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.
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12$\begingroup$ 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. $\endgroup$– BrunoCommented Aug 31, 2011 at 16:53
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4$\begingroup$ 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. $\endgroup$– hamed hassaniCommented Sep 1, 2011 at 6:26
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4$\begingroup$ 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. $\endgroup$– Michael RenardyCommented Aug 15, 2013 at 14:25
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1$\begingroup$ 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 $\endgroup$– David WhiteCommented Aug 15, 2013 at 16:12
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3$\begingroup$ This is a surprisingly nice question, but this isn't clear until actually attempting to solve it. I think it deserves some effort! $\endgroup$– user14166Commented Apr 12, 2015 at 22:30
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1 Answer
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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].
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8$\begingroup$ 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. $\endgroup$ Commented Sep 1, 2011 at 6:17