# 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 $M=\sup f-\inf f$. 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 $(M+1)\rho$ 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, 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 $x>6$.