diff --git a/src/dg/complex/derivative.tex b/src/dg/complex/derivative.tex index ab89dbd..60b68c2 100644 --- a/src/dg/complex/derivative.tex +++ b/src/dg/complex/derivative.tex @@ -280,63 +280,4 @@ (5) $\Rightarrow$ (3): By the equivalence of the prior points, for any $\phi \in E^*$, $\phi \circ f$ satisfies (3). By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, $f$ also satisfies (3). \end{proof} -\begin{proposition} -\label{proposition:existence-curves} - Let $K \subset \complex$ be compact and $U \in \cn_\complex(K)$, then there exists closed rectifiable curves $\seqf{\gamma_j}$ such that for any separated locally convex space $E$, $f \in H(U; E)$, and $z_0 \in K$, - \[ - f(z) = \sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz - \] -\end{proposition} -\begin{proof}[Proof, {{\cite[Proposition VIII.1.1]{ConwayComplex}}}. ] - Via \hyperref[fattening]{proposition:distance-compact}, let $V \in \cn_\complex^o(K)$ precompact with $\ol V \subset U$. Identify $\complex = \real^2$, then since $V$ is precompact, there exists $\delta > 0$ and $\seqf{(x_j, y_j)} \subset U$ such that: - \begin{enumerate} - \item For each $1 \le j \le n$, $R_j = [x_j, x_j + \delta] \times [y_j, y_j + \delta] \subset U$. - \item $\bigcup_{j = 1}^n R_j \supset V$. - \item $x_j, y_j = 0 \mod \delta$. - \end{enumerate} - - In other words, $V$ is covered with a grid of squares with side-length $\delta$ and corners $\seqf{(x_j, y_j)}$. Now, for each $1 \le j \le n$ and $t \in [0, 1]$, let - \begin{align*} - \gamma_{j, \rightarrow}(t) &= (1 - t)(x_j, y_j) + t(x_j + \delta, y_j) \\ - \gamma_{j, \uparrow}(t) &= (1 - t)(x_j + \delta, y_j) + t(x_j + \delta, y_j + \delta) \\ - \gamma_{j, \leftarrow}(t) &= (1 - t)(x_j + \delta, y_j + \delta) + t(x_j, y_j + \delta) \\ - \gamma_{j, \downarrow}(t) &= (1 - t)(x_j, y_j + \delta) + t(x_j, y_j) \\ - \end{align*} - - and $\gamma_j = \gamma_{j, \downarrow} \cdot \gamma_{j, \leftarrow} \cdot \gamma_{j, \uparrow} \cdot \gamma_{j, \rightarrow}$ be their concatenation, then for each $z \in U \setminus \partial R_j$, - \[ - \int_{\gamma_j} \frac{f(z)}{z - z_0}dz = \begin{cases} - f(z) &z \in R_j^o \\ - 0 &z \in U \setminus R_j - \end{cases} - \] - - by \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula}. Let $\seqf[N]{\mu_j}$ be an enumeration of - \[ - \bigcup_{j = 1}^n \bracs{\gamma_{j, \downarrow}, \gamma_{j, \leftarrow}, \gamma_{j, \uparrow}, \gamma_{j, \rightarrow}} - \] - - then for each $z_0 \in \bigcup_{j = 1}^n R_j^o$ and $f \in H(U; E)$, - \[ - f(z) = \sum_{j = 1}^N \int_{\mu_j}\frac{f(z)}{z - z_0}dz - \] - - From here, it is sufficient to eliminate line segments that intersect $K$ and ensure that the remaining segments form a collection of loops. Let $1 \le j \le k \le N$, then $(\mu_j, \mu_k)$ is \textit{redundant} if for each $t \in [0, 1]$, $\mu_j(t) = \mu_k(1 - t)$. If $(\mu_j, \mu_k)$ are redundant, then - \begin{enumerate}[start=3] - \item There exists no $1 \le l \le N$ with $l \ne j, k$ such that $(\mu_j, \mu_l)$ or $(\mu_k, \mu_l)$ is redundant. - \item $\int_{\mu_j}\frac{f(z)}{z - z_0}dz + \int_{\mu_k}\frac{f(z)}{z - z_0}dz = 0$. - \end{enumerate} - - so every line segment either cannot form a redundant pair, or forms a unique one. - - By relabeling, let $\bracs{\mu_j|1 \le j \le m}$ such that for each $1 \le j \le m$, there exists no $1 \le k \le n$ such that $(\mu_j, \mu_k)$ is redundant. By (5), for each $z_0 \in \bigcup_{j = 1}^n R_j^o$ and $f \in H(U; E)$, - \[ - f(z) = \sum_{j = 1}^m \int_{\mu_j}\frac{f(z)}{z - z_0}dz - \] - - Let $1 \le j \le N$ such that $\mu_j([0, 1]) \cap K \ne \emptyset$. Since $V \in \cn_\complex(K)$, there exists $1 \le k \le N$ such that $(\mu_j, \mu_k)$ are redundant by (2) and (3). Therefore $\bigcup_{j = 1}^m \mu_j([0, 1]) \cap K = \emptyset$. Since $\bigcup_{j = 1}^n R_j^o$ is dense in $K$, the above also holds for all $z \in K$. - - Finally, let $1 \le j \le m$. Since $\mu_j$ does not form a redundant pair, $\mu_j(1)$ intersects at most two distinct squares by (3). In which case, there must exist $1 \le k \le m$ such that $\mu_k(0) = \mu_j(1)$. Therefore $\seqf[m]{\mu_j}$ forms a collection of closed rectifiable curves. -\end{proof} - diff --git a/src/dg/complex/index.tex b/src/dg/complex/index.tex index 877c7ba..2e67f04 100644 --- a/src/dg/complex/index.tex +++ b/src/dg/complex/index.tex @@ -3,8 +3,9 @@ \input{./derivative.tex} +\input{./sphere.tex} \input{./space.tex} \input{./log.tex} \input{./entire.tex} - +\input{./runge.tex} diff --git a/src/dg/complex/runge.tex b/src/dg/complex/runge.tex new file mode 100644 index 0000000..61e4eac --- /dev/null +++ b/src/dg/complex/runge.tex @@ -0,0 +1,170 @@ +\section{Runge's Theorem} +\label{section:runge} + +\begin{proposition} +\label{proposition:existence-curves} + Let $K \subset \complex$ be compact and $U \in \cn_\complex(K)$, then there exists closed rectifiable curves $\seqf{\gamma_j}$ such that for any separated locally convex space $E$, $f \in H(U; E)$, and $z_0 \in K$, + \[ + f(z) = \sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz + \] +\end{proposition} +\begin{proof}[Proof, {{\cite[Proposition VIII.1.1]{ConwayComplex}}}. ] + Via \hyperref[fattening]{proposition:distance-compact}, let $V \in \cn_\complex^o(K)$ precompact with $\ol V \subset U$. Identify $\complex = \real^2$, then since $V$ is precompact, there exists $\delta > 0$ and $\seqf{(x_j, y_j)} \subset U$ such that: + \begin{enumerate} + \item For each $1 \le j \le n$, $R_j = [x_j, x_j + \delta] \times [y_j, y_j + \delta] \subset U$. + \item $\bigcup_{j = 1}^n R_j \supset V$. + \item $x_j, y_j = 0 \mod \delta$. + \end{enumerate} + + In other words, $V$ is covered with a grid of squares with side-length $\delta$ and corners $\seqf{(x_j, y_j)}$. Now, for each $1 \le j \le n$ and $t \in [0, 1]$, let + \begin{align*} + \gamma_{j, \rightarrow}(t) &= (1 - t)(x_j, y_j) + t(x_j + \delta, y_j) \\ + \gamma_{j, \uparrow}(t) &= (1 - t)(x_j + \delta, y_j) + t(x_j + \delta, y_j + \delta) \\ + \gamma_{j, \leftarrow}(t) &= (1 - t)(x_j + \delta, y_j + \delta) + t(x_j, y_j + \delta) \\ + \gamma_{j, \downarrow}(t) &= (1 - t)(x_j, y_j + \delta) + t(x_j, y_j) \\ + \end{align*} + + and $\gamma_j = \gamma_{j, \downarrow} \cdot \gamma_{j, \leftarrow} \cdot \gamma_{j, \uparrow} \cdot \gamma_{j, \rightarrow}$ be their concatenation, then for each $z \in U \setminus \partial R_j$, + \[ + \int_{\gamma_j} \frac{f(z)}{z - z_0}dz = \begin{cases} + f(z) &z \in R_j^o \\ + 0 &z \in U \setminus R_j + \end{cases} + \] + + by \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula}. Let $\seqf[N]{\mu_j}$ be an enumeration of + \[ + \bigcup_{j = 1}^n \bracs{\gamma_{j, \downarrow}, \gamma_{j, \leftarrow}, \gamma_{j, \uparrow}, \gamma_{j, \rightarrow}} + \] + + then for each $z_0 \in \bigcup_{j = 1}^n R_j^o$ and $f \in H(U; E)$, + \[ + f(z) = \sum_{j = 1}^N \int_{\mu_j}\frac{f(z)}{z - z_0}dz + \] + + From here, it is sufficient to eliminate line segments that intersect $K$ and ensure that the remaining segments form a collection of loops. Let $1 \le j \le k \le N$, then $(\mu_j, \mu_k)$ is \textit{redundant} if for each $t \in [0, 1]$, $\mu_j(t) = \mu_k(1 - t)$. If $(\mu_j, \mu_k)$ are redundant, then + \begin{enumerate}[start=3] + \item There exists no $1 \le l \le N$ with $l \ne j, k$ such that $(\mu_j, \mu_l)$ or $(\mu_k, \mu_l)$ is redundant. + \item $\int_{\mu_j}\frac{f(z)}{z - z_0}dz + \int_{\mu_k}\frac{f(z)}{z - z_0}dz = 0$. + \end{enumerate} + + so every line segment either cannot form a redundant pair, or forms a unique one. + + By relabeling, let $\bracs{\mu_j|1 \le j \le m}$ such that for each $1 \le j \le m$, there exists no $1 \le k \le n$ such that $(\mu_j, \mu_k)$ is redundant. By (5), for each $z_0 \in \bigcup_{j = 1}^n R_j^o$ and $f \in H(U; E)$, + \[ + f(z) = \sum_{j = 1}^m \int_{\mu_j}\frac{f(z)}{z - z_0}dz + \] + + Let $1 \le j \le N$ such that $\mu_j([0, 1]) \cap K \ne \emptyset$. Since $V \in \cn_\complex(K)$, there exists $1 \le k \le N$ such that $(\mu_j, \mu_k)$ are redundant by (2) and (3). Therefore $\bigcup_{j = 1}^m \mu_j([0, 1]) \cap K = \emptyset$. Since $\bigcup_{j = 1}^n R_j^o$ is dense in $K$, the above also holds for all $z \in K$. + + Finally, let $1 \le j \le m$. Since $\mu_j$ does not form a redundant pair, $\mu_j(1)$ intersects at most two distinct squares by (3). In which case, there must exist $1 \le k \le m$ such that $\mu_k(0) = \mu_j(1)$. Therefore $\seqf[m]{\mu_j}$ forms a collection of closed rectifiable curves. +\end{proof} + +\begin{lemma} +\label{lemma:rational-curve-approximation} + Let $\gamma \in C([a, b]; \complex)$ be a rectifiable curve, $K \subset \complex$ such that $K \cap \gamma([a, b]) = \emptyset$, $f \in C(\gamma([a, b]); \complex)$, and $\eps > 0$, then there exists $R \in \complex(z)$ such that: + \begin{enumerate} + \item $R \in H(\complex \setminus \gamma([a, b]); \complex)$. + \item For each $z_0 \in K$, + \[ + \abs{\int_{\gamma} \frac{f(z)}{z - z_0}dz - R(z)} < \eps + \] + \end{enumerate} +\end{lemma} +\begin{proof}[Proof, {{\cite[Lemma VIII.1.5]{ConwayComplex}}}. ] + Since the mapping + \[ + \varphi: K \times [a, b] \to \complex \quad (z_0, t) \mapsto \frac{f \circ \gamma(t)}{\gamma(t) - z} + \] + + is continuous, it is uniformly continuous by \autoref{proposition:uniform-continuous-compact}. Hence the mappings $\bracs{\varphi(z_0, \cdot)|t \in [a, b]}$ are uniformly equicontinuous. Thus there exists $\delta > 0$ such that for each $s, t \in [a, b]$ with $|s - t| \le \delta$, + \[ + \abs{\frac{f \circ \gamma(s)}{\gamma(s) - z_0} - \frac{f \circ \gamma(t)}{\gamma(t) - z_0}} < \eps + \] + + for all $z_0 \in K$. + + Let $(P = \seqfz{t_j}) \in \scp([a, b])$ with $\sigma(t) < \delta$, and + \[ + R(z) = \sum_{j = 1}^n f \circ \gamma(t_j)\frac{\gamma(t_j) - \gamma(t_{j-1})}{\gamma(t_j) - z} + \] + + then $R \in \complex(z) \cap H(\complex \setminus \gamma([a, b]); \complex)$, and for each $z_0 \in K$, + \begin{align*} + \abs{\int_\gamma \frac{f(z)}{z - z_0}dz - R(z)} &\le + \sum_{j = 1}^n \int_{t_{j-1}}^{t_j}\abs{\frac{f \circ \gamma(t)}{\gamma(t) - z_0} - \frac{f \circ \gamma(t_j)}{\gamma(t_j) - z}}\gamma(dt) \\ + &\le \eps \norm{\gamma}_{\text{var}} + \end{align*} +\end{proof} + +\begin{lemma}[Pole Pushing] +\label{lemma:pole-pushing} + Let $K \subset \complex$ be compact and $P \subset \complex_\infty \setminus K$ such that $P$ intersects every connected component of $\complex_\infty \setminus K$, then for every $a \in \complex \setminus K$ and $\eps > 0$, there exists $R \in \complex(z)$ such that: + \begin{enumerate} + \item $R \in H(\complex_\infty \setminus P; \complex)$. + \item For every $z \in K$, $|R(z) - 1/(z - a)| < \eps$. + \end{enumerate} +\end{lemma} +\begin{proof}[Proof, {{\cite[Lemma III.1.10]{ConwayComplex}}}. ] + First suppose that $\infty \not\in P$. Let $A \subset \complex_\infty \setminus K$ be the collection of elements such that the lemma holds\footnote{Under the identification that $1/(z - \infty) = 0$.}. + + Let $a_0 \in \complex \setminus K$, then there exists $V \in \cn_\complex(a)$ compact such that $V \cap K = \emptyset$. In which case, the function + \[ + K \times V \to \complex \quad (z, a) \mapsto \frac{1}{z - a} + \] + + is continuous, and hence uniformly continuous by \autoref{proposition:uniform-continuous-compact}. Thus $1/(z - a) \to 1/(z - a_0)$ uniformly on compact sets as $a \to a_0$, and if $a_0 \in \ol{A}$, then $a_0 \in A$ as well. Therefore $A$ is a closed subset of $\complex_\infty \setminus K$. + + Now, let $a_0 \in A \cap \complex$, then by \autoref{proposition:distance-compact}, $r = d(a_0, K) > 0$. Let $a \in B_\complex(a_0, r)$, then for each $z \in K$, + \begin{align*} + \frac{1}{z - a} &= \frac{1}{z - a_0}\frac{z - a_0}{z - a} = \frac{1}{z - a_0}\braks{\frac{z - a}{z - a_0}}^{-1} \\ + &= \frac{1}{z - a_0}\braks{1 - \frac{a - a_0}{z - a_0}}^{-1} + \end{align*} + + Since $\sup_{z \in K}|a - a_0|/|z - a_0| < 1$, + \[ + \frac{1}{z - a} = \frac{1}{z - a_0}\sum_{n = 0}^\infty \braks{\frac{a - a_0}{z - a_0}}^n + \] + + where the convergence is uniform on $K$. Therefore $B_\complex(a_0, r) \subset A$ as well. + + Finally, for each $a \in \complex \setminus K$ with $|a| > \sup_{z \in K}|z|$, + \[ + \frac{1}{z - a} = -\frac{1}{a\paren{1 - \frac{z}{a}}} = -\frac{1}{a}\sum_{n = 0}^\infty \braks{\frac{z}{a}}^n + \] + + where the convergence is uniform on $K$, so there exists a neighbourhood of $\infty$ that is contained in $A$. + + Since $P \cup \bracs{\infty} \subset A$, and $A$ is an open and closed subset of $C_\infty \setminus K$, $A$ is a union of connected components of $C_\infty \setminus K$ by \autoref{lemma:union-connected-components}. Given that $P$ intersects every connected component of $C_\infty \setminus K$, the lemma holds for all $a \in \complex \setminus