Proposition 6

Let A be an involutive Banach algebra and let $\lambda$ be a continuous positive linear form on A. Let $(E_1, x_1, \varphi_1)$ and $(E_2, x_2, \varphi_2)$ be Hilbertian realizations of $\lambda$. Suppose that $(E_1, x_1, \varphi_1)$ is a cyclic Hilbertian realization.

a) There exists a unique continuous linear mapping $u$ from $E_1$ into $E_2$ which is a morphism of representations of $\varphi_1$ into $\varphi_2($I, p. 11) and which satisfies $u(\varphi_1(a)x_1) =\varphi_2(a)x_2$ for every $a\in A$;

b) The linear mapping $u$ is isometric;

c) If $(E_2, x_2, \varphi_2)$ is cyclic, then $u$ is an isomorphism;

d) If $(E_2, x_2, \varphi_2)$ is cyclic and if A is unital, then $u(x_1) =x_2$.

For $j\in \{1,2\}$, define $\gamma_j: A\rightarrow E_j$ by $\gamma_j(a) =\varphi_j(a)x_j$ for every $a\in A$. By definition, the pairs $(E_j, \gamma_j)$ are Hilbertian realizations of the universally positive kernel $f$ on A defined by $(a, b)\mapsto \lambda (a^*b)$. The Hilbertian realization $(E_1, \gamma_1)$ is cyclic; by prop. 1 of V, p. 434, there exists therefore a unique continuous linear mapping $u: E_1\rightarrow E_2$ such that $\gamma_2=u\circ \gamma_1$, and this mapping is isometric. Moreover, if $(E_2, x_2, \varphi_2)$ is also cyclic, then $u$ is an isomorphism. To prove a), b) and c), it is enough to prove that $u$ is a morphism of representations of $\varphi_1$ into $\varphi_2$.

Let $a\in A$. For every $b\in A$, we have

therefore the continuous linear mappings $u\circ \varphi_1(a)$ and $\varphi_2(a)\circ u$ coincide on the subspace of $E_1$ generated by the image of $\gamma_1$; this subspace is dense in $E_1$ by hypothesis, whence $u\circ \varphi_1(a) =\varphi_2(a)\circ u$.

Finally, let us prove d). Suppose therefore that A is unital and that $(E_2, x_2, \varphi_2)$ is cyclic. There exists a cyclic Hilbertian realization $(E_3, x_3, \varphi_3)$ of $\lambda$ such that $\varphi_3$ is a unital morphism (prop. 5). Let $j\in \{1,2\}$. Applying the foregoing to $(E_j, x_j, \varphi_j)$ and $(E_3, x_3, \varphi_3)$, we see that there exists an isometric isomorphism $\widetilde{u}_j$ of $E_j$ into $E_3$ such that $\widetilde{u}_j\circ \varphi_j(a) =\varphi_3(a)\circ \widetilde{u}_j$ for every $a\in A$. Taking $a= 1_A$, we see that $\varphi_j$ is unital. We then have $x_2=\varphi_2(1_A)x_2=u(\varphi_1(1_A)x_1) =u(x_1)$.