Update

1 files changed, 54 insertions, 59 deletions

diff --git a/pages/operator-algebras/banach-algebras/index.md b/pages/operator-algebras/banach-algebras/index.md
index 2fb8f03..3335d78 100644
--- a/pages/operator-algebras/banach-algebras/index.md
+++ b/pages/operator-algebras/banach-algebras/index.md
@@ -47,25 +47,23 @@ We say that $\mathcal{A}$ is an *unital* Banach algebra, if $\mathcal{A}$ contai
 
 It is easy to see that a Banach algebra has at most one unit.
 
-{: .proposition-title #neumann-series }
-> Proposition (Neumann Series)
->
-> Let $\mathcal{A}$ be a unital Banach algebra
-> and let $x \in \mathcal{A}$ satisfy $\norm{x} < 1$.
-> Then $\mathbf{1}-x$ is invertible
-> and the inverse is given by the series
-> 
-> $$
-> (\mathbf{1}-x)^{-1} = \sum_{n=0}^{\infty} x^n,
-> $$
->
-> which converges absolutely in norm.
-> Moreover, we have the estimate
-> 
-> $$
-> \norm{(\mathbf{1}-x)^{-1}} \le \frac{1}{1 - \norm{x}}.
-> $$
-> {: .katex-display .mb-0 }
+{% proposition Neumann Series %}
+Let $\mathcal{A}$ be a unital Banach algebra
+and let $x \in \mathcal{A}$ satisfy $\norm{x} < 1$.
+Then $\mathbf{1}-x$ is invertible,
+and the inverse is given by the series
+
+$$
+(\mathbf{1}-x)^{-1} = \sum_{n=0}^{\infty} x^n,
+$$
+
+which converges absolutely in norm.
+Moreover, we have the estimate
+
+$$
+\norm{(\mathbf{1}-x)^{-1}} \le \frac{1}{1 - \norm{x}}.
+$$
+{% endproposition %}
 
 {% proof %}
 Since the Banach algebra norm is submultiplicative,
@@ -91,18 +89,17 @@ The estimate follows from $\norm{s} \le \sum \norm{x}^n = 1 / (1 - \norm{x})$.
 
 ## The Spectrum
 
-{: .definition-title }
-> Definition (Spectrum, Resolvent Set)
->
-> 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 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 $R_{\lambda} = (\lambda - x)^{-1}$. \
->   The mapping $R : \rho(x) \to \mathcal{A}$, $\lambda \mapsto R_{\lambda}$, is called *resolvent map*.
+{% definition Spectrum, Resolvent Set %}
+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 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 $R_{\lambda} = (\lambda - x)^{-1}$. \
+  The mapping $R : \rho(x) \to \mathcal{A}$, $\lambda \mapsto R_{\lambda}$, is called *resolvent map*.
+{% enddefinition %}
 
 {% theorem %}
 Suppose $x$ is an element of a unital Banach algebra $\mathcal{A}$.
@@ -122,7 +119,7 @@ $$
 
 {% 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).
@@ -165,7 +162,7 @@ We assume that $\sigma(x)$ is empty
 and derive a contradiction.
 Observe that the resolvent map $R$ is defined on the whole complex plane.
 By [this corollary](#resolvent-map-is-analytic), $R$ is analytic, hence entire.
-Analytic functions are countinuous;
+Analytic functions are continuous;
 therefore $R$ is bounded on the compact disk $\abs{\lambda} \le 2 \norm{x}$.
 For $\abs{\lambda} > 2 \norm{x}$ we may expand $R_{\lambda}$ into a [Neumann series](#neumann-series),
 
@@ -188,13 +185,12 @@ $$
 This shows that $R$ is a bounded entire function. Now
 [Liouville's Theorem](/pages/complex-analysis/one-complex-variable/cauchys-integral-formula.html#liouvilles-theorem)
 (for vector-valued functions) implies that $R$ is constant.
-This is contradictiory because XXX
+This is contradictory because XXX
 {% endproof %}
 
-{: .theorem-title }
-> Gelfand–Mazur Theorem
->
-> Every Banach algebra in which all nonzero elements are invertible is isometrically isomorphic to $\CC$.
+{% theorem * Gelfand–Mazur Theorem %}
+Every Banach algebra in which all nonzero elements are invertible is isometrically isomorphic to $\CC$.
+{% endtheorem %}
 
 {% proof %}
 For any Banach algebra $A$,
@@ -220,40 +216,39 @@ include:
 - $\mathcal{A}$ is a division algebra.
 - The underlying ring of $\mathcal{A}$ is a field.
 
-{: .theorem-title }
-> Spectral Radius Formula
->
-> For every Banach algebra element $x$ the spectral radius is given by
-> 
-> $$
-> r(x) = \lim_{n \to \infty} \norm{x^n}^{1/n}.
-> $$
-> {: .katex-display .mb-0 }
+{% theorem * Spectral Radius Formula %}
+For every Banach algebra element $x$ the spectral radius is given by
+
+$$
+r(x) = \lim_{n \to \infty} \norm{x^n}^{1/n}.
+$$
+{% endtheorem %}
 
 ## Gelfand’s Theory
 
-Proposition
+{% proposition %}
 Let $\mathcal{A}$ be a unital commutative Banach algebra.
 If $\phi$ is a nonzero multiplicative linear functional on $\mathcal{A}$,
 then its kernel $\ker \phi$ is a maximal ideal in $\mathcal{A}$.
 Every maximal ideal $\mathcal{I}$ in $\mathcal{A}$ is of the form
 $I = \ker \phi$ for some nonzero multiplicative linear functional $\phi$ on $\mathcal{A}$.
+{% endproposition %}
 
-In other words, the mapping $\phi \mapsto \ker \phi$ is gives a bijection
+In other words, the mapping $\phi \mapsto \ker \phi$ gives a bijection
 between the sets of nonzero multiplicative linear functionals and maximal ideals.
 
+{% definition %}
+The *Gelfand space* $\Gamma_{\mathcal{A}}$ of a unital commutative Banach algebra $\mathcal{A}$
+is the set of maximal ideals of $\mathcal{A}$; its topology is inherited from
+the weak* topology on the dual of $\mathcal{A}$ via the correspondence described above.
+{% enddefinition %}
 
-Definition
+{% definition %}
 The *maximal ideal space* $\mathcal{M}_{\mathcal{A}}$ of a unital commutative Banach algebra $\mathcal{A}$
 is the set of maximal ideals of $\mathcal{A}$; its topology is inherited from
-the weak* topology on the dual of $\mathcal{A}$ via the correspondece described above.
-
-Proposition
-The *maximal ideal space* of a unital commutative Banach algebra is a compact Hausdorff space.
-
-{% definition bla, blubb %}
-a
-b
+the weak* topology on the dual of $\mathcal{A}$ via the correspondence described above.
 {% enddefinition %}
 
-
+{% proposition %}
+The *Gelfand space* of a unital commutative Banach algebra is a compact Hausdorff space.
+{% endproposition %}