How can I prove the following Liouville theorem without using the mean value property?



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> 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.
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The proof that I know indeed uses the mean value property for harmonic functions.




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From the comments: is it rigorous to do it like this: $-\Delta u = 0 \implies \int\_{\mathbb R^n} |\nabla u|^2 = 0$ (integrating by parts, hence $u$ is constant? It seems to easy, probably I'm missing something.



Prove Liouville theorem without using mean value property
Below we will prove, by purely energy methods, the following sharper statement: Let $u$ be a harmonic function which satisfies $$\liminf\_{r \to \infty} \frac1{r^2} \frac{1}{|B\_r|}\int\_{B\_r} |u|^2 = 0.$$ Then $u$ is constant.


All Liouville theorems of this sort are soft/qualitative versions of harder/more quantitative regularity statements. In this case, the Liouville statement follows from the interior gradient $L^\infty$ bound for harmonic functions (as we will see below). So another version of your question could be: Is there an energy methods proof of the interior $L^\infty$ bound for harmonic functions which in particular does not use the mean value property?


The classical energy methods proof of the pointwise estimates for harmonic functions uses only two ingredients: the Caccioppoli inequality (the most basic energy estimate), and the Sobolev inequality (because you have to get pointwise bounds from $L^2$ bounds somehow!).


The Caccioppoli inequality says that
\begin{equation\*}
\int\_{B\_r}
|\nabla u|^2
\leq 
\frac{C}{r^2} \int\_{B\_{2r}} u^2 \,.
\end{equation\*}
You get this by testing the equation with $\varphi^2u$ where $\varphi$ is an appropriate cutoff function.


Iterating the Caccioppoli inequality many times, we get
\begin{equation\*}
\int\_{B\_r} |\nabla^k u|^2
\leq 
\frac{C}{r^{2k}} \int\_{B\_{2^kr}} u^2 \,.
\end{equation\*}
If $k > 1+\frac d2$, then we can the Sobolev inequality to get
\begin{equation\*}
\| \nabla u \|\_{L^\infty(B\_r)}^2 
\leq 
\frac{C}{r^2} 
\frac{1}{|B\_r|}
\int\_{B\_{2^kr}} u^2\,.
\end{equation\*}
Now send $r\to \infty$ along a good subsequence for the liminf, and you discover that $\nabla u=0$, which means $u$ is constant.


