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

Question

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.

Answer 1

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.

Answer 2

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.