Schwarz lemma

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.

Lemma statement

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

The Schwarz lemma states that under these circumstances $$|f(z)| le |z| for all $$z in D, and $$|f'(0)| le 1.

Moreover, if the equality $$|f(z)| = |z| holds for any $$z e 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 .

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 "D"_"r" be a closed disc within "D" with radius "r". By the maximum modulus principle,

:$$ |g(z)| = frac le frac{1}{r}

for all "z" in "D"_"r" and all "z"_"r" on the boundary of "D"_"r". As "r" approaches 1 we get |"g"("z")| &le; 1.

Moreover, if there exists a "z"₀ in "D" such that "g"("z"₀) = 1. Then, applying the 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.

chwarz-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 $$fcolon D o D be holomorphic. Then, for all $$z_1,z_2in D,

:$$left|frac{f(z_1)-f(z_2)}{1-overline{f(z_1)}f(z_2)} ight
le frac{left|z_1-z_2 ight{left|1-overline{z_1}z_2 ight

and, for all $$zin D

:$$frac{left|f'(z) ight{1-left|f(z) ight|^2} lefrac{1}{1-left|z ight|^2}.

The expression

:$$d(z_1,z_2)= anh^{-1}left(frac{left|z_1-z_2 ight{left|1-overline{z_1}z_2 ight ight)

is the distance of the points $$z_1,z_2 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 $$fcolonmathbb{H} omathbb{H} be holomorphic. Then, for all $$z_1,z_2in mathbb{H},

:$$left|frac{f(z_1)-f(z_2)}{overline{f(z_1)}-f(z_2)} ight
le frac{left|z_1-z_2 ight{left|overline{z_1}-z_2 ight.

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 $$D. Then, the map $$Wcirc fcirc 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 $$W, we get the desired result. Also, for all $$zinmathbb{H},

:$$frac{left|f'(z) ight{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.

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 $$frac{z-z_0}{overline{z_0}z-1} where $$|z_0|<1 maps the unit circle to itself. Fix $$z_1 and define the Möbius transformations $$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 $$M(z_1)=0 and the Möbius transformation is invertible, the composition $$phi(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))} ight| le |z|.

Now calling $$z_2=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)} ight| le left|frac{z_1-z_2}{1-overline{z_1}z_2} ight|.

To prove the second part of the theorem, we just let $$z_2 tend to $$z_1.

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.

References

* Jurgen Jost, "Compact Riemann Surfaces" (2002), Springer-Verlag, New York. ISBN 3-540-43299-X "(See Section 2.3)"
*cite book | author = S. Dineen | title = The Schwarz Lemma | publisher = Oxford | year = 1989 | id=ISBN 0-19-853571-6