Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If one considers $C^1([a,b])$, then immediately one has $f$ is uniformly continuous for all $f\in C^1([a,b])$ since $[a,b]$ is compact in ${\mathbb R}$. When it comes to an open interval, things may be different. For example $f(x)=1/x$, $f\in C^1((0,1))$ but $f$ is not uniformly continuous on $(0,1)$. One property of $f$ is that both $f$ and $f'$ are unbounded on $(0,1)$, which, for example, is different from that of $g(x)=\sqrt{x}$.

Here comes my first question:

  • Is the following statement true?

    For $f\in C^1((a,b))$, $f$ is uniformly continuous if and only if $f$ is bounded on $(a,b)$.

It seems that one may give the "only if" part from the answer to this question. For the "if" part, I can't come up with a counterexample. What's more,

  • if we consider $C((a,b))$, then what would be the relationship between uniform continuity and boundedness of these functions?
share|improve this question

1 Answer 1

up vote 3 down vote accepted

No. A counterexample is $x\mapsto \sin\frac{1}{x}$ on $(0,1)$. It is bounded and $C^\infty$ on this interval, but not uniformly continuous.

You should be able to prove that $f$ is uniformly continuous on $(a,b)$ if and only if it is continuous and $\lim_{x\to a_+} f(x)$ and $\lim_{x\to b_-} f(x)$ exist (and are finite).

share|improve this answer
    
with the hypothesis given above I'm tempted to argue like this: Let $\epsilon>0$. Since $f\in C'((a,b))$, there is a bound $M>0$ for $f'$. Take $\delta=\epsilon/M$ then by them MVT: $|f(x)-f(y)|=|f'(c)||x-y|\leq M|x-y|$. But your example clearly shows that the theorem don't holds. What is the mistake in my proof? –  leo Sep 25 '11 at 16:09
    
@leo: I don't think "there is necessarily a bound $M$ for $f'$" since the interval is not closed. –  Jack Sep 25 '11 at 16:15
    
@Jack: I understand that "a function $f$ of class $C^1$" as $f'$ exist and is continuous in the domain (here $(a,b)$), then continuity implies boundedness. Am I wrong? –  leo Sep 25 '11 at 16:18
    
@leo, continuity in an open interval does not imply boundedness, as Jack's example $x\mapsto 1/x$ on $(0,1)$ shows. –  Henning Makholm Sep 25 '11 at 16:21
    
yes it is true, thanks for the enlightening –  leo Sep 25 '11 at 16:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.