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

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.

• If I remember correctly, the answer is 'when the integral of the $1$-form $\phi$ along any two paths joining $p$ and $q$ has a value $F(p,q)=-F(q,p)$ that depends only on the endpoints $p$ and $q$'. I think that there is also an answer based on $d\phi=0$ where the exterior derivative $d$ is suitably extended to the appropriate Sobolev spaces. Nov 14, 2014 at 12:04
• Ah. Thankyou for that! Can you suggest any relevant references?
– DCM
Nov 14, 2014 at 13:43
• Proposition 4.3.9 on page 334 in Horváth's Topological Vector Spaces and Distributions gives $\partial_i f_j=\partial_j f_i$ 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 $\Omega$ ?
– TaQ
Nov 14, 2014 at 13:44
• I will certainly investigate Horvath's book. Thanks!
– DCM
Nov 14, 2014 at 13:47
• @DCM, your question is essentially the same question as the one you linked to, except with one level of regularity less. As the discussion there shows, this is a rather non-trivial question and the best available information seems fairly recent. So, if this question as posed is really of importance to you, you should look into that literature as cited there by Jochen and Pedro. Nov 14, 2014 at 15:33

$\def\ssp{\kern.4mm}$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 $f$ .