1

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?

Community
  • 1

1 Answers1

2

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).

hmakholm left over Monica
  • 286,031
  • 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. –  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. – hmakholm left over Monica Sep 25 '11 at 16:21
  • yes it is true, thanks for the enlightening – leo Sep 25 '11 at 16:25