ker(f) ⊆ Sing(A) for all Multiplicative Linear Functionals f

# For Algebras with Unit A - ker(f) ⊆ Sing(A) for all Multiplicative Linear Functionals f

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

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

We will now prove that if $\mathfrak{A}$ is an algebra with unit and $f$ is a multiplicative linear functional on $\mathfrak{A}$ then $f(x) = 0$ implies that $x$ is singular.

Proposition 1: Let $\mathfrak{A}$ be an algebra with unit. If $f$ is a multiplicative linear functional on $\mathfrak{A}$ then $\ker (f) \subseteq \mathrm{Sing}(\mathfrak{A})$. |

**Proof:**Let $f$ be a multiplicative linear functional. Let $x \in \ker (f)$ and suppose instead that $x \not \in \mathrm{Sing}(\mathfrak{A})$. Then $x \in \mathrm{Inv}(\mathfrak{A})$. So $xx^{-1} = 1_{\mathfrak{A}} = x^{-1}x$ where $1_{\mathfrak{A}}$ denotes the unit in $\mathfrak{A}$. Applying $f$ gives us that:

\begin{align} \quad 0 = 0f(x^{-1}) = f(x)f(x^{-1}) = f(xx^{-1}) = f(1_{\mathfrak{A}}) = 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 $y \in \mathfrak{A}$ we have that $y1_{\mathfrak{A}} = y = 1_{\mathfrak{A}}y$, so $f(y) = f(y)f(1_{\mathfrak{A}}) = f(1_{\mathfrak{A}})f(y) = 0$, thus, $f$ is the zero functional - a contradiction since by definition $f$ is not identically zero.

- Thus the assumption that $x \not \in \mathrm{Sing}(\mathfrak{A})$ is false. So $x \in \mathrm{Sing}(\mathfrak{A})$ showing that $\ker (f) \subseteq \mathrm{Sing}(\mathfrak{A})$. $\blacksquare$