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Let $D$ be the unit disc centered at the origin and f olomorphic in the disc. Show that

$2|f'(0)|\leq sup_ {z,w\in D}|f(z)-f(w)|$

Furthermore, we have equality if and only if f is linear.

I only know Cauchy's estimation for derivatives:

$f^{(n)}(0)\leq \frac{n!}{r^n}Max_{|z|=r} |f|$ for $0<r<1$

but i've no idea how to apply this to this case.

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The Cauchy estimates are kind of useless here. Try starting from the definition of the complex derivative. – jmracek Nov 13 '12 at 13:06
up vote 1 down vote accepted

Cauchy's derivative-integral formula:

$$f'(0) = \frac{1}{2\pi \imath} \int_{|\xi|=r} \frac{f(\xi)}{\xi^2} \, d\xi$$

where $0<r<1$. Thus

$$2f'(0)= \frac{1}{2\pi \imath} \int_{|\xi|=r} \frac{f(\xi)-f(-\xi)}{\xi^2} \, d\xi \\ \Rightarrow 2|f'(0)| \leq \frac{1}{2\pi \imath} \cdot \frac{2\pi r}{r^2} \cdot \sup\{|f(\xi)-f(-\xi)|; |\xi|=r\} \\ \leq \frac{1}{r} \cdot \sup\{|f(z)-f(w)|; z,w \in D\}$$

Now let $r \to 1$...

Clearly the equality holds if $f$ is linear. On the other hand: Since $f(z) = \sum_{n \geq 0} a_n \cdot z^n$ we have

$$\sup_{w,z \in D} |f(z)-f(w)| = \sup_{w,z \in D} \left| f'(0) \cdot (z-w)+\sum_{n \geq 2} a_n \cdot (z^n-w^n) \right| \stackrel{!}{=} 2|f'(0)|$$ and this can only hold if $a_n =0$ for all $n \geq 2$.

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