Past exam of the analysis course of the University of Oxford / 2011 / 2 / c / Solution

Let $Z=\overline{\mathrm{span}}\{x_n:n\ge 1\}$. If $Z\ne X$, part (a) provides a nonzero bounded functional annihilating $Z$; normalize it to have norm one. Density of the sequence in the dual unit ball gives an index $n$ with $\Vert f_n-f\Vert <1/4$. Since $f(x_n)=0$ and $\Vert x_n\Vert \le 1$,But the hypothesis and reverse triangle inequality givea contradiction. Therefore the span is dense.

If $X^{\prime }$ is separable, its unit ball has a countable dense set. For each nonzero $f_n$, the definition of its norm permits choosing $x_n$ in the unit ball with $|f_n(x_n)|\ge \Vert f_n\Vert /2$; choose zero for a zero functional. The preceding result gives a dense countable span over $\mathbb{Q}$, or over $\mathbb{Q}+i\mathbb{Q}$ in the complex case. Thus $X$ is separable.

The converse fails: ${\ell }^1$ is separable, since finite rational-coordinate sequences are dense, but $({\ell }^1{)}^{\prime }\cong {\ell }^{\infty }$ is not separable. The dual identification sends a bounded sequence $a$ to $x\mapsto {\sum }_n{a}_n{x}_n$, isometrically; conversely a bounded functional is determined by its values on the coordinate vectors and hence has this form by truncation. The uncountable family of zero-one sequences in ${\ell }^{\infty }$ has pairwise distance one, which no countable dense set can approximate within a radius smaller than $1/2$.