[MSE-SOLVED] Subfunction is functorial!!!!

[JmsNxn]

Yes! That was it; I'm glad I don't have to go searching for it. I'm gonna spend some time fiddling with,

  I expected to blow your mind dropping those analogies... xD but you are cold as ice. I mean, those regularities are fire... something magic.

\(
(f,x^{\overline{z}}) = \Gamma(z) \frac{d^{-z}}{dw^{-z}}\vartheta\\
(Hf,x^{\overline{z}}) = \Gamma(z) \frac{d^{-z}}{dw^{-z}}H\vartheta\\
\)
...
We are assuming that \( |f(z)| < M \) is bounded for \( \Re(z) > 0 \). We are assuming it takes \( \mathbb{R}^+ \to \mathbb{R}^+ \); and additionally that \( f : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C}_{\Re(z) > 0} \). This is enough for these transforms to converge.

Mmhh but what  \( x^{\overline{z}} \) is? I can't follow properly.

if \( z = a+ib \) then \( \overline{z} = a-ib \) then \( x^{\overline{z}} = x ^{a-ib} \). Sorry, I should've been clearer.

Then this is just a strange way of writing the differintegral,

\(
(f,g) = \int_0^\infty f(x)\overline{g(x)}\,dx/x\\
(f,x^{\overline {z}}) = \int_0^\infty f(x)x^{z-1}\,dx\\
\)

Which is; as we've talked about before; the Kernel of the differintegral. And I wanna see if we can do something to the kernel to get the \( \uparrow \) operator somehow. Not too sure if this is possible; as it may imply linearity if done too obviously. And we know it isn't linear.

I'm just trying to think of ways we can make \( (\frac{d^{z}}{dw^z}|_{w=0} H)^n \) for complex \( n \); where for natural n it just makes a hyperoperation-chain.


....

I apologize for not being so surprised. I had learned all of that before, I just needed a reminder of its exact shape