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?
Your condition is a special case of Hölder continuity. If $\alpha = 1$, it is usually called Lipschitz instead of Hölder. I'll give some hints.
I. Suppose $\alpha = 1 + \epsilon$ for $\epsilon > 0$. Then $\left|\frac{f(x) - f(y)}{x-y}\right| \leq |x-y|^\epsilon$. If we take the limit as $y \to x$, what does this say about the derivative of $f$ at $x$?
II. Consider $f(x) = |x|$.
III. Let $\epsilon > 0$, and $\delta = \epsilon^{\frac{1}{\alpha}}$. If $|x-y| < \delta$, then what can be said about |f(x) - f(y)|?
If $f: I \rightarrow \Bbb R$ is a function for which exists $c, \alpha > 0$ such that $$|fx - fy| \leq c|x - y|^\alpha$$ then $f$ is said to be Hölder continuous. If $\alpha = 1$, then we say $f$ is Lipschitz continuous. This way, way, every Lipschitz function is a Hölder function. Now, if $\alpha > 1$, $f$ is constant because it's derivative is zero, see that: $$ 0 < |fx - fy| \leq |x-y|^\alpha \\ 0 < \frac{|fx - fy|}{|x - y|} \leq c|x - y|^{\alpha - 1}$$ Since $\alpha - 1 > 0$, making the difference $x - y \to 0$, we get that $f' \equiv 0$, hence $f$ is constant. If $f$ is differentiable, then $f$ is continuous.
However, we can say something stronger. Every Hölder continuous function is uniformly continuous. Let $\epsilon > 0$, choose $\delta = \sqrt[\alpha]{\frac{\epsilon}{c}}$, and $x,y \in I$ such that $|x - y| < \delta$. Then we get $$|fx - fy| \leq c|x-y|^\alpha < c\sqrt[\alpha]{\dfrac{\epsilon}{c}}^\alpha = \epsilon$$
For the counter-example, $f(x) = |x|$ will do, as @user71352 said.
Here is a question a bit related to Hölder continuity, with a great answer, and lot of references.
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 $y$.
