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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 $f$.

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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
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
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 $n=1$.

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

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