Another way to continuum sum!

[JmsNxn]

Just for curiosity, but in other words the operator \( \mathcal {L} \) conjugates the differential operator and the difference operator?

That is the same as

\( \Delta \circ \mathcal {L}=\mathcal {L} \)\( \circ {D} \)

So \( \mathcal {L} \) is like a the solution of an "abel functional equation for operators"...


Maybe if you can find an operator \( \mathcal {V} \) such that

\( \Sigma \circ \mathcal {V}=\mathcal {V} \)\( \circ {D} \)

where \( \Sigma f(x):=f(S(f^{-1}(x))) \)

(aka an operator that conujugates the differential operator and the subfunction operator)
you could conjugate the fractional differentiation of a function by \( \mathcal {V} \) and obtain a fractional iteration of the subfunction operator:

\( \Sigma^{\circ\sigma} \circ \mathcal {V}=\mathcal {V} \)\( \circ {D^{\circ\sigma}} \)

\( \Sigma^{\circ\sigma}=\mathcal {V} \)\( \circ {D^{\circ\sigma}} \circ \mathcal {V^{-1}} \)

Do you think it is possible to find shuch \( \mathcal {V} \)?

Yep, but it's not linear and it requires a lot more sophistication to it than in the form I have. And before you get ahead of yourself, it looks as though it WILL NOT work on hyper operators. Unless some very magical theorems fall from the air. It's hard enough getting it to work for tetration let alone all functions.