How to prove $\sin(1/x)$ is not uniformly continuous

How do I go about proving $f(x)=\sin(1/x)$ is not uniformly continuous?

(Or: different question, but same intention* how do I prove that $x\sin(x)$ is not uniformly continuous)

*I'm trying to grasp how one would prove $f$ is not uniformly continuous for functions other than the simple $x^n$. I have seen one technique being to set an $\epsilon$ and set $x, y$ in the form of $\delta$ (e.g. $\delta/2$, etc.) then subsequently proving that $f(x)-f(y)\ge\epsilon$

Solution 1:

Ultimately a very brief solution could be given to this problem, but I decided to write in some detail how you might approach it.

You want to negate the following: $$\forall \varepsilon>0,\exists\delta>0,\forall x,y, |x-y|<\delta\implies|f(x)-f(y)|<\varepsilon.$$

You can write out what that negation is rather mechanically, swapping universal and existential quantifiers, until you finally negate the implication by ensuring that $|x-y|<\delta$ and $|f(x)-f(y)|\geq \varepsilon$. See here for a discussion of dissecting analysis problems like this by Tim Gowers.

That is, you want to prove:

$$\exists \varepsilon>0,\forall\delta>0,\exists x,y,\text{ such that } |x-y|<\delta\text{ and }|f(x)-f(y)|\geq\varepsilon.$$

Before giving the final argument, it is a good idea to experiment in a "backwards" fashion; think about where you want to end up and how you can get there. Roughly, the conclusion "$|x-y|<\delta\text{ and }|f(x)-f(y)|\geq\varepsilon$" will be saying that $x$ and $y$ will be close while $f(x)$ and $f(y)$ will stay a distance $\varepsilon$ away. The property of $\sin(1/x)$ that allows this to happen is that it oscillates like crazy between $1$ and $-1$ over smaller and smaller intervals of $x$ values. So there will be "nearby" $x$ and $y$ such that $f(x)=-1$ and $f(y)=1$. The distance between the function values here is $2$, while the distance between the input values can be arbitrarily small. This leads to the conclusion that $\varepsilon = 2$ will be a sufficient choice.

Next, with $\varepsilon$ fixed at $2$, and $\delta>0$ arbitrary but fixed, you need to show that there are $x$ and $y$ with $|x-y|<\delta$ and $|f(x)-f(y)|\geq 2$. As indicated above, the last part can be achieved by ensuring that $f(x)=-1$ and $f(y)=1$. For what $x$ and $y$ is it true that $\sin(1/x)=-1$ and $\sin(1/y)=1$? Use what you know about the sine function to answer this question (I'll leave this to you). Notice that the choices of such $x$ and $y$ get arbitrarily close to $0$, and in particular you can choose such $x$ and $y$ with $0<x,y<\delta$, which implies that $|x-y|<\delta$.

Largely the same approach applies to $x\sin(x)$, except that its reason for not being uniformly continuous changes. Now the problem is where $x$ gets very large, and the rate of oscillation doesn't change, but the amplitude does. A hint is to consider the function values at $x=2\pi n$ and $y=2\pi n + c$, where $c>0$ is "small", as $n$ goes to infinity.

Solution 2:

Choose two sequences $T_n = \frac{1}{n}$ and $S_n = \frac{1}{n+\pi}$. Their difference goes to zero, as $n$ goes to infinity. But $|f(S_n)-f(T_n)|=2|\cos n|$. Thus, can't find sigma for any epsilon greater than zero (from definition).