Let $f$ be a real function such that such that $|f(x)-f(y)|\leq \sqrt {|x-y|}, \forall x,y\in\Bbb R.$ Does this condition imply that $f$ will be differentiable ? If Lipschitz order is greater than $1$ then function is constant so differentiable . If Lipschitz order is equal to $1$ then $ |x|$ is counterexample for differentiablity . For this question I have no idea . Please suggest me a counterexample or proof of differentiablity hint . Thank you.


1 Answer 1


$f(x) = \min(\epsilon,|x|)$ works for any $\epsilon$ small enough. Its (globally) $1$-Lipschitz continuous with constant $1$, therefore also (locally) $1/2$-Lipschitz continuous, and as you said, not differentiable. The following annoying work and the point of using $\epsilon$ to cutoff the function is to get global control on the seminorm $\frac{|f(x)-f(y)|}{\sqrt{|x-y|}}$ which is already well behaved for $x,y$ sufficiently small.

For $|x|<1/2$, $|y|<1/2$, then $|x-y|<1$, so $$ |f(x)-f(y)|\le |x-y|\le \sqrt{|x-y|}.$$ If $|x|<\epsilon$ but $|y|\ge 2\epsilon$, then $|x-y|\ge \epsilon$, and (so long as $\epsilon <1$) $$|f(x)-f(y)| = ||x|-\epsilon|=\epsilon -|x| \le \epsilon \le \sqrt{\epsilon} \le \sqrt{|x-y|}. $$ The case $|y|<\epsilon$ and $|x|\ge 2\epsilon$ is treated similarly.

If $|x|\ge \epsilon,|y|\ge \epsilon$ , then $|f(x)-f(y)| = 0 \le \sqrt{|x-y|}$.

You can check that if $\epsilon<1/4$, then $$\{|x|<1/2,|y|<1/2\}\cup \{ |x|<\epsilon,|y|\ge 2\epsilon\} \cup \{ |y|<\epsilon,|x|\ge 2\epsilon\}\cup \{|x|\ge \epsilon , |y|\ge \epsilon\} = \mathbb R^2, $$ enter image description here so we're done.

PS $\alpha$-Lipschitz functions are also (more commonly?) called $\alpha$-Hölder continuous functions.

PPS The same proof with the obvious modifications gives $\sup_{x\neq y} \frac{|f(x)-f(y)|}{|x-y|^\alpha}\le 1$, for all $0<\alpha < 1$.

  • $\begingroup$ thank for giving solutions. My expectations was some simple counterexample. $\endgroup$
    – neelkanth
    Aug 24, 2020 at 15:42

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.