Tangential Sobolev spaces

Question

• Let $\Omega \subset \mathbf R^n$ be a smooth domain and define $U_s=\{x\in\Omega \mid d(x,\partial \Omega)<s\}$;
• let $f\in W^{1,p}(\Omega)∩W_{\mathrm{loc}} ^{2,p}(\Omega)$;
• let $v$ be the unit normal to $\Omega$ and consider $v$ to be smooth with bounded derivatives.

Using the notation: $$ \begin{split} \nabla_N f &=\nabla f\cdot v \in L^p (\Omega)\\ \nabla_T f &=\nabla f-v\nabla_N f \in (L^p (\Omega))^n \end{split} $$ Then we have: $f\in W^{2,p} (\Omega)\iff \nabla_N f\in W^{1,p} (U_s)$ and $\nabla_T f\in (W^{1,p} (U_s ))^n$

this theorem is found in "Mathematical Tools for the Navier-Stokes Equations and Related Models Study of the Incompressible" Theorem III.3.14

The forward implication “$\Longrightarrow$” it's obvious to me, but the reverse one is harder. Can somebody please explain it ?

Here is my attempt:

Assume $\nabla_N f\in W^{1,p} (U_s)$ and $\nabla_T f\in (W^{1,p} (U_s ))^n$: then $\nabla f\in W^{1,p} (U_s )$ since $$ \nabla f=\nabla_T f+v\nabla_N f $$ and both $\nabla_T f$ and $v\nabla_N f$ are smooth enough. Now obviously $$ \nabla f\in W^{1,p} (U_s )\implies f\in W^{2,p} (U_s ). $$ Now assume further that \Omega Is bounded, if this is the case then we can find a compact $K$ such that $\Omega\setminus U_s\subset K\subset \Omega$; for instance a compact can be the closure of all the points in $\Omega$ at distance from the boundary greater or equal to $s/2$ . Then $$ f\in W_{loc}^{2,p} (\Omega)\implies f\in W^{2,p} (K). $$ Now $$ f\in W^{2,p} (K)\cap W^{2,p}(U_s )\implies f\in W^{2,p} (\Omega) $$ However the assumption $\Omega$ bounded is not present in the original theorem, and I can't see how to proceed.

Answer 1

The assumption that $\Omega$ is bounded is in fact required. (So your attempt is the correct proof, once you fix the omission in the statement.)

Counterexample: let $\Omega$ be the upper half plane. Let $\phi$ be a smooth function of compact support on the reals. Let

$$ f(x,y) = \phi(x) \frac{1}{(1+y)^m} \sin(y^k) $$

for $m,k\in \mathbb{N}$ to be determined. Certainly in a band $U_s$ around the $x$-axis you have that $f\in W^{2,p}(U_s)$.

• In order for $f$ to be in $W^{1,p}(\Omega)$, it suffices that $p(k-1-m) < -1$.
• But if $p[(k-1)^2 - m] \geq -1$, while it is true that $f\in W^{2,p}_{loc}(\Omega)$, we don't have $W^{2,p}(\Omega)$ due to the slow decay as $y\to \infty$.

So for example, if we set $k = 10$ and $m = 20$, we get a counterexample for every $p\in [1,\infty)$.