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I know that $1/x$ is unbounded on $(0,5)$ (for example) and that since it is unbounded, it is not uniformly continuous.

Does a function have to be bounded to be uniformly continuous? I don't think one exists.

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The function $f(x) = x$ is unbounded on $\mathbb{R}$, but uniformly continuous on $\mathbb{R}$. The function $f(x) = \sqrt{x}$ is another interesting example.

Perhaps you meant to ask something like, if $I$ is a bounded interval (not necessarily closed) and $f: I \to \mathbb{R}$ is uniformly continuous, then is $f$ bounded? The answer to this is yes. Find $\delta > 0$ such that for $|x - y| < \delta$, $|f(x) - f(y)| < 1$. Then by partitioning the interval $I$ up into a finite number of pieces smaller than $\delta$, you can show $f$ is bounded.

The same holds true if $I$ is any bounded set, not just an interval.

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  • $\begingroup$ why 1, not $\epsilon$? $\endgroup$ Commented May 11, 2014 at 23:04
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    $\begingroup$ @Jack I picked $\epsilon = 1$ because it doesn't matter what epsilon you use. You just need to show $f$ is bounded, and picking a smaller $\epsilon$ won't help you get a tighter bound. $\endgroup$ Commented May 11, 2014 at 23:05
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Just to add that a bounded derivative is sufficient for an (obviously differentiable) function $ f:\mathbb R \rightarrow \mathbb R$ to be uniformly-continuous, but a necessary condition for the limit as $ x\rightarrow \infty $ to be finite (when the limit $f'(x)$ exists as $x \rightarrow \infty$) is that lim$_{x\rightarrow \infty}f'(x)=0. $ So, to make your statement true, you need to add these two conditions: the limit of $f'(x)$ exists as you go to $\infty$ , and it is $0$.

EDIT: As pointed out by timur, these conditions are necessary, but not sufficient, i.e., we need $f'(x) \rightarrow 0$ for the limit to be finite, a.k.a., for f to be bounded . I suspect that adding a condition on $|f''(x)| \rightarrow \infty$ is sufficient, i.e., the function decreases fast-enough to not become unbounded.

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  • $\begingroup$ It seems not: For example consider $\sqrt{x}$. $\endgroup$
    – timur
    Commented May 12, 2014 at 1:15
  • $\begingroup$ @timur: What do you mean? its derivative is unbounded, and the function itself is unbounded. I don't see how it is a counter to anything I posted. $\endgroup$
    – user99680
    Commented May 12, 2014 at 1:18
  • $\begingroup$ The derivative goes to $0$ as $x\to\infty$, no? $\endgroup$
    – timur
    Commented May 12, 2014 at 1:19
  • $\begingroup$ @timur: good point, let me think for a bit. I guess the best I can say for now is that these are all necessary conditions. $\endgroup$
    – user99680
    Commented May 12, 2014 at 1:25
  • $\begingroup$ @timur: I think adding the condition that $f''(x)>0$ is sufficient, or, even better, $<-\infty <M<f'(x)<0$ and $f''(x)\rightarrow \infty $? $\endgroup$
    – user99680
    Commented May 12, 2014 at 1:35

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