uniform continuity of $f$ on $[0,\infty)$ when $\lim_{x\rightarrow\infty}f(x)/x=1$.

Question

Prove or disprove : If $f$ is continuous on $[0,\infty)$ such that $\lim_{x\rightarrow\infty}\frac{f(x)}{x}=1$, then $f$ is uniformly continuous on $[0,\infty)$

I think I have an issue of interpretation the given hypothesis. On top of that, this becomes more harder when it is asked to consider the validity of statement. I do not even know where to begin. Any trick and idea to deal with this kind of problem? Please help...

Answer 1

So, here is what I would do, that is, how I would tackle this problem (since I think this is the question and not really weather this statement is true or false):

• First: This extra condition gives you an idea of how the functions behaves at infinity, so it is a growth condition.

• Now, since you also have to decide weather this statement is true or not, you could try to come up with a counterexample. Usually (at least in my experience) these are either the obvious ones or so involved that one rather looses hope and will just try to proof this statement.

• Now, you should write down the definition of uniformly continuous and also that of continuos (since $f$ is given to be that) and then play around to maybe see weather you can work this growth condition into a proof.

• It also might be helpful to rewrite this extra condition in a different way, that is, what does $\lim_{x\to\infty} \frac{f(x)}{x} =1$ actually mean?

• Another useful thing, is to think about what do you know about the connection of continuous and uniformly continuous functions. That is every uniformly continuous function is continuous and the converse is usually false, but under which extra asumptions is it actually true.

I hope this might be helpful.

Answer 2

What happens when the graph of $f$ stays relatively close to the line $y=x$, but oscilates more and more rapidly around it as $x$ tends to $\infty$?

Answer 3

Let $f(x) = x + \sin(nx)$ for $x \in [2\pi n, 2 \pi (n+1)]$, $n = 0, 1, 2, \dots$

We have $\sin(n 2\pi n)= \sin (2 \pi n^2) = 0$ and $\sin(n 2\pi (n+1))= \sin (2 \pi n(n+1)) = 0$, thus $f$ is well-defined and continuous. Since $\lvert \sin(nx) \rvert \le 1$ we see that $\lim_{x\to \infty} \frac{f(x)}{x} = 1$.

We have $f(2 \pi n) = 2 \pi n$ and $f(2 \pi n + \frac{\pi}{2n}) = 2 \pi n + 1$ which shows that $f$ is not uniformly continuous.