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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?

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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.

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    $\begingroup$ One more reason to become familiar with capacities, thank you :) $\endgroup$ – Dreipunkt Jun 25 '16 at 16:05

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