Conditions for a function to be increasing close to 0 Suppose $f:[0,1]\to[0,1)$ is a continuous function such that $f(0)=0$ and $f(x)>0$ for $x\neq 0$. Under what natural conditions there exists $\epsilon>0$ such that $f$ is strictly increasing over $[0,\epsilon]$? 
For example, if $f$ is differentiable over [0,1] and $\partial_+ f(0)\neq 0$ (the right-derivative is non-zero), then such $\epsilon$ must exist, by here. Could we make the same conclusion if the differentiability assumption is removed (so, instead of differentiability, we have that  $\partial_+ f(0)$ is defined and is positive).
 A: If $f$ is not strictly increasing on $[0,\epsilon]$, there are $a,b$ with $0 \le a < b \le \epsilon$ and $f(a) \ge f(b)$.  The  maximum value of $f$ on $[a,b]$ is attained at some $c < b$, and then the upper right-hand Dini derivative $$f'_+(c) = \limsup_{x \to c+} \frac{f(x) - f(c)}{x-c} \le 0$$
Thus a sufficient condition for $f$ to be strictly increasing on $[0,\epsilon)$ is that $f'_+(x) > 0$ for all $x \in [0, \epsilon)$.  On the other hand, a necessary condition for $f$ to be increasing on $[0,\epsilon)$ is that $f'_-(x) \ge 0$ for all $x \in [0,\epsilon)$, where $f'_-$ is the lower right-hand Dini derivative $$f'_-(x) = \liminf_{t \to x+} \frac{f(t) - f(x)}{t - x}$$ 
A: Take $f(x) = \frac{1}{2}x  +x^2 \sin{\frac{1}{2x}} $ and $f(0)=0$
From the inequality $\sin{\frac{1}{2x}} \leq \frac{1}{2x} $  we can conclude that $f \ge 0$ one $[0,1]$.  Also  $f' (0+) = \frac{1}{2}$ and $f' (x) = \frac{1}{2} - \cos{\frac{1}{2x}} + 2x \sin{\frac{1}{2x}} $ for all $x >0$. 
Observe that $f'$ oscillates with positive and negative values as $x \to 0^+.$ This shows $f$ can not be increasing on any interval $[0, \epsilon].$
This function satisfies all conditions in question but not the claim!
