91Ó°ÊÓ

Consider the space \(\ell^{1}\), both as a Banach space and as the locally convex topological space with the weak topology induced by its dual space \(\ell^{\infty}\); cf. E 2.3.2. Show that every weakly convergent sequence in \(\ell^{1}\) is norm convergent. Explain why this fact does not imply that the two topologies coincide. Hint: If \(x_{n} \rightarrow 0\) weakly, but \(\left\|x_{n}\right\|_{1} \geq \varepsilon>0\) for infinitely many \(n\), we can find (passing if necessary to a subsequence) an increasing sequence of numbers \(a(n)\), with \(a(0)=0\), such that $$ \sum_{k=1}^{a(n-1)}\left|x_{n}(k)\right| \leq n^{-1}, \quad \sum_{k=a(n-1)+1}^{a(n)}\left|x_{n}(k)\right| \geq\left\|x_{n}\right\|_{1}-2 n^{-1} $$ for every \(n\). Define \(x\) in \(\ell^{\infty}\) by \(x(k)=\operatorname{sign} \overline{x_{n}(k)}\) for \(a(n-1)

(Uniformly convex spaces.) A Banach space \(\mathfrak{X}\) is uniformly convex if whenever we have sequences \(\left(x_{n}\right)\) and \(\left(y_{n}\right)\) in \(\mathfrak{X}\), with \(\left\|x_{n}\right\|=\left\|y_{n}\right\|=1\) for all \(n\) and \(\left\|\frac{1}{2}\left(x_{n}+y_{n}\right)\right\| \rightarrow 1\), then \(\left\|x_{n}-y_{n}\right\| \rightarrow 0 .\) Show in this case that every closed, convex subset of \(\mathfrak{X}\) has a uniquely determined element with smallest norm.