de Rham theorem for tempered distributions I am wondering if the following statement holds.

If $u\in \mathscr{S}'$ satisfies $\left< u,\Phi\right>=0$ for all $\Phi \in \mathscr{S}$ with $\mathrm{div}\, \Phi=0$, then there exists $p\in \mathscr{S}'$ such that $u=\nabla p$ in $\mathscr{S}'$. Here <,> denotes the dual pairing.

It is well known that if we replace $\mathscr{S}$ with $\mathscr{D}$, then this is well-known theorem due to de Rham.
It is hard for me to find the analog version of tempered distribution.
Thank you very much for your time.
 A: This works. As explained in my comment, we need only show that if $p\in\mathcal D'(\mathbb R^n)$ and $\nabla p\in\mathcal S'$ (vector valued), then $p\in\mathcal S'$.
The condition for a function $\varphi$ to be a divergence of a vector field is $\int\varphi=0$; see here. So if we fix a $\varphi_0\in\mathcal D$ with $\int\varphi_0=1$ and $\varphi\in\mathcal S$ is arbitrary, then we can write $\varphi=\textrm{div }\Phi + c\varphi_0$, for some $\Phi\in\mathcal S$, and with $c=\int\varphi$.
Let $a=(p,\varphi_0)$. We then obtain
$$
(p,\varphi)=(p,\textrm{div }\Phi) + a \int\varphi = -(\nabla p, \Phi) + a\int\varphi .
$$
This shows that $p$ is well defined and continuous on $\mathcal S$, so is a tempered distribution. (Strictly speaking, I was too lazy here to actually show the continuity explicitly and I more relied on the meta-principle that anything that has a natural definition as a functional on $\mathcal S$ is continuous. To do it properly, we'd have to establish that $\Phi$ can be made to depend continuously on $\varphi$ in the topology of $\mathcal S$.)
