Suppose that \(f\) is holomorphic on the punctured disk \(0 < |z| < 2\).
Prove that if there is a real constant \(C\) such that \(|f(z)| \leq C\), then \[ \int_{|z|<1} |f'(z)|^2\,dz < \infty. \]
What happens when \(|f|\) is unbounded?
Solution
The integral is taken with respect to area. Write \(f(z)=\sum_{n\in\ZZ}a_nz^n\), the Laurent series on \(0<\abs z<2\).
1. For \(0<r<1\), \(\int_{r<\abs z<1}\abs{f'}^2\,dA=\pi\sum_{n\ne0}n\abs{a_n}^2(1-r^{2n})\).
In polar coordinates, orthogonality of \(e^{i(n-1)\theta}\) on each circle gives \(\int_0^{2\pi}\abs{f'(\rho e^{i\theta})}^2\,d\theta=2\pi\sum_nn^2\abs{a_n}^2\rho^{2n-2}\), and \(\int_r^1 2\pi n^2\rho^{2n-1}\,d\rho=\pi n(1-r^{2n})\) for \(n\ne0\). Each term \(n(1-r^{2n})\) is nonnegative, for \(n<0\) as well.
2. (A) If \(\abs f\le C\), the integral is finite.
By the removable singularity theorem \(a_n=0\) for \(n<0\). Then the terms of step 1 increase to \(\pi n\abs{a_n}^2\) as \(r\to0\), and \(\pi\sum_{n>0}n\abs{a_n}^2\le\pi\sum_nn^2\abs{a_n}^2=\int_0^{2\pi}\abs{f'(e^{i\theta})}^2\,\frac{d\theta}2<\infty\), since \(f'\) is continuous on \(\abs z=1\).
3. (B) If \(\abs f\) is unbounded near \(0\), the integral is \(\boxed{\infty}\).
The singularity is then not removable, so \(a_n\ne0\) for some \(n<0\), and its term in step 1 is \(\abs n\abs{a_n}^2\pi(r^{-2\abs n}-1)\to\infty\) as \(r\to0\). For example \(f(z)=1/z\) gives \(\int_{\abs z<1}\abs z^{-4}\,dA=2\pi\int_0^1r^{-3}\,dr=\infty\).
4. Q.E.D.
Steps 2 and 3 answer parts (A) and (B).