# Is a compactly supported function on a locally compact Hausdorff space uniformly continuous?

Consider the following fragment from Folland's book "A course in abstract harmonic analysis":

Let $$G$$ be a locally compact Hausdorff group.

I want to prove that the representation $$\pi: G \to B(L^2(\mu))$$ defined by $$(\pi(x)f)(s) = f(x^{-1}s)$$ is strongly continuous. I.e. if $$f \in L^2(\mu)$$ and $$\{g_\alpha\} \to 1$$, we must show that $$\int_G |f(g_\alpha^{-1}s)-f(s)|^2 d \mu(s) = \|\pi(g_\alpha)f-f\|_2^2 \to 0.$$

My attempt: A routine density argument shows that it is sufficient to prove this for $$f \in C_c(S)$$.

In the hope of accomplishing this, I try to prove the following lemma (I try to mimique the proof of proposition 2.42, as suggested by Folland):

Lemma: Let $$X$$ be a locally compact Hausdorff space and $$G$$ a locally compact Hausdorff topological group together with a continuous action $$G\times X \to X$$. If $$f \in C_c(X)$$, then the following property is satisfied:

For every $$\epsilon >0$$, there exists an open subset $$V\subseteq G$$ containing $$1$$ such that for all $$g \in V$$, we have $$\sup_{x \in X}|f(g^{-1}x)-f(x)| <\epsilon$$.

Given this lemma, I can finish, but the classical proof (where $$X=G$$ and we consider the translation action) does not generalise to general actions.

To prove the lemma, I tried to modify the proof of proposition 2.6 in Folland's book, but it becomes clear that this proof no longer works, though I still hope that there exists some modification that saves the argument.

• What is your question: To prove the Lemma or the uniform continuity of a compactly supported function? Feb 3, 2022 at 20:41
• @PaulFrost I want to prove that this representation is continuous. How this is achieved is immaterial to me, just wanted to sketch my attempt. Feb 3, 2022 at 20:46
• Because $f$ has compact support, $x \in X$ really can be changed to $x \in K$ for some compact $K.$ For every such $x \in K$ you have a neighbourhood $V$ where the inequality you want happens due to continuity. Now cover $K$ by finitely many such neighbourhoods. Is this not a good approach? Feb 3, 2022 at 22:51
• @WilliamM. I definitely tried this, but it didn't work out (basically because the term $g^{-1}x$ does annoying things). Could you provide more details? Perhaps write an answer? Feb 3, 2022 at 22:57
• I was probably thinking in the case where $G$ is compact. Feb 3, 2022 at 23:37

Here is a proof of the Lemma in the OP based an a nice auxiliary result by Lynn H. Loomis.

Lemma (Loomis): Suppose $$X,Y$$ are locally compact Hausdorff (l.c.H) spaces and $$Z$$ a topological space, and let $$f:X\times Y\rightarrow Z$$ be a continuous function. If $$K\subset X$$ is compact and $$U\subset Z$$ open, then $$W=\{y\in Y: f(x,y)\in U\ \text{for all}\ x\in K\}$$ is open in $$Y$$.

Proof: Fix $$y_0\in W$$. We show that there is an open neighborhood $$V_{y_0}$$ of $$y_0$$ fully contained in $$W$$. The continuity of $$f$$ implies that for each $$x\in K$$, there are open neighborhoods $$A_x\subset X$$ containing $$x$$ and $$B_x$$ containing $$y_0$$ such that $$f(A_x\times B_x)\subset U$$. By compactness of $$K$$, there are finite $$x_1,\ldots, x_m\in K$$ such that $$K\subset \bigcup^m_{j=1}A_{x_j}$$. The set $$V_{y_0}=\bigcap^m_{j=1}B_{x_j}$$ is open, contains $$y_0$$. Suppose $$y\in V_{y_0}$$, and let $$x\in K$$. Choose $$A_{x_j}$$ containing $$x$$. It follows that $$(x,y)\in A_{x_j}\times V_{y_0}\subset A_{x_j}\times B_{x_j}$$ and so, $$f(x,y)\in U$$.

As in the OP, let $$X$$ be a l.c.H space and $$G$$ a l.c.H topological group that acts continuous on $$X$$, and let $$f\in\mathcal{C}_c(X)$$, ale let $$K=\operatorname{supp}(f)$$. Recall that an action can be seen as a continuous function $$\Phi:G\times X\rightarrow X$$ such that $$\Phi(1,x)=x$$ for all $$x\in X$$ and $$\Phi(gh,x)=\Phi(g,\Phi(h, x))$$ for all $$g,h\in G$$ and $$x\in X$$. It is usual to denote the group action as $$g\cdot x:= \Phi(g,x)$$.

Choose any symmetric neighborhood $$U$$ of $$1\in G$$ with compact closure. Define $$F:X\times G\rightarrow \mathbb{R}$$ by $$F(x,g)= f(g^{-1}\cdot x)-f(x)=f(\Phi(g^{-1},x))-f(x)$$ This is a continuous function. Notice that $$\overline{U}\cdot K:=\Phi(\overline{U}\times K)\subset X$$ is compact. The Lemma above shows that $$W=\{g\in G: |F(x,g)|<\varepsilon,\,\text{for all}\, x\in \overline{U}\cdot K\}$$ is an open subset of $$G$$ which contains $$1$$; hence, $$V:=(W\cap W^{-1})\cap U$$ is a symmetric open neighborhood of $$1$$.

Let $$s\in V$$, and let $$x\in X$$. If $$x\in \overline{U}\cdot X$$, then by definition of $$W$$, $$|f(s^{-1}\cdot x)-f(x)|=|F(x,s)|<\varepsilon$$ If $$x\notin \overline{U}\cdot K$$, then $$g^{-1}\cdot x\notin K$$ for all $$g\in V$$. Indeed, if $$g^{-1}\cdot x\in K$$ for some $$g\in V$$, then $$x=ss^{-1}\cdot x=s\cdot(s^{-1}\cdot x)\in U\cdot K$$ which is a contradiction. This means that $$|f(s^{-1}\cdot x)-f(x)|=|f(s^{-1}\cdot x)-f(1\cdot x)|=0<\varepsilon$$

This proves the Lemma in the OP.

• Nice answer! While reading through it, I corrected some typos/made some improvements. I hope you don't mind (otherwise you can rollback). Feb 4, 2022 at 13:19