%% LyX 2.2.3 created this file.  For more info, see http://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\documentclass[oneside,czech,english,reqno]{amsbook}
\usepackage[T1]{fontenc}
\usepackage[latin2,latin9]{inputenc}
\usepackage{babel}
\usepackage{mathrsfs}
\usepackage{amstext}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{wasysym}
\usepackage[unicode=true]
 {hyperref}

\makeatletter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Textclass specific LaTeX commands.
\numberwithin{section}{chapter}
\numberwithin{figure}{section}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands.
% README!
% - in the case you are using this source code, please at least CITE
%   in your work this document as online source with url
%   http://www.karlin.mff.cuni.cz/~prazak/vyuka/Pdr2/
%   and the authors Radim Cajzl and Dalibor Pra¸ák
% - this document was created using LyX
%   (graphical TeX editor, see http://www.lyx.org)
%   Corresponding LaTeX source is AUTOMATICALLY GENERATED and therefore
%   not designed to be easily readable.
%   You are strongly advised to make any edits directly using LyX.

%\parindent 0pt                                                                                                                                                                 
%\usepackage[scale=0.8]{geometry}            

\def\dx{\,dx}

% underbrace which correctly sets scriptstyle where appropriate
\makeatletter
\let\saved@underbrace\underbrace
\renewcommand*\underbrace[1]{\@ifnextchar_{\ub@with{#1}}{\ub@without{#1}}}
\def\ub@with#1_#2{\mathpalette\underbrace@i{{#1}{_{#2}}}}
\newcommand*\ub@without[1]{\mathpalette\underbrace@i{{#1}{}}}
\newcommand*\underbrace@i[2]{\underbrace@ii#1#2}
\newcommand*\underbrace@ii[3]{\saved@underbrace{#1#2}#3}
\makeatother

\parindent 0pt                                                                                                                                                                 
\usepackage[scale=0.8]{geometry}

\makeatletter
\renewcommand{\@maketitle}{%
  \thispagestyle{empty}%
  \begingroup \topskip\z@skip
  \centering
  \begingroup
    \LARGE\bfseries
    \vglue 1in
    {\Huge\@title}\par\vspace{24pt}%
    \vglue 0.5 in
    \def\and{\par\medskip}\centering
    \mdseries\authors\par\bigskip
  \endgroup
  \vfill
  \begin{center}
  MFF UK, 2016
  \end{center}
  \endgroup
  \newpage
}

\makeatother

\begin{document}

\title{Partial differential equations II}

\author{Radim Cajzl}

\author{Dalibor Pra¸ák}
\maketitle

\chapter*{Introduction}

These lecture notes are a transcript of the course ``PDE2'', given
by the second author at MFF UK during spring 2016. All the material
presented here is well-known. Precise wording of theorems and their
proofs were inspired by various sources we list in the bibliography.

An up-to-date version of these notes, along with supplementary material,
can be found at the webpage of the course: \texttt{\href{http://www.karlin.mff.cuni.cz/~prazak/vyuka/Pdr2/}{http://www.karlin.mff.cuni.cz/$\sim$prazak/vyuka/Pdr2/}.}

Feel free to write an e-mail to the authors in the case you would
like to share some ideas or report found errors in the text. The authors'
email adresses are \texttt{\href{mailto:}{radim.cajzl@gmail.com}}
and \texttt{\href{mailto:}{prazak@karlin.mff.cuni.cz}}.

Special thanks go to Zdenek Mihula for providing detailed solutions
to selected exercises. 

\tableofcontents{}

\chapter{Vector\textendash valued functions}

Vector\textendash valued \dots{} values in infinite\textendash dimensional
Banach space.

\subsection*{Notation:}

we consider:
\begin{itemize}
\item $u\left(t\right):I\to X$, $I=\left[0,T\right]$ is time interval,
$X$ is Banach space, 
\item $\left\Vert u\right\Vert _{X}$ is norm in $X$
\item $X^{*}$ is dual space $X^{*}$, $\left\langle x^{*},x\right\rangle _{X^{*},X}$
is notation of duality
\item scalar case: $X=\mathbb{R}$
\end{itemize}

\section{Vector\textendash valued integrable functions (Bochner Integral)}

\subsection*{Def.: {[}Simple, weakly and strongly measurable function{]}}

Function $u\left(t\right):I\to X$ is called
\begin{description}
\item [{simple}] if $u\left(t\right)=\sum_{j=1}^{N}\chi_{A_{j}}\left(t\right)x_{j}$,
where $A_{j}\subset I$ are (Lebesgue in $\mathbb{R}$) measurable
and $x_{j}\in X$
\item [{(strongly)~measurable}] if there exist simple functions $u_{n}\left(t\right)$
such that $u_{n}\left(t\right)\to u\left(t\right)$, $n\to\infty$
for a. e. $t\in I$.
\item [{weakly~measurable}] if the (scalar) function $t\mapsto\left\langle x^{*},u\left(t\right)\right\rangle $
is measurable in $I$ for $\forall x^{*}\in X^{*}$.
\end{description}

\subsection*{Remark:}
\begin{enumerate}
\item Strongly meas. $\implies$ weakly meas.
\item $u\left(t\right)$ simple $\iff$ $u\left(t\right)$ is measurable
and $u\left(I\right)\subset X$ is finite.
\end{enumerate}

\subsection*{Theorem 1.1 {[}Pettis characterization of measurability{]}}

Function $u\left(t\right):I\to X$ is measurable iff $u\left(t\right)$
is weakly measurable and moreover $\exists N\subset I$ such that
$\lambda\left(N\right)=0$ and $u\left(I\backslash N\right)$ is separable
(i. e., $u\left(t\right)$ is ``essentially separably\textendash valued'').

\subsection*{Corollary}
\begin{enumerate}
\item For $X$ separable: weakly measurable $\iff$ strongly measurable.
\item $u_{n}\left(t\right)$ is measurable, $u_{n}\left(t\right)\to u\left(t\right)$
a. e. $\implies$ $u\left(t\right)$ is measurable.
\item $u\left(t\right)$ continuous $\implies$ $u\left(t\right)$ measurable
\end{enumerate}

\subsubsection*{Proof:}

$u_{n}\left(t\right)$ measurable $\stackrel{\mbox{th.1.1}}{\implies}\exists N_{n}\subset I$,
$\lambda\left(N_{n}\right)=0$. Set $u\left(I\backslash N_{n}\right)=M_{n}\subset X$
is separable, also $\exists N_{0}\subset I$, $\lambda\left(N_{0}\right)$
such that $u_{n}\left(t\right)\to u\left(t\right)$ for $\forall t\in I\backslash N_{0}$.
Set $N=\cup_{j=0}^{\infty}N_{j}$, then $\lambda\left(N\right)=0$
and moreover $u\left(I\backslash N\right)\subseteq\overline{\cup_{n}M^{n}}^{X}$
\dots{} separability is preserved under countable unions and closures
$\implies$ $u\left(t\right)$ is essentially separably\textendash valued.

Is $u\left(t\right)$ weakly measurable? Consider $x^{*}\in X^{*}$fixed,
then the mapping $t\mapsto\left\langle x^{*},u_{n}\left(t\right)\right\rangle $
is measurable and $\left\langle x^{*},u_{n}\left(t\right)\right\rangle \to\left\langle x^{*},u\left(t\right)\right\rangle $
as $n\to\infty$ for a. e. $t\in I$, the mapping $t\mapsto\left\langle x^{*},u\left(t\right)\right\rangle $
is measurable by scalar (Lebesgue) theory.

$\left(3\right)$ HW1

\subsection*{Def.: {[}Bochner integral{]}}

function $u\left(t\right):I\to X$ is called Bochner integrable, if
there exists $u_{n}\left(t\right)$ simple such that $\int_{I}\left\Vert u\left(t\right)-u_{n}\left(t\right)\right\Vert dt\to0$,
$n\to\infty$. The integral (Bochner integral) is defined as follows
\begin{enumerate}
\item $\int_{I}u\left(t\right)dt=\sum_{j=1}^{N}\lambda\left(A_{j}\right)x_{j}$,
if $u\left(t\right)$ is simple
\item $\int_{I}u\left(t\right)dt=\lim_{n\to\infty}\int_{I}u_{n}\left(t\right)dt$,
if $u\left(t\right)$ is integrable with $u_{n}\left(t\right)$ simple
from the definition above.
\end{enumerate}

\subsection*{Remark:}

It is possible to show that the definiton is correct, i. e., independent
on the choice of $A_{j}$, $x_{j}$, $u_{n}\left(t\right)$. Also
it is possible to prove that $\left\Vert \int_{I}u\left(t\right)dt\right\Vert _{X}\leq\int\left\Vert u\left(t\right)\right\Vert _{X}dt$.

\subsection*{Theorem 1.2 {[}Bochner characterization of measurability{]}}

Function $u\left(t\right):I\to X$ is Bochner integrable iff $u\left(t\right)$
is measurable and $\int_{I}\left\Vert u\left(t\right)\right\Vert dt<\infty$.

\subsection*{Theorem 1.3 {[}Lebesgue{]}}

Let $u_{n}\left(t\right):I\to X$ be measurable, $u_{n}\left(t\right)\to u\left(t\right)$
as $n\to\infty$ for a. e. $t\in I$, let there $\exists g\left(t\right):I\to\mathbb{R}$
integrable and such that $\left\Vert u_{n}\left(t\right)\right\Vert \leq g\left(t\right)$
for $\forall n$ and a. e. $t\in I$. Then $\int_{I}u_{n}\left(t\right)dt\to\int_{I}u\left(t\right)dt$,
moreover $\int_{I}\left\Vert u_{n}\left(t\right)-u\left(t\right)\right\Vert dt\to0$
for $n\to\infty$ and $u$ is integrable.

\subsection*{Recall:}

for scalar $u\left(t\right):I\to\mathbb{R}$ we say that $t\in I$
is a Lebesgue point if $\lim_{h\to0}\frac{1}{2h}\int_{-h}^{h}\left|u\left(t+s\right)-u\left(t\right)\right|ds=0$.
\textbf{Lebesgue theorem:} $u\left(t\right):I\to\mathbb{R}$ is integrable
$\implies$ a. e. $t\in I$ is a Lebesgue point.

\subsection*{Def.: {[}Lebesgue point of function{]}}

We say that $t\in I$ is a Lebesgue point of $u\left(t\right):I\to X$
if 
\[
\frac{1}{2h}\int_{-h}^{h}\left\Vert u\left(t+s\right)-u\left(t\right)\right\Vert ds\to0\mbox{ as h\ensuremath{\to}0 .}
\]
 

\subsection*{Theorem 1.4 {[}On Lebesgue points a. e.{]}}

If $u\left(t\right):I\to X$ is integrable, then a. e. $t\in I$ is
a Lebesgue point.

\subsection*{Remark}

Let $u\left(t\right):I\to X$ be integrable.
\begin{enumerate}
\item Set $U\left(x\right)=\int_{t_{0}}^{t}u\left(t\right)dt$, $t_{0}\in I$
fixed $\stackrel{\mbox{HW1}}{\implies}$ $U'\left(t\right)=u\left(t\right)$
for every Lebesgue point of $u\left(t\right)$, in particular a. e.
\item (regularization kernels) Let $\psi_{0}\left(t\right):\mathbb{R}\to\mathbb{R}$
be bounded, zero outside interval $\left[-1,1\right]$ and such that
$\int_{-1}^{1}\psi_{0}\left(s\right)ds=1$ (\& possibly other regularizations).
\begin{figure}
\includegraphics[width=0.5\textwidth]{img/k1_reg_kernel}

\caption{Regularization kernel}
\end{figure}
\\
$\psi_{n}\left(t\right)=n\psi_{0}\left(nt\right)$, $u*\psi_{n}\left(t\right)=\int_{\mathbb{R}}u\left(t-s\right)\psi_{n}\left(s\right)ds$
for $t\in\left[\frac{1}{n},T-\frac{1}{n}\right]$ or set $u\left(t\right)=0$
outside $I$.\\
Claim: $u*\psi_{n}\left(t\right)\to u\left(t\right)$ if $t\in I$
is Lebesgue point of $u$. Proof: 
\begin{eqnarray*}
u*\psi_{n}\left(t\right)-u\left(t\right) & = & \int_{\mathbb{R}}u\left(t-s\right)\psi_{n}\left(s\right)ds-u\left(t\right)\\
 & = & \int_{-\frac{1}{n}}^{\frac{1}{n}}\left[u\left(t-s\right)-u\left(t\right)\right]\psi_{n}\left(s\right)ds
\end{eqnarray*}
\[
\implies\left\Vert u\psi_{n}\left(t\right)-u\left(t\right)\right\Vert \leq c\underbrace{n}_{h^{-1}}\int_{\underbrace{-\frac{1}{n}}_{-h}}^{\frac{1}{n}}\left\Vert u\left(t-s\right)u\left(t\right)\right\Vert ds\stackrel{\left(*\right)}{\to}0,\ n\to\infty
\]
where $\left(*\right)$ holds by definition of Lebesgue points.
\end{enumerate}

\subsubsection*{Proof of th. 1.4:}

By th. 1.1 there $\exists N_{0}\subset I$, $\lambda\left(N_{0}\right)=0$
such that $u\left(I\backslash N_{0}\right)\subseteq\overline{\left\{ x_{1},x_{2},\ldots,x_{k},\ldots\right\} }^{X}$.
Set $\varphi_{k}=\left\Vert u\left(t\right)-x_{k}\right\Vert $ \dots{}
scalar, integrable. By scalar version of the theorem $\exists N_{k}\subset I$,
$\lambda\left(N_{k}\right)=0$ such that $\varphi_{k}\left(t\right)$
has Lebesgue point at every $t\in I\backslash N_{k}$. Set $N=\cup_{k=0}^{\infty}N_{k}$,
then $\lambda\left(N\right)=0$ and $\forall t\in I\backslash N$
is Lebesgue point of $u\left(t\right)$, i. e., $\frac{1}{2h}\int_{-h}^{h}\left\Vert u\left(t+s\right)-u\left(t\right)\right\Vert ds\to0$,
$h\to0$. Fix $t\in I\backslash N$, $\varepsilon>0$ arbitrary. Observe:
$\exists k$ such that $\left\Vert u\left(t\right)-x_{k}\right\Vert \leq\varepsilon$.
Then 
\begin{eqnarray*}
\frac{1}{2h}\int_{-h}^{h}\left\Vert u\left(t+s\right)-u\left(t\right)\pm x_{k}\right\Vert ds & \leq\\
\underbrace{\frac{1}{2h}\int_{-h}^{h}\underbrace{\left\Vert u\left(t+s\right)-x_{k}\right\Vert }_{\varphi\left(t+s\right)}ds}_{I\left(h\right)}+\underbrace{\frac{1}{2h}\int_{-h}^{h}\underbrace{\left\Vert u\left(t\right)-x_{k}\right\Vert }_{\leq\varepsilon}ds}_{\leq\varepsilon\ \left(*\right)}
\end{eqnarray*}
\[
I\left(h\right)=\frac{1}{2h}\int_{-h}^{h}\varphi_{k}\left(t+s\right)ds\to\underbrace{\varphi_{k}\left(t\right)}_{<\varepsilon\makebox[0pt][l]{\mbox{ by choice of \ensuremath{k}}}},\ h\to0
\]
 since $t$ is L. point of $\varphi\left(t\right)$.

$\left(*\right)<2\varepsilon$ for $h$ small enough.

\subsection*{Def.: {[}Spaces $L^{p}\left(I;X\right)${]}}

for $p\in[1,\infty)$ we set 
\begin{eqnarray*}
L^{p}\left(I;X\right) & = & \left\{ u\left(t\right):I\to X\mbox{ measurable, }\int_{I}\left\Vert u\left(t\right)\right\Vert _{X}^{p}dt<\infty\right\} \\
L^{\infty}\left(I;X\right) & = & \bigg\{ u\left(t\right):I\to X\mbox{ measurable, essentially bounded,}\\
 &  & \quad\mbox{i. e., }\exists c<\infty\mbox{ s. t. }\left\Vert u\left(t\right)\right\Vert _{X}<c\mbox{ a. e.}\bigg\}
\end{eqnarray*}

\subsection*{Remark:}

These are Banach spaces with the usual norm and the convention that
$u\left(t\right)$, $\widetilde{u}\left(t\right)$ are identified
if $u\left(t\right)=\widetilde{u}\left(t\right)$ a. e. in $I$. If
$X$ is Hilbert with scalar product $\left(\cdot,\cdot\right)_{X}$,
then $L^{2}\left(I;X\right)$ is Hilbert with scalar product $\left(u,v\right)_{L^{2}\left(I;X\right)}=\int_{I}\left(u\left(t\right),v\left(t\right)\right)_{X}dt$.

\subsection*{Note:}

$L^{1}\left(I;X\right)$ is space of Bochner integrable functions,
$L^{p}\left(I;X\right)\subset L^{q}\left(I;X\right)$ for $p\geq q$
(since $I$ is bounded).

\subsection*{Lemma 1.1 {[}Approximation and density in $L^{p}\left(I;X\right)${]}}

Let $p\in[1,\infty)$. Then
\begin{enumerate}
\item Simple functions are dense in $L^{p}\left(I;X\right)$.
\item Functions of the form $u\left(t\right)=\sum_{j=1}^{N}\varphi_{j}\left(t\right)x_{j}$,
$\varphi_{j}\left(t\right)\in C_{c}^{\infty}\left(I,\mathbb{R}\right)$
are dense in $L^{p}\left(I;X\right)$
\item If the space $Y$ is dense in $X$, then $C_{c}^{\infty}\left(I,Y\right)$
are dense in $L^{p}\left(I;X\right)$.
\item Let $\psi_{n}\left(t\right)$ be regularizing kernels, let $u\left(t\right)\in L^{p}\left(I;X\right)$
be extended by $0$ outside of $I$. Then $u*\psi_{n}\left(t\right)\to u\left(t\right)$
in $L^{p}\left(I;X\right)$ for $n\to\infty$.
\end{enumerate}

\subsection*{Remark:}

\begin{eqnarray*}
C\left(I,X\right) & = & \left\{ u\left(t\right):I\to X\mbox{ continuous}\right\} \\
C^{1}\left(I,X\right) & = & C\left(I,X\right)\cap\left\{ u'\left(t\right)\in C\left(I,X\right)\right\} \\
C_{c}\left(I,X\right) & = & C\left(I,X\right)\cap\left\{ u\left(t\right)=0\mbox{ on }\left[0,\delta\right]\cup\left[T-\delta,T\right]\mbox{ for some \ensuremath{\delta}}\right\} 
\end{eqnarray*}

\subsubsection*{Proof:}
\begin{enumerate}
\item $u\left(t\right)\in L^{p}\left(I;X\right)$ given $\stackrel{?}{\implies}$$\exists u_{n}\left(t\right)$
simple such that $\int_{I}\left\Vert u_{n}\left(t\right)-u\left(t\right)\right\Vert _{X}^{p}dt\to0$.

We know $u\left(t\right)$ is measurable, hence $\exists\widetilde{u}_{n}\left(t\right)$
simple such that $\widetilde{u}_{n}\left(t\right)\to u\left(t\right)$
a. e. in $I$. Set $u_{n}=\begin{cases}
\widetilde{u}_{n}\left(t\right) & \quad\mbox{if \ensuremath{\left\Vert \widetilde{u}_{n}\left(t\right)\right\Vert \leq1+\left\Vert u\left(t\right)\right\Vert }}\\
0 & \quad\mbox{otherwise}
\end{cases}$. Clearly $u_{n}\left(t\right)\to u\left(t\right)$, but $\left\Vert u_{n}\left(t\right)\right\Vert \leq1+\left\Vert u\left(t\right)\right\Vert $
for a. e. 
\begin{eqnarray*}
\underbrace{\left\Vert u_{n}\left(t\right)-u\left(t\right)\right\Vert ^{p}}_{\smash{\to0\mbox{ s. v.}}} & \leq & \left(1+2\left\Vert u\left(t\right)\right\Vert \right)^{p}\\
 & \leq & c_{p}\left(1+\left\Vert u\left(t\right)\right\Vert ^{p}\right)=g\left(t\right)\mbox{\makebox[0pt][l]{ \dots\ integrable}}
\end{eqnarray*}
By scalar Lebesgue theorem we are done.
\item by $\left(1\right)$ we know: $u\left(t\right)=\sum_{j=1}^{N}\chi_{A_{j}}\left(t\right)x_{j}$
are dense in $L^{p}\left(I;X\right)$. Scalar theory: $\chi_{A_{j}}\left(t\right)$
can be approximated in $L^{p}\left(I;\mathbb{R}\right)$ by $\varphi_{j}\left(t\right)\in C_{c}^{\infty}\left(I,\mathbb{R}\right)$.
\item Functions $u\left(t\right)=\sum_{j=1}^{N}\varphi_{j}\left(t\right)x_{j}$,
$\varphi_{j}\left(t\right)\in C_{c}^{\infty}\left(I,\mathbb{R}\right)$,
$x_{j}\in X$ can be approximated in $L^{p}\left(I;X\right)$ by functions
$u\left(t\right)=\sum_{j=1}^{N}\underbrace{\varphi_{j}\left(t\right)y_{j}}_{\smash{\makebox[0pt][l]{\ensuremath{\in C_{c}^{\infty}\left(I;Y\right),\mbox{ obviously}}}}}$,
where $\varphi_{j}\in C_{c}^{\infty}\left(I,\mathbb{R}\right)$, $y_{j}\in Y$.
\item $\varepsilon\geq0$ given, arbitrary. Find $v\left(t\right)\in C_{c}^{\infty}\left(I,X\right)$
such that $\left\Vert u\left(t\right)-v\left(t\right)\right\Vert _{L^{p}\left(I;X\right)}<\varepsilon$.
Then 
\[
u*\psi_{n}\left(t\right)\pm v\left(t\right)=\left[u-v\right]*\psi_{n}\left(t\right)+v*\psi_{n}\left(t\right)=f_{n}\left(t\right)+g_{n}\left(t\right)
\]
\begin{itemize}
\item $f_{n}\left(t\right)$: use Young inequality: $\left\Vert \varphi_{1}*\varphi_{2}\right\Vert _{L^{r}}\leq\left\Vert \varphi_{1}\right\Vert _{L^{p}}\left\Vert \varphi_{2}\right\Vert _{L^{q}}$,
in particular for $p=r$, $q=1$: $\left\Vert \varphi_{1}*\varphi_{2}\right\Vert _{L^{p}}\leq\left\Vert \varphi_{1}\right\Vert _{L^{p}}\cdot\left\Vert \varphi_{2}\right\Vert _{L^{1}}$.
Set $\varphi_{1}=u-v$, then $\left\Vert u-v\right\Vert _{L^{p}\left(I;X\right)}<\varepsilon$,
$\varphi_{2}=\psi_{n}$, $\left\Vert \psi_{n}\right\Vert _{L^{1}}=1$.
$\implies\left\Vert f_{n}\left(t\right)\right\Vert _{L^{p}\left(I;X\right)}<\varepsilon$.
\item $g_{n}\left(t\right)=v*\psi_{n}\left(t\right)\rightrightarrows v\left(t\right)$
in $I$, hence $v*\psi\left(t\right)\to v\left(t\right)$ in $L^{p}\left(I;X\right)$,
hence $\left\Vert g_{n}\left(t\right)-v\left(t\right)\right\Vert _{L^{p}\left(I;X\right)}<\varepsilon$
for $n$ large enough.
\end{itemize}
\end{enumerate}
\[
u*\psi_{n}\left(t\right)-u\left(t\right)=\underbrace{f_{n}\left(t\right)}_{\to0}+\underbrace{g_{n}\left(t\right)-v\left(t\right)}_{{\mbox{\ensuremath{\left\Vert \cdot\right\Vert }}\makebox[0pt][l]{<\ensuremath{\varepsilon}}\atop \mbox{for \ensuremath{n} large}}}\ +\ \underbrace{v\left(t\right)-u\left(t\right)}_{\mbox{\ensuremath{\left\Vert \cdot\right\Vert }}\makebox[0pt][l]{<\ensuremath{\varepsilon}}},
\]
therefore $\left\Vert u*\psi_{n}\left(t\right)-u\left(t\right)\right\Vert <3\varepsilon$
for $n$ large.

\subsection*{Remark}
\begin{enumerate}
\item Corollary: $X$ separable $\implies$$L^{p}\left(I;X\right)$ is separable
for $p\in[1,\infty)$.
\item None of these is true for $p=\infty$.
\item More about $L^{p}\left(\ldots\right)$ can be found below (dual space,
geometry).
\end{enumerate}

\section{$AC$ functions and weak time derivative}

\subsection*{Def.:}

$u\left(t\right):I\to X$ is absolutely continuous ($AC\left(I,X\right)$)
$\iff$ for $\forall\varepsilon>0\ \exists\delta>0$: $\forall$ disjoint
finite system $\left(\alpha_{j},\beta_{j}\right)\subset I$ it holds
that if $\sum_{j}\left|\alpha_{j}-\beta_{j}\right|<\delta$ then $\sum_{j}\left\Vert u\left(\alpha_{j}\right)-u\left(\beta_{j}\right)\right\Vert <\varepsilon$.

\subsection*{Theorem 1.5 {[}Derivative of $AC$ function{]}}

Let $u\left(t\right)\in AC\left(I,X\right)$, let $X$ be reflexive
and separable. Then $u'\left(t\right)=\lim_{h\to0}\frac{1}{h}\left(u\left(t+h\right)-u\left(t\right)\right)$
for a. e. $t\in I$, $u'\left(t\right)\in L^{1}\left(I;X\right)$
and $u\left(t_{2}\right)-u\left(t_{1}\right)=\int_{t_{1}}^{t_{2}}u'\left(s\right)ds$
for $\forall t_{1},t_{2}\in I$.

\subsection*{Recall:}

We assume scalar version $\left(X=\mathbb{R}\right)$ is already proven.
\begin{itemize}
\item $I=\left[0,T\right]$
\item $X$ is Banach space.
\end{itemize}

\subsection*{Recall: (Eberlein\textendash Śmulian theorem)}

$X$ is reflexive, $\left\Vert u_{n}\right\Vert $ is bounded $\implies$
$\exists$ weakly convergent subsequence $\widetilde{u}_{n}$, i.
e., $\exists u\in X$ such that $\widetilde{u}_{n}\rightharpoonup u$
meaning $\left\langle x^{*},\widetilde{u}_{n}\right\rangle \to\left\langle x^{*},u\right\rangle $
for $\forall x^{*}\in X^{*}$ fixed.

\subsection*{Recall:}

norm is weakly lower-semicontinuous: 
\[
u_{n}\rightharpoonup u\implies\left\Vert u\right\Vert \leq\liminf_{n\to\infty}\left\Vert u_{n}\right\Vert 
\]

\subsubsection*{Proof of th. 1.5:}

\textbf{Step 1:} set 
\[
V\left(t\right)=\mbox{var}\left(u\left(t\right),0,t\right)\stackrel{\mbox{by def.}}{=}\sup_{D}\sum_{j=1}^{N}\left\Vert u\left(\tau_{j}\right)-u\left(\tau_{j-1}\right)\right\Vert 
\]
 where the supremum is taken over all partitions $D$ of the interval
$\left[0,t\right]$, i. e., $D:0=\tau_{0}<\tau_{1}<\ldots<\tau_{n}=t$.

observe: $V\left(t\right)\geq0$, $V\left(t\right)$ is non-decreasing,
even AC. 
\[
\left\Vert \frac{u\left(t+h\right)-u\left(t\right)}{h}\right\Vert \leq\frac{V\left(t+h\right)-V\left(t\right)}{h}\to V'\left(t\right)\in\mathbb{R}\mbox{ for a. e. \ensuremath{t\in I\backslash N_{0}}},
\]
where $\lambda\left(N_{0}\right)=0$. Hence $\left\Vert \frac{u\left(t+h\right)-u\left(t\right)}{h}\right\Vert $
is bounded as $h\to0$, for a. e. $t\in I$.

\textbf{Step 2:} $X^{*}$ is separable, let $\left\{ x_{1}^{*},x_{2}^{*},\ldots\right\} $
be some countable dense set in $X^{*}$. Define auxiliary functions
$\varphi_{k}\left(t\right)=\left\langle x_{k}^{*},u\left(t\right)\right\rangle :I\to\mathbb{R}$,
$\varphi_{k}$ are $AC$. Therefore by scalar case $\varphi'\left(t\right)\in\mathbb{R}$,
$\exists$ for $\forall t\in I\backslash N_{k}$, $\lambda\left(N_{k}\right)=0$
and $\varphi_{k}\left(t_{2}\right)-\varphi_{k}\left(t_{1}\right)=\int_{t_{1}}^{t_{2}}\varphi'\left(s\right)ds\ \forall t_{1},t_{2}$.
For $t\in I\backslash\bigcup_{k>0}N_{k}$ define $v\left(t\right)\in X$
as follows: take some $h_{n}\to0$ such that $\frac{u\left(t+h_{n}\right)-u\left(t\right)}{h_{n}}\rightharpoonup v\left(t\right)$
(by Step 1 \& Eberlein\textendash Śmulian).

observe: $v\left(t\right)$ is independent of the sequence $h_{n}$,
hence $\frac{u\left(t+h\right)-u\left(t\right)}{h}\rightharpoonup v\left(t\right)$,
$h\to0$.

Assume: $\frac{u\left(t+\widetilde{h}_{n}\right)-u\left(t\right)}{\widetilde{h}_{n}}\rightharpoonup\widetilde{v}\left(t\right)\in X$.
But then 
\begin{eqnarray*}
\left\langle x_{k}^{*},\frac{u\left(t+h_{n}\right)-u\left(t\right)}{h_{n}}\right\rangle  & \to & \left\langle x_{k}^{*},v\left(t\right)\right\rangle \\
\Vert\quad\quad\quad\quad\quad &  & \quad\Vert\\
\frac{1}{h_{n}}\left(\varphi_{k}\left(t+h_{n}\right)-\varphi_{k}\left(t\right)\right) & \to & \varphi_{k}'\left(t\right),
\end{eqnarray*}
the same holds with $\widetilde{h}_{n}$: $\left\langle x_{k}^{*},\widetilde{v}\left(t\right)\right\rangle =\varphi_{k}'$$\left(t\right)=\left\langle x^{*},v\left(t\right)\right\rangle $,
therefore $\left\langle x_{k}^{*},\widetilde{v}\left(t\right)-v\left(t\right)\right\rangle =0$
for $\forall k$, by density of $x_{k}^{*}$ in $X$ we have $\widetilde{v}\left(t\right)=v\left(t\right)$.

\textbf{Step 3:} $v\left(t\right)\in L^{1}\left(I,X\right)$, i. e.,
\begin{enumerate}
\item measurable \dots{} see excercises, HW1
\item $\int_{I}\left\Vert v\left(t\right)\right\Vert dt<\infty$ \dots{}
weak lower semicontinuity
\end{enumerate}
\begin{eqnarray*}
\int_{I}\left\Vert v\left(t\right)\right\Vert dt & \leq & \int_{I}\liminf_{n\to\infty}\left\Vert \frac{1}{h_{n}}\left(v\left(t+h_{n}\right)-v\left(t\right)\right)\right\Vert dt\\
 & \leq & \int_{I}\liminf_{n\to\infty}\frac{1}{h_{n}}\left(V\left(t+h_{n}\right)-V\left(t\right)\right)dt\\
\mbox{we know} & = & \int_{I}\lim_{n\to\infty}\frac{1}{h_{n}}\left(V\left(t+h_{n}\right)-V\right)dt\\
 & = & \int_{I}V'\left(t\right)dt<\infty
\end{eqnarray*}
since $V\left(t\right)$ is $AC$.

\textbf{Step 4:} $u\left(t_{2}\right)-u\left(t_{1}\right)=\int_{t_{1}}^{t_{2}}v\left(s\right)ds$
for $\forall t_{1},t_{2}\in I$.\\
We have: 
\[
\left\langle x_{k}^{*},u\left(t_{2}\right)\right\rangle -\left\langle x_{k}^{*},u\left(t_{1}\right)\right\rangle =\int_{t_{1}}^{t_{2}}\underbrace{\left\langle x_{k}^{*},v\left(s\right)\right\rangle }_{\varphi_{k}'\left(s\right)}ds=\left\langle x_{k}^{*},\int_{t_{1}}^{t_{2}}v\left(s\right)ds\right\rangle 
\]
We conclude by density of $x_{k}^{*}$ in $X^{*}$.

\textbf{Step 5:} $\frac{u\left(t+h\right)-u\left(t\right)}{h}\to v\left(t\right)$
strongly for a. e. $t\in I$ \dots{} consequence of Th. 1.4 and step
4.

\subsection*{Notation:}

$D\left(I\right)=C_{c}^{\infty}\left(I,\mathbb{R}\right)$ \dots{}
scalar test-functions.

\subsection*{Lemma 1.2 {[}Weak characterization of constant and zero function{]}}

Let $u\left(t\right)\in L^{1}\left(I;X\right)$.
\begin{enumerate}
\item If $\int_{I}u\left(t\right)\varphi\left(t\right)dt=0$ for $\forall\varphi\left(t\right)\in D\left(I\right)$
then $u\left(t\right)=0$ a. e. in $I$.
\item If $\int_{I}u\left(t\right)\varphi'\left(t\right)dt=0$ for $\forall\varphi\left(t\right)\in D\left(I\right)$
then $\exists x_{0}\in X$ such that $u\left(t\right)=x_{0}$ a. e.
in $I$.
\end{enumerate}

\subsubsection*{Proof:}

$\left(1\right)$: Set $u_{n}\left(t\right)=u*\psi_{n}\left(t\right)$,
where $\psi_{n}\left(t\right)=n\psi_{0}\left(nt\right)$, $\psi_{0}\left(t\right)$
is convolution kernel such that $\psi_{0}\in D\left(\left[-1,1\right]\right)$.

$PS=\int_{\mathbb{R}}u\left(s\right)\underbrace{\psi_{n}\left(t-s\right)}_{\mbox{\ensuremath{\in}}\makebox[0pt][l]{\ensuremath{\,D\left(I\right)} for \ensuremath{n} large enough and \ensuremath{t\in\left[0,T\right]} fixed}}ds=0$
by assumptions.

\begin{figure}
\includegraphics[width=0.6\textwidth]{img/psi_n}

\caption{Regularization function $\psi_{n}\left(t-s\right)$}

\end{figure}

On the other hand: $LS=u_{n}\left(t\right)\to u\left(t\right)$ a.
e. (by th. 1.4)

$\left(2\right)$ $u_{n}\left(t\right)$ as above: $u_{n}'\left(t\right)=\frac{d}{dt}\int_{\mathbb{R}}u\left(s\right)\psi_{n}\left(t-s\right)ds\stackrel{\makebox[0pt][c]{ex.}}{=}\int_{\mathbb{R}}u\left(s\right)\underbrace{\psi_{n}'\left(t-s\right)}_{-\varphi'\left(s\right)}ds=0$,
$t\in\left(0,T\right)$ fixed, $n$ large enough, where $\varphi\left(s\right)=\psi_{n}\left(t-s\right)\in D\left(I\right)$.
Hence $u_{n}\left(t\right)$ is smooth, $u_{n}'\left(t\right)=0$,
therefore $u_{n}\left(t\right)=x_{n}$ everywhere. But $u_{n}\left(t\right)\to u\left(t\right)$
a. e. in $I$, which concludes the proof.

\subsection*{Lemma 1.3 {[}Equivalent definition of weak derivative{]}}

Let $u\left(t\right)$, $g\left(t\right)\in L^{1}\left(I;X\right)$.
Then the following are equivalent:
\begin{enumerate}
\item $\exists x_{0}\in X$ such that $u\left(t\right)=x_{0}+\int_{0}^{t}g\left(s\right)ds$
for a. e. $t\in I$.
\item $\int_{I}u\left(t\right)\varphi'\left(t\right)dt=-\int_{I}g\left(t\right)\varphi\left(t\right)dt$
for $\forall\varphi\left(t\right)\in D\left(I\right)$.
\item $\frac{d}{dt}\left\langle x^{*},u\left(t\right)\right\rangle =\left\langle x^{*},g\left(t\right)\right\rangle $
in the sense of distributions in $\left(0,T\right)$ for $\forall x^{*}\in X^{*}$
fixed.
\end{enumerate}

\subsubsection*{Proof:}
\begin{itemize}
\item[$\left(2\right)\iff\left(3\right)$]  recall: $\left(3\right)$ means $\int_{I}\left\langle x^{*},u\left(t\right)\right\rangle \varphi'\left(t\right)dt=-\int_{I}\left\langle x^{*},g\left(t\right)\right\rangle \varphi\left(t\right)dt$
for $\forall\varphi\in D\left(t\right)$. \\
It follows that $\left\langle x^{*},\int_{I}u\left(t\right)\varphi'\left(t\right)dt\right\rangle =-\left\langle x^{*},\int g\left(t\right)\varphi\left(t\right)dt\right\rangle $
for $\forall x^{*}\in X^{*}$ and $\forall\varphi\in D\left(I\right)$.
Therefore $\int_{I}u\left(t\right)\varphi'\left(t\right)dt=\int g\left(t\right)\varphi\left(t\right)dt\iff\left(2\right)$
holds.
\item[$\left(1\right)\implies\left(3\right)$]  take $x^{*}\in X^{*}$ arbitrary, apply to $\left(1\right)$: 
\[
\underbrace{\left\langle x^{*},u\left(t\right)\right\rangle }_{\mbox{\ensuremath{=}}\makebox[0pt][l]{\ensuremath{\,\xi\left(t\right)}}}=\left\langle x^{*},x_{0}\right\rangle +\int_{0}^{t}\underbrace{\left\langle x^{*},g\left(s\right)\right\rangle }_{\mbox{\ensuremath{\eta\left(s\right)}}}ds=\widetilde{\xi}\left(t\right)
\]
clearly $\widetilde{\xi}\left(t\right)\in AC$, $\widetilde{\xi}\left(t\right)=\xi\left(t\right)$
a. e. Take $\varphi\left(t\right)\in D\left(I\right)$: one easily
verifies that $\widetilde{\xi}\left(t\right)\varphi\left(t\right)\in AC$
and $\left(\widetilde{\xi}\left(t\right)\varphi\left(t\right)\right)'=\widetilde{\xi}\left(t\right)\varphi\left(t\right)+\widetilde{\xi}\left(t\right)\varphi'\left(t\right)$
a. e. Integrate over $\left[0,T\right]$, note that $\varphi\left(0\right)=\varphi\left(T\right)=0$:
$0=\int_{I}\underbrace{\widetilde{\xi}\left(t\right)}_{\mbox{\ensuremath{=}}\makebox[0pt][l]{\ensuremath{\,\eta\left(t\right)} a. e.}}\varphi\left(t\right)+\underbrace{\widetilde{\xi}\left(t\right)}_{\mbox{\ensuremath{=}}\makebox[0pt][l]{\ensuremath{\xi\left(t\right)} a. e.}}\varphi'\left(t\right)dt$.
Therefore $\left(3\right)$ holds.
\item[$\left(2\right)\implies\left(1\right)$]  set $\widetilde{u}\left(t\right)=\int_{0}^{t}g\left(s\right)ds$
\dots{} $\widetilde{u}\left(t\right)\in AC\left(I,X\right)$, $\widetilde{u}'\left(t\right)=g\left(t\right)$
a. e.\\
Take $\varphi\left(t\right)\in D\left(I\right)$ arbitrary \dots{}
$\widetilde{u}\left(t\right)\varphi\left(t\right)\in AC\left(I,X\right)$,
by remark below Th. 1.4 it holds that 
\begin{eqnarray*}
\left(\widetilde{u}\left(t\right)\varphi\left(t\right)\right)' & = & \widetilde{u}'\left(t\right)\varphi\left(t\right)+\widetilde{u}\left(t\right)\varphi'\left(t\right)\quad\Bigg/\int_{I}dt\\
\implies\quad0 & = & \int_{I}g\left(t\right)\varphi\left(t\right)dt+\int_{I}\widetilde{u}\left(t\right)\varphi'\left(t\right)dt,
\end{eqnarray*}
 by $\left(2\right)$ we have: $\int_{I}g\left(t\right)\varphi\left(t\right)dt+\int_{I}u\left(t\right)\varphi'\left(t\right)dt=0$.\\
Subtract: $\int_{I}\left[u\left(t\right)-\widetilde{u}\left(t\right)\right]\varphi'\left(t\right)=0$
$\forall\varphi\left(t\right)\in D\left(I\right)$ $\stackrel{\mbox{L. 1.2, 2}}{\implies}$
$\exists x_{0}\in X$ s. t. $u\left(t\right)-\widetilde{u}\left(t\right)=x_{0}$
a. e. in $I$.\\
But $\widetilde{u}\left(t\right)=\int_{0}^{t}g\left(s\right)ds$,
hence $\left(1\right)$ holds.
\end{itemize}

\subsection*{Def.: {[}Weak derivative of $u\left(I\right):\to X${]}}

Let $u\left(t\right)$, $g\left(t\right)\in L^{1}\left(I;X\right)$.
We say that $g\left(t\right)$ is weak time derivative of $u\left(t\right)$,
if one (and hence any) of the assertions of L. 1.3 holds. We write
$\frac{d}{dt}u\left(t\right)=g\left(t\right)$. We then define $W^{1,p}\left(I;X\right)=\left\{ u\left(t\right)\in L^{p}\left(I;X\right),\frac{d}{dt}u\left(t\right)\in L^{p}\left(I;X\right)\right\} $.

\section{Geometry and duality in $L^{p}\left(I;X\right)$ spaces}

\subsection*{Recall: {[}Reflexive space, canonical embedding $X\to X^{**}${]}}

mapping $J:X\to\left(X^{*}\right)^{*}$ is isometric into. For $X$
reflexive, $J$ is onto.

$x\mapsto Jx$, where $\left\langle Jx,y\right\rangle _{X^{**},X^{*}}=\left\langle y,x\right\rangle _{X^{*},X}$
for $\forall y\in X^{*}$. Eberlein\textendash Śmulian: $X$ reflexive,
$\left\{ u_{n}\right\} \subset X$ bounded $\implies$ $\exists$
subseq. $\widetilde{u}_{n}$: $\exists u\in X$ s. t. $\widetilde{u}_{n}\rightharpoonup u$
(weak convergence, meaning $\left\langle y,\widetilde{u}_{n}\right\rangle \to\left\langle y,u\right\rangle $
for $\forall y\in X^{*}$).

\subsection*{Def.: {[}Strictly and uniformly convex space{]}}

$X$ is:
\begin{description}
\item [{strictly~convex}] $\left\Vert x\right\Vert ,\left\Vert y\right\Vert \leq1$,
$x\ne y$ $\implies$ $\left\Vert \frac{x+y}{2}\right\Vert <1$.
\item [{uniformly~convex}] $\forall\varepsilon>0$ $\exists\delta>0$
s. t. $\left\Vert x\right\Vert ,\left\Vert y\right\Vert \leq1$, $\left\Vert x-y\right\Vert \geq\varepsilon$
$\implies$$\left\Vert \frac{x+y}{2}\right\Vert <1-\delta$.
\end{description}

\subsection*{Theorem 1.6 {[}Relationship between weak and strong convergence in
UC spaces{]}}

Let $X$ be uniformly convex, let $x_{n}\rightharpoonup x$ and let
$\left\Vert x_{n}\right\Vert \to\left\Vert x\right\Vert $. Then $x_{n}\to x$.

\subsubsection*{Proof:}

WLOG: $x\ne0$, set $y_{n}=\frac{x_{n}}{\left\Vert x_{n}\right\Vert }$,
$y=\frac{x}{\left\Vert x\right\Vert }$, clearly $y_{n}\rightharpoonup y$,
$\left\Vert y_{n}\right\Vert \to\left\Vert y\right\Vert =1$. Assume
for contradiction $y_{n}\nrightarrow y$, hence $\exists\varepsilon>0$,
$\exists$ subseq $\widetilde{y}_{n}$ such that $\left\Vert \widetilde{y}_{n}-y\right\Vert \geq\varepsilon$
for $\forall n$. But $\widetilde{y}_{n}\rightharpoonup y=\frac{y+y}{2}\leftharpoonup\frac{y_{n}+y}{2}$
and since the norm is weakly lower semicontinuous, we have $\left\Vert y\right\Vert \leq\liminf_{n}\left\Vert \frac{y_{n}+y}{2}\right\Vert \stackrel{\mbox{\ensuremath{X} unif. conv.}}{\leq}1-\delta<1$,
by which we have obtained a contradiction.

\subsection*{Remark:}

general principle: $x_{n}\rightharpoonup x$, $A\left(x_{n}\right)\rightharpoonup A\left(x\right)$
for suitable $A$$\implies x_{n}\to x$.

\subsection*{Remark:}
\begin{itemize}
\item $X$ Hilbert $\implies$ $X$ uniformly convex,
\item $X$ uniformly convex $\implies$ $X$ reflexive,
\item $L^{p}\left(\Omega\right)$ is uniformly convex for $p\in\left(1,+\infty\right)$.
\end{itemize}

\subsection*{Theorem 1.7}

Let $X$ be uniformly convex, $p\in\left(1,+\infty\right)$. Then
$L^{p}\left(I;X\right)$ is uniformly convex.

\subsection*{Recall:}
\begin{itemize}
\item $p,p'\in[1,\infty)$ are Hölder conjugate $\iff\frac{1}{p}+\frac{1}{p'}=1$
\item Hölder inequality: $\int_{\Omega}\left|u\left(x\right)v\left(x\right)\right|\dx\leq\left(\int_{\Omega}\left|u\left(x\right)\right|^{p}\dx\right)^{\frac{1}{p}}\left(\int_{\Omega}\left|v\left(x\right)\right|^{p'}\right)^{\frac{1}{p'}}$
\item consequence: $v\left(x\right)\in L^{p'}\left(\Omega\right)$ \dots{}
$F_{v}\in\left(L^{p}\left(\Omega\right)\right)^{*}:u\left(\cdot\right)\mapsto\int_{\Omega}u\left(x\right)v\left(x\right)\dx$
for $\forall u\in L^{p}\left(\Omega\right)$.
\end{itemize}

\subsection*{Theorem 1.8 {[}Hölder inequality{]}}

Let $u\left(t\right)\in L^{p}\left(I;X\right)$, $v\left(t\right)\in L^{p'}\left(I;X^{*}\right)$,
$p,p'$ Hölder conjugate. Then the mapping $t\mapsto\left\langle v\left(t\right),u\left(t\right)\right\rangle _{X^{*},X}$
is measurable and 
\[
\int_{I}\left|\left\langle v\left(t\right),u\left(t\right)\right\rangle \right|dt\leq\left(\int_{I}\left\Vert u\left(t\right)\right\Vert _{X}^{p}\right)^{\frac{1}{p}}\left(\int_{I}\left\Vert v\left(t\right)\right\Vert _{X^{*}}^{p'}\right)^{\frac{1}{p'}}.
\]

\subsubsection*{Proof:}

$u\left(t\right)$, $v\left(t\right)$ are strongly measurable, $PS<\infty$.
Then $\exists u_{n}\left(t\right):I\to X$ simple such that $u_{n}\left(t\right)\to u\left(t\right)$
and $\exists v_{n}\left(t\right):I\to X$ simple such that $v_{n}\left(t\right)\to v\left(t\right)$
a. e. in $X,X^{*}$. Therefore the mapping $t\mapsto\left\langle v_{n}\left(t\right),u_{n}\left(t\right)\right\rangle $
is simple and measurable. Since $\left\langle v_{n}\left(t\right),u_{n}\left(t\right)\right\rangle \to\left\langle v\left(t\right),u\left(t\right)\right\rangle $
a. e. ($\left\langle \cdot,\cdot\right\rangle $ is continuous), the
mapping $t\mapsto\left\langle v\left(t\right),u\left(t\right)\right\rangle $
is measurable. Finally: $\left|\left\langle v\left(t\right),u\left(t\right)\right\rangle \right|\leq\left\Vert v\left(t\right)\right\Vert _{X^{*}}\left\Vert u\left(t\right)\right\Vert _{X}$
and by scalar Hölder inequality we are done.

\subsection*{Corollary:}

$v\left(t\right)\in L^{p'}\left(I;X^{*}\right)\implies F_{v}:u\mapsto\int_{I}\left\langle v\left(t\right),u\left(t\right)\right\rangle _{X^{*},X}$
is well defined for $\forall u\left(t\right)\in L^{p}\left(I;X\right)$.

\subsection*{Theorem 1.9 {[}Characterization of $L^{p}\left(I,X\right)$ dual{]}}

Let $X$ be reflexive, separable, let $p\in[1,+\infty)$. Denote $\mathscr{X}=L^{p}\left(I;X\right)$.
Then for any $F\in\mathscr{X}^{*}$ there is $v\left(t\right)\in L^{p'}\left(I,X^{*}\right)$
such that 
\[
\left\langle F,u\left(t\right)\right\rangle _{\mathscr{X}^{*},\mathscr{X}}=\int_{I}\left\langle v\left(t\right),u\left(t\right)\right\rangle _{X^{*},X}\,dx\quad\forall u\left(t\right)\in\mathscr{X}.
\]
 Moreover, $v\left(t\right)$ is uniquely defined and $\left\Vert v\left(t\right)\right\Vert _{L^{p'}\left(I;X\right)}=\left\Vert F\right\Vert _{\mathscr{X}^{*}}$.

\subsubsection*{Proof.}

\textbf{Step 1:} For $\tau\in I$ and $x\in X$ denote $u_{\tau,x}(t)=\chi_{[0,\tau]}(t)x$.
Clearly $u_{\tau,x}(t)\in\mathscr{X}$ and $x\mapsto\left\langle F,u_{\tau,x}(\cdot)\right\rangle _{\mathscr{X}^{*},\mathscr{X}}$
is linear continuous. Hence there is $g(\tau)\in X^{*}$ such that
$\left\langle F,u_{\tau,x}(\cdot)\right\rangle _{\mathscr{X}^{*},\mathscr{X}}=\left\langle g(\tau),x\right\rangle _{X^{*},X}$
for all $x\in X$. 

We will show there is $v(t)\in L^{p'}(I;X^{*})$ such that 
\begin{align}
g(\tau) & =\int_{0}^{\tau}v(t)\,dt\label{T1.9g1}\\
\left\Vert v(t)\right\Vert _{L^{p'}(I;X^{*})} & \le\left\Vert F\right\Vert _{\mathscr{X}^{*}}\label{T1.9g2}
\end{align}
Observe that with this we are done: (\ref{T1.9g1}) implies that 
\[
\left\langle F,u_{\tau,x}(\cdot)\right\rangle _{\mathscr{X}^{*},\mathscr{X}}=\left\langle \int_{0}^{\tau}v(t)\,dt,x\right\rangle _{X^{*},X}=\int_{I}\left\langle v(t),u_{\tau,x}(t)\right\rangle _{X^{*},X}\,dt
\]
By linearity we have 
\begin{equation}
\left\langle F,u(\cdot)\right\rangle _{\mathscr{X}^{*},\mathscr{X}}=\int_{I}\left\langle v(t),u(t)\right\rangle \,dt\label{T1.9eq}
\end{equation}
for any $u(t)=\sum_{j}\chi_{(\alpha_{j},\beta_{j}]}(t)x_{j}$. But
such functions are dense in $\mathscr{X}$, to which (\ref{T1.9eq})
extends, using continuity of $F$ on the left, and Hölder inequality
on the right. Furthermore, it now follows from (\ref{T1.9eq}) that
\[
\left\Vert F\right\Vert _{\mathscr{X^{*}}}=\sup_{\left\Vert u(t)\right\Vert _{\mathscr{X}}=1}\left\langle F,u(\cdot)\right\rangle _{\mathscr{X^{*},X}}=\sup_{\left\Vert u(t)\right\Vert _{\mathscr{X}}=1}\int_{I}\left\langle v(t),u(t)\right\rangle _{X^{*},X}\,dt\le\left\Vert v(t)\right\Vert _{L^{p'}(I;X^{*})}
\]
by Hölder inequality again. Together with (\ref{T1.9g2}) we obtain
$\left\Vert F\right\Vert _{\mathscr{X^{*}}}=\left\Vert v(t)\right\Vert _{L^{p'}(I;X^{*})}$;
this also implies that $v(t)$ is uniquely defined.

\medskip{}
\textbf{Step 2:} Towards proving (\ref{T1.9g1}), we first show that
$g(t):I\to X^{*}$ is absolutely continuous. Let $(\alpha_{j},\beta_{j})\subset I$
be disjoint. It follows from reflexivity of $X$ that there exist
$x_{j}\in X$ with $\left\Vert x_{j}\right\Vert =1$ such that $\left\Vert g(\beta_{j})-g(\alpha_{j})\right\Vert _{X^{*}}\left\langle g(\beta_{j})-g(\alpha_{j}),x_{j}\right\rangle _{X^{*},X}$. 

On the other hand 
\[
\left\langle g(\beta_{j})-g(\alpha_{j}),x_{j}\right\rangle _{X^{*},X}=\left\langle F,u_{\beta_{j},x_{j}}(\cdot)-u_{\alpha_{j},x_{j}}(\cdot)\right\rangle _{\mathscr{X^{*},X}}=\left\langle F,\chi_{(\alpha_{j},\beta_{j}]}(\cdot)x_{j}\right\rangle _{\mathscr{X}^{*},\mathscr{X}}
\]
Hence 
\begin{equation}
\begin{aligned}\sum_{j}\left\Vert g(\beta_{j})-g(\alpha_{j})\right\Vert _{X^{*}} & =\left\langle F,\sum_{j}\chi_{(\alpha_{j},\beta_{j}]}(\cdot)x_{j}\right\rangle _{\mathscr{X}^{*},\mathscr{X}}\\
 & \le F_{\mathscr{X}^{*}}\left\Vert \sum_{j}\chi_{(\alpha_{j},\beta_{j}]}(t)x_{j}\right\Vert _{\mathscr{X}}=\left\Vert F\right\Vert _{\mathscr{X^{*}}}\left(\sum_{j}(\beta_{j}-\alpha_{j})\right)^{\frac{1}{p}}
\end{aligned}
\label{T1.9ac}
\end{equation}
Obviously, this implies $g(t)\in AC(I;X^{*})$.

\medskip{}
\textbf{Step 3:} By previous step and the fact that $g(0)=0$, we
see that (\ref{T1.9g1}) holds with some $v(t)\in L^{1}(I;X^{*})$.
It remains to establish (\ref{T1.9g2}). 

If $p=1$, observe that (\ref{T1.9ac}) implies $g(t)$ is lipschitz,
and in particular $v(t)=g'(t)$ a.~e. is essentially bounded by $\left\Vert F\right\Vert _{\mathscr{X}^{*}}$.
In other words, (\ref{T1.9g2}) holds with $p'=\infty$ as required. 

If $p\in(1,\infty)$, one can proceed as follows. As in Step 1 we
use linearity to deduce (\ref{T1.9eq}) for all $u(t)=\sum_{j}\chi_{(\alpha_{j},\beta_{j}]}(t)x_{j}$.
We now just have $v(t)\in L^{1}(I;X^{*})$, so the density argument
only extends to $u(t)\in L^{\infty}(I;X)$. 

We need one more limiting argument: set 
\[
v_{n}(t)=\begin{cases}
v(t),\quad\textrm{ if }\left\Vert v(t)\right\Vert _{X^{*}}\le n\\
0,\quad\textrm{otherwise}
\end{cases}
\]
and $u_{n}(t)=z(t)\left\Vert v_{n}(t)\right\Vert _{X^{*}}^{p'-1}$,
where $z(t)\in X$ are such that $\left\Vert z(t)\right\Vert _{X}=1$
and $\left\langle v(t),z(t)\right\rangle _{X^{*},X}=\left\Vert v(t)\right\Vert _{X^{*}}$.
Now $u_{n}(t)$ are essentially bounded, and $\left\Vert u_{n}(t)\right\Vert _{X}^{p}=\left\langle v(t),u_{n}(t)\right\rangle _{X^{*},X}=\left\Vert v_{n}(t)\right\Vert _{X^{*}}^{p'}$.
Plugging $u_{n}(t)$ into (\ref{T1.9eq}) gives, after a simple manipulation,
that 
\[
\left(\int_{I}\left\Vert v_{n}(t)\right\Vert _{X^{*}}^{p'}\,dt\right)^{\frac{1}{p'}}\le\left\Vert F\right\Vert _{\mathscr{X^{*}}}
\]
Since $\left\Vert v_{n}(t)\right\Vert _{X^{*}}\nearrow\left\Vert v(t)\right\Vert _{X^{*}}$,
estimate (\ref{T1.9g2}) follows by Levi theorem. 

\subsection*{Corollary:}

$X$ reflective, separable, $p\in\left(1,+\infty\right)$ $\implies$
$L^{p}\left(I;X\right)$ is reflexive and separable. Hence any sequence
bounded in $L^{p}\left(I;X\right)$ has a weakly convergent subsequence.

\section[More on weakly differentiable functions]{More on weakly differentiable functions:\protect \\
extensions, approximations, embeddings}

\subsection*{Recall: }

$u\left(t\right),g\left(t\right)\in L^{1}\left(I;X\right)$ \dots{}
$g\left(t\right)$ is a weak derivative of $u\left(t\right)$ $\left(\frac{d}{dt}u\left(t\right)=g\left(t\right)\right)$
iff (by L. 1.3) $\exists x_{0}\in X$ such that $u\left(t\right)=x_{0}+\int_{0}^{t}g\left(s\right)ds$
for a. e. $t\in I=\left[0,T\right]$ or $\iff\int_{I}u\left(t\right)\varphi'\left(t\right)dt=-\int_{I}g\left(t\right)\varphi\left(t\right)dt$
for $\forall\varphi\left(t\right)\in D\left(I\right)=C_{c}^{\infty}\left(I,\mathbb{R}\right)$.

