The question I would be asking is roughly : do the smooth Sobolev functions defined on an open bounded domain extend to smooth Sobolev functions on the Euclidean space ?

For detail :

Fix $p \geq 1. $ By the word 'smooth', I will always mean $C^{\infty}$ .Let $U, V$ be two bounded, connected open domains in $ \mathbb{R}^n $ with smooth boundary, so that $\bar{U}\subset V $. Let $u \in W^{1,p}(U) \cap C^0(\bar{U}) $. In many PDE books ( for example , L.C. Evans' "Partial Differential Equations", P. 254 ) there are standard methods described to produce an extension $Eu$ of $u$ such that :

$Eu \in W^{1,p}(\mathbb{R}^n), support (Eu) \subset V, Eu = u $ a.e.on $U$ . The method is first to treat the case where $u \in W^{1,p}(U) \cap C^0(\bar{U}) \cap C^{\infty}(\bar{U}) $ ,locally flatten the boundary $ \partial{U} $, extend $u$ by higher order reflection across the boundary, using partition of unity and then approach arbitrary $u \in W^{1,p}(U) $ by $C^{\infty}(\bar{U})$ functions in $W^{1,p}(U)$ -norm. We do not even need $ u \in C^0(\bar{U}) $ in this proof .

I understand that if $u \in C^{\infty}(\bar{U}) $as well. , then $Eu \in C^\infty(\mathbb{R}^n)$, but is the same true if $u \in C^{\infty}(U) \cap C^0(\bar{U}) $ only, not $C^{\infty}(\bar{U})$ ? This is my main question.