Lemma 1

Let F be a Hilbert space, K a compact group, and s ↦ U(s) a representation of K in the group of invertible elements of $\mathscr{L}(F; F)$, continuous for the topology of pointwise convergence. There exists a nondegenerate positive hermitian form $\varphi$ on F such that

for all $s \in K, x \in F, y \in F$, and such that the topological vector space structure of F defined by $\varphi$ (TVS, V, §1, No. 3) is identical to the original structure of F.

Let $\alpha$ be a Haar measure on K. For any $x, y$ in F, the mapping $s \mapsto (U(s)x|U(s)y)$ is continuous. Set

It is immediate that $\varphi(x, y)$ is a sesquilinear form on F. Since the set of endomorphisms $U(s)$ is compact in $\mathscr{L}_s(F; F)$, there exists a constant M such that $\|U(s)\| \leq M$ for all $s \in K$. For every $x \in F$, we therefore have

whence the inequalities

which shows that $\varphi$ is positive and nondegenerate, and that the norm $\varphi(x, x)^{1/2}$ is equivalent to the norm $\|x\|$. Finally, for all $t \in K$,

When the field of scalars is $\mathbf{H}$, one argues exactly as before on replacing everywhere the function $s \mapsto (U(s)x|U(s)y)$ by the function $s \mapsto \Phi(sx, sy)$ defined on G, with values in $\mathbf{H}$. This completes the proof of the proposition.