Update

1 files changed, 23 insertions, 19 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.