0
$\begingroup$

Give an example of a continuous function $f: \mathbb R\to \mathbb R$ which is not uniformly continuous on $(1,3)$. I'm not looking for someone to do this for me or anything but I feel completely stuck. I can't even find a jumping off point.

I understand what qualifies a function as continuous and uniformly continuous. What I'm having trouble with (and I feel silly for admitting it) is being able to product a function that is not uniformly continuous at the given interval, (1,3).

I've done the internet searches but I'm having trouble relating what I'm finding to my given interval. I'm sorry if that doesn't narrow it down enough, but feeling stuck from the very start doesn't give me a whole lot to shave off.

$\endgroup$
2
  • $\begingroup$ Where are you stuck, what have your approaches been so far? Do you have an idea of how to try to solve it? Have you consulted any sources? If so, please post those in your question. Otherwise, a quick web search may give you some ideas. $\endgroup$
    – JMCF125
    Dec 5, 2013 at 20:53
  • 2
    $\begingroup$ In fact, as you've stated the conditions, no such function exists. You might double check with whoever you got the problem from. Otherwise, you might try proving that every function which is continuous on $\mathbb{R}$ is uniformly continuous on $(1,3)$ (or any other bounded set). $\endgroup$ Dec 5, 2013 at 21:34

2 Answers 2

1
$\begingroup$

Someone has given you an impossible task. If a function $f$ with domain $\mathbb{R}$ is continuous, then $f$ is continuous on the compact interval $[1,3]$, and a function that is continuous on $[1,3]$ is uniformly continuous on $[1,3]$, hence it is uniformly continuous on any subset of $[1,3]$, such as $(1,3)$.

The fact that if $f$ is continuous on $[1,3]$, then it must be uniformly continuous on $[1,3]$, is not too hard to prove. One way to prove it is by contradiction. Assume $f$ is continuous but not uniformly continuous on $[1,3]$, use the definition of uniform continuity and the compactness of $[1,3]$, and eventually you will get a contradiction.

Undoubtedly this is proven in scores of real analysis textbooks, but I don't know of one offhand, and I don't know if proof by contradiction is the most popular or the best way to do it.

$\endgroup$
0
$\begingroup$

Here's a hint:

Can $f$ be uniformly continuous on an interval $I$ if there is a limit point, $a$, of $I$ for which $\lim_{x\rightarrow a}f(x)=\pm\infty$?

$\endgroup$
1
  • 1
    $\begingroup$ Also, if a function is continuous on a compact interval then it is uniformly continuous on that compact interval. Hence your constructed function cannot be continuous on $[1,3]$. $\endgroup$
    – Eric
    Dec 5, 2013 at 20:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .