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I can't see why the following could not be a counterexample.

Let $f:[-1;1]\to\mathbb{R}$ defined as $f(x) = \sqrt[3]{x}$. It's a continuous function that goes from a compact set to $\mathbb{R}$ but it's not uniformly continuous.

I really appreciate any kind of help.

Alp Uzman
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Davide Modesto
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    Why do you think it's not uniformly continuous? – Rob Arthan Apr 16 '20 at 20:46
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    Very possibly you are accidentally asking more than the definition/characterization of "uniformly continuous". For example, yes, we can see that the derivative blows up at $0$. So that precludes an easy proof of uniform continuity (using boundedness of a continuous derivative on a compact set)... but that failure of proof mechanism does not mean that the conclusion does not hold... – paul garrett Apr 16 '20 at 22:16

1 Answers1

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A continuous function on compact space is uniformly continuous.

Then a function $f:X\to Y$ is uniformly continuous on $X$ if and only if:

$$(\forall\epsilon>0)(\exists\delta>0)(\forall x_1,x_2\in X): |x_1-x_2|<\delta \implies |f(x_1)-f(x_2)|<\epsilon.$$

We will show that $f:[-1;1]\to\mathbb{R}$ defined by the rule $f(x) = \sqrt[3]{x}$ is uniformly continuous.

Let $\epsilon>0$. Use the inequality $|\sqrt[3]{x} - \sqrt[3]{y}\big| \leq \sqrt[3]{|x-y|}$, to continue the proof.

Salech Alhasov
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  • $|\sqrt[3]{x} - \sqrt[3]{y}\big| \leq \sqrt[3]{|x-y|} < \sqrt[3]{\delta} $ so $\delta = \epsilon^3$ should work, right? – Davide Modesto Apr 17 '20 at 09:41
  • @DavideModesto, yes. You have shown the existence of $\delta>0$, given $\epsilon$. See the simulation here: https://en.wikipedia.org/wiki/Uniform_continuity – Salech Alhasov Apr 17 '20 at 11:08
  • I have problem. If I consider $x = -y$, from the inequality you gave me follows that $2 \leq \sqrt[3]{2}$ which is false. – Davide Modesto Apr 17 '20 at 14:53
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    @DavideModesto, yes, you are right. I didn't mention a crucial condition: $x$ and $y$ should have the same sign. Hence the proof splits into two: showing that $f$ is uniformly continuous on [-1,0] and on $[0,1]$, then use the following fact that discussed here: https://math.stackexchange.com/questions/217683/if-f-is-unformly-continuous-on-two-sets-show-that-f-is-also-uniformly-continuou – Salech Alhasov Apr 17 '20 at 16:37