6
$\begingroup$

Suppose $f$ is a holomorphic map on the unit disc. Let $d$ be the diameter of the image of $f$. If $ 2|f'(0)|=d $, please show that $f$ is a linear function.

I think maybe I can prove it by using Schwarz lemma, but I failed.

$\endgroup$
  • $\begingroup$ Is it meant so that the image of $f$ is contained in the unit disk? $\endgroup$ – Berci Mar 24 '13 at 13:45
8
$\begingroup$

Historical note: this was proved by Landau and Toeplitz in 1907. A modern source, with many generalizations, is the paper Area, capacity and diameter versions of Schwarz's Lemma by Robert B. Burckel, Donald E. Marshall, David Minda, Pietro Poggi-Corradini, Thomas J. Ransford.

I divided the proof into several parts, so you can decide to stop before reading further.

Opening scene

Let's normalize things so that $f'(0)=1$ and $d=2$. The first thing that comes to mind is to introduce $g(z)=(f(z)-f(-z))/2$, apply the equality case of the Schwarz lemma, and conclude that $g(z)\equiv z$. So, $f$ contains no odd powers above linear. But we know nothing about even powers.

Commercial break

A healthy reaction is to try to find a counterexample of the form $f(z)=z+\epsilon z^2$. Nope, does not work. For this function the differences $|f(z)-f(-z)|$ do not realize the diameter of the image: they are not even stationary points. We get more from $|f(z')-f(-z)|$ with $z'$ is close to $z$.

This suggests a variational argument, differentiating with respect to $z$ lying on the unit circle (so far, illegally):
$$\begin{align}|f(z+dz)-f(-z)| &= |f(z)+f'(z)\,dz-f(-z)| = |2z+f'(z)\,dz|\\ &= 2|1+f'(z)\, dz/z| = 2+ 2\operatorname{Re}( f'(z)\, dz/z) +o(|dz|)\end{align}$$ Since $|f(z+dz)-f(-z)|$ is not to exceed $2$, it follows that $\operatorname{Re}( f'(z)\, dz/z)=0$. Here $dz/z$ is imaginary because $dz$ is tangent to the unit circle at $z$. Hence, $\operatorname{Im} f'(z)=0$. Since $\operatorname{Im} f'$ vanishes on the unit circle, it vanishes identically, and $f$ is indeed linear.

Your ad could be here

Now we try to legalize the above in the usual way, taking $0<r<1$ and considering $f_r(z)=r^{-1}f(rz)$, which is smooth on the unit circle. Of course $f_r'(0)=1$, but what can we say about the image of the unit disk $\mathbb D$ under $f_r$? If it also has diameter $2$, then the variational argument applies to $f_r$ and we are done. Let's see how the function $D(r):=\operatorname{diam} f_r(\mathbb D)$ depends on $r$: $$ D(r) = \sup_{\theta \in [0,2\pi]} \sup_{|z|=1}\, |f_r(e^{i\theta}z) - f_r(z)| = \sup_{\theta \in [0,2\pi]} \sup_{|z|=r}\, \left|\frac{f( e^{i\theta}z) - f (z)}{z}\right| $$ For any fixed $\theta$, the function $\dfrac{f( e^{i\theta}z) - f (z)}{z}$ is holomorphic in the unit disk. By the maximum principle, the supremum of its modulus over $|z|=r$ is a nondecreasing function of $r$. Taking the supremum over $\theta$, we conclude that $D$ is a nondecreasing function of $r$.

Exit music

Since $D(0+)=2=D(1)$, it follows that $D(r)\equiv 2$, and this is the final piece of the puzzle.

$\endgroup$
  • $\begingroup$ This is very nice, +1! $\endgroup$ – Malik Younsi Mar 25 '13 at 20:03
  • 1
    $\begingroup$ Why do you have $D(0+)=2$? $\endgroup$ – Bach Sep 28 '19 at 18:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.