Let $X$ be a topological affine space over $\mathbb C$, with no additional assumptions. Let $X^*$ denote its dual space of continuous affine functionals $X \to \mathbb C$, equipped with the weak-$*$ topology.

Let $\mathbb P$ be a finite-variance Radon measure on $X$, meaning that each functional is square-integrable with respect to $\mathbb P$. Where $\mathbb E$ denotes the expectation operator for $\mathbb P$, define the variance by $\operatorname{var}_{\mathbb P}(\psi) := \mathbb E(|\psi|^2) - |\mathbb E(\psi)|^2$. Let $d_{\mathbb P}(\psi, \varphi) := \sqrt{\operatorname{var}_{\mathbb P}(\psi - \varphi)}$ be the induced pseudo-metric on the dual space.

Is the $d_{\mathbb P}$-topology a refinement of the weak-$*$ topology on $X^*$? In particular, is every closed ball (with respect to $d_{\mathbb P}$) a closed set in the weak-$*$ topology?


Tom LaGatta: "Is the $d_{\mathbb P}$-topology a refinement of the weak-$*$ topology on $X^*$?"

I would suspect not generally. Possibly a counterexample could be constructed by taking $X=\mathbb C^{\kern1mm\mathbb N_0}$ with the product topology. Then $X^*=\mathbb C^{\kern1mm(\mathbb N_0)}$, the space of sequences with only a finite number of nonzero terms, equipped with the largest vector space topology which is not metrizable. As $\mathbb P$ one would take the product measure of the Gaussians on $\mathbb C$. If the $d_{\mathbb P}$-topology is a vector space topology, then it cannot be a refinement of the topology of $X^*$. (The preceding is just an idea. I have not thought through the details.)


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