Approximation of bounded Sobolev functions in $L^\infty$

399 Views Asked by At

I'm trying to understand the problem of my former question in detail, and the crucial point (at least in my attempts to solve the problem) seems to be the following:

Let $\Omega\subset \mathbb R^d$ be a bounded domain with smooth boundary. Is there a $p>2$ such that $W^{1,p}_0(\Omega)\cap L^\infty(\Omega)$ is dense in $W^{1,2}_0(\Omega)\cap L^\infty(\Omega)$ with respect to the $L^\infty$-norm?

Is there a proof or a nice counter-example or at least a book or paper which deals with such questions?

1

There are 1 best solutions below

1
On BEST ANSWER

A counterexample for $d=2$: let $\Omega$ be the disk $\{x:\|x\|<\exp(-\exp(\pi))\}$, and $$ f(x) = \sin \log \log \frac{1}{\|x\|} $$ This function is in $W_0^{1,2}(\Omega)\cap L^\infty(\Omega)$ (relevant calculations here) but has a discontinuity at $0$, and moreover cannot be made continuous by redefining it on a set of measure zero.

On the other hand, any function in $W^{1,p}(\Omega)$ for $p>2$ is continuous (by Morrey-Sobolev) and so the limit of such functions in $L^\infty$ must have a continuous representative.


I don't have a counterexample in higher dimensions, but suspect the same problem occurs there: Sobolev functions can be made continuous after removing a set of small capacity, and the relevant capacity depends on the exponent $p$. So it seems that the functions in $W^{1,p}$ are continuous in too many places to approximate the functions in $W^{1,2}$ uniformly.