3

Let be $u\in W^{1,2}(U)$, where $U\subset \Bbb R^n$ is an bounded open. Define $$\sup_{\partial U} u_+:=\inf \{\ l\in \Bbb R: (u_+-l)_+\in W_0^{1,2}(U)\}.$$why the set is not empty?If $k\in \Bbb R$ is such that $$\sup_{\partial U} u_+\le k<\sup_U u$$ then why $(u-k)_+\in W^{1,2}_0(U)?$

The question is easy if $u\in C^1(\overline{U})$ but why it is right in general?

Domenico Vuono
  • 2,672
  • I'd expect that $\sup_{\partial U} u_+$ coincides with the essential supremum of $u|{\partial U}$ (the trace), which can be infinite. In other words, I'd guess that the set may be empty and in this case we put $\sup{\partial U} u_+ = \infty$. – Michał Miśkiewicz Jun 14 '21 at 21:32
  • And $k$ how can i choose ? – Domenico Vuono Jun 15 '21 at 07:37
  • @DomenicoVuono why don't you tell us where the statement comes from? Maybe you do not need to say that $\sup_{\partial U} u_+<\infty$ for every $u\in W^{1,2}(U)$ (which, as Michał Miśkiewicz pointed out, is not true). The second part of the statement is a standard exercise in Sobolev spaces, for every $k\in \mathbb R$: it suffices to show that $u_+\in W^{1,2}$ when $u$ is. The $0$-boundary condition then holds by definition of $k$, since $u-k\leq 0$ on $\partial U$. The condition $k<\sup_U u$ is just there to grant that $(u-k)_+\neq 0$ imo. – AlephBeth Jun 19 '21 at 16:16

0 Answers0