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 $f$.

  • $\begingroup$ actually I still have a doubt about your notation: what is $\|\cdot\|_{L_2(\Omega)}$ ? $\endgroup$ Commented Nov 20, 2010 at 9:19
  • $\begingroup$ I have editted the post to answer your comment. This is the standard notation and definition, right? $\endgroup$
    – alext87
    Commented Nov 20, 2010 at 16:33
  • $\begingroup$ 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? $\endgroup$
    – Deane Yang
    Commented Nov 20, 2010 at 16:56
  • $\begingroup$ 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 $f$. $\endgroup$
    – Deane Yang
    Commented Nov 20, 2010 at 16:57
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Nov 21, 2010 at 12:19

1 Answer 1


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 $n=1$.

  • $\begingroup$ Thank you very much. After doing some reading I now understand your answer. Thanks! :D $\endgroup$
    – alext87
    Commented Nov 22, 2010 at 10:01

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.