I'll remind readers of something additional that is very valuable, the function:
\[
\mathcal{B} g(x) : \mathbb{R}_{<0} \to \mathbb{R}_{<0}\\
\]
And has the expansion:
\[
\mathcal{B} g(-x) = \sum_{k=1}^\infty g_k \frac{(-x)^k}{k!}\\
\]
For \( |x| < \delta\), so it is analytic in a neighborhood of \(x =0\). This means we can write:
\[
\int_0^\infty e^{-t} \mathcal{B} g(-tx)\,dt \sim \sum_{k=1}^\infty (-1)^k g_k \cdot (-x)^k \, \text{as}\,x\to 0\\\\
\]
But, this also equals the iterated function: so that:
\[
\int_0^\infty e^{-t}\mathcal{B} g(tx)\,dt = g(x)\\
\]
For \(\Re(x) < 0\). This effectively shows, that having the existence of the half iterate, is enough for an integral construction. What's even more fascinating, is that:
\[
\mathcal{L} \mathcal{B} g(x) = g(1/x)\\
\]
Where \(\mathcal{L}\) is the Laplace transform. And this construction is solely dependent on \(g\) being bounded in a closed half plane.
There's a much more general theorem at play here. Which is that, if the half plane is invariant, and \(g\) is a contraction mapping on that half-plane--we get a Borel summable expression for said iterate. Adititionally it is nearly completely decided by an asymptotic expansion at \(0\).
If we have an asymptotic series:
\[
g(z) \sim \sum_{k=1}^\infty g_k z^k\\
\]
And we are guaranteed that \(g : \mathcal{H} \to \mathcal{H}\) where \(g\) is bounded, and additionally bounded on the closure of \(\mathcal{H}\) for \(\mathcal{H}\) a half plane. We can do all the heavy lifting from this last result. So this result is certainly definable on a larger domain.
It can probably be done for all fractional iterates about a parabolic point; but I'm not sure how to drop the Half plane requirement. And I think the fact that our original \(g\) sent a half plane to itself is very important--especially in how we define asymptotics in the Mellin transform.
But most (all?) parabolic points (probably just roots of unity multipliers) have an asymptotic series at said fixed point for the iterations. The question is, can they construct an iteration in a half plane from there, (and only within the attracting petal, the repelling petal would require us to look at the inverse).
If there exists a stable iteration \(f^{\circ t_0}(z) : \mathcal{H} \to \mathcal{H}\) with a parabolic point on its boundary (or it's conjugate similar to some iteration like this). I think we can Borel sum much much broader functions.
------------------------------------
I should also, further prove one thing from above.
When I wrote:
I realize it might not be clear to you guys how I showed this will grow like \(c^N N!\). The first thing you can note is that:
\[
|g^{(N)}(x)| \le N! |g(x)| p(x)^N\\
\]
for some \(p(x)\). The value \(p(x) \to 0\) as \(x \to \infty\) because the radius of convergence continues to grow when we center about \(x\). So the integral ends up looking like:
\[
F(z-N) \approx \frac{N!}{(z-1)(z-2)\cdots (z-N)}\\
\]
Where:
\[
\Gamma(z) = \lim_{N\to\infty} \frac{N! N^z}{z(z+1)(z+2) \cdots (z+N)}\\
\]
Which, this approximation for \(F(z-N)\) diverges just faster than converging to the gamma function, but of which our function will experience really similar growth. And then the \(\frac{1}{\Gamma(1+N-z)}\) will level everything out. I can further justify this if you'd like, and additionally I can find some sources related to Ramanujan summation and the lot which uses all these techniques.
