Added more complex analysis.

src/dg/complex/space.tex

@@ -0,0 +1,40 @@
+\section{Spaces of Holomorphic Functions}
+\label{section:holomorphic-function-space}
+
+\begin{definition}[Space of Holomorphic Functions]
+\label{definition:holomorphic-function-space}
+    Let $E$ be a complete separated locally convex space over $\complex$ and $U \subset \complex$ be open, then $H(U; E)$ is the \textbf{space of $E$-valued holomorphic functions on $U$}, equipped with the topology of uniform convergence on compact sets. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:holomorphic-complete}
+    Let $E$ be a complete separated locally convex space over $\complex$ and $U \subset \complex$ be open, then $H(U; E)$ is complete. 
+\end{proposition}
+\begin{proof}
+    By \hyperref[corollary:cauchy-estimate]{Cauchy's estimate}, uniform convergence on compact sets is equivalent to uniform convergence of derivatives of all orders on compact sets. Since $U$ is locally compact, uniform limits of holomorphic functions are holomorphic by \autoref{theorem:differentiable-uniform-limit}.
+\end{proof}
+
+\begin{theorem}[Montel]
+\label{theorem:montel}
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open, and $\cf \subset H(U; E)$, then the following are equivalent:
+    \begin{enumerate}[label=(B\arabic*)]
+        \item $\cf$ is equicontinuous, and for each $x \in U$, $\cf(x) = \bracs{f(x)|f \in \cf}$ is bounded.
+        \item $\cf$ is bounded in $H(U; E)$.
+    \end{enumerate}
+
+    and the following are equivalent:
+    \begin{enumerate}[label=(C\arabic*)]
+        \item $\cf$ is precompact in $H(U; E)$. 
+        \item $\cf$ is bounded in $H(U; E)$, and for each $x \in U$, $\cf(x) = \bracs{f(x)|f \in \cf}$ is precompact.  
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (B1) $\Rightarrow$ (B2): Let $K \subset U$ be compact and $V \in \cn_E(0)$ be circled. Since $\cf$ is equicontinuous, for each $x \in K$, there exists $U_x \in \cn_U(x)$ such that $f(y) - f(x) \in V$ for all $y \in U_x$ and $f \in \cf$. By compactness of $K$, there exists $\seqf{x_j} \subset K$ such that $K \subset \bigcup_{j = 1}^n U_{x_j}$. Since $B = \bigcup_{j = 1}^n \cf(x_j)$ is bounded, there exists $\lambda > 0$ such that $\lambda V \supset B$. In which case, 
+    \[
+    (\lambda + 1)V \supset B + V \supset \bigcup_{x \in K}^n \cf(x)
+    \]
+
+    (B2) $\Rightarrow$ (B1): By \hyperref[Cauchy's Estimate]{corollary:cauchy-estimate}, $\bracsn{Df|f \in \cf}$ is also uniformly bounded on every compact set. Thus $\cf$ is equicontinuous. 
+
+    (C1) $\Leftrightarrow$ (C2): By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}.
+\end{proof}