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\title{UFA, 2021--2022\\ zkou\v{s}kov\'e p\'isemky}
\author{Marek C\'uth}

\begin{document}
\noindent\begin{center}\textbf{\'Uvod do funkcion\'aln\'i anal\'yzy, ZS 2021-2022\\Zad\'an\'i p\'isemn\'e \v{c}\'asti zkou\v{s}ky - term\'in 6.1.}\end{center}

 \bigskip
 
\begin{example}[8 bod\r{u}] Na prostoru $X  = L_\infty([0,1]) \oplus_2 L_3([0,1])$ nad t\v{e}lesem komplexn\'ich \v{c}\'isel definujme funkcion\'al $\varphi:X\to \mathbb C$ p\v{r}edpisem
\[
\varphi(f,g):=\int_0^1 \big(f(t)+g(t)\big) e^{-it} \d t,\quad f\in L_\infty([0,1]),\; g\in L_3([0,1]).
\]
Uka\v{z}te, \v{z}e $\varphi\in X^*$ a ur\v{c}ete normu $\|\varphi\|$.
\end{example}

\begin{example}[15 bod\r{u}] V tomto p\v{r}\'ikladu jsou uva\v{z}ujte komplexn\'i Banachovy prostory. Nech\v{t} je d\'an oper\'ator $T:C([0,1])\to c_0$ p\v{r}edpisem
\[
Tf = \big(f(\tfrac{1}{n}) - f(0)\big)_{n=1}^\infty,\quad f\in C([0,1]).
\]
\begin{enumerate}[(i)]
\item Doka\v{z}te, \v{z}e $T$ je dob\v{r}e definovan\'y spojit\'y line\'arn\'i oper\'ator a ur\v{c}ete hodnotu $\|T\|$. (4 body)
\item Vyj\'ad\v{r}ete du\'aln\'i oper\'ator $T^*:\ell_1\to M([0,1])$ pomoc\'i reprezentace du\'al\r{u} klasick\'ych prostor\r{u}. (6 bod\r{u})
\item Ur\v{c}ete, zda je oper\'ator $T$ kompaktn\'i. (5 bod\r{u})
%\item Nalezn\v{e}te oper\'ator $S:c_0\to C([0,1])$ spl\v{n}uj\'ic\'i $TS = I$ a odvo\v{d}te, \v{z}e $c_0$ je izomorfn\'i komplementovan\'emu podprostoru $C([0,1])$.
\end{enumerate}
\end{example}

\begin{example}[27 bod\r{u}] V tomto p\v{r}\'ikladu jsou uva\v{z}ujte re\'aln\'e Banachovy prostory. Nech\v{t} je d\'an oper\'ator $T:L_2([0,1])\to L_2([0,1])$ p\v{r}edpisem
\[
Tf(t) = \int_0^1 \min(s,t) f(s) \d s,\qquad f\in L_2([0,1]).
\]
\begin{enumerate}[(i)]
    \item Uka\v{z}te, \v{z}e $T$ je spojit\'y line\'arn\'i oper\'ator, kterĂ˝ je kompaktnĂ­. (10 bod\r{u})\\
    
    \emph{Hint: Uva\v{z}ujte oper\'atory $A:L_2([0,1])\to C([0,1])$ a $B:C([0,1])\to L_2([0,1])$, kde $A$ je d\'ano stejn\'ym p\v{r}edpisem jako $T$ a $B$ je ``identita'' (p\v{r}esn\v{e}ji: $Af(t)=\int_0^1 \min(t,s)f(s) \d s$ a $Bf = [f]$). Uka\v{z}te, \v{z}e $A$ a $B$ jsou line\'arn\'i, spojit\'e  a s pomoc\'i Arzela-Ascoliho v\v{e}ty uka\v{z}te, \v{z}e $A$ je kompaktn\'i. Na z\'av\v{e}r pou\v{z}ijte identitu $T = B\circ A$.}\\
    
    \item Uka\v{z}te, \v{z}e $T$ je samoadjungovan\'y. (2 body)
    \item Uka\v{z}te, \v{z}e pro $f\in C([0,1])$ plat\'i $Tf(0)=0$, $(Tf)'(1)=0$ a $(Tf)''(t) = -f(t)$. (3 body)
    \item Uka\v{z}te, \v{z}e kdykoliv $f\in L_2([0,1])$, $\lambda\in \mathbb R$ a $Tf=\lambda f$, pak $f\in C^2([0,1])$ (2 body).
    \item Nalezn\v{e}te $\sigma(T)$ a $\sigma_p(T)$ a uka\v{z}te, \v{z}e vlastn\'i podprostory p\v{r}\'islu\v{s}n\'e nenulov\'ym vlastn\'im \v{c}\'isl\r{u}m jsou jednodimenzion\'aln\'i prostory generovan\'e funkcemi $\sin((\tfrac{\pi}{2}+k\pi)t)$ pro $k\in \mathbb N\cup \{0\}$. (6 bod\r{u})\\
    
    \emph{Pom\r{u}cka: Pokud to budete pot\v{r}ebovat, m\r{u}\v{z}ete bez v\'ypo\v{c}tu pou\v{z}\'ivat n\'asleduj\'ic\'i vzorce platn\'e pro ka\v{z}d\'e $\alpha\in\mathbb R\setminus\{0\}$ a ka\v{z}d\'e $t\in[0,1]$:
    \[
    \int_0^1 \min(s,t) \sin(\alpha s) \d s = \frac{1}{\alpha^2} \sin(\alpha t) - \frac{t}{\alpha}\cos(\alpha),
    \]
    \[
    \int_0^1 \min(s,t) \cos(\alpha s) \d s = \frac{1}{\alpha^2} \Big(\cos(\alpha t) - 1\Big) + \frac{t}{\alpha}\sin(\alpha).
    \]}
    
    \item Ur\v{c}ete p\v{r}esnou hodnotu $\|T\|$. (2 body)
    \item Z Hilbert-Schmidtovy v\v{e}ty nalezn\v{e}te diagonalizaci oper\'atoru $T$, tj. posloupnost re\'aln\'ych \v{c}\'isel $(\lambda_n)_{n=1}^\infty$ a posloupnost funkc\'i $(f_n)_{n=1}^\infty$ spl\v{n}uj\'ic\'i, \v{z}e
    \[
     Tf = \sum_{n=1}^\infty \lambda_n \Big(\int_0^1 f(t)f_n(t) \d t\Big) f_n,\qquad
f\in L_2([0,1]).    \]
    (2 body)
\end{enumerate}
\end{example}

\newpage

\textbf{NĂˇstin Ĺ™eĹˇenĂ­}

\textbf{PĹ™Ă­klad 1}: UvaĹľujme funkcionĂˇly $\varphi_1:L_\infty\to\mathbb C$ a $\varphi_2:L_3\to\mathbb C$ definovanĂ© pĹ™edpisem
\[
\varphi_1(f) = \int_0^1 f(t) e^{-it} \mathrm{d} t,\quad \varphi_2(f) = \int_0^1 f(t) e^{-it} \mathrm{d} t.
\]
Pak pro $f\in L_\infty$ mĂˇme
\[
|\varphi_1(f)|\leq \int_0^1 |f(t)| \leq \|f\|_\infty
\]
a tedy $\varphi_1\in (L_\infty)^*$ a $\|\varphi_1\|\leq 1$. Nav\'ic, pro $f(t)=e^{it}$, $t\in [0,1]$ m\'ame $|\varphi_1(f)| = \int_0^1 1 \mathrm{d} t = 1 = \|f\|_{\infty}$ a tedy $\|\varphi_1\| = 1$. Dle reprezentace $(L_3)^* = L_{3/2}$ mĂˇme $\varphi_2\in (L_3)^*$ a $\|\varphi_2\| = \|e^{-it}\|_{3/2} = 1$.

KoneÄŤnÄ›, mĂˇme $\varphi(f,g) = \varphi_1(f) + \varphi_2(g)$ a tedy dle reprezentace $(L_\infty\oplus_2 L_3)^* = (L_\infty)^*\oplus_2 L_{3/2}$ mĂˇme $\varphi\in X^*$ a $\|\varphi\| = \|(\|\varphi_1\|,\|\varphi_2\|)\|_2 = \sqrt{1^2 + 1^2} = \sqrt{2}$.\\[10pt]

\textbf{PĹ™Ă­klad 2}: Pro $f\in C([0,1])$ mĂˇme $f(\tfrac{1}{n})\to f(0)$, proto $Tf(n)\to 0$ a tedy $Tf\in c_0$. DĂˇle platĂ­ 
\[|Tf(n)|\leq |f(\tfrac{1}{n})| + |f(0)|\leq 2\|f\|_\infty,\]
a tedy $T$ je spojitĂ˝ lineĂˇrnĂ­ operĂˇtor (linearita je zĹ™ejmĂˇ) a $\|T\|\leq 2$. Pro $f(t) = 1-2t$ platĂ­ $\|Tf\| \geq |f(1) - f(0)| = 2 = 2\|f\|_\infty$ a tedy $\|T\|=2$.

Pro kaĹľdĂ© $f\in C([0,1])$ a $x\in \ell_1$ platĂ­
\[T^*x(f) = x(Tf) = \sum_{n=1}^\infty x_n Tf(n) = \sum_{n=1}^\infty x_n\big(f(\tfrac{1}{n}) - f(0)\big).\]
ProtoĹľe $\sum_{n=1}^\infty \|x_n\big(\delta(\tfrac{1}{n}) - \delta(0)\big)\|\leq \sum_{n=1}^\infty 2|x_n| \leq 2\|x\|_1 &lt;\infty$, je $\mu:=\sum_{n=1}^\infty x_n\big(\delta(\tfrac{1}{n}) - \delta(0)\big)$ dobĹ™e definovanĂ˝ prvek prostoru $M([0,1])$ a 
\[\mu(f)  = \sum_{n=1}^\infty x_n\big(\delta(\tfrac{1}{n}) - \delta(0)\big)(f) = \sum_{n=1}^\infty x_n\big(f(\tfrac{1}{n}) - f(0)\big),\quad f\in C([0,1])\]
tedy dle pĹ™edchozĂ­ho vĂ˝poÄŤtu mĂˇme $T^*x = \mu = \sum_{n=1}^\infty x_n\big(\delta(\tfrac{1}{n}) - \delta(0)\big)$ pro kaĹľdĂ© $x\in \ell_1$.

