[proofplan] We first show that every character on $A$ is unital and contractive, so $\Delta(A)$ is contained in the weak-* compact closed unit ball of $A^*$. We then express unitality and multiplicativity as weak-* closed equations inside that unit ball. Thus $\Delta(A)$ is a weak-* closed subset of a weak-* compact set. Finally, the weak-* topology on $A^*$ is Hausdorff, and hence the [subspace topology](/page/Subspace%20Topology) on $\Delta(A)$ is Hausdorff. [/proofplan]

[step:Show that every character is unital and contractive]Let $\tau\in\Delta(A)$. Since $\tau$ is nonzero, there exists $a_0\in A$ such that $\tau(a_0)\neq 0$. Multiplicativity gives \begin{align*}\tau(a_0)=\tau(1_Aa_0)=\tau(1_A)\tau(a_0).\end{align*} Since $\tau(a_0)\neq 0$, it follows that $\tau(1_A)=1$. We next prove that $|\tau(a)|\leq \|a\|_A$ for every $a\in A$. Fix $a\in A$. If $\lambda\in\mathbb C$ satisfies $|\lambda|>\|a\|_A$, then \begin{align*}\|\lambda^{-1}a\|_A=|\lambda|^{-1}\|a\|_A<1.\end{align*} By [citetheorem:8545], applied in the unital Banach algebra $A$ to $\lambda^{-1}a$, the element $1_A-\lambda^{-1}a$ is invertible. Multiplying by the nonzero scalar $\lambda$, the element $\lambda 1_A-a$ is invertible in $A$. Suppose for contradiction that $|\tau(a)|>\|a\|_A$. Set $\lambda:=\tau(a)$. Then $\lambda 1_A-a$ is invertible by the preceding paragraph, while \begin{align*}\tau(\lambda 1_A-a)=\lambda\tau(1_A)-\tau(a)=\lambda-\lambda=0.\end{align*} If $b\in A$ denotes the inverse of $\lambda 1_A-a$, multiplicativity gives \begin{align*}1=\tau(1_A)=\tau((\lambda 1_A-a)b)=\tau(\lambda 1_A-a)\tau(b)=0.\end{align*} a contradiction. Hence $|\tau(a)|\leq \|a\|_A$ for every $a\in A$, so $\tau$ is bounded and $\|\tau\|_{A^*}\leq1$.[/step]

[guided]We begin by proving the standard boundedness property of characters rather than taking it as a black box. Let $\tau\in\Delta(A)$. By definition, $\tau:A\to\mathbb C$ is a nonzero complex-linear multiplicative functional. Because it is nonzero, there is an element $a_0\in A$ with $\tau(a_0)\neq 0$. Applying multiplicativity to $1_Aa_0=a_0$ gives \begin{align*}\tau(a_0)=\tau(1_Aa_0)=\tau(1_A)\tau(a_0).\end{align*} The scalar $\tau(a_0)$ is nonzero, so cancellation in $\mathbb C$ gives $\tau(1_A)=1$. Now fix an element $a\in A$. We want to show $|\tau(a)|\leq \|a\|_A$. The useful idea is to prove that every scalar $\lambda$ with $|\lambda|>\|a\|_A$ lies outside the possible value $\tau(a)$. If $|\lambda|>\|a\|_A$, then \begin{align*}\|\lambda^{-1}a\|_A=|\lambda|^{-1}\|a\|_A<1.\end{align*} The hypotheses of [citetheorem:8545] are satisfied because $A$ is a unital Banach algebra and $\lambda^{-1}a\in A$ has norm strictly less than $1$. Therefore $1_A-\lambda^{-1}a$ is invertible in $A$. Multiplication by the nonzero scalar $\lambda$ preserves invertibility, so $\lambda 1_A-a$ is invertible. Assume, toward a contradiction, that $|\tau(a)|>\|a\|_A$. Define the scalar $\lambda:=\tau(a)$. By the preceding paragraph, $\lambda 1_A-a$ is invertible. On the other hand, using complex linearity of $\tau$ and the already proved identity $\tau(1_A)=1$, we get \begin{align*} \tau(\lambda 1_A-a)=\lambda\tau(1_A)-\tau(a)=\lambda-\lambda=0. \end{align*} Let $b\in A$ be the inverse of $\lambda 1_A-a$. Multiplicativity now gives \begin{align*} 1=\tau(1_A)=\tau((\lambda 1_A-a)b)=\tau(\lambda 1_A-a)\tau(b)=0. \end{align*} This contradiction proves $|\tau(a)|\leq \|a\|_A$ for every $a\in A$. Hence $\tau$ is bounded and $\|\tau\|_{A^*}\leq 1$, which is the contractive estimate needed to place $\tau$ in the closed unit ball of $A^*$.[/guided]

