Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 6 / 2 / iv / Solution

True. Choose a norm-dense sequence $(x_n)$ in the unit ball of the separable normed vector space $E$. On the dual unit ball $B^{\prime }$, defineEach summand is at most $2^{1-n}$, so the series converges uniformly. Symmetry and the triangle inequality follow term by term. If $d(\varphi ,\psi )=0$, the two continuous functionals agree on the dense sequence, hence on the unit ball and then on all of $E$. Thus $d$ is a metric.

A relative weak-star neighbourhood controls finitely many evaluations. Conversely, to make $d$ small, first choose a tail whose sum is small and then control the finitely many evaluations at $x_1,\dots ,x_N$. This proves that each metric ball is weak-star open. For the other direction, let $x\in E$ and $ϵ>0$. Approximate $x/\Vert x\Vert$ by $x_n$ when $x\ne 0$. Since $\Vert \varphi -\psi \Vert \le 2$ on $B^{\prime }$,The second term can be made smaller than $ϵ/2$ by choosing $n$; the first is smaller than $ϵ/2$ whenever $d(\varphi ,\psi )$ is sufficiently small, because $|\varphi (x_n)-\psi (x_n)|\le 2^{n}d(\varphi ,\psi )$. Every evaluation is therefore metric-continuous. These two inclusions show that $d$ induces exactly the weak-star topology, proving weak-star metrizability of the dual ball without a completeness assumption on $E$.