Analysis Problem of the Day 25

Today’s problem appeared as Problem 8 on the UCLA Fall 2023 Analysis Qual:

Problem 8: For $z_1, z_2 \in \mathbb{D},$ define

$\Delta(z_1,z_2)=\left|\frac{z_1-z_2}{1-\overline{z_2}z_1}\right|.$

a) If for $\alpha \in \mathbb{D},$ $g: \mathbb{D} \to \mathbb{D}$ is the Möbius transformation $g(z) = \frac{z-\alpha}{1-\overline{\alpha} z},$ show that $\Delta(f(z_1),f(z_2)) = \Delta(z_1,z_2)$ for all $z_1,z_2 \in \mathbb{D}.$

b) Show that if $f: \mathbb{D} \to \mathbb{D}$ is holomorphic, then $\Delta(f(z_1),f(z_2)) \leq \Delta(z_1,z_2)$ for all $z_1,z_2 \in \mathbb{D}.$

c) Characterize all $f$ such that equality in b) holds for at least one pair $z_1,z_2.$

Solution: a) Since we are dealing with functions on the unit disc, it is helpful to remember the characterization of automorphisms of the unit disc – those are precisely maps of the form

$h(z) = e^{i\theta} \frac{z-\alpha}{1-\overline{\alpha}z}$

for $\alpha \in \mathbb{D}.$ We will prove that a) holds for all functions of the above form. Note that the constant factor out front cancels in the definition of $\Delta$ since one is taking magnitudes. Thus, it suffices to prove the statement for all functions of the form $g$ given in a). Given an automorphism $h_\alpha$ of the unit disc that maps a given point $\alpha$ to the origin, there is a unique map of the form in a) (which we denote by $g_\alpha$ ) that corresponds to this automorphism. Moreover, $\Delta(z_1,z_2) = |g_{z_2}(z_1)|.$ Thus,

$\Delta(h_\alpha(z_1),h_\alpha(z_2)) = \Delta(g_\alpha(z_1),g_\alpha(z_2))=|g_{g_\alpha(z_2)}(g_\alpha(z_1))|.$

As a function of $z_1,$ the function inside the modulus is a composition of automorphisms and therefore $\Delta(h_\alpha(z_1),h_\alpha(z_2))=|g_\beta(z_1)|$ for some $\beta \in \mathbb{D}.$ But notice that $\Delta(h_\alpha(z_2),h_\alpha(z_2))=0,$ so $\beta = z_2.$ Thus,

$\Delta(h_\alpha(z_1),h_\alpha(z_2)) = |g_{z_2}(z_1)| = \Delta(z_1,z_2)$

for all maps $h$ as above.

b) Since we want to establish an inequality for maps on the unit disc, we are immediately reminded of the Schwarz lemma. In order to apply the lemma, we need to map the origin to itself, which can be done by an appropriate composition with Möbius transformations. Namely, as a function of $z_1,$ since $\Delta(f(z_1),f(z_2)) = g_{f(z_2)}(f(z_1)),$ we have that $z_1 \to \Delta(f(\cdot),f(z_2)) \circ g^{-1}_{z_2}$ maps $0 \to z_2 \to f(z_2) \to 0,$ as required. Thus, by the Schwarz lemma,

$|g_{f(z_2)} \circ f \circ g^{-1}_{z_2}(z_1)| \leq |z_1|,$

i.e. $\Delta(f(z_1),f(z_2)) \leq |g_{z_2}(z_1)| = \Delta(z_1,z_2)$ for any $z_1,z_2 \in \mathbb{D}.$

c) Recall that equality in the Schwarz lemma holds if and only if the function is of the form $e^{i\theta} z.$ Thus, if equality holds for at least one pair $z_1 \not = z_2,$ then $g_{f(z_2)} \circ f \circ g^{-1}_{z_2}(z_1) = e^{i\theta} z,$ i.e. $f = g_{f(z_2)}^{-1}(e^{i\theta} g_{z_2}(z_1)).$ This is an automorphism of the unit disc and therefore takes the form $h$ above. By a), this implies that if equality holds for at least one pair of distinct points, then equality holds for all points and the map $f$ is necessarily an automorphism of the unit disc of the form $h$ above.

Remark: The map $\Delta$ is a metric on $\mathbb{D}$ (and in general an arbitrary Riemann surface) known as the hyperbolic metric.

Leave a Reply Cancel reply