[step:Embed the character space in a weak-* compact ball]Define the closed unit ball of the [dual space](/page/Dual%20Space) by $K:=\{\lambda\in A^*: \|\lambda\|_{A^*}\leq 1\}$. By the previous step, $\Delta(A)\subseteq K$. We use the [Banach-Alaoglu theorem](/page/Banach-Alaoglu%20Theorem) in the following precise form: for every normed complex [vector space](/page/Vector%20Space) $X$, the closed unit ball of $X^*$ is compact in the weak-* topology $\sigma(X^*,X)$. Its hypotheses apply with $X=A$, since a unital complex Banach algebra is in particular a normed complex vector space. Therefore $K$ is compact for the weak-* topology $\sigma(A^*,A)$.[/step]

[guided]The previous step gives a [uniform norm](/page/Uniform%20Norm) bound on all characters, and this is the point where compactness enters. Define the closed unit ball of the dual space by \begin{align*} K:=\{\lambda\in A^*: \|\lambda\|_{A^*}\leq 1\}. \end{align*} For every $\tau\in\Delta(A)$, the contractive estimate $\|\tau\|_{A^*}\leq 1$ proved above gives $\tau\in K$. Hence \begin{align*} \Delta(A)\subseteq K. \end{align*} We now invoke the Banach-Alaoglu theorem in the precise form needed here: if $X$ is a normed complex vector space, then the closed unit ball of $X^*$ is compact in the weak-* topology $\sigma(X^*,X)$. This theorem applies with $X=A$ because a unital complex Banach algebra is, after forgetting multiplication, a normed complex vector space. The closed unit ball of $A^*$ is exactly the set $K$ defined above. Therefore $K$ is compact for the weak-* topology $\sigma(A^*,A)$.[/guided]

[step:Express unitality and multiplicativity as weak-* closed equations]For each $a\in A$, define the point-evaluation map \begin{align*} \operatorname{ev}_a:A^*\to\mathbb C,\qquad \lambda\mapsto \lambda(a). \end{align*} By definition of the weak-* topology $\sigma(A^*,A)$, each map $\operatorname{ev}_a$ is weak-* continuous. Define \begin{align*} F_1:=\{\lambda\in K:\lambda(1_A)=1\}. \end{align*} Then $F_1=\operatorname{ev}_{1_A}^{-1}(\{1\})\cap K$, so $F_1$ is weak-* closed in $K$ because $\{1\}$ is closed in $\mathbb C$. For each ordered pair $(a,b)\in A\times A$, define \begin{align*} M_{a,b}:K\to\mathbb C,\qquad \lambda\mapsto \lambda(ab)-\lambda(a)\lambda(b). \end{align*} The map $M_{a,b}$ is weak-* continuous because it is built from the weak-* continuous maps $\operatorname{ev}_{ab}$, $\operatorname{ev}_a$, and $\operatorname{ev}_b$ using scalar multiplication and subtraction in $\mathbb C$. Hence \begin{align*} F_{a,b}:=\{\lambda\in K:\lambda(ab)=\lambda(a)\lambda(b)\}=M_{a,b}^{-1}(\{0\}) \end{align*} is weak-* closed in $K$. Now set \begin{align*} F:=F_1\cap\bigcap_{(a,b)\in A\times A}F_{a,b}. \end{align*} Arbitrary intersections of closed sets are closed, so $F$ is weak-* closed in $K$.[/step]

