Lemma 3

Let A be a unital involutive Banach algebra admitting an approximate unit (I, p. 120, Definition 7). Let $\lambda$ be a continuous positive linear form on A.

a) For every $a$ in A, one has $\lambda (a^*) =\overline{\lambda(a)}$ and $|\lambda (a)|^2\leqslant \|\lambda \|\lambda (a^*a)$;

b) Let $\widetilde{A}$ be the involutive algebra obtained from A by adjunction of a unit element, and let $e$ be its unit element. There exists a continuous positive linear form $\widetilde{\lambda}$ on $\widetilde{A}$ extending $\lambda$ and such that $\widetilde{\lambda}(e) =\|\lambda \|$;

c) For all $a$ and $b$ in A, one has $|\lambda (b^*ab)|\leqslant \|a\|\lambda (b^*b)$.

Let us prove a). Let $\mathfrak{F}$ be an approximate unit of A. Let $a\in A$. Using Lemma 2, a) and the definition of an approximate unit, one finds

$\lambda (a^*) =$ lim$_{f,\mathfrak{F}}\lambda (f a^*) =$ lim$_{f,\mathfrak{F}}\lambda (af^*) =$ lim$_{f,\mathfrak{F}}\lambda ((f a^*)^*) =\overline{\lambda(a)}$,

whence (loc. cit.)

$|\lambda (a)|^2=$ lim$_{f,\mathfrak{F}}|\lambda (f a)|^2\leqslant \lambda (a^*a)$ lim sup$_{f,\mathfrak{F}}\lambda (f f^*)\leqslant \|\lambda \|\lambda (a^*a)$,

since $\mathfrak{F}$ is a filter on the unit ball of A.

To prove b), one may suppose that $\lambda$ is not zero, then that $\|\lambda \|= 1$. For $a\in A$ and $z\in \mathbf{C}$, let us put $\widetilde{\lambda}(a+z\cdot e) =\lambda (a) +z$. The mapping $\widetilde{\lambda}$ is a continuous linear form on $\widetilde{A}$ which extends $\lambda$ and satisfies $\widetilde{\lambda}(e) = 1$. It is positive: for all $a\in A$ and $z\in \mathbf{C}$, by a), we calculate

Finally, let $b$ be an element of A. The mapping $a\mapsto \widetilde{\lambda}(b^*ab)$ is a positive linear form on the unital involutive Banach algebra $\widetilde{A}$. It is therefore continuous of norm equal to its value at $e$ (Lemma 2, b)), which is equal to $\lambda (b^*b)$, whence assertion c).