Added more complex analysis.

src/dg/complex/derivative.tex

@@ -234,7 +234,7 @@
         \[
         f(z_0) = \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z - z_0}dz
         \]
-        \item (\textbf{Analyticity}) For each $z_0 \in U$, there exists $r > 0$ and $\seq{a_n} \subset E$ such that for each $z \in B(z_0, r)$, 
+        \item (\textbf{Analyticity}) For each $z_0 \in U$ and $r > 0$ with $\ol{B(0, r)} \subset U$, there exists $\seq{a_n} \subset E$ such that for each $z \in B(z_0, r/2)$, 
         \[
         f(z) = \sum_{n = 0}^\infty a_n(z - z_0)^n
         \]
@@ -257,7 +257,7 @@
     
     Let
     \[
-    g(z) = \sum_{k = 0}^\infty \frac{1}{k!} D^kf(z_0)(z - z_0)^n
+    g(z) = \sum_{k = 0}^\infty \frac{1}{k!} D^kf(z_0)(z - z_0)^k
     \]
 
     then by \hyperref[Cauchy's Estimate]{corollary:cauchy-estimate}, for any $k \in \natz$ and continuous seminorm $[\cdot]_E: E \to [0, \infty)$, 
@@ -281,3 +281,4 @@
 \end{proof}
 
 
+

src/dg/complex/entire.tex

@@ -0,0 +1,44 @@
+\section{Entire Functions}
+\label{section:entire-holomorphic}
+
+\begin{definition}[Entire Function]
+\label{definition:entire-function}
+    Let $E$ be a complete separated locally convex space over $\complex$, then an \textbf{entire $E$-valued function} is an $E$-valued holomorphic function on $\complex$.
+\end{definition}
+
+\begin{proposition}
+\label{proposition:entire-power-series}
+    Let $E$ be a complete separated locally convex space over $\complex$, and $f \in H(\complex; E)$, then for any $z_0 \in \complex$ and $z \in \complex$,
+    \[
+    f(z) = \sum_{k = 0}^\infty \frac{1}{k!} D^kf(z_0)(z - z_0)^k
+    \]
+
+    where the radius of convergence of the series is infinite. 
+\end{proposition}
+\begin{proof}
+    See (4) of \autoref{definition:complex-analytic}. 
+\end{proof}
+
+\begin{theorem}[Liouville]
+\label{theorem:liouville}
+    Let $E$ be a complete separated locally convex space over $\complex$ and $f \in H(\complex; E)$. If $f$ is bounded, then $f$ is constant. 
+\end{theorem}
+\begin{proof}
+    Let $x_0 \in \complex$ and $[\cdot]_E: [E, \infty)$ be a continuous seminorm on $E$, then by \hyperref[Cauchy's Estimate]{corollary:cauchy-estimate},
+    \[
+    [Df(z_0)]_E \le \frac{1}{r} \sup_{z \in \ol{B(z_0, r)}}[f(z)]_E
+    \]
+
+    for all $r > 0$. Therefore $Df = 0$, and $f$ is constant. 
+\end{proof}
+
+\begin{theorem}[Fundamental Theorem of Algebra]
+\label{theorem:fundamental-theorem-of-algebra}
+    Let $p \in \complex[z]$ be a non-constant polynomial, then there exists $z \in \complex$ such that $f(z) = 0$. 
+\end{theorem}
+\begin{proof}
+    Let $p \in \complex[z]$ such that $p(z) \ne 0$ for all $z \in \complex$. Let $f = 1/p$, then $f \in H(\complex; \complex) \cap C_0(\complex; \complex)$, so $f$ is bounded. By \hyperref[Liouville's Theorem]{theorem:liouville}, $f$ and thus $p$ is constant. 
+\end{proof}
+
+
+

src/dg/complex/index.tex

@@ -3,6 +3,8 @@
 
 
 \input{./derivative.tex}
+\input{./space.tex}
 \input{./log.tex}
+\input{./entire.tex}
 
 

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}