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Need a little help in the following:

Let $f(z)$ analytic function on $D = \{z\in\mathbb C: |z| < 1\}$. Define $\displaystyle d = \sup_{z,w \in D} |f(z) - f(w)|$.

Prove that $|f'(0)| \leq \frac{d}{2}$.

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1 Answer

Hint(s):

  1. Put $g(z)=f(z)-f(-z)$. What is the link between $g'(0)$ and $f'(0)$?
  2. Apply the Cauchy integral formula to the circle $C(0,r)$ for $0<r<1$ to get $|g'(0)|\leq \frac dr$.
  3. Conclude.

Note that $\frac 12$ is the best constant we can hope; take $f(z)=z$ to see that, and you can, by the same method, get a bound for the odd derivatives at $0$.

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Could you write for me explicitly the step $2$? Please, $g(0)=\frac{1}{2\pi i}\int_{|z|=r}\frac{g(z)}{z}dz$ am I right? –  Bunuelian Trick Apr 30 '13 at 14:58
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Yes, you are right. Hence $g'(0)=\frac 1{2\pi i}\int_{\{|z|=r\}}\frac{g(z)}{z^2}dz$, and $|g(z)|\leqslant d$. –  Davide Giraudo Apr 30 '13 at 15:03
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