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Let us consider the identity function $$f:(\mathbb{R},d)\to (\mathbb{R},d_{usual}) $$ $$f:x \to x$$ Here we are considering $d(x,y)=|(x)^3-(y)^3|$

Is the function $f$ uniformly continuous on closed and bounded interval?

I am looking for an example of function $f$ which is not uniformly continuous on a closed and bounded interval but it is continuous.

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  • $\begingroup$ Any continuous function on a closed bounded interval is uniformly continuous. $\endgroup$ Feb 24, 2022 at 9:06
  • $\begingroup$ That is in R under usual metric.What if the metric is different? $\endgroup$
    – Antimony
    Feb 24, 2022 at 9:07
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    $\begingroup$ Any continuous function on a compact metric space into any metric space is uniformly continuous. $\endgroup$ Feb 24, 2022 at 9:12
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    $\begingroup$ If you are using the metric $d$ on the domain also you should specify it. Also, a closed and bounded interval is compact in the metric $d$ also. $\endgroup$ Feb 24, 2022 at 9:15
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    $\begingroup$ No. It doesn't. On a closed and bounded interval $x^{1/3}$ is continuous hence also uniformly continuous. That makes your map uniformly continuous. $\endgroup$ Feb 24, 2022 at 11:35

1 Answer 1

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In this answer I assume that $\tan^{-1}$ denotes the arctangent $\arctan$.

No, your example does not work, because $\arctan$ is Lipschitz continuous. Indeed for all $x, y \in \Bbb R$ $$|\arctan x - \arctan y| \le |x-y|$$

this implies that the identity map $f: \Bbb R \to ( \Bbb R , d)$ is Lipschitz (hence uniformly continuous).

Anyway, if you want to build an example of a continuous function which is not uniformly continous on a closed and bounded interval, you are trying to find a counterexample to Heine-Cantor theorem. In particular you need to find a distance defined on the real line with the property that closed and bounded intervals are not compact.

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  • $\begingroup$ Is there a way to modify the example? $\endgroup$
    – Antimony
    Feb 24, 2022 at 9:08
  • $\begingroup$ How about $(R,d_{usual}) \to (R,d)$ where $d(x,y)=|x^3-y^3|$. Will this work if I take $f(x)=x$? $\endgroup$
    – Antimony
    Feb 24, 2022 at 9:10
  • $\begingroup$ This is an interesting question. I don't think so, but now I have to leave, sorry $\endgroup$
    – Crostul
    Feb 24, 2022 at 9:15

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