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\begin{document}
\title{Problem set 6}
\author{Due date: Apr 16}
\maketitle
\begin{enumerate}
\item Suppose that $X$ is a Banach space with the norm topology. Show that the Ball X is norm closed if and only if it is 
weakly closed
\item Sequential non-compactness of $\ell^{\infty}(\N)^{\ast}$. Let $b_{n}: \ell^{\infty}(\N) \to \R$ be the functions defined by
$$
b_{n} (x) = x_{n} \, \quad 
$$ 
where $x = (x_{1},x_{2},\ldots, x_{n},x_{n+1},\ldots)$. Show that $b_{n}$ does not have any subsequence that converges pointwise, where pointwise convergence in this setup means that for every fixed $x\in\ell^{\infty}(\N)$, $b_{n}(x)$ is Cauchy.
\item Suppose that $X$ is a reflexive Banach space, $M$ is a closed linear subspace of $X$ and that $x_{0} \in X \setminus M$, then there is a point $y_{0} \in M$ such that $\|x_{0}-y_{0} \| = \text{dist}(x_{0},M)$. In order to prove this result show that, the norm on $X$ is lower semicontinuous for the weak topology, and the norm of $X^{\ast}$ is lower semicontinuous for the weak-${\ast}$ topology. Further show that a lower semicontinuous function on a compact set must achieve its minimum. Note that a function $f:X\to \R$ is lower semicontinuous at $x_{0}$ in a topological space if there exists an open neighborhood $U$ of $x_{0}$ such that 
$$
f(x) \geq f(x_{0}) - \eps \quad \forall x\in U
$$
\item Show that a sequence $f_{n} \in C[0,1]$ equipped with the supremum norm converges weakly if and only if it is
norm bounded and pointwise convergent. Give an example of a sequence in the space which converges weakly but not in norm.
\end{enumerate}
\end{document}