The question is: if $f(z)$ is analytic and $|f(z)|\leq M$ for $|z|\leq r$, find an upper bound for$|f^{(n)}(z)|$ in $|z|\leq\frac{r}{2}$.

My attempt: Since $f(z)$ is analytic where $|z|\leq r$, we know that $$f^{(n)}(z)=\frac{n!}{2\pi i}\int_{|z|=r}\frac{f(w)}{(w-z)^{n+1}}dw,$$ and $f(z)$ is bounded by $M$. We know that $\bigg|\int_Cf(z)dz\bigg|\leq\max|f(z)|\cdot\text{(length of C)}$, so $$\bigg|f^{(n)}(z)\bigg|=\bigg|\frac{n!}{2\pi i}\int_{|z|=r}\frac{f(w)}{(w-z)^{n+1}}dw\bigg|\leq n!\cdot M_n\cdot2\pi r,$$ where $M_n:=\max|\frac{f(w)}{(w-z)^{n+1}}|$, for a fixed $z$.

Is this correct? Just finding an upper bound seems like it shouldn't be too difficult, but I feel that I didn't do it correctly. Also, on a somewhat related note, since $f(z)$ is analytic, and its derivatives are analytic, shouldn't $\int_{|z|=r}f^{(n)}(z)dz=0$?

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    $\begingroup$ $|w-z| \geq r/2$ for $|z|<r/2$. Also $2 \pi $ should be cancelled. Then you don't need $M_n$. You can just use $r/2$ and $M$ to do it. $\endgroup$ – k99731 Feb 27 '16 at 1:39
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    $\begingroup$ For "$\int_{|z|=r}f^{(n)}(z)dz=0$?", yes. But you want to bound $f^{(n)}(z)$, instead of its integral. You can expect there are a lot of negative and positive part, so turn out it becomes $0$ after summing up. $\endgroup$ – k99731 Feb 27 '16 at 1:47
  • $\begingroup$ Oh, that makes sense, thank you. And what am I canceling $2\pi$ out with? And how does that get rid of $M_n$? $\endgroup$ – Matt G Feb 27 '16 at 1:50
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    $\begingroup$ Because $$|\frac{f(w)}{(w-z)^{n+1}}| \leq \frac{M}{(r/2)^{n+1}}$$. After converting it to $M$ and $r$, it becomes independent to $w$, so you can pull it out. Also, $$\int_{|z|=r} 1 dw = 2 \pi r$$. $\endgroup$ – k99731 Feb 27 '16 at 1:53

You are on the right track, but the estimate is not enough.

Fix a $z$ in the Disk of $r$. Consider disk $\gamma$ centered at $z$: $$ |w-z|=\frac{r-|z|}{\rho} $$ where $\rho>1$ so that it is completely inside the Disk of $r$. Then \begin{align} \bigg|f^{(n)}(z)\bigg|&=\bigg|\frac{n!}{2\pi i}\int_{|z|=r}\frac{f(w)}{(w-z)^{n+1}}dw\bigg| \\ &=\bigg|\frac{n!}{2\pi i}\int_{\gamma}\frac{f(w)}{(w-z)^{n+1}}dw\bigg| \\ &\leqslant\frac{n!}{2\pi}\int_{\gamma}\frac{M}{|w-z|^{n+1}}|dw| \\ &=\frac{M n!}{2\pi}\int_{0}^{2\pi}\frac{1}{|w-z|^{n+1}}\frac{r-|z|}{\rho}d\theta \\ &=\frac{M n!}{2\pi}\int_{0}^{2\pi}\frac{\rho^n}{(r-|z|)^{n}}d\theta \\ &=\frac{M n!\rho^n}{(r-|z|)^{n}} \end{align} So in $|z|\leq\frac{r}{2}$, there is $$ \bigg|f^{(n)}(z)\bigg|\leqslant\frac{M n!\rho^n}{(r-r/2)^{n}}=\frac{M n!(2\rho)^n}{r^{n}} $$

  • $\begingroup$ What is $\rho$? $\endgroup$ – Matt G Feb 27 '16 at 17:10
  • $\begingroup$ $ρ$ is any real >1, which is to make sure that $\gamma$ is within disk of r. $\endgroup$ – Math Wizard Feb 27 '16 at 22:12

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