# I can compute a function $F(x)$ such that $F(x(1/2-x))= F(x)/2$, It is analytic on a filled Julia set.

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.

Questions:

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.