Talk:Lipschitz continuity

The definition
The definition here seems to be restricted to R. Other definitions are in higher spaces than R? is this true? http://planetmath.org/encyclopedia/LipschitzCondition.htmlUser A1 06:22, 27 September 2006 (UTC)
 * The definition is not restricted to R, Lipshitz functions are defineded on any metric space, in the section on metric spaces in this article. Oleg Alexandrov (talk) 15:12, 27 September 2006 (UTC)

Could it be useful to talk about the familiar metric space R^n as a special case?128.8.94.54 (talk) 11:40, 17 September 2008 (UTC)
 * Yes, it could. --A r m y 1 9 8 7 ! ! ! 12:02, 17 September 2008 (UTC)
 * Actually I see nothing in the "Real numbers" subsection which cannot apply to any other metric space (replacing |x &minus; y| with d(x, y). I'm going to merge them, mentioning real numbers as an example. --A r m y 1 9 8 7 ! ! ! 12:13, 17 September 2008 (UTC)

The following sentence is not clear to me: "Any such K is referred to as a Lipschitz constant for the function ƒ. The smallest constant is sometimes called the (best) Lipschitz constant; however in most cases the latter notion is less relevant." I would remove the part from "; however..." as we are just giving the definition here. 64.134.134.33 (talk) —Preceding undated comment added 01:31, 27 June 2010 (UTC).

In the first paragraph of the definition section it is said, that the function f is called Lipschitz continuous if the following theorem holds; which doesn't thus have to be valid for all Lipschitz continuous functions. Later in the 3rd paragraph the same condition (excluding the case x1 = x2) it is stated as a biimplication ("if and only if"). What should the reader now believe? (Téleo (talk) 13:26, 17 May 2016 (UTC))


 * This is a usual mathematical jargon; in a definition, "X is called A if R" means "X is called A if and only if R" (just because otherwise it would not be a definition); see If and only if. Boris Tsirelson (talk) 16:19, 26 May 2016 (UTC)

I've added a subtitle for Real-valued functions, to make it more easily readable and understandable, specifically by people not used to metric spaces. I'm new to Wikipedia, so unsure whether this was a good idea or if it was executed well, so feel free to expand upon it. 7dare (talk) 19:21, 13 January 2017 (UTC)


 * Nice. But you did not indicate the end of the special case. Thus I try a different design. Boris Tsirelson (talk) 19:33, 13 January 2017 (UTC)

I included a phrase to indicate the usage of the phrase 'L-Lipschitz' for a function that is Lipschitz continuous with Lipschitz constant L, as described here: [] and used for example here (see Assumption 1): []. Not sure if my edit was accepted but I would appreciate if someone more knowledgeable than me can confirm this usage and review the edit. Thanks, Billtubbs (talk) 15:42, 26 September 2021 (UTC)


 * Yes, “L-Lipschitz” is a standard and very common short form for “Lipschitz of constant L”, both in written and spoken mathematical language (and “L-Lip” if you want to be even shorter). In this article a short form for “Lipschitz of constant L” doesn’t seem necessary because the phrase is not repeated too many times. However, the meaning of L-Lipschitz is an information that shouldn’t be missing so I think you did well.pm a 06:50, 28 September 2021 (UTC)

Property of bilipschitz functions
The following text seems tautological:
 * Every bilipschitz function is injective. A bilipschitz function is the same thing as a Lipschitz bijection whose inverse function is also Lipschitz.

In other words, if we define a bilipschitz function as a bijection that is Lipschitz and has a Lipschitz inverse, then it is trivially injective. Ideas? Haseldon 21:18, 9 November 2006 (UTC)
 * But this is not how bilipschits functions were defined in the article. The definition was:

If there exists a $$K \ge 1$$ with


 * $$\frac{1}{K}d(x,y) \le d'(f(x), f(y)) \le K d(x, y)$$

then f is called bilipschitz.

