Let $\Omega\subset \mathbb{R}^d$ be an open and bounded domain with Lipschitz smooth boundary. Let $\delta>0$ and 
$
\Omega\_\delta = \{ x\in\Omega : \inf\_{y\in\partial\Omega} \left\|x-y\right\|\_{L\_2(\Omega)}\geq \delta \}
$


Is there a $\hat{\delta}>0$ such that $\forall 0<\delta<\hat{\delta}$ the space $\Omega\_{\delta}$ has a Lipschitz smooth boundary?


This statement seems like it should be true. I would really appreciate some help. 


Thanks in advance. 



Shrinking a Lipschitz smooth domain.
My answer is **Yes**. Of course, I presume that you assume $\Omega$ is on one side only of its boundary.


The boundary $\partial\Omega$ is compact. By assumptions, it has an atlas with finitely many charts, each one corresponding to a piece $\Gamma\_j$ which is the graph of a Lipschitz function: in appropriate coordinates, $\Gamma\_j$ is given by $x\_d=\phi\_j(x\_1,\ldots,x\_{d-1})$.


Let us choose $\delta$ smaller than one tenth of
$$\delta\_0:=\min\_{x\in\bar\Omega}\\,\max\_j\{d(x,\partial\Gamma\_j)\}>0$$


Let $\bar x$ be on the boundary of $\Omega\_\delta$, not on $\partial\Omega$. Let $j$ be such that $B(\bar x;9\delta)\cap\partial\Omega\subset\Gamma\_j$. There exists a point $\bar y\in\partial\Gamma\_j$ such that $d(\bar x,\bar y)=\delta$. Wlog, we have $\bar y=0$ and the graph is locally $x\_d=\phi\_j(x\_1,\ldots,x\_{d-1})$. Let $M$ be the Lipschitz constant of $\phi\_j$. Locally, the boundary of $\Omega\_\delta$ is the upper envellop of the balls $B(y;\delta)$ with $y\in \Gamma\_j$; as a matter of fact, for $x\in B(\bar x;\delta)$, the closest points of $\partial\Omega$ to $x$ belong to $\Gamma\_j$. It is easy to see that it is the graph of a function $\phi\_j^\delta$ whose Lipschitz constant is at most $M$.


