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Is $f: \mathbb{R} \to \mathbb{R}: x \mapsto x^2\cos x$ uniformly continuous?

My attempt:

My intuition doesn't tell me anything on this one. I tried standard procedures: its derivative is not bounded, hence not Lipschitz.

I tried to find equivalent sequences with non equivalent image sequences, but I couldn't find any.

This made me believe that the function is uniformly continuous, but I don't know how to prove it.

I tried to do this by the definition but I don't know how to handle the expression

$$|x² \cos(x) - y² \cos(x)|$$ given that $|x-y| < \delta$ for an appropriate delta.

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  • $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. $\endgroup$ – Clement C. Dec 4 '17 at 2:07
  • $\begingroup$ I didn't have time to check all the answers in detail, but I will check them in the future and then I'll mark the answer that helped me the most! Thanks for your concern though! $\endgroup$ – user370967 Dec 4 '17 at 12:48
  • $\begingroup$ In fact, I was hoping someone provided two simple equivalent sequences whose image sequences are not equivalent. $\endgroup$ – user370967 Dec 4 '17 at 12:50
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    $\begingroup$ Possibly useful for intuiuton: math.stackexchange.com/questions/2540090/… $\endgroup$ – Ethan Bolker Dec 9 '17 at 18:54
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An explicit counterexample:

Consider the sequence $(x_n)_n$ defined by $x_n \stackrel{\rm def}{=} 2\pi n$. Assuming uniform continuity, we have, for $\varepsilon = 1$, some small $\delta>0$ (wlog $\delta < 2\pi$) such that in particular $$ \lvert f(x_n+\delta) - f_n(x_n+\delta) \rvert \leq 1 $$ for all $n\geq 0$. However, $$ f(x_n) = x_n^2 \cos(x_n) = x_n^2 $$ but $$\begin{align} f(x_n+\delta) &= (x_n+\delta)^2\cos(x_n+\delta) = (x_n+\delta)^2\cos(2\pi n+\delta) = (x_n+\delta)^2\cos(\delta) \\&= x_n^2\cos(\delta)+(2\delta x_n + \delta^2)\cos(\delta) = x_n^2\cos(\delta) + o(x_n^2) \end{align}$$ so that $$\lvert f(x_n+\delta)-f(x_n)\rvert = (1-\cos(\delta))x_n^2 + o(x_n^2) \xrightarrow[n\to\infty]{}\infty $$ contradicting the assumption.

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    $\begingroup$ This is more convincing answer (at least for beginners) +1 $\endgroup$ – Paramanand Singh Nov 30 '17 at 7:09
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No, because $x^2$ has an unbounded derivative and there are arbitrarily large values of $x$ where $\cos(x)$ has its derivative bounded away from zero.

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    $\begingroup$ “cos has arbitrarily large values” ?! What do you mean? $\endgroup$ – Lierre Nov 29 '17 at 20:18
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    $\begingroup$ @Lierre "There are arbitrary large values where the derivative [of $\cos$] is bounded away from zero." $\endgroup$ – Clement C. Nov 29 '17 at 23:40
  • $\begingroup$ Thanks - I did not say exactly what I wanted to. I've made a change which I hope makes it clearer. $\endgroup$ – marty cohen Nov 29 '17 at 23:56

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