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\section{States}
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\label{section:cstar-states}
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\begin{definition}[State]
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\label{definition:cstar-state}
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Let $A$ be a unital $C^*$-algebra and $\phi \in A^*$, then $\phi$ is a \textbf{state} if $\phi$ is positive and $\dpn{1, \phi}{A} = 1$.
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The set of states $S(A) \subset A^*$ of $A$ equipped with the weak* topology is the \textbf{state space} of $A$.
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\end{definition}
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\begin{lemma}
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\label{lemma:cstar-state-cauchy-schwarz}
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Let $A$ be a unital $C^*$-algebra and $\phi \in A^*$ be a positive linear functional, then the mapping
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\[
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A \times A \to \complex \quad (x, y) \mapsto \dpn{x, y}{\phi} := \dpn{y^*x, \phi}{A}
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\]
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is a pseudo inner product. In particular, for any $x, y \in A$,
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\[
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|\dpn{y^*x, \phi}{A}|^2 = |\dpn{x, y}{\phi}|^2 \le \dpn{x, x}{\phi} \cdot \dpn{y, y}{\phi}
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\]
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\end{lemma}
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\begin{proof}
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By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}.
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\end{proof}
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\begin{definition}[Pure State]
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\label{definition:pure-state}
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Let $A$ be a unital $C^*$-algebra and $\phi \in S(A)$, then $\phi$ is a \textbf{pure state} if $\phi$ is an extreme point of $S(A)$. The set $P(A)$ is the collection of all pure states of $A$.
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\end{definition}
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\begin{proposition}
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\label{proposition:state-space-compact-convex}
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Let $A$ be a unital $C^*$-algebra, then $S(A)$ is a compact convex set, and $S(A)$ is the weak*-closed convex hull of $P(A)$.
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\end{proposition}
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\begin{proof}
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Since the evaluation map is weak* continuous and
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\[
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S(A) = \bracs{\phi \in A^*|\dpn{1, \phi}{A} = 1} \cap \bigcap_{\substack{x \in A \\ x \ge 0}}\bracs{\phi \in A^*|\dpn{x, \phi}{A} \ge 0}
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\]
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the state space is an intersection of convex and weak*-closed sets, so it is closed and convex.
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By \autoref{theorem:cstar-positive-algebraic}, $S(A) \subset \ol{B_{A^*}(0, 1)}$, which is weak* compact by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. Therefore $S(A)$ is compact by \autoref{proposition:compact-extensions}.
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By the \hyperref[Krein-Milman Theorem]{theorem:krein-milman}, $S(A)$ is the weak*-closed convex hull of $P(A)$.
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\end{proof}
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\begin{proposition}
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\label{proposition:multiplicative-pure-state}
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Let $A$ be a unital $C^*$-algebra, then:
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\begin{enumerate}
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\item $\Omega(A) \subset P(A)$.
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\item If $A$ is commutative, then $\Omega(A) = P(A)$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): Let $\phi \in \Omega(A)$. By \autoref{proposition:multiplicative-unit}, $\norm{\phi}_{A^*} = \dpn{1, \phi}{A} = 1$. Thus $\phi$ is a state by \autoref{theorem:cstar-positive-algebraic}, and $\Omega(A) \subset S(A)$.
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Let $\psi, \rho \in S(A)$ and $t \in (0, 1)$ such that $\phi = (1 - t)\psi + t\rho$, then for each $x \in \ker(\phi)$, $x^*x \in \ker(\phi)$ as well. As $t \ne 0$, $x^*x \in \ker(\psi)$ and $x^*x \in \ker(\rho)$. By the \hyperref[Cauchy-Schwarz inequality]{lemma:cstar-state-cauchy-schwarz},
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\[
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|\dpn{x, \psi}{A}|^2 = |\dpn{1^*x, \psi}{A}|^2 \le \dpn{1, \psi}{A} \cdot \dpn{x^*x, \psi}{A} = 0
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\]
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Likewise, $\dpn{x, \rho}{A} = 0$ as well. Hence $\ker(\psi), \ker(\rho) \supset \ker(\phi)$. Thus there exist scalars $\alpha, \beta \in \complex$ such that $\phi = \alpha \psi = \beta \rho$. However, since $\phi, \psi, \rho \in S(A)$, $\alpha = \beta = 1$, and $\phi = \psi = \rho$. Therefore $\phi$ is a pure state.
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(2): Using the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, identify $A$ with $C(\Omega(A); \complex)$ and $S(A)$ as Radon probability measures on $\Omega(A)$.
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Let $\cm = \bracs{t\mu|\mu \in S(A), t \in [0, 1]}$. By \autoref{proposition:space-of-measures-extreme-points}, the extreme points of $\cm$ are the delta masses $\bracs{\delta_x|x \in \Omega(A)}$, and possibly $0$. For any $\mu \in S(A)$, $\nu, \rho \in \cm$, and $t \in (0, 1)$, $\mu = (1 - t)\nu + t\rho$ implies that $\nu(\Omega(A)) = \rho(\Omega(A)) = 1$, and $\nu, \rho \in S(A)$ as well. Thus the extreme points of $S(A)$ are exactly the delta masses $\bracs{\delta_x|x \in \Omega(A)}$, which correspond to $\Omega(A)$ itself.
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\end{proof}
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\begin{theorem}[Extension of States]
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\label{theorem:cstar-pure-state-extension}
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Let $A$ be a unital $C^*$-algebra, $B \subset A$ be a $C^*$-subalgebra with $1_A \in B$, and $\phi \in S(B)$, then
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\begin{enumerate}
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\item There exists $\Phi \in S(A)$ such that $\Phi|_B = \phi$.
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\item If $\phi \in P(B)$, then there exists $\Phi \in P(A)$ such that $\Phi|_B = \phi$.
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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(1): By \autoref{theorem:cstar-positive-algebraic}, $\norm{\phi}_{B^*} = \dpn{1_A, \phi}{B}$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach}, there exists $\Phi \in A^*$ such that $\Phi|_B = \phi$ and $\norm{\Phi}_{A^*} = \norm{\phi}_{B^*} = \dpn{1_A, \Phi}{A}$. Thus \autoref{theorem:cstar-positive-algebraic} implies that $\Phi \in S(A)$.
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(2): Let $E(\phi) = \bracs{\Phi \in S(A)|\Phi|_B = \phi}$ be the collection of all extensions of $\phi$, then $E(\phi)$ is a weak*-closed convex subset of $S(A)$. By (1), $E(\phi)$ is non-empty, and as such admits an extreme point $\Phi$ by the \hyperref[Krein-Milman Theorem]{theorem:krein-milman}.
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Let $\psi, \rho \in S(A)$ and $t \in (0, 1)$ such that $\Phi = (1 - t)\psi + t\rho$. In which case, $\phi = (1 - t)\psi|_B + t\rho|_B$. Since $\phi \in P(B)$, $\phi = \psi|_B = \rho|_B$, so $\psi, \rho \in E(\phi)$. As $\Phi$ is an extreme point of $E(\phi)$, $\Phi = \psi = \rho$. Therefore $\Phi \in P(A)$.
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\end{proof}
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\begin{theorem}
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\label{theorem:cstar-state-existence}
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Let $A$ be a unital $C^*$-algebra, $x \in A$, and $\lambda \in \sigma_A(x)$, then there exists $\phi \in S(A)$ such that $\dpn{x, \phi}{A} = \lambda$.
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Theorem 13.7]{Zhu}}}. ]
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Let $B = \text{span}\bracs{x, 1}$. For each $\alpha x + \beta \in B$, let $\dpn{\alpha x + \beta, \phi_0}{B} = \alpha \lambda + \beta$. Since $\sigma_A(1) = \bracs{1}$, $\phi_0 \in B^*$ is a well-defined linear functional with $\dpn{x, \phi_0}{B} = \lambda$ and $\dpn{1, \phi_0}{B} = 1$.
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In addition, for each $\alpha x + \beta \in B$, $\alpha \lambda + \beta \in \sigma_A(\alpha x + \beta)$ by \autoref{proposition:commutative-spectrum-gymnastics}, and
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\[
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|\alpha \lambda + \beta| \le [\alpha x + \beta]_{sp} \le \norm{\alpha x + \beta}_A
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\]
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Thus $\norm{\phi_0}_{B^*} = \dpn{1, \phi_0}{B} = 1$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach}, there exists $\phi \in A^*$ such that $\phi|_B = \phi_0$ and $\norm{\phi}_{A^*} = \norm{\phi_0}_{B^*}$. In which case, $\dpn{x, \phi}{A} = \lambda$ and $\dpn{1, \phi}{A} = \norm{\phi}_{A^*} = 1$. By \autoref{theorem:cstar-positive-algebraic}, $\phi$ is positive and hence a state.
