Kernels of Mult. Functs. are Subsets of Sing(X) in a Ban. Alg. with Unit

# Kernels of Multiplicative Functionals are Subsets of Sing(X) in a Banach Algebra with Unit

Recall from the Multiplicative Linear Functionals on a Banach Algebra page that if $X$ is a Banach algebra then a linear functional $f$ on $X$ is said to be multiplicative if for all $x, y \in X$ we have that:

(1)
\begin{align} \quad f(xy) = f(x)f(y) \end{align}

We will now prove that if $X$ is a Banach algebra with unit and $f$ is a multiplicative linear functional then $f(x) = 0$ implies that $x$ is singular.

 Proposition 1: Let $X$ be a Banach algebra with unit. If $f$ is a multiplicative linear functional on $X$ then $\ker (f) \subseteq \mathrm{Sing}(X)$.
• Proof: Let $f$ be a multiplicative linear functional. Let $x \in \ker (f)$ and suppose instead that $x \not \in \mathrm{Sing}(X)$. Then $x \in \mathrm{Inv}(X)$. So $xx^{-1} = 1_X = x^{-1}x$ where $1_X$ denotes the unit in $X$. Applying $f$ gives us that:
(2)
\begin{align} \quad 0 = 0f(x^{-1}) = f(x)f(x^{-1}) = f(xx^{-1}) = f(1_X) = f(x^{-1}x) = f(x^{-1})f(x) = f(x^{-1})0 = 0 \end{align}
• So $f(1_X) = 0$. But then for any $y \in X$ we have that $y1_X = y = 1_Xy$, so $f(y) = f(y)f(1_X) = f(1_X)f(y) = 0$, thus, $f$ is the zero functional - a contradiction since by definition $f$ is not identically zero.
• Thus the assumption that $x \not \in \mathrm{Sing}(X)$ is false. So $x \in \mathrm{Sing}(X)$ showing that $\ker (f) \subseteq \mathrm{Sing}(X)$. $\blacksquare$