VyĹˇetĹ™eme nynĂ­ kompaktnost operĂˇtoru $T$. Pro $f_n(t) = \max\{0,1-(n+1)|1-nt|\}$ platĂ­ Ĺľe $\{f_n\neq 0\} \subset [\tfrac{1}{n+1},\tfrac{1}{n-1}]$ a $\|f_n\| = f_n(\tfrac{1}{n}) = 1$, tedy $\|Tf_n-Tf_m\|\geq |(Tf_n-Tf_m)(\tfrac{1}{n})| = |1-0|=1$ pro $n\neq m$. Tedy $T(B_{C([0,1])})$ obsahuje nekoneÄŤnou $1$-separovanou mnoĹľinu a $T$ proto nenĂ­ kompaktnĂ­.\\[10pt]

\textbf{PĹ™Ă­klad 3}: PĹ™ednÄ› si uvÄ›domĂ­me, Ĺľe pro $f\in L_2([0,1])$ je $\|f\|_1\leq \|1\|_2\cdot \|f\|_2 = \|f\|_2 &lt;\infty$ a tedy $L_2([0,1])\subset L_1([0,1])$. DĂˇle $Af$ je spojitĂˇ funkce pro kaĹľdĂ© $f\in L_2([0,1])$ dle vÄ›ty o integrĂˇlu zĂˇvislĂ©m na parametru (integrovatelnĂˇ majoranta $\min(s,t)f(s)$ je napĹ™Ă­klad $|f(s)|\in L_1$). NavĂ­c, platĂ­
\[
|Af(t)|\leq \int_0^1 |\min(s,t)f(s)| \mathrm{d} s\leq \int_0^1 |f(s)| \mathrm{d} s = \|f\|_1\leq \|f\|_2,\quad t\in [0,1],\; f\in L_2([0,1]),
\]
a tedy $A:L_2([0,1])\to C([0,1])$ je spojitĂ˝ lineĂˇrnĂ­ operĂˇtor a $\|A\|\leq 1$. DĂˇle pro $f\in B_{L_2([0,1])}$ a $t,t'\in[0,1]$ platĂ­
\[\begin{split}
|Af(t) - Af(t')| &amp; \leq \int_0^1 |\min(s,t)-\min(s,t')||f(s)| \mathrm{d} s\\ &amp;  \leq |t-t'|\int_0^1 |f(s)| \mathrm{d} s \leq |t-t'|\|f\|_1\leq |t-t'|\|f\|_2\leq |t-t'|
\end{split},\]
kde vyuĹľĂ­vĂˇme faktu Ĺľe $|\min(s,t)-\min(s,t')|\leq |t-t'|$ (coĹľ je snadnĂ© ovÄ›Ĺ™it). Tedy kaĹľdĂˇ $Af$ je $1$-LipschitzovskĂˇ a dle Arzela-Ascoliho vÄ›ty tak dostĂˇvĂˇme, Ĺľe $A$ je kompaktnĂ­ operĂˇtor.

DĂˇle pro $f\in C([0,1])$ platĂ­
\[
\|Bf\|_2 = \|f\|_2 \leq \sqrt{\int_0^1 \|f\|_\infty^2} \leq \|f\|_\infty,
\]
a tedy $B$ je spojitĂ˝ lineĂˇrnĂ­ operĂˇtor. ProtoĹľe zĹ™ejmÄ› $T = B\circ A$, je $T$ sloĹľenĂ­m kompaktnĂ­ho operĂˇtoru se spojitĂ˝m operĂˇtorem a tedy $T$ je kompaktnĂ­ operĂˇtor.

OvÄ›Ĺ™me nynĂ­, Ĺľe $T$ je samoadjungovanĂ˝ operĂˇtor. Pro $f,g\in L_2$ mĂˇme
\[
\langle Tf,g\rangle = \int_0^1 \int_0^1 \min(s,t) f(s) \mathrm{d} s g(t) \mathrm{d} t = \int_0^1 f(s) \int_0^1 \min(s,t) g(t) \mathrm{d} t \mathrm{d} s = \langle f,Tg\rangle
\]
a tedy $T$ je samoadjungovanĂ˝ (ve vĂ˝poÄŤtu jsme pouĹľili Fubiniovu vÄ›tu).

Pokud je $f\in C([0,1])$, pak mĂˇme $Tf(0) = \int_0^1 \min(s,0)f(s) = \int_0^1 0 = 0$ a takĂ©
\[
Tf(t) = \int_0^t s f(s) \mathrm{d} s + \int_t^1 t f(s) \mathrm{d} s,
\]
a tedy podle "Newton Leibnitzovy formule" pro spojitĂ© funkce mĂˇme
\[Tf'(t) = tf(t) + \Big(t\int_t^1 f(s) \mathrm{d} s\Big)'(t) = tf(t) + \int_t^1 f(s) \mathrm{d} s - tf(t) = \int_t^1 f(s) \mathrm{d} s\]
a speciĂˇlnÄ› tak $(Tf)'(1) = 0$ a takĂ©
\[Tf''(t) = \Big(\int_t^1 f(s) \mathrm{d} s\Big)'(t) = -f(t),\quad t\in [0,1].
\]

VĂ˝Ĺˇe jsme jiĹľ ovÄ›Ĺ™ili, Ĺľe pro $f\in L_2([0,1])$ je $Tf = Af$ spojitĂˇ funkce a tedy pokud $Tf=\lambda f$, pak $f$ je takĂ© spojitĂˇ funkce a dle pĹ™edchozĂ­ho pak $(Tf)'' = -f$ a tedy $Tf\in C^2([0,1])$.

Pokud je tedy splnÄ›no $Tf = \lambda f$, pak po dvojĂ­m zderivovĂˇnĂ­ dostĂˇvĂˇme $\lambda f'' = -f$. Pokud je $\lambda = 0$, pak nutnÄ› $f=0$ a tedy $\lambda\notin \sigma_p(T)$. Pro $\lambda\neq 0$ musĂ­me vyĹ™eĹˇit diferenciĂˇlnĂ­ rovnici $f'' + \tfrac{1}{\lambda} f = 0$ s podmĂ­nkami $f(0) = 0$ a $f'(1) = 0$. Pokud je $\lambda&lt;0$, pak fundamentĂˇlnĂ­ systĂ©m Ĺ™eĹˇenĂ­ je tvoĹ™en funkcemi $\{e^{t/\sqrt{|\lambda|}},e^{-t/\sqrt{|\lambda|}}\}$ a tedy $f(t) = Ae^{t/\sqrt{|\lambda|}} + Be^{-t/\sqrt{|\lambda|}}$, po dosazenĂ­ podmĂ­nky $f(0)=0$ dostĂˇvĂˇme Ĺľe $A=-B$ a po dosazenĂ­ podmĂ­nky $f'(1)=0$ dostĂˇvĂˇme $A=B$, tedy $A=B=0$ a proto $f=0$. Tedy $\sigma_p(T)\subset (0,\infty)$. Pro $\lambda &gt; 0$ je fundamentĂˇlnĂ­ systĂ©m Ĺ™eĹˇenĂ­ je tvoĹ™en funkcemi $\{\sin(t/\sqrt{\lambda}),\cos(t/\sqrt{\lambda})\}$ a tedy
\[f(t) = A\sin(t/\sqrt{\lambda}) + B\cos(t/\sqrt{\lambda})\]
a po dosazenĂ­ podmĂ­nky $f(0)=0$ dostĂˇvĂˇme, Ĺľe $B=0$. Tedy $f(t) = A\sin(t/\sqrt{\lambda})$ a po dosazenĂ­ podmĂ­nky $f'(1)=0$ dostĂˇvĂˇme, Ĺľe buÄŹ $f=0$ (takovĂ© Ĺ™eĹˇenĂ­ ale nehledĂˇme) nebo $\cos(1/\sqrt{\lambda})=0$, tedy $\tfrac{1}{\sqrt{\lambda}} = \tfrac{\pi}{2} + k\pi$ pro nÄ›jakĂ© $k\in\mathbb{N}\cup \{0\}$. DĂˇle je snadnĂ© ovÄ›Ĺ™it, Ĺľe pro kaĹľdĂ© $k\in \mathbb{N}\cup \{0\}$ je \[T\Big(\sin\big((\tfrac{\pi}{2} + k\pi)t\big)\Big) = \tfrac{1}{(\tfrac{\pi}{2} + k\pi)^2} \sin\big((\tfrac{\pi}{2} + k\pi)t\big)\] a tedy $\sigma_p(T) = \{\tfrac{1}{(\tfrac{\pi}{2} + k\pi)^2}\colon k\in \mathbb{N}\cup \{0\}\}$ a pĹ™Ă­sluĹˇnĂ© vlastnĂ­ podprostory jsou jednodimenzionĂˇlnĂ­ prostory generovanĂ© funkcemi $\sin\big((\tfrac{\pi}{2} + k\pi)t\big)$. ProtoĹľe $T$ je kompaktnĂ­ operĂˇtor, mĂˇme $\sigma(T) = \{0\}\cup \sigma_p(T)$.

ProtoĹľe $T$ je kompaktnĂ­, samoadjungovanĂ˝ a $\sigma(T)\subset [0,\infty)$, mĂˇme $\|T\| = \max\{\lambda\colon \lambda\in\sigma_p(T)\} = \tfrac{4}{\pi^2}$.

KoneÄŤnÄ›, dle Hilbert-Schmidtovy vÄ›ty mĂˇme
\[
Tf = \sum_{n=0}^\infty \lambda_n \Big(\int_0^1 f(t)f_n(t) \d t\Big) f_n,
\]
kde $\lambda_n = \tfrac{1}{(\tfrac{\pi}{2} + n\pi)^2}$ jsou nenulovĂˇ vlastnĂ­ ÄŤĂ­sla  a $f_n= \tfrac{\sin\big((\tfrac{\pi}{2} + k\pi)t\big)}{\|\sin\big((\tfrac{\pi}{2} + k\pi)t\big)\|_2}$ jsou pĹ™Ă­sluĹˇnĂ© znormovanĂ© vlastnĂ­ vektory.

\newpage

\noindent\begin{center}\textbf{\'Uvod do funkcion\'aln\'i anal\'yzy, ZS 2021-2022\\Zad\'an\'i p\'isemn\'e \v{c}\'asti zkou\v{s}ky - term\'in 17.1.}\end{center}

 \bigskip

\begin{example}[13 bod\r{u}] V re\'aln\'em Hlbertov\v{e} prostoru $\ell_2$ je d\'an dvoudimenzion\'aln\'i podprostor $Y$ generovan\'y body $(2^{-n})_{n=1}^\infty\in\ell_2$ a $(3^{-n})_{n=1}^\infty\in \ell_2$. Najd\v{e}te ortonorm\'aln\'i b\'azi $Y$ a s pomocĂ­ vzorce pro ortogonĂˇlnĂ­ projekci najd\v{e}te nejbli\v{z}\v{s}\'i bod v $Y$ k bodu $e_1 = (1,0,0,\ldots)\in \ell_2$.

\end{example}

\begin{example}[23 bod\r{u}] V tomto p\v{r}\'ikladu uva\v{z}ujte re\'aln\'e Banachovy prostory. Nech\v{t} je d\'an oper\'ator $T:L_1([0,2])\to L_1([0,2])$ p\v{r}edpisem
\[
Tf(t) = \int_0^1 f(s) \mathrm{d} s + f(t),\quad f\in L_1([0,2]).
\]
\begin{enumerate}[(i)]
    \item UkaĹľte, Ĺľe $T$ je spojitĂ˝ lineĂˇrnĂ­ operĂˇtor. (4 body)
    \item Ur\v{c}ete hodnotu $\|T\|$. (3 body)
    \item NaleznÄ›te $\sigma_p(T)$ a $\sigma(T)$. (6 bod\r{u})
    \item UrÄŤete, zda je $T$ isomorfismus a zda je $T$ kompaktnĂ­ operĂˇtor. (2 body)
    \item Vyj\'ad\v{r}ete du\'aln\'i oper\'ator $T^*$ pomoc\'i reprezentace du\'al\r{u} klasick\'ych prostor\r{u}. (8 bod\r{u})
\end{enumerate}
\end{example}

\begin{example}[14 bod\r{u}] \v{R}e\v{s}te n\'asleduj\'ic\'i \'ulohy.
\begin{enumerate}[(i)]
    \item UrÄŤete Fourierovu transformaci funkce $e^{-a|x|}$ pro kaĹľdĂ© $a&gt;0$. (5 bod\r{u})
    \item Nalezn\v{e}te funkci $h\in L_1(\mathbb{R})$ spl\v{n}uj\'ic\'i
    \[\widehat{h}(t) = \frac{1}{(t^2 + 1)(t^2+4)} ,\quad t\in\mathbb{R}.\qquad \text{(4 body)}\]
    (Hint: S pomoc\'i v\'ysledku z (i) vyj\'ad\v{r}ete funkci na lev\'e stran\v{e} rovnosti jako line\'arn\'i kombinaci funkc\'i tvaru $\widehat{e^{-a|x|}}(t)$.)
    \item Ur\v{c}ete, \v{c}emu je rovno $\hat{\hat{h}}(t)$ pro $t\in \mathbb{R}$. (2 body)\\
    (Hint: aplikujte V\v{e}tu o inverzi)
    \item Aplikujte p\v{r}edchoz\'i body a odvo\v{d}te pro ka\v{z}d\'e $x\in \mathbb{R}$ hodnotu integr\'alu
    \[
    \int_{0}^\infty \frac{\cos(tx)}{(t^2 + 1)(t^2+4)} \mathrm{d} t.\qquad \text{(3 body)}
    \]
    (Hint: vyj\'ad\v{r}ete cosinus ve tvaru exponenci\'aly a zadan\'y integr\'al ve tvaru "$const \int e^{-itx} \widehat{h}(t) \mathrm{d} t = const \cdot \hat{\hat{h}}(x)$".)
\end{enumerate}
\end{example}

