Assume that we have a function $f : [0, 1] \rightarrow [0, 1]$ which is differentiable, $f(0) = 0$ and $f(1)=1$. Can we deduce that there exists an open sub-interval in $[0, 1]$ in which $f$ is strictly increasing over that?
Note that here, continuity of $f^{\prime}$ is not mentioned.
The answer is no, a counterexample (and discussion/history around it) can be found in
K. C. Ciesielski - "Monsters in Calculus", page 4.
The basic idea is that if one picks the Pompeiu function $h$, there exists a dense set $G$ of translates such that for each $t \in G$ the zeroes of the derivatives of $h(x)$ and $h(x-t)$ do not coincide. Then, it is easy to conclude that the derivative of $h(x)-h(t-x)$ is positive on a dense subset and negative on another dense subset of $\mathbb R$.
Edit simplified the next
Let $g(x)=h(x)-h(t-x)$. Then, since $g$ is not constant, there exists some $a <b$ such that $g(a) \neq g(b)$.
Then for the right values of $\alpha, \beta$ $$f(x)= \alpha g( a+ (b-a)x) +\beta$$ satisfies $f(0)=0$ and $f(1)=1$.
