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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?.

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Good luck with that, when $n\geq 2$. –  YangMills Nov 16 '12 at 14:30
    
I would rather construct counterexamples. The Sobolev embedding theorem is known to be quite sharp. –  Alexander Shamov Nov 16 '12 at 15:03
    
Sorry, I meant good luck when $n\geq 4$, since $H^2=W^{2,2}$ does not in general embed into $L^\infty$. –  YangMills Nov 16 '12 at 19:30

1 Answer 1

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.

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