Equiv. Crit. for a Linear Funct. to be Multiplicative in a Banach Algebra

# Equivalent Criterion for a Linear Functional to be Multiplicative in a Banach Algebra

Recall from the Jordan 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 a Jordan functional if $f(x^2) = [f(x)]^2$ for all $x \in \mathfrak{A}$. We proved that if $\mathfrak{A}$ is a Banach algebra then every Jordan functional is multiplicative. We will use this result in the theorem below

 Theorem 1: Let $\mathfrak{A}$ be a Banach algebra with unit and let $f$ be a linear functional on $\mathfrak{A}$. Then the following statements are equivalent: 1) $f(1_{\mathfrak{A}}) = 1$ and $\ker (f) \subseteq \mathrm{Sing}(\mathfrak{A})$. 2) $f(x) \in \mathrm{Sp}(\mathfrak{A}, x)$ for all $x \in \mathfrak{A}$. 3) $f$ is multiplicative.
• Proof: $1) \Rightarrow 2)$ Suppose that $f(1_{\mathfrak{A}}) = 1$ and $\ker (f) \subseteq \mathrm{Sing}(\mathfrak{A})$. Let $x \in \mathfrak{A}$. We want to show that $x - f(x)1_{\mathfrak{A}} \in \mathrm{Sing}(\mathfrak{A})$. But observe that $x - f(x)1_{\mathfrak{A}} \in \ker (f)$ since $f(x - f(x)1_{\mathfrak{A}}) = 0$. But $\ker (f) \subseteq \mathrm{Sing}(\mathfrak{A})$ by hypothesis. Hence $x = f(x)1_{\mathfrak{A}} \in \mathrm{Sing}(\mathfrak{A})$ showing that $f(x) \in \mathrm{Sp}(\mathfrak{A}, x)$ for each $x \in \mathfrak{A}$.
• $2) \Rightarrow 1)$ Suppose that $f(x) \in \mathrm{Sp}(\mathfrak{A}, x)$ for all $x \in \mathfrak{A}$. Then $x - f(x)1_{\mathfrak{A}} \in \mathrm{Sing}(\mathfrak{A})$ for all $x \in \mathfrak{A}$.
• Suppose $x \in \ker (f)$. Then $f(x) = 0$. From above $x - f(x)1_{\mathfrak{A}} \in \mathrm{Sing}(\mathfrak{A})$ says that $x \in \mathrm{Sing}(\mathfrak{A})$. Thus $\ker (f) \subseteq \mathrm{Sing}(\mathfrak{A})$. Furthermore, we have that $f(1_{\mathfrak{A}}) \in \mathrm{Sp}(\mathfrak{A}, 1_{\mathfrak{A}})$ which says that $1_{\mathfrak{A}} - f(1)1_{\mathfrak{A}} \in \mathrm{Sing}(\mathfrak{A})$, that is, $[1 - f(1)]1_{\mathfrak{A}} \in \mathrm{Sing}(\mathfrak{A})$. We know that $1_{\mathfrak{A}} \in \mathrm{Inv}(\mathfrak{A})$ and that $\lambda 1_{\mathfrak{A}} \in \mathrm{Inv}(\mathfrak{A})$ for all $\lambda \neq 0$. Thus $[1 - f(1)]1_{\mathfrak{A}} \in \mathrm{Sing}(\mathfrak{A})$ implies $f(1) = 1$.
• $3) \Rightarrow 1)$ Suppose that $f$ is multiplicative linear functional. Then certainly $f(1_{\mathfrak{A}}) = 1$. To see why, note that since $f$ is multiplicative we have that $f(1_{\mathfrak{A}}) = f(1_{\mathfrak{A}}1_{\mathfrak{A}}) = f(1_{\mathfrak{A}})f(1_{\mathfrak{A}})$. So $f(1_{\mathfrak{A}}) = 0$ or $f(1_{\mathfrak{A}}) = 1$. But if $f(1_{\mathfrak{A}}) = 0$ then $f$ is identically the zero functional - a contradiction. Thus $f(1_{\mathfrak{A}}) = 1$.
• $1) \Rightarrow 3)$ Suppose that $f(1_{\mathfrak{A}}) = 1$ and $\ker(f) \subseteq \mathrm{Sing}(\mathfrak{A})$. Let $x \in \mathfrak{A}$ be such that $\| x \| \leq 1$ and let $\lambda \in \mathbb{C}$ be such that $|\lambda| > 1$. Then:
(1)
\begin{align} \quad r \left ( \frac{1}{\lambda} x \right ) \leq \left \| \frac{1}{\lambda} x \right \| = \frac{1}{|\lambda|} \| x \| < 1 \end{align}
