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Let $E$ be a normed $\mathbb R$-vector space, $\sigma(E',E)$ and $c(E',E)$ denote the weak* topology and the topology of compact convergence on $E'$, respectively, $\mu$ be a probability measure on $E$ and $$\hat\mu:E'\to\mathbb C\setminus\{0\}\;,\;\;\varphi\mapsto\int\mu({\rm d}x)e^{{\rm i}\langle x,\:\varphi\rangle}$$ denote the characteristic function of $\mu$.

As I have shown in this answer, $\hat\mu$ is $c(E',E)$-continuous as long as we assume that $\mu$ is sufficiently regular$^1$. However, isn't it $\sigma(E',E)$-continuous as well? In fact, if $\varphi_n\to\varphi$ wrt $\sigma(E',E)$ it holds $$\forall x\in E:\langle x,\varphi_n\rangle\to\langle x,\varphi\rangle\tag1$$ and hence $$\int\mu({\rm d}x)e^{{\rm i}\langle x,\:\varphi_n\rangle}\to\int\mu({\rm d}x)e^{{\rm i}\langle x,\:\varphi\rangle}\tag2$$ by the dominated convergence theorem.

Now, since $\sigma(E',E)\subseteq c(E',E)$, couldn't we immediately infer from this result that $\hat\mu$ is $c(E',E)$-continuous? (And hence the $c(E',E)$-continuity holds without any regularity assumption on $\mu$.)

$^1$ I've assumed thightness of $\mu$ in my answer. So, $E$ being complete and separable would be sufficient.

0xbadf00d
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  • I am not sure whether the weak$^{\ast}$ dual of a normed space is always sequential. So I think you need to take a converging net there and dominated convergence won't apply – G. Chiusole Apr 28 '21 at 21:31
  • Indeed, for infinite dimensional reflexive Banach spaces this is never the case. see here – G. Chiusole Apr 28 '21 at 21:33
  • @G.Chiusole Thank you for pointing that out. So, it seems like we cannot apply the dominated convergence, but that doesn't necessarily mean that $(2)$ could still hold for arbitrary nets $\varphi_n$. Do you have an idea (or a counterexample)? – 0xbadf00d Apr 29 '21 at 05:12
  • @G.Chiusole Continuity is at least claimed in this paper (at the beginning of page 886), but without a proof, and here (below Definition 1.49), again without a proof. – 0xbadf00d Apr 29 '21 at 05:15
  • The paper you linked seems to consider the strong dual of the Banach space i.e. with the operator norm topology. The second reference also considers the strong topology, which is stronger than compact convergence – G. Chiusole Apr 29 '21 at 07:21
  • @G.Chiusole Might be true; the second reference is better. – 0xbadf00d Apr 29 '21 at 09:09

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