Proof of Cauchy inequality: $\vert f''(z_0)\vert \leq \frac{2}{r^2}\max\limits_{\partial D_r} \vert f\vert$

Let $G\subset\mathbb{C}$ a domain, $f$ holomorphic on $G$, $r>0, z_0\in G$ and $D_r=D_r(z_0)\subset\subset G$. Prove this inequality: $$\vert f''(z_0)\vert \leq \frac{2}{r^2}\max\limits_{\partial D_r} \vert f\vert$$

Obviously this is the case $n=2$ of the Cauchy inequalities. In the lecture we've proved these two inequalities: \begin{align*} (1)\quad \vert f(z_0)\vert &\leq \max\limits_{\partial D_r(z_0)} \vert f\vert \\ (2)\quad \vert f'(z_0) \vert &\leq \frac4{r}\max\limits_{\partial D_r(z_0)} \vert f\vert \end{align*}

So I think it's to easy to derivate both sides and get the result. I tried this.

For $z\in \overline{D_\frac{r}{2}(z_0)}$ one can say with Cauchy integral formula: $$f''(z_0)=\frac{1}{\pi i} \oint\limits_{|z|=\frac{r}{2}} \frac{f(\zeta)}{\left(\zeta- z\right)^3}d\zeta$$

For $\zeta \in \partial D_r(z_0)$ is $\vert\zeta -z\vert \geq \frac{r}{2} \Rightarrow$ $$\vert f''(z_0)\vert \leq \frac{1}{\pi}\cdot 2\pi r \max\limits_{\partial D_r(z_0)}\left\vert\frac{f(\zeta)}{\left(\zeta- z\right)^3}\right\vert=\frac{8}{r^2}\max\limits_{\partial D_r(z_0)} \vert f\vert$$

Then I couldn't get the right factor. Where's my mistake? Thank you!

• For one thing, you've got the Cauchy formula all messed up (too many conflicting zeds). I don't know how much it matters, but perhaps it does. – user228113 Jun 29 '17 at 9:46
• Why too many $z$? – jacmeird Jun 29 '17 at 9:48
• Actually, not "too many zeds", but rather "misplaced zeds". Have a look at this with $z_0=z_0$, $z=\zeta$ and $a=z$ and see the difference. – user228113 Jun 29 '17 at 9:58

Why $r/2$ ??. We have
$f''(z_0)=\frac{1}{\pi i} \oint\limits_{|\zeta-z_0|=r} \frac{f(\zeta)}{\left(\zeta- z_0\right)^3}d\zeta$.
Let $M=\max\limits_{\partial D_r(z_0)} \vert f\vert$.
For $|\zeta-z_0|=r$:
$\frac{|f(\zeta)|}{\left|\zeta- z_0\right|^3} \le \frac{M}{r^3}$. Hence
$|f''(z_0)| \le \frac{1}{\pi} 2 \pi r \frac{M}{r^3}=\frac{2M}{r^2}$.