Question:
Let u be a harmonic function in $\mathbb{R}^n$ and suppose that
$$\int_{\mathbb{R}^n}|u(x)|^2dx\lt\infty.$$
Show that $u\equiv0$.
My attempt:
$\forall B(x,r)\subset\mathbb{R}^n, \text{we have...}$
$$0\le|u(x)|=|\unicode{x2a0d}_{B(x,r)}u(y)dy|\le(\unicode{x2a0d}_{B(x,r)}|u(y)|^2dy)^{\frac{1}{2}}(\unicode{x2a0d}_{B(x,r)}1dy)^{\frac{1}{2}}=(\unicode{x2a0d}_{B(x,r)}|u(y)|^2dy)^{\frac{1}{2}}.$$
$\text{Since it is clear that:}$
$$\lim_{r\to\infty}\int_{B(x,r)}|u(x)|^2dx=\int_{\mathbb{R}^n}|u(x)|^2dx\lt\infty\quad and\quad \lim_{r\to\infty}\frac{1}{|B(x,r)|}=\lim_{r\to\infty}\frac{1}{\alpha(n)r^n}\\=\frac{1}{\alpha(n)}\lim_{r\to\infty}\frac{1}{r^n}=0$$
$$\therefore \lim_{r\to\infty}|u(x)|\le\lim_{r\to\infty}(\unicode{x2a0d}_{B(x,r)}|u(y)|^2dy)^{\frac{1}{2}}=(\frac{1}{|B(x,r)|}\int_{B(x,r)}|u(y)|^2dy)^{\frac{1}{2}}=0$$
$\text{Hence, we find, via the Sandwich Principle, that } u\equiv0$
Help!
Here is my solution to the given problem, I am unsure if this checks out as a rigorous solution any help or even alternative solutions would be greatly appreciated.