3
$\begingroup$

Is there a holomorphic function on the unit disk $f$ such that $\left|f^{\left(n\right)}\left(0\right)\right|\geq n^{2n}$ for all $n\geq0$? Prove or disprove.

I disproved it with the following claim, but I'm unsure of it since it would work for $n^n$ as well, and I'm not sure why the original question asked for $n^2n$:

Suppose f is holomorphic on the unit disk and take the disk of radius $\frac{1}{2}$, $\frac{1}{2}\mathbb{D}$, Cauchy's integral formula tells us that for $\gamma=\partial\frac{1}{2}\mathbb{D}$,

$$\left|f^{\left(n\right)}\left(0\right)\right|=\left|\frac{n!}{2\pi i}\int_{\gamma}\frac{f\left(z\right)}{z^{n+1}}dz\right|$$

As $\frac{1}{2}\mathbb{D}$ is compact, $\left|f\right|$ must be bounded on it by some number $M$, and then $$\left|\int_{\gamma}\frac{f\left(z\right)}{z^{n+1}}d z\right|\leq\left|\max_{\gamma}\left|\frac{f}{z^{n+1}}\right|\right|\cdot L\left(\gamma\right)\leq M\cdot L\left(\gamma\right)$$

where as $\gamma$ is the circle of radius $\frac{1}{2}$, we know it's length is $\pi$, and we get that

$$\left|\frac{n!}{2\pi i}\int_{\gamma}\frac{f\left(z\right)}{z^{n+1}}d z\right|\leq\frac{n!}{2\pi}\left|M\pi\right|=\frac{n!\cdot M}{2}$$

where using Stirling's approximation, we see that

$$\lim_{n\to\infty}\frac{\left(n!\cdot M\right)/2}{n^{2n}}=(M/2)\lim_{n\to\infty}\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}}{n^{2n}}=(M/2)\lim_{n\to\infty}\frac{\sqrt{2\pi n}}{n^{n}e^{n}}\to0$$

showing that for some $n$, it must be that $f^{\left(n\right)}\left(0\right)\leq\frac{n!\cdot M}{2}<n^{2n}$, contradicting the original claim.

$\endgroup$
  • 1
    $\begingroup$ Yes, this proof looks fine. Well done. $\endgroup$ – Chappers Jan 4 '17 at 15:16
  • $\begingroup$ can't you just use the Taylor expansion of $f$ around $x=0$? $\endgroup$ – Yon Teh Jan 28 '17 at 11:48
  • $\begingroup$ @Nescio : $max_{\gamma} |\dfrac {f}{z^{n+1}}|=M.2^{n+1}$ as over $\gamma $ , $|z|^{n+1}=(1/2)^{n+1}$ $\endgroup$ – user228169 Mar 19 '17 at 13:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.