Semi-Simple Commutative Banach Algebras

# Semi-Simple Commutative Banach Algebras

Recall from The Radical of a Commutative Banach Algebra page that if $X$ is a commutative Banach algebra over $\mathbb{C}$ then the radical of $X$ is defined to be the set:

(1)\begin{align} \quad \mathrm{Rad}(X) = \bigcap_{f \in \Phi_X} \ker (f) \end{align}

We proved that $\mathrm{Rad}(X)$ is nonempty (as it contains $0$) and that:

(2)\begin{align} \quad \mathrm{Rad}(X) = \{ x \in X : r(x) = 0 \} \end{align}

We give a special name to commutative Banach algebras $X$ over $\mathbb{C}$ whose radical contains ONLY the $0$.

Definition: Let $X$ be a commutative Banach algebra over $\mathbb{C}$. $X$ is said to be Semi-Simple if $\mathrm{Rad}(X) = \{ 0 \}$. |

The following proposition gives criteria for when a commutative Banach algebra $X$ over $\mathbb{C}$ is semi-simple.

Proposition 1: Let $X$ be a commutative Banach algebra. The following statements are equivalent:1) $X$ is semi-simple.2) $\Phi_X$ separates the points of $X$.3) The Gelfand representation $x \to \hat{x}$ is a monomorphism of $X$ into $C_0(\Phi_X)$.4) The spectral radius $r$ defined on $X$ is an algebra norm on $X$. |

*Recall that if $X$ is a commutative complex Banach algebra then $r(x) = \| \hat{x} \|_{\infty} = \sup_{f \in \Phi_X} |\hat{x}(f)|$. We use this throughout the proof below.*

**Proof:**$1) \Rightarrow 2)$ Let $X$ be semi-simple. Then $\mathrm{Rad}(X) = \{ 0 \}$. Let $x, y \in X$ be such that $x \neq y$. Suppose instead that for every $f \in \Phi_X$ we have that $f(x) = f(y)$. Then by linearity of $f$ we have that $f(x - y) = 0$, so $x - y \in \ker (f)$ for every $f \in \Phi_X$, implying that $x - y \in \mathrm{Rad}(X)$. But $\mathrm{Rad}(X) = \{ 0 \}$, so $x - y = 0$, a contradiction. Thus there must exist an $f \in \Phi_X$ such that $f(x) \neq f(y)$, i.e., $\Phi_X$ separates the points of $X$.

- $2) \Rightarrow 3)$ Let $\Phi_X$ separate the points of $X$. We have already proven that the Gelfand representation $x \to \hat{x}$ is a homomorphism of $X$ to $C_0(\Phi_X) =$ when $X$ has a unit. When $X$ does not have a unit, we have proven already that $\Phi_X$ is compact and that the Gelfand representation is a homomorphism of $X$ to $C(\Phi_X)$. But since $\Phi_X$ is compact, $C(\Phi_X) = C_0(\Phi_X)$.

- In either case, all that remains to show is that the Gelfand representation is injective. Let $x, y \in X$ and suppose that $\hat{x} = \hat{y}$. Then for all $f \in \Phi_X$ we have that $\hat{x}(f) = \hat{y}(f)$, i.e., for all $f \in \Phi_X$ we have that $f(x) = f(y)$. If $x \neq y$ then $\Phi_X$ cannot separate the points of $X$. Thus we must have that $x = y$, and so the Gelfand representation is a monomorphism of $X$ to $C_0(\Phi_X)$.

- $3) \Rightarrow 4)$ Suppose the Gelfand representation $x \to \hat{x}$ is a monomorphism of $X$ into $C_0(\Phi_X)$. We want to show that the spectral radius is a norm on $X$. First observe $r : X \to [0, \infty)$.

**1. Showing that $r(x) = 0$ if and only if $x = 0$:**Suppose $r(x) = 0$. Then $0 = r(x) = \|\hat{x} \|_{\infty} = \sup_{f \in \Phi_X} |\hat{x}(f)|$, so that $\hat{x}(f) = 0$ for all $f \in \Phi_X$. But also $\hat{0}(f) = 0$ for all $f \in \Phi_X$. Since the Gelfand representation is a monomorphism (and hence injective) and $\hat{x} = \hat{0}$ we have that $x = 0$. On the other hand if we suppose that $x =0$ then $r(0) = \| \hat{0} \|_{\infty} = \sup_{f \in \Phi_X} |\hat{0}(f)| = \sup_{f \in \Phi_X} |f(0)| = 0$. So we conclude that $r(x) = 0$ if and only if $x = 0$.

**2. Showing that $r(\alpha x) = |\alpha| r(x)$:**Let $\alpha \in \mathbb{C}$, $x \in X$. Then:

\begin{align} \quad r(\alpha x) = \| \alpha x \|_{\infty} = \sup_{f \in \Phi_X} |\hat{\alpha x} (f)| = \sup_{f \in \Phi_X} |f(\alpha x)| = |\alpha| \sup_{f \in \Phi_X} |f(x)| = |\alpha| \sup_{f \in \Phi_X} |\hat{x}(f)| = |\alpha| \| x \|_{\infty} = |\alpha| r(x) \end{align}

**3. Showing that $r(x + y) \leq r(x) + r(y)$:**Let $x, y \in X$. Then:

\begin{align} \quad \quad r(x + y) = \| x + y \|_{\infty} = \sup_{f \in \Phi_X} |\hat{x + y}(f) | = \sup_{f \in \Phi_X} |f(x + y)| \leq \sup_{f \in \Phi_X} |f(x)| + \sup_{f \in \Phi_X} |f(y)| = \sup_{f \in \Phi_X} |\hat{x}(f)| + \sup_{f \in \Phi_X} |\hat{y}(f)| = \| x \|_{\infty} + \| y \|_{\infty} = r(x) + r(y) \end{align}

**4. Showing that $r(xy) \leq r(x)r(y)$:**Let $x, y \in X$. Then:

\begin{align} \quad \quad r(xy) = \| xy \|_{\infty} = \sup_{f \in \Phi_X} | \hat{xy}(f)| = \sup_{f \in \Phi_X} |f(xy)| \leq \sup_{f \in \Phi_X} |f(x)| \cdot \sup_{f \in \Phi_X} |f(y)| = \sup_{f \in \Phi_X} |\hat{x}(f)| \cdot \sup_{f \in \Phi_X} |\hat{y}(f)| = \| x \|_{\infty} \| y \|_{\infty} = r(x)r(y) \end{align}

- Thus the spectral radius $r$ is an algebra norm on $X$.

- $4) \Rightarrow 1)$ Suppose that the spectral radius $r$ is an algebra norm on $X$. Then in particular, $r(x) = 0$ if and only if $x = 0$. We have proven already that $\mathrm{Rad}(X) = \{ x \in X : r(x) = 0 \}$ at the proposition referenced at the top of this page. Therefore we see that $\mathrm{Rad}(X) = \{ 0 \}$, i.e., $X$ is semi-simple. $\blacksquare$