From a058df31639cc8cc7cd5e6b5efacba94e1087ee0 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Mon, 18 May 2026 18:06:02 -0400 Subject: [PATCH] Added more complex analysis. --- src/dg/complex/derivative.tex | 5 ++-- src/dg/complex/entire.tex | 44 +++++++++++++++++++++++++++++++++++ src/dg/complex/index.tex | 2 ++ src/dg/complex/space.tex | 40 +++++++++++++++++++++++++++++++ 4 files changed, 89 insertions(+), 2 deletions(-) create mode 100644 src/dg/complex/entire.tex create mode 100644 src/dg/complex/space.tex diff --git a/src/dg/complex/derivative.tex b/src/dg/complex/derivative.tex index e7fd834..777915d 100644 --- a/src/dg/complex/derivative.tex +++ b/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} + diff --git a/src/dg/complex/entire.tex b/src/dg/complex/entire.tex new file mode 100644 index 0000000..60c240a --- /dev/null +++ b/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} + + + diff --git a/src/dg/complex/index.tex b/src/dg/complex/index.tex index 9a6467a..877c7ba 100644 --- a/src/dg/complex/index.tex +++ b/src/dg/complex/index.tex @@ -3,6 +3,8 @@ \input{./derivative.tex} +\input{./space.tex} \input{./log.tex} +\input{./entire.tex} diff --git a/src/dg/complex/space.tex b/src/dg/complex/space.tex new file mode 100644 index 0000000..f8c2283 --- /dev/null +++ b/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}