Let $n=1,2,3,...$. For any $x \in \mathbb{R}^n$ and $r>0$ is every open ball $B_r(x):=\{y\in\ \mathbb{R}^n:\|y-x\|<r\}$ a bounded Lipschitz domain?
I am having troubles understanding Lipschitz domains. My teacher said that a bounded Lipschitz domain is a domain whose boundary is the graph of a Lipschitz function. I really don't understand the meaning of that claim. I am reading the Wikipedia definition and now I am really confused.
Can somebody help me with this problem? I'll appreciate it so much.
Try to look at the $n=2$ situation: the unit circle looks locally like the graph of the function $\sqrt{1-x^2}$, which is Lipschitz on $(-1/2,1/2)$.
That is, if you take a point $x_0$ on the circle, an open ball with small radius at the centered at that point, then the intersection of the circle with the ball is a after rotation and translation part of the graph of $\sqrt{1-x^2}$. Moreover, points near $x_0$ inside/outside the domain are below/above the graph.