Con. of a Homo. from a Com. Ban. Alg. to a Semi-Sim. Com. Ban. Alg.

# Continuity of a Homomorphism from a Commutative Banach Algebra to a Semi-Simple Commutative Banach Algebra

Recall from the Semi-Simple Commutative Banach Algebras page that a commutative Banach algebra is said to be semi-simple if:

(1)\begin{align} \quad \mathrm{Rad}(X) = \{ 0 \} \end{align}

If $X$ and $Y$ are commutative Banach algebras and furthermore, if $Y$ is semi-simple, then every homomorphism $\alpha : X \to Y$ is continuous as we prove below.

Theorem 1: Let $X$ be a commutative Banach algebra over $\mathbb{C}$ and let $Y$ be a semi-simple commutative Banach algebra over $\mathbb{C}$. If $\alpha$ is a homomorphism from $X$ to $Y$ then $\alpha$ is continuous on $X$. |

*Recall that a homomorphism $\alpha$ from $X$ to $Y$ is a linear operator from $X$ to $Y$ such that $\alpha (xy) = \alpha(x) \alpha(y)$ for all $x, y \in X$ and that every multiplicative linear functional on $X$ is a nonzero homomorphism from $X$ to $\mathbb{C}$.*

**Proof:**Let $\alpha : X \to Y$ be a homomorphism. Let $(x_n)$ be a sequence in $X$ that converges to $0$ and suppose that $(\alpha(x_n))$ converges to some $y \in Y$. We aim to show that $y = \alpha (0) = 0$ to apply the closed graph theorem.

- Let $f \in \Phi_Y$. Observe that since $\alpha$ is a homomorphism $\alpha$ is by definition a linear operator from $X$ to $Y$. $f$ is also a linear functional on $Y$, so the composition $f \circ \alpha : X \to \mathbb{C}$ is a linear functional on $X$. We first show that $\alpha \circ f$ is further a multiplicative linear functional on $X$.

- Let $x_1, x_2 \in X$. Since $\alpha$ is a homomorphism and $f$ is multiplicative we have that:

\begin{align} \quad (f \circ \alpha)(x_1x_2) = f(\alpha(x_1x_2)) = f(\alpha(x_1)\alpha(x_2)) = f(\alpha(x_1))f(\alpha(x_2)) = (f \circ \alpha)(x_1) (f \circ \alpha)(x_2) \end{align}

- Thus $\alpha \circ f$ is a multiplicative linear functional and so $\alpha \circ f \in \Phi_X$. Therefore $\alpha \circ f$ is continuous by the theorem on the Multiplicative Linear Functionals on a Banach Algebra page. Since $\alpha \circ f$ is continuous on $X$ and $f$ is continuous on $Y$ we have that:

\begin{align} \quad f(y) = f \left ( \lim_{n \to \infty} \alpha(x_n) \right ) = \lim_{n \to \infty} (f \circ \alpha)(x_n) = (f \circ \alpha) \left ( \lim_{n \to \infty} x_n \right )= (f \circ \alpha)(0) = 0 \end{align}

- Therefore $f(y) = 0$ for all $f \in \Phi_Y$, that is, $\hat{y}(f) = 0$ for every $f \in \Phi_Y$.

- Since $Y$ is a semi-simple commutative Banach algebra we have by the theorem on the Semi-Simple Commutative Banach Algebras page that the Gelfand representation $y \to \hat{y}$ is a monomorphism of $X$ into $C_0(\Phi_Y)$. We know that $\hat{0}(f) = 0$ for all $f \in \Phi_Y$ and so $\hat{y} = \hat{0}$. Since $y \to \hat{y}$ is a monomorphism (and hence injective), we have that $y = 0$.

- Thus whenever $(x_n)$ converges to $0$ in $X$ and $(\alpha(x_n))$ converges to $y$ in $Y$ we have that $y = 0$. Thus by the corollary to The Closed Graph Theorem, $\alpha$ is continuous. $\blacksquare$