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!

  • 2
    $\begingroup$ For related ideas, look for "modulus of continuity". $\endgroup$
    – GEdgar
    Jun 27 '21 at 14:38
  • $\begingroup$ Thanks. Do you know if the proof I have provided below is sound? $\endgroup$
    – WhiteLake
    Jun 27 '21 at 14:50

I think I managed to figure it out.

  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
    $\begingroup$ +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. $\endgroup$ Jun 28 '21 at 2:06

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