When is a `1-form' with continuous coefficients exact?

Question

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.

Answer 1

$\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 $f$ .