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

Table of Contents

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:

(1)

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

(2)

\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:

(3)

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

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

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