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 $\mathfrak{A}$ is said to be semi-simple if:

(1)

If $\mathfrak{A}$ and $\mathfrak{B}$ are commutative Banach algebras and furthermore, if $\mathfrak{B}$ is semi-simple, then every homomorphism $\alpha : \mathfrak{A} \to \mathfrak{B}$ is continuous as we prove below.

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

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

• Proof: Let $\alpha : \mathfrak{A} \to \mathfrak{B}$ be a homomorphism. Let $(x_n)$ be a sequence in $\mathfrak{A}$ that converges to $0$ and suppose that $(\alpha(x_n))$ converges to some $y \in \mathfrak{B}$. We aim to show that $y = \alpha (0) = 0$ to apply the closed graph theorem.
• Let $f \in \Phi_{\mathfrak{B}}$. Observe that since $\alpha$ is a homomorphism, $\alpha$ is by definition a linear operator from $\mathfrak{A}$ to $\mathfrak{B}$. $f$ is also a linear functional on $\mathfrak{B}$, so the composition $f \circ \alpha : \mathfrak{A} \to \mathbb{C}$ is a linear functional on $\mathfrak{A}$. We first show that $\alpha \circ f$ is further a multiplicative linear functional on $\mathfrak{A}$.
• Let $x_1, x_2 \in \mathfrak{A}$. Since $\alpha$ is a homomorphism and $f$ is multiplicative we have that:
(2)
\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 $\alpha \circ f \in \Phi_{\mathfrak{A}}$. 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 $\mathfrak{A}$ and $f$ is continuous on $\mathfrak{B}$ we have that:
(3)
\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 $f \in \Phi_{\mathfrak{B}}$, that is, $\hat{y}(f) = 0$ for every $f \in \Phi_{\mathfrak{B}}$.
• Since $\mathfrak{B}$ 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 $\mathfrak{B}$ into $C_0(\Phi_{\mathfrak{B}})$. We know that $\hat{0}(f) = 0$ for all $f \in \Phi_{\mathfrak{B}}$ 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 $\mathfrak{A}$ and $(\alpha(x_n))$ converges to $y$ in $\mathfrak{B}$ we have that $y = 0$. Thus by the corollary to The Closed Graph Theorem, $\alpha$ is continuous. $\blacksquare$