The Carrier Space of a Commutative Banach Algebra
 Table of Contents

# The Carrier Space of a Commutative Banach Algebra

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

(1)
\begin{align} \quad \Phi_{\mathfrak{A}} \subset \Phi_\mathfrak{A}^{\infty} \subseteq \mathfrak{A}^* \end{align}

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

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

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

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

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

We begin with some properties of the carrier space for a commutative Banach algebra $\mathfrak{A}$.

 Proposition 1: Let $\mathfrak{A}$ be a commutative Banach algebra. Then: a) $\Phi_{\mathfrak{A}}$ has one-point compactification $\Phi_{\mathfrak{A}}^{\infty}$. b) If $\mathfrak{A}$ has a unit then $\Phi_{\mathfrak{A}}$ is compact. c) $\Phi_{\mathfrak{A}}$ is locally compact and Hausdorff.
• Proof of a) As mentioned above, if $f \in \Phi_{\mathfrak{A}}^{\infty}$. Then $\| f \| \leq 1$ and so $\Phi_{\mathfrak{A}}^{\infty}$ is contained in $B_{\mathfrak{A}^*}$ where $B_{\mathfrak{A}^*}$ denotes the closed unit ball of $\mathfrak{A}^*$. Recall from Alaoglu's Theorem that $B_{\mathfrak{A}^*}$ is weak-* compact. So to show that $\Phi_{\mathfrak{A}}^{\infty} \subseteq B_{\mathfrak{A}^*}$ is weak-* compact it is sufficient to prove that $\Phi_{\mathfrak{A}}^{\infty}$ is weak-* closed (this is because every closed subset of a compact space is also compact).
• If $\Phi_{\mathfrak{A}}^{\infty} = \mathfrak{A}^*$ then we are done.
• Otherwise, suppose that $\Phi_{\mathfrak{A}}^{\infty} \neq \mathfrak{A}^*$. We show that $\Phi_{\mathfrak{A}}^{\infty}$ is weak-* closed by showing that its complement $\mathfrak{A}^* \setminus \Phi_{\mathfrak{A}}^{\infty}$ is weak-* open, i.e., by showing that each $f \in \mathfrak{A}^* \setminus \Phi_{\mathfrak{A}^*}$ has an open neighbourhood fully contained in $\mathfrak{A}^* \setminus \Phi_{\mathfrak{A}^*}$.
• Let $f \in \mathfrak{A}^* \setminus \Phi_{\mathfrak{A}}^{\infty}$. Then $f$ is a linear functional but is NOT multiplicative. So for some $x, y \in \mathfrak{A}$ 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 $g \in \mathfrak{A}^* \setminus \Phi_{\mathfrak{A}}^{\infty}$ so that $V \subset \mathfrak{A}^* \setminus \Phi_{\mathfrak{A}}^{\infty}$ to conclude that $\mathfrak{A}^* \setminus \Phi_{\mathfrak{A}}^{\infty}$ 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 $g \in \mathfrak{A}^* \setminus \Phi_{\mathfrak{A}}^{\infty}$.
• Thus $g \in \mathfrak{A}^* \setminus \Phi_{\mathfrak{A}}^{\infty}$. So we have constructed an open neighbourhood of $f$ fully contained in $\mathfrak{A}^* \setminus \Phi_{\mathfrak{A}}^{\infty}$. Since $f$ was chosen arbitrarily in $\mathfrak{A}^* \setminus \Phi_{\mathfrak{A}}^{\infty}$, this shows that $\mathfrak{A}^* \setminus \Phi_{\mathfrak{A}}^{\infty}$ is weak-* open, so $\Phi_{\mathfrak{A}}^{\infty}$ is a weak-* closed subset of the weak-* compact space $\mathfrak{A}^*$, i.e., $\Phi_{\mathfrak{A}}^{\infty}$ is compact with respect to the $\mathfrak{A}$-topology. $\blacksquare$
• Proof of b) Suppose that $\mathfrak{A}$ has a unit $1_{\mathfrak{A}}$. Recall that then if $f \in \Phi_{\mathfrak{A}}$ then $f(1_{\mathfrak{A}}) = 1$. Therefore we see that:
(8)
\begin{align} \quad \Phi_{\mathfrak{A}} = \{ f \in \Phi_{\mathfrak{A}}^{\infty} : f(1_{\mathfrak{A}}) = 1 \} \end{align}
• So $\Phi_{\mathfrak{A}}$ is a closed subset of the $\mathfrak{A}$-topology compact space $\Phi_{\mathfrak{A}}^{\infty}$ which shows that $\Phi_{\mathfrak{A}}$ is compact with respect to the $\mathfrak{A}$ topology. $\blacksquare$
• Proof of c) By (a) we have that $\Phi_{\mathfrak{A}}^{\infty}$ is compact with respect to the $\mathfrak{A}$-topology. For any $f \in \Phi_{\mathfrak{A}}$, by definition, $f$ is not identically zero so there exists an $x_0 \in \mathfrak{A}$ with $f(x_0) \neq 0$. Let:
(9)
\begin{align} \quad N(f, x_0) = \left \{ g \in \Phi_{\mathfrak{A}}^{\infty} : |g(x_0)| \geq \frac{1}{2} |f(x_0)| \right \} \end{align}
• The above set is a closed subset of $\Phi_{\mathfrak{A}}^{\infty}$ and is thus compact in $\Phi_{\mathfrak{A}}$ with respect to the $\mathfrak{A}$-topology; contains $f$, and does not contain the zero functional. So $\Phi_{\mathfrak{A}}$ is locally compact.
• Lastly we know that the dual space $\mathfrak{A}^*$ is Hausdorff. Since $\Phi_{\mathfrak{A}}$ is a subspace of $\mathfrak{A}^*$ we have that $\Phi_{\mathfrak{A}}$ is also Hausdorff. $\blacksquare$
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