Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $f: \Omega\rightarrow \mathbb{R}$ where $\Omega\subset\mathbb{R}^d$ is bounded with lipschitz smooth boundary. Further suppose that $f\in\mathcal{H}^{\tau}(\Omega)$, $\tau>0$ and that $f$ decays rapidly to $0$ on the boundary.

Let $$ \Omega_{\delta} = \{ x\in\Omega : \inf_{y\in\partial\Omega} \left\|x-y\right\|_2 > \delta \} .$$ Where $\delta>0$ is small enough to preserve smoothness in the boundary of $\Omega_{\delta}$. See: Shrinking a Lipschitz smooth domain.

Are there any known bounds on $\left\|f\right\|_{L_2(\Omega\setminus\Omega_{\delta})}$? i.e. bounding $f$ near the boundary of $\Omega$.

Note: $ \left\|f\right\|_{L_2(\Omega)}= \left(\int_{\Omega} |f|^2\right)^{\frac{1}{2}} $

I ideally would like some bound of the form: Given $f\in\mathcal{H}^\tau(\Omega)$, $\tau>d/2$ which is zero on the boundary and $\delta$ sufficiently small then:

$\left\|f\right\|_{L_2(\Omega\setminus\Omega_{\delta})} \leq C\delta^\alpha\left\|f\right\|_{L_2(\Omega)}$ with $\alpha>0$ as large as possible (hopefully $\alpha=1$) and $C$ is a constant not depending on $\delta$ or $f$.

share|improve this question
    
actually I still have a doubt about your notation: what is $\|\cdot\|_{L_2(\Omega)}$ ? –  Pietro Majer Nov 20 '10 at 9:19
    
I have editted the post to answer your comment. This is the standard notation and definition, right? –  alext87 Nov 20 '10 at 16:33
    
I think Pietro is referring to the norm applied to $x-y$ in the second paragraph. The notation is normally used for the $L_2$ norm of a function only. What do you mean when you apply it to a vector? –  Deane Yang Nov 20 '10 at 16:56
    
And what do you want to bound the $L_2$ norm of $f$ by? You can make the norm as large as you want by rescaling $f$. –  Deane Yang Nov 20 '10 at 16:57
1  
OK, thank you for clarifying; it's as I had guessed. I've added the exponent 1/2 that was missing in the definition of the $L^2$ norm of $f$. Note that the last inequality you wrote is trivially and universally true with no hypothesis, taking $C=1$. Feel free to ask if something is not clear to you about the answer below. –  Pietro Majer Nov 21 '10 at 12:19

1 Answer 1

up vote 3 down vote accepted

For sure $\|f\|_ {L^2(\Omega\setminus\Omega_\delta)}=o(1)$ as $\delta\to0$ for any $f\in L^2(\Omega)$ (this, even if $\Omega$ was not bounded). For $f\in L^\infty(\Omega)$ you have $\|f\|_ {L^2(\Omega\setminus\Omega_\delta)}=O(\delta)$, for the Lebesgue measure of $\Omega\setminus\Omega_\delta$ is bounded by $\delta\mathcal{H}^{n-1}(\partial\Omega),$ as a consequence of the coarea formula applied to the distance function, or directly, on the lines of Denis Serre's construction in the linked answer. For the same reason, if $f$ is Hölder continuous of exponent $0\leq \alpha\le1$ (for instance, it is in a Sobolev space in the hypothesis of the Morrey-Sobolev embedding) and vanishes on $\partial\Omega$, you have $\|f\|_ {L^2(\Omega\setminus\Omega_\delta)}=O(\delta^{1+\alpha}).$ Finally, it is not completely clear what you mean exactly by "rapidly decaying to 0", but certainly any bound on $|f|$ on $\Omega\setminus\Omega_\delta$ gives a bound on the norm as said, and in few words, everything is like in the case $n=1$.

share|improve this answer
    
Thank you very much. After doing some reading I now understand your answer. Thanks! :D –  alext87 Nov 22 '10 at 10:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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