From 777f9d3fd8caf56e6bc6999a4b05379307d0733f Mon Sep 17 00:00:00 2001
From: Justin Gassner <justin.gassner@mailbox.org>
Date: Tue, 12 Sep 2023 07:36:33 +0200
Subject: Initial commit

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 pages/operator-algebras/c-star-algebras/states.md | 43 +++++++++++++++++++++++
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+---
+title: States
+parent: C*-Algebras
+grand_parent: Operator Algebras
+nav_order: 1
+# cspell:words
+---
+
+# {{ page.title }}
+
+{: .definition-title }
+> Definition (State, State Space)
+>
+> A norm-one [positive linear functional]( {% link pages/operator-algebras/c-star-algebras/positive-linear-functionals.md %} ) on a C\*-algebra is called a *state*.\
+> The *state space* $S(\mathcal{A})$ of a C\*-algebra $\mathcal{A}$ is the set of all its states.
+
+Note that $S(\mathcal{A})$ is a subset of the unit ball in the dual space of $\mathcal{A}$.
+
+{: .corollary }
+> A linear functional $\omega$ on a C\*-algebra is a state
+> if and only if $\omega(\mathbf{1}) = 1 = \norm{\omega}$.
+
+{: .proposition }
+> The state space of a C\*-algebra is convex and weak\* compact.
+
+{% proof %}
+First, we show convexity.
+Let $\omega_0, \omega_1$ be states on $\mathcal{A}$ and let $t \in (0,1)$.
+Consider the convex combination $\omega = (1-t)\omega_0 + t\omega_1$.
+Clearly, $\omega$ is linear and $\omega(\mathbf{1}) = 1$.
+By the triangle inequality, $\norm{\omega} \le 1$.
+It follows from the lemma above that $\omega$ lies in $S(\mathcal{A})$. This proves that $S(\mathcal{A})$ is convex.
+
+Next we show weak\* compactness. Since $S(\mathcal{A})$ is contained
+in the closed unit ball in the dual of $\mathcal{A}$,
+which is weak\* compact by the
+[Banach–Alaoglu Theorem]({% link pages/functional-analysis-basics/banach-alaoglu-theorem.md %}),
+it will suffice to show that $S(\mathcal{A})$ is weak\* closed.
+Let $(\omega_i)$ be a net of states that weak\* converges to some bounded linear functional $\omega$ on $\mathcal{A}$.
+This means that $\omega_i(x) \to \omega(x)$ for every $x \in \mathcal{A}$.
+For all $i$ we have $\omega_i(x) \ge 0$ for $x \ge 0$ and $\omega_i(\mathbf{1}) = 1$; hence $\omega(x) \ge 0$ for $x \ge 0$ and $\omega(\mathbf{1}) = 1$. Thus $\omega$ is again a state.
+This shows that the state space is weak* closed, completing the proof.
+{% endproof %}
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