Talk:Lipschitz continuity

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[edit] 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)

[edit] 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)

[edit] 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)