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Divergent integral


Let u satisfy u \in \mathcal{C}^2(\Bbb{R}^n),\ \Delta u=0 on \Bbb{R}^n. Show that the integral \displaystyle \int_{\Bbb{R}^n}u^2 dx is convergent if and only if u \equiv 0.
PHD 6201

Proof: Suppose there exists x_0 \in \Bbb{R}^n with u(x_0) \neq 0. Apply the mean value property for the harmonic function u and get

u(x_0)=\frac{1}{\omega_n \rho^{n-1}} \int_{ \partial B(x_0,\rho)} u(x) d \sigma, where \omega_n is the surface of the n-dimensional unit ball. By Cauchy Schwarz inequality we get

u^2(x_0) \leq (\frac{1}{\omega_n \rho^{n-1}})^2 \int_{ \partial B(x_0,\rho)} u^2(x) d \sigma \int_{ \partial B(x_0,\rho)} d\sigma, which means that
\int_{ \partial B(x_0,\rho)} u^2(x) d \sigma \geq \omega_n \rho^{n-1}u^2(x_0).

We finish by whe following inequality: \int_{\Bbb{R}^n}u^2(x)dx \geq \int_0^r \int_{ \partial B(x_0,\rho)} u^2(x) d \sigma d \rho \geq \frac{\omega_n}{n}u^2(x_0)r^n. The inequality is true for any r>0, and therefore, for r \to \infty we see that the integral is divergent. This means that the given integral is convergent if and only if u \equiv 0.

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