Possible Duplicate:
First derivative bounded by supremum of difference of values in disc
Let $f$ be holomorphic in the disk $D_1(0)$ and let $d=\operatorname{diam}(f(D_1(0))$. I want to show that
$2|f'(0)|\leq d$
I have the following:
Let $\gamma$ be the circle of radius $r<1$ traversed counter-clockwise. Let $-\gamma$ be the same circle traversed clockwise.
By Cauchy's integral formula for derivatives, we have that
$f'(0)=\frac{1}{2\pi i}\int_\gamma \frac{f(\theta)}{\theta^2}d\theta=-\frac{1}{2\pi i}\int_{-\gamma}\frac{f(\theta)}{\theta^2}d\theta=-\frac{1}{2\pi i}\int_\gamma \frac{f(-\theta)}{\theta^2}d\theta$
So
$\begin{align} 2f'(0) &=\frac{1}{2\pi i}\int_\gamma \frac{f(\theta)-f(-\theta)}{\theta^2} \; d\theta\\ \end{align}$
and by the standard estimate
$\begin{align} 2|f'(0)| &=|\frac{1}{2\pi i}\int_\gamma \frac{f(\theta)-f(-\theta)}{\theta^2}d\theta|\\ &\leq \frac{1}{2\pi}\max_{\theta \in \gamma}\{|\frac{f(\theta)-f(-\theta)}{\theta^2}|\}2\pi r\\ &\leq r\max_{\theta \in \gamma}\{|\frac{f(\theta)-f(-\theta)}{\theta^2}|\} \end{align}$ which is almost what I want, but not quite.
We haven't discussed the maximum modulus principle in class, so I can't use that. Any suggestions as to how to finish this up?
Thanks.