First, some background: I wanted to prove that, if $f$ is a measurable function such that $\nabla f\in L^p\_\text{loc}(\mathbb R^n)$, then $f\in L^p\_\text{loc}(\mathbb R^n)$, $p\in(1,\infty)$. This is proven, for instance, in the book Sobolev Spaces by Vladimir Maz'ya, but I don't like the proof there. I was thinking about the possibility of a proof in the following outline: Fix a bounded open set $\Omega\subset\mathbb R^n$ and $\phi\in C\_c^{\infty}(\Omega)$. Use the notation $\mathbf{f}=(f,f,\cdots,f)$ with $n$ components. Then, in the sense of distributions we observe
$$
|(\mathbf{f},\nabla\phi)|=|(\nabla f,\phi)|\leq\Vert\nabla f\Vert\_{L^p(\Omega)}\Vert\phi\Vert\_{L^q(\operatorname{supp}\phi)}\leq C\Vert\nabla\phi\Vert\_{L^q(\Omega)},
$$
where in the last line we used Holder's inequality and then the Sobolev inequality. The above estimate implies that the vector function $\mathbf{f}$ is an element of the dual space of the subspace in $L^p(\Omega)^n$ consisting of those vector fields which are conservative, i.e. those vector fields $\mathbf{r}$ for which there exists a function $\phi$ such that $\mathbf{r}=\nabla\phi$. Of course, this is not enough to show that $\mathbf{f}\in L^p(\Omega)^n$, which is what is desired to show, but I think that, if I could characterize the elements of the space that $\mathbf{f}$ has been shown to live in, then I could finish the proof somehow. 


Of course, the Helmholtz-Hodge decomposition states that at this point it suffices for me to check that $\mathbf{f}$ is a bounded linear functional on the subspace of $L^p(\Omega)^n$ consisting of divergence-free vector fields, but it's not immediately clear how to prove this though. Anyway, my main question is:


**What is the dual space of the subspace of $L^p(\Omega)^n$ consisting of vectors $\mathbf{r}$ for which there exists a function $\phi$ such that $\mathbf{r}=\nabla\phi$?**



What is the dual space of $L^p$(conservative vector fields on a bounded set)?
**Here is a short, elementary, and self-contained proof of the result you wanted to prove.** It is similar to the one given in Maz'ya's book, but simpler.
For a related post see: <https://mathoverflow.net/a/297392/121665>



> 
> If $f\in L^1\_{\rm loc}$ or even if $f$ is a distribution and $\nabla
> f\in L^p$, then $f\in L^p\_{\rm loc}$.
> 
> 
> 


This follows from the fact proven below. For $1\leq p\leq \infty$ define
$$
L^{1,p}(\Omega)=\{ f\in \mathcal{D}'(\Omega):\, \nabla f\in L^p(\Omega)\},
\quad
W^{1,p}(\Omega)=\{ f\in L^p(\Omega):\, \nabla f\in L^p(\Omega)\}.
$$
Similarly we define spaces $L^{1,p}\_{\rm loc}$ and $W^{1,p}\_{\rm loc}$.



> 
> **Theorem.** *$L\_{\rm loc}^{1,p}(\Omega)\subset W^{1,p}\_{\rm loc}(\Omega)$ for $1\leq p\leq\infty$.*
> 
> 
> 


**Remark.** By induction the result generalizes to spaces with higher order derivatives.


In the proof we will need the following well known result.


**Lemma.** *If $u\in W^{1,1}(\mathbb{R}^{n})$ then $$ u(x)= \frac{1}{n\omega\_{n}}\int\_{\mathbb{R}^{n}}\frac{(x-y)\cdot \nabla u(y)}{|x-y|^{n}}\, dy \quad \text{a.e.,} $$ where $\omega\_n$ denotes the volume of the unit ball.*


*Proof.* By a density argument it suffices to prove it for $u\in C\_0^\infty(\mathbb{R}^n)$. Let
$s\in S^{n-1}$ (unit sphere). We have 
$$
u(x)=-\int\_{0}^{\infty}\frac{d}{dr}u(x+rs)\, dr=
-\int\_{0}^{\infty}Du(x+rs)\cdot s\, dr.
$$
Taking the average with respect to $s\in S^{n-1}$ we get
(recall that the volume of $S^{n-1}$ equals $n\omega\_n$)
$$
u(x)=-\frac{1}{n\omega\_{n}}\int\_{S^{n-1}}\int\_{0}^{\infty}Du(x+rs)\cdot
s\, dr\, ds 
$$
and the lemma follows after substituting
$x+rs=y$, so $dy=r^{n-1}drds$, $s=(y-x)/|y-x|$,
and $drds=|x-y|^{(1-n)}dy$. $\Box$


The lemma can be seen as the
integral representation of the Dirac $\delta$ distribution. 



