# Bounding near the boundary for a Sobolev function.

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.

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.
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? – alext87 Dec 3 '10 at 11:24
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. – Pietro Majer Dec 3 '10 at 13:48