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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?

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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).

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    yes it is true, thanks for the enlightening2011-09-25