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

Let $G$ be a locally compact Hausdorff group.

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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.

Any help/hint/suggestion/answer is highly appreciated! Thanks in advance for any help!

  • $\begingroup$ What is your question: To prove the Lemma or the uniform continuity of a compactly supported function? $\endgroup$
    – Paul Frost
    Feb 3, 2022 at 20:41
  • $\begingroup$ @PaulFrost I want to prove that this representation is continuous. How this is achieved is immaterial to me, just wanted to sketch my attempt. $\endgroup$
    – Andromeda
    Feb 3, 2022 at 20:46
  • $\begingroup$ 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? $\endgroup$
    – William M.
    Feb 3, 2022 at 22:51
  • $\begingroup$ @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? $\endgroup$
    – Andromeda
    Feb 3, 2022 at 22:57
  • $\begingroup$ I was probably thinking in the case where $G$ is compact. $\endgroup$
    – William M.
    Feb 3, 2022 at 23:37

1 Answer 1


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.

  • 1
    $\begingroup$ Nice answer! While reading through it, I corrected some typos/made some improvements. I hope you don't mind (otherwise you can rollback). $\endgroup$ Feb 4, 2022 at 13:19

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