Let $E$ be a Banach space with norm $|| \cdot ||$, and let $E^{\ast}$ be the subspace of bounded linear functionals of $\textrm{Hom}_{\mathbb{C}}(E,\mathbb{C})$. What is the "topology of uniform convergence on compact sets?" Is this the same as the norm topology on $E^{\ast}$?
This topology is mentioned in Bourbaki, Differential and Analytic Manifolds.
Best Answer
It's also known as compact-open topology, in this case specialized to linear functionals. A basis of neighborhoods of $0$ in this topology is given by sets $$ U_{K,r} = \{f\in E^*: \sup_K|f|<r\},\quad K\subset E \text{ compact}, r>0 $$
When $E$ is finite-dimensional, it's the same as norm topology (because the unit ball of $E$ is compact, and also because all topologies on a finite-dimensional TVS are equivalent). From now on I assume $E$ infinite-dimensional.
Since one-point sets are compact, the compact-open topology on $E^*$ is at least as strong as weak* topology.
Claim: On bounded subsets of $E^*$ the compact-open topology agrees with weak* topology.
This claim is proved below. In particular, it implies that compact-open topology is strictly weaker than the norm topology: for example, the unit ball of $E^*$ is compact in it (by the Banach-Alaoglu theorem), unlike in the norm topology on $E^*$.
Since every weak*-convergent sequence is bounded in the norm (by the UBP), it follows that a sequence in $E^*$ converges in the compact-open topology if and only if it is weak*-convergent. This does not mean the two topologies are the same; sequences are not enough to detect that.
Aside: when applied to operators on a Hilbert space, the compact-open topology is quite complicated, as the MathOverflow discussion indicates.
Proof of the claim. Let $U$ be a compact-open neighborhood of $0$, meaning there is a compact set $K\subset E$ and $r>0$ such that $U=\{f\in E^*: \sup_K|f|<r\}$. Also fix $M>0$ and restrict attention to functionals within the ball $B_M = \{f\in E^*:\|f\|\le M\}$. Since $K$ is compact, there is a finite-dimensional subspace $P$ such that $K$ is contained in the $r/(2M)$-neighborhood of $P$. Since $K$ is bounded, this means $K$ is contained in the $r/(2M)$-neighborhood of some bounded set $Q\subset P$.
Pick a basis $\{b_k\}$ for $P$. If $|f(b_k)|<\epsilon$ for some sufficiently small $\epsilon>0$ and for every $k$, then $\sup_Q |f|<r/2$. Using the fact that $K$ is close to $Q$, and that $\|f\|\le M$, conclude that $\sup_K |f| < r$. Conclusion: the weak*-neighborhood of $0$, defined by $$ V = \{f\in E^*: |f(b_k)|<\epsilon \ \forall k\} $$ satisfies $V\cap B_M\subset U\cap B_M$, proving the claim. $\quad\Box$