# Talk:Lipschitz continuity

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## 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. -- 12:02, 17 September 2008 (UTC)
Actually I see nothing in the "Real numbers" subsection which cannot apply to any other metric space (replacing |xy| with d(x, y). I'm going to merge them, mentioning real numbers as an example. -- 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).

## 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 ≤ 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.) -- 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. —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)