diff options
Diffstat (limited to 'pages/operator-algebras')
3 files changed, 37 insertions, 26 deletions
diff --git a/pages/operator-algebras/banach-algebras/index.md b/pages/operator-algebras/banach-algebras/index.md index 3335d78..9d70df8 100644 --- a/pages/operator-algebras/banach-algebras/index.md +++ b/pages/operator-algebras/banach-algebras/index.md @@ -80,7 +80,7 @@ $$ (\mathbf{1}-x) s_n = s_n (\mathbf{1}-x) = \mathbf{1} - x^{n+1}. $$ -In the limit $n \to \infty$ we obtain $(\mathbf{1}-x) s = s (\mathbf{1}-x) = \mathbf{1}$, +In the limit $n \to \infty$ we obtain $(\mathbf{1}-x) s = s (\mathbf{1}-x) = \mathbf{1}$, because multiplication in a Banach algebra is continuous, and because $y^n \to 0$ when $\norm{y} < 1$. This proves that $s$ is the inverse of $\mathbf{1}-x$. @@ -94,10 +94,12 @@ Suppose $x$ is an element of a unital Banach algebra $\mathcal{A}$. {: .mb-0 } {: .my-0 } -- The *spectrum* of $x$ is the set $\sigma(x) = \lbrace\lambda \in \CC : x - \lambda$ is not invertible in $\mathcal{A}\rbrace$. \ +- The *spectrum* of $x$ is the set + $\sigma(x) = \lbrace\lambda \in \CC : x - \lambda$ is not invertible in $\mathcal{A}\rbrace$. \ The elements of $\sigma(x)$ are called *spectral values* of $x$. - The *resolvent set* of $x$ is the set $\rho (x) = \CC \setminus \sigma(x)$. \ - For $\lambda \in \rho(x)$ the *resolvent* of $x$ is the algebra element $R_{\lambda} = (\lambda - x)^{-1}$. \ + For $\lambda \in \rho(x)$ the *resolvent* of $x$ is + the algebra element $R_{\lambda} = (\lambda - x)^{-1}$. \ The mapping $R : \rho(x) \to \mathcal{A}$, $\lambda \mapsto R_{\lambda}$, is called *resolvent map*. {% enddefinition %} @@ -119,7 +121,7 @@ $$ {% proof %} Let $\lambda$ be in the resolvent set of $x$. -Then $\lambda - x$ is invertible, and we have for all $\mu \in \CC$ +Then $\lambda - x$ is invertible, and we have for all $\mu \in \CC$ $$ \mu - x = \bigl(\mathbf{1} - (\lambda - \mu) (\lambda - x)^{-1}\bigr) (\lambda - x). @@ -138,20 +140,22 @@ the claimed formula for its inverse follows by an application of the rule $(ab)^{-1} = b^{-1} a^{-1}$ for invertible $a,b \in \mathcal{A}$. {% endproof %} -{: .corollary #resolvent-set-is-open #spectrum-is-closed } -> The resolvent set $\rho(x)$ is open and the spectrum $\sigma(x)$ is closed. +{% corollary %} +The resolvent set $\rho(x)$ is open and the spectrum $\sigma(x)$ is closed. +{% endcorollary %} -{: .corollary #resolvent-map-is-analytic } -> Suppose $x$ is an element of a unital Banach algebra $\mathcal{A}$. -> The resolvent map -> -> $$ -> R : \rho(x) \longrightarrow \mathcal{A}, \quad \lambda \longmapsto R_{\lambda} = (\lambda - x)^{-1}, -> $$ -> -> is (strongly) analytic. +{% corollary %} +Suppose $x$ is an element of a unital Banach algebra $\mathcal{A}$. +The resolvent map + +$$ +R : \rho(x) \longrightarrow \mathcal{A}, \quad \lambda \longmapsto R_{\lambda} = (\lambda - x)^{-1}, +$$ + +is (strongly) analytic. +{% endcorollary %} - --- +--- {: .proposition #spectrum-is-not-empty } > Suppose $x$ is an element of a unital Banach algebra. @@ -169,7 +173,7 @@ For $\abs{\lambda} > 2 \norm{x}$ we may expand $R_{\lambda}$ into a [Neumann ser $$ R_{\lambda} = (\lambda - x)^{-1} -= \lambda^{-1} (\mathbf{1} - \lambda^{-1} x)^{-1} += \lambda^{-1} (\mathbf{1} - \lambda^{-1} x)^{-1} = \lambda^{-1} \sum_{n=0}^{\infty} (\lambda^{-1} x)^n, $$ @@ -177,7 +181,7 @@ and make the estimate $$ \norm{R_{\lambda}} -\le \abs{\lambda}^{-1} (1 - \norm{\lambda^{-1} x})^{-1} +\le \abs{\lambda}^{-1} (1 - \norm{\lambda^{-1} x})^{-1} = (\abs{\lambda} - \norm{x})^{-1} < \norm{x}^{-1}. $$ @@ -197,7 +201,7 @@ For any Banach algebra $A$, the mapping $\varphi : \CC \to A$, $\lambda \mapsto \lambda \mathbf{1}$, is linear, multiplicative and isometric, hence injective. Let $x$ be any element of $A$. -Since its +Since its [spectrum is not empty](/pages/operator-algebras/banach-algebras/index.html#spectrum-is-not-empty), there must exist a complex number $\lambda$ such that $x - \lambda \mathbf{1}$ is not invertible. diff --git a/pages/operator-algebras/c-star-algebras/positive-linear-functionals.md b/pages/operator-algebras/c-star-algebras/positive-linear-functionals.md index ea15f87..bdf7b3d 100644 --- a/pages/operator-algebras/c-star-algebras/positive-linear-functionals.md +++ b/pages/operator-algebras/c-star-algebras/positive-linear-functionals.md @@ -22,7 +22,9 @@ A norm-one positive linear functional on a $C^*$-algebra is called a *state*. {% enddefinition %} {% definition State Space %} -The *state space* of a $C^*$-algebra $\mathcal{A}$, denoted by $S(\mathcal{A})$, is the set of all states of $\mathcal{A}$. +The *state space* of a $C^*$-algebra $\mathcal{A}$, +denoted by $S(\mathcal{A})$, +is the set of all states of $\mathcal{A}$. {% enddefinition %} Note that $S(\mathcal{A})$ is a subset of the unit ball in the dual space of $\mathcal{A}$. 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 - |