3

The following question comes from Evans' PDEs book (Chapter 5, exercise 13).

Give an example of an open set $U \subset \mathbb{R}^n$ and a function $u \in W^{1,\infty}(U)$ such that $u$ is not Lipschitz continuous on $U$. (Hint: Take $U$ to be the open unit disk in $\mathbb{R}^2$, with a slit removed).

A example of such a function was given here. Specifically,

Let $U$ be the plane with the slit along negative $x$-axis. Using polar coordinates $r,\theta$ with, $\pi<\theta<\pi$, define $u(r,\theta)=r\theta$.

I have a couple of questions:

  1. What does a "slit" mean here? In the above example, does it just mean $y \neq 0$ for negative $x$?
  2. How does this exercise not contradict Evans' theorem, where $W^{1,\infty}(U) = C^{1,0}(U)$?

For reference, here is the theorem:

(Characterization of $W^{1,\infty}$). Let $U$ be open and bounded, with $\partial U$ of class $C^1$. Then $u: U \rightarrow \mathbb{R}$ is Lipschitz continuous iff $u \in W^{1,\infty}(U)$.

Why does removing a "slit" from the unit disk make $\partial U$ badly behaved? Perhaps I'm misinterpreting what $C^1$ boundary means. Is there is layman's definition of $C^1$ boundary?

Pedro
  • 19,965
  • 9
  • 70
  • 138

1 Answers1

2

  1. A "slit" is in line with your number $1$. In particular, the "slit" is the "cut" that you will make to $U$: $\{(x,y): y= 0, x > 0\}$.

  2. A $C^1$ (resp. something else) boundary for $U$ (in $\mathbb{R}^{n+1})$ in Evans depends on how $U$ sits in $\mathbb{R}^{n+1}$ (It is not a property of $U$ as a topological space, e.g.). He defines it so that the idea is as follows: There are local coordinates given by a $C^1$ function such that the boundary is given by the $x_{1}$-coordinate and the inside of $U$ corresponds to the region above the $x_1$-axis. In particular, in this example with the "slit" removed, no matter how you try to draw local coordinates at the point $(1/2, 0)$, you cannot make the region above the $x_1$-axis correspond to the region in $U$: because $U$ lies above and below the $x_1$ axis.

user357980
  • 1,894