Added the holomorphic functional calculus.

src/dg/complex/runge.tex

@@ -3,10 +3,17 @@
 
 \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 $z_0 \in K$, 
-    \[
-    f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz
-    \]
+    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 $E$, 
+    \begin{enumerate}
+        \item For every $f \in H(U; E)$ and $z_0 \in K$, 
+        \[
+        f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz
+        \]
+        \item For every $V \in \cn_\complex(U)$, $f \in H(V; E)$, and $z_0 \in V \setminus U$, 
+        \[
+        \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz = 0
+        \]
+    \end{enumerate}
 \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:
@@ -60,6 +67,7 @@
     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: