Wikipedia:Reference desk/Archives/Mathematics/2016 September 21

= September 21 =

Uniform continuity
I'm struggling with the definition: $$f:A\to\R$$ is uniformly continuous if $$\forall\ \varepsilon>0\ \exist\ \delta>0:\forall\ x_1,x_2\in A:|x_1-x_2|<\delta\ \Rightarrow\ \Big|f(x_1)-f(x_2)\Big|<\varepsilon$$.

Is this definition close or somehow "equivalent" to the Cauchy sequence definition, were per any given $$\varepsilon>0$$ there exists a $$N$$ s.t. per any 2 indexes $$m,n>N\ \Rightarrow\ |a_m-a_n|<\varepsilon$$ ?

יהודה שמחה ולדמן (talk) 00:24, 21 September 2016 (UTC)
 * Hmm, I guess there's a very vague analogy there, but I think the better answer is "not really". A sequence is a Cauchy sequence if eventually all elements of the sequence are close to each other.  A function is uniformly continuous if you can establish uniform bounds to how close two x values have to be to make two y values close.
 * It's probably easiest to think about if you consider examples of functions that aren't uniformly continuous. For example, if you look at $$y=\frac{1}{x}$$ on the half-open interval (0,1].  Say you want to guarantee that two y values are within 1 of each other.  By picking x values close to 0, you can get two x values that are as close as you like to each other, but for which their corresponding y values differ by at least 1.
 * Partly this is because the derivative of the function goes to infinity as x goes to 0. But that's not the full story, because every continuous function on a compact interval is uniformly continuous, even if its derivative goes to &infin; at some point. --Trovatore (talk) 01:16, 21 September 2016 (UTC)
 * Which one of these 2 claims below does $$\forall\ x_1,x_2\in A:|x_1-x_2|<\delta\ \Rightarrow\ \Big|f(x_1)-f(x_2)\Big|<\varepsilon$$ mean:
 * all $$x_1,x_2\in A$$ that fulfill $$|x_1-x_2|<\delta$$ so we get $$\Big|f(x_1)-f(x_2)\Big|<\varepsilon$$ ?
 * all $$x_1,x_2\in A$$ need to fulfill $$|x_1-x_2|<\delta$$ so we get $$\Big|f(x_1)-f(x_2)\Big|<\varepsilon$$ ?
 * יהודה שמחה ולדמן (talk) 08:00, 21 September 2016 (UTC)
 * I'm not sure I follow the question. Going by my best guess as to what you mean, then I'd say option 1.  But it would be better if you worked it out yourself.  Start with a concrete example of a uniformly continuous function, and another concrete example of one that's not uniformly continuous, and test your formulations against them.  If you need examples, I gave you an example above of a continuous function that's not uniformly continuous.
 * For an example of a uniformly continuous function, try $$f(x)=\sqrt x$$ on the interval [0,1]. (I made it not completely trivial by arranging for the derivative to go to &infin; at the left endpoint.) --Trovatore (talk) 08:08, 21 September 2016 (UTC)


 * Not my area of expertise but perhaps you can think Cauchy sequences as uniformly continuous functions of a particular uniform space defined on the the natural numbers. Let Pn denote {x∈N: x≥n} and define an entourage on N to be any subset of N×N which contains Δ and Pn×Pn for some n. The axioms for a uniform space hold and a sequence is Cauchy iff it's a uniformly continuous map from N with the above uniform structure to the metric space with the standard uniform structure on a metric space. --RDBury (talk) 07:04, 22 September 2016 (UTC)
 * Wow, nicely done! I haven't checked all the details, but assuming you're right, I'm impressed. --Trovatore (talk) 07:30, 22 September 2016 (UTC)
 * (And I suppose I would be remiss not to mention that it also vindicates the original poster's insight.) --Trovatore (talk) 08:04, 22 September 2016 (UTC)
 * A bit more pedestrian: equip N with the induced metric from the set $$\{1/n|n\in\mathbb N\}$$. Then a sequence is Cauchy iff it is a uniformly continuous map with respect to this metric.   Sławomir Biały  (talk) 10:31, 22 September 2016 (UTC)
 * That's good too (and easier to check). --Trovatore (talk) 18:13, 22 September 2016 (UTC)