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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 $f\in\mathcal{H}^{\tau}(\Omega)$, $\tau>\frac{d}{2}$ (i.e. $f$ is continuous) and $f$ is zero 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..

For sufficiently small $\delta>0$ it is true that:

$\left\|f\right\|_{L_2(\Omega\setminus\Omega_{\delta})} \leq C\delta^\alpha\left\|f\right\|_{L_2(\Omega)}$ with $\alpha\geq 1$ and $C$ is a constant not depending on $\delta$ or $f$.

If this is not possible what is the largest $\alpha\in(0,1)$ so that the above inequality holds.

Thanks in advance.

Note this is almost identical to my previous post Bounding a smooth function near the boundary but presented in a clearer fashion. I would be grateful for any comments or help.

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I don't believe your inequality holds. Your proposed inequality depends only on the $L_2$ norm of $f$. Since the Sobolev space is dense in $L_2$, if your inequality were true, it would also extend to any $L_2$ function as well. I am also fairly sure it is easy to construct counterexamples to your inequality. You need to use the Sobolev norm on the right side of the inequality and not just the $L_2$ norm.

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  • $\begingroup$ Thanks, okay I was trying to show too much. If I change $L_2$ to $\mathcal{H}^{\tau}$ in the inequality and $\alpha$ to $1+\tau$ do you think the inequality could hold? $\endgroup$
    – alext87
    Commented Dec 3, 2010 at 11:24
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    $\begingroup$ Yes, as soon as $C\delta^\alpha < 1$, a counterexample is simply any $C^\infty$ function with support in $\Omega\setminus\Omega_\delta$. To understand what can be true, think of the 1-dimensional case. $\endgroup$
    – Pietro Majer
    Commented Dec 3, 2010 at 13:48
  • $\begingroup$ To be honest, I don't have time to play with this, but it seems to me that you can figure out what's true by simply alternating between looking for simple counterexamples (and, as Pietro points out, the 1-d situation is a good place to start) and figuring out what you can prove using the fundamental theorem of calculus and the Holder inequality. There's nothing more subtle than that going on here. Just assume both the domain and function are smooth. If you get anything, it extends to the general case by continuity. $\endgroup$
    – Deane Yang
    Commented Dec 3, 2010 at 13:58

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