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$f: D\to \mathbb C$ is holomorphic such that $\sup_{u,v\in D}|f(v)-f(u)|=d$, then $|f'(0)|=d/2\implies f$ is linear.

Here, let's suppose that $D$ is a closed unit disk.

By Cauchy's integral formula, since $f(z)$ and $f(-z)$ are both holomorphic, we have for every circle $C_r$ centered at $0$ and with radius $r<1,$ $$2|f'(0)|=\frac 1{2\pi}|\int_{C_r}\frac{f(z)-f(-z)}{z^2}|\le d/r$$

Since $r$ can be brought arbitrarily close to $1$, we have $2|f'(0)|\le d$.

We have the equality in case $f(z)=a+bz$. But I'm not sure how to show the converse i.e., if $2|f'(0)|= d$, then $f$ is linear.

Koro
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  • I've answered this here using a convexity argument and there is another solution too: https://math.stackexchange.com/questions/3619008/2-leftf0-right-sup-z-omega-in-mathbbd-leftfz-f-omega-right/3621526#3621526 – Conrad Feb 28 '23 at 21:38
  • This is Schwartz lemma applied to $f/d/2$ – Salcio Mar 01 '23 at 04:41
  • @Salcio: Using Schwarz lemma, we only have $f(z)-f(-z)=adz$, where a is such that |a|=1. How do we go from here? – Koro Mar 01 '23 at 06:03
  • @Salcio: I think I see it now. So $f(z)= \sum_{n=0}^\infty a_n z^n\implies f(z)-f(-z)= \sum_{n=0}^\infty a_n (1-(-1)^n) z^n= ad z\implies a_n=0, \forall n>1$. It follows that $f(z)= a_0 +a_1 z$. Is this correct? Thanks a lot. – Koro Mar 01 '23 at 06:10
  • @Conrad: Thanks a lot. – Koro Mar 01 '23 at 06:10
  • Oh no. It concluded only that $a_n=0$ for odd $n$. I'm not sure how to go from here. – Koro Mar 01 '23 at 06:14
  • @Conrad: Ahh, I think it can be fixed as follows: I apply SL on $(f(z)- f(-iz))\sqrt 2/d$ and get a similar result as above. But this time for n=2, we have $a_2 (1-(-i)^2)z^2= 2a_2$ so $a_2$ should be zero. Then, we similarly, by SL on $f(z)-f(-i^{1/2}z))\times $something, show that $a_4=0$ and continue this process inductively to show that $a_n$=0 for even n>0. – Koro Mar 01 '23 at 06:33

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