 Solution

A reflexive Banach space is one for which $J_X:X\to X^{\ast \ast }$ is onto. If $X$ is reflexive, every $\theta \in X^{\ast \ast }$ has the form $\theta (f)=f(x)$ for some $x\in X$. Thus the functionals defining $\sigma (X^{\ast },X^{\ast \ast })$ and $\sigma (X^{\ast },X)$ are precisely the same, proving equality of the topologies.

Conversely, assume those topologies coincide. For any $\theta \in X^{\ast \ast }$, the linear functional $f\mapsto \theta (f)$ on $X^{\ast }$ is then weak-star continuous. Its continuity at zero gives finitely many $x_1,\dots ,x_m\in X$ and $ϵ>0$ such thatIf all these evaluations are zero, arbitrary scalar multiples of $f$ still satisfy the inequalities, forcing $\theta (f)=0$. Hence $\theta$ factors through the finite-dimensional evaluation map $f\mapsto (f(x_1),\dots ,f(x_m))$. A linear functional on its image extends to ${\mathbb{R}}^m$, so for suitable scalars $a_j$,Thus $\theta$ lies in $J_X(X)$. Since it was arbitrary, $X$ is reflexive. We have proved