[Math] Holder continuous but not Lipschitz

continuitymetric-spacesreal-analysis

Is there a function that is Holder continuous but not Lipschitz continuous?

Best Answer

From wiki,

The function $f(x) = x^\beta$ (with $\beta \le 1$) defined on $[0, 1]$ serves as a prototypical example of a function that is $C^{0,\alpha}$ Hölder continuous for $0 < \alpha \le \beta$, but not for $\alpha>\beta$. In particular, if we choose $\beta<1$, it's not Lipschitz.

Do you know how to check this statement?