Proposition 12

Let E be a complex Hilbert space and $\widetilde{\pi}$ an involutive algebra morphism from $L^1(G)$ into $\mathscr{L}(E)$. If the representation $\widetilde{\pi}$ is non-degenerate, then there exists a unique unitary representation $\pi$ of G in E such that $\widetilde{\pi}(f) =\pi (f)$ for every $f\in L^1(G)$.

The uniqueness follows from INT, VIII, p. 139, § 2, n$^o7$, cor. 3 of lemma 4.

Let us prove the existence of $\pi$. We have $\|\widetilde{\pi}\|\leqslant 1$ by Prop. 2 of I, p. 104. Let F be the subspace of E generated by the vectors of the form $\widetilde{\pi}(f)x$ for $f$ in $L^1(G)$ and $x$ in E; it is dense in E since the representation $\widetilde{\pi}$ is non-degenerate.

For every compact neighbourhood V of $e$, let $\varphi_V$ be a positive continuous function with support contained in V such that $\int\varphi_V\mu= 1$. Let $\mathfrak{B}$ be a base of the filter of compact neighbourhoods of $e$.

Let $g$ be in G and $f$ in $L^1(G)$. We have $(\varphi_V*\varepsilon_g)*f\rightarrow \varepsilon_g*f$ in $L^1(G)$ along the filter of sections of $\mathfrak{B}$ (INT, VIII, p. 172, § 4, n$^o7$, Prop. 20), hence $\widetilde{\pi}(\varphi_V*\varepsilon_g)\widetilde{\pi}(f)$ converges to $\widetilde{\pi}(\varepsilon_g*f)$ in $\mathscr{L}(E)$. This implies that $\widetilde{\pi}(\varphi_V*\varepsilon_g)$ converges simply along the filter of sections of $\mathfrak{B}$ to a linear mapping $\pi (g)$ of F into F; this mapping is continuous by EVT, III, p. 26, Cor. 3, since $\|\widetilde{\pi}(\varphi_V*\varepsilon_g)\|\leqslant 1$ for every compact neighbourhood V of $e$. There therefore exists a unique endomorphism of E which induces $\pi (g)$ by passing to subspaces. We again denote this endomorphism by $\pi (g)$; we have $\|\pi (g)\|\leqslant 1$.

Let $f\in L^1(G)$. Since $\widetilde{\pi}(\varphi_V*\varepsilon_g)\widetilde{\pi}(f)$ converges to $\widetilde{\pi}(\varepsilon_g*f)$, we have $\pi (g)\widetilde{\pi}(f) =\widetilde{\pi}(\varepsilon_g*f)$.

For $g=e$, this relation shows that $\pi (e)$ is the identity on F, hence $\pi (e) = 1_E$. Let $g_1$ and $g_2$ be in G. We have

whence $\pi (g_1)\pi (g_2) =\pi (g_1g_2)$ on F, and therefore on E. This proves that the mapping $g\mapsto \pi (g)$ is a representation of G in E. Since $\|\pi (g)\|\leqslant$ 1 and $\|\pi (g)^{-1}\|$ = $\|\pi (g^{-1})\|\leqslant$ 1, the endomorphisms $\pi (g)$ of E are isometric, hence unitary (EVT, V, p. 40, Prop. 3).

Let $f\in L^1(G)$ and $x\in E$. The mapping $g\mapsto \varepsilon_g*f$ of G into $L^1(G)$ is continuous (INT, VIII, p. 136, § 2, n$^o5$ and p. 144, § 3, n$^o2$, formula (5)), and $\widetilde{\pi}$ is continuous, hence the mapping $g\mapsto \widetilde{\pi}(\varepsilon_g*f)x=\pi (g)\widetilde{\pi}(f)x$ is continuous at $g=e$. Since F is dense in E, the representation $\pi$ is unitary (Lemma 4 of V, p. 380).

Let $f_1$ and $f_2$ be in $L^1(G)$. By INT, VIII, p. 127, §1, n$^o5$, Prop. 7, we have the relation

in $L^1$(G), whence

using INT, VI, p. 9, § 1, n$^o1$, prop. 1. It follows that

for every $f\in L^1(G)$. The proposition is proved.