Let $f: \Bbb R \to \Bbb R$ be such that $|f(x) -f(y)| \le |x - y |^2 \ \forall \ x,\ y \in \Bbb R$. How can I show that $f$ is constant?

Question

Let $f: \Bbb R \to \Bbb R$ be such that $|f(x) -f(y)| \le |x - y |^2 \ \forall \ x,\ y \in \Bbb R$.

How can I show that $f$ is constant? To me the statement looks an awful lot like something having to do with continuity/uniform continuity, but the fact that $|x - y|$ is being raised to the power of $2$ is confusing me.. I also recognize that the problem doesn't say anything about continuity, it's just what it looks like to me. Any tips?

Answer 1

$$\left| \frac{f(x)-f(y)}{x-y}\right|\leq |x-y| \Rightarrow \forall x\; \exists f'(x) \mbox{ and } f'(x) = 0$$