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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.

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    I've posted some history on questions regarding continuous selections of deltas [here](http://math.stackexchange.com/a/1628855/23611).2016-01-27

1 Answers 1

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The function $\delta$ may be defined as follows. For every $x\geqslant 0$, let $F(x)=\sup\{|f(z)|\,;|z|\leqslant x\}$. Then $\delta(0)=0$ and, for every $t\gt0$, $\delta(t)=\sup\{x\gt0\,;\forall z\lt x, F(z)\lt t\}$. One sees that the question really concerns the nondecreasing function $F$ defined on $[0,+\infty)$.

The function $\delta:t\mapsto\delta(t)$ is defined on $[0,+\infty)$ and continuous on the left on $(0,+\infty)$. Let $t\geqslant0$. The function $\delta$ is continuous on the right at $t$ if and only if $F$ is strictly increasing on the right at $t$, that is, if and only if $F(s)\gt F(t)$ for every $s\gt t$.

Now, $F$ is always nondecreasing and $F$ is increasing if and only if the function $g$ defined on $[0,+\infty)$ by $g(x)=\max\{|f(x)|,|f(-x)|\}$ for every $x\geqslant 0$, is increasing.

The corresponding necessary and sufficient condition in the general case is that $g$ is increasing where, for every $x\geqslant0$, $g(x)=\sup\{|f(z)|\,;\|z\|=x\}$.