Extending to a Lipschitz function

Asked Mar 13 '21 at 11:31

Active Mar 13 '21 at 11:43

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Let $(X, d)$ be a metric space, and let $A, B \subset X$ be two open sets such that $d(A, B) \geq 1$. Suppose we have a function $f$ defined only on $A \cup B$ such that $f|_A \equiv 1$ and $f|_B \equiv -1$. My question is, can we extend $f$ as a Lipschitz function $f : X \to \mathbb{R}$ such that the Lipschitz norm of $f$ is $\leq 2$?

My intuitive guess is that the answer is yes, and $f$ could be defined in terms of some convenient distance function, but I cannot quite put my finger on it. I am also unsure whether $X$ needs to be compact for this to happen. Any help will be appreciated.

edited Mar 13 '21 at 11:43

Henno Brandsma

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asked Mar 13 '21 at 11:31

SMS

1

$f(x)=\frac {d(x,B)-d(x,A)} {d(x,A)+d(x,B)}$ is Lipschitz, $1$ on $A$ and $-1$ in $B$ but I am not sure if the Lipschitz constant is at most $2$. Try to work out the constant. – Kavi Rama Murthy Mar 13 '21 at 11:39
This might help. – Lázaro Albuquerque Mar 13 '21 at 19:41
This is excellent, truly nothing new under the sun! But of course, KaviRamaMurthy's solution was also working (essentially a variant of the one you gave). – SMS Mar 14 '21 at 06:26

Extending to a Lipschitz function

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