5

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?

Arthur
  • 199,419
BenL
  • 981
  • Use curly braces, like x^{1/3} to make the entire fraction appear in the exponent: $x^{1/3}$. – Arthur Mar 25 '17 at 14:04
  • 1
    You mean $\sqrt[3]{x}$ instead of $x^{1/3}$? – ntt Mar 25 '17 at 14:05
  • I am aware. I typed this on an iPhone. I went back and fixed it. See my edits above. – BenL Mar 25 '17 at 14:06
  • @nnt are you asking if I prefer one way of writing the same number to another way writing the same number? If that's your question then yes, I do have a preference. My preference is the way I wrote it;) – BenL Mar 25 '17 at 14:07
  • 3
    What @ntt meant was that strictly speaking, fractional exponents are undefined for negative numbers, even though the corresponding root exists. This is because $\frac13=\frac26$, and you should, at all times, be able to switch between the two. This conflicts with $x^{2/6}$ being ill-defined for negative $x$, so instead of creating a theory where rational numbers are treated differently from how they're usually treated, we ban the combination of non-integer exponents and negative bases. – Arthur Mar 25 '17 at 14:10

1 Answers1

6

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.

uniquesolution
  • 18,541
  • I apologize, I should have specified; I am not admitted the concept of a derivative. That said, I appreciate the idea of using a closed interval to cross the origin, as in $\mathbb{R}$ closed sets are compact, and continuous functions on compact metric spaces are uniformly continuous. Then I can just patch together the rest of the domain. Thanks! – BenL Mar 25 '17 at 14:17
  • Hi. I understand the whole proof properly, but which simple argument axactly can I use to show, that the funcion is uniformly continous in R ? – martina Jun 23 '19 at 11:23