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\usepackage{amsmath, amsthm, amsfonts, amssymb}
\usepackage{enumerate}
\input{../common/math.tex}
%\input{../common/thm_en.tex}
\newtheorem{thm}{Exercise}
%opening
\title{Functional analysis problem sheet II}
\author{Max Klinger and Kilian Klebes}
\begin{document}
\maketitle
\thm{
Let $f: \RR\rightarrow \RR$ be bounded and measurable, then $\exists$ step function $f_n : \RR \rightarrow \RR$ with finitely many levels and which converge uniformly to f.
}
\proof{
Since f is bounded, we know $\exists a** 0$ and cover [a,b] with balls of that diameter, e.g. choose $y_{n,k} =a +k\epsilon$. Since [a,b] is a compact subset of $\RR$, we can choose a finite subcovering of this cover and still have all of the set covered, that means in effect that only finitely many $y_{n,k}$ (in our case $n= \lceil\frac{b-a}{\epsilon}\rceil$ are necessary). Now define $A_{n,k}= f^{-1}([y_{n,k},y_{n,k}+\epsilon])$ and $f_\epsilon = \sum^{n-1}_{k=0} y_{n,k} \1_{A_{n,k}}(x)$. These $f_n$ have n and thus finitely many levels. $\norm{f-f_n}_\infty = \sup_{x\in \RR} \abs{(f-f_n)(x)}\leq \diam{A_{n,k}}\leq \epsilon$ since $\forall x \in \RR \; \exists A_{n,k}$ such that $f(x) \in A_{n,k}$ and $\diam(A_{n,k}) = \epsilon$, providing an upper bound on the difference between the two functions.
}
\thm{
Let $l^2(\ZZ^d) = \{\phi: \ZZ^d =\rightarrow \CC: \sum_{x\in \ZZ^d}\abs{\phi(x)}^2 < \infty \}$ and $\bigtriangleup: l^2(\ZZ^d) \rightarrow l^2(\ZZ^d), \phi\mapsto \bigtriangleup(\phi)$ where $\bigtriangleup(\phi)(x) = \sum_{\stackrel{x\in \ZZ^d}{\abs{x-y}=1}}(\phi(x)-\phi(y))$
}
\proof[Solution]{
\begin{enumerate}[5.1)]
\item On the lattice $\ZZ^d$ there is always one element left and right of every x fulfilling $\abs{x-y} =1$ along every basis element, i. e. there are 2d elements.
The operator can thus be expressed as follows: $\bigtriangleup = \sum^{d}_{k=1} 2\id -R_k -L_k$, where $R_k, L_k$ is a right/left shift in the kth component of the argument.
\\Check: $$(\bigtriangleup\phi)(x) = \sum^d_{k=1} 2\phi(x) -\phi(x+e_k)-\phi(x-e_k) =$$ $$=\sum^d_{k=1} (\phi(x) -\phi(x+e_k)) +\sum^d_{k=1} (\phi(x) -\phi(x-e_k)) =$$ $$= \sum_{\stackrel{x\in \ZZ^d}{y-x=1}}(\phi(x)-\phi(y)) + \sum_{\stackrel{x\in \ZZ^d}{x-y=1}}(\phi(x)-\phi(y)) =$$ $$=\sum_{\stackrel{x\in \ZZ^d}{\abs{x-y}=1}}(\phi(x)-\phi(y))$$
\item $$\bigtriangleup^* = (\sum^{d}_{k=1} 2\id -R_k -L_k)^* = \sum^{d}_{k=1} 2\id^* -R_k^* -L_k^* = \sum^{d}_{k=1} 2\id - L_k -R_k = \bigtriangleup$$
$$\norm{\bigtriangleup} = \norm{\sum^{d}_{k=1} 2\id -R_k -L_k} \leq \sum^{d}_{k=1} 2\norm{\id} -\norm{R_k} -\norm{L_k} = \sum^d_{k=1} 4 = 4d$$
To see the other direction consider $\delta: x\mapsto \left\{\begin{array}{cl} 1,& x = 0\\ 0, & else\end{array}\right.$ Then $L \delta = 0$ as well as $R\delta$ thus $\scalprod{\bigtriangleup\phi}{\bigtriangleup\phi} = (\sum_{x\in\ZZ^d} 2\phi)^2 = 4d\norm{\phi}$ and therefore $\norm{\bigtriangleup}= 4d$.
%the following sequence:\\ $\phi: x\mapsto \left\{\begin{array}{cl} e^{x_i^2}\hat{x}_i,& x_i\text{ ungerade }\\ 0, & x_i\text{ gerade}\end{array}\right.$ Since now all components of $\phi$ except the ith are 0 the norm of $\bigtriangleup\phi$ is $(2id-L-R)(\phi)(2id-L-R)(\phi) = 4\phi -2L\phi -2R\phi -2L\phi -2R\phi +L^2\phi +R^2\phi +2LR\phi = 6 \phi -4L\phi - 4R\phi + L^2\phi +R^2 \phi$. L\phi as well as
\item It is clear that for the scalar product as given for $l^2(\ZZ^2)$ the optimum under the constraint that $\eta,\phi$ are normed lies with $\phi=\eta$, thus $\scalprod{\phi}{L\phi} \leq \norm{\phi}^2$ and an analougus statement holds for $\scalprod{\phi}{R\phi}$. From this we can conclude $\scalprod{\phi}{\bigtriangleup\phi} = 2 \norm{\phi} -\scalprod{\phi}{L\phi} -\scalprod{\phi}{R\phi} \leq 0$. From 5.2) we know $\norm{\bigtriangleup}=4d$, thus $\bigtriangleup\leq 4d$. and thus the statement holds
\item This is clear from Corollary 1.3 from the lecture and 5.3) since $M(\bigtriangleup) = 4d$ and $m(\bigtriangleup) =0$
\item Too annoying.
\end{enumerate}
}
\end{document}
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