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\begin{document}
\begin{center}
{\bf \large Math 702 Problem Set 9} \vs \\
{\bf due Friday 4/25/2014} \vs \vs
\end{center}

\noindent
Unless specified otherwise, all vector spaces can be real or complex, every $L^p$ space is equipped with the corresponding $L^p$-norm $(1 \leq p \leq \infty)$, and the space $\sC[a,b]$ is equipped with the sup norm. \vs \vs

\noindent
{\bf Problem 1.} Suppose $X,Y$ are normed spaces, $L_n \to  L$ in $\sL(X,Y)$ and $x_n \to x$ in $X$. Show that $L_n(x_n) \to L(x)$ in $Y$. \vs

\noindent
{\bf Problem 2.} Suppose $X$ is a normed space, $Y$ is a Banach space, $\{ L_n \}$ is a bounded sequence in $\sL(X,Y)$, and $\{ L_n(x) \}$ converges for every $x$ in a dense subset of $X$. \vs
\begin{enumerate}
\item[(i)]
Show that $L(x)=\lim_{n \to \infty} L_n(x)$ exists for every $x \in X$, and $L \in \sL(X,Y)$. \vs

\item[(ii)]
Show by an example that the convergence $L_n \to L$ in $\sL(X,Y)$ may not hold. \vs 
\end{enumerate}
(Hint for (ii): Look for a sequence $\{ L_n \}$ of bounded linear functionals on $\sC[0,1]$ with $\| L_n \|=1$ such that $L_n(f) \to 0$ for every $f$.) \vs 

\noindent
{\bf Problem 3.} Consider the linear functionals $\Phi, \Psi$ on $\sC[-1,1]$ defined by 
\begin{align*}
\Phi(f) & =\int_{-1}^1 x f(x) \, dx \\
\Psi(f) & = \int_0^1 f(x) \, dx - \int_{-1}^0 f(x) \, dx. 
\end{align*}
Show that $\Phi, \Psi$ are bounded and find their operator norms. How would the answers change if we regarded $\Phi, \Psi$ as linear functionals on $L^1[-1,1]$? \vs

\noindent
{\bf Problem 4.} {\it The topology of pointwise convergence in $\sC[0,1]$ is not normable:} Prove that there is no norm $\| \cdot \|$ on $\sC[0,1]$ with respect to which $\| f_n-f \| \to 0$ iff $f_n(x) \to f(x)$ for every $x \in [0,1]$. \vs

\noindent
{\bf Problem 5.} Suppose $(X, \| \cdot \|)$ is a normed space and $Y$ is a proper closed subspace of $X$. Prove the following statements: \vs
\begin{enumerate}
\item[(i)]
$\| x+Y \| = \inf_{y \in Y} \| x + y \|$ defines a norm on the quotient space $X/Y$. \vs

\item[(ii)]
The natural projection $p:X \to X/Y$ is a bounded linear map with $\| p \|=1$. \vs

\item[(iii)]
If $X$ is a Banach space, so are $Y$ and $X/Y$, and the natural projection $p$ is an open map. \vs

\item[(iv)]
If $Y$ and $X/Y$ are Banach spaces, so is $X$. \vs 
\end{enumerate}

\noindent
{\bf Problem 6.} It will be convenient to call a norm $\| \cdot \|$ on a vector space $X$ a {\bit Banach norm} if $(X,\| \cdot \|)$ is a Banach space. \vs
\begin{enumerate}

\item[(i)]
Show that if $\| \cdot \|$ and $\| \cdot \|'$ are Banach norms on $X$ and $\| x \| \leq C \| x \|'$ for some constant $C>0$, then $\| \cdot \|$ and $\| \cdot \|'$ are equivalent norms. \vs

\item[(ii)]
Critique the following ``proof'' that any two Banach norms $\| \cdot \|$ and $\| \cdot \|'$ on a vector space $X$ are equivalent: Define $\| x \|'' = \| x \| + \| x \|'$, which is a Banach norm on $X$ since every Cauchy sequence in $\| \cdot \|''$ is Cauchy in both $\| \cdot \|$ and $\| \cdot \|'$. Since $\| x \| \leq \| x \|''$ and $\| x \|' \leq \| x \|''$, part (i) shows that $\| \cdot \|$ and $\| \cdot \|'$ are equivalent to $\| \cdot \|''$, hence equivalent to each other. \vs
\end{enumerate}
(Hint for (i): Use the open mapping theorem. With regards to (ii), one can show that every infinite dimensional Banach space admits non-equivalent Banach norms.) \vs