K$. +\end{proof} + +\begin{theorem}[Runge] +\label{theorem:runge} + Let $K \subset \complex$ be compact and $P \subset C_\infty \setminus K$ such that $P$ intersects every connected component of $C_\infty \setminus K$, then $\complex(x) \cap H(C_\infty \setminus P; \complex)$ is dense in $H(K; \complex)$ with respect to the uniform topology. +\end{theorem} +\begin{proof} + Let $U \in \cn_\complex(K)$, $f \in H(U; \complex)$, and $\eps > 0$. By \autoref{proposition:existence-curves}, there exists closed rectifiable curves $\seqf{\gamma_j}$ in $U \setminus V$ such that for each $z_0 \in K$, + \[ + f(z_0) = \sum_{j = 1}^n \int_{\gamma_j}\frac{f(z)}{z - z_0}dz + \] + + Let $T$ be the union of the images of $\seqf{\gamma_j}$, then by \autoref{lemma:rational-curve-approximation}, there exists $\seqf{R_j} \subset \complex(z) \cap H(\complex \setminus T; \complex)$ such that for each $z_0 \in K$ and $1 \le j \le n$, + \[ + \abs{\int_{\gamma_j}\frac{f(z)}{z - z_0}dz - R_j(z)} < \frac{\eps}{n} + \] + + so + \[ + \abs{f(z_0) - \sum_{j = 1}^n R_j(z)} < \eps + \] + + for all $z_0 \in K$. By the \hyperref[pole pushing lemma]{lemma:pole-pushing}, there exists $S \in \complex(z) \cap H(\complex_\infty \setminus P)$ such that $|S(z_0) - \sum_{j = 1}^n R_j(z_0)| < \eps$ for all $z_0 \in K$. +\end{proof} + +\begin{corollary} +\label{corollary:runge-rational-approximation} + Let $K \subset \complex$ be compact, then $\complex(z) \cap H(K; \complex)$ is dense in $H(K; \complex)$. +\end{corollary} + + + diff --git a/src/dg/complex/sphere.tex b/src/dg/complex/sphere.tex new file mode 100644 index 0000000..b8d5754 --- /dev/null +++ b/src/dg/complex/sphere.tex @@ -0,0 +1,16 @@ +\section{The Riemann Sphere} +\label{section:riemann-sphere} + + +\begin{definition}[Extended Complex Plane] +\label{definition:extended-complex-plane} + Let $\complex$ be the complex plane, then its one-point compactification $\complex_\infty = \complex \sqcup \bracs{\infty}$ is the \textbf{extended complex plane}. +\end{definition} + +\begin{definition}[Holomorphic on $\complex_\infty$] +\label{definition:holomorphic-on-sphere} + Let $E$ be a separated locally convex space and $f \in C(\complex_\infty; E)$, then $f$ is \textbf{holomorphic at $\infty$} if $z \mapsto f(z^{-1})$ (under the identification that $1/0 = \infty$) is holomorphic at $0$. +\end{definition} + + + diff --git a/src/topology/main/connected.tex b/src/topology/main/connected.tex index 5c1e8e9..dc11bde 100644 --- a/src/topology/main/connected.tex +++ b/src/topology/main/connected.tex @@ -45,3 +45,13 @@ \begin{proof} Let $C$ be the union of all connected sets that contain $A$, then $C$ is connected by \autoref{proposition:connected-union}, and is the maximum connected set containing $A$ by definition. \end{proof} + +\begin{lemma} +\label{lemma:union-connected-components} + Let $X$ be a topological space and $A \subset X$ be both open and closed, then $A$ is a union of connected components of $X$. +\end{lemma} +\begin{proof} + Let $C \subset X$ be a connected component, then $C \cap A$ and $C \setminus A$ are both open. Since $C$ is connected, either $C \cap A = \emptyset$ and $C \subset A^c$, or $C \setminus A = \emptyset$ and $C \subset A$. +\end{proof} + +