Show $\sqrt[3]{x}$ is or isn't uniformly continuous. I'm pretty sure $f(x)=\sqrt[3]{x}$ is uniformly continuous but having a hard time proving it.  If I let $\varepsilon > 0$ and try to get $\delta$ so that $|f(x)-f(y)| = |\sqrt[3]{x} - \sqrt[3]{y}|<\varepsilon$ I don't see how to relate this to an arbitrary $|x-y|<\delta$.  
I thought about $|\sqrt[3]{x}-\sqrt[3]{y}|^3 < \varepsilon^3$ is 
$$|x-y+(3\sqrt[3]{xy^2}-3\sqrt[3]{x^2y})|$$ 
but I don't see a way forward from here.
I know the function behaves significantly differently on $(0,1)$ and $(-\infty, -1) \cup (1,\infty)$ so I might want to break this into two proofs, where since $x^{1/3}$ is point-wise continuous then it's continuous on the compact set $[-1,1]$ and all that remains is to prove it on the rest---although I'm not seeing how to do that.
(However, this worries me:  I don't see a theorem in Rudin about ``A function uniformly continuous on a set $A$ and a set $B$ is therefore uniformly continuous on $A\cup B$ where $A\cup B$ is [some relevant property here, like being an interval or something].''  So I'm not sure how secure this proof strategy is.)
 A: It is on $[0,1]$ since a continuous function on a closed bounded interval is uniformly continuous.  For $x\ge 1$, 
$$|f'(x)| = 1/3x^{-2/3} \le 1/3.$$
Invoke the mean value theorem to get unform continuity on $[1,\infty)$. Now spread the wealth to the whole line by the oddness of this function.
A: A classical theorem (one that hopefully has come up in your real analysis class) is that a real function that is continuous on a closed, bounded interval is uniformly continuous. Thus for any big $N$, we know that $x^{1/3}$ is uniformly continuous on $[-N,N]$, and so the problem region should be recognized as $x\rightarrow\infty$. In particular, hopefully you have seen a proof that $f(x)=x^2$ is not uniformly continuous on $\mathbb{R}$; is there a way you can adapt this proof to suit the case when $f(x)=x^{1/3}$? No.
Away from zero, on the set $(-\infty,-N]\cup[N,\infty)$ you should be able to prove uniform continuity.
In short, yes, if a function is uniformly continuous on $[a,b]$ as well as $[b,c]$, then it will be uniformly continuous on the interval $[a,c]$. Think about how you might prove this, and indeed will generalize to any finite number of intervals...
A: Regarding your worry about whether uniform continuity on $[a,b]$ and $[b,c]$ is enough to conclude uniform continuity on $[a,c]$, I think this detail is important enough to merit a proof.
Given $\epsilon > 0$, there exist $\delta_1 > 0$ and $\delta_2 > 0$ such that $|f(x)-f(y)| < \epsilon/2$ provided that either of these conditions is met:


*

*$x,y \in [a,b]$ and $|x-y| < \delta_1$

*$x,y \in [b,c]$ and $|x-y| < \delta_2$


But what if $x\in [a,b]$ and $y \in [b,c]$? Let's suppose this is the case. Let $\delta = \min(\delta_1, \delta_2)$. Suppose that $|x-y| < \delta$. Then $b$ lies between $x$ and $y$, so $|x-b| < \delta \leq \delta_1$. Therefore, $|f(x) - f(b)| < \epsilon/2$. Similarly, $|b-y| < \delta \leq \delta_2$, so $|f(b) - f(y)| < \epsilon / 2$. (Note that we have exploited the fact that $b$ is contained in both intervals.) Therefore, by the triangle inequality,
$$|f(x) - f(y)| \leq |f(x) - f(b)| + |f(b) - f(y)| < \epsilon$$
so now all cases are covered, so the "spreading the wealth" technique of establishing uniform continuity on a finite number of adjacent intervals is legit.
A: Any function $x^\alpha,\enspace 0\le\alpha\le 1$ is uniformly continuous on $\mathbf R$ because :


*

*it is continuous, hence it is uniformly continuous on any bounded closed interval.

*its derivative is bounded from above by $\alpha$ on $[1+\infty)$, hence it is Lipschitz on $[1,+\infty)$.

*it is an odd function.

A: The function $f(x) := \sqrt[3]{x}$ is uniformly continuous in $\mathbb{R}$ because of the inequality:
$$\big| \sqrt[3]{x} - \sqrt[3]{y}\big| \leq \sqrt[3]{|x-y|}\; ,$$
which holds for every $x,y\in \mathbb{R}$.
In general, if $\alpha \in ]0,1]$, the function $f(x):=|x|^\alpha$ is uniformly continuous in $\mathbb{R}$ because of the inequality:
$$\big| |x|^\alpha - |y|^\alpha \big| \leq |x-y|^\alpha\; ,$$
which express the fact that $f$ is Hölder-continuous.