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\end{proof}
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\begin{corollary}
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\label{corollary:cstar-positive-weakstar-dense}
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Let $A$ be a unital $C^*$-algebra, then:
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\begin{enumerate}
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\item For each $x \in A$, $x = 0$ if and only if $\dpn{x, \phi}{A} = 0$ for all $\phi \in P(A)$.
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\item The linear span of $P(A)$ is weak*-dense in $A^*$.
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\end{enumerate}
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\end{corollary}
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\begin{proof}
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(1): Let $x \in A$ such that $\dpn{x, \phi}{A} = 0$ for all $\phi \in P(A)$. First suppose that $x$ is self-adjoint. By \autoref{theorem:cstar-state-existence}, $\sigma_A(x) = \bracs{0}$, and $\norm{x}_A = [x]_{sp} = 0$ by \autoref{theorem:c-star-normal-spectral-radius}.
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Now suppose that $x$ is arbitrary. In this case, for each $\phi \in P(A)$,
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\[
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0 = \text{Re}(\dpn{x, \phi}{A}) = \dpn{\text{Re}(x), \phi}{A}
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\]
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because $\phi$ is Hermitian. Similarly, $\dpn{\text{Im}(x), \phi}{A} = 0$ as well. Thus $\text{Re}(x) = \text{Im}(x) = 0$, and $x = 0$ as well.
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(2): Since the linear span of $P(A)$ separates points in $A$, it is weak*-dense in $A^*$ by \autoref{lemma:duality-dense}.
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\end{proof}
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\begin{corollary}
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\label{corollary:cstar-positive-property-probe}
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Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then\footnote{The crude bound seems kind of tragic, but it wouldn't be true otherwise. }
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\begin{align*}
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\sigma_A(x) &\subset \bracs{\dpn{x, \phi}{A}|\phi \in P(A)} \\
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&\subset \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} = \ol{\text{Conv}}(\sigma_A(x))
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\end{align*}
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In particular,
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\begin{enumerate}
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\item $x$ is self-adjoint if and only if $\dpn{x, \phi}{A} \in \real$ for all $\phi \in P(A)$.
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\item $x$ is positive if and only if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in P(A)$.
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\item There exists $\phi \in P(A)$ such that $\norm{x}_A = |\dpn{x, \phi}{A}|$.
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\end{enumerate}
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\end{corollary}
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\begin{proof}
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Let $\lambda \in \sigma_A(x)$. By \autoref{proposition:gelfand-transform-gymnastics}, there exists $\phi \in \Omega(A[x])$ such that $\dpn{x, \phi}{A[x]} = \lambda$. By \autoref{proposition:multiplicative-pure-state}, $\phi \in P(A[x])$. The \hyperref[pure state extension theorem]{theorem:cstar-pure-state-extension} implies that there exists $\Phi \in P(A)$ such that $\Phi|_{A[x]} = \phi$. Thus $\Phi$ is a pure state with $\dpn{x, \Phi}{A} = \lambda$, and $ \sigma_A(x) \subset \bracs{\dpn{x, \Phi}{A}|\Phi \in P(A)}$.
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Let $\Phi \in S(A)$ and $\phi = \Phi|_{A[x]}$, then $\phi \in S(A[x])$ as well. By the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, the \hyperref[Spectral Theorem]{theorem:spectral-c-star}, and the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, $\phi$ takes the form of a Radon probability measure $\mu$ on $\sigma_A(x)$. In which case,
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\[
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\dpn{x, \Phi}{A} = \dpn{x, \phi}{A[x]} = \int_{\sigma_A(x)}\lambda \mu(d\lambda) \in \ol{\text{Conv}}(\sigma_A(x))
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\]
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Finally, since $S(A)$ is compact and convex by \autoref{proposition:state-space-compact-convex},
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\begin{align*}
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\bracs{\dpn{x, \phi}{A}|\phi \in S(A)} &= \ol{\text{Conv}}(\bracs{\dpn{x, \phi}{A}|\phi \in P(A)}) \\
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&\subset \ol{\text{Conv}}(\sigma_A(x))
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\end{align*}
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by \autoref{proposition:compact-extensions} and \autoref{proposition:closure-of-image}.
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(1), (2): By \autoref{corollary:spectrum-characterisation-iff}.
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\end{proof}
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Bokuan Li