Added some spectrum gruntwork.

src/op/banach/spectrum.tex

@@ -77,7 +77,7 @@
     Let $x \in A$. By \autoref{proposition:spectrum-non-empty}, there exists $\lambda \in \sigma_A(x)$. Since $\lambda \cdot 1 - x$ is not invertible and every non-zero element of $A$ is invertible, $x = \lambda \cdot 1$. Therefore the mapping $\complex \to A$ defined by $\lambda \mapsto \lambda \cdot 1$ is an isometric isomorphism. 
 \end{proof}
 
-\begin{proposition}[Spectral Radius Formula]
+\begin{proposition}[Beurling's Spectral Radius Formula]
 \label{proposition:spectral-radius-hadamard}
     Let $A$ be a unital Banach algebra and $x \in A$, then $[\cdot]_{sp} = \limsup_{n \to \infty}\normn{x^n}_A^{1/n}$. 
 \end{proposition}
@@ -128,7 +128,7 @@
 
 \begin{proposition}
 \label{proposition:commutative-spectrum-gymnastics}
-    Let $A$ be a unital Banach algebra and $x, y \in A$ with $x = y$, then
+    Let $A$ be a commutative unital Banach algebra and $x, y \in A$ with $x = y$, then
     \begin{enumerate}
         \item $\sigma_A(x + y) \subset \sigma_A(x) + \sigma_A(y)$.
         \item $\sigma_A(xy) \subset \sigma_A(x)\sigma_A(y)$.