\subsection*{Remark:}
\begin{itemize}
\item by the above results: $u\left(t\right)$ is weakly differentiable
$\iff$ $\exists\widetilde{u}\left(t\right)\in AC$ such that $\widetilde{u}\left(t\right)=u\left(t\right)$,
$\widetilde{u}'\left(t\right)=\frac{d}{dt}u\left(t\right)$ a. e.
\item in applications we will often have $u\left(t\right)\in L^{p}\left(I;Y\right)$,
$\frac{d}{dt}u\left(t\right)\in L^{q}\left(I;Z\right)$. This means:
$\exists X$ such that $Y,Z\subset X$ (often $Y\subset Z=X$). Then
$\frac{d}{dt}u\left(t\right)=g\left(t\right)$ in $L^{1}\left(I;X\right)$
and moreover $u\left(t\right)\in L^{p}\left(I;Y\right)$, $g\left(t\right)\in L^{q}\left(I;Z\right)$. 
\end{itemize}

\subsection*{Lemma 1.4 {[}Weak derivative of product and convolution{]}}

Let $u\left(t\right):I\to X$ be weakly differentiable. Then:
\begin{enumerate}
\item If $\eta\left(t\right):I\to\mathbb{R}$ is $C^{\infty}$,\footnote{Holds for lipschitz, proven for $C^{\infty}$.}
then $u\left(t\right)\eta\left(t\right):I\to X$ is weakly differentiable
and $\frac{d}{dt}\left(u\left(t\right)\eta\left(t\right)\right)=\left(\frac{d}{dt}u\left(t\right)\right)\eta\left(t\right)+u\left(t\right)\eta'\left(t\right)$.
\item If $\psi\left(t\right)\in D\left(I\right)$, then $u*\psi\left(t\right):I\to X$
is smooth and moreover $\left(u*\psi\right)'\left(t\right)=\left(\frac{d}{dt}u\right)*\psi\left(t\right)$,
whenever $t-\mbox{supp}\,\psi\subset\left(0,T\right)$.
\end{enumerate}

\subsubsection*{Proof:}
\begin{enumerate}
\item $\varphi\left(t\right)\in D\left(I\right)$ given:
\begin{eqnarray*}
\int_{I}\left[u\left(t\right)\eta\left(t\right)\right]\varphi'\left(t\right)dt & = & \int_{I}u\left(t\right)\underbrace{\eta\left(t\right)\varphi'\left(t\right)}_{{\mbox{\ensuremath{\Vert}}\makebox[0pt][l]{\ (smooth)}\atop \makebox[0pt][c]{\ensuremath{\left(\eta\left(t\right)\varphi\left(t\right)\right)'-\eta'\left(t\right)\varphi\left(t\right)}}}}dt\\
 & = & \underbrace{\int_{I}u\left(t\right)\underbrace{\left(\eta\left(t\right)\varphi\left(t\right)\right)'}_{\mbox{\ensuremath{\mbox{\ensuremath{\in}\,}}\makebox[0pt][l]{\ensuremath{\,D\left(I\right)}}}}dt}_{\mbox{\ensuremath{=}}\makebox[0pt][l]{\ensuremath{-\int_{I}g\left(t\right)\eta\left(t\right)\varphi\left(t\right)dt,\ g\left(t\right)=\frac{d}{dt}u\left(t\right)}}}-\int_{I}u\left(t\right)\eta'\left(t\right)\varphi\left(t\right)dt\\
 & = & -\int_{I}\left[\frac{d}{dt}u\left(t\right)\eta\left(t\right)+u\left(t\right)\eta'\left(t\right)\right]\varphi\left(t\right)dt
\end{eqnarray*}
\item extend $u\left(s\right)=0$ for $s\notin I$ (not neccessary if $t-\mbox{supp}\,\psi\subset I$).
Let $t\in I$ be fixed.
\begin{eqnarray*}
u*\psi\left(t\right) & = & \int_{\mathbb{R}}u\left(t-s\right)\psi\left(s\right)ds\stackrel{\mbox{subst.}}{=}\int_{\mathbb{R}}u\left(s\right)\psi\left(t-s\right)ds\\
 & = & \int_{I}u\left(s\right)\psi\left(t-s\right)ds\\
\makebox[0pt][r]{we know: }\left(u*\psi\right)'\left(t\right) & = & \int_{I}u\left(s\right)\underbrace{\psi'\left(t-s\right)}_{\mbox{\ensuremath{=}}\makebox[0pt][l]{\ensuremath{\,-\varphi'\left(s\right)}, where \ensuremath{\varphi\left(s\right)=\psi\left(t-s\right)}}}ds\\
 & = & \int_{I}g\left(s\right)\varphi\left(s\right)ds=\int_{I}\frac{d}{dt}u\left(s\right)\psi\left(t-s\right)=PS
\end{eqnarray*}
\end{enumerate}

\subsection*{Theorem 1.10 {[}Extension operator for weak derivative{]}}

Let $u\left(t\right)\in L^{p}\left(I,Y\right)$, $\frac{d}{dt}u\left(t\right)\in L^{q}\left(I;Z\right)$,
where $I=\left[0,T\right]$. Denote $I_{\varDelta}=\left[-\varDelta,T+\varDelta\right]$
with $\varDelta>0$. Then there is $Eu\left(t\right)\in L^{p}\left(I_{\varDelta};Y\right)$,
$\frac{d}{dt}Eu\in L^{q}\left(I_{\varDelta},Z\right)$ and such that
$Eu\left(t\right)=u\left(t\right)$, $\frac{d}{dt}Eu\left(t\right)=\frac{d}{dt}u$
for a. a. $t\in I$. Moreover, $\left\Vert Eu\left(t\right)\right\Vert _{L^{p}\left(I_{\Delta},Y\right)}\leq C\left\Vert u\left(t\right)\right\Vert _{L^{p}\left(I_{\Delta},Y\right)}$
and $\left\Vert \frac{d}{dt}Eu\left(t\right)\right\Vert _{L^{q}\left(I_{\Delta},Y\right)}\leq C\left\Vert \frac{d}{dt}u\left(t\right)\right\Vert _{L^{q}\left(I_{\Delta},Y\right)}$
for suitable $C>0$.

\subsubsection*{Proof:}

\textbf{Trick:} use even extension as on Fig. \ref{fig:Ev-ext}: $\widetilde{u}\left(t\right)=\begin{cases}
u\left(t\right) & \quad t\in\left[0,T\right]\\
u\left(-t\right) & \quad t\in[-T,0)
\end{cases}$.

\begin{figure}
\includegraphics[width=0.7\textwidth]{img/k1_even_ext}

\caption{\label{fig:Ev-ext}Even extension}
\end{figure}

Claim: $\widetilde{u}\left(t\right)$ is weakly differentiable and
$\frac{d}{dt}\widetilde{u}\left(t\right)=\widetilde{g}\left(t\right)$
in $\left[-T,T\right]$,\\
where $\widetilde{g}\left(t\right)=\begin{cases}
g\left(t\right) & \quad t\in\left[0,T\right]\\
-g\left(-t\right) & \quad t\in\left[-T,0\right)
\end{cases}$ with $g\left(t\right)=\frac{d}{dt}u\left(t\right)$.

Note: $t\in\left[-T,0\right]\implies$
\begin{eqnarray*}
\widetilde{u}\left(t\right) & \stackrel{\mbox{L. 1.3, 1}}{=} & x_{0}+\int_{0}^{-t}g\left(s\right)ds=\underbrace{x_{0}+\int_{0}^{T}g\left(s\right)ds}_{\widetilde{x}_{0}\in X}-\underbrace{\int_{-t}^{T}g\left(s\right)ds}_{\mbox{subst. }\mbox{\ensuremath{s=-\tau}}}\\
 & = & \widetilde{x}_{0}-\int_{-T}^{t}\widetilde{g}\left(-\tau\right)d\tau=\widetilde{x}_{0}+\int_{-T}^{t}\widetilde{g}\left(s\right)ds,
\end{eqnarray*}
hence we have shown: $\widetilde{u}\left(t\right)=\widetilde{x}_{0}+\int_{-T}^{t}\widetilde{g}\left(s\right)ds$,
$t\in\left[-T,0\right)$, but also $u\left(t\right)=x_{0}+\int_{0}^{t}g\left(s\right)ds$,
therefore $\widetilde{u}\left(t\right)=\widetilde{x}_{0}+\int_{-T}^{t}\widetilde{g}\left(s\right)ds$,
$t\in\left[-T,T\right]$. Clearly $\left\Vert \widetilde{u}\left(t\right)\right\Vert _{L^{p}\left(\left[-T,2T\right];Y\right)}=3\left\Vert u\right\Vert _{L^{p}\left(I;Y\right)}$,
$\left\Vert \frac{d\widetilde{u}}{dt}\right\Vert _{L^{q}\left(\left[-T,2T\right],Z\right)}=3\left\Vert \frac{d}{dt}u\right\Vert _{L^{q}\left(I;Z\right)}$.

\subsection*{Remarks:}
\begin{itemize}
\item extension operator $u\mapsto Eu$ is linear,
\item we can have $Eu\left(t\right)=0$ for $t$ outside $I_{\varDelta/2}$
(multiply by cutoff function $\eta\left(t\right)$: L. 1.4,1).
\end{itemize}

\subsection*{Theorem 1.11 {[}Smooth approximation of weakly differentiable functions{]}}

Let $u\left(t\right)\in L^{p}\left(I;Y\right)$, $\frac{d}{dt}u\left(t\right)\in L^{q}\left(I;Z\right)$.
Then there exist functions $u_{n}\left(t\right)\in C^{1}\left(I;Y\right)$
such that $u_{n}\left(t\right)\to u\left(t\right)$ in $L^{p}\left(I;Y\right)$,
$u_{n}'\left(t\right)\to\frac{d}{dt}u\left(t\right)$ in $L^{q}\left(I;Z\right)$.

\subsubsection*{Proof:}

Set $u_{n}\left(t\right)=Eu*\psi_{n}\left(t\right)$, where $\psi_{n}\left(t\right)=n\psi_{0}\left(nt\right)$,
$\psi_{0}\left(t\right)\in C_{c}^{\infty}\left(\mathbb{R},\mathbb{R}\right)$,
$\int_{\mathbb{R}}\psi_{0}\left(t\right)dt=1$, $\psi_{0}\left(t\right)=0$
outside $\left[-1,1\right]$. $Eu\left(t\right)$ is defined for $t\in\left[-\varDelta,T+\varDelta\right]$
\dots{} from Th. 1.10, set $Eu\left(t\right)=0$ elsewhere.

Observe: $u_{n}\left(t\right)\in C^{1}\left(I,Y\right)$ \dots{} ex.
1.3.1, $u_{n}\left(t\right)\to Eu\left(t\right)$, $n\to\infty$ in
$L^{p}\left(I_{\varDelta},Y\right)$ \dots{} L 1.1,4 and $Eu\left(t\right)=u\left(t\right)$
in $\left[0,T\right]$, hence $u_{n}\left(t\right)\to u\left(t\right)$
in $L^{p}\left(I;Y\right)$. Moreover, $u_{n}'\left(t\right)=\left(\frac{d}{dt}Eu\right)*\psi_{n}\left(t\right)\to\left(\frac{d}{dt}Eu\left(t\right)\right)*\psi_{n}$
in $L^{q}\left(I;Z\right)$ by L. 1.1,4 and $\frac{d}{dt}Eu\left(t\right)=\frac{d}{dt}u\left(t\right)$
in $I$, hence $u_{n}'\left(t\right)\to\frac{d}{dt}u\left(t\right)$
in $L^{q}\left(I;Z\right)$.

\subsection*{Def.: {[}Gelfand triple{]}}

Let $X$ be separable, reflexive, densely embedded into a Hilbert
space $H$. Then, by Gelfand triple we mean $X\subset H=H^{*}\subset X^{*}$.

\subsubsection*{Commentary:}

Riesz: $H=H^{*}$ (identification): $\forall y\in H^{*}$ $\exists!\widetilde{y}\in H$
such that $\left\langle y,x\right\rangle _{H^{*},H}=\left(\widetilde{y},x\right)_{H}$
$\forall x\in H$, where $\left(\cdot,\cdot\right)_{H}$ is scalar
product in $H$. $X\subset H\implies H^{*}\subset X^{*}$, in particular
$X\subset X^{*}$ via the embedding: $x_{0}\mapsto x_{0}^{*}$, where
$\left\langle x_{0}^{*},x\right\rangle _{X^{*},X}=\left(x_{0},x\right)_{H}$
$\forall x\in X$. Can be shown: injective and on the dense set, see
ex. 3.3. 

\subsubsection*{Application:}

we will often work with $u\left(t\right)\in L^{p}\left(I;X\right)$,
$\frac{d}{dt}u\left(t\right)\in L^{q}\left(I;X^{*}\right)$.

\subsection*{Lemma 1.5 {[}Weak representative for $W^{1,p}\left(I;X\right)${]}}

$W^{1,p}\left(I;X\right)\subset C\left(I,X\right)$ in the sense of
representative: for any $u\left(t\right)\in W^{1,p}\left(I,X\right)$
there is $\widetilde{u}\left(t\right)\in C\left(I,X\right)$ such
that $\widetilde{u}\left(t\right)=u\left(t\right)$ a. e. and $\left\Vert \widetilde{u}\left(t\right)\right\Vert _{C\left(I,X\right)}\leq c\left\Vert u\left(t\right)\right\Vert _{W^{1,p}\left(I;X\right)}$.

\subsection*{Remark:}

One even has $W^{1,p}\left(I;X\right)\hookrightarrow C^{0,\alpha}\left(I,X\right)$,
where $\alpha=1-\frac{1}{p}$ (HW 2).

\subsubsection*{Proof:}

\textbf{Step 1:} $u\left(t\right)\in C^{1}\left(I,X\right)$ \dots{}
fix $t\in I$, WLOG let $t\in\left[\frac{T}{2},T\right]$, where $I=\left[0,T\right]$.
Then $u\left(t\right)=u\left(\tau\right)+\int_{\tau}^{t}\frac{d}{ds}u\left(s\right)ds$
for $\tau\in\left[0,\frac{t}{2}\right]$. Therefore

\begin{eqnarray*}
\left\Vert u\left(t\right)\right\Vert  & \leq & \left\Vert u\left(\tau\right)\right\Vert +\int_{I}\left\Vert \frac{d}{dt}u\left(s\right)\right\Vert ds\quad/\int_{0}^{T/2}d\tau\\
\frac{T}{2}\left\Vert u\left(t\right)\right\Vert  & \leq & \int_{0}^{\frac{T}{2}}\left\Vert u\left(\tau\right)\right\Vert d\tau+\frac{T}{2}\int_{I}\left\Vert \frac{d}{dt}u\left(s\right)\right\Vert ds\\
 & \leq & c\left\Vert u\left(t\right)\right\Vert _{W^{1,1}\left(I;X\right)}\leq\widetilde{c}\left\Vert u\right\Vert _{W^{1,p}\left(I;X\right)}
\end{eqnarray*}
\[
\implies\left\Vert u\left(t\right)\right\Vert _{C\left(I;X\right)}\leq c\left\Vert u\left(t\right)\right\Vert _{W^{1,p}\left(I;X\right)}\quad\left(\#\right)
\]
 $c=c\left(T\right)$ for any $u\left(t\right)\in C^{1}\left(I,X\right)$

\textbf{Step 2:} $u\left(t\right)\in W^{1,p}\left(I;X\right)$ arbitrary.
Th. 1.11 $\implies$ $\exists u_{n}\left(t\right)\in C^{1}\left(I;X\right)$,
$u_{n}\left(t\right)\to u\left(t\right)$ in $W^{1,p}\left(I;X\right)$.
Hence $\left\{ u_{n}\left(t\right)\right\} $ is cauchy in $W^{1,p}\left(I;X\right)$
and we apply $\left(\#\right)$ to $u_{n}\left(t\right)-u_{m}\left(t\right)$
$\implies$ $\left\{ u_{n}\left(t\right)\right\} $ is cauchy in $C\left(I,X\right)$,
hence $u_{n}\left(t\right)\rightrightarrows\widetilde{u}\left(t\right)$,
where $\widetilde{u}\left(t\right)\in C\left(I,X\right)$ is the continuous
representative.

\subsection*{Theorem 1.12 {[}Continuous representative for $L^{p}\left(I;X\right)$
with $L^{p'}\left(I,X^{*}\right)$ derivative{]}}

Let $X\subset H=H^{*}\subset X^{*}$ be Gelfand triple. Let $u\left(t\right)\in L^{p}\left(I;X\right)$,
$\frac{d}{dt}u\left(t\right)\in L^{p'}\left(I;X^{*}\right)$, where
$p,p'$ are Hölder conjugate. Then
\begin{enumerate}
\item $u\left(t\right)\in C\left(I,H\right)$ in the sense of representative:\\
$\exists\widetilde{u}\left(t\right)\in C\left(I,H\right)$ such that
$\left\Vert \widetilde{u}\right\Vert _{C\left(I,H\right)}\leq c\left(\left\Vert u\right\Vert _{L^{p}\left(I;X\right)}+\left\Vert \frac{d}{dt}u\right\Vert _{L^{p'}\left(I;X^{*}\right)}\right)$
and $\widetilde{u}\left(t\right)=u\left(t\right)$ a. e. in $I$.
\item $t\mapsto\left\Vert u\left(t\right)\right\Vert _{H}^{2}$ is weakly
differentiable with $\frac{d}{dt}\left\Vert u\left(t\right)\right\Vert _{H}^{2}=2\left\langle \frac{d}{dt}u\left(t\right),u\left(t\right)\right\rangle _{X^{*},X}$
a. e. in $I$, in particular, $\left\Vert \widetilde{u}\left(t_{2}\right)\right\Vert _{H}^{2}=\left\Vert \widetilde{u}\left(t_{1}\right)\right\Vert _{X}^{2}+\int_{t_{1}}^{t_{2}}2\left\langle \frac{d}{dt}u\left(s\right),u\left(s\right)\right\rangle _{X^{*},X}ds$
where $\widetilde{u}\left(t\right)$ is the continuous representative.
\end{enumerate}

\subsubsection*{Proof:}
\begin{enumerate}
\item \textbf{Step 1:} $u\left(t\right)\in C^{1}\left(I,X\right)$. \textbf{Trick:}
$u\left(t\right)=\underbrace{u\left(t\right)\theta\left(t\right)}_{u_{1}\left(t\right)}+\underbrace{u\left(t\right)\left(1-\theta\left(t\right)\right)}_{u_{2}\left(t\right)}$,
$\theta\left(t\right):I\to\mathbb{R}$ is smooth, $\theta\left(0\right)=0$,
$\theta\left(T\right)=1$, e. g. $\theta\left(t\right)=\frac{t}{T}.$
\begin{eqnarray*}
\left\Vert u_{1}\left(t\right)\right\Vert _{H}^{2} & = & \left(u_{1}\left(t\right),u_{1}\left(t\right)\right)_{H}-\underbrace{\left(u_{1}\left(0\right),u_{1}\left(0\right)\right)_{H}}_{0}\\
 & = & \int_{0}^{t}\frac{d}{ds}\left(u_{1}\left(s\right),u_{1}\left(s\right)\right)ds\\
 & = & \int_{0}^{t}2\Big(\underbrace{u_{1}'\left(s\right)}_{\makebox[0pt][r]{\ensuremath{u'\left(s\right)\theta\left(s\right)+u\left(s\right)\theta'\left(s\right)\quad\quad\quad\quad}}},\underbrace{u_{1}\left(s\right)}_{\makebox[0pt][l]{\ensuremath{\quad u\left(s\right)\theta\left(s\right)}}}\Big)ds\\
 & = & 2\int_{0}^{t}\left(u'\left(s\right),u\left(s\right)\right)_{H}\theta^{2}\left(s\right)+\left(u\left(s\right),u\left(s\right)\right)_{H}\theta'\left(s\right)\theta\left(s\right)ds\\
\left({\mbox{Gelfand emb.}\atop \iota:X\to X^{*}}\right) & = & 2\int_{0}^{t}\left\langle \iota u'\left(s\right),u\left(s\right)\right\rangle _{X^{*},X}\theta\left(s\right)+\left\langle \iota u\left(s\right),u\left(s\right)\right\rangle _{X^{*},X}\theta'\left(s\right)\theta\left(s\right)ds\\
 & \leq & C\int_{I}\left|\left\langle \iota u'\left(s\right),u\left(s\right)\right\rangle \right|+\left|\left\langle \iota u\left(s\right),u\left(s\right)\right\rangle \right|ds\\
\noalign{\mbox{"generalized scalar product": Gelfand: \ensuremath{\left\langle \iota u,v\right\rangle _{X^{*},X}=\left(u,v\right)_{H}} \dots\ use Hölder}}\\
 & \leq & C\left(\left\Vert \iota u'\right\Vert _{L^{p'}\left(I;X\right)}\left\Vert u\right\Vert _{L^{p}\left(I;X\right)}\right)+\underbrace{\left\Vert \iota u\right\Vert _{L^{\infty}\left(I;X^{*}\right)}}_{\mbox{\ensuremath{\stackrel{\mbox{L. 1.1}}{\leq}}}\mbox{\,\ensuremath{C\left\Vert u\right\Vert _{W^{1,1}\left(I;X^{*}\right)}}}}\left\Vert u\right\Vert _{L^{1}\left(I;X\right)}\\
 & \leq & C\left(\left\Vert u\right\Vert _{L^{p}\left(I;X\right)}+\left\Vert \frac{d}{dt}\iota u\right\Vert _{L^{p'}\left(I;X\right)}\right)^{2}
\end{eqnarray*}
And therefore part 1 holds for $u\left(t\right)\in C^{1}\left(I,X\right)$,
namely we have: \\
$\left(\sun\right)$ $\sup_{t\in\left[0,T\right]}\left\Vert u\left(t\right)\right\Vert _{H}\leq c\left(\left\Vert u\right\Vert _{L^{p}\left(I;X\right)}+\left\Vert \frac{d}{dt}u\right\Vert _{L^{p}\left(I;X^{*}\right)}\right)$.
\\
\textbf{Step 2:} let $u\left(t\right)\in L^{p}\left(I;X\right)$ with
$\frac{d}{dt}u\left(t\right)\in L^{p'}\left(I;X^{*}\right)$ be given.
By Th. 1.11 there exists $u_{n}\in C^{1}\left(I;X\right)$ such that
$u_{n}\left(t\right)\to u\left(t\right)$ in $L^{p}\left(I;X\right)$
and $\iota u_{n}'\left(t\right)\to\frac{d}{dt}u\left(t\right)$ in
$L^{p'}\left(I;X^{*}\right)$. Apply $\left(\sun\right)$ to $u_{n}\left(t\right)-u_{m}\left(t\right)\in C^{1}\left(I,X\right)$,
hence $RHS$ is small for $m,n\geq n_{0}$ for $n_{0}$ large enough.
Therefore $LHS$ is small $\implies\left(B.C.\right)$ for $u_{n}\left(t\right):I\to H$
$\implies\exists\widetilde{u}\left(t\right)\in C\left(I,H\right)$
such that $u_{n}\left(t\right)\rightrightarrows\widetilde{u}\left(t\right)$
in $I$. Also: $u_{n}\left(t\right)\to u\left(t\right)$ a. e. in
$X$, hence $u\left(t\right)=\widetilde{u}\left(t\right)$ a. e.,
where $\widetilde{u}\left(t\right)$ is the representative.
\item Let $u_{n}\left(t\right)$ be smooth approximations: 
\begin{eqnarray*}
\frac{d}{dt}\left\Vert u_{n}\left(t\right)\right\Vert _{H}^{2} & = & \frac{d}{dt}\left(u_{n}\left(t\right),u_{n}\left(t\right)\right)=2\left(u_{n}'\left(t\right),u_{n}\left(t\right)\right)\\
 & = & 2\left\langle \iota u_{n}'\left(t\right),u_{n}\left(t\right)\right\rangle _{X^{*},X}
\end{eqnarray*}
For $\forall t_{1},t_{2}\in I$ we have 
\begin{eqnarray*}
\left\Vert u_{n}\left(t_{2}\right)\right\Vert _{H}^{2} & = & \left\Vert u_{n}\left(t_{1}\right)\right\Vert _{H}^{2}+2\int_{t_{2}}^{t_{1}}\left\langle \iota u_{n}'\left(s\right),u_{n}\left(s\right)\right\rangle _{X^{*},X}ds\\
 & \downarrow\makebox[0pt][l]{n\ensuremath{\to\infty}}\\
\left\Vert \widetilde{u}\left(t_{2}\right)\right\Vert _{H}^{2} & = & \left\Vert \widetilde{u}\left(t_{1}\right)\right\Vert _{H}^{2}+2\int_{t_{2}}^{t_{1}}\left\langle \iota u'\left(s\right),u\left(s\right)\right\rangle _{X^{*},X}ds
\end{eqnarray*}
by Hölder inequality (ex.: $u_{n}\left(t\right)\to u\left(t\right)$
in $L^{p}\left(I;X\right)$, $v_{n}\left(t\right)\to v\left(t\right)$
in $L^{p'}\left(I;X^{*}\right)$ which implies the convergence $\int_{I}\left\langle v_{n}\left(s\right),u_{n}\left(s\right)\right\rangle ds\to\int_{I}\left\langle v\left(s\right),u\left(s\right)\right\rangle ds$,
see ex. 2.5.
\end{enumerate}

\subsection*{Notation:}
\begin{description}
\item [{embedding}] $X\hookrightarrow Y$: $X\subset Y$, $\exists c>0$
such that $\left\Vert u\right\Vert _{Y}\leq c\left\Vert u\right\Vert _{X}$
\item [{compact~embedding}] $X\hookrightarrow\hookrightarrow Y$: $X\hookrightarrow Y$
and for any $\left\{ u_{n}\right\} \subset X$ bounded there is a
strongly convergent (in $Y$) subsequence.
\end{description}

\subsection*{Remarks:}
\begin{itemize}
\item Th. 1.12: $\left\{ u\left(t\right)\in L^{p}\left(I;X\right),\frac{d}{dt}u\left(t\right)\in L^{p'}\left(I;X\right)\right\} \hookrightarrow C\left(I,H\right)$
in the sense of representative.
\item Gelfand triple: $X\hookrightarrow H\stackrel{\sim}{=}H^{*}\hookrightarrow X^{*}$,
also $X\hookrightarrow X^{*}$, more precisely: the embedding of $X$
into $X^{*}$ is given by $\iota X\to X^{*}$, where $\left\langle \iota u,v\right\rangle _{X^{*},X}=\left(u,v\right)_{H}$,
$u,v\in X$, because: $\frac{d}{dt}u\left(t\right)=g\left(t\right)\iff\int_{I}\iota u\left(t\right)\varphi'\left(t\right)dt=-\int g\left(t\right)\varphi\left(t\right)dt$
$\forall\varphi\left(t\right)\in D\left(I\right)$.
\end{itemize}

\subsection*{Lemma 1.6 {[}Ehrling lemma{]}}

Let $Y\hookrightarrow\hookrightarrow X\hookrightarrow Z$. Then for
any $a>0$ there is $C>0$ such that $\left\Vert u\right\Vert _{X}\leq\alpha\left\Vert u\right\Vert _{Y}+C\left\Vert u\right\Vert _{Z}$.

\subsubsection*{Proof:}

for contradiction assume that $\forall n\ \exists u_{n}$ such that
$\left\Vert u_{n}\right\Vert _{X}>a\left\Vert u_{n}\right\Vert _{Y}+n\left\Vert u_{n}\right\Vert _{Z}$,
WLOG $\left\Vert u_{n}\right\Vert _{X}=1$, hence $1>a\left\Vert u_{n}\right\Vert _{Y}+n\left\Vert u_{n}\right\Vert _{Z}$.
Therefore $\left\{ u_{n}\right\} $ is bounded in $Y\implies\exists$subsequence
$\widetilde{u}_{n}\to u\in X$, $\left\Vert u\right\Vert _{X}=1$.
But: $\frac{1}{n}>\left\Vert u_{n}\right\Vert _{Z}$, $u_{n}\to0$
in $Z$. Therefore $u=0$ in $Z$ and thus we have obtained a contradiction:
$u=0$ in $Z$ $\implies$ $u=0$ in $X$ since $X\hookrightarrow Z$

\subsection*{Theorem 1.13 {[}Aubin\textendash Lions lemma{]}}

Let $Y\hookrightarrow\hookrightarrow X\hookrightarrow Z$, where $X$,
$Z$ are reflexive and separable. Let $p,q\in\left(1,\infty\right)$.
Then for any sequence $u_{n}\left(t\right)$ bounded in $L^{p}\left(I;Y\right)$
such that $\frac{d}{dt}u_{n}\left(t\right)$ are bounded in $L^{q}\left(I;Z\right)$
there is a subsequence converging strongly in $L^{p}\left(I;X\right)$.

(Briefly: $\left\{ u\left(t\right)\in L^{p}\left(I,Y\right),\ \frac{d}{dt}u\left(t\right)\in L^{q}\left(I;Z\right)\right\} \hookrightarrow\hookrightarrow L^{p}\left(I;X\right)$.)

\subsubsection*{Proof:}

\textbf{Step 1:} $L^{p}\left(I;Y\right)$, $L^{q}\left(I;Z\right)$
are reflexive, by Eberlein\textendash Śmulian $\exists$ subsequence
such that $u_{n}\left(t\right)\rightharpoonup u\left(t\right)$ in
$L^{p}\left(I;Y\right)$ and $\frac{d}{dt}u_{n}\left(t\right)\rightharpoonup g\left(t\right)$
in $L^{q}\left(I;Z\right)$ and $\frac{d}{dt}u\left(t\right)=g\left(t\right)$.
We will show: $u_{n}\left(t\right)\to u\left(t\right)$ in $L^{p}\left(I;X\right)$.
WLOG $u\left(t\right)=0$ \dots{} otherwise subtract the limit (see
ex.~2.1). 

Subproblem: $u_{n}\left(t\right)\rightharpoonup0$ in $L^{p}\left(I;Y\right)$
and $\frac{d}{dt}u_{n}\left(t\right)\rightharpoonup0$ in $L^{q}\left(I;Z\right)$
$\implies$ $u_{n}\left(t\right)\to0$ in $L^{p}\left(I;X\right)$.

\textbf{Step 2:} $u_{n}\to0$ (strongly) in $Z$ for any $t\in I$
fixed. \textbf{Trick:} $u_{n}\left(t\right)=u_{n}\left(t+s\right)-\int_{t}^{t+s}\frac{d}{dt}u_{n}\left(\sigma\right)d\sigma$,
$s\in\left(t,t+\delta\right)$. Integrate $\int_{0}^{\delta}\cdot ds$,
multiply by $\frac{1}{\delta}$: 
\[
u_{n}\left(t\right)=\underbrace{\frac{1}{\delta}\int_{0}^{\delta}u_{n}\left(t+s\right)ds}_{I_{1n}}-\underbrace{\frac{1}{\delta}\int_{0}^{\delta}\int_{t}^{t+s}u_{n}\left(\sigma\right)d\sigma ds}_{I_{2n}}
\]

$I_{1n}$: recall: $x_{n}\rightharpoonup0$ in $X$, $L:X\to Y$ linear
continuous $\implies Lx_{n}\rightharpoonup0$ in $Y$ (see ``Properties
of weak convergence'' in appendices). Observe: $u\mapsto\frac{1}{\delta}\int_{0}^{\delta}u\left(t+s\right)ds$
is linear continuous $L^{p}\left(I;Y\right)\to Y\hookrightarrow\hookrightarrow Z$.
Recall: $x_{n}\rightharpoonup0$ in $Y$, $Y\hookrightarrow\hookrightarrow Z\implies x_{n}\to0$
in $Z$. Hence $I_{1n}\to0$ in $Z$.

$I_{2n}$: 
\begin{eqnarray*}
\int_{0}^{\delta}\left(\int_{t}^{t+s}\frac{d}{dt}u_{n}\left(\sigma\right)d\sigma\right)ds & \stackrel{\makebox[0pt][c]{ Fub.}}{=} & \int_{t}^{t+\delta}\left(\int_{\sigma-t}^{\sigma}\frac{d}{dt}u_{n}\left(\sigma\right)ds\right)d\sigma\\
 & = & \int_{t}^{t+\delta}\left(\sigma-\sigma+t\right)\frac{d}{dt}u_{n}\left(\sigma\right)d\sigma
\end{eqnarray*}
\begin{eqnarray*}
\implies\left\Vert I_{2n}\right\Vert _{Z} & \leq & \int_{t}^{t+\delta}\left\Vert \frac{d}{dt}u_{n}\left(\sigma\right)\right\Vert _{Z}d\sigma\\
\left(\mbox{Hölder}\right) & \leq & \left(\int_{t}^{t+\delta}\left\Vert \frac{d}{dt}u_{n}\left(\sigma\right)\right\Vert _{Z}^{q}d\sigma\right)^{\frac{1}{q}}\left(\int_{_{t}}^{t+\delta}1^{q'}d\sigma\right)^{\frac{1}{q'}}\\
 & \leq & \left\Vert \frac{d}{dt}u_{n}\right\Vert _{L^{q}\left(I;Z\right)}\delta^{1-\frac{1}{\delta}}\leq K\delta^{1-\frac{1}{\delta}}
\end{eqnarray*}
where $1-\frac{1}{q}>0$ since $q>1$.

Therefore for $\varepsilon>0$ given choose $\delta>0$ small such
that $\left\Vert I_{2n}\right\Vert _{Z}\leq\frac{\varepsilon}{2}$
for $\forall n$. Then choose $n_{0}$ such that $\left\Vert I_{1n}\right\Vert _{Z}<\frac{\varepsilon}{2}$
if $n>n_{0}$ $\implies$$\left\Vert u_{n}\left(t\right)\right\Vert _{Z}<\varepsilon$
for $n>n_{0}$.

\textbf{Step 3:} $u_{n}\left(t\right)\to0$ in $L^{p}\left(I;X\right)$,
$\varepsilon>0$ given. Use L. 1.6: $\left\Vert u_{n}\left(t\right)\right\Vert _{X}\leq a\left\Vert u_{n}\left(t\right)\right\Vert _{Y}+C\left\Vert u_{n}\left(t\right)\right\Vert _{Z}$.
$\Delta$-inequality: $\left\Vert u_{n}\right\Vert _{L^{p}\left(I;X\right)}\leq a\left\Vert u_{n}\right\Vert _{L^{p}\left(I;Y\right)}+c\left\Vert u_{n}\right\Vert _{L^{p}\left(I;Z\right)}$.
Choose $a=\frac{\varepsilon}{2k}$, where $\left\Vert u_{n}\right\Vert _{L^{p}\left(I;Y\right)}\leq K$.
But: $\left\Vert u_{n}\left(t\right)\right\Vert _{L^{p}\left(I;Z\right)}=\left(\int_{I}\left\Vert u_{n}\left(t\right)\right\Vert _{Z}^{p}dt\right)^{\frac{1}{p}}\to0$
by Lebesgue. Since $\left\Vert u_{n}\left(t\right)\right\Vert _{Z}\to0$
by step 2 and $\left\Vert u_{n}\left(t\right)\right\Vert _{Z}\leq C$
by embedding $W^{1,1}\left(I;Z\right)\hookrightarrow C\left(I,Z\right)$
is $u_{n}$ is bounded in $W^{1,1}\left(I;Z\right)$, hence second
term $<\frac{\varepsilon}{2}$ for $n$ large, therefore $\left\Vert u_{n}\right\Vert _{L^{p}\left(I;X\right)}\leq\varepsilon$
for $n$ large.

\chapter{Parabolic second order equations}

\begin{eqnarray*}
\left(P1\right)\quad\partial_{t}u-\mbox{div}\,a\left(\nabla u\right)+f\left(u\right) & = & h\left(t,x\right),\quad\left(t,x\right)\in I\times\Omega\\
\left(P2\right)\quad u & = & u_{0},\quad t=0,\ x\in\Omega\\
\left(P3\right)\quad u & = & 0,\quad x\in\partial\Omega
\end{eqnarray*}
where $\nabla u=\left(\frac{\partial u}{\partial x_{1}},\ldots,\frac{\partial u}{\partial x_{2}}\right)$;
operators $\Delta$ and $\mbox{div}$ are defined as usual. $u=u\left(t,x\right)$
is unknown solution, $h=h\left(t,x\right)$ is right\textendash hand
side, $u_{0}=u_{0}\left(x\right)$ is initial condition. $h$ and
$u_{0}$ define the data of the equation.

\begin{figure}
\includegraphics[width=0.9\textwidth]{img/k2_cylinder}

\caption{Initial and boundary condition in spacetime cylinder.}

\end{figure}

\subsection*{Assumptions}
\begin{itemize}
\item[$\left(A1\right)$] $\Omega\in\mathbb{R}^{n}$ is bounded domain (i. e., open, connected)
with regular (Lipschitz) boundary.
\item[$\left(A2\right)$] $a\left(\xi\right):\mathbb{R}^{n}\to\mathbb{R}^{n}$, $a\left(0\right)=0$,
$\left|a\left(\xi_{1}\right)-a\left(\xi_{2}\right)\right|\leq\alpha_{1}\left|\xi_{1}-\xi_{2}\right|$,
$\left(a\left(\xi_{1}\right)-a\left(\xi_{2}\right)\right)\cdot\left(\xi_{1}-\xi_{2}\right)\ge\alpha_{0}\left|\xi_{1}-\xi_{2}\right|^{2}$
$\forall\xi_{1},\xi_{1}\in\mathbb{R}^{n}$.
\item[$\left(A3\right)$] $f\left(z\right):\mathbb{R}\to\mathbb{R}$, $\left|f\left(z_{1}\right)-f\left(z_{2}\right)\right|\leq\ell\left|z_{1}-z_{2}\right|$,
$\forall z_{1},z_{2}\in\mathbb{R}$.
\end{itemize}

\subsection*{Remark:}
\begin{itemize}
\item special case: $a\left(\xi\right)=\xi$, $f=0$ \dots{} heat equation
$\partial_{t}u-\Delta u=h\left(t,x\right)$.
\end{itemize}

\subsection*{Plan:}
\begin{enumerate}
\item well\textendash posedness ($\exists!$ weak solution)
\item regularity (strong solution for smooth data)
\item further properties
\end{enumerate}

\subsection*{Recall:}
\begin{itemize}
\item Sobolev spaces $W^{1,2}=\left\{ u\in L^{2}\left(\Omega\right):\nabla u=v\in L^{2}\left(\Omega\right)\right\} $,
where $\nabla u$ is weak derivative, i. e., $-\int_{\Omega}u\frac{\partial\varphi}{\partial x_{j}}\dx=\int_{\Omega}v_{j}\varphi\dx$
$\forall\varphi\in D\left(\Omega\right),\forall j$.\\
$\left\Vert u\right\Vert _{W^{1,2}\left(\Omega\right)}=\left\Vert u\right\Vert _{L^{2}\left(\Omega\right)}+\left\Vert \nabla u\right\Vert _{L^{2}\left(\Omega\right)}$
\item $W_{0}^{1,2}\left(\Omega\right)=\left\{ u\in W^{1,2}\left(\Omega\right),u=0\mbox{ on }\partial\Omega\mbox{ in the sense of traces}\right\} =\overline{C_{c}^{\infty}\left(\Omega\right)}^{W^{1,2}\left(\Omega\right)}$
\item $W^{-1,2}\left(\Omega\right)=\left(W_{0}^{1,2}\left(\Omega\right)\right)^{*}$
\end{itemize}

\subsection*{Facts:}
\begin{itemize}
\item compact embedding: $W^{1,2}\left(\Omega\right)\hookrightarrow\hookrightarrow L^{2}\left(\Omega\right)$
\item Poincaré inequality: $\left\Vert u\right\Vert _{L^{2}\left(\Omega\right)}\leq c_{p}\left\Vert \nabla u\right\Vert _{L^{2}\left(\Omega\right)}$
for $\forall u\in W_{0}^{1,2}\left(\Omega\right)$.\\
Corollary: norms $\left\Vert u\right\Vert _{W^{1,2}\left(\Omega\right)}$
and $\left\Vert \nabla u\right\Vert _{L^{2}\left(\Omega\right)}$
are equivalent on $W_{0}^{1,2}\left(\Omega\right)$.
\item $W^{1,2}\left(\Omega\right)$ is reflexive, separable, $W_{0}^{1,2}\left(\Omega\right)$
is a closed subspace of $W^{1,2}\left(\Omega\right)$ hence also reflexive,
separable.
\item $W_{0}^{1,2}\left(\Omega\right)$, $W^{1,2}\left(\Omega\right)$ are
dense in $L^{2}\left(\Omega\right)$.
\end{itemize}

\subsection*{Notation:}

we will write $L^{2}$, $W^{1,2}$, $W_{0}^{1,2}$ instead of $L^{2}\left(\Omega\right)$,
$\left\Vert u\right\Vert _{2}$, $\left\Vert u\right\Vert _{1,2}$,
$\left\Vert u\right\Vert _{-1,2}$ instead of $\left\Vert u\right\Vert _{L^{2}\left(\Omega\right)}$
etc.\\
$\left(\cdot,\cdot\right)$ is scalar product in $L^{2}\left(\Omega\right)$,
i. e., $\left(f,g\right)=\int_{\Omega}\underbrace{f\left(x\right)}_{\!\!\makebox[0pt][l]{both functions in \ensuremath{\mathbb{R}^{n}} or \ensuremath{\mathbb{R}}}}\cdot\underbrace{g\left(x\right)}\dx$,
$\left\langle \cdot,\cdot\right\rangle $ is duality of $W^{-1,2}$
and $W_{0}^{1,2}$.

Gelfand triple: $W_{0}^{1,2}\hookrightarrow L^{2}\stackrel{\sim}{=}\left(L^{2}\right)^{*}\hookrightarrow W^{-1,2}$
with the embedding $\iota:L^{2}\hookrightarrow L^{2}:\left\langle \iota u,v\right\rangle =\left(u,v\right)$,
$\forall v\in L^{2}$. 

\subsection*{Recall:}

stationary problem: 
\begin{eqnarray*}
-\mbox{div}a\left(\nabla u\right) & = & h,\quad x\in\Omega,h\in W^{-1,2}\\
u & = & 0,\quad x\in\partial\Omega
\end{eqnarray*}
weak solution: $u\in W_{0}^{1,2}$ such that: $\int_{\Omega}a\left(\nabla u\left(x\right)\right)\cdot\nabla v\left(x\right)\dx=\left\langle h,v\right\rangle $
for $\forall v\in W_{0}^{1,2}$
\begin{itemize}
\item by nonlinear Lax\textendash Milgram theorem: $\forall h\in W^{-1,2}$
$\exists!u\in W_{0}^{1,2}$ weak solution
\item we introduce operator $A:W_{0}^{1,2}\to W^{-1,2}$ by $\left\langle A\left(u\right),v\right\rangle =\left(a\left(\nabla u\right),v\right)$.
By Lax\textendash Milgram: $A:W_{0}^{1,2}\to W^{-1,2}$ is continuous,
1-to-1.
\end{itemize}

\subsection*{Remark:}

if $a\left(\xi\right)=\xi$, then $\left\langle A\left(u\right),v\right\rangle =\left(\nabla u,\nabla v\right)$,
i. e., $A\left(u\right)=-\Delta u$ (weak laplacian).

\subsection*{Assumption:}

$h\left(t\right)\in L^{2}\left(I;W^{-1,2}\right)$, where $h\left(t\right)=h\left(t,x\right)$
is the $RHS$ in $\left(P1\right)$.

\subsection*{Def.: {[}Weak solution of parabolic equation{]}}

\addcontentsline{toc}{section}{Definition of weak solution}

function $u\left(t\right)\in L^{2}\big(I;W_{0}^{1,2}\big)$ is called
a weak solution to $\left(P1\right)$\textendash $\left(P3\right)$
provided that:
\[
\frac{d}{dt}\left(u\left(t\right),v\right)+\left(a\left(\nabla u\left(t\right)\right),\nabla v\right)+\left(f\left(u\left(t\right)\right),v\right)=\left\langle h\left(t\right),v\right\rangle 
\]
 in the sense of distributions in $\left(0,T\right)$ for $\forall v\in W_{0}^{1,2}$.
(``Infinite system of ODEs.'')

\subsection*{Remark: }

Expanding the definition of $\frac{d}{dt}$ we get
\[
\frac{d}{dt}y\left(t\right)=g\left(t\right)\iff\int_{I}y\left(t\right)\varphi'\left(t\right)=-\int_{I}g\left(t\right)\varphi\left(t\right)dt\ \forall\varphi\left(t\right)\in D\left(I\right)
\]
\begin{eqnarray*}
 & \implies & \left(*\right)\ -\int_{I}\left(u\left(t\right),v\right)\varphi'\left(t\right)dt+\int_{I}\left(a\left(\nabla u\left(t\right)\right),\nabla v\right)\varphi dt\\
 &  & +\int_{I}\left(f\left(u\left(t\right)\right),v\right)\varphi\left(t\right)dt=\int_{I}\left\langle h\left(t\right),v\right\rangle \varphi\left(t\right)dt\\
 &  & \quad\mbox{for }\forall v\in W_{0}^{1,2},\forall\varphi\left(t\right)\in D\left(I\right)
\end{eqnarray*}

by expanding further $\left(\cdot,\cdot\right)$ we get
\begin{eqnarray*}
 &  & -\iint_{I\times\Omega}u\left(t,x\right)v\left(x\right)\varphi'\left(t\right)dt\dx+\iint_{I\times\Omega}a\left(\nabla u\left(t,x\right)\right)\cdot\nabla v\left(x\right)\varphi\left(t\right)dt\dx\\
 &  & +\iint_{I\times\Omega}f\left(u\left(t,x\right)\right)v\left(x\right)\varphi\left(t\right)dt\dx=\int_{I}\left\langle h\left(t\right),v\right\rangle \varphi\left(t\right)dt\\
 &  & \quad\mbox{for }\forall v\in W_{0}^{1,2}
\end{eqnarray*}

\subsection*{Note:}
\begin{itemize}
\item multiply $\left(P1\right)$ by $v\left(x\right)\varphi\left(t\right)$,
integrate by parts \dots{} formally we obtain the equation above.
\item definiton makes sense: all integrals are finite (Hölder \& $\left(A1\right)$,
$\left(A2\right)$ \dots{} see ex. 3)
\end{itemize}

\subsection*{Lemma 2.1 {[}Properties of weak solution of parabolic equation{]}}

Let $u\left(t\right)\in L^{2}\left(I;W_{0}^{1,2}\right)$ be weak
solution. Then
\begin{enumerate}
\item $u\left(t\right)$ is weakly differentiable as a function $I\to W^{-1,2}$
with 
\[
\left(**\right)\quad\frac{d}{dt}u\left(t\right)+Au\left(t\right)+\iota f\left(u\left(t\right)\right)=h\left(t\right),
\]
in particular $\frac{d}{dt}u\left(t\right)\in L^{2}\left(I,W^{-1,2}\right)$,
\item $u\left(t\right)\in C\left(I,L^{2}\right)$ in the sense of representative,
\item $t\mapsto\left\Vert u\left(t\right)\right\Vert _{2}^{2}$ is weakly
differentiable with
\[
\left(***\right)\quad\frac{d}{dt}\left\Vert u\left(t\right)\right\Vert _{2}^{2}+\left\langle A\left(u\left(t\right)\right),u\left(t\right)\right\rangle +\left(f\left(u\left(t\right)\right),u\left(t\right)\right)=\left\langle h\left(t\right),u\left(t\right)\right\rangle \ \mbox{a. e. in }I
\]
\end{enumerate}

\subsubsection*{Proof:}
\begin{enumerate}
\item with embedding $\iota:W_{0}^{1,2}\to W^{-1,2}$ from Gelfand triple:
$\left(u\left(t\right),v\right)=\left\langle \iota u\left(t\right),v\right\rangle $,
$\left(f\left(u\left(t\right)\right),v\right)=\left\langle \iota f\left(u\left(t\right)\right),v\right\rangle $,
by definition: $\left(a\left(\nabla u\left(t\right)\right),\nabla v\right)=\left\langle A\left(u\left(t\right)\right),v\right\rangle $.
But then $\left(*\right)$ is rewritten as 
\[
\left\langle \underbrace{\int_{I}-\iota u\left(t\right)\varphi'\left(t\right)+A\left(u\left(t\right)\right)\varphi\left(t\right)+\iota f\left(u\left(t\right)\right)\varphi\left(t\right)-h\left(t\right)\varphi\left(t\right)dt}_{{\mbox{\ensuremath{\Vert}}\atop \mbox{0}}\makebox[0pt][l]{ since eq. holds \ensuremath{\forall\varphi\left(t\right)\in D\left(I\right)}, \ensuremath{\forall v\in W_{0}^{1,2}} }},v\right\rangle =0.
\]
Therefore $\left(**\right)$ is proven.\\
$\frac{d}{dt}u\left(t\right)\in L^{2}\left(I,W^{-1,2}\right)$ follows
from $\left(**\right)$ and the fact that $A\left(u\left(t\right)\right)$,
$f\left(u\left(t\right)\right)$, $h\left(t\right)$ belong to this
space.
\item Apply Th. 1.12,1 with $X=W_{0}^{1,2}$, $X^{*}=W^{-1,2}$, $H=L^{2}$,
$p=p'=2$.
\item Apply Th. 1.12,2 with $X=W_{0}^{1,2}$, $X=W_{0}^{1,2}$, $X^{*}=W^{-1,2}$,
$H=L^{2}$, $p=p'=2$ \dots{} $\frac{d}{dt}\left\Vert u\left(t\right)\right\Vert _{2}^{2}=2\left\langle \frac{d}{dt}u\left(t\right),u\left(t\right)\right\rangle \stackrel{\mbox{use }\left(**\right)}{=}\text{\dots}$
\end{enumerate}

\subsection*{Remark:}
\begin{itemize}
\item we will always work with continuous representative, hence $u\left(t\right)$
is well-defined for all $t\in I$, in particular initial condition
$u\left(0\right)=u_{0}$ makes sense,
\item also, $t\to\left\Vert u\left(t\right)\right\Vert _{2}^{2}$ is $AC$
(Th. 1.12).
\end{itemize}

\subsection*{Theorem 2.1 {[}Uniqueness of weak solution of parabolic equation{]}}

\addcontentsline{toc}{section}{Uniqueness of weak solution}

Weak solution is unique.

\subsubsection*{Proof:}

Let $u\left(t\right),v\left(t\right)\in L^{2}\left(I;W_{0}^{1,2}\right)$
be weak solutions and assume $u\left(0\right)=v\left(0\right)$. Set
$w\left(t\right)=u\left(t\right)-v\left(t\right)$. Goal: $w\left(t\right)=0$
$\forall t\in I$.

By L. 2.1: $w\left(t\right)$ is weakly differentiable with $\frac{d}{dt}w\left(t\right)+A\left(u\left(t\right)\right)-A\left(v\left(t\right)\right)+f\left(u\left(t\right)\right)-f\left(v\left(t\right)\right)=0$
in $W^{-1,2}$ for a. a. $t\in I$. Apply $\left\langle \cdot,w\left(t\right)\right\rangle $,
by Th. 1.12:

\[
\underbrace{\left\langle \frac{d}{dt}w\left(t\right),w\left(t\right)\right\rangle }_{{\mbox{\makebox[0pt][r]{Th. 1.12 }\ensuremath{\Vert}}\atop \makebox[0pt][c]{\ensuremath{\frac{1}{2}\frac{d}{dt}\left\Vert w\left(t\right)\right\Vert _{2}^{2}}}}}+\underbrace{\left\langle A\left(u\left(t\right)\right)-A\left(v\left(t\right)\right),w\right\rangle }_{\left(t_{2}\right)}+\underbrace{\left(f\left(u\left(t\right)\right)-f\left(v\left(t\right)\right),w\left(t\right)\right)}_{\left(t_{3}\right)}=0
\]
\[
\left(t_{2}\right):\quad\int_{\Omega}\underbrace{\left(a\left(\nabla u\left(t\right)\right)-a\left(\nabla v\left(t\right)\right)\right)\left(\nabla u\left(t\right)-\nabla v\left(t\right)\right)}_{\mbox{\ensuremath{\geq}}\makebox[0pt][l]{\ensuremath{\,0} by \ensuremath{\left(A1\right)}}}\dx\geq0
\]
\begin{eqnarray*}
 & \left(t_{3}\right): & \quad\Bigg|\int_{\Omega}\underbrace{\left(f\left(u\left(t\right)\right)-f\left(v\left(t\right)\right)\right)}_{\leq\makebox[0pt][l]{\,\ensuremath{\ell}\ensuremath{\ensuremath{\left|u\left(t\right)-v\left(t\right)\right|}}\mbox{ by }\ensuremath{\left(A2\right)}}}\left(u\left(t\right)-v\left(t\right)\right)\dx\Bigg|\\
 &  & \leq\int_{\Omega}\ell\left|w\left(t\right)\right|^{2}\dx=\ell\left\Vert w\left(t\right)\right\Vert _{2}^{2}
\end{eqnarray*}
Therefore 
\begin{eqnarray*}
\frac{1}{2}\frac{d}{dt}\left\Vert w\left(t\right)\right\Vert _{2}^{2} & = & \underbrace{-\left(t_{2}\right)}_{\mbox{\ensuremath{\leq0}}}+\underbrace{\left(-\left(t_{3}\right)\right)}_{\mbox{\ensuremath{\leq}}\makebox[0pt][l]{\ensuremath{\,\left|\left(t_{3}\right)\right|}}}\\
 & \leq & 0+\ell\left\Vert w\left(t\right)\right\Vert _{2}^{2}
\end{eqnarray*}

$\implies\frac{d}{dt}\left\Vert w\left(t\right)\right\Vert _{2}^{2}\leq2\ell\left\Vert w\left(t\right)\right\Vert _{2}^{2}$
for a. a. $t\in I$ $\stackrel{\mbox{L. 2.2}}{\implies}\left\Vert w\left(t\right)\right\Vert _{2}^{2}\leq\left\Vert w\left(0\right)\right\Vert _{2}^{2}e^{2\ell t}$
for $\forall t$, therefore $w\left(t\right)=0$ in $I$.

\subsection*{Lemma 2.2 {[}Gronwall lemma{]}}

Let $y\left(g\right)$, $g\left(t\right)$ be nonnegative (scalar)
functions, $y\left(t\right)$ continuous, $g\left(t\right)$ integrable.
Let $y\left(t\right)\leq K+\int_{0}^{t}g\left(s\right)y\left(s\right)ds$
for $\forall t\in I$. Then $y\left(t\right)\leq K\exp\left(\int_{0}^{t}g\left(s\right)ds\right)$
for $\forall t\in I$.

\subsubsection*{Proof:}

see ex. 3.

\subsection*{Goal:}

for $\forall u_{0}\in L^{2}$, $h\left(t\right)\in L^{2}\left(I;W^{-1,2}\right)$
$\exists u\in L^{2}\big(I;W_{0}^{1,2}\big)$ weak solution such that
$u\left(0\right)=u_{0}$.

\subsection*{Def: {[}Monotone and hemi-continuous operator{]}}

$\mathscr{X}$ is Banach space, let $A:\mathscr{X}\to\mathscr{X}^{*}$
be (nonlinear) operator, A is
\begin{description}
\item [{monotone}] $\left\langle A\left(u\right)-A\left(v\right),u-v\right\rangle \geq0$
$\forall u,v\in\mathscr{X}$.
\item [{hemicontinuous}] $t\mapsto A\left(u+tv\right)$ is continuous $\left(\mathbb{R}\to\mathscr{X}\right)$
for $\forall u,v\in\mathscr{X}$ fixed.
\end{description}

\subsection*{Lemma 2.3 {[}Minty's trick{]}}

Let $A:\mathscr{X}\to\mathscr{X^{*}}$ be monotone, hemicontinuous.
Let $u_{n}\rightharpoonup u$ in $\mathscr{X}$, $A\left(u_{n}\right)\rightharpoonup\alpha$
in $\mathscr{X}^{*}$. Let $\limsup_{n\to\infty}\left\langle A\left(u_{n}\right),u_{n}\right\rangle \leq\left\langle \alpha,u\right\rangle $.
Then $\alpha=A\left(u\right)$, i. e., $A\left(u_{n}\right)\rightharpoonup A\left(u\right)$.

