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Spring 2017 - Problem 11

harmonic functions

Let $1 \leq p < \infty$ and let $U\p{z}$ be a harmonic function on the complex plane $\C$ such that

\iint_{\R \times \R} \abs{U\p{x + iy}}^p \,\diff{x} \,\diff{y} < \infty.

Prove $U\p{z} = 0$ for all $z = x + iy \in \C$ .

Solution.

Let $q$ be such that $\frac{1}{p} + \frac{1}{q} = 1$ . Since $U$ is harmonic everywhere, we may apply the mean value property on any $B\p{z, R}$ with $R > 0$ to get

\begin{aligned} \abs{U\p{z}} = \abs{\frac{1}{\pi R^2} \int_{B\p{z,R}} U\p{w} \,\diff{A}} &\leq \frac{1}{\pi R^2} \int_{B\p{z,R}} \abs{U\p{w}} \,\diff{A} \\ &\leq \frac{\norm{U}_{L^p}}{\pi R^2} \norm{\chi_{B\p{z,R}}}_{L^q} \\ &\leq \frac{\norm{U}_{L^p}}{\pi R^2} m\p{B\p{z, R}}^{1/q} \\ &= \frac{\norm{U}_{L^p}}{\pi R^2} \pi^{1/q} R^{2/q} \\ &= \frac{\norm{U}_{L^p}}{\pi^{1-\p{1/q}} \p{R^2}^{1-\p{1/q}}} \\ &= \frac{\norm{U}_{L^p}}{\pi^{1/p} R^{2/p}}, \end{aligned}

which tends to $0$ as $R \to \infty$ , since $p \neq \infty$ . Thus, as $z$ was arbitrary, we have $U\p{z} = 0$ everywhere.