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.