Proposition 13

Let $\varphi \in L^{\infty}(G)$. Denote by $\lambda_{\varphi}$ the continuous linear form $f\mapsto \langle f, \varphi \rangle$ on $L^1(G)$. Then $\varphi$ is the class of a function of positive type on G if and only if $\lambda_{\varphi}$ is a positive continuous linear form on $L^1(G)$.

Suppose that $\lambda_{\varphi}$ is a positive continuous linear form on $L^1(G)$. Let $(E, x,\widetilde{\pi})$ be a cyclic Hilbert realization of $\lambda$ (prop. 5 of V, p. 438).

Let $\pi$ be a unitary representation of G in E such that $\pi (f) =\widetilde{\pi}(f)$ for every $f\in L^1(G)$ (prop. 12). Then $\lambda_{\varphi}(f) =\langle x|\pi (f)x\rangle$ for every $f\in L^1(G)$.

Since

for every $f\in L^1$(G), we get

for every $f\in L^1(G)$ (INT, VI, p. 9, § 1, n$^o1$, prop. 1). Thus, $\varphi$ is the class in $L^{\infty}(G)$ of the function on G defined by $g\mapsto \langle x|\pi (g)x\rangle$, which is a function of positive type on G (prop. 9 of V, p. 443).

Conversely, let $\varphi \in$ Pos(G). Let $f\in \mathscr{K}(G)$. Then we have

(INT, VII, p. 19, § 1, n$^o3$, formula (22)). The continuous function on $G\times G$ defined by $(z, y)\mapsto \varphi (y)f(z)f(zy)$ is bounded and

according to INT, V, p. 94, § 8, n$^o3$, prop. 8. We therefore deduce from the Lebesgue-Fubini theorem (INT, V, p. 96, § 8, n$^o4$, th. 1) that

This implies that $\lambda_{\varphi}(f^**f)\geqslant 0$ by Theorem 1, (i) of V, p. 432 applied to the measure induced by $\mu$ on the support of $f$ (INT, IV, p. 186, § 5, n$^o7$, def. 4). By continuity it follows that $\lambda_{\varphi}(f^**f)\geqslant 0$ for every $f\in L^1(G)$.

The spaces $L^{\infty}(G)$ and $\mathscr{C}_b(G)$ are endowed with their topologies as Banach spaces, whose norm is denoted by $f\mapsto \|f\|_{\infty}$. Here we shall call weak topology on $L^{\infty}$(G), $\mathscr{C}_b(G)$ or Pos(G) the topology induced by the weak topology $\sigma (L^{\infty}(G),L^1(G))$.

Since $L^{\infty}(G)$ is identified with the dual of $L^1(G)$ (INT, V, p. 61, §5, n$^o8$, th. 4), every closed ball of $L^{\infty}(G)$ is compact for the weak topology (EVT, III, p. 17, cor. 3).