Let $f:\mathbb{R} \to \mathbb{R} $ be a function such that there exists an $M>0$ such that $|f(x)-f(y)|\le M|x-y|^2$ for all $x, y\in \mathbb{R}$. Show that $f$ is constant.
I know that this gives
${|f(x)-f(y)| \over |x-y|}\le M|x-y|$
so
$|f'(y)|\le M|x-y|$.
I also know that to prove something is constant then you need to show that $f(x)=f(y) \forall x,y \in \mathbb{R}$ with $f'(x)=0 \forall x \in \mathbb{R}$ but I cannot decide what to use and how to do this, using the mean value theorem or the definition of the derivative or something else.
