Proposition 1

Let $f$ be a universally positive kernel on X. Let $(E_1, g_1)$ and $(E_2, g_2)$ be Hilbert realizations of $f$. Suppose that $(E_1, g_1)$ is a cyclic Hilbert realization. There then exists a unique continuous linear mapping $u: E_1\rightarrow E_2$ such that $g_2=u\circ g_1$. This mapping is isometric. If $(E_2, g_2)$ is also cyclic, then $u$ is an isomorphism.

The uniqueness of $u$ results from the fact that the image of $g_1$ is total in $E_1$. Let $F =\mathbf{C}^{(X)}$ and let $(e_x)_{x\in X}$ be the canonical basis of F. For $j= 1$ and $j= 2$, denote by $u_j$ the linear mapping from F into $E_j$ determined by $u_j(e_x) =g_j(x)$, and denote by $F_j$ its image; the space $F_1$ is dense in $E_1$.

Let $t=\sum t_xe_x$ be an element of F. One has

Consequently, there exists an isometric linear mapping $v$ from $F_1$ into $F_2$ such that $u_2=v\circ u_1$, and in particular $g_2(x) =v(g_1(x))$ for all $x\in X$. Since the image of $g_1$ is total in $E_1$, this mapping extends to an isometric linear mapping $u$ from $E_1$ into $E_2$ such that $g_2=u\circ g_1$.

By Lemma 8 of I, p. 107, the image of $u$ is closed in $E_2$. If $(E_2, g_2)$ is a cyclic Hilbertian realization, the image of $u$ is also dense in $E_2$ and $u$ is therefore an isomorphism.