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\begin{center}
\textbf{201B, Winter '11, Professor John Hunter}

\textbf{Homework 8 Solutions}

\end{center}

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\begin{enumerate}

\item[1.] A sequence of bounded linear operators $A_n \in \mathcal{B}(\mathcal{H})$ on a Hilbert space $\mathcal{H}$ is said to converge to an operator $A \in \mathcal{H}$: {\it uniformly} if $A_n \rightarrow A$ with respect to the operator norm on $\mathcal{B}(\mathcal{H})$; {\it strongly} if $A_n x \rightarrow Ax$ strongly in $\mathcal{H}$ for every $x \in \mathcal{H}$. 
\begin{enumerate}
\item Give an example of a sequence of operators that converges strongly but not uniformly.
\begin{proof}
\emph{Remember we did this in 201A. Solution due to Eric!}\\
$\bullet $ Let $T_n = T^n$ where $T$ is the left shift operator on $\ell^2 (\mathbb{Z})$ given by 
 $$T(x_1,x_2,\ldots) = (x_2,x_3,\ldots).$$ 
 First, we will show that $T_n$ converges strongly to $0$. Then we will show that it doesn't converge uniformly to zero. \\
Pick any sequence $x = (x_1,x_2,\ldots)$ in $\ell^2(\mathbb{Z})$; we have that 
\begin{equation*}
x_n\rightarrow 0 \; \mbox{as} \;  n\rightarrow \infty .
\end{equation*}
Remember that the $\ell^2 (\mathbb{Z})$-norm of $T_n x$ is  given as follows:
\[||T_n x|| =  \left(\sum_{i=n}^{\infty} x_n^2\right)^{1/2} < \infty , \]
which monotonically approaches $0$ as $n\rightarrow \infty$. \\
So, $T_n$ converges strongly to the zero operator.\\ 

$\bullet$ If $T_n$ converged uniformly, it would have to agree with the strong limit we have found above i.e., it suppose to be $0$.  We can calculate the norm of $||T_n||$ by first noting that clearly $||T_n|| \leq 1$. To prove that indeed the norm of $T_{n}$ is $1$, we do the usual trick: if we take any sequence $s_n \in \ell^2(\mathbb{T})$ that begins with $n$ zeros then $||T_n s_n|| = ||s_n||$, which implies that $||T_n||\geq 1$. 
Thus, $||T_n|| = 1$ for all $n \in \mathbb{N}$.\\

Therefore we can conclude that, $T_n$ does not converge uniformly since we cannot have 
 $$||T_n||=1 \rightarrow 0.$$  \\
\
\end{proof}

\item Give an example of a sequence of operators that converges weakly but not strongly. 
\begin{proof}
\emph{Remember we did this in 201A. Solution due to Eric!}\\
$\bullet$ Let $S_n = S^n$ where $S$ is the right shift operator on $\ell^2(\mathbb{Z})$ given by $$S(x_1,x_2,\ldots) = (0,x_1,x_2,\ldots).$$ Our proof here will mirror the structure of the argument in part ($a$). First, we show that $S_n$ converges weakly to zero. Secondly, we will show that it cannot converge strongly to zero, implying that it must not converge strongly at all.\\

$\bullet$ Given any bounded linear functional $\phi$ on $\ell^2(\mathbb{Z})$ and any $x\in \ell^2(\mathbb{Z})$, we can linearly decompose the action of $\phi$ on $x$ as the action of the components $\phi_i$ of $\phi$ on the components $x_i$ of $x$ given by $\phi_i(x_j) = a_ix_j$ and find
\begin{equation*}
\begin{aligned}
\phi(S_n x) = &\phi(0,\ldots,0,x_1,x_2,\ldots) \\
=& \sum_{i=n+1}^{\infty} \phi_i(x_{i-n})\\
=& \sum_{i=n+1}^{\infty} a_ix_{i-n} \\
\leq & \left(\sum_{j=1}^{\infty} x_j^2\right)^{(1/2)} \left(\sum_{i=n+1}^{\infty} a_i^2\right)^{(1/2)},
\end{aligned}
\end{equation*}
which goes to zero since $a_i \rightarrow 0$ as $n\rightarrow \infty$. \\

Thus, $S_n$ converges weakly to 0. However, it is clear from the definition that $||S_n x|| = ||x||$ for all $n \in \mathbb{N}$. Therefore, we cannot have 
\begin{equation*}
S_n x \rightarrow 0 \; \mbox{ as} \; n\rightarrow \infty . 
\end{equation*}
Hence, $S_n$ converges weakly to $0$, but not strongly.
\\
\
\end{proof}
\end{enumerate}

