$f(x)$ is uniformly continuous on $(0, \infty).$

Question

Is the function $f(x) = x^2 + x$ uniformly continuous on $(0, \infty)?$ Justify your answer.

Attempt: Let $x,y \in (0,\infty).$ Let $\varepsilon>0.$ Then, if $|x-y|<\delta,$

$$|x^2+x-y^2-y| \le |x^2-y^2|+|x-y| = |x-y||x+y|+|x-y|<\delta|x+y|+\delta$$

Here I don't know how to deal with $|x+y|$. Could you give some hint?

I am also wondering if the theorem (if $f$ is continuous on the closed interval, $f$ is uniformly continuous on that interval) is still applicable on $[0, \infty).$

Thank you in advance.

Answer 1

No, the theorem is not still applicable on $[0,+\infty)$.

If $n\in\mathbb N$, then$$\lim_{x\to\infty}f\left(x+\frac1n\right)-f(x)=\lim_{x\to\infty}\frac{2x}n+\frac1n+\frac1{n^2}=+\infty$$Can you take it from here?

Answer 2

Suppose that $f$ is uniformly continuous on $(0, \infty)$. Then there is $\delta>0$ such that

$$|f(x)-f(x+ \delta/2)|<1$$

for all $x>0$.

An easy computation shows that $|f(x)-f(x+ \delta/2)|=x \delta+ \frac{\delta^2}{4}+\frac{\delta}{2}$.

Hence it results that $x \delta+ \frac{\delta^2}{4}+\frac{\delta}{2}<1$ for all $x>0$, which is absurd.

Answer 3

For any $a> 0$, you have this

$$f(x + a) - f(x) = (x+a)^2 + x + a - (x^2 + x) = 2xa + a^2 + a.$$

As $x\to\infty$, $f(x+a) - f(x) \to \infty$. This debars uniform continuity on $(0,\infty).$