The Radical of a Commutative Banach Algebra

The Radical of a Commutative Banach Algebra

Definition: Let $X$ be a commutative Banach algebra over $\mathbb{C}$. The Radical of $X$ is defined to be the set $\displaystyle{\mathrm{Rad} (X) = \bigcap_{f \in \Phi_X} \ker (f)}$.

Note that if $f \in \Phi_X$ 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 $0 \in \mathrm{Rad}(X)$, so $\mathrm{Rad}(X)$ is never empty.

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

Proposition 1: Let $X$ be a commutative Banach algebra over $\mathbb{C}$. Then $\mathrm{Rad} (X) = \{ x \in X : r(x) = 0 \}$.
  • Proof: Let $x \in \mathrm{Rad}(X)$. Then $x \in \ker (f)$ for every $f \in \Phi_X$, that is, $f(x) = 0$ for all $f \in \Phi_X$. Thus:
(1)
\begin{align} \quad r(x) = \| \hat{x} \|_{\infty} = \sup_{f \in \Phi_X} |\hat{x}(f)| = \sup_{f \in \Phi_X} |f(x)| = 0 \end{align}
  • So $r(x) = 0$, hence $\mathrm{Rad}(X) \subseteq \{ x \in X : r(x) = 0 \}$.
  • Now let $x \in \{ x \in X : r(x) = 0 \}$ so that $r(x) = 0$. Thus:
(2)
\begin{align} \quad 0 = r(x) = \| \hat{x} \|_{\infty} = \sup_{f \in \Phi_X} |\hat{x}(f)| = \sup_{f \in \Phi_X} |f(x)| \end{align}
  • So $|f(x)| = 0$ for all $f \in \Phi_X$ implying $x \in \ker (f)$ for all $f \in \Phi_X$. Thus $x \in \mathrm{Rad}(X)$ so $\mathrm{Rad}(X) \supseteq \{ x \in X : r(x) = 0 \}$. $\blacksquare$
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