• Since $\mathfrak{A}$ is a Banach algebra with unit and since $r \left ( \frac{1}{\lambda} x \right ) < 1$ we have by the theorem on the Invertibility of 1 - a When r(a) < 1 in a Banach Algebra with Unit page that $1_{\mathfrak{A}} - \frac{1}{\lambda} x \in \mathrm{Inv}(\mathfrak{A})$. So $1_{\mathfrak{A}} - \frac{1}{\lambda} x \not \in \mathrm{Sing}(\mathfrak{A})$. Since $\ker (f) \subseteq \mathrm{Sing}(\mathfrak{A})$. Using this and the hypothesis that $f(1_{\mathfrak{A}}) = 1$ and we get that:
(2)
\begin{align} \quad f \left (1_{\mathfrak{A}} - \frac{1}{\lambda}x \right ) \neq 0 \\ \quad f(1_{\mathfrak{A}}) - \frac{1}{\lambda} f(x) \neq 0 \\ \quad 1 - \frac{1}{\lambda} f(x) \neq 0 \end{align}
• Rearranging the above equation shows that for all $x \in \mathfrak{A}$ with $\| x \| \leq 1$ we have that $f(x) \neq \lambda$ for any $\lambda \in \mathbb{C}$ with $|\lambda| > 1$. Thus for all $x \in \mathfrak{A}$ with $\| x \| \leq 1$ we have that $|f(x)| \leq 1$. So $f$ is bounded and $\| f \| = \sup_{\| x \| \leq 1} \{ |f(x)| \} \leq 1$.
• For each $x \in \mathfrak{A}$ let $F : \mathbb{C} \to \mathbb{C}$ be defined for all $z \in \mathbb{C}$ by:
(3)
\begin{align} \quad F(z) & := f(\exp(zx)) \\ & = f \left ( \sum_{n=0}^{\infty} \frac{z^nx^n}{n!} \right ) \\ & = f \left ( 1_{\mathfrak{A}} + \sum_{n=1}^{\infty} \frac{z^nx^n}{n!} \right ) \\ & = 1 + f \left ( \sum_{n=1}^{\infty} \frac{z^nx^n}{n!} \right ) \\ & = 1 + \sum_{n=1}^{\infty} \frac{z^n}{n!} f(x^n) \quad (\dagger) \end{align}
• The function $F$ is an entire function and furthermore we see that for all $z \in \mathbb{C}$:
(4)
\begin{align} \quad |F(z)| \leq 1 + \left | \sum_{n=1}^{\infty} \frac{z^n}{n!} f(x^n) \right \| \leq 1 + \sum_{n=1}^{\infty} \frac{|z|^n}{n!} |f(x^n)| \overset{(*)} \leq \| x \| + \sum_{n=1}^{\infty} \frac{|z|^n}{n!} \| x \|^n = \exp(|z| \| x \|) \end{align}
• Where the inequality at $(*)$ comes from the fact that $f$ is bounded with $\| f \| \leq 1$ so that $|f(x)| \leq \| x \|$.
• Note note that:
(5)
\begin{align} \quad \left \| 1 - \exp (zx) \right \| = \left \| 1 - \left [ 1 + \sum_{n=1}^{\infty} \frac{z^nx^n}{n!} \right ] \right \| = \left \| \sum_{n=1}^{\infty} \frac{z^nx^n}{n!} \right \| \leq \sum_{n=1}^{\infty} \frac{|z|^n}{n!} \| x \|^n \leq \sum_{n=1}^{\infty} \frac{|z|^n}{n!} \end{align}
• With this it can be shown that there exists an $\alpha \in \mathbb{C}$ for which:
(6)
\begin{align} \quad F(z) &= \exp (\alpha z) \\ &= 1 + \sum_{n=1}^{\infty} \frac{z^n}{n!} \alpha^n \quad (\dagger \dagger) \end{align}
• Comparing $(\dagger)$ and $(\dagger \dagger)$ shows us that for each $n \in \mathbb{N}$:
(7)
\begin{align} \quad f(x^n) = \alpha^n \end{align}
• In particular, when $n = 1$ we get that $f(x) = \alpha$, and when $n = 2$ we get that $f(x^2) = \alpha^2$. So for all $x \in \mathfrak{A}$ we have that:
(8)
\begin{align} \quad f(x^2) = \alpha^2 = [f(x)]^2 \end{align}
• So $f$ is a Jordan functional. But every Jordan functional on a Banach algebra is multiplicative, so $f$ is multiplicative. $\blacksquare$

We can formulate a similar theorem when $\mathfrak{A}$ is a Banach algebra and is NOT assumed to have a unit.

 Theorem 2: Let $\mathfrak{A}$ be a Banach algebra (with or without unit) and let $f$ be a linear functional on $\mathfrak{A}$. Then $f$ is a multiplicative linear functional if and only if $f(x) \in \mathrm{Sp}(\mathfrak{A}, x)$ for every $x \in \mathfrak{A}$.