I have been studying a function $F(x)$ obeying $F(p(x))=F(x)/2$. I did numerical work for $p(x)=x(1/2-x)$, although a similar functional equation could be solved for any polynomial with an attractive fixed point at $0$. It has a power series about $0$, which seems to have a radius of convergence of $1/2$, which makes sense because $p(x)$ has a repulsive fixed point at $-1/2$. We have $F(p^n(x))=F(x)/2^n$, F(x) is differentiable so we can define $d p^n(x)/dn$ by differentiating both sides and turn the discrete mapping $x \mapsto p(x)$ into a vector field (infinitesimal diffeomorphisms). I compute $F(x)$ by computing $y=p^n(x)$ with n sufficiently large that the power series is accurate for $F(y)$ then put $F(x)=2^nF(y)$. $F(x)=0$ for all $0$'s of $p^n(x)=0$ and any integer $n$. It is singular on the boundary of the basin of the fixed point, which in my case is the Julia set.


1)Is this known/interesting ?

2) It seems to me that there is no way to extend $F(x)$ beyond the basin of $0$. Is this correct.

3)The inverse function to $F(x)$ would be interesting, but I have not had time to study it.


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