Let $\Omega$ (possibly unbounded) be a domain of $\mathbb{R}^{n}$ with smooth boundaries. Consider $$\gamma_{D}^{0}:C^{\infty}(\overline{\Omega})\to C^{\infty}(\partial\Omega)\\\gamma_{0}^{D}u=u|_{\partial\Omega}$$ and $$\gamma_{N}^{0}:C^{\infty}(\overline{\Omega})\to C^{\infty}(\partial\Omega)\\\gamma_{0}^{N}u=\frac{\partial u}{\partial\nu}|_{\partial\Omega}$$ Extending both to $H^{1}(\Omega)$ leads to the operators $\gamma_{D}:H^{1}(\Omega)\to H^{1/2}(\partial\Omega)$ and $\gamma_{N}:H^{1}(\Omega)\to H^{-1/2}(\partial\Omega)$. What happens if I use $H^{2}(\Omega)$ as domain for the 2 operators instead? Does the neumann trace stay in $L^2$ in this case?
Edit: I imagine that you "lose" half derivative for Dirichlet and one and a half for Neumann. But, does this hold for general domains?
Edit2: In the article A Proof of the Trace Theorem of Sobolev Spaces on Lipschitz Domains by Ding it is reported that for $s>3/2$ the Dirichlet trace $\gamma_{D}:H^{s}(\Omega)\to H^{s-1/2}(\partial\Omega)$is not continuous. But instead is continuous as an operator acting between $H^{s}(\Omega)$ and $H^{1}(\partial\Omega)$.
I am not sure if this result is valid for unbounded Lipschitz domains (Theorem 1 of the paper refers to Bounded sets, while Theorems 2 and 3 do not). If it is the case, does that mean that the Neumann trace is then continuous in $L^{2}(\partial\Omega)$?
If $\Omega$ is uniformly $C^{2,1}$, then $$\gamma^0_D:H^2(\Omega)\to H^{3/2}(\partial\Omega)$$ and $$\gamma^0_N:H^2(\Omega)\to H^{1/2}(\partial\Omega)$$ are both continuous. It's Theorem 18.51 in the book Leoni "A first course in Sobolev spaces," AMS, second edition, 2017.
The definition of uniformly $C^{2,1}$ is due to Stein. In particular, a half-space, or the supergraph of a function $f:\mathbb{R}^{d-1}\to\mathbb{R}$ of class $C^{2,1}$ is a domain with uniformly $C^{2,1}$ boundary, but you can have more general unbounded domains.
The problem in Ding is that for Lipschitz domains, it is difficult to define $H^{s}(\partial\Omega)$. If $$\Omega=\{x=(x',x_d)\in \mathbb{R}^{d-1}\times \mathbb{R}:\, x_d>f(x')\}$$ and you have a smooth function $u:C^\infty(\overline{\Omega})$, then its trace should be $v(x'):=u(x',f(x'))$, $x'\in\mathbb{R}^{d-1}$. If $f$ is only Lipschitz, you can take the first derivatives of $v$ but not the second or fractional derivatives of an order bigger than one.