Is map of "deltas" to points on closed interval continuous?

I am having trouble properly coming up with an answer to the following:

Let $$f: [0, 1] \to \mathbb{R}$$ be a continuous function, using the usual (Cauchy) characterization of continuity. Define the set $$D_{\epsilon} (x) = \{d \in \mathbb{R}: \forall x' \in [0, 1]:|x'-x| < d \implies |f(x')-f(x)| < \epsilon\}$$ for all $$x \in [0,1]$$, and let $$\delta(x) = \sup{D_{\epsilon}}$$. Does the fact that $$f(x)$$ is continuous on $$[0,1]$$ imply that the map $$\delta(x)$$ is continuous too?

I have tried to derive a contradiction, by showing that if some points in the interval are close enough, but their $$\delta$$ differs by a finite amount, one of the $$\delta$$ is not actually the supremum of the set D for one of them, but I didn't manage to take my argument too far.

Any help is much appreciated!

• For related ideas, look for "modulus of continuity". Jun 27 '21 at 14:38
• Thanks. Do you know if the proof I have provided below is sound? Jun 27 '21 at 14:50

1. $$f: [0, 1] \to \mathbb{R}$$ is continuous, if: $$\forall \epsilon > 0: \forall x \in [0, 1]: \exists \delta > 0: |x'-x| < \delta \implies |f(x')-f(x)| < \epsilon$$.
2. If the map $$\delta_{\epsilon}(x)$$ is continous, this means that: $$\forall E>0: \forall x \in [0, 1]: \exists d > 0: |x'-x| < d \implies |\delta_{\epsilon}(x') - \delta_{\epsilon}(x)| < E$$.
3. Suppose the map $$\delta_{\epsilon}(x)$$ were not continuous. This would mean that there is some $$x_0 \in [0,1]$$ such that for any $$d>0$$, there is some $$x \in (x_0 - d, x_0 + d)$$ for which $$|\delta_{\epsilon}(x)-\delta_{\epsilon}(x_0)| > E$$.
4. Now fix $$E$$ and $$x_0$$. If 3. is true, then we can choose some $$x_1$$ arbitrarily close to $$x_0$$ such that $$|\delta_{\epsilon}(x_1)-\delta_{\epsilon}(x_0)| > E$$, WLOG drop the absolute value sign. Then, by definition of our function $$\delta_{\epsilon}(x)$$, there is some $$x' \in [x_0 + \delta_{\epsilon}(x_0), x_1 + \delta_{\epsilon}(x_1))$$ such that $$|f(x') - f(x_0)| > \epsilon$$.
5. By the triangle inequality, $$|f(x') - f(x_0)| \leq |f(x')-f(x_1)| + |f(x_1) - f(x_0)|$$. Since $$x'$$ lies in $$(x_1 - \delta_{\epsilon}(x_1), x_1 + \delta_{\epsilon}(x_1))$$, $$|f(x') - f(x_1)|< \epsilon$$. Also, since $$f$$ is continous, we can always choose $$d$$ to make $$|f(x_1) - f(x_0)|$$ arbitrarily small. This implies that $$|f(x')-f(x_0)| < \epsilon$$, contradicting what was deduced in 3.
6. So the map $$d_{\epsilon}(x)$$ is continous.
Remark: this then also means that the function we could define from the $$d$$-s is continous, and so on ad infinitum.
• +1. I would have said basically the same, but by contradiction: If $D_{\epsilon}$ were discontinuous at $x$, there would be sequences $(b_n)_n$ and $(c_m)_m$ converging to $x$ with $D_{\epsilon}(b_n)\to B$ and $D_{\epsilon}(c_n)\to C>B.$ But if $b_n, c_m$ are close enough to $x,$ and hence close enough to each other, and if $D_{\epsilon}(b_n), D_{\epsilon}(c_m)$ are close enough to $B,C$ respectively, this leads to a contradiction. Jun 28 '21 at 2:06