Show that $ x^{(1/3)}$ is uniformly continuous on $\mathbb{R}$

Question

I've proved it when restricted to the positive and negative portions of the domain (by assuming the x and y in consideration are both positive or negative, respectively). I cannot for the life of me figure out how to prove it when x (assumed to be >y) is positive and y negative. My intuition is that it must be uniformly continuous, because it seems like $\delta < \epsilon ^3$ should be a limit on how much the function can change over a given interval; that said, I can't prove that $ |x - y| < \epsilon ^3 \Rightarrow |x^{1/3}-y^{1/3}| < \epsilon$ when the domain crosses the origin. Help?

Answer 1

Let $f(x)=x^{1/3}$. On $[-1,1]$ your function is uniformly continuous, as is any continuous function on a closed and bounded interval in $\mathbb{R}$. On the complement of $[-1,1]$, the derivative of your function is bounded. Indeed, $|\frac{1}{3x^{2/3}}|<\frac{1}{3}$ for every $|x|>1$. As a result, for every $x,y$ outside of $[-1,1]$ (*), using the mean value theorem; $$|f(x)-f(y)|=|f'(\xi)||x-y|\leq |x-y|/3$$ where $\xi$ is a suitable point between $x$ and $y$. It follows that $f$ is uniformly continuous on the complement of $[-1,1]$, too, and so it is uniformly continuous in $\mathbb{R}$ (by a simple argument).

(*) Strictly speaking, we should treat $x,y$ in $(1,\infty)$ and $(-\infty,-1)$ separately: we don't want $x>1$ and $y<-1$. Nonetheless, the idea should be clear.