Does there exist an unbounded function that is uniformly continuous?
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.
$\endgroup$2 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$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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