Oleg Alexandrov (talk) 04:08, 10 November 2006 (UTC)

A bilipschitz function need not be either surjective or injective. However, if a bilipschitz function is bijective, its inverse function is, in fact, bilipschitz. It's even satisfied by the same constant (just switch objects with their images and rearrange the inequalities.) Jwuthe2 (talk) 03:59, 8 October 2010 (UTC)


 * This is wrong. A bilipschitz function is always injective.  I hope you agree also that a bilipschitz function is surjective onto its image (as the article currently implies).  Furthermore, the characterization of bilipschitz function as a Lipschitz injection with Lipschitz inverse is the basic motivation for even being interested in bilipschitz mappings in the first place.   Sławomir Biały  (talk) 12:01, 4 April 2011 (UTC)

uniform Lipschitz condition
The text currently states: --- A function f, defined on [a,b], is said to satisfy a uniform Lipschitz condition of order α > 0 on [a,b] if there exists a constant M > 0 such that

| f(x) − f(y) | < M | x − y | ^α

for all x and y in [a,b]. --- which appears to be the same as Hölder continuity. It also appears to be a misuse of the term `uniform', which should mean `independent of x and y', i.e. not locally Lipschitz. Agreed? Jorn74 (talk) 22:11, 18 May 2008 (UTC)
 * I agree that "uniform Lipschitz condition of order α " appears the same as Hölder continuity. I think the "uniform" part in the article is right, there is nothing local in that definition. Oleg Alexandrov (talk) 01:53, 19 May 2008 (UTC)

I ran into this article looking for the more general "Lipschitz condition" which, unlike "Lipschitz uniform condition" or "Lipschitz continuity", does not even assume the function is differentiable everywhere. Robert Seeley's"An Introduction to Fourier Series and Integrals", for example, defines it for a continuous and piecewise-differentiable function g(x) as |g(x) - g(y)| <= M|x-y| for all x and y (in the domain of g).

This definition and its link with Lipschitz continuity should be mentioned in this article, this section seems to be a good place for it. 68.164.80.180 (talk) 17:59, 10 July 2014 (UTC)

Can K be less than 1 in short maps?
The article metric map says that a map is metric if $$d_{Y}(f(x),f(y)) \leq d_{X}(x,y).$$ This definition includes contractions, whereas the one in this article doesn't, if K needs to be "the smallest such constant". To make this definition be equivalent to the one in metric map, "K = 1" should be "K &le; 1". (Also, depending on whether K must be "the smallest such constant" or not, the recent addition "0 < K" in the definition of contraction either has the effect of excluding constant functions from the definition, or is completely useless.) -- A r m y 1 9 8 7 ! ! !  10:21, 27 September 2008 (UTC)

stronger?
The first paragraph states Lipschitz continuity condition is "stronger than regular continuity". is this true? need an example in the example section. Jackzhp (talk) 22:03, 18 May 2009 (UTC)
 * It's definitely stronger. Consider $$x \mapsto \sqrt{x}$$ where $$x \in [0,\infty)$$; this function is continuous over its domain, but it is not Lipschitz continuous as the slope goes to infinity as $$x \to 0$$. In fact, this case is already in the examples section. Further down in the page, it also says that "Every Lipschitz continuous map is uniformly continuous, and hence a fortiori continuous." This uniform continuity is the essence of the "magic" behind all of the nice features of Lipschitz continuous functions. &mdash;TedPavlic (talk) 12:39, 19 May 2009 (UTC)

Lipschitz Constant and Examples
There should be more examples and in the definition it should state somewhere that in the R^2 case the inequality is simply |f(y)-f(x)| is equal to or less than K|x-y|. Also, the Lipschitz constant should be discussed more, explicitly stating how to find it and that it cannot be less than 0 or infinity.--Gustav Ulsh Iler (talk) 20:52, 23 October 2009 (UTC)

The definition Iler just gave is for the Lipschitz condition, not for Lipschitz continuity. The Lipschitz condition is defined for piecewise differentiable functions, so is more general than Lipschitz continuity.

