A functional recursion problem..do you have any idea? I have a problem which is related to algebra and polynomials. I would be very grateful if any of you could give a hand to 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 help.
 A: This isn't an answer, but it at least provides a possible path forward towards a proof. The general idea is to write $f_n(x)$ as a fourier series over the interval $x\in(0,1)$. As was noted in the comments, the choice of $f_0(x)$ does not seem to change the fact that $f_n(x)$ is monotone on the interval $[0,1/2]$ and $[1/2,1]$ (at least for continuous $f_0(x)$). Thus, I propose the following proposition:
If $f_0(x)$ is a continuous function on $[0,1]$ and $f_n(x)$ is defined as
$$f_n(x)=\frac{1}{2}(f_{n-1}(x^2)+f_{n-1}((1-x)^2)),$$
then both the following are true:
$$f_n(x)\to c_n g(x) +d$$
(where $c_n$ is a real number depending on $n$ and $f_0(x)$, while $d$ is a fixed real depending only on $f_0(x)$) and this function is eventually monotone on the intervals $[0,1/2]$ and $[1/2,1]$. We can think of $c_n$ and $d$ as scaling factors that arise due to the initial choice of $f_0(x)$. For example, if we select $f_0(x)=\sqrt{x}$, then
$$f_0(x)=\sqrt{x}$$
$$f_n(x)=1/2\text{ for }n\geq 1$$
which implies $c_n=0$ and $d=1/2$.
In order to show this proposition is at least semi-valid, note that for any choice of continuous $f_0(x)$ on $[0,1]$ we have
$$f_n(x)=\frac{1}{2}(f_{n-1}(x^2)+f_{n-1}((1-x)^2))$$
$$=\frac{1}{2}(f_{n-1}((1-x)^2)+f_{n-1}(x^2))$$
$$=\frac{1}{2}(f_{n-1}((1-x)^2)+f_{n-1}((1-(1-x))^2))=f_n(1-x)$$
(for $n\geq 1$). Thus, except for possibly $n=0$, $f_n(x)$ is symmetric about the line $x=1/2$. Of course, this implies $f_n(0)=f_n(1)$.
Now, note that over the interval $x\in [0,1]$ we can write 
$$f_n(x)=a_0+\sum_{i=1}^\infty a_i \cos(2\pi i x)+\sum_{i=1}^\infty b_i \sin(2\pi i x)$$
where
$$a_0=\int_0^1f_n(x)dx$$
$$a_i=2\int_0^1f_n(x)\cos(2\pi i x)dx\text{ for }i\geq 1$$
$$b_i=2\int_0^1f_n(x)\sin(2\pi i x)dx\text{ for }i\geq 1.$$
Now, notice that $b_i=0$ (at least for $n\geq 1$) as
$$b_i=2\int_0^1f_n(x)\sin(2\pi i x)dx=2\int_{-1/2}^{1/2}f_n(x-1/2)\sin(2\pi i x-\pi i)dx=0$$
as $f_n(x-1/2)$ is an even function (this follows as it is symmetric around $x=1/2$) and $\sin(2\pi i x-\pi i)$ is an odd function. Then $f_n(x)$ simplifies to 
$$f_n(x)=a_0+\sum_{i=1}^\infty a_i \cos(2\pi i x).$$
Now, if we add the further stipulation that $f_0(x)$ is piecewise-smooth on $[0,1]$, we can differentiate this term by term as $f_n(x)$ is continuous on $[0,1]$, piecewise-smooth on $[0,1]$, and $f_n(0)=f_n(1)$ for (see here for details) to get
$$f_n^{'}(x)=-2\pi \sum_{i=1}^\infty a_i i\sin(2\pi i x).$$
Now, if we could prove that this function is monotone on $(0,1/2)$ (as $f_n(x)$ is symmetric about $x=1/2$), we would have our proof. Numerical evidence would suggest that this is the case at large enough $n$ regardless of $f_0(x)$. Unfortunately, this seems just as difficult as the original problem, but maybe someone with more experience/knowledge about fourier series can expand upon this approach.
One thing I did look at was considering the function $g(x)$ as defined in the proposition (as the constants $c_n$ and $d$ do not affect the monotinicity of the derivative). Then one would simply have to prove 
$$g^{'}(x)=-4\pi \sum_{i=1}^\infty  i\sin(2\pi i x)\int_0^1g(y)\cos(2\pi i y)dy$$
is monotonic where $g(x)$ is any function that satisfies
$$g(x)=\frac{1}{2}(g(x^2)+g((1-x)^2))$$
on $[0,1]$. This seems reasonable as in the limit as $i$ goes to infinity,
$$\int_0^1g(y)\cos(2\pi i y)dy=O\left(\frac{1}{i^2}\right)$$
seems to hold for all continuous functions (not just $g(x)$). This is promising as on $(0,1)$
$$\sum_{i=1}^{\infty}\frac{\sin(2\pi i x)}{i}=\pi(1/2-x)$$
which is clearly monotonic.
