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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 $\mathbf{r}=\nabla\phi$?

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  • $\begingroup$ If you write \operatorname{supp}\phi rather than \text{supp }\phi then you don't need to add spacing manually, and moreover, the spacing is context-dependent, so the space to the right of $\operatorname{supp}$ is different in $\operatorname{supp}\varphi$ and $\operatorname{supp}(\varphi). \qquad$ $\endgroup$ – Michael Hardy Mar 29 '18 at 4:08
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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 $K_{i}=n^{-1}\omega_{n}^{-1}x_{i}|x|^{-n}$.

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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 $\eta K_i\in L^1$. $\Box$

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