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Let $\Omega\subset\mathbb{R}^N$ be a bounded domain of class $C^2$. Take $u\in W_0^{1,p}(\Omega)$ for $p\in (1,\infty)$ and assume that $$\frac{\partial u}{\partial \nu}(x)<0,\ \forall\ x\in \partial\Omega,$$

where $\nu$ is the exterior unitary normal vector through the boundary of $\Omega$ and I am assuming that the normal derivative does exist in the classical sense. For $\delta>0$, define $$\Omega_\delta=\{x\in \Omega:\ \operatorname{dist}(x,\partial\Omega)<\delta\},$$

where $\operatorname{dist}$ is the distance function. My question is the following:

Can we find $\delta>0$ and a sequence $u_n\in W_0^{1,p}(\Omega)$ such that $u_n\to u$ in $W^{1,p}(\Omega_\delta)$ and the functions $u_n$ are $p$-subharmonic in $\Omega_\delta$, that is, $$\int _{\Omega_\delta}|\nabla u_n|^{p-2}\nabla u_n\nabla \varphi\le 0,\ \forall\ \varphi\in C_0^\infty(\Omega_\delta).$$

Any idea is appreciated. I have nothing for this problem by now.

Tomás
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  • For $u \in W_0^{1,p}(\Omega)$ the trace of the normal derivative $\partial u / \partial \nu$ is not well-defined. – gerw May 23 '16 at 06:30
  • What do you mean by "exists in the classical sense"? That $u$ is $C^1$ in a neighborhood of the boundary? – gerw May 23 '16 at 16:51
  • I mean that $$\lim_{h\to 0^+}\frac{u(x-h\nu)-u(x)}{h}>0.$$ – Tomás May 23 '16 at 17:51
  • But for $p \le N$, the point values of $u$ are not well-defined. – gerw May 23 '16 at 19:00
  • You are right @gerw. I have to think more on this question. Thank you for yoru time. – Tomás May 26 '16 at 10:42
  • No problem. Does it help that you can define a normal derivative for $u \in H_0^1(\Omega)$ with $\Delta u \in L^2(\Omega)$? – gerw May 26 '16 at 13:51
  • Interesting, I didn't knew it. Can you give me references for it? Is it true also for the $p$-Laplacian? I mean: $u\in W_0^{1,p}(\Omega)$ with $\Delta_p u\in L^p (\Omega)$. – Tomás May 26 '16 at 13:58
  • Let $E : H^{1/2}(\partial\Omega) \to H^1(\Omega)$ a right-inverse of the trace map. Then, you define $\partial u /\partial \nu \in (H^{1/2}(\Omega))^*$ via $\partial u/\partial \nu (v) := \int_\Omega E(v) , \Delta u + \nabla E(v) \cdot \nabla u,\mathrm{d}x$ for $v \in H^{1/2}(\partial\Omega)$. This formula is motivated by integration by parts. I think it also works for $u \in W_0^{1,p}(\Omega)$, but you need $\Delta u$ (and not the $p$-Laplacian). I do not have a reference by hand, sorry. – gerw May 26 '16 at 14:07
  • Is your definition independente of $E$? – Tomás May 26 '16 at 14:15
  • I think so. You might also check http://mathoverflow.net/a/109765/32507 – gerw May 26 '16 at 18:24

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