In the Examples section, the first bullet is:

(f is beta-Holder and 0 < alpha <= beta <=1) implies (f is alpha-Holder)

If f is the identity on the Reals the f is Lipschitz but f is not Holder of exponent one half.

Is some extra restriction on the domain required here?

Several problems with the definitions in this page

I totally agree with the above remark. I mentioned in the section "examples" that the identity is not 1/2-Hölder according to the definition. This might indicate that one also wants to speak about locally Hölder, or indicate what happens when the metric is changed (to a bounded metric on R, for instance).

One virtually always works with Holder continuous functions over bounded sets, otherwise the Holder spaces are very weird. In fact if you consider Holder continuous functions over unbounded domains, you end up in the situation where differentiable functions do not belong to any Holder spaces for Holder coefficients less than one. These function spaces are pretty much useless so no one uses them. Paul Laroque (talk) 02:02, 1 April 2010 (UTC)[reply]

Also, it is not natural to define $C^{1,\alpha }$ on an open set by requiring that the derivative be bounded on the open set. Someone has to open an authoritative book and check what the common usages are. Bdmy (talk) 09:37, 20 December 2008 (UTC)--Bdmy (talk) 19:28, 19 January 2009 (UTC)[reply]

Indeed there are several possibilities on these functional spaces and norms. The choice depends on the use. --pma (talk) 09:38, 6 September 2009 (UTC)[reply]

absolute vs. norm vs. metric

can the absolute to be extended to any metric? Jackzhp (talk) 21:13, 18 May 2009 (UTC)[reply]

What you want is not very clear. Try the RefDesk/Maths.--pma (talk) 09:27, 6 September 2009 (UTC)[reply]

I guess, he's asking whether you can define Hölder continuity on general metric spaces or general normed spaces, not just R or R^n. --David Pal (talk) 22:02, 4 March 2011 (UTC)[reply]

then the answer is yes of course; check also modulus of continuity --pm a 13:48, 24 March 2014 (UTC) [reply]

Introductory section could be clearer

The introductory section includes this passage:

" We have the following chain of inclusions for functions over a compact subset of the real line

Continuously differentiable ⊆Lipschitz continuous ⊆ α-Hölder continuous ⊆ uniformly continuous ⊆ continuous

where 0 < α ≤ 1. "

But the subset notation is incompatible with the adjectives Continuously differentiable, Lipschitz continuous, α-Hölder continuous', uniformly continuous, continuous, which do not mean sets. Plus it is confusing.

Much clearer would be the sequence of implications

continuously differentiable ⇒ Lipschitz continuous ⇒ α-Hölder continuous' ⇒ uniformly continuous ⇒ continuous,

where 0 < α ≤ 1.

whose meaning is readily understood.Daqu (talk) 06:42, 1 October 2014 (UTC)[reply]

Bad writing

The introductory section begins as follows:

"In mathematics, a real or complex-valued function f on d-dimensional Euclidean space satisfies a Hölder condition, or is Hölder continuous, when there are real constants C ≥ 0, α > 0, such that

|f(x)-f(y)|\leq C\|x-y\|^{\alpha }

for all x and y in the domain of f. More generally, the condition can be formulated for functions between any two metric spaces. The number α is called the exponent of the Hölder condition."

But there is no reason to assume that the exponent α in the displayed equation is unique.

In fact, it is far from unique in all cases.

Therefore it is entirely inappropriate (as in the last quoted sentence) to refer to "The number α".

Unclear statement

The introductory section contains this statement:

"We have the following chain of strict inclusions for functions defined on a closed and bounded interval [a, b] of the real line with a < b :

Continuously differentiable ⊂ Lipschitz continuous ⊂ α-Hölder continuous ⊂ uniformly continuous ⊂ continuous,

where 0 < α ≤ 1."

This statement neglects to state how the number α is quantified:

Is this true for all α in (0, 1] ?

Or instead does there merely exist some α in (0, 1] for which it is true?

I hope someone knowledgeable about this subject can clarify this.