First we recall the statement of *Schwarz’s lemma*, which is a basic result in complex analysis. Let be the unit disk in .

**Schwarz’s lemma.** Let . If for all and , then , and for all . If or for some , then for some with .

To prove this, consider the holomorphic function and apply the maximum principle. (See any textbook for details.) Note that is maximized if and only if is a conformal self-map of (with the normalization ). If we do not specify the value of , we get Pick’s lemma.

**Pick’s lemma**. Suppose is holomorphic. Then

for any . Equality holds for some if and only if is a conformal self-map of (and in that case equality holds everywhere).

Pick’s lemma leads naturally to the *hyperbolic metric* on .

Schwarz’s lemma has the following useful Corollary called the *Principle of Subordination*. It makes clear the role of conformal maps in Schwarz’s lemma. First we need a notation. If is holomorphic (or merely continuous) on and , we define

.

**Principle of Subordination.** Let . Suppose that , is one-to-one, and , then and for all .

*Proof*. Since is one-to-one, is in fact a conformal map from to . Let be the inverse of . Then is holomorphic and maps into with since . By Schwarz’s lemma, we have

and

for all . By the chain rule, . So the first inequality gives .

Fix . The second inequality implies that

.

It follows that

.

Note that the second last equality follows from the maximum principle.

As an exercise, formulate the condition for equality. Here we give a simple example.

**Example.** Let be the right half plane. Let be holomorphic. We claim that .

*Proof*. Write , where . Define . Then is a conformal map from to with . Clearly . By the principle of subordination, we have . By direct calculation, and the result follows.

Using this estimate, we can prove the following interesting result.

**Example (a prelim problem).** Let be holomorphic. Then .

*Proof*. Since omits , there exists such that . Note that

and .

Since , we have . It follows that maps into . Let . By the above result, we have

.

It follows that

.

Maximizing the last expression over , we have

.

More examples will be added later.

Nice post. Admittedly I have never thought about hyperbolic metric this way ( I am making my remark here for my own benefit). Let me try to figure out what you mean by the Pick’s lemma gives rise to the hyperbolic metric. Perhaps this is somewhat trivial and is well-known to many of you.

In most textbooks, the hyperbolic metric is given first and then it is deduced that the isometry group is exactly the group of all conformal self maps of the disk i.e. disk-preserving holomorphic and antiholomorphic functions. Suppose we want to reverse this process and want to find a (hyperbolic) metric which is preserved by the conformal self maps. We observe from the equality case of the Pick’s lemma that for to be a conformal self map,

i.e.

Thus we find that the hyperbolic metric is given by (up to a positive factor)

Interestingly, the other form of Pick’s lemma is given by where , with equality holds if and only if it is a conformal self map. (This can be seen by , where and putting . ) Writing it out, it is

with equality hold iff is a conformal self map. It suggests (without integrating) that the

hyperbolic distanceis actually given by (up a scaling). HOWEVER, it is different from the usual definition , I can’t think if any way to “see” (except integrating) that the correct definition is the later one instead of the former one (actually I haven’t checked if the former one really satisfies the triangle inequality). Can anyone help?