Let $\mathbb{H}$ denote the upper half plane of $\mathbb{C}$, i.e. \begin{equation*} \mathbb{H}=\{z \in \mathbb{C}: Im(z)> 0\} \end{equation*} Suppose $f:\mathbb{H}\to\mathbb{H}$ is analytic. Prove that

  1. $\Big|\frac{f(z)-f(w)}{f(z)-\overline{f(w)}}\Big|\le \Big|\frac{z-w}{z-\overline w}\Big|$, for all $z,w\in\mathbb{H}$.
  2. If $f\in Aut(\mathbb{H})$, then the equality in the above formula holds for any $z,w\in\mathbb{H}$.
  3. If the equality in the above formula holds for any one pair $z_0\ne w_0\in\mathbb{H}$, then $f\in Aut(\mathbb{H})$.

Hint: Recall that for any $z_0\in\mathbb{H}$, $h_{z_0}(z)=\frac{z-z_0}{z-\overline{z_0}}\in Iso (\mathbb{H},\mathbb{D})$. Fix $w\in\mathbb{H}$ and apply Schwarz lemma to $g=h_{f(w)}\circ f\circ h_{w}^{-1}$.

Proof (attempt)

Following the hint, I let $h_w(z)=\frac{z-w}{z-\overline{w}}$ and solved for its inverse, which is $h_w^{-1}(z)=\frac{\overline{w}z-w}{z-1}$. Then we have \begin{align*} g(z)&=\left(h_{f(w)}\circ f\circ h_{w}^{-1}\right)(z) \\ &=\frac{f\left(\frac{\overline{w}z-w}{z-1}\right)-f(w)}{f\left(\frac{\overline{w}z-w}{z-1}\right)-\overline{f(w)}} \end{align*} which looks like quite a mess. However, I see that \begin{align*} g(0)&=\frac{f\left(\frac{0-w}{0-1}\right)-f(w)}{f\left(\frac{0-w}{0-1}\right)-\overline{f(w)}} \\ &=\frac{f\left(w\right)-f(w)}{f\left(w\right)-\overline{f(w)}} \\ &=0 \end{align*} which is a requirement of the Schwarz lemma, which, applying to $g$, says that \begin{align*} |g(z)|\le |z| \end{align*} i.e. \begin{align*} \left|\left(h_{f(w)}\circ f\circ h_{w}^{-1}\right)(z)\right| \le |z| \end{align*} or \begin{align*} \left|\frac{f\left(\frac{\overline{w}z-w}{z-1}\right)-f(w)}{f\left(\frac{\overline{w}z-w}{z-1}\right)-\overline{f(w)}}\right| \le |z| \end{align*} but I don't see the inequality I'm looking for. Any help with this one? Thank you very much!


1 Answer 1


Let $z'=f(z)$ and $w'=f(w)$. Use a Mobius transformation $\phi$ to map conformally $\mathbb H$ to the unit disk so that $z$ maps to $0$, and $w$ maps to a point $\phi(w)=\tilde w$ of the disk. Similarly, use another Mobius transformation $\psi$ to map $z'$ to $0$ and $w'$ to a point $\psi(w')=\tilde w'$ of $\mathbb D$. Consider the composition $\psi\circ f\circ \phi^{-1}: \mathbb D \to \mathbb D$ that maps $0$ to $0$ by construction. Schwarz's lemma now applies to yield $$|\psi \circ f\circ \phi^{-1} (\zeta)| \leq |\zeta|$$ for all $\zeta \in \mathbb D$. Using $\zeta=\tilde w= \phi(w)$ we obtain $$ |\psi \circ f(w)|\leq |\phi(w)|$$ If you use the exact formulas of $\phi$ and $\psi $ then you obtain the desired $$\frac{|z'-w'|}{|z'-\overline {w'}|}\leq \frac{|z-w|}{|z-\overline w|} $$

If $f\in Aut(\mathbb H)$ then $\psi \circ f\circ \phi^{-1}$ is an autmorphism of $\mathbb D$ that maps $0$ to $0$, so it is a rotation and $|\psi \circ f\circ \phi^{-1}(\zeta)|=|\zeta|$ which implies that we have equality in the above inequality.

Conversely, if we have equality, then $|\psi \circ f\circ \phi^{-1}(\zeta)|=|\zeta|$ for $\zeta=\phi(w)$, so the equality case in Schwarz's lemma implies that $\psi \circ f\circ \phi^{-1}(\zeta)=e^{i\theta}\zeta$, and in fact $f$ is an automorphism of $\mathbb H$ as a composition of conformal maps.

  • $\begingroup$ Outstanding answer, and very well explained. Thank you very much for your help! $\endgroup$ Commented Apr 22, 2015 at 15:55
  • $\begingroup$ I thought the standard conformal mapping from $H$ to $D$ was $\frac{z-i}{z+i}$, but this is not the one being used here. With this mapping it's such a hassle to work out the compositions that I've given up, would you care to share the conformal mappings that you chose? $\endgroup$
    – Mike
    Commented May 25, 2016 at 10:19
  • $\begingroup$ Read the hint in the OP. $\endgroup$
    – Dimitris
    Commented May 25, 2016 at 16:39

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