Graph (or manifold) Lipschitz satisfy the sphere (ball) condition?

Question

Consider $\varphi: U\subset \mathbb{R}^{n-1}\to \mathbb{R}$ a Lipschitz function and $\Omega=Graph(\varphi)$, i.e.,
$$\Omega=\{x=(x_1,...,x_n)\in U\times\mathbb{R};x_n=\varphi(x_1,...,x_{n-1})\}.$$

The set $\Omega$ satisfies the Sphere Condition?

Sphere Condition means that existe $r>0$ (uniform) such that for each $z\in\Omega$, there is $\xi$ satisfying $z\in \partial B_r(\xi)$ and $B_r(\xi)\cap\Omega=\emptyset$.

More generally, if $\Omega$ is only Lipschitz manifold (not necessarily graphic), $\Omega$ satisfies the Sphere Condition?

Answer 1

This is not true. The Lipschitz condition allows for corners, which are bad for the sphere condition. In two dimensions, a simple counterexample is $\varphi(x) = |x_1|-|x_2|$ pictured below.

A sphere containing $(0,0,0)$ has to be either above or below the graph; due to symmetry we may assume it's above it. Then its intersection with the plane $x_2=0$ must be the single point $(0,0,0)$; for otherwise it would be a circle, and a circle cannot stay above the graph of $x_3=|x_1|$. Therefore, the center of the sphere is $(0,r,0)$ where $r$ is its radius. But then the sphere contains the point $(0, r-r/\sqrt{2}, -r/\sqrt{2})$ which is below the graph of $\varphi$.

Here is another example, which works in all dimensions starting with $1$: $\varphi = x_1 \sin \log |x_1|$. This is a Lipschitz function ($\varphi'$ is bounded), but the graph of $\varphi$ meets both $x_n= |x_1|$ and $x_n=-|x_1|$ infinitely often. This rules out a sphere touching the origin from either above or below. The example works in higher dimensions with the same $\varphi$: it just does not depend on $x_2,\dots,x_{n-1}$. The only thing I don't like about this example is that it's impossible to see what goes on from looking at an actual plot of $\varphi$.

But $\varphi = x_1 \sin (10 \log |x_1|)$, which is also Lipschitz, makes a better illustration.

Answer 2

Neither the category of semi-algebraic it is valid, as done by the user @user127096, namely $\varphi(x,y)=|x|-|y|$ is semi-algebraic.

Answer 3

A sharp answer to your question

Theorem: A domains $\Omega\subset \Bbb R^d$ satisfies the both interior and exterior sphere condition if and only if $\partial \Omega$ is $C^{1,1}$. See Theorem 1.0.9, here

Counter-example of $C^{1,\alpha}$ domain not satisfying the interior sphere condition

Consider $\Omega=\{(,)\in \Bbb R^2:>||^{1+\alpha}\},$ with $0\leq \alpha<1$. This domain is $C^{1,\alpha}$ smooth. Because for $ f(x)= |x|^{1+\alpha},$ we have $f'(x)=(\alpha+1) |x|^{\alpha-1}x$ which is $\alpha$-H"older continuous. In fact it can be shown that \begin{align} |f'(x)- f'(y)|= \alpha||x|^{\alpha-1}x-|y|^{\alpha-1}y|\leq (\alpha+1)2^{1-\alpha} |x-y|^{\alpha}\,. \end{align}
Let $ B= B((x_0, y_0), r)$ , $r>0$ be a ball touching the boundary at $(0,0)$ and such that $ \overline{B}\subset \Omega$. We have that $(0,0)\in \partial \Omega\cap B((x_0, y_0), r)$ then necessarily \begin{align*} \text{$x_0^2 +y_0^2=r^2$ and $y_0>|x_0|^{1+\alpha}$}\implies y_0= \sqrt{r^2-x_0}. \end{align*} Moreover, since $\partial B\subset \Omega$ in particular the bottom part $\partial B_-= \{ (x,y): y=y_0- \sqrt{r^2-(x-x_0)^2} \}$ is contained in $\Omega$. That is, for all $(x,y)\in \partial B_-$ with $|x-x_0|\leq r$ we have \begin{align*} y=y_0- \sqrt{r^2-(x-x_0)^2}= \sqrt{r^2-x_0}- \sqrt{r^2-(x-x_0)^2}\qquad\text{and}\qquad y>|x|^{1+\alpha} \end{align*} That is for all $|x-x_0|\leq r$ we get that \begin{align*} \sqrt{r^2-x_0^2}- \sqrt{r^2-(x-x_0)^2}>|x|^{1+\alpha} \end{align*} Letting $x\to x_0$ it follows that \begin{align*} r^2-x_0^2\geq (r+|x_0|^{1+\alpha})^2\Longleftrightarrow -x_0^2\geq 2r|x_0|^{1+\alpha}+|x_0|^{2(1+\alpha)} \end{align*} Which is possible only if $x_0=0$ so that $y_0=r$ that is $(x_0, y_0)= (0,r)$. In this case, for $0<|x|<r$, the above inequality becomes \begin{align*} r^2+|x|^{2(1+\alpha)} -2r|x|^{1+\alpha}= (r-|x|^{1+\alpha})^2> r^2-x^2, \quad \text{ i.e., }\quad |x|^{2\alpha} -2r|x|^{\alpha-1}+1)>0. \end{align*} Given that $ |x|^{2\alpha} -2r|x|^{\alpha-1}+1\xrightarrow{x\to0}-\infty$ as $\alpha-1<0$, there is $0<|x|<r$ such that \begin{align*} -1>( |x|^{2\alpha} -2r|x|^{\alpha-1}+1)>0. \end{align*} This is impossible and hence the point $(0,0)$ does not satisfies the interior sphere condition. However $(0,0)$ satisfies the exterior sphere condition.