Show that the characteristic function of a finite signed measure on a normed vector space is uniformly continuous

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 $$$$\begin{split}\left|\hat\mu(\varphi_1)-\hat\mu(\varphi_2)\right|&\le\underbrace{\int_{K\cap\tilde K}\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|{\rm d}\left|\mu\right|}_{<\:\varepsilon}\\&\;\;\;\;\;\;\;\;\;\;\;\;+\int_{K\cap\tilde K^c}\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|{\rm d}\left|\mu\right|\\&\;\;\;\;\;\;\;\;\;\;\;\;+\underbrace{\int_{K\cap\tilde K}\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|{\rm d}\left|\mu\right|}_{<\:2\varepsilon}\end{split}\tag4$$$$ for all $$\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 $$K$$):

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.

• I think this requires separability of $E$ (or that $\mu$ has separable support). – Kavi Rama Murthy Oct 24 '20 at 11:56

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 $$M=\sup \{\|x\|:x \in K\}$$.

• (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? – 0xbadf00d Oct 24 '20 at 13:05
• Of course integrability is good enough. Support is the smallest closed set of full measure .@0xbadf00d – Kavi Rama Murthy Oct 24 '20 at 13:09
• Is this definition equivalent to the one given here: en.wikipedia.org/wiki/Support_(measure_theory)#Definition? – 0xbadf00d Oct 24 '20 at 13:13
• @0xbadf00d Yes, they are equivalent. – Kavi Rama Murthy Oct 25 '20 at 1:15
• 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 $x'-y'$. – 0xbadf00d Dec 17 '20 at 11:24

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 $$$$\begin{split}|\hat\mu(\varphi_1)-\hat\mu(\varphi_2)|&\le\int\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|{\rm d}|\mu|\\&=\underbrace{\int_K\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|{\rm d}|\mu|}_{<\:\frac13\varepsilon}+\underbrace{\int_{K^c}\underbrace{\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|}_{\le\:2}{\rm d}|\mu|}_{<\:\frac23\varepsilon}<\varepsilon\end{split}\tag7$$$$ for all $$\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

• 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? – Giuseppe Negro Dec 20 '20 at 11:31
• @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. – 0xbadf00d Dec 20 '20 at 11:37
• Ooh right, that makes sense, sorry for the silly question. – Giuseppe Negro Dec 20 '20 at 12:26
• @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. – 0xbadf00d Dec 20 '20 at 15:33
• I had noticed. I am reading your edited answer right now. Thank you for the editing. – Giuseppe Negro Dec 20 '20 at 15:34