[guided]We now convert the algebraic definition of a character into topological equations. For each $a\in A$, define \begin{align*} \operatorname{ev}_a:A^*\to\mathbb C,\qquad \lambda\mapsto \lambda(a). \end{align*} The weak-* topology $\sigma(A^*,A)$ is precisely the topology of pointwise convergence on $A$, so each $\operatorname{ev}_a$ is continuous by definition. First encode the condition that a functional is unital. Define \begin{align*} F_1:=\{\lambda\in K:\lambda(1_A)=1\}. \end{align*} This is the inverse image of the [closed set](/page/Closed%20Set) $\{1\}\subset\mathbb C$ under the continuous map $\operatorname{ev}_{1_A}$, intersected with $K$. Therefore $F_1$ is weak-* closed in $K$. Next encode multiplicativity. Fix $a,b\in A$. We need the equation $\lambda(ab)=\lambda(a)\lambda(b)$ to define a closed condition on $\lambda$. Define the map \begin{align*} M_{a,b}:K\to\mathbb C,\qquad \lambda\mapsto \lambda(ab)-\lambda(a)\lambda(b). \end{align*} The first term $\lambda\mapsto\lambda(ab)$ is $\operatorname{ev}_{ab}$, hence weak-* continuous. The two maps $\lambda\mapsto\lambda(a)$ and $\lambda\mapsto\lambda(b)$ are also weak-* continuous, and multiplication $\mathbb C\times\mathbb C\to\mathbb C$ is continuous in the usual topology. Thus $\lambda\mapsto\lambda(a)\lambda(b)$ is weak-* continuous, and subtracting continuous complex-valued functions preserves continuity. Hence $M_{a,b}$ is weak-* continuous. The multiplicativity equation for this fixed pair $(a,b)$ is exactly the zero set of $M_{a,b}$: \begin{align*} F_{a,b}:=\{\lambda\in K:\lambda(ab)=\lambda(a)\lambda(b)\}=M_{a,b}^{-1}(\{0\}). \end{align*} Since $\{0\}$ is closed in $\mathbb C$, the set $F_{a,b}$ is weak-* closed in $K$. Finally, a functional is multiplicative for all pairs $a,b\in A$ precisely when it lies in every $F_{a,b}$. Combining this with the unital condition gives \begin{align*} F:=F_1\cap\bigcap_{(a,b)\in A\times A}F_{a,b}. \end{align*} This is an arbitrary intersection of weak-* closed subsets of $K$, and arbitrary intersections of closed sets are closed. Therefore $F$ is weak-* closed in $K$.[/guided]

[step:Identify the closed set with the character space]We claim that $F=\Delta(A)$. If $\tau\in\Delta(A)$, then the first step gives $\tau\in K$ and $\tau(1_A)=1$, while multiplicativity gives $\tau(ab)=\tau(a)\tau(b)$ for all $a,b\in A$. Thus $\tau\in F$. Conversely, if $\lambda\in F$, then $\lambda\in A^*$ is complex-linear, satisfies $\lambda(1_A)=1$, and satisfies $\lambda(ab)=\lambda(a)\lambda(b)$ for all $a,b\in A$. Since $\lambda(1_A)=1$, $\lambda$ is nonzero. Hence $\lambda\in\Delta(A)$. Therefore $F=\Delta(A)$, and $\Delta(A)$ is weak-* closed in $K$.[/step]

