Grönwall area theorem

Remarks

For $r > 1$, the image of the circle $\abs{z} = r$ is a Jordan curve, and the bounded region $E_r$ it encloses is the complement of $g(\set{\abs{z} > r})$. Green's formula gives the area of $E_r$ as $\frac{1}{2i}\oint \bar{w}\, dw$ over this curve, which works out to $\pi\bigl(r^2 - \sum_{n \ge 1} n \abs{b_n}^2 r^{-2n}\bigr)$. This area is nonnegative, and letting $r \to 1$ gives the inequality. Equality holds exactly when the complement of $g(\set{\abs{z} > 1})$ has area $0$. Example: $g(z) = z + 1/z$ maps $\set{\abs{z} > 1}$ onto $\C \setminus [-2, 2]$, and $b_1 = 1$ gives equality. In particular $\abs{b_1} \le 1$. Applied to $g(z) = 1/\sqrt{f(1/z^2)}$ for $f$ in the class $S$ of normalised univalent functions, it gives Bieberbach's inequality $\abs{a_2} \le 2$.

Going deeper

For a student

The area theorem is the starting point of the coefficient theory of univalent functions. Through the substitution $f \mapsto 1/\sqrt{f(1/z^2)}$, it gives $\abs{a_2} \le 2$ for $f \in S$, hence the Koebe quarter theorem and the distortion theorems. The Grunsky inequalities on the Grunsky matrix refine it to a statement about all the coefficients of $\log\frac{g(z) - g(\zeta)}{z - \zeta}$. The equality case shows that the bound $\abs{b_1} \le 1$ is attained exactly by maps onto the complement of a segment.