Conditions needed for a unique root to also be a "clear-cut" root

Question

Suppose $f:[0,1] \longrightarrow [-1,1]$ is a continuous function that has a unique root $r_{0} \in (0,1)$.

I want $r_{0}$ to be a ``clear-cut root" (not sure what to call it) in the following sense: For any $\varepsilon > 0$, there exists a $\delta_{\epsilon} > 0$ such that $|t - r_{0}| > \varepsilon$ implies that $|f(t)| \geq \delta_{\epsilon}$.

My question is: are there any standard smoothness conditions on $f$ (such as f is Lipschitz continuous) which guarantee that whenever $r_{0} \in (0,1)$ is a unique root of $f$, it is also a ``clear-cut root"?

Answer 1

Every continuous $f$ satisfies that condition.

$|f|$ has a minimum on the compact set $[0, 1] \setminus (r_0-\epsilon, r_0+\epsilon) \, ,$ and that minimum must be positive if $r_0$ is the only zero of $f$ in $[0, 1]$.

Answer 2

You might want $f$ to be twice continuously differentiable with a non-zero derivative in the root. Then $$ |f(r_0+h)|=\Bigl|f'(r_0)h+\tfrac12f''(\xi)h^2\Bigr|\ge |h|·\Bigl(|f'(r_0)|-\tfrac12·|h|·M_2\Bigr) $$ where $M_2$ is a bound for the second derivative. This gives you reasonably good control on the function values for $h\approx 0$.