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

Put $T=R\circ J_Z:Z\to X^{\ast }$. Then $(Tz)(x)=x(z)$ and $\Vert Tz\Vert \le \Vert z\Vert$. We must prove $T$ onto.

Take $f\in B_{X^{\ast }}$. Part (iv) supplies $\theta \in B_{Z^{\ast \ast }}$ with $R\theta =f$. Goldstine's theorem gives a net $z_{\alpha }\in B_Z$ with $J_Z{z}_{\alpha }\to \theta$ weak-star. By continuity of $R$, $T{z}_{\alpha }\to f$ in $\sigma (X^{\ast },X)$. Reflexivity of $X$ makes this the weak topology on $X^{\ast }$ by part (iii). Thus $B_{X^{\ast }}$ is contained in the weak closure of $T(B_Z)$.

The reverse containment holds because $T(B_Z)\subseteq B_{X^{\ast }}$, and the latter ball is weakly closed: it is the intersection of the closed evaluation constraints $|g(x)|\le 1$ for $x\in B_X$. The set $T(B_Z)$ is convex, so part (ii), now on the Banach space $X^{\ast }$, givesDensity of the ball gives exact surjectivity by a convergent correction argument. For nonzero $f\in X^{\ast }$, choose $z_1$ with $\Vert z_1\Vert \le \Vert f\Vert$ and $\Vert f-T{z}_1\Vert \le \Vert f\Vert /2$. Inductively approximate the current residual $r_{n-1}$ by $T{z}_n$ with $\Vert z_n\Vert \le \Vert r_{n-1}\Vert$ and $\Vert r_{n-1}-T{z}_n\Vert \le \Vert r_{n-1}\Vert /2$. If a residual is zero, stop.

The residual norms are bounded by $2^{-n}\Vert f\Vert$, while ${\sum }_n\Vert z_n\Vert \le 2\Vert f\Vert$. Completeness of $Z$ gives $z={\sum }_n{z}_n\in Z$, and continuity of $T$ gives $Tz=f$. For $f=0$ use $z=0$. ThereforeThis is evaluation onto the dual of a reflexive dual subspace. The correction argument is needed: a dense image alone would not imply surjectivity.