Hey, Bo. I was just as surprised this would work as you.
The main property I used, was mentioned by David E speyer; which is that:
\[
g(z) : \mathbb{C}/[0,\infty) \to \mathbb{C}\\
\]
And which for our purposes was reduced to :
\[
g(z) : \{\Re(z) < 0\} \to \{\Re(z) < 0\}\\
\]
We asked that \(g(z)\) is bounded on this domain as well. And additionally we asked that \(g(i\mathbb{R})\) was bounded.
From there, we asked that \(g(0) = 0\), and additionally had an asymptotic expansion at zero.
This gave us \(F(z)\) is meromorphic for \(\Re(z) < 0\), has simple poles with simple residues \((-1)^k g_k\) when \(z = -k\); And ensured the decay in the imaginary argument \(F(x+iy) = O(e^{-\frac{\pi}{2}|y|})\).
From here, we are just taking a Fourier transform/Inverse mellin transform of \(F(x+iy)/\Gamma(1-x-iy)\).
The really hard part, I'm not sure about when generalizing, is that \(g^{(N)}(x)\) looks like \(M N! x^{-N}\). This should definitely happen in some form in the general case, but it might be a difficult argument to follow through generally. I kind of drew from fractional calculus here. Which is that if a function is differintegrable; it must grow at worst like the Gamma function. This is probably the only weak point in my proof; but it is correct. It's just not as clean as everything else.
So to answer your question. Can you make a half iterate \(g(g(x)) = xe^x\) that satisfies all of the above? If so, yeah it could work. I think \(e^z-1\) was kind of a perfect storm though. You're going to have to do more legwork for other functions.
EDIT: Personally, I would say that we can always construct a function \(g\) on a petal of \(f\) about a parabolic fixed point; and then we can map that domain to the Left half plane with the fixedpoint on the boundary--whereupon you just do everything I did above; but on a different space of functions.
I mean upto change of variables; I believe this is gneralizable.
The main property I used, was mentioned by David E speyer; which is that:
\[
g(z) : \mathbb{C}/[0,\infty) \to \mathbb{C}\\
\]
And which for our purposes was reduced to :
\[
g(z) : \{\Re(z) < 0\} \to \{\Re(z) < 0\}\\
\]
We asked that \(g(z)\) is bounded on this domain as well. And additionally we asked that \(g(i\mathbb{R})\) was bounded.
From there, we asked that \(g(0) = 0\), and additionally had an asymptotic expansion at zero.
This gave us \(F(z)\) is meromorphic for \(\Re(z) < 0\), has simple poles with simple residues \((-1)^k g_k\) when \(z = -k\); And ensured the decay in the imaginary argument \(F(x+iy) = O(e^{-\frac{\pi}{2}|y|})\).
From here, we are just taking a Fourier transform/Inverse mellin transform of \(F(x+iy)/\Gamma(1-x-iy)\).
The really hard part, I'm not sure about when generalizing, is that \(g^{(N)}(x)\) looks like \(M N! x^{-N}\). This should definitely happen in some form in the general case, but it might be a difficult argument to follow through generally. I kind of drew from fractional calculus here. Which is that if a function is differintegrable; it must grow at worst like the Gamma function. This is probably the only weak point in my proof; but it is correct. It's just not as clean as everything else.
So to answer your question. Can you make a half iterate \(g(g(x)) = xe^x\) that satisfies all of the above? If so, yeah it could work. I think \(e^z-1\) was kind of a perfect storm though. You're going to have to do more legwork for other functions.
EDIT: Personally, I would say that we can always construct a function \(g\) on a petal of \(f\) about a parabolic fixed point; and then we can map that domain to the Left half plane with the fixedpoint on the boundary--whereupon you just do everything I did above; but on a different space of functions.
I mean upto change of variables; I believe this is gneralizable.