The Radical of a Commutative Banach Algebra

Definition: Let $\mathfrak{A}$ be a commutative Banach algebra. The Radical of $\mathfrak{A}$ is defined to be the set $\displaystyle{\mathrm{Rad} (\mathfrak{A}) = \bigcap_{f \in \Phi_{\mathfrak{A}}} \ker (f)}$.

Note that if $f \in \Phi_\mathfrak{A}$ then $f(0) = 0$. To see why, note that $f(0) = f(0 + 0) = f(0) + f(0)$, so that $f(0) = 0$. Therefore we see that $0 \in \mathrm{Rad}(\mathfrak{A})$, so $\mathrm{Rad}(\mathfrak{A})$ is never empty.

The following proposition tells us exactly what the radical of a commutative Banach algebra is.

Proposition 1: Let $\mathfrak{A}$ be a commutative Banach algebra. Then $\mathrm{Rad} (\mathfrak{A}) = \{ x \in \mathfrak{A} : r(x) = 0 \}$.

• Proof: Let $x \in \mathrm{Rad}(\mathfrak{A})$. Then $x \in \ker (f)$ for every $f \in \Phi_{\mathfrak{A}}$, that is, $f(x) = 0$ for all $f \in \Phi_{\mathfrak{A}}$. Thus:

(1)

\begin{align} \quad r(x) = \| \hat{x} \|_{\infty} = \sup_{f \in \Phi_{\mathfrak{A}}} |\hat{x}(f)| = \sup_{f \in \Phi_{\mathfrak{A}}} |f(x)| = 0 \end{align}

• So $r(x) = 0$, hence $\mathrm{Rad}(\mathfrak{A}) \subseteq \{ x \in {\mathfrak{A}} : r(x) = 0 \}$.

• Now let $x \in \{ x \in \mathfrak{A} : r(x) = 0 \}$ so that $r(x) = 0$. Thus:

(2)

\begin{align} \quad 0 = r(x) = \| \hat{x} \|_{\infty} = \sup_{f \in \Phi_{\mathfrak{A}}} |\hat{x}(f)| = \sup_{f \in \Phi_{\mathfrak{A}}} |f(x)| \end{align}

• So $|f(x)| = 0$ for all $f \in \Phi_{\mathfrak{A}}$ implying $x \in \ker (f)$ for all $f \in \Phi_{\mathfrak{A}}$. Thus $x \in \mathrm{Rad}(\mathfrak{A})$ so $\mathrm{Rad}(\mathfrak{A}) \supseteq \{ x \in \mathfrak{A} : r(x) = 0 \}$. $\blacksquare$