Bounding a smooth function near the boundary 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 $f$. 
 A: 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 $n=1$.