> 
> **Corollary.**
> *$\sum\_{i=1}^{n}\partial K\_{i}/\partial x\_{i}=\delta$, where $K\_{i}=n^{-1}\omega\_{n}^{-1}x\_{i}|x|^{-n}$.*
> 
> 
> 


$\Box$


*Proof of the theorem.*
Suppose that $u\in L^{1,p}(\Omega)$. Let
$V\Subset V\_{\varepsilon}\Subset\Omega$, where 
$V\_{\varepsilon}=\{ x\, |\, {\rm dist}\,(x,V)<\varepsilon\}$ and let
$\varphi\in C\_0^\infty(\Omega)$, $\varphi|\_{V\_{\varepsilon}}\equiv 1$. It suffices to
prove that $w=\varphi u\in L^{p}(V)$. In other words it suffices to prove
that the distribution $\varphi u$ when restricted to $V$ 
can be represented by a certain $L^{p}(V)$ function.


Let $\eta\in C\_0^\infty(B^{n}(0,\varepsilon))$,
$\eta|\_{B^{n}(0,\varepsilon/2)}\equiv 1$. According to the corollary we have
$$
\sum\_{i=1}^{n}\frac{\partial(\eta K\_{i})}{\partial x\_{i}}=
\sum\_{i=1}^{n}\eta\frac{\partial K\_{i}}{\partial x\_{i}}+\xi=\delta+\xi,
$$
where 
$\xi\in C\_0^\infty(B^{n}(0,\varepsilon))$. 
The fact that $\xi$ is smooth follows from the observation 
$K\_{i}\partial\eta/\partial x\_{i}\in C\_0^\infty(B^{n}(0,\varepsilon))$.
Now
$$
w+\xi\*w=(\delta+\xi)\*w=\sum\_{i=1}^{n}\frac{\partial(\eta K\_{i})}{\partial x\_{i}}\*w 
=\sum\_{i=1}^{n}(\eta K\_{i})\*\frac{\partial w}{\partial x\_{i}}.
$$
By properties of the convolution, the distribution $\xi\*w$ is
smooth in $\mathbb{R}^{n}$ and hence it belongs to $L^{p}(V)$ (after
being restricted to $V$). Therefore it remains to show that
$$
(\eta K\_{i})\*\frac{\partial w}{\partial x\_{i}}\in L^{p}(V)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\*)
$$
Since ${\rm supp}\,\eta K\_{i}\subset B^{n}(0,\varepsilon)$, the 
restriction of the distribution 
$(\eta K\_{i})\*(\partial w/\partial x\_{i})$ to $V$ does not depend on the
behavior of $\partial w/\partial x\_{i}$ outside $V\_{\varepsilon}$, but
$\partial w/\partial x\_{i}=\partial u/\partial x\_{i}$ in
$\mathcal{D}'(V\_{\varepsilon})$ (as $\varphi=1$ in $V\_{\varepsilon}$) so 
$$
(\eta K\_{i})\*\frac{\partial w}{\partial x\_{i}}=
(\eta K\_{i})\*\frac{\partial u}{\partial x\_{i}}
\quad
\text{in $\mathcal{D}'(V)$.}
$$
Now (\*) follows since 
$$
\Vert(\eta K\_i)\*\frac{\partial u}{\partial x\_i}\Vert\_p\leq\Vert\eta K\_i\Vert\_1
\Vert\frac{\partial u}{\partial x\_i}\Vert\_p\leq C \Vert\frac{\partial u}{\partial x\_i}\Vert\_p.
$$
Indeed, $\eta$ has compact support and $|K\_i|\leq C|x|^{1-n}$ so $\eta K\_i\in L^1$.
$\Box$