\newpage

\textbf{NĂˇstin Ĺ™eĹˇenĂ­}

\textbf{PĹ™Ă­klad 4}: Nejprve nalezneme ortonormĂˇlnĂ­ bĂˇzi $Y$. Zvolme $\widetilde{x_1}:=(2^{-n})_{n=1}^\infty$, pak
\[
\|\widetilde{x_1}\|_2 = \sqrt{\sum_{n=1}^\infty 4^{-n}} = \sqrt{\frac{1/4}{1-1/4}} = \sqrt{1/3}.
\]
PoloĹľme tedy $x_1:=\sqrt{3}\cdot (2^{-n})_{n=1}^\infty$. UvaĹľujme dĂˇle
\[\begin{split}
\widetilde{x_2} &amp; = (3^{-n})_{n=1}^\infty - \langle (3^{-n})_{n=1}^\infty, x_1\rangle x_1 = (3^{-n})_{n=1}^\infty - 3\Big(\sum_{n=1}^\infty 2^{-n}3^{-n}\Big)(2^{-n})_{n=1}^\infty\\
&amp; = (3^{-n})_{n=1}^\infty - 3\frac{1/6}{1-1/6}(2^{-n})_{n=1}^\infty = (3^{-n} - \tfrac{3}{5}2^{-n})_{n=1}^\infty.
\end{split}\]
Pak $\{x_1,\widetilde{x_2}\}$ je ortogonĂˇlnĂ­ bĂˇze $Y$ a zbĂ˝vĂˇ znormovat prvek $\widetilde{x_2}$. MĂˇme
\[\begin{split}
\|\widetilde{x_2}\|_2^2 &amp; = \sum_{n=1}^\infty (3^{-n} - \tfrac{3}{5}2^{-n})^2 = \sum_{n=1}^\infty 9^{-n} - \frac{6}{5}\sum_{n=1}^\infty 6^{-n} + \frac{9}{25}\sum_{n=1}^\infty 4^{-n}\\ &amp; = \frac{1/9}{1-1/9} - \frac{6}{5}\frac{1/6}{1-1/6} + \frac{9}{25}\frac{1/4}{1-1/4} = \frac{1}{8} - \frac{6}{5.5} + \frac{3}{25} = \frac{25-24}{8.25} = \frac{1}{200},
\end{split}\]
tedy pro $x_2 = \tfrac{\widetilde{x_2}}{\|\widetilde{x_2}\|_2} = 10\sqrt{2}\widetilde{x_2}$ dostĂˇvĂˇme, Ĺľe ON bĂˇze $Y$ je $\{x_1,x_2\}$.

OrtogonĂˇlnĂ­ projekce na $Y$ je pak dĂˇna pĹ™edpisem
\[
P_Y(x) = \langle x,x_1\rangle x_1 + \langle x,x_2\rangle x_2,\quad x\in\ell_2,
\]
a tedy nejbliĹľsĂ­ bod v $Y$ k bodu $e_1$ je bod
\[\begin{split}
\langle e_1,x_1\rangle x_1 + \langle e_1,x_2\rangle x_2 &amp; = x_1(1)x_1 + x_2(1)x_2 = \tfrac{3}{2}\widetilde{x_1} + 200(\tfrac{1}{3}-\tfrac{3}{10})\widetilde{x_2}\\ &amp; = \tfrac{3}{2}\widetilde{x_1} + \tfrac{20}{3}\widetilde{x_2} = \big((\tfrac{3}{2}-\tfrac{20}{5})2^{-n} + \tfrac{20}{3}3^{-n}\big)_{n=1}^\infty = \big(-\tfrac{5}{2^{n+1}} + \tfrac{20}{3^{n+1}}\big)_{n=1}^\infty
\end{split}.\]

\textbf{PĹ™Ă­klad 5}: Je snadnĂ© si rozmyslet, Ĺľe $T$ je lineĂˇrnĂ­. DĂˇle pro kaĹľdĂ© $f\in L_1([0,2])$ mĂˇme
\[
\|Tf\|_1 = \int_0^2 \Big|f(t) + \int_0^1 f(s) \d s\Big|\d t \leq \|f\|_1 + \int_0^2 \int_0^1 |f(s)| \d s\d t \leq \|f\|_1 + \int_0^2 \|f\|_1 \d t  = 3\|f\|_1,
\]
tedy $T$ je spojitĂ˝ a $\|T\|\leq 3$.

UvaĹľujme nynĂ­ $f = \chi_{[0,1]}$. Pak $\|f\|_1 = 1$ a 
\[\|Tf\|_1 = \int_0^2\Big|\int_0^1 1 \d s + f(t)\Big| \d t = \int_0^2 1 + \chi_{[0,1]}(t) \d t = 2 + \int_0^2 \chi_{[0,1]}(t) \d t = 3,\]
tedy $\|T\|=3$ a normy se nabĂ˝vĂˇ.

UvÄ›domme si nynĂ­, Ĺľe $T = I + S$, kde $S:L_1([0,2])\to L_1([0,2])$ je koneÄŤnÄ›-dimenzionĂˇlnĂ­ operĂˇtor danĂ˝ pĹ™edpisem $S(f)(t) = \int_0^1 f(s) \d s$ (tedy v oboru hodnot $S$ jsou pouze konstantnĂ­ funkce, tj. obor hodnot $S$ je jednodimenzionĂˇlnĂ­).

Pro urÄŤenĂ­ $\sigma_p(S)$ Ĺ™eĹˇme rovnici $Sf = \lambda f$. Pokud je $\lambda = 0$, pak napĹ™Ă­klad $f = \chi_{[0,1/2]} - \chi_{1/2,1}$ je nenulovĂ˝m Ĺ™eĹˇenĂ­m tĂ©to rovnice a tedy $0\in \sigma_p(S)$. Pokud $\lambda\neq 0$, pak $f = Sf/\lambda$ a tedy $f$ musĂ­ bĂ˝t konstantnĂ­ funkce, dejme tomu $f(t) = c$, $t\in [0,2]$. Pak ale $\lambda c = \lambda f(t)$ a zĂˇroveĹ $Sf(t) = c$ a tedy nutnÄ› nenulovĂ© Ĺ™eĹˇenĂ­ rovnice $Sf = \lambda f$ existuje, prĂˇvÄ› kdyĹľ $\lambda = 1$. Celkem tedy $\sigma_p(S) = \{0,1\}$ a protoĹľe $S$ je kompaktnĂ­ (neboĹĄ je koneÄŤnÄ›-dimenzionĂˇlnĂ­), dostĂˇvĂˇme $\sigma_p(S) = \sigma(S) = \{0,1\}$.

Pro kaĹľdĂ© $\lambda\in \mathbb{R}$ mĂˇme $T-\lambda I = S - (\lambda - 1)I$ a tedy snadno dostĂˇvĂˇme, Ĺľe $\sigma_p(T) = \sigma(T) = \{1,2\}$.

ProtoĹľe $0\notin \sigma(T)$, je operĂˇtor $T$ prostĂ˝ a na, tedy isomorfismus a proto nenĂ­ kompaktnĂ­.

Pro urÄŤenĂ­ duĂˇlnĂ­ho operĂˇtoru si uvÄ›domĂ­me, Ĺľe $T^* = S^* + I^* = S^* + I$ a tedy bude 
staÄŤit urÄŤit adjungovanĂ˝ operĂˇtor $S^*$. S pouĹľitĂ­m reprezentace duĂˇlĹŻ klasickĂ˝ch prostorĹŻ dostĂˇvĂˇme, Ĺľe pro kaĹľdĂ© $f\in L_\infty([0,2])$ je $S^*f$ jedinĂˇ funkce z $L_\infty([0,2])$ splĹujĂ­cĂ­ pro kaĹľdĂ© $g\in L_1([0,2])$
\[
\int_0^1 g(t)S^*f(t) \d t = S^*f(g) = f(Sg) = \int_0^2 f(s) \int_0^1 g(t) \d t \d s = \int_0^2 g(t) \chi_{[0,1]}(t) \int_0^2 f(s) \d s \d t,
\]
 a protoĹľe zĹ™ejmÄ› $t\mapsto \chi_{[0,1]}(t) \int_0^2 f(s) \d s$ je omezenĂˇ mÄ›Ĺ™itelnĂˇ funkce, dostĂˇvĂˇme Ĺľe $S^*f(t) = \chi_{[0,1]}(t) \int_0^2 f(s) \d s$, $t\in [0,2]$. Tedy \[T^*f(t) = f(t) + \chi_{[0,1]}(t) \int_0^2 f(s) \d s, \quad t\in [0,2].\]
 
 \textbf{PĹ™Ă­klad 6}: OznaÄŤme $f_a(x) = e^{-a|x|}$ pro $x\in \mathbb{R}$ a $a&gt;0$. Pak pro kaĹľdĂ© $t\in \mathbb{R}$ s pouĹľitĂ­m sudosti funkce cosinus dostĂˇvĂˇme
 \[
 \widehat{f_a}(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-a|x|}(\cos(tx) - i\sin(tx)) \d x = \frac{2}{\sqrt{2\pi}} \int_0^\infty e^{-ax}\cos(tx) \d x,
 \]
a pokud dvakrĂˇt aplikujeme metodu per partes, dostaneme pro $t\neq 0$
\[
I = \int_0^\infty e^{-ax}\cos(tx) \d x = \ldots = \frac{a}{t^2} - \frac{a^2}{t^2}I,
\]
z ÄŤehoĹľ spoÄŤteme Ĺľe $I = \tfrac{a}{a^2 + t^2}$ a tedy
\[\widehat{f_a}(t) = \frac{2a}{(a^2 + t^2)\sqrt{2\pi}}, \quad t\neq 0\]
 a protoĹľe funkce $\widehat{f_a}$ je spojitĂˇ, vzorec platĂ­ i pro $t=0$.
 
 NynĂ­ zkusĂ­me najĂ­t $A,B\in\mathbb{R}$ splĹujĂ­cĂ­
 \[
 \frac{1}{(t^2+1)(t^2+4)} = \frac{A}{t^2+1} + \frac{B}{t^2 + 4}.
 \]
 Po pĹ™evedenĂ­ na spoleÄŤnĂ©ho jmenovatele snadno zjistĂ­me, Ĺľe rovnost vĂ˝Ĺˇe platĂ­ pro $A = \tfrac{1}{3} = -B$. Tedy,
 \[
 \frac{1}{(t^2+1)(t^2+4)} = \frac{1}{3(t^2+1)} - \frac{1}{3(t^2 + 4)} = \frac{\sqrt{2\pi}}{2}\Big(\tfrac{1}{3}\widehat{f_1}(t) - \tfrac{1}{6}\widehat{f_2}(t)\Big),\quad t\in \mathbb{R}
 \]
 a tak pro $h(x) = \frac{\sqrt{2\pi}}{6}(e^{-|x|} - \tfrac{1}{2}e^{-2|x|})$, $x\in \mathbb{R}$ dostĂˇvĂˇme, Ĺľe 
 \[\widehat{h}(t) = \frac{1}{(t^2+1)(t^2+4)},\quad t\in \mathbb{R}.\]
 
 ProtoĹľe $h\in L_1(\mathbb{R})$ a $\widehat{h}\in L_1(\mathbb{R})$ (srovnĂˇnĂ­m u nekoneÄŤna s funkcĂ­ $\tfrac{1}{t^2}$) a protoĹľe funkce $h$ je zĹ™ejmÄ› spojitĂˇ, dle VÄ›ty o inverzi dostĂˇvĂˇme Ĺľe $\hat{\hat{h}}(t) = h(-t) = h(t)$ pro kaĹľdĂ© $t\in\mathbb{R}$.
 
 KoneÄŤnÄ›, s pouĹľitĂ­m pĹ™edchozĂ­ch bodĹŻ a sudosti funkce cosinus dostĂˇvĂˇme
 \[\begin{split}
 \int_{0}^\infty \frac{\cos(tx)}{(t^2 + 1)(t^2+4)} \mathrm{d} t &amp; = \frac{1}{2}\int_{-\infty}^\infty \frac{e^{-itx}}{(t^2 + 1)(t^2+4)} \mathrm{d} t = \frac{1}{2}\int_{-\infty}^\infty e^{-itx}\widehat{h}(t) \mathrm{d} t \\ 
 &amp; = \frac{\sqrt{2\pi}}{2} \hat{\hat{h}}(x) = \frac{\sqrt{2\pi}}{2} h(x) = \frac{\pi}{6}(e^{-|x|} - \tfrac{1}{2}e^{-2|x|}), \quad x\in \mathbb{R}.
 \end{split}\]
 
 \newpage
 \noindent\begin{center}\textbf{\'Uvod do funkcion\'aln\'i anal\'yzy, ZS 2021-2022\\Zad\'an\'i p\'isemn\'e \v{c}\'asti zkou\v{s}ky - term\'in 24.1.}\end{center}

 \bigskip
 
 \begin{example}[11 bod\r{u}] V tomto p\v{r}\'ikladu uva\v{z}ujte re\'aln\'e Banachovy prostory. Nech\v{t} je d\'an oper\'ator $T:c_0\to L_3([0,1])$ p\v{r}edpisem
\[
Tx(t) = \sum_{n=1}^\infty \frac{x_n}{n^2} \exp(-nt),\quad x\in c_0.
\]
\begin{enumerate}[(i)]
    \item UkaĹľte, Ĺľe $T$ je dob\v{r}e definovan\'y spojitĂ˝ lineĂˇrnĂ­ operĂˇtor. (5 bod\r{u})
    \item Vyj\'ad\v{r}ete du\'aln\'i oper\'ator $T^*$ pomoc\'i reprezentace du\'al\r{u} klasick\'ych prostor\r{u}. (6 bod\r{u})
\end{enumerate}
\end{example}

\begin{example}[21 bod\r{u}] V tomto p\v{r}\'ikladu uva\v{z}ujte re\'aln\'e Banachovy prostory. Nech\v{t} je d\'an oper\'ator $T:C_0(\mathbb{R})\to C_0(\mathbb{R})$ p\v{r}edpisem
\[
Tf(t) = \frac{f(|t|)}{t^2 + 1},\quad t\in \mathbb{R}.
\]
\begin{enumerate}[(i)]
    \item UkaĹľte, Ĺľe $T$ je spojitĂ˝ lineĂˇrnĂ­ operĂˇtor. (2 body)
    \item Ur\v{c}ete $\|T\|$. (2 body)
    \item NaleznÄ›te $\sigma_p(T)$. (5 bod\r{u})
    \item NaleznÄ›te $\sigma(T)$. (10 bod\r{u})
    \item UrÄŤete, zda je $T$ isomorfismus a zda je $T$ kompaktnĂ­ operĂˇtor. (2 body)
\end{enumerate}
\end{example}

\begin{example}[18 bod\r{u}] V tomto p\v{r}\'ikladu uva\v{z}ujeme konvoluci vzhledem k m\'i\v{r}e z definice Fourierovy transformace, tj. $\mu = \tfrac{1}{\sqrt{2\pi}}\lambda$. Nech\v{t} jsou d\'any funkce $f_n = \chi_{[-n,n]}$, $n\in \mathbb{N}$.
\begin{enumerate}[(i)]
    \item Spo\v{c}t\v{e}te hodnotu $(f_n * f_1)(0)$, $(f_n * f_1)(x)$ pro $|x|&gt;n+1$ a doka\v{z}te, \v{z}e $\|f_n * f_1\|_\infty\leq \tfrac{2}{\sqrt{2\pi}}$ pro ka\v{z}d\'e $n\in\mathbb{N}$. (3 body)
    \item Doka\v{z}te, \v{z}e
    \[\widehat{f_n * f_1}(x) = \frac{2}{\pi}\frac{\sin(xn)\sin(x)}{x^2},\quad x\in \mathbb{R}\setminus\{0\}.\]
    (4 body)\\
    (Hint: pou\v{z}ijte vztah mezi konvoluc\'i a Fourierovou transformac\'i)
    \item Polo\v{z}me $g_n = \widehat{f_n * f_1}$ pro ka\v{z}d\'e $n\in \mathbb{N}$. Uka\v{z}te, \v{z}e $g_n\in L_1(\er)$, $n\in\mathbb{N}$. (2 body)
    \item Doka\v{z}te, \v{z}e $\|g_n\|_{L_1}\to \infty$. (5 bod\r{u})\\
    (Hint: Uka\v{z}te, \v{z}e dokonce $\int_0^1 |g_n(t)| \mathrm{d} t \to \infty$. K tomu \'u\v{c}elu nejprve doka\v{z}te nerovnost $\sin t\geq  t \sin 1$, $t\in [0,1]$ a tuto nerovnost pak pou\v{z}ijte spole\v{c}n\v{e} se zn\'am\'ym faktem, \v{z}e $\int_1^\infty \tfrac{|\sin x|}{x} \mathrm{d} x = \infty$.)
    \item S pomoc\'i v\v{e}ty o inverzi uka\v{z}te, \v{z}e $\|\widehat{g_n}\|_\infty \leq \tfrac{2}{\sqrt{2\pi}}$ pro ka\v{z}d\'e $n\in\mathbb{N}$. (2 body)
    \item Doka\v{z}te, \v{z}e Fourierova transformace jako\v{z}to zobrazen\'i z $L_1(\mathbb{R})$ do $C_0(\mathbb{R})$ nen\'i surjektivn\'i. (2 body)\\
    (Hint: pouĹľijte vĂ˝Ĺˇe zjiĹˇtÄ›nĂ© vlastnosti funkcĂ­ $g_n$ a jeden z dĹŻsledkĹŻ VÄ›ty o otevĹ™enĂ©m zobrazenĂ­.)
\end{enumerate}
\end{example}

\newpage
\textbf{NĂˇstin Ĺ™eĹˇenĂ­}

\textbf{PĹ™Ă­klad 7}: Pro kaĹľdĂ© $x\in c_0$ a $t\in [0,1]$ platĂ­
\[
|Tx(t)| \leq \sum_{n=1}^\infty \tfrac{|x_n|}{n^2}\exp(-nt)\leq \|x\|_\infty\cdot \sum_{n=1}^\infty \tfrac{1}{n^2} &lt; \infty,
\]
tedy $Tf(t)$ je dobĹ™e definovanĂ© reĂˇlnĂ© ÄŤĂ­slo a navĂ­c
\[
\|Tx\|_3^3 = \int_0^1 |Tx(t)|^3 \d t \leq \|x\|_\infty^3\cdot \Big(\sum_{n=1}^\infty \tfrac{1}{n^2}\Big)^3 \int_0^1 1 \d t = \|x\|_\infty^3\cdot \Big(\sum_{n=1}^\infty \tfrac{1}{n^2}\Big)^3,
\]
tedy $\|Tx\|_3\leq \|x\|_\infty\cdot \sum_{n=1}^\infty \tfrac{1}{n^2}$. NynĂ­ je snadnĂ© si uvÄ›domit, Ĺľe $T$ je spojitĂ˝ lineĂˇrnĂ­ operĂˇtor a $\|T\|\leq \sum_{n=1}^\infty \tfrac{1}{n^2}$.

S pouĹľitĂ­m reprezentace duĂˇlĹŻ klasickĂ˝ch prostorĹŻ dostĂˇvĂˇme, Ĺľe pro kaĹľdĂ© $f\in L_{3/2}([0,1])$ je $T^*f$ posloupnost z $\ell_1 = (c_0)^*$ a protoĹľe mnoĹľina $\{e_n\colon n\in\N\}$ je lineĂˇrnÄ› hustĂˇ v $c_0$, je $T^*f$ jedinĂˇ posloupnost z $\ell_1$ splĹujĂ­cĂ­ pro kaĹľdĂ© $n\in\N$
\[
T^*f(n) = T^*f(e_n) = f(Te_n) = \int_0^1 f(t) (Te_n)(t) \d t = \int_0^1 \frac{f(t)\exp(-nt)}{n^2} \d t.
\]
Celkem tedy dostĂˇvĂˇme, Ĺľe $T^*f = \Big(\int_0^1 \frac{f(t)\exp(-nt)}{n^2} \d t\Big)_{n=1}^\infty$.

\textbf{PĹ™Ă­klad 8}: Pro kaĹľdĂ© $f\in C_0(\er)$ je zĹ™ejmÄ› $Tf$ spojitĂˇ funkce a $\lim_{t\to \pm \infty}Tf(t) = 0$, tedy $Tf\in C_0(\er)$. Je snadnĂ© si uvÄ›domit, Ĺľe operĂˇtor $T$ je lineĂˇrnĂ­ a Ĺľe platĂ­
\[
\|Tf\|_\infty\leq \|f\|_\infty\cdot \|\tfrac{1}{t^2+1}\|_\infty  = \|f\|_\infty,
\]
tedy $T$ je spojitĂ˝ lineĂˇrnĂ­ operĂˇtor a $\|T\|\leq 1$. Na druhou stranu, pro libovolnou $f\in C_0(\er)$ splĹujĂ­cĂ­ $\|f\|_\infty = 1 = f(0)$ (napĹ™Ă­klad $f(t) = \tfrac{1}{t^2+1}$) mĂˇme $\|Tf\|_\infty \geq |Tf(0)| = 1$, tedy $\|T\|=1$.

