If $p$ is a distribution, what is the meaning of the claim $\nabla p\in L^p(\Omega)^d$ 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 $q=\infty$?]
 A: This is standard notation: if $p \in \mathcal D ' (\Omega)$, then $\nabla p \in \mathcal D ' (\Omega, \Bbb R^d)$ is a vector-valued distribution (the terminology is misleading, because it suggests that a vector-valued distribution has vectors as values, which it does not; the term is quite natural, though, in that a vector-valued distribution is a distribution on a space of vector-valued test functions) given by $\langle \nabla p, \varPhi \rangle = - \langle p, \text{div } \varPhi \rangle$.
The definition is quite natural if you think about it for a moment: if $p$ is in fact a smooth function and $F$ a smooth vector-valued map with compact support (i.e. a vector-valued test function), then $\nabla p$ can be viewed as a vector of distributions, and its value on $F$ is (using $\partial _i$ for $\frac {\partial} {\partial x_i}$)
$$\langle \nabla p, F \rangle = \sum _i \langle (\partial_i p), F_i \rangle = \sum _i \int (\partial_i p) F_i = \sum _i \int \partial_i (p F_i) - \sum _i \int p \partial_i F_i = \\
\sum_i 0 - \int p \sum _i \partial_i F_i = - \int p \ \text{div} F = - \langle p, \text{div } F \rangle ,$$
where $\int \partial_i (p F_i) = 0$ because $pF_i$ has compact support in the variable $x_i$ (because $F$ has).
On the other hand, the space $L^q (\Omega)^d$ is the space of vector-valued Lebesgue $q$-integrable functions, i.e. $L^q (\Omega)^d = L^q (\Omega, \Bbb R^d) = \{ V : \Omega \to \Bbb R^d \text{ measurable } \mid \int _\Omega \| V \|^q < \infty \}$, with equality understood as equality almost everywhere (as usual with Lebesgue spaces).
With all this clarified, the notation $\nabla p \in L^q (\Omega) ^d$ means that $\nabla p$ is a vector-valued distribution on $\Bbb R^d$ such that there exist $V \in L^q (\Omega)^d$ with $\nabla p = V$. The equality makes sense because, as with usual distributions, the space $L^q (\Omega) ^d$ embeds naturally in $\mathcal D ' (\Omega, \Bbb R^d)$.
