$f:[0,1]\to\mathbb R$ satisfies the following conditions:
$$f(0)=0,$$ $$f'(x)\neq0 \ \text{for any } \{x: f(x)=0\}.$$
What is the necessary and sufficient condition for $\exists \epsilon$ such that $f(x)$ has no root on the interval $(0,\epsilon)$?
Claim 1: Continuity is not a sufficient condition.
Claim 2: Almost everywhere differentiablity of $f$ is not sufficient
Claim 3: Absolute continuity is sufficient
For the Claim 1, at a first glance, continuity seems sufficient: since $f'(x)\neq 0$, $\{x|f(x)=0\}$ is a discrete set; we then pick the root next to $0$, denote it as $\epsilon$. However, without countability, the operation of "pick the next point" does not make sense. In fact, let $f$ be a Weierstrass function, it is possible that $\forall\epsilon>0$, $f(x)$ has uncountable many roots on $(0,\epsilon)$.
For the Claim 2, it is still possible that $\forall\epsilon>0$, $f(x)$ has roots on $(0,\epsilon)$.
Not sure about Claim 3, though...
I think $f$ being differentiable is sufficient but not necessary.
It fine to interpret $f'(0)\neq 0$ as $f(x)$ is differentiable at $0$; any intuition on this $f(0)$ differentiable case will be a good answer. Though I was more interested in another case: $f(x)$ is not differentiable at $0$ and the subderivative $f'(0)\neq 0$.