What's the fastest way to tell if a function is uniformly continuous or not? I have my real analysis final tomorrow and there are multiple choice questions.  I'm wondering about a fast way to tell if a function is uniformly continuous or not.  I know and understand the definition of uniform continuity, and I understand its difference from continuity, but I'm realizing I don't really know how to tell if a function is uniformly continuous or not (on a given interval or on R).
One of my classmates suggested that a function is NOT uniformly continuous if its derivative diverges in the given interval.  Is this true?  Can I just think of the graph of the function and if its slope does not eventually settle to some point, is it not uniformly continuous?
Thanks in advance for any help regarding how to approach these kinds of questions!
 A: Some common situations:
A continuous function $f$ is uniformly continuous if


*

*$f$ is a map from a closed bounded interval

*if $f$ is a map from any compact set

*If $f$ is differentiable and has a bounded derivative, usually... (see comments on this one.  If the domain is the union of finitely intervals that have no common border though, this is true).


A function $f$ is not uniformly continuous if


*

*$f$ is not continuous

*$f$ has a vertical asymptote (consider $1/x$ on $(0,1)$)

*$f:[a,\infty)\to \mathbb{R}$ is differentiable, and $|f'|$ grows without bound as $x \to  \pm \infty$.


Outside of that, I'd say just go back to the definition.
I hope you find this helpful.
A: Probably not as useful as the sufficient conditions listed above, but the necessary and sufficient condition for a function $f$ to be uniformly continuous is that whenever you have two sequences $(x_n)$ and $(y_n)$ such that $d(x_n,y_n)$ goes to $0$ (here $d$ stands for the distance, which is a notion pertaining to the general theory of metric spaces), then the same happens to $d\bigl(f(x_n),f(y_n)\bigr)$. Note that is not required the sequences $(x_n),(y_n)$ to have some "nice" behavior (convergence, Cauchy condition, boundedness, etc.) other than their corresponding terms come closer.
A: One more condition that may be intuitively/visually compelling: If $U \subset \mathbf{R}^n$ is bounded and $f:U \to \mathbf{R}$ is continuous, then $f$ is uniformly continuous on $U$ iff there exists a continuous extension of $f$ to the closure of $U$.
(Consequently, a bounded, continuous, monotone function on an interval of reals is uniformly continuous (even if the domain is unbounded); $f(x) = \sin(1/x)$ is not uniformly continuous on $(0, 1)$; $\operatorname{sgn}(x) = x/|x|$ is not uniformly continuous on $\mathbf{R}\setminus\{0\}$, but is uniformly continuous on $\mathbf{R}\setminus[0, \delta]$ for every $\delta > 0$, etc.)
