Area theorem (conformal mapping)

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In the mathematical theory of conformal mappings, the area theorem gives an inequality satisfied by the power series coefficients of certain conformal mappings. The theorem is called by that name, not because of its implications, but rather because the proof uses the notion of area.

Statement[edit]

Suppose that f is analytic and injective in the punctured open unit disk \mathbb D\setminus\{0\} and has the power series representation


f(z)= \frac 1z + \sum_{n=0}^\infty a_n z^n,\qquad z\in \mathbb D\setminus\{0\},

then the coefficients a_n satisfy


\sum_{n=0}^\infty n|a_n|^2\le 1.

Proof[edit]

The idea of the proof is to look at the area uncovered by the image of f. Define for r\in(0,1)

\gamma_r(\theta):=f(r\,e^{-i\theta}),\qquad \theta\in[0,2\pi].

Then \gamma_r is a simple closed curve in the plane. Let D_r denote the unique bounded connected component of \mathbb C\setminus\gamma[0,2\pi]. The existence and uniqueness of D_r follows from Jordan's curve theorem.

If D is a domain in the plane whose boundary is a smooth simple closed curve \gamma, then


\mathrm{area}(D)=\int_\gamma x\,dy=-\int_\gamma y\,dx\,,

provided that \gamma is positively oriented around D. This follows easily, for example, from Green's theorem. As we will soon see, \gamma_r is positively oriented around D_r (and that is the reason for the minus sign in the definition of \gamma_r). After applying the chain rule and the formula for \gamma_r, the above expressions for the area give


\mathrm{area}(D_r)= \int_0^{2\pi} \Re\bigl(f(r e^{-i\theta})\bigr)\,\Im\bigl(-i\,r\,e^{-i\theta}\,f'(r e^{-i\theta})\bigr)\,d\theta = -\int_0^{2\pi} \Im\bigl(f(r e^{-i\theta})\bigr)\,\Re\bigl(-i\,r\,e^{-i\theta}\,f'(r e^{-i\theta})\bigr)d\theta.

Therefore, the area of D_r also equals to the average of the two expressions on the right hand side. After simplification, this yields


\mathrm{area}(D_r) = -\frac 12\, \Re\int_0^{2\pi}f(r\,e^{-i\theta})\,\overline{r\,e^{-i\theta}\,f'(r\,e^{-i\theta})}\,d\theta,

where \overline z denotes complex conjugation. We set a_{-1}=1 and use the power series expansion for f, to get


\mathrm{area}(D_r) = -\frac 12\, \Re\int_0^{2\pi} \sum_{n=-1}^\infty
\sum_{m=-1}^\infty
m\,r^{n+m}\,a_n\,\overline{a_m}\,e^{i\,(m-n)\,\theta}\,d\theta\,.

(Since \int_0^{2\pi} \sum_{n=-1}^\infty\sum_{m=-1}^\infty m\,r^{n+m}\,|a_n|\,|a_m|\,d\theta<\infty\,,
the rearrangement of the terms is justified.) Now note that \int_0^{2\pi} e^{i\,(m-n)\,\theta}\,d\theta is 2\pi if n= m and is zero otherwise. Therefore, we get


\mathrm{area}(D_r)= -\pi\sum_{n=-1}^\infty n\,r^{2n}\,|a_n|^2.

The area of D_r is clearly positive. Therefore, the right hand side is positive. Since a_{-1}=1, by letting r\to1, the theorem now follows.

It only remains to justify the claim that \gamma_r is positively oriented around D_r. Let r' satisfy r<r'<1, and set z_0=f(r'), say. For very small s>0, we may write the expression for the winding number of \gamma_s around z_0, and verify that it is equal to 1. Since, \gamma_t does not pass through z_0 when t\ne r' (as f is injective), the invariance of the winding number under homotopy in the complement of z_0 implies that the winding number of \gamma_r around z_0 is also 1. This implies that z_0\in D_r and that \gamma_r is positively oriented around D_r, as required.

Uses[edit]

The inequalities satisfied by power series coefficients of conformal mappings were of considerable interest to mathematicians prior to the solution of the Bieberbach conjecture. The area theorem is a central tool in this context. Moreover, the area theorem is often used in order to prove the Koebe 1/4 theorem, which is very useful in the study of the geometry of conformal mappings.

References[edit]