Why is $\phi_\mu(x)= \int_{\widehat{G}} \xi(x)d\mu(\xi)$ a continuous map?

Question

Let $G$ be a locally compact Hausdorff abelian group. Let $\widehat{G}$ be its dual group, consisting of the unitary characters $G \to \mathbb{T}$. If $\mu \in M(\widehat{G})$ (= complex Radon measures on $\widehat{G}$), define $$\phi_\mu: G \to \mathbb{C}: x\mapsto\int_{\widehat{G}} \xi(x) d\mu(\xi).$$ Note that this integral exists because $$\int_{\widehat{G}}|\xi(x)|d|\mu|(\xi)= |\mu|(\widehat{G}) < \infty.$$

Why is the map $\phi_\mu$ continuous? I.e. assume that we have the net convergence $x_\alpha \to x$ in $G$. Then we should be able to show that $$\int_\widehat{G} \xi(x_\alpha) d \mu(\xi)\to \int_\widehat{G} \xi(x)d\mu(\xi).$$

I tried to estimate $$\left|\int_\widehat{G} \xi(x_\alpha)d\mu(\xi)-\int_\widehat{G} \xi(x)d\mu(\xi)\right|\le \int_\widehat{G}|\xi(x_\alpha)-\xi(x)| d|\mu|(\xi).$$ Now, I am not sure how to proceed. Any hints/help are highly appreciated!

Reference: Folland's book "A course in abstract harmonic analysis", second edition p103.

Answer 1

Let $z_0 \in G$ and $\varepsilon > 0$ be given. By inner regularity, one can find a compact set $K \subset \hat G$ such that $|\mu|(K^c) \le \varepsilon$. The mapping $(z,\xi) \longmapsto \xi(z)$ is continuous because $G$ is LCH and $\hat G$ is equipped with the compact-open topology. Then, for every $\xi_0 \in K$ there are open sets $U \ni z_0$ and $V\ni \xi_0$ such that $|\xi(z)-\xi_0(z_0)| \le \varepsilon$ whenever $z \in U$ and $\xi \in V$.

Such open sets $V$ cover $K$, so by compactness there are $\xi_1,\ldots,\xi_n \in K$ and open sets $U_i \ni z_0$, $V_i \ni \xi_i$ such that $$ K \subseteq \bigcup_{i=1}^n V_i ~~\text{ and }~~ \forall z\in U_i,\forall \xi \in V_i ~,~ |\xi(z)-\xi_i(z_0)| \le \varepsilon $$ Now let $U = \bigcap_{i=1}^n U_i$ and observe that for any $z\in U$ and $\xi \in K$, we have $$ |\xi(z)-\xi(z_0)| \le |\xi(z)-\xi_i(z_0)| + |\xi_i(z_0)-\xi(z_0)| \le 2\varepsilon $$where $i$ is chosen so that $\xi \in V_i$. Finally, whenever $z \in U$, we have $$ \begin{split} |\phi_\mu(z)-\phi_\mu(z_0)| &\le \int_{K} |\xi(z)-\xi(z_0)| \,|\mu|(\mathrm d\xi) + \int_{K^c} |\xi(z)-\xi(z_0)| \,|\mu|(\mathrm d\xi)\\ &\le 2\varepsilon |\mu|(K) + 2|\mu|(K^c)\\ &\le 2\varepsilon \|\mu\|_\text{TV} + 2\varepsilon \end{split} $$ The last quantity can be as small as we want, hence $\phi_\mu$ is continuous at $z_0$.

Answer 2

The first relevant point to observe is that the topology of the dual of a locally compact abelian group $G$ is defined as the topology of uniform convergence on compact sets. In other words, a net $\{\xi _\alpha \}_\alpha $ in $\widehat G$ converges to some $\xi $ in $\widehat G$ iff $$ \sup_{x\in K}|\xi _\alpha (x)-\xi (x)| \to 0 \qquad\qquad{(\dagger)} $$ for every compact subset $K\subseteq G$.

Next we need to focus on the precise meaning of Pontryagin duality, namely the result according to which $\, \widehat {\!\widehat G}$ coincides with $G$. To begin with, this says that every character on $\widehat G$ is of the form $$ \Phi_x: \xi \in \widehat G \mapsto \xi (x)\in \mathbb T, $$ for some $x\in G$, so that the map $$ \Phi: x\in G \mapsto \Phi_x\in \, \widehat {\!\widehat G} $$ turns out to be a group isomorphism. Pontryagin duality does not stop there as it also states that $\Phi$ is a topological homeomorphism! Interpreting the latter fact from the point of view of the definition of the topology on the dual group $\, \widehat {\!\widehat G}$, we see that a net $\{x_\alpha \}_\alpha $ in $G$ converges to some $x$ in $G$ iff $$ \sup_{\xi\in K}|\xi (x_\alpha )-\xi (x)| \to 0 $$ for every compact subset $K\subseteq \widehat G$. It is perhaps a good idea to contrast this with $(\dagger)$, paying special attention to the precise position of the subscript $\alpha $!

This said, given $\varepsilon >0$, and using the regularity of the measure $\mu $, and hence also of $|\mu |$, choose some compact $K\subseteq \widehat G$ such that $|\mu |(\widehat G\setminus K)<\varepsilon $. Next, choose some $\alpha _0$ such that for all $\alpha \geq \alpha _0$, one has that $$ \sup_{\xi\in K}|\xi (x_\alpha )-\xi (x)| <\varepsilon . $$ For all such $\alpha $ one then has that $$ \left|\int_{\widehat{G}} \xi(x_\alpha)d\mu(\xi)-\int_{\widehat{G}} \xi(x)d\mu(\xi)\right|\le \int_{\widehat{G}}|\xi(x_\alpha)-\xi(x)| d|\mu|(\xi) = $$$$ = \int_{K}|\xi(x_\alpha)-\xi(x)| d|\mu|(\xi) + \int_{\widehat{G}\setminus K}|\xi(x_\alpha)-\xi(x)| d|\mu|(\xi) \leq $$$$\leq \varepsilon |\mu |(K)+ 2 |\mu |(\widehat G \setminus K) \leq \varepsilon |\mu |(K)+ 2 \varepsilon . $$