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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 $u \equiv 0$.