\subsubsection*{Proof:}

\[
\left\langle A\left(u_{n}\right)-A\left(v\right),u_{n}-v\right\rangle \geq0\quad\ensuremath{v\in\mathscr{X}}\mbox{ to be specified later.}
\]
\begin{eqnarray*}
\left\langle A\left(u_{n}\right),u_{n}\right\rangle  & \geq & \left\langle A\left(u_{n}\right),v\right\rangle +\left\langle A\left(v\right),u_{n}-v\right\rangle \quad\Bigg/\limsup_{n\to\infty}\\
\limsup_{n\to\infty}\left\langle A\left(u_{n}\right),u_{n}\right\rangle  & \geq & \left\langle \alpha,v\right\rangle +\left\langle A\left(v\right),u-v\right\rangle 
\end{eqnarray*}
using last assumptions:
\begin{eqnarray*}
\left\langle \alpha,u\right\rangle  & \geq & \left\langle \alpha,v\right\rangle +\left\langle A\left(v\right),u-v\right\rangle \\
\left\langle \alpha-A\left(v\right),u-v\right\rangle  & \geq & 0\quad\mbox{trick: }v=u\pm\lambda w,\lambda>0,w\in\mathscr{X}
\end{eqnarray*}
\begin{eqnarray*}
\implies\left\langle \alpha-A\left(u\pm\lambda v\right),\mp\lambda w\right\rangle  & \geq & 0\\
\implies\big<\alpha-\underbrace{A\left(u\pm\lambda v\right)}_{{\mbox{\ensuremath{\downarrow}\makebox[0pt][l]{ by hemicontinuity}}\atop \mbox{\ensuremath{A\left(u\right)}}}},w\big> & = & 0\\
\implies\left\langle \alpha-A\left(u\right),w\right\rangle  & = & 0\quad\mbox{for }\forall w\in\mathscr{X}\\
\implies\alpha-A\left(u\right) & = & 0
\end{eqnarray*}

\subsection*{Theorem 2.2 {[}Compactness of weak solutions of parabolic equation{]}}

\addcontentsline{toc}{section}{Compactness of weak solutions}

Let $u_{n}\left(t\right)\in L^{2}\left(I;W_{0}^{1,2}\right)$ be weak
solutions such that $u_{n}\left(0\right)\to u_{0}$ in $L^{2}$. Then
there is a subsequence $\widetilde{u}_{n}\left(t\right)\rightharpoonup u\in L^{2}\left(I;W_{0}^{1,2}\right)$
such that $u\left(t\right)$ is a weak solution with $u\left(0\right)=u_{0}$.

\subsubsection*{Proof:}

\textbf{Step 1:} a priori estimates

\[
\underbrace{\frac{d}{dt}u_{n}\left(t\right)}_{\left(t_{1}\right)}+\underbrace{A\left(u_{n}\left(t\right)\right)}_{\left(t_{2}\right)}+\underbrace{f\left(u_{n}\left(t\right)\right)}_{\left(t_{3}\right)}=\underbrace{h\left(t\right)}_{\left(t_{4}\right)}\quad\mbox{a. e. in \ensuremath{I},\ensuremath{\quad\Bigg/\left\langle \cdot,u_{n}\left(t\right)\right\rangle }}
\]
\begin{itemize}
\item[$\left(t_{1}\right)$]  $\left\langle \frac{d}{dt}u_{n}\left(t\right),u_{n}\left(t\right)\right\rangle =\frac{1}{2}\frac{d}{dt}\left\Vert u_{n}\left(t\right)\right\Vert _{2}^{2}$
by Th. 1.12, L. 2.1.
\item[$\left(t_{2}\right)$ ] below we omit $\left(t\right)$ (argument of functions).
\begin{eqnarray*}
\left\langle A\left(u_{n}\right),u_{n}\right\rangle  & = & \int_{\Omega}a\left(\nabla u_{n}\right)\cdot\nabla u_{n}\dx\\
 & = & \int_{\Omega}\Big(a\left(\nabla u_{n}\right)-\underbrace{a\left(\nabla0\right)}_{=0}\Big)\left(\nabla u_{n}-\nabla0\right)\dx\\
 & \stackrel{\makebox[0pt][c]{\ensuremath{\left(A1\right)}}}{\geq} & \int_{\Omega}\alpha\left|\nabla u_{n}-\nabla0\right|^{2}\dx\\
 & = & \alpha\left\Vert \nabla u_{n}\right\Vert _{2}^{2}\stackrel{\mbox{Poincaré}}{\geq}c_{1}\left\Vert u_{n}\right\Vert _{1,2}^{2}
\end{eqnarray*}
\item[$\left(t_{3}\right)$] 
\begin{eqnarray*}
\left(f\left(u_{n}\right),u_{n}\right) & = & \int_{\Omega}f\left(u_{n}\right)u_{n}\dx\\
 & = & \int_{\Omega}\underbrace{\left(f\left(u_{n}\right)-f\left(0\right)\right)\left(u_{n}-0\right)}_{\mbox{\ensuremath{\left|\cdot\right|}}\makebox[0pt][l]{<\ensuremath{\ell\left|u_{n}\right|} by \ensuremath{\left(A2\right)}}}\dx+f\left(0\right)\int_{\Omega}u_{n}\dx
\end{eqnarray*}
\[
\implies\left|\left(t_{3}\right)\right|\leq\left\Vert u_{n}\right\Vert _{1}+f\left(0\right)\cdot\left\Vert u_{n}\right\Vert _{1}\stackrel{\left(*\right)}{\leq}
\]
but $\left\Vert u_{n}\right\Vert _{1}\stackrel{\mbox{\ensuremath{\Omega} bounded}}{\leq}c\left\Vert u_{n}\right\Vert _{2}$.
Young: $\left\Vert u_{n}\right\Vert _{2}=\left\Vert u_{n}\right\Vert _{2}\cdot1\leq\frac{1}{2}\left\Vert u_{n}\right\Vert _{2}^{2}+\frac{1}{2}$
\[
\stackrel{\left(*\right)}{\leq}c_{3}\left(1+\left\Vert u_{n}\right\Vert _{2}^{2}\right)\quad\mbox{where \ensuremath{c_{3}=c_{3}\left(\Omega,\ell,\ldots\right)}}
\]
\item[$\left(t_{4}\right)$] 
\[
\left|\left(t_{4}\right)\right|=\left|\left\langle h\left(t\right),u_{n}\right\rangle \right|\leq\left\Vert h\right\Vert _{-1,2}\left\Vert u_{n}\right\Vert _{1,2}\stackrel{\mbox{Young}}{\leq}\frac{c_{1}}{2}\left\Vert u_{n}\right\Vert _{1,2}^{2}+\frac{1}{2c_{1}}\left\Vert h\right\Vert _{-1,2}^{2}
\]
 Then:
\[
\underbrace{\left\langle \frac{d}{dt}u_{n}\left(t\right),u_{n}\left(t\right)\right\rangle }_{\mbox{=}\makebox[0pt][l]{\ensuremath{\,\frac{1}{2}\frac{d}{dt}\left\Vert u_{n}\left(t\right)\right\Vert _{2}^{2}}}}+\underbrace{\left\langle A\left(u_{n}\left(t\right)\right),u_{n}\left(t\right)\right\rangle }_{\mbox{\ensuremath{\geq}}\makebox[0pt][l]{\ensuremath{\,c_{1}\left\Vert u_{n}\right\Vert _{1,2}^{2}}}}+\underbrace{\left\langle f\left(u_{n}\left(t\right)\right),u_{n}\left(t\right)\right\rangle }_{\mbox{\ensuremath{\leq}}\makebox[0pt][l]{\ensuremath{\,c_{3}\left(1+\left\Vert u_{n}\right\Vert _{2}^{2}\right)}}}\ =\ \underbrace{\left\langle h\left(t\right),u_{n}\left(t\right)\right\rangle }_{\mbox{\ensuremath{\le}}\makebox[0pt][l]{\ensuremath{\,\frac{c_{1}}{2}\left\Vert u_{n}\right\Vert _{1,2}^{2}+\frac{1}{2c_{1}}\left\Vert h\right\Vert _{-1,2}^{2}}}}
\]
 
\begin{eqnarray*}
\frac{1}{2}\frac{d}{dt}\left\Vert u_{n}\left(t\right)\right\Vert _{2}^{2}+c_{1}\left\Vert u_{n}\left(t\right)\right\Vert _{1,2}^{2} & \leq\\
c_{3}\left(1+\left\Vert u_{n}\left(t\right)\right\Vert _{2}^{2}\right)+\frac{c_{1}}{2}\left\Vert u_{n}\left(t\right)\right\Vert _{1,2}^{2}+\frac{1}{2c_{1}}\left\Vert h\left(t\right)\right\Vert _{-1,2}^{2} &  & \Bigg/\ 2\int_{0}^{t}ds
\end{eqnarray*}
\begin{eqnarray*}
\implies\left\Vert u_{n}\left(t\right)\right\Vert _{2}^{2}+c_{1}\int_{0}^{t}\left\Vert u_{n}\left(s\right)\right\Vert _{1,2}^{2}ds & \leq & \left\Vert u_{n}\left(0\right)\right\Vert _{2}^{2}+2c_{3}t+\\
\frac{1}{c_{1}}\int_{0}^{t}\left\Vert h\left(s\right)\right\Vert _{-1,2}^{2}ds+2c_{3}\int_{0}^{t}\left\Vert u_{n}\left(s\right)\right\Vert _{2}^{2}ds &  & \mbox{for \ensuremath{\forall t\in I}.}
\end{eqnarray*}
Set $Y\left(t\right)=\left\Vert u_{n}\left(t\right)\right\Vert _{2}^{2}+c_{1}\int_{0}^{t}\left\Vert u_{n}\left(s\right)\right\Vert _{1,2}^{2}ds$,
let $K>0$ be such that $\left\Vert u_{n}\left(0\right)\right\Vert _{2}^{2}+2c_{3}t+\frac{1}{c_{1}}\int_{0}^{t}\left\Vert h\left(s\right)\right\Vert _{1,2}^{2}ds\leq K$
for $\forall t\in I$, $\forall n$. Hence we obtain $Y\left(t\right)\leq K+\int_{0}^{t}2c_{3}Y\left(s\right)ds$
for $\forall t\in I$. By L. 2.2 (Gronwall): $Y\left(t\right)\leq Ke^{2c_{3}t}$
for $\forall t\in I=\left[0,T\right]$. Therefore $Y\left(t\right)$
is bounded in $I$ (independently of $n$) and so $Yu_{n}\left(t\right)$
is bounded in $L^{\infty}\left(I;L^{2}\right)$ and $L^{2}\left(I;W_{0}^{1,2}\right)$.
Moreover $\frac{d}{dt}u_{n}\left(t\right)$ are bounded in $L^{2}\left(I;W^{-1,2}\right)$
\dots{} follows from the equation (ex. 3).
\end{itemize}

\subsection*{Step 2:\textmd{ convergent subsequences: $\exists$ subsequence (denote
by $u_{n}\left(t\right)$) such that $u_{n}\rightharpoonup u\left(t\right)$
in $L^{2}\left(I;W_{0}^{1,2}\right)$ by Eberlein\textendash Śmulian
and $\frac{d}{dt}u_{n}\left(t\right)\rightharpoonup g\left(t\right)$
in $L^{2}\left(I;W^{-1,2}\right)$. By HW2 we know $g\left(t\right)=\frac{d}{dt}u\left(t\right)$.
By Th. 1.13 (with $Y=W_{0}^{1,2}$, $X=L^{2}$, $Z=W^{-1,2}$, $p=q=2$)
it holds that $u_{n}\left(t\right)\to u\left(t\right)$ in $L^{2}\left(I;L^{2}\right)$.}}

\subsection*{\textmd{Moreover: $u_{n}\left(t\right)\rightharpoonup u\left(t\right)$
in $L^{2}$ for $\forall t\in I$ fixed (representatives).}}
\begin{itemize}
\item \textit{Proof:} $u_{n}\left(t\right)\rightharpoonup u\left(t\right)$
in $W=\left\{ u\left(t\right)\in L^{2}\left(I;W_{0}^{1,2}\right),\ \frac{d}{dt}u\left(t\right)\in L\left(I;W^{-1,2}\right)\right\} $,
but the mapping $u\left(\cdot\right)\mapsto u\left(t\right)$, is
continuous and linear for $t\in I$ fixed as $W\to L^{2}$ by Th.
1.12. The proof is completed since the weak convergence is preserved
by continuous mappings, see ``Properties of weak convergence'' in
appendices.
\end{itemize}
\textbf{Step 3:} passage to the limit.

\begin{eqnarray*}
\frac{d}{dt}u_{n}\left(t\right)+A\left(u_{n}\left(t\right)\right)+f\left(u_{n}\left(t\right)\right) & = & h\left(t\right)\mbox{ in }L^{2}\left(I;W^{-1,2}\right)\\
\downharpoonright\quad\quad\quad\quad\quad\downharpoonright\quad\quad\quad\quad\downharpoonright\quad\quad\\
\frac{d}{dt}u\left(t\right)\quad\quad+\alpha\left(t\right)\quad+f\left(u\left(t\right)\right)\ \  & = & h\left(t\right)
\end{eqnarray*}
\begin{itemize}
\item easy to show: $f\left(u_{n}\left(t\right)\right)\to f\left(u\left(t\right)\right)$
in $L^{2}\left(I;L^{2}\right)$, hence also $\rightharpoonup$ in
$L^{2}\left(I;W^{-1,2}\right)$ (ex. 3).
\item observe: $A\left(u_{n}\left(t\right)\right)$ is bounded in $L^{2}\left(I;W^{-1,2}\right)$,
hence $A\left(u_{n}\right)\rightharpoonup\alpha\left(t\right)$ in
$L^{2}\left(I;W^{-1,2}\right)$ for some $\alpha\left(t\right)$ after
taking a subsequence.
\end{itemize}
It remains to show that $A\left(u\left(t\right)\right)=\alpha\left(t\right)$
for a. a. $t\in I$.

Apply L. 2.3 (Minty's trick): , $X=L^{2}\left(I,W_{0}^{1,2}\right),$$X^{*}=L^{2}\left(I,W^{-1,2}\right)$,
$\tilde{A}:X\to X*$: $u\left(t\right)\mapsto A\left(u\left(t\right)\right)$.
easy to see: $\tilde{A}$ monotone and hemicontinuous (in fact continuous,
HW3). Need to show: (This completes the proof.) 

\[
\limsup_{n\to\infty}\underbrace{\int_{I}\left\langle A\left(u_{n})t\right),u_{n}\left(t\right)\right\rangle _{W^{-1,2},W_{0}^{1,2}}dt}_{\left\langle \tilde{A}\left(u_{n}\right),u_{n}\right\rangle _{X^{*},X}}\leq\int_{I}\left\langle \alpha\left(t\right),u\left(t\right)\right\rangle _{W^{-1,2},W_{0}^{1,}}dt\quad\left(*\right)
\]

By step 1: $\frac{d}{dt}\left\Vert u_{n}\right\Vert _{2}^{2}+\left\langle A\left(u_{n}\left(t\right)\right),u_{n}\left(t\right)\right\rangle +\left(f\left(u_{n}\left(t\right)\right),u_{n}\left(t\right)\right)=\left\langle h\left(t\right),u_{n}\left(t\right)\right\rangle /\int_{0}^{T}dt$,
rearange:
\[
\int_{I}\left\langle A\left(u_{n}\left(t\right)\right),u_{n}\left(t\right)\right\rangle dt=\int_{I}\left\langle h\left(t\right),u_{n}\left(t\right)\right\rangle dt-\int_{I}\left(f\left(u_{n}\left(t\right)\right),u_{n}\left(t\right)\right)dt+\frac{1}{2}\left\Vert u_{n}\left(0\right)\right\Vert ^{2}-\frac{1}{2}\left\Vert u_{n}\left(T\right)\right\Vert ^{2}
\]
\[
\limsup_{n\to\infty}\int_{I}\left\langle A\left(u_{n}\left(t\right)\right),u_{n}\left(t\right)\right\rangle dt=\limsup_{n\to\infty}\left(J_{1n}+J_{2n}+J_{3n}+J_{4n}\right)\leq\sum_{k=1}^{4}\limsup_{n\to\infty}J_{kn}
\]

\[
\limsup_{n\to\infty}J_{1n}=\lim_{n\to\infty}\int_{I}\left\langle h\left(t\right),u_{n}\left(t\right)\right\rangle dt=\int_{I}\left\langle h\left(t\right),u\left(t\right)\right\rangle dt\mbox{ as }u_{n}\rightharpoonup u\mbox{, in }L^{2}\left(I,W_{0}^{1,2}\right)
\]

\[
\limsup_{n\to\infty}J_{2n}=\lim_{n\to\infty}\int_{I}\left(f\left(u_{n}\left(t\right)\right),u_{n}\left(t\right)\right)dt=\int_{I}\left\langle \iota f\left(u\left(t\right)\right),u\left(t\right)\right\rangle dt\mbox{ as }u_{n}\to u\mbox{, in }L^{2}\left(I,L^{2}\right)
\]
\[
\limsup_{n\to\infty}J_{3n}=\lim_{n\to\infty}\frac{1}{2}\left\Vert u_{n}\left(0\right)\right\Vert _{2}^{2}=\left\Vert u\left(0\right)\right\Vert _{2}^{2}
\]
\[
\limsup_{n\to\infty}J_{4n}=\limsup_{n\to\infty}\left(-\frac{1}{2}\left\Vert u_{n}\left(T\right)\right\Vert _{2}^{2}\right)=-\liminf_{n\to\infty}\left(\frac{1}{2}\left\Vert u_{n}\left(T\right)\right\Vert _{2}^{2}\right)\leq-\frac{1}{2}\left\Vert u\left(T\right)\right\Vert _{2}^{2}
\]
 because $u_{n}\left(T\right)\rightharpoonup u\left(T\right)$ in
$L^{2}$ and the norm is weakly lower-semicontinuous, i. e. $\left\Vert u\left(T\right)\right\Vert _{2}\leq\liminf_{n\to\infty}\left\Vert u_{n}\left(T\right)\right\Vert _{2}^ {}$.

Finally: $\limsup_{n\to\infty}\int_{I}\left\langle A\left(u_{n}\left(t\right)\right),u_{n}\left(t\right)\right\rangle dt\leq\int_{I}\left\langle h\left(t\right),u\left(t\right)\right\rangle dt+\int_{I}\left(f\left(u\left(t\right)\right),u\left(t\right)\right)dt+\frac{1}{2}\left\Vert u\left(0\right)\right\Vert _{2}^{2}-\frac{1}{2}\left\Vert u\left(T\right)\right\Vert _{2}^{2}$.
But test eq. for $u\left(t\right)$ (three underbraces above), apply
$\left\langle \cdot,u\left(t\right)\right\rangle $:
\[
\frac{1}{2}\frac{d}{dt}\left\Vert u\left(t\right)\right\Vert _{2}^{2}+\left\langle \alpha\left(t\right),u\left(t\right)\right\rangle +\left(f\left(t\right),u\left(t\right)\right)=\mbox{\ensuremath{\left\langle h\left(t\right),u\left(t\right)\right\rangle \quad/\int_{I}dt}}
\]
obtain: 
\[
\int_{I}\left\langle \alpha\left(t\right),u\left(t\right)\right\rangle dt=\int_{I}\left\langle h\left(t\right),u\left(t\right)\right\rangle dt-\int_{I}\left(f\left(u\left(t\right)\right),u\left(t\right)\right)dt+\frac{1}{2}\left\Vert u\left(0\right)\right\Vert _{2}^{2}-\frac{1}{2}\left\Vert u\left(T\right)\right\Vert _{2}^{2}
\]
by comparing the RHS we get the $\left(*\right)$, which completes
the proof.

\subsection*{Remark: }

$u\left(t\right)$ is w. s. $\iff\frac{d}{dt}\iota u\left(t\right)=\mathcal{F}\left(t,u\left(t\right)\right),$
where $\mathcal{F}\left(t,u\right):I\times W_{0}^{1,2}\to W^{-1,2}$,
where 
\[
\mathcal{F}\left(t,u\right)=-\mathcal{A}\left(u\right)-\iota f\left(u\right)+h\left(t\right)\iff
\]
\[
\frac{d}{dt}\left(u\left(t\right),v\right)=\mathcal{F}\left\langle t,u\left(t\right)\right\rangle \mbox{ for }\forall v\in W_{0}^{1,2}\mbox{ fixed}
\]

easy to show $\iff\frac{d}{dt}\left(u\left(t\right),w_{j}\right)=\left\langle \mathcal{F}\left(t,u\left(t\right)\right),w_{j}\right\rangle ,$
where $w_{1},w_{2},\dots$ are dense in $W_{0}^{1,2}$ \dots{} infinite
(countable) system of ODEs. Good choice of basis $w_{j}$ is useful,
it makes things nice.

\subsection*{Recall {[}Weak formulation of laplacian in $W_{0}^{1,2}$''{]}:}

Eigenvalue problem for Dirichlet laplacian:
\begin{eqnarray*}
-\Delta u & = & \lambda u\quad\mbox{in }\Omega\\
u & = & 0\mbox{ on }\partial\Omega
\end{eqnarray*}

weak formulation: find $u\in W_{0}^{1,2}$ s. t. $\left(\nabla u,\nabla v\right)=\lambda\left(u,v\right)$
for $\forall v\in W_{0}^{1,2}$. We know: $\exists\lambda_{j}>0,\lambda_{j}\to\infty,w_{j}\in W_{0}^{1,2}$
pairs of eigenvalue/eigenfunction s t. $\left\{ w_{j}\right\} $ are
complete ON basis of $L^{2}$ and also complete ON basis of $W_{0}^{1,2}$
\dots{} Hilbert space with scalar product $\left(\left(\cdot,\cdot\right)\right)=\left(\nabla u,\nabla v\right)$
\dots{} (due to Poincaré it gives an equivalent norm on $W_{0}^{1,2}$)

\subsection*{Notation:}

$P_{n}$ \dots{} projection $L^{2}\to\mbox{span}\left\{ w_{1},\dots,w_{n}\right\} $.
Clearly $\left\Vert P_{n}\right\Vert =1$, but also $\left\Vert P_{n}\right\Vert =1$
with respect to $W_{0}^{1,2}$ norm, see ex. 4.2.

\subsection*{Theorem 2.3 {[}Existence of weak solutions of parabolic equation{]} }

\addcontentsline{toc}{section}{Existence of weak solutions}

Let $u_{0}\in L^{2}$, $h\left(t\right)\in L^{2}\left(I,W^{-1,2}\right).$
Then there is $u\left(t\right)\in L^{2}\left(I,W_{0}^{1,2}\right)$
a weak solution such that $u\left(0\right)=u_{0}$.

\subsubsection*{Proof:}

\textbf{Step 0:} approximating solutions $u^{N}\left(t\right)=\sum_{j=1}^{N}c_{j}^{N}\left(t\right)w_{j}$.
($^{N}$ is index), $w_{j}$ laplace eigenfunctions, $c_{j}^{N}\left(t\right):I\to\mathbb{R}$
are AC s. t. 
\[
\left(P_{N}\right)\quad\frac{d}{dt}c_{j}^{N}\left(t\right)=\left\langle \mathcal{F}\left(t,u^{N}\left(t\right)\right),w_{j}\right\rangle \quad j=1,\dots,N
\]
\dots{} system of $N$ ODEs.

Note: $c_{j}^{N}\left(t\right)=\left(u^{N}\left(t\right),w_{j}\right)\text{ \dots\ since}$
$w_{j}$ are ON in $L^{2}.$ Hence $\left(P_{N}\right)\iff\frac{d}{dt}\left(u^{N}\left(t\right),w_{j}\right)=\left\langle \mathcal{F}\left(t,u^{N}\left(t\right),w_{j}\right)\right\rangle ,j=1\dots N$$\iff P_{N}\left(\frac{d}{dt}u^{N}\left(t\right)-\mathcal{F}\left(t,u\left(t\right)\right)\right)=0$.
$c_{j}^{N}\left(t\right)$ solves $\left(P_{N}\right)$ with init.
cond. $c_{j}^{N}=\left(u_{0},w_{j}\right),\forall j=1,\dots,N$$\iff u^{N}\left(0\right)=P_{N}u_{0}$
by ON of $w_{j}$, hence $u^{N}\left(0\right)\to u_{0}=u\left(0\right)$
in $L^{2}$.

Easy to see: (HW3): RHS of $\left(P_{N}\right)$ satisfies Carathéodory
conditions $\implies\exists c_{j}^{N}\left(t\right)\in AC\left(I\right)$.
Hence $u^{N}\left(t\right)$ are well-defined in $I$.

\textbf{Step 1:} a priori estimates (independent of $N$ fixed)
\[
\left(P_{N}\right)\quad\frac{d}{dt}c_{j}^{N}\left(t\right)+\left\langle \mathcal{A}\left(u^{N}\left(t\right),w_{j}\right)\right\rangle +\left(f\left(u^{N}\left(t\right)\right),w_{j}\right)=\left\langle h\left(t,w_{j}\right)\right\rangle \quad\Bigg/\cdot c_{j}^{N}\left(t\right),\sum_{j=1}^{N}
\]
\begin{itemize}
\item first term: $\sum_{j=1}^{N}\left(\frac{d}{dt}c_{j}^{N}\right)c_{j}^{N}=\frac{1}{2}\frac{d}{dt}\underbrace{\sum_{j=1}^{N}\left(c_{j}^{N}\right)^{2}}_{\mbox{=}\makebox[0pt][l]{\ensuremath{\left\Vert u^{N}\right\Vert _{2}^{2}}}}=\frac{1}{2}\frac{d}{dt}\left\Vert u^{N}\right\Vert _{2}^{2}$,
since $w_{j}\in L^{2}$ are ON, the sum is finite and $c_{j}^{N}\in AC$.
\item second term: $\left\langle \mathcal{A}\left(u^{N}\right),u^{N}\right\rangle =\int_{\Omega}a\left(\nabla u^{N}\right)\nabla u^{N}\stackrel{\left(A1\right)}{\geq}\alpha_{0}\left\Vert \nabla u^{N}\right\Vert _{2}^{2}\geq c_{1}\left\Vert u^{N}\right\Vert _{1,2}^{2}$
as in Th. 2.2.
\item third term: $\left|\left(f\left(u^{N}\left(t\right)\right),u^{N}\left(t\right)\right)\right|\leq\ldots\leq c_{2}\left(1+\left\Vert u^{N}\left(t\right)\right\Vert _{2}^{2}\right)$
as in Th. 2.2
\item fourth (last) term: $\left|\left\langle h\left(t\right),u^{N}\left(t\right)\right\rangle \right|\leq\ldots\leq\frac{c_{1}}{2}\left\Vert u^{N}\left(t\right)\right\Vert _{1,2}^{2}+\frac{1}{2c_{1}}\left\Vert h\left(t\right)\right\Vert _{1,2}^{2}$.
\end{itemize}
Finally: $\frac{d}{dt}\left\Vert u^{N}\left(t\right)\right\Vert _{2}^{2}+c_{1}\left\Vert u^{N}\left(t\right)\right\Vert _{1,2}^{2}\leq c_{3}+c_{2}\left\Vert u^{N}\left(t\right)\right\Vert _{2}^{2}$.

By Gronwall: $u^{N}\left(t\right)$ bounded in $L^{2}\left(I,W_{0}^{1,2}\right)$
and $L^{\infty}\left(I,L^{2}\right)$.

We now need: $\frac{d}{dt}\iota u^{N}\left(t\right)$ bounded in $L^{2}\left(I,W^{-1,2}\right)$
\dots{} use dual characterization of norm:
\[
\left\Vert \frac{d}{dt}\iota u^{N}\left(t\right)\right\Vert _{-1,2}=\sup_{v\in W_{0}^{1,2},\left\Vert v\right\Vert =1}\left\langle \frac{d}{dt}\iota u^{N}\left(t\right),v\right\rangle _{W^{-1,2},W_{0}^{1,2}}
\]
but for $v$ fixed we have: 
\begin{eqnarray*}
\left\langle \frac{d}{dt}\iota u^{N}\left(t\right),v\right\rangle  & \stackrel{\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\makebox[0pt][l]{\ensuremath{P_{N}u^{N}=u^{N}} due to the previous equality}}{=}\\
\left\langle \iota P_{N}\frac{d}{dt}u^{N}\left(t\right),v\right\rangle  & = & \left(P_{N}\frac{d}{dt}u^{N}\left(t\right),v\right)=\left(\frac{d}{dt}u^{N}\left(t\right),P_{N}v\right)=\left\langle \iota\frac{d}{dt}u^{N}\left(t\right),P_{N}v\right\rangle \\
 & = & \left\langle \mathcal{F}\left(t,u^{N}\left(t\right)\right),P_{N}v\right\rangle \\
 & \leq & \left\Vert \mathcal{F}\left(t,u^{N}\left(t\right)\right)\right\Vert _{-1,2}\underbrace{\left\Vert P_{N}v\right\Vert _{1,2}}_{\leq\makebox[0pt][l]{\ensuremath{\,1}\mbox{ because \ensuremath{\left\Vert v\right\Vert _{1,2}=1} and \ensuremath{\left\Vert P_{N}\right\Vert _{\mathcal{L}\left(W_{0}^{1,2}\right)}=1}}}}\leq\left\Vert \mathcal{F}\left(t,u^{N}\left(t\right)\right)\right\Vert _{-1,2}\\
 & \leq & C\left(\left\Vert u^{N}\left(t\right)\right\Vert _{1,2}+\left\Vert h\left(t\right)\right\Vert _{-1,2}\right)\mbox{ by (HW 3)}
\end{eqnarray*}
taking $\sup$: $\left\Vert \frac{d}{dt}u^{N}\left(t\right)\right\Vert _{L^{2}\left(I,W^{-1,2}\right)}\leq C$
independent of $N$.

\textbf{Step 2:} $\exists$ convergent subsequences (as in Thm. 2.
2) s. t. 
\begin{eqnarray*}
u^{N}\left(t\right) & \rightharpoonup & u\left(t\right)\mbox{ in }L^{2}\left(I,W_{0}^{1,2}\right)\\
\frac{d}{dt}u^{N}\left(t\right) & \rightharpoonup & \frac{d}{dt}u\left(t\right)\mbox{ in }L^{2}\left(I,W^{-1,2}\right)\\
u^{N}\left(t\right) & \to & u\left(t\right)\mbox{ in }L^{2}\left(I,L^{2}\right)\\
\mathcal{A}\left(u^{N}\left(t\right)\right) & \rightharpoonup & \alpha\left(t\right)\mbox{ in }L^{2}\left(I,W^{-1,2}\right)\\
u^{N} & \rightharpoonup & u\left(t\right)\mbox{ in }L^{2}\mbox{ for any \ensuremath{t\in I} fixed}
\end{eqnarray*}

\textbf{Step 3:} limit passage $\left(P_{N}\right)\to\left(P_{I}\right)$
\[
\left(P_{N}\right)\quad\frac{d}{dt}\left(u^{N},w_{j}\right)=\left\langle \mathcal{\mathcal{F}}\left(t,u^{N}\left(t\right)\right),w_{j}\right\rangle \mbox{ for }j\leq N
\]
\begin{eqnarray*}
\iff-\int_{I}\left(u^{N}\left(t\right),w_{j}\right)\varphi'\left(t\right)dt+\int_{I}\left\langle \mathcal{A}\left(u^{N}\left(t\right)\right),w_{j}\right\rangle \varphi\left(t\right)dt\\
+\int_{I}\left(f\left(u^{N}\left(t\right)\right),w_{j}\right)\varphi\left(t\right)dt=\int_{I}\left\langle h\left(t\right),w_{j}\right\rangle \varphi\left(t\right)dt
\end{eqnarray*}
 for $\forall\varphi\left(t\right)\in D\left(I\right)$. Fix $j$,
let $N\to\infty$.
\begin{eqnarray*}
\implies-\int_{I}\left(u\left(t\right),w_{j}\right)\varphi'\left(t\right)dt+\int_{I}\left\langle \left(\alpha\left(t\right)\right),w_{j}\right\rangle \varphi\left(t\right)dt\\
+\int_{I}\left(f\left(u\left(t\right)\right),w_{j}\right)\varphi\left(t\right)dt=\int_{I}\left\langle h\left(t\right),w_{j}\right\rangle \varphi\left(t\right)dt
\end{eqnarray*}

This holds for $\forall j$, by density of $w_{j}$ in $W_{0}^{1,2}$
we can replace $w_{j}$ by $v\in W_{0}^{1,2}$ arbitrary.
\[
\implies\frac{d}{dt}\iota u\left(t\right)+\alpha\left(t\right)+\iota f\left(u\left(t\right)\right)=h\left(t\right)\mbox{ in }L^{2}\left(I,W^{-1,2}\right)
\]
It remains to show, that $\mathcal{A}\left(u\left(t\right)\right)=\alpha\left(t\right)$,
same argument as in Th. 2.2. This finishes the proof.

\subsection*{Remark:}

If $\mathcal{A}$ is linear (e. g. heat equation), then $\mathcal{A}\left(u_{n}\left(t\right)\right)\rightharpoonup\mathcal{A}\left(u\left(t\right)\right)$
is automatic.

\subsection*{Recall:}

maximum principle: $\partial_{t}u-\Delta u=h\left(t\right)$, $u\left(0\right)\leq0$,
$h\left(t\right)\leq0$ then $u\left(t\right)\leq0$ for $\forall t\geq0$.
Classical argument: let $\left(t_{0},x_{0}\right)$ be the point of
maximum, then $\partial_{t}u\left(t\right)=0$, $\frac{\partial^{2}}{\partial x^{2}}u\leq0$
at $\left(t_{0},x_{0}\right)$, then contradiction: $\partial_{t}u-\Delta u\geq0$
up to some details.

\subsection*{Problem:}

it requires $u\left(t\right)\in C\left(I\times\bar{\Omega}\right)\cap C_{t}^{1}\cap C_{x}^{2}$.
Therefore this is not available for weak solution.

\glqq weak argument``: $\frac{d}{dt}\int_{\Omega}u^{+}\left(t\right)\dx\leq0$.
(Then $u^{+}=0\implies u^{+}\left(t\right)=0\forall t\geq0$. How
to prove $\frac{d}{dt}\int_{\Omega}u^{+}\left(t\right)\dx\leq0$?)

Observe $u^{+}=\psi\left(u\right)$ where $\psi\left(z\right)=\begin{cases}
z & ;z\geq0\\
0 & ;z<0
\end{cases}$. Key point: $\psi$ is convex. Convex funcions have nice behaviour
with respect to diffusion.

\[
\frac{d}{dt}\int_{\Omega}\psi\left(u\right)\dx=\int_{\Omega}\psi'\left(u\right)\underbrace{\partial_{t}u}_{\Delta u\ldots}\dx=\int_{\Omega}\psi'\left(u\right)\Delta u\dx\stackrel{PP}{=}\int_{\Omega}\underbrace{-\psi''\left(u\right)}_{\makebox[0pt][c]{infinite but with good sign}}\left(\nabla u\cdot\nabla u\right)\dx
\]
This is the basic idea, now let us do it rigorously.

\subsection*{Lemma 2.4 {[}Weak derivative of compound functions{]} }

Let $\psi\left(z\right):\mathbb{R}\to\mathbb{R}$, with $\psi'$,
$\psi''$ bounded.
\begin{enumerate}
\item if $u\in W_{0}^{1,2}$, then $\psi\left(u\right)\in W_{0}^{1,2}$,
$\nabla\psi\left(u\right)=\psi'\left(u\right)\nabla u$ in the weak
sense and moreover $u\mapsto\psi\left(u\right)$ is continuous $W_{0}^{1,2}\to W_{0}^{1,2}$.
\item if $u\left(t\right)\in L^{2}\left(I,W_{0}^{1,2}\right)$, $\frac{d}{dt}u\left(t\right)\in L^{2}\left(I,W^{-1,2}\right)$
then $\frac{d}{dt}\int_{I}\psi\left(u\left(t\right)\right)\dx=\left\langle \frac{d}{dt}u\left(t\right),\psi'\left(u\left(t\right)\right)\right\rangle $
for a. e. $t\in I$.
\end{enumerate}

\subsubsection*{Proof:}
\begin{enumerate}
\item given $u\in W_{0}^{1,2}$ there $\exists u_{n}\in C_{c}^{\infty}$,
$u_{n}\to u$ in $W^{1,2}$, let $\varphi\in C_{c}^{\infty}$ be fixed.
\[
\int_{\Omega}\psi\left(u_{n}\right)\frac{d\varphi}{dx_{j}}dx=-\int_{\Omega}\psi'\left(u_{n}\right)\nabla u_{n}\varphi dx\mbox{ by parts, all is smooth, zero boundary}
\]
Take limit $n\to\infty$:
\[
\int_{\Omega}\psi\left(u\right)\frac{\partial\varphi}{\partial x_{j}}dx=-\int_{\Omega}\psi'\left(u\right)\nabla u\varphi dx
\]
hence $\nabla\psi\left(u\right)=\psi'\left(u\right)\nabla u$ weakly.\\
Continuity: mapping $u\mapsto\psi\left(u\right)$ is continous $W_{0}^{1,2}\to W_{0}^{1,2}$?
We want to show, that if $u_{n}\to u$ in $W^{1,2}$, i. e., $u_{n}\to u$
in $L^{2}$ and $\nabla u_{n}\to\nabla u$ in $L^{2}$, then $\psi\left(u_{n}\right)\to\psi\left(u\right)$
in $W^{1,2}$, i. e., $\psi\left(u_{n}\right)\to\psi\left(u\right)$
in $L^{2}$ and $\psi'\left(u_{n}\right)\nabla u_{n}\to\psi'\left(u\right)\nabla u$.
Routine argument (Lebesgue): $\psi'$, $\psi''$ bounded, hence $\left|\psi\left(z\right)\right|\leq C\left(1+\left|z\right|\right)$.
\item Let first $u\left(t\right)$ be smooth, say $u\left(t\right)\in C^{1}\left(I,L^{2}\right)$,
$\psi\in C^{2}$ Take $t_{1}$, $t_{2}\in I$ fixed.
\[
u\left(t_{1}\right)-u\left(t_{2}\right)=\int_{t_{1}}^{t_{2}}\frac{d}{dt}u\left(t\right)dx\mbox{ in \ensuremath{L^{2}}}
\]
 hence $u\left(t_{2},x\right)-u\left(t_{1},x\right)=\int_{t_{1}}^{t_{2}}\frac{d}{dt}u\left(t,x\right)dx$
for a. e. $x\in\Omega$.\\
We know: $\frac{d}{dt}u\left(t,x\right)\in C\left(I,L^{2}\right)\hookrightarrow L^{1}\left(I,L^{1}\right)=L^{1}\left(I\times\Omega\right)=L^{1}\left(\Omega\times I\right)$,
Fubini \dots{} see ex. 2.2\textbf{.}\\
Hence for a. e. $x\in\Omega$ the following is true: $\frac{d}{dt}u\left(t,x\right)\in L^{1}\left(I\right)\implies t\mapsto u\left(t\right)\in AC$$\implies t\mapsto\psi\left(u\left(t,x\right)\right)\in AC$
and $\frac{d}{dt}\psi\left(u\left(t,x\right)\right)=\psi'\left(u\left(t,x\right)\right)\frac{d}{dt}u\left(t,x\right)$.
Finally:
\[
\psi\left(u\left(t_{2},x\right)\right)-\psi\left(u\left(t_{1},x\right)\right)=\int_{t_{1}}^{t_{2}}\psi\left(u\left(t,x\right)\right)\frac{d}{dt}u\left(t,x\right)dt\quad\mbox{for a. e. \ensuremath{x\in\Omega}}
\]
Integrate $\int_{\Omega}\dx$:
\[
\int_{\Omega}\psi\left(u\left(t_{2}\right)\right)\dx-\int_{\Omega}\psi\left(u\left(t_{1}\right)\right)\dx=\int_{t_{1}}^{t_{2}}\underbrace{\int_{\Omega}\psi'\left(u\left(t\right)\right)\frac{d}{dt}u\left(t\right)\dx}_{\left(\frac{d}{dt}u\left(t\right),\psi'\left(u\left(t\right)\right)\right)}dt=\int_{t_{1}}^{t_{2}}\left\langle \iota\frac{d}{dt}u\left(t\right),\psi'\left(u\left(t\right)\right)\right\rangle dt
\]
If $u\left(t\right)\in L^{2}\left(I,W_{0}^{1,2}\right)$ is arbitrary,
$\frac{d}{dt}u\left(t\right)\in L^{2}\left(I,W^{-1,2}\right)$ \dots{}
$\exists u_{n}\in C^{1}\left(I,W_{0}^{1,2}\right)$ s. t. $u_{n}\to u\left(t\right)$
in $L^{2}\left(I,W_{0}^{1,2}\right)$, $\iota\frac{d}{dt}u_{n}\left(t\right)\to\frac{d}{dt}u\left(t\right)$
in $L^{2}\left(I,W^{-1,2}\right)$ and $u_{n}\to u\left(t\right)$
in $L^{2}$ for all $t\in I$ (continuous representative).\\
By the above: 
\[
\int_{\Omega}\psi\left(u_{n}\left(t_{2}\right)\right)\dx-\int_{\Omega}\psi\left(u_{n}\left(t_{1}\right)\right)\dx=\int_{t_{1}}^{t_{2}}\left\langle \iota\frac{d}{dt}u_{n}\left(t\right),\psi'\left(u_{n}\left(t\right)\right)\right\rangle dt
\]
let $n\to\infty$ (by Lebesgue: routine limit passage (ex. 4)): 
\[
\int_{\Omega}\psi\left(u\left(t_{2}\right)\right)\dx-\int_{\Omega}\psi\left(u\left(t_{1}\right)\right)\dx=\int_{t_{1}}^{t_{2}}\left\langle \frac{d}{dt}u\left(t\right),\psi'\left(u\left(t\right)\right)\right\rangle dt
\]
\end{enumerate}

\subsection*{Def.: {[}Nonnegative functions in spaces $L^{p}$, $W_{0}^{1,2}$,
$W^{-1,2}${]}}

If $v\in L^{2}$ or $W_{0}^{1,2}$, then $v\geq0$ (or $v\leq0$)
means $v\left(x\right)\geq0$ (or $v\left(x\right)\leq0$) for a.
e. $x\in\Omega$.

If $h\in W^{-1,2}$ then $h\geq0$ (or $h\leq0$) means $\left\langle h,v\right\rangle \geq0$
(or $\left\langle h,v\right\rangle \leq0$) for all $v\in W_{0}^{1,2}$,
$v\geq0$.

\subsection*{Theorem 2.4 {[}Maximum principle for weak solution of parabolic equation{]}}

\addcontentsline{toc}{section}{Maximum principle for weak solutions}

Let $u\left(t\right)\in L^{2}\left(I,W_{0}^{1,2}\right)$ be w. s.,
let $u\left(0\right)\leq0$, $h\left(t\right)\leq0$ for a. e. $t\in I$,
$f\left(\cdot\right)\geq0$. Then $u\left(t\right)\leq0$ for a. e.
$t\in I$.

\subsubsection*{Proof:}

take $\psi_{\varepsilon}:\mathbb{R}\to\mathbb{R}$ s. t. $0\leq\psi_{\varepsilon}\leq z^{+}$
, $\psi_{\varepsilon}\left(z\right)\nearrow z^{+}$, $\varepsilon\to0^{+}$,
$\psi'_{\varepsilon}$, $\psi''_{\varepsilon}\geq0$ bounded.

\begin{figure}
\includegraphics[width=0.4\textwidth]{img/k2_aux_fun}\caption{Graph of $z^{+}$ and $\psi_{\varepsilon}$.}
\end{figure}
\[
\frac{d}{dt}\iota u\left(t\right)+\mathcal{A}\left(u\left(t\right)\right)+\iota f\left(u\left(t\right)\right)=h\left(t\right)\mbox{ in \ensuremath{W^{1,2},} take }\Big<\cdot,\underbrace{\psi'_{\varepsilon}\left(u\left(t\right)\right)}_{\makebox[0pt][c]{in \ensuremath{W_{0}^{1,2}} by L. 2.4}}\Big>
\]
first term: $\left\langle \frac{d}{dt}\iota u\left(t\right),\psi'_{\varepsilon}\left(u\left(t\right)\right)\right\rangle =\frac{d}{dt}\int_{\Omega}\psi_{\varepsilon}\left(u\left(t\right)\right)dt$
by L2.4\\
second term: $\left\langle \mathcal{A}\left(u\left(t\right)\right),\psi'_{\varepsilon}\left(u\left(t\right)\right)\right\rangle =a\left(\nabla u,\nabla\psi'_{\varepsilon}\left(u\right)\right)\stackrel{L2.4}{=}\left(a\left(\nabla u\right),\psi''_{\varepsilon}\left(u\right)\nabla u\right)=\int_{\Omega}\underbrace{a\left(\nabla u\right)\cdot\nabla u}_{\geq0\mbox{ by }\left(A1\right)}\underbrace{\psi''_{\varepsilon}}_{\geq0}\dx\geq0$.\\
third term$\left\langle \iota f\left(u\right),\psi'_{\varepsilon}\left(u\right)\right\rangle =\int_{\Omega}f\left(u\right)\psi'_{\varepsilon}\left(u\right)\geq0$\\
fourth term:$\left\langle h\left(t\right),\psi'_{\varepsilon}\left(u\right)\right\rangle \leq0$\\
together:$\frac{d}{dt}\int_{\Omega}\psi_{\varepsilon}\left(u\left(t\right)\right)\dx\leq0$,
take $\int_{0}^{\tau}dt$, $\tau\in I$: $\int_{\Omega}\psi_{\varepsilon}\left(u\left(\tau\right)\right)\dx\leq\int_{\Omega}\psi_{\varepsilon}\left(u\left(0\right)\right)\dx$,
take $\varepsilon\to0^{+}$: $\int_{\Omega}u^{+}\left(\tau\right)\dx\leq\int_{\Omega}u^{+}\left(0\right)\dx$,
hence $u^{+}\left(0\right)=0$ hence $u^{+}\left(\tau\right)=0$ for
all $\tau$.

\subsection*{Theorem 2.5 {[}Strong solution of heat equation{]}}

\addcontentsline{toc}{section}{Existence of strong solution to the heat equation}

Let $u\left(t\right)\in L^{2}\left(I,W_{0}^{1,2}\right)$ be w. s.
to the heat equation:

\[
\frac{d}{dt}u-\Delta u+f\left(u\right)=h\left(t\right)
\]
Let $u\left(0\right)\in W_{0}^{1,2}$, $h\left(t\right)\in L^{2}\left(I,L^{2}\right)$.
Then $u\left(t\right)\in L^{\infty}\left(I,W_{0}^{1,2}\right)\cap L^{2}\left(I,W^{2,2}\right)$,
$\frac{d}{dt}u\left(t\right)\in L^{2}\left(I,L^{2}\right)$.

\subsection*{Remark:}

usually for w. s. $u\left(0\right)\in L^{2}$, $h\left(t\right)\in L^{2}\left(I,W^{-1,2}\right)$
and $u\left(t\right)\in L^{\infty}\left(I,L^{2}\right)\cap L^{2}\left(I,W_{0}^{1,2}\right)$,
$\frac{d}{dt}u\left(t\right)\in L^{2}\left(I,W_{0}^{-1,2}\right)$.
Strong solution: one derivative better in spatial derivative

\subsubsection*{Proof:}

\textbf{Step 1:} (formal) \dots{} multiply by $\frac{d}{dt}u$, $\int_{\Omega}\dx$:
\begin{itemize}
\item first term: $\int_{\Omega}\frac{d}{dt}u\cdot\frac{d}{dt}u\dx$=$\left\Vert \frac{d}{dt}u\right\Vert _{2}^{2}$
\item second term: $-\int_{\Omega}\Delta u\frac{d}{dt}u\dx=\int_{\Omega}\nabla u\cdot\nabla\left(\frac{d}{dt}u\right)\dx=\frac{1}{2}\frac{d}{dt}\left\Vert \nabla u\right\Vert _{2}^{2}$
\item third term: $\left|\left(f\left(u\right),\frac{d}{dt}u\right)\right|\leq\left\Vert f\left(u\right)\right\Vert _{2}\cdot\left\Vert \frac{d}{dt}u\right\Vert _{2}\stackrel{\mbox{Young}}{\leq}\left\Vert f\left(u\right)\right\Vert _{2}+\frac{1}{4}\left\Vert \frac{d}{dt}u\right\Vert _{2}^{2}$
(Young: $ab\leq\frac{\varepsilon}{2}a^{2}+\frac{1}{2\varepsilon}b^{2}$,
here $\varepsilon=2$)
\item fourth term: $\left|\left(h\left(t\right),\frac{d}{dt}u\left(t\right)\right)\right|\leq\left\Vert h\left(t\right)\right\Vert _{2}\cdot\left\Vert \frac{d}{dt}u\right\Vert \leq\left\Vert h\left(t\right)\right\Vert _{2}^{2}+\frac{1}{4}\left\Vert \frac{d}{dt}u\right\Vert _{2}^{2}$.
\end{itemize}
All together: $\frac{1}{2}\left\Vert \frac{d}{dt}u\right\Vert _{2}^{2}+\frac{1}{2}\frac{d}{dt}\left\Vert \nabla u\right\Vert _{2}^{2}\leq\left\Vert h\left(t\right)\right\Vert _{2}^{2}+\left\Vert f\left(u\right)\right\Vert _{2}^{2}$
integrate $2\int_{0}^{t}dt$:
\[
\int_{0}^{t}\left\Vert \frac{d}{dt}u\left(s\right)\right\Vert _{2}^{2}ds+\left\Vert \nabla u\left(t\right)\right\Vert _{2}^{2}\leq\underbrace{2\int_{0}^{t}\left\Vert h\left(s\right)\right\Vert _{2}^{2}+\left\Vert f\left(u\left(s\right)\right)\right\Vert {}_{2}^{2}ds}_{\leq\makebox[0pt][l]{\ensuremath{\,K}}}+\overbrace{\left\Vert \nabla u\left(0\right)\right\Vert _{2}^{2}}^{\makebox[0pt][c]{ok, \ensuremath{u\left(0\right)\in W_{0}^{1,2}}}}\quad\forall t\in I
\]
 where $K$ is independent of $t\in I$ because $h\left(t\right)\in L^{2}\left(I;L^{2}\right)$,
$u\left(t\right)\in L^{2}\left(I;L^{2}\right)$ and $f$ is lipschitz.
Take $\sup_{t\in I}$: $\frac{d}{dt}u\left(t\right)\in L^{2}\left(I,L^{2}\right)$,
$\left\Vert \nabla u\left(t\right)\right\Vert _{2}\in L^{\infty}\left(I\right)$,
by Poincaré ($\left\Vert \nabla u\right\Vert \approx\left\Vert u\right\Vert _{1,2}$
in $W_{0}^{1,2}$): $u\in L^{\infty}\left(I,W_{0}^{1,2}\right)$.

\textbf{Step 2:} rigorous proof: let $u^{N}\left(t\right)=\sum_{j=1}^{N}c_{j}^{N}\left(t\right)w_{j}$
be approximations from Thm. 2.3., i. e., $\frac{d}{dt}c_{j}^{N}+\left(\nabla u^{N}\left(t\right),\nabla w_{j}\right)+\left(f\left(u^{N}\left(t\right)\right),w_{j}\right)=\left(h\left(t\right),w_{j}\right)$
in $I,j=1,\ldots,N$, $c_{j}^{N}\left(0\right)=\left(u\left(0\right),w_{j}\right)$.
Multiply by $\frac{d}{dt}c_{j}^{N}$, $\sum_{j=1}^{N}$:
\begin{itemize}
\item first term: note that $\frac{d}{dt}u^{N}\left(t\right)=\sum_{j=1}^{N}\frac{d}{dt}c_{j}^{N}\left(t\right)w_{j}$,
hence $\sum_{j=1}^{N}\left(\frac{d}{dt}c_{j}^{N}\right)^{2}=\left\Vert \frac{d}{dt}u^{N}\right\Vert _{2}^{2}$,
since $w_{j}\in L^{2}$ are ON,
\item second term: $\Big(\nabla u^{N},\nabla\underbrace{\sum_{j=1}^{N}\frac{d}{dt}c_{j}^{N}w_{j}}_{\frac{d}{dt}u^{N}}\Big)=\left(\nabla u^{N},\frac{d}{dt}\nabla u^{N}\right)=\frac{1}{2}\frac{d}{dt}\left\Vert \nabla u^{N}\right\Vert _{2}^{2}$,
\item third term: $\left|\left(f\left(u^{N}\right),\frac{d}{dt}u^{N}\right)\right|\stackrel{\mbox{C.S.}}{\leq}\left\Vert f\left(u^{N}\right)\right\Vert _{2}\cdot\left\Vert \frac{d}{dt}u^{N}\right\Vert _{2}\stackrel{\mbox{Young}}{\leq}\left\Vert f\left(u^{N}\right)\right\Vert _{2}^{2}+\frac{1}{4}\left\Vert \frac{d}{dt}u^{N}\right\Vert _{2}^{2}$,
\item fourth term: $\left|\left(h\left(t\right),\frac{d}{dt}\left(u^{N}\right)\right)\right|\leq\ldots\leq\left\Vert h\left(t\right)\right\Vert _{2}^{2}+\frac{1}{4}\left\Vert \frac{d}{dt}u^{N}\right\Vert _{2}^{2}$.
\end{itemize}
Together as before: $\frac{1}{2}\left\Vert \frac{d}{dt}u^{N}\right\Vert _{2}^{2}+\frac{1}{2}\frac{d}{dt}\left\Vert \nabla u^{N}\right\Vert _{2}^{2}\leq\left\Vert h\left(t\right)\right\Vert _{2}^{2}+$$\left\Vert f\left(u^{N}\right)\right\Vert _{2}^{2}$,
take $\int_{I}ds$:
\[
\int_{0}^{t}\left\Vert \frac{d}{dt}u^{N}\left(s\right)\right\Vert _{2}^{2}+\left\Vert \nabla u^{N}\left(t\right)\right\Vert _{2}^{2}\leq2\int_{0}^{t}\left\Vert h\left(s\right)\right\Vert _{2}^{2}+\left\Vert f\left(u^{N}\left(s\right)\right)\right\Vert _{2}^{2}ds+\left\Vert \nabla u^{N}\left(0\right)\right\Vert _{2}^{2}
\]
Observe: RHS $\leq K$ independent of $t\in I$ and $N$: beacuse
$h\left(t\right)\in L^{2}\left(I,L^{2}\right)$, $u^{N}\left(t\right)$
is bounded in $L^{\infty}\left(I,L^{2}\right)$, $f\left(\cdot\right)$
lipschitz and $u^{N}\left(0\right)=P_{N}u\left(0\right)$, where $P_{N}:L^{2}\to\mbox{span}\left\{ w_{1},\ldots,w_{N}\right\} $
but we also know $\left\Vert P_{N}u\right\Vert _{1,2}\leq\left\Vert u\right\Vert _{1,2}$.
Hence $\left\Vert u^{N}\left(0\right)\right\Vert _{1,2}$ is bounded.

By argument of step 1: $u^{N}\left(t\right)$ bounded in $L^{\infty}\left(I,W_{0}^{1,2}\right)$
and $\frac{d}{dt}u^{N}\left(t\right)$ is bounded in $L^{2}\left(I,L^{2}\right)$.
Therefore w. s. $u\left(t\right)$ constructed in Thm 2.3 belongs
to these spaces. But we have uniqueness (Th. 2.1), therefore every
solution has this regularity.

\textbf{Step 3:} $u\left(t\right)\in L^{2}\left(I,W^{2,2}\right)$:

Recall elliptic regularity for laplacian: if $-\Delta u=F$ weakly
(i. e. $\left(\nabla u,\nabla v\right)=\left(F,v\right)$ for $\forall v\in W_{0}^{1,2}$),
where $F\in L^{2}$, then $u\in W^{2,2}$ and $\left\Vert u\right\Vert _{2,2}\leq C_{R}\left\Vert F\right\Vert _{2}$,
$C_{R}$ depends only on $\Omega$.

\begin{eqnarray*}
-\int_{I}\left(u\left(t\right),v\right)\varphi'\left(t\right)dt & + & \int_{I}\left(\nabla u\left(t\right),\nabla v\right)\varphi\left(t\right)dt+\int_{I}\left(f\left(u\left(t\right)\right),v\right)\varphi\left(t\right)dt\\
 &  & =\int_{I}\left(h\left(t\right),v\right)\varphi\left(t\right)dt\quad\forall v\in W_{0}^{1,2}\ \forall\varphi\left(t\right)\in D\left(I\right)
\end{eqnarray*}
but $\frac{d}{dt}u\left(t\right)=g\left(t\right)\in L^{2}\left(I,L^{2}\right)$
by step 2, hence first term$=\int_{I}\left(g\left(t\right),v\right)\varphi\left(t\right)dt$.
Therefore:

\[
\int_{I}\left(\nabla u\left(t\right),\nabla v\right)\varphi\left(t\right)dt=\int_{I}\left(F\left(t\right),v\right)\varphi\left(t\right)dt\quad\forall v,\varphi\left(t\right)
\]
where $F\left(t\right)=h\left(t\right)-f\left(u\left(t\right)\right)-g\left(t\right)$.
By standard argument: $\left(\nabla u\left(t\right),\nabla v\right)=\left(F\left(t\right),v\right)$
$\forall v\in W_{0}^{1,2}$ and for a. e. $t\in I$.

Now use elliptic regularity: $\left\Vert u\left(t\right)\right\Vert _{1,2}\leq C_{R}\left\Vert F\left(t\right)\right\Vert _{2}$.
But $F\left(t\right)\in L^{2}\left(I,L^{2}\right)$ hence $u\left(t\right)\in L^{2}\left(I,W^{2,2}\right)$.

