Distributional derivatives are locally integrable implies the distribution is also locally integrable?

Question

Let $T$ be a distribution on $\mathbb{R}^n$ such that there are functions $f_1,\ldots,f_n \in L^1_\text{loc}(\mathbb{R}^n)$ so that $\dfrac{\partial T}{\partial x_j} = f_j, \forall j=1,\ldots,n. $

My question: Is it true that $T$ is indeed a function in $L^1_\text{loc}(\mathbb{R}^n)$, i.e. there is $u\in L^1_\text{loc}(\mathbb{R}^n$) such that $$T(\phi) = \int_{\mathbb{R}^n} u \phi ,\; \forall \phi\in C_0^\infty (\mathbb{R}^n)?$$

When $n=1$, the answer is YES, and a proof can be found here. But I could not adapt this since I am not sure how to use the absolute continuity mentioned in one dimension in the case $n>1.$

I asked this question also at https://math.stackexchange.com.

Answer 1

This is just to complement Bazin's answer in the case where $T$ does not have a compact support.

Consider $T_\epsilon=\rho_\epsilon *T \in C^\infty (\mathbb R^n)$ which converges to $T$ as a distribution (in the weak sense) and $D_i T_\epsilon $ converges to $f_i$ in $L^1_{loc}$. Fix a ball $B_R$, $0 \leq \psi \in C_c^\infty (B_R)$, $\psi \not \equiv 0$ and use Poincarè inequality in the following form $$ \|u\|_{L^1(B_R)} \leq C \left ( \|\nabla u\|_{L^1(B_R)}+|\int_{B_R} u\psi| \right ) $$ to show that $T_\epsilon$ is Cauchy in $L^1(B_R)$ to conclude that $T \in L^1(B_R)$. The Poincarè inequality above is easily proved by contradiction, using the compactness of the embedding of $W^{1,1}(B_R)$ into $L^1(B_R)$.

Answer 2

Yes, this works. I'll discuss the case $n=2$. I hope this can be generalized, but to be honest, I haven't thought that through properly.

Let $u(x,y)=\int_0^x f_1(s,y)\, ds$, $v(x,y)=\int_0^y f_2(x,t)\, dt$; note that these functions only change on a null set if a different representative of $f_j$ is chosen and in particular the distributions induced by $u,v$ are well defined. To avoid trouble later on, fix representatives once and for all.

We have $T_x=u_x$ in $\mathcal D'$. This implies that $$ (T,\varphi) = (u,\varphi) + \int_{-\infty}^{\infty} S(\varphi(x,\cdot))\, dx \quad\quad\quad\quad (1) $$ for some $S\in\mathcal D'(\mathbb R)$. See Hormander I, Theorem 3.1.4'.

Now consider test functions of product form $\varphi(x,y)=g(x)h(y)$. If we fix one of the factors, say $h$, then the composed map $g \mapsto (U,gh)$ gives us a distribution on $\mathbb R$ for any $U\in\mathcal D'(\mathbb R^2)$. We have $$ -(T_y,gh)=(T,gh')=(u,gh')+S(h')\int g =-(v_y, gh)=(v,gh') . $$ The third and fifth expressions, thought of as one-dimensional distributions acting on $g$, are functions. Hence $$ S(h')=\int (v(x,y)-u(x,y))h'(y)\, dy $$ for almost every $x$. Here the exceptional null set may depend on $h'$. However, since a countable set of $h$'s suffices to determine $S$ on the codimension $1$ subspace of derivatives $h'$, it is in fact true that $v(x,y)-u(x,y)=w(y)+c(x)$ for almost every $x$. See here. Hence $S'=-w$ and thus also $S$ itself are functions. Now we can go back to (1) to confirm that $T$ is a function also.

Answer 3

If $T$ is a distribution on $\mathbb R^n$ such that $\nabla T$ belongs to $L^1_{\text{loc}}$, the isoperimetric inequality implies that $$ T\in L^\frac{n}{n-1}_{\text{loc}}. \tag{1} $$ To prove this, you need only to multiply $T$ by a smooth compactly supported function $\chi$, so that $u=\chi T$ is a distribution with compact support and with gradient in $L^1(\mathbb R^n)$. You may then define $ v_\epsilon=u\ast \rho_\epsilon $ where $\rho_\epsilon$ is a standard mollifier and you get that $v_\epsilon$ is smooth and compactly supported so that $$ c_n\Vert v_\epsilon\Vert_{L^\frac{n}{n-1}(\mathbb R^n)}\le \Vert \nabla v_\epsilon\Vert_{L^1(\mathbb R^n)}, $$ which is essentially enough to get that $u$ belongs to $L^\frac{n}{n-1}(\mathbb R^n)$ and thus (1).

Answer 4

Here's another way to get the same conclusion, maybe a tiny bit more elementary than the answers above. First, the question can be localized replacing $T$ with $T\theta$ for some arbitrary bump function $\theta$ : I assume now that $T$ is compactly supported with $\nabla T=F\in L^1(\mathbf{R}^n)$.

For a any smooth function $\varphi$ and $x,h\in\mathbf{R}^n$ there holds $$ \tau_h \varphi - \varphi = \int_0^1 (\tau_{sh}\nabla \varphi,h)\, \mathrm{d}s$$ where $( \cdot,\cdot)$ is the euclidean inner-product and $\tau_h \varphi(x) = \varphi(x+h)$. A direct bracket/duality computation gives in our case $$\tau_h T - T = \int_0^1 ( \tau_{sh} F,h) \,\mathrm{d} s:= G_h,$$ where one checks that $G_h$ is a well-defined $L^1(\mathbf{R}^d)$ element. Up to now, this is more or less equivalent to what have done Christian Remling above. Now, since $T$ has compact support, say $K$, we fix a test function $\psi$ supported in $K$ and we pick $h$ large enough so that the support of $\tau_h T$ does not meet $K$ : we have $\langle T,\psi \rangle = \langle T-\tau_h T,\psi\rangle = -\langle G_h,\psi\rangle$ and the conclusion follows.