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Diffstat (limited to 'pages/operator-algebras/c-star-algebras/states.md')
| -rw-r--r-- | pages/operator-algebras/c-star-algebras/states.md | 17 |
1 files changed, 11 insertions, 6 deletions
diff --git a/pages/operator-algebras/c-star-algebras/states.md b/pages/operator-algebras/c-star-algebras/states.md index 29cf5f5..a483915 100644 --- a/pages/operator-algebras/c-star-algebras/states.md +++ b/pages/operator-algebras/c-star-algebras/states.md @@ -8,7 +8,9 @@ nav_order: 1 # {{ page.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*.\ +A norm-one +[positive linear functional](/pages/operator-algebras/c-star-algebras/positive-linear-functionals.html) +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. {% enddefinition %} @@ -30,20 +32,23 @@ 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. +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 %}), +[Banach–Alaoglu Theorem](/pages/functional-analysis-basics/banach-alaoglu-theorem.html), 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 $\mathcal{A}$. +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. +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 %} TODO: state space is nonempty TODO: pure states - |