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

The continuous dual space $X^{\prime }$ is the vector space of bounded scalar-valued linear functionals, with operator normThis makes it a normed space, and in fact a Banach space whether or not $X$ is complete.

The norm-preserving Hahn-Banach theorem states that a bounded linear functional $f_0$ on any linear subspace $M\subseteq X$ extends to a bounded linear functional $f$ on all of $X$ with $f{|}_M=f_0$ and $\Vert f\Vert =\Vert f_0\Vert$. The subspace need not be closed. This holds over both the real and complex scalar fields, using the complex version when necessary.

Put $d=\mathrm{dist}(x,Z)$. If $d=0$, the zero functional works. If $d>0$, then $x\notin \overline{Z}$, and on $M=Z+\mathrm{span}\{x\}$ defineThe representation is unique, so this is linear. For $a\ne 0$,and the bound also holds for $a=0$. Thus $\Vert f_0\Vert \le 1$. The Hahn-Banach theorem extends it to $f\in X^{\prime }$ with the same bound, provingNo closedness assumption on $Z$ was required; distance is unchanged on taking its closure.

For the centered ball, the Hahn-Banach theorem supplies a norm-one real functional with $f(y)=\Vert y\Vert >r$. Hence $f(z)\le r$ on the ball, and any $t\in (r,\Vert y\Vert )$ separates it strictly from $y$.

For the general center, apply the same construction to $y-x$: choose $\Vert f\Vert =1$ and $f(y-x)=\Vert y-x\Vert$. ThenFor example,gives $f(z)<t<f(y)$ for every $z\in B$. The strict gap uses exactly the assumption that $y$ lies outside the closed ball.

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 $1/2$.