Hey, Tommy!
Not a problem. I'll explain again. I made an edit to the post above after you had already posted. I'll put it here.
TO clarify.
The correct way to say it, is if \(F\) is holomorphic in the right half plane:
\[
|F(z)| \le e^{\rho|\Re(z)| + \tau|\Im(z)|}\,\,\text{for}\,\,\tau,\rho>0\,\,\tau < \pi/2\\
\]
Then; as you call the gamma interpolation, is equivalent to \(F\). So that \(F(z)\) is fully determined by its behaviour on the naturals.
Suppose that:
\[
EF(n) = F(n+1)\\
\]
And let's also suppose that:
\[
EF(z)\,\,\text{has the same bounds as above}\\
\]
Then:
\[
EF(z) = F(z+1)\\
\]
This is because:
\[
EF(z) - F(z+1) = \frac{d^{z-1}}{dw^{z-1}}\Big{|}_{w=0} \sum_{k=0}^\infty \left(EF(k+1) - F(k+2)\right)\frac{w^k}{k!} = \frac{d^{z-1}}{dw^{z-1}} 0 = 0\\
\]
This works for composition, because composition is a linear operator. Write \(Ez = g(z)\) locally about fixed points with real positive multipliers (it works with complex multipliers but it's very tricky). Then the above "gamma interpolation" of \(g^{\circ n}\) just produces the standard Schroder iteration about that fixed point. This is apparent without doing any work at all. the function \(g^{\circ z}\), using the Schroder iteration and keeping the multiplier real positive, satisfies the above bounds I wrote, and is holomorphic in a right half plane. So we just use Ramanujan's master theorem. Absolutely it satisfies the equation.
Now Applying this logic to the above, if \(\alpha \uparrow^s z\) is in this Ramanujan space (across \(s\)), as well as \(\alpha \uparrow^{s-1}\left(\alpha \uparrow^s z\right)\). Then, their difference is certainly in this ramanujan space. Then:
\[
\alpha \uparrow^s z - \alpha \uparrow^{s-1}\left(\alpha \uparrow^s z\right) = \frac{d^{s-2}}{dw^{s-2}}\Big{|}_{w=0}\sum_{k=0}^\infty \left(\alpha \uparrow^{k+2} z - \alpha \uparrow^{k+1}\left(\alpha \uparrow^{k+2} z\right)\right)\frac{w^k}{k!} = \frac{d^{z-1}}{dw^{z-1}} 0 = 0\\
\]
So since they agree on the naturals, they agree continuously. The downside is that we need to ensure that:
\[
\alpha \uparrow^{s-1}\left(\alpha \uparrow^s z\right)\\
\]
Is in this Ramanujan space. The closest I ever got to showing this, was showing that:
\[
\alpha \uparrow^{s-1} \alpha = \alpha \uparrow^s 2\\
\]
This is actually manageable, to show this isn't too out there. But even then, it requires showing that the "gamma interpolation" works for hyper-operators. Where last I looked at it, there was a lemma that escaped being proven which held it together... I'd have to go over my old notes, But I believe I reduced the problem into showing that the following converges:
\[
F(s) = \frac{d^{s-2}}{dw^{s-2}}\sum_{n=0}^\infty \alpha \uparrow^{n+2} \infty \frac{w^n}{n!}\\
\]
So if you can interpolate the fixed points of the bounded hyperoperators, you can interpolate the hyperoperators themselves. This is very doable, because as you increase \(s\) you get closer and closer to \(\alpha\).
Not a problem. I'll explain again. I made an edit to the post above after you had already posted. I'll put it here.
TO clarify.
The correct way to say it, is if \(F\) is holomorphic in the right half plane:
\[
|F(z)| \le e^{\rho|\Re(z)| + \tau|\Im(z)|}\,\,\text{for}\,\,\tau,\rho>0\,\,\tau < \pi/2\\
\]
Then; as you call the gamma interpolation, is equivalent to \(F\). So that \(F(z)\) is fully determined by its behaviour on the naturals.
Suppose that:
\[
EF(n) = F(n+1)\\
\]
And let's also suppose that:
\[
EF(z)\,\,\text{has the same bounds as above}\\
\]
Then:
\[
EF(z) = F(z+1)\\
\]
This is because:
\[
EF(z) - F(z+1) = \frac{d^{z-1}}{dw^{z-1}}\Big{|}_{w=0} \sum_{k=0}^\infty \left(EF(k+1) - F(k+2)\right)\frac{w^k}{k!} = \frac{d^{z-1}}{dw^{z-1}} 0 = 0\\
\]
This works for composition, because composition is a linear operator. Write \(Ez = g(z)\) locally about fixed points with real positive multipliers (it works with complex multipliers but it's very tricky). Then the above "gamma interpolation" of \(g^{\circ n}\) just produces the standard Schroder iteration about that fixed point. This is apparent without doing any work at all. the function \(g^{\circ z}\), using the Schroder iteration and keeping the multiplier real positive, satisfies the above bounds I wrote, and is holomorphic in a right half plane. So we just use Ramanujan's master theorem. Absolutely it satisfies the equation.
Now Applying this logic to the above, if \(\alpha \uparrow^s z\) is in this Ramanujan space (across \(s\)), as well as \(\alpha \uparrow^{s-1}\left(\alpha \uparrow^s z\right)\). Then, their difference is certainly in this ramanujan space. Then:
\[
\alpha \uparrow^s z - \alpha \uparrow^{s-1}\left(\alpha \uparrow^s z\right) = \frac{d^{s-2}}{dw^{s-2}}\Big{|}_{w=0}\sum_{k=0}^\infty \left(\alpha \uparrow^{k+2} z - \alpha \uparrow^{k+1}\left(\alpha \uparrow^{k+2} z\right)\right)\frac{w^k}{k!} = \frac{d^{z-1}}{dw^{z-1}} 0 = 0\\
\]
So since they agree on the naturals, they agree continuously. The downside is that we need to ensure that:
\[
\alpha \uparrow^{s-1}\left(\alpha \uparrow^s z\right)\\
\]
Is in this Ramanujan space. The closest I ever got to showing this, was showing that:
\[
\alpha \uparrow^{s-1} \alpha = \alpha \uparrow^s 2\\
\]
This is actually manageable, to show this isn't too out there. But even then, it requires showing that the "gamma interpolation" works for hyper-operators. Where last I looked at it, there was a lemma that escaped being proven which held it together... I'd have to go over my old notes, But I believe I reduced the problem into showing that the following converges:
\[
F(s) = \frac{d^{s-2}}{dw^{s-2}}\sum_{n=0}^\infty \alpha \uparrow^{n+2} \infty \frac{w^n}{n!}\\
\]
So if you can interpolate the fixed points of the bounded hyperoperators, you can interpolate the hyperoperators themselves. This is very doable, because as you increase \(s\) you get closer and closer to \(\alpha\).