Let $\,D\subseteq\mathbb{R}^{n-1}$ be a convex bounded domain. Let$A:D\to(0,\infty)$ be a Lipschitz continuous function. Let $\,\Omega\,$ be bounded domain in $\,\mathbb{R}^{n}\,$ of the form \begin{equation} \Omega:=\left\{(x',x_{n})\in\mathbb{R}^{n}|\,x'\in D,\,0\leqslant x_{n}<A(x')\right\}. \end{equation}

Let $u\in C^{1}(\overline{\Omega})$ be such that \begin{equation} u=0\quad\mbox{on}\quad \overline{\Omega} \cap\{x_{n}=0\}. \end{equation} QUESTION: Let $p>1$ and let $\,\mathbb{R}_{+}^{n}=\{x\in\mathbb{R}^{n}|\,x_{n}>0\}.\,$ Given $\,\epsilon>0,\,$ does there exist a function $\,\tilde{u}\in W_{0}^{1,p}(\mathbb{R}_{+}^{n})\,$such that \begin{equation} \tilde{u}|_{\Omega}=u, \mbox{and}\quad \int_{\mathbb{R}_{+}^{n}\setminus\Omega}|\nabla\tilde{u}|^{p}\,dx<\epsilon. \end{equation}

  • $\begingroup$ Suppose Omega is, say, a square or cube, and u(x) = x_n. Can you solve your problem in this or other simplified situation? $\endgroup$ – Tommi Feb 23 '15 at 11:59
  • $\begingroup$ Somehow this reminds me about the definition of $p$-capacity of the set $\Omega$. I suspect the geometry of the set $\Omega$ plays a role here. $\endgroup$ – Juhana Siljander Mar 16 '15 at 20:33

If you want to prove that the answer is no, you can relax the problem, and simplify it. Choose $u=x_n$, $p=2$ and minimise $$ I:=\int_{R^n_{+}} |\nabla v - 1_\Omega e_n|^2 $$ Clearly, if what you are looking for was possible, then there would be a bounded sequence $v_n$ in $W^{1,2}_0$ such that $I(v_n)\leq\frac{1}{n}$, and therefore the minimum would be 0. But you can solve that problem (take $\Omega$ to be a cube or half a disk as suggested in the comments), and find what the minimum is : it is not zero by Liouville's Theorem for example (but you can compute it explicitely for the nice geometries mentioned above).


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