# Prove that a function is not uniformly continuous

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...

Thanks!

Take $$a_n=n$$ and $$b_n=n+\frac{1}{n}$$ to see what will happen!
• $\big( f(a_n) - f(b_n)\big) \rightarrow 2$... Very nice! Commented Jun 23, 2019 at 4:32
• 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 Commented 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.