# Weak divergence implies weak differentiability of components?

Suppose $\Omega$ is an open set in $\Bbb{R}^N$ and $\sigma : \Omega \to \Bbb{R}^N$ is a field with all components belonging to $L^2(\Omega)$.

We say that $\sigma$ has weak divergence if there exists a function $w \in L^2(\Omega)$ such that for all $\varphi \in C_c^\infty (\Omega)$ we have

$$\int_\Omega \sigma \cdot \nabla \varphi=-\int_\Omega w \varphi.$$

My question is:

Can we establish a result of the form: if $\sigma$ has weak divergence then each component of $\sigma$ is weakly differentiable?

The idea is that I've seen this technique in proving that if a function $u$ is $H^1(\Omega)$ and it satisfies some convenient weak condition then $\nabla u$ has a weak divergence and therefore $u \in H^2(\Omega)$. The book where I've seen this technique is aimed for engineers, and therefore it is not very rigorous. That is why I've asked this question.

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If I understood your comment at the end what was actually shown was that a function $u\in H^1(\Omega)$ satisfied $\Delta u=f\in L^2(\Omega)$ weakly. This is a standard consequence of "elliptic regularity". What you are asking can't possibly be true since you have no control over most of the derivatives of each component of $\sigma$. –  Rbega Oct 23 '12 at 17:14
That is what I was thinking, but I wanted another confirmation. –  Beni Bogosel Oct 23 '12 at 17:35
as Rbega says: if it is weakly differentiable (i.e., $\sigma\in H^1(\Omega)$, then it clearly also has a weak divergence, but (generally) not viceversa. i do not even see why elliptic regularity applies here. what one can generally say is that if $u\in H^1(\Omega)$ is a weak solution of $\Delta u=f\in L^2(\Omega)$, then $u\in H^2_{loc}(\Omega)$. –  Delio Mugnolo Oct 23 '12 at 19:25

So $\sigma=\sum_{1\le j\le n}\sigma_j(x)\frac{\partial}{\partial x_j}$ is a vector field with distributions coefficients $\sigma_j$ and divergence in $L^2$: $$\sum_{1\le j\le n}\frac{\partial \sigma_j}{\partial x_j}\in L^2.$$ If I understand your question correctly, you ask if this implies that each $\sigma_j$ belongs to $L^2$. Of course not since you can choose all $\sigma_j=0$ for $j\ge 2$ and $\sigma_1$ to be any distribution in the variables $(x_2,\dots, x_n)$.
Note nevertheless that when $n=2$ and $\text{div} \sigma=0$, then there exists a function $\psi(x_1,x_2)$ such that $$\sigma_1=\frac{\partial \psi}{\partial x_2},\quad \sigma_2=-\frac{\partial \psi}{\partial x_1}.$$ If $\sigma$ happens to be continuous (resp. locally $L^2$), then $\psi$ is locally Lipschitz continuous (resp. locally $H^1)$.