Given $z, w \in D$ (unit disk, open), what are the functions (analytic, from unit disk to unit disk) $f$ that maximize the norm of $f(z)-f(w)$?

My attempt: We have that $$|f(z)-f(w)| \leq |f(z)|+|f(w)| \leq |z|+|w| \tag{1}$$ using the triangle inequality and Schwarz lemma. Moreover, $$|f(z)| \leq |z| \text{ and } |f(w)| \leq |w|. \tag{2}$$

Now, I found the challenge to be deciding which inequality in (1) to maximize. In the collinear case (for $z$, $w$ and $0$), by Schwarz lemma, (2) are equalities only if $f(z) = az$ for $z$ unimodular. Now, this would also make the first inequality in (1) an equality. Thus, we found the class of functions maximizing the distance.

Now I do not fully understand what to do/which inequality to maximize in the non-collinear case (for $0$, $z$ and $w$). A unimodular constant multiple would preserve the distance, which intuitively does not seem like a max to me. My intuitions are probably wrong. Any help is appreciate!

  • 1
    $\begingroup$ Do you assume that $f(0)=0?$ If not, Schwarz lemma cannot imply $|f(z)|\le |z|$. $\endgroup$ – 23rd Nov 30 '12 at 3:29

There is a fractional linear transformation $\varphi$ that preserves $D$ such that $\varphi(z) = -\varphi(w)$. If $F: D \to D$ is analytic, consider the function $g(\zeta) = \dfrac{F(\zeta) - F(-\zeta)}{2 \zeta}$, with $g(0) = F'(0)$. Applying the maximum modulus principle to $g$, conclude that $|g(\zeta)| \le 1$. Thus $|F(\zeta) - F(-\zeta)| \le 2 |\zeta|$ (which is attained by the identity function). Taking $F = f \circ \varphi^{-1}$, we get $$|f(z) - f(w)| = |F(\varphi(z)) - F(-\varphi(z))| \le 2 |\varphi(z)| = |\varphi(z) - \varphi(w)|$$

  • $\begingroup$ Is it true that those specific fractional linear transformation is the only one who attain the maximum distance(between all the analytic functions)? $\endgroup$ – GAJO May 16 '14 at 14:39
  • 1
    $\begingroup$ Suppose $|f(z) - f(w)| = |\varphi(z) - \varphi(w)|$, i.e. $|g(z)| = 1$. The maximum modulus principle then implies that $g(\zeta)$ is a constant $c$ of absolute value $1$, i.e. $F(\zeta) - F(-\zeta) = 2 c \zeta$, and $F'(0) = c$. Schwarz's lemma then implies that $F(\zeta) = c \zeta$, i.e. $f(s) = c \varphi(s)$. $\endgroup$ – Robert Israel May 16 '14 at 15:03
  • $\begingroup$ but in order to use Schwarz's lemma we need to reqiure F(0) = 0 $\endgroup$ – GAJO May 18 '14 at 7:44
  • $\begingroup$ Oops, yes. But consider this. Take any $\omega$ with $|\omega| = 1$, and a sequence $z_n \to \omega$ with $z_n \in D$. Some subsequence $F(z_{n_k})$ has a limit $L$ with $|L|\le 1$. Then $F(-z_{n_k}) \to L - 2 c \omega$, and $|L - 2 c \omega| \le 1$ as well. But since $|c\omega| = 1$ this implies $L = c \omega$. From that you can conclude $\lim_{\zeta \to \omega} F(\zeta) = c \omega$ for all $\omega$ with $|\omega|=1$, and using the Cauchy integral formula that $F(\zeta) = c \zeta$ for $\zeta \in D$. $\endgroup$ – Robert Israel May 19 '14 at 19:32

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.