\subsection*{Remarks:}
\begin{enumerate}
\item local regularity if $u\left(t\right)$ is w. s., $u\left(0\right)\in L^{2}$,
$h\left(t\right)\in L^{2}\left(I,L^{2}\right)$, but $u\left(t\right)\in L^{2}\left(I,W_{0}^{1,2}\right)$$\implies u\left(\tau\right)\in W_{0}^{1,2}$
for a. e. $\tau\in\left(0,T\right)$ $\implies u\left(t\right)\in L^{\infty}\left(\tau,T;W_{0}^{1,2}\right)\cap L^{2}\left(\tau,T,W_{0}^{1,2}\right)$,
$\frac{d}{dt}u\left(t\right)\in L^{2}\left(\tau,T,L^{2}\right)$ for
arbitrarily small $\tau>0$.,
\item even better data imply even better solutions.
\end{enumerate}

\chapter{Hyperbolic second order equations}

\begin{eqnarray*}
\left(H1\right)\quad\partial_{tt}u-\Delta u+\alpha\partial_{t}u+f\left(u\right) & = & h\left(t\right),\quad t\in I,x\in\Omega\\
\left(H2\right)\quad u & = & u_{0},\quad t=0,x\in\Omega\\
\left(H3\right)\quad\partial_{t}u & = & u_{1,},\quad t=0,x\in\Omega\\
\left(H4\right)\quad u & = & 0,\quad t\in I,x\in\partial\Omega
\end{eqnarray*}

\subsection*{Assumptions:}

$\left(A\right)\quad\Omega\subset\mathbb{R}^{n}$ bounded domain,
$\partial\Omega$ regular (lipschitz), $f\left(z\right):\mathbb{R}\to\mathbb{R}$,
$\left|f\left(z_{1}\right)-f\left(z_{2}\right)\right|\leq l\left|z_{1}-z_{2}\right|$,
$\forall z_{1},z_{2}\in\mathbb{R}$, $\alpha\in\mathbb{R}$.

\subsection*{Def.: {[}Weak solution of hyperbolic equation{]}}

\addcontentsline{toc}{section}{Definition of weak solution}

Function $u\left(t\right)\in L^{\infty}\left(I,W_{0}^{1,2}\right)$
with $\frac{d}{dt}u\in L^{\infty}\left(I,L^{2}\right)$ is called
weak solution of the problem $\left(H1\right)$, iff 
\[
\frac{d^{2}}{dt^{2}}\left(u\left(t\right),v\right)+\left(\nabla u\left(t\right),\nabla v\right)+\alpha\left(\frac{d}{dt}u\left(t\right),v\right)+\left(f\left(u\left(t\right)\right),v\right)=\left(h\left(t\right),v\right)
\]
 in the sense of distributions in $\left(0,T\right)$, for any $v\in W_{0}^{1,2}$
fixed.

\subsection*{Remarks: }
\begin{enumerate}
\item define $A:W_{0}^{1,2}\to W^{-1,2}$ by $\left\langle Au,v\right\rangle =\left(\nabla u,\nabla u\right)\forall v\in W_{0}^{1,2}$,
($A=-\Delta$), $u\left(t\right)$ is w. s. $\stackrel{\mbox{as in L2.1}}{\iff}\frac{d^{2}}{dt^{2}}\iota u\left(t\right)+Au\left(t\right)+\alpha\iota\frac{d}{dt}u\left(t\right)+\iota f\left(u\left(t\right)\right)=\iota h\left(t\right)$
in $W^{-1,2}$ or by d'Alambert: $\frac{d}{dt}u\left(t\right)=v\left(t\right)$
in $L^{2}$, 
\[
\frac{d}{dt}\iota v\left(t\right)+Au\left(t\right)+\alpha v\left(t\right)+\iota f\left(u\left(t\right)\right)=\iota h\left(t\right)\mbox{ in }W_{0}^{-1,2}
\]
\item continuous representatives: clearly $u\left(t\right)\in W^{1,2}\left(I,L^{2}\right)\hookrightarrow C\left(I,L^{2}\right)$,
$\frac{d}{dt}u\left(t\right)=v\left(t\right)\in W^{1,2}\left(I,W^{-1,2}\right)\hookrightarrow C\left(I,W^{-1,2}\right)$.
We will show better continuity later.\\
I. e.: $u\left(t\right)\in L^{\infty}\left(I,W_{0}^{1,2}\right)\hookrightarrow L^{2}\left(I,L^{2}\right),\frac{d}{dt}u\left(t\right)\in L^{2}\left(I,L^{2}\right)\implies u\left(t\right)\in C^{0,\frac{1}{2}}\left(I,L^{2}\right)$
by L. 1.5 ex. 2.1,\\
$\frac{d}{dt}u\in L^{2}\left(I,L^{2}\right)\stackrel{\iota}{\hookrightarrow}L^{2}\left(I,W^{-1,2}\right)$,
$\frac{d^{2}}{dt^{2}}u\left(t\right)=-Au\left(t\right)-\alpha\frac{d}{dt}u\left(t\right)-\iota f\left(u\left(t\right)\right)+h\left(t\right)\in L^{2}\left(I,W^{-1,2}\right)$,
therefore $\frac{d}{dt}u\left(t\right)\in C^{0,\frac{1}{2}}\left(I,W^{-1,2}\right)$.
\end{enumerate}
However (see ex. 4.1): if $X\hookrightarrow Z$, and $u\left(t\right)\in L^{\infty}\left(I,X\right)\cap C\left(I,Z\right)\implies u\left(t\right)\in C\left(I,X^{\mbox{weak}}\right)$.
Using this $u\left(t\right)\in C\left(I,\left(W_{0}^{1,2}\right)^{\mbox{weak}}\right)$,
$\frac{d}{dt}u\left(t\right)\in C\left(I,\left(L^{2}\right)^{\mbox{weak}}\right)$.

We even have have $u\left(t\right)\in C\left(I,W_{0}^{1,2}\right)$
and $\frac{d}{dt}u\left(t\right)\in C\left(I,L^{2}\right)$ for $u\left(t\right)$
weak solution. See the last remark of this chapter.

\subsection*{Def.:}

$u\left(t\right)\in C\left(I,X^{\mbox{weak}}\right)\iff t\mapsto\left\langle f,u\left(t\right)\right\rangle $
is continuous for $\forall f\in X^{*}$ fixed.

\subsection*{Remark:}

proving uniqueness the same way as for solution of parabolic equation:
let $u_{1}$, $u_{2}$ be weak solutions, $u_{1}\left(0\right)=u_{2}\left(0\right)$,
$\partial_{t}u_{1}\left(0\right)=\partial_{t}u_{2}\left(0\right)$,
set $u\left(t\right)=u_{1}-u_{2}$ \dots{} is $u\left(t\right)=0\ \forall t\in I$?

Subtract equations for $u_{1}$, $u_{2}$: 
\[
\partial_{tt}u-\Delta u+\alpha\partial_{t}u+f\left(u_{1}\right)-f\left(u_{2}\right)=0/\mbox{formally }\partial_{t}u,\int_{\Omega}\dx
\]
\begin{itemize}
\item first term: $\int_{\Omega}\underbrace{\partial_{tt}u\partial_{t}ud}_{\frac{1}{2}\partial_{t}\left(\partial_{t}u\right)^{2}}x=\frac{1}{2}\frac{d}{dt}\left\Vert \partial_{t}u\right\Vert _{2}^{2}$,
\item second term: $\int_{\Omega}\nabla u\cdot\nabla\left(\partial_{t}u\right)\dx=\frac{1}{2}\frac{d}{dt}\left\Vert \nabla u\right\Vert _{2}^{2}$,
\item third term: $\alpha\left\Vert \partial_{t}u\right\Vert _{2}^{2}$,
\item fourth term: $\left|\left(f\left(u_{1}\right)-f\left(u_{2}\right),\partial_{t}u\right)\right|\leq l\left\Vert u\right\Vert _{2}\left\Vert \partial_{t}u\right\Vert _{2}\leq\ldots$
\end{itemize}
Together: $\frac{d}{dt}\left(\frac{1}{2}\left\Vert \partial_{t}u\right\Vert _{2}^{2}+\frac{1}{2}\left\Vert \nabla u\right\Vert _{2}^{2}\right)\leq c\left(\left\Vert \partial_{t}u\right\Vert _{2}^{2}+\left\Vert u\right\Vert _{2}^{2}\right)$,
use Gronwall lemma.

\subsection*{Problem:}

we can not do this for w. s., since $\partial_{tt}u\in L^{2}\left(I,W^{-1,2}\right),\partial_{t}u\in L^{2}\left(I,L^{2}\right)$,
and that is not enough for first and second term above.

\subsection*{Remark:}
\begin{itemize}
\item Goal: existence and (more precisely ``or'') uniqueness of solution.
The solution could be weak, strong, classical or on the other hand
very weak, in measures, in distributions. The uniqueness is easy to
prove for stronger solutions, whereas the existence is easier for
weaker ones.
\item Parabolic equation: 
\[
\frac{d}{dt}u+\mathcal{A}\left(u\right)+\ldots\quad/\cdot u,\int_{\Omega}\dx
\]
\[
\implies\frac{1}{2}\frac{d}{dt}\left\Vert u\right\Vert _{2}^{2}+\underbrace{\left\langle \mathcal{A}\left(u\right),u\right\rangle }_{\geq c_{1}\left\Vert u\right\Vert _{1,2}^{2}}+\ldots\quad/\int_{0}^{\tau}dt
\]
\[
\left\Vert u\left(\tau\right)\right\Vert _{2}^{2}+\int_{0}^{\tau}\left\Vert u\left(t\right)\right\Vert _{1,2}^{2}dt\leq C=C\left(u_{0},h\left(t\right),\Omega\right)
\]
$\implies$ a priori estimates: $u\left(t\right)\in L^{\infty}\left(I,L^{2}\right)\cap L^{2}\left(I,W^{1,2}\right)$,
$\frac{d}{dt}u\left(t\right)\in L^{2}\left(I,W^{-1,2}\right)$ (formal)\\
a priori estimates + approximation (Galerkin: $u^{N}\left(t\right)=\sum_{j=1}^{N}c_{j}^{N}\left(t\right)w_{j}$)
$\implies$ existence of a solution in the class of solutions above.
A priori estimates are valid ($u\left(t\right)$ is admissible test
function) $\implies$ easy proof of uniquenes: (a priori estimates
for difference of solutions $w=u-v$, Gronwall).
\item hyperbolic equation: 
\begin{eqnarray*}
\frac{d^{2}}{dt^{2}}u+Au+\alpha\frac{d}{dt}u+f\left(u\right) & = & h\left(t,x\right)\\
u\left(0\right) & = & u_{0}\\
\frac{d}{dt}u\left(0\right) & = & u_{1}
\end{eqnarray*}
assumptions: $A=-\Delta$ (weak), $f\left(\cdot\right)$ globally
lipschitz, $\alpha\in\mathbb{R}$, $h\left(t,x\right)\in L^{2}\left(I,L^{2}\right)$,
$u_{0}\in W_{0}^{1,2}$, $u_{1}\in L^{2}$.\\
Formal a priori estimates: use $\frac{d}{dt}u$ as a test functions,
the first and second term yield: $\frac{d}{dt}E\left[u\right]$, where
$E\left[u\right]=\frac{1}{2}\left(\left\Vert \frac{d}{dt}u\right\Vert _{2}^{2}+\left\Vert \nabla u\right\Vert _{2}^{2}\right)$
is energy (sensible defined for physicists). Integrate $\int_{0}^{\tau}dt\implies$$E\left[u\left(\tau\right)\right]\leq C=C\left(\mathrm{data}\right)$$\implies u\in L^{\infty}\left(I,W_{0}^{1,2}\right)$,
$\frac{d}{dt}u\left(t\right)\in L^{\infty}\left(I,L^{2}\right)$.
Problem: this can not be done, since $\frac{d^{2}}{dt^{2}}u\left(t\right)\in L^{\infty}\left(I,W^{-1,2}\right)$
is not admissible test function.
\end{itemize}

\subsection*{Theorem 3.1 {[}Uniqueness of weak solution of hyperbolic equation{]}}

\addcontentsline{toc}{section}{Uniqueness of weak solution}

The weak solution is uniquely defined by the initial conditions.

\subsubsection*{Proof:}

let $u\left(t\right)$, $v\left(t\right)$ be weak solutions, $u\left(0\right)=v\left(0\right)$,
$\frac{d}{dt}u\left(0\right)=\frac{d}{dt}v\left(0\right)$, denote
$w\left(t\right)=u\left(t\right)-v\left(t\right)$. $w\left(t\right)$
satisfies:
\begin{align*}
\frac{d^{2}}{dt^{2}}\iota w+Aw\left(t\right)+\alpha\iota\frac{d}{dt}w+\iota\left(f\left(u\right)-f\left(v\right)\right) & =0\mbox{}\\
\noalign{\centerline{\text{(weakly in \ensuremath{I}: equation in \ensuremath{W^{-1,2}}, \ensuremath{w\left(0\right)=0}, \ensuremath{\frac{d}{dt}w\left(0\right)=0}})}}
\end{align*}
\textbf{Trick:} $\int_{0}^{t}ds$: four terms:
\begin{enumerate}
\item $\int_{0}^{t}\frac{d^{2}}{dt^{2}}\iota wds=\left[\frac{d}{dt}\iota w\right]_{0}^{t}=\frac{d}{dt}\iota w\left(t\right)$
\item $\int_{0}^{t}Aw\left(s\right)ds=A\underbrace{\int_{0}^{t}w\left(s\right)ds}_{\makebox[0pt][c]{\mbox{denote }W\ensuremath{\left(t\right)\in W_{0}^{1,2}}}}=AW\left(t\right)\in W^{-1,2}$
\item $\alpha\int_{0}^{t}\frac{d}{dt}\iota w\left(s\right)ds=\alpha\left[\iota w\left(s\right)\right]_{0}^{t}$
\item $\iota\int_{0}^{t}f\left(u\left(s\right)\right)-f\left(v\left(s\right)\right)ds$
\end{enumerate}
Together:
\[
\frac{d}{dt}\iota w+AW+\alpha W+\iota\int_{0}^{t}f\left(u\left(s\right)\right)-f\left(v\left(s\right)\right)ds=0\quad\mbox{weakly in \ensuremath{I} with values in \ensuremath{W^{-1,2}}, \ensuremath{\big<\cdot,\underbrace{w}_{\in\makebox[0pt][l]{\ensuremath{W_{0}^{1,2}}}}}\ensuremath{\big>}}
\]
\begin{enumerate}
\item $\left\langle \frac{d}{dt}\iota w,w\right\rangle =\left(\frac{d}{dt}w,w\right)=\frac{1}{2}\frac{d}{dt}\left\Vert w\right\Vert _{2}^{2}$\\
Remark: special case of Th. 1.12: $u\left(t\right)\in L^{2}\left(I,H\right)$,
$\frac{d}{dt}u\in L^{2}\left(I,H\right)$$\implies\frac{d}{dt}\left\Vert u\left(t\right)\right\Vert _{H}^{2}=2\left(\frac{d}{dt}u\left(t\right),u\left(t\right)\right)_{H}$
weakly. Apply for $H=L^{2}$.
\item Apply previous remark for $H=W_{0}^{1,2}$: 
\[
\left\langle AW,w\right\rangle =\left(\nabla W,\nabla w\right)=\underbrace{\Big(\Big(W,\overbrace{w}^{\mbox{\ensuremath{=}}\makebox[0pt][l]{\ensuremath{\frac{d}{dt}W}}}\Big)\Big)}_{\makebox[0pt][r]{\mbox{scalar product in \ensuremath{W_{0}^{1,2}}}}}\stackrel{\text{remark}}{=}\frac{1}{2}\frac{d}{dt}\underbrace{\left\Vert W\right\Vert _{1,2}^{2}}_{\makebox[0pt][l]{norm induced by\emph{} \ensuremath{\left(\left(\cdot,\cdot\right)\right)}}}=\frac{1}{2}\frac{d}{dt}\left\Vert \nabla W\right\Vert _{2}^{2}
\]
\item $\left|\alpha\left\langle \iota w,w\right\rangle \right|=\left|\alpha\left(w,w\right)\right|\leq\left|\alpha\right|\left\Vert w\right\Vert _{2}^{2}$
\item 
\begin{align*}
\left|\left\langle \iota\int_{0}^{t}f\left(u\left(s\right)\right)-f\left(v\left(s\right)\right)ds,w\left(t\right)\right\rangle \right| & =\left|\int_{0}^{t}\left(f\left(u\left(s\right)\right)-f\left(v\left(s\right)\right)ds,w\left(t\right)\right)\right|\\
 & \leq\int_{0}^{t}\underbrace{\left\Vert f\left(u\left(s\right)\right)-f\left(v\left(s\right)\right)\right\Vert _{2}}_{\leq\ell\left\Vert w\left(s\right)\right\Vert _{2}}\cdot\left\Vert w\left(t\right)\right\Vert ds\\
 & \stackrel{\mbox{\ensuremath{f} lipsch.}}{\leq}\ell\int_{0}^{t}\left\Vert w\left(s\right)\right\Vert _{2}\left\Vert w\left(t\right)\right\Vert ds\\
 & \stackrel{\mbox{Young}}{\leq}\int_{0}^{t}\frac{\ell^{2}}{2}\left\Vert w\left(s\right)\right\Vert ds+\frac{t}{2}\left\Vert w\left(t\right)\right\Vert _{2}^{2}
\end{align*}
$ $
\end{enumerate}
Together: $\frac{1}{2}\frac{d}{dt}\Big(\underbrace{\left\Vert w\left(t\right)\right\Vert _{2}^{2}+\left\Vert \nabla W\left(t\right)\right\Vert _{2}^{2}}_{\mbox{denote by \ensuremath{Y\left(t\right)}}}\Big)\leq\underbrace{\left(\left|\alpha\right|+\frac{t}{2}\right)}_{\leq c_{1}}\underbrace{\left\Vert w\left(t\right)\right\Vert _{2}^{2}}_{\leq Y\left(t\right)}+\frac{l^{2}}{2}\int_{0}^{t}\left\Vert w\left(s\right)\right\Vert _{2}^{2}ds$,
$/\int_{0}^{\tau}dt$, we know: $Y\left(0\right)=0$ due to the initial
conditions:
\[
Y\left(\tau\right)\leq c_{1}\int_{0}^{\tau}Y\left(t\right)ds+\frac{l^{2}}{2}\underbrace{\int_{0}^{\tau}\int_{0}^{t}\left\Vert w\left(s\right)\right\Vert _{2}^{2}dsdt}_{\mbox{\ensuremath{\leq}}\makebox[0pt][l]{\ensuremath{\,c\int_{0}^{\tau}\left\Vert w\left(s\right)\right\Vert _{2}^{2}ds\mbox{ \dots\ Fubini}}}}
\]

So $Y\left(\tau\right)\leq c_{3}\int_{0}^{\tau}Y\left(s\right)ds$
for $\forall\tau\in I$, Gronwall: $Y\left(\tau\right)\leq0$ pro
$\forall\tau\in I$, i. e. $w\left(t\right)=0$ a. e. in $I$. The
proof is done.

The trick works due to linearity of all problematic terms and the
integration which increases regularity.

\subsection*{Lemma 3.1 {[}Testing hyperbolic equation by the derivative{]}}

\addcontentsline{toc}{section}{Time derivative as a test function}

Let $u\left(t\right)\in L^{2}\left(I,W_{0}^{1,2}\right)$, $\frac{d}{dt}u\left(t\right)\in L^{2}\left(I,L^{2}\right)$,
let $\frac{d^{2}}{dt^{2}}\iota u+Au=\iota H\left(t\right)$ weakly
in $I$, where $H\left(t\right)\in L^{2}\left(I,L^{2}\right)$. Then
$t\mapsto E\left[u\left(t\right)\right]$ is weakly differentiable
and $\frac{d}{dt}E\left[u\left(t\right)\right]=\left(H\left(t\right),\frac{d}{dt}u\left(t\right)\right)$
for a. e. $t\in I$.

\subsubsection*{Proof:}

molifying by convolution: $\psi_{n}\left(t\right)=n\psi_{0}\left(nt\right)$,
$\psi_{0}\left(t\right)$ \dots{} convolution (smooth) molifier. Define
$u_{n}\left(t\right)=u*\psi_{n}\left(t\right)\in C^{\infty}\left(J,W_{0}^{1,2}\right)$,
where $J=\left(\delta,T-\delta\right)$, $\delta>0$ fixed, $\frac{1}{n}<\delta$,
here $u*\psi_{n}\left(t\right)=\int_{\mathbb{R}}u\left(s\right)\psi_{n}\left(t-s\right)ds=\int_{\mathbb{R}}u\left(t-s\right)\psi_{n}\left(t\right)ds$
makes sense.

We know: $u_{n}''\left(t\right)=u*\psi_{n}''\left(t\right)=\int_{\mathbb{R}}u\left(s\right)\psi_{n}''\left(t-s\right)ds$.

The equation (weakly): $\int_{I}\iota u\left(s\right)\varphi''\left(s\right)ds+\int_{I}Au\left(s\right)\varphi\left(s\right)ds=\int_{I}\iota H\left(s\right)\varphi\left(s\right)ds$
for $\forall\varphi\left(s\right)\in C_{c}^{\infty}\left(I\right)$
choose $\varphi\left(s\right)=\psi_{n}\left(t-s\right)$, $t\in J$
fixed, i. e..:
\begin{align*}
\left\langle \cdot,u_{n}'\left(t\right)\right\rangle /\quad\iota u''_{n}\left(t\right)+Au_{n}\left(t\right) & =\iota H_{n}\left(t\right)\quad\mbox{in \ensuremath{J}}\\
\noalign{\centerline{\text{(holds classically)}}}
\end{align*}
where $H_{n}\left(t\right)=H\left(t\right)*\psi_{n}\left(t\right)$.
Thus we get:
\begin{eqnarray*}
\frac{d}{dt}E\left[u_{n}\left(t\right)\right] & = & \left\langle \iota H_{n}\left(t\right),\frac{d}{dt}u_{n}\left(t\right)\right\rangle =\left(H_{n}\left(t\right),\frac{d}{dt}u_{n}\left(t\right)\right)\quad\Bigg/\int_{0}^{\tau}dt\\
E\left[u_{n}\left(\tau\right)\right] & = & E\left[u_{n}\left(0\right)\right]+\int_{0}^{\tau}\left(H_{n}\left(t\right),\frac{d}{dt}u_{n}\left(t\right)\right)dt\quad;n\to\infty\\
E\left[u\left(\tau\right)\right] & = & E\left[u\left(0\right)\right]+\int_{0}^{\tau}\left(H\left(t\right),\frac{d}{dt}u\left(t\right)\right)dt
\end{eqnarray*}
since $u_{n}\left(t\right)\to u\left(t\right)$ in $W_{0}^{1,2}$,
$\frac{d}{dt}u_{n}\left(t\right)\to\frac{d}{dt}u_{n}\left(t\right)$
v $L^{2}$ for a. e. $t\in I$; $H_{n}\left(t\right)\to H\left(t\right)$
v $L^{2}\left(I,L^{2}\right)$, by which the proof is done.

\subsection*{Corollary:}

uniqueness of weak solution (other proof): apply L. 3.1 to $w\left(t\right)=u\left(t\right)-v\left(t\right)$,
where $u\left(t\right)$ and $v\left(t\right)$ are solutions. I.
e. $\frac{d^{2}}{dt^{2}}\iota w+Aw=\iota\underbrace{\left(-\alpha w+f\left(v\right)-f\left(u\right)\right)}_{\mbox{denote by }H\left(t\right)\in L^{2}\left(I,L^{2}\right)}$,
we get: $\frac{d}{dt}E\left[u\left(t\right)\right]=\left(-w+\left(f\left(u\right)-f\left(v\right)\right),\frac{d}{dt}w\right)\leq cE\left[w\right]$
(Young, routine estimates). Gronwall: $E\left[w\left(\tau\right)\right]\leq E\left[w\left(0\right)\right]e^{c\tau}$
for $\forall\tau\in I$. Uniqueness, moreover, continuous dependence
on initial condition.

\subsection*{Theorem 3.2: {[}Existence of weak solution for hyperbolic equation{]}}

\addcontentsline{toc}{section}{Existence of weak solution}

Let $u_{0}\in W_{0}^{1,2}$, $u_{1}\in L^{2}$, $h\left(t\right)\in L^{2}\left(I,L^{2}\right)$.
Then there exists $u\left(t\right)\in L^{\infty}\left(I,W_{0}^{1,2}\right)$
such that $\frac{d}{dt}u\left(t\right)\in L^{\infty}\left(I,L^{2}\right)$
is weak solution of $\left(H1\right)$ with initial conditions $u\left(0\right)=u_{0}$,
$\frac{d}{dt}u\left(0\right)=u_{1}$ in the sense of continuous representatives.

\subsubsection*{Proof:}

\textbf{Step 1:} Galerkin approximation: $u^{N}\left(t\right)=\sum_{j=1}^{N}c_{j}^{N}\left(t\right)w_{j}$,
$w_{j}\in W_{0}^{1,2}$ are laplace eigenfunctions, $c_{j}^{N}\left(t\right):I\to\mathbb{R}$,
solve the system of equations
\begin{eqnarray*}
\left(H_{N}\right)\quad\frac{d^{2}}{dt^{2}}\left(\iota u^{N}\left(t\right),w_{j}\right) & + & \left\langle Au^{N}\left(t\right),w_{j}\right\rangle +\alpha\left(\frac{d}{dt}u^{N}\left(t\right),w_{j}\right)\\
+\left(f\left(u_{n}\left(t\right)\right),w_{j}\right) &  & =\left(h\left(t\right),w_{j}\right)\quad\mbox{v }I,\forall j=1,\dots,N
\end{eqnarray*}
Initial conditions: $u^{N}\left(0\right)=P_{N}u_{0}$, $\frac{d}{dt}u^{N}\left(0\right)=P_{N}u_{1}$.

Observe: $\left(u^{N}\left(t\right),w_{j}\right)=c_{j}^{N}\left(t\right)$,
$\left\langle Au_{N}\left(t\right),w_{j}\right\rangle =\left(\nabla u^{N}\left(t\right),\nabla w_{j}\right)=\lambda_{j}\underbrace{\left(u^{N}\left(t\right),w_{j}\right)}_{c_{j}^{N}\left(t\right)}$.
Therefore 
\begin{eqnarray*}
\left(H_{N}\right)\quad\partial_{tt}c_{j}^{N}+\lambda_{j}c_{j}^{N}+\alpha\frac{d}{dt}c_{j}^{N}+\mathcal{F}_{j}\left(t,c_{1}^{N},\dots,c_{N}^{N}\right) & = & 0\\
c_{j}^{N}\left(0\right) & = & \left(u_{0},w_{j}\right)\\
\frac{d}{dt}c_{j}^{N}\left(0\right) & = & \left(u_{1},w_{j}\right)
\end{eqnarray*}
where $\mathcal{F}_{j}\left(t,c_{1}^{N},\dots,c_{N}^{N}\right)$=$\left(f\left(\sum_{l=1}^{N}c_{l}^{N}w_{l}\right),w_{j}\right)-\left(h\left(t\right),w_{j}\right)$.
I. e., we have a system of nonlinear equations. The nonlinearity of
$\mathcal{F}$ satisfies Carathéodory conditions (lipschitz with respect
to $c_{k}^{N}$, integrable in $t$) $\implies$ therefore there exists
exactly one solution $c_{j}^{N}\left(t\right),\frac{d}{dt}c_{j}^{N}\in AC\left(I\right)$.

\textbf{Step 2:} estimates independent of $N$. $\left(H_{N}\right)\cdot\frac{d}{dt}c_{j}^{N}$,
$\sum_{j=1}^{N}$ (formally analogous to multiplication of equation
by time derivative and integration $\int_{\Omega}dx$). We get five
terms:
\begin{enumerate}
\item $\left(\frac{d^{2}}{dt^{2}}u^{N},\frac{d}{dt}u^{N}\right)=\frac{1}{2}\frac{d}{dt}\left\Vert \frac{d}{dt}u^{N}\right\Vert _{2}^{2}$
\item $\left\langle Au^{N},\frac{d}{dt}u^{N}\right\rangle =\left(\nabla u^{N},\nabla\left(\frac{d}{dt}u^{N}\right)\right)=\frac{1}{2}\frac{d}{dt}\left\Vert \nabla u^{N}\right\Vert _{2}^{2}$.
Together with the first term gives $\frac{d}{dt}E\left[u^{N}\right]$
\item $\left|\alpha\left(\frac{d}{dt}u^{N},\frac{d}{dt}u^{N}\right)\right|\leq\left|\alpha\right|\cdot\left\Vert \frac{d}{dt}u^{N}\right\Vert _{2}^{2}$
\item 
\begin{eqnarray*}
\left|\left(f\left(u^{N}\right),\frac{d}{dt}u^{N}\right)\right| & \leq & \left\Vert f\left(u^{N}\right)\right\Vert _{2}\left\Vert \frac{d}{dt}u^{N}\right\Vert _{2}\stackrel{\mbox{Ex. 3.2}}{\leq}c\left(1+\left\Vert u^{N}\right\Vert _{2}\right)\cdot\left\Vert \frac{d}{dt}u^{N}\right\Vert _{2}\\
 & \stackrel{\mbox{Poincaré}}{\leq} & c_{1}\left(1+\left\Vert \nabla u^{N}\right\Vert _{2}\right)\cdot\left\Vert \frac{d}{dt}u^{N}\right\Vert _{2}\\
 & \stackrel{\mbox{Young}}{\leq} & c_{3}\Bigg(1+\underbrace{\left\Vert \nabla u^{N}\right\Vert _{2}^{2}+\left\Vert \frac{d}{dt}u^{N}\right\Vert _{2}^{2}}_{2E\left[u^{N}\right]}\Bigg)
\end{eqnarray*}
\item $\left|\left(h\left(t\right),\frac{d}{dt}u^{N}\right)\right|\leq\left\Vert h\left(t\right)\right\Vert _{2}\left\Vert \frac{d}{dt}u^{N}\right\Vert _{2}\leq\frac{1}{2}\left\Vert h\left(t\right)\right\Vert _{2}^{2}+\underbrace{\frac{1}{2}\left\Vert \frac{d}{dt}u^{N}\right\Vert _{2}^{2}}_{\leq E\left[u^{N}\right]}$
\end{enumerate}
Together: 
\[
\frac{d}{dt}E\left[u^{N}\left(t\right)\right]\leq c_{4}E\left[u^{N}\left(t\right)\right]+\frac{1}{2}\left\Vert h\left(t\right)\right\Vert _{2}^{2}\quad\Bigg/\int_{0}^{\tau}dt
\]
 
\[
E\left[u^{N}\left(\tau\right)\right]\leq\underbrace{E\left[u^{N}\left(0\right)\right]+\frac{1}{2}\int_{o}^{\tau}\left\Vert h\left(t\right)\right\Vert _{2}^{2}}_{\leq\makebox[0pt][l]{\ensuremath{\,K}\mbox{\ independent of \ensuremath{N} (\ensuremath{\sun})}}}+c_{4}\int_{0}^{\tau}E\left[u^{N}\left(t\right)\right]dt,
\]
where $K$ is independent of $N$ due to $h\left(t\right)\in L^{2}\left(I,L^{2}\right)$
and Gronwall: $E\left[u^{N}\left(t\right)\right]\leq K$ , $\forall N\in\mathbb{N},\forall\tau\in I$,
i. e., $u^{N}\left(t\right)$ is bounded in $L^{\infty}\left(I,W_{0}^{1,2}\right)$,
$\frac{d}{dt}u^{N}\left(t\right)$ bounded in $L^{\infty}\left(I,L^{2}\right)$.
The estimate $\left(\sun\right)$ holds since $h\left(t\right)\in L^{2}\left(I,L^{2}\right)$
and $E\left[u^{N}\left(0\right)\right]=\frac{1}{2}\left\Vert \nabla P_{N}u_{0}\right\Vert _{2}^{2}+\frac{1}{2}\left\Vert P_{N}u_{1}\right\Vert _{2}^{2}\leq\frac{1}{2}\left\Vert \nabla u_{0}\right\Vert _{2}^{2}+\frac{1}{2}\left\Vert u_{1}\right\Vert _{2}^{2}$.

We claim: $\iota\frac{d^{2}}{dt^{2}}u^{N}\left(t\right)$ is bounded
in $L^{2}\left(I,W^{-1,2}\right)$. Use: $\left\Vert \iota\frac{d^{2}}{dt^{2}}u^{N}\left(t\right)\right\Vert _{-1,2}=\sup_{{w\in W_{0}^{1,2}\atop \left\Vert w\right\Vert =1}}\left\langle \iota\frac{d^{2}}{dt^{2}}u^{N}\left(t\right),w\right\rangle $:

\begin{eqnarray*}
\left\langle \iota\frac{d^{2}}{dt^{2}}u^{N}\left(t\right),w\right\rangle  & = & \left(\frac{d^{2}}{dt^{2}}u^{N}\left(t\right),w\right)=\left(\frac{d^{2}}{dt^{2}}P_{N}u^{N}\left(t\right),w\right)\\
 & = & \left(\frac{d^{2}}{dt^{2}}u^{N}\left(t\right),P_{N}w\right)=\left\langle \iota\frac{d^{2}}{dt^{2}}u^{N}\left(t\right),P_{N}w\right\rangle 
\end{eqnarray*}
Therefore $\left|\left\langle \iota\frac{d^{2}}{dt^{2}}u^{N},w\right\rangle \right|=\left|\left(\frac{d^{2}}{dt^{2}}u^{N}\left(t\right),P_{N}w\right)\right|\stackrel{*}{=}$

Aside:
\[
\left(H_{N}\right):\quad\Big<\iota\frac{d^{2}}{dt^{2}}u^{N}\left(t\right)+Au^{N}\left(t\right)+\alpha\iota\frac{d}{dt}u^{n}\left(t\right)+\iota f\left(u^{N}\left(t\right)\right)-\iota h\left(t\right),\underbrace{w_{j}}_{\makebox[0pt][c]{generate \ensuremath{P_{N}}}}\Big>=0\quad\forall j=1,\dots,N.
\]
Therefore $\iff P_{N}\left(\iota\frac{d^{2}}{dt^{2}}u^{N}\left(t\right)+Au^{N}\left(t\right)+\alpha\iota\frac{d}{dt}u^{n}\left(t\right)+\iota f\left(u^{N}\left(t\right)\right)-h\left(t\right)\right)=0$
and $\left\langle \iota\frac{d^{2}}{dt^{2}}u^{N}\left(t\right),P_{N}w\right\rangle =\left\langle -Au^{N}+\alpha\iota\frac{d}{dt}u^{N}+\iota f\left(u^{N}\right)-\iota h\left(t\right),P_{N}w\right\rangle $

$\stackrel{*}{=}\left|\left\langle \iota\frac{d^{2}}{dt^{2}}u,w\right\rangle \right|\leq\left\Vert -Au^{N}\left(t\right)-\alpha\iota\frac{d}{dt}u^{N}\left(t\right)-\iota f\left(u^{N}\left(t\right)\right)+\iota h\left(t\right)\right\Vert _{-1,2}\underbrace{\left\Vert P_{N}w\right\Vert _{1,2}}_{\mbox{\ensuremath{\leq}}\makebox[0pt][l]{\ensuremath{\,\left\Vert w\right\Vert _{1,2}=1}}}$. 

Use the estimates $\left\Vert Au^{N}\left(t\right)\right\Vert _{-1,2}\leq\left\Vert \nabla u^{N}\left(t\right)\right\Vert _{2}$,
$\left\Vert \iota\frac{d}{dt}u^{N}\left(t\right)\right\Vert _{-1,2}\leq c\left\Vert \frac{d}{dt}u^{N}\left(t\right)\right\Vert $.

Together: $\left\Vert \iota\frac{d^{2}}{dt^{2}}u^{N}\left(t\right)\right\Vert _{-1,2}\leq c\Big(1+\underbrace{\left\Vert \nabla u^{N}\left(t\right)\right\Vert _{2}+\left\Vert \frac{d}{dt}u^{N}\left(t\right)\right\Vert _{2}+\left\Vert u^{N}\left(t\right)\right\Vert _{2}+\left\Vert h\left(t\right)\right\Vert _{2}}_{\mbox{bounded in }L^{2}\left(I\right)\mbox{ independently of \ensuremath{N}}}\Big)$.

\textbf{Step 3:} passage to the limit: Eberlein\textendash Śmuljan:
$\exists u\left(t\right)$ t. ¸. $u^{N}\left(t\right)\stackrel{*}{\rightharpoonup}u\left(t\right)$
weakly-{*} v $L^{\infty}\left(I,W_{0}^{1,2}\right)$, $\frac{d}{dt}u^{N}\left(t\right)\stackrel{*}{\rightharpoonup}\frac{d}{dt}u\left(t\right)$
weakly-{*} in $L^{\infty}\left(I,L^{2}\right)$, due to Aubin\textendash Lions
lemma $u^{N}\left(t\right)\to u\left(t\right)$ strongly in $L^{2}\left(I,L^{2}\right)$
and therefore $f\left(u^{N}\left(t\right)\right)\to f\left(u\left(t\right)\right)$
strongly in $L^{2}\left(I,L^{2}\right)$ (see ex.~3.2).

$\left(H_{N}\right)\cdot\varphi\left(t\right)\in C_{c}^{\infty}\left(I\right),\int_{I}dt$:

first term: $\left(\frac{d^{2}}{dt^{2}}u^{N}\left(t\right),w_{j}\right)\varphi\left(t\right)$\dots $2\times$by
parts, we get:
\begin{align*}
\int_{I}\left(u^{N}\left(t\right),w_{j}\right)\varphi''\left(t\right)dt+\int_{I}\left(\nabla u^{N}\left(t\right),\nabla w_{j}\right)\varphi dt+\alpha\int_{I}\left(\frac{d}{dt}u^{N}\left(t\right),w_{j}\right)dt+\int_{I}\left(f\left(u^{N}\right),w_{j}\right)\varphi\left(t\right)dt & =\\
\int_{I}\left(h\left(t\right),w_{j}\right)\varphi\left(t\right)dt
\end{align*}
. $u^{N}\to u$ due to the convergences above we can take the limit
in every term, holds for $\forall w_{j}$ since $w_{j}$ are dense
it holds for arbitrary $w\in W_{0}^{1,2}$, i. e., $u\left(t\right)$
is a weak solution.

\textbf{Step 4:} the initial conditions are satisfied (for continuous
representatives): let us consider the space $\mathcal{W}=\left\{ u\left(t\right)\in L^{2}\left(I,L^{2}\right),\frac{d}{dt}u\left(t\right)\in L^{2}\left(I,L^{2}\right)\right\} =W^{1,2}\left(I,L^{2}\right)$.
We know: $u^{N}\left(t\right)\rightharpoonup u\left(t\right)$ v $\mathcal{W}$.
Also: the operator $\tau:\mathcal{W}\to L^{2}$, $u\left(\cdot\right)\mapsto\tilde{u}\left(\tau\right)$
is continuous (L. 1.5), i. e. $u^{N}\left(\tau\right)\rightharpoonup\tilde{u}\left(\tau\right)$
in $L^{2}$ for $\forall\tau\in I$. We also know that $u^{N}\left(0\right)=P_{N}u_{0}\to u_{0}$
strongly in $N\to\infty$. Therefore $\tilde{u}\left(0\right)=u_{0}$.
For $u_{1}=\frac{d}{dt}\widetilde{u}\left(0\right)$ in $W^{-1,2}$
analogously. Thus the theorem is proven.

\subsection*{Theorem 3.3. {[}Strong solution of hyperbolic equation{]}}

\addcontentsline{toc}{section}{Existence of strong solution}

Let $u\left(t\right)$ be a weak solution of $\left(H1\right)$, let
$\Delta u_{0}+h\left(0\right)-f\left(u\left(0\right)\right)\in L^{2}$,
$u_{1}\in W_{0}^{1,2}$, $\frac{d}{dt}h\left(t\right)\in L^{2}\left(I,L^{2}\right)$
and $f\left(z\right):\mathbb{R}\to\mathbb{R}$ is smooth with bounded
derivative, let $\partial\Omega\in C^{2}$. Then $u\left(t\right)\in L^{\infty}\left(I,W^{2,2}\right)$,
$\frac{d}{dt}u\left(t\right)\in L^{\infty}\left(I,W_{0}^{1,2}\right)$
and $\frac{d^{2}}{dt^{2}}u\left(t\right)\in L^{\infty}\left(I,L^{2}\right)$.

\subsubsection*{Proof: }

(formally): $\frac{d^{2}}{dt^{2}}u+Au+\alpha\frac{d}{dt}u=h\left(t\right)-f\left(u\left(t\right)\right)\quad\Big/\frac{d}{dt}$,
denote $v=\frac{d}{dt}u$. We get: $\frac{d^{2}}{dt^{2}}v+Av+\alpha\frac{d}{dt}v=\underbrace{\frac{d}{dt}h\left(t\right)-f'\left(u\left(t\right)\right)v\left(t\right)}_{\mbox{denote by }H\left(t\right)\in L^{2}\left(I,L^{2}\right)}$,
$v\left(0\right)=\frac{d}{dt}u\left(0\right)=u_{1}\in W_{0}^{1,2}$,
$\frac{d}{dt}v\left(0\right)=\frac{d^{2}}{dt^{2}}u\left(0\right)\stackrel{\mbox{equation for \ensuremath{u\left(t\right)}}}{=}\Delta u\left(0\right)+h\left(0\right)-f\left(u\left(0\right)\right)-\alpha\frac{d}{dt}u\left(0\right)\in L^{2}$.
Estimates (see step 2 in Th. 3.2) for $v\left(t\right)$ $\implies v\left(t\right)\in L^{\infty}\left(I,W_{0}^{1,2}\right)$,
$\frac{d}{dt}v\left(t\right)\in L^{\infty}\left(I,L^{2}\right)$.
It remains to show $u\left(t\right)\in L^{\infty}\left(I,W_{2,2}\right)$:
we want $\left\Vert u\left(t\right)\right\Vert _{2,2}\in L^{\infty}\left(I\right)$
\dots{} again by the equation: $-\Delta u\left(t\right)=\underbrace{h\left(t\right)-f\left(u\left(t\right)\right)-\alpha\frac{d}{dt}u\left(t\right)-\frac{d^{2}}{dt^{2}}u\left(t\right)}_{\mbox{denote by }F\left(t\right)}$.
By previous estimates we know (for $h\left(t\right)$ due to L. 1.5)
we know that $F\left(t\right)\in L^{\infty}\left(I,L^{2}\right)$.
By elliptic regularity: $\left\Vert u\left(t\right)\right\Vert _{2,2}\leq c_{R}\left\Vert F\left(t\right)\right\Vert _{2}$,
where $c_{R}$ depends on $\partial\Omega$, we need $C^{2}$ regularity
of boundary. More correct proof: for $u^{N}\left(t\right)$ suitable
approximations (i. e., Galerkin). 

\subsection*{Remarks:}
\begin{itemize}
\item If $u\left(t\right)$ is a weak solution of $\left(H1\right)$ and
it holds that $u\left(t\right)\in L^{\infty}\left(I,W^{2,2}\right)$,
$\frac{d}{dt}u\left(t\right)\in L^{\infty}\left(I,W_{0}^{2,2}\right)$,
then $u\left(t\right)\in W^{2,2}$, $\frac{d}{dt}u\left(t\right)\in W_{0}^{1,2}$
for \textbf{every} $t\in I$, moreover, we have weak continuity with
respect to $t$ (see ex. 4.1).
\item For parabolic equation (for simplicity, equation of heat): let $\frac{d}{dt}u-\Delta u=0$,
$u\left(0\right)=u_{0}\in L^{2}$ (and not better), $\partial\Omega$
smooth.\\
We know that there exists a weak solution $u\left(t\right)\in L^{2}\left(I,W_{0}^{1,2}\right)$$\implies\exists t>0$
(arbitrarily small) s. t. $u\left(t_{0}\right)\in W_{0}^{1,2}$. Let
us use Th. 2.5: $u\left(t\right)\in L^{\infty}\left(t_{0},T;W_{0}^{1,2}\right)\cap L^{2}\left(t_{0},T;W^{2,2}\right)$.
$\implies\exists t_{1}>t_{0}$ (arbitrarily close) s. t. $u\left(t_{1}\right)\in W^{2,2}$.
We can arbitrarily increase the regularity as long as the boundary
is sufficiently smooth. Therefore the regularity of weak solution
increases in time.
\item Hyperbolic equation, for simplicity: let $\frac{d^{2}}{dt^{2}}u-\Delta u+\alpha\frac{d}{dt}u=0$,
$\partial\Omega$ be smooth, $u\left(0\right)\in W_{0}^{1,2}$, $\frac{d}{dt}u\left(0\right)\in L^{2}$
and not better.\\
We claim: $\nexists\tau>0$ s. t. $u\left(\tau\right)\in W^{2,2}$,
$\frac{d}{dt}u\left(\tau\right)\in W_{0}^{1,2}$. Proof: time reversal
in the wave equation: $u\left(t\right)$ is a weak solution in $\left[0,\tau\right]$
$\iff$$\hat{u}\left(t\right)=u\left(\tau-t\right)$ is a weak solution
in $\left[0,\tau\right]$, where $\alpha\leftarrow-\alpha$ (not significant
for previous theory). Therefore the regularity can't be increased:
if $u\left(\tau\right)\in W^{2,2}$, $\frac{d}{dt}u\left(\tau\right)\in W_{0}^{1,2}$$\implies\hat{u}\left(0\right)\in W^{2,2}$,
$\frac{d}{dt}\hat{u}\left(0\right)\in W^{2,2}$, due to Th. 3.3 and
following remark we have $\hat{u}\left(t\right)\in W^{2,2}$, $\frac{d}{dt}u\left(t\right)\in W_{0}^{1,2}$
for $\forall t$. Especially for $t=\tau$ we have a contradiction
with non\textendash smoothness of $u\left(0\right)$ and $\frac{d}{dt}u\left(0\right)$.
\end{itemize}

\subsection*{Notation: }
\begin{itemize}
\item $B\left(x_{0},r\right)=\left\{ x\in\mathbb{R}^{n},\left|x-x_{0}\right|<R\right\} $, 
\item $e=e\left(t,x\right)=\frac{1}{2}\left|\frac{d}{dt}u\left(t,x\right)\right|^{2}+\frac{1}{2}\left|\nabla u\left(t,x\right)\right|^{2}$,
where $u=u\left(t,x\right)$ is a weak solution. Remark: $E\left[u\left(t\right)\right]=\int_{\Omega}e\left(t,x\right)<dx$,
in particular $e\left(t,x\right)\in L^{1}\left(I\times\Omega\right)$.
\end{itemize}

\subsection*{Theorem 3.4. {[}The wave principle{]}}

\addcontentsline{toc}{section}{Finite speed of propagation}

Let $u\left(t\right)$ be a weak solution of the wave equation $\frac{d^{2}}{dt^{2}}u-\Delta u+\alpha\frac{d}{dt}u=0$,
where $\alpha\geq0$. Let $x_{0}\in\Omega$, $\tau\in I$ be such
that $B\left(x_{0},\tau\right)\subset\Omega$. Then $\int_{B\left(x_{0},\tau-t\right)}e\left(t,x\right)dx\leq\int_{B\left(x_{0},\tau\right)}e\left(0,x\right)dx$
for $\forall t\in\left[0,\tau\right]$.

\begin{figure}
\includegraphics[width=0.6\textwidth]{img/k3_wave_principle}

\caption{Wave principle illustration.}
\end{figure}

\subsection*{Corollary {[}Finite speed of propagation{]}}

$u\left(t\right)$ is a weak solution, $u\left(0\right)=0$, $\frac{d}{dt}u\left(0\right)=0$
in $B\left(x_{0},\tau\right)$ $\implies u=0$ in $\bigcup_{t\in\left[0,\tau\right]}B\left(x_{0},\tau-t\right)$.

\subsection*{Remark: }

analogous inequality holds for difference of solutions (linearity
of the equation).

\subsubsection*{Proof:}

\textbf{Step 1:} let $u\left(t,x\right)$ be smooth. Define auxiliary
function $p\left(t\right)=\int_{B\left(t\right)}e\left(t,x\right)dx$,
where $B\left(t\right)=B\left(x_{0},\tau-t\right)$. Auxiliary calculation
(see ``Time derivative of integral over time dependent domain''
in Appendices): $p'\left(t\right)=\frac{d}{dt}\int_{B\left(t\right)}e\left(t,x\right)dx=\int_{B\left(t\right)}\partial_{t}e\left(t,x\right)dx-\int_{\partial B\left(t\right)}e\left(t,x\right)dS\left(x\right)$.

Goal: $p'\left(t\right)\leq0$ (then we are done) 
\[
\partial_{t}e\left(t,x\right)=\partial_{t}\left(\frac{1}{2}\left(\partial_{t}u\right)^{2}+\frac{1}{2}\left|\nabla u\right|^{2}\right)=\partial_{tt}u\cdot\partial_{t}u+\partial_{t}\nabla u\cdot\nabla u.
\]
So we obtain: 
\[
p'\left(t\right)=\int_{B\left(t\right)}\partial_{tt}u\cdot\partial_{t}u+\underbrace{\partial_{t}\nabla u\cdot\nabla u}_{\text{by parts}}dx-\int_{\partial B\left(t\right)}edS\left(x\right).
\]

by parts (Gauss theorem): $\int_{\partial B\left(t\right)}\left(\partial_{t}u\right)\nabla u\cdot\underbrace{\nu}_{\makebox[0pt][c]{outer normal}}dS\left(x\right)=\int_{B\left(t\right)}\nabla\cdot\left(\left(\partial_{t}u\right)\nabla u\right)dx=\int_{B\left(t\right)}\nabla\partial_{t}u\cdot\nabla u+\left(\partial_{t}u\right)\Delta u\dx$.

Therefore $p'\left(t\right)=\int_{B\left(t\right)}\partial_{tt}u\cdot\partial_{t}u-\Delta u\cdot\partial_{t}u\dx+\int_{\partial B\left(t\right)}-e+\left(\partial_{t}u\right)\nabla u\cdot\nu dS\left(x\right)=I_{1}+I_{2}$.\\
ad $I_{1}$: by the equation we have: $\partial_{tt}u-\Delta u=-\alpha\partial_{t}u$,
tj. $I_{1}=\int_{B\left(t\right)}-\alpha\left(\partial_{t}u\right)^{2}\dx\leq0$
since $\alpha\geq0$.\\
ad $I_{2}$: $\left|\left(\partial_{t}u\right)\nabla u\cdot\nu\right|\leq\left|\partial_{t}u\right|\left|\nabla u\right|\underbrace{\left|\nu\right|}_{=1}=\left|\partial_{t}u\right|\left|\nabla u\right|\stackrel{\mbox{Hölder}}{\leq}\frac{1}{2}\left|\partial_{t}u\right|^{2}+\frac{1}{2}\left|\nabla u\right|^{2}=e$,
therefore $I_{2}\leq0$, since the integrand is $\leq0$. Therefore
we are done for smooth solution.

\textbf{Step 2:} $u\left(t\right)$ is an arbitrary weak solution
\dots{} suitable approximation $u^{n}\left(t\right)$: sufficiently
smooth so the first step is valid and at the same time $u^{n}\left(t\right)\to u\left(t\right)$
sufficiently strongly so the claim is preserved. Use Galerkin approximation
$u^{N}\left(t\right)$ from Th. 3.2. 

Observe: $u^{N}\left(t\right)=\sum_{j=1}^{N}c_{j}^{N}\left(t\right)w_{j}$
solves the equation: $\left(H_{N}\right)\quad\frac{d^{2}}{dt^{2}}c_{j}^{N}+\lambda_{j}c_{j}^{N}+\alpha\frac{d}{dt}c_{j}^{N}=0$,
i. e., $c_{k}^{N}\in C^{\infty}\left(I\right)$ since it solves the
ODE with constant coefficients. Moreover, $w_{j}\in C^{\infty}\left(\Omega\right)$
due to local regularity of laplacian: $-\Delta w_{j}=\lambda_{j}w_{j}$
($PS\in L^{2}\implies LS\in W^{2,2}\implies PS\in W^{2,2}$ etc.).
Moreover, it holds that: $\frac{d^{2}}{dt^{2}}u^{N}-\Delta u^{N}+\alpha\frac{d}{dt}u^{N}=0$,
i. e., the approximation exactly solves the equation. (Moreover, $c_{j}^{N}\left(t\right)w_{j}$
exactly satisfies the equation for $\forall j$ fixed.) Therefore
the conclusion of theorem holds for $u^{N}$.

Passage to the limit: we know that $\nabla u^{N}\left(t\right)\rightharpoonup\nabla u\left(t\right)$,
$\frac{d}{dt}u^{N}\left(t\right)\rightharpoonup\frac{d}{dt}u\left(t\right)$
in $L^{2}$, moreover $\nabla u^{N}\left(0\right)\to\nabla u\left(0\right)$,
$\frac{d}{dt}u^{N}\left(0\right)\to\frac{d}{dt}u\left(0\right)$ v
$L^{2}$. $t\in\left[0,\tau\right]$ is fixed, by weak lower semicontinuity
we have: 
\begin{align*}
\int_{B\left(t\right)}\frac{1}{2}\left|\frac{d}{dt}u\left(t,x\right)\right|^{2}+\frac{1}{2}\left|\nabla u\left(t,x\right)\right|^{2}\,dx & \leq\liminf_{N\to\infty}\int_{B\left(t\right)}\frac{1}{2}\left|\partial_{t}u^{N}\left(t,x\right)\right|^{2}+\frac{1}{2}\left|\nabla u^{N}\left(t,x\right)\right|^{2}\,dx\\
 & \stackrel{\mbox{Step 1 for \ensuremath{u^{N}\left(t\right)}}}{\leq}\liminf_{N\to\infty}\int_{B\left(0\right)}\frac{1}{2}\left|\partial_{t}u^{N}\left(0,x\right)\right|+\frac{1}{2}\left|\nabla u^{N}\left(0,x\right)\right|^{2}\,dx\\
 & =\int_{B\left(0\right)}\frac{1}{2}\left|\frac{d}{dt}u\left(0,x\right)\right|^{2}+\frac{1}{2}\left|\nabla u\left(0,x\right)\right|^{2}\,dx
\end{align*}
.

\subsection*{Remark:}

corollary of Th. 3.3: strong continuity of weak solutions: let $u\left(t\right)$
be a weak solution of $\left(H1\right)$, $u_{0}\in W_{0}^{1,2}$,
$u_{1}\in L^{2}$, let $\partial\Omega\in C^{2}$, where $f\left(z\right)$
is smooth with bounded derivative, $\frac{d}{dt}h\left(t\right)\in L^{2}\left(I,L^{2}\right)$.
Then we claim $u\left(t\right)\in C\left(I,W_{0}^{1,2}\right)$, $\frac{d}{dt}u\left(t\right)\in C\left(I,L^{2}\right)$.

\subsubsection*{Proof:}

choose $u_{0}^{n}\in W^{2,2}$, $u_{1}^{n}\in W_{0}^{1,2}$ s. t.
$u_{0}^{n}\to u_{0}$ strongly in $W_{0}^{1,2}$ and $u_{1}^{n}\to u_{1}$
strongly in $L^{2}$. Let $u^{n}\left(t\right)$ be a strong solution
for more smooth initial conditions $u_{1}^{n}$, $u_{0}^{n}$. Therefore
$u^{n}\left(t\right)$ is at the same time a strong solution (Th.
3.3). At the same time it holds that $u^{n}\left(t\right)\rightharpoonup u\left(t\right)$
in the sense of Th. 3.3.

\textbf{Trick:} the equation for $u^{n}\left(t\right)-u^{m}\left(t\right),$
test by $\frac{d}{dt}\left(u^{n}\left(t\right)-u^{m}\left(t\right)\right)$.
(Standard estimates + Gronwall) \dots{} we obtain: $E\left[u^{m}\left(t\right)-u^{n}\left(t\right)\right]\leq c_{0}\cdot E\left[u^{m}\left(0\right)-u^{n}\left(0\right)\right]$.
Initial conditions $u^{n}\left(0\right)$, $\frac{d}{dt}u^{n}\left(0\right)$
are cauchy in $W_{0}^{1,2}$ resp. $L^{2}$, i. e., $u^{n}\left(t\right)$,
$u^{m}\left(t\right)$ are cauchy in $C\left(I,W_{0}^{1,2}\right)$
resp. $C\left(I,L^{2}\right)$. By completeness \& continuity of strong
solution we obtain the conclusion.

\chapter{Semigroup theory}

Up to now we considered evolution PDEs: $\frac{d}{dt}u-\Delta u+\cdots$.
The rest of lectures is more close to functional analysis.

\section{Homogeneous equation}

\subsection*{Motivation:}

Let us have the equation $x'=ax$, where $x\left(0\right)=1$ \dots{}
the solution is $e^{at}$, an exponential function. Generalization:
$a\leftarrow A\in\mathbb{R}^{n\times n}$: $x'=Ax$, $x\left(0\right)=x_{0}$
the solution is $e^{tA}x_{0}$, a matrix exponential function. Goal:
generalization to general Banach space, the study of equations of
the type 
\[
\left(4.1\right)\quad\frac{d}{dt}x=Ax,\ x\left(0\right)=x_{0},\ x\in X,
\]
where $X$ is a Banach space, $A:X\to X$ is a linear operator, e.
g., $A=\Delta$. How to define a general exponential function $e^{tA}$?
The power series is suitable only for bounded operators. Problem:
$\Delta$ is unbounded operator, $\sum_{n=0}^{\infty}\frac{t^{n}A^{n}}{n!}$
in general does not make sense. Remark: $-\Delta:W_{0}^{1,2}\to W^{-1,2}$
is bounded, but in different spaces.

\subsection*{Idea:}

$A$ is ``well unbounded'' (i. e., bounded from above), then $e^{tA}$
will be possible to define for $t>0$.

