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.
 A: $f(x) = \min(\epsilon,|x|)$ 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 $0<\alpha < 1$.
