Suppose that $f:(a,b) \to \Bbb R$ satisfies $|f(x)-f(y)|=M|x-y|^\alpha$ for some $\alpha > 1$, some $M \geq 0$, an all $(x,y)$ in $(a,b)$. Prove $f$ is constant on $(a,b)$
I have absolutely no idea how to go about this....Full proof would be exceedingly helpful.
see that $\frac{\mid f(x) - f(y) \mid}{\mid x - y \mid} = M \mid x -y \mid ^ {a-1}$ . now use the definition of derivative to show f has derivative zero.