TITLE: Prove that f is differentiable in $\mathbb{R}$
QUESTION [6 upvotes]: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ some function that all $x$ and $y$ in $\mathbb{R}$ satisfies: $$\left|f(x)-f(y)\right| \le (x-y)^2$$

Prove that f is differentiable in any point in $\mathbb{R}$.
Prove that f is constant.

REPLY [7 votes]: I would like to point out that it is possible to prove the assertion without proving differentiability in the first place.
For each x we have 
$$|f(x)-f(0)| = \left|\sum_{k=1}^n f\left(\frac{k}{n} \cdot x\right)-f\left(\frac{k-1}{n}\cdot x\right)\right|$$
Thus by the triangle inequality
$$|f(x)-f(0)| \le \sum_{k=1}^n \left|f\left(\frac{k}{n}\cdot x\right)-f\left(\frac{k-1}{n}\cdot x\right)\right|$$
Since for each $k$
$$\left|f\left(\frac{k}{n}\cdot x\right)-f\left(\frac{k-1}{n}\cdot x\right)\right| \le \left|\frac{k}{n}\cdot x-\frac{k-1}{n}\cdot x\right|^2 = \left|\frac{x}{n}\right|^2$$ 
It follows that 
$$|f(x)-f(0)| \le \sum_{k=1}^n \frac{x^2}{n^2} = \frac{x^2}{n}$$
And thus
$$|f(x)-f(0)|\le \limsup_{n\rightarrow\infty} \frac{x^2}{n} = 0$$
Hence for all $x$
$$f(x)=f(0)$$
Thus f is constant. It follows that f is differentiable everywhere ;)