Definition 27.1.6 (Holomorphic).label Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open, and $f \in C(U; E)$, then the following are equivalent:
- (1)
(Complex Differentiability) $f \in C^{1}(U; E)$.
- (2)
(Cauchy-Riemann Equations) Under the identification of $C = \real^{2}$, $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\in C(U; E)$ and
\[\frac{\partial f}{\partial x}= i\frac{\partial f}{\partial y}\] - (3)
(Cauchy’s Integral Formula) For each $z_{0} \in U$, $r > 0$ such that $\ol{B(z_0, r)}\subset U$, and closed rectifiable path $\gamma \in C([a, b]; U)$ homotopic to $\omega_{z_0, r}$ on $U \setminus \bracs{z_0}$,
\[f(z_{0}) = \frac{1}{2\pi i}\int_{\gamma} \frac{f(z)}{z - z_{0}}dz\] - (4)
(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 $z \in B(z_{0}, r/2)$,
\[f(z) = \sum_{n = 0}^{\infty} a_{n}(z - z_{0})^{n}\]where the radius of convergence of the series is at least $r$.
- (5)
(Weak Holomorphy) For each $\phi \in E^{*}$, $\phi \circ f$ satisfies the above.
If the above holds, then $f$ is holomorphic/complex analytic on $U$.
Proof. (1) $\Leftrightarrow$ (2): Lemma 27.1.1.
(1) + (2) $\Rightarrow$ (3): See Cauchy’s Integral Formula.
(3) $\Rightarrow$ (4): By Cauchy’s Integral Formula, $f \in C^{\infty}(U; E)$ where for each $k \in \natz$,
Let
then by Cauchy’s Estimate, for any $k \in \natz$ and continuous seminorm $[\cdot]_{E}: E \to [0, \infty)$,
Thus $[D^{k}f(z_{0})/k!]_{E} \le C/r^{k}$ for all $k \in \natz$, and the radius of convergence of $g$ is at least $r$.
Let $z \in B(z_{0}, r/2)$, $s = |z - z_{0}|$, and $n \in \natp$, then by Taylor’s Formula and Cauchy’s Estimate,
which tends to $0$ as $n \to \infty$.
(4) $\Rightarrow$ (1): By Theorem 26.7.3.
(5) $\Rightarrow$ (3): By the equivalence of the prior points, for any $\phi \in E^{*}$, $\phi \circ f$ satisfies (3). By the Hahn-Banach Theorem, $f$ also satisfies (3).$\square$
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