[proofplan] We define the left and right regular operators on $L^2(G,\mu)$ and first check that the definitions do not depend on the chosen representative of an $L^2$ class. Haar invariance gives unitarity of each translation operator. A direct pointwise computation on measurable representatives then proves that the left and right regular operators commute, and the same computation verifies the product law for the resulting $G\times G$ action. Strong continuity follows by approximating an arbitrary $L^2$ function by a [continuous function](/page/Continuous%20Function) and using [uniform continuity](/page/Uniform%20Continuity) of continuous functions on the compact group $G$. [/proofplan] [step:Define the translation operators on $L^2(G,\mu)$ and verify well-definedness] Let $g,h\in G$. For a Borel measurable representative \begin{align*} f:G\to\mathbb C \end{align*} of an element of $L^2(G,\mu)$, define the measurable maps \begin{align*} \ell_g f:G\to\mathbb C,\qquad x\mapsto f(g^{-1}x) \end{align*} and \begin{align*} r_h f:G\to\mathbb C,\qquad x\mapsto f(xh). \end{align*} The maps $x\mapsto g^{-1}x$ and $x\mapsto xh$ are diffeomorphisms of $G$, hence Borel measurable. If $f_1=f_2$ $\mu$-almost everywhere, let \begin{align*} N:=\{x\in G:f_1(x)\neq f_2(x)\}. \end{align*} Then $\mu(N)=0$. By left invariance of Haar measure, \begin{align*} \mu(gN)=\mu(N)=0. \end{align*} Since $\{x\in G:f_1(g^{-1}x)\neq f_2(g^{-1}x)\}=gN$, it follows that $\ell_g f_1=\ell_g f_2$ $\mu$-almost everywhere. Since $\mu$ is a normalized left Haar measure on the compact Lie group $G$, [citetheorem:8829] implies that $\mu$ is right invariant. Applying that right invariance to the Borel set $N$ and the element $h^{-1}\in G$ gives \begin{align*} \mu(Nh^{-1})=\mu(N)=0. \end{align*} Since $\{x\in G:f_1(xh)\neq f_2(xh)\}=Nh^{-1}$, it follows that $r_h f_1=r_h f_2$ $\mu$-almost everywhere. Thus $L_g$ and $R_h$ are well-defined on $L^2(G,\mu)$ by \begin{align*} L_g[f]=[\ell_g f],\qquad R_h[f]=[r_h f]. \end{align*} [guided] The only subtlety in defining $L_g$ and $R_h$ is that an element of $L^2(G,\mu)$ is an equivalence class of functions, not a single pointwise function. We therefore start with a Borel measurable representative \begin{align*} f:G\to\mathbb C \end{align*} and define \begin{align*} \ell_g f:G\to\mathbb C,\qquad x\mapsto f(g^{-1}x) \end{align*} and \begin{align*} r_h f:G\to\mathbb C,\qquad x\mapsto f(xh). \end{align*} These are measurable because left and right translations are diffeomorphisms of the Lie group $G$, hence Borel measurable maps. Now suppose $f_1$ and $f_2$ represent the same $L^2$ class. This means that the disagreement set \begin{align*} N:=\{x\in G:f_1(x)\neq f_2(x)\} \end{align*} has Haar measure zero. For the left translation, the new disagreement set is exactly $gN$, because \begin{align*} f_1(g^{-1}x)\neq f_2(g^{-1}x)\iff g^{-1}x\in N\iff x\in gN. \end{align*} Left invariance of Haar measure gives \begin{align*} \mu(gN)=\mu(N)=0. \end{align*} Therefore $\ell_g f_1=\ell_g f_2$ $\mu$-almost everywhere. For the right translation, the disagreement set is $Nh^{-1}$, since \begin{align*} f_1(xh)\neq f_2(xh)\iff xh\in N\iff x\in Nh^{-1}. \end{align*} Because $G$ is compact and $\mu$ is a normalized left Haar measure, [citetheorem:8829] applies and says that $\mu$ is right invariant: for every Borel set $A\subset G$ and every $a\in G$, $\mu(Aa)=\mu(A)$. We use this with $A=N$ and $a=h^{-1}$, obtaining \begin{align*} \mu(Nh^{-1})=\mu(N)=0. \end{align*} Thus $r_h f_1=r_h f_2$ $\mu$-almost everywhere. This proves that the formulas \begin{align*} L_g[f]=[\ell_g f],\qquad R_h[f]=[r_h f] \end{align*} are independent of the representative $f$. [/guided] [/step] [step:Prove that each left and right translation operator is unitary] Let $f\in L^2(G,\mu)$ and choose a Borel measurable representative, again denoted $f:G\to\mathbb C$. Using left invariance of $\mu$ under the substitution $y=g^{-1}x$, we obtain \begin{align*} \|L_g f\|_{L^2(G,\mu)}^2=\int_G |f(g^{-1}x)|^2\,d\mu(x)=\int_G |f(y)|^2\,d\mu(y)=\|f\|_{L^2(G,\mu)}^2. \end{align*} Using right invariance of $\mu$ under the substitution $y=xh$, which holds for compact groups by [citetheorem:8829], we likewise obtain \begin{align*} \|R_h f\|_{L^2(G,\mu)}^2=\int_G |f(xh)|^2\,d\mu(x)=\int_G |f(y)|^2\,d\mu(y)=\|f\|_{L^2(G,\mu)}^2. \end{align*} The inverse operators are $L_{g^{-1}}$ and $R_{h^{-1}}$, respectively: on representatives, $L_gL_{g^{-1}}=L_e=L_{g^{-1}}L_g$ and $R_hR_{h^{-1}}=R_e=R_{h^{-1}}R_h$. Hence $L_g$ and $R_h$ are unitary operators on $L^2(G,\mu)$. [/step] [step:Compute both orders of left and right regular translation] Let $f:G\to\mathbb C$ be a Borel measurable representative of an element of $L^2(G,\mu)$, and let $x\in G$. By the definitions of $L_g$ and $R_h$, \begin{align*} (L_gR_hf)(x)=(R_hf)(g^{-1}x)=f(g^{-1}xh). \end{align*} Similarly, \begin{align*} (R_hL_gf)(x)=(L_gf)(xh)=f(g^{-1}xh). \end{align*} Thus $L_gR_hf=R_hL_gf$ pointwise for every Borel measurable representative $f$. Passing from representatives to $\mu$-equivalence classes gives $L_gR_h=R_hL_g$ as operators on $L^2(G,\mu)$. [/step] [step:Verify the product law for the $G\times G$ action] Define \begin{align*} \rho:G\times G\to GL(L^2(G,\mu)),\qquad \rho(g,h):=L_gR_h. \end{align*} For $(g,h),(g',h')\in G\times G$ and a Borel measurable representative $f:G\to\mathbb C$, the commutation established above gives \begin{align*} \rho(g,h)\rho(g',h')f=L_gR_hL_{g'}R_{h'}f=L_gL_{g'}R_hR_{h'}f. \end{align*} For every $x\in G$, \begin{align*} (L_gL_{g'}R_hR_{h'}f)(x)=f(g'^{-1}g^{-1}xhh')=f((gg')^{-1}xhh'). \end{align*} Therefore \begin{align*} \rho(g,h)\rho(g',h')=\rho(gg',hh'). \end{align*} Also $\rho(e,e)$ is the identity operator on $L^2(G,\mu)$, where $e\in G$ is the identity element. Hence $\rho$ is a representation of $G\times G$. [/step] [step:Prove strong continuity of the product representation] Let $(g_k,h_k)_{k=1}^{\infty}$ be a sequence in $G\times G$ converging to $(g,h)\in G\times G$, and let $f\in L^2(G,\mu)$. Since $G$ is a compact Lie group, it is a compact [Hausdorff space](/page/Hausdorff%20Space), and normalized Haar measure is a finite regular Borel measure on $G$. Let $C(G;\mathbb C)$ denote the complex [vector space](/page/Vector%20Space) of continuous functions from $G$ to $\mathbb C$. Hence the standard density theorem for continuous functions in $L^2$ over a compact Hausdorff space with finite regular Borel measure gives that $C(G;\mathbb C)$ is dense in $L^2(G,\mu)$. Fix $\varepsilon>0$. Choose $u\in C(G;\mathbb C)$ such that \begin{align*} \|f-u\|_{L^2(G,\mu)}<\varepsilon. \end{align*} The function $u$ is uniformly continuous on the [compact space](/page/Compact%20Space) $G$. Because the map $G\times G\times G\to G$, $(a,b,x)\mapsto a^{-1}xb$, is continuous and $G$ is compact, $u(g_k^{-1}xh_k)\to u(g^{-1}xh)$ uniformly in $x\in G$. Hence \begin{align*} \|\rho(g_k,h_k)u-\rho(g,h)u\|_{L^2(G,\mu)}\to 0. \end{align*} Using unitarity of every $\rho(a,b)$ already proved for $a,b\in G$, we obtain \begin{align*} \|\rho(g_k,h_k)f-\rho(g,h)f\|_{L^2(G,\mu)}\leq 2\|f-u\|_{L^2(G,\mu)}+\|\rho(g_k,h_k)u-\rho(g,h)u\|_{L^2(G,\mu)}. \end{align*} Taking the limit superior in $k$ and then using the arbitrary choice of $\varepsilon>0$ gives $\rho(g_k,h_k)f\to\rho(g,h)f$ in $L^2(G,\mu)$. Since $G\times G$ is first countable as a Lie group, sequential continuity is continuity. Thus $\rho:G\times G\to GL(L^2(G,\mu))$ is strongly continuous in the strong operator topology. [/step] [step:Conclude that the product action is a strongly continuous unitary representation] For every $(g,h)\in G\times G$, the operator $\rho(g,h)=L_gR_h$ is a product of the unitary operators $L_g$ and $R_h$. Hence $\rho(g,h)$ is unitary. Its action on representatives is precisely \begin{align*} (\rho(g,h)f)(x)=f(g^{-1}xh). \end{align*} Together with the product law and strong continuity proved above, the displayed formula defines a strongly continuous unitary representation of $G\times G$ on $L^2(G,\mu)$. [/step]