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 $K$?


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 $f(x)=x$).

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

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