Edit: This is sort of a prima facie thing in fractional calculus. If we are able to make a Mellin expansion this good, the worst this integral can explode is \(O(N!)\). I'll try to find an old paper I remember that covered these asymptotic expansions--including the theorems I used above. It's somewhere out there.
\[
\mathcal{B} g(x) : \mathbb{R}_{<0} \to \mathbb{R}_{<0}\\
\]
And has the expansion:
\[
\mathcal{B} g(-x) = \sum_{k=1}^\infty g_k \frac{(-x)^k}{k!}\\
\]
For \( |x| < \delta\), so it is analytic in a neighborhood of \(x =0\). This means we can write:
\[
\int_0^\infty e^{-t} \mathcal{B} g(-tx)\,dt \sim \sum_{k=1}^\infty (-1)^k g_k \cdot (-x)^k \, \text{as}\,x\to 0\\\\
\]
But, this also equals the iterated function: so that:
\[
\int_0^\infty e^{-t}\mathcal{B} g(tx)\,dt = g(x)\\
\]
For \(\Re(x) < 0\). This effectively shows, that having the existence of the half iterate, is enough for an integral construction. What's even more fascinating, is that:
\[
\mathcal{L} \mathcal{B} g(x) = g(1/x)\\
\]
Where \(\mathcal{L}\) is the Laplace transform. And this construction is solely dependent on \(g\) being bounded in a closed half plane.
There's a much more general theorem at play here. Which is that, if the half plane is invariant, and \(g\) is a contraction mapping on that half-plane--we get a Borel summable expression for said iterate. Adititionally it is nearly completely decided by an asymptotic expansion at \(0\).
If we have an asymptotic series:
\[
g(z) \sim \sum_{k=1}^\infty g_k z^k\\
\]
And we are guaranteed that \(g : \mathcal{H} \to \mathcal{H}\) where \(g\) is bounded, and additionally bounded on the closure of \(\mathcal{H}\) for \(\mathcal{H}\) a half plane. We can do all the heavy lifting from this last result. So this result is certainly definable on a larger domain.
It can probably be done for all fractional iterates about a parabolic point; but I'm not sure how to drop the Half plane requirement. And I think the fact that our original \(g\) sent a half plane to itself is very important--especially in how we define asymptotics in the Mellin transform.
But most (all?) parabolic points (probably just roots of unity multipliers) have an asymptotic series at said fixed point for the iterations. The question is, can they construct an iteration in a half plane from there, (and only within the attracting petal, the repelling petal would require us to look at the inverse).
If there exists a stable iteration \(f^{\circ t_0}(z) : \mathcal{H} \to \mathcal{H}\) with a parabolic point on its boundary (or it's conjugate similar to some iteration like this). I think we can Borel sum much much broader functions.
------------------------------------
I should also, further prove one thing from above.
When I wrote:
(08/21/2022, 12:17 AM)JmsNxn Wrote: ...
And iterating this, we get:
\[
F(z-N) = \frac{1}{(z-1)(z-2)\cdots (z-N)}\int_0^\infty g^{(N)}(x)x^{z-1}\,dx\\
\]
And this clearly has Gamma function growth, where by we must have the appropriate decay.
I realize it might not be clear to you guys how I showed this will grow like \(c^N N!\). The first thing you can note is that:
\[
|g^{(N)}(x)| \le N! |g(x)| p(x)^N\\
\]
for some \(p(x)\). The value \(p(x) \to 0\) as \(x \to \infty\) because the radius of convergence continues to grow when we center about \(x\). So the integral ends up looking like:
\[
F(z-N) \approx \frac{N!}{(z-1)(z-2)\cdots (z-N)}\\
\]
Where:
\[
\Gamma(z) = \lim_{N\to\infty} \frac{N! N^z}{z(z+1)(z+2) \cdots (z+N)}\\
\]
Which, this approximation for \(F(z-N)\) diverges just faster than converging to the gamma function, but of which our function will experience really similar growth. And then the \(\frac{1}{\Gamma(1+N-z)}\) will level everything out. I can further justify this if you'd like, and additionally I can find some sources related to Ramanujan summation and the lot which uses all these techniques.
Edit: This is sort of a prima facie thing in fractional calculus. If we are able to make a Mellin expansion this good, the worst this integral can explode is \(O(N!)\). I'll try to find an old paper I remember that covered these asymptotic expansions--including the theorems I used above. It's somewhere out there.