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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.

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  • $\begingroup$ This is on a complete, perhaps noncompact, Riemannian manifold, and $u$ and $\Phi$ are tangent vector fields, correct? $\endgroup$
    – Ben McKay
    Commented May 9, 2022 at 9:37
  • $\begingroup$ @BenMcKay Yes. I am working on $\mathbb{R}^n$, but I am not sure what it means the `tangent vector fields' since I am not familiar with DG. I want to find an exact reference for this fact. $\endgroup$
    – Will Kwon
    Commented May 9, 2022 at 14:14
  • $\begingroup$ $u=\nabla p$ for some $p\in \mathcal D'$ by the fact you quote, so a more direct version of your question is: Let $p\in\mathcal D'$. Does $\nabla p\in\mathcal S'$ imply that $p\in\mathcal S'$? $\endgroup$
    – Christian Remling
    Commented May 9, 2022 at 14:42

1 Answer 1

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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 $\mathcal S$.)

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  • $\begingroup$ Thank you for your answer. In fact, I got a similar proof as yours, but it was a bit tricky to show the assertion given in the second paragraph. To show the assertion, I mimicked the proof of Wang's paper (math.fsu.edu/~wxm/paper/Wang_pressure.pdf), the Proposition given on page 2. I will try to fix or find references if it is possible. Thank you very much. $\endgroup$
    – Will Kwon
    Commented May 10, 2022 at 3:59
  • $\begingroup$ @WillKwon: Yes, this is exactly what is required here if you want to do it carefully. $\endgroup$
    – Christian Remling
    Commented May 10, 2022 at 16:10

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