Right translation is a continuous action for fixed $f\in L^2(G)$

Question

Let $G$ be a compact Lie group, and consider the space $L^2(G)$ with the Haar measure. For $f\in L^2(G)$, denote $L_gf$ by the map which sends $x$ to $f(xg)$. I want to show that the map $$ g\mapsto L_gf $$ for fixed $f\in L^2(G)$ is continuous. I was thinking of giving an $\epsilon-\delta$ proof (since I have a lack of experience in this topic). So I want to argue that if $g$ is close enough to some fixed $g_0$, then $$ \int_G (L_gf-L_{g_0}f)^2<\epsilon. $$ However, I'm not exactly sure how to tackle this, as we don't have continuity of $f$. Maybe there are some results I am unaware of, which could make this proof straightforward? Any help is appreciated.

Answer 1

The result is true in general for $1\le p\le \infty. $ We use the density of the continuous functions in $L^p(G)$ together with the translation invariance of the Haar measure, $\mu$:

Without loss of generality, $g_0=e.$ Take an $h\in C(G)$ such that $\|h-f\|_p<\epsilon. \ $ There is an open set $U\ni e$ such that $\|h(xg)-h(x)\|_p<\epsilon$ whenever $xgx^{-1}\in U.$ This is possible because $h$ is uniformly continuous $^{*}$ and $\mu(G)=1.$ Then, we have

$\|f(xg)-f(x)\|_p=\|f(xg)-h(xg)+h(xg)-h(x)+h(x)-f(x)\|_p\leq\\ \|f(xg)-h(xg)\|_p+\|h(xg)-h(x)\|_p+\|h(x)-f(x)\|_p<3\varepsilon$

$^{*}$ Here is a sketch of the proof that $h$ is uniformly continuous:

$\underline{\text{Preliminaries}}$

$h: G\to \mathbb C$ is (left) uniformly continuous on $G$ if (and only if) for any $\epsilon > 0$ there exists an open neighborhood $V$ of $e$ such that $|h(g)-h(g')|<\epsilon$ for any $g,g'\in G$ such that $g^{-1}g'\in V.$ Equivalently,

$\tag1 |h(gv)-h(v)|<\epsilon\ \forall g\in G,v\in V$

The multiplication

$\tag2 \varphi:G\times G\to G:(g,g')\mapsto gg'$

is continuous, and

$\tag3\varphi_g:G\to G:g'\mapsto gg'$

is a homeomorphism for each $g\in G$ (in fact, if $G$ is a Lie group, then $\varphi_g$ is a diffeomorphism with respect to its manifold structure).

$\underline{\text{The proof}}$

Let $\epsilon>0.$ Using $(3)$, for each $g\in G$ there is a neighborhood $U_g\ni e$ such that $|h\circ \varphi_g(u)-h\circ \varphi_g(e)|=|h(gu)-h(ge)|=|h(gu)-h(g)|<\frac{\epsilon}{2}$ as soon as $u\in U_g.$ Now, from $(2)$ it follows that there is a neighborhood $V_g\ni e$ such that

$\tag4 V_gV_g\subseteq U_g$

$G$ is compact, so there is a $\{g_i\}^n_{i=1}$ such that $G=\bigcup^n_{i=1} g_iV_{g_i}$. Set $V:=\bigcap^n_{i=1}V_{g_i}$, which is open and non-empty and suppose $g\in G.$ Then, from this and $(4)$, we have for some $1\le i \le n,$ $g\in g_iV_{g_i}\subseteq g_iU_{g_i}$ so $g=g_iu$ for some $u\in U_{g_i}$ and so

$\tag 5|h(g)-h(g_i)|=|h(g_iu)-h(g_i)|<\frac{\epsilon}{2}$

And if $v\in V,$ then $gv\in g_iV_{g_i}V\subseteq g_iV_{g_i}V_{g_i}\subseteq g_iU_{g_i}$ so $gv=g_iu$ for some $u\in U_{g_i}$, which implies that

$\tag 6|h(gv)-h(g_i)|=|h(g_iu)-h(g_i)|<\frac{\epsilon}{2}$

To finish, note that $(5)$ and $(6)$ combine to give $(1).$