The Carrier Space of a Commutative Banach Algebra

The Carrier Space of a Commutative Banach Algebra

Recall from the Multiplicative Linear Functionals on a Banach Algebra page that if $X$ is a Banach algebra then every multiplicative linear functional $f$ on $X$ is bounded and $\| f \| \leq 1$. The set of all multiplicative linear functionals on $X$ is denoted by $\Phi_X$, and the set of all multiplicative linear functionals on $X$ and the zero functional on $X$ is denoted by $\Phi_X^{\infty}$. Also recall that the dual space $X^*$ is the collection of all bounded linear functionals on $X$. Thus we have that:

(1)
\begin{align} \quad \Phi_X \subset \Phi_X^{\infty} \subseteq X^* \end{align}

We can consider the weak-* topology on $X^*$. Recall the weak-* topology on $X^*$ is the topology induced by the collection $\hat{X} = \{\hat{x} : x \in X \} \subseteq X^{**}$ of linear functionals on $X^*$ where for each $x \in X$ we defined $\hat{x} : X^* \to \mathbb{C}$ for all $f \in X^*$ by $\hat{x}(f) = f(x)$.

We noted that a sequence $(f_n)$ in $X^*$ is said to weak-* converge to $f \in X^*$ if for all $x \in X$ we have that $\lim_{n \to \infty} \hat{x}(f_n) = \hat{x}(f)$, that is, for all $x \in X$ we have that:

(2)
\begin{align} \quad \lim_{n \to \infty} f(x_n) = f(x) \end{align}

We can consider both $\Phi_X$ and $\Phi_X^{\infty}$ as subspaces of $(X^*, \mathrm{weak}-*)$. When we do, we give the spaces a special name.

Definition: Let $X$ be a commutative Banach algebra. Equip the dual space $X^*$ with the weak-* topology. The $X$-Topology on $\Phi_X$ (or $\Phi_X^{\infty}$) is the subspace topology on $\Phi_X$ (or $\Phi_X^{\infty}$). The Carrier Space for $X$ is the space $\Phi_X$ (or sometimes $\Phi_X^{\infty}$) equipped with the $X$-topology.

We begin with some properties of the carrier space for a commutative Banach algebra $X$.

