Supposing that D is a bounded Lipschitz domain (and not smooth) in $\mathbb{R}^d$. From what I know, it is known that the trace operator is well-defined and continuous from $H^s(D)$ to $H^l(\partial D)$ when $l=s-1/2$ and $ 1/2<s<3/2$. My question is what happens when $s>3/2$, is the above result true? Also I am interested in versions concerning more general spaces like Besov or Tribel-Lizorkin. Any answer or reference will be appreciated.
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$\begingroup$ Should be l=s-1/2. $\endgroup$– EddyCommented Jun 26, 2017 at 14:37
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2 Answers
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The problem here is with the definition of $H^\ell (\partial D)$. Typically, $\partial D$ is a manifold and you can define this set by local charts. In order to define $H^\ell (\partial D)$ with $0 < \ell \le 1$, it is sufficient to have bi-Lipschitz charts, whereas for $1 < \ell \le 2$ you would typically need diffeomorphisms with Lipschitz derivatives ($\mathcal{C}^{1, 1}$).
Going further, you also need to know how the boundary is attached to the domain, and this is the point of defining a Lipschitz domain by requiring the boundary to be a graph of Lipschitz function instead of say a Lipschitz manifold.
answered Jul 14, 2017 at 8:14
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For Sobolev spaces $W_p^l(D)$, $l\in \mathbb N$, $1<p<\infty$, traces on a Lipschitz boundary $\partial D$ is considered in O. V. Besov, V. P. Il'in, S. M. Nikol'skii, Integral representations of functions and imbedding theorems, $\S20$, Theorem 20.12. Traces are completely characterised in the sense of necessary and sufficient conditions for a system of functions $\{f^{(\alpha)}\}$ on $\partial D$ to be traces of some function $f\in W_p^l(D)$ and its derivatives up to a certain order.
