Geodesic
Representation of functions by generalized exponential series
Sbornik. Mathematics, Tome 62 (1989) no. 2, pp. 491-505 
A. F. Leont'ev. Representation of functions by generalized exponential series. Sbornik. Mathematics, Tome 62 (1989) no. 2, pp. 491-505. http://geodesic.mathdoc.fr/item/SM_1989_62_2_a9/
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     author = {A. F. Leont'ev},
     title = {Representation of functions by generalized exponential series},
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     year = {1989},
     volume = {62},
     number = {2},
     language = {en},
     url = {http://geodesic.mathdoc.fr/item/SM_1989_62_2_a9/}
}
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Voir la notice de l'article provenant de la source Math-Net.Ru

Let $f(z)=\sum_0^\infty\frac{a_k}{k!}z^k$ be an entire function of exponential type, $\gamma(t)=\sum_0^\infty\frac{a_k}{t^{k+1}}$; let the singular points of $\gamma(t)$ lie in the disk $|t|\leqslant1$, let $t=1$ be a singular point of $\gamma(t)$. By definition, $f\in A_0$ if every function $\Phi(z)$ that is analytic in a convex domain $D$, $0\in D$, can be represented in the form $\Phi(z)=\sum_1^\infty c_kf(\lambda_kz )$ with $\lim_{k\to \infty}\frac{\ln k}{\lambda_k}=0$. It was established previously that if the singular points of $\gamma(t)$ and of $\gamma_1(t)=\sum_0^\infty\frac1{a_kt^{k+1}}$ lie on $[0,1]$, then $f\in A_0$. The following is now established: under the stated conditions, $f(z)$ is a function of completely regular growth in the half-plane $\operatorname{Re}z\geqslant0$; if $f\in A_0$ and $f(z)$ is of completely regular growth in $\operatorname{Re}z\geqslant0$, then the singular points of $\gamma(t)$ and of $\gamma_1(t)$ lie on $[0,1]$. Bibliography: 8 titles.

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