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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!

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Do you assume that $f(0)=0?$ If not, Schwarz lemma cannot imply $|f(z)|\le |z|$. – 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 $$|f(z) - f(w)| = |F(\varphi(z)) - F(-\varphi(z))| \le 2 |\varphi(z)| = |\varphi(z) - \varphi(w)|$$

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Is it true that those specific fractional linear transformation is the only one who attain the maximum distance(between all the analytic functions)? – GAJO May 16 '14 at 14:39
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. $f(s) = c \varphi(s)$. – Robert Israel May 16 '14 at 15:03
but in order to use Schwarz's lemma we need to reqiure F(0) = 0 – GAJO May 18 '14 at 7:44
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 $\zeta \in D$. – Robert Israel May 19 '14 at 19:32

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