my question is as follows: Given a bounded, simply connected Lipschitz domain $\Omega\subset\mathbb{R}^{3}$ (so in particular open) and a suitably smooth function $u\colon\Omega\to\mathbb{R}^{3}$, assume that for any $\varepsilon>0$ there a suitably smooth function $v_{\varepsilon}\colon\Omega_{\varepsilon}\to\mathbb{R}$ such that $\nabla v_{\varepsilon}=u$ on $\Omega_{\varepsilon}:=\{x\in\Omega\colon\text{dist}(x,\partial\Omega)>\varepsilon\} $ whenever $\Omega_{\varepsilon}$ itself is simply connected. I would be interested in whether this suffices to conclude that $u=\nabla v$ on $\Omega$ for some suitable $v$, and how $v$ can be obtained from the $v_{\varepsilon}$'s. Clearly, one should somehow send $\varepsilon\searrow 0$, but isn't this somehow pointless -- I mean, does $\Omega_{\varepsilon}$ need to simply connected for some $\varepsilon>0$ at all? If $\Omega$ is simply connected only, then clearly not -- however, might there be any hope invoking the Lipschitz condition?
Best & thanks, Jim
Given any $p\in\Omega$ there is an $\epsilon>0$ such that $p\in\Omega_\epsilon$. From $u=\nabla v_\epsilon$ in $\Omega_\epsilon$ it then follows that ${\rm curl}\,u\equiv0$ in the neighborhood of $p$. Since $p\in\Omega$ was arbitrary we can conclude that $u$ has vanishing curl throughout $\Omega$, and as $\Omega$ is simply connected this implies that $u=\nabla v$ for a potential $v:\>\Omega\to{\mathbb R}$.
If you want to work with the $v_\epsilon$ note that these are uniquely determined up to an additive constant. By requiring $v_\epsilon(p_0)=0$ for some "origin" $p_0\in\Omega$ you then can be sure that $v_{\epsilon'}\restriction\Omega_\epsilon=v_\epsilon$ when $\epsilon'<\epsilon$, so that the convergence is obvious.