continuation of fix A to fix B ?

[tommy1729]

Just a thought.
it can be extended to the family of schroder function by
\[\sigma_f(z) = \int_{\mathbb{R}}{h(k)g(f^k(z))\mathrm{d}k}\] (and thus on measurable set in the way that k has some kinda shift-invariance), where \(h\) is any function that fits \(h(k-1)=s\,h(k)\), \(s\) is the multiplier, and \(g\) is any function.

Hey, Leo

This is entirely correct and what I was hinting at, but I didn't bother to go full Lebesgue!

The correct way to discuss this is:

\[
F_c (z) = \int_{\mathbb{R}}f^{\circ k}(z) \,d\mu ( c)\\
\]

And we would choose a (hopefully quasi Borel) measure \(\mu( c)\); which only depends on \(c\) and \(k\). Which would then give us (hopefully) all possible solutions.

This would say, for all \(\mu( c)\) such that:

\[
F_c(f(z)) = cF_c(z)\\
\]

The value:

\[
\sum_{k=-\infty}^\infty c^k f^{\circ k}(z)\\
\]

Would belong to this class--where there is some \(\mu^*\) which satisfies this (has moles/atoms at \(k \in \mathbb{Z}\)). I believe the official notation would be: \(\mu^*( c) = \sum_{n\in\mathbb{Z}} c^n \delta(k-n)\) (I'm a little out of date on my distribution knowledge, lol )

But, let's play your game, Leo! I was avoiding this discussion. But let's write:

\[
\int_{-\infty}^\infty f^{\circ k}(z) c^k \,dk\\
\]

It turns out this is the "total" holomorphic function to satisfy Tommy's identity. And the only other thing that can is a Fourier transform. What we need is an "indicator" function \(\chi\) that is translationally invariant. This means that \(\chi(k+1) = \chi(k)\). And \(\chi : \mathbb{R} \to \mathbb{C}\). So that:

\[
\int_{-\infty}^\infty f^{\circ k}(z) c^k \chi(k) \,dk \\
\]

Are all solutions. This is assuming that we're considering the ideal case of \(f(z) = \sqrt{2}^z\)--and the standard iteration. So tommy's solution, in and of itself, is the indicator function \(\chi_{\mathbb{Z}}\) on the integers. We can defs choose other indicators though. But they will always be discrete.

This is the super fun part. Imagine we write:

\[
F[\chi_c](z)= \int_{-\infty}^\infty f^{\circ k}(z) c^k \chi_c(k) dk
\]

Now, let's write the standard fourier analysis approach...

\[
F[\chi_c](z) = \sum_{j=-\infty}^\infty b_j( c) F[e^{-2\pi i j k}](z)\\
\]

Which happens because every element of Tommy's space of solutions, can be found as the integral:

\[
F[\chi_c](z)= \int_{-\infty}^\infty f^{\circ k}(z) c^k \chi_c(k) dk
\]

And each:

\[
\chi_c(k) = \sum_{j=-\infty}^\infty b_j(c ) e^{-2 \pi i j k}\\
\]

BUT!!!!!!!!!!!!!!

This only happens in a distributional sense. In a technical sense, the value \(\mu^*(c )\) cannot be represented as this. When I write:

\[
F[\chi_{\mathbb{Z}}](z) = \int_{-\infty}^\infty f^{\circ k}(z) c^k \chi_{\mathbb{Z}}(k) dk = \sum_{k=-\infty}^\infty f^{\circ k}(z)c^k\\
\]

We do not mean that the naive numerical approximation methods work. What we mean is that this converges in a distributional sense. This means we have to introduce an \(L^2\) space. In this case, we can take something standard, like the standard Fourier Analysis space \(L^2(\mathbb{R})\). And from this, we have the inner product:

\[
(f,g) = \int_{-\infty}^\infty f(k)\overline{g(k)}\,dk\\
\]

And from the extension of the Hilbert space (using Von Neumann magic), we can define:

\[
F[\chi_c](z) = (f^{\circ k}(z) c^k, \chi_c(k))\\
\]

And as a distribution; we can Fourier decompose \(\chi_c\), even though it may not be obvious....

Tommy's \(\chi_{\mathbb{Z}}\) would be the same thing as a "square wave". It can be approximated with things underneath the integral....

correct.