Theorem 27.1.4 (Cauchy’s Integral Formula).label Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open, $z_{0} \in U$, $r > 0$ such that $\ol{B(z_0, r)}\subset U$, $\gamma \in C([a, b]; \complex)$ be a closed, rectifiable path homotopic to $\omega_{z_0, r}$ on $U \setminus \bracs{z_0}$, and $f \in C^{1}(U; E)$, then
- (1)
$\int_{\gamma} f = 0$.
- (2)
$f(z) = \frac{1}{2\pi i}\int_{\gamma} \frac{f(z)}{z - z_{0}}dz$.
More over, for any $g \in C(U; E)$ that satisfies (2) for all $z_{0} \in U$, $r > 0$ with $\ol{B(z_0, r)}\subset U$, closed rectifiable curve $\gamma \in C([a, b]; \complex)$ homotopic to $\omega_{z_0, r}$ on $U \setminus \bracs{z_0}$,
- (3)
$g \in C^{\infty}(U; E)$, where for each $k \in \natz$,
\[D^{k}g(z_{0}) = \frac{k!}{2\pi i}\int_{\gamma}\frac{g(z)}{(z - z_{0})^{k+1}}dz\]
Proof. By Theorem 27.1.2 and the change of variables formula, for any $g \in C^{1}(U \setminus \bracs{z_0}; E)$,
(1): Since $f \in C(U; E)$, $f$ is bounded on $\ol{B(z_0, r)}$, so for any $s \in (0, r)$,
As $E$ is locally convex,
(2): Since $f \in C(U; E)$,
(3): Suppose inductively that (3) holds for $k \in \natz$. For sufficiently small $h \in \complex$,
Therefore $g \in C^{k+1}(U; E)$ with
$\square$
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