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Fall 2014 - Problem 10

harmonic functions

Let us introduce a vector space $\mathcal{B}$ defined as follows,

\mathcal{B} = \set{\func{u}{\C}{\C},\ u \text{ is holomorphic and } \iint_\C \abs{u\p{x + iy}}^2 e^{-\p{x^2+y^2}} \,\diff{x} \,\diff{y} < \infty}.

Show that $\mathcal{B}$ becomes a complete space when equipped with the norm

\norm{u}^2 = \iint_\C \abs{u\p{x + iy}}^2 e^{-\p{x^2+y^2}} \,\diff{x} \,\diff{y}.

Solution.

Observe that $\mathcal{B}$ is the subspace of holomorphic functions in $L^2\p{\mu}$ , where

\mu\p{A} = \iint_A e^{-\p{x^2 + y^2}} \,\diff{x} \,\diff{y},

which is a finite measure on $\C$ . Note also that $0 \leq e^{-\p{x^2 + y^2}} \leq 1$ . Since $\mathcal{B}$ is a subspace of a complete space, we only need to show that it is closed in $L^2$ .

Suppose $\set{u_n}_n \subseteq \mathcal{B}$ is a sequence such that $u_n \to u$ in $L^2$ . If we show that $u_n$ converges locally uniformly, then we're done.

Let $R > 0$ and let $K = \cl{B\p{0, R}}$ . For each $z \in K$ , we have

u_n\p{z} = \frac{1}{2\pi i} \int_{\partial B\p{z,r}} \frac{u_n\p{\zeta}}{\zeta - z} \,\diff\zeta = \frac{1}{2\pi} \int_0^{2\pi} u_n\p{z + re^{i\theta}} \,\diff\theta,

so taking absolute values, multiplying both sides by $r$ and integrating from $0$ to $1$ ,

\begin{aligned} \abs{u_n\p{z}} &\leq \frac{1}{\pi} \iint_{B\p{z,1}} \abs{u_n\p{x + iy}} \,\diff{x} \,\diff{y} \\ &= \frac{1}{\pi} \iint_{B\p{z,1}} \frac{\abs{u_n\p{x + iy}} e^{-\sqrt{x^2 + y^2}}}{e^{-\sqrt{x^2 + y^2}}} \,\diff{x} \,\diff{y} \\ &= \frac{1}{\pi e^{\sqrt{R+1}}} \iint_{B\p{z,1}} \abs{u_n\p{x + iy}} e^{-\sqrt{x^2 + y^2}} \,\diff{x} \,\diff{y} \\ &\leq \frac{\mu\p{B\p{z, 1}}}{\pi e^{\sqrt{R+1}}} \p{\iint_\C \abs{u_n\p{x + iy}}^2 e^{-\p{x^2 + y^2}} \,\diff{x} \,\diff{y}}^{1/2} && \p{\text{Cauchy-Schwarz}} \\ &\leq \frac{\mu\p{\C}}{\pi e^{\sqrt{R+1}}} \norm{u_n}_{L^2\p{\mu}}. \end{aligned}

Thus, $\set{u_n}_n$ is uniformly bounded on $K$ . Since any compact set is contained in a compact ball, we may apply Montel's theorem. Hence, we get a locally uniformly convergent subsequence $\set{u_{n_k}}_k$ on $\C$ . Passing to another sequence if necessary, $\set{u_{n_k}}_k$ also converges almost everywhere to $u$ , since it converges to $u$ in $L^2$ . By local uniform convergence, $u$ must also be holomorphic, so $\mathcal{B}$ is a closed subspace of $L^2\p{\mu}$ , hence complete.