Show that $f(x) = x\cos^3(x)$ is not uniformly continuous on $\mathbb{R}$

Question

Show that $f(x) = x\cos^3(x)$ is not uniformly continuous on $\mathbb{R}$.

I tried $x_n = \pi/2 + n\pi$ and $y_n = \pi/2 + n\pi + 1/(n\pi)$.

Since $\cos^3(x) = \frac14 (\cos(3x)+3\cos x)$, $f(x_n) = 0$. So, $$\left|\frac34(\pi/2 + n\pi + 1/(n\pi))\cos(\pi/2 + n\pi + 1/(n\pi)) + \frac14(\pi/2 + n\pi + 1/(n\pi))\cos(3\pi/2 + n\pi + 1/(n\pi))\right| > \frac12|(\pi/2 + n\pi + 1/(n\pi))\sin(1/(n\pi))| > \frac12 n \pi\sin(1/(n\pi)) \to \frac12$$ so it is not uniformly continuous.

Is this correct?

Answer 1

Producing two sequences $(x_n)_{n\geq1}$ and $(y_n)_{n\geq1}$ with $y_n-x_n\to0$ and $|f(y_n)-f(x_n)|\geq\epsilon_0$ is a good idea, in principle. But I'm afraid there is some error in your calculations. Note that $$f'(x)=\cos^2 x(\cos x -3x\sin x)$$ is very small near the odd multiples of ${\pi\over2}$. Therefore it will be difficult to detect large differences $|f(y_n)-f(x_n)|$ in the neighborhood of such points. But at the odd multiples of ${\pi\over4}$ the derivative $|f'(x)|$ has order of magnitude $|x|$. The mean value theorem then should allow you to prove that you can achieve $|f(y)-f(x)|\geq1$ for suitably selected points $x$, $y$ arbitrarily close to each other.