Application of expansions of entire functions in series of exponentials

The equality
\begin{equation} \varlimsup_{r\to\infty}\frac{\ln|F(re^{i\varphi})|}{r^\rho}=\varlimsup_{r\to\infty}\frac{\ln\Phi(re^{i\varphi})}{r^\rho} \end{equation}
is established for those values of $\varphi$ for which the left-hand side is nonnegative. Here $F(z)=\sum_1^\infty a_ke^{\lambda_kz}$, $\Phi(z)=\sum_1^\infty |a_ke^{\lambda_kz}|$, $\rho>1$. It is assumed that the $\lambda_k$ ($k\geqslant1$) are the zeros of an entire function $L(\lambda)\in[\rho_1,0]$ ($1/\rho+1/\rho_1=1$), that
$$ \lim_{k\to\infty}\frac1{|\lambda_k|^{\rho_1}}\ln\biggl|\frac1{L'(\lambda_k)}\biggr|=0 $$
and that the right-hand side of (1) is finite. It follows from this result that the indicator $h_F(\varphi)$ of $F(z)$ is determined by the moduli of the coefficients $a_k$.
The equation
\begin{equation} \sum_0^\infty c_k F^{(k)}(z)=f(z)\qquad\biggl(\sum_0^\infty c_k\lambda^k=L(\lambda)\biggr) \end{equation}
is also considered. Let $0<H(\varphi)<\infty$ and let $H(\varphi)r^\rho$ be a convex function of $z=re^{i\varphi}$. If $h_f(\varphi)\leqslant H(\varphi)$ ($h_f(\varphi)$ is the indicator of $f(z)$ for order $\rho$) then equation (2) has a solution with $h_F(\varphi)\leqslant H(\varphi)$. It is shown by using the results stated above that there are not always solutions of (2) satisfying the condition $h_F(\varphi)\leqslant h_f(\varphi)$.
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