Is it true that a smooth, harmonic functions with compactly supported gradient is trivial?

Question

Let $u \in C^{\infty}(\mathbb{R}^n)$ be such that $\nabla u \in C^{\infty}_c(\mathbb{R}^3; \mathbb{R}^3)$ and $\Delta u=0$. I wish to prove or disprove that $u = c$ for some $c \in \mathbb{R}^n$; note the statement would be true with $c=0$ (by Liouville classical theorem) if $u$ itself is compactly supported.

I had the idea of exploiting a representation formula like:

$$ u(x) = \frac{1}{n \omega_n} \int_{\mathbb{R}^3} \nabla u(y) \cdot \frac{x-y}{|x-y|^n} dy $$

together with an integration by parts in order to get the Laplacian. Unfortunately, the latter is only valid if $u \in W^{1,1}(\mathbb{R}^n)$ and here we don't have the hypothesis $u \in L^1(\mathbb{R}^n)$. I am stuck here, but at same time I cannot produce a counterexample.

Answer 1

Note that the gradient has compact support, thus $u$ is constant outside a compact set. Hence, there exists a constant $c$ such that $v:=u-c$ has compact support. But $v$ is harmonic and thus vanishes. This implies that $u$ is constant.

Answer 2

I think that you can prove that the function is constant, but not necessarily zero: since $\nabla u$ has compact support, there exists $R>0$ such that $\nabla u \equiv 0$ in $\mathbb{R}^{n} \setminus \overline{B(0;R)}$, and then $u \equiv c$ ($c \in \mathbb{R}$ is a constant) in $\mathbb{R}^{n} \setminus \overline{B(0;R)}$. From this, $$ |u(x)| \leq \max \left \lbrace c, \Bigg| \max_{x \in \overline{B(0,R)}}u \Bigg|\right \rbrace, $$ that is, $u$ is bounded. By Liouville theorem, you obtain that $u$ is constant.

Answer 3

Differentiating the equation we note $$0=\partial_i\Delta u = \Delta \partial_i u$$ which means that the components of $\nabla u$ are themselves harmonic. Since they are compactly supported, they vanish identically, and so $u$ is constant.