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](https://mathoverflow.net/questions/180451/poincare-lemma-for-non-smooth-differentiable-forms). 



When is a `1-form' with continuous coefficients exact?
$\def\ssp{\kern.4mm}
$Here is a sketch of proof of sufficiency of $d\ssp f=0$ , i.e. of $\partial\_i f\_j=\partial\_j f\_i$ (in the distributional sense) assuming that the case where $f$ is $C^1$ is known, for which I refer e.g. to the Poincaré lemma in §V.5 on pages 124−125 in Serge Lang's *Differential Manifolds*, Springer 1988.


Fix any $x\_0\in\Omega$ , and define $u$ by $\Omega\owns x\mapsto\int\_{\,0}^{\,1}f(x\_0+t\,(x-x\_0))\cdot(x-x\_0)\,{\rm d\,}t$ . For fixed $x\in\Omega$ to prove that $d\ssp u(x)=f(x)$ , take a smooth "cut-off" function $\chi$ which has value $1$ up to "sufficiently" near to the boundary of $\Omega$ and which has value $0$ "very near" to the boundary. Further, take a "smooth bump" $\varphi$ at the origin with integral $1$ whose support is "very small". Let $\Omega\_0$ be a convex open set with compact closure included in the interior of $\chi^{-1}[\{1\}]$ and $x\_0,x\in\Omega\_0$ and $\Omega\_0-{\rm supp\,}\varphi\subset\chi^{-1}[\{1\}]$ . Then letting $g=(\chi\cdot f)\*\varphi$ and defining $v$ by $\Omega\owns x\mapsto\int\_{\,0}^{\,1}g\ssp(x\_0+t\,(x-x\_0))\cdot(x-x\_0)\,{\rm d\,}t$ , by the $C^1$ or even the smooth case, noting that $\partial\_i g\_j=(\chi\cdot\partial\_i f\_j)\*\varphi=(\chi\cdot\partial\_j f\_i)\*\varphi=\partial\_j g\_i$ holds on $\Omega\_0$ , we have $\partial\_i v=g\_i=(\chi\cdot f\_i)\*\varphi$ on $\Omega\_0$ . Taking here in place of $\varphi$ the function $\varphi\_n:z\mapsto n^d\ssp\varphi(n\ssp z)$ , and letting $n\to\infty$ , we get $\partial\_i u=f\_i$ pointwise on $\Omega\_0$ since both convergences $g\_i\ssp(\varphi\_n)\to f\_i$ and $v\ssp(\varphi\_n)\to u$ as $n\to\infty$ are uniform on $\Omega\_0$ . As $x\in\Omega$ here is arbitrarily fixed, we get $d\ssp u=f$ on $\Omega$ which further extends to the boundary by the assumed uniform continuity of $f$ .


