I'm trying to solve the following question:
If $u$ is harmonic in $\mathbb{R}^N$ e $\nabla u \in L^2(\mathbb{R}^N)$, then $u$ is constant.
I've found this solution, which I posted below, but I have one question about the problem itself and one about the argument of the solution.
Solution:
First of all, you should point out what definition of harmonic function you use; I will use the fact that a continuous function $u$ which satisfies the mean property is harmonic (it is equivalent to $u \in C^2$ and $\Delta u =0$). I remark also the fact that an harmonic function is necessarily $C^{\infty}$ (indeed, analytic) - you can easily see this by using mollifiers.
If $u$ is harmonic, then $\partial_{x_i}u$ is harmonic for every $i=1,\ldots , N$. It follows that $\nabla u$ has the mean property: using Cauchy-Schwarz, $$ \vert \nabla u(x) \vert =\left\vert \frac{1}{\omega_N R^N}\int_{B(x,R)}\nabla u(y)dy \right\vert \leq \frac{1}{\omega_N R^N}\int_{B(x,R)} \vert \nabla u(y) \vert dy\leq \frac{\sqrt{\omega_N R^N}}{\omega_N R^N}\left(\int_{B(x,R)}|\nabla u(y)|^2 dy\right)^{1/2}\leq\frac{||\nabla u||_2}{\sqrt{\omega_N R^N}} $$
Let $R \to + \infty$ and you get $\nabla u(x)=0$, which is the claim. Hope this helps (but I am afraid of misunderstanding the question).
My questions:
- $\nabla u: \mathbb{R}^N \to \mathbb{R}^N$, but if a function $g \in L^2(\mathbb{R}^N)$, then $g:\mathbb{R}^N \to \mathbb{R}$, so how $\nabla u \in L^2(\mathbb{R}^N)$?
- If $\nabla u$ has the mean-value property, then $$ |\nabla u(x)| = \left\vert\left(\frac{\partial u}{\partial x_1}, \dots, \frac{\partial u}{\partial x_N} \right)\right\vert = \frac{1}{|B(x,r)|} \left\vert\left(\int_{B(x,r)}\frac{\partial u(y)}{\partial x_1}~dy, \dots, \int_{B(x,r)}\frac{\partial u(y)}{\partial x_N}~dy\right)\right\vert = \frac{1}{B(x,r)}\left\vert\int_{B(x,r)}\nabla u~dy \right\vert. $$ How is that last equality possible? Does the integral of a function $g: \mathbb{R}^N \to \mathbb{R}^N$ equal the integrals of its components?