When I was in the shower this morning a question went through my head about continuity of a function at a point. The simplest formulation of this question is:
Let $f : \mathbb{R} \to \mathbb{R}$ be an unbounded continuous function with $f(0) = 0$. Define $\delta_f : (0, \infty) \to (0, \infty)$ by $\delta_f(\varepsilon) = \sup \{ \delta > 0\, :\, |x|< \delta \Rightarrow |f(x)| < \varepsilon \}$ Under what conditions is $\delta_f$ a continuous function of $\varepsilon$?
The answer to this probably involves monotonicity; for example, it seems that $\delta_f$ is continuous whenever $f$ is strictly monotone. I'd like to find (if possible) the weakest condition on $f$ to make $\delta_f$ continuous.
My hunch is that the answer is that $\delta_f$ is continuous if and only if $|f| : \mathbb{R} \to [0,\infty)$ is strictly monotone, but I await counterexamples with open arms.
More generally, the question can be formulated as follows:
Let $f : V \to W$ be a continuous unbounded function between normed spaces with $f(0_V) = 0_W$. Define $\delta_f : (0, \infty) \to (0, \infty)$ by $\delta_f(\varepsilon) = \sup \{ \delta > 0 \, :\, \lVert x \rVert < \delta \Rightarrow \lVert f(x) \rVert < \varepsilon \}$ Under what conditions is $\delta_f$ a continuous function of $\varepsilon$?
My ultimate goal is to prove that $\delta_f$ is continuous if and only if $\lVert f \rVert : V \to [0, \infty)$ is strictly monotone, or to find a counterexample.