Gronwall's Area Theorem
   HOME

TheInfoList



OR:

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 f:\mathbf\to\mathbb from the unit disk \mathbf onto a subset of the complex plane contains the disk whose center is f(0) and whose radius is , f'(0), /4.
The theorem is named after
Paul Koebe Paul Koebe (15 February 1882 – 6 August 1945) was a 20th-century German mathematician. His work dealt exclusively with the complex numbers, his most important results being on the uniformization of Riemann surfaces in a series of four papers in ...
, who conjectured the result in 1907. The theorem was proven by
Ludwig Bieberbach Ludwig Georg Elias Moses Bieberbach (; 4 December 1886 – 1 September 1982) was a German mathematician and Nazi. Biography Born in Goddelau, near Darmstadt, he studied at Heidelberg and under Felix Klein at Göttingen, receiving his doctorat ...
in 1916. The example of the Koebe function shows that the constant 1/4 in the theorem cannot be improved (increased). A related result is the
Schwarz lemma In mathematics, the Schwarz lemma, named after Hermann Amandus Schwarz, is a result in complex analysis about holomorphic functions from the open unit disk to itself. The lemma is less celebrated than deeper theorems, such as the Riemann mapp ...
, and a notion related to both is
conformal radius In mathematics, the conformal radius is a way to measure the size of a simply connected planar domain ''D'' viewed from a point ''z'' in it. As opposed to notions using Euclidean distance (say, the radius of the largest inscribed disk with center '' ...
.


Grönwall's area theorem

Suppose that :g(z) = z +b_1z^ + b_2 z^ + \cdots is univalent in , z, >1. Then :\sum_ n, b_n, ^2 \le 1. In fact, if r > 1, the complement of the image of the disk , z, >r is a bounded domain X(r). Its area is given by : \int_ dx\,dy = \int_\overline\,dz = \int_\overline\,dg=\pi r^2 - \pi\sum n, b_n, ^2 r^. Since the area is positive, the result follows by letting r decrease to 1. The above proof shows equality holds if and only if the complement of the image of g has zero area, i.e. Lebesgue measure zero. This result was proved in 1914 by the Swedish mathematician
Thomas Hakon Grönwall Thomas Hakon Grönwall or Thomas Hakon Gronwall (January 16, 1877 in Dylta bruk, Sweden – May 9, 1932 in New York City, New York) was a Swedish mathematician. He studied at the University College of Stockholm and Uppsala University and comple ...
.


Koebe function

The Koebe function is defined by :f(z)=\frac=\sum_^\infty n z^n Application of the theorem to this function shows that the constant 1/4 in the theorem cannot be improved, as the image domain f(\mathbf) does not contain the point z=-1/4 and so cannot contain any disk centred at 0 with radius larger than 1/4. The rotated Koebe function is :f_\alpha(z)=\frac=\sum_^\infty n\alpha^ z^n with \alpha a complex number of absolute value 1. The Koebe function and its rotations are '' schlicht'': that is, univalent (analytic and one-to-one) and satisfying f(0)=0 and f'(0)=1.


Bieberbach's coefficient inequality for univalent functions

Let : g(z) = z + a_2z^2 + a_3 z^3 + \cdots be univalent in , z, <1. Then :, a_2, \le 2. This follows by applying Gronwall's area theorem to the odd univalent function : g(z^)^= z - a_2 z^ + \cdots. Equality holds if and only if g is a rotated Koebe function. This result was proved by
Ludwig Bieberbach Ludwig Georg Elias Moses Bieberbach (; 4 December 1886 – 1 September 1982) was a German mathematician and Nazi. Biography Born in Goddelau, near Darmstadt, he studied at Heidelberg and under Felix Klein at Göttingen, receiving his doctorat ...
in 1916 and provided the basis for his celebrated conjecture that , a_n, \leq n, proved in 1985 by
Louis de Branges Louis de Branges de Bourcia (born August 21, 1932) is a French-American mathematician. He is the Edward C. Elliott Distinguished Professor of Mathematics at Purdue University in West Lafayette, Indiana. He is best known for proving the long-stan ...
.


Proof of quarter theorem

Applying an affine map, it can be assumed that :f(0)=0,\,\,\, f^\prime(0)=1, so that : f(z) = z + a_2 z^2 + \cdots . If w is not in f(\mathbf), then :h(z)= = z +(a_2+w^) z^2 + \cdots is univalent in , z, <1. Applying the coefficient inequality to f and h gives : , w, ^ \le , a_2, + , a_2 + w^, \le 4, so that : , w, \ge .


Koebe distortion theorem

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. Let f(z) be a univalent function on , z, <1 normalized so that f(0)=0 and f'(0)=1 and let r=, z, . Then :\le , f(z), \le : \le , f^\prime(z), \le : \le \left, z\ \le with equality if and only if f is a Koebe function : f(z) =.


Notes


References

* * * * * * * *{{cite book, last=Rudin , first=Walter , authorlink=Walter Rudin , year=1987 , title=Real and Complex Analysis , series=Series in Higher Mathematics , publisher=McGraw-Hill , edition=3 , isbn=0-07-054234-1 , mr=924157


External links

* Koebe 1/4 theorem a
PlanetMath
Theorems in complex analysis