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\begin{document}

\centerline {\bf Math 523b - Take home midterm}

\bigskip

\noindent {\bf Due Monday, March 25 at 10 am}  

\noindent {\bf Rules:}
You may not talk to anyone about the exam. If the problem is 
unclear you can ask me for clarification, but I will not give hints.
You may refer to your class notes, your homework solutions 
and Folland. But you cannot use any other books or the web.

\noindent {\bf Exposition} is important. I will take off points if 
your solution is not clearly explained or contains irrelevant arguments. 

\bigskip

\begin{enumerate} 

\item 
Let $\e$ be the set of even polynomials which we consider as 
functions on $[0,1]$. For $1 \le p \le \infty$
this is a subspace of $L^p([0,1],\borel,m)$
where $m$ is Lebesgue measure and $\borel$ are the Borel sets.
Determine (with proof) the closure of $\e$ in 
$L^p$ for $1 \le p \le \infty$. 

\item Let $(X,\M,\mu)$ be a measure space. We abbreviate 
$L^p(X,\M,\mu)$ as $L^p$. Let $1 <p,r < \infty$. 
Fix a function $f \in L^p$ and define a linear operator $T$ by 
$T g = f g $, where $g \in L^r$. Find a condition on $p,q,r$ that
implies $T g \in L^q$ for all $g \in L^r$ and prove that $T$ is 
a bounded linear operator from $L^r$ to $L^q$ when this condition holds.

\item Let $\ps$ be the set of power series $\sum_{n=0}^\infty c_n x^n$
such that 
$$
\sum_{n=0}^\infty (n+1) |c_n| \le 1
$$ 
It is easy to show that such a series converges 
uniformly on $[0,1]$ and so defines
a continuous function. So $\ps \subset C([0,1])$. 
(You don't have to show this). 
Prove that the closure of $\ps$ in $C([0,1])$ is compact. 
The norm is the usual sup norm.

\item 
Let $1<p<\infty$ and let $q$ be the conjugate exponent given 
by $1/p+1/q=1$. 
Let $b_n$ be a real valued sequence such that 
$\sum_{n=1}^\infty a_n  b_n$ converges for all real sequences $a_n$
in $l^p$. Prove that $b_n$ is in $l^q$. 
Hint: consider the linear functionals $f_N$ on $l^p$ defined by 
$$
f_N(\{a_n\}_{n=1}^\infty)= \sum_{n=1}^N b_n a_n
$$

\newpage

\item Consider $L^2=L^2([0,\infty),\borel,m)$ where $m$ is Lebesgue measure
and $\borel$ is the Borel sets.
Let 
$$
V_n = \{ f \in L^2 : f=0 \, a.e. \, on \, (n,\infty)\}
$$
($f=0 \, a.e. \, on \, (n,\infty)$ means that 
$m(\{x \in (n,\infty):f(x) \neq 0 \})=0$. )
It is not hard to show $V_n$ is closed. You may assume this fact without
proving it. Let $P_n$ be the orthogonal projection onto $V_n$. 
\begin{enumerate}
\item Find an explicit formula for $P_n$, i.e., given $ f \in L^2$
what is $P_n f$?
\item It is natural to expect that $P_n$ converges to 
the identity $I$ in some sense. 
Here are two possible senses in which it could converge.
\begin{enumerate}
\item $||P_n-I|| \ra 0$. (Here the norm is the operator norm.)
\item $P_n$ converges to $I$ in the strong operator topology, i.e., 
for all $f \in L^2$, $||P_n f - f|| \ra 0$. (Here the norm is the norm 
in $L^2$.)
\end{enumerate}
For each of these sense, either prove or disprove that $P_n$ converges to 
$I$ in this sense. 

\end{enumerate}

\end{enumerate}

\end{document}

\end
 

\item  In a homework problem you showed that the dual of $c_0$, 
the space of sequences that converge to zero with the sup norm, 
is isometric to $l^1$. Now let $c$ denoted the space of real-valued 
sequences that converge to some finite real number. 
Given $y=\{y_n\}_{n=1}^\infty$ we can still define a linear
functional $f_y$ by 
$$
f_y(\{x_n\}_{n=1}^\infty)=\sum_{n=1}^\infty x_n y_n
$$
and the same argument as for $c_0$ proves this is a bounded linear
functional with norm equal to the $l^1$ norm of $y$. 
So we have an isometry from $l^1$ {\bf into} $c^*$.
\begin{enumerate} 
\item Prove that this isometry is not onto.
\item Prove that there is a linear function $f_\infty \in c^*$ so 
that every linear functional in $c^*$ can be written as 
$f_y + a f_\infty$ for some $y \in l^1$ and some real number $a$. 
(So the range of the isometry has co-dimension 1.)
\end{enumerate} 

