Hausdorff extension of Lipschitz functions

The extension appeared in the previous post and is very simple: given a closed set $A\subset X$ and a bounded function $f\colon A\to\mathbb R$ , we define $\displaystyle F(x)=\inf_{a\in A} \left(f(a)+\frac{d(x,a)}{d(x,A)}-1\right)$ for all $x\in X\setminus A$ . It is easy to see that $F(x)\to f(c)$ as $x\to c\in A$ . Indeed, the boundedness of $f$ implies that the infimum is actually taken only over the set $A_{x,M}=\lbrace a\in A\colon d(x,a)\le M\,d(x,A)\rbrace$ where $M=1+\sup f-\inf f$ . As the distance $\rho=d(x,c)$ tends to 0, the set $A_{x,M}$ shrinks at the same rate: indeed, $A_{x,M}$ is contained in the ball of radius $M\rho$ with center $c$ . Thus, $F$ furnishes a continuous extension of $f$ whenever $f$ is continuous on $A$ .

Now suppose that $f$ is Lipschitz on $A$ . Will the extension be Lipschitz? Let’s consider a one-dimensional example with $X=\mathbb R$ , $A=[0,5]$ , and $f(x)=x$ on $A$ . When $x<5$ , the extension is

$\displaystyle F(x)=\inf_{a\in [0,5]} \left(a+\frac{x-a}{x-5}-1\right)=\frac{5}{x-5}+\inf_{a\in [0,5]} a\,\left(1-\frac{1}{x-5}\right)$

When $5<x<6$ , the infimum is at $a=5$ , hence $F(x)=1$ . When $x> 6$ , it shifts to $a=0$ , which yields $F(x)=\frac{5}{x-5}$ . Here is the plot:

While the original function $f$ had Lipschitz constant 1, the extension has Lipschitz constant 5 (attained to the right of the transition point $x=6$ ). We can make this progressively worse by replacing 5 with larger numbers. So, the Lipschitz constant of the extension does not admit a bound in terms of $\mathrm{Lip}\,(f)$ alone. The oscillation $M=\sup f-\inf f$ enters the game as well… although it’s not entirely clear that it must. In the example above, there is no apparent reason for $F$ to decrease so quickly (or at all) when $x>6$ .

3 thoughts on “Hausdorff extension of Lipschitz functions”

Should the a_{x,M} have M = 1 + sup f – inf f?

Calculus7 says:

2016-07-18 at 17:10

You are right. And then M+1 can be replaced by M below. Edited.

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1. t says:
  
  2016-07-19 at 01:17
  
  Thank. These notes definitely helped me to get better sense of the construction.

Hausdorff extension of Lipschitz functions

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3 thoughts on “Hausdorff extension of Lipschitz functions”

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