Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
Do you assume that $f(0)=0?$ If not, Schwarz lemma cannot imply $|f(z)|\le |z|$. –  23rd Nov 30 '12 at 3:29
add comment

1 Answer

There is a fractional linear transformation $\varphi$ that preserves $D$ such that for some $x \in D$, $\varphi(z) = -\varphi(w)$. 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 $$|f(z) - f(w)| = |F(\varphi(z)) - F(-\varphi(z))| \le 2 |\varphi(z)| = |\varphi(z) - \varphi(w)|$$

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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