ker(f) is a Two-Sided Ideal of A

# For Algebras A - ker(f) is a Two-Sided Ideal of A for all Multiplicative Linear Functionals f

Proposition 1: Let $\mathfrak{A}$ be a Banach algebra and let $f$ be a multiplicative linear functional on $\mathfrak{A}$. Then $\ker(f)$ is a two-sided ideal of $\mathfrak{A}$ and moreover, if $\ker (f)$ is modular and $u$ is a modular unit for $\ker (f)$ then $f(u) = 1$. |

**Proof:**Let $J = \ker (f)$. Let $x \in \mathfrak{A}$ and $j \in J$ so that $f(j) = 0$. Since $f$ is multiplicative we have that:

\begin{align} \quad f(xj) = f(x)f(j) = f(x)0 = 0 \end{align}

- Thus $xj \in J$ showing that $\mathfrak{A}J \subseteq J$. Similarly:

\begin{align} \quad f(jx) = f(j)f(x) = 0f(x) = 0 \end{align}

- Thus $jx \in J$ showing that $J\mathfrak{A} \subseteq J$. So $J = \ker (f)$ is a two-sided ideal of $\mathfrak{A}$.

- Now suppose that $\ker (f)$ is modular and let $u$ be a modular unit for $\ker (f)$. Then $(1 - u)\mathfrak{A} \subseteq \ker (f)$ and $\mathfrak{A}(1 - u) \subseteq \ker (f)$. So for all $x \in \mathfrak{A}$ we have that $x - ux \in \ker (f)$ and $x - xu \in \ker(f)$, so:

\begin{align} \quad 0 = f(x - ux) = f(x) - f(u)f(x) = [1 - f(u)]f(x) \end{align}

- And also:

\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 $x_0 \in \mathfrak{A}$ with $f(x_0) \neq 0$. Plugging $x_0$ into both of the above equalities shows us that $f(u) = 1$. $\blacksquare$