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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 $K$?

Guillaume F.
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1 Answers1

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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 $f(x)=x$).