# Function $f$ such that $|f(x)-f(y)|\leq \sqrt {|x-y|}, \forall x,y\in\Bbb R.$

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.

$$f(x) = \min(\epsilon,|x|)$$ 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 $$\{|x|<1/2,|y|<1/2\}\cup \{ |x|<\epsilon,|y|\ge 2\epsilon\} \cup \{ |y|<\epsilon,|x|\ge 2\epsilon\}\cup \{|x|\ge \epsilon , |y|\ge \epsilon\} = \mathbb R^2,$$ 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 $$0<\alpha < 1$$.

• thank for giving solutions. My expectations was some simple counterexample. Aug 24, 2020 at 15:42