Koebe quarter theorem

In complex analysis, a branch of mathematics, the Koebe 1/4 theorem states that the image of an injective analytic function f:\mathbb D\to\mathbb C from the unit disk \mathbb D 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, who conjectured the result in 1907. The theorem was proven by Ludwig Bieberbach in 1914. The Koebe function ƒ(z) = z/(1 − z)2 shows that the constant 1/4 in the theorem cannot be improved.

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

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Gronwall's area theorem

Suppose that

g(z) = z %2Bb_1z^{-1} %2B b_2 z^{-2} %2B \cdots

is univalent in  |z| > 1. Then

\sum_{n\ge 1} 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_{X(r)} dxdy = {1\over 2i} \int_{\partial X(r)}\, \overline{z}\,dz = -{1\over 2i}\int_{|z|=r} \,\overline{g}\,dg={1\over 2\pi r^2} -{1\over 2\pi}\sum n |b_n|^2 r^{2n}.

Since the area is positive, the result follows by letting r decrease to 1. The above proof shows quality 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 Gronwall.

Bieberbach's coefficient inequality for univalent functions

Let

 g(z) = z %2B a_2z^2 %2B a_3 z^3 %2B \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^2)^{-1/2}= z^{-1} -{1\over 2} a_2 z %2B \cdots.

Equality holds if and only if g is a rotation of the Koebe function.

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

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 %2B a_2 z^2 %2B \cdots .

If w is not in f(\mathbb D), then

h(z)={wh(z)\over w-h(z)} = z %2B(a_2%2Bw^{-1}) z^2 %2B \cdots

is univalent in  |z|<1.

Applying the coefficient inequality to f and h gives

 |w|^{-1} \le |a_2| %2B |a_2 %2B w^{-1}|\le 4,

so that

 |w|\ge {1\over 4}.

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.[1]

Let f(z) be a univalent function on |z| < 1 normalized so that f(0) = 0 and f'(0) = 0 and let r = |z|. Then

{r \over (1%2Br)^2}\le |f(z)|\le {r\over (1-r)^2}
{1-r\over (1%2Br)^3} \le |f^\prime(z)| \le {1%2Br\over (1-r)^3}
{1-r\over 1%2Br} \le \left|z{f^\prime(z)\over f(z)}\right| \le {1%2Br\over 1-r}

with equality if and only if f is a Koebe function

 f(z) ={z\over(1-e^{i\theta}z)^2}.

Notes

  1. ^ Pommerenke 1975, p. 21-22

References