Pro urÄŤenĂ­ $\sigma_p(T)\subset [-1,1]$ Ĺ™eĹˇme pro $\lambda\in [-1,1]$ rovnici
\begin{equation}\label{eq:pointSpectrum}
\lambda f(t) = Tf(t) = \frac{f(|t|)}{1+t^2}.
\end{equation}
Pokud existuje $t_0&gt;0$ splĹujĂ­cĂ­ $f(t_0)\neq 0$, pak ze spojitosti $f$ najdeme otevĹ™enĂ˝ interval $I\subset (0,\infty)$ na kterĂ©m je funkce $f$ nenulovĂˇ a po vydÄ›lenĂ­ rovnice vĂ˝Ĺˇe ÄŤĂ­slem $f(t)$ na intervalu $I$ dostĂˇvĂˇme $\lambda = \frac{1}{1+t^2}$, $t\in I$. To ale nenĂ­ moĹľnĂ©, neboĹĄ $\lambda$ je konstanta na intervalu $I$. Tedy kaĹľdĂ© Ĺ™eĹˇenĂ­ $f$ rovnice \eqref{eq:pointSpectrum} splĹuje Ĺľe $f|_{(0,\infty)}\equiv 0$ a tedy $\lambda f(t) = 0$, $t\in \er$. Pokud je $\lambda\neq 0$, pak $f\equiv 0$ a tedy $\lambda \notin \sigma_p(T)$. Pokud je $\lambda = 0$, pak napĹ™Ă­klad $f(t) = \max\{0,1-|t+1|\}$ je nenulovĂ˝m Ĺ™eĹˇenĂ­m a tedy $\sigma_p(T) = \{0\}$.

Pro urÄŤenĂ­ $\sigma(T)$ si nejprve uvÄ›domme, Ĺľe $\sigma(T)\subset B(0,\|T\|) = [-1,1]$. Zafixujme $g\in C_0(\er)$ a Ĺ™eĹˇme pro $\lambda\in [-1,1]$ rovnici
\begin{equation}\label{eq:spectrum}
    g(t) = Tf(t) - \lambda f(t) = \frac{f(|t|)}{1+t^2} - \lambda f(t).
\end{equation}
Pro $t\geq 0$ tak dostĂˇvĂˇme $g(t) = f(t)(\tfrac{1}{t^2 + 1} - \lambda)$ a tedy $f(t) = g(t)\Big(\tfrac{1}{t^2 + 1} - \lambda\Big)^{-1}$ pro $t\in [0,\infty)\cap \{t\colon \tfrac{1}{1+t^2} \neq \lambda\}$. UvÄ›domme si, Ĺľe $\operatorname{Rng}(\tfrac{1}{t^2 + 1}) = (0,1]$ a $\tfrac{1}{t^2+1} = 1$ prĂˇvÄ› kdyĹľ $t=0$, a tedy pokud $\lambda\in (0,1)$, pak existuje $t_0 &gt; 0$ splĹujĂ­cĂ­ $\tfrac{1}{1+t_0^2} = \lambda$ a protoĹľe funkce $g$ je omezenĂˇ, dostĂˇvĂˇme pak
\[
\lim_{t\to t_0} |f(t)| = \lim_{t\to t_0} \Big|g(t)\Big(\tfrac{1}{t^2 + 1} - \lambda\Big)^{-1}\Big| = \infty
\]
a proto $f$ nenĂ­ omezenĂˇ funkce, speciĂˇlnÄ› $f\notin C_0(\er)$ a tedy neexistuje v $C_0(\er)$ Ĺ™eĹˇenĂ­ rovnice \eqref{eq:spectrum} a proto $\sigma(T)\supset (0,1)$. ProtoĹľe $\sigma(T)$ je uzavĹ™enĂˇ mnoĹľina, dostĂˇvĂˇme $[0,1]\subset \sigma(T)$. Na druhou stranu, zvolme $\lambda\in [-1,0)$. Dle vĂ˝Ĺˇe zjiĹˇtÄ›nĂ©ho kaĹľdĂ© Ĺ™eĹˇenĂ­ rovnice \eqref{eq:spectrum} splĹuje $f(t) = g(t)\Big(\tfrac{1}{t^2 + 1} - \lambda\Big)^{-1}$ pro $t\in [0,\infty)$. Pro $t&lt;0$ zase s pouĹľitĂ­m pĹ™edchozĂ­ho dostĂˇvĂˇme 
\[g(t) = \frac{f(-t)}{1+t^2} - \lambda f(t) = \frac{g(-t)}{1+t^2}\Big(\frac{1}{t^2 + 1} - \lambda\Big)^{-1} - \lambda f(t) = \frac{g(-t)}{1 - \lambda(1+t^2)} - \lambda f(t),\]
a tedy jedinĂ˝m Ĺ™eĹˇenĂ­m rovnice \eqref{eq:spectrum} je funkce
\[
f(t) = \begin{cases} g(t)\Big(\tfrac{1}{t^2 + 1} - \lambda\Big)^{-1} &amp; t\geq 0,\\
\frac{1}{\lambda}\Big(\frac{g(-t)}{1 - \lambda(1+t^2)} - g(t)\Big) &amp; t&lt;0.
\end{cases}
\]
ProtoĹľe $\lambda\notin\operatorname{Rng}(\tfrac{1}{t^2+1})$, je $f$ spojitĂˇ funkce na $\er\setminus\{0\}$ a snadno spoÄŤteme, Ĺľe $\lim_{t\to 0^+} f(t) = f(0) = g(0)(1-\lambda)^{-1}$, $\lim_{t\to 0^-} f(t) = \tfrac{g(0)}{\lambda}(\tfrac{1}{1-\lambda} - 1)$ a tak snadno nahlĂ©dneme Ĺľe $\lim_{t\to 0^+} f(t) = f(0) = \lim_{t\to 0^-} f(t)$, tedy $f$ je spojitĂˇ funkce na $\er$. NavĂ­c, snadno spoÄŤteme Ĺľe
\[
\lim_{t\to\infty} f(t) = (0-\lambda)^{-1}\lim_{t \to\infty} g(t) = 0
\]
a
\[
\lim_{t\to -\infty} f(t) = \tfrac{1}{\lambda} \Big(0\cdot \lim_{t\to \infty} g(t) - \lim_{t\to -\infty} g(t)\Big) = 0,
\]
a tedy $f\in C_0(\er)$ je Ĺ™eĹˇenĂ­m rovnice \eqref{eq:spectrum} a $\lambda\notin \sigma(T)$. Celkem tedy $\sigma(T) = [0,1]$.

ProtoĹľe $0\in \sigma_p(T)$, operĂˇtor $T$  nenĂ­ prostĂ˝ a tedy nenĂ­ isomorfismus. ProtoĹľe spektrum operĂˇtoru $T$ nenĂ­ spoÄŤetnĂ©, operĂˇtor $T$ nenĂ­ kompaktnĂ­.

\textbf{PĹ™Ă­klad 9}: Pro kaĹľdĂ© $x\in \er$ mĂˇme
\[
(f_n * f_1)(x) = \frac{1}{\sqrt{2\pi}}\int_{-n}^n \chi_{[-1,1]}(x-y) \d y = \frac{1}{\sqrt{2\pi}}\int_{[-n,n]\cap [x-1,x+1]} \d y = \frac{1}{\sqrt{2\pi}} \lambda([-n,n]\cap [x-1,x+1]).
\]
Tedy, $(f_n * f_1)(0) = \frac{2}{\sqrt{2\pi}}$, $(f_n * f_1)(x) = 0$ pro $|x|&gt;n+1$ a koneÄŤnÄ›
\[\|f_n * f_1\|_\infty\leq \frac{1}{\sqrt{2\pi}} \sup_{x\in \er} \lambda([x-1,x+1]) = \frac{2}{\sqrt{2\pi}},\qquad n\in \N.\]

DĂˇle platĂ­ $\widehat{f_n * f_1} = \widehat{f_n}\cdot \widehat{f_1}$. SpoÄŤtÄ›me tedy nejprve $\widehat{f_n}$ (ve vĂ˝poÄŤtu nĂ­Ĺľe pouĹľĂ­vĂˇme sudost funkce cosinus a lichost funkce sinus):
\[
\widehat{f_n}(t) = \frac{1}{\sqrt{2\pi}}\int_{-n}^n e^{-itx} \d x = \frac{2}{\sqrt{2\pi}}\int_{0}^n \cos(tx) \d x = \frac{2}{\sqrt{2\pi}}\Big[\tfrac{\sin(tx)}{t}\Big]_0^n = \frac{2\sin(nt)}{t\sqrt{2\pi}},\quad t\neq 0.
\]
DostĂˇvĂˇme tak Ĺľe pro $x\neq 0$ platĂ­ $\widehat{f_n * f_1}(x) = \widehat{f_n}(x) \widehat{f_1}(x) = \frac{2\sin(nx)}{x\sqrt{2\pi}} \frac{2\sin(x)}{x\sqrt{2\pi}} = \frac{2}{\pi}\frac{\sin(nx)\sin(x)}{x^2}$.

ProtoĹľe Fourierova transformace je zobrazenĂ­ do $C_0(\er)$, je funkce $|g_n|=|\widehat{f_n * f_1}|$ spojitĂˇ a tedy je integrovatelnĂˇ, prĂˇvÄ› kdyĹľ je integrovatelnĂˇ "u $\pm \infty$". ProtoĹľe $|g_n(x)|\leq \tfrac{2}{\pi x^2}\in L_1(\er\setminus [-1,1])$, ze srovnĂˇvacĂ­ho kritĂ©ria vidĂ­me, Ĺľe $|g_n|$ je integrovatelnĂˇ "u $\pm \infty$" a tedy $g_n\in L_1(\er)$.

OdhadnÄ›me nynĂ­ $\|g_n\|_{L_1}$. PĹ™ednÄ› metodami z prvnĂ­ho semestru matematickĂ© analĂ˝zy ovÄ›Ĺ™Ă­me, Ĺľe $\sin t\geq t\sin 1$ pro $t\in [0,1]$ (plyne napĹ™Ă­klad z toho, Ĺľe funkce $\sin t$ je konkĂˇvnĂ­ na $[0,1]$). Pak odhadneme
\[\begin{split}
\|g_n\|_1 &amp; \geq \frac{2}{\pi}\int_0^1 \frac{|\sin(nx)\sin(x)|}{x^2} \d x  = \frac{2}{\pi}\int_0^n \frac{n|\sin(t)|\sin(\tfrac{t}{n})}{t^2} \d t \geq \frac{2\sin 1}{\pi}\int_0^n \frac{|\sin(t)|}{t} \d t\\
&amp; \geq \frac{2\sin 1}{\pi}\int_1^n \frac{|\sin(t)|}{t} \d t \stackrel{n\to\infty}{\to} \frac{2\sin 1}{\pi}\int_0^\infty \frac{|\sin(t)|}{t} \d t = \infty,
\end{split}\]
tedy $\|g_n\|_1\to \infty$.

ProtoĹľe $\widehat{f_n * f_1} = g_n\in L_1$, z vÄ›ty o inverzi dostĂˇvĂˇme $\widehat{g_n}(x) = \hat{\hat{f_n * f_1}}(x) = (f_n * f_1)(-x)$ s.v. a ze spojitosti funkce $\widehat{g_n}$ tak dostĂˇvĂˇme, Ĺľe $\|\widehat{g_n}\|_\infty \leq \|f_n * f_1\|$ a s pouĹľitĂ­m jiĹľ dokĂˇzanĂ©ho bodu (i) tak platĂ­ $\|\widehat{g_n}\|_\infty\leq \tfrac{2}{\sqrt{2\pi}}$ pro kaĹľdĂ© $n\in\N$.

VĂ­me, Ĺľe Fourierova transformace je prostĂ© lineaĂˇrnĂ­ zobrazenĂ­ $L_1$ do $C_0(\er)$. Kdyby bylo surjektivnĂ­, pak by se jednalo o isomorfismus (dle jednoho z dĹŻsledkĹŻ VÄ›ty o otevĹ™enĂ©m zobrazenĂ­). Ale to by byl spor neboĹĄ vĂ˝Ĺˇe bylo ukĂˇzĂˇno, Ĺľe $\|\widehat{g_n}\|_\infty\leq \tfrac{2}{\sqrt{2\pi}}$ kdeĹľto $\|g_n\|_1\to \infty$, tedy nemĹŻĹľe existovat konstanta $c$ splĹujĂ­cĂ­ $\|g_n\|_1\leq c\|g_n\|_\infty$, $n\in \N$.