 \item[2.] A subset $E$ of a vector space $X$ is said to be convex if 
\[\lambda x + (1-\lambda)y \in E \quad \forall x,y \in E, 0\leq \lambda \leq 1.\]
\begin{enumerate}
\item Show that a strongly closed, convex subset of a Hilbert space is weakly closed.
\begin{proof}
If $x_{n} \rightharpoonup x$, where $\left\lbrace  x_{n}\right\rbrace _{n} \subset E $, by \emph{Mazur's theorem}, there is a sequence of  $\left\lbrace  y_{n}\right\rbrace _{n} \subset E $ of finite convex combination of $\left\lbrace  x_{n}\right\rbrace _{n} $ such that $y_{n}\rightarrow x.$ Note that the sequence $\left\lbrace  y_{n}\right\rbrace _{n}  $ is realy a subset of $E$ because we were given that $E$ is convex! So $x\in E$, because $E$ is strongly closed. Therefore, we conclude that $E$ is weakly closed.

\end{proof}
\item Show that every strongly closed, convex subset of a Hilbert space contains a point of minimum norm.
\begin{proof}
Suppose $E$ is a closed convex set in a Hilbert space $\hs$. Let
\begin{equation*}
d=\inf_{x\in E}\|x\|.
\end{equation*}

$\bullet$ If $d=0$, then we can find a sequence $\left\lbrace x_{n}\right\rbrace _{n}$ so that $$\lim_{n}\left\|x_{n}\right\|=0.$$ Then we find $\left\lbrace x_{n}\right\rbrace _{n}$ is convergent to $0$. Therefore $0$ is a limit point of $E$. Since $E$ is closed we find $0\in E$. If $\norm x=0$, then $x=0$. Hence $0$ is the unique minimum point when $d=0$.\\

$\bullet$ Suppose $d>0$. Then we can find a sequence $\left\lbrace x_{n}\right\rbrace _{n}$ so that $\lim_{n}\left\|x_{n}\right\|=d$. \\
By the parallelogram's law, we find
\begin{equation}
\label{1}
\left\|x_{n}-x_{m}\right\|^{2}=2\left\|x_{n}\right\|^{2}+2\left\|x_{m}\right\|^{2}-4\left\|\frac{x_{n}+x_{m}}{2}\right\|^{2}.
\end{equation}
Since $E$ is convex, $x_{n}\in E$ for all $n\in\mb N$, then we find $(x_{n}+x_{m})/2\in E$. By the definition of $d$, we see that
\begin{equation}
\label{2}
d^{2}\leq \left\|\frac{x_{n}+x_{n}}{2}\right\|^{2}.
\end{equation}
(\ref{1}) and (\ref{2}) imply
\begin{equation*}
\left\|x_{n}-x_{m}\right\|^{2}\leq 2\left\|x_{n}\right\|^{2}+2\left\|x_{m}\right\|^{2}-4d^{2}.
\end{equation*}

Since $\lim_{n\to\infty}\left\|x_{n}\right\|=d$, (or $\lim_{n\to\infty}\|x_{n}\|^{2}=d^{2}$), then for each $\epsilon>0$, we can find $N_{\epsilon}>0$ so that $n\geq N_{\epsilon}$,
\begin{equation*}
0\leq \left\|x_{n}\right\|^{2}-d^{2} <\frac{\epsilon^{2}}{4},
\end{equation*}
which implies that whenever $n,m\geq N_{\epsilon}$, we have
\begin{equation*}
\left\|x_{n}-x_{m}\right\|^{2} < \left(2d^{2}+\frac{\epsilon^{2}}{2}\right)+\left(2d^{2}+\frac{\epsilon^{2}}{2}\right)-4d^{2}=\epsilon^{2}.
\end{equation*}

We proved that $\left\lbrace x_{n}\right\rbrace _{n}$ is a Cauchy sequence.\\

$\bullet$ Since $\hs$ is a Hilbert space, then we can find $x\in \hs$ so that $$\lim_{n\to\infty}x_{n}=x.$$ We find $x$ is a limit point of $E$. Since $E$ is closed, then $x\in E$. \\
We also have $$\lim_{n\to\infty}\|x_{n}\|=\|x\|=d,$$
and this is given by the continuity of the norm. Hence $x$ is indeed a minimum point.\\

$\bullet$ Suppose $y$ is another point with $\|y\|=d$. Again, using the parallelogram's law, we find
\begin{equation*}
\begin{aligned}
0\leq \|x-y\|^{2}=&2\|x\|^{2}+2\|y\|^{2}-4\left\|\frac{x+y}{2}\right\|^{2}\\
=&4d^{2}-4\left\|\frac{x+y}{2}\right\|^{2}\\
\leq& 4d^{2}-4d^{2}=0.
\end{aligned}
\end{equation*}

We find $\|x-y\|=0$, which implies that $x=y$. We conclude that the minimum point is unique.
\end{proof}

\end{enumerate}
\end{enumerate}
\end{document}