Let u(x) a harmonic function in $\mathbb{R}^n$ such as: \begin{equation} \int_{\mathbb{R}^n}|u(x)|dx =K< \infty \end{equation} Show thtat $u(x)=0$, $\forall x \in \mathbb{R}^n$.
Using the property of volumetric mean I get that, for a sphere $B(a,r) \subset \mathbb{R}^n$ the value of $u(a)$ is: \begin{equation} u(a)=\frac{1}{\omega_nr^{n-1}}\int\limits_{B(a,r)}u(x)dx \end{equation}
By the fact that $u(a)\leq |u(a)|$ we get the following inequalities \begin{equation} u(a)\leq |u(a)|=\bigg|\frac{1}{\omega_nr^{n-1}}\int\limits_{B(a,r)}u(x)dx\bigg|\leq \frac{1}{\omega_nr^{n-1}}\int\limits_{B(a,r)}|u(x)|dx\leq \frac{1}{\omega_nr^{n-1}}\int\limits_{\mathbb{R}^n}|u(x)|dx= \frac{1}{\omega_nr^{n-1}}K \end{equation}
As $r\to \infty $ we get that $|u(a)|\leq K\lim\limits_{r \to \infty} \frac{1}{\omega_nr^{n-1}}=0 $.
So $|u(a)|\leq 0\implies |u(a)|= 0 \implies u(a)=0$.
Since $a$ is arbitrary we get that \begin{equation} u(x)=0 , \ \ \ \ \forall x \in \mathbb{R}^n \end{equation}
I'm not so secure that this argumentation is valid. If it is wrong can anybody tell me where?