\newpage

\noindent\begin{center}\textbf{\'Uvod do funkcion\'aln\'i anal\'yzy, ZS 2021-2022\\Zad\'an\'i p\'isemn\'e \v{c}\'asti zkou\v{s}ky - term\'in 31.1.}\end{center}

 \bigskip

\begin{example}[14 bod\r{u}]
V tomto p\v{r}\'ikladu uva\v{z}ujte re\'aln\'e Banachovy prostory. Nech\v{t} je d\'an oper\'ator $T:C([0,1])\to L_2([0,1],\sin t \; \mathrm{d}\lambda)$ p\v{r}edpisem
\[
Tf(t) = \sqrt{t}f(t),\quad f\in C([0,1]).
\]
(p\v{r}ipome\v{n}me, \v{z}e m\'irou $\sin t \; \mathrm{d}\lambda$ rozumĂ­me borelovskou m\'iru $\mu$, pro kterou plat\'i $\int_0^1 f(t) \mathrm{d} \mu(t) = \int_0^1 f(t) \sin t \; \mathrm{d} t$ pro kaĹľdou $f\in C([0,1])$)
\begin{enumerate}[(i)]
    \item UkaĹľte, Ĺľe $T$ je spojitĂ˝ lineĂˇrnĂ­ operĂˇtor. (4 body)
    \item Vyj\'ad\v{r}ete du\'aln\'i oper\'ator $T^*$ pomoc\'i reprezentace du\'al\r{u} klasick\'ych prostor\r{u}. (5 bod\r{u})
    \item UvaĹľujme podprostor \[Z = \{f\in C([0,1])\colon f|_{[0,\tfrac{1}{2}]} = 0\}\subset C([0,1]).\] Je $T|_Z$ isomorfismus? (5 bod\r{u})
\end{enumerate}
\end{example}

\begin{example}[27 bod\r{u}] V tomto p\v{r}\'ikladu uva\v{z}ujte komplexn\'i Banachovy prostory. Nech\v{t} je d\'an oper\'ator $T:\ell_1\to \ell_1$ p\v{r}edpisem
\[
T(x) = (\tfrac{2}{1}x_2,-\tfrac{1}{2}x_1,\tfrac{3}{2}x_4,-\tfrac{2}{3}x_3,\ldots,\tfrac{n+1}{n}x_{2n},-\tfrac{n}{n+1}x_{2n-1},\ldots),\qquad x\in \ell_1.
\]
\begin{enumerate}[(i)]
    \item UkaĹľte, Ĺľe $T$ je spojitĂ˝ lineĂˇrnĂ­ operĂˇtor. (3 body)
    \item UrÄŤete $\|T\|$. (2 body)
    \item NaleznÄ›te $\sigma_p(T)$ a pro kaĹľdĂ© $\lambda\in \sigma_p(T)$ naleznÄ›te alespoĹ jeden nenulovĂ˝ vlastnĂ­ vektor. (10 bod\r{u})
    \item NaleznÄ›te $\sigma(T)$. (10 bod\r{u})
    \item UrÄŤete, zda je $T$ isomorfismus a zda je $T$ kompaktnĂ­ operĂˇtor. (2 body)
\end{enumerate}
\end{example}

\begin{example}[9 bod\r{u}] Nech\v{t} je d\'ana funkce $f(x) = (|x|-3)\chi_{[-3,3]}(x)$, $x\in\mathbb R$.
\begin{enumerate}[(i)]
    \item NaleznÄ›te Fourierovu transformaci $\widehat{f}(t)$, $t\in\mathbb R$. (5 bod\r{u})
    \item Aplikujte p\v{r}edchoz\'i bod a s pomoc\'i Plancherelovy v\v{e}ty odvo\v{d}te hodnotu integr\'alu
    \[\int_{-\infty}^\infty \Big(\frac{\cos(3t)-1}{t^2}\Big)^2 \mathrm{d} t.\]
    (4 body)
\end{enumerate}
\end{example}
\newpage

\textbf{NĂˇstin Ĺ™eĹˇenĂ­}

\textbf{PĹ™Ă­klad 10}: Pro kaĹľdou $f\in C([0,1])$ platĂ­
\[
\|Tf\|_{L_2}^2 = \int_0^1 t f^2(t) \sin t \d t\leq \|f\|_\infty^2 \int_0^1 t\sin t \d t\leq \sin 1 \|f\|_\infty^2,
\]
tedy $Tf\in L_2([0,1],\sin t \d \lambda)$ a $T$ je dobĹ™e definovanĂ˝ spojitĂ˝ lineĂˇrnĂ­ operĂˇtor splĹujĂ­cĂ­ $\|T\|\leq \sqrt{\sin 1}$.

S pouĹľitĂ­m reprezentace duĂˇlĹŻ klasickĂ˝ch prostorĹŻ dostĂˇvĂˇme, Ĺľe pro kaĹľdĂ© $f\in L_{2}([0,1], \sin t \d \lambda)$ je $T^*f\in M([0,1])$ jedinĂˇ znamĂ©nkovĂˇ mĂ­ra splĹujĂ­cĂ­ pro kaĹľdĂ© $g\in C([0,1])$
\[
    \int_0^1 g(t) \d (T^*f)(t) = T^*f(g) = g(Tf) = \int_0^1 \sqrt{t}f(t)g(t) \sin t \d t,
\]
a tedy vidĂ­me, Ĺľe $T^*f = \sqrt{t}f(t) \sin t \d \lambda$.

Pro $n\in\N$, $n\geq 2$ uvaĹľujme funkce $f_n(x):=\max\{1-n|x-1|,0\}\in S_{Z}$, $x\in [0,1]$. Pak $\{f_n\neq 0\}\subset (1-1/n,1+1/n)$ a tedy
\[
\|Tf_n\|_{L_2}^2 = \int_{1-\tfrac{1}{n}}^1 tf_n(t)^2\sin t \d t\leq \int_{1-\tfrac{1}{n}}^1 \sin 1 \d t = \frac{\sin 1}{n} \to 0,
\]
a proto $T|_Z$ nenĂ­ isomorfismus.

\textbf{PĹ™Ă­klad 11}: Pro kaĹľdĂ© $x\in \ell_1$ platĂ­ 
\[
\|Tx\|_1 = \sum_{n=1}^\infty |\tfrac{n+1}{n}x_{2n}| + |\tfrac{n}{n+1}x_{2n-1}|\leq 2^3\sum_{n=1}^\infty |x_{2n}| + |x_{2n-1}| \leq 2\|x\|_3,
\]
tedy $Tx\in \ell_1$ a snadno pak nahlĂ©dneme, Ĺľe $T$ je spojitĂ˝ lineĂˇrnĂ­ operĂˇtor a $\|T\|\leq 2$. Pro $e_2\in S_{\ell_1}$ dĂˇle platĂ­ $\|T(e_2)\|_1 = \|2e_1\|_1 = 2$, tedy $\|T\|=2$.

Pro urÄŤenĂ­ $\sigma_p(T)\subset B_{\mathbb{C}}(0,2)$ Ĺ™eĹˇme pro $\lambda\in B_{\mathbb{C}}(0,2)$ rovnici $\lambda x = Tx$.
PorovnĂˇnĂ­m jednotlivĂ˝ch souĹ™adnic vidĂ­me, Ĺľe $x\in\ell_1$ je Ĺ™eĹˇenĂ­m tĂ©to rovnice, prĂˇvÄ› kdyĹľ pro kaĹľdĂ© $n\in\N$ je splnÄ›na nĂˇsledujĂ­cĂ­ soustava rovnic
\[
\lambda x_{2n-1} = \tfrac{n+1}{n}x_{2n}\quad \&amp; \quad \lambda x_{2n} = -\tfrac{n}{n+1}x_{2n-1}.
\]
Pokud je $\lambda=0$, pak dostĂˇvĂˇme $x_{2n}=x_{2n-1}=0$ a tedy nenulovĂ© Ĺ™eĹˇenĂ­ neexistuje, a proto $0\notin \sigma_p(T)$. Pro $\lambda\neq 0$ musĂ­me  pro kaĹľdĂ© $n\in\N$ Ĺ™eĹˇit soustavu rovnic danou maticĂ­
\[
\left(\begin{array}{cc|c}  
 \lambda &amp; -\tfrac{n+1}{n} &amp; 0\\  
 \tfrac{n}{n+1} &amp; \lambda &amp; 0  
\end{array}\right) \sim \left(\begin{array}{cc|c}  
 \lambda &amp; -\tfrac{n+1}{n} &amp; 0\\  
 0 &amp; 1+\lambda^2 &amp; 0  
\end{array}\right)
\]
a tedy pokud mĂˇ soustava rovnic nenulovĂ© Ĺ™eĹˇenĂ­, musĂ­ bĂ˝t $\lambda^2+1  = 0$ (tj. $\lambda = \pm i$). Tedy $\sigma_p(T)\subset \{\pm i\}$. Pro $\lambda = \pm i$ lehko spoÄŤteme, Ĺľe nenulovĂ˝m Ĺ™eĹˇenĂ­m rovnice $\lambda x=Tx$ je napĹ™Ă­klad vektor $x = (\mp i2,1,0,0,0,\ldots)$ (nebo jakĂ˝koliv jinĂ˝ vektor, pro kterĂ˝ platĂ­ $\lambda x_{2n-1} = \tfrac{n+1}{n}x_{2n}$, $n\in\N$).

