Let $U \subset \mathbb R^3$ be an open, bounded and connected set with a $C^2-$regular boundary $\partial U$. I'm trying to understand the following implication:

If $f\in W^{2-1/2,2}(U)$ then $f{\vert}_{\partial U} \in W^{1,2}(\partial U)(*)$

So, I'm aware of this theorem:

General Trace Theorem: if $f\in W^{1-1/p,p}(\partial \Omega)$, then there exists a function $f \in W^{1,p}(\Omega)$ such that $f{\vert}_{\partial \Omega}=f$

QUESTION: Is the above theorem still valid if we replace $\partial \Omega$ with $U$ and $\Omega$ with $\partial U$ so that $(*)$ makes sense? If not, is there any other way to deduce $(*)$?

Any help is appreciated. Thanks in advance!

EDIT: regularity in $(*)$ fixed

  • $\begingroup$ You're asking about whether you can extend a function from the boundary into the domain? $\endgroup$ – Ian Nov 13 '18 at 15:52
  • $\begingroup$ I think that you have the regularities reversed. Anyway, see here. $\endgroup$ – Giuseppe Negro Nov 13 '18 at 15:55
  • $\begingroup$ @Ian no... I am trying to understand the implication in $(*)$. There the extension is from the domain to boundary... $\endgroup$ – kaithkolesidou Nov 13 '18 at 15:56
  • $\begingroup$ The given theorem is about extending from the inside of the domain to the boundary. Switching the two around would amount to extending a function given on the boundary to the domain, unless I am misunderstanding your intended meaning. $\endgroup$ – Ian Nov 13 '18 at 15:57
  • $\begingroup$ @GiuseppeNegro What do you mean by reversed regularities? $\endgroup$ – kaithkolesidou Nov 13 '18 at 16:00

$\newcommand{\R}{\mathbb{R}}$ The implication (*) is invalid, and for obvious reasons. The trace $f|_{\partial U}$ is kind of a restriction of $f$, so you cannot expect it to have more derivatives than $f$!

To have an example, choose your favorite function $g \in W^{1/2,2}(\R)$ that doesn't belong to $W^{1,2}(\R)$. Then consider \begin{align*} U &= \{ (x,y,z) \in \R^3 : z > 0 \}, \\ f(x,y,z) & = g(x,y). \end{align*} Then $f$ is locally in $W^{1/2,2}$, but its restriction to the boundary $\partial U = \R^2 \times \{0\}$ is simply $g$, which is not in $W^{1,2}$.

Of course, one can modify this example to have a bounded domain.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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