Theorem 4 (Gelfand–Raikov)

For every $x\not =e$ in G, there exists an irreducible unitary representation $\pi$ of G such that $\pi (x)$ is not the identity endomorphism of the space of $\pi$.

Let $x\in G$ be such that $\pi (x)$ is the identity for every $\pi \in \widehat{G}$. It follows that $\varphi (x) =\varphi (e) = 1$ for every extreme point $\varphi$ of Pos$_1(G)$ (prop. 11 of V, p. 444), hence for every $\varphi \in$ Pos$_1(G)$ by prop. 14 since the linear form $\varphi \mapsto \varphi (e)$ is continuous on Pos$_1(G)$ endowed with the weak topology (th. 3). But if $x\not =e$, there exists $f\in L^2(G)$ of norm 1 such that $\langle f|\boldsymbol{\gamma }_G(x)f\rangle = 0$(V, p. 406, lemma 3). Since the left-hand side of this equality is of the form $\varphi (x)$, where $\varphi \in$ Pos$_1$(G), this is a contradiction.

If G is not locally compact, it is not always true that the irreducible unitary representations of G separate the points of G.