\noindent
{\bf Problem 7.} Recall that $\ell^1$ and $\ell^{\infty}$ are the vector spaces of all complex-valued sequences $x=\{ x_i \}$ for which the norms 
$$
\| x \|_1 = \sum_{i=1}^{\infty} |x_i| \qquad \text{and} \qquad 
\| x \|_{\infty} = \sup_{i} |x_i| 
$$
are finite. Let $c_0$ be the subspace of $\ell^{\infty}$ consisting of all $x=\{ x_i \}$ for which $\lim_{i \to \infty} x_i =0$. Prove the following statements: \vs
\begin{enumerate}
\item[(i)]
Every $x \in \ell^\infty$ defines an $\hat{x} \in (\ell^1)^\ast$ by 
$$
\hat{x}(y) = \sum_{i=1}^{\infty} x_i y_i,
$$
and $\| \hat{x} \|=\| x \|_\infty$. Moreover, every element of $(\ell^1)^\ast$ is of the form $\hat{x}$ for some $x \in \ell^\infty$. Thus, the map $x \mapsto \hat{x}$ is an isometric isomorphism between $\ell^\infty$ and $(\ell^1)^\ast$. \vs 

\item[(ii)] 
Every $x \in \ell^1$ defines an $\hat{x} \in (c_0)^\ast$ by 
$$
\hat{x}(y) = \sum_{i=1}^{\infty} x_i y_i,
$$
and $\| \hat{x} \|=\| x \|_1$. Moreover, every element of $(c_0)^\ast$ is of the form $\hat{x}$ for some $x \in \ell^1$. Thus, the map $x \mapsto \hat{x}$ is an isometric isomorphism between $\ell^1$ and $(c_0)^\ast$. \vs    

\item[(iii)]
Every $x \in \ell^1$ defines an $\hat{x} \in (\ell^\infty)^\ast$ by the same formula as in (ii), and again $\| \hat{x} \|=\| x \|_1$. However, there are non-trivial elements of $(\ell^\infty)^\ast$ which vanish on $c_0$, so the map $x \mapsto \hat{x}$ from $\ell^1$ to $(\ell^{\infty})^\ast$ is not surjective. \vs
\end{enumerate}
(Hint: The space $S$ of sequences $\{ z_i \}$ where $z_i=0$ for all but finitely many $i$ is dense in both $\ell^1$ and $c_0$, so every element of $(\ell^1)^\ast$ or $(c_0)^\ast$ is determined by its values on $S$.) \vs

\newpage
\noindent
{\bf Problem 8.} 
\begin{enumerate}
\item[(i)]
Show that $\ell^1$ and $c_0$ are separable Banach spaces but $\ell^{\infty}$ is not. \vs 

\item[(ii)]
Show, however, that every separable Banach space is isometric to a subspace of $\ell^\infty$. In other words, if $(X, \| \cdot \|)$ is a separable Banach space, there is an injective linear map $L: X \to \ell^{\infty}$ such that $\| L(x) \|_{\infty}= \| x \|$ for every $x \in X$. \vs
\end{enumerate}
(Hint for (ii): Take a countable dense set $\{ x_n \}$ in $X$ and find $f_n \in X^\ast$ such that $\| f_n \|=1$ and $f_n(x_n)=\| x_n \|$. Now manufacture $L$ using the $f_n$.) \vs   

\noindent
{\bf Problem 9.} A sequence $\{ x_n \}$ in a normed space $X$ {\bit converges weakly} to $x \in X$, written as $x_n \wto x$, if $f(x_n) \to f(x)$ for every $f \in X^\ast$. Prove the following statements: \vs
\begin{enumerate}
\item[(i)]
A sequence has at most one weak limit, and $x_n \to x$ implies $x_n \wto x$. \vs

\item[(ii)]
If $x_n \wto x$, then $\| x \| \leq \liminf_{n \to \infty} \| x_n \|$. \vs

\item[(iii)]
If $x_n \wto x$, then $\{ x_n \}$ is a bounded sequence. \vs
\end{enumerate}
(Hint for (iii): Let $\hat{x}_n$ and $\hat{x}$ be the corresponding elements of $X^{\ast\ast}$, so $\hat{x}_n(f) \to \hat{x}(f)$ for every $f \in X^\ast$. Apply the uniform boundedness principle to the collection $\{ \hat{x}_n \}$.) \vs 

\noindent 
{\bf Problem 10.} In this problem, you can use the isomorphisms $(c_0)^\ast \cong \ell^1$ and $(\ell^1)^\ast \cong \ell^{\infty}$ of problem 7. \vs
\begin{enumerate}
\item[(i)]
Find a sequence $\{ x_n \}$ in $c_0$ such that $x_n \wto 0$ but $\| x_n \|_{\infty}=1$ for all $n$. This shows that a weakly convergent sequence is not necessarily convergent. \vs

\item[(ii)]
Prove the following special property of $\ell^1$: If $x_n \wto x$ in $\ell^1$, then $x_n \to x$.  
\end{enumerate}

\end{document}

