Function is defined on the whole real line and $|f(x) -f(y)| \leq |x-y|^\alpha$, then.... Given: $f(x)$ is defined on $\mathbb{R}$  and  $|f(x) -f(y)|  \le |x-y|^\alpha$. Which of the following statements are true?
I. If $\alpha > 1$, then $f(x)$ is constant.
II. If $\alpha = 1$, then $f(x)$  is differentiable.
III. $0 < \alpha < 1$, then $f(x)$ is continuous.
Answer: I $-$ true, II $-$ false, III $-$ true.
I wonder how this result was obtained. Maybe somebody can give some explanations?

Hints: For (I) $0\le\vert\frac{f(x)-f(y)}{x-y}\rvert\le\vert x-y\rvert^{\alpha-1}$
For (II) Think about the function $x\mapsto\lvert x\rvert$
For (III) $0\le\lvert f(x)-f(y)\rvert\le\lvert x-y\rvert^{\alpha}$ let $x$ tend to $y$.
