Proposition 10

Let $\varphi$ be a function of positive type on G and let $(\varrho_1, x_1)$ and $(\varrho_2, x_2)$ be Hilbertian realizations of $\varphi$, the Hilbertian representation $(\varrho_1, x_1)$ being cyclic. There exists a unique isometric G-morphism $u$ of $\varrho_1$ into $\varrho_2$ such that $u(x_1) =x_2$. If $(\varrho_2, x_2)$ is also cyclic, then $u$ is an isomorphism.

For $1\leqslant j\leqslant 2$, let us denote by $E_j$ the space of $\varrho_j$ and by $\gamma_j$ the function on G defined by $\gamma_j(g) =\varrho_j(g)x_j$ for every $g\in G$. The pairs $(E_1, \gamma_1)$ and $(E_2, \gamma_2)$ are Hilbertian realizations of the function of positive type $(g, h)\mapsto \varphi (g^{-1}h)$, and $(E_1, \gamma_1)$ is a cyclic Hilbertian realization. By Prop. 1 of V, p. 434, there exists a unique isometric linear mapping $u: E_1\rightarrow E_2$ such that $\gamma_2=u\circ \gamma_1$. In particular, one has $x_2=\gamma_2(e) =u(\gamma_1(e)) =u(x_1)$. Moreover, for all $g$ and $h$ in G, one has

and since the set of elements $\gamma_1(h)$ for $h\in G$ is total in $E_1$, this means that $u$ is a G-morphism. By loc. cit., it is an isometric isomorphism if $(\varrho_2, x_2)$ is also cyclic.