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    Area formula for a univalent power series on the disk

    P-BKS80-19

    Area
    Prelim
    Seen at
    BERKELEY
    Years
    1980
    Status
    draft

    Let \[ f(z)=\sum_{n=0}^{\infty}c_nz^n \] be analytic and one-to-one on the unit disk \(\mathbb D\), mapping it onto a domain \(G\) of area \(A\). Prove that \[ A=\pi\sum_{n=1}^{\infty}n|c_n|^2. \]

    Solution

    For \(0<r<1\), put \[ D_r\coloneqq\{z\in\CC:\abs{z}<r\}, \qquad G_r\coloneqq f(D_r), \qquad A_r\coloneqq\operatorname{area}(G_r). \]

    1. For every \(0<r<1\),\[ A_r=\int_{D_r}\abs{f'(z)}^2\,dA(z). \]

    An injective holomorphic function has nonvanishing derivative, so the restriction \[ f:D_r\longrightarrow G_r \] is a real \(C^1\) diffeomorphism. For a holomorphic map, the determinant of the real Jacobian is \[ J_f(z)=\abs{f'(z)}^2. \] The change-of-variables formula therefore gives the displayed identity.

    2. For every \(0<r<1\),\[ A_r = \pi\sum_{n=1}^{\infty}n\abs{c_n}^2r^{2n}. \]

    Differentiating the power series gives \[ f'(z)=\sum_{n=1}^{\infty}n c_n z^{n-1}. \] For each fixed \(0\le\rho<1\), orthogonality of the functions \(e^{i(n-1)\theta}\) on \([0,2\pi]\) gives \[ \int_0^{2\pi} \abs{f'(\rho e^{i\theta})}^2\,d\theta = 2\pi\sum_{n=1}^{\infty} n^2\abs{c_n}^2\rho^{2n-2}. \] Hence step 1 and polar coordinates give \[ \begin{aligned} A_r &= \int_0^r\int_0^{2\pi} \abs{f'(\rho e^{i\theta})}^2 \rho\,d\theta\,d\rho\\ &= 2\pi\sum_{n=1}^{\infty} n^2\abs{c_n}^2 \int_0^r\rho^{2n-1}\,d\rho\\ &= \pi\sum_{n=1}^{\infty} n\abs{c_n}^2r^{2n}. \end{aligned} \] The interchange with the radial integral is valid because all terms in the resulting series are nonnegative.

    3. As \(r\uparrow1\),\[ A_r\longrightarrow A \]and\[ \sum_{n=1}^{\infty}n\abs{c_n}^2r^{2n} \longrightarrow \sum_{n=1}^{\infty}n\abs{c_n}^2. \]

    The disks \(D_r\) increase to the unit disk, so the sets \(G_r=f(D_r)\) increase to \(G\). Continuity from below of planar area gives \(A_r\to\operatorname{area}(G)=A\).

    For each \(n\), the term \[ n\abs{c_n}^2r^{2n} \] increases to \(n\abs{c_n}^2\) as \(r\uparrow1\). Monotone convergence for the counting measure on the positive integers therefore gives the second limit.

    4. Therefore\[ \boxed{ A=\pi\sum_{n=1}^{\infty}n\abs{c_n}^2 }. \]

    Take \(r\uparrow1\) in the identity from step 2 and use both limits in step 3.

    5. Q.E.D.

    Step 4 is the required area formula.