If $p$ is a distribution such that $\nabla p$ is regular, then $p$ must be regular too

Let

• $d\in\mathbb N$
• $\Omega\subseteq\mathbb R^d$ be open
• $\mathcal D(\Omega):=C_c^\infty(\Omega)$
• $q\ge 1$

Each $f\in L^1_{\text{loc}}(\Omega)$ can be identified with $\langle f\rangle\in\mathcal D'(\Omega)$, $$\langle f\rangle(\phi):=\langle\phi,f\rangle_{L^2(\Omega)}\;\;\;\text{for }\phi\in\mathcal D(\Omega)$$ and each $g\in L^1_{\text{loc}}(\Omega)^d$ can be identified with $\langle g\rangle:\mathcal D(\Omega)^d\to\mathbb R$, $$\langle g\rangle(\phi):=\sum_{i=1}^d\langle g_i\rangle(\phi_i)\;\;\;\text{for }\phi\in\mathcal D(\Omega)^d\;.$$ If $p\in \mathcal D'(\Omega)$, then $$\frac{\partial p}{\partial x_i}(\phi):=-p\left(\frac{\partial\phi}{\partial x_i}\right)\;\;\;\text{for }i\in\left\{1,\ldots,d\right\}\text{ and }\phi\in\mathcal D(\Omega)$$ and $$\nabla p(\phi):=\sum_{i=1}^d\frac{\partial p}{\partial x_i}(\phi_i)\;\;\;\text{for }\phi\in\mathcal D(\Omega)^d\;.$$

Let $p\in\mathcal D'(\Omega)$ with $$\nabla p=\langle g\rangle$$ for some $g\in L^q(\Omega)^d$. Can we show that there is some $f\in L_{\text{loc}}^q(\Omega)$ with $p=\langle f\rangle$?

• Maybe I am confused with your notations. But your request is not equivalent to requiring that the function $f$ is the weak derivative of $g$? as I understand it, this is just a definition. – Andrew Jul 3 '16 at 15:10