# Least area of the image of a simply connected domain under holomorphic maps

I found the following statement in the book "The Kernel Function and Conformal mapping" at page 23-24 by Stefan Bergman:

Let $\Omega$ be a bounded, simply connected domain in $\mathbb{C}.$ Let $p\in\Omega.$ Now consider the family $\mathcal{F}:=\{f\in\mathcal{O}(\Omega,\mathbb{C}):f(p)=0,\,f'(p)=1\}$. Given $f\in\mathcal{F}$ let $D_f$ denote the range of $f.$ Then $D_f$ has least area in the case in which $D_f$ is a disc with center at the origin.

I'm not able to see why this is true. Please help to understand the above.

## 1 Answer

First of all, let's be clear about what exactly Bergman claims. It is well known2 that the minimum area is attained in the case in which $D$ is a circle with center at the origin.

Explanation: Bergman considers the range to be a Riemann surface, which can be multi-sheeted over the plane. In plain/plane terms, he counts the area of the range with multiplicities: a part covered twice counts twice, etc. The area of range with multiplicities is the integral of Jacobian determinant over the domain -- this can be taken as definition.

As Daniel Fischer pointed out to me, we can assume that the domain is itself a disk centered at $0$ by pre-composing $f$ with a Riemann map. I will normalize the Riemann map so that its derivative at $0$ is $1$; thus, the disk is not necessarily of unit radius. Let $R$ be its radius.

So, now we have $f:\{z: |z|<R\}\to D_f$ with $f'(0)=1$. Write $f(z)=\sum_{n=0}^\infty a_nz^n$. The derivative is $f'(z)=\sum_{n=1}^\infty na_nz^{n-1}$. Viewed as a map on $\mathbb R^2$, $f$ has the Jacobian determinant $|f'(z)|^2$. The area of $D_f$ is the integral of Jacobian over $\{|z|<R\}$. Note that $$|f'(z)|^2=f'(z)\overline{f'(z)}= \sum_{m=1}^\infty\sum_{n=1}^\infty mn a_m\overline{a_n}z^{m-1} \bar z^{n-1}$$ Integrating in polar coordinates, we find that when $m\ne n$, the term $z^{m-1} \bar z^{n-1}$ integrates to zero over every circle, because it is a multiple of $e^{i(m-n)\theta}$. When $m=n$, the term is $n^2|a_n|^2|z|^{2n-2}$, which is nonnegative. Conclusion: the area iz minimized by setting all coefficients with $n>1$ to $0$.

For completemess, this is what you get after integration: $$\operatorname{area}(D_f)= \pi \sum_{n=1}^\infty n |a_n|^2 R^{2n}$$

One may ask the same question without counting multiplicities: that is, minimize the area of $D_f$ considered as a subset of $\mathbb C$. The answer remains the same: the minimum is attained when $D_f$ is a disk. However, the proof I know is substantially more involved: it relies on the fact that capacity decreases under symmetrization. A story for another time...

• There's a small problem in that the domain of $f$ is $\Omega$, not (necessarily) the unit disk. That can be treated by composing with a biholomorphic mapping $\varphi \colon \mathbb{D}\to \Omega$ with $\varphi(0) = p$ and $\varphi'(0) > 0$. There's another problem, which I don't see how to treat, namely that it is not given that $f$ is conformal, so how do we treat the case of a non-injective $f$? And does the conclusion even hold then? – Daniel Fischer Dec 23 '13 at 15:51
• @DanielFischer Actually, my answer (with your correction) still works for the claim that is actually made by Bergman (found on Google Books). – Post No Bulls Dec 23 '13 at 17:01
• Yes. Counting sheets, the proof indeed remains simple. Nice job. – Daniel Fischer Dec 23 '13 at 17:05
• Got your idea, Thanks. But to understand I'll take time having also in my mind the concerns (above) that Daniel had. Thanks again. – Abelvikram Dec 23 '13 at 17:22
• I think the above answer is revised having the points that Daniel made? – Abelvikram Dec 23 '13 at 17:27