Spring 2019 - Problem 12

Solution.

1. Suppose $\set{f_n}_n \subseteq \mathcal{H}$ converges to $f \in L^2\p{\mu}$ and let $K = \cl{B\p{0, R}}$. By the mean value property, for any $z \in K$, we have

   $$\begin{aligned} \abs{f_n\p{z}} &\leq \frac{1}{\pi} \int_{B\p{z, 1}} \abs{f_n\p{w}} \,\diff\lambda\p{w} \\ &\leq \frac{1}{\pi} \lambda\p{B\p{z, 1}}^{1/2} \p{\int_{B\p{z,1}} \abs{f_n\p{w}}^2 \,\diff\mu\p{w}}^{1/2} \\ &\leq \frac{\lambda\p{\D}^{1/2}}{\pi} e^{\p{R+1}^2} \sup_n\,\norm{f_n}_{L^2\p{\mu}} \\ &< \infty. \end{aligned}$$

   In other words, $\set{f_n}_n$ is a normal family, so it admits a subsequence which converges locally uniformly. Hence, $f$ is a locally uniform limit of entire functions, so $f$ itself is entire, i.e., $f \in \mathcal{H}$, so $\mathcal{H}$ is closed.

2. We follow the hint. Notice that

   $$\begin{aligned} \norm{e_n}_{L^2\p{\mu}}^2 &= \frac{1}{\pi n!} \int_\C z^{2n} e^{-\abs{z}^2} \,\diff\lambda\p{z} \\ &= \frac{1}{\pi n!} \int_0^\infty \int_0^{2\pi} r^{2n+1} e^{-r^2} \,\diff{\theta} \,\diff{r} \\ &= \frac{2}{n!} \int_0^\infty r^{2n} \cdot re^{-r^2} \,\diff{r} \\ &= \left. -\frac{r^{2n} e^{-r^2}}{n!} \right\rvert_0^\infty + \frac{2n}{n!} \int_\C r^{2n-1} e^{-r^2} \,\diff{r} \\ &= \norm{e_{n-1}}_{L^2\p{\mu}}^2. \end{aligned}$$

   When $n = 0$,

   $$\norm{e_0}_{L^2\p{\mu}}^2 = 2 \int_0^\infty re^{-r^2} \,\diff{r} = \left. -e^{-r^2} \right\rvert_0^\infty = 1,$$

   so by induction, $\norm{e_n}_{L^2\p{\mu}} = 1$ for all $n \geq 0$. Continuing, if $n \neq m$, we h ave

   $$\begin{aligned} \inner{e_n, e_m}_{L^2\p{\mu}} &= \frac{1}{\pi\p{n! \cdot m!}^{1/2}} \int_\C z^n \p{\conj{z}}^m e^{-\abs{z}^2} \,\diff\lambda\p{z} \\ &= \frac{1}{\pi\p{n! \cdot m!}^{1/2}} \int_0^\infty r^{n+m+1} e^{-r^2} \int_0^{2\pi} e^{i\p{n-m}\theta} \,\diff\theta \,\diff{r} \\ &= 0, \end{aligned}$$

   since the inner integral vanishes by orthogonality. Thus, $\set{e_n}_n$ is indeed an orthonormal system on $\mathcal{H}$. To show that it is a basis, suppose $\inner{f, e_n}_{L^2\p{\mu}} = 0$ for all $n \geq 0$. Then if we write $f\p{z} = \sum_{n=0}^\infty a_nz^n$, we get

   $$\begin{aligned} 0 &= \inner{f, e_n}_{L^2\p{\mu}} \\ &= \frac{1}{\p{\pi n!}^{1/2}} \int_\C \sum_{m=0}^\infty a_m z^m \p{\conj{z}}^n \,\diff\mu\p{z} \\ &= \frac{1}{\p{\pi n!}^{1/2}} \int_0^\infty \int_0^{2\pi} \sum_{m=0}^\infty a_m r^{n+m+1} e^{i\p{m-n}\theta} e^{-r^2} \,\diff{\theta} \,\diff{r} \\ &= \frac{1}{\p{\pi n!}^{1/2}} \int_0^\infty \sum_{m=0}^\infty a_m r^{n+m+1} e^{-r^2} \int_0^{2\pi} e^{i\p{m-n}\theta} \,\diff{\theta} \,\diff{r} && \p{\text{absolute convergence}} \\ &= \frac{2\pi a_n}{\p{\pi n!}^{1/2}} \int_0^\infty r^{2n+1} e^{-r^2} \,\diff{r} \\ &= \p{\pi n!}^{1/2} a_n. \end{aligned}$$

   Thus, if $f$ is orthogonal to all of the $e_n$, then it is $0$, so $\set{e_n}_n$ is a basis for $\mathcal{H}$. By Fubini-Tonelli, we have

   $$\begin{aligned} \sum_{n=0}^\infty \int_\C \abs{f\p{w}} \frac{\abs{z\conj{w}}^n}{n!} \,\diff\mu\p{w} &= \sum_{n=0}^\infty \int_\C \abs{f\p{w}} \frac{\abs{z\conj{w}}^n}{n!} \,\diff\mu\p{w} \\ &= \int_\C \abs{f\p{w}} \sum_{n=0}^\infty \frac{\abs{z\conj{w}}^n}{n!} \,\diff\mu\p{w} \\ &= \int_\C \abs{f\p{w}} e^{\abs{z\conj{w}}} \,\diff\mu\p{w} \\ &\leq \norm{f}_{L^2\p{\mu}} \p{\int_\C e^{\abs{z\conj{w}}-\abs{w}^2} \,\diff\lambda\p{w}}^{1/2} \\ &< \infty. \end{aligned}$$

   Hence, we may apply Fubini's theorem in the following to interchange the sum and integral:

   $$\begin{aligned} f\p{z} &= \sum_{n=0}^\infty \inner{f, e_n} e_n\p{z} \\ &= \sum_{n=0}^\infty \frac{z^n}{\pi n!} \int_\C f\p{w} \p{\conj{w}}^n \,\diff\mu\p{w} \\ &= \frac{1}{\pi} \sum_{n=0}^\infty \int_\C f\p{w} \frac{\p{z\conj{w}}^n}{n!} \,\diff\mu\p{w} \\ &= \frac{1}{\pi} \int_\C f\p{w} \sum_{n=0}^\infty \frac{\p{z\conj{w}}^n}{n!} \,\diff\mu\p{w} \\ &= \frac{1}{\pi} \int_\C f\p{w} e^{z\conj{w}} \,\diff\mu\p{w}, \end{aligned}$$

   which was what we wanted to show.