Lemma 4

Let A be an involutive normed algebra. For every continuous linear form $f$ on A, one has $\|f^*\|=\|f\|$. If moreover $f$ is hermitian, then $\|f\|=\|f|A_h\|$.

The first assertion follows from the definitions. For the second, let $g$ denote the restriction of $f$ to $A_h$. One has $\|f\|\geqslant \|g\|$. Let us prove the converse inequality. For every $\varepsilon >$ 0, there exists $x\in$ A such that $\|x\|\leqslant 1$ and $|f(x)|\geqslant \|f\| -\varepsilon$. Multiplying $x$ by a complex number of modulus 1, we may suppose $f(x)\geqslant 0$. Then the element $\frac{1}{2}(x+x^*)$ belongs to $A_h$ and has norm $\leqslant 1$. One has

$|(1_*)|$ 1 $*$

2 2

hence $\|g\|\geqslant \|f\| -\varepsilon$. It follows that $\|g\|\geqslant \|f\|$.

In the sequel, hermitian continuous linear forms on A and real continuous linear forms on $A_h$ will be identified.