Question

Let $IR$ be the set of real numbers and $f:IR\to IR$ be such that for all $x,y\in IR$, $\left|f\right(x)-f(y\left)\right|\le {|x-y|}^3$. Prove that $f$ is a constant function.

Answer 1

Given $\left|f\right(x)-f(y\left)\right|\le {|x-y|}^3$

$\Rightarrow \underset{x\to y}{lim}\frac{\left|f\right(x)-f(y\left)\right|}{|x-y|}\le \underset{x\to y}{lim}{|x-y|}^2$

$\Rightarrow \left|f^{\prime }\right(x\left)\right|\le 0$

$\Rightarrow f^{\prime }\left(x\right)=0\Rightarrow f\left(x\right)=$ constant