# Prove that $\left.F\right|_{\left\{ϕ∈C_c^∞(Ω,ℝ^d):∇⋅ϕ=0\right\}}=0⇔∃p∈C_c^∞(Ω)$ with $F=∇p$, for all $F∈H_0^1(Ω,ℝ^d)'$

Let

• $d\in\mathbb N$
• $\Omega\subseteq\mathbb R^d$ be open
• $\langle\;\cdot\;,\;\cdot\;\rangle$ denote the inner product on $L^2(\Omega,\mathbb R^d)$
• $\mathcal D:=C_c^\infty(\Omega,\mathbb R^d)$, $$H:=\overline{\mathcal D}^{\langle\;\cdot\;,\;\cdot\;\rangle_H}\color{blue}{\stackrel{\text{def}}=H_0^1(\Omega,\mathbb R^d)}$$ with $$\langle\phi,\psi\rangle_H:=\langle\phi,\psi\rangle+\underbrace{\sum_{i=1}^d\langle\nabla\phi_i,\nabla\psi_i\rangle}_{=:\;(\phi,\:\psi)}\;\;\;\text{for }\phi,\psi\in\mathcal D$$ and $$\mathfrak D:=\left\{\phi\in\mathcal D:\nabla\cdot\phi=0\right\}$$

By a well-known theorem, $$\left.F\right|_{\mathfrak D}=0\;\Leftrightarrow\;\exists p\in C_c^\infty(\Omega)':F=\nabla p\tag 1$$ for all $F\in\mathcal D'$. Now, the author of a seminar paper that I'm reading states (in the proof of Theorem 9) that $(1)$ implies that $$\left\{F\in H':\left.F\right|_{\mathfrak D}=0\right\}=\left\{F\in H'\mid\exists p\in C_c^\infty(\Omega):F=\nabla p\right\}\;.\tag 2$$

Why?

It's clear to me that $$\langle v\rangle_H:=\left.\langle\;\cdot\;,v\rangle_H\right|_{\mathcal D}\in\mathcal D'\;\;\;\text{for all }v\in H\tag 2$$ and the associated map $H\to\mathcal D'$ is injective. Moreover, since $H\cong H'$, $$f:=\left.F\right|_{\mathcal D}\in\mathcal D'\;\;\;\text{for all }F\in H'\tag 3$$ and, again, the associated map $H'\to\mathcal D'$ is injective. However, the only thing I'm able to deduce from $(1)$ is that $$\left.F\right|_{\mathfrak D}=0\;\Leftrightarrow\;\exists p\in C_c^\infty(\Omega)':\left.F\right|_{\mathfrak D}=\nabla p\tag 4$$ for all $F\in H'$.

So, how do we obtain $(2)$? Or is this impossible and we need to interpret $(2)$ in some special sense?

• Just remembered the en.wikipedia.org/wiki/Helmholtz_decomposition theorem, I'm 99% sure it is the solution – reuns Jul 6 '16 at 19:13
• @user1952009 I'm aware of the decomposition theorem (it is proven in the referenced paper), but I don't see how it should be relevant here. – 0xbadf00d Jul 6 '16 at 21:00