\subsection*{Notation: {[}Unbounded operator{]}}
\begin{itemize}
\item $X$ \dots{} Banach space with respect to $\left\Vert \cdot\right\Vert $.
\item $\mathcal{L}\left(X\right)=\left\{ L:X\to X\mbox{ is linear continuous operator}\right\} $
is a Banach space, $\left\Vert L\right\Vert _{\mathcal{L}\left(X\right)}=\sup_{{x\in X\atop \left\Vert x\right\Vert =1}}\left\Vert Lx\right\Vert $,
\item Unbounded operator is the couple $\left(A,D\left(A\right)\right)$,
where $D\left(A\right)\subset\subset X$ is a subspace (domain of
definition of $A$), $A:D\left(A\right)\to X$ is linear.
\end{itemize}

\subsection*{Def.: {[}Semigroup, $c_{0}$-semigroup{]}}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ $c_0$ semigroup}

The function $S\!\left(t\right):[0;\infty)\to\mathcal{L}\left(X\right)$
is called a semigroup, iff
\begin{enumerate}
\item $S\!\left(0\right)$ is identity
\item $S\!\left(t\right)S\!\left(s\right)=S\!\left(t+s\right)$, $\forall t,s\geq0$
\end{enumerate}
If moreover 
\begin{enumerate}
\item[\foreignlanguage{czech}{\inputencoding{latin2}$\left(3\right)$}] $S\!\left(t\right)x\to x$, $t\to0^{+}$ for $\forall x\in X$ fixed,
we call $S\,\left(t\right)$ a $c_{0}$-semigroup.
\end{enumerate}

\subsection*{Remark:}
\begin{itemize}
\item $c_{0}$-semigroup \dots{} abstract exponential. Possible definitions
of standard exponential: either a solution of $x'=ax$, $x\left(0\right)=1$
or a power series $e^{at}=\sum_{n=0}^{\infty}\frac{\left(at\right)^{n}}{n!}=\lim_{n\to\infty}\left(1+\frac{at}{n}\right)^{n}$,
or a solution of functional equation: $f\left(x+y\right)=f\left(x\right)f\left(y\right)$
+ continuity and $f\left(\cdot\right)$ is nonzero. Then \glqq $S\!\left(t\right)=e^{ta}$``,
$c_{0}$-semigroup is a suitable candidate for exponential.
\item stronger assumption $\left(3'\right)$ $\left\Vert S\!\left(t\right)-I\right\Vert _{\mathcal{L}\left(X\right)}\to0$,
$t\to0^{+}$ (so called uniform continuity) implies $S\!\left(t\right)=e^{tA}$
for some linear continuous operator $A$, see ex. 5.1.
\end{itemize}

\subsection*{Lemma 4.1 {[}Exponential estimates, continuity in time of $c_{0}$-semigroup{]}}

Let $S\!\left(t\right)$ be a $c_{0}$-semigroup in $X$. Then
\begin{enumerate}
\item $\exists M\geq1$, $\omega\geq0$ s. t. $\left\Vert S\!\left(t\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq M\cdot e^{\omega t}$
for $\forall t\geq0$. 
\item $t\mapsto S\!\left(t\right)x$ is continuous $[0,\infty)\to X$ for
$\forall x\in X$ fixed.
\end{enumerate}

\subsubsection*{Proof:}
\begin{enumerate}
\item we claim: $\exists M\geq1$, $\exists\delta>0$ s. t. $\left\Vert S\!\left(t\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq M$,
$\forall t\in\left[0,\delta\right]$: by contradiction: if not, then
$\exists t_{n}\to0^{+}$ s. t. $\left\Vert S\!\left(t_{n}\right)\right\Vert _{\mathcal{L}\left(X\right)}\to+\infty$,
but $S\!\left(t_{n}\right)x\to x$ for $\forall x\in X$ fixed due
to the part $\left(3\right)$ of semigroup definition and so $\left\Vert S\!\left(t_{n}\right)x\right\Vert $
is bounded. That is a contradiction to the principle of uniform boundednes,
see functional analysis (a set of operators is bounded in operator
norm iff $\left\Vert S\!\left(t_{n}\right)x\right\Vert $ is bounded
for $\forall x$).\\
Set $\omega=\frac{1}{\delta}\ln M$, i. e.. $M=e^{\omega\delta}$,
then for $t\geq0$ arbitrary it holds that $t=n\delta+\varepsilon$,
$\varepsilon\in[0,\delta)$, $n\in\mathbb{N}$. Then $\left\Vert S\!\left(t\right)\right\Vert _{\mathcal{L}\left(X\right)}=\big\Vert S\big(\underbrace{\delta+\delta+\cdots+\delta}_{\mbox{\ensuremath{n\times}}}+\varepsilon\big)\big\Vert_{\mathcal{L}\left(X\right)}=\left\Vert S\!\left(\delta\right)\cdots S\!\left(\delta\right)S\!\left(\varepsilon\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq\left\Vert S\!\left(\delta\right)\right\Vert _{\mathcal{L}\left(X\right)}^{n}\left\Vert S\!\left(\varepsilon\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq M\cdot\underbrace{M^{n}}_{\mbox{\ensuremath{=}}\makebox[0pt][l]{\ensuremath{\,e^{\omega n\delta}}}}\leq M\cdot e^{\omega t}$.
\item continuity: in $0^{+}$ we have due to part $\left(3\right)$ of definiton
of semigroup. Continuity (from the right and from the left) in $t>0$
remains:\\
Continuity from the right: $S\!\left(t+h\right)x=S\!\left(t\right)\underbrace{S\!\left(h\right)x}_{\to x}\to S\!\left(t\right)x$,
$h\to0^{+}$ due to the property $\left(3\right)$, $S\!\left(t\right)\in\mathcal{L}\left(X\right)$.\\
Continuity from the left: (WLOG $h<t$) $S\!\left(t-h\right)x-\underbrace{S\!\left(t\right)x}_{S\!\left(t-h\right)S\!\left(h\right)x}=S\!\left(t-h\right)\left[x-S\!\left(h\right)x\right]$.
Estimate: $\left\Vert S\!\left(t-h\right)x-S\!\left(t\right)x\right\Vert \leq\underbrace{\left\Vert S\!\left(t-h\right)\right\Vert _{\mathcal{L}\left(X\right)}}_{\mbox{\ensuremath{\leq}}\makebox[0pt][l]{\,\ensuremath{Me^{\omega t}\mbox{, independent of \ensuremath{h} due to the first part}}}}\overbrace{\left\Vert x-S\!\left(h\right)x\right\Vert }^{\mbox{\ensuremath{\to}}\makebox[0pt][l]{\ensuremath{\,0} due to \ensuremath{\left(3\right)}}}$,
$h\to0^{+}$.
\end{enumerate}

\subsection*{Def.: {[}Generator of a semigroup{]}}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Generator of semigroup, properties}

An unbounded operator $\left(A,D\left(A\right)\right)$ is called
a generator of semigroup $S\!\left(t\right)$ iff 
\[
Ax=\lim_{h\to0^{+}}\frac{1}{h}\left(S\!\left(h\right)x-x\right),\ D\left(A\right)=\left\{ x\in X,\lim_{h\to0^{+}}\frac{1}{h}\left(S\!\left(h\right)x-x\right)\mbox{ exists v \ensuremath{X}}\right\} .
\]

\subsection*{Remark:}

it is easy to show that the operator defined by this formula is linear
and $D\left(A\right)\subset X$ is a linear subspace.

\subsection*{Theorem 4.1 {[}Basic properties of a generator{]}}

Let $\left(A,D\left(A\right)\right)$ be a generator of $S\!\left(t\right)$,
a $c_{0}$-semigroup in $X$. Then:
\begin{enumerate}
\item $x\in D\left(A\right)\implies S\!\left(t\right)x\in D\left(A\right)$
for $\forall t\geq0$,
\item $x\in D\left(A\right)\implies AS\!\left(t\right)x=S\!\left(t\right)Ax=\frac{d}{dt}S\!\left(t\right)x$
for $\forall t\ge0$ (in $t=0$ only from the right),
\item $x\in X$, $t\ge0\implies\int_{0}^{t}S\!\left(s\right)xds\in D\left(A\right)$,
$A\left(\int_{0}^{t}S\!\left(s\right)xds\right)=S\!\left(t\right)x-x$.
\end{enumerate}

\subsubsection*{Proof:}
\begin{enumerate}
\item $x\in D\left(A\right)$, $t\geq0$ given, $\underbrace{\frac{1}{h}\big(\overbrace{S\!\left(h\right)S\!\left(t\right)}^{\mbox{\ensuremath{=}}\makebox[0pt][l]{\ensuremath{\,S\!\left(h+t\right)=S\!\left(t\right)S\!\left(h\right)}due to \ensuremath{\left(2\right)}}}x-S\!\left(t\right)x\big)}_{(*)}\stackrel{?}{\to}y\implies S\!\left(t\right)x\in D\left(A\right),AS\!\left(t\right)x=y$
\[
\left(*\right)=\frac{1}{h}\left(S\!\left(t\right)S\!\left(h\right)x-S\!\left(t\right)x\right)=S\!\left(t\right)\underbrace{\left(\frac{1}{h}\left(S\!\left(h\right)x-x\right)\right)}_{\mbox{\ensuremath{\to}}\makebox[0pt][l]{\ensuremath{Ax}}}\to S\!\left(t\right)Ax
\]
\item $x\in D\left(A\right)$ \dots{} $AS\!\left(t\right)x=S\!\left(t\right)Ax$
see part 1, $\frac{d}{dt}S\!\left(t\right)x=S\!\left(t\right)Ax$
from the right for $\forall t\geq0$, see first part $\left(\frac{1}{h}\left(S\!\left(t+h\right)x-S\!\left(t\right)x\right)\to S\!\left(t\right)Ax\right)$,
$h\to0^{+}$.\\
From the left? $\frac{S\!\left(t-h\right)x-S\!\left(t\right)x}{-h}\to S\!\left(t\right)Ax$
as $h\to0^{+}$ for $t>0$ fixed? $\frac{S\!\left(t-h\right)x-S\!\left(t\right)x}{-h}=S\!\left(t-h\right)\left[\frac{x-S\!\left(h\right)x}{-h}\right]-S\!\left(t\right)Ax=S\!\left(t-h\right)\left[\frac{x-S\!\left(h\right)x}{-h}\right]-S\!\left(t-h\right)S\!\left(h\right)Ax\stackrel{?}{\to0}$:
$\underbrace{S\!\left(t-h\right)}_{\makebox[0pt][r]{L. 4.1, 1: bounded in }\left\Vert \cdot\right\Vert _{\mathcal{L}\left(X\right)}\!\!\!\!\!\!\!\!\!\!\!\!}\Big\{\underbrace{\left[\frac{S\!\left(h\right)x-x}{h}\right]}_{\mbox{\ensuremath{\to}}\makebox[0pt][l]{\ensuremath{Ax}}}-\underbrace{S\!\left(h\right)Ax}_{\mbox{\ensuremath{\to}}\makebox[0pt][l]{\ensuremath{Ax} due to \ensuremath{\left(3\right)}}}\Big\}\to0$
as in L. 4.1, 2.
\item Denote $y=\int_{0}^{t}S\!\left(s\right)xds$, $x\in X$ for $t>0$
fixed. 
\begin{eqnarray*}
\frac{1}{h}\left(S\!\left(h\right)y-y\right) & = & \frac{1}{h}\Bigg(\underbrace{S\!\left(h\right)\int_{0}^{t}S\!\left(s\right)xds}_{\mbox{\ensuremath{\int_{0}^{t}S\!\left(s+h\right)xds}}\mbox{\makebox[0pt][l]{, subst. \textendash\ a shift by \ensuremath{h}}}}-\int_{0}^{t}S\!\left(s\right)xds\Bigg)\\
 & = & \frac{1}{h}\left(\int_{h}^{t+h}S\!\left(s\right)xds-\int_{0}^{t}S\!\left(s\right)xds\right)\\
 & = & \frac{1}{h}\int_{t}^{t+h}S\!\left(s\right)xds-\frac{1}{h}\int_{0}^{h}S\!\left(s\right)xds\underbrace{\stackrel{h\to0^{+}}{\to}}_{\makebox[0pt][c]{derivative of continuous integrand (L. 4.1) w. r. t. upper bound }}S\!\left(t\right)x-\overbrace{S\!\left(0\right)x}^{\mbox{\ensuremath{=}}\makebox[0pt][l]{\,x}}
\end{eqnarray*}
 Therefore $y\in D\left(A\right)$, $Ay=S\!\left(t\right)x-x$, which
was to be proven.
\end{enumerate}

\subsection*{Remark:}

The theorem states:
\begin{enumerate}
\item $D\left(A\right)$ is invariant w. r. t. $S\!\left(t\right)$.
\item $S\!\left(t\right)$, $A$ commute in v $D\left(A\right)$, moreover
$t\mapsto S\!\left(t\right)x$ is a classical solution of $\frac{d}{dt}x=Ax$,
$x\left(0\right)=x_{0}$, if $x\in D\left(A\right)$.
\end{enumerate}

\subsection*{Def.: {[}Closed operator{]}}

We say that an unbounded operator $\left(A,D\left(A\right)\right)$
is closed, iff: $u_{n}\in D\left(A\right)$, $u_{n}\to u$, $Au_{n}\to v\implies u\in D\left(A\right)$
and $Au=v$.

\subsection*{Remark:}

it is easy to show that $\left(A,D\left(A\right)\right)$ is closed
$\iff D\left(A\right)$ is complete (i. e., Banach) with respect to
the norm $\left\Vert u\right\Vert +\left\Vert Au\right\Vert $, the
so-called graph norm.

\subsection*{Remark:}

unbounded, but closed operators: natural property of derivative in
different function spaces, examples:
\begin{enumerate}
\item $\mathscr{X}=L^{1}\left(I,X\right)$, $A:u\left(t\right)\mapsto\frac{d}{dt}u\left(t\right)$,
$D\left(A\right)=W^{1,1}\left(I,X\right)$ \dots{} see chap. 1: statement
(see ex.~2.1): $u_{n}\left(t\right)\in W^{1,1}\left(I,X\right),\ u_{n}\left(t\right)\to u\left(t\right)$
v $L^{1}\left(I,X\right)$, $\frac{d}{dt}u_{n}\left(t\right)\to g\left(t\right)$
v $L^{1}\left(I,X\right)$ $\implies u\left(t\right)\in W^{1,1}\left(I,X\right)$,
$\frac{d}{dt}u\left(t\right)=g\left(t\right)$. This is equivalent
to closedness of $\left(A,D\left(A\right)\right)$.
\item $X=C^{1}\left(\left[0,1\right]\right)$ \dots{} theorem from analysis:
$f_{n}\left(t\right)\in C^{1}\left(\left[0,1\right]\right)$, $f_{n}\left(t\right)\rightrightarrows f\left(t\right)$
v $\left[0,1\right]$, $\frac{d}{dt}f_{n}\left(t\right)\rightrightarrows\frac{d}{dt}g\left(t\right)$
v $\left[0,1\right]$ $\implies f\left(t\right)\in C^{1}\left(\left[0,1\right]\right)$,
$\frac{d}{dt}f\left(t\right)=g\left(t\right)$. That is equivalent
to closedness of ``$\frac{d}{dt}$'' in $C\left(\left[0,1\right]\right)=X$
with the definition domain $C^{1}\left(\left[0,1\right]\right)$.
\end{enumerate}

\subsection*{Theorem 4.2 {[}Density and closedness of generator{]}}

Let $\left(A,D\left(A\right)\right)$ be a generator of a $c_{0}$-semigroup
$S\!\left(t\right)$ in $X$. Then $D\left(A\right)$ is dense in
$X$ and $\left(A,D\left(A\right)\right)$ is closed. 

\subsubsection*{Proof:}

\textbf{Density} \dots{} $x\in X$ given. $x=\lim_{h\to0^{+}}\underbrace{\frac{1}{h}\int_{0}^{h}S\!\left(s\right)xds}_{\mbox{\ensuremath{\in}}\makebox[0pt][l]{\ensuremath{\,D\left(A\right)}\mbox{ dle V.4.1.3}}}$
(continuity of integrand), i. e., we have elements from the definition
domain which approximate given element.

\textbf{Closedness} \dots{} $x_{n}\in D\left(A\right)$ given, $x_{n}\to x$,
$Ax_{n}\to y\stackrel{?}{\implies}x\in D\left(A\right),$ $Ax=y$.
Observe: $s\mapsto S\!\left(s\right)x_{n}$ is $C^{1}$ since $\frac{d}{dt}S\!\left(s\right)x_{n}=S\!\left(s\right)Ax_{n}$
due to Th.4.1, 2 and due to Newton-Leibnitz we have $S\!\left(h\right)x_{n}-S\!\left(0\right)x_{n}=\int_{0}^{h}\frac{d}{ds}S\!\left(s\right)x_{n}ds=\int_{0}^{h}S\!\left(s\right)Ax_{n}ds$.
Take a limit $n\to\infty$. $LHS\to S\!\left(h\right)x-x$, $RHS$
\dots{} exchange of $\lim$ and $\int$: $Ax_{n}\to y$, therefore
$\left\Vert S\!\left(s\right)Ax-S\!\left(s\right)y\right\Vert \leq\underbrace{\left\Vert S\!\left(s\right)\right\Vert _{\mathcal{L}\left(X\right)}}_{\mbox{\makebox[0pt][c]{bounded independently of \ensuremath{s\in\mbox{\ensuremath{\left[0,h\right]}}}}}}\left\Vert Ax_{n}-y\right\Vert $,
uniform convergence. Therefore by the limit we obtain $\frac{1}{h}\left(S\!\left(h\right)x-x\right)=\frac{1}{h}\int_{0}^{h}S\!\left(s\right)yds$,
take $h\to0^{+}$. $RHS\to y$ (continuity of integrand), i. e., $LHS\to y$
or in other words $x\in D\left(A\right)$, $Ax=y$, which was to be
proven.

\subsection*{Remark:}

Theorem 4.2, proof of closedness: $S\!\left(h\right)x_{n}-x_{n}=\int_{0}^{h}S\!\left(s\right)Ax_{n}ds$,
$x_{n}\in D\left(A\right)$ is needed. In Th. 4.1, 3 we already have
$S\!\left(h\right)x-x=A\left(\int_{0}^{h}S\!\left(s\right)xds\right)$,
$x\in X$. Wouldn't it be possible to shift $A$ into the integral
straight away?

Problem: does it hold that $A\left(\int_{I}f\left(s\right)ds\right)=\int_{I}Af\left(s\right)ds$?
For continuous operators $A$ it holds, see ex. 1.1. For closed operators
$A$ it is possible to prove, see ex. 5.3. It is not possible to use
this argument above as we are just proving the closedness of generator
$A$.

\subsection*{Remark:}
\begin{itemize}
\item Th. 4.1:$\left(A,D\left(A\right)\right)$ is a generator of a semigroup
$S\!\left(t\right)$$\implies\forall x_{0}\in D\left(A\right)$ is
$x\left(t\right)=S\!\left(t\right)x_{0}$ a classical solution of
(4.1).
\item Key problem: $\left(A,D\left(A\right)\right)$ given, $\stackrel{?}{\implies}\exists c_{0}$-sg.
$S\!\left(t\right)$ s. t. $A$ is a generator of $S\!\left(t\right)$.
\end{itemize}

\subsection*{Lemma 4.2 {[}Uniqueness of a semigroup{]}}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Uniqueness of the semigroup}

Let $S\!\left(t\right)$, $\tilde{S}\left(t\right)$ be $c_{0}$-semigroups
which have the same generator. Then $S\!\left(t\right)=\tilde{S}\left(t\right)$
for $\forall t>0$.

\subsubsection*{Proof:}

\textbf{Trick:} $y\left(t\right)=S\!\left(T-t\right)\widetilde{S}\left(t\right)x$,
$x\in D\left(A\right)$, check $y\left(t\right)\in C\left(\left[0,T\right],X\right)$,
$y'\left(t\right)=0\ \forall t\in\left(0,T\right)$ $\implies y\left(T\right)=\widetilde{S}\left(T\right)y\left(0\right)=S\!\left(T\right)x$.
$D\left(A\right)$ is dense in $X$ (Th. 4.2)

\subsection*{Def.: {[}Resolvent, resolvent set, spectrum{]}}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Definintion: resolvent, resolvent set, spectrum}

Let $\left(A,D\left(A\right)\right)$ be an unbounded operator. We
define
\begin{description}
\item [{resolvent~set}] $\rho\left(A\right)=\left\{ \lambda;\lambda I-A\to X\text{ is one-to-one}\right\} \subset\mathbb{R}$
(generally can be considered a subset of $\mathbb{C}$), 
\item [{resolvent}] $R\left(\lambda,A\right)=\left(\lambda I-A\right)_{-1}:X\to D\left(A\right),\lambda\in\rho\left(A\right)$,
\item [{spectrum}] $\sigma\left(A\right)=\left\{ \lambda\in\mathbb{C},\lambda I-A\mbox{ is not invertible}\right\} $.
Equivalently $\sigma\left(A\right)=\mathbb{C}\backslash\rho\left(A\right)$.
\end{description}

\subsection*{Remark:}
\begin{itemize}
\item $\left(A,D\left(A\right)\right)$ is closed $\implies$$R\left(\lambda,A\right)\in\mathcal{L}\left(X\right)$,
since $A$ is closed $\iff D\left(A\right)$ is Banach with graph
norm $\left\Vert x\right\Vert +\left\Vert Ax\right\Vert $. Moreover
by the closedness is t $A:D\left(A\right)\to X$ continuous. Banach
theorem on open mapping: inversion is continuous, i. e., $R\left(\lambda,A\right):X\to D\left(A\right)$
is continuous.
\item the following relations hold: \\
\begin{eqnarray*}
\left(i\right)\quad AR\left(\lambda,A\right)x & = & \lambda R\left(\lambda,A\right)x-x\ \forall x\in X,\\
\left(ii\right)\quad R\left(\lambda,A\right)Ax & = & \lambda R\left(\lambda,A\right)x-x\ \forall x\in D\left(A\right),\\
\left(iii\right)\quad R\left(\lambda,A\right)x-R\left(\mu A\right)x & = & \left(\mu-\lambda\right)R\left(\lambda,A\right)R\left(\mu,A\right)x\ \forall x\in X,
\end{eqnarray*}
where $\left(iii\right)$ is so-called resolvent identity.
\end{itemize}

\subsubsection*{Proof of (i):}

$AR\left(\lambda,A\right)x=\left[\left(A-\lambda I\right)+\lambda I\right]R\left(\lambda,A\right)x=\underbrace{-\left(\lambda I-A\right)R\left(\lambda,A\right)X}_{-x}+\lambda R\left(\lambda,A\right)x$.\\
The other relations are proven similarly, (i)-(ii)$\implies AR\left(\lambda,A\right)x=RA\left(\lambda,A\right)x,\ x\in D\left(A\right)$.
Heuristics: $R\left(\lambda,A\right)=\frac{1}{\lambda-A}$.

\subsection*{Lemma 4.3 {[}Formula for the resolvent by Laplace transform{]}}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Resolvent and Laplace transform}

Let $\left(A,D\left(A\right)\right)$ be a generator of $c_{0}$-semigroup
$S\!\left(t\right)$; let $\left\Vert S\!\left(t\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq Me^{\omega t}$.
Then $\lambda\in\rho\left(A\right)$ for $\forall\lambda>\omega$
and the resolvent can be expressed as $R\left(\lambda,A\right)x=\int_{0}^{\infty}e^{-\lambda t}S\!\left(t\right)xdt$,
$\left\Vert R\left(\lambda,A\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq\frac{M}{\lambda-\omega}$.

\subsubsection*{Proof:}

WLOG: $\omega=0$, (see ex. 6.2) since $c_{0}$-sg. generated by $\left(A,D\left(A\right)\right)$$\iff\widetilde{S}\left(t\right)=e^{-\omega t}S\!\left(t\right)$
is $c_{0}$-sg. generated by $\left(\widetilde{A},D\left(\widetilde{A}\right)\right)$
where $\widetilde{A}=A-\omega I$, $D\left(\widetilde{A}\right)=D\left(A\right)$.
Moreover $R\left(\lambda,\widetilde{A}\right)=R\left(\lambda+\omega,A\right)$.

Therefore $\left\Vert S\!\left(t\right)\right\Vert _{\mathcal{L}\left(X\right)}\le M$,
$\lambda>0\stackrel{?}{\implies}\lambda\in\rho\left(A\right)$. Denote
$\widetilde{R}x=\int_{0}^{\infty}e^{-\lambda t}S\!\left(t\right)xdt$
(Laplace transform of semigroup $S\!\left(t\right)$), $x\in X$,
$\lambda>0$ fixed. Integral defined: integrand continuous (L. 4.1),
$\left\Vert \mbox{integrand}\right\Vert \leq e^{-\lambda t}M\left\Vert x\right\Vert \in L^{1}\left(0,\infty\right)$,\\
$\left\Vert \widetilde{R}x\right\Vert \leq\int_{0}^{\infty}e^{-\lambda t}M\left\Vert x\right\Vert dt=\frac{M}{\lambda}\left\Vert x\right\Vert $,
i. e., $\widetilde{R}\in L\left(X\right)$, $\left\Vert \widetilde{R}\right\Vert _{\mathcal{L}\left(X\right)}\leq\frac{M}{\lambda}$.

We will show that $\widetilde{R}x\in D\left(A\right)$: 
\[
\frac{1}{h}\left[S\!\left(h\right)-I\right]\widetilde{R}x=\frac{1}{h}\Bigg[\int_{0}^{\infty}e^{-\lambda t}\underbrace{S\!\left(h\right)S\!\left(t\right)}_{S\!\left(h+t\right)}x-e^{-\lambda t}S\!\left(t\right)x\,dt\Bigg],
\]
Substution in first integral: $\int_{h}^{\infty}e^{-\lambda\left(t-h\right)}S\!\left(t\right)x$,
$\pm\int_{0}^{h}e^{-\lambda\left(t-h\right)}S\!\left(t\right)x$,
together: 
\[
=\frac{e^{\lambda h}-1}{h}\underbrace{\int_{0}^{\infty}e^{-\lambda t}S\!\left(t\right)x\,dt}_{\widetilde{R}x}-\frac{e^{\lambda h}}{h}\int_{0}^{h}e^{-\lambda t}S\!\left(t\right)x\,dt
\]
 Take $h\to0^{+}$: $\to\lambda\widetilde{R}x-x$.

I. e., $\widetilde{R}x\in D\left(A\right)$, $A\widetilde{R}x=\lambda\widetilde{R}x-x$,
$\forall x\in X$, in other words $\left(\lambda I-A\right)\widetilde{R}x=x$,
i. e. $\lambda I-A:D\left(A\right)\to X$ is onto.

Is injective? Let $x\in D\left(A\right)$ be fixed, 
\begin{align*}
A\widetilde{R}x & =A\left(\int_{0}^{\infty}e^{-\lambda t}S\!\left(t\right)x\right)\stackrel{\mbox{Ex. 5.4, \ensuremath{A} closed (Th. 4.2.)}}{=}\int_{0}^{\infty}A\left(e^{-\lambda t}S\!\left(t\right)x\right)dt\\
\left(\mbox{exchange op. and sg: Th. 4.1, 2}\right) & =\int_{0}^{\infty}e^{-\lambda t}S\!\left(t\right)Axdt=\widetilde{R}Ax
\end{align*}
, i. e. $A\widetilde{R}=\widetilde{R}A$ in $D\left(A\right)$ $\implies\widetilde{R}\left(\lambda I-A\right)x=\lambda\widetilde{R}x-\widetilde{R}AX\stackrel{A\widetilde{R}x=\lambda\widetilde{R}x-x}{=}x$,
i. e., $\forall x\in D\left(A\right):\widetilde{R}\left(\lambda I-A\right)x=x$
\dots{} $\lambda I-A$ is injective! I. e., $\widetilde{R}=R\left(\lambda,A\right)$,
the proof is done.

\subsection*{Def.: {[}Semigroup of contractions{]}}

We say that $S\!\left(t\right)$ is a semigroup of contractions, if
$\left\Vert S\!\left(t\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq1$,
$\forall t\ge0$

\subsection*{Theorem 4.3 {[}Hille\textendash Yosida (for contractions){]}}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Hille--Yosida theorem for contractions}

Let $\left(A,D\left(A\right)\right)$ be an unbounded operator. Then
it is equivalent:
\begin{enumerate}
\item $\exists c_{0}$-semigroup of contractions, which is generated by
$\left(A,D\left(A\right)\right)$.
\item $\left(A,D\left(A\right)\right)$ is closed, $D\left(A\right)$ is
dense in $X$, $\lambda\in\rho\left(A\right)$ for $\forall\lambda>0$
and it holds that $\left\Vert R\left(\lambda,A\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq\frac{1}{\lambda}$.
\end{enumerate}

\subsubsection*{Proof:}
\begin{itemize}
\item[$\left(1\right)\implies\left(2\right)$]  we already have proven: Th. 4.2 (density and closedness), L. 4.3
(resolvent, $M=1$, $\omega=0$).
\item[$\left(2\right)\implies\left(1\right)$]  Yosida's approximation: $A_{n}=nAR\left(n,A\right),n\in\mathbb{N}$.
Strategy: $S\!\left(t\right)=\lim_{n\to\infty}e^{tA_{n}}x$\\
\textbf{Step 1:} we claim that $A_{n}=n^{2}R\left(n,A\right)-nI$,
i. e., $A_{n}\in\mathcal{L}\left(X\right)$, $nR\left(n,A\right)x\to x$,
$n\to\infty$ for $\forall x\in X$ fixed, $A_{n}x\to Ax$, $n\to\infty$,
$\forall x\in D\left(A\right)$.
\begin{itemize}
\item[$\left(i\right)$] $AR\left(n,A\right)=nR\left(n,A\right)-I$, $\Big/\cdot n$$\implies\underbrace{nAR\left(n,A\right)}_{A_{n}}=n^{2}R\left(n,A\right)-nI\in\mathcal{L}\left(X\right)$,
since $R\left(n,A\right)\in\mathcal{L}\left(X\right)$.
\item[$\left(ii\right)$] convergence principle (ex. 1.4, 1): $F_{n}:X\to X$ linear, $\left\Vert F_{n}\right\Vert _{\mathcal{L}\left(X\right)}\leq C$,
$F_{n}y\to y$ for $\forall y\in S$ fixed, $S$ dense in $X$ $\implies F_{n}x\to x$
for $\forall x\in X$.\\
Apply $F_{n}=nR\left(n,A\right)$, $\left\Vert F_{n}\right\Vert _{\mathcal{L}\left(X\right)}\leq1$
due to $\left\Vert R\left(\lambda,A\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq\frac{1}{\lambda}$
$\forall\lambda>0$, $S=D\left(A\right)$ dense in $X$ (assumption),
$y\in D\left(A\right):nR\left(n,A\right)y=y+\underbrace{AR\left(n,A\right)y}_{\underbrace{R\left(n,A\right)}_{\left\Vert \cdot\right\Vert _{\mathcal{L}\left(X\right)}\leq\frac{1}{n}}\underbrace{Ay}_{\mbox{fixed}}}\to y$,
$n\to\infty$.
\item[$\left(iii\right)$] $A_{n}x=nAR\left(n,A\right)x=nR\left(n,A\right)Ax\to Ax$ due to
(ii)
\end{itemize}
\item \textbf{Step 2:} approximation of semigroup $S_{n}\left(t\right)=e^{tA_{n}}=\sum_{k=0}^{\infty}\frac{t^{k}A_{n}^{k}}{k!}\in\mathcal{L}\left(X\right)$,
since $A_{n}\in\mathcal{L}\left(X\right)$ due to step 1. At the same
time we know: $A_{n}=\underbrace{n^{2}R\left(n,A\right)-nI}_{\mbox{are commutative}}$
$\implies e^{tA_{n}}=e^{tn^{2}R\left(n,A\right)-tnI}=e^{-nt}\cdot\underbrace{e^{n^{2}tR\left(n,A\right)}}_{\left(*\right)}$.
$\left\Vert \left(*\right)\right\Vert _{\mathcal{L}\left(X\right)}=\left\Vert \sum_{k=0}^{\infty}\frac{\left(nt\right)^{k}\left(nR\left(n,A\right)\right)^{k}}{k!}\right\Vert _{\mathcal{L}\left(X\right)}\leq\sum_{k=0}^{\infty}\frac{\left(nt\right)^{k}}{k!}\underbrace{\left\Vert nR\left(n,A\right)\right\Vert _{\mathcal{L}\left(X\right)}^{k}}_{\leq1}\leq e^{nt}$.
I. e., we have $\left\Vert S_{n}\left(t\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq1$
for $\forall n$, in other words $S_{n}\left(t\right)$ are contractions.\\
\textbf{Step 3:} existence of $\lim_{n\to\infty}S_{n}\left(t\right)x$,
for $\forall x\in X$ fixed. For now, let $x\in D\left(A\right)$.
\textbf{}\\
\textbf{Trick:} let $t$ be fixed.
\begin{eqnarray*}
S_{n}\left(t\right)x-S_{m}\left(t\right)x & = & \left[S_{n}\left(t-s\right)S_{m}\left(s\right)x\right]_{s=0}^{s=t}=\left[e^{\left(t-s\right)A_{n}}e^{sA_{m}}x\right]_{s=0}^{t}\\
 & = & \int_{0}^{t}\frac{d}{ds}\left[e^{\left(t-s\right)A_{n}}e^{sA_{m}}x\right]ds\\
 & = & \int_{0}^{t}-A_{n}e^{\left(t-s\right)A_{n}}e^{sA_{m}}x+e^{\left(t-s\right)A_{n}}A_{m}e^{sA_{m}}xds=\left(\sun\right)
\end{eqnarray*}
We know: $A_{n}A_{m}=A_{m}A_{n}$ \dots{} is implied by expression
of resolvent ($R\left(n,A\right)$, $R\left(m,A\right)$ are commutative).\\
\[
\left(\sun\right)=\int_{0}^{t}\underbrace{e^{\left(t-s\right)A_{n}}e^{sA_{m}}}_{\underbrace{S_{n}\left(t-s\right)S_{m}\left(s\right)}_{\mbox{\ensuremath{\left\Vert \cdot\right\Vert }}\makebox[0pt][l]{\ensuremath{_{\mathcal{L}\left(x\right)}\leq1}}}}\left[A_{m}x-A_{n}x\right]ds
\]
 
\[
\implies\left\Vert S_{n}\left(t\right)x-S_{m}\left(t\right)x\right\Vert \leq t\left\Vert A_{m}x-A_{n}x\right\Vert 
\]
$x\in D\left(A\right)\implies A_{n}x$, $A_{m}x\to Ax$, $A_{n}x-A_{m}x\to0$
$\implies\left\{ S_{n}\left(t\right)x\right\} $ satisfy B. C. condition
uniformly with respect to $t\in\left[0,T\right]$, which is bounded
interval, i. e., there exists $S_{n}\left(t\right)\stackrel{loc}{\rightrightarrows}S\!\left(t\right)x$
defined as this uniform limit. The convergence principle (density
of $D\left(A\right),$$\left\Vert S_{n}\left(t\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq1$:
$S\left(t\right)$ is clearly a $c_{0}$-semigroup of contractions)
$\implies$ $\exists$ limit for $\forall x\in X$.\\
\textbf{Step 4:} $S\!\left(t\right)$ is generated by the original
$\left(A,D\left(A\right)\right)$. Denote the generator of $S\!\left(t\right)$
by $\left(\widetilde{A},D\left(\widetilde{A}\right)\right)$, let
$x\in D\left(A\right)$. 
\[
S_{n}\left(t\right)x-x=\int_{0}^{t}\frac{d}{ds}S_{n}\left(s\right)xds=\int_{0}^{t}A_{n}S_{n}\left(s\right)xds=\int_{0}^{t}\underbrace{S_{n}\left(s\right)}_{\mbox{\ensuremath{\rightrightarrows}}\makebox[0pt][l]{\ensuremath{\,S\!\left(s\right)}}}\overbrace{A_{n}x}^{\mbox{\ensuremath{\to}}\makebox[0pt][l]{\ensuremath{\,Ax}}}ds
\]
For $n\to\infty$: $LHS\to S\!\left(t\right)x-x=\int_{0}^{t}S\!\left(s\right)A\left(x\right)ds$,
i. e., $\frac{1}{h}\left(S\!\left(h\right)x-x\right)=\frac{1}{h}\int_{0}^{t}S\!\left(s\right)Axds\to Ax$,
$h\to0^{+}$, in other words $x\in D\left(\widetilde{A}\right)$,
$\widetilde{A}x=Ax$, i. e., we have $D\left(A\right)\subset D\left(\widetilde{A}\right)$,
$\widetilde{A}x=Ax$ for $x\in D\left(A\right)$.\\
Opposite implication: choose $\lambda>0$ arbitrarily: 
\[
\lambda\in\underbrace{\rho\left(A\right)}_{\mbox{\makebox[0pt][c]{assumption}}}\cap\overbrace{\rho\left(\widetilde{A}\right)}^{\mbox{L.4.3: \makebox[0pt][l]{\ensuremath{\left\Vert S\!\left(t\right)\right\Vert _{\mathcal{L}\left(X\right)}\le1=Me^{\omega t},M=1,\omega=0}} }}
\]
 and so $\lambda I-\widetilde{A}=\underbrace{\lambda I-A\mbox{ in D\ensuremath{\left(A\right)} }}_{\mbox{injective, onto}}$,
therefore $RHS=\lambda I-\widetilde{A}:D\left(\widetilde{A}\right)\to X$
is onto, injective: $D\left(A\right)\subset D\left(\widetilde{A}\right),\lambda\in\rho\left(\widetilde{A}\right)$.
\end{itemize}

\subsection*{Remarks: }
\begin{enumerate}
\item Generalized Hille\textendash Yosida theorem: $\left(A,D\left(A\right)\right)$
generates $c_{0}$-semigroup, satisfying the estimate $\left\Vert S\!\left(t\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq Me^{\omega t}\iff$
is closed, densely defined and it holds that $\lambda\in\rho\left(A\right)$
for $\forall\lambda>\omega$ and $\left\Vert R^{n}\left(\lambda,A\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq\frac{M}{\left(\lambda-\omega\right)^{n}}\ \forall n\in\mathbb{N}$.\\
Proof: reduce to Th. 4.3: we know $\omega\ne0$ \dots{} $\widetilde{S}=e^{-\omega t}S\!\left(t\right)$,
$M\neq1$ \dots{} by equivalent norm in $X$.
\item Lumer\textendash Phillips theorem: $\left(A,D\left(A\right)\right)$
generates $c_{0}$-semigroup of contractions $\iff$ it is closed
and densely defined and $\left\Vert \lambda x-Ax\right\Vert \geq\lambda\left\Vert x\right\Vert \ \forall x\in X$
for $\forall\lambda>0$ and $\exists\lambda_{0}>0$ s. t. $\lambda_{0}I-A:D\left(A\right)\mapsto X$
is onto.\\
Proof: reduce to Th. 4.3: 
\[
\big\Vert\underbrace{\left(\lambda I-A\right)x}_{y}\big\Vert\geq\lambda\big\Vert\underbrace{x}_{\makebox[0pt][c]{\ensuremath{R\left(\lambda,A\right)y}}}\big\Vert\iff\left\Vert R\left(\lambda,A\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq\frac{1}{\lambda}.
\]
 
\item Heuristic remarks:
\begin{itemize}
\item key condition for generator of semigroup of contractions is in some
sense ``$A\leq0$'': $\left\Vert \lambda x-Ax\right\Vert \geq\lambda\left\Vert x\right\Vert $\dots $\big\Vert x\underbrace{-\frac{A}{\lambda}}_{\mbox{\ensuremath{\approx}}\makebox[0pt][l]{\ensuremath{\,\geq0}}}x\big\Vert\geq\left\Vert x\right\Vert $
\item in the proof of Th. 4.3 we have used the approximations $A_{n}x\to Ax$;
$A_{n}x=n^{2}R\left(n,A\right)-nI$, $nR\left(n,A\right)x\to x$.
For quick check and as clue to remember: $R\left(n,A\right)=\frac{1}{n-A}$$\implies nR\left(n,A\right)=\frac{n}{n-A}=\frac{1}{1-\frac{A}{n}}\to1$;
$A_{n}=\frac{n^{2}}{n-A}-n=\frac{n^{2}-n\left(n-A\right)}{n-A}=\frac{nA}{n-A}=\frac{A}{1-\frac{A}{n}}\to A$.
\end{itemize}
\end{enumerate}

\subsection*{Remark: }

$\left(4.1\right)$ \dots{} $\frac{d}{dt}x=Ax$, $x\left(0\right)=x_{0}$
\dots{} if $S\!\left(t\right)$ is $c_{0}$-semigroup, generated by
$A$$\implies$$S\!\left(t\right)x_{0}$ \dots{} we have reasonable
concept of solution\dots{}
\begin{itemize}
\item $x_{0}\in D\left(A\right)$ \dots{} even a classical solution (Th.
4.1, 2), 
\item $x_{0}\in X\backslash D\left(A\right)$ \dots{} $S\!\left(t\right)x_{0}=\lim_{n\to\infty}S\!\left(t\right)x_{n}$
where $x_{n}\in D\left(A\right),\ x_{n}\to x_{0}$ uniformly in $\left[0,T\right]$.
\end{itemize}

\section{Nonhomogeneous equation}

\[
\left(4.2\right)\quad\frac{d}{dt}u=Au+f\left(t\right),\ u\left(0\right)=u_{0}\in X
\]
where $f\left(t\right):I\to X$ is integrable, $I=\left[0,T\right]$,
$\left(A,D\left(A\right)\right)$ is unbounded operator.

\subsection*{Def.: {[}Classical and strong solution of nonhomogeneous task{]}}

A function $u\left(t\right)$ is called:
\begin{description}
\item [{classical~solution~of~$\left(4.2\right)$}] if $u\left(t\right)\in C^{1}\left(I,X\right)\cap C\left(I,D\left(A\right)\right)$
satisfies the equation $\left(4.2\right)$ for $\forall t\in I$,
\item [{strong~solution~of~$\left(4.2\right)$}] if $u\left(t\right)\in W^{1,1}\left(I,X\right)\cap L^{1}\left(I,D\left(A\right)\right)$
satisfies the equation $\left(4.2\right)$ for a. e. $t\in I$.
\end{description}

\subsection*{Remarks }
\begin{enumerate}
\item Classical solution is a strong one \dots{} classical derivative is
a weak derivative, continuity implies integrability. $\exists u$
classical solution, neccessarily $f\in C\left(I,X\right)$.
\item $u\left(t\right)$ is a strong solution $\iff$ $u\left(t\right)\in L^{1}\left(I,D\left(A\right)\right)$
and $u\left(t\right)=u_{0}+\int_{0}^{t}Au\left(s\right)+f\left(s\right)ds$
for a. e. $t\in I$.\\
Proof: L 1.3, moreover $RHS$ \dots{} $AC$ representative of $u\left(t\right)\in W^{1,1}$.
\item $u\left(t\right)\in C\left(I,D\left(A\right)\right)$ means $t\mapsto u\left(t\right)$
is continuous with respect to the graph norm $D\left(A\right)$: $\left\Vert u\right\Vert +\left\Vert Au\right\Vert $.
$\left(A,D\left(A\right)\right)$ is closed $\iff$$D\left(A\right)$
is complete (Banach) in graph norm. It is easy to show that if $\left(A,D\left(A\right)\right)$
is closed, then $u\left(t\right)\in C\left(I,D\left(A\right)\right)\iff u\left(t\right)\in C\left(I,X\right),\ u\left(t\right)\in D\left(A\right)$
for $\forall t\in I$ and $Au\left(t\right)\in C\left(I,X\right)$.
\end{enumerate}

\subsection*{Lemma 4.4 {[}Hille theorem on characterization of $L^{1}\left(I;D\left(A\right)\right)${]}}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Hille's characterization of $L^1(I;D(A))$}

Let $\left(A,D\left(A\right)\right)$ be closed. Then $u\left(t\right)\in L^{1}\left(I,D\left(A\right)\right)\iff u\left(t\right)\in L^{1}\left(I,X\right)$,
$u\left(t\right)\in D\left(A\right)$ for a. e. $t$, $Au\left(t\right)\in L^{1}\left(I,X\right)$.
Moreover in a such case, $\int_{I}u\left(t\right)dt\in D\left(A\right)$,
$A\left(\int_{I}u\left(t\right)dt\right)=\int_{I}Au\left(t\right)dt$.

\subsubsection*{Proof:}

``$\implies$'' and moreover \dots{} see ex. 5.3.

``$\impliedby$'': observe: for $u_{1}\left(t\right):I\to X_{1}$,
$u_{2}\left(t\right):I\to X_{2}$ integrable and $X_{1}$, $X_{2}$
Banach spaces is $\left(u_{1}\left(t\right),u_{2}\left(t\right)\right):I\to X_{1}\times X_{2}$
integrable, where we consider the norm $\left\Vert \left(u_{1},u_{2}\right)\right\Vert _{X_{1}\times X_{2}}=\left\Vert u_{1}\right\Vert _{X_{1}}+\left\Vert u_{2}\right\Vert _{X_{2}}$
and $\int_{I}\left(u_{1}\left(t\right),u_{2}\left(t\right)\right)dt=\left(\int_{I}u_{1}\left(t\right)dt,\int_{I}u_{2}\left(t\right)dt\right)$
\dots{} easy to prove by expansion of definitions.

Apply this observation to $u_{1}\left(t\right)=u\left(t\right)$,
$u_{2}\left(t\right)=Au\left(t\right)$, $x_{1}=x_{2}=x$ \dots{}
$u_{1}$,$u_{2}:I\to D\left(A\right)$. 

\subsection*{Def.: {[}Mild solution{]}}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Definition: mild solution}

Let $\left(A,D\left(A\right)\right)$ be a generator of $c_{0}$-semigroup
$S\!\left(t\right)$. Then the function $u\left(t\right)$, satisfiying
$u\left(t\right)=S\!\left(t\right)u_{0}+\int_{0}^{t}S\!\left(t-s\right)f\left(s\right)ds$,
$\forall t\in I$ is called a mild solution of $\left(4.2\right)$.

\subsection*{Remark:}

motivation: ``variation of constans'' since $S\!\left(t\right)=e^{tA}$.
Integral is finite, therefore $u\left(t\right)\in C\left(I,X\right)$
\dots{} see next lemma, trivially $\exists!$ mild solution.

\subsection*{Lemma 4.5 {[}Properties of convolution of semigroup{]}\protect \\
}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Properties of semigroup convolutions}

Let $S\!\left(t\right)$ be a $c_{0}$-semigroup, let $v\left(t\right)=\int_{0}^{t}S\!\left(t-s\right)f\left(s\right)ds$.
Then:
\begin{enumerate}
\item $f\left(t\right)\in L^{1}\left(I,X\right)\implies v\left(t\right)\in C\left(I,X\right)$
\item $f\left(t\right)\in C^{0,1}\left(I,X\right)\implies v\left(t\right)\in C^{0,1}\left(I,X\right)$
\item $f\left(t\right)\in C^{1}\left(I,X\right)\implies v\left(t\right)\in C^{1}\left(I,X\right)$
and $v'\left(t\right)=S\!\left(t\right)f\left(0\right)+\int_{0}^{t}S\!\left(t-s\right)f'\left(s\right)ds$
$\forall t\in I$.
\end{enumerate}

\subsubsection*{Proof:}

$v\left(t\right)=\int_{0}^{t}S\!\left(t-s\right)f\left(s\right)ds$,
$S\!\left(t-s\right)f\left(s\right)=g\left(s\right)\in L^{1}\left(\left(0,t\right),X\right)$
(excercise: use measurability and integrability of $f$, $S\!\left(t-s\right)\in\mathcal{L}\left(X\right)$,
$\left\Vert S\!\left(t\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq c_{0}$
\dots{} L. 4.1, $t\in I=\left[0,T\right]$) \dots{} well defined for
$f\left(t\right)\in L^{1}\left(I,X\right)$. 
\begin{enumerate}
\item continuity: let $t,t+h\in I$, WLOG $h>0$. 
\begin{eqnarray*}
v\left(t+h\right)-v\left(t\right) & = & \int_{0}^{t+h}S\!\left(t+h-s\right)f\left(s\right)ds-\int_{0}^{t}S\!\left(t-s\right)f\left(s\right)ds\\
 & = & \int_{0}^{h}S\!\left(t+h-s\right)f\left(s\right)ds+\underbrace{\int_{h}^{t+h}S\!\left(t+h-s\right)f\left(s\right)ds}_{\mbox{subst. \ensuremath{s\to s+h}}}\\
 &  & -\int_{0}^{t}S\!\left(t-s\right)f\left(s\right)ds\\
 & = & \int_{0}^{h}S\!\left(t+h-s\right)f\left(s\right)ds+\int_{0}^{t}S\!\left(t-s\right)\left[f\left(s+h\right)-f\left(s\right)\right]ds
\end{eqnarray*}
\[
\implies\left\Vert v\left(t+h\right)-v\left(t\right)\right\Vert \leq C_{0}\int_{0}^{h}\left\Vert f\left(s\right)\right\Vert ds+C_{0}\int_{0}^{T}\left\Vert f\left(s+h\right)-f\left(s\right)\right\Vert ds.
\]
Set $f\left(t\right)=0$ outside $I=\left[0,T\right]$, we know: $h\to0$:
$RHS\to0$: $f\left(s+h\right)\to f\left(s\right)$ in $L^{1}$ for
$h\to0$ $\implies LHS\to0$.
\item lipschitz continuity: use the last estimate once more, where $f\left(t\right)\in C^{0,1}\left(I,X\right)\implies\left\Vert f\left(t\right)\right\Vert _{X}\leq C_{1}$,
$\left\Vert f\left(t+h\right)-f\left(t\right)\right\Vert \leq C_{2}\left|h\right|$,
together: $\left\Vert v\left(t+h\right)-v\left(t\right)\right\Vert \leq C_{0}C_{1}h+C_{0}TC_{2}h=C_{3}h$.
\item differentiability: 
\[
\frac{1}{h}\left[v\left(t+h\right)-v\left(t\right)\right]=\underbrace{\frac{1}{h}\int_{0}^{h}S\!\left(t+h-s\right)f\left(s\right)ds}_{I_{1}\left(h\right)}+\underbrace{\int_{0}^{t}S\!\left(t-s\right)\left[\frac{f\left(s+h\right)-f\left(s\right)}{h}\right]ds}_{I_{2}\left(h\right)}
\]
 Let $t\in[0,T)$ be fixed, $h\to0^{+}$: then $I_{1}\left(h\right)\to S\!\left(t\right)f\left(0\right)$
\dots{} continuity of integrand (with respect to all variables, see
ex. 6.4), $I_{2}\left(h\right)\to\int_{0}^{t}S\!\left(t-s\right)f'\left(s\right)ds$
\dots{} since $\left\Vert S\!\left(t-s\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq C_{0}$,
we know: $\frac{f\left(s+h\right)-f\left(s\right)}{h}\to\frac{d}{dt}f\left(s\right)=f'\left(s\right)$
a. e. (see ex. 5.4, exchange of limit and integral: Lebesgue theorem,
bounded majorant) \dots{} derivative from the right, derivative from
the left is analogous $t\in(0,T]$. Together: $v'\left(t\right)=S\!\left(t\right)f\left(0\right)+\int_{0}^{t}S\!\left(t-s\right)f'\left(s\right)ds$
$\forall t\in I$ and that is a continuous function of time $t$ (property
of $c_{0}$-semigroup + first part for $f'$ instead of $f$), therefore
$v\left(t\right)\in C^{1}\left(I,X\right)$.
\end{enumerate}

\subsection*{Corollary:}

mild solution is neccessarily continuous.

\subsection*{Lemma 4.6 {[}Equivalent definition of mild solution{]} }

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Equivalent definition of mild solution}

A function $u\left(t\right)$ is a mild solution of $\left(4.2\right)$$\iff$$u\left(t\right)\in C\left(I,X\right)$,
$\int_{0}^{t}u\left(s\right)ds\in D\left(A\right)$, $u\left(t\right)=u_{0}+A\left(\int_{0}^{T}u\left(s\right)ds\right)+\int_{0}^{T}f\left(s\right)ds$
for $\forall t\in I$.

\subsubsection*{Proof:}

\textbf{Step 1:}
\begin{itemize}
\item[``$\implies$'']  let $u\left(t\right)=S\!\left(t\right)u_{0}+\int_{0}^{t}S\!\left(t-s\right)f\left(s\right)ds=u_{1}\left(t\right)+u_{2}\left(t\right)$
be a mild solution. (Remark: $u_{1}$ and $u_{2}$ are mild solutions
of $\left(4.2\right)$ for $f\left(t\right)=0$ and $u_{0}=0$, respectively.)\\
ad $u_{1}$: $\int_{0}^{t}u_{1}\left(s\right)ds=\int_{0}^{t}S\!\left(s\right)u_{0}ds\stackrel{\mbox{Th. 4.1, 3}}{\in}D\left(A\right)$
and $A\left(\int_{0}^{t}S\!\left(s\right)u_{0}ds\right)=S\!\left(t\right)u_{0}-u_{0}$.\\
ad $u_{2}$: consider $A\left(\int_{0}^{t}u_{2}\left(s\right)ds\right)=$
\dots{} 
\begin{eqnarray*}
\int_{0}^{t}u_{2}\left(s\right)ds & = & \underbrace{\int_{0}^{t}\left(\int_{0}^{s}S\!\left(s-\sigma\right)f\left(\sigma\right)d\sigma\right)ds}_{\mbox{denote by \ensuremath{I_{1}}}}\\
\mbox{\ensuremath{\left(\mbox{Fubini: }\ensuremath{0<\sigma<s<t}\right)}} & = & \underbrace{\int_{0}^{t}\underbrace{\left(\int_{\sigma}^{t}S\!\left(s-\sigma\right)f\left(\sigma\right)ds\right)}_{\mbox{denote by \ensuremath{w\left(\sigma\right)}}}d\sigma}_{\mbox{designate \ensuremath{I_{2}}}}
\end{eqnarray*}
 $w\left(\sigma\right)=\int_{0}^{t-\sigma}S\!\left(s\right)\underbrace{f\left(\sigma\right)}_{\makebox[0pt][c]{fixed in \ensuremath{X}}}ds\in D\left(A\right)$.
$Aw\left(\sigma\right)=\underbrace{S\!\left(t-\sigma\right)f\left(\sigma\right)}_{\mbox{\ensuremath{\in}}\makebox[0pt][l]{\ensuremath{\,L^{1}\left(I,X\right)}}}-f\left(\sigma\right)$
\dots{} due to Th. 4.1, 3. Due to L. 4.4 we know: $A\left(\int_{0}^{t}u_{2}\left(s\right)ds\right)=\int_{0}^{t}Aw\left(\sigma\right)d\sigma=\int_{0}^{t}S\!\left(t-\sigma\right)f\left(\sigma\right)-f\left(\sigma\right)d\sigma=u_{2}\left(t\right)-\int_{0}^{t}f\left(s\right)ds$.\\
It is neccessary to check Fubini theorem for $\left(s,\sigma\right)\mapsto S\!\left(s-\sigma\right)f\left(\sigma\right)\in X$:
we know that $I_{1}$, $I_{2}$ exist, it remains to show: $\left\langle f,I_{1}\right\rangle =\left\langle f,I_{2}\right\rangle \ \forall f\in X^{*}$,
i. e., $\int_{0}^{t}\int_{0}^{s}\left\langle f,S\!\left(s-\sigma\right)f\left(\sigma\right)\right\rangle =\int_{0}^{t}\int_{\sigma}^{t}\left\langle f,S\!\left(s-\sigma\right)f\left(\sigma\right)\right\rangle $
\dots{} holds true due to (scalar) Fubini theorem.\\
together $\int_{0}^{t}u_{1}\left(s\right)+u_{2}\left(s\right)ds\in D\left(A\right)$,
\[
A\left(\int_{0}^{t}u_{1}\left(s\right)+u_{2}\left(s\right)ds\right)=S\!\left(t\right)u_{0}-u_{0}+\underbrace{\int_{0}^{t}S\!\left(t-s\right)f\left(s\right)ds}_{u_{2}\left(t\right)}-\int_{0}^{t}f\left(s\right)ds
\]
 which was to be proven. 
\item[``$\impliedby$'']  it suffices to show: the conditions in alternative definiton of
mild solution are satisfied at most by one function $u\left(t\right)$
(since the mild solution satisfies these conditions due to previous
implication). Let $u\left(t\right)$, $\widetilde{u}\left(t\right)$
satisfy the conditions. Denote $w\left(t\right)=u\left(t\right)-\widetilde{u}\left(t\right)$,
therefore: $w\left(t\right)\in C\left(I,X\right)$, $\int_{0}^{t}w\left(s\right)ds\in D\left(A\right)$
and $w\left(t\right)=A\left(\int_{0}^{t}w\left(s\right)ds\right)$
$\forall t\in I$. We will show that $w\left(t\right)=0$ in $I$.
Denote $U\left(t\right)=\int_{0}^{t}w\left(s\right)ds$, we know:
$U\left(0\right)=0$, $\frac{d}{dt}U\left(t\right)=w\left(t\right)$
(due to the continuity of $w\left(t\right)$), $U\left(t\right)\in D\left(A\right)$,
$AU\left(t\right)=w\left(t\right)$. Therefore $U\left(t\right)$
is a classical solution of $\frac{d}{dt}u=Au$, $u\left(0\right)=0$.
Therefore it is zero, see the proof of L. 4.2 (see ex. 6.4, auxiliary
function $y\left(t\right)=S\!\left(T-t\right)U\left(t\right)$, zero
derivative).
\end{itemize}

\subsection*{Corollaries: }
\begin{enumerate}
\item Strong solution is a mild solution: proof: strong solution $u\left(t\right)=u_{0}+\int_{0}^{t}Au\left(s\right)ds+\int_{0}^{t}f\left(s\right)ds$
for a. e. $t\in I$, $u\left(t\right)\in L^{1}\left(I,D\left(A\right)\right)$,
or more precisely its $AC$ representative (due to L. 4.4 we can ``pull''
$A$ outside the integral.
\item There is at most one classical and/or strong solution. Proof: we know
classical$\implies$strong$\implies$mild, which is unique. (Up to
sets of zero measure for the strong solution.)
\end{enumerate}

\subsection*{Remarks: }
\begin{enumerate}
\item The mild solution is not neccessarily a neither strong or classical
solution. E. g., consider $\left(4.2\right)$ for $u_{0}=0$, $f\left(t\right)=S\!\left(t\right)x$,
where $x\in X$ is such that $S\!\left(t\right)x\notin D\left(A\right)$
for any $t>0$. I. e., the mild solution $u\left(t\right)=S\!\left(t\right)0+\int_{0}^{t}\underbrace{S\!\left(t-s\right)S\!\left(s\right)}_{S\!\left(t\right)}xds=tS\!\left(t\right)x\notin D\left(A\right)$,
$t>0$ can be neither strong nor classical solution. 
\item Every mild solution is a (locally uniform) limit of very nice classical
solutions.\\
D\r{u}kaz: let $u\left(t\right)$ be a mild solution, i. e., $u\left(t\right)=u_{0}+A\left(\int_{0}^{t}u\left(s\right)ds\right)+\int_{0}^{t}f\left(s\right)ds$,
$t\in I$. Define $u_{h}\left(t\right)=\frac{1}{h}\int_{t}^{t+h}u\left(s\right)ds$,
$t\in\left[0,T-\Delta\right]$, $\Delta>0$, $h\in\left(0,\Delta\right)$.
We know: $u_{h}\left(t\right)\rightrightarrows u\left(t\right)$ in
$\left[0,T-\Delta\right]$ by the continuity of $u\left(t\right)$.
Equation for $u_{h}\left(t\right)$? $\frac{d}{dt}u_{h}\left(t\right)=\frac{1}{h}\left[u\left(t+h\right)-u\left(t\right)\right]=\frac{1}{h}\left[A\left(\int_{t}^{t+h}u\left(s\right)ds\right)+\int_{t}^{t+h}f\left(s\right)ds\right]=Au_{h}\left(t\right)-f_{h}\left(t\right)$,
where $f_{h}\left(t\right)=\frac{1}{h}\int_{t}^{t+h}f\left(s\right)ds$,
$u_{h}\left(0\right)\in D\left(A\right)$.
\end{enumerate}

\subsection*{Theorem 4.4 {[}Regularity of mild solution{]}}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Regularity of mild solution}
\begin{enumerate}
\item Let $u_{0}\in D\left(A\right)$, $f\left(t\right)\in C^{1}\left(I,X\right)$.
Then the mild solution is classical.
\item Let $u_{0}\in D\left(A\right)$, $f\left(t\right)\in C^{0,1}\left(I,X\right)$,
moreover let $X$ be reflexive. Then the mild solution is strong.
\end{enumerate}

\subsubsection*{Proof:}

$u\left(t\right)=S\!\left(t\right)u_{0}+\int_{0}^{t}S\!\left(t-s\right)f\left(s\right)ds=u_{1}\left(t\right)+v\left(t\right)$,
$u_{1}\left(t\right)$ is classical solution (and therefore a strong
one) for $f\left(t\right)\equiv0$, it remains to show: $v\left(t\right)$
is a classical (strong) solution $\left(4.2\right)$ for $u_{0}=0$.
Let $f\left(t\right)\in C^{0,1}\left(I,X\right)$. Then due to L.
4.5, 2 we know: $v\left(t\right)\in C^{0,1}\left(I,X\right)\subset W^{1,1}\left(I,X\right)$,
moreover $\exists\frac{d}{dt}v\left(t\right)\in X$ for a. e. $t\in I$
(see Chapter 1: Th. 1.5, use reflexivity).

