# Alternative proof of Liouville theorem for harmonic functions

From Prove Liouville theorem without using mean value property the following question arises: To prove the Liouville theorem

If $$u$$ is harmonic on $$\mathbb{R}^n$$ and $$\int_{\mathbb{R}^n}|\nabla u|^2 dx \leq C$$ for some $$C > 0$$, then $$u$$ is constant.

is it possible to use the following reasoning?

Fix a non-negative smooth function $$\phi$$ which is identically $$1$$ on $$B_1$$ and vanishes identically outside $$B_2$$. Pick $$M$$ so $$|\Delta \phi| \leq M$$. Set $$\phi_R(x)=\phi(x/R)$$. We have $$|\nabla \phi_R(x)|\leq M/R$$ and $$|\Delta \phi_R(x)|\leq M/R^2$$, then estimate $$\int_{\mathbb R} \phi_R |\nabla u|^2$$ in a way such that the right hand side goes to zero as $$R \to \infty$$.

Unfortunately, I'm not able to complete this proof, but it seems that it should work.

$$|\nabla u|$$ is subharmonic because $$\ |\nabla u|=\sup_{x\in S_p} |\partial_x u|$$,let us write $$\nabla u=v$$. $$\int_{R^{n}}|v|^{2} d x=\int_{0}^{+\infty} \int_{\partial B_{r}(0)}|v(r, \theta)|^{2} d \theta d r$$ by AM-GM inequality and $$\|\cdot \|$$ is a concave function and possion kernel expression, we have for $$0\leq r_1\leq r_2<\infty$$, $$\frac{1}{V(\partial B_{r_{2}}(b))}\int_{\partial B_{r_{2}}(b)}|v(r, \theta)|^{2} d \theta \geqslant \frac{1}{V(\partial B_{r_{1}}(b))}\int_{\partial B_{r_{1}(0)}}|v(r, \theta)|^{2} d \theta$$ So $$\int_{\mathbb{R}^{2}}|v|^{2} d x \geqslant \int_{0}^{+\infty}|v(0)|^{2} d r \quad \quad (*)$$ In this argument 0 can change to any point $$p\in \mathbb{R}^2$$, so we proved $$\forall p\in \mathbb{R}^2$$, $$v(p)=0$$, so $$v \equiv 0, \quad u \equiv \text { constant }$$.
And you can take a $$\phi_{R}$$ Polished $$(*)$$ and prove $$\int_{\mathbb{R}} \phi_{R}|\nabla u|^{2}<\infty$$ imitate this.