Talk:Modulus of continuity

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Why ω : [ 0 , ∞ ] → [ 0 , ∞ ] {\displaystyle \omega :[0,\infty ]\to [0,\infty ]} rather than ω : [ 0 , ∞ ) → [ 0 , ∞ ] {\displaystyle \omega :[0,\infty )\to [0,\infty ]} ? We're told in the remarks section that infinity is needed as an element of the range. But it's not clear why it's needed in the domain. Jowa fan (talk) 02:57, 14 June 2012 (UTC)[reply]

Actually it is not needed, and several textbooks just take [ 0 , ∞ ) {\displaystyle [0,\infty )} as domain in the definition of a mdc. The choice [ 0 , ∞ ] {\displaystyle [0,\infty ]} is just to make it more natural the composition rule (see the second stated elementary fact).--pma 12:14, 14 June 2012 (UTC)[reply]

The "Examples of Use" section uses the "modulus of a partition P". I think many readers, even many mathematicians, will not know what this means. P is a finite set, and notation such as "|P|" is often used for the cardinality of P, which in this case is n + 1, but that makes no sense here. I think the definition of "modulus of a partition", or a link to a definition, or at least a reference (if no link can be found) should be given, or the example should be removed. Gsspradlin (talk) 22:24, 7 November 2013 (UTC)[reply]

For many domains that most users will actually use, a finite-valued modulus of continuity always exists (as in the second sentence of the whole article, in which the domain is a real interval). The definition of modulus of continuity does not require finitivity, so the curious reader will wonder why one allows a modulus of continuity to be infinite. The end of the "Remarks" section contains the statement "However, the situation is different for uniformly continuous functions defined on compact or convex subsets of normed spaces", which, while better than nothing, is vague, and it would be better to have a definitive statement that actually asserts something. I am quite sure that if the domain is a convex subset of a normed linear space, this is possible. I am also quite sure this is possible if the domain is a metric space whose metric is Lipschitz (most readers, including probably most mathematicians, won't know what this means, so a link or reference should be included). Unfortunately, I don't have a reference at hand. Given the brevity of this article, links would be more appropriate than proofs. The only counterexample I have seen (where no finite modulus of continuity exists for a uniformly continuous function) involved a noncompact domain, so it is conceivable to me that if the domain is compact, then any uniformly continuous function admits a finite modulus of continuity. In this case, I don't know this for sure. There may be an easy, elementary proof (I haven't tried). Any mention of compactness should be backed up with a definite statement with an accompanying reference, or omitted. Gsspradlin (talk) 15:12, 9 November 2013 (UTC)[reply]

Well, the reason why we allow the value infinity for moduli of continuity is that it makes true the statement: "f is uniformly continuous if and only if it has a modulus of continuity". The fact is that sometimes one does not have sufficient information on the domain of a uniformly continuous map. The "vague sentence" at the end of the Remarks section that you quoted serves as introduction to the following section, where everything is explained in detail. Yes, if the domain of a uniformly continuous map is convex there is a real-valued module of continuity for it, indeed even concave (read the article). "A metric space whose metric is Lipschitz" makes no or little sense (every metric is 1-Lipschitz wrto itself; what is a non-Lipschitz metric then?). It is obvious that a continuous function on a compact metric space has a bounded modulus of continuity (as explained; read the article). In conclusion, I think most readers may be happy with the introductory part, while curious or interested readers can find any explanations in the subsequent sections. (Of course, these readers are supposed to have a minimal knowledge of the English language, in particular, mathematical terms such as "convex", "compact", "function"; there is no need of a reference of these terms although we may link them with other articles; yet it does not seem the case) pma 00:58, 19 November 2013 (UTC)[reply]

The common definition of the modulus of continuity for a uniformly continuous function is | f ( x ) − f ( y ) | < ε {\displaystyle |f(x)-f(y)|<\varepsilon } whenever | x − y | < ω ( ε ) {\displaystyle |x-y|<\omega (\varepsilon )} (as given on page 34 of Bishop's foundations of constructive analysis for example), for any constructive interval. It's also relatively straightforward to see that defining this way makes a lot more sense in actual proofs. The article seems to define it as the inverse of Bishop's operation, which looks really weird to me Saolof (talk) 13:10, 8 October 2021 (UTC)[reply]

The definition in the lead section seems to confuse the metric for the function itself: I → R {\displaystyle I\to \mathbb {R} } is properly the signature of the metric d I {\displaystyle d_{I}} , whereas continuity is defined for a function f : I → I {\displaystyle f:I\to I} . Moreover it doesn't mention anywhere the fact that I {\displaystyle I} (or the object ( X , d X ) {\displaystyle (X,d_{X})} that the "Formal definition" section discusses without asserting any axioms) is a metric space, which is fundamental to the notion of uniform continuity. Including infinity in the metric is also incorrect, since having it as a value contradicts the notion that ω {\displaystyle \omega } is a modulus of uniform continuity:

• If ω ( ϵ ) = ∞ {\displaystyle \omega (\epsilon )=\infty } , then d ( x , y ) < ω ( ϵ ) {\displaystyle d(x,y)<\omega (\epsilon )} for all x , y {\displaystyle x,y} , contradicting the definition that it is the maximum variance (and not merely an upper bound on variance). Thus f {\displaystyle f} is continuous (because ω {\displaystyle \omega } is still defined to be monotonic) but not uniformly continuous (because the variance is unbounded).
• Alternatively it could be that the space is disconnected, meaning d ( x , y ) = ∞ {\displaystyle d(x,y)=\infty } for some x , y {\displaystyle x,y} . But this contradicts the definition of metric space, which restricts d {\displaystyle d} to the reals (so it can't take on a value of ∞ {\displaystyle \infty } ).
• Zero is necessary: let f ( x ) = 1 {\displaystyle f(x)=1} for all x {\displaystyle x} (which is trivially uniformly continuous), then d ( f ( x ) , f ( y ) ) < ϵ {\displaystyle d(f(x),f(y))<\epsilon } for all 0}"> ϵ > 0 {\displaystyle \epsilon >0} , so we require ω ( ϵ ) = 0 {\displaystyle \omega (\epsilon )=0} .

It's possibly fine if we relax the notion to arbitrary continuity, but that is not what the article (or, as far as I know, the literature) describes. I've slapped {{accuracy}} on the article. But this is really a subtopic of uniform continuity and probably not worthy of a separate article. Hairy Dude (talk) 17:22, 28 November 2023 (UTC)[reply]

I think that metric spaces are linked in the introduction. It never occurred for me for a moment that d_X is not metric for X. Moreover the topic of why omega is defined to take infinity as an input is given on this talk page. Search for "Infinity not needed in domain?". No one is claiming that a metric should take on infinite values.

I agree that that more explicit definitions for the symbols use may improve clarity. But I was not confused by them. And frankly I have been unable to locate any factual errors.

As for "The definition given does not match the one common in the literature", this is a matter of perceptive. The entire field of approximation theory essentially adopts the definition in this article. This fact alone makes the article worthy of existing independently of uniform continuity. Radiodont2 (talk) 15:01, 7 March 2024 (UTC)[reply]