Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The trace theorem says that the restriction of a $W^{1,p}(\Omega)$ function $u$, $Tu$ belongs to $W^{1-1/p,p}(\partial\Omega)$ if $\Omega$ satisfies some smooth condition, for example, $\Omega$ is convex. Now my question is the inverse of the Trace Theorem. Suppose $\Omega$ is convex, and $\phi\in W^{1-1/p,p}(\partial\Omega)$, is there exists a fucntion $\Phi\in W^{1,p}(\Omega)$ with $\|\Phi\|\leq C\|\phi\|?$

Dees the extension theorem is related to this question? But usually the extension Theorem talks about the extension from a domain to the whole space.

share|improve this question
    
This is very closely related to my question here: mathoverflow.net/questions/70740/image-of-the-trace-operator –  Willie Wong May 10 '12 at 8:01
add comment

2 Answers

At least if $p=2$ (Hilbert case), the answer is yes. Not only that, but there exist explicit right inverses, that is linear bounded operators $L:W^{1/2}(\partial\Omega)\rightarrow W^{1,2}(\Omega)$ such that $T\circ L$ is the identity. Since such operators are pseudo-differential, it is likely that they are bounded for $p\ne2$ as well, and thus satisfy again $T\circ L={\rm id}$.

share|improve this answer
add comment

I think your question is answered in

Article (JonWal1978) Jonsson, A. & Wallin, H. A Whitney extension theorem in $L_p$ and Besov spaces Ann. Inst. Fourier (Grenoble), 1978, 28, vi, 139-192

and

Article (Marsch1987) Marschall, J. The trace of Sobolev-Slobodeckij spaces on Lipschitz domains Manuscripta Math., 1987, 58, 47-65

Theorem 2 of the latter paper states that if $\Omega$ is a Lipschitz domain, with $s,p$ satisfying some inequalities (as usual), then $W^{s,p}(\Omega)$ traces to (as a surjection) some Besov space on the boundary $\partial\Omega$ and that the trace operator has bounded linear right inverse if $s-1/p$ is not an integer.

share|improve this answer
add comment

Your Answer

 
discard

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.