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I am thinking of the possible validity of a statement like this:

given any positive mapping $\delta$ on $[a,b]$, there exists a partition $\, a=a_0<a_1<\cdots<a_n=b \,$ of $ \,[a,b] \,$ so that $$a_r - a_{r-1} < \delta (a_{r-1})\qquad(r=1,\ldots,n)$$ Jean Mawhin says it is not true at the end of his contribution in Fulvia Skof (Ed.) Giuseppe Peano between Mathematics and Logic (2011).

I cannot understand if the reason is banal or not, so I need your help to know why it is not true. Thank you in advance.

(the question is tied to the need of the so-called straddle lemma in the introduction of the Henstock-Kurzweil integral)

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A counterexample should illustrate. Consider the interval $[0,2]$ define your function $\delta:[0,2]\to R^+$ by requiring that $\delta(x)< 1-x$ for all $x\in [0,1]$ and any other values you wish elsewhere. If an interval $[a_{r-1},a_r]$ in the first half of the interval satisfies $a_r-a_{r-1}<\delta(a_{r-1})$ then $a_r< a_{r-1} + \delta(a_{r-1})<1$. So you never reach the point $1$ in your partition.

But that does not mean that there is no left-tagged partitions possible if you make suitable adjustments. John Hagood came up with a very nice idea for this and we used it to integrate Dini derivatives of continuous functions. If you care to pursue see

MR2202919 (2006i:26010) Hagood, John W. ; Thomson, Brian S.
Recovering a function from a Dini derivative. Amer. Math. Monthly 113 (2006), no. 1, 34--46. Download link here.

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  • $\begingroup$ @ B.S.Thomson Pedagogically speaking, it is a pity that one has to introduce an alternate definition of derivative just at the start of the theory in order to prove that every derivative is HK-integrable. If the derivative is continuous there is no need. $\endgroup$ – Tony Piccolo Oct 30 '15 at 23:28
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    $\begingroup$ I don't understand the comment. The HK integrates all ordinary derivatives quite trivially. In fact it is easier to prove this than to prove that a continuous derivative is Riemann integrable. Maybe you are referring to something Mawhin said that I am not familiar with. As to integrating the Dini derivative of a continuous function, that is a different matter since it might not even be Lebesgue integrable. $\endgroup$ – B. S. Thomson Oct 30 '15 at 23:33
  • $\begingroup$ @ B.S.Thomson I refer to the second "activity" in my previous question where it is used the fact that a continuous derivative is the derivative of a uniformly differentiable function. If the statement in my post were valid, the algebra would be the same for every derivative. Not being so, it is better to consider arbitrarily tagged partitions from the start. $\endgroup$ – Tony Piccolo Oct 31 '15 at 22:52
  • $\begingroup$ @ B.S.Thomson Mawhin's writing deals with a theorem of Peano: at the end he says that the proof could be more direct if the above statement about left-tagged partitions were valid. $\endgroup$ – Tony Piccolo Nov 1 '15 at 6:27
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    $\begingroup$ Thanks Tony. That is interesting and the history is quite curious. I would expect that you can make John Hagood's lemma do the job for you and you can give a suitable proof. Peano was claiming the existence of any partition but he didn't think of it being finer than some $\delta(x)$. If he had (in 1884) he would get priority over Cousin for it. As to your comment about "continuous ... no need." Well yes--if all derivatives were continuous the theory gets quite easy. But they aren't. $\endgroup$ – B. S. Thomson Nov 1 '15 at 23:44
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The answer to Tony's question is in the negative, but with a slight adjustment we can do this more positively. Here is Tony's motivation, quoting from his previous post on the subject:

Peano's Exercise: Let $f$ be defined on $[a,b]$ and there differentiable. Show that for every $ϵ>0 $ there exists a partition $ a=a_0<a_1<...<a_n=b $ of $[a,b] $ so that$$ \left|\frac{f(a_{i+1})-f(a_i)}{a_{i+1}-a_i}-f'(a_i) \right|<\epsilon\qquad(i=0,...,n-1)$$

This proof was left as an exercise to the Belgian mathematician P. Gilbert by G. Peano in a quarrel (1884) about a mistake made by C. Jordan in his Cours d'analyse vol.1 (1882). According to Peano, Jordan's proof of the mean value inequality theorem presented a fallacious argument: Gilbert did not agree.

Naturally, Tony who is aware and perhaps even a fan of the Cousin Covering lemma thought that there must be a proof that uses that lemma or something similar. Certainly it looks so. But we need a modification.

Piccolo-Cousin Covering Lemma: Let $\cal C$ be a collection of closed subintervals of $[a,b]$ with the following two properties:

(a) For every $a\leq x < b$ there is a $\delta(x)>0$ so that $[x,x+t] \in {\cal C}$ for all $0<t<\delta(x)$.

(b) For every $a<x\leq b$ there is at least one interval $[c,x]\in {\cal C}$.

Then ${\cal C}$ contains a partition of $[a,b]$.

We call it the "Piccolo" lemma in honor of Tony or, perhaps, because we are thinking of this a "little" version of the Cousin lemma, little (piccolo) because it assumes so little about what is happening on the left at each point. Of course, if you assume less you get less: here we have a partition of $[a,b]$ but not of every subinterval of $[a,b]$.

Solution of Peano's Exercise using Piccolo coverings: Define the collection $${\cal C}=\left\{[u,v]: \left|\frac{f(v)-f(u)}{v-u} - f'(u)\right| <\epsilon \right\}$$ Just verify that $\cal C$ satisfies the two conditions of the lemma. The condition (a) is quite evident. The condition (b) follows from the mean-value theorem but is elementary (Tony's other post shows how). By the lemma there is a partition that satisfies Peano's requirements.

As Fermat once said (roughly), I believe I have valid proofs of these statements but StackExchange allows too few characters to add them here.

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  • $\begingroup$ Thank you for the "naming", Brian. How do you prove the lemma ? $\endgroup$ – Tony Piccolo Nov 7 '15 at 6:30
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    $\begingroup$ Good name for it, but probably not your real name. Here is a sketch: (i) define $S$ as the set of all $s$ such that $\cal C$ contains a partition of $[s,b]$, (ii) show $S\not=\emptyset$, (iii) define $\hat{s}=\inf S$, (iv) show that $\hat{s}\in S$, and (iv) show $\hat{s}=a$. For other or better proofs maybe post this as a new question with appropriate background information. I found your posts on these ideas entertaining and perhaps others do too. $\endgroup$ – B. S. Thomson Nov 7 '15 at 18:48
  • $\begingroup$ Thanks for your suggestion. Do you think it is possible to prove "by bisection" ? $\endgroup$ – Tony Piccolo Nov 7 '15 at 19:29
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    $\begingroup$ "By bisection?" Probably not, because the conclusion is that there is a partition of $[a,b]$, but not necessarily a partition of many other subintervals. For example there is a partition of $[a,b]$ but under these assumptions there may be no partition of $[c,b]$ for any value $a<c<b$. $\endgroup$ – B. S. Thomson Nov 7 '15 at 20:07

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