MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
If $p > n$, then this won't work, because $W^{1,p} \subset C^{\alpha}$ for some $\alpha > 0$. So if $u$ is not Holder continuous on $\overline{U}$, it cannot be extended in $W^{1,p}$ beyond $\overline{U}$. – Deane Yang Apr 21 '12 at 21:58
@ Prof. Deane Yang : Thank you. If we assume $ p > n $, then is there any condition on $u$ that would make sure that the smooth Sobolev function $u$ on $ U $ can be extended to $Eu \in C^{\infty}(\mathbb{R}^n) \cap W^{1,p}(\mathbb{R}^n)$. Is Holder continuity of $u$ or that $ u $ is Lipchitz enough for this ? Actually, I am looking at this problem as a part of another problem, and there I have $ p > 2, u $ is Lipchitz upto the boundary, because its derivatives are bounded in the interior. – Analysis Now Apr 21 '12 at 22:37
For my case, $n = 2$. – Analysis Now Apr 21 '12 at 22:40

To my taste, the cleanest approach to the extension problem is contained in Stein's 1970 book "Singular integrals and differentiability properties of functions". For a bounded open set with Lipschitz boundary, he constructs a 'universal' extension operator which, when applied to a $W^{k,p}$ function on the open set, produces a $W^{k,p}$ function on the whole space, with support in a fixed neighbourhood, and suitable bounds on the norm. The method also works if the set is unbounded, provided some uniform bounds at infinity are assumed on the coordinate patches at the boundary. It is not necessary at any step to assume continuity or smoothness up to the boundary.

As Pietro mentions in his answer, the extension property is blatantly false if the boundary contains inward cusps, but I understand from your question that your boundary is smooth hence at least Lipschitz.

share|cite|improve this answer

Extension is a well studied topic in the theory of Sobolev spaces, included in any treatise on the topic. Have a look e.g. to Adams' Sobolev spaces, Chapter 4 (Interpolation and Extension theorems). For positive results, you need geometric properties of domains.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.