论文标题
通过距离函数和Lorentz空间来表征零迹线的功能
Characterization of functions with zero traces via the distance function and Lorentz spaces
论文作者
论文摘要
考虑一个常规域$ω\ subset \ mathbb {r}^n $，让$ d（x）= \ operatatorName {dist}（x，x，\partialΩ）$。表示$ l^{1，\ infty} _a（ω）$具有绝对连续的icoasinorms的$ l^{1，\ infty}（ω）$的功能空间。该集基本上小于$ l^{1，\ infty}（ω）$，但与此同时，基本上大于所有$ l^{1，q}（ω）$，$ q \ in [1，\ infty）$的联合。 1980年代末的一个经典结果指出，（1，\ infty）$和$ m \ in \ mathbb {n} $，$ u $，$ u $属于sobolev space $ w^{m，p} _0 _0（ω）$，并且仅当$ u/d^m \ in l^p（in l^p（p（p）） l^p（ω）$。在随之而来的几十年中，几位作者花了大量精力来放松表征条件。最近，事实证明，$ u \ in w^{m，p} _0（ω）$，仅当$ u/d^m \ in l^1（ω）$和$ \ weft | \ nabla^m u \ r \ right | \ in l^p（ω）$中。在本文中，我们表明，对于$ n \ geq1 $和$ p \ in（1，\ infty）$，我们在w^{1，p} _0（ω）$中具有$ u \，并且仅当$ u/d \ in l^{1，\ infty} _a（Ω）$ u/d \ in l^{1，\ undy} _a（ω）$ wept | weft | wept | wept | \ w |此外，我们提出了一个反例，该例子表明，在放松条件后，$ u/d \ in l^{1，\ infty} _a（ω）$ to $ u/d \ in l^{1，\ infty}（ω）$等价不再持有。
Consider a regular domain $Ω\subset \mathbb{R}^N$ and let $d(x)=\operatorname{dist}(x,\partialΩ)$. Denote $L^{1,\infty}_a(Ω)$ the space of functions from $L^{1,\infty}(Ω)$ having absolutely continuous quasinorms. This set is essentially smaller than $L^{1,\infty}(Ω)$ but, at the same time, essentially larger than a union of all $L^{1,q}(Ω)$, $q\in[1,\infty)$. 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 $W^{m,p}_0(Ω)$ if and only if $u/d^m\in L^p(Ω)$ and $\left|\nabla^m u\right|\in L^p(Ω)$. During the consequent decades, several authors have spent considerable effort in order to relax the characterizing condition. Recently, it was proved that $u\in W^{m,p}_0(Ω)$ if and only if $u/d^m\in L^1(Ω)$ and $\left|\nabla^m u\right|\in L^p(Ω)$. In this paper we show that for $N\geq1$ and $p\in(1,\infty)$ we have $u\in W^{1,p}_0(Ω)$ if and only if $u/d\in L^{1,\infty}_a(Ω)$ and $\left|\nabla u\right|\in L^p(Ω)$. Moreover, we present a counterexample which demonstrates that after relaxing the condition $u/d\in L^{1,\infty}_a(Ω)$ to $u/d\in L^{1,\infty}(Ω)$ the equivalence no longer holds.
