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First derivative bounded by supremum of difference of values in disc

Let $f$ be holomorphic in the disk $D_1(0)$ and let $d=\operatorname{diam}(f(D_1(0))$. I want to show that

$$2|f'(0)|\leq d$$

I have the following:

Let $\gamma$ be the circle of radius $r<1$ traversed counter-clockwise. Let $-\gamma$ be the same circle traversed clockwise.

By Cauchy's integral formula for derivatives, we have that

$$f'(0)=\frac{1}{2\pi i}\int_\gamma \frac{f(\theta)}{\theta^2}d\theta=-\frac{1}{2\pi i}\int_{-\gamma}\frac{f(\theta)}{\theta^2}d\theta=-\frac{1}{2\pi i}\int_\gamma \frac{f(-\theta)}{\theta^2}d\theta$$

So

$$\begin{align} 2f'(0) &=\frac{1}{2\pi i}\int_\gamma \frac{f(\theta)-f(-\theta)}{\theta^2} \; d\theta\\ \end{align}$$

and by the standard estimate

$$\begin{align} 2|f'(0)| &=|\frac{1}{2\pi i}\int_\gamma \frac{f(\theta)-f(-\theta)}{\theta^2}d\theta|\\ &\leq \frac{1}{2\pi}\max_{\theta \in \gamma}\{|\frac{f(\theta)-f(-\theta)}{\theta^2}|\}2\pi r\\ &\leq r\max_{\theta \in \gamma}\{|\frac{f(\theta)-f(-\theta)}{\theta^2}|\} \end{align}$$ which is almost what I want, but not quite.

We haven't discussed the maximum modulus principle in class, so I can't use that. Any suggestions as to how to finish this up?

Thanks.

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marked as duplicate by Davide Giraudo, sdcvvc, William, wentaway, Matt N. Sep 3 '12 at 10:21

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3 Answers

up vote 1 down vote accepted

Notice that $|f(\theta)-f(-\theta)| \leq |f(\theta)-f(\phi)| \leq d$ where $\phi$ is another boundary point on the circle. The inequality holds simply because we can only have increased the distance by taking a more arbitrary point.

It follows that $r\max_{\theta \in \gamma}\{|\frac{f(\theta)-f(-\theta)}{\theta^2}|\} \leq \frac{rd}{r^2} = d/r$ which is true for all $r<1$ so taking the limit as $r$ approaches $1$ you get your result (since $2|f'(0)| \leq d/r$ for all $r$ means that it is less than or equal to all values greater than $d$, and therefore less than or equal to $d$ itself).

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In fact you are almost done. Put $g(z)=f(z)-f(-z)$, which is still holomorphic, then $$2f'(0)=g'(0)=\frac 1{2\pi i}\int_{C(0,r)}\frac{f(\xi)-f(-\xi)}{\xi^2}d\xi$$ so for each $0<r<1$, using the fact that $|\xi|=r$ on $C(0,r)$ $$2|f'(0)|\leq \frac 1{2\pi}\int_{C(0,r)}\frac{|f(\xi)-f(-\xi)|}{|\xi|^2}d\xi=\frac 1{2\pi r^2}d2\pi r=\frac dr,$$ and we get the result letting $r$ converging to $1$.

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Another solution by using Schwarz Lemma.

Suppose $d\ne 0$. Let $z_0$ be a point in $D_1(0)$ such that $|f(z) - f(z_0)|\le d/2$ for all $z \in D_1(0)$. Let $h$ be a biholomorphic function of $D_1(0)$ mapping $c := \frac{f(z_0) -f(0)}{d/2} $ to $0$, for example, $$h:z \mapsto \frac{z-c}{1-z\bar{c}}.$$

Set $g : z \mapsto h\left(\frac{f(z_0) -f(z)}{d/2} \right)$. By Schwarz Lemma applying to $g$, one has $\left|h'(c)f'(0)\right|\le d/2$. Since $\left|h'(c) \right| = 1/(1-|c|^2) \ge 1$, one has $2|f'(0)|\le d$.

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The Scwarz Lemma is a direct corollary of the maximum modulus principal. So if you can't use the latter, it stands to reason that the former hasn't been covered and is similarly "out of bounds" :) Nice answer nonetheless. –  davin Feb 21 '12 at 22:55
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