[proofplan] We prove convexity directly from the defining two conditions for a state: positivity and normalization at the unit. For compactness, we first place every state in the closed unit ball of $A^*$ by the norm formula for positive functionals. Banach-Alaoglu gives weak* compactness of that ball, and it remains only to prove that the state space is weak* closed. The closedness follows because normalization and positivity are both expressed as weak* closed pointwise conditions. [/proofplan]

[step:Check that convex combinations of states are states]Let $\rho_0,\rho_1\in S(A)$, and let $t\in[0,1]$. Define the linear functional $\rho_t:A\to\mathbb C$ by \begin{align*} \rho_t(a) = t\rho_1(a)+(1-t)\rho_0(a) \end{align*} for every $a\in A$. For every $a\in A$, positivity of $\rho_0$ and $\rho_1$ gives $\rho_t(a^*a)=t\rho_1(a^*a)+(1-t)\rho_0(a^*a)\ge 0$. Thus $\rho_t$ is positive. Also, $\rho_t(1_A)=t\rho_1(1_A)+(1-t)\rho_0(1_A)=t+(1-t)=1$. Therefore $\rho_t\in S(A)$, so $S(A)$ is convex.[/step]

[guided]We must show that the line segment between any two states remains inside the state space. Let $\rho_0,\rho_1\in S(A)$ and let $t\in[0,1]$. Define the functional $\rho_t:A\to\mathbb C$ by $\rho_t(a)=t\rho_1(a)+(1-t)\rho_0(a)$ for every $a\in A$. Because $\rho_0$ and $\rho_1$ are linear functionals and $t,1-t\in\mathbb R$, the functional $\rho_t$ is linear. To verify positivity, take an arbitrary element $a\in A$. Since $\rho_0$ and $\rho_1$ are states, they are positive, so $\rho_0(a^*a)\ge 0$ and $\rho_1(a^*a)\ge 0$. Since $t\ge 0$ and $1-t\ge 0$, their weighted sum is also nonnegative: $\rho_t(a^*a)=t\rho_1(a^*a)+(1-t)\rho_0(a^*a)\ge 0$. Thus $\rho_t$ is positive. It remains to check the normalization at the unit. Since $\rho_0(1_A)=1$ and $\rho_1(1_A)=1$, we have $\rho_t(1_A)=t\rho_1(1_A)+(1-t)\rho_0(1_A)=t+(1-t)=1$. Therefore $\rho_t$ is a positive linear functional normalized by $\rho_t(1_A)=1$, hence $\rho_t\in S(A)$. This proves that $S(A)$ is convex.[/guided]

[step:Place the state space inside the weak* compact closed unit ball] Let $B_{A^*}:=\{\rho\in A^*:\|\rho\|_{A^*}\le 1\}$ be the closed unit ball of the dual [Banach space](/page/Banach%20Space) $A^*$. If $\rho\in S(A)$, then $\rho$ is a positive linear functional on the unital $C^*$-algebra $A$, and $\rho(1_A)=1$. By the Norm Formula for Positive Functionals (TEMP-14), $\|\rho\|_{A^*}=\rho(1_A)=1$. Hence $S(A)\subset B_{A^*}$. By the [Banach-Alaoglu Theorem](/page/Banach-Alaoglu%20Theorem) applied to the normed space $A$, the closed unit ball $B_{A^*}$ is compact for the [weak* topology](/page/Weak*%20Topology) $\sigma(A^*,A)$ (citing a result not yet in the wiki: Banach-Alaoglu Theorem). [/step]

[step:Express the state space as a weak* closed subset of the unit ball] For each $b\in A$, define the evaluation map $\operatorname{ev}_b:(A^*,\sigma(A^*,A))\to\mathbb C$ by $\operatorname{ev}_b(\rho)=\rho(b)$ for every $\rho\in A^*$. By the definition of the weak* topology $\sigma(A^*,A)$, each map $\operatorname{ev}_b$ is continuous. The normalization condition is weak* closed because $\{\rho\in A^*:\rho(1_A)=1\}=\operatorname{ev}_{1_A}^{-1}(\{1\})$, and $\{1\}$ is closed in $\mathbb C$. For each $a\in A$, define $P_a:=\{\rho\in A^*:\rho(a^*a)\in[0,\infty)\}$. Since $[0,\infty)$ is closed in $\mathbb C$ and $P_a=\operatorname{ev}_{a^*a}^{-1}([0,\infty))$, each $P_a$ is weak* closed. A linear functional $\rho:A\to\mathbb C$ is positive exactly when $\rho(a^*a)\ge 0$ for every $a\in A$. Therefore $S(A)=\{\rho\in A^*:\rho(1_A)=1\}\cap\bigcap_{a\in A}P_a$. This is an intersection of weak* closed subsets of $A^*$, so $S(A)$ is weak* closed in $A^*$. [/step]

[step:Conclude weak* compactness] From the previous steps, $S(A)\subset B_{A^*}$ and $S(A)$ is weak* closed in $A^*$. Hence $S(A)$ is weak* closed as a subset of the weak* [compact space](/page/Compact%20Space) $B_{A^*}$. Therefore $S(A)$ is weak* compact. Together with the convexity proved above, this shows that $S(A)$ is a weak* compact convex subset of $A^*$. [/step]