On equivalence of compound operators to the operator of differentiation (in Ukrainian)

Let $A_R$ ($0 < R < \infty$) be a space of all univalent and analytic in the disc $|z| < R$ functions with uniform convergence topology and $D = d/dz$ be a usual differentiation operator in the $A_R.$ There is also considered the operator $P : (Pf)(z) = f(-z)$ ($\forall f \in A_R$) and $L = \alpha(D) + \beta(D)P,$ where $\alpha(\lambda)$ and $\beta(\lambda)$ are given entire function from the class $[1,0]$ respectively for $R \ll \infty$. Under such assumptions the opearator $L$ of this form is linear continuous mapping of the space $A_R$ into itself, wchich commute with the operator $D^2$ in the space. Moreover, the operator of this form (they are called compound) exhaust the set of linear continuous operators, which commutes with $D^2$ and their matrices in power basis $\{z^n\}_{n=0}^\infty$ of the space $A_R$ are uppertriangular. The goal of this paper is to find conditions of existence such isomorphism $\tau$ of the space $A_R$ onto itself that $TL = DT$ (the property is denited by $L \sim D$).