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@@ -280,5 +280,63 @@
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(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).
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\end{proof}
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\begin{proposition}
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\label{proposition:existence-curves}
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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 $z_0 \in K$,
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\[
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f(z) = \sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz
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\]
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\end{proposition}
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\begin{proof}[Proof, {{\cite[Proposition VIII.1.1]{ConwayComplex}}}. ]
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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:
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\begin{enumerate}
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\item For each $1 \le j \le n$, $R_j = [x_j, x_j + \delta] \times [y_j, y_j + \delta] \subset U$.
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\item $\bigcup_{j = 1}^n R_j \supset V$.
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\item $x_j, y_j = 0 \mod \delta$.
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\end{enumerate}
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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
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\begin{align*}
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\gamma_{j, \rightarrow}(t) &= (1 - t)(x_j, y_j) + t(x_j + \delta, y_j) \\
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\gamma_{j, \uparrow}(t) &= (1 - t)(x_j + \delta, y_j) + t(x_j + \delta, y_j + \delta) \\
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\gamma_{j, \leftarrow}(t) &= (1 - t)(x_j + \delta, y_j + \delta) + t(x_j, y_j + \delta) \\
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\gamma_{j, \downarrow}(t) &= (1 - t)(x_j, y_j + \delta) + t(x_j, y_j) \\
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\end{align*}
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and $\gamma_j = \gamma_{j, \downarrow} \cdot \gamma_{j, \leftarrow} \cdot \gamma_{j, \uparrow} \cdot \gamma_{j, \rightarrow}$ be their concatenation, then for each $z \in U \setminus \partial R_j$,
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\[
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\int_{\gamma_j} \frac{f(z)}{z - z_0}dz = \begin{cases}
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f(z) &z \in R_j^o \\
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0 &z \in U \setminus R_j
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\end{cases}
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\]
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by \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula}. Let $\seqf[N]{\mu_j}$ be an enumeration of
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\[
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\bigcup_{j = 1}^n \bracs{\gamma_{j, \downarrow}, \gamma_{j, \leftarrow}, \gamma_{j, \uparrow}, \gamma_{j, \rightarrow}}
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\]
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then for each $z_0 \in \bigcup_{j = 1}^n R_j^o$ and $f \in H(U; E)$,
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\[
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f(z) = \sum_{j = 1}^N \int_{\mu_j}\frac{f(z)}{z - z_0}dz
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\]
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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
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\begin{enumerate}[start=3]
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\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.
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\item $\int_{\mu_j}\frac{f(z)}{z - z_0}dz + \int_{\mu_k}\frac{f(z)}{z - z_0}dz = 0$.
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\end{enumerate}
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so every line segment either cannot form a redundant pair, or forms a unique one.
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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 $f \in H(U; E)$,
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\[
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f(z) = \sum_{j = 1}^m \int_{\mu_j}\frac{f(z)}{z - z_0}dz
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\]
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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 $z \in K$.
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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.
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\end{proof}
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Bokuan Li