I am wondering if the following statement holds.



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> 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.
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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.



de Rham theorem for tempered distributions
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.](https://mathoverflow.net/questions/112180/which-test-functions-are-the-divergence-of-a-vector-field%3F) 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 $\mathcal S$.)


