if $|f(x)-f(y)|\le |x-y|^{\sqrt 2}$ then is $f$ a constant function?

Question

if $f: \mathbb{R}\to \mathbb{R}$ satisfies $$|f(x)-f(y)|\le |x-y|^{\sqrt{2}}$$ for all $x,y\in \mathbb{R}$ ,then is f increasing ,decreasing or constant?
in my view ,it is clear that $|f(x)-f(y)|$ is not zero so it is not constant. And since the rate of change is positive so the function would be increasing.Is my solution correct or wrong?

Answer 1

Consider the more general case: $|f(x)-f(y)|\le |x-y|^{1+\varepsilon}$, with $\varepsilon>0$.

Then $f$ is differentiable with $f'=0$ because $$ \lim_{x\to x_0} \left|\frac{f(x)-f(x_0)}{x-x_0}\right| \le \lim_{x\to x_0}|x-x_0|^{\varepsilon} = 0 $$

So, $\sqrt2$ is a red herring. Its only relevant property is $\sqrt2>1$.