Pro urÄŤenĂ­ $\sigma(T)$ zvolme $\lambda\in B(0,2)\setminus\{\pm i\}$, $y\in \ell_1$ a Ĺ™eĹˇme rovnici $\lambda x - Tx = y$ pro $x\in\ell_1$. Pak $x\in\ell_1$ je Ĺ™eĹˇenĂ­m tĂ©to rovnice, prĂˇvÄ› kdyĹľ pro kaĹľdĂ© $n\in\N$ je splnÄ›na soustava rovnic
\[
\lambda x_{2n-1} - \tfrac{n+1}{n}x_{2n} = y_{2n-1}\quad \&amp; \quad \lambda x_{2n} + \tfrac{n}{n+1}x_{2n-1} = y_{2n}.
\]
Pokud je $\lambda = 0$, pak rovnici Ĺ™eĹˇĂ­ $x = (2y_2,-\tfrac{1}{2}y_1,\ldots,\tfrac{n+1}{n}y_{2n},-\tfrac{n}{n+1}y_{2n-1},\ldots) = Ty\in\ell_1$, a tedy $0\notin \sigma(T)$. Pro $\lambda\neq 0$ musĂ­me  pro kaĹľdĂ© $n\in\N$ Ĺ™eĹˇit soustavu rovnic danou maticĂ­
\[\begin{split}
\left(\begin{array}{cc|c}  
 \lambda &amp; -\tfrac{n+1}{n} &amp; y_{2n-1}\\  
 \tfrac{n}{n+1} &amp; \lambda &amp; y_{2n}
\end{array}\right) &amp; \sim \left(\begin{array}{cc|c}  
 \lambda\tfrac{n}{n+1} &amp; -1 &amp; \tfrac{n}{n+1}y_{2n-1}\\  
 -\lambda\tfrac{n}{n+1} &amp; -\lambda^2 &amp; - \lambda y_{2n} 
\end{array}\right)\\
&amp; \sim \left(\begin{array}{cc|c}  
 \lambda\tfrac{n}{n+1} &amp; -1 &amp; \tfrac{n}{n+1}y_{2n-1}\\  
 0 &amp; -(1+\lambda^2) &amp; \tfrac{n}{n+1}y_{2n-1} - \lambda y_{2n} 
\end{array}\right)\\
&amp; \sim \left(\begin{array}{cc|c}  
-(1+\lambda^2)\lambda\tfrac{n}{n+1} &amp; 0 &amp; -\tfrac{n\lambda^2}{n+1}y_{2n-1} - \lambda y_{2n}\\  
 0 &amp; -(1+\lambda^2) &amp; \tfrac{n}{n+1}y_{2n-1} - \lambda y_{2n} 
\end{array}\right)\\
&amp; \sim \left(\begin{array}{cc|c}  
(1+\lambda^2) &amp; 0 &amp; \lambda y_{2n-1} + \tfrac{n+1}{n}y_{2n}\\  
 0 &amp; (1+\lambda^2) &amp; -\tfrac{n}{n+1}y_{2n-1} + \lambda y_{2n} 
\end{array}\right)
\end{split}\]
a tedy Ĺ™eĹˇenĂ­ rovnice je dĂˇno vektorem $x$, pro nÄ›jĹľ platĂ­ (pouĹľĂ­vĂˇme, Ĺľe $\lambda\notin\{0,\pm i\}$)
\[
x_{2n-1} = \tfrac{\lambda}{1+\lambda^2}y_{2n-1} + \tfrac{n+1}{n(1+\lambda^2)}y_{2n},\qquad
x_{2n} = - \tfrac{n}{(n+1)(1+\lambda^2)} y_{2n-1} + \frac{\lambda}{1+\lambda^2}y_{2n},
\]
coĹľ je vsutku prvek $\ell_1$, neboĹĄ
\[\begin{split}
|x_{2n-1}| + |x_{2n}| &amp; \leq \tfrac{|\lambda|}{|1+\lambda^2|}|y_{2n-1}| + \tfrac{2}{|1+\lambda^2|}|y_{2n}| + \tfrac{1}{|1+\lambda^2|}|y_{2n-1}| + \tfrac{|\lambda|}{|1+\lambda^2|}|y_{2n}|\\ 
&amp; \leq \tfrac{2}{|1+\lambda^2|}(2|y_{2n-1}| + 2|y_{2n}|) = \tfrac{4}{|1+\lambda^2|}(|y_{2n-1}| + |y_{2n}|),
\end{split}\]
a tedy $\|x\|_1\leq \tfrac{4}{|1+\lambda^2|}\|y\|_1$. Celkem tak dostĂˇvĂˇme, Ĺľe $\sigma(T) = \sigma_p(T)=\{\pm i\}$.

ProtoĹľe $0\notin \sigma(T)$, operĂˇtor $T$ je isomorfismus a tedy nenĂ­ kompaktnĂ­.

\textbf{PĹ™Ă­klad 12:} Pro $t\in\er$ platĂ­ (pouĹľĂ­vĂˇme, Ĺľe funkce $f(x)$ je sudĂˇ, a tedy $f(x)\sin x$ je lichĂˇ a $f(x)\cos x$ je sudĂˇ)
\[
\widehat{f}(t) = \frac{1}{\sqrt{2\pi}}\int_\er  f(x)e^{-itx}\d x = \frac{2}{\sqrt{2\pi}}\int_0^\infty f(x)\cos tx \d x = \frac{2}{\sqrt{2\pi}}\int_0^3 (x-3)\cos tx \d x
\]
a s pouĹľitĂ­m metody per partes pro $t\neq 0$ pak dostĂˇvĂˇme
\[
\widehat{f}(t) = \frac{2}{\sqrt{2\pi}}\Big(\tfrac{1}{t}\Big[(x-3)\sin tx\Big]_{x=0}^3 - \tfrac{1}{t}\int_0^3 \sin tx \d x\Big) = \frac{2}{t\sqrt{2\pi}}\Big[\frac{\cos tx}{t}\Big]_{x=0}^3 = \frac{2}{\sqrt{2\pi}}\frac{\cos(3t) - 1}{t^2}
\]
a pro $t=0$ mĂˇme
\[
\widehat{f}(0) = \frac{2}{\sqrt{2\pi}}\int_0^3 (x-3) \d x = \frac{2}{\sqrt{2\pi}}(\tfrac{9}{2} - 9) = -\frac{9}{\sqrt{2\pi}}.
\]

ProtoĹľe $f\in L_1\cap L_2$, aplikacĂ­ Plancherelovy vÄ›ty mĂˇme $\|\widehat{f}\|_2^2 = \|f\|_2^2$ a tedy
\[\begin{split}
\int_\er \Big(\frac{\cos(3t)-1}{t^2}\Big)^2 \d t &amp; = \frac{2\pi}{4}\int_\er \big(\widehat{f}(t)\big)^2 \d t = \frac{\pi}{2} \|f\|_2^2 = \frac{\pi}{2} \int_{-3}^3 (|x|-3)^2 \d x = \pi \int_0^3 (x-3)^2 \d x \\
&amp; = \pi \Big[\frac{(x-3)^3}{3}\Big]_0^3 = 9\pi.
\end{split}\]

\newpage

\noindent\begin{center}\textbf{\'Uvod do funkcion\'aln\'i anal\'yzy, ZS 2021-2022\\Zad\'an\'i p\'isemn\'e \v{c}\'asti zkou\v{s}ky - term\'in 7.2.}\end{center}

 \bigskip

\begin{example}[10 bod\r{u}] Na prostoru $X  = c_0 \oplus_\infty C([0,1])$ nad t\v{e}lesem re\'aln\'ych \v{c}\'isel definujme funkcion\'al $\varphi:X\to \mathbb R$ p\v{r}edpisem
\[
\varphi(x,f):=\sum_{n=1}^\infty \frac{x_n}{3^n} + \int_0^1 t f(t^2) \d t + f(0),\quad x\in c_0,\; f\in C([0,1]).
\]
Uka\v{z}te, \v{z}e $\varphi\in X^*$ a ur\v{c}ete normu $\|\varphi\|$.
\end{example}

\begin{example}[25 bod\r{u}] V tomto p\v{r}\'ikladu uva\v{z}ujte re\'aln\'e Banachovy prostory. Nech\v{t} je d\'an Hilbert\r{u}v prostor $H = L_2([0,1], t\;\mathrm{d} t)$ (p\v{r}ipome\v{n}me, \v{z}e m\'irou $t \; \mathrm{d} t$ rozumĂ­me borelovskou m\'iru $\mu$, pro kterou plat\'i $\int_0^1 f(t) \mathrm{d} \mu(t) = \int_0^1 f(t) t \; \mathrm{d} t$ pro kaĹľdou $f\in C([0,1])$). A\v{t} jsou dĂˇny funkce $f_1, f_2\in H$ p\v{r}edpisem \[f_1(t)=\sqrt{2},\quad f_2(t) = 6t-4, \quad t\in [0,1].\]
Nech\v{t} oper\'ator $T:H\to H$ je d\'an p\v{r}edpisem
\[
Tf(t) = f(t) + \langle f,f_1\rangle + \langle f,f_2\rangle t,\quad f\in L_2([0,1]).
\]
\begin{enumerate}[(i)]
    \item UkaĹľte, Ĺľe $\{f_1,f_2\}$ je ortonorm\'aln\'i syst\'em v $H$. (3 body)
    \item UkaĹľte, Ĺľe $T$ je spojitĂ˝ lineĂˇrnĂ­ operĂˇtor. (3 body)
    \item NaleznÄ›te $\sigma_p(T)$ a $\sigma(T)$. (10 bod\r{u})
    \item UrÄŤete, zda je $T$ isomorfismus a zda je $T$ kompaktnĂ­ operĂˇtor. (2 body)
    \item Vyj\'ad\v{r}ete hilbertovsky adjungovan\'y oper\'ator $T^\star$. (7 bod\r{u})
\end{enumerate}
\end{example}

\begin{example}[15 bod\r{u}] Nech\v{t} je d\'ana funkce $f(x) = e^{ix-3|x|}$, $x\in\mathbb R$.
\begin{enumerate}[(i)]
    \item NaleznÄ›te Fourierovu transformaci $\widehat{f}(t)$, $t\in\mathbb R$. (7 bod\r{u})\\
    (Hint: mÄ›lo by vĂˇm vyjĂ­t "const. $\tfrac{1}{t^2-2t+10}$")
    \item Uka\v{z}te, \v{z}e $\widehat{f}\in L_1(\mathbb{R})$ a ur\v{c}ete, \v{c}emu je rovno $\hat{\hat{f}}(t)$ pro $t\in \mathbb{R}$. (2 body)
    \item Aplikujte p\v{r}edchoz\'i body a odvo\v{d}te pro ka\v{z}d\'e $x\in \mathbb{R}$ hodnotu integr\'alu
    \[
    \int_{-\infty}^\infty \frac{\sin(tx)}{t^2 -2t + 10} \mathrm{d} t.
    \]
    (6 bod\r{u})\\
    (Hint: uv\v{e}domte si, \v{z}e $\sin(tx)$ lze napsat s pomoc\'i imagin\'arn\'i \v{c}\'asti $e^{-itx}$ a  vyj\'ad\v{r}ete zadan\'y integr\'al ve tvaru "$const \operatorname{Im}\big(\int e^{-itx} \widehat{f}(t) \mathrm{d} t\big) = const \cdot \operatorname{Im}\big(\hat{\hat{f}}(x)\big)$".)
\end{enumerate}
\end{example}

\newpage

\textbf{NĂˇstin Ĺ™eĹˇenĂ­}

\textbf{PĹ™Ă­klad 13}: UvaĹľujme funkcionĂˇly $\varphi_1:c_0 \to\mathbb R$ a $\varphi_2:C([0,1])\to\mathbb R$ definovanĂ© pĹ™edpisem
\[
\varphi_1(x) = \sum_{n=1}^\infty \frac{x_n}{3^n},\quad \varphi_2(f) = \int_0^1 tf(t^2) \d t + f(0).
\]
Pak pro $f\in C([0,1])$ mĂˇme
\[
|\varphi_2(f)|\leq \int_0^1 |tf(t^2)| \d t  + |f(0)|\leq \|f\|_\infty\int_0^1 t \d t + \|f\|_\infty = \frac{3}{2}\|f\|_\infty
\]
a tedy $\varphi_1\in (C([0,1]))^*$ a $\|\varphi_2\|\leq \tfrac{3}{2}$. Nav\'ic, pro $f(t)=1$, $t\in [0,1]$ m\'ame $|\varphi_2(f)| = \int_0^1 t \mathrm{d} t + 1 = \tfrac{3}{2} = \tfrac{3}{2}\|f\|_{\infty}$ a tedy $\|\varphi_2\| = \tfrac{3}{2}$. Dle reprezentace $(c_0)^* = \ell_1$ mĂˇme $\varphi_1\in (c_0)^*$ a $\|\varphi_1\| = \|(3^{-n})\|_{1} = \sum_{n=1}^\infty 3^{-n} = \tfrac{1}{2}$.

KoneÄŤnÄ›, mĂˇme $\varphi(f,g) = \varphi_1(f) + \varphi_2(g)$ a tedy dle reprezentace $(c_0\oplus_\infty C([0,1]))^* = (c_0)^*\oplus_1 (C([0,1]))^*$ mĂˇme $\varphi\in X^*$ a $\|\varphi\| = \|(\|\varphi_1\|,\|\varphi_2\|)\|_1 = \tfrac{1}{2} + \tfrac{3}{2} = 2$.

