2

Let $\Omega\subset \mathbb{R}^N$ be a domain with $N\ge 2$. Let $K\subset \Omega$ be a compact set and take $u:\overline{\Omega}\to\mathbb{R}$ such that $u$ is Lipschitz and $u=1$ in $\partial K$.

Assume that $\operatorname{int}{K}\ne\emptyset$ and conseider the new function $v:\overline{\Omega}\to \mathbb{R}$ defined by

$$v(x) = \left\{ \begin{array}{rl} 1 &\mbox{ if $x\in \operatorname{int}{K}$} \\ u(x) &\mbox{ otherwise} \end{array} \right. $$

Is $v$ a Lipschitz function?

Tomás
  • 22,559

2 Answers2

3

Here is another approach:

Choose $x,y$. If both lie in $K^\circ$ or neither lie in $K^\circ$ then since $1,u$ are Lipschitz we see that $v$ satisfies a Lipschitz condition.

So, suppose $x \in K^\circ$, $y \in \bar{\Omega} \setminus K^\circ$. Let $\phi(t) = x+t(y-x)$, and let $T=\sup_{t \in [0,1]} \{ t | \phi(t) \in K^\circ \}$. Since $K$ is closed, we see that $T \in (0,1)$, and continuity shows that $v(\phi(T)) = u(\phi(T))=1$.

Then \begin{eqnarray} |v(x)-v(y)| &\le & |v(x)-v(\phi(T))|+ |v(\phi(T))-v(y)| \\ &=& |v(\phi(T))-v(y)| \\ &=& |u(\phi(T))-u(y)| \\ &\le& L \|\phi(T)-y\| \\ &=& L (1-T) \|x-y\| \\ &\le& L \|x-y\| \end{eqnarray}

copper.hat
  • 172,524
0

Yes, even with the same constant. Let $L$ be the Lipschitz constant of $u$. Then $$|u(x)-1|\le L\operatorname{dist}(x,K),\quad x \in \Omega\setminus K$$ The function $w(x)=L\operatorname{dist}(x,K)$ is also $L$-Lipschitz. Setting $$v = 1 + \min(\max(u-1,-w),w)$$ we conclude that $v$ is $L$-Lipschitz. This is the same $v$ as in your question.

user127096
  • 9,683
  • I confess to you that I really tried to write $v$ as you have did, however, with no result. I wonder if some day I will tackle some problems easily as you do. Anyway, thank you again. – Tomás Mar 26 '14 at 17:58
  • Sorry to make another question, but although I can see that your aprroach does work, on the other hand, I can't figure out the intuition behind the idea. Maybe you could provide me with some general abstract framework which contains this example, or maybe just some tricks? – Tomás Mar 26 '14 at 18:16
  • 1
    @Tomás You know the truncation trick: $\min(\max( u, a),b))$ truncates the function so that the values of new function are between $a$ and $b$. More generally, we can truncate between two functions. They should be Lipschitz, they should be both equal to $1$ on $K$, and away from $K$ they should give $u$ enough room so that it does not get truncated there. The functions $1\pm L\operatorname{dist}(x,K)$ do the job. – user127096 Mar 26 '14 at 18:26