Multiplicative Linear Functionals on a Banach Algebra

# Multiplicative Linear Functionals on a Banach Algebra

Definition: Let $\mathfrak{A}$ be a normed algebra over $\mathbf{F}$. A linear functional $f : \mathfrak{A} \to \mathbf{F}$ is said to be Multiplicative if $f$ is not the zero functional and $f(xy) = f(x)f(y)$ for all $x, y \in \mathfrak{A}$. The Set of All Multiplicative Linear Functionals on $\mathfrak{A}$ is denoted $\Phi_{\mathfrak{A}}$. The Set of All Multiplicative Linear Functionals on $\mathfrak{A}$ AND the Zero Linear Functional is denoted $\Phi_{\mathfrak{A}}^{\infty}$. |

*Note that a multiplicative linear functional is simply a nonzero homomorphism of $\mathfrak{A}$ to $\mathbf{F}$.*

We begin by proving a very important result regarding multiplicative linear functionals on a Banach algebra.

Proposition 1: Let $\mathfrak{A}$ be a Banach algebra over $\mathbf{F}$. If $f : \mathfrak{A} \to \mathbf{F}$ is a multiplicative linear functional then $f$ is bounded and $\| f \| \leq 1$. |

**Proof:**Let $f$ be a multiplicative linear functional. We will show that for all $x \in \mathfrak{A}$ with $\| x \| < 1$ we have that $f(x) \neq 1$.

- Suppose instead that there exists an $x \in \mathfrak{A}$ with $\| x \| < 1$ such that $f(x) = 1$. Let $\displaystyle{y = \sum_{n=1}^{\infty} x^n}$. Observe that $\displaystyle{\sum_{n=1}^{\infty} \| x^n \| \leq \sum_{n=1}^{\infty} \| x \|^n < \infty}$ since $\| x \| < 1$. Since $\mathfrak{A}$ is a Banach space, the absolute convergence of $y$ implies that $y \in \mathfrak{A}$. Furthermore observe that:

\begin{align} \quad x +xy = x + x \sum_{n=1}^{\infty} x^n = x + \sum_{n=2}^{\infty} x^n = \sum_{n=1}^{\infty} x^n = y \end{align}

- Therefore $y = x + xy$. Applying $f$ to both sides of this equation yields:

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

- Which is a contradiction. So for all $x \in \mathfrak{A}$ with $\| x \| < 1$ we have that $f(x) \neq 1$.

- We now show that if $x \in \mathfrak{A}$ is such that $\| x \| < 1$ then $|f(x)| \leq 1$. Let $x \in \mathfrak{A}$ with $\| x \| < 1$ and suppose that $|f(x)| > 1$. Then $\| x \| < |f(x)|$. Consider the vector $\frac{x}{f(x)}$. Then:

\begin{align} \quad \left \| \frac{x}{f(x)} \right \| = \frac{\| x \|}{|f(x)|} < 1 \end{align}

- So by the proof from earlier we must have that:

\begin{align} \quad 1 = f \left ( \frac{x}{f(x)} \right ) \neq 1 \end{align}

- Again a contradiction. Thus if $x \in \mathfrak{A}$ is such that $\| x \| < 1$ then $|f(x)| \leq 1$.

- Finally, if $x \in \mathfrak{A}$ is such that $\| x \| \leq 1$ then $|f(x)| \leq 1$. To see why, let $0 < \alpha < 1$. Then $\| \alpha x \| < 1$ and so $\alpha |f(x)| = |f(ax)| \leq 1$. Thus $|f(x)| \leq \frac{1}{\alpha}$ for all $0 < \alpha < 1$ implying that $|f(x)| \leq 1$. So:

\begin{align} \quad \| f \| = \sup_{\| x \| \leq 1} \{ |f(x)| \} \leq 1 \quad \blacksquare \end{align}

Proposition 2: Let $\mathfrak{A}$ be a Banach algebra over $\mathbf{F}$ with unit. If $f : \mathfrak{A} \to \mathbb{C}$ is a multiplicative linear functional then $f(1) = 1$ and so $\| f \| = 1$. |

**Proof:**We have that $1 = 1 \cdot 1$ and so:

\begin{align} \quad f(1) = f(1 \cdot 1) = f(1)f(1) \end{align}

- So either $f(1) = 0$ or $f(1) = 1$. If $f(1) = 0$ then for all $x \in \mathfrak{A}$ we have that $f(x) = f(1 \cdot x) = f(1)f(x) = 0$, which cannot happen since $f$ is not the zero functional. Thus $f(1) = 1$. Thus by Proposition 1 we have that:

\begin{align} \quad \| f \| = \sup_{\| x \| \leq 1} \{ |f(x)| \} = 1 \end{align}