Central continuous approximate identities

$C(G)$ is dense in $L^2(G)$ by [L2], because a compact Lie group is a compact LCH space and $C_c(G)=C(G)$ there.

For a decreasing sequence $r_n\downarrow 0$ put $b_n(x):=\max \{0,1-d(x,e)\mathrm{/}{r}_n\}$; it is continuous, nonnegative, supported in the ball of radius $r_n$, central and inversion invariant by the bi-invariance of $d$, and its integral is positive because it is positive on the nonempty open ball of radius $r_n$; setting $k_n:=b_n\mathrm{/}{\int }_G{b}_n$ gives $k_n\ge 0$, continuous, central, inversion invariant, of integral one, with support shrinking to $\{e\}$.

For $k\in C(G)$ and $H\in L^2(G)$, $T_{k}H$ is continuous: for $x,x^{\mathrm{\prime }}\in G$ one has $\mathrm{\mid }{T}_{k}H(x)-T_{k}H(x^{\mathrm{\prime }})\mathrm{\mid }\le \mathrm{\parallel }k(x^{-1}\cdot )-k(x^{\mathrm{\prime }-1}\cdot ){\mathrm{\parallel }}_2\mathrm{\parallel }H{\mathrm{\parallel }}_2$ by Cauchy–Schwarz, and the first factor tends to $0$ as $x^{\mathrm{\prime }}\to x$ by uniform continuity of $k$.

Both regular representations are strongly continuous: given $H\in L^2(G)$ and $\epsilon >0$, choose $c\in C(G)$ with $\mathrm{\parallel }H-c{\mathrm{\parallel }}_2<\epsilon$ by step 1.1; since $c$ is uniformly continuous on $G$ and $G$ is compact, for $x$ close to $e$ one has $\mathrm{\mid }c(x^{-1}y)-c(y)\mathrm{\mid }<\epsilon$ for all $y$, so $2\epsilon +\mathrm{\parallel }{L}_{x}c-c{\mathrm{\parallel }}_2\ge \mathrm{\parallel }{L}_{x}H-H{\mathrm{\parallel }}_2$ tends to $0$ as $x\to e$; the same argument applies to $R_x$.

For $f\in C(G)$ and $x\in G$, $(T_{k_n}f)(x)-f(x)={\int }_G{k}_n(u)(f(xu)-f(x))du$ because $\int k_n=1$, so $\mathrm{\mid }{T}_{k_n}f(x)-f(x)\mathrm{\mid }\le {\sup }_{u\in \mathrm{supp}{k}_n}\mathrm{\mid }f(xu)-f(x)\mathrm{\mid }\to 0$ uniformly in $x$ by uniform continuity of $f$ and the shrinking supports; hence $T_{k_n}f\to f$ uniformly.

For $H\in L^2(G)$, $\mathrm{\parallel }{T}_{k_n}H-H{\mathrm{\parallel }}_2\le {\sup }_{u\in \mathrm{supp}{k}_n}\mathrm{\parallel }{R}_{u}H-H{\mathrm{\parallel }}_2$, again because $T_{k_n}H=\int k_n(u)R_{u}Hdu$ and $\int k_n=1$; by strong continuity (step 2.1) and the shrinking supports this tends to $0$.

Let $V\subseteq L^2(G)$ be closed and invariant under left translations, let $k$ be central, and let $H\in V$. Centrality gives $k(x^{-1}y)=k(y{x}^{-1})$. With $v=y{x}^{-1}$ and bi-invariance of Haar measure, $T_{k}H(x)={\int }_{G}k(v)H(vx)dv={\int }_{G}k(v)(L_{v^{-1}}H)(x)dv$. The map $v\mapsto L_{v^{-1}}H$ is continuous into $V$ by step 2.1, so this integral is an $L^2$-limit of finite linear combinations of elements of $V$; closedness gives $T_{k}H\in V$. If a closed subspace is invariant under conjugation, the same Riemann-sum argument applied to $g\mapsto H(g^{-1}\cdot g)$ gives stability under conjugation averaging; if it is invariant under inversion, applying the inversion operator preserves it by hypothesis.