Show that $x\sin(x)$ is not uniformly continuous on $\Bbb R$

Question

For one of my homework problems, I'm supposed to show that $f(x)=x\sin(x)$ is not uniformly continuous on the real numbers, but this doesn't seem right to me.

I know that if a function is continuous on a closed and bounded interval then it is uniformly continuous on that interval. I also know that f is continuous on $\Bbb R$. It seems like a simple induction would show that f is UC on $\Bbb R$, so what's going on?

Please don't actually tell me how to solve the problem, but if you can tell me what I'm missing I would appreciate it.

Answer 1

Hint: what happens with the derivative of this function for large $x$? What can we say when we try to apply the UC definition and look at $x=2\pi k$, $k\in \Bbb N$?

Answer 2

Any continuous function on a compact interval is uniformly continuous.

This does not extend to $\mathbb{R}$ by induction, as $\mathbb{R}$ is not compact.

If it is uniformly continuous then for all $\epsilon>0$ there is some $\delta >0$ such that $|f(x)-f(y)| < \epsilon$ whenever $|x-y| < \delta$.

So, you need to show that there is some $\epsilon >0$ such that for all $\delta>0$ there is some $x,y$ with $|x-y| < \delta$ such that $|f(x)-f(y)| \ge \epsilon$.

In your particular example, you can pick an arbitrary $\epsilon>0$.

Hint: Look at how quickly the function changes in the neighbourhood of a zero crossing.

Answer 3

We shall assume $x \sin(x)$ is uniformly continuous on $\mathbb{R}$ and reach a contradiction. Let $\varepsilon>0$ be given. Since $x \sin(x)$ is uniformly continuous on $\mathbb{R}$ there is $\delta>0$ so that for all $k \in \mathbb{N}$ we have

\begin{equation} \begin{split} | 2\pi \big(k +\frac{\delta}{2}\big) \sin\big(2\pi \big(k +\frac{\delta}{2}\big)\big) - 2\pi \big(k -\frac{\delta}{2}\big) \sin\big(2\pi \big(k -\frac{\delta}{2}\big)\big)| &= 4 \pi k \sin \big(\frac{\delta}{2}\big) \\ &<\varepsilon. \end{split} \end{equation}

However by the Archimedean property, we know that there is a positive integer $\hat{k}$ so that $4 \pi \hat{k} \sin \big(\frac{\delta}{2}\big)> \varepsilon$. This is a contradiction, and so the result is proved.

It should not be hard to see the above equality. Just do a little work to get from the left side to the right side of it.