Slight adjustments in known results.

src/op/example/hardy.tex

@@ -23,12 +23,11 @@
         \item If $\fU \to x_0 \in \ol{D}$, then $f \in A(D)$, $\phi_{\fU}(f) = f(z_0)$. 
         \item $\phi_{\fU}$ is a multiplicative linear functional on $H^\infty(D)$. 
     \end{enumerate}
-    
 \end{proposition}
 \begin{proof}
     Let $f \in H^\infty(D)$, then by \autoref{proposition:imagefilterbase}, $f(\fU)$ is an ultrafilter base. Since $f$ is bounded, $f(\fU)$ converges to exactly one element of $\complex$. Hence the limit is well-defined. 
 
-    (2): By \autoref{proposition:operator-space-completeness}. 
+    (2): By \autoref{definition:multiplicative-linear-functional-space}. 
 \end{proof}