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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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    If you define $g(\delta) = \sup_{|x| \le \delta} |f(x)|$ then $g$ is continuous and non decreasing. I think $\delta_f$ is continuous if and only if $g$ is strictly increasing (in which case $\delta_f = g^{-1}$).2012-01-22
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    If you take $f(x) = x$, then $\delta_f(\epsilon) = \epsilon$ but $|f| : x \mapsto |x|$ is not monotone (whereas the function $g$ I defined in my previous comment is).2012-01-22
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    if $f$ is monotonic but constant on an interval you already have an example where $\delta$ is not continous (take $\varepsilon = f(x)$ for some $x$ from the constancy interval). E.g. $f(x)= x$ for $x<1$, $f(x)=1$ for $1<= x <= 2$ and strictly linearly increasing for $x>2$. You can explicitly calculate $\delta$ for this $f$. So I think it is fruitless to search for a weaker condition than strict monotonicity. Analogous examples are easy to construct in the more general case you are asking about.2012-01-22
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    @JoelCohen: Of course, it should have been obvious that (for $V \ne \mathbb{R}$) then notion of a function $f : V \to (0,\infty)$ being monotone doesn't make any sense. Thanks!2012-01-22
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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

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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 $x\geqslant0$, $g(x)=\sup\{|f(z)|\,;\|z\|=x\}$.