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 18:16
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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]. 

