0
$\begingroup$

Let $\Omega $ be a bounded open subset of $R^n,\; n\ge 1.$

I m asking about the existence of a subregion $\omega\subset \Omega$ such that the map $y\to y|_\omega $ from $H^2(\Omega)$ into $L^\infty(\omega)$ is continuous?.

$\endgroup$
3
  • $\begingroup$ Good luck with that, when $n\geq 2$. $\endgroup$
    – YangMills
    Nov 16, 2012 at 14:30
  • $\begingroup$ I would rather construct counterexamples. The Sobolev embedding theorem is known to be quite sharp. $\endgroup$ Nov 16, 2012 at 15:03
  • $\begingroup$ Sorry, I meant good luck when $n\geq 4$, since $H^2=W^{2,2}$ does not in general embed into $L^\infty$. $\endgroup$
    – YangMills
    Nov 16, 2012 at 19:30

1 Answer 1

2
$\begingroup$

I think it would help to specify what you mean by subregion $\omega \subset \Omega$. In general, it is absolutely true for $\omega$ a lower dimensional object. For example, when $n=2$ and $u \in H^1(\Omega)$, you can expect that for most (Lebesgue almost every) slices the restriction $u|_\omega$ is absolutely continuous (and therefore continuous and bounded).

If you want a subregion of full measure, then the dimension and exponent are important. A nice example is a function like $|x|^a$, whose derivative is $a|x|^{a-1}$ and second derivative is $a(a-1)|x|^{a-2}$. Computing the $L^2$ norm of this second derivative near zero, we find that we require

$\int_0^1 r^{2(a-2)}\;r^{n-1}dr <\infty$

which is equivalent to saying $2(a-2)+n-1>-1$, so that we should have $a>2-\frac{n}{2}$. Therefore, if $n > 4$, the function $|x|^{-\epsilon}$ is in $H^2(B)$ but not in $L^\infty(B)$. Then, you can define

$u(x):= \sum_n \frac{1}{2^n} |x-x_n|^{-\epsilon}$,

where $(x_n)_n$ is a dense set and obtain a function which is unbounded in every open set and still in $H^2$, by the above calculation.

Also, YangMills comment is right, since we see when $n=4$ it should still be possible, but one should use logarithms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.