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Let $D \subset \mathbb{C}$ be the open unit disk, $f:D \to \mathbb{C}$ a holomorphic function and $d=\sup \{|f(z)-f(w)|:z,w\in D \}$. How can I show that $|f'(0)|\leq \frac{d}{2}$?

I tried using Schwarz's lemma on the function $g:D \to \mathbb{C}$ defined by $g(z)=\frac{f(z)-f(0)}{d}$, but I am only getting $|f'(0)|\leq d$.

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By Cauchy's integral formula, $$ f'(0) = \frac{1}{2\pi i} \int_{|z|=r} \frac{f(z)}{z^2}\,dz = \frac{1}{2\pi i} \int_{|z|=r} \frac{-f(-z)}{z^2}\,dz, $$ for any $r < 1$. Hence $$ f'(0) = \frac12 \cdot \frac{1}{2\pi i} \int_{|z|=r} \frac{f(z)-f(-z)}{z^2}\,dz $$ and consequently (by the standard estimation lemma) $$ |f'(0)| \le \frac12 \cdot \frac{1}{2\pi} \cdot 2\pi r \cdot \max_{|z|=r} \left|\frac{f(z)-f(-z)}{z^2} \right| \le \frac{d}{2r}. $$ Let $r \nearrow 1$ to get the desired inequality.

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  • $\begingroup$ Can we prove that the equality holds if and only if f is a linear function? $\endgroup$ – Yixuan Huang May 2 at 15:51

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