[proofplan] We first prove that the state space is nonempty by extending the canonical state on the scalar subalgebra $\mathbb C1_A$ to all of $A$. Then we use the [compactness and convexity of the state space](/theorems/8559) in the weak-* topology. Finally, Krein-Milman applies because the weak-* topology is locally convex, giving that $S(A)$ is the weak-* closed convex hull of its extreme points; by the definition of pure state, these extreme points are precisely the pure states. [/proofplan]

[step:Extend the scalar state to obtain a state on $A$]Let $B:=\mathbb C1_A\subset A$. Since $A$ is unital and nonzero, $1_A\neq 0$, so every element of $B$ has a unique form $\lambda 1_A$ with $\lambda\in\mathbb C$. Define \begin{align*} \rho_0:B&\to\mathbb C \end{align*} by \begin{align*} \rho_0(\lambda 1_A)=\lambda. \end{align*} The subspace $B$ is a unital $C^*$-subalgebra of $A$ with the same unit. If $b=\lambda 1_A\in B$ is positive, then $\lambda\in[0,\infty)$, hence $\rho_0(b)=\lambda\ge 0$. Also $\rho_0(1_A)=1$, so $\rho_0$ is a state on $B$. By [citetheorem:8561] applied to the unital inclusion $B\subset A$, there exists a state $\rho:A\to\mathbb C$ such that $\rho|_B=\rho_0$. Hence $S(A)\neq\varnothing$.[/step]

[guided]The purpose of this step is to prove that there is at least one state before applying any compactness or convexity theorem to the state space. Let \begin{align*} B:=\mathbb C1_A\subset A. \end{align*} Because $A$ is assumed nonzero and unital, its unit satisfies $1_A\neq 0$. Therefore the representation of an element of $B$ as $\lambda 1_A$ is unique: if $\lambda 1_A=\mu 1_A$, then $(\lambda-\mu)1_A=0$, and since $1_A\neq 0$, we get $\lambda=\mu$. Define the functional \begin{align*} \rho_0:B&\to\mathbb C \end{align*} by \begin{align*} \rho_0(\lambda 1_A)=\lambda. \end{align*} This map is linear because scalar multiplication and addition in $B$ are inherited from $A$. We verify that it is a state on $B$. First, $B$ is a unital $C^*$-subalgebra of $A$ with unit $1_A$. Second, if $b=\lambda 1_A\in B$ is positive, then the spectrum of $b$ in the scalar algebra is contained in $[0,\infty)$, so $\lambda\in[0,\infty)$. Thus \begin{align*} \rho_0(b)=\lambda\ge 0. \end{align*} So $\rho_0$ is positive. Finally, \begin{align*} \rho_0(1_A)=1, \end{align*} and therefore $\rho_0$ is a state on $B$. Now [citetheorem:8561] applies: its hypotheses require a unital $C^*$-algebra $A$, a unital $C^*$-subalgebra $B\subset A$ with the same unit, and a state on $B$. We have verified each of these conditions. Hence there exists a state $\rho:A\to\mathbb C$ extending $\rho_0$. In particular, at least one state on $A$ exists, so \begin{align*} S(A)\neq\varnothing. \end{align*}[/guided]

[step:Use weak-* compactness and convexity of the state space] By [citetheorem:8559], the state space $S(A)$ is a weak-* compact convex subset of $A^*$. Since the weak-* topology on $A^*$ is the topology of pointwise convergence on $A$, it is a locally convex topology. Thus $S(A)$ is a nonempty compact convex subset of the locally convex [topological vector space](/page/Topological%20Vector%20Space) $(A^*,w^*)$. [/step]

[step:Apply Krein-Milman to identify the weak-* closed convex hull]We apply the Krein-Milman theorem to the nonempty compact convex set $S(A)\subset (A^*,w^*)$. The theorem gives \begin{align*} S(A)=\overline{\operatorname{conv}(\operatorname{Ext} S(A))}^{\,w^*}. \end{align*} Here $\operatorname{Ext} S(A)$ denotes the set of extreme points of $S(A)$, and the closure is taken in the weak-* topology.[/step]

[guided]We now use the geometric structure of $S(A)$. The Krein-Milman theorem applies to compact convex subsets of locally convex topological vector spaces. The ambient space is $A^*$ equipped with the weak-* topology. For each $a\in A$, define the seminorm \begin{align*} p_a:A^*&\to[0,\infty) \end{align*} by \begin{align*} p_a(\rho):=|\rho(a)|. \end{align*} The weak-* topology is generated by the family $(p_a)_{a\in A}$, hence it is locally convex. By [citetheorem:8559], applied to the unital $C^*$-algebra $A$, the set $S(A)$ is weak-* compact and convex in $A^*$. By the first step it is also nonempty. Therefore the hypotheses of the Krein-Milman theorem are satisfied for the set $S(A)\subset (A^*,w^*)$. Its conclusion is that a compact convex set is the closed convex hull of its extreme points. In this setting, that says exactly \begin{align*} S(A)=\overline{\operatorname{conv}(\operatorname{Ext} S(A))}^{\,w^*}. \end{align*} The superscript $w^*$ records that the closure is taken in the weak-* topology, not in the norm topology of $A^*$.[/guided]

[step:Identify extreme points with pure states and conclude existence] By the definition of pure state, a state on $A$ is pure precisely when it is an extreme point of the convex set $S(A)$. Hence $\operatorname{Ext} S(A)$ is the set of pure states on $A$, and the displayed formula is the asserted formula. It remains only to see that pure states exist. If $\operatorname{Ext} S(A)=\varnothing$, then \begin{align*} \operatorname{conv}(\operatorname{Ext} S(A))=\varnothing, \end{align*} so its weak-* closure is also empty. This contradicts the formula above and the already proved fact that $S(A)\neq\varnothing$. Therefore $\operatorname{Ext} S(A)\neq\varnothing$, and $A$ has at least one pure state. [/step]