Talk:Monotonic function

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I know I'm not the most mathematically knowledgeable person here, but I think I may have found an error in this article. It says that a monotonic function can only have countably many discontinuities in its domain. Isn't the devil's staircase (the one related to the Cantor set) a counterexample to this? Thanks for any clarifications you can give! -- Oliver Pereira 02:12 Nov 29, 2002 (UTC)

Sorry, that's not what I meant to say. The devil's staircase is continuous, of course! But if you turn it on its side, don't you get a function (okay, specifying the values on the straight line bits in some arbitrary way) from [0, 1] to [0, 1] which is monotonic and which has an uncountable number of discontinuities? -- Oliver Pereira 02:27 Nov 29, 2002 (UTC)

Nope, okay, the set of discontinuities is countable, of course. Why don't I think before posting these stupid questions??? -- Oliver Pereira 02:31 Nov 29, 2002 (UTC)

Contents 1 Derivative 2 nondecreasing 3 clarification 4 Isotone 5 Article needs references 6 Proof of only countable discontinuities?

[edit] Derivative

Is there any name for this theorem? Tosha 20:21, 19 Feb 2005 (UTC)

if f is a monotonic functions defined on an interval I, then f is differentiable almost everywhere on I, i.e. the set of numbers x in I such that f is not differentiable in x has Lebesgue measure zero.

[edit] nondecreasing

if u r going to have nondecreasing redirect here, then you should better define nondecreasing. inpired by the shape of the graph? can you show the graph then??--Jaysscholar 23:21, 12 October 2005 (UTC)

• Non-decreasing is defined on the page. Is your issue with the spelling nondecreasing vs. non-decreasing? My personal preference is nondecreasing, but I'm sure many opinions differ. Ah, maybe I see your point. The definition relies on the definition of monotone above, which is a bit far away. I don't see an easy improvement, but I agree that the current definition is hard to read locally.

--Erik Demaine 00:55, 13 October 2005 (UTC)

[edit] clarification

does this definition mean that the equation's slope is always positive or negative (no points of inflection)? The definition is unclear (similarly so in my calculus textbook, unfortunately...). --anon

There can be points of inflection. Positive slope is about first derivative, and points of inflection are about second derivative, so they are not mutually exclusive. Example:

f(x) = x³ + (x − 1)³.

First derivative is always > 0, while second derivative is 6(2x-1) which is 0 at x=1/2, so there are points of inflection. Oleg Alexandrov (talk) 19:12, 21 February 2006 (UTC)

Actually the answer is that under the definition of a monotonic function, ie a function f such that for all , constant functions f(x)=c for some number c are also monotonic. For example, for the constant function f(x)=2, if then because f(x) = f(y) = 2. Thus f(x)=2 is monotonic, and has a constant slope of 0 at all points in its domain.

A typical related definition is that a function f is called strictly monotonic if for all x > y f(x) > f(y). Such functions can still have points of inflection (eg f(x) = x³ is strictly monotonic with an inflection point at x=0). However, they can't be constant over an interval. Dugwiki 19:00, 6 November 2006 (UTC)

[edit] Isotone

Should Isotone really redirect here? I'm looking at a different definition in Bhatia's "Matrix Analysis" - if x is majorized by y, then f(x) is submajorized by f(y). For instance, the entropy function (from Rⁿ₊ to R₊) is isotone. A5 18:59, 6 March 2006 (UTC)

[edit] Article needs references

I noticed by chance while reading the article that it doesn't have any citations. References will be needed. Especially handy would be references to proofs of some of the statements in the article, such as the section that implies that all monotone functions have both left and right limits at every point in their domain, and that only having jump discontinuities implies only having a countable number of discontinuities. (I'm sure the statements are true, but it would be informative to reference proofs to that effect.) Dugwiki 18:07, 31 October 2006 (UTC)

[edit] Proof of only countable discontinuities?

Ok, for the heck of it I was trying to prove that a monotonic function can only have countably many discontinuities, but I'm stuck. :/ I'm sure it's true, but I haven't figured out how to formally prove it. Anybody have a nice little proof of it, just out of curiosity? Dugwiki 22:56, 1 November 2006 (UTC)

To follow up, I think a possible pseudo proof might be something like this:

Let f is a monotonic function on the reals, and let D be the set of jump discontinuities in the domain of f. Assume that D is an uncountable set. Then it follows (I think) that there must exist a point d in D such that for some e there is a finite interval (d-e,d+e) that contains an uncountable subset of D. (I suspect this part is true but am not sure of the proof.)

