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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!

user114014
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    Caution. $\sqrt{x}$ is uniformly continuous on $(0,\infty)$, but its derivative is not bounded. Depending on what is meant with "its derivative diverges in the given interval", the criterion might be wrong. But if the derivative exists and is bounded, the function certainly is uniformly continuous. – Daniel Fischer Dec 09 '13 at 23:45
  • If you have a continuous function from a compact set. That's a nice one. – LASV Dec 09 '13 at 23:48
  • To add to Daniel Fischer's comment: bounded derivatives implies the function is Lipschitz, and Lipschitz continuity implies uniform continuity, but uniform continuity does not imply Lipschitz continuity. – Patterns_43 Dec 10 '13 at 00:00
  • @DanielFischer - so if the derivative is bounded, it is uniformly continuous. But the converse is false, so if it is uniformly continuous, the derivative is not necessarily bounded? – user114014 Dec 10 '13 at 00:42
  • That's right. The boundedness of the derivative is sufficient, but not necessary. – Daniel Fischer Dec 10 '13 at 00:46

3 Answers3

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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.

Ben Grossmann
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  • "If $f$ is differentiable and has a continuous, bounded derivative, and the domain is an interval (or, in $\mathbf{R}^n$, is convex)". :) (Otherwise there are examples sign as $x/|x|$ on $\mathbf{R}\setminus{0}$, or the polar angle $\theta$ on a slit annulus, or....) – Andrew D. Hwang Dec 10 '13 at 00:34
  • Thank you, this was pretty helpful! Two questions. First, by "f is a map from a closed bounded interval", you mean like if I = [a,b] and f:I --> R, right? Second, you mentioned vertical asymptotes - does this mean that tan(x) is not uniformly continuous on R, but it is uniformly continuous wherever cos(x) ≠ 0? – user114014 Dec 10 '13 at 00:39
  • @user114014 first question: that's exactly what I mean. Second question: note that $\tan(x)$ is not even continuous on $\Bbb R$, so it can't be uniformly continuous. What I mean though is that if you look at $\tan(x)$ on an open interval such as $(-\pi/2,\pi/2)$ where $\tan(x)$ is continuous but has an asymptote on the boundary, you get something that is not uniformly continuous. – Ben Grossmann Dec 10 '13 at 00:48
  • @user86418 Fair point! Never thought of that. – Ben Grossmann Dec 10 '13 at 00:50
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    But if you use a smaller closed interval such as [$ -\pi/2 + \epsilon, pi/2 - \epsilon$] you retrieve your uniform continuity. On the smaller closed interval the derivative is bounded; on the entire open interval the function does have vertical asymptotes and cannot be uniformly continuous. Re Dan Fisher's example of $\sqrt x on [0,1], a continuous function on a closed, bounded interval is uniformly continuous -- even tho in this case the derivative does become unbounded. – Betty Mock Dec 10 '13 at 00:54
  • In item 3, you don't need continuity of derivative (mean value theorem does not require it). On the other hand "the complement of the domain of $f$ contains no isolated points" is not enough: consider $1/(x-c)$ restricted to the complement of Cantor set $C$, where $c\in C$. I think here it is reasonable to limit the consideration to intervals; people rarely ask about uniform continuity on disconnected sets. – Post No Bulls Dec 10 '13 at 00:57
  • @PostNoBills thanks for that. What a strange rabbit-hole I've uncovered... – Ben Grossmann Dec 10 '13 at 01:04
  • +1 Finally, a direct answer to the point. – Felix Marin Sep 12 '20 at 20:58
  • Just to be clear, you're asserting that $f:A \to \mathbb{R}$ is differentiable with a bounded derivative, i.e $|f'(x)|, \forall x \in A$ and $A$ is union of finitely many intervals with no common border, then $f$ is uniformly continuous. You don't need a continuous derivative, right? So $A= (1,2)\cup(3,4)$ would work but not $A=(1,2)\cup(2,3)$ wouldn't because they share a border $2$? – William Aug 24 '22 at 19:58
  • @William That is what I was saying, and indeed continuity of the derivative is unnecessary. – Ben Grossmann Aug 24 '22 at 20:11
  • Thank you. I got confused by the comments. Also in your last point, I think there's a mistake? You say $|f'(x)|\to ∞$ but you write $x \to ±∞$ instead of just $x \to ∞$ (see domain). Were you trying to say maybe, that given $a\in \mathbb{R}$ for differentiable $f:[a,∞)\to R$, $|f'(x)| →+∞$ as $x \to ∞$ and for differentiable $f:(-∞,a] \to R$, $|f'|\to +∞$ as $x \to -∞$? – William Aug 24 '22 at 20:38
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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.

Matemáticos Chibchas
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  • Yeah, isn't that just the definition of uniform continuity? – user114014 Dec 10 '13 at 01:11
  • @user114014 Hmmm, yes and no, I mean, the usual definition of uniform continuity is that using $\epsilon$ and $\delta$ in the same way as the definition of continuity, except that you specify that the value of $\delta$ depends only on $\epsilon$, not on the point of the domain. In other words, the same value of $\delta$ works for every point in the domain, which is not the case for plain continuity. – Matemáticos Chibchas Dec 10 '13 at 01:13
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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.)

Andrew D. Hwang
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