# 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.

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