Inequality relating diameter of the image of a holomorphic function on the unit disk to the derivative at 0. 
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. 
 A: Notice that $|f(\theta)-f(-\theta)| \leq |f(\theta)-f(\phi)| \leq d$ where $\phi$ is another boundary point on the circle. The inequality holds simply because we can only have increased the distance by taking a more arbitrary point.
It follows that $r\max_{\theta \in \gamma}\{|\frac{f(\theta)-f(-\theta)}{\theta^2}|\} \leq \frac{rd}{r^2} = d/r$ which is true for all $r<1$ so taking the limit as $r$ approaches $1$ you get your result (since $2|f'(0)| \leq d/r$ for all $r$ means that it is less than or equal to all values greater than $d$, and therefore less than or equal to $d$ itself).
A: In fact you are almost done. Put $g(z)=f(z)-f(-z)$, which is still holomorphic, then $$2f'(0)=g'(0)=\frac 1{2\pi i}\int_{C(0,r)}\frac{f(\xi)-f(-\xi)}{\xi^2}d\xi$$ so for each $0<r<1$, using the fact that $|\xi|=r$ on $C(0,r)$
$$2|f'(0)|\leq \frac 1{2\pi}\int_{C(0,r)}\frac{|f(\xi)-f(-\xi)|}{|\xi|^2}d\xi=\frac 1{2\pi r^2}d2\pi r=\frac dr,$$
and we get the result letting $r$ converging to $1$.
A: Another solution by using Schwarz Lemma.
Suppose $d\ne 0$. Let $z_0$ be a point in $D_1(0)$ such that $|f(z) - f(z_0)|\le d/2$ for all $z \in D_1(0)$. Let $h$ be a biholomorphic function of $D_1(0)$ mapping $c := \frac{f(z_0) -f(0)}{d/2} $ to $0$, for example,
$$h:z \mapsto \frac{z-c}{1-z\bar{c}}.$$
Set $g : z \mapsto h\left(\frac{f(z_0) -f(z)}{d/2} \right)$. By Schwarz Lemma applying to $g$, one has $\left|h'(c)f'(0)\right|\le d/2$. Since $\left|h'(c) \right| = 1/(1-|c|^2) \ge 1$, one has $2|f'(0)|\le d$.
