Jump to content

Koebe quarter theorem

From Wikipedia, the free encyclopedia
(Redirected from Bieberbach's inequality)

In complex analysis, a branch of mathematics, the Koebe 1/4 theorem states the following:

Koebe Quarter Theorem. The image of an injective analytic function from the unit disk onto a subset of the complex plane contains the disk whose center is and whose radius is .

The theorem is named after Paul Koebe, who conjectured the result in 1907. The theorem was proven by Ludwig Bieberbach in 1916. The example of the Koebe function shows that the constant in the theorem cannot be improved (increased).

A related result is the Schwarz lemma, and a notion related to both is conformal radius.

Grönwall's area theorem

[edit]

Suppose that

is univalent in . Then

In fact, if , the complement of the image of the disk is a bounded domain . Its area is given by

Since the area is positive, the result follows by letting decrease to . The above proof shows equality holds if and only if the complement of the image of has zero area, i.e. Lebesgue measure zero.

This result was proved in 1914 by the Swedish mathematician Thomas Hakon Grönwall.

Koebe function

[edit]

The Koebe function is defined by

Application of the theorem to this function shows that the constant in the theorem cannot be improved, as the image domain does not contain the point and so cannot contain any disk centred at with radius larger than .

The rotated Koebe function is

with a complex number of absolute value . The Koebe function and its rotations are schlicht: that is, univalent (analytic and one-to-one) and satisfying and .

Bieberbach's coefficient inequality for univalent functions

[edit]

Let

be univalent in . Then

This follows by applying Gronwall's area theorem to the odd univalent function

Equality holds if and only if is a rotated Koebe function.

This result was proved by Ludwig Bieberbach in 1916 and provided the basis for his celebrated conjecture that , proved in 1985 by Louis de Branges.

Proof of quarter theorem

[edit]

Applying an affine map, it can be assumed that

so that

In particular, the coefficient inequality gives that . If is not in , then

is univalent in .

Applying the coefficient inequality to gives

so that

Koebe distortion theorem

[edit]

The Koebe distortion theorem gives a series of bounds for a univalent function and its derivative. It is a direct consequence of Bieberbach's inequality for the second coefficient and the Koebe quarter theorem.[1]

Let be a univalent function on normalized so that and and let . Then

with equality if and only if is a Koebe function

Notes

[edit]
  1. ^ Pommerenke 1975, pp. 21–22

References

[edit]
  • Bieberbach, Ludwig (1916), "Über die Koeffizienten derjenigen Potenzreihen, welche eine schlichte Abbildung des Einheitskreises vermitteln", S.-B. Preuss. Akad. Wiss.: 940–955
  • Carleson, L.; Gamelin, T. D. W. (1993), Complex dynamics, Universitext: Tracts in Mathematics, Springer-Verlag, pp. 1–2, ISBN 0-387-97942-5
  • Conway, John B. (1995), Functions of One Complex Variable II, Berlin, New York: Springer-Verlag, ISBN 978-0-387-94460-9
  • Duren, P. L. (1983), Univalent functions, Grundlehren der Mathematischen Wissenschaften, vol. 259, Springer-Verlag, ISBN 0-387-90795-5
  • Gronwall, T.H. (1914), "Some remarks on conformal representation", Annals of Mathematics, 16: 72–76, doi:10.2307/1968044
  • Nehari, Zeev (1952), Conformal mapping, Dover, pp. 248–249, ISBN 0-486-61137-X
  • Pommerenke, C. (1975), Univalent functions, with a chapter on quadratic differentials by Gerd Jensen, Studia Mathematica/Mathematische Lehrbücher, vol. 15, Vandenhoeck & Ruprecht
  • Rudin, Walter (1987). Real and Complex Analysis. Series in Higher Mathematics (3 ed.). McGraw-Hill. ISBN 0-07-054234-1. MR 0924157.
[edit]