In the Adams and Fournier book on Sobolev spaces, they mention a bounded domain with a locally Lipschitz boundary (meaning the boundary is a finite union of graphs of Lipschitz functions) satisfies the cone condition (meaning there is a cone $C$ such that for every $x$ in the domain, there is a cone $C_x$ with its origin at $x$ that is congruent to $C$).
- Is there an intuitive explanation for why this must hold?
- A link to a source with a proof would be very helpful!
Here is a sketch of the main argument of a proof.
Let $f:\mathbb R^{n-1} \to \mathbb R$ be Lipschitz with modulus $L$. Then due to the Lipschitz continuity of $f$, we have that $$ f(y) \le f(x) + L \|x-y\|_2 $$ for all $y$ close to $x$. That is, we can fit the cone $$\{ (x,t): \ \|x\| \le t \}$$ inside the set $$ \{ (x,t) : \ f(x) \le t\}, $$ which is the epigraph of $f$ and locally models the boundary of the domain.
For some intuition, consider the function $$ f(x) = \sqrt{ |x| }. $$ Because of the infinite slope at zero, we cannot fit any non-trivial cone (one with non-empty interior) inside the epigraph of $f$ at $x=0$.
Of course, the Lipschitz-ness is not necessary. The function $$ f(x) = \sqrt{ \max(x,0) } $$ is not Lipschitz, but the epigraph satisfies the cone condition. This is kind of cheating, because upon rotating the graph of $f$ it becomes the graph of a Lipschitz function.
Also the double brick domain (two 3d bricks on top of each other, one is rotated 90 degrees against the other) is not Lipschitz but satisfies the cone condition.