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Let $E$ be a normed $\mathbb R$-vector space, $\mu$ be a finite signed measure on $(E,\mathcal B(E))$ and $$\hat\mu:E'\to\mathbb C\;,\;\;\;\varphi\mapsto\int\mu({\rm d}x)e^{{\rm i}\varphi}$$ denote the characteristic function of $\mu$.

Replying to a previous formulation of this question, Kavi Rama Murthy has shown that if $E$ is complete and separable and $\mu$ is nonnegative, then $\hat\mu$ is uniformly continuous.

It is easy to see that his proof still works in the general case as long as we are assuming that $\mu$ is tight$^1$, i.e. $$\forall\varepsilon>0:\exists K\subseteq E\text{ compact}:|\mu|(K^c)<\varepsilon\tag1.$$

Taking a closer look at the proof, I've observed the following: Let $\langle\;\cdot\;,\;\cdot\;\rangle$ denote the duality pairing between $E$ and $E'$ and $$p_x(\varphi):=|\langle x,\varphi\rangle|\;\;\;\text{for }\varphi\in E'$$ for $x\in E$. By definition, the weak* topology $\sigma(E',E)$ on $E'$ is the topology generated by the seminorm family $(p_x)_{x\in E}$.

Now, if $K\subseteq E$ is compact, $$p_K(\varphi):=\sup_{x\in K}p_x(\varphi)\;\;\;\text{for }\varphi\in E'$$ should be a seminorm on $E'$ as well. And if I'm not missing something, the topology generated by $(p_K:K\subseteq E\text{ is compact})$ is precisely the topology $\sigma_c(E',E)$ of compact convergence on $E'$.

What Kavi Rama Murthy has shown is that, since $\mu$ is tight, for all $\varepsilon>0$, there is a compact $K\subseteq E$ and a $\delta>0$ with $$|\hat\mu(\varphi_1)-\hat\mu(\varphi_2)|<\varepsilon\;\;\;\text{for all }\varphi_1,\varphi_2\in E'\text{ with }p_K(\varphi_1-\varphi_2)<\delta\tag2.$$

Question: Are we able to conclude that $\hat\mu$ is $\sigma_c(E',E)$-continuous?

EDIT:

In order to conclude that $\hat\mu$ is (uniformly) $\sigma_c(E',E)$-continuous, we need to that $(2)$ holds for $K$ replaced by an arbitrary compact $\tilde K\subseteq E$. Given $\varepsilon>0$, we can show $(2)$ by choosing the compact subset $K\subseteq E$ such that $$|\mu|(K^c)<\varepsilon\tag3.$$

We may then write \begin{equation}\begin{split}\left|\hat\mu(\varphi_1)-\hat\mu(\varphi_2)\right|&\le\underbrace{\int_{K\cap\tilde K}\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|{\rm d}\left|\mu\right|}_{<\:\varepsilon}\\&\;\;\;\;\;\;\;\;\;\;\;\;+\int_{K\cap\tilde K^c}\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|{\rm d}\left|\mu\right|\\&\;\;\;\;\;\;\;\;\;\;\;\;+\underbrace{\int_{K\cap\tilde K}\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|{\rm d}\left|\mu\right|}_{<\:2\varepsilon}\end{split}\tag4\end{equation} for all $\varphi_1,\varphi_2\in E'$ with $p_{\tilde K}(\varphi_1-\varphi_2)<\delta$, where $$\delta:=\frac\varepsilon{\left\|\mu\right\|},$$ but I have no idea how we can control the second integral.

EDIT 2

A "proof" of this claim can be (found in Linde's Probability in Banach Spaces), but I have no idea why this proof is correct, since he is concluding the continuity immediately from $(2)$ (for a single $K$):

enter image description here enter image description here

Maybe we need to assume that $\mu$ is even Radon, i.e. that for all $B\in\mathcal (E)$, there is a compact $C\subseteq E$ with $C\subseteq B$ and $|\mu|(B\setminus C)<\varepsilon$. The author is actually imposing this assumption, but he obviously doesn't make use of it in his proof (he would need to consider an arbitrary compact $\tilde K\subseteq E$, as I did above).


$^1$ On a complete separable metric space, every finite signed measure is tight.

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  • $\begingroup$ I think this requires separability of $E$ (or that $\mu $ has separable support). $\endgroup$ – Kavi Rama Murthy Oct 24 '20 at 11:56
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Partial answer: I will give a proof assuming that $E$ is separable. Of course this will give a proof when $E$ is not separable but $\mu$ has separable support.

