A geometric proof of Koebe 1/4 theorem
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The famous Koebe one-quarter theorem gives a sharp bound on the size of the image of univalent functions locally. The standard proof of this theorem which can be found in most complex analysis books (e.g. Rudin1 chapter 14) relies on the area theorem and a rather unnatural conformal change of variables. I found another proof of the theorem from the book of Hubbard2 which is, in my opinion, more fascinating and geometric, relying on covering spaces and conformal moduli.
Let’s denote the open disk centered at $a \in \mathbb{C}$ of radius $r>0$ by $\mathbb{D}(a,r)$. The unit disk will be especially denoted by $\mathbb{D}$. We’d also denote the round annulus $\{ r < \vert z \vert < R \}$ by $\mathbb{A}(r,R)$.
Koebe 1/4 Theorem: Let $f: \mathbb{D} \to \mathbb{C}$ be a univalent map. If $f(0)=0$ and $\vert f’(0)\vert =1$, then the image $f(\mathbb{D})$ contains the disk $\mathbb{D}(0,\frac{1}{4})$.
There are a few comments I would like to make.
- The constant $\frac{1}{4}$ is sharp. This value is achieved by the Koebe function $ K(z)=\frac{z}{(z-1)^2}$ which conformally maps $\mathbb{D}$ onto the slit plane $\mathbb{C}\backslash (-\infty,-\frac{1}{4}]$. In fact, this function serves as the extremal function in many other results in geometric function theory.
- By affine change of variables, the result can easily be generalised as follows: the image of any univalent map $f$ on a disk $\mathbb{D}(a,r)$ must contain the disk $\mathbb{D}(f(a), \frac{r}{4}\vert f’(a)\vert)$.
Alright, let’s go straight to the proof. Let $f: \mathbb{D} \to \mathbb{C}$ be a univalent map such that $f(0)=0$ and $\vert f’(0)\vert =1$. Define the map $g(z) := (1+\delta)f(z)$ for some arbitrarily small $\delta>0$; clearly, $g(0) = 0$ and $\vert g’(0) \vert > 1$. Let’s assume for a contradiction that $g(\mathbb{D})$ does not contain some point $w$ in the circle $\{ \vert z \vert = \frac{1}{4}\}$.
As a rational map, the Koebe function $K(z)$ is a double branched covering map of the Riemann sphere $\mathbb{P}^1$ onto itself with critical points $\pm 1$ and critical values $K(1) = \infty$ and $K(-1)=-\frac{1}{4}$. The unique non-trivial deck transformation of this covering is $\frac{1}{z}$ since $K(\frac{1}{z}) \equiv K(z)$. By composition with some rotation, we can assume that $w=-\frac{1}{4}$, so that the image $g(\mathbb{D})$ of $g$ contains no critical values of $K$. We can then lift $g$ to two distinct meromorphic functions $g_0, g_{\infty} : \mathbb{D} \to \mathbb{P}$ such that $K \circ g_j = g$ and $g_j(0) = j$ for $j=\{0,\infty\}$. The two lifts are related by the equation $g_\infty = 1/g_0$ and their images are disjoint.
By Taylor series about $0$, we can deduce that for any $\epsilon >0$, for sufficiently small $r>0$, $g_0(\mathbb{D}(0,r))$ contains the disk $\mathbb{D}(0, r(1-\epsilon)\vert g’(0) \vert)$. Since $g_\infty = 1/g_0$, the image $g_\infty(\mathbb{D}(0,r))$ also contains the domain $\{ \vert z \vert > \frac{1}{r(1-\epsilon)\vert g’(0) \vert} \}$. In overall, we can conclude that the (disjoint) annuli $A_j := g_j(\mathbb{A}(r, 1))$, $j \in \{0,\infty\}$ are contained in the bigger annulus
\[A := \mathbb{A} \left( r(1-\epsilon)\vert g'(0) \vert, \frac{1}{r(1-\epsilon)\vert g'(0) \vert}\right).\]The annuli $A$, $A_0$ and $A_\infty$ are all homotopic to each other as each of them separates $0$ and $\infty$. At this point, we’d like to consider the family of curves joining the inner and outer boundary components of each of the annuli $A$, $A_0$ and $A_\infty$. (Look back at this post for the theory of extremal lengths.) We calculate the moduli of each of these annuli:
\[\text{mod}(A_0) = \text{mod}(A_\infty) = \frac{1}{2\pi} \log\frac{1}{r}, \quad \text{mod}(A) = \frac{1}{2\pi} \log\frac{1}{r^2(1-\epsilon)^2\vert g'(0) \vert^2}.\]By the series law, we then have the inequality $\text{mod}(A_0) + \text{mod}(A_\infty) \leq \text{mod}(A)$ which then gives us:
\[\frac{2}{2\pi} \log\frac{1}{r} \leq \frac{1}{2\pi} \log\frac{1}{r^2(1-\epsilon)^2\vert g'(0) \vert^2}.\]Upon simplification, we obtain the inequality $\vert g’(0) \vert \leq \frac{1}{1-\epsilon}$ for all small $\epsilon >0$. This implies $\vert g’(0) \vert \leq 1$, which is then a contradiction.
References
1: W. Rudin. Real and Complex Analysis. McGraw-Hill, Series in Higher Mathematics, 3rd edition, 1987.
2: J. H. Hubbard. Teichmüller Theory and Applications to Geometry, Topology, and Dynamics. Vol. 1. Matrix Editions, Ithaca, NY, 2nd edition, 2006.