Then for each jump discontinuity within that interval, there is an associated non-zero "distance" jumped which is the difference between the right hand and left hand limits at the discontinuity. Consider the set of all such jump distances within the above finite interval (d-e,d+e). Since f is monotonic, and this is a finite subinterval of the domain, it follows that the righthand limit of f at (d-e) must be less than the lefthand limit of f at d+e. And, in fact, the righthand limit at d+e must be at least as large as the lefthand limit at d-e plus the sum of all jump distances for all discontinuities between the two points. This total amount must be finite, because if it were unbounded in either direction then since f is monotone it would not be able to cover values beyond the unbounded direction (ie if the distance grows unbounded toward the point d+e, then f(y) could not exist for y>(d+e) since f(y) must be greater than all function values within the interval (d-e,d+e)).

But an uncountable sum series of positive values can only converge if at most a countable number of elements in the series are non-zero. Since D is uncountable, the uncountable sum of its elements' jump distances can therefore likewise only converge to a finite total distance if only a countable number of those jump distances are non-zero. However, if the jump distance is equal to zero, then the lefthand and righthand limits are equal, and there is by definition no jump discontinuity at that point. Thus there can only be at most a countable number of discontinuities within (d-e,d+e), a contradiction with the above initial assumptions that (d-e,d+e) contained an uncountable set of discontinuities. Therefore, by contradiction, there are no finite intervals in the domain of f with an uncountable number of discontinuities, and thus the set of all discontinuities D can not be uncountable.

That's about the best I can do at the moment. If anyone can clean up the above "proof" or has something neater, please feel free. Dugwiki 20:37, 2 November 2006 (UTC)

If you have access to a copy of Riesz and Nagy's book 'Functional Analysis', it is worth reading - they discus lots of ideas around this area, with some excellent proofs. Madmath789 20:51, 2 November 2006 (UTC)

Thanks. I don't have access to that book, unfortunately, but I'll keep that in mind if I'm at Borders. Dugwiki 21:59, 2 November 2006 (UTC)

Oh, also, if you can pinpoint the proofs in that book, you might want to include them in the article as footnotes for references. The article is currently uncited. Dugwiki 22:01, 2 November 2006 (UTC)

As a follow-up, I thought it about this some more Friday night and came up with what I think is a reasonable proof of the corollary I mentioned regarding a point with an uncountable number of neighbors in an uncountable subset of the reals. Here's how this bit of original research goes:

Lemma: For all uncountable subsets S of the reals, there exists at least one element s of S such that for all k>0 there are an uncountable number of elements of S in the interval [s-k, s+k].

Proof: Let S be a given uncountable subset of the reals, and let k be any given strictly positive real number. Consider the partition of the reals defined by all subintervals of the form [nk, (n+2)k] for all integers n. In other words, a partition of the reals of intervals of width 2k including the interval [0,2k] and expanding outward in both directions.

This partition covers the reals, and hence includes all elements of S in its combined intervals. The number of intervals in this partition is countable, though, so since S is uncountable there must exist at least one interval in the partition that contains an uncountable number of elements in S. The reason is that if every interval contained only a countable number of elements of S, then the combined union of all those elements would be a countable number of elements from a countable number of intervals, which is at most a countable union.

Therefore, by the definition of the partition outlined above, there exists at least one number n such that the interval [nk, (n+2)k] contains an uncountable subset of S. This interval can be rewritten as [(n+1)k - k, (n+1)k + k], and thus meets the criteria of an interval of the form desired in the lemma. (qed)

With that lemma now proven, it follows that the total "jump continuity distance" within a finite interval can only be finite number if there are at most a countable number of strictly positive distances jumped. Thus no finite interval in the domain of a monotonic function can have an uncountable number of jump discontinuities, and therefore by the lemma above there can only be at most a countable number of total discontinuities on a real monotonic function's total domain.

I'm not sure how this proof compares to ones in the textbook reference mentioned previously, but it was certainly fun to play around with last week. :) Dugwiki 18:20, 6 November 2006 (UTC)