Tempered distributions

Question

Let $P$ be a vector whose components are polynomials in $\mathbb{R}^n$ and harmonic. it's true that exists a polynomial T that $\nabla T = P$?

I think this has something to be with Fourier transform, but I'm not sure.

Answer 1

In general, if $\Omega\subset\mathbb{R}^n$ is simply connected and $\vec F=(F_1,\dots,f_n)\colon\Omega\to\mathbb{R}^n$ is a vector field of class $C^1$, then $\vec F$ is a gradient if and only if $$ \frac{\partial F_i}{\partial x_j}=\frac{\partial F_j}{\partial x_i},\quad1\le i,j\le n.\tag{*} $$ If the $F_i$ are polynomials, then they are $C^\infty$ on $\mathbb{R}^n$, which is simply connected. In order to be a gradient, they must satisfy (*). In that case, the potential function is in fact a polynomial.