# If $F∈H^{-1}(Ω,ℝ^d)$ and $∃p∈\mathcal D'(Ω):\left.F\right|_{\mathcal D(Ω,ℝ^d)}=∇p$, then $∃\overline p∈H^{-1}(Ω):F=∇\overline p$

Let

• $d\in\mathbb N$
• $\Omega\subseteq\mathbb R^d$ be open
• $\mathcal D(\Omega):=C_c^\infty(\Omega)$ and $\mathcal D(\Omega,\mathbb R^d):=C_c^\infty(\Omega,\mathbb R^d)$
• $H^{-1}(\Omega):=H_0^1(\Omega)'$ and $H^{-1}(\Omega,\mathbb R^d):=H_0^1(\Omega,\mathbb R^d)'$

Let me cite a well-known theorem:

Let $u\in\mathcal D'(\Omega,\mathbb R^d)$ $\Rightarrow$ $$\left.u\right|_{\mathfrak D(\Omega)}=0\;\Leftrightarrow\;\exists p\in\mathcal D'(\Omega):u=\nabla p\tag 1$$ where $$\mathfrak D(\Omega):=\left\{\phi\in\mathcal D(\Omega,\mathbb R^d):\nabla\cdot\phi=0\right\}\;.$$

Now, if $F\in H^{-1}(\Omega,\mathbb R^d)$, then $$f:=\left.F\right|_{\mathcal D(\Omega,\:\mathbb R^d)}\in\mathcal D'(\Omega,\mathbb R^d)\;.$$ And since $\mathfrak D(\Omega)\subseteq\mathcal D(\Omega,\mathbb R^d)$, we obtain $$\left.f\right|_{\mathfrak D(\Omega)}=\left.F\right|_{\mathfrak D(\Omega)}\;.$$ Thus, $$\left.F\right|_{\mathfrak D(\Omega)}=0\;\Leftrightarrow\;\exists p\in\mathcal D'(\Omega):\left.F\right|_{\mathcal D(\Omega,\:\mathbb R^d)}=\nabla p\tag 2$$ by $(1)$.

I would like to conclude that $$\left.F\right|_{\mathfrak D(\Omega)}=0\;\Leftrightarrow\;\exists\overline p\in H^{-1}(\Omega):F=\nabla\overline p\;.\tag 3$$

Is that possible (at least the direction "$\Rightarrow$")? And if so, how?

• There is something that I don't get: this result is like a Poincaré lemma for distributions. The essential ingredient in Poincaré's lemma (which, at its core, is a topological result) the the simple-connectedness of $\Omega$. Are you sure, then, that you don't have to add some topological condition on $\Omega$ (such as, in decreasing order of generality: convexity, star-shapedness, simply-connectedness)? – Alex M. Jul 8 '16 at 8:32
• @AlexM. I'm quite sure. The only crucial claim is $(1)$, but you can find a proof of $(1)$ for arbitrary (open) $\Omega$. – 0xbadf00d Jul 8 '16 at 9:41