Prove that if f is an element of Lip a and a>1 then f is constant

Question

We have that a function $f: [a,b]\mapsto\Bbb R$ satisfies a Lipschitz condition of order $a > 0$ if there is some positive constant $M$ so that $$|f(x_1)-f(x_2)| \leq M|x_1-x_2|^a \qquad \forall x_1,x_2 \in [a,b]$$ and $\operatorname{Lip}(a)$ denoted the set of all functions satisfying a Lipschitz condition of order $a$. So I have to prove that if $f$ is an element of $\operatorname{Lip}(a)$ and $a>1$ then $f$ is constant.

Answer 1

Hint

We have

$$\left|\frac{f(x)-f(y)}{x-y}\right|\le M|x-y|^{a-1}$$

so what do you conclude if you pass to the limit $x\to y$?