Let
- $d\in\mathbb N$
- $\Omega\subseteq\mathbb R^d$ be open
- $\mathcal D(\Omega):=C_c^\infty(\Omega)$
- $q\ge 1$
I've seen the following Lemma (without a proof) in a paper and don't understand how I need to interpret it:
Let $p\in\mathcal D'(\Omega)$ with $\nabla p\in L^q(\Omega)^d$ $\Rightarrow$ $p\in L_{\text{loc}}^q(\Omega)$.
By definition, $$\nabla p(\Phi)\stackrel{\text{def}}=\sum_{i=1}^d\frac{\partial p}{\partial x_i}(\Phi_i)\stackrel{\text{def}}=-\sum_{i=1}^dp\left(\frac{\partial\Phi_i}{\partial x_i}\right)\;\;\;\text{for all }\Phi\in\mathcal D(\Omega)^d\;.\tag 1$$
I know that each $f\in\mathcal L^1_{\text{loc}}(\Omega)$ can be identified with $\langle f\rangle\in\mathcal D'(\Omega)$, $$\langle f\rangle(\phi):=\langle\phi,f\rangle_{L^2(\Omega)}\;\;\;\text{for }\phi\in\mathcal D(\Omega)\;.\tag 2$$ I understand that this identification is the meaning of $L^1_{\text{loc}}(\Omega)\subseteq\mathcal D'(\Omega)$. By $(1)$ and $(2)$, we see that $$\nabla\langle f\rangle(\Phi)=-\langle\nabla\cdot\Phi,f\rangle_{L^2(\Omega)}\;\;\;\text{for all }\Phi\in\mathcal D(\Omega)^d\tag 3\;.$$
However, even with $(3)$, I'm not able to make sense of $\nabla p\in L^q(\Omega)^d$. So, what is meant?
[As a secondary question, where can I find a proof of the Lemma and does the Lemma even hold for $q=\infty$?]