Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 6 / 4 / Solution

The weak topology $\sigma (E,E^{\ast })$ is the coarsest topology making every continuous linear functional in $E^{\ast }$ continuous. A basic neighborhood of $x$ imposes finitely many inequalities $|f_j(z-x)|<\epsilon$, and a sequence converges weakly precisely when each $f(x_n)$ converges to $f(x)$. The Hahn-Banach theorem makes the dual separate points, so the weak topology is Hausdorff.

Here is a proof of the requested weak subsequence property, the compact-to-sequential direction of the Eberlein-Šmulian theorem. Let $Y$ be the closed linear span of the countable set $\{x_n:n\ge 1\}$. It is a separable Banach space. The Hahn-Banach theorem separates points outside a closed vector subspace from that subspace, so $Y$ is weakly closed in $E$. Therefore $C=K\cap Y$ is weakly compact.

Choose a countable dense set $(u_j)$ in the unit sphere of $Y$, and choose norming functionals $f_j\in Y^{\ast }$ with $\Vert f_j\Vert =1$ and $f_j(u_j)=1$. They separate points: for $y\ne 0$, a sufficiently close $u_j$ to $y/\Vert y\Vert$ gives $|f_j(y)|>0$. This is a countable norming family. Extend the functionals to $E$ by the Hahn-Banach theorem. The evaluation mapis weakly continuous and injective. Its domain is compact and its target Hausdorff, so it is a homeomorphism onto its image: images of closed subsets are compact, hence closed. A countable product of real or complex lines is metrizable, for example by ${\sum }_j{2}^{-j}\min (1,|a_j-b_j|)$. Thus $C$ has a compact metrizable weak topology. Compact metric spaces are sequentially compact, giving a subsequence with $\boxed{x_{n_j}\rightharpoonup x\in C\subseteq K}$. If $Y=\{0\}$ the result is immediate. We did not assert that a separable Banach space has a separable dual or that its weak topology is generally metrizable.