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 $f_0(x) =x(1x)$ and for $n \geq 0$ define $$f_{n+1}(x) =\frac 12 (f_n(x^2) + f_n((1x)^2)).$$ 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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closed as too localized by José FigueroaO'Farrill, Gjergji Zaimi, Felipe Voloch, Emil Jeřábek, Andy Putman Aug 31 '11 at 18:16This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question. 


First, notice that $f_n(x)=f_n(1x).$ Therefore, $f_{n+1}(x)=1/2(f_n(x^2)+f_n(1(1x)^2))=1/2(f_n(x^2)+f_n(2xx^2))$, but both x^2, 2xx^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]. 

