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.



Weak divergence implies weak differentiability of components?
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 $H^1)$.


For the general statement that you mention, note that the gradient operator is elliptic, which is not the case of the divergence operator.


