- Assume \(f(z)=\sum_{n=0}^{\infty} c_{n} z^{n}\) converges in \(|z|<R\). Show that for \(r<R\),
\[ \frac{1}{2 \pi} \int_{0}^{2 \pi}\left|f\left(r e^{i \theta}\right)\right|^{2} d \theta=\sum_{n=0}^{\infty}\left|c_{n}\right|^{2} r^{2 n} \]
- Deduce Liouville’s theorem from (a).
Solution
Computing the LHS: \[ \int_{[0, 2\pi]} \abs{f(re^{i\theta})}^2 \dtheta &= \int_{[0, 2\pi]} f(re^{i\theta}) \bar{f(re^{i\theta}) } \dtheta \\ &= \int_{[0, 2\pi]} \sum_{k\geq 0} c_k r^k e^{ik\theta} \sum_{j\geq 0} \bar{c_j} r^j e^{-ij\theta} \dtheta \\ &= \int_{[0, 2\pi]} \sum_{k,j\geq 0} c_k\bar{c_j} r^{k+j} e^{i(k-j)\theta} \dtheta \\ &= \sum_{k,j\geq 0} c_k\bar{c_j} r^{k+j} \int_{[0, 2\pi]} e^{i(k-j)\theta} \dtheta \\ &= \sum_{k,j\geq 0} c_k\bar{c_j} r^{k+j} \chi_{i=j}\cdot 2\pi \\ &= \sum_{k\geq 0} c_k\bar{c_k} r^{2k} \cdot 2\pi \\ &= 2\pi \sum_{k\geq 0}\abs{c_k}^2 r^{2k} ,\] where we’ve used that the series converges uniformly in its radius of convergence to commute sums and integrals.
Now supposing \(\abs{f(z)}\leq M\) for all \(z\in \CC\), if \(f\) is entire then \(\sum_{k\geq 0} c_k z^k\) converges for all \(r\), so \[ \sum_{k\geq 0} \abs{c_k}^2 r^{2k} = {1\over 2\pi }\int_{[0, 2\pi]} \abs{f(re^{i\theta})}^2 \dtheta \leq {1\over 2\pi}\int_{[0, 2\pi]} M^2 \dtheta = M^2 .\] Thus for all \(r\), \[ \abs{c_0}^2 + \abs{c_1}^2 r^2 + \abs{c_2}^2 r^{4} + \cdots \leq M^2 ,\] and taking \(r\to\infty\) forces \(\abs{c_1}^2 = \abs{c_2}^2 = \cdots = 0\). So \(f(z) = c_0\) is constant.