Is every Lipschitz continuous function is holder continuous with exponent $\in (0,1)$?

This seems to be true,but I haven't found such a conclusion in any textbook.


1 Answer 1


For every bounded Lipschitz-function, this is true. To see this, note that for $|x-y| \leq 1$, we have

$$ |f(x) - f(x)| \leq C \cdot |x-y| = C\cdot |x-y|^\alpha \cdot |x-y|^{1-\alpha} \leq C\cdot |x-y|^\alpha $$

and for $|x-y| > 1$, we have

$$ |f(x) - f(y)| \leq|f(x)| + |f(y)| \leq 2K \leq 2K \cdot |x-y|^\alpha. $$

As any Lipschitz function on a bounded domain is bounded, the claim holds in particular for Lipschitz functions on bounded domains.

But for unbounded functions, this is not true in general. For example, $f : \Bbb{R} \to \Bbb{R}, x\mapsto x$ is trivially Lipschitz, but

$$ \frac{|f(n) - f(0)|}{|n-0|^\alpha} = n^{1-\alpha} \to \infty \text{ for } n \to \infty, $$

so that $f$ is not $\alpha$ Hölder-continuous for any $\alpha \in (0,1)$.


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