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 $X$ is a Banach algebra then a linear functional $f$ on $X$ is said to be a Jordan functional if $f(x^2) = [f(x)]^2$ for all $x \in X$. We proved that every Jordan functional is multiplicative. We will use this result in the theorem below

Theorem 1: Let $X$ be a Banach algebra with unit and let $f$ be a linear functional on $X$. Then the following statements are equivalent:
1) $f(1_X) = 1$ and $\ker (f) \subseteq \mathrm{Sing}(X)$.
2) $f(x) \in \mathrm{Sp}(X, x)$ for all $x \in X$.
3) $f$ is multiplicative.
  • Proof: $1) \Rightarrow 2)$ Suppose that $f(1_X) = 1$ and $\ker (f) \subseteq \mathrm{Sing}(X)$. Let $x \in X$. We want to show that $x - f(x)1_X \in \mathrm{Sing}(X)$. But observe that $x - f(x)1_X \in \ker (f)$ since $f(x - f(x)1_X) = 0$. But $\ker (f) \subseteq \mathrm{Sing}(X)$ by hypothesis. Hence $x = f(x)1_X \in \mathrm{Sing}(X)$ showing that $f(x) \in \mathrm{Sp}(X, x)$ for each $x \in X$.
  • $2) \Rightarrow 1)$ Suppose that $f(x) \in \mathrm{Sp}(X, x)$ for all $x \in X$. Then $x - f(x)1_X \in \mathrm{Sing}(X)$ for all $x \in X$.
  • Suppose $x \in \ker (f)$. Then $f(x) = 0$. From above $x - f(x)1_X \in \mathrm{Sing}(X)$ says that $x \in \mathrm{Sing}(X)$. Thus $\ker (f) \subseteq \mathrm{Sing}(X)$. Furthermore, we have that $f(1_X) \in \mathrm{Sp}(X, 1_X)$ which says that $1_X - f(1)1_X \in \mathrm{Sing}(X)$, that is, $[1 - f(1)]1_X \in \mathrm{Sing}(X)$. We know that $1_X \in \mathrm{Inv}(X)$ and that $\lambda 1_X \in \mathrm{Inv}(X)$ for all $\lambda \neq 0$. Thus $[1 - f(1)]1_X \in \mathrm{Sing}(X)$ implies $f(1) = 1$.
  • $3) \Rightarrow 1)$ Suppose that $f$ is multiplicative linear functional. Then certainly $f(1_X) = 1$. To see why, note that since $f$ is multiplicative we have that $f(1_X) = f(1_X1_X) = f(1_X)f(1_X)$. So $f(1_X) = 0$ or $f(1_X) = 1$. But if $f(1_X) = 0$ then $f$ is identically the zero functional - a contradiction. Thus $f(1_X) = 1$.
  • $1) \Rightarrow 3)$ Suppose that $f(1_X) = 1$ and $\ker(f) \subseteq \mathrm{Sing}(X)$. Let $x \in X$ 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 $X$ 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 - x When r(x) < 1 in a Banach Algebra page that $1_X - \frac{1}{\lambda} x \in \mathrm{Inv}(X)$. So $1_X - \frac{1}{\lambda} x \not \in \mathrm{Sing}(X)$. Since $\ker (f) \subseteq \mathrm{Sing}(X)$. Using this and the hypothesis that $f(1_X) = 1$ and we get that:
(2)
\begin{align} \quad f \left (1_X - \frac{1}{\lambda}x \right ) \neq 0 \\ \quad f(1_X) - \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 $x \in X$ with $\| x \| \leq 1$ we have that $f(x) \neq \lambda$ for any $\lambda \in \mathbb{C}$ with $|\lambda| > 1$. Thus for all $x \in X$ 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 $x \in X$ 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_X + \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 $x \in X$ 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 $X$ is a Banach algebra and is NOT assumed to have a unit.

Theorem 2: Let $X$ be a Banach algebra and let $f$ be a linear functional on $X$. Then $f$ is a multiplicative linear functional if and only if $f(x) \in \mathrm{Sp}(X, x)$ for every $x \in X$.
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