= Solution
Put $T=R\circ J_Z:Z\to X^*$. Then $(Tz)(x)=x(z)$ and $\|Tz\|\leq\|z\|$. We must prove $T$ onto.
Take $f\in B_{X^*}$. Part (iv) supplies $\theta\in B_{Z^{**}}$ with $R\theta=f$. <Goldstine theorem>[Goldstine's theorem] gives a net $z_\alpha\in B_Z$ with $J_Zz_\alpha\to\theta$ weak-star. By <continuity> of $R$, $Tz_\alpha\to f$ in $\sigma(X^*,X)$. Reflexivity of $X$ makes this the <weak topology> on $X^*$ by part (iii). Thus $B_{X^*}$ is contained in the weak <closure> of $T(B_Z)$.
The reverse containment holds because $T(B_Z)\subseteq B_{X^*}$, and the latter ball is weakly closed: it is the intersection of the closed evaluation constraints $|g(x)|\leq1$ for $x\in B_X$. The set $T(B_Z)$ is convex, so part (ii), now on the <Banach space> $X^*$, gives
$$
\boxed{\overline{T(B_Z)}^{\|\cdot\|}=B_{X^*}.}
$$
Density of the ball gives exact surjectivity by a convergent correction argument. For nonzero $f\in X^*$, choose $z_1$ with $\|z_1\|\leq\|f\|$ and $\|f-Tz_1\|\leq\|f\|/2$. Inductively approximate the current residual $r_{n-1}$ by $Tz_n$ with $\|z_n\|\leq\|r_{n-1}\|$ and $\|r_{n-1}-Tz_n\|\leq\|r_{n-1}\|/2$. If a residual is zero, stop.
The residual <norms> are bounded by $2^{-n}\|f\|$, while $\sum_n\|z_n\|\leq2\|f\|$. Completeness of $Z$ gives $z=\sum_nz_n\in Z$, and <continuity> of $T$ gives $Tz=f$. For $f=0$ use $z=0$. Therefore
$$
\boxed{\forall f\in X^*\ \exists z\in Z:\quad x(z)=f(x)\text{ for every }x\in X.}
$$
This is <evaluation onto the dual of a reflexive dual subspace>. The correction argument is needed: a dense image alone would not imply surjectivity.
Back to article page
