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I am encountering a geometrical question that intuitively seems obvious but I have a lack of argument to prove or disprove it in a rigorous manner.

Let $\Omega\subset\Bbb R^d$ be an open bounded (may be connected just to make it simpler)

For $\delta>0$ small enough we define the shrunken version of $\Omega$ as by $$\Omega_\delta= \{x\in \Omega~: \operatorname{dist}(x,\partial \Omega)>\delta\}$$

Basically, for $\delta$ small enough, $\Omega$ and $\Omega_\delta$ have a similar shape. For instance if $\Omega= B(0,1)$ is the unit ball then, $\Omega_\delta=B(0,\delta) =\delta B(0,1)$ is the ball centered at 0 with radius $\delta$.

Definition An open set $\Omega$ is said to satisfy the Volum Density Condition(VDC) if there exists a constant $\kappa>0$ such that for all $x\in \partial \Omega$ and $r\in (0,1)$ $$|\Omega\cap B(x,r)|\geq \kappa r^d.$$

An open set $\Omega$ is said to be of class $C^k$ if for every $x\in \partial \Omega$ there exists $r>0$ and a mapping $\gamma: \Bbb R^{d-1}\to\Bbb R$ such that

$$ \Omega\cap B(x,r)= \{x=(x',x_d)\in B(x,r)~: x_d>\gamma(x')\}$$

A natural question would be: Does $\Omega_\delta$ inherit the regularity properties of $\Omega$? Precisely,

1)If $\Omega$ is $C^k$ do we have that $\Omega_\delta$ is also $C^k$?

2) If $\Omega$ satisfies the Volum Density Condition(VDC) does $\Omega_\delta$ also satisfies the Volum Density Condition(VDC)?

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    $\begingroup$ Question 1 has a clear answer: yes if $k \geqslant 2$ and $\delta > 0$ is small enough. This is given in most textbooks on PDE's, I suppose, I can search for an exact reference if you like. However, if $\delta > 0$ is too large, then smoothness of $\Omega_\delta$ may deteriorate: consider, say, a unit ball and $\delta = 1$. Also, if $k = 1$, then $\Omega_\delta$ may become highly irregular even when $\delta > 0$ is small. An example: $\Omega$ a region above the graph of $y = |x| / \log(1/|x|)$ near $x = 0$. $\endgroup$ – Mateusz Kwaśnicki Jun 7 at 21:02
  • $\begingroup$ @MateuszKwaśnicki Please I would like to see does reference. Do you have an idea about what could happen with VDC condition? I technically like that condition very much. $\endgroup$ – Guy Fsone Jun 7 at 21:06
  • $\begingroup$ Of course, large $\delta $ are not interesting at all. $\endgroup$ – Guy Fsone Jun 7 at 21:08
  • $\begingroup$ In the example of Mateusz Kwasnicki, the "highly irregular" means "having an outward cusp", and such things do not satisfy VDC (while $C^1$ domains to, of course). $\endgroup$ – Kostya_I Jun 7 at 21:24
  • $\begingroup$ @GuyFsone: Section 14.6 in Gilbarg and Trudinger has what you need. $\endgroup$ – Mateusz Kwaśnicki Jun 7 at 21:35

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