[guided]We must check that the closed set $F$ constructed from equations is exactly the character space, not merely a related closed set. First let $\tau\in\Delta(A)$. By definition of $\Delta(A)$, the map $\tau:A\to\mathbb C$ is a nonzero complex-linear multiplicative functional. The first step proved two additional facts for every such $\tau$: it belongs to $K$ and satisfies $\tau(1_A)=1$. Its multiplicativity gives \begin{align*} \tau(ab)=\tau(a)\tau(b) \end{align*} for every $a,b\in A$. Thus $\tau$ satisfies the defining condition for $F_1$ and every defining condition for $F_{a,b}$, so $\tau\in F$. Conversely, let $\lambda\in F$. Since $F\subseteq K\subseteq A^*$, the map $\lambda:A\to\mathbb C$ is a bounded complex-linear functional. The condition $\lambda\in F_1$ says \begin{align*} \lambda(1_A)=1. \end{align*} In particular, $\lambda$ is nonzero. Also, membership in every $F_{a,b}$ says that for all $a,b\in A$, \begin{align*} \lambda(ab)=\lambda(a)\lambda(b). \end{align*} Therefore $\lambda$ is a nonzero multiplicative complex-linear functional on $A$, so $\lambda\in\Delta(A)$. We have proved both inclusions $\Delta(A)\subseteq F$ and $F\subseteq\Delta(A)$, hence $F=\Delta(A)$. Since the previous step proved that $F$ is weak-* closed in $K$, it follows that $\Delta(A)$ is weak-* closed in $K$.[/guided]

[step:Conclude compactness and Hausdorffness]Since $K$ is weak-* compact and $\Delta(A)$ is weak-* closed in $K$, the subspace $\Delta(A)$ is compact. It remains to verify the Hausdorff property. Let $\lambda,\mu\in A^*$ with $\lambda\neq\mu$. Then there exists $a\in A$ such that $\lambda(a)\neq\mu(a)$. Because $\mathbb C$ is Hausdorff, there are disjoint open sets $U,V\subset\mathbb C$ with $\lambda(a)\in U$ and $\mu(a)\in V$. The weak-* open sets $\operatorname{ev}_a^{-1}(U)$ and $\operatorname{ev}_a^{-1}(V)$ separate $\lambda$ and $\mu$ in $A^*$. Thus $A^*$ is Hausdorff in the weak-* topology. Therefore its subspace $\Delta(A)$ is Hausdorff. Combining compactness and Hausdorffness, $\Delta(A)$ is a compact [Hausdorff space](/page/Hausdorff%20Space) in the weak-* topology.[/step]

[guided]The topological conclusion comes from compactness and separation. We have defined \begin{align*} K:=\{\lambda\in A^*: \|\lambda\|_{A^*}\leq 1\} \end{align*} and proved that $K$ is compact in the weak-* topology $\sigma(A^*,A)$. We have also proved that $\Delta(A)=F$, where $F$ is an intersection of weak-* closed subsets of $K$. Therefore $\Delta(A)$ is weak-* closed in $K$. A closed subset of a compact [topological space](/page/Topological%20Space) is compact, so $\Delta(A)$ is compact in the subspace topology inherited from $\sigma(A^*,A)$. It remains to prove that this topology is Hausdorff. Let $\lambda,\mu\in A^*$ with $\lambda\neq\mu$. Since $\lambda$ and $\mu$ are distinct functions from $A$ to $\mathbb C$, there exists $a\in A$ such that \begin{align*} \lambda(a)\neq\mu(a). \end{align*} The complex plane $\mathbb C$ is Hausdorff in its usual topology, so there exist disjoint open sets $U,V\subset\mathbb C$ with $\lambda(a)\in U$ and $\mu(a)\in V$. By definition of the weak-* topology, the point-evaluation map \begin{align*} \operatorname{ev}_a:A^*\to\mathbb C,\qquad \nu\mapsto \nu(a) \end{align*} is weak-* continuous. Hence $\operatorname{ev}_a^{-1}(U)$ and $\operatorname{ev}_a^{-1}(V)$ are weak-* open subsets of $A^*$. They are disjoint because $U\cap V=\varnothing$, and they contain $\lambda$ and $\mu$, respectively. Thus $A^*$ is Hausdorff in the weak-* topology. Every subspace of a Hausdorff space is Hausdorff, so $\Delta(A)$ is Hausdorff. Combining compactness and Hausdorffness proves that $\Delta(A)$ is a compact Hausdorff space in the weak-* topology.[/guided]