Theorem 1

Let $f\in \mathscr{C}(X\times X)$. The following conditions are equivalent:

(i) For every compact subset Y of X and every positive measure $\mu$ on Y, the endomorphism of $L^2(Y, \mu)$ defined by the kernel $f|(Y\times Y)$ (Def. 1 of III, p. 29) is positive; in other words, one has

for every $h\in \mathscr{L}^2(Y, \mu)$;

(ii) For every integer $n\in \mathbf{N}$, every family $(x_i)_{0\leqslant i\leqslant n}$ in X and every family $(t_i)_{0\leqslant i\leqslant n}$ of complex numbers, one has

(iii) There exist a complex Hilbert space E and a continuous mapping $g: X\rightarrow E$ with total image such that $f(x, y) =\langle g(x)|g(y)\rangle$ for all $x$ and $y$ in X;

(iv) There exist a complex Hilbert space E and a continuous mapping $g: X\rightarrow E$ such that $f(x, y) =\langle g(x)|g(y)\rangle$ for all $x$ and $y$ in X.

It is apparent that (iii) implies (iv), and one sees that (i) implies (ii) by considering the discrete measure $\mu$ which is the image of the counting measure on $\{0, . . . , n\}$ under the mapping $i\mapsto x_i$ and the function $h$ such that

$x_i=x$

Let us prove that (iv) implies (i). Suppose that there exist a complex Hilbert space E and a continuous mapping $g: X\rightarrow E$ such that $f(x, y) =\langle g(x)|g(y)\rangle$ for every $(x, y)\in X\times X$. Let Y be a compact subset of $X,\mu$ a positive measure on Y and $h\in \mathscr{L}^2(Y, \mu)$. One has

by INT, V, p. 97, § 8, no$^o4$, Prop. 9.

Finally, let us prove that (ii) implies (iii). Let $\widetilde{E}$ be the space of complex measures with finite support on X. For $\mu_1$ and $\mu_2$ in $\widetilde{E}$, put

The sesquilinear form resulting on $\widetilde{E}$ is a positive Hermitian form. In fact, let $\mu\in \widetilde{E}$; there exist a finite family $(x_i)_{0\leqslant i\leqslant n}$ in X and complex numbers $(t_i)_{0\leqslant i\leqslant n}$ such that $\mu=\sum^n_{i=0}t_i\varepsilon_{x_i}$. One then has

by hypothesis. Define $\widetilde{g}: X\rightarrow \widetilde{E}$ by $\widetilde{g}(x) =\varepsilon_x$. The image of the mapping $\widetilde{g}$ generates $\widetilde{E}$. Moreover, for every $(x, y)\in X\times X$, one has on the one hand $f(x, y) =\langle \widetilde{g}(x)|\widetilde{g}(y)\rangle$ and on the other hand

which implies that $\widetilde{g}$ is continuous since $f$ is continuous.

Let E be the separated-completed Hilbert space of $\widetilde{E}$ (EVT, V, p. 8, cor. of the prop. 4) and $g: X\rightarrow E$ the composition of $\widetilde{g}$ and of the canonical mapping $\widetilde{E}\rightarrow E$. Then the mapping $g$ is continuous, its image is total in E, and one has $f(x, y) =\langle g(x)|g(y)\rangle$ for all $(x, y)\in X\times X$.

The method used to prove that (ii) implies (iii) is called the Gelfand–Naimark–Segal construction.