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Let $\Omega\subset\mathbb{R}^N$ be an open, bounded and connected Lipschitz domain. Is it true that we can find some $R>0$ such that any $N$-dimensional open ball $B(x,r)$ with $r\leq R$ that intersects $\Omega$ has the property that $\Omega\cap B(x,r)$ is a open, bounded and connected Lipschitz domain too?

I don't know any reference that specify those facts and I decided to ask here.

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EDIT: I look at the solutions and see that the problem is that we lose connectedness.

Question: If the intersection between an open ball and a Lipschitz domain is connected, then we can say that the intersection is a Lipschitz domain?

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    $\begingroup$ It might fail to be connected. $\endgroup$
    – Teri
    Commented Jul 9 at 11:48
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    $\begingroup$ Isn't it now just the definition of a Lipschitz domain? The boundary locally looks like the graph of a Lipschitz function. By compactness of the closure pick the minimal radius.. $\endgroup$
    – Dmitrii Korshunov
    Commented Jul 9 at 12:37
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    $\begingroup$ It's not even true for the complement of a right-angled sector, isn't it? (You can restrict to $r=1$ by scaling invariance and see that connectedness may fail.) $\endgroup$
    – Martin Hairer
    Commented Jul 9 at 13:28
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    $\begingroup$ I can't see how you lose connectedness by choosing a small radius? $\endgroup$
    – Bogdan
    Commented Jul 9 at 13:33
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    $\begingroup$ @Bogdan Take its center on the bisector of the missing wedge in such a way that it intersects both sides but doesn't contain the origin. $\endgroup$
    – Martin Hairer
    Commented Jul 9 at 19:31

1 Answer 1

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In general, no such $R$ exists.

Let $\Omega=\left\{x\in (-1,1)^2:\ \min\{x_1,x_2\}<0\right\}$, which is an open, bounded, connected Lipschitz domain. For any $r\in(0,1)$, we can find a ball of radius $r$ whose intersection with $\Omega$ is not connected, which will show that no $R$ has the desired property.

There exists some $\gamma\in\left( \tfrac1{\sqrt2}r,\ r \right)$. Let $B$ denote the open ball of radius $r$ around $(\gamma,\gamma)$. Then $B\cap\Omega$ is not connected. It's easy to visualize this, but let me prove it.

Let $D_1:=\{x\in\mathbb R^2:\ x_1>x_2\}$, let $D_2:=\{x\in\mathbb R^2:\ x_1<x_2\}$, and let $D_0:=\{x\in\mathbb R^2:\ x_1=x_2\}=\mathbb R^2\setminus(D_1\cup D_2)$. Observe, for small enough $\epsilon>0$, we know $D_1\cap B\cap\Omega$ contains the element $(\gamma,-\epsilon)$ and $D_2\cap B\cap\Omega$ contains the element $(-\epsilon,\gamma)$. Meanwhile, any $x\in D_0\cap\Omega$ has $\Vert(\gamma,\gamma)-x\Vert_2=\sqrt2(\gamma-x)>\sqrt2\gamma>r$ so that $x\notin B$. Thus $D_1\cap B\cap\Omega$ and $D_2\cap B\cap\Omega$ form a nontrivial partition of $B\cap\Omega$ into open sets.

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  • $\begingroup$ I think the question asks it for balls centered on the boundary (although that's not what's written), in which case answer is yes. Without this constraint, the answer is yes iff the set has positive reach I think. $\endgroup$
    – alesia
    Commented Jul 9 at 18:10
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    $\begingroup$ You're right—even though the question does not say so, the picture depicts this special situation. $\endgroup$
    – e.lipnowski
    Commented Jul 9 at 18:35
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    $\begingroup$ But even if you further require that the ball be centered on the boundary, the intersection can be disconnected if you let the center of the ball vary with $r$. For instance, say $\Omega=\left\{x\in(-1,1)^2:\ \min\{x_1,x_2\}<0 \text{ or } x_2>x_1\right\}$. Then for any $r\in(0,1)$, you can find a $\gamma\in(r,1)$ such that the distance between $(\gamma,0)$ and the line $\{x\in\mathbb R^2:\ x_1=x_2\}$ is strictly less than $r$. But then $(\gamma,0)$ is a boundary point, and the ball of radius $r$ centered at $(\gamma,0)$ has non-connected intersection with $\Omega$. $\endgroup$
    – e.lipnowski
    Commented Jul 9 at 20:08
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    $\begingroup$ @TimCarson The derivative of $xcos(\pi/x)$ is unbounded, so I don't beleive that this is a Lipschitz domain. $\endgroup$
    – Bogdan
    Commented Jul 10 at 5:55
  • $\begingroup$ @e.lipnowski Thanks for the solution! I add a question in the final of my post. Do you think it's true? $\endgroup$
    – Bogdan
    Commented Jul 10 at 7:26

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