Some people may have thought it too obvious to mention, but Iler is right that the article should point out explicitly that the Lipschitz constant must be in between 0 and infinity. Otherwise the definition is simply incorrect. 68.164.80.180 (talk) 18:04, 10 July 2014 (UTC)

should something about Norton's Dome be added to this page?
should something about Norton's Dome be added to this page? — Preceding unsigned comment added by 72.83.192.20 (talk) 03:15, 1 November 2012 (UTC)


 * Note: we now have a (very stubby) article on Norton's dome that needs to be filled out into a proper article. -- The Anome (talk) 17:51, 5 August 2014 (UTC)

One-sided Lipschitz
The example of an one-sided Lipschitz function does not make sense when using the definition in the section; F(x) is scalar, which makes the left-hand side of the definition vector times scalar, thus a vector, while the right-hand side is a norm of a vector. Also, the image of F(x) should be specified, for example to be from R^d to R^d as in "Uniqueness and weak stability for multi-dimensional transport equations with one-sided Lipschitz coefficient" by Bouchut et al. — Preceding unsigned comment added by Bojeryd91 (talk • contribs) 16:46, 30 October 2014 (UTC)

Wording in Introduction
Should the every be changed to any in the opening sentence? "here exists a definite real number such that, for every pair of points on the graph of this function". Every seems to indicate some sort of pairs that already exist, while any is a bit more clear that it can be talking about any 2 point on the graph. I'm not familiar with mathematical terminology so I'm not comfortable making a change. 69.172.155.110 (talk) 04:32, 5 June 2016 (UTC)


 * No, "any" is not recommended in mathematics, since it is somewhat ambiguous: means not always "every", but also "some". Boris Tsirelson (talk) 04:58, 5 June 2016 (UTC)

Contraction
I don't think a function is called a contraction unless it maps a metric space to itself; 0 ≤ K < 1 is not sufficient. The page should be updated to reflect this

Merudo (talk) 12:57, 17 June 2018 (UTC)

Where is square root uniformly continuous?
Am I missing something?

The section "Continuous functions that are not (globally) Lipschitz continuous" says that square root is not Lipschitz on [0,1] (true) but that it is uniformly continuous. It seems to me that if zero is included in the interval, then it is not uniformly continuous either. The cited article [Robbin[ says in example 15 that The function square root is uniformly continuous on the set S = (0,infinity). Note that zero is NOT included.

Update: Sorry, I think I was missing something. Although in example 15, the set S does not include zero, when I actually read the proof I think it works even if zero is included. I don't know how to delete this comment, only update it.

Programmer in Chief (talk) 17:38, 21 January 2019 (UTC)


 * Yes, the square root IS uniformly continuous on the whole $$[0,\infty).$$ Just because $$\sqrt x \le \sqrt{x+a} \le \sqrt x + \sqrt a.$$ Boris Tsirelson (talk) 20:13, 21 January 2019 (UTC)

Inclusion chain
We have the following chain of strict inclusions for functions over a closed and bounded (i.e. compact) non-trivial interval of the real line


 * Continuously differentiable ⊂ Lipschitz continuous ⊂ α-Hölder continuous ⊂ uniformly continuous = continuous

On closed interval of real line, closed and bounded implies compact due to Heine–Borel theorem and uniformly continuous = continuous due to Heine–Cantor theorem? Should this be mentioned? Should the more general statement that uniform continuity implies continuity for any metric space be included? Wqwt (talk) 07:16, 4 April 2019 (UTC)


 * Yes, this is a good idea. For now, this chain "oscillates": someone replaces the last equality with strict inclusion, then someone restores it. This happens during years. A clarification would help. Boris Tsirelson (talk) 10:54, 4 April 2019 (UTC)

Hi, there is also writen that we assume that alpha is between zero and one, then it should be α-Hölder continuous ⊂ Lipschitz continuous or not? Coohy (talk) 20:37, 12 April 2022 (UTC)