Characterization of functions with zero traces via the distance function and Lorentz spaces (2209.13486v2)

Abstract: Consider a regular domain $\Omega \subset \mathbb{R}^N$ and let $d(x)=\operatorname{dist}(x,\partial\Omega)$. Denote $L^{1,\infty}_a(\Omega)$ the space of functions from $L^{1,\infty}(\Omega)$ having absolutely continuous quasinorms. This set is essentially smaller than $L^{1,\infty}(\Omega)$ but, at the same time, essentially larger than a union of all $L^{1,q}(\Omega)$, $q\in[1,\infty)$. A classical result of late 1980's states that for $p\in (1,\infty)$ and $m \in \mathbb{N}$, $u$ belongs to the Sobolev space $W^{m,p}_0(\Omega)$ if and only if $u/d^m\in L^p(\Omega)$ and $\left|\nabla^m u\right|\in L^p(\Omega)$. During the consequent decades, several authors have spent considerable effort in order to relax the characterizing condition. Recently, it was proved that $u\in W^{m,p}_0(\Omega)$ if and only if $u/d^m\in L^1(\Omega)$ and $\left|\nabla^m u\right|\in L^p(\Omega)$. In this paper we show that for $N\geq1$ and $p\in(1,\infty)$ we have $u\in W^{1,p}_0(\Omega)$ if and only if $u/d\in L^{1,\infty}_a(\Omega)$ and $\left|\nabla u\right|\in L^p(\Omega)$. Moreover, we present a counterexample which demonstrates that after relaxing the condition $u/d\in L^{1,\infty}_a(\Omega)$ to $u/d\in L^{1,\infty}(\Omega)$ the equivalence no longer holds.