Question on Sobolev extension onto boundary

Question

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

You're asking about whether you can extend a function from the boundary into the domain? — Ian, Nov 13 '18 at 15:52
I think that you have the regularities reversed. Anyway, see here. — Giuseppe Negro, Nov 13 '18 at 15:55
@Ian no... I am trying to understand the implication in $(*)$. There the extension is from the domain to boundary... — kaithkolesidou, Nov 13 '18 at 15:56
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. — Ian, Nov 13 '18 at 15:57
I mean that $1-1/2$ goes with $\partial U$ and $1$ goes with $U$, as in the statement of the trace theorem which you recalled. — Giuseppe Negro, Nov 13 '18 at 16:03
@Ian Well I am a bit puzzled this is why I may stated my question incorrectly. All I try is to understand why this implication (*) is valid. — kaithkolesidou, Nov 13 '18 at 16:04
@GiuseppeNegro exactly! So I'm wondering if this could be opposite. Otherwise I can't see why (*) holds — kaithkolesidou, Nov 13 '18 at 16:06

score 0 · Answer 1 · answered Nov 13 '18 at 18:51

$\newcommand{\R}{\mathbb{R}}$ 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.

Question on Sobolev extension onto boundary

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