Haar measure on locally compact group

Question

Please I need a help to solve two problems in the book of principles of Harmonic analysis of Deitmar and Echterhoff Exercise 1.4 Let $G$ be a locally compact group with Haar measure $\mu$, and let $S\subset G$ be a measurable subset with $0< \mu (S) < \infty$. Show the map $x \mapsto \mu(S \cap xS)$ from $G$ to $\mathbb R$ is continuous. They give me a hint to write the map as $\langle 1_S,L_{x^{-1}}1_S \rangle$ and then use the Cauchy Schwarz inequality. In Exercise 1.5 Show that the set $K$ of all $k\in G$ with $\mu(S \cap kS)$=$\mu(S)$ is a closed subgroup of $G$.

Answer 1

Fix a measurable subset $S$ with $0<\mu(S)<\infty$. Since continuity on locally compact topological group is determined at the identity $e\in G$, it suffices to check that the map $x\mapsto \mu(S\cap xS)$ is continuous at $x=e$. Let $f$ be a compactly supported continuous function such that $\left\|f-1_{S}\right\|_{L^{2}(G)}<\epsilon$, given $\epsilon>0$. Since $G$ is locally compact, there exists a nnbhd $V\ni e$ which is compact and symmetric. Observe that for any $x\in V$, $L_{x}f$ has compact support $xK\subset VK$, where $K:=\text{supp}(f)$. Since $f$ is uniformly continuous on $G$, there exists an open symmetric nbhd $W\ni e$ such that $y^{-1}x\in W$ or $xy^{-1}\in W$ $\Rightarrow \left|f(x)-f(y)\right|<\epsilon/(\mu(VK))^{1/2}$. Set $U:=W\cap V$ to obtain a symmetric nbhd of $e$. For $x\in U$, uniform continuity gives $\left|L_{x}f(y)-f(y)\right|<\epsilon/(\mu(VK))^{1/2}$ for all $y\in G$.

Observe that $1_{S\cap xS}=1_{S}1_{xS}=1_{S}L_{x}1_{S}$, where $L_{x}1_{S}(y)=1_{S}(x^{-1}y)$. By Cauchy-Schwarz,

$$\begin{array}{lcl}\displaystyle\left|\mu(S)-\mu(S\cap xS)\right|&=&\displaystyle\left|\int_{G}1_{S}(y)\left[\overline{1_{S}}(y)-\overline{1_{xS}}(y)\right]d\mu(y)\right|\\[2 em]&\leq&\displaystyle\left(\int_{G}\left|1_{S}(y)dy\right|^{2}d\mu(y)\right)^{1/2}\left(\int_{G}\left|1_{S}(y)-1_{xS}(y)\right|^{2}d\mu(y)\right)^{1/2}\end{array}$$

Using the triangle inequality we have that the second factor above is $\leq$ $$\left\|1_{S}-f\right\|_{L^{2}(G)}+\left\|f-L_{x}f\right\|_{L^{2}(G)}+\left\|L_{x}f-1_{xS}\right\|_{L^{2}(G)}\leq\epsilon+\left\|f-L_{x}f\right\|_{L^{2}(G)}+\left\|L_{x}f-1_{xS}\right\|_{L^{2}(G)}$$

By left-invariance of $\mu$, $\left\|L_{x^{-1}}f-1_{xS}\right\|_{L^{2}(G)}<\epsilon$. $$\left\|f-L_{x}f\right\|_{L^{2}(G)}^{2}\leq\int_{G}\left|f(y)-L_{x}f(y)\right|^{2}d\mu(y)\leq\int_{G}\left[\epsilon/(\mu(VK))^{1/2}\right]^{2}\cdot 1_{VK}(y)d\mu(y)=\epsilon^{2}$$

Putting all these results together, we obtain that $$\forall x\in U, \qquad \left|\mu(S)-\mu(S\cap xS)\right|\leq 3\mu(S)^{1/2}\epsilon$$ This completes the proof.

For the second exercise, it suffices to show that if $(x_{\alpha})_{\alpha\in\Lambda}\subset K$ is a net converging to some $x\in G$, then $x\in K$. But since the map $y\mapsto \mu(S\cap yS)$ is continuous, as shown above, we see that

$$\mu(S)=\mu(S\cap x_{\alpha}S)\rightarrow\mu(S\cap xS),$$

whence $\mu(S\cap xS)=\mu(S)$ by uniqueness of limits. So $x\in K$.