It is an interesting fact that if support of $\mu$ exists in the sense that there is a smallest closed set of full measure then it is necessarily separable. [This requires Axiom of Choice]

Under this hypothesis it is known that $\mu$ is tight. Ref. Convergence of Probability Measures by Billingsley.

Let $\epsilon >0$ and choose a compact set $K$ such that $\mu (K^{c}) <\epsilon$. Then $$|\phi (x')-\phi (y')|$$ $$ \leq \int |e^{i \langle x', x \rangle}-e^{i \langle x', x \rangle}| d\mu (x)$$ $$\leq \int_K |e^{i \langle x', x \rangle}-e^{i \langle x', x \rangle}| d\mu (x)+2\epsilon.$$ So $$|\phi (x')-\phi (y')| \leq \|x'-y'\|\int_K \|x|| d\mu(x)+2\epsilon<3\epsilon$$ if $$\|x'-y'\| <\frac {\epsilon} {M\mu(E)}$$ where $$M=\sup \{\|x\|:x \in K\}$$.

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  • $\begingroup$ (a) Thank you for your answer. I'll need to check the details, but do you agree that the claim is generally true (i.e. even for nonseparable $E$) using $(3)$ assuming that $\mu$ has a finite first moment? (b) Could you provide the definition of the "support" you're using? $\endgroup$ – 0xbadf00d Oct 24 '20 at 13:05
  • $\begingroup$ Of course integrability is good enough. Support is the smallest closed set of full measure .@0xbadf00d $\endgroup$ – Kavi Rama Murthy Oct 24 '20 at 13:09
  • $\begingroup$ Is this definition equivalent to the one given here: en.wikipedia.org/wiki/Support_(measure_theory)#Definition? $\endgroup$ – 0xbadf00d Oct 24 '20 at 13:13
  • $\begingroup$ @0xbadf00d Yes, they are equivalent. $\endgroup$ – Kavi Rama Murthy Oct 25 '20 at 1:15
  • $\begingroup$ Please take a look at my latest edit. I've noticed from your proof that we only need to control $|x'-y'|$ on the compact set $K$; not the whole operator norm of $x'-y'$. $\endgroup$ – 0xbadf00d Dec 17 '20 at 11:24
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Hopefully, I didn't make any stupid mistake, but I think I figured out why the argument in the excerpt is correct.

First of all, let's establish a common understanding of the definitions:

Definition 1

  1. If $(E,\tau)$ is a topological space, then $$\mathcal N_\tau(x):=\{N:N\text{ is a }\tau\text{-neighborhood of }x\}\;\;\;\text{for }x\in E.$$
  2. If $(E_i,\tau_i)$ is a topological vector space, then $f:E_1\to E_2$ is called uniformly $(\tau_1,\tau_2)$-continuous if $$\forall N\in\mathcal N_{\tau_2}(0):\exists M\in N_{\tau_1}(0):\forall x,y\in E_1:x-y\in M\Rightarrow f(x)-f(y)\in N.$$
  3. If $(E_i,\tau_i)$ is a topological vector space, then $\mathcal F\subseteq E_2^{E_1}$ is called uniformly $(\tau_1,\tau_2)$-equicontinuous if $$\forall N\in\mathcal N_{\tau_2}(0):\exists M\in N_{\tau_1}(0):\forall f\in\mathcal F:\forall x,y\in E_1:x-y\in M\Rightarrow f(x)-f(y)\in N.$$

Definition 2: If $(E,\mathcal E)$ is a measurable space, then $$\mathcal M(E,\mathcal E):=\{\mu:\mu\text{ is a finite signed measure on }(E,\mathcal E)\}.$$ If $\mu\in\mathcal M(E,\mathcal E)$, then $|\mu|$ denotes the total variation of $\mu$. Thte total variation norm $\left\|\;\cdot\;\right\|$ on $\mathcal M(E,E)$ is defined by $$\left\|\mu\right\|:=|\mu|(E)\;\;\;\text{for }\mu\in\mathcal M(E,\mathcal E).$$ If $E$ is a Hausdorff space, then $\mathcal F\subseteq\mathcal M(E):=\mathcal M(E,\mathcal B(E))$ is called tight if $$\forall\varepsilon>0:\exists K\subseteq E\text{ compact}:\sup_{\mu\in\mathcal F}|\mu|(K^c)<\varepsilon.$$

Now, it is important to remember the following fact:

Lemma 1: If $(X,\tau)$ is a topological vector space and $p$ is a seminorm on $X$, then

  1. $p$ is $\tau$-continuous;
  2. $p$ is $\tau$-continuous at $0$;
  3. $U_p:=\{x\in X:p(x)<1\}$ is a $\tau$-neighborhood of $0$

are equivalent.

We are ready to establish the following result:

Theorem 1: If $\mathcal F\subseteq\mathcal M(E)$ be $\left\|\;\cdot\;\right\|$-bounded and tight, then $\{\hat\mu:\mu\in\mathcal F\}$ is uniformly $\sigma_c(E',C)$-equicontinuous.

ProofI: Let $\varepsilon>0$. Since $\mathcal F$ is $\left\|\;\cdot\;\right\|$-bounded, $$c:=\sup_{\mu\in\mathcal F}\left\|\mu\right\|<\infty.$$ And since $\mathcal F$ is tight, there is a compact $K\subseteq E$ with $$\sup_{\mu\in\mathcal F}|\mu|(K^c)<\frac\varepsilon3.\tag5$$ Assume $c\ne0$. Then $$\delta:=\frac\varepsilon{3c}$$ is well-defined. Let $$N:=\{\varphi\in E':p_K(\varphi)<\delta\}.$$ Now, $$\int_K\underbrace{\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|}_{\le\:|\varphi-1-\varphi_2|}{\rm d}|\mu|\le\left\|\mu\right\|p_K(\varphi_1-\varphi_2)<\frac\varepsilon3\tag6$$ and hence \begin{equation}\begin{split}|\hat\mu(\varphi_1)-\hat\mu(\varphi_2)|&\le\int\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|{\rm d}|\mu|\\&=\underbrace{\int_K\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|{\rm d}|\mu|}_{<\:\frac13\varepsilon}+\underbrace{\int_{K^c}\underbrace{\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|}_{\le\:2}{\rm d}|\mu|}_{<\:\frac23\varepsilon}<\varepsilon\end{split}\tag7\end{equation} for all $\mu\in\mathcal F$ and $\varphi_1,\varphi_2\in E'$ with $p_K(\varphi_1-\varphi_2)<\delta$; i.e. $$\forall\mu\in\mathcal F:\forall\varphi_1,\varphi_2\in E':\varphi_1-\varphi_2\in N\Rightarrow\hat\mu(\varphi_1)-\hat\mu(\varphi_2)\in B_\varepsilon(0)\tag8.$$

By definition of $\sigma_c(E',E)$, the seminorm $p_K$ is $\sigma_c(E',E)$-continuous. Thus, by Lemma 1, $$N=\delta U_{p_K}\in\mathcal N_{\sigma_c(E',\:E)}(0)\tag9$$ and hence we should have shown the claim.

Remark: I would highly appreciate any confirmation of my proof or any hint to a mistake in the comment section below

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  • $\begingroup$ I don't understand something in equation (7). Since $\lvert d\mu\rvert$ is a positive measure, the map $\phi\mapsto \int1-\Re e^{{\rm i}(\varphi)}\:{\rm d}|\mu|$ is a map of $E'$ into $\mathbb R$. How can it be $\sigma(E', E)$ continuous? $\endgroup$ – Giuseppe Negro Dec 20 '20 at 11:31
  • $\begingroup$ @GiuseppeNegro $\sigma(E',E)$ is the topology on the domain (which is $E'$) of this map. Its codomain (which is $\mathbb R$) is equipped with the Euclidean topology $\tau$. So, being $\sigma_c(E',E)$-continuous means more precisely that it is being $(\sigma_c(E',E),\tau)$-continuous. $\endgroup$ – 0xbadf00d Dec 20 '20 at 11:37
  • $\begingroup$ Ooh right, that makes sense, sorry for the silly question. $\endgroup$ – Giuseppe Negro Dec 20 '20 at 12:26
  • $\begingroup$ @GiuseppeNegro Please note that I've cleaned the proof and improved the shown result. Maybe you can let me know whether you agree or something is still not clear. $\endgroup$ – 0xbadf00d Dec 20 '20 at 15:33
  • $\begingroup$ I had noticed. I am reading your edited answer right now. Thank you for the editing. $\endgroup$ – Giuseppe Negro Dec 20 '20 at 15:34

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