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this idea is a draft and perhaps somewhat farfetched ...

consider a non-linear real entire function f(x) strictly rising on the reals with no more than 2 conjugate fixpoints at R + oo i and R - oo i.

also f(x) is uniquely invertible on the line segment connecting the 2 conjugate fixpoints L and L*.

let R be the real value of L , and x be real then f(2x) i = f((x-R)i + R)

now consider the integral from R - oo i till R + oo i of f(z) dz = C_0

let C_n be the integral after substitution z = f^[1/2^n](x).

now consider G = sum over +integers n : abs(C_n - C_0)

G = 0 implies a uniqueness condition on the superfunction of f(x).

G = 0 is equivalent to df/dx f^[1/2^n](x) >= 0 and also equivalent to

df/dq f^[q](x) >= 0.

how to compute or prove a sum like abs(C_n - C_0) ?

i have considered taking the continuum sum of C_x - C_0 or a contour integral around it and then trying to prove the period of one of those ...

this also resembles fourrier transforms but i dont know how to use transforms on this ...

it would be nice if we could convert this problem to tetration , however it might be problematic ; half-iterates tend not to agree on both fixpoints , the mapping to another function might not be topologically conjugate , an integral on a riemann sphere can be closed and its equivalent on the complex plane might not ( + oo i IS - oo i at the sphere).

however this is a draft and conjugate fixpoints are special cases so there might be progress possible with some " luck ".

regards

tommy1729