Prove that if $u:\mathbb{R}^d \to \mathbb{R}$ is harmonic with integrable square then $u\equiv 0$.

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Prove that if $u:\mathbb{R}^d \to \mathbb{R}$ is harmonic and $$\int_{\mathbb{R}^d}u^2(x)dx=M<\infty$$ then $u\equiv 0$.

I'm not sure what the intended method is because there are no solutions, but I'd like to know if my proof is correct.

Since $u$ is harmonic, $u^2$ is subharmonic, hence by the mean value property, for any $x\in\mathbb{R}^d$ and $r>0$ we have

$$u(x)^2\leq \frac{1}{|B_r(x)|}\int_{B_r(x)}u(y)^2dy\leq \frac{M}{|B_r(x)|}=\frac{M}{r^d |B_1(0)|}\to 0$$ as $r\to\infty$. Therefore $u\equiv 0$.