Is there a transformation $\mathcal{T}$ of maps $\mathbb{R}_{{\geq}0}^{n} \rightarrow \mathbb{R}_{{\geq}0}^{n}$ with the following property?

If a map $F : \mathbb{R}_{{\geq}0}^{n} \rightarrow \mathbb{R}_{{\geq}0}^{n}$ is smooth and order-reversing with respect to the product order and possesses a unique fixed point $\omega$, then
  • The point $\omega$ is the unique fixed point of ${\mathcal{T}}\!F$; and
  • For some (known) point $a \in \mathbb{R}_{{\geq}0}^{n}$, the sequence of iterates of ${\mathcal{T}}\!F$ starting at $a$ converges to the (unknown) point $\omega$.

For more on this question, please refer to the 3-page PDF document at this address: http://math.gillesgnacadja.info/files/FixedPointAlgo_OPEN.html. I would have liked to post everything here but I could not find a way to save and preview the question before posting it. The content of the document is as follows.

  1. The Question
  2. Why this Question?
  3. The Trivial Case $n = 1$
  4. Where Does This Question Come From?
  5. Satisfying the Hypotheses of the Question
  6. An Analogous Question
  • $\begingroup$ What do you mean by "order-reversing" if $n>1$? $\endgroup$ – fedja Dec 13 '10 at 2:47
  • $\begingroup$ My guess is the product order, although it might be lexicographic order. $\endgroup$ – user5810 Dec 13 '10 at 3:25
  • $\begingroup$ Thanks to fedja for the question and to Ricky Demer for the answer. The order is indeed the product order. I added the precision in the question. $\endgroup$ – Gilles Gnacadja Dec 13 '10 at 16:58

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.