Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I have a nice problem which is related to algebra and polynomials or even operator theory. I would be very grateful if any of you could solve it. Here is the problem: Consider the function $f_0(x) =x(1-x)$ and for $n \geq 0$ define $$f_{n+1}(x) =\frac 12 (f_n(x^2) + f_n((1-x)^2)).$$ Now, looking more closely at $f_0(x)$, we see that it is increasing on $[0,\frac 12]$ and decreasing on $[\frac 12, 1]$ . The problem is to prove that such a property holds for all the $f_n$'s. More precisely, prove that each $f_n (x)$ is increasing on $x \in [0,\frac 12]$ and decreasing on $x \in [\frac 12, 1]$ . I would be very thankful if any of you could come up with a solution.

share|improve this question
Two remarks: First remark, your title is the worst title we can imagine... It is not informative at all! Second remark, I do not think your question is appropriate for mathoverflow since it is not a research-level question (post on math.stackexchange instead). To help you a little bit with your question though, $f_0$, thus each $f_n$, is smooth so you can consider its derivative, and you can have a recurrence relation on the derivatives which should help you. –  Bruno Aug 31 '11 at 16:53
Thanks Bruno, Sorry to bother you with the title, I do not have a lot of experience in posting problems. In fact this is a research problem and I am not aware of any place that a solution to it exists. I am sure that the claim is true, since I have simulated the recursion for large values of $n$ and it is true. Many thanks again. –  hamed hassani Sep 1 '11 at 6:26
I find $$f_{n+1}'(x)=xf_n'(x^2)-(1-x)f_n'((1-x)^2).$$ But I reach no obvious conclusion from that, since for $x<1/2$, $(1-x)^2$ may be less or greater than 1/2. I do not think this problem is as easy as people have presumed. Voting to reopen. –  Michael Renardy Aug 15 '13 at 14:25
I'll vote to reopen if he changes the title and edits to include this information about the recursion which he posted in a comment –  David White Aug 15 '13 at 16:12
This is a surprisingly nice question, but this isn't clear until actually attempting to solve it. I think it deserves some effort! –  Alex Amenta Apr 12 at 22:30

1 Answer 1

First, notice that $f_n(x)=f_n(1-x).$ Therefore, $f_{n+1}(x)=1/2(f_n(x^2)+f_n(1-(1-x)^2))=1/2(f_n(x^2)+f_n(2x-x^2))$, but both x^2, 2x-x^2 are increasing functions on [0, 1/2]. So, by induction, you conclude that $f_n$ is increasing on [0, 1/2]. Now, by symmetry, $f_n$ is decreasing on [1/2, 1].

share|improve this answer
Hi, Many thanks for your comment. The problem is that the range of $2x-x^2$ over $[0, \frac 12]$ is $[0, \frac 34]$. As a result what you mentioned does not work here. –  hamed hassani Sep 1 '11 at 6:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.