\textbf{Trick:} $A\left(v\left(t\right)\right)=?$ 
\begin{eqnarray*}
\frac{1}{h}\left[S\!\left(h\right)-I\right]v\left(t\right) & = & \frac{1}{h}\Big[\underbrace{S\!\left(h\right)\int_{0}^{t}S\!\left(t-s\right)f\left(s\right)ds}_{\left(*\right)}-\int_{0}^{t}S\!\left(t-s\right)f\left(s\right)ds\Big]\\
 & = & \frac{1}{h}\left[v\left(t+h\right)-v\left(t\right)\right]-\frac{1}{h}\int_{t}^{t+h}S\!\left(t+h-s\right)f\left(s\right)ds,
\end{eqnarray*}
\begin{eqnarray*}
\mbox{since }\left(*\right) & = & \int_{0}^{t}S\!\left(t+h-s\right)f\left(s\right)ds\pm\int_{t}^{t+h}S\!\left(t+h-s\right)f\left(s\right)ds\\
 & = & v\left(t+h\right)-\int_{t}^{t+h}S\!\left(t+h-s\right)f\left(s\right)ds.
\end{eqnarray*}

Take $h\to0^{+}$: $RHS\to\frac{d}{dt}v\left(t\right)-f\left(t\right)$
\dots{} due to the existence of derivative a. e. and the continuity
of integrand. ($S(\underbrace{t+h-s}_{\to0})f\left(s\right)=\varphi\left(h,s\right)$
\dots{} continuous, $\varphi\left(0,0\right)=f\left(t\right)$).

I. e., for a. e. $t$: $v\left(t\right)\in D\left(A\right)$, $Av\left(t\right)=\frac{d}{dt}v\left(t\right)-f\left(t\right)\in L^{1}\left(I,X\right)$,
L. 4.4$\implies v\left(t\right)\in L^{1}\left(I,D\left(A\right)\right)$,
which was to be proven.

Part one is proven similarly, the derivative $Av\left(t\right)=\frac{d}{dt}v\left(t\right)-f\left(t\right)$
is evaluated everywhere, L. 4.5, 3, we will get a classical equation,
$v\left(t\right)$ is continuous.

\begin{figure}
\includegraphics[width=0.6\textwidth]{img/k4_triangle}

\caption{Relations between semigroup, generator and resolvent.}

\end{figure}

\subsection*{Remarks: }

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Remarks on exponential formulas}

Exponential formula $A\to S\!\left(t\right)=\text{\quotedblbase}e^{tA}\text{\textquotedblright}$.
\begin{enumerate}
\item $A\in\mathcal{L}\left(X\right)$ \dots{} $S\!\left(t\right)=e^{tA}=\sum_{k=0}^{\infty}\frac{t^{k}}{k!}A^{k}\in\mathcal{L}\left(X\right)$
\dots{} exactly all uniformly continuous semigroups, i. e., $S\!\left(t\right)\to I$
v $\mathcal{L}\left(X\right)$, $t\to0^{+}$ (see ex. 5).
\item Hille\textendash Yosida (Th. 4.3): $S\!\left(t\right)=\lim_{n\to\infty}e^{-nt}e^{t\underbrace{n^{2}R\left(n,A\right)}_{\in\mathcal{L}\left(X\right)}}$
\dots{} ``Yosida approximation''.
\item Statement: $S\!\left(t\right)$ is a $c_{0}$-semigroup, $\implies$$S\!\left(t\right)u=\lim_{h\to0^{+}}e^{tA_{h}u}$
\dots{} holds for $\forall u\in X$ fixed, uniformly with respect
to$t\in\left[0,T\right]$, where $A_{h}=\frac{1}{h}\left[S\!\left(h\right)-I\right]$.\\
An example of application: ex. 5.4: shift semigroup: $S\!\left(t\right):u\left(x\right)\mapsto u\left(x+t\right)$,
$u\left(x\right)\in L^{2}\left(\mathbb{R}\right)=X$, generator: $A=\frac{d}{dx}\ \left(\mbox{weakly}\right)$,
$D\left(A\right)=W^{1,2}\left(\mathbb{R}\right)$.\\
$LHS$: $\left[S\!\left(t\right)u\right]\left(x\right)=u\left(x+t\right)$,
$RHS$: $A_{h}=\frac{1}{h}\left[S\!\left(h\right)-I\right]:u\left(x\right)\mapsto\frac{u\left(h+h\right)-u\left(x\right)}{h}=d_{h}u\left(x\right)$.
\\
Then $\left(e^{tA_{h}}u\right)\left(x\right)=\sum_{k=0}^{\infty}\frac{t^{k}}{k!}\left(d_{h}\right)^{k}u\left(x\right)$,
where $d_{h}^{2}u\left(x\right)=d_{h}\left(d_{h}u\left(x\right)\right)=\frac{1}{h}\left(d_{h}\left(u+h\right)-d_{h}u\left(x\right)\right)=\frac{1}{h^{2}}\left(u\left(x+2h\right)+u\left(x\right)-2u\left(x+h\right)\right)$.\\
Together: $u\left(x+t\right)=\lim_{h\to0^{+}}\sum_{k=0}^{\infty}\frac{t^{k}}{k!}\left(d_{h}^{k}u\right)\left(x\right)$
\dots{} Taylor (formally: if $d_{h}^{k}u\left(x\right)\to\left(\frac{d}{dx}\right)^{k}u\left(x\right)$,
then we get an approximation by Taylor expansion, rigorously for $u\left(x\right)$
smooth.)
\item Statement: $S\!\left(t\right)$ is a $c_{0}$-semigroup generated
by $\left(A,D\left(A\right)\right)$ , then $S\!\left(t\right)u=\lim_{n\to\infty}\left(I-\frac{t}{n}A\right)^{-n}u$
for $\forall u\in X$ fixed. \\
\\
\textbf{Remarks:}
\begin{itemize}
\item an analogy of formula from basic analysis: $e^{at}=\lim_{n\to\infty}\left(1-\frac{t}{n}a\right)^{-n}$
\item expression on $LHS$ makes sense: $\left(I-\frac{t}{n}A\right)^{-1}=\left(\frac{t}{n}\left(\frac{n}{t}I-A\right)\right)^{-1}=\frac{n}{t}R\left(\frac{n}{t},A\right)$
\dots{} well defined for $n$ sufficiently large (see L. 4.3)
\end{itemize}
\end{enumerate}

\subsection*{Def.:}

Let $S\!\left(t\right)$ be a $c_{0}$-semigroup on $X$ generated
by $\left(A,D\left(A\right)\right)$. We say that $S\!\left(t\right)$
is differentiable, if $S\!\left(t\right)u_{0}\in D\left(A\right)$
for $\forall t>0$ and $\forall u_{0}\in X$.

\subsection*{Remark:}

It is easy to show that $S\!\left(t\right)u_{0}\in D\left(A\right)\iff\lim_{h\to0^{+}}\frac{1}{h}\left(S\!\left(t+h\right)-S\!\left(t\right)\right)u_{0}$
exists for$\forall t>0$, $u_{0}\in X$ and therefore $\lim\dots=AS\!\left(t\right)u_{0}$.
$\implies S\!\left(t\right)u_{0}$ satisfies $\left(4.1\right)$ classically
for $\forall t>0$ \dots{} ``mollifying property''.

\subsection*{Examples:}
\begin{enumerate}
\item Shift semigroup $S\!\left(t\right):u\left(x\right)\mapsto u\left(x+t\right)$
is not differentiable since $S\!\left(t\right)u_{0}\in D\left(A\right)=W^{1,2}\left(\mathbb{R}\right)\iff u_{0}\in W^{1,2}\left(\mathbb{R}\right)$.
\item Heat semigroup is differentiable (see below and ex.~6.3).
\end{enumerate}

\subsection*{Remark:}

the heat semigroup is differentiable. 

\subsubsection*{Proof:}

goal: $\mathcal{S}\left(t\right)u_{0}=\sum_{j}\underbrace{e^{-\lambda_{j}t}u_{j}}_{\widetilde{u}_{j}}w_{j}\in D\left(\Delta\right)=W_{0}^{1,2}\cap W^{2,2}$
for $t>0$, $u_{0}\in L^{2}$ arbitrary. Due to statement 6.3.2, b)
$\iff$ $\sum_{j}\lambda_{j}^{2}\widetilde{u}_{j}^{2}<\infty$ $\iff\sum_{j}\underbrace{\lambda_{j}^{2}e^{-2\lambda_{j}t}}_{\leq c_{t}}u_{j}^{2}<\infty$
\dots{} which follows by the followin:
\begin{itemize}
\item $\sum_{j}u_{j}^{2}<\infty$
\item $\lambda\mapsto\lambda^{2}e^{-2\lambda t}$ is bounded with respect
to $\lambda\in[0,\infty)$, $\lambda_{j}>0$.
\end{itemize}

\subsection*{Remark:}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Applications of semigroup theory to heat and wave equation}

\begin{eqnarray*}
\left(H\!E\right)\quad\frac{d}{dt}u & = & \Delta u\\
u\left(0\right) & = & u_{0}
\end{eqnarray*}
solution from the point of view of chapter 4: define an operator $\Delta:D\left(\Delta\right)\to L^{2}$.
We want to show that it is a generator of$c_{0}$-semigroup (moreover,
a semigroup of contractions) on $L^{2}$. Due to Th. 4.3 (Hille\textendash Yosida)
it suffices (and at the same time we know that it is neccessary) to
check that:
\begin{itemize}
\item $\Delta$ is densely defined and closed.
\item $\lambda I-\Delta:D\left(\Delta\right)\to L^{2}$ is injective and
onto for $\forall\lambda>0$
\item $\big\Vert\underbrace{\left(\lambda I-\Delta\right)^{-1}}_{R\left(\lambda,\Delta\right)}f\big\Vert\leq\frac{1}{\lambda}\left\Vert f\right\Vert _{2}$
\end{itemize}
(voluntary excercise), see below for the wave equation.

\subsection*{Example:}

\begin{eqnarray*}
\left(W\!E\right)\quad\frac{d^{2}}{dt^{2}}u & = & \Delta u\\
u\left(0\right) & = & u_{0}\\
\frac{d}{dt}u\left(0\right) & = & u_{1}
\end{eqnarray*}
we want to use the semigroup theory. How to deal with $\frac{d^{2}}{dt^{2}}$:
d'Alembert: $U=\left(\begin{array}{c}
u\\
v
\end{array}\right)$, $v=\frac{d}{dt}u$ \dots{} $\left(W\!E\right)$ is transformed to
$\frac{d}{dt}U=AU$, $A=\left(\begin{array}{cc}
0 & I\\
\Delta & 0
\end{array}\right)$. $X=W_{0}^{1,2}\times L^{2}$, $u_{0}\in W_{0}^{1,2}$, $v\in L^{2}$,
$D\left(A\right)=\underbrace{D\left(\Delta\right)}_{=W_{0}^{1,2}\cap W^{2,2}}\times W_{0}^{1,2}$. 

Goal: $A$ is a generator of $c_{0}$-semigroup of contractions on
$X$ \dots{} by Hille\textendash Yosida.
\begin{itemize}
\item density: $D\left(A\right)$ is dense in $X$ \dots{} $W_{0}^{1,2}\cap W^{2,2}$
is dense in $W_{0}^{1,2}$ and $W_{0}^{1,2}$ is dense in $L^{2}$
\dots{} known from previous courses on PDE.
\item closedness: $U_{n}\in D\left(A\right)$, $AU_{n}=F_{n}$, $U_{n}\to U$,
$F_{n}\to F$ in $X$, then $U\in D\left(A\right)$ and $AU=F$.\\
component-wise: $U_{n}=\left(\begin{array}{c}
u_{n}\\
v_{n}
\end{array}\right)$, $F_{N}=\left(\begin{array}{c}
f_{n}\\
g_{n}
\end{array}\right)$, therefore $\left(W\!E\right)\iff{v_{n}=f_{n}\mbox{ \dots\ in \ensuremath{W_{0}^{1,2}}}\atop \Delta u_{n}=g_{n}\mbox{ \dots\ in \ensuremath{L^{2}}}}$.
For $n\to\infty$: the first equation: $v=f\mbox{ }$ \dots{} in $W_{0}^{1,2}$,
second equation: \textbf{trick:} a weak formulation $-\left(\nabla u_{n},\nabla w\right)=\left(g_{n},v\right)$,
$n\to\infty$: $-\left(\nabla u,\nabla w\right)=\left(g,w\right)$.
Use elliptic regularity: $-\Delta u=-g$ weakly, $g\in L^{2}$, elliptic
regularity $\implies u\in W^{2,2}$.
\item resolvent: we need: $\forall\lambda>0,\forall F\in X\ \exists!U\in D\left(A\right):\ \left(\lambda I-A\right)U=F$,
$\left(U=R\left(\lambda,A\right)F\right)$, we want to show that $\left\Vert U\right\Vert _{X}\leq\frac{1}{\lambda}\left\Vert F\right\Vert _{X}$.
Component-wise: $F=\left(\begin{array}{c}
f\\
g
\end{array}\right)\in W_{0}^{1,2}\times L^{2}$\dots $\exists!U=\left(\begin{array}{c}
u\\
v
\end{array}\right)\in W_{0}^{1,2}\cap W^{2,2}\times W_{0}^{1,2}$, i. e.: 
\begin{eqnarray*}
\lambda u-v & = & f\quad/\lambda\cdot\mbox{, add to second equation}\\
\lambda v-\Delta u & = & g
\end{eqnarray*}
\textbf{Trick:} 
\begin{eqnarray*}
\iff\quad\quad\quad\quad v & = & \lambda u-f\\
\lambda^{2}u-\Delta u & = & g+\lambda f
\end{eqnarray*}
There exists only one solution of second equation $u\in W_{0}^{1,2}\cap W^{2,2}$
(Lax-Milgram \& elliptic regularity \dots{} weak formulation, $RHS\in L^{2}$)
first equation: $v\in W_{0}^{1,2}$.\\
the ``correct norm'' on $X=W_{0}^{1,2}\times L^{2}$ is $\left\Vert \left(u,v\right)\right\Vert =\left(\left\Vert \nabla u\right\Vert _{2}^{2}+\left\Vert v\right\Vert _{2}^{2}\right)^{\frac{1}{2}}$.
\textbf{Trick:} 
\begin{eqnarray*}
\lambda u-v & = & f\quad/-\Delta u,\int_{\Omega}\\
\lambda v-\Delta u & = & g\quad/v,\int_{\Omega}
\end{eqnarray*}
 
\begin{eqnarray*}
\lambda\left\Vert \nabla u\right\Vert _{2}^{2}-\left(\nabla v,\nabla u\right) & = & \left(\nabla f,\nabla u\right)\\
\lambda\left\Vert v\right\Vert _{2}^{2}+\left(\nabla u,\nabla v\right) & = & \left(g,v\right)
\end{eqnarray*}
 sum: 
\begin{eqnarray*}
\lambda\left(\left\Vert \nabla u\right\Vert _{2}^{2}+\left\Vert v\right\Vert _{2}^{2}\right) & \leq & \left\Vert \nabla f\right\Vert _{2}\left\Vert \nabla u\right\Vert _{2}+\left\Vert g\right\Vert _{2}\left\Vert v\right\Vert _{2}.\\
 & \stackrel{\mbox{\makebox[16pt][r]{C.\textendash S. in \ensuremath{\ensuremath{\mathbb{R}^{2}}}}}}{\le} & \left(\left\Vert \nabla f\right\Vert _{2}^{2}+\left\Vert g\right\Vert _{2}^{2}\right)^{\frac{1}{2}}\left(\left\Vert \nabla u\right\Vert _{2}^{2}+\left\Vert \nabla v\right\Vert _{2}^{2}\right)^{\frac{1}{2}}.
\end{eqnarray*}
Therefore $\lambda\left(\left\Vert \nabla u\right\Vert _{2}^{2}+\left\Vert v\right\Vert _{2}^{2}\right)^{\frac{1}{2}}\leq\left(\left\Vert \nabla f\right\Vert _{2}^{2}+\left\Vert g\right\Vert _{2}^{2}\right)^{\frac{1}{2}}$.
\end{itemize}

\subsection*{Example:}

\begin{eqnarray*}
\left(4.2\right)\quad\frac{d}{dt}u & = & Au+f\left(t\right)\\
u\left(0\right) & = & u_{0}
\end{eqnarray*}
$f\left(t\right):I\to X$ and $u_{0}\in X$ are given.

\subsection*{Recall:}

Mild solution: $u\left(t\right)=S\!\left(t\right)u_{0}+\int_{0}^{t}S\!\left(t-s\right)f\left(s\right)ds$,
$t\in I$, where $S\!\left(t\right)$ is a $c_{0}$-semigroup genereated
by $A$.

Th. 4.4: $u_{0}\in D\left(A\right)$, $f\left(t\right)\in C^{0,1}$
or $C^{1}$, then the mild solution is strong or classical, respectively.

\subsection*{Example:}

$\frac{\partial}{\partial t}u=\frac{\partial}{\partial x}u+f\left(t,x\right)$,
$t>0$, $x\in\mathbb{R}$, $u=u_{0}$ for $t=0$, $x\in\mathbb{R}$.
Transformation to $\left(4.2\right)$ \dots{} $A=\frac{d}{dx}$, $D\left(A\right)=W^{1,2}\left(\mathbb{R}\right)$,
$u_{0}\in X=L^{2}\left(\mathbb{R}\right)$, $f\left(t,x\right):I\leftrightarrow L^{2}\left(\mathbb{R}\right)$,
$t\mapsto f\left(t,\cdot\right)$. We know: $\left(A,D\left(A\right)\right)$
is generator of shift $c_{0}$-semigroup: $\left[S\!\left(t\right)u\right]=u\left(x+t\right)$.

Mild solution: $u\left(t\right)=\left[S\!\left(t\right)u_{0}\right]\left(x\right)+\int_{0}^{t}\underbrace{\left[S\!\left(t-s\right)f\left(s,\cdot\right)\right]\left(x\right)}_{f\left(s,x+t-s\right)}ds=u_{0}\left(x+t\right)+\int_{0}^{t}f\left(s,x+t-s\right)ds\in L^{2}\left(\mathbb{R}\right)$.

\subsection*{Remark:}

formally (i. e., classically for $u_{0}$, $f$ smooth enough) it
is a solution \dots{} (voluntary excercise), but the mild solution
makes sense even for $u_{0}\in L^{2}\left(\mathbb{R}\right)$, $f\left(t\right):I\to L^{2}$
integrable (and not better).

\addtocontents{toc}{\protect\newpage}

\part*{Excercises}

\subsection*{Ex 1.1}

Assume that $u(t)\in C(I,X)$ (always $I=[0,T]$) and prove the following
propositions:
\begin{enumerate}
\item $u(I)\subset X$ is compact, and $u(t):I\to X$ is uniformly continuous.
\item Prove that $u(t):I\to X$ is strongly measurable (a) using the Pettis
theorem and (b) directly from the definition
\item Prove that $u(t):I\to X$ is Bochner integrable (a) using the Bochner
theorem and (b) directly from the definition
\item Show that $\int_{I}u(t)\,dt=\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{n}u(jT/n)$ 
\end{enumerate}

\subsection*{Ex 1.2}
\begin{enumerate}
\item Let $u(t):I\to X$ be Bochner integrable, let $F:X\to Y$ be linear,
continuous. Then $Fu(t):I\to Y$ is Bochner integrable, and 
\[
F\left(\int_{I}u(t)\,dt\right)=\int_{I}F\big(u(t)\big)\,dt
\]
In particular: $\langle x^{*},\int_{I}u(t)\,dt\rangle=\int_{I}\langle x^{*},u(t)\rangle\,dt$
for any $x^{*}\in X^{*}$.
\item Prove the following version of Fatou's lemma: let $u_{n}(t):I\to X$
are (strongly) measurable, and $u_{n}(t)$ converge weakly to $u(t)$
for a.e. $t\in I$. Then $u(t)$ is measurable and $\int_{I}\|u(t)\|\,dt\le\liminf_{n\to\infty}\int_{I}\|u_{n}(t)\|\,dt$.
In particular, if $\int_{I}\|u_{n}(t)\|\,dt$ are bounded, then $u(t)$
is integrable.
\end{enumerate}

\subsection*{Ex 1.3}

Let $\psi_{0}(t):\mathbb{R}\to\mathbb{R}$ be convolution kernel,
i.e. $\psi_{0}(t)$ is bounded, zero outside $[-1,1]$ and $\int_{-1}^{1}\psi_{0}(t)\,dt=1$.
Let $u(t):I\to X$ be given, and assume that $u(t)$ is extended by
zero outside of $I$. Let $\psi_{n}(t)=n\psi_{0}(nt)$ and finally
let $u_{n}(t)=u*\psi_{n}(t)=\int_{\mathbb{R}}u(t-s)\psi_{n}(s)\,ds$.
\begin{enumerate}
\item Show that $u_{n}(t)\in C(I,X)$ and if $\psi_{0}\in C^{1}$, then
also $u_{n}(t)\in C^{1}(I,X)$ and $u'_{n}(t)=u*\psi'_{n}(t)$.
\item Show that norm of $u_{n}(t)$ is not larger than the norm of $u(t)$
in the spaces $C(I,X)$, $L^{p}(I,X)$.
\item Show that if $u(t)\in C(J,X)$ for some $J$ strictly larger than
$I$, then $u_{n}(t)\rightrightarrows u(t)$ in $C(I,X)$, for $n\to\infty$. 
\end{enumerate}

\subsection*{Ex 1.4}
\begin{enumerate}
\item Prove the following Convergence Principle: Let $F_{n}:X\to X$ be
a sequence of linear operators such that the norms $\|F_{n}\|$ are
bounded independently of $n$. Let there be a dense $S\subset X$
such that $F_{n}v\to v$ as $n\to\infty$ for any $v\in S$. Then
$F_{n}u\to u$ as $n\to\infty$ for any $u\in X$.
\item Apply the Convergence Principle to prove part 4 of Lemma 1.1.
\end{enumerate}

\subsection*{Ex 2.1}

Let $X$ be reflexive, separable.
\begin{enumerate}
\item Let $p\in(1,\infty)$. Show that any $u(t)\in W^{1,p}(I;X)$ has a
$\alpha$-Hölder continuous representative, with $\alpha=1-1/p$.
\item Show that $W^{1,\infty}(I;X)=C^{0,1}(I;X)$ (the space of Lipschitz
functions), in the sense of representative.
\item Let $u_{n}(t)$ be weakly differentiable, and let $u_{n}(t)\rightharpoonup u(t)$,
$\frac{d}{dt}u_{n}(t)\rightharpoonup g(t)$ (weakly) in $L^{1}(I;X)$.
Then $u(t)$ is weakly differentiable, with $\frac{d}{dt}u(t)=g(t)$.
\item Let $u_{n}(t)$ are bounded in $L^{p}(I;Y)$, $\frac{d}{dt}u_{n}(t)$
are bounded in $L^{q}(I;Z)$, where $p$, $q\in(1,\infty)$ and $Y$,
$Z$ are reflexive, separable. Then there is a subsequence so that
$\tilde{u}_{n}(t\rightharpoonup u(t)$, $\frac{d}{dt}\tilde{u}_{n}(t)\rightharpoonup g(t)$
in the respective spaces, and $\frac{d}{dt}u(t)=g(t)$.
\end{enumerate}

\subsection*{Ex 2.2}
\begin{enumerate}
\item Prove that $L^{p}(I;L^{p}(\Omega))=L^{p}(I\times\Omega)$ if $p\in[1,\infty)$,
but $L^{\infty}(I;L^{\infty}(\Omega))\subsetneq L^{\infty}(I\times\Omega)$.
\item Prove that if $u_{n}\rightharpoonup u$ in $L^{p}(I;L^{q}(\Omega))$,
and $\Omega\subset\mathbb{R}^{n}$ is open, bounded, then 
\[
\int_{I\times\Omega}u_{n}(t,x)\psi(t,x)\,dtdx\to\int_{I\times\Omega}u(t,x)\psi(t,x)\,dtdx
\]
for any (say) bounded, measurable function $\psi(t,x)$.
\end{enumerate}

\subsection*{Ex 2.3}

Let $H$ be a Hilbert space. Show that $H$ is uniformly convex. Show
directly that if $u_{n}\rightharpoonup u$ and $\left\Vert u_{n}\right\Vert _{H}\to\left\Vert u\right\Vert _{H}$,
then $u_{n}\to u$.

\subsection*{{*} Ex 2.4}

Let $u_{n}(t)$ be bounded in $L^{p}(I;X)$, where $p\in(1,\infty]$,
and $X$ be reflexive, separable. Prove that there is a weakly convergent
($*$-weak if $p=\infty$) subsequence, using only Theorem 1.9 and
separability of $L^{p'}(I;X^{*})$.

\subsection*{Ex 2.5}

Let $u_{n}(t)\rightharpoonup u(t)$ in $L^{p}(I;X)$, $v_{n}(t)\to v(t)$
in $L^{p'}(I;X^{*})$, where $p$, $p'$ are Hölder conjugate. Prove
that $\int_{I}\left\langle u_{n}(t)v_{n}(t)\right\rangle _{X,X^{*}}\,dt\to\int_{I}\left\langle u(t),v(t)\right\rangle _{X,X^{*}}\,dt$.

\subsection*{Ex 3.1}

Let $u(t)\in L^{2}(I;W_{0}^{1,2})$, $g(t)\in L^{2}(I;W^{-1,2})$
and $u_{0}\in L^{2}$. Then the following are equivalent: 
\begin{enumerate}
\item[(i)] $\frac{d}{dt}u(t)=g(t)$ and $u(0)=u_{0}$ (in the sense of representative) 
\item[(ii)] for any $v\in W_{0}^{1,2}$, $\varphi\in C_{c}^{\infty}((-\infty,T))$
one has 
\[
-\int_{I}(u(t),v)\varphi'(t)\,dt=\int_{I}\left\langle g(t),v\right\rangle \varphi(t)\,dt+(u_{0},v)\varphi(0)
\]
\end{enumerate}

\subsubsection*{Corollary:}

$u\left(t\right)\in L^{2}\left(I,W_{0}^{1,2}\right)$ is w. s. with
$u\left(0\right)=u_{0}$ (for representative) iff
\begin{align*}
\int_{I}\left(u\left(t\right),v\right)\varphi'\left(t\right)dt+ & \int_{I}\left(a\left(\nabla u\left(t\right)\right),v\right)\varphi\left(t\right)dt+\int_{I}\left(f\left(u\left(t\right)\right),v\right)\varphi dt\\
 & =\int_{I}\left\langle h\left(t\right),v\right\rangle +\left(u_{0},v\right)\varphi\left(0\right)
\end{align*}

\subsection*{Ex 3.2}

Recall the notation and assumptions from Chapter 2: let $f(z):\mathbb{R}\to\mathbb{R}$,
$a(\xi):\mathbb{R}^{n}\to\mathbb{R}^{n}$ be Lipschitz continuous.
Let the operators $\mathcal{A}:W_{0}^{1,2}\to W^{-1,2}$ and $\mathcal{F}:I\times W_{0}^{1,2}\to W^{-1,2}$
be defined as 
\begin{align*}
\left\langle \mathcal{A}(u),v\right\rangle  & =\int_{\Omega}a(\nabla u(x))\cdot\nabla v(x)\dx\\
\mathcal{F}(t,u) & =-\mathcal{A}(u)-\iota f(u)+h(t)
\end{align*}
where $h(t)\in L^{2}(I;W^{-1,2})$ is a fixed function.
\begin{enumerate}
\item Prove that $u\mapsto f(u)$ is Lipschitz as operator $L^{2}\to L^{2}$,
and also $u(t)\mapsto f(u(t))$ is Lipschitz as operator $L^{2}(I;L^{2})\to L^{2}(I;L^{2}))$.
N. B. do not forget to verify that $f(u)$ and $f(u(t))$ are \emph{measurable}
in the appropriate sense.
\item Deduce that also $u(t)\mapsto\iota f(u(t))$ is Lipschitz as operator
$L^{2}(I;W_{0}^{1,2})\to L^{2}(I;W^{-1,2})$.
\item Prove that $u\to\mathcal{A}(u)$ is Lipschitz continuous as operator
$W_{0}^{1,2}(\Omega)\to W^{-1,2}(\Omega)$.
\item Show that $\left\Vert \mathcal{F}(t,u)\right\Vert _{-1,2}\le c(1+\left\Vert u\right\Vert _{1,2}+\left\Vert h(t)\right\Vert _{-1,2})$
with some constant only depending on the nonlinearities $a(\cdot)$
and $f(\cdot)$.
\end{enumerate}

\subsection*{Ex 3.3}

Let $W_{0}^{1,2}\hookrightarrow L^{2}\hookrightarrow W^{-1,2}$ be
the Gelfand triple, with the embedding $\iota:W_{0}^{1,2}\to W^{-1,2}$. 
\begin{enumerate}
\item Observe that due to the Poincaré inequality, $W_{0}^{1,2}$ is a Hilbert
space with the scalar product $\left(\left(u,v\right)\right)=\int_{\Omega}\nabla u\left(x\right)\cdot\nabla v\left(x\right)\dx$.
\item By Riesz theorem, any $f\in W^{-1,2}$ can be represented by some
$u_{f}\in W_{0}^{1,2}$ so that 
\[
\left\langle f,v\right\rangle =\left(\left(u_{f},v\right)\right)\quad\forall v\in W_{0}^{1,2}
\]
\item Show that by a Green formula $\left(\left(u,v\right)\right)=\left(-\Delta u,v\right)$
for any $v\in W_{0}^{1,2}$ and $u\in C_{c}^{\infty}$.
\item Combine that above with the density of $C_{c}^{\infty}$ in $W_{0}^{1,2}$
to show that for any $f\in W^{-1,2}$ there exists smooth functions
$u_{n}$ such that $\iota u_{n}\to f$.
\item Finally, show that $\iota:W_{0}^{1,2}\to W^{-1,2}$ is \emph{injective}.
\end{enumerate}
These are the reasons why no symbol ``$\iota$'' is normally employed
and $\left(\cdot,\cdot\right)_{L^{2}}$ is simply seen as a generalization
of $\left\langle \cdot,\cdot\right\rangle _{W^{-1,2},W_{0}^{1,2}}$
without further notational ado.

\subsection*{Ex 4.1}

Let $X$ be reflexive, separable, $X\hookrightarrow Z$, and let $u(t)\in L^{\infty}(I;X)\cap C(I;Z)$.
Show that $u(t)\in X$ for \emph{all} $t\in I$ and moreover, $t\mapsto u(t)\in X$
is weakly continuous.

\subsection*{Ex 4.2}

Let $w_{j}$ be the eigenfunctions of $-\Delta u=\lambda u$ with
zero Dirichlet b.c. Let $P_{N}$ be the ON projection (in $L^{2}$)
on the space $\text{span}\,\{w_{1},\dots,w_{N}\}$. Clearly $P_{N}$
is continuous $L^{2}\to L^{2}$ with norm 1.
\begin{enumerate}
\item Show that $P_{N}$ is also continuous $W_{0}^{1,2}\to W_{0}^{1,2}$
with norm 1, if $W_{0}^{1,2}$ is taken as a Hilbert space with scalar
product $(\!(u,v)\!)=(\nabla u,\nabla v)$.
\item Show that $\left\Vert P_{N}u\right\Vert _{2,2}\le c\left\Vert u\right\Vert _{2,2}$
for any $u\in W_{0}^{1,2}\cap W^{2,2}$ (assume $\partial\Omega$
sufficiently regular).
\end{enumerate}

\subsection*{Ex 4.3}

Let $\psi(z):\mathbb{R}\to\mathbb{R}$ be smooth function with a bounded
derivative. Show that $u_{n}\to u$ in $W^{1,2}$ implies $\psi(u_{n})\to\psi(u)$
in $W^{1,2}$.

\textbf{Ex 4.4.} {[}d'Alembert's transform{]}. Let $u(t):I\to X$,
$g(t):I\to X$ be integrable functions. Then the following assertions
are equivalent:

(i) $\frac{d^{2}}{dt^{2}}u(t)=g(t)$ weakly, i.e. 
\[
\int_{I}u(t)\varphi''(t)\,dt=\int_{I}g(t)\varphi(t)\,dt\qquad\forall\varphi(t)\in C_{c}^{\infty}(I)
\]

(ii) there is $v(t):I\to X$ integrable such that $\frac{d}{dt}u(t)=v(t)$
and $\frac{d}{dt}v(t)=g(t)$ weakly in $I$.

\subsection*{Ex 5.1}

Let $S(t)$ be a $c_{0}$-semigroup in $X$. Show that the following
are equivalent:
\begin{itemize}
\item[(1)] $S(t)=e^{tA}$ for some $A\in\mathcal{L}(X)$
\item[(2)] $S(t)$ is uniformly continuous, i.e. $S(t)\to I$ in $\mathcal{L}(X)$
for $t\to0+$
\end{itemize}

\subsection*{Ex 5.2}

Let $u(t)\in L^{2}(I;W_{0}^{1,2})\cap C(I;L^{2})$ be the (unique)
weak solution to the heat equation 
\[
\frac{d}{dt}u-\Delta u=0,\qquad u(0)=u_{0}
\]
Verify that the solution operators $S(t):u_{0}\mapsto u(t)$ form
a $c_{0}$-semigroup in $L^{2}$.

\subsection*{Ex 5.3}

Let $(A,\mathcal{D}(A))$ be an unbounded operator in $X$, which
is closed, and let $\mathcal{D}(A)$ be dense in $X$.
\begin{enumerate}
\item Let $v'(t)=\lim_{h\to0}\frac{1}{h}(v(t+h)-v(t)$ be the classical
derivative in $X$. Assuming that $u'(t)$ and $(Au)'(t)$ exist,
show that $u'(t)\in\mathcal{D}(A)$ and $A(u'(t))=(Au)'(t)$.
\item Assume that $u(t):I\to\mathcal{D}(A)$ be Bochner integrable, where
$\mathcal{D}(A)$ is equipped with the graph-norm $\left\Vert u\right\Vert _{X}+\left\Vert Au\right\Vert _{X}$. 

Show that both $u(t):I\to X$ and $Au(t):I\to X$ are Bochner integrable,
and $A\big(\int_{I}u(t)\,dt)=\int_{I}Au(t)\,dt$ .
\end{enumerate}

\subsection*{Ex 5.4}

Let $X=L^{2}(\mathbb{R})$ and define the ``shift'' operators $S(t):X\to X$
by $S(t):f(x)\mapsto f(x+t)$.
\begin{enumerate}
\item Verify that $S(t)$ form a $c_{0}$-semigroup
\item Show that $\left\Vert S(t)-I\right\Vert _{\mathcal{L}(X)}=2$ for
any $t>0$, hence the semigroup is not uniformly continuous
\item Prove that if $f(x)\in W^{1,2}(\mathbb{R})$, then $\frac{1}{h}(S(h)f(x)-f(x))\to\frac{d}{dx}f(x)$
in $L^{2}(\mathbb{R})$, as $h\to0^{+}$.
\item Prove conversely that if $f(x)$, $g(x)\in L^{2}(\mathbb{R})$ are
such that $\frac{1}{h}(S(h)f(x)-f(x))\to g(x)$ in $L^{2}(\mathbb{R})$,
as $h\to0^{+}$, then $f(x)\in W^{1,2}(\mathbb{R})$ and $\frac{d}{dx}f(x)=g(x)$
\item Observe that the above assertions imply that the generator of $S(t)$
is the operator $A:f(x)\mapsto\frac{d}{dx}f(x)$ with the domain of
definition $\mathcal{D}(A)=W^{1,2}(\mathbb{R})$.
\end{enumerate}

\subsection*{Ex 6.1}

Consider space $\ell^{2}$ of sequences $\{u_{j}\}$ with the norm
$\big(\sum_{j}u_{j}^{2}\big)^{1/2}$. 

Let $\lambda_{j}$ be real numbers. Define operator $A:D\left(A\right)\subset\ell^{2}\to\ell^{2}$
as 
\[
A:\{u_{j}\}\mapsto\{\lambda_{j}u_{j}\},\ \qquad D\left(A\right)=\left\{ \{u_{j}\}\in\ell^{2};\ \sum_{j}\lambda_{j}^{2}u_{j}^{2}<\infty\right\} 
\]
\begin{enumerate}
\item Observe that $D\left(A\right)=\ell^{2}$ and $A\in\mathcal{L}(\ell^{2})$
$\iff$ the sequence $\{\lambda_{j}\}$ is bounded.
\item Show that $(A,D\left(A\right))$ is closed, and $D\left(A\right)$
is dense in $\ell^{2}$.
\item Assume that $\lambda_{j}\le\omega$ for all $j$. Deduce that any
$\lambda>\omega$ belongs to the resolvent set $\rho(A)$. Write an
explicit formula for the resolvent $R(\lambda,A)$.
\item By Hille-Yosida theorem, $A$ is a generator of a $c_{0}$-semigroup
in $\ell^{2}$, which satisfies $\left\Vert S(t)\right\Vert _{\mathcal{L}(\ell^{2})}\le e^{\omega t}$,
$t\ge0$. Compute $S(t)$ explicitely.
\end{enumerate}

\subsection*{Ex 6.2}

Let $S(t)$ be a $c_{0}$-semigroup in $X$, satisfying $\left\Vert S(t)\right\Vert _{L\left(X\right)}\le Me^{\omega t}$,
$t\ge0$. 

Let $\tilde{S}(t)=e^{-\omega t}S(t)$. Prove that:
\begin{enumerate}
\item $\tilde{S}(t)$ is also a $c_{0}$-semigroup, and $\left\Vert \tilde{S}(t)\right\Vert _{L\left(X\right)}\le M$,
$t\ge0$.
\item Let $(A,D\left(A\right))$, $\tilde{A},\mathcal{D}(\tilde{A}))$ be
the generators of $S(t)$, $\tilde{S}(t)$, respectively. Show that
$D\left(A\right)=\mathcal{D}(\tilde{A})$ and $\tilde{A}=A-\omega I$.
\end{enumerate}

\subsection*{{*}Ex. 6.3}

Let $\Omega\subset\mathbb{R}^{d}$ be a bounded domain with smooth
boundary. Let $\lambda_{j}$, $w_{j}$ be the eigenvalues and eigenfunctions
of the Dirichlet laplacian, with $\left\Vert w_{j}\right\Vert _{2}=1$.

By Parseval's theorem, we have $u_{0}=\sum_{j}u_{j}w_{j}$, where
$u_{j}=(u,w_{j})$ and the sum converges in $L^{2}$.
\begin{enumerate}
\item Show that $u(t)=\sum_{j}e^{-\lambda_{j}t}u_{j}w_{j}$ is a weak solution
to the heat equation $\frac{d}{dt}u-\Delta u=0$, $u(0)=u_{0}$.
\item Show that $\sum_{j}\lambda_{j}u_{j}^{2}<\infty$ if and only if $u_{0}\in W_{0}^{1,2}$.
Show that $\sum_{j}\lambda_{j}^{2}u_{j}^{2}<\infty$ if and only if
$u_{0}\in W_{0}^{1,2}\cap W^{2,2}$.
\item Observe that the ``heat semigroup'' (see Exercise 5.2) can thus
be identified with a multiplicative semigroup (as in Exercise 6.1).
It follows that the generator of heat semigroup is the operator $\Delta=\sum_{i=1}^{d}\frac{\partial^{2}u}{\partial x_{i}^{2}}$,
with derivatives understood weakly on the domain of definition $\mathcal{D}(\Delta)=W_{0}^{1,2}\cap W^{2,2}$.
\end{enumerate}

\subsection*{Ex 6.4}

Let $S(t)$ be a $c_{0}$-semigroup on $X$.
\begin{enumerate}
\item Show that the map $(t,x)\mapsto S(t)x$ is \emph{jointly} continuous
$[0,\infty)\times X\to X$.
\item Assuming that $\tilde{S}(t)$ is another $c_{0}$-semigroup, show
that $y(t)=S(T-t)\tilde{S}(t)x$ is continuous $[0,T]\to X$, where
$T>0$ and $x\in X$ are fixed.
\item Assuming finally that $S(t)$ and $\tilde{S}(t)$ have the same generator,
show that $y'(t)=0$ for any $t\in(0,T)$ if $x\in D\left(A\right)$.
\item Deduce that $S(t)=\tilde{S}(t)$ for all $t\ge0$, thus establishing
Lemma 4.2.
\end{enumerate}

\section*{Hints}

\subsection*{Ex 1.1}
\begin{enumerate}
\item Follows from compactness of $I$ just as in the scalar case $X=\mathbb{R}$.
\item (a) compact implies separable, and continuous scalar is measurable;
(b) set $u_{n}(t)=u(jT/n)$ for $t\in[(j-1)T/n,jT/n]$ \textendash{}
these are simple functions and $u_{n}(t)\rightrightarrows u(t)$ thanks
to uniform continuity

parts 3 and 4 use very similar ideas
\end{enumerate}

\subsection*{Ex 1.2}

1. Let $u_{n}(t)$ be simple functions from the definition of $\int_{I}u(t)\,dt$.
Then $Fu_{n}(t)$ are simple\dots{} In particular: set $Y=\mathbb{R}$.

2. Use the fact that a separable set can be enlarged to a closed and
convex (hence weakly closed) set. Use weak lower semicontinuity of
the norm and scalar version of Fatou's lemma.

\subsection*{Ex 1.3}

1. Rewrite $u_{n}(t)=\int_{\mathbb{R}}u(s)\psi_{n}(t-s)\,ds$ and
show that usual theorems about dependence of integral on parameter
apply. (In fact for $\psi_{0}$ smooth enough, the dependence on $t$
is uniform, so the exchange of integral and limit is trivial.)

2. For $p=1$, this follows by Fubini's theorem.

3. Use uniform continuity of $u(t)$ on the neighborhood of $I$.

\subsection*{Ex 1.4}

1. Fix $u\in X$, and let $\varepsilon>0$ arbitrary be given. Pick
$v\in S$ such that $\|u-v\|<\varepsilon$. Write $F_{n}u-u=F_{n}(u-v)+\big(F_{n}v-v\big)+(v-u)$
and show that each term is estimated by (a multiple of) $\varepsilon$
if $n$ is large enough.

2. Set $X=L^{p}(I;X)$, $F_{n}u=u*\psi_{n}$ and $S=C_{c}(I;X)$.
Use the results of Ex 1.3.2 and 1.3.3. 

\subsection*{Ex 2.1}
\begin{enumerate}
\item By Lemma 1.5, there is a continuous representative $\tilde{u}(t)$
such that $\tilde{u}(t_{1})-\tilde{u}(t_{2})=\int_{t_{1}}^{t_{2}}\frac{d}{dt}u(s)\,ds$.
Estimate the integral using the Hölder inequality.
\item Inclusion $\subset$ is as above. For the converse, note that Lipschitz
function is absolutely continuous, and its derivative is $L^{\infty}$,
cf. Theorem 1.5.
\item Explain (in detail), that weak convergence is enough to pass in the
definition of the weak derivative.
\item Use Eberlein-Śmulian and the previous problem.
\end{enumerate}

\subsection*{Ex 2.3}
\begin{enumerate}
\item Use $\left\Vert \frac{x+y}{2}\right\Vert _{H}^{2}+\left\Vert \frac{x-y}{2}\right\Vert _{H}^{2}=\left\Vert x\right\Vert _{H}^{2}+\left\Vert y\right\Vert _{H}^{2}$.
\item Write $\left\Vert u-u_{n}\right\Vert _{H}^{2}=\left\Vert u\right\Vert _{H}^{2}-2(u,u_{n})_{H}+\left\Vert u_{n}\right\Vert _{H}^{2}$.
\end{enumerate}

\subsection*{Ex 2.4}

Let $v_{n}(t)$ be a countable dense set in $L^{p'}(I;X^{*})$ \dots{}

\subsection*{Ex 2.5}

Add and subtract $\left\langle u_{n}\left(t\right),v\left(t\right)\right\rangle $.

\subsection*{Ex 4.1.}

$\exists K>0$, $N\subset I$ s.t. $\lambda(N)=0$ and $\left\Vert u(t)\right\Vert _{X}\le K$
for all $t\in I\setminus N$. Approximate $t_{0}\in N$ with $t_{n}\to t_{0}$,
$t_{n}\in I\setminus N$ to show that $\left\Vert u(t_{0})\right\Vert _{X}\le K$.
Prove continuity by contradiction, using uniqueness of limits in $Z$.

\subsection*{Ex 4.2}
\begin{itemize}
\item[$\left(i\right)$] Rewrite $P_{N}$ as ON (in $W_{0}^{1,2}$ w.r. to $(\!(\cdot,\cdot)\!)$)
projection
\item[$\left(ii\right)$] Show that $P_{N}(-\Delta u)=-\Delta P_{N}u$; use elliptic regularity
for the laplacian
\end{itemize}

\subsection*{Ex 4.3}

In view of Lemma 2.4, it is enough to show that $u_{n}\to u$, $\nabla u_{n}\to\nabla u$
in $L^{2}$ implies $\psi'(u_{n})\nabla u_{n}\to\psi'(u)\nabla u$
in $L^{2}$. By taking a subsequence we can in the first step assume
$u_{n}\to u$ a.e. Show further by contradiction (and step one) that
convergence takes place even without taking a subsequence.

\subsection*{Ex. 5.3.2}

Let $u_{n}(t)$ be simple functions and $u_{n}(t)\to u(t)$ in the
norm of $\mathcal{D}(A)$ for a.e. $t\in I$, \dots{}

\subsection*{Ex. 5.4. }
\begin{itemize}
\item[2.] Consider suitable $f(x)\in L^{2}(\mathbb{R})$ with compact support. 
\item[3.] Working with AC representative, we have $f(x+h)-f(x)=\int_{0}^{h}g(x+s)\,ds$,
where $g=\frac{d}{dx}f$. Deduce that $\frac{1}{h}\big(f(x+h)-f(x)\big)$
can be written as convolution of $g$ with suitable kernels, and use
Lemma 1.1, part 4. 
\item[4.] Let $\varphi(x)\in C_{c}^{\infty}(\mathbb{R})$ be given test function
and $h>0$ be fixed. Prove that 
\[
\int_{\mathbb{R}}\frac{f(x+h)-f(x)}{h}\varphi(x)\dx=\int_{\mathbb{R}}f(x)\frac{\varphi(x-h)-\varphi(x)}{h}\dx
\]
Using the assumptions, show that you can take the limit $h\to0+$
on both sides, to obtain that $\frac{d}{dx}f(x)=g(x)$ in the sense
of weak derivative. 
\end{itemize}

\subsection*{Ex. 6.1}

in part 2, for density, consider $\mathcal{C}\subset D\left(A\right)$
consisting of $\{u_{j}\}$ with only finitely many nonzero $u_{j}$,
for 3. $R(\lambda,A):\{v_{j}\}\mapsto\{\frac{1}{\lambda-\lambda_{j}}v_{j}\}$,
in 4. use Yosida approximation.

Alternatively, one can guess that the answer $S(t):\{u_{j}\}\mapsto\{e^{\lambda_{j}t}u_{j}\}$
and then verify that $S(t)$ is a $c_{0}$-semigroup on $\ell^{2}$
and $A$ is its generator. So $S(t)$ must be the sought-for semigroup
by Lemma 4.2.

For taking limits in $\ell^{2}$, you can use the following version
of Lebesgue's theorem (where sum is seen as an integral): if $\sum_{j}|a_{j}|<\infty$
and $b_{j}(t)$ are bounded independently of $j$ and $t$, and $b_{j}(t)\to\beta_{j}$,
then $\sum_{j}b_{j}(t)a_{j}\to\sum_{j}\beta_{j}a_{j}$.

\subsection*{Ex. 6.4}

for 1, use part 1 of Lemma 4.1; in writing $y(t+h)-y(t)$, add $\pm S(T-(t+h))\tilde{S}(t)x$
and use joint continuity and definition of the generator, and Theorem
4.1; in 4. $S(T)x=\tilde{S}(T)x$ if $x\in D\left(A\right)$ and use
density.

\section*{Solutions}

\subsection*{Ex. 1.2 }

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Bochner integral, convolution}
\begin{enumerate}
\item Let us first prove that $Fu(t)$ is Bochner integrable. Since $u(t)$
is Bochner integrable, there exist simple functions $u_{n}(t)=\sum_{j=1}^{k(n)}\chi_{A_{j}^{n}}(t)x_{j}^{n}$
such that $u_{n}(t)\rightarrow u(t)$ for a. e. $t\in I$ and $\int_{I}\left\Vert u(t)-u_{n}(t)\right\Vert _{x}\,dt\rightarrow0$.
Denote $s_{n}(t)=\sum_{j=1}^{k(n)}\chi_{A_{j}^{n}}(t)F(x_{j}^{n})$.
Then $s_{n}$ are clearly simple. By linearity of $F$ we have $s_{n}(t)=F(u_{n}(t))$.
Since $F$ is also continuous, it holds that 
\[
\lim_{n\rightarrow\infty}s_{n}(t)=F\left(\lim_{n\rightarrow\infty}u_{n}(t)\right)=Fu(t)
\]
for a. e. $t\in I$. By continuity and linearity of $F$ we have ($\left\Vert F\right\Vert <\infty$)
\[
\int_{I}\left\Vert Fu(t)-Fu_{n}(t)\right\Vert _{Y}\,dt\leq\left\Vert F\right\Vert \int_{I}\left\Vert u(t)-u_{n}(t)\right\Vert _{X}\,dt\rightarrow0.
\]
Thus we see that $Fu(t)$ is Bochner integrable.

It follows from the definition of Bochner integral and continuity
and linearity $F$ that
\begin{align*}
F\left(\int_{I}u(t)\,\mathrm{dt}\right) & =F(\lim_{n\rightarrow\infty}\int_{I}u_{n}(t)\,dt)=\lim_{n\rightarrow\infty}F\left(\int_{I}u_{n}(t)\,dt\right)\\
 & =\lim_{n\rightarrow\infty}F\left(\sum_{j=1}^{k(n)}\lambda(A_{j}^{n})x_{j}^{n}\right)=\lim_{n\rightarrow\infty}\sum_{j=1}^{k(n)}\lambda(A_{j}^{n})F(x_{j}^{n})\\
 & =\lim_{n\rightarrow\infty}\int_{I}s_{n}(t)\,dt=\int_{I}Fu(t)\,dt.
\end{align*}
In particular, if we choose $Y=\mathbb{R}$, we obtain for every $x^{*}\in X^{*}$
\[
\langle x^{*},\int_{I}u(t)\,dt\rangle=\int_{I}\langle x^{*},u(t)\rangle\,dt.
\]
\item We begin by proving the measurability of $u(t)$. Let $x^{*}\in X^{*}$
be arbitrary. Since $u_{n}(t)$ are measurable and therefore weakly
measurable, the scalar functions $\langle x^{*},u_{n}(t)\rangle$
are measurable. But since $u_{n}(t)\rightharpoonup u(t)$ for a. e.
$t\in I$, we have 
\[
\langle x^{*},u_{n}(t)\rangle\rightarrow\langle x^{*},u(t)\rangle
\]
for a. e. $t\in I$. Scalar function $\langle x^{*},u(t)\rangle$
is for a. e. $t\in I$ pointwise limit of sequence of scalar measurable
functions and therefore it is a scalar measurable function. As $x^{*}\in X^{*}$
was chosen arbitrarily, the function $u(t)$ is weakly measurable.

Since $u_{n}(t)$ are measurable and therefore $\lambda$-separably
valued, there exist Lebesgue null sets $N_{n}$ such that $u_{n}(I\setminus N_{n})$
are separable sets. Moreover, there exists Lebesgue null null set
$N_{0}$ such that $u_{n}(t)\rightharpoonup u(t)$ for a. e. $t\in I\setminus N_{0}$.
Denote 
\[
N=\bigcup\limits _{k=0}^{\infty}N_{k},
\]
which is also a Lebesgue null set. Therefore $u_{n}(I\setminus N)$
are separable sets and for every $t\in I\setminus N$ we have that
$u_{n}(t)\rightharpoonup u(t)$. As a countable union of separable
sets is a separable set, the union 
\[
\bigcup\limits _{n=1}^{\infty}u_{n}(I\setminus N)
\]
is a separable set and thus $\mathrm{span}\left(\bigcup\limits _{n=1}^{\infty}u_{n}(I\setminus N)\right)$
is a separable subspace. Since subspace is clearly a convex subset,
it follows easily that 
\[
u(t)\in\overline{\bigcup\limits _{n=1}^{\infty}u_{n}(I\setminus N)}^{_{w}}\subseteq\overline{\mathrm{span}(\bigcup\limits _{n=1}^{\infty}u_{n}(I\setminus N))}^{_{w}}=\overline{\mathrm{span}(\bigcup\limits _{n=1}^{\infty}u_{n}(I\setminus N))}^{\left\Vert \cdot\right\Vert _{X}}
\]
for every $t\in I\setminus N$. Therefore $u(t)$ is also $\lambda$-separably
valued and measurable by the Pettis theorem.

