When do the weak-star and compact convergence (compact-open) topology coincide on the dual of a Banach space?

Question

In Measures Which Agree on Balls by Hoffmann-Jørgenson, it is claimed on page 323 that for an arbitrary Banach space $E$, if $\pi$ is the topology on $E^*$ of uniform convergence on compact subsets of $E$, then $\pi$ coincides with $\sigma(E^*,E)$ on bounded subsets of $E^*$. Why is this true?

Answer 1

Weak$^*$-convergence of a net means pointwise convergence on singletons, and therefore (uniform) convergence finite sets. Uniform convergence on compact subsets means exactly that. Since finite sets are compact, the latter is stronger than the former. If the net is bounded, they are equivalent notions. Suppose we have a net $(x_\lambda^*)$ which is weak$*$-convergent to $x^*$ and $\sup_\lambda \|x^*_\lambda\|<R<\infty$. By weak$^*$ lower semicontinuity of the norm, $\|x^*\|\leqslant R$. Fix a compact subset $K$ of $E$ and $\epsilon>0$. Let $F\subset K$ be a finite $\epsilon/3R$ net. There exists $\lambda_0$ such that for each $\lambda_0\leqslant \lambda$, and each $y\in F$, $|x^*(y)-x^*_\lambda(y)|<\epsilon/3$. For $x\in K$, there exists $y\in F$ such that $\|x-y\|\leqslant \epsilon/3$. Then for $\lambda_0\leqslant \lambda$, \begin{align*} |x^*(x)-x^*_\lambda(x)| & \leqslant |x^*(x)-x^*(y)| + |x^*(y)-x^*_\lambda(y)|+|x^*_\lambda(y)-x^*_\lambda(x)| \\ & < R\|x-y\|+\epsilon/3+R\|x-y\|<\epsilon.\end{align*}

Without boundedness, $\|x-y\|$ being small doesn't get you anywhere.