Suppose $f \colon \mathbb{D} → \mathbb{C}$ is holomorphic. Then I want to show that the diameter $$d=\sup _{z, w∈\mathbb{D}} |f (z) − f (w)|$$
of the image of $f$ satisfies $2|f′(0)| ≤ d$ and that equality holds precisely when $f$ is linear, this is $f(z) = a_0 +a_1z$.
Then I did the following:
Using Cauchy's integral formula we get that,
$$2|f′(0)|=\left|\frac{1}{2πi}∫_{|z|=r}\frac{f(z)−f(−z)}{z^2}\,dz\right|.$$
Then I parametrize the circle, so arrive at:
$$\frac{1}{2πi}∫_{0}^{2 \pi}\frac{f(Re^{i \theta})−f(−Re^{i \theta})}{R^2e^{2i \theta}}iRe^{i \theta}\,d\theta.$$
So given that $R=1$ and that the length of the curve is $2 \pi$ we arrive to this inequality
$$\left|\frac{1}{2πi}\int_{0}^{2 \pi}\frac{f(Re^{i \theta})−f(−Re^{i \theta})}{R^2e^{2i \theta}}iRe^{i \theta}\,d\theta\right|< \sup \left|f(e^{i \theta})−f(−e^{i \theta})\right|$$
and for the equality we see that if $f$ is linear then $2f'(0)=2a_0$ and $d=\sup|a_1(z-w)|=a_1$, but the thing there is no reason to have that $a_0=a_1.$
So Am I right in the first approach (or how can I fix it)? Also, how Can I get the equality?
Thanks a lot in advance.
