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<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:

Large Lipschitz constant

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.

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