Let's take \(z \in U\), such that \(f^{\circ k}(z) \to 2\) and \(f^{\circ -k}(z) \to 4\), as \(k\to\infty\)--where \(f(z) = \sqrt{2}^z\). The set \(U\) is simply connected domain, and subject to all our "double fixed point talk". Then taking the regular iteration for \(f:U \to U\). This means that \(f^{\circ s} : U \to U\) for \(\Re s > 0\).
Then:
\[
\widehat{f^{\circ s}}(\xi,z) = \int_{-\infty+it}^{\infty+it} \left( f^{\circ s}(z) -2\right) e^{-2\pi i \xi s}\,ds\\
\]
Applications of \(f\) become applications of a wave multiplier \(e^{-2\pi i \xi}\). Which is the equation:
\[
\widehat{f^{\circ s}}(\xi,f^{\circ q}(z)) = e^{2\pi i \xi q} \widehat{f^{\circ s}}(\xi,z)
\]
We can prove this in a very strong sense. The heavy lifting is handled by Henryk, and Dmitri. The paper they wrote on the Four super exponentials of \(\sqrt{2}^z\) proves everything we need. I'd put a lot of the credit to Henryk--because I had written integrals similar before. But Henryk has paved a way to absolutely, rigorously, prove it.
Then:
\[
\widehat{f^{\circ s}}(\xi,z) = \int_{-\infty+it}^{\infty+it} \left( f^{\circ s}(z) -2\right) e^{-2\pi i \xi s}\,ds\\
\]
Applications of \(f\) become applications of a wave multiplier \(e^{-2\pi i \xi}\). Which is the equation:
\[
\widehat{f^{\circ s}}(\xi,f^{\circ q}(z)) = e^{2\pi i \xi q} \widehat{f^{\circ s}}(\xi,z)
\]
We can prove this in a very strong sense. The heavy lifting is handled by Henryk, and Dmitri. The paper they wrote on the Four super exponentials of \(\sqrt{2}^z\) proves everything we need. I'd put a lot of the credit to Henryk--because I had written integrals similar before. But Henryk has paved a way to absolutely, rigorously, prove it.