By weak lower semicontinuity of the norm and by the scalar version
of Fatou's lemma it holds that 
\[
\int_{I}\left\Vert u(t)\right\Vert \,dt\leq\int_{I}\liminf_{n\rightarrow\infty}\left\Vert u_{n}(t)\right\Vert \,dt\leq\liminf_{n\rightarrow\infty}\int_{I}\left\Vert u_{n}(t)\right\Vert \,dt.
\]

Moreover, if the sequence $\int_{I}\left\Vert u_{n}(t)\right\Vert \,dt$
is bounded, it follows easily that\\
$\int_{I}\left\Vert u(t)\right\Vert \,dt\leq\liminf_{n\rightarrow\infty}\int_{I}\left\Vert u_{n}(t)\right\Vert \,dt<\infty$,
therefore $u(t)$ is integrable by the Bochner theorem.
\end{enumerate}

\subsection*{Ex. 1.3}

Convolution kernel: $\psi_{0}\left(t\right):\mathbb{R}\to\mathbb{R}$,
bounded, measurable, nonnegative, $\psi_{0}\left(t\right)=0$ outside
$\left[-1,1\right]$, $\int_{\mathbb{R}}\psi_{0}\left(t\right)dt=1$.
Define $\psi_{n}\left(t\right)=n\psi_{0}\left(nt\right)$, $u_{n}\left(t\right)=u*\psi_{n}\left(t\right)=\int_{\mathbb{R}}u\left(t-s\right)\psi_{n}\left(s\right)ds=\int_{\mathbb{R}}\psi_{n}\left(t-s\right)u\left(s\right)ds$,
we set $u\left(t\right)=0$ outside of $I$.
\begin{enumerate}
\item $u\left(t\right)\in L^{1}\left(I;X\right)$, $\psi_{0}\left(t\right)\in C^{1}\implies u_{n}\left(t\right)\in C^{1}\left(I,X\right)$
and $u_{n}'$$\left(t\right)=u*\psi_{n}'\left(t\right)$ for any $n$
fixed.\\
Is $u_{n}\left(t\right)$ continuous? Let $t\in I$ be fixed. $h_{j}\to0\implies u_{n}\left(t+h_{j}\right)\to u_{n}\left(t\right)$,
but $u_{n}\left(t+h_{j}\right)-u_{n}\left(t\right)=\int_{I}\underbrace{\left[\psi_{n}\left(t+h_{j}-s\right)-\psi_{n}\left(t-s\right)\right]u\left(s\right)}_{v_{j}\left(s\right)}ds$.
By Th. 1.3 (Lebesgue) we have: $v_{j}\left(s\right)\to0$, $j\to\infty$
for a. e. $s\in\mathbb{R}$ fixed and $\left\Vert v_{j}\left(s\right)\right\Vert \le\left(\left|\psi_{n}\left(\ldots\right)\right|+\left|\psi_{n}\left(\ldots\right)\right|\right)\left\Vert u\left(s\right)\right\Vert \le c\left\Vert u\left(s\right)\right\Vert \in L^{1}\left(I;\mathbb{R}\right)$.\\
Is $u$ differentiable? 
\begin{align*}
u_{n}'\left(t\right) & =\frac{d}{dt}\int_{I}\psi_{n}\left(t-s\right)u\left(s\right)ds\\
 & =\lim_{h\to0}\int_{I}\underbrace{\left[\frac{\psi_{n}\left(t+h-s\right)-\psi_{n}\left(t-s\right)}{h}\right]}_{\text{\ensuremath{\to}}\makebox[0pt][l]{\ensuremath{\,\psi_{n}'\left(t-s\right)} for \ensuremath{\forall s} fixed}}u\left(s\right)ds\\
\noalign{\centerline{\text{\ensuremath{\left|\left[\ldots\right]\right|\le C} \dots\ Lagrange mean value theorem \ensuremath{\left(C=\sup\left|\psi_{n}'\right|\right)}}}}\\
 & =\int_{I}\psi_{n}'\left(t-s\right)u\left(s\right)ds
\end{align*}
\item $\left\Vert u_{n}\left(t\right)\right\Vert _{\mathcal{F}}\leq\left\Vert u\left(t\right)\right\Vert _{\mathcal{F}}$
$\forall n$, where $\mathcal{F}=C\left(I,X\right)$ or $L^{p}\left(I;X\right)$.\\
For $L^{p}$:
\begin{itemize}
\item $p=1$: 
\begin{align*}
\left\Vert u_{n}\left(t\right)\right\Vert _{L^{1}\left(I;X\right)} & =\int_{I}\left\Vert u*\psi_{n}\left(t\right)\right\Vert _{X}dt=\int_{I}\left\Vert \int_{\mathbb{R}}u\left(s\right)\psi_{n}\left(t-s\right)ds\right\Vert _{X}dt\\
 & \leq\int_{I}\left(\int_{\mathbb{R}}\left\Vert u\left(s\right)\right\Vert _{X}\psi_{n}\left(t-s\right)ds\right)dt\\
 & \stackrel{\makebox[0pt][c]{Fubini}}{=}\ \ \int_{\mathbb{R}}\left(\int_{\mathbb{R}}\left\Vert u\left(s\right)\right\Vert _{X}\psi_{n}\left(t-s\right)dt\right)ds\\
 & =\int_{\mathbb{R}}\left\Vert u\left(s\right)\right\Vert \underbrace{\left(\int_{\mathbb{R}}\psi_{n}\left(t-s\right)dt\right)}_{=\makebox[0pt][l]{\ensuremath{\,1}}}ds\\
 & =\int_{\mathbb{R}}\left\Vert u\left(s\right)\right\Vert ds=\left\Vert u\left(\cdot\right)\right\Vert _{L^{1}\left(I;X\right)}
\end{align*}
\item for $p=\infty$ easy (same as $C\left(I,X\right))$.
\item $p\in\left(1,\infty\right):$ Hölder inequality: $\left(\frac{1}{p}+\frac{1}{p'}=1\right)$
\[
\int_{\Omega}\left|fg\right|\leq\left(\int_{\Omega}\left|f\right|^{p}\right)^{\frac{1}{p}}\left(\int_{\Omega}\left|g\right|^{p'}\right)^{\frac{1}{p'}}.
\]
\textbf{Trick:} 
\begin{align*}
\left\Vert u*\psi_{n}\left(t\right)\right\Vert  & \leq\int_{\mathbb{R}}\left\Vert u\left(s\right)\right\Vert \psi_{n}\left(t-s\right)ds\\
 & =\int\underbrace{\left\Vert u\left(s\right)\right\Vert \left(\psi_{n}\left(t-s\right)\right)^{\frac{1}{p}}}_{f\left(s\right)}\underbrace{\left(\psi_{n}\left(t-s\right)\right)^{1-\frac{1}{p}}}_{g\left(s\right)}ds\\
 & \stackrel{\makebox[0pt][c]{Hölder}}{\leq}\ \ \left(\int_{\mathbb{R}}\left\Vert u\left(s\right)\right\Vert ^{p}\psi_{n}\left(t-s\right)ds\right)^{\frac{1}{p}}\underbrace{\left(\int_{\mathbb{R}}\left(\psi_{n}\left(t-s\right)\right)ds\right)^{\frac{1}{p'}}}_{=\makebox[0pt][l]{\ensuremath{\,1}}}
\end{align*}
$\implies\left\Vert u*\psi_{n}\left(t\right)\right\Vert ^{p}\leq\int_{\mathbb{R}}\left\Vert u\left(s\right)\right\Vert ^{p}\psi_{n}\left(t-s\right)ds$,
now as before: $\int_{\mathbb{R}}dt$, Fubini 
\[
\implies\int_{\mathbb{R}}\left\Vert u*\psi_{n}\left(t\right)\right\Vert ^{p}dt\leq\int_{\mathbb{R}}\left\Vert u\left(s\right)\right\Vert ^{p}ds.
\]
\end{itemize}
\item If $u\left(t\right)\in C\left(J,X\right)$, $J$ compact larger than
$I$, then $u_{n}\left(t\right)\rightrightarrows u\left(t\right)$
in $I$. 
\begin{align*}
\left\Vert u_{n}\left(t\right)-u\left(t\right)\right\Vert  & =\left\Vert \int_{\mathbb{R}}u\left(s\right)\psi_{n}\left(t-s\right)ds-u\left(t\right)\right\Vert \\
 & \le\int_{\mathbb{R}}\left\Vert u\left(s\right)-u\left(t\right)\right\Vert \underbrace{\psi_{n}\left(t-s\right)}_{\ne\makebox[0pt][l]{\ only if \ensuremath{\left|t-s\right|\le\frac{1}{n}}}}ds
\end{align*}
($t\in I$, we only consider $s\in\left[-\frac{1}{n},T+\frac{1}{n}\right]\subset J$
for $n$ large). Uniform continuity of $u\left(t\right)$ on $J$:
$\varepsilon>0$ given $\implies\exists\delta>0$ s. t. $\left\Vert u\left(t\right)-u\left(s\right)\right\Vert \le\varepsilon$
for $\forall t,s\in J$ s. t. $\left|t-s\right|<\delta$. Take n s.
t. $\frac{1}{n}\le\delta$.
\[
\left\Vert u_{n}\left(t\right)-u\left(t\right)\right\Vert <\int_{\mathbb{R}}\varepsilon\psi_{n}\left(t-s\right)ds=\varepsilon
\]
\end{enumerate}

\subsection*{Ex. 2.1}
\begin{enumerate}
\item[1.] Since $u(t)\in W^{1,p}\left(I;X\right)$, there exists a continuous
representative $\tilde{u}(t)\in C(I;X)$. For $r,s\in I$ we have
\begin{align*}
\left\Vert \tilde{u}(s)-\tilde{u}(r)\right\Vert _{X} & =\left\Vert \int_{r}^{s}\frac{d}{dt}u(t)\,dt\right\Vert _{X}\leq\left|\int_{r}^{s}\left\Vert \frac{d}{dt}u(t)\right\Vert _{X}\,dt\right|\\
 & \stackrel{\text{Hölder}}{\leq}\left|s-r\right|^{\frac{1}{p'}}\left\Vert \frac{d}{dt}u(t)\right\Vert _{L^{p}\left(I;X\right)}\\
 & \leq\left|s-r\right|^{\frac{1}{p'}}\left\Vert u(t)\right\Vert _{W^{1,p}\left(I;X\right)},
\end{align*}
where $\frac{1}{p'}=1-\frac{1}{p}$.

Since $r,s\in I$ were chosen arbitrary, we have 
\[
\sup\limits _{\substack{r,s\in I\\
r\neq s
}
}\frac{\left\Vert \tilde{u}(s)-\tilde{u}(r)\right\Vert _{x}}{\left|s-r\right|^{1-\frac{1}{p}}}\leq\left\Vert u(t)\right\Vert _{W^{1,p}\left(I;X\right)}.
\]

Therefore $\tilde{u}(t)$ is $\alpha$-Hölder for $\alpha=1-\frac{1}{p}$
.
\item[3.] Denote $\Upsilon=L^{1}(I;X)$. Choose arbitrary fixed test function
$\varphi\in\mathcal{D}(I)$ and define 
\begin{align*}
\Phi(v(t)) & =\int_{I}v(t)\varphi'(t)\,dt,\\
\Psi(v(t)) & =-\int_{I}v(t)\varphi(t)\,dt,\quad v(t)\in\Upsilon.
\end{align*}
We see that $\Phi$ and $\Psi$ are well\textendash defined linear
operators from $\Upsilon$ to $X$ directly by the definition. Moreover,
they are continous since 
\[
\left\Vert \Phi(v(t))\right\Vert _{X}\leq\int_{I}\left\Vert v(t)\right\Vert _{X}\left|\varphi'(t)\right|\,\mathrm{dt}\stackrel{\text{Hölder}}{\leq}\left\Vert v(t)\right\Vert _{\Upsilon}\left\Vert \varphi'(t)\right\Vert _{L^{\infty}(I;\mathbb{R})},
\]
where $\left\Vert \varphi'(t)\right\Vert _{L^{\infty}(I;\mathbb{R})}<\infty$
as $\varphi$ is a test function and therefore $\varphi'$ is a continuous
compactly supported function. Similar arguments apply for proving
the continuity of $\Psi$. We thus have that $\Phi,\Psi\in\mathcal{L}(\Upsilon,X)$.
By the definition of weak derivative we have 
\[
\Phi(u_{n}(t))=\int_{I}u_{n}(t)\varphi'(t)\,dt=-\int_{I}\frac{d}{dt}u_{n}(t)\varphi(t)\ \mathrm{dt}=\Psi(\frac{d}{dt}u_{n}(t)).
\]
We now have that $\Phi(u_{n}(t))\rightharpoonup\Phi(u(t))$ and $\Psi\left(\frac{d}{dt}u_{n}(t)\right)\rightharpoonup\Psi(g(t))$
in $X$. By uniqueness of the weak limit it holds that $\Phi(u(t))=\Psi(g(t))$
and thus 
\[
\int_{I}u(t)\varphi'(t)\,dt=\Phi(u(t))=\Psi(g(t))=-\int_{I}g(t)\varphi(t)\,dt,
\]
which means that $u(t)$ is weakly differentiable and $\frac{d}{dt}u(t)=g(t)$
since $\varphi\in\mathcal{D}(I)$ was chosen arbitrarily. 
\item[5.] We can write
\begin{align*}
\int_{I}\langle v_{n}(t),u_{n}(t)\rangle_{X^{*},X}\,dt=\int_{I}\langle v_{n}(t)-v(t),u_{n}(t)\rangle_{X^{*},X}\,dt+\int_{I}\langle v(t),u_{n}(t)\rangle_{X^{*},X}\,dt.
\end{align*}
If we denote $\Upsilon=L^{p}(I;X)$, $\Upsilon^{*}=\left(L^{p}(I;X)\right)^{*}$
and denote by $F_{n}\in\Upsilon^{*}$ the unique element of dual $\Upsilon$
corresponding to the $v_{n}(t)\in L^{p'}(I;X^{*})$ (and by $F\in\Upsilon^{*}$
the element corresponding to $v(t)\in L^{p'}(I;X^{*})$) by canonical
dual mapping, then we have 
\[
\int_{I}\langle v(t),u_{n}(t)\rangle_{X^{*},X}\,dt=\langle F,u_{n}(t)\rangle_{\Upsilon^{*},\Upsilon}\stackrel{n\rightarrow\infty}{\longrightarrow}\langle F,u(t)\rangle_{\Upsilon^{*},\Upsilon}=\int_{I}\langle v(t),u(t)\rangle_{X^{*},X}\,dt,
\]
as $u_{n}(t)\rightharpoonup u(t)$ in $\Upsilon$. Moreover, we have
that 
\begin{align*}
\left|\int_{I}\langle v_{n}(t)-v(t),u_{n}(t)\rangle_{X^{*},X}\ \mathrm{dt}\right| & \leq\int_{I}\left\Vert v_{n}(t)-v(t)\right\Vert _{X^{*}}\left\Vert u_{n}(t)\right\Vert _{X}\,dt\\
 & \stackrel{\text{Hölder}}{\leq}\left\Vert v_{n}(t)-v(t)\right\Vert _{L^{p'}(I;X^{*})}\left\Vert u_{n}(t)\right\Vert _{L^{p}(I;X)}\stackrel{n\rightarrow\infty}{\longrightarrow}0,
\end{align*}
since the sequence $\left\{ u_{n}(t)\right\} $ is weakly convergent
in $L^{p}(I;X)$ and therefeore bounded in $L^{p}(I;X)$ and $v_{n}\longrightarrow v$
in $L^{p'}(I;X^{*})$. Now it is easily seen that $\int_{I}\langle v_{n}(t),u_{n}(t)\rangle_{X^{*},X}\,dt\stackrel{n\rightarrow\infty}{\longrightarrow}\int_{I}\langle v(t),u(t)\rangle_{X^{*},X}\,dt$. 
\end{enumerate}

\subsection*{Ex. 2.2}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Relation of $L^p\left(I;L^p(\Omega)\right)$ and $L^p(I\times\Omega)$} 

We will show only $L^{p}\left(I;L^{p}\left(\Omega\right)\right)=L^{p}\left(I\times\Omega\right)$
for $p\in[1,\infty)$:
\begin{itemize}
\item Let $u=0$ in $L^{p}\left(I\times\Omega\right)\iff u\left(t,x\right)=0$
for a. e. $\left(t,x\right)\in I\times\Omega$. By Fubini we know
that this is equivalent to $u\left(t,x\right)=0$ a. e. $x\in\Omega$
for a. e. $t\in I$. Therefore $u\left(t,\cdot\right)=0$ in $L^{2}\left(\Omega\right)$
for a. e. $t\in I$.\\
Moreover $N\subset I\times\Omega$ is zero set $\iff$ for a. e. $t\in I$
the set $N_{t}=\left\{ x\in\Omega,\left(t,x\right)\in N\right\} $
is zero set. Therefore $u=0$ in $L^{p}\left(I\times\Omega\right)$
$\iff$ $u\left(t\right)=0$ in $L^{2}\left(I;L^{p}\left(\Omega\right)\right)$
\item $\left\Vert u\right\Vert _{L^{p}\left(I\times\Omega\right)}^{p}=\int_{I\times\Omega}\left|u\left(t,x\right)\right|^{p}dtdx\stackrel{\text{Fubini}}{=}\int_{I}\underbrace{\left(\int_{\Omega}\left|u\left(t,x\right)\right|^{p}dx\right)}_{\left\Vert u\left(t,\cdot\right)\right\Vert _{L^{p}\left(\Omega\right)}^{p}}dt=\left\Vert u\left(t\right)\right\Vert _{L^{p}\left(I;L^{p}\left(\Omega\right)\right)}^{p}$.
\item measurability: let $u\left(t,x\right)\in L^{p}\left(I\times\Omega\right)$.
Then $\exists u_{n}\left(t,x\right)\in C_{c}^{\infty}\left(I\times\Omega\right)$
such that $u_{n}\left(t,x\right)\to u\left(t,x\right)$ in $L^{p}\left(I\times\Omega\right)$.
$u_{n}\left(t\right):I\to L^{p}\left(\Omega\right)$ are continuous,
therefore strongly measurable.\\
Therefore $\int_{I}\left\Vert u_{n}\left(t\right)-u\left(t\right)\right\Vert _{L^{p}\left(\Omega\right)}dt\to0$
($u_{n}\left(t\right)$ is Cauchy since $u_{n}\left(t,x\right)$ is
Cauchy in $L^{p}\left(I\times\Omega\right)$), hence $u_{n}\left(t\right)\to u$
a. e. in $L^{p}\left(I;L^{p}\left(\Omega\right)\right)$
\end{itemize}

\subsection*{Ex. 3.1}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Integral formulation of weak derivative with initial condition} 
\begin{itemize}
\item Let 1. be true. By part 1: $\frac{d}{dt}\iota u\left(t\right)=g\left(t\right)$
in $I$ and the function $\iota u\left(t\right):I\to W^{-1,2}$ is
weakly differentiable. By Lemma 1.3 $\implies\exists\tilde{\tilde{u}}\left(t\right)\in AC\left(I,W^{-1,2}\right)$
s. t. $\tilde{\tilde{u}}\left(t\right)=\iota u\left(t\right)$ a.
e. in $I$. But also (by Thm. 1.12) $\exists\tilde{u}\left(t\right)\in C\left(I,L^{2}\right)$
s. t. $u\left(t\right)=\tilde{u}\left(t\right)$ a. e. Hence clearly
$\tilde{\tilde{u}}\left(t\right)=\iota\tilde{u}\left(t\right)$ for
all $t\in I$ and $\iota u\left(0\right)=u_{0}.$ Take $v\in W_{0}^{1,2}$,
$\varphi\left(t\right)\in C_{c}^{\infty}\left(\left(-\infty,T\right)\right)$,
clearly $t\mapsto\underbrace{\left\langle \iota\tilde{u}\left(t\right),v\right\rangle \varphi\left(t\right)}_{x\left(t\right)}$
is $AC$. $x'\left(t\right)=\frac{d}{dt}\left(\left\langle \iota u\left(t\right),v\right\rangle \varphi\left(t\right)\right)=\left\langle \frac{d}{dt}\iota u\left(t\right),v\right\rangle \varphi\left(t\right)+\left\langle \iota\tilde{u}\left(t\right),v\right\rangle \varphi'\left(t\right)$
a. e., write $\underbrace{x\left(T\right)}_{\text{\ensuremath{=}}\makebox[0pt][l]{\ensuremath{\,0}}}-\underbrace{x\left(0\right)}_{\text{\ensuremath{=}}\makebox[0pt][l]{\ensuremath{-\left\langle \iota\tilde{u}\left(0\right),v\right\rangle \varphi\left(0\right)}}}=\overbrace{\int_{I}x'\left(t\right)dt}^{\makebox[0pt][r]{\ensuremath{\text{\ensuremath{\int_{I}\left\langle g\left(t\right),v\right\rangle \varphi\left(t\right)}+\ensuremath{\left\langle \iota\tilde{u}\left(t\right),v\right\rangle \varphi}'\ensuremath{\left(t\right)}dt}}}}$
since $x\left(t\right)\in AC$ we have $\left(u_{0},v\right)=\int_{I}\left\langle g\left(t\right),v\right\rangle \varphi\left(t\right)+\left(\tilde{u}\left(t\right),v\right)\varphi'\left(t\right)dt$,
hence 2. holds.
\item Let 2. be true: in particular for $\forall\varphi\left(t\right)\in C_{C}^{\infty}\left(\left(0,T\right)\right)$
we have 
\[
-\int_{I}\left(u\left(t\right),v\right)\varphi'\left(t\right)dt=\int_{I}\left\langle g\left(t\right),v\right\rangle \varphi\left(t\right)dt,\forall v\in W_{0}^{1,2}
\]
$\iff\frac{d}{dt}\iota u\left(t\right)=g\left(t\right)$, hence we
have first part of 1, hence by the above argument (integration by
parts):\\
compare with 2. $\implies\left(\tilde{u}\left(0\right),v\right)\varphi\left(0\right)=\left(u_{0},v\right)\varphi\left(0\right)$,
where $v\in W_{0}^{1,2}$ is arbitrary, $W_{0}^{1,2}$ is dense in
$L^{2}$ $\implies u_{0}=\tilde{u}_{0}$.
\end{itemize}

\subsection*{Ex. 3.2}
\begin{enumerate}
\item Let $u\in L^{2}$. Since $f$ is $l$-Lipschitz continuous and in
particular continuous, $f(u)$ is measurable scalar function as ``continuous$\circ$measurable''
is a measurable scalar function. For $u,v\in L^{2}$ we have 
\begin{align*}
\left\Vert f(u)-f(v)\right\Vert _{2}^{2} & =\int_{\Omega}\left|f(u(x))-f(v(x))\right|^{2}\,dx\\
 & \leq l^{2}\int_{\Omega}\left|u(x)-v(x)\right|^{2}\,dx=l^{2}\left\Vert u-v\right\Vert _{2}^{2},
\end{align*}
therefore the mapping $u\in L^{2}\mapsto f(u)\in L^{2}$ is $l$-Lipschitz.

Now let $u(t)\in L^{2}\left(I;L^{2}\right)$. In particular, $u(t)$
is strongly measurable and therefore there exist simple functions
$u_{n}(t)$ such that $u_{n}(t)\rightarrow u(t)$ in $L^{2}$ for
a. e. $t\in I$. Denote by $T_{f}:L^{2}\longrightarrow L^{2}$ the
(nonlinear) operator defined by the formula $T_{f}(u)=f(u)$. This
operator is Lipschitz and therefore continuous due to previous part.
Thus $T_{f}\circ u_{n}(t)\rightarrow T_{f}\circ u(t)=f(u(t))$ in
$L^{2}$ for a. e. $t\in I$. But $T_{f}\circ u_{n}(t)$ are simple
functions (they clearly map their definition domain to finitely many
values), which means that $f(u(t))$ is strongly measurable mapping.
For $u(t),v(t)\in L^{2}\left(I;L^{2}\right)$, using previous parts,
we have 
\begin{align*}
\left\Vert f(u(t))-f(v(t))\right\Vert _{L^{2}\left(I;L^{2}\right)}^{2} & =\int_{I}\left\Vert f(u(t))-f(v(t))\right\Vert _{2}^{2}\,dt\\
 & \leq l^{2}\int_{I}\left\Vert u(t)-v(t)\right\Vert _{2}^{2}\ \mathrm{dt}=l^{2}\left\Vert u(t)-v(t)\right\Vert _{L^{2}\left(I;L^{2}\right)}^{2},
\end{align*}
thus the mapping $u(t)\in L^{2}\left(I;L^{2}\right)\mapsto f(u(t))\in L^{2}\left(I;L^{2}\right)$
is $l$-Lipschitz.
\item Let $u(t),v(t)\in L^{2}\left(I;W_{0}^{1,2}\right)$. Then in particular
(for $t\in I$ fixed) we have $f(u(t)),f(v(t))\in L^{2}$, as $f$
is Lipschitz and $u(t),v(t)\in L^{2}$. Fix arbitrary $w\in W_{0}^{1,2},\left\Vert w\right\Vert _{1,2}\leq1$.
Then we obtain 
\begin{align*}
\left|\left(f(u(t))-f(v(t)),w\right)\right| & \leq\int_{\Omega}\left|f(u(t)(x))-f(v(t)(x))\right|\left|w(x)\right|\,dx\\
 & \leq l\int_{\Omega}\left|u(t)(x)-v(t)(x)\right|\left|w(x)\right|\,dx\\
 & \leq l\left\Vert u(t)-v(t)\right\Vert _{2}\left\Vert w\right\Vert _{2}\leq l\left\Vert u(t)-v(t)\right\Vert _{2},
\end{align*}
hence
\begin{equation}
\left\Vert \iota f(u(t))-\iota f(v(t))\right\Vert _{-1,2}\leq l\left\Vert u(t)-v(t)\right\Vert _{2}.\label{e3.2ineq1}
\end{equation}
Using (\ref{e3.2ineq1}), we have 
\begin{align*}
\left\Vert \iota f(u(t))-\iota f(v(t))\right\Vert _{L^{2}\left(I;W^{-1,2}\right)}^{2} & =\int_{I}\left\Vert \iota f(u(t))-\iota f(v(t))\right\Vert _{-1,2}^{2}\,dt\\
 & \leq l^{2}\int_{I}\left\Vert u(t)-v(t)\right\Vert _{2}^{2}\,dt\leq l^{2}\int_{I}\left\Vert u(t)-v(t)\right\Vert _{1,2}^{2}\,dt\\
 & =l^{2}\left\Vert u(t)-v(t)\right\Vert _{L^{2}\left(I;W_{0}^{1,2}\right)}^{2},
\end{align*}
which means that the mapping $u(t)\in L^{2}\left(I;W_{0}^{1,2}\right)\mapsto\iota f(u(t))\in L^{2}\left(I;W^{-1,2}\right)$
is $l$-Lipschitz.
\item Let $u,v\in W_{0}^{1,2}$. Choose $w\in W_{0}^{1,2},\left\Vert w\right\Vert _{1,2}\leq1$
arbitrary fixed. Then 
\begin{align*}
\left|\langle\mathcal{A}(u)-\mathcal{A}(v),w\rangle\right| & \leq\int_{\Omega}\left|a(\nabla u(x))-a(\nabla v(x))\right|\left|\nabla w(x)\right|\,dx\\
 & \leq\alpha_{1}\int_{\Omega}\left|\nabla u(x)-\nabla v(x)\right|\left|\nabla w(x)\right|\,dx\\
 & \leq\alpha_{1}\left\Vert \nabla u-\nabla v\right\Vert _{2}\left\Vert \nabla w\right\Vert _{2}\leq\alpha_{1}\left\Vert \nabla u-\nabla v\right\Vert _{2}\\
 & \leq\alpha_{1}\left\Vert u-v\right\Vert _{1,2},
\end{align*}
thus $\left\Vert \mathcal{A}(u)-\mathcal{A}(v)\right\Vert _{-1,2}\leq\alpha_{1}\left\Vert u-v\right\Vert _{1,2}$.
Therefore the mapping $u\in W_{0}^{1,2}\mapsto\mathcal{A}(u)\in W^{-1,2}$
is $\alpha_{1}$-Lipschitz.
\item Let $t\in I$ and $u\in W_{0}^{1,2}$. Fix arbitrary $v\in W^{1,2},\left\Vert v\right\Vert _{1,2}\leq1$.
Then (we assumed that $a\left(0\right)=0$ and $\left|\Omega\right|<\infty$)
\begin{align*}
 & \left|\int_{\Omega}a(\nabla u(x))\cdot\nabla v(x)\,dx+(f(u),v)\right|=\left|\int_{\Omega}a(\nabla u(x))\cdot\nabla v(x)\,dx+\int_{\Omega}f(u(x))v(x)\,dx\right|\\
 & \leq\int_{\Omega}\left|a(\nabla u(x))\right|\left|\nabla v(x)\right|\,dx+\int_{\Omega}\left|f(u(x))\right|\left|v(x)\right|\,dx\\
 & =\int_{\Omega}\left|a(\nabla u(x))-a(o)\right|\left|\nabla v(x)\right|\,dx+\int_{\Omega}\left|f(u(x))\right|\left|v(x)\right|\,dx\\
 & \leq\alpha_{1}\int_{\Omega}\left|\nabla u(x)\right|\left|\nabla v(x)\right|\,dx+\int_{\Omega}\left|f(u(x))\right|\left|v(x)\right|\,dx\\
 & \leq\alpha_{1}\left\Vert \nabla u\right\Vert _{2}\left\Vert \nabla v\right\Vert _{2}+\int_{\Omega}\left|f(u(x))\right|\left|v(x)\right|\,dx\\
 & \leq\alpha_{1}\left\Vert \nabla u\right\Vert _{2}+\int_{\Omega}\left|f(u(x))-f(0)\right|\left|v(x)\right|\,dx+\left|f(0)\right|\int_{\Omega}\left|v(x)\right|\,dx\\
 & \leq\alpha_{1}\left\Vert \nabla u\right\Vert _{2}+l\int_{\Omega}\left|u(x)\right|\left|v(x)\right|\,dx+\left|f(0)\right|\left|\Omega\right|^{\frac{1}{2}}\\
 & \leq\alpha_{1}\left\Vert \nabla u\right\Vert _{2}+l\left\Vert u\right\Vert _{2}+\left|f(0)\right|\left|\Omega\right|^{\frac{1}{2}}.
\end{align*}
Therefore we have proven that 
\begin{equation}
\left\Vert -\mathcal{A}(u)-\iota f\left(u\right)\right\Vert _{-1,2}\leq\max\{\alpha_{1},l,\left|f(0)\right|\left|\Omega\right|^{\frac{1}{2}}\}(1+\left\Vert u\right\Vert _{1,2}).\label{e3.2ineq2}
\end{equation}
By the inequality (\ref{e3.2ineq2}) it easily follows that 
\begin{align*}
\left\Vert \mathcal{F}(t,u)\right\Vert _{-1,2} & =\left\Vert -\mathcal{A}(u)-\iota f(u)+h(t)\right\Vert _{-1,2}\leq\left\Vert -\mathcal{A}(u)-\iota f(u)\right\Vert _{-1,2}+\left\Vert h(t)\right\Vert _{-1,2}\\
 & \leq\max\{1,\alpha_{1},l,\left|f(0)\right|\left|\Omega\right|^{\frac{1}{2}}\}(1+\left\Vert u\right\Vert _{1,2}+\left\Vert h(t)\right\Vert _{-1,2}),
\end{align*}
which concludes the proof.
\end{enumerate}

\subsection*{Ex. 3.3}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Gelfand triple and embedding $L^2$ in $W^{-1,2}$} 

Let $W_{0}^{1,2}\hookrightarrow L^{2}\hookrightarrow W^{-1,2}$ be
the Gelfand triple, with the embedding $\iota:W_{0}^{1,2}\to W^{-1,2}$. 
\begin{enumerate}
\item[$\left(1\right)$] By Poincaré inequality, the norms $\left\Vert \cdot\right\Vert :u\mapsto\left\Vert \nabla u\right\Vert _{2}$
and $\left\Vert u\right\Vert _{1,2}$ are equivalent on $W_{0}^{1,2}$.
Therefore the mapping $\left(\left(\cdot,\cdot\right)\right):W_{0}^{1,2}\times W_{0}^{1,2}\to\mathbb{R}$
defined as $\left(\left(u,v\right)\right)=\left(\nabla u,\nabla v\right)$
is a scalar product on $W_{0}^{1,2}$ and $\left(W_{0}^{1,2},\left(\left(\cdot,\cdot\right)\right)\right)$
is a Hilbert space.
\item[$\left(4\right)$] Let $f\in W^{-1,2}=\left(W_{0}^{1,2}\right)^{*}$ be given. By Riesz
theorem, there exists $u_{f}\in W_{0}^{1,2}$ so that 
\[
\left\langle f,v\right\rangle =\left(\left(u_{f},v\right)\right)\quad\forall v\in W_{0}^{1,2}
\]
By density of $C_{c}^{\infty}$ in $W_{0}^{1,2}$, there exists a
sequence $u_{n}\in C_{c}^{\infty}$ such that $u_{n}\to u_{f}$ in
$W_{0}^{1,2}$, hence $\left\langle f,v\right\rangle =\lim_{n\to\infty}\left(\left(u_{n},v\right)\right)$.
Now we can write 
\[
\left(\left(u_{n},v\right)\right)=\int_{\Omega}\nabla u_{n}\left(x\right)\nabla v\left(x\right)dx=\int_{\Omega}\Delta u_{n}\left(x\right)v\left(x\right)dx=\left(-\Delta u_{n},v\right)=\Big<\iota\underbrace{\left(-\Delta u_{n}\right)}_{\mbox{\ensuremath{f_{n}}}\makebox[0pt][r]{\ensuremath{\in C_{c}^{\infty}\subset L^{2}}}},v\Big>
\]
hence $f=\lim_{n\to\infty}\iota f_{n}$ (limit in $W^{-1,2}$), $f_{n}\in L^{2}$
\end{enumerate}

\subsection*{Ex. 3.4}
\begin{itemize}
\item[$4.$]  Let $f\in W^{-1,2}$ be given. $W^{-1,2}=\left(W_{0}^{1,2}\right)^{*}$,
$W_{0}^{1,2}$ is a Hilbert space with scalar product $\left(\left(u,v\right)\right)=\left(\nabla u,\nabla v\right)$.
By Riesz theorem we know: $\exists u_{f}\in W_{0}^{1,2}$ s. t. $\left\langle f,v\right\rangle =\left(\left(u_{f},v\right)\right)=\left(\nabla u_{f},\nabla v\right)\forall v\in W_{0}^{1,2}$.
\\
We know: $\exists u_{n}\in C_{c}^{\infty}$ s. t. $u_{n}\to u_{f}$
in $W_{0}^{1,2}$ and therefore $\left\langle f,v\right\rangle =\lim_{n\to\infty}\left(\nabla u_{n},\nabla v\right)$,
but due to Green's theorem we have $\left(\nabla u,\nabla v\right)=\int_{\Omega}\nabla u_{n}\cdot\nabla vdx=-\int_{\Omega}\underbrace{\Delta u_{n}}_{=-f_{n}}vdx$$=\left(f_{n},v\right)=\left\langle \iota f_{n},v\right\rangle $.
We have shown: $\left\langle f,v\right\rangle =\lim_{n\to\infty}\left\langle \iota f_{n},v\right\rangle $,
i. e., $\iota f_{n}\rightharpoonup f$ weakly (for $v\in W_{0}^{1,2}$
fixed). The same holds even strongly: $\left\Vert \iota f_{n}-f\right\Vert =\sup_{{v\in W_{0}^{1,2}\atop \left\Vert v\right\Vert \leq1}}\left\langle \iota f_{n}-f,v\right\rangle =\left(\nabla u_{n}-\nabla u,v\right)\leq\left\Vert \nabla u_{n}-\nabla u\right\Vert _{2}\to0$,
i. e., the convergence is strong. Therefore $\overline{\left(\iota L^{2}\right)}^{W^{-1,2}}=W^{-1,2}$.
\item[$5.$]  it suffices to show that $\iota u=0$ in $W^{-1,2}\implies u=0$
in $L^{2}$: $\iota u=0\iff\left\langle \iota u,v\right\rangle =0$
for $\forall v\in W_{0}^{1,2}$, i. e., $\left(u,v\right)=0\forall v\in W_{0}^{1,2}$
but $W_{0}^{1,2}$ is dense in $L^{2}$$\implies\left(u,v\right)=0\ \forall v\in L^{2}$
$\implies u=0$ in $L^{2}$. Follow by linearity.
\end{itemize}

\subsection*{Ex. 4.1}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Continuity of projection to laplacian eigenvalues} 

Let $u(t)\in L^{\infty}(I;X)\cap C(I;Z)$. I. e., there exists a measurable
subset $N\subseteq I,\left|N\right|=0$ such that for $\forall t\in I\setminus N$
it holds that $u(t)\in X$ and $\left\Vert u(t)\right\Vert _{X}\leq\left\Vert u(t)\right\Vert _{L^{\infty}\left(I;X\right)}=M<\infty$.
We begin by proving that this holds for every $t\in I$. Choose $t_{0}\in N$
arbirtrary fixed. As $\left|N\right|=0$ (and in particular $N$ does
not contain any nondegenerate interval), there exists $t_{n}\in I\setminus N$
such that $t_{n}\rightarrow t_{0}$. For every $n\in\mathbb{N}$ we
thus have $u(t_{n})\in X$ and $\left\Vert u(t_{n})\right\Vert _{X}\leq M$.
As $X$ is reflexive, we can choose a subsequence $\left\{ s_{n}\right\} \subseteq\left\{ t_{n}\right\} $
such that $u(s_{n})\rightharpoonup x_{0}\in X$ in $X$. Since $X\hookrightarrow Z$,
we also have $u(s_{n})\rightharpoonup x_{0}$ in $Z$. But $u(t)\in C(I;Z)$,
therefore $u(s_{n})\rightarrow u(t_{0})$ in $Z$ as $s_{n}\rightarrow t_{0}$.
By uniqueness of weak limit it follows that $u(t_{0})=x_{0}\in X$
and by weak lower semicontinuity of the norm we have $\left\Vert u(t_{0})\right\Vert _{X}\leq\liminf u(s_{n})\leq M$.
We have thus shown that 
\begin{align}
u(t)\in X & ,\nonumber \\
\left\Vert u(t)\right\Vert _{X}\leq M & ,\quad\forall t\in I.\label{e4.1ineq1}
\end{align}

Now we proceed to prove that the mapping $t\in I\mapsto u(t)\in X$
is weakly continuous. I. e., we need to show that for arbitrary $\varphi\in X^{*}$
the mapping $t\in I\mapsto\langle\varphi,u(t)\rangle_{X^{*},X}$ is
continuous. Fix $\varphi\in X^{*}$. By (\ref{e4.1ineq1}) we have
that for every $t\in I$ it holds that $\langle\varphi,u(t)\rangle_{X^{*},X}\in\left[-\left\Vert \varphi\right\Vert _{X^{*}}M,\left\Vert \varphi\right\Vert _{X^{*}}M\right]$.
Assume for contradiction that $t\in I\mapsto\langle\varphi,u(t)\rangle_{X^{*},X}$
is not continuous. Therefore there exists $t_{0}\in I$ and a sequence
$t_{n}\in I$ such that $t_{n}\rightarrow t_{0}$. But $\langle\varphi,u(t_{n})\rangle_{X^{*},X}\not\rightarrow\langle\varphi,u(t_{0})\rangle_{X^{*},X}\in\left[-\left\Vert \varphi\right\Vert _{X^{*}}M,\left\Vert \varphi\right\Vert _{X^{*}}M\right]$.
Thus there exists a chosen subsequence $\left\{ s_{n}\right\} \subseteq\left\{ t_{n}\right\} $
such that $\langle\varphi,u(s_{n})\rangle_{X^{*},X}\rightarrow c_{0}\in\left[-\left\Vert \varphi\right\Vert _{X^{*}}M,\left\Vert \varphi\right\Vert _{X^{*}}M\right]\setminus\left\{ \langle\varphi,u(t_{0})\rangle_{X^{*},X}\right\} $.
But by similar reasoning as in first part we know that we can choose
a subsequence $\left\{ r_{n}\right\} \subseteq\left\{ s_{n}\right\} $
such that $u(r_{n})\rightharpoonup u(t_{0})$ in $X$. Therefore,
in particular $\langle\varphi,u(r_{n})\rangle_{X^{*},X}\rightarrow\langle\varphi,u(t_{0})\rangle_{X^{*},X}$.
This contradicts our assumptions as $\langle\varphi,u(r_{n})\rangle_{X^{*},X}\rightarrow c_{0}\neq\langle\varphi,u(t_{0})\rangle_{X^{*},X}$.
The mapping $t\in I\mapsto\langle\varphi,u(t)\rangle_{X^{*},X}$ is
therefore continuous. It is obvious that $t\in I\mapsto u(t)\in X$
is weakly continuous as $\varphi\in X^{*}$ was chosen arbitrarily. 

\subsection*{Ex. 4.2}
\begin{enumerate}
\item For $u\in W_{0}^{1,2}$ we have (we know that the eigenvalues $\lambda_{j}$
are positive) 
\[
P_{N}u=\sum_{j=1}^{N}(u,w_{j})w_{j}=\sum_{j=1}^{N}\frac{1}{\lambda_{j}}(\nabla u,\nabla w_{j})w_{j}=\sum_{j=1}^{N}((u,\frac{w_{j}}{\sqrt{\lambda_{j}}}))\frac{w_{j}}{\sqrt{\lambda_{j}}}.
\]
But the set $\left\{ \frac{w_{1}}{\sqrt{\lambda_{1}}},\dots,\frac{w_{N}}{\sqrt{\lambda_{N}}}\right\} $
is orthonormal in ON v $W_{0}^{1,2}$ with the scalar product $((\cdot,\cdot))$
as $(\nabla w_{j},\nabla w_{l})=\delta_{jl}\lambda_{j}$. We have
expressed $P_{N}$ as (nontrivial) orthonormal projection on $\text{span}\,\left\{ \frac{w_{1}}{\sqrt{\lambda_{1}}},\dots,\frac{w_{N}}{\sqrt{\lambda_{N}}}\right\} $,
therefore the mapping $P_{N}:W_{0}^{1,2}\longrightarrow W_{0}^{1,2}$
is continuous with norm equal to $1$ if $W_{0}^{1,2}$ is considered
Hilbert space with the scalar product $((\cdot,\cdot))$. 
\item Add extra assumption $\partial\Omega\in C^{2}$ for we need to use
theorem about elliptic regularity. Application of this theorem to
eigenfunctions of laplacian for Dirichlet problem gives that $w_{j}\in W_{0}^{1,2}\cap W^{2,2}$
and in particular $\Delta w_{j}=-\lambda_{j}w_{j}$ a. e. in $\Omega$.
Now we use second part of this theorem about estimate of $W^{2,2}$
norm and we obtain that for every $v\in W^{2,2}$ it holds that 
\begin{equation}
\left\Vert v\right\Vert _{2,2}\leq C_{R}\left(\left\Vert v\right\Vert _{2}+\left\Vert \Delta v\right\Vert _{2}\right),\label{e4.2ineq2}
\end{equation}
where $C_{R}$ is independent of $v$. Let $u\in W_{0}^{1,2}\cap W^{2,2}$.
Since $\frac{\partial u}{\partial x_{i}}\in W^{1,2}$ and $w_{j}\in W_{0}^{1,2}$,
we can use integration by parts for Sobolev functions and obtain 
\[
\int_{\Omega}\frac{\partial u}{\partial x_{i}}(x)\frac{\partial w_{j}}{\partial x_{i}}(x)\,dx=-\int_{\Omega}\frac{\partial^{2}u}{\partial x_{i}^{2}}(x)w_{j}(x)\,dx.
\]
Therefore
\begin{align*}
(\nabla u,\nabla w_{j}) & =\sum_{i=1}^{n}\int_{\Omega}\frac{\partial u}{\partial x_{i}}(x)\frac{\partial w_{j}}{\partial x_{i}}(x)\,dx\\
 & =-\sum_{i=1}^{n}\int_{\Omega}\frac{\partial^{2}u}{\partial x_{i}^{2}}(x)w_{j}(x)\,dx\\
 & =-(\Delta u,w_{j}),
\end{align*}
by which it follows that 
\begin{align*}
\Delta P_{N}u & =\sum_{j=1}^{N}(u,w_{j})\Delta w_{j}=\sum_{j=1}^{N}-\lambda_{j}(u,w_{j})w_{j}\\
 & =\sum_{j=1}^{N}-(\nabla u,\nabla w_{j})w_{j}=\sum_{j=1}^{N}(\Delta u,w_{j})w_{j}\\
 & =P_{N}\Delta u,
\end{align*}
a. e. in $\Omega$. We have thus shown that 
\begin{equation}
\Delta P_{N}u=P_{N}\Delta u,\label{e4.4eq1}
\end{equation}
a. e. in $\Omega$. Combining (\ref{e4.2ineq2}), (\ref{e4.4eq1})
and the fact that $P_{N}$ is orthonormal projection $L^{2}\longrightarrow L^{2}$,
we obtain that 
\begin{align*}
\left\Vert P_{N}u\right\Vert _{2,2} & \leq C_{R}(\left\Vert P_{N}u\right\Vert _{2}+\left\Vert \Delta P_{N}u\right\Vert _{2})=C_{R}(\left\Vert P_{N}u\right\Vert _{2}+\left\Vert P_{N}\Delta u\right\Vert _{2})\\
 & \leq C_{R}(\left\Vert u\right\Vert _{2}+\left\Vert \Delta u\right\Vert _{2})\leq C_{R}\left\Vert u\right\Vert _{2,2}.
\end{align*}
\end{enumerate}

\subsection*{Ex. 4.3}

By Lemma 2.3 we know that (for this part of the lemma the boundedness
of first derivative suffices) $\psi(u)\in W^{1,2}$ for $u\in W^{1,2}$
and that $\nabla\psi(u)=\psi'(u)\nabla u$. Let $u_{n}\rightarrow u$
in $W^{1,2}$. The goal is to prove that $\psi(u_{n})\rightarrow\psi(u)$
in $W^{1,2}$. Obviously $\psi(u_{n})\rightarrow\psi(u)$ in $L^{2}$
since 
\begin{align*}
\int_{\Omega}\left|\psi(u_{n}(x))-\psi(u(x))\right|^{2}\,dx\leq\left\Vert \psi'\right\Vert _{\infty}^{2}\int_{\Omega}\left|u_{n}(x)-u(x)\right|^{2}\,dx\rightarrow0,
\end{align*}
as $\left\Vert \psi'\right\Vert _{\infty}<\infty$ and $u_{n}\rightarrow u$
v $L^{2}$. It remains to show that 
\begin{equation}
\nabla\psi(u_{n})\rightarrow\nabla\psi(u)\label{e4.3goal1}
\end{equation}
in $L^{2}$. In first step we will prove (\ref{e4.3goal1}) with an
additional assumption: 
\[
u_{n}(x)\rightarrow u(x)
\]
a. e. in $\Omega$. Then it holds that 
\[
\left\Vert \psi'(u_{n})\nabla u_{n}-\psi'(u)\nabla u\right\Vert _{2}\leq\left\Vert \psi'(u_{n})\nabla u_{n}-\psi'(u_{n})\nabla u\right\Vert _{2}+\left\Vert \psi'(u_{n})\nabla u-\psi'(u)\nabla u\right\Vert _{2},
\]
since
\[
\left\Vert \psi'(u_{n})\nabla u_{n}-\psi'(u_{n})\nabla u\right\Vert _{2}\leq\left\Vert \psi'\right\Vert _{\infty}\left\Vert \nabla u_{n}-\nabla u\right\Vert _{2}\rightarrow0,
\]
as $\left\Vert \psi'\right\Vert _{\infty}<\infty$ and $\nabla u_{n}\rightarrow\nabla u$
in $L^{2}$ and moreover we have that 
\[
\left\Vert \psi'(u_{n})\nabla u-\psi'(u)\nabla u\right\Vert _{2}^{2}=\int_{\Omega}\left|\psi'(u_{n}(x))\nabla u(x)-\psi'(u(x))\nabla u(x)\right|^{2}\,dx\rightarrow0,
\]
because $\left|\psi'(u_{n}(x))\nabla u(x)-\psi'(u(x))\nabla u(x)\right|^{2}\rightarrow0$
a. e. in $\Omega$ and 
\[
\left|\psi'(u_{n}(x))\nabla u(x)-\psi'(u(x))\nabla u(x)\right|^{2}\leq4\left\Vert \psi'\right\Vert _{\infty}^{2}\left|\nabla u(x)\right|^{2}\in L^{1},
\]
so we can use Lebesgue dominated convergence theorem. We have thus
proved that $\left\Vert \psi'(u_{n})\nabla u_{n}-\psi'(u)\nabla u\right\Vert _{2}\rightarrow0$,
which concludes the proof of (\ref{e4.3goal1}) za with an additional
assumption

It remains to prove the general case. Assusme for contradiction that
$\nabla\psi(u_{n})\not\rightarrow\nabla\psi(u)$ in $L^{2}$. Therefore
there exists a chosen subsequence $\left\{ v_{n}\right\} \subseteq\left\{ u_{n}\right\} $
such that 
\[
\left\Vert \psi'(v_{n})\nabla v_{n}-\psi'(u)\nabla u\right\Vert _{2}\rightarrow c\in(0,\left\Vert \psi'\right\Vert _{\infty}(2\left\Vert \nabla u\right\Vert _{2}+1)].
\]
But since $v_{n}\rightarrow u$ in $L^{2}$ and $L^{2}\hookrightarrow L^{1}$
(we have $\left|\Omega\right|<\infty$), we can choose a subsequence
$\left\{ z_{n}\right\} \subseteq\left\{ v_{n}\right\} $ such that
$z_{n}(x)\rightarrow u(x)$ a. e. in $\Omega$. Due to the first step
it holds that $\left\Vert \nabla\psi(z_{n})-\nabla\psi(u)\right\Vert _{2}\rightarrow0$,
which contradicts our assumptions as $\left\Vert \nabla\psi(z_{n})-\nabla\psi(u)\right\Vert _{2}\rightarrow c\neq0$.
Therefore (\ref{e4.3goal1}) holds, which remained to prove.

\subsection*{Ex. 5.1}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Characterization of semigroups with bounded generators} 

$S\left(t\right)$ is a $c_{0}$-semigroup in $X$. The following
statements are equivalent:
\begin{enumerate}
\item[\foreignlanguage{czech}{\inputencoding{latin2}$\left(1\right)$}] $S\left(t\right)=e^{tA}$, kde $A\in\mathcal{L}\left(X\right)$
\item[\foreignlanguage{czech}{\inputencoding{latin2}$\left(2\right)$}] $S\left(h\right)\to I$ v $\mathcal{L}\left(X\right)$, $h\to0^{+}$.
\end{enumerate}

\subsubsection*{Remark:}

(2) \dots{} uniform continuity: $\left\Vert S\left(h\right)-I\right\Vert _{\mathcal{L}\left(X\right)}\to0\implies\sup_{\left\Vert x\right\Vert \leq1}\left\Vert S\left(h\right)x-x\right\Vert \to0$,
$h\to0^{+}$ ($\implies$3. $S\left(h\right)x\to x$, $h\to0^{+}$,
$\forall x\in X$ fixed).
\begin{itemize}
\item[$\left(1\right)\implies\left(2\right)$]  let $S\left(t\right)=e^{tA}=\sum_{n=0}^{\infty}\frac{t^{n}A^{n}}{n!}$
\dots{} absolutely convergent in $\mathcal{L}\left(X\right)$ for
$\forall t\in\mathbb{C}$, the proof is similar as in the case of
$A\in\mathbb{C}^{n\times n}$ since $\mathcal{L}\left(X\right)$ is
complete and$\left\Vert AB\right\Vert _{\mathcal{L}\left(X\right)}\le\left\Vert A\right\Vert _{\mathcal{L}\left(X\right)}\left\Vert B\right\Vert _{\mathcal{L}\left(X\right)}$.
Moreover $\frac{d}{dt}S\left(t\right)=AS\left(t\right)$, as a mapping
$t\in\mathbb{C}\mapsto S\left(t\right)\in\mathcal{L}\left(X\right)$
(differentiation temr by term), in particular, $\frac{1}{h}\left(S\left(x\right)-x\right)\to Ax$,
$\forall x\in X$, i. e., we have shown that if $\left(\tilde{A},D\left(A\right)\right)$
is a generator of $S\left(t\right)=e^{tA}$ then $D\left(\tilde{A}\right)=X$,
$\tilde{A}x=Ax$ for $\forall x\in X$.
\item[$\left(2\right)\implies\left(1\right)$]  let $S\left(t\right)$ be a semigroup and $S\left(h\right)\to I$,
$h\to0^{+}$ in $\mathcal{L}\left(X\right)$. Denote the generator
of $S\left(t\right)$ by $\left(A,D\left(A\right)\right)$. We will
show that $D\left(A\right)=X$, $A\in\mathcal{L}\left(X\right)$ $\implies$
the proof is done, define $\tilde{S}\left(t\right)=e^{tA}$ in the
sense of $\left(1\right)$. We know that $A$ generates $\tilde{S}\left(t\right)$,
due to L. 4.2 we have: $S\left(t\right)$ and $\tilde{S}\left(t\right)$
are generated by the same operator $\implies$ $S\left(t\right)=\tilde{S}\left(t\right)$,
i. e., $S\left(t\right)=e^{tA}$. \textbf{Trick:} $S\left(s\right)\to I$,
$s\to0^{+}\implies\underbrace{\frac{1}{\tau}\int_{0}^{\tau}S\left(s\right)ds}_{B_{\tau}}\to I,\ \tau\to0^{+}$
$\implies B_{\tau}=I-\underbrace{\left(I-B_{\tau}\right)}_{\mbox{small}}$
is an invertible operator: ($\left(I-Q\right)^{-1}=\sum_{n=0}^{\infty}Q^{n}$
if $\left\Vert Q\right\Vert _{\mathcal{L}\left(X\right)}<1$). $\frac{S\left(h\right)-I}{h}B_{\tau}=\frac{1}{h}\left(S\left(h\right)\frac{1}{\tau}\int_{0}^{\tau}S\left(s\right)ds-\int_{0}^{\tau}S\left(s\right)ds\right)$
move $S\left(h\right)$ inside the integral in the first term, combine
to $S\left(h+s\right)$ and obtain $\frac{1}{\tau}\left(\int_{\tau}^{\tau+h}S\left(s\right)ds-\frac{1}{h}\int_{0}^{h}S\left(s\right)ds\right)$,
$h\to0^{+}:\frac{1}{\tau}\left(S\left(\tau\right)-I\right)$ since
$\frac{1}{h}\left(S\left(h\right)-I\right)B_{\tau}\to\frac{1}{\tau}\left(S\left(\tau\right)-I\right)$
in $\mathcal{L}\left(X\right)$, $h\to0^{+}$, $\frac{1}{h}\left(S\left(h\right)-I\right)=\frac{1}{h}\left(S\left(h\right)-I\right)B_{\tau}B_{\tau}^{-1}\to\frac{1}{\tau}\left(S\left(\tau\right)-I\right)B_{\tau}^{-1}=A\in\mathcal{L}\left(X\right)$.
Therefore $\frac{1}{h}\left(S\left(h\right)x-x\right)\to Ax$ for
$\forall x\in X$ fixed, i. e., $A$ is a bounded generator of $S\left(t\right)$.
\end{itemize}

\subsection*{Ex. 5.2}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Semigroup generated by the heat equation} 

Heat equation (see Chap. 2) $\frac{d}{dt}u-\Delta u=0$, $t>0$, $u\left(0\right)=u_{0}$.
Th. 2.3 $\implies\forall u_{0}\in L^{2}\exists!u\left(t\right)\in L^{2}\left(I,W_{0}^{1,2}\right)\cap C\left(I,L^{2}\right)$,
where $I=\left[0,T\right]$ is arbitrary. Denote $S\left(t\right)u_{0}=u\left(t\right)$
(operator which maps the initial condition to solution at the time
$t$).

We claim: $S\left(t\right)$ is a $c_{0}$-semigroup in $L^{2}$.
\begin{itemize}
\item $S\left(t+s\right)=S\left(t\right)S\left(s\right)$ \dots{} concatenation
of solutions (the solution at time $t$ is considered to be initial
condition for solution at time $s$). 
\item $S\left(t\right)\in\mathcal{L}\left(X\right)$ \dots{} linearity of
equation, 
\item $\left\Vert S\left(t\right)\right\Vert _{\mathcal{L}\left(L^{2}\right)}\leq1$,
$\forall t\ge0$. (Use difference of two solutions $w=u-v$ as a test
function $\implies\frac{1}{2}\frac{d}{dt}\left\Vert w\left(t\right)\right\Vert _{2}^{2}+\left\Vert \nabla w\right\Vert _{2}^{2}=0$
$\implies\left\Vert w\left(t\right)\right\Vert _{2}^{2}\leq\left\Vert w\left(0\right)\right\Vert _{2}^{2}$)
\item Third property: $t\mapsto S\left(t\right)u_{0}=u\left(t\right)$ is
continuous (follows from $u\left(t\right)\in C\left(I;L^{2}\right)$). 
\end{itemize}
? generator $\left(A,D\left(A\right)\right)$: auxiliary idea: let
$u_{0}\in W_{0}^{1,2}$ (better initial condition) Th. 2.5$\implies u\left(t\right)\in L^{\infty}\left(I,W_{0}^{1,2}\right)$$\implies$
the mapping $t\mapsto u\left(t\right)$ is continuous into the weak
topology $W_{0}^{1,2}$ (see ex. 4.1). On the other hand: by the weak
formulation we know: $\left(u\left(t\right),v\right)=\left(u_{0},v\right)-\int_{0}^{t}\nabla u\left(s\right)\cdot\nabla vds$
$\forall t\ge0$, $v\in W_{0}^{1,2}$. Moreover: $\left(\frac{S\left(t\right)u_{0}-u_{0}}{t},v\right)=\frac{1}{t}\int_{0}^{t}\underbrace{\nabla u\left(s\right)\nabla v}_{\makebox[0pt][c]{continuous integrand}}ds$.
Take the limit $t\to0^{+}$: $\frac{S\left(t\right)u_{0}-u_{0}}{h}\to-\tilde{A}u_{0}$,
where $\tilde{A}$ is the weak laplacian from chapter 2: $\left\langle \tilde{A}u,v\right\rangle =\int_{\Omega}\nabla u\cdot\nabla vdx$.
I. e., $\Delta$ in some form is the generator. Later we will show
that $D\left(A\right)=W_{0}^{1,2}\cap W^{2,2}$, $A:u\mapsto\Delta u\in L^{2}$.

