Non uniform continuity of $f(x)=x^3$ on the interval $[10,\infty)$

Question

How to show that $f(x)=x^3$ is not uniform continuous on the interval $[10,\infty)$?

I am well aware that $f(x)^3$ is not uniform continuous on the set of real numbers but and I believe on $[10,\infty)$ it is also not uniform continuous. However, how do I show it's not for a given set? I have seen proofs of it on the reals but not with a given interval. How do I use the interval to help me decide it's not uniform continuous?

Answer 1

One shortcut... if the function is continuous and differentiable over a domain, and the derivative is bounded for all x in the domain, then then the function is uniformly continuous over that domain.

If the domain is unbounded e.g. $[3,\infty)$ the derivative of $f(x)=x^3$ goes to infinity as $x$ goes to infinity, and $f(x)$ is not uniformly continuous.

But over the interval $[0, 10^{10})$ it is.

However, it is possible for the derivative to be undefined / unbounded and the function to still be uniformly continuous.

e.g. $f(x) = \sqrt x$

$\lim_\limits {x\to 0^+} f'(x) = \infty$

Yet, $f(x)$ is uniformly continous over $[0,\infty)$

If you want to be safe, always check against the definition.

The function is uniformly continuous if:

$\forall \epsilon>0,\exists \delta>0: \forall x,y \in [10,\infty), |x-y|<\delta \implies |x^3 - y^3|<\epsilon$

And therefore is not uniformly continuous if

$\forall \delta > 0, \exists \epsilon > 0, \exists x,y\in [10,\infty): |x-y|<\delta$ and $|x^3 - y^3|>\epsilon$

$|x^3 - y^3| = |x-y||x^2 +xy + y^2|$

For any $\epsilon, \delta$ we can choose $x = \min(\frac 32 \sqrt {\frac {\epsilon}{\delta}},10), y = x+ \frac {\delta}{2}$

Answer 2

Assume by way of contradiction that it is uniformly continuous. Then for $\epsilon = 1$, $\exists \delta > 0$ such that $|x - y| < \delta \Rightarrow |f(x) - f(y)| = |x^3 - y^3| < 1$ for all $x,y \in [10,\infty)$. Note that $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$.

We can choose $y = x + \delta/2$ so that $|x - y| = \delta/2 < \delta$. Observe, \begin{align} |f(x) - f(y)| &= |x - y||x^2 + xy + y^2|\\ &= \frac{\delta}{2}|x^2 + x(x + \delta/2) + (x + \delta/2)^2|\\ &= \frac{\delta}{2}|3x^2 + \frac{3\delta}{2}x + \frac{\delta^2}{4}| < 1 \end{align} a contradiction since $\frac{\delta}{2}|3x^2 + \frac{3\delta}{2}x + \frac{\delta^2}{4}|$ is unbounded for a fixed $\delta$ (all non-constant polynomials are unbounded). Hence $f(x)$ is not uniformly continuous.