Proposition 3

Let A be a stellar algebra and let $\widetilde{A}$ be the involutive algebra deduced from A by adjunction of a unit element. There exists a unique norm on $\widetilde{A}$ extending that of A and making $\widetilde{A}$ into a stellar algebra.

The uniqueness of such a norm results from prop. 2. Let us now prove its existence. Let $\widetilde{e}$ denote the unit element of $\widetilde{A}$.

Suppose first that A possesses a unit element $e$. The product of the normed involutive algebras A and $\mathbf{C}(\widetilde{e}-e)$ is identified with $\widetilde{A}$ and is a stellar algebra (example 5). The norm on $\widetilde{A}$ extends that of A, whence the assertion.

Suppose henceforth that A does not possess a unit element. For every $x\in \widetilde{A}$, let $\boldsymbol{\gamma }_x$ be the multiplication operator $y\mapsto xy$ from A into A, and set $\|x\|_{\widetilde{A}}=\|\boldsymbol{\gamma }_x\|$. The mapping $x\mapsto \|x\|_{\widetilde{A}}$ is a seminorm on $\widetilde{A}$. For all $x$ and $x'$ in $\widetilde{A}$, one has $\|xx'\|_{\widetilde{A}}\leqslant \|x\|_{\widetilde{A}}\|x'\|_{\widetilde{A}}$. Moreover, by lemma 7, one has $\|x\|_{\widetilde{A}}=\|x\|$ for all $x\in A$.

Let us show that the mapping $x\mapsto \|x\|_{\widetilde{A}}$ is a norm on $\widetilde{A}$. Let $\lambda \in \mathbf{C}$ and $x\in A$ be such that $\|\lambda \widetilde{e}+x\|_{\widetilde{A}}$ = 0. If $\lambda \not= 0$, the condition $(\lambda \widetilde{e}+x)y= 0$ for every $y\in A$ implies that $-\lambda^{-1}x$ is a left unit element in A. Analogously, the element $-\overline{\lambda}^{-1}x^*$ is a right unit element. Thus the algebra A would then possess a unit element, contrary to the assumption. Hence $\lambda = 0$. But then $0 =\|x\|_{\widetilde{A}}=$ $\|x\|$, and therefore $x= 0$.

Since A is complete and of codimension 1 in $\widetilde{A}$, the space $\widetilde{A}$ endowed with the norm $x\mapsto \|x\|_{\widetilde{A}}$ is complete. To conclude, it is therefore sufficient to show that one has $\|x\|^2_{\widetilde{A}}\leqslant \|x^*x\|_{\widetilde{A}}$ for every $x\in \widetilde{A}$ (Lemma 6). One may suppose that $\|x\|_{\widetilde{A}}= 1$. For every real number $r <1$, there therefore exists $y\in A$ such that $\|y\|=\|y^*\|\leqslant 1$ and $\|xy\|^2\geqslant r$. Since $xy\in A$, one has

It follows that $\|x^*x\|_{\widetilde{A}}\geqslant 1$, and therefore the involutive algebra $\widetilde{A}$ endowed with the norm $x\mapsto \|x\|_{\widetilde{A}}$ is a star algebra.