\subsection*{Ex. 5.3}
\begin{enumerate}
\item Since $(Au)'(t)$ exists, $Au(t+h)$ is well\textendash defined for
small $h$, therefore there exists $\delta>0$ such that $u(t+h)\in D(A)$
for $0<\left|h\right|<\delta$ . For these $h$ we thus have 
\[
\frac{u(t+h)-u(t)}{h}\in D(A)\longrightarrow u'(t),\quad h\rightarrow0.
\]
By linearity of $A$ we have 
\[
A\left(\frac{u(t+h)-u(t)}{h}\right)=\frac{Au(t+h)-Au(t)}{h}\longrightarrow(Au)'(t),\quad h\rightarrow0.
\]
By closedness of operator $A$ it follows that $u'(t)\in D(A)$ and
$(Au)'(t)=A(u'(t))$. 
\item Denote $\left\Vert u\right\Vert _{D(A)}=\left\Vert u\right\Vert _{X}+\left\Vert Au\right\Vert _{X}$
for $u\in D(A)$. As $A$ is closed, the space $\left(D(A),\left\Vert \cdot\right\Vert _{D\left(A\right)}\right)$
is Banach so we can use the theory of Bochner integral. Since $u(t)\in L^{1}(I;D(A))$,
there exist simple functions $u_{n}(t):I\longrightarrow D(A)$ such
that $u_{n}(t)\longrightarrow u(t)$ in $D(A)$ for a. e. $t\in I$.
A fortiori we have $u_{n}(t)\longrightarrow u(t)$ in $X$ for a.
e. $t\in I$, since clearly $\left(D(A),\left\Vert \cdot\right\Vert _{D\left(A\right)}\right)\hookrightarrow\left(X,\left\Vert \cdot\right\Vert _{X}\right)$.
Therefore $u(t):I\longrightarrow X$ is strongly measurable. Moreover
$\int_{I}\left\Vert u(t)\right\Vert _{X}\,dt\leq\int_{I}\left\Vert u(t)\right\Vert _{D\left(A\right)}\,dt<\infty$
and so $u(t)\in L^{1}(I;X)$ due to Bochner theorem. By linearity
of $A$ the functions $Au_{n}(t):I\longrightarrow X$ are simple and
moreover 
\[
\left\Vert Au_{n}(t)-Au(t)\right\Vert _{X}\leq\left\Vert u_{n}(t)-u(t)\right\Vert _{D\left(A\right)}\longrightarrow0
\]
for a. e. $t\in I$. Therefore $Au(t):I\longrightarrow X$ is strongly
measurable. Clearly $\int_{I}\left\Vert Au(t)\right\Vert _{X}\,dt\leq\int_{I}\left\Vert u(t)\right\Vert _{D\left(A\right)}\,dt<\infty$,
thus $Au(t)\in L^{1}(I;X)$.

We know that
\[
\int_{I}u_{n}(t)\,dt\in D(A)\longrightarrow\int_{I}u(t)\,dt,
\]
and by linearity of $A$ and the fact that $u_{n}(t)$ are simple
functions we have 
\[
A\int_{I}u_{n}(t)\,dt=\int_{I}Au_{n}(t)\,dt\longrightarrow\int_{I}Au(t)\,dt.
\]
By closedness of $A$ it therefore holds that $\int_{I}u(t)\,dt\in D(A)$
and $A\int_{I}u(t)\,dt=\int_{I}Au(t)\,dt$ which concludes the proof.
\end{enumerate}

\subsection*{Ex. 5.4}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Semigroup of shift operators} 
\begin{enumerate}
\item $S\left(t\right)$ form a $c_{0}$-semigroup:
\begin{enumerate}
\item[$\left(i\right)$] $S\left(t\right)S\left(s\right)=S\left(t+s\right)$, $S\left(0\right)=I$
\dots{} clear
\item[$\left(ii\right)$] continuity: we know $f\left(x+h\right)\to f\left(x\right)$, $h\to0^{+}$
in $L^{2}\left(\mathbb{R}\right)$. By Ex.~1.4 we know: $S\left(h\right)f\to f$,
$h\to0^{+}$, moreover $\left\Vert S\left(h\right)\right\Vert \leq1$
independently of $\forall h$. 
\end{enumerate}
\item Fix $t>0$. Obviously for arbitrary $f\in B_{L^{2}}$ it holds that
\[
\left\Vert f(x+t)-f(x)\right\Vert _{2}\leq\left\Vert f(x+t)\right\Vert _{2}+\left\Vert f\right\Vert _{2}=2\left\Vert f\right\Vert _{2}\leq2.
\]
Thus
\begin{equation}
\left\Vert S(t)-I\right\Vert _{\mathcal{L}(X)}\leq2.\label{e5.4ineq1}
\end{equation}
To saturate the operator norm it is easily seen that we are looking
for functions from unit ball for which $f(x+t)$ is ``almost equal''
to $-f(x)$. Define 
\[
f_{1}(x)=\frac{1}{\sqrt{2t}}\left(\chi_{(0,t)}(x)-\chi_{(-t,0)}(x)\right).
\]
Then clearly 
\[
\left\Vert f_{1}\right\Vert _{2}=1
\]
and
\[
f_{1}(x+t)=-f_{1}(x),\text{ for }x\in(-t,0).
\]
Hence 
\begin{align*}
\left\Vert f_{1}(x+t)-f_{1}(x)\right\Vert _{2} & =\frac{1}{\sqrt{2t}}\sqrt{\int_{-2t}^{-t}1\,dx+\int_{-t}^{0}4\,dx+\int_{0}^{t}1\,dx}=\sqrt{\frac{1+4+1}{2}}\\
 & =\sqrt{3}.
\end{align*}
Define 
\[
f_{2}(x)=\frac{1}{\sqrt{4t}}\left(\chi_{(-t,0)\cup(t,2t)}(x)-\chi_{(-2t,-t)\cup(0,t)}(x)\right).
\]
Clearly
\[
\left\Vert f_{2}\right\Vert _{2}=1
\]
and 
\[
f_{2}(x+t)=-f_{2}(x),\text{ for }x\in(-2t,t).
\]
Thus 
\begin{align*}
\left\Vert f_{2}(x+t)-f_{2}(x)\right\Vert _{2} & =\frac{1}{\sqrt{4t}}\sqrt{\int_{-3t}^{-2t}1\,dx+\int_{-2t}^{t}4\,dx+\int_{t}^{2t}1\,dx}=\sqrt{\frac{1+3\cdot4+1}{4}}\\
 & =\sqrt{\frac{7}{2}}.
\end{align*}
In general, define 
\begin{align*}
f_{n}(x) & =\frac{1}{\sqrt{2tn}}(\chi_{(-(n-1)t,-(n-2)t)\cup(-(n-3)t,-(n-4)t)\cup\dots\cup((n-1)t,nt)}(x)\\
 & -\chi_{(-nt,-(n-1)t)\cup(-(n-2)t,-(n-3)t)\cup\dots\cup((n-2)t,(n-1)t)}(x)).
\end{align*}
Then obviously 
\begin{equation}
\left\Vert f_{n}\right\Vert _{2}=1\label{e5.4eq1}
\end{equation}
and 
\[
f_{n}(x+t)=-f_{n}(x),\text{ pro }x\in(-nt,(n-1)t),
\]
so 
\begin{align}
\left\Vert f_{n}(x+t)-f_{n}(x)\right\Vert _{2} & =\frac{1}{\sqrt{2nt}}\sqrt{\int_{-\left(n+1\right)t}^{-nt}1\,dx+\int_{-nt}^{(n-1)t}4\,dx+\int_{(n-1)t}^{nt}1\,dx}\nonumber \\
 & =\sqrt{\frac{1+4(2n-1)+1}{2n}}=\sqrt{\frac{4n-1}{n}}\longrightarrow\sqrt{4}=2.\label{e5.4eq2}
\end{align}
Therefore by (\ref{e5.4eq1}) and (\ref{e5.4eq2}) we see that 
\[
\left\Vert S(t)-I\right\Vert _{\mathcal{L}(X)}\geq2,
\]
which togehter with (\ref{e5.4ineq1}) gives the desired result: 
\[
\left\Vert S(t)-I\right\Vert _{\mathcal{L}(X)}=2.
\]
\item Consider the regularization kernels 
\begin{align*}
\psi_{0}(t) & =\chi_{(-1,0)}(t),\\
\psi_{h}(t) & =\frac{1}{h}\psi_{0}\left(\frac{t}{h}\right),\quad h>0.
\end{align*}
We know that $\mathrm{spt}\psi_{h}\subseteq[-h,0]$, and for $f\in L^{2}$
it holds that $(f*\psi_{h})(x)\longrightarrow f(x)$ in $L^{2}$.
Let $f\in W^{1,2}$. We can assume that $f$ is absolutely continuous
and we will obtain, denoting $g=\frac{d}{dx}f\in L^{2}$, the following:
\begin{align*}
\frac{S(h)f(x)-f(x)}{h} & =\frac{f(x+h)-f(x)}{h}=\frac{\int_{x}^{x+h}g(s)\,ds}{h}\\
 & =\frac{\int_{-h}^{0}g(x-s)\,ds}{h}=\frac{\int_{-h}^{0}g(x-s)\chi_{(-h,0)}(s)\,ds}{h}\\
 & =\frac{\int_{-h}^{0}g(x-s)\chi_{(-1,0)}(\frac{s}{h})\,ds}{h}=\frac{\int_{-h}^{0}g(x-s)\psi_{0}(\frac{s}{h})\,\mathrm{ds}}{h}\\
 & =\int_{-h}^{0}g(x-s)\psi_{h}(s)\,ds=\int_{\mathbb{R}}g(x-s)\psi_{h}(s)\,ds\\
 & =(g*\psi_{h})(x)\longrightarrow g(x)=\frac{d}{dx}f(x)
\end{align*}
in $L^{2}$ for $h\rightarrow0_{+}$, as was to be shown.
\item Choose an arbitrary, fixed test function $\varphi\in D(\mathbb{R})$.
We have 
\begin{align*}
\int_{\mathbb{R}}\frac{f(x+h)-f(x)}{h}\varphi(x)\,dx & =\frac{1}{h}\left[\int_{\mathbb{R}}f(x+h)\varphi(x)\,dx-\int_{\mathbb{R}}f(x)\varphi(x)\,dx\right]\\
 & =\frac{1}{h}\left[\int_{\mathbb{R}}f(x)\varphi(x-h)\,dx-\int_{\mathbb{R}}f(x)\varphi(x)\,dx\right]\\
 & =\int_{\mathbb{R}}f(x)\underbrace{\frac{\varphi(x-h)-\varphi(x)}{h}}_{\to\,\makebox[0pt][l]{\ensuremath{\frac{d}{dt}\varphi=\varphi'} by \ensuremath{\left(3\right)}}}\,dx.
\end{align*}
The proof is completed since the test function $\varphi$ was chosen
arbitrarily and therefore$\frac{d}{dx}f(x)=g(x)\in L^{2}$ and $f\in W^{1,2}$.
\end{enumerate}

\subsection*{Ex. 6.1}
\begin{enumerate}
\item Let $D(A)=\ell^{2}$ and let $A\in\mathcal{L}(\ell^{2})$. Therefore
there exists $C>0$ such that 
\[
\left\Vert Au\right\Vert _{2}\leq C\left\Vert u\right\Vert _{2}
\]
for every $u\in\ell^{2}$. In particular by the choice $u=\mathbf{e}_{j}$
we obtain 
\[
\left|\lambda_{j}\right|=\sqrt{\sum_{k=1}^{\infty}\lambda_{k}^{2}u_{k}^{2}}=\left\Vert A\mathbf{e}_{j}\right\Vert _{_{2}}\leq C\left\Vert \mathbf{e}_{j}\right\Vert _{2}=C,
\]
and thus the sequence $\left\{ \lambda_{j}\right\} $ is bounded.

On the other hand, let $\left\{ \lambda_{j}\right\} $ be bounded.
Then clearly for every $u\in\ell^{2}$ we have 
\[
\left\Vert Au\right\Vert _{2}=\sqrt{\sum_{k=1}^{\infty}\lambda_{k}^{2}u_{k}^{2}}\leq\left\Vert \left\{ \lambda_{j}\right\} \right\Vert _{\infty}\sqrt{\sum_{k=1}^{\infty}u_{k}^{2}}=\left\Vert \left\{ \lambda_{j}\right\} \right\Vert _{\infty}\left\Vert u\right\Vert _{2}<\infty.
\]
Hence $D(A)=\ell^{2}$ and $A\in\mathcal{L}(\ell^{2})$, the linearity
is evident.
\item Let $u^{n}\in D(A)\longrightarrow u$ and $Au^{n}\longrightarrow v$
in $\ell^{2}$. We will choose a rapidly convergent subsequence which
converges pointwise. I. e., we will choose $\left\{ w^{n}\right\} \subseteq\left\{ u^{n}\right\} $
such that 
\begin{equation}
w_{j}^{n}\longrightarrow u_{j},\quad n\rightarrow\infty\label{e6.1eq1}
\end{equation}
and
\begin{equation}
\left(Aw^{n}\right)_{j}=\lambda_{j}w_{j}^{n}\longrightarrow v_{j},\quad n\rightarrow\infty\label{e6.1eq2}
\end{equation}
for every $j\in\mathbb{N}$. By (\ref{e6.1eq1}) we thus have that
\[
\lambda_{j}w_{j}^{n}\longrightarrow\lambda_{j}u_{j},\quad n\rightarrow\infty,
\]
which along with (\ref{e6.1eq2}) gives the result 
\[
\lambda_{j}u_{j}=v_{j},
\]
for every $j\in\mathbb{N}$, therefore $u\in D(A)$ and $Au=v$. Therefore
the operator $A$ is closed.

The operator is moreover densely defined as it is sufficient to consider
finitely supported sequences, which are dense in $\ell^{p}$ for $p\in[1,\infty)$
(in particular for $p=2$). Finitely supported sequences are clearly
contained in the definition domain of $A$ and since they are dense
in $\ell^{2}$, a fortiori $D(A)$ is dense in $\ell^{2}$. 
\item Moreover, let $\lambda_{j}\leq\omega$ and $\lambda>\omega$. The
goal is to prove that $\lambda\in\rho(A)$, i. e., the operator $\lambda I-A:D(A)\longrightarrow\ell^{2}$
is bijective. Let $\lambda u-Au=\lambda v-Av$ for $u,v\in D(A)$.
Therefore for every $j\in\mathbb{N}$ we have that 
\[
\lambda u_{j}-\lambda_{j}u_{j}=\lambda v_{j}-\lambda_{j}v_{j},
\]
which means that $u_{j}=v_{j}$ as $\lambda-\lambda_{j}\neq0$. Hence
$u=v$, i. e., the operator $\lambda I-A$ is injective.

Let $v\in\ell^{2}$ be given. We are looking for $u\in D(A)$ such
that $\lambda u-Au=v$. The goal is to prove that for every $j\in\mathbb{N}$
it holds that 
\[
\lambda u_{j}-\lambda_{j}u_{j}=v_{j},
\]
thus the only candidate for the argument of $A$ is 
\begin{equation}
u=\left\{ \frac{v_{j}}{\lambda-\lambda_{j}}\right\} .\label{e6.1eq3}
\end{equation}
It remains to show that this $u$ is in the definition domain of $A$.
Clearly $u\in\ell^{2}$, since 
\[
\sum_{j=1}^{\infty}\frac{v_{j}^{2}}{(\lambda-\lambda_{j})^{2}}\leq\frac{1}{(\lambda-\omega)^{2}}\sum_{j=1}^{\infty}v_{j}^{2}=\frac{1}{(\lambda-\omega)^{2}}\left\Vert v\right\Vert _{2}^{2}<\infty.
\]
Moreover
\[
\sum_{j=1}^{\infty}\frac{\lambda_{j}^{2}v_{j}^{2}}{(\lambda-\lambda_{j})^{2}}\leq\max\left\{ 1,\frac{\omega^{2}}{(\lambda-\omega)^{2}}\right\} \sum_{j=1}^{\infty}v_{j}^{2}<\infty,
\]
where the inequality $\frac{\lambda_{j}^{2}}{(\lambda-\lambda_{j})^{2}}\leq\max\left\{ 1,\frac{\omega^{2}}{(\lambda-\omega)^{2}}\right\} $
can be easily obtained by analysis of the function $f(x)=\frac{x^{2}}{(\lambda-x)^{2}}$
on the interval $(-\infty,\omega]$. Therfore $u\in D(A)$, hence
the operator$\lambda I-A$ is onto.

From this it may be concluded that $\lambda\in\rho(A)$ and by (\ref{e6.1eq3})
we know that $R(\lambda,A)v=\left\{ \frac{v_{j}}{\lambda-\lambda_{j}}\right\} $
for $\lambda>\omega$ and $v\in\ell^{2}$. 
\item Since the convergence in $\ell^{2}$ in particular means that we are
able to choose a pointwise convergent subsequence, it is easy to guess
a candidate for $S\left(t\right)$:
\begin{equation}
S(t)u=\left\{ \mathrm{e}^{\lambda_{j}t}u_{j}\right\} \label{e6.1def1}
\end{equation}
for $u\in\ell^{2}$ and $t\in[0,\infty)$. Operator defined in this
way is clearly linear and bounded on $\ell^{2}$, since 
\[
\sum_{j=1}^{\infty}\mathrm{e}^{2\lambda_{j}t}u_{j}^{2}\leq\mathrm{e}^{2\omega t}\sum_{j=1}^{\infty}u_{j}^{2},
\]
therefore $S(t)\in\mathcal{L}(\ell^{2})$. Moreover, it is evident
that $S(0)=\mathrm{I}$ and $S(t+s)=S(t)S(s)$ for $s,t\geq0$. Fix
$u\in\ell^{2}$. We want to show that
\begin{equation}
S(t)u\longrightarrow u,\quad t\rightarrow0_{+}.\label{e6.1eq4}
\end{equation}
Clearly 
\begin{equation}
\mathrm{e}^{\lambda_{j}t}u_{j}\longrightarrow u_{j},\quad t\rightarrow0_{+}\label{e6.1eq5}
\end{equation}
for every $j\in\mathbb{N}$ and for $0<t<1$ we have 
\begin{equation}
(\mathrm{e}^{\lambda_{j}t}u_{j}-u_{j})^{2}\leq2(\mathrm{e}^{2\lambda_{j}t}u_{j}^{2}+u_{j}^{2})\leq2(\mathrm{e}^{2\omega t}u_{j}^{2}+u_{j}^{2})\leq2(\mathrm{e}^{2\left|\omega\right|}u_{j}^{2}+u_{j}^{2}),\label{e6.1eq6}
\end{equation}
where $\left\{ 2(\mathrm{e}^{2\left|\omega\right|}u_{j}^{2}+u_{j}^{2})\right\} \in\ell^{1}$,
hence by (\ref{e6.1eq5}), (\ref{e6.1eq6}) and Lebesgue dominated
convergence theorem (using the arihmetic measure on $\mathbb{N}$)
the equation (\ref{e6.1eq4}) holds. Thus $S(t)$ is a $c_{0}-$semigroup.

It suffices to show that $(A,D(A))$ is a generator of this semigroup.
Let $u\in D(A)$. Clearly 
\begin{equation}
\frac{\mathrm{e}^{\lambda_{j}h}u_{j}-u_{j}}{h}\longrightarrow\lambda_{j}u_{j},\quad h\rightarrow0_{+}\label{e6.1eq7}
\end{equation}
for every $j\in\mathbb{N}$. Since 
\[
\frac{\mathrm{e}^{\lambda_{j}h}-1}{h}\leq\frac{\mathrm{e}^{\omega h}-1}{h}\longrightarrow\omega,\ h\rightarrow0_{+},
\]
there exists $h_{0}>0$ (independent on $j$) such that for everry
$h\in(0,h_{0})$ it holds that 
\[
\frac{\mathrm{e}^{\lambda_{j}h}-1}{h}\leq2\left|\omega\right|.
\]
Using the inequality $\mathrm{e}^{x}\geq1+x$ we have 
\[
\frac{1-\mathrm{e}^{\lambda_{j}h}}{h}\leq\frac{-\lambda_{j}h}{h}\leq\left|\lambda_{j}\right|.
\]
Combining these steps together we obtain that for every $j\in\mathbb{N}$
and every $h\in(0,h_{0})$ it holds that
\begin{equation}
\left|\frac{\mathrm{e}^{\lambda_{j}h}-1}{h}\right|\leq\max\left\{ \left|\omega\right|,\left|\lambda_{j}\right|\right\} \leq\left|\omega\right|+\left|\lambda_{j}\right|.\label{e6.1eq8}
\end{equation}
Using (\ref{e6.1eq8}) we get that for $h\in(0,h_{0})$ the following
holds 
\[
\left|\frac{\mathrm{e}^{\lambda_{j}h}-1}{h}u_{j}-\lambda_{j}u_{j}\right|^{2}\leq2\left(\left|\frac{\mathrm{e}^{\lambda_{j}h}-1}{h}\right|^{2}u_{j}^{2}+\lambda_{j}^{2}u_{j}^{2}\right)\leq2\left(2\left(\omega^{2}+\lambda_{j}^{2}\right)u_{j}^{2}+\lambda_{j}^{2}u_{j}^{2}\right),
\]
where $\left\{ 5\lambda_{j}^{2}u_{j}^{2}+4\omega^{2}u_{j}^{2}\right\} \in\ell^{1}$
as $u\in D(A)$, which together with (\ref{e6.1eq7}) and Lebesgue
dominated convergence theorem yields 
\[
\frac{S(h)u-u}{h}\longrightarrow Au,\quad h\rightarrow0_{+}.
\]
It remains to show that $D(A)=\left\{ u\in\ell^{2};\lim\limits _{h\rightarrow0_{+}}\frac{S(h)u-u}{h}\in\ell^{2}\right\} $.\\
We already know that $D(A)\subseteq\left\{ u\in\ell^{2},\lim\limits _{h\rightarrow0_{+}}\frac{S(h)u-u}{h}\in\ell^{2}\right\} $.
To prove the opposite inclusion, consider 
\[
u\in\left\{ v\in\ell^{2},\lim\limits _{h\rightarrow0_{+}}\frac{S(h)v-v}{h}\in\ell^{2}\right\} 
\]
and let $\lim\limits _{h\rightarrow0_{+}}\frac{S(h)u-u}{h}=v\in\ell^{2}$.
In particular, for $h_{n}=\frac{1}{n}$ it holds that $\frac{S(h_{n})u-u}{h_{n}}\longrightarrow v$.
Choose pointwise converging subsequence, i. e., choose a subsequence
$\left\{ \tau_{n}\right\} \subseteq\left\{ h_{n}\right\} $ such that
for every $j\in\mathbb{N}$ we have 
\[
\frac{\mathrm{e}^{\tau_{n}\lambda_{j}}u_{j}-u_{j}}{\tau_{n}}\longrightarrow v_{j},\quad n\rightarrow\infty,
\]
but 
\[
\frac{\mathrm{e}^{\tau_{n}\lambda_{j}}u_{j}-u_{j}}{\tau_{n}}\longrightarrow\lambda_{j}u_{j},\quad n\rightarrow\infty,
\]
hence $v_{j}=\lambda_{j}u_{j}$. I. e., $\left\{ \lambda_{j}u_{j}\right\} =v\in\ell^{2}$,
therefore $u\in D(A)$.

We have thus proven (two $c_{0}$-semigroups generated by the same
operator are identic, see L. 4.2) that $S(t)$ defined by (\ref{e6.1def1})
is clearly a $c_{0}$-semigroup generated by the operator $(A,D(A))$. 
\end{enumerate}

\subsection*{Ex. 6.2}

Let $S\left(t\right)$ be a $c_{0}$-semigroup generated by $\left(A,D\left(A\right)\right)$
such that $\left\Vert S\left(t\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq Me^{\omega t}$.
Denote $\widetilde{S}\left(t\right)=e^{-\omega t}S\left(t\right)$.
Then: 
\begin{enumerate}
\item $\widetilde{S}\left(t\right)$ is a semigroup, $\left\Vert \widetilde{S}\left(t\right)\right\Vert _{\mathcal{L}\left(X\right)}\leq M$
$\forall t\ge0$
\item If $\left(\widetilde{A},D\left(\widetilde{A}\right)\right)$ generates
$\widetilde{S}\left(t\right)$, then $D\left(A\right)=D\left(\widetilde{A}\right)$,
$\widetilde{A}=A-\omega I$
\item $\rho\left(\widetilde{A}\right)=\rho\left(A\right)-\omega$, $R\left(\lambda,\widetilde{A}\right)=R\left(\lambda+\omega,A\right)$
$\forall\lambda\in\rho\left(\widetilde{A}\right)$.
\end{enumerate}

\subsubsection*{Proof:}

1., 3. \dots{} easy (expand)

2. it suffices to show: $x\in D\left(A\right)\implies x\in D\left(\widetilde{A}\right)$
and $\widetilde{A}x=Ax-\omega x$:

$\frac{1}{h}\left[\widetilde{S}\left(h\right)x-x\right]=\frac{1}{h}\left[e^{-\omega h}S\left(h\right)x-x\pm S\left(h\right)x\right]=\underbrace{\frac{e^{-\omega h}-1}{h}}_{\to-\omega}\underbrace{S\left(h\right)x}_{\to x}+\underbrace{\frac{1}{h}\left[S\left(h\right)x-x\right]}_{\to Ax}$

\subsection*{Ex. 6.3}

Heat semigroup $\mathcal{S}\left(t\right):L^{2}\to L^{2}$, $u_{0}\mapsto u\left(t\right)$
\dots{} weak solution (in the sense of chap.~2) of the heat equation
$\left(HE\right)\quad\frac{d}{dt}u-\Delta u=0,\ u|_{t=0}=0,\ u|_{\partial\Omega}=0,\ x\in\Omega$
in the space $L^{2}\left(I,W_{0}^{1,2}\right)\cap C\left(I,L^{2}\right)$.
What is the generator of $\mathcal{S}\left(t\right)$?

Recall: we know: $\exists\lambda_{j}>0$, $w_{j}\in W_{0}^{1,2}$
\dots{} eigenfunctions of laplacian, i. e., $-\Delta w_{j}=\lambda_{j}w_{j}$
weakly in $\Omega$: $\left(\nabla w_{j},\nabla v\right)=\lambda_{j}\left(w_{j},v\right)$
$\forall v\in W_{0}^{1,2}$. Moreover $\left\{ w_{j}\right\} $ form
a complete ON basis of $L^{2}$, i. e., $\forall u\in L^{2}$ can
be expressed as $u=\sum_{j=1}^{\infty}u_{j}w_{j}$, where $u_{j}=\left(u,w_{j}\right)$;
Parseval: $\left\Vert u\right\Vert _{L^{2}}^{2}=\sum_{j}u_{j}^{2}$.

\subsection*{Statement 6.3.2:}
\begin{itemize}
\item[(a)]  $u_{0}\in W_{0}^{1,2}\iff\sum_{j}\lambda_{j}u_{j}^{2}<\infty$,
in particular, $\sum_{j}\lambda_{j}u_{j}=\left\Vert \nabla u_{0}\right\Vert _{2}^{2}$.
\item[(b)]  $u_{0}\in W_{0}^{1,2}\cap W^{2,2}\iff\sum_{j}\lambda_{j}^{2}u_{j}^{2}<\infty$,
in particular, $\sum_{j}\lambda_{j}u_{j}=\left\Vert \Delta u_{0}\right\Vert _{2}^{2}$

\subsubsection*{Proof of (b):}
\begin{itemize}
\item[``$\impliedby$'']  let $u_{0}\in W_{0}^{1,2}$, $\sum\lambda_{j}^{2}u_{j}^{2}<\infty$,
where $u_{j}=\left(u_{0},w_{j}\right)$ $\stackrel{?}{\implies}u_{0}\in W^{2,2}$.
Denote $u_{N}=P_{N}u_{0}=\sum_{j=1}^{N}u_{j}w_{j}$: we know: $u_{N}\to u_{0}$
v $L^{2}.$ Moreover: $-\Delta u_{N}=\sum_{j=1}^{N}u_{j}\left(-\Delta u_{j}\right)=\underbrace{\sum_{j=1}^{N}\lambda_{j}u_{j}w_{j}}_{\mbox{denote by }z_{N}}$
weakly. We know: $\left\Vert -\Delta u_{N}\right\Vert _{2}^{2}=\left\Vert z_{N}\right\Vert _{2}^{2}=\left\Vert \sum_{j=1}^{N}\lambda_{j}^{2}u_{j}w_{j}\right\Vert _{2}^{2}=\sum_{j=1}^{N}\lambda_{j}^{2}w_{j}^{2}\le\sum_{j=1}^{\infty}\lambda_{j}^{2}u_{j}^{2}=c<\infty$.
We will use elliptic regularity: $\left\Vert u_{N}\right\Vert _{2,2}\le C_{R}\left\Vert z_{N}\right\Vert _{2}\le C_{R}\sqrt{c}$,
where $C_{R}$ depends only on $\Omega$. Therefore $\left\{ u_{N}\right\} \subset W^{2,2}$
is bounded $\implies\exists$subsequence $\widetilde{u}_{N}\rightharpoonup\widetilde{u}\in W_{0}^{1,2}$,
therefore $\widetilde{u}_{N}\rightharpoonup\widetilde{u}\in L^{2}$
\item[``$\implies$'']  let $u_{0}\in W_{0}^{1,2}\cap W^{2,2}\stackrel{?}{\implies}\sum_{j=1}^{\infty}\lambda_{j}^{2}u_{j}^{2}<\infty$
and is equal to $\left\Vert \Delta u_{0}\right\Vert _{2}^{2}=\sum_{j=1}^{n}\frac{\partial^{2}u}{\partial x_{j}^{2}}$
(weak derivative).
\end{itemize}
\textbf{Recall:} Gauss\textendash Green theorem: $\left(\nabla u,\nabla w\right)=\left(-\Delta u,w\right)$,
$\forall u\in W^{2,2}$, $w\in W_{0}^{1,2}$.

\textit{Proof:} $0=\int_{\partial\Omega}w\left(\nabla u\cdot n\right)dS=\int_{\Omega}\underbrace{\mbox{div}\left(w\nabla u\right)}_{\nabla w\cdot\nabla u+w\Delta u}dx=\int_{\Omega}\nabla w\cdot\nabla udx+\int_{\Omega}w\Delta udx$,
\dots{} $u$, $w$ is smooth and follow by density in given spaces.

Apply to $u=u_{0}\in W^{2,2}$, $w=w_{j}\in W_{0}^{1,2}$ and obtain:
$\left(\nabla u_{0},\nabla w_{j}\right)=\left(-\Delta u_{0},w_{j}\right)=z_{j}$.
$LS=\lambda_{j}\left(u_{0},w_{j}\right)$ since $-\Delta w_{j}=\lambda_{j}w_{j}$
weakly, $u_{0}\in W_{0}^{1,2}$ ``test function''. Parseval: $\sum z_{j}^{2}=\left\Vert -\Delta u_{0}\right\Vert _{2}^{2}$,
at the same time we know that $\sum z_{j}^{2}=\sum\lambda_{j}^{2}u_{j}^{2}$.

\subsection*{Statement 6.3.1:}

$u_{0}\in L^{2}$ are given, $u_{0}=\sum_{j}u_{j}w_{j}\implies\sum_{j}e^{-\lambda_{j}t}u_{j}w_{j}$
is a weak solution of $\left(HE\right)$.

\subsection*{Corollary:}

the generator of heat semigroup is $\underbrace{\Delta}_{=\sum_{j=1}^{n}\frac{\partial^{2}}{\partial x_{i}^{2}}}$
with the definition domain $W_{0}^{1,2}\cap W^{2,2}$.

\subsubsection*{Proof of corollary:}

the statement implies that $\left(\mathcal{S}\left(t\right),L^{2}\right)\leftrightarrow\left(\widetilde{\mathcal{S}}\left(t\right),\ell^{2}\right)$,
where $\widetilde{\mathcal{S}}\left(t\right)$ is a semigroup generated
by the operator $\widetilde{A}:\left\{ u_{j}\right\} \mapsto\left\{ \lambda_{j}u_{j}\right\} $,
we know (see ex. 6.1 and ex. 6.3): $D\left(\widetilde{A}\right)=\left\{ \sum_{j}\lambda_{j}^{2}u_{j}^{2}<\infty\right\} $
$\implies$ $D\left(A\right)=W_{0}^{1,2}\cap W^{2,2}$.

\subsubsection*{Proof of statement 6.3.1:}

we will show $\left(i\right)$ $u\left(t\right)\in L^{2}\left(I,W_{0}^{1,2}\right)$,
$\left(ii\right)$ $\frac{d}{dt}\left(u\left(t\right),v\right)+\left(\nabla u\left(t\right),\nabla w\right)=0$
weakly in $\left[0,T\right]$ for $\forall w\in W_{0}^{1,2}$ fixed.
\begin{itemize}
\item[$\left(i\right)$] $\left\Vert \nabla u\left(t\right)\right\Vert _{2}^{2}\stackrel{\mbox{6.3.2, a)}}{=}\sum_{j}\lambda_{j}e^{-2\lambda_{j}t}u_{j}^{2}$
since $u\left(t\right)=\sum_{j}e^{-\lambda_{j}t}u_{j}w_{j}$. 
\[
\int_{0}^{T}\left\Vert \nabla u\left(t\right)\right\Vert _{2}^{2}dt=\int_{0}^{T}\sum_{j}\lambda_{j}e^{-2\lambda_{j}t}u_{j}^{2}dt=\sum_{j}\underbrace{\int_{0}^{T}\lambda_{j}e^{-2\lambda_{j}t}u_{j}^{2}dt}_{\left[\frac{e^{-2\lambda_{j}t}}{-2}\right]_{0}^{T}u_{j}^{2}}\le\frac{1}{2}\left\Vert u_{0}\right\Vert _{2}^{2}
\]
Remark: from chapter 2 we know: $\frac{1}{2}\left\Vert u\left(T\right)\right\Vert _{2}^{2}+\int_{0}^{T}\left\Vert \nabla u\left(t\right)\right\Vert _{2}^{2}dt=\frac{1}{2}\left\Vert u_{0}\right\Vert _{2}^{2}$.
\item[$\left(ii\right)$] due to the density it suffices to consider $w=w_{k}$, where $k=1,2,\dots$
is arbitrary fixed. Then $\left(u\left(t\right),w\right)=\left(\sum_{j}e^{-\lambda_{j}t}u_{j}w_{j},w_{k}\right)\stackrel{\mbox{ON}}{=}e^{-\lambda_{k}t}u_{k}$.
$\left(\nabla u\left(t\right),\nabla w\right)=\left(\nabla\sum_{j}e^{-\lambda_{j}t}u_{j}w_{j},\nabla w_{k}\right)=$\\
$\sum_{j}e^{-\lambda_{j}t}u_{j}\underbrace{\left(\nabla w_{j},\nabla w_{k}\right)}_{\mbox{\ensuremath{=}}\makebox[0pt][l]{\ensuremath{\,\lambda_{k},k=1;0\text{\ensuremath{\mbox{ otherwise }\text{\dots\ OG v \ensuremath{W_{0}^{1,2}}}}}}}}$
since the sum converges also in $W_{0}^{1,2}$ \dots $=\lambda_{k}e^{-\lambda_{k}t}u_{k}$.
\end{itemize}
\end{itemize}

\subsection*{Ex. 6.4}

\addcontentsline{toc}{section}{\ \ \ \ \ \ \ \ Uniqueness of semigroup w.r.t. generator} 
\begin{enumerate}
\item Let $\varepsilon>0$ be given and $(t_{0},x_{0})\in[0,\infty)\times X$.
Since the mapping $t\in[0,\infty)\mapsto S(t)x_{0}$ is continuous,
there exists $\delta>0$ such that 
\begin{equation}
\left\Vert S(t)x_{0}-S(t_{0})x_{0}\right\Vert _{X}\leq\frac{\varepsilon}{2}\label{e6.4ineq1}
\end{equation}
for every $t\in(t_{0}-\delta,t_{0}+\delta)\cap[0,\infty)$. Moreover
we know that there exists constats $M\geq1$ and $\omega\geq0$ such
that 
\begin{equation}
\left\Vert S(t)x-S(t)x_{0}\right\Vert _{X}\leq M\mathrm{e}^{\omega t}\left\Vert x-x_{0}\right\Vert _{X}\label{e6.4ineq2}
\end{equation}
for every $t\geq0$. For $(t,x)\in((t_{0}-\delta,t_{0}+\delta)\cap[0,\infty))\times B\left(x_{0},\frac{\varepsilon}{2M\mathrm{e}^{\omega(t_{0}+\delta)}}\right)$
we obtain by (\ref{e6.4ineq1}) and (\ref{e6.4ineq2}) the following:
\begin{align*}
\left\Vert S(t)x-S(t_{0})x_{0}\right\Vert _{X} & \leq\left\Vert S(t)x-S(t)x_{0}\right\Vert _{X}+\left\Vert S(t)x_{0}-S(t_{0})x_{0}\right\Vert _{X}\\
 & \leq M\mathrm{e}^{\omega t}\left\Vert x-x_{0}\right\Vert _{X}+\frac{\varepsilon}{2}\leq M\mathrm{e}^{\omega(t_{0}+\delta)}\left\Vert x-x_{0}\right\Vert _{X}+\frac{\varepsilon}{2}\\
 & \leq\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,
\end{align*}
thus the mapping $(t,x)\in[0,\infty)\times X\mapsto S(t)x$ is continuous. 
\item Fix $T>0$ and $x\in X$. Fix an arbitrary $t_{0}\in[0,T]$. We know
that 
\[
\tilde{S}(t_{0}+h)x\longrightarrow\tilde{S}(t_{0})x,\quad h\rightarrow0,
\]
and by the first part we know that the mapping $(t,x)\in[0,\infty)\times X\mapsto S(t)x$
is continuous, hence 
\[
S(T-(t_{0}+h))(\tilde{S}(t_{0}+h)x)\longrightarrow S(T-t_{0})\tilde{S}(t_{0})x,\quad h\rightarrow0,
\]
and the mapping $y(t)$ is continuous in $t_{0}$, which was chosen
arbitrarily (in the case of $t_{0}=0$ or $t_{0}=T$ it suffices to
consider limit from the right or from the left, respectively). 
\item Denote by $(A,D(A))$ the common generator of $S(t)$ and $\tilde{S}(t)$.
Let $x\in D(A)$ and $t\in(0,T)$. The goal is to prove that $y'(t)=0$.
At first, compute the derivative from the right: 
\[
\frac{\tilde{S}(t+h)x-\tilde{S}(t)x}{h}=\tilde{S}(t)\left(\frac{\tilde{S}(h)x-x}{h}\right)\longrightarrow\tilde{S}(t)Ax,\quad h\rightarrow0_{+},
\]
as $\frac{\tilde{S}(h)x-x}{h}\longrightarrow Ax$ for $h\rightarrow0_{+}$
($x\in D(A)$) and $S(t)\in\mathcal{L}(X)$. Using the first part:
\begin{align}
 & \frac{S(T-(t+h))\tilde{S}(t+h)x-S(T-(t+h))\tilde{S}(t)x}{h}\nonumber \\
 & =S(T-(t+h))\left(\frac{\tilde{S}(t+h)x-\tilde{S}(t)x}{h}\right)\longrightarrow S(T-t)\tilde{S}(t)Ax,\quad h\rightarrow0_{+}.\label{e6.4eq1}
\end{align}
By properties of generator ($x\in D(A)$) we know that $\tilde{S}(t)x\in D(A)$,
so 
\[
\frac{\tilde{S}(t)x-S(h)\tilde{S}(t)x}{h}\longrightarrow-A\tilde{S}(t)x,\quad h\rightarrow0_{+},
\]
and once more by using the continuity from the first part we get 
\begin{align}
 & \frac{S(T-(t+h))\tilde{S}(t)x-S(T-t)\tilde{S}(t)x}{h}\nonumber \\
 & =S(T-(t+h))\left(\frac{\tilde{S}(t)x-S(h)\tilde{S}(t)x}{h}\right)\longrightarrow-S(T-t)A\tilde{S}(t)x,\quad h\rightarrow0_{+}.\label{e6.4eq2}
\end{align}
By combining (\ref{e6.4eq1}) and (\ref{e6.4eq2}) we thus obtain
\begin{align*}
 & \frac{y(t+h)-y(t)}{h}\\
 & =\frac{S(T-(t+h))\tilde{S}(t+h)x-S(T-(t+h))\tilde{S}(t)x}{h}\\
 & +\frac{S(T-(t+h))\tilde{S}(t)x-S(T-t)\tilde{S}(t)x}{h}\\
 & \longrightarrow S(T-t)\tilde{S}(t)Ax-S(T-t)A\tilde{S}(t)x=0,\quad h\rightarrow0_{+},
\end{align*}
since $\tilde{S}(t)Ax=A\tilde{S}(t)x$ from the properties of generator
it follows that $y'(t_{+})=0$.

For the derivative from the left let us write 
\begin{align*}
 & \frac{y(t-h)-y(t)}{-h}\\
 & =\frac{S(T-(t-h))\tilde{S}(t-h)x-S(T-(t-h))\tilde{S}(t)x}{-h}\\
 & +\frac{S(T-(t-h))\tilde{S}(t)x-S(T-t)\tilde{S}(t)x}{-h}\\
 & =S(T-(t-h))\left(\tilde{S}(t-h)\frac{\tilde{S}(h)x-x}{h}\right)\\
 & -S(T-t)\left(\frac{S(h)\tilde{S}(t)x-\tilde{S}(t)x}{h}\right)
\end{align*}
and then it is sufficient to apply similar reasoning.
\item Fix an arbitrary $T>0$ and $x\in D(A)$. By the second part we know
that the mapping $y(t)$ is continuous on $[0,T]$. By the third part
we know that $y'(t)=0$ for $t\in(0,T)$. Therefore $y(t)$ is constant
on $[0,T]$, which in particular means that 
\[
\tilde{S}(T)x=y(T)=y(0)=S(T)x.
\]
As $T>0$ and $x\in D(A)$ were chosen arbitrarily, we have derived
that
\[
\tilde{S}(t)x=S(t)x
\]
for every $t\geq0$ (for $t=0$ it holds trivially ) and $x\in D(A)$.
Since the generator of $c_{0}$-semigroup is densely defined, it follows
by the continuity of operators $S(t),\tilde{S}(t)$ ($t$ fixed),
that 
\[
\tilde{S}(t)x=S(t)x
\]
for $x\in X$, which concludes the proof.
\end{enumerate}
\begin{thebibliography}{10}
\bibitem{barbu}Barbu, V.: \emph{Nonlinear semigroups and differential
equations in Banach spaces}, Editura Academiei, 1976.

\bibitem{cerny}\v{C}erný, R.: \emph{PDR}, Lecture notes, MFF UK,
2015.

\bibitem{evans}Evans, E.C.: \emph{Partial differential equations,}
AMS 2010.

\bibitem{fasangova}Fa¨angová, E.: \emph{Attractor for a beam equation
with weak damping}, Diploma thesis, MFF UK, 1994.

\bibitem{gajewski}Gajewski, Gröger, Zacharias\emph{: Nichtlineare
Operatorgleichungen und Operatordifferentialgleichungen}, Berlin,
1974

\bibitem{kreuter}Kreuter, M.: \emph{Sobolev spaces of vector-valued
functions}, Diploma thesis, Ulm University, 2015.

\bibitem{lukes}Luke¨, J.: \emph{Zápisky z FA}, Lecture notes, MFF
UK, 2015.

\bibitem{vNeerven}van Neerven, J.: \emph{Stochastic Evolution Equations},
Lecture notes, ISEM, 2010.

\bibitem{pazy}Pazy, A.: \emph{Semigroups of linear operators and
applications to partial differential equation}, Springer, 1983.

\bibitem{showalter}Showalter, R. E.: \emph{Monotone operators in
Banach space and nonlinear partial differential equations}, AMS, 1997 

\bibitem{MR1886588}Arendt, W. and Batty, Ch. and Hieber, M. and Neubrander,
F.: \emph{Vector-valued Laplace transforms and Cauchy problems}, Monographs
in Mathematics, vol. 96, Birkhäuser Verlag, Berlin, 2001.
\end{thebibliography}

\part*{Appendices}

\chapter*{Properties of weak convergence}

Let $u_{n}\rightharpoonup u$ in $X$ Then the following statements
hold:
\begin{enumerate}
\item $Lu_{n}\rightharpoonup Lu$ in $Y$ if $L:X\to Y$ is linear continuous
\item $u_{n}\rightharpoonup u$ in $Y$ if $X\hookrightarrow Y$
\begin{enumerate}
\item $u_{n}\to u$ in $Y$, if $X\hookrightarrow\hookrightarrow Y$
\end{enumerate}
\end{enumerate}

\subsubsection*{Proof:}
\begin{enumerate}
\item goal: $\left\langle g,Lu_{n}\right\rangle _{Y^{*},Y}\to\left\langle g,Lu\right\rangle _{Y^{*},Y}$
for $\forall g\in Y*$ fixed. Observe: mapping $f:u\mapsto\left\langle g,Lu\right\rangle _{Y^{*},Y}\in X^{*},X\to\mathbb{R}$
(is usually denoted $\underbrace{L^{*}g}_{=f}$: $L^{*}:Y^{*}\to X^{*}$).
\\
Conclusion follows as $u_{n}\rightharpoonup u$, hence $\left\langle f,u_{n}\right\rangle \to\left\langle f,u\right\rangle $.
\item special case of 1: $X\hookrightarrow Y$ means: $Id:X\to Y$ is continuous.
\item $u_{n}\rightharpoonup u$ in $X$ $\implies\left\Vert u_{n}\right\Vert _{X}$
is bounded (Banach-Steinhaus), hence $X\hookrightarrow\hookrightarrow Y$
implies $\exists$ subsequence $\underbrace{\tilde{u}_{n}\to u}_{\implies\tilde{u}_{n}\rightharpoonup u\mbox{ in }Y}$
in $Y$, on the other hand $\tilde{u}_{n}\rightharpoonup u$ in $Y$,
hence $u=\tilde{u}$, because the weak limit is unique (Hahn-Banach).\\
We need more: $u_{n}\to u$ without taking subsequences. By contradiction:\\
$u_{n}\nrightarrow u$, hence $\exists\varepsilon>0\exists$subsequence
$\tilde{\tilde{u}}_{n}$ s. t. $\left\Vert \tilde{\tilde{u}}_{n}-u\right\Vert _{Y}\geq\varepsilon\forall n$,
we can assume $\tilde{\tilde{u}}_{n}\stackrel{(\rightharpoonup)}{\to}\tilde{\tilde{u}}$
in $Y$, clearly $\left\Vert \tilde{\tilde{u}}-u\right\Vert _{Y}\geq\varepsilon,$
but on the other hand $\tilde{\tilde{u}}_{n}\rightharpoonup u$ in
$Y$ \dots{} contradiction.
\end{enumerate}

\chapter*{Time derivative of integral over time-dependent domain}

Let $e\left(t,x\right)$ be smooth.
\[
\frac{d}{dt}\int_{B\left(x_{0},\tau-t\right)}e\left(t,x\right)dx=\int_{B\left(x_{0},\tau-t\right)}\stackrel{\left(1\right)}{\partial_{t}e\left(t,x\right)}dx-\int_{\partial B\left(x_{0},\tau-t\right)}\stackrel{\left(2\right)}{e\left(t,x\right)}\,dS_{x}
\]

\subsubsection*{Proof:}

WLOG $x_{0}=0$, \textbf{TRICK:} substitution $x=\left(\tau-t\right)y$,
$dx=\left(\tau-t\right)^{n}dy$, $y\in B\left(0,1\right)$, where
$n$ is dimension. 
\[
\int_{B\left(0,\tau-t\right)}e\left(t,x\right)dx\stackrel{\text{subst.}}{=}\int_{B\left(0,1\right)}e\left(t,\left(\tau-t\right)y\right)\cdot\left(\tau-t\right)^{n}dx
\]
\begin{align*}
LHS & =\int_{B\left(0,1\right)}\partial_{t}e\left(t,\left(\tau-t\right)y\right)\cdot\left(\tau-t\right)^{n}+e\left(t,\left(\tau-t\right)y\right)\cdot n\left(\tau-t\right)^{n-1}+\nabla_{x}e\left(t,\left(\tau-t\right)y\right)\cdot\left(\tau-t\right)^{n}y\,dy\\
\text{subst. }y\to x & =\int_{B\left(0,\tau-t\right)}\underbrace{\partial_{t}e\left(t,x\right)}_{\left(1\right)}+\nabla_{x}e\left(t,x\right)\cdot\frac{x}{\tau-t}+\frac{n}{\tau-t}e\left(t,x\right)dx
\end{align*}
The two remaining terms result in $\left(2\right)$ since 
\begin{align*}
\int_{\partial B\left(x_{0},\tau-t\right)}e\left(t,x\right)\,dS_{x} & =\int_{\partial B\left(x_{0},\tau-t\right)}\left[e\left(t,x\right)\frac{x}{\tau-t}\right]\cdot\nu\,dS_{x}\\
 & =\int_{B\left(x_{0},\tau-x\right)}\text{div}_{x}\bigg[e\left(t,x\right)\underbrace{\frac{x}{\tau-t}}_{\mbox{\ensuremath{\nu}}\makebox[0pt][l]{ \dots\ outer normal}}\bigg]dx\\
\left(\text{Gauss}\right) & =\int_{\partial B\left(x_{0},\tau-t\right)}e\left(t,x\right)\underbrace{\left|\nu\right|^{2}}_{1}\,dS_{x}
\end{align*}

\chapter*{Regularity of the heat equation}

Consider the following problem:
\begin{eqnarray*}
\frac{d}{dt}u-\Delta u & = & F\left(t\right)\\
u\left(0\right) & = & u_{0}
\end{eqnarray*}

\textbf{Th. 2.3}: $u_{0}\left(t\right)\in L^{2}$, $F\left(t\right)\in L^{2}\left(I,W^{-1,2}\right)$$\implies\exists\mbox{w. s. }u\left(t\right)\in L^{\infty}\left(I,L^{2}\right)\cap L^{2}\left(I,W_{0}^{1,2}\right)$,
$\frac{d}{dt}u\left(t\right)\in L^{2}\left(I,W^{-1,2}\right)$

\textbf{Th. 2.5}: $u_{0}\left(t\right)\in W_{0}^{1,2}$, $F\left(t\right)\in L^{2}\left(I,L^{2}\right)\implies u\left(t\right)\in L^{\infty}\left(I,W_{0}^{1,2}\right)\cap L^{2}\left(I,W^{2,2}\right)$,
$\frac{d}{dt}u\left(t\right)\in L^{2}\left(I,L^{2}\right)$.

\textbf{Th. R}: let $u_{0}\left(t\right)\in W^{2m+1,2}$, $\frac{d^{k}}{dt^{k}}F\left(t\right)\in L^{2}\left(I,W^{2m-2k,2}\right)$,
$k=0,\ldots,m$ and let the compatibility conditions $\left(C_{0}\right)\ldots\left(C_{m}\right)$
hold: 
\begin{eqnarray*}
\left(C_{0}\right)\quad g_{0} & = & u_{0}\in W_{0}^{1,2}\\
\left(C_{1}\right)\quad g_{1} & = & F\left(0\right)+\Delta g_{0}\in W_{0}^{1,2}\\
 & \vdots\\
\left(C_{m}\right)\quad g_{m} & = & \frac{d^{m-1}}{dt^{m-1}}F\left(0\right)+\Delta g_{m-1}\in W_{0}^{1,2}
\end{eqnarray*}
then $\frac{d^{k}}{dt^{k}}u\left(t\right)\in L^{2}\left(I,W^{2m+2-2k,2}\right)$,
$k=0,\ldots,m+1$.

\subsubsection*{Remarks concerning the proof of Theorem R}
\begin{itemize}
\item $m=0$: assumptions: $u_{0}\in W_{0}^{1,2}$, $F\left(t\right)\in L^{2}\left(I,L^{2}\right)$,
$\left(C_{0}\right)$: $u_{0}\in W_{0}^{1,2}$. Conlusion: for $k=0$:
$u\left(t\right)\in L^{2}\left(I,W^{2,2}\right)$, $k=1$:$\frac{d}{dt}u\left(t\right)\in L^{2}\left(I,\underbrace{W^{0,2}}_{=L^{2}}\right)$,
i. e., Th. 2.5, already proven.
\item $m=1$: differentiate the equation with respect to time and use Th.
2.5 for the derivative. Denote $v=\frac{d}{dt}u$: 
\begin{eqnarray*}
\frac{d^{2}}{dt^{2}}u+\frac{d}{dt}\left(-\Delta u\right) & = & \frac{d}{dt}F\left(t\right)\\
\frac{d}{dt}v-\Delta v & = & \frac{d}{dt}F\left(t\right)\\
v\left(0\right) & = & \frac{d}{dt}u\left(0\right)\stackrel{\mbox{the equation}}{=}F\left(0\right)+\Delta u\left(0\right)
\end{eqnarray*}
assumptions: $\left(k=1\right)$ $\frac{d}{dt}F\left(t\right)\in L^{2}\Big(I,\underbrace{W^{0,2}}_{L^{2}}\Big)$,
$\left(C_{1}\right)$: $F\left(0\right)+\Delta u\left(0\right)=v\left(0\right)\in W_{0}^{1,2}$,
due to Th. 2.5. we have: $v\left(t\right)\in L^{2}\left(I,W^{2,2}\right)$,
$\frac{d}{dt}v\left(t\right)\in L^{2}\left(I,L^{2}\right)$. We want:
$\frac{d^{k}}{dt^{k}}u\in L^{2}\left(I,W^{4-2k,2}\right)$ for $k=0,1,2$.
For $k=1,2$ it is already done due to Th. 2.5. Remains to show for
$k=0$: $u\left(t\right)\in L^{2}\left(I,W^{4,2}\right)$ \dots{}
obtain from the equation for $u\left(t\right)$ and th regularity
of laplacian: $-\Delta u\left(t\right)=\underbrace{F\left(t\right)-\frac{d}{dt}u\left(t\right)}_{\in W^{2,2}}$.
I. e., $\Delta u\left(t\right)\in W^{2,2}$$\stackrel{\mbox{regularity of laplacian}}{\implies}u\left(t\right)\in W^{4,2}$
and $\left\Vert u\left(t\right)\right\Vert _{4,2}\leq c\left(\left\Vert F\left(t\right)-\frac{d}{dt}u\left(t\right)\right\Vert _{2,2}\right)$.
\item for $m\in\mathbb{N}$ general follow by induction, the induction step
is always Th. 2.5. Rigorous proof: approximation (e. g., Galerkin).
\end{itemize}

\subsection*{Remark}

Why do we need the compatibility conditions? We can not prescribe
the initial and boundary conditions completely arbitrarily. The compatibility
conditions mean correspondence of boundary and initial conditions
for $t=0$, $x\in\partial\Omega$, where these two conditions overlap.

Other equations:
\begin{enumerate}
\item Linear elliptic operator $\mathcal{A}\left(u\right)=-\text{div}\left(a\left(x\right)\nabla u\right)+b\left(x\right)\nabla u+c\left(x\right)u$
\dots{} the same techniques work for reasonable (e. g. smooth) $a\left(x\right),b\left(x\right),c\left(x\right)$
\dots{} See Evans: PDE.
\item Nonlinear operator: more difficult to prove, but similar technique
also works.
\item Other terms in equation: $\partial_{t}u-\Delta u+f\left(u\right)=h\left(t\right)$
rewrite as: $\partial_{t}u-\Delta u=\underbrace{h\left(t\right)-f\left(u\right)}_{=F\left(t\right)}$,
here additional regularity of $f\left(\cdot\right)$ is needed.
\end{enumerate}

Niemytskii operators: $u\left(x\right)\mapsto f\left(u\left(x\right)\right):$$L^{p}\left(\Omega\right)\to L^{p}\left(\Omega\right)$,
$f\left(z\right):\mathbb{R}\to\mathbb{R}$ continuous, typically $\left|f\left(z\right)\right|\leq c\left(1+\left|z\right|^{\rho}\right)$,
$p>0$.
\begin{itemize}
\item measurability $f\left(u\left(x\right)\right)$ \dots{} easy to show
by Lebesgue theory,
\item integrability: $f\left(u\left(x\right)\right)$: $u\left(x\right)\in L^{p}\left(\Omega\right)\implies f\left(u\left(x\right)\right)\in L^{q}\left(\Omega\right)$
for $\forall q$ s. t. $\rho q\leq p$,\\
proof:$\int_{\Omega}\left|f\left(u\right)\right|^{q}dx\leq\int_{\Omega}\left(c\left(1+\left|u\right|^{\rho}\right)\right)^{q}dx\leq c_{1}\int_{\Omega}1+\left|u\right|^{\rho q}dx<\infty$
(estimate $\left(a+b\right)^{q}\leq c_{q}\left(a^{q}+b^{q}\right)$
for $\forall a,b,q\geq0$),
\item actually we even have that the mapping $u\mapsto f\left(u\right)$
is continuous $L^{p}\to L^{q}$ as long as the assumptions above hold.
(See web pages \texttt{\href{http://www.karlin.mff.cuni.cz/~prazak/vyuka/Pdr2/}{http://www.karlin.mff.cuni.cz/$\sim$prazak/vyuka/Pdr2/}.})
\end{itemize}

\end{document}
