# Let $f$ be holomorphic and bounded, prove $\left|f'(0)\right| \leq \frac{\text{diam}(f(\mathbb{D}))}{2}$ [duplicate]

For it, use that $$\oint_{|z|=r} \frac{f(z)-f(-z)}{z^2} dz = 4\pi i f'(0) ; \hspace{1cm}\text{ with } 0<r<1$$ Also remember that diam$(A)=\text{sup}\{\left|z-w\right| : z,w\in A\}$ for A bounded, $\mathbb{D}$ is the unit disc and $f:\mathbb{D}\rightarrow\mathbb{C}$.

First, observe that because changing the starting point of a path does not affect the value of an integral, we may write $\partial D(0,1):[0,1]\rightarrow\mathbb{C}$ as $\partial D(0,1)(\theta)=-e^{i\theta}$ so that: $$\int_{\partial D(0,1)}\frac{f(w)}{w^2}\,dw=\int_0^{2\pi}\frac{f(-e^{i\theta})}{e^{2i\theta}}(-e^{i\theta})\,dw=-\int_{\partial D(0,1)}\frac{f(-w)}{w^2}\,dw$$ By Cauchy's integral formula, we have that $f'(0)=\frac{1}{2\pi i}\int_{\partial D(0,1)}\frac{f(w)}{w^2}\,dw$ so that by the above $2f'(0)=\frac{1}{2\pi i}\int_{\partial D(0,1)}\frac{f(w)-f(-w)}{w^2}\,dw$. Thus: $$|f'(0)|=|\frac{1}{4\pi i}\int_{\partial D(0,1)}\frac{f(w)-f(-w)}{w^2}\,dw|=\frac{1}{4\pi}\max_{w\in\partial D(0,1)}\frac{|f(w)-f(-w)|}{|w^2|}||\partial D(0,1)||=\\\frac{1}{2}\max_{w\in\partial D(0,1)}|f(w)-f(-w)|\leq\frac{1}{2}\max_{z,w\in\overline{D}(0,1)}|f(z)-f(w)|$$
• These computations are not valid since the function $f$ is not defined on $\partial D(0,1)$. This is why one introduces the $r$-business. – m.s Nov 24 '16 at 21:44
Using the formula you wrote we obtain for every $0 < r < 1$ the estimate $$|4 \pi f´(0)| \leq 2 \pi r\text{diam}{(f(\{z:|z| < r\}})) \leq 2 \pi r \text{diam}{(f(\mathbb{D}}))\,$$ which means that for every $0 < r < 1$ we have $$| f´(0)| \leq r \frac{\text{diam}{(f(\mathbb{D}}))}{2}$$ This implies the claim (take for instance $r_n = 1-1/n$ an let $n \rightarrow \infty$ in the inequality).