\textbf{PĹ™Ă­klad 14}: ProtoĹľe
\[
\langle f_1,f_2\rangle = \int_0^1 \sqrt{2}(6t-4) t \d t = \sqrt{2} \Big[6\tfrac{t^3}{3} - 4\tfrac{t^2}{2}\Big]_{t=0}^1 = 0,
\]
je $\{f_1,f_2\}$ ortogonĂˇlnĂ­ systĂ©m a protoĹľe
\[
\|f_1\|^2 = \langle f_1,f_1\rangle = \int_0^1 2 t \d t = 1,\quad \|f_2\|^2 = \langle f_2,f_2\rangle =\int_0^1 (6t-4)^2 t \d t = 4\int_0^1 9t^3 - 12t^2 + 4 t \d t = 1,
\]
je tento systĂ©m dokonce ortonormĂˇlnĂ­.

OznaÄŤme $S:L_2([0,1], t \d t)\to L_2([0,1], t \d t)$ operĂˇtor danĂ˝ pĹ™edpisem $Sf(t) = \langle f,f_1 \rangle + \langle f, f_2\rangle t$, pak pro $f\in L_2$ s pouĹľitĂ­m Cauchy-Schwartzovy nerovnosti mĂˇme
\[
|Sf(t)|\leq \|f\|_2 \|f_1\|_2 + \|f\|_2 \|f_2\|_2 t = \|f\|_2 ( 1 + t),\quad  t\in [0,1],
\]
a tedy
\[
\|Sf\|_2^2\leq \|f\|_2^2 \int_0^1 (1+t)^2 t \d t \leq  4 \|f\|_2^2 \int_0^1 \d t = 4 \|f\|_2^2.
\]
Snadno tak nahlĂ©dneme, Ĺľe $S$ je spojitĂ˝ lineĂˇrnĂ­ operĂˇtor, $\|S\|\leq 2$ a protoĹľe $T = I + S$, je $T$ takĂ© spojitĂ˝ lineĂˇrnĂ­ operĂˇtor (a $\|T\|\leq 3$).

UvÄ›domme si nynĂ­, Ĺľe $T = I + S$, kde $S$ je koneÄŤnÄ›-dimenzionĂˇlnĂ­ operĂˇtor definovanĂ˝ vĂ˝Ĺˇe (v oboru hodnot $S$ jsou funkce tvaru $a + bt$, $t\in [0,1]$, tj. obor hodnot $S$ je dvoudimenzionĂˇlnĂ­ a je generovĂˇn funkcemi $f_1$ a $f_2$ neboĹĄ $\operatorname{Span}\{1,t\} = \operatorname{Span}\{f_1,f_2\}$).

Pro urÄŤenĂ­ $\sigma_p(S)$ Ĺ™eĹˇme rovnici $Sf = \lambda f$. Pokud je $\lambda = 0$, pak libovolnĂˇ nenulovĂˇ funkce $f\in \{f_1,f_2\}^\perp$ splĹuje Ĺľe $Sf = 0$ a tedy $0\in \sigma_p(S)$ (takovĂˇ $f$ zĹ™ejmÄ› existuje, neboĹĄ v opaÄŤnĂ©m pĹ™Ă­padÄ› by $\{f_1,f_2\}$ byla ortonormĂˇlnĂ­ bĂˇze $H$ a celĂ˝ prostor $H$ by byl koneÄŤnÄ›-dimenzionĂˇlnĂ­, coĹľ zĹ™ejmÄ› nenĂ­ neboĹĄ napĹ™Ă­klad $t^2\in H\setminus \operatorname{span}\{f_1,f_2\}$). Pokud $\lambda\neq 0$, pak $f = Sf/\lambda\in \operatorname{Rng} S = \operatorname{Span}\{f_1,f_2\}$ a tedy $f$ musĂ­ bĂ˝t tvaru $f = af_1 + bf_2$ pro nÄ›jakĂ© konstanty $a,b\in \mathbb R$. Pak ale pro s.v. $t\in [0,1]$ platĂ­
\[
Sf(t) = \langle af_1 + bf_2,f_1\rangle + \langle af_1 + bf_2,f_2\rangle t = a + bt,\]
a zĂˇroveĹ
\[
\lambda f(t) = \lambda (af_1(t) + b f_2(t)) = \lambda(\sqrt{2} a + b (6t-4)) = \lambda(\sqrt{2}a - 4b + 6bt),
\]
tedy $\lambda\neq 0$ je vlastnĂ­m ÄŤĂ­slem, prĂˇvÄ› kdyĹľ existuje nenulovĂ© Ĺ™eĹˇenĂ­ soustavy rovnic
\[
a = \lambda(\sqrt{2}a - 4b) , \quad b=6\lambda b.\]
To existuje pouze pokud $\lambda = \tfrac{1}{6}$ (pak soustavu Ĺ™eĹˇĂ­ napĹ™Ă­klad $(a,b) = (\tfrac{-4}{6-\sqrt{2}},1)$), nebo pokud $b=0$ a $\lambda = \tfrac{1}{\sqrt{2}}$ (pak soustavu Ĺ™eĹˇĂ­ napĹ™Ă­klad $(a,b) = (1,0)$). Celkem tedy dostĂˇvĂˇme, Ĺľe $\sigma_p(S) = \{0,\tfrac{1}{6},\tfrac{1}{\sqrt{2}}\}$. ProtoĹľe operĂˇtor $S$ je koneÄŤnÄ›-dimenzionĂˇlnĂ­, je takĂ© kompaktnĂ­ a tedy $\sigma_p(S) = \sigma(S) = \{0,\tfrac{1}{6},\tfrac{1}{\sqrt{2}}\}$.

Pro kaĹľdĂ© $\lambda\in \mathbb{R}$ mĂˇme $T-\lambda I = S - (\lambda - 1)I$ a tedy snadno dostĂˇvĂˇme, Ĺľe $\sigma_p(T) = \sigma(T) = \{1,\tfrac{7}{6},1+ \tfrac{1}{\sqrt{2}}\}$.

ProtoĹľe $0\notin \sigma(T)$, je operĂˇtor $T$ prostĂ˝ a na, tedy isomorfismus a proto nenĂ­ kompaktnĂ­.

Pro urÄŤenĂ­ hilbertovsky adjungovanĂ©ho operĂˇtoru si uvÄ›domĂ­me, Ĺľe $T^\star = S^\star + I^\star = S^\star + I$ a tedy bude staÄŤit urÄŤit adjungovanĂ˝ operĂˇtor $S^\star$. Pro kaĹľdĂ© $g\in L_2$ je $S^\star g$ jedinĂˇ funkce z $L_2$ splĹujĂ­cĂ­ pro kaĹľdĂ© $f\in L_2$ rovnici $\langle Sf,g\rangle = \langle f,S^\star g\rangle$, kde
\[
\langle f,S^\star g\rangle = \int_0^1 t f(t) S^\star g(t) \d t.
\]
a
\[\begin{split}
\langle Sf,g\rangle &amp; = \int_0^1 tg(t)(\langle f,f_1\rangle + \langle f,f_2\rangle t) \d t = \int_0^1 t g(t) \int_0^1 f(s)(f_1(s)+tf_2(s))s \d s \d t\\
&amp; \stackrel{FUBINI}{=} \int_0^1 s f(s) \int_0^1 tg(t)f_1(s) + t^2g(t)f_2(s)\d t \d s\\
&amp; = \int_0^1 s f(s) \Big(\langle 1, g\rangle f_1(s) + \langle t, g\rangle f_2(s)\Big) \d s
\end{split}\]
a tedy vidĂ­me, Ĺľe $S^\star g(s) = \langle 1, g\rangle f_1(s) + \langle t, g\rangle f_2(s)$, $s\in [0,1]$ (jednĂˇ se opravdu o funkci z $L_2$ neboĹĄ to je lineĂˇrnĂ­ kombinace funkcĂ­ $f_1$ a $f_2$).

Tedy, pro $g\in L_2$ mĂˇme $T^\star g(s) = g(s) + \langle 1, g\rangle f_1(s) + \langle t, g\rangle f_2(s)$, $s\in [0,1]$.

\textbf{PĹ™Ă­klad 15}: Pro $t\in\mathbb R$ s pouĹľitĂ­m lichosti funkce sinus a sudosti funkce cosinus dostĂˇvĂˇme
\[
\widehat{f}(t) = \frac{1}{\sqrt{2\pi}}\int_{\er} e^{ix-3|x|}e^{-itx} \d x = \frac{1}{\sqrt{2\pi}}\int_{\er} e^{-3|x|}e^{-i(t-1)x} \d x = \frac{2}{\sqrt{2\pi}} \int_{0}^\infty e^{-3x}\cos((t-1)x) \d x
\]
a pokud dvakrĂˇt aplikujeme metodu per partes, dostaneme pro $t\neq 1$
\[
I = \int_0^\infty e^{-3x}\cos((t-1)x) \d x = \ldots = \frac{3}{(t-1)^2} - \frac{9}{(t-1)^2}I,
\]
z ÄŤehoĹľ spoÄŤteme Ĺľe $I = \tfrac{3}{9 + (t-1)^2}$ a tedy
\[
\widehat{f}(t) = \frac{6}{(9 + (t-1)^2)\sqrt{2\pi}} = \frac{6}{(t^2-2t+10)\sqrt{2\pi}}, \quad t\neq 1
\]
a protoĹľe funkce $\widehat{f}$ je spojitĂˇ, vzorec platĂ­ i pro $t=1$.

ProtoĹľe funkce $\widehat{f}$ je spojitĂˇ, je integrovatelnĂˇ, prĂˇvÄ› kdyĹľ je integrovatelnĂˇ "u $\pm \infty$". ProtoĹľe $|\widehat{f}(t)|\leq \tfrac{6}{(t-1)^2}\in L_1(\er\setminus [-10,10])$, ze srovnĂˇvacĂ­ho kritĂ©ria vidĂ­me, Ĺľe $\widehat{f}(t)$ je integrovatelnĂˇ "u $\pm \infty$" a tedy $\widehat{f}\in L_1(\er)$. ProtoĹľe funkce $f$ je spojitĂˇ, dle vÄ›ty o inverzi tak mĂˇme, Ĺľe $\hat{\hat{f}}(t) = f(-t)$ pro $t\in\er$.

KoneÄŤnÄ›, s pouĹľitĂ­m pĹ™edchozĂ­ch bodĹŻ a toho Ĺľe $\sin(tx) = -\operatorname{Im}(e^{-itx})$ dostĂˇvĂˇme pro $x\in\er$
\[\begin{split}
\int_{-\infty}^\infty \frac{\sin(tx)}{t^2 -2t + 10} \mathrm{d} t &amp; = -\frac{\sqrt{2\pi}}{6}\int_{\er} \widehat{f}(t) \operatorname{Im}(e^{-itx}) \d t = -\frac{\sqrt{2\pi}}{6}\operatorname{Im}\Big(\int_{\er} \widehat{f}(t) e^{-itx} \d t\Big)\\
&amp; = -\frac{\sqrt{2\pi}}{6}\operatorname{Im}(\sqrt{2\pi}\hat{\hat{f}}(x)) 
 = -\frac{2\pi}{6}\operatorname{Im}(f(-x)) = -\frac{\pi}{3}\operatorname{Im}(e^{-ix-3|x|})\\
 &amp; = \frac{\pi}{3}e^{-3|x|}\sin(x).
\end{split}\]
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