Mult. Functs. on Modular Units of Ker(f) in a Banach Algebra

# Multiplicative Linear Functionals Applied to Modular Units of Ker(f) in a Banach Algebra

 Proposition 1: Let $X$ be a Banach algebra and let $f$ be a multiplicative linear functional on $X$. Then $\ker(f)$ is a two-sided ideal of $X$ and moreover, if $\ker (f)$ is modular and $u$ is a modular unit for $\ker (f)$ then $f(u) = 1$.
• Proof: We have already shown that $\ker (f)$ is a two-sided ideal of $X$ but we will prove it again for completeness. Let $J = \ker (f)$. Let $x \in X$ and $j \in J$ so that $f(j) = 0$. Since $f$ is multiplicative we have that:
(1)
\begin{align} \quad f(xj) = f(x)f(j) = f(x)0 = 0 \end{align}
• Thus $xj \in J$ showing that $XJ \subseteq J$. Similarly:
(2)
\begin{align} \quad f(jx) = f(j)f(x) = 0f(x) = 0 \end{align}
• Thus $jx \in J$ showing that $JX \subseteq J$. So $J = \ker (f)$ is a two-sided ideal of $X$.
• Now suppose that $\ker (f)$ is modular and let $u$ be a modular unit for $\ker (f)$. Then $X - uX \subseteq \ker (f)$ and $X - Xu \subseteq \ker (f)$. So for all $x \in X$ we have that $x - ux \in \ker (f)$ and $x - xu \in \ker(f)$, so:
(3)
\begin{align} \quad 0 = f(x - ux) = f(x) - f(u)f(x) = [1 - f(u)]f(x) \end{align}
• And also:
(4)
\begin{align} \quad 0 = f(x - xu) = f(x) - f(x)f(u) = f(x)[1 - f(u)] \end{align}
• Since $f$ is multiplicative, $f$ is not identically the zero functional. So there exists an $x_0 \in X$ with $f(x_0) \neq 0$. Plugging $x_0$ into both of the above equalities shows us that $f(u) = 1$. $\blacksquare$