# What is the dual space of $L^p$(conservative vector fields on a bounded set)?

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

• 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 $\operatorname{supp}(\varphi). \qquad$ – Michael Hardy Mar 29 '18 at 4:08

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