Let $\Omega$ be a convex, bounded open subset of $\mathbb{R}^d$, and let $C^1(\bar \Omega)$ be usual space of continuous functions on $\bar \Omega$ which are $C^1$ in $\Omega$ and whose partials in $\Omega$ are bounded and uniformly continuous in $\Omega$. My question is about the natural differentiation operator $D:C^1(\bar \Omega)\longrightarrow C(\bar \Omega)^d$, given by

$$ Du=(\partial_1u,\dots,\partial_du), $$

where $\partial_i u$ is the unique extension to $\bar\Omega$ of the partial derivative of $u$ in the $e_i$ direction. My question is about the failure of surjectivity of $D$ when $d>1$: for which $f=(f_1,\dots,f_d)\in C(\bar \Omega)^d$ is there a $u\in C^1(\bar \Omega)$ such that $Du=f$?

There are (I think) some obvious necessary conditions concerning the relationship between the distributional derivatives of the $f_i$s if $f$ is to be `$D$ exact', and some restriction needs to be in effect to take care of what has to happen with line integrals. However, this question is a little outside my usual area of operation, so I thought I'd see if anyone know the answer. It occurs to me that there may be a connection with this question: Poincare lemma for non-smooth differentiable forms.

Topological Vector Spaces and Distributionsgives $\partial_i f_j=\partial_j f_i$ as a necessary and sufficient condition in the case $\Omega=\mathbb R\kern.4mm^n$ . Maybe the proof can be adapted to the case of bounded convex $\Omega$ ? $\endgroup$6more comments