Proposition 9

Let A be a Banach algebra, let B be a commutative Banach algebra without radical, Every morphism of the underlying algebra of A into the underlying algebra of B is continuous.

Let $h: A\rightarrow B$ be an algebra morphism and let $(a, b)\in A\times B$ be a point adherent to the graph Γ of $h$. Let $\chi \in \mathsf{X}'(B)$. The function $x\mapsto \chi (h(x))$ from A into $\mathbf{C}$ is an algebra homomorphism, hence is continuous (I, p. 29, th. 1). The map from $A\times B$ into $\mathbf{C}$ given by $(x, y)\mapsto \chi (h(x))-\chi (y)$ is therefore continuous; it is zero on Γ, hence zero at $(a, b)$. Thus one has $\chi (h(a)) =\chi (b)$ for every $\chi \in \mathsf{X}'(B)$. Since B is without radical, one has $h(a) =b$. Thus the graph of $h$ is closed and therefore $h$ is continuous (EVT, I, p. 19, cor. 5).