Schwarz lemma

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In mathematics, the Schwarz lemma, named after Hermann Amandus Schwarz, is a result in complex analysis about holomorphic functions defined on the open unit disk.

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[edit] Lemma statement

Let D = \{z : | z | < 1\}\ be the open unit disk in the complex plane C. Let f : D \to \overline D be a holomorphic function with f(0) = 0\ .

The Schwarz lemma states that

| f(z) | \le | z | for all z \in D, and

| f'(0) | \le 1,

and if the equality | f(z) |=| z |\ holds for any z \ne 0\, or | f'(0) | = 1\, then f\ is a rotation: f(z) = az\, with |a| = 1\ .

This lemma is less celebrated than stronger theorems, such as the Riemann mapping theorem, which it helps to prove; however, it is one of the simplest results capturing the "rigidity" of holomorphic functions. No similar result exists for real functions, of course.

To prove the lemma, one applies the maximum modulus principle to the function f(z)/z\ .

[edit] Proof

Let

g(z) = \frac{f(z)}{z}.\,

The function g(z) is holomorphic in D since f(0) = 0 and f is holomorphic. Let Dr be a closed disc within D with radius r. By the maximum modulus principle,

|g(z)| = \frac{|f(z)|}{|z|} \le \frac{|f(z_r)|}{|z_r|} \le \frac{1}{r}

for all z in Dr and all zr on the boundary of Dr. As r approaches 1 we get |g(z)| \le 1 .

Moreover, if there exists a z0 in D such that g(z0) = 1. Then, applying maximum modulus principle to g, we obtain that g is constant, hence f(z) = kz, where k is constant and | k | = 1. This is also the case if | f'(0) | = 1.

[edit] Schwarz-Pick theorem

A variant of the Schwarz lemma can be stated that is invariant under analytic automorphisms on the unit disk, i.e. bijective holomorphic mappings of the unit disc to itself. This variant is known as the Schwarz-Pick theorem (after Georg Pick):

Let f\colon D \to D be holomorphic. Then, for all z_1,z_2\in D,

\left|\frac{f(z_1)-f(z_2)}{1-\overline{f(z_1)}f(z_2)}\right| \le \frac{\left|z_1-z_2\right|}{\left|1-\overline{z_1}z_2\right|}

and, for all z\in D

\frac{\left|f'(z)\right|}{1-\left|f(z)\right|^2} \le \frac{1}{1-\left|z\right|^2}.

The expression

d(z_1,z_2)=\tanh^{-1}\left(\frac{\left|z_1-z_2\right|}{\left|1-\overline{z_1}z_2\right|}\right)

is the distance of the points z1,z2 in the Poincaré metric, i.e. the metric in the Poincaré disc model for hyperbolic geometry in dimension two. The Schwarz-Pick theorem then essentially states that a holomorphic map of the unit disk into itself decreases the distance of points in the Poincaré metric. If equality holds throughout in one of the two inequalities above (which is equivalent to saying that the holomorphic map preserves the distance in the Poincaré metric) , then f must be an analytic automorphism of the unit disc, given by a Möbius transformation mapping the unit disc to itself.

An analogous statement on the upper half-plane \mathbb{H} can be made as follows:

Let f\colon\mathbb{H}\to\mathbb{H} be holomorphic. Then, for all z_1,z_2\in \mathbb{H},

\left|\frac{f(z_1)-f(z_2)}{\overline{f(z_1)}-f(z_2)}\right| \le \frac{\left|z_1-z_2\right|}{\left|\overline{z_1}-z_2\right|}.

This is an easy consequence of the Schwarz-Pick theorem mentioned above: One just needs to remember that the Cayley transform W(z) = (z − i) / (z + i) maps the upper half-plane \mathbb{H} conformally onto the unit disc D. Then, the map W\circ f\circ W^{-1} is a holomorphic map from D onto D. Using the Schwarz-Pick theorem on this map, and finally simplifying the results by using the formula for W, we get the desired result. Also, for all z\in\mathbb{H},

\frac{\left|f'(z)\right|}{\mbox{Im }f(z)} \le \frac{1}{\mbox{Im }(z)}.

If equality holds for either the one or the other expressions, then f must be a Möbius transformation with real coefficients. That is, if equality holds, then

f(z)=\frac{az+b}{cz+d}

with a,b,c,d being real numbers, and ad − bc > 0.

[edit] Proof of Schwarz-Pick theorem

The proof of the Schwarz-Pick theorem follows from Schwarz's lemma and the fact that a Möbius transformation of the form \frac{z-z_0}{\overline{z_0}z-1} where | z0 | < 1 maps the unit circle to itself. Fix z1 and define the Möbius transformations M(z)=\frac{z_1-z}{1-\overline{z_1}z} and \phi(z)=\frac{f(z_1)-z}{1-\overline{f(z_1)}z}.

Since M(z1) = 0 and the Möbius transformation is invertible, the composition φ(f(M − 1(z))) maps 0 to 0 and the unit disk is mapped into itself. Thus we can apply Schwarz's lemma, which is to say

|\phi(f(M^{-1}(z)))|=\left|\frac{f(z_1)-f(M^{-1}(z))}{1-\overline{f(z_1)}f(M^{-1}(z))}\right| \le |z|.

Now calling z2 = M − 1(z) (which will still be in the unit disk) yields the desired conclusion

\left|\frac{f(z_1)-f(z_2)}{1-\overline{f(z_1)}f(z_2)}\right| \le \left|\frac{z_1-z_2}{1-\overline{z_1}z_2}\right|.

To prove the second part of the theorem, we just let z2 tend to z1.

[edit] Further generalizations and related results

The Schwarz-Ahlfors-Pick theorem provides an analogous theorem for hyperbolic manifolds.

De Branges' theorem, formerly known as the Bieberbach Conjecture, is an important extension of the lemma, giving restrictions on the higher derivatives of f at 0 in case f is injective; that is, conformal.

The Koebe 1/4 theorem provides a related estimate in the case that f is conformal.

[edit] References

This article incorporates material from Schwarz lemma on PlanetMath, which is licensed under the GFDL.