Fall 2016 - Problem 7

Solution.

1. Let $\delta = \frac{1}{2}d\p{z_0, \D^\comp}$ so that $\cl{B\p{z_0, \delta}} \subseteq \D$. Then by the mean value property and Cauchy-Schwarz,

   $$\abs{f\p{z_0}} \leq \frac{1}{\pi\delta^2} \iint_{B\p{z_0,\delta}} \abs{f\p{z}} \,\diff{A}\p{z} \leq \frac{m\p{\D}^{1/2}}{\pi\delta^2} \norm{f}_{L^2},$$

   so $L_{z_0}$ is a bounded linear operator.

2. Recall that

   $$f\p{z_0} = \frac{1}{\pi} \int_\D \frac{f\p{z}}{\p{1 - \conj{z}z_0}^2} \,\diff{A}\p{z}.$$

   From this, we immediately see that $g\p{z} = \frac{1}{\p{1 - z\conj{z_0}}^2}$ represents $L_{z_0}$: clearly $g$ is holomorphic, and it is bounded on $\D$ since its only pole is at $\frac{z_0}{\abs{z_0}^2}$ which is outside $\D$. To complete the proof, it remains to prove this formula.

   Write

   $$f\p{z} = \sum_{n=0}^\infty a_nz^n \quad\text{and}\quad \frac{1}{\p{1 - \conj{z}z_0}^2} = \sum_{n=0}^\infty \p{n + 1}\p{\conj{z}z_0}^n.$$

   Observe also that if $n \neq m$,

   $$\int_\D z^n \conj{z}^m \,\diff{A}\p{z} = \int_0^1 \int_0^{2\pi} r^{n+m+1} e^{in\theta} e^{-im\theta} \diff\theta \,\diff{r} = 0.$$

   If $n = m$, then

   $$\int_\D z^n \conj{z}^m \,\diff{A}\p{z} = \frac{\pi}{n + 1}.$$

   Hence, because all series converge locally uniformly, we have

   $$\begin{aligned} \frac{1}{\pi} \int_\D \frac{f\p{z}}{1 - \conj{z}z_0} \,\diff{A}\p{z} &= \frac{1}{\pi} \int_\D \p{\sum_{n=0}^\infty a_nz^n} \p{\sum_{m=0}^\infty \p{m+1}\p{\conj{z}z_0}^m} \,\diff{A}\p{z} \\ &= \frac{1}{\pi} \sum_{n=0}^\infty \sum_{m=0}^\infty a_nz_0^m\p{m + 1} \int_\D z^n\conj{z}^m \,\diff{A}\p{z} \\ &= \frac{1}{\pi} \sum_{n=0}^\infty a_nz_0^n\p{n + 1} \frac{\pi}{n + 1} \\ &= \sum_{n=0}^\infty a_n z_0^n \\ &= f\p{z_0}, \end{aligned}$$

   which completes the proof.