I came across the following task
Suppose $f(x)$ to be defined on $[0, 1]$ and $f(0) = 0$ also $\forall \ x, y: 0 \leq x \leq y \leq 1 \rightarrow f(y) - f(x) \leq (y-x)^2$. Then there is a function $f(x)$ that is discontinuous on $[0, 1]$
I can't imagine this situation because I thought that if for all, arbitrarily closely located to each other, points, the values of the function differ even less it means a uniform continuity which entails a continuity. But it seems I was wrong. Could you please help me to think up a such function? Thanks in advance for any hints!