Lebesgue integrable functions and relation with the distance function Let $\Omega\subset\mathbb{R}^{N}$ be a smooth bounded domain, define
$\Omega_{\rho}:=\{x\in\Omega:d(x,\partial\Omega)<\rho\}$, and let $f\in
L^{p}(\Omega)$ for some $p>N$. My question is:
Is it true that there exists $\rho_{0}>0$ small such that $\left\vert f(x)\right\vert \leq Cd(x,\partial\Omega)^{\alpha
}$ $a.e. x \in$ $\Omega_{\rho_{0}}$, for some $C>0$ and some
$\alpha>-\frac{1}{p}$ ? If not, is it valid for some other $\alpha$ ?
Thanks in advance!
Uriel
 A: No. I don't think you have that inequality. Take $N=2$ and say that the boundary of $\Omega$ has a flat part, say the segment $(-1,1)\times \{-\rho_0\}$. Then $\Omega_{\rho_0}$ has a flat part at $(-1/2,1/2)\times \{0\}$. Consider any $a>0$ and take $f(x,y)=\frac{1}{x^{a/p}}$ for $0<y<x^b$ and $0<x<1$, where $b>a$ and $f=0$ everywhere else. 
Then $$\int_\Omega f^pdxdy=\int_0^1 \frac{1}{x^{a}}\int_0^{x^b}1\,dydx
=\int_0^1 x^{b-a}\,dx<\infty.$$
Near $(0,0)$ you have $d((x,y),\partial\Omega)\ge \frac{\rho}2$ but $f$ is unbounded near $(0,0)$.
A: Let $(a,b)$ be an interval and let $0<p<\infty.$ Then there is no rate of growth function on $(a,b)$ that a function $f\in L^p(a,b)$ must satisfy. In other words, if $g:(a,b) \to (0,\infty)$ is continuous, then there exists a continuous $f\in L^p(a,b)$ such that $f(x)/g(x)$ is unbounded on $(a,b).$
Proof: Infinitely many disjoint triangular spikes of huge heights and extremely small bases.
Well ... maybe this will be downvoted, but surely somewhere on MSE this needs to be said.
