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Fall 2015 - Problem 9

normal families

Let $\set{f_j}_j$ be a sequence of entire functions such that, writing $z = x + iy$ , we have

\iint_\C \abs{f_j\p{z}}^2 e^{-\abs{z}^2} \,\diff{x} \,\diff{y} \leq C, \quad j = 1, 2, \ldots,

for some constant $C > 0$ . Show that there exists a subsequence $\set{f_{j_k}}_k$ and an entire function $f$ such that we have

\iint_\C \abs{f_{j_k}\p{z} - f\p{z}}^2 e^{-2\abs{z}^2} \,\diff{x} \,\diff{y} \to 0, \quad k \to \infty.

Solution.

Let $R > 0$ and $K = \cl{B\p{0, R}}$ . Then for $z \in \cl{B\p{0, 1+R}}$ , we get $e^{\abs{z}^2} \leq e^{\p{1+R}^2}$ and so

\begin{aligned} \iint_{\cl{B\p{0,1+R}}} \abs{f_j\p{z}}^2 \,\diff{x} \,\diff{y} &= \iint_{\cl{B\p{0,1+R}}} \abs{f_j\p{z}}^2 e^{-\abs{z}^2} e^{\abs{z}^2} \,\diff{y} \\ &\leq Ce^{R^2}. \end{aligned}

Notice that for any $z \in K$ , we have $\cl{B\p{z, 1}} \subseteq \cl{B\p{0, 1+R}}$ , so by the mean value property,

\begin{aligned} \abs{f_j\p{z}} &= \frac{1}{\pi} \abs{\iint_{B\p{z,1}} f_j\p{x + iy} \,\diff{x} \,\diff{y}} \\ &\leq \frac{1}{\pi} \iint_{\cl{B\p{0,1+R}}} \abs{f_j\p{x + iy}} \,\diff{x} \,\diff{y} \\ &\leq \frac{1}{\pi} \p{\iint_{\cl{B\p{0,1+R}}} \abs{f_j\p{x + iy}}^2 \,\diff{x} \,\diff{y}}^{1/2} m\p{B\p{z, 1}}^{1/2} \\ &\leq \frac{m\p{\D}^{1/2}}{\pi} Ce^{R^2}. \end{aligned}

Thus, $\set{f_j}_j$ is a normal family, so there exists a subsequence $\set{f_{j_k}}_k$ which converges locally uniformly to an entire function $f$ . By Fatou's lemma, $f$ satisfies

\iint_\C \abs{f\p{z}}^2 e^{-\abs{z}^2} \,\diff{x} \,\diff{y} \leq \liminf_{k\to\infty} \iint_\C \abs{f_{j_k}\p{z}}^2 e^{-\abs{z}^2} \,\diff{x} \,\diff{y} \leq C.

To complete the proof, let $\epsilon > 0$ . If $R > 0$ ,

\begin{aligned} \iint_{B\p{0,R}^\comp} \abs{f_{j_k}\p{z} - f\p{z}}^2 e^{-2\abs{z}^2} \,\diff{z} &\leq e^{-R^2} \iint_{B\p{0,R}^\comp} \abs{f_{j_k}\p{z} - f\p{z}}^2 e^{-\abs{z}^2} \,\diff{z} \\ &\leq 2Ce^{-R^2}, \end{aligned}

so these go to $0$ as $R \to \infty$ , uniformly in $k$ . Let $R$ be large enough so that

\iint_{B\p{0,R}^\comp} \abs{f_{j_k}\p{z} - f\p{z}}^2 e^{-2\abs{z}^2} \,\diff{z} < \epsilon.

Thus, because $\cl{B\p{0, R}}$ is compact, there exists $N \in \N$ so that $\abs{f_{j_k}\p{z} - f\p{z}} < \epsilon$ for all $z \in \cl{B\p{0, R}}$ , for all $k \geq N$ . Hence,

\begin{aligned} \iint_\C \abs{f_{j_k}\p{z} - f\p{z}}^2 e^{-2\abs{z}^2} \,\diff{x} \,\diff{y} &= \iint_{B\p{0,R}} \abs{f_{j_k}\p{z} - f\p{z}}^2 e^{-2\abs{z}^2} \,\diff{x} \,\diff{y} + \iint_{B\p{0,R}^\comp} \abs{f_{j_k}\p{z} - f\p{z}}^2 e^{-2\abs{z}^2} \,\diff{x} \,\diff{y} \\ &\leq \epsilon^2 \int_0^R \int_0^{2\pi} re^{-2r^2} \,\theta \,\diff{r} + \epsilon \\ &= \frac{\pi\epsilon^2\p{1 - e^{-2R^2}}}{4} + \epsilon \\ &\leq \frac{\pi\epsilon^2}{4} + \epsilon, \end{aligned}

which completes the proof.