The L1 action of a strongly continuous unitary representation

Fix $f\in L^1(K,\mu ;\mathbb{C})$ and $v\in H$: the orbit map $k\mapsto \pi (k)v$ is continuous on the compact space $K$ and hence strongly measurable, because for each $n$ the compact set $\pi (K)v$ is covered by finitely many balls of radius $1\mathrm{/}n$ whose open preimages disjointify to Borel sets on which the values at chosen centres define a measurable simple function within $1\mathrm{/}n$ of $\pi (\cdot )v$; choosing a measurable representative of $f$, [F3] applied to its real and imaginary parts provides scalar simple functions $s_n$ with $s_n\to f$ pointwise, so the simple products $s_n{t}_n$ converge pointwise to $f(k)\pi (k)v$, which is therefore strongly measurable by [F2]; since ${\int }_K\mathrm{\parallel }f(k)\pi (k)v\mathrm{\parallel }d\mu (k)=\mathrm{\parallel }v\mathrm{\parallel }\mathrm{\parallel }f{\mathrm{\parallel }}_1<\mathrm{\infty }$, the criterion [F4] makes $k\mapsto f(k)\pi (k)v$ Bochner integrable, and $\pi (f)v:={\int }_{K}f(k)\pi (k)vd\mu (k)$ is well defined and unchanged when $f$ or $v$ is changed on a null set; the integral is linear in the integrand, so $v\mapsto \pi (f)v$ is linear, and the norm inequality gives $\mathrm{\parallel }\pi (f)v\mathrm{\parallel }\le {\int }_K\mathrm{\mid }f(k)\mathrm{\mid }\mathrm{\parallel }\pi (k)v\mathrm{\parallel }d\mu (k)=\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }v\mathrm{\parallel }$, so by [F10] the operator $\pi (f)$ is bounded with $\mathrm{\parallel }\pi (f)\mathrm{\parallel }\le \mathrm{\parallel }f{\mathrm{\parallel }}_1$.

For every bounded linear functional $S:H\to \mathbb{C}$ the commutation theorem [F4] gives $S(\pi (f)v)={\int }_{K}f(k)S(\pi (k)v)d\mu (k)$, and taking $S=⟨\cdot ,w⟩$ gives the pairing formula $⟨\pi (f)v,w⟩={\int }_{K}f(k)⟨\pi (k)v,w⟩d\mu (k)$, which applied to $x\mapsto \pi (x)\pi (g)v$ also gives $⟨\pi (x)\pi (g)v,w⟩={\int }_{K}g(y)⟨\pi (x)\pi (y)v,w⟩d\mu (y)$; combining the two, $⟨\pi (f)\pi (g)v,w⟩={\int }_K{\int }_{K}f(x)g(y)⟨\pi (xy)v,w⟩d\mu (y)d\mu (x)$ for $f,g\in C(K)$ by [F7], since the integrand is bounded and $f,g$ are integrable on the probability space, and the substitution $z=xy$ with left invariance [F6] turns this into ${\int }_K(f\ast g)(z)⟨\pi (z)v,w⟩d\mu (z)=⟨\pi (f\ast g)v,w⟩$ by [F5]. Both sides of $\pi (f\ast g)=\pi (f)\pi (g)$ are bounded bilinear in $(f,g)$ with norm at most $\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1$ by [F5] and step 1.1, and they agree on the dense subset $C(K)\times C(K)$ by [F1], so the identity holds for all $f,g\in L^1(K)$; this is the first identity of (1).

For $v,w\in H$, the pairing formula of step 2.1 gives $⟨\pi (f{)}^{\ast }v,w⟩=⟨v,\pi (f)w⟩=\overline{{\int }_{K}f(x)⟨\pi (x)w,v⟩d\mu (x)}={\int }_K\overline{f(x)}⟨v,\pi (x)w⟩d\mu (x)={\int }_K\overline{f(x)}⟨\pi (x^{-1})v,w⟩d\mu (x)$, and substituting $y=x^{-1}$, which preserves $\mu$ by [F6], turns this into ${\int }_K\overline{f(y^{-1})}⟨\pi (y)v,w⟩d\mu (y)={\int }_K{f}^{\ast }(y)⟨\pi (y)v,w⟩d\mu (y)=⟨\pi (f^{\ast })v,w⟩$ because ${\mathrm{\Delta }}_K\equiv 1$ gives $f^{\ast }(y)=\overline{f(y^{-1})}$ by [F5]; since $v,w$ are arbitrary, $\pi (f^{\ast })=\pi (f{)}^{\ast }$, which is the second identity of (1).

By the pairing formula of step 2.1 and [F8], $⟨\pi (k)\pi (f)v,w⟩=⟨\pi (f)v,\pi (k{)}^{-1}w⟩={\int }_{K}f(x)⟨\pi (x)v,\pi (k{)}^{-1}w⟩d\mu (x)={\int }_{K}f(x)⟨\pi (k)\pi (x)v,w⟩d\mu (x)={\int }_{K}f(x)⟨\pi (kx)v,w⟩d\mu (x)$, and substituting $z=kx$ with left invariance [F6] gives ${\int }_{K}f(k^{-1}z)⟨\pi (z)v,w⟩d\mu (z)=⟨\pi (\lambda (k)f)v,w⟩$; similarly $⟨\pi (f)\pi (k)v,w⟩={\int }_{K}f(x)⟨\pi (x)\pi (k)v,w⟩d\mu (x)={\int }_{K}f(x)⟨\pi (xk)v,w⟩d\mu (x)$ and substituting $z=xk$ gives ${\int }_{K}f(z{k}^{-1})⟨\pi (z)v,w⟩d\mu (z)=⟨\pi (\rho (k{)}^{-1}f)v,w⟩$ because $\rho (k{)}^{-1}f(z)=f(z{k}^{-1})$ on the compact group; as $v,w$ are arbitrary, the two covariance identities of (2) follow.

For $f\in C(K)$ with ${\int }_{K}fd\mu =1$ the linearity of the Bochner integral gives $\pi (f)v-v={\int }_{K}f(k)(\pi (k)v-v)d\mu (k)$, so the norm inequality yields $\mathrm{\parallel }\pi (f)v-v\mathrm{\parallel }\le {\int }_K\mathrm{\mid }f(k)\mathrm{\mid }\mathrm{\parallel }\pi (k)v-v\mathrm{\parallel }d\mu (k)$; if $v\mathrm{\ne }0$, continuity of $k\mapsto \pi (k)v$ at the identity gives an open identity neighbourhood $U$ with $\mathrm{\parallel }\pi (k)v-v\mathrm{\parallel }<\mathrm{\parallel }v\mathrm{\parallel }\mathrm{/}2$ for $k\in U$, and [F9] applied to $\{e\}\subseteq U$ provides a nonnegative continuous $g$ with $g(e)>0$ and vanishing outside $U$, so $f:=g\mathrm{/}{\int }_{K}gd\mu$ is nonnegative continuous with integral one and vanishing outside $U$, whence $\mathrm{\parallel }\pi (f)v-v\mathrm{\parallel }<\mathrm{\parallel }v\mathrm{\parallel }\mathrm{/}2$ and $\pi (f)v\mathrm{\ne }0$; this proves (3). The Axiom of Choice is consumed through the normalized Haar probability and the cited Bochner, convolution and density suppliers; the computations above are choice-free apart from those inputs.