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Note, there is a similar question here: Show $\sqrt[3]{x}$ is or isn't uniformly continuous.

I am asking my question anyway because the one here does not ask about the Sequential Criterion for Absence of Uniform continuity

$ f(x)=\sqrt[3]{x}$ is continuous on $[0,\infty)$, note $|f(x)-f(c) | \rightarrow |\sqrt[3]{x}-\sqrt[3]{c}| <\epsilon$

$|\sqrt[3]{x}-\sqrt[3]{c}|= |(x-c)(\sqrt[3]{x^2}+\sqrt[3]{xc}+\sqrt[3]{c^2})| $ If we insist on $\delta < 1$, then we ensure that $x$ is within $(c-1,c+1)$

Then $(\sqrt[3]{x^2}+\sqrt[3]{xc}+\sqrt[3]{c^2})| $ $\geq$ $3\sqrt[3]{c^2}$, giving us $|(x-c)(3\sqrt[3]{c^2})|<\epsilon$.

So $\delta = \dfrac{\epsilon} {3\sqrt[3]{c^2}}$

Is it uniformly continuous on $[0,\infty)$?

Here I use the Sequential Criterion for Absence of Uniform Continuity which states:

A function $f: A \rightarrow \mathbb{R}$ fails to be uniformly continuouson $A$ if and only if $\exists \epsilon_0 >0$ and two sequences $(x_n), (y_n) \in A$ satisfying $|x_n - y_n| \rightarrow 0$ but $|f(x_n) - f(y_n)| \geq \epsilon_0$

I let $x_n=n, y_n=n+\frac{1}{n}$ and observe that $|x_n - y_n| \rightarrow 0$ but

$$|f(x_n) - f(y_n)| = \left| \left( \sqrt[3]{n}-\sqrt[3]{\left(n+\frac{1}{n} \right) }\, \right) \left\{ \frac{\sqrt[3]{n^2} + \sqrt[3]{n\left( n + \frac 1 n \right) + \sqrt[3]{\left( n + \frac 1 n \right)^2}}}{\sqrt[3]{n^2} + \sqrt[3]{ n \left( n + \frac 1 n \right) + \sqrt[3]{\left( n + \frac 1 n \right)^2}}} \right\} \right| $$

$$= \frac{|n-(n + \frac{1}{n})|}{\sqrt[3]{n^2}+\sqrt[3]{n(n+\frac{1}{n})}+\sqrt[3]{(n+\frac{1}{n}})^2},$$

which goes to $0$ and thus $f$ is uniformly continuous on $[0,\infty)$.

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  • $\begingroup$ Do you want an answer to your title, or a response to the errors in your derivation that the cube root function is uniformly continuous on its domain? $\endgroup$ – Eric Towers May 6 '16 at 22:54
  • $\begingroup$ If one is allowed to use some theory, our function is uniformly continuous on $[0,1]$, and for the rest one can use the Mean Value Theorem. $\endgroup$ – André Nicolas May 6 '16 at 22:58
  • $\begingroup$ You falsified incorrectly the thesis of your long-name criterion. In fact, the same exact argument could be done with $A=(0,\infty)$, $f=\ln$ and the same two sequences. But, of course, $\ln$ is not uniformly continuous on $(0,\infty)$. In fact, I would not use that criterion, because the negation of the part after the " if and only if " requires you to check a large number of sequences and, truth to be told, it would appear in a form wich is not be the best phrasing of uniform continuity. $\endgroup$ – user228113 May 6 '16 at 23:04
  • $\begingroup$ @G.Sassatelli I did not falsify anything, but I believe I may have used it incorrectly. Thanks to Eric Towers. $\endgroup$ – Jabernet May 6 '16 at 23:09
  • $\begingroup$ @AndréNicolas Technically I can't use the Mean Value Theorem here, it is in the section of the book prior to discussion of the Mean Value Theorem. $\endgroup$ – Jabernet May 6 '16 at 23:10
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Regarding uniform continuity...

You have shown that there is a single pair of sequences for which $|x_n - y_n| \rightarrow 0$ and not $|f(x_n) - f(y_n)| \geq \epsilon_0$. You have not shown this for any other sequence. Since you need to show this for any possible pair of sequences, you have only begun your task.

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  • $\begingroup$ Okay, that makes a bit more sense. I can use it to show something is not uniformly continuous in this manner, but if I want to use it the other way I have to show it is true for all sequences. $\endgroup$ – Jabernet May 6 '16 at 23:06
  • $\begingroup$ @Jabernet : Agreed. $\endgroup$ – Eric Towers May 6 '16 at 23:10

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