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First, see this link on the alternative characterizations of analytic functions. I want to prove a version of 3) for complex-analytic functions. In particular:

If $f$ is a complex-analytic function on an open set $D \subset \mathbb{C}$, then for every compact set $K \subset D$ there exists a constant $C$ such that for every $x \in K$ and every non-negative integer $k$ the following estimate holds: $$ \left| \frac{d^k f}{dx^k}(x) \right| \leq C^{k+1} k!$$

Using Cauchy's estimates on individual points gives a "local" version of the estimate. How do I choose a single $C$ that will work for all $x \in K$? I'm pretty sure the compactness of $K$ will play a vital role. Can I surround $K$ by a precompact neighborhood $N$ contained in $D$, then choose $C$ as the reciprocal of the distance between $K$ and the boundary of $N$?

I believe the proof of the real-analytic case might give me some insights but I haven't seen a satisfactory proof yet for the real-analytic case; if you can point me to some reference, I'll appreciate it.

I found a "proof" of the real-analytic case here but I wasn't satisfied (see my comments underneath the answer). Basically, I don't understand the transition from the Cauchy's estimates being true for centers of the sets in the finite cover to being true for all of $x$ in the compact set.

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There exists $r>0$ such that $\cup_{a\in K}B(a,r)$ is contained in a larger compact set $K_1\subset D.$ Let $M= \sup_{K_1}|f|.$ Then for all $a\in K,$

$$|f^{(k)}(a)| \le k!\frac{M}{r^k} = k!\left( \frac{M^{1/k}}{r}\right)^{k}$$

by Cauchy's estimates. This will give your result.

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  • $\begingroup$ This is the solution I had in mind but I wasn't sure about the existence of such an $r$. I would appreciate it if you gave more details on how to choose or why there is such an $r$. Then I can accept your answer. Thanks. $\endgroup$ – Alex Strife Jul 14 '15 at 1:33
  • $\begingroup$ The distance from $K$ to $D^c$ is positive. Any $r$ less than this distance will work. $\endgroup$ – zhw. Jul 14 '15 at 1:42
  • $\begingroup$ In the case $D = \mathbb{C}$, $D^C$ is empty... so I believe we should first surround $K$ by a bounded open set contained in $D$ and consider the complement of that instead. $\endgroup$ – Alex Strife Jul 23 '15 at 5:28
  • $\begingroup$ I have already completed the solution though. Thanks! $\endgroup$ – Alex Strife Jul 23 '15 at 5:30

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