Proposition 5

Let A be an involutive Banach algebra admitting an approximate unit. Every continuous positive linear form on A admits a cyclic Hilbertian realization $(E, x, \varphi )$. If A is unital, then there exists such a realization in which the morphism $\varphi$ is unital.

One may suppose that A is a unital algebra (Lemma 3, b)).

Let $\lambda$ be a continuous positive linear form on A. Let $(E, g)$ be a cyclic Hilbertian realization of the universally positive kernel $f$ on A defined by $f(a, b) =\lambda (a^*b)$ (Prop. 4 and Theorem 1). The mapping $g$ is continuous, its image is total in E, and one has $\lambda (a^*b) =\langle g(a)|g(b)\rangle$ for every $(a, b)\in A^2$. Put $x=g(1)\in E$.

The mapping $g$ of A into E is linear: in fact, for $(a, b, c)\in A^3$ and $(s, t)\in \mathbf{C}^2$, we have

whence the assertion, since the image of $g$ is total in E.

In particular, the image F of $g$ is a dense vector subspace of E. The kernel of $g$ is a left ideal of A: if $g(b) = 0$, then for all $a$ and $c$ in A, we have

therefore $g(ab) = 0$ since F is dense in E.

Let $a\in A$. Since the kernel of $g$ is a left ideal of A, there exists a linear mapping $\widetilde{\varphi}(a) : F\rightarrow F$ such that $\widetilde{\varphi}(a)(g(b)) =g(ab)$ for every $b\in A$. Moreover, for every $b\in A$, we have

(lemma 3, c)), therefore $\widetilde{\varphi}(a)$ is continuous and of norm $\leqslant \|a\|$. Consequently, there exists a unique endomorphism $\varphi (a)\in \mathscr{L}(E)$ which induces $\widetilde{\varphi}(a)$ by passing to the subspaces.

Let $a$ and $b$ be in A. We have

for every $c$ in A, therefore $\varphi (ab) =\varphi (a)\circ \varphi (b)$ since F is dense in E.

The mapping $\varphi$ of A into $\mathscr{L}(E)$ is linear: in fact, for every $(a, b)\in A^2$ and every $(s, t)\in \mathbf{C}^2$, we have

for every $c\in A$, whence $\varphi (sa+tb) =s\varphi (a) +t\varphi (b)$ since F is dense in E. Since $\|\varphi (b)\|\leqslant \|b\|$ for every $b\in A$, the mapping $\varphi$ is continuous. We also have $\varphi (1) = 1_E$ since $\varphi (1)(g(a)) =g(a)$ for every $a\in A$ and F is dense in E.

Lastly, let $a,b,c$ be in A. We have

whence $\varphi (a^*) =\varphi (a)^*$ since F is dense in E.

In conclusion, the mapping $\varphi$ is a continuous morphism of involutive algebras of A into $\mathscr{L}(E)$ and $g(a) =\varphi (a)x$ for every $a\in A$; consequently, the triplet $(E, x, \varphi )$ is a cyclic Hilbertian realization of $\lambda$.