Proposition 1: Let $X$ be a commutative Banach algebra. Then:
a) $\Phi_X$ has one-point compactification $\Phi_X^{\infty}$.
b) If $X$ has a unit then $\Phi_X$ is compact.
c) $\Phi_X$ is locally compact and Hausdorff.
  • Proof of a) As mentioned above, if $f \in \Phi_X^{\infty}$. Then $\| f \| \leq 1$ and so $\Phi_X^{\infty}$ is contained in $B_{X^*}$ where $B_{X^*}$ denotes the closed unit ball of $X^*$. Recall from Alaoglu's Theorem that $B_{X^*}$ is weak-* compact. So to show that $\Phi_X^{\infty} \subseteq B_{X^*}$ is weak-* compact it is sufficient to prove that $\Phi_X^{\infty}$ is weak-* closed (this is because every closed subset of a compact space is also compact).
  • If $\Phi_X^{\infty} = X^*$ then we are done.
  • Otherwise, suppose that $\Phi_X^{\infty} \neq X^*$. We show that $\Phi_X^{\infty}$ is weak-* closed by showing that its complement $X^* \setminus \Phi_X^{\infty}$ is weak-* open, i.e., by showing that each $f \in X^* \setminus \Phi_{X^*}$ has an open neighbourhood fully contained in $X^* \setminus \Phi_{X^*}$.
  • Let $f \in X^* \setminus \Phi_X^{\infty}$. Then $f$ is a linear functional but is NOT multiplicative. So for some $x, y \in X$ we have that $f(x)f(y) \neq f(xy)$. Define:
(3)
\begin{align} \quad \delta := |f(x)f(y) - f(xy)| > 0 \end{align}
  • Let:
(4)
\begin{align} \quad \epsilon = \min \left \{ \frac{1}{2} \delta^{1/2}, \frac{1}{4} \left ( 1 + |f(x)| + |f(y)| \right )^{-1}\delta \right \} > 0 \end{align}
  • And consider the following open neighbourhood $V$ of $f$:
(5)
\begin{align} \quad V = \left \{ g : |g(x) - f(x)| < \epsilon, |g(y) - f(y)| < \epsilon, |g(xy) - f(xy)| < \epsilon \right \} \end{align}
  • Suppose that $g \in V$ so that $|g(x) - f(x)| < \epsilon$, $|g(y) - f(y)| < \epsilon$, and $|g(xy) - f(xy)| < \epsilon$. We want to show that $g \in X^* \setminus \Phi_X^{\infty}$ so that $V \subset X^* \setminus \Phi_X^{\infty}$ to conclude that $X^* \setminus \Phi_X^{\infty}$ is weak-* open. We have that:
(6)
\begin{align} \quad |g(x)g(y) - f(x)f(y)| &\leq |g(x)g(y) [- g(x)f(y) - f(x)g(y) + f(x)f(y) + g(x)f(y) + f(x)g(y) - f(x)f(y)] - f(x)f(y) | \\ & \leq |[g(x)g(y) - g(x)f(y) - f(x)g(y) + f(x)f(y)] + g(x)f(y) + f(x)g(y) - f(x)f(y) - f(x)f(y) | \\ & \leq |[g(x) - f(x)][g(y) - f(y)] + [g(x)f(y) -f(x)f(y)] + [f(x)g(y) - f(x)f(y)] | \\ & \leq |g(x) - f(x)||g(y) - f(y)| + |f(y)| |g(x) - f(x)| + |f(x)||g(y) - f(y)| \\ & < \epsilon^2 + |f(y)| \epsilon + |f(x)| \epsilon \\ & < \epsilon^2 + \epsilon[|f(y)| + |f(x)|] \\ & < \left ( \frac{1}{2} \delta^{1/2} \right )^2 + \frac{1}{4}\left (1 + |f(x)| + |f(y)| \right )^{-1} \delta \left [ |f(y)| + |f(x)| \right ] \\ & < \frac{1}{4} \delta + \frac{1}{4} \frac{|f(x)| + |f(y)|}{1 + |f(x)| + |f(y)|} \delta \\ & < \frac{1}{4} \delta + \frac{1}{4} \delta \\ & < \frac{1}{2} \delta \end{align}
  • Now we have also assumed above that
(7)
\begin{align} \quad |g(xy) - f(xy)| < \epsilon \leq \frac{1}{4} \delta \end{align}
  • Therefore $|g(xy) - g(x)g(y)| \geq \frac{1}{4} \delta$ showing that $g$ is not multiplicative. (See the image below). So $g \in X^* \setminus \Phi_X^{\infty}$.
Screen%20Shot%202018-06-18%20at%208.01.31%20PM.png
  • Thus $g \in X^* \setminus \Phi_X^{\infty}$. So we have constructed an open neighbourhood of $f$ fully contained in $X^* \setminus \Phi_X^{\infty}$. Since $f$ was chosen arbitrarily in $X^* \setminus \Phi_X^{\infty}$, this shows that $X^* \setminus \Phi_X^{\infty}$ is weak-* open, so $\Phi_X^{\infty}$ is a weak-* closed subset of the weak-* compact space $X^*$, i.e., $\Phi_X^{\infty}$ is compact with respect to the $X$-topology. $\blacksquare$
  • Proof of b) Suppose that $X$ has a unit $1_X$. Recall that then if $f \in \Phi_X$ then $f(1_X) = 1$. Therefore we see that:
(8)
\begin{align} \quad \Phi_X = \{ f \in \Phi_X^{\infty} : f(1_X) = 1 \} \end{align}
  • So $\Phi_X$ is a closed subset of the $X$-topology compact space $\Phi_X^{\infty}$ which shows that $\Phi_X$ is compact with respect to the $X$ topology. $\blacksquare$
  • Proof of c) By (a) we have that $\Phi_X^{\infty}$ is compact with respect to the $X$-topology. For any $f \in \Phi_X$, by definition, $f$ is not identically zero so therre exists an $x_0 \in X$ with $f(x_0) \neq 0$. Let:
(9)
\begin{align} \quad N(f, x_0) = \left \{ g \in \Phi_X^{\infty} : |g(x_0)| \geq \frac{1}{2} |f(x_0)| \right \} \end{align}
  • The above set is a closed subset of $\Phi_X^{\infty}$ and is thus compact in $\Phi_X$ with respect to the $X$-topology; contains $f$, and does not contain the zero functional. So $\Phi_X$ is locally compact.
  • Lastly we know that the dual space $X^*$ is Hausdorff. Since $\Phi_X$ is a subspace of $X^*$ we have that $\Phi_X$ is also Hausdorff. $\blacksquare$
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