Show that: $f:\mathbb{R} \rightarrow \mathbb{R}$ where $f(x) = x^2 + x$ is not uniformly continuous.

To prove that, I need to show that there are $x,y \in \mathbb{R}$ such that $\exists \epsilon > 0$, $\forall \delta > 0$ and it follows that $|x-y| \leq \delta$ but $|f(x) - f(y)| > \epsilon$.

First suppose that thee following statements are true:

  • $x > \frac{1}{2\delta}$
  • $\epsilon = 1$
  • $y = x + \delta$

Now let $\delta > 0$.

Clearly $|x-y| \leq \delta$ because $|x-y|=|x-(x+\delta)| = |\delta| = \delta \leq \delta$. Now see that $$ \begin{align*} |f(x)-f(y)|&=|x^2 + x - (x+\delta)^2 - (x+\delta)|\\ &=|2x\delta + \delta^2 + \delta|\\ &=2x\delta + \delta^2 + \delta\\ &>2x\delta = 2 > 1 = \epsilon \end{align*} $$

Terefore $f(x)$ is no uniformly continuous.

Can someone please check my proof? Also, it's a proved theorem that if $a_n,b_n$ are functions in the domain of $f$ such that $(a_n - b_n) \rightarrow 0$ and $f(a_n) - f(b_n) \nrightarrow 0$ then $f$ is not uniformly continuous. Can someone please show me an example of $a_n$ and $b_n$ for that case?

I tried to prove some not uniformly continuous functions using the previous theorem but it's not trivial (at least for me), to find those sequences...



2 Answers 2


Take $a_n=n$ and $b_n=n+\frac{1}{n}$ to see what will happen!

  • $\begingroup$ How did you think about them? $\endgroup$
    – Bruno Reis
    Jun 23, 2019 at 4:28
  • $\begingroup$ $\big( f(a_n) - f(b_n)\big) \rightarrow 2 $... Very nice! $\endgroup$
    – Bruno Reis
    Jun 23, 2019 at 4:32
  • 1
    $\begingroup$ The same sequences is a familiar counterexample for the non uniform continuity of $x^2$. This sequence will also work for your function also. That why I pick this sequences $\endgroup$ Jun 23, 2019 at 4:32

In your own estimate $|f(y)-f(x)| \ge 2x\delta$, with $y=x+\delta$. So choose the sequence so that $x \delta$ is never close to $0$, so $\delta=\frac1n$, $x=n$ is a good option, and sort of the simplest. This suggests taking $a_n = n$ and $b_n = n+\frac1n$ as the sequences. Or $2n,2n+\frac{1}{n}$ etc. Lots of options.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .