Added citation.

src/op/banach/spectrum.tex

@@ -65,7 +65,7 @@
 \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}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Theorem 1.8]{FollandHarmonic}}}. ]
     Let $r = \limsup_{n \to \infty}\normn{x^n}_A^{1/n}$, then for any $\lambda \in \complex$ with $|\lambda| > r$, the series $\sum_{n = 0}^\infty \lambda^{-n-1}x^n$ converges absolutely, and to the inverse of $(x - \lambda)$. Therefore $r \ge [\cdot]_{sp}$. 
 
     Let $D = \bracs{\lambda \in \complex|\ |\lambda| > [x]_{sp}}$, then since the series $\sum_{n = 0}^\infty \lambda^{-n-1}x^n$ is the expansion of $R_x$ at infinity, which is defined on $D$, it must converge on $D$. Let $\phi \in A^*$, then by the preceding discussion, the series $\sum_{n = 0}^\infty \lambda^{-n-1}\dpn{x^n, \phi}{A}$ converges on $D$. Thus for any $\lambda \in D$,