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 \( \mathcal {V} \)?
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 \( \mathcal {V} \)?
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)