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I am trying to prove the following statement.

Given $U \subseteq \mathbb{C}$ open, a compact set $K \subset U$, and $j \in \mathbb{N}$, show that there exists a constant $C > 0$ such that for any holomorphic $f : U \to \mathbb{C}$ and $z \in K$ we have \begin{equation*} \lvert f^{(j)}(z) \rvert \leq C \sup_{w \in U} \lvert f(w) \rvert. \end{equation*}

Here is my attempt. In what follows, $D(P,r)$ and $\overline{D(P, r)}$ will denote the open and closed disks of radius $r$ centered at $P$, respectively.

Proof attempt. Let $z \in K \subset U$. Since $U$ is open, we can find $r_z > 0$ such that $\overline{D(z, r_z)} \subseteq U$. Cover $K$ with $\{D(z, r_z) \mid z \in K\}$; since $K$ is compact, we can find finitely many $z_i \in K$ such that $\{D(z_i, r_i) \mid 1 \leq i \leq n, r_i := r_{z_i}\}$ covers $K$. Now let $f : U \to \mathbb{C}$ be holomorphic and let $z \in K$. Then $z \in D(z_i, r_i)$ for some $1 \leq i \leq n$, so by Cauchy's integral formula and a bound on the path integral we have \begin{equation*} \lvert f^{(j)}(z) \vert = \left \vert \frac{j!}{2 \pi i} \oint_{\partial D(z_i, r_i)} \frac{f(w) \ dw}{(w - z)^{j + 1}} \right \vert \leq \frac{j!}{2 \pi} \sup_{w \in \partial D(z_i, r_i)} \left \vert \frac{f(w)}{(w - z)^{j+1}} \right \vert \cdot 2 \pi r_i. \end{equation*} We get \begin{equation*} \left \vert w - z \right \vert = \vert (w - z_i) - (z - z_i) \vert \geq \vert \vert w - z_i \vert - \vert z_i - z \vert \vert = \vert r_i - \vert z_i - z \vert \vert \end{equation*}
for $w \in \partial D(z_i, r_i)$, so \begin{equation*} \vert f^{(j)}(z) \vert \leq r_i \cdot j! \sup_{w \in \partial D(z_i, r_i)} \left \vert \frac{f(w)}{(w - z)^{j+1}} \right \vert \leq r_i j! \frac{\sup_{w \in \partial D(z_i, r_i)} \left \vert f(w) \right \vert}{\vert r_i - \vert z_i - z \vert \vert^{j+1}} \leq \frac{j!}{r_i^j} \frac{1}{\left \vert 1 - \frac{\left \vert z_i - z\right \vert}{r_i}\right \vert^{j+1}} \sup_{w \in U} \vert f(w) \vert. \end{equation*}

This is as far as I could go. The triangle inequality and the bound I have on $z_i - z$ work against me: I can bound it above by $r_i$ and below by $0$, and both are useless with the reciprocal. If I could get a bound $M_{i,j}$ for \begin{equation*} \frac{1}{\left \vert 1 - \frac{\left \vert z_i - z \right \vert}{r_i}\right \vert^{j+1}} \end{equation*} that did not depend on $z$ then I could take $C = \max \{M_{i,j} \cdot j!/r_i^j \mid 1 \leq i \leq n\}$ and the statement would follow. However, I see no way of getting a useful bound.

Any suggestion or hint would be greatly appreciated.

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1 Answer 1

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Cover $K$ by the disks $D(z,\frac {r_z} 2)$ instead of $D(z,r_z)$. Then you will have $ {|z-z_i|} <\frac {r_i} 2$ which gives $|1-\frac {|z-z_i|} {r_i}|^{j+1} \geq (\frac1 2)^{j+1}$.

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  • $\begingroup$ Dang. Thank you. $\endgroup$ Sep 20, 2020 at 15:37

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