Lemma 2

Let A be an involutive Banach algebra and $\lambda$ a positive linear form on A.

a) For all $a$ and $b$ in A, one has $\lambda (a^*b) =\overline{\lambda(b^*a)}$ and

b) If A is unital, then the linear form $\lambda$ is continuous and its norm is equal to $\lambda (1)$.

The mapping $(a, b)\mapsto \lambda (a^*b)$ is a positive Hermitian form on A; it therefore satisfies $\lambda (a^*b) =\lambda (b^*a)$ and $|\lambda (a^*b)|^2\leqslant \lambda (a^*a)\lambda (b^*b)$ for all $a$ and $b$ in A (EVT, V, p. 2, remark and p. 3, prop. 2).

Let us prove b). Let $a\in A$ be a hermitian element of norm $<1$. The spectral radius of $a$ is less than $\|a\|$, hence the spectrum of $a$ is contained in the open unit disc D of $\mathbf{C}$ (Theorem 1 of I, p. 24). Let $f$ be a holomorphic function on D such that $f(z)^2= 1-z$ for all $z\in D$ (Lemma 1 of V, p. 435). Applying the holomorphic functional calculus to the element $a$ and to the function $f$, the element $b=f(a)$ satisfies $b^2= 1-a($I, p. 74, Theorem 5). By Prop. 3 of V, p. 435, one moreover has $f(a)^*=f^*(a^*) =f(a)$, hence $b$ is hermitian. We then have $\lambda (1-a) =\lambda (b^*b)\geqslant 0$, whence $\lambda (a)\leqslant \lambda (1)$.

Now let $b\in A$ be of norm $<1$. The element $b^*b$ is hermitian and $\|b^*b\|<1$, hence applying a) with $a= 1$, we find

with equality if $b= 1$. The linear form $\lambda$ is therefore continuous, and its norm is equal to $\lambda (1)$. Assertion b) is proved.