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In the following link, it says that Lipschitz domain plus or minus small ball may not be a Lipschitz domian.

Therefore, I'm woundering that $C^{1,1}$ domain plus ro minus small ball is a Lipschitz domain or not. This is a simply question and has been answered by Hajlasz.

his question is rather straightforward and has already been addressed by Hajłasz.

To clarify, let me restate the question: would the boundary ball of $C^{1,1}$ domain be Lipschitz, precisely,

For any $x \in \partial \Omega$, would there always exists a $r_x$, such that $B(x,r_x) \cap \Omega$ is a Lipschitz domain?

Let $\Omega\subseteq \mathbb{R}^n$ be a $C^{1,1}$ domain with compact boundary. Just to be precise, this means that there are finitely many cylinders $U_1,U_2,\ldots, U_k$ which cover $\partial \Omega$ and such that for each $j$ the set $U_j\cap \Omega$ is (or can be rotated to be) the super-level set of a $C^{1,1}$ function (differentiable funciton with Lipschitz derivative) $\varphi_j$.

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Edit: This is the answer to the previous version of the problem, before it was reformulated.

No. If your domain is the union of two externally tangent balls or the difference of two internally tangent balls, then it is not $C^{1,1}$. Not even Lipschitz. The boundary is not a graph of a function near the point where the balls are tangent. Perhaps you incorrectly formulated the question.

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  • $\begingroup$ Sorry, I restate the question now. Thank you. $\endgroup$ Commented Sep 28 at 2:35

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