Question: Let $f$ be an analytic function in the unit disc $D={z\in C: |z|<1}$. Consider a point $z_0\in D$. Show that there must be a positive integer n such that the n-th derivative of $f$ at $z_0$ satisfies $|f^{(n)}(z_0)| \leq n!n^n$.

Thoughts so far: In class, we discussed power series and their usefulness in proving a variety of facts of complex analysis. Just by rearranging the inequality we see that $\frac{|f^{(n)}(z_0)|}{n!} \leq n^n$. Clearly $\frac{|f^{(n)}(z_0)|}{n!}$ is a term of the Taylor series expansion for $f(z)$, but the $n^n$ term is annoying the royal bananas out of me as I cannot think of a way to get that. I am considering using the ratio ($\lim_{n\to\infty} \frac{\frac{f^{(n)}(z_0)}{n!}}{\frac{f^{(n-1)}(z_0)}{(n-1)!}} = \lim_{n\to \infty} \frac{f^{(n)}(z_0)}{nf^{n-1}(z_0)} \leq 1 \implies \lim_{n\to\infty} \frac{f^{(n)}(z_0)}{f^{n-1}(z_0)} \leq n$) or n root test ($\lim_{n\to\infty} \sqrt[n]{\frac{f^{(n)}(z_0)}{n!}} \leq 1$) for convergence since analyticity implies convergence of the series, but I cannot seem to coax the $n^n$ out of any of my attempts. Also, I am uncertain whether I am using the notion of analyticity and power series correctly.

Thank you in advance for any help that you may provide.

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    $\begingroup$ Why are you talking about $n^2$ when the inequality you have to prove (for some $n$) is $|f^{(n)}(z_0)| \le n! n^n$? $\endgroup$ – Niccolò Oct 20 '12 at 22:57
  • $\begingroup$ That was an unfortunate typo. Thank you for pointing it out. Should be fixed now. $\endgroup$ – ABC Bach Oct 20 '12 at 23:47
  • $\begingroup$ There's still an 'n^n' that needs to be converted to math type. :) $\endgroup$ – Haskell Curry Oct 20 '12 at 23:56

According to Cauchy's estimates, if $ f $ is a holomorphic function on $ D(a;R) $ and $ |f(z)| \leq M $ for all $ z \in D(a;R) $, then $$ \forall n \in \mathbb{N}: \quad |{f^{(n)}}(a)| \leq \frac{n! M}{R^{n}}. $$ Now, let $ z_{0} \in D(0;1) $. Choose $ R > 0 $ small enough so that $ D(z_{0};R) \subseteq D(0;1 - \epsilon) $ for some $ \epsilon \in (0,1) $. This guarantees that the given $ f $ is bounded on $ D(z_{0};R) $. Hence, suppose that $ |f(z)| \leq M $ for all $ z \in D(z_{0};R) $. Choose $ n \in \mathbb{N} $ large enough so that $ M \leq (nR)^{n} $. Then by Cauchy's estimates, $$ |{f^{(n)}}(z_{0})| \leq \frac{n! M}{R^{n}} \leq n! M \cdot \frac{n^{n}}{M} = n! n^{n}. $$ I hope that this works.

  • $\begingroup$ Thank you for your response. Quite an elegant answer. $\endgroup$ – ABC Bach Oct 21 '12 at 0:11

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