0
$\begingroup$

Suppose $f$ is a Lipschitz function $(X, d_X) \to (Y, d_Y)$ with Lipschitz constant $K$.

Question. If $d_X'$ and $d'_Y$ are metrics that are topologically equivalent to $d_X$ and $d_Y$, is $f$ Lipschitz continuous with the same Lipschitz constant $K$?

$\endgroup$
1
$\begingroup$

The answer is very much negative.

The metric $d_Y'=2d_Y$ is topologically equivalent to $d_Y$, but the Lipschitz constant is twice as much with respect to it.

The metric $d_Y'=\sqrt{d_Y}$ is also topologically equivalent to $d_Y$, but there's a good chance the map is no longer Lipschitz $d_X\to d_Y'$ (for example, consider $X=Y=\mathbb{R}$ with the standard metric, and $f(x)=x$).

$\endgroup$
  • $\begingroup$ Interesting! I had no idea that Lipschitz continuity cannot be expressed in terms of topology! $\endgroup$ – Guillaume F. Jun 17 '18 at 12:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.