Are compactly supported continuous functions dense in weighted Sobolev spaces?

44 Views Asked by At

I have a question, here in this paper, the author define a weighted Sobolev space $W^{m,p}(D,w)$, with $1 \leq m < \infty$, $1 \leq p < \infty$, as a normed space of locally summable, $m$ times weakly differentiable functions $f : D → \mathbb{R}$ ($D ⊂ R^n$). $$ || f ||_{W^{m,p} (D,w) } = \left( \int_D |f(x)|^pw(x)dx \right)^{1/p}+ \sum_{ |\alpha|= m} \left( \int_D |D^{\alpha}f(x)|^pw(x)dx \right)^{1/p}.$$ with $D^{\alpha}f$ is the weak derivative of order $\alpha$ of the function $f$. He states the following density result:

Theorem 1. Let $D ⊂ R^n$ be an open set and let a weight $w$ satisfy the $A_p$-condition, i.e. $w$ is locally integrable and

$${\displaystyle \left({\frac {1}{|B|}}\int _{B}w (x)\,dx\right)\left({\frac {1}{|B|}}\int _{B}w (x)^{{\frac {1}{p-1}}}\,dx\right)^{p-1}\leq C<\infty ,}$$ where $|B|$ is the Lebesgue measure of the ball $B$. Then $W^{m,p}(D,w), 1 \leq m < \infty, 1 < p < \infty$ is a Banach space. Smooth functions of the class $W^{m,p}(D, w)$ are dense in $W^{m,p}(D, w)$.

I want to understand what's the problem with $W^{1, \infty}(D,w)$, wouldn't it also be Banach? For example if $D = \mathbb{R}^n$. We can also change the density conclusion to the class of smooth functions with compact support instead of just smooth ones. As this paper is from 2009, if anyone knows of any more recent advances that address my question, I would appreciate the reference.