Hadamard Compositions of Gelfond–Leont’ev Derivatives

Abstract

For analytic functions $f_j(z)={\displaystyle \sum _{n=0}^{\infty }}{a}_{n,j}{z}^n$, $1\le j\le p$, the notion of a Hadamard composition ${(f_1\ast \dots \ast f_p)}_m={\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \sum _{k_1+\cdots +k_p=m}}{c}_{k_1\dots k_p}{a}_{n,1}^{k_1}·\dots ·a_{n,p}^{k_p}\right)z^n$ of genus m is introduced. The relationship between the growth of the Gelfond–Leont’ev derivative of the Hadamard composition of functions $f_j$ and the growth Hadamard composition of Gelfond–Leont’ev derivatives of these functions is studied. We found conditions under which these derivatives and the composition have the same order and a lower order. For the maximal terms of the power expansion of these derivatives, I describe behavior of their ratios.

1. Introduction

Let

$$f(z)=\sum _{n=0}^{\infty }{a}_n{z}^n$$

(1)

and

$$f_j(z)=\sum _{n=0}^{\infty }{a}_{n,j}{z}^n,1\le j\le p$$

(2)

be analytic functions. As in [1], I say that the function is similar to the Hadamard composition of the functions $f_j$ if $a_n=w(a_{n,1},\dots ,a_{n,p})$ for all n, where $w:{\mathbb{C}}^p\to \mathbb{C}$ is a continuous function. Clearly, if $p=2$ and $w(a_{n,1},a_{n,2})=a_{n,1}{a}_{n,2}$, then $f=(f_1\ast f_2)$ is [2] the Hadamard composition (product) of the functions $f_1$ and $f_2$. Obtained by J. Hadamard, the properties of this composition find the applications [3,4] in the theory of the analytic continuation of the functions represented by a power series.

Here, I consider the case when w is a homogeneous polynomial. Recall that a polynomial is named homogeneous if all monomials with nonzero coefficients have the identical degree. A polynomial $P(x_1,\dots ,x_p)$ is homogeneous to the degree m if, and only if, $P(t{x}_1,\dots ,t{x}_p)=t^{m}P(x_1,\dots ,x_p)$ for all t from the field above that a polynomial is defined. Function (1) is called a Hadamard composition of genus $m\ge 1$ of functions (2) if $a_n=P(a_{n,1},\dots ,a_{n,p})$, where

$$P(x_1,\dots ,x_p)=\sum _{k_1+\cdots +k_p=m}{c}_{k_1\dots k_p}{x}_1^{k_1}·\dots ·x_p^{k_p},k_j\in {\mathbb{Z}}_{+}$$

is a homogeneous polynomial of degree $m\ge 1$ with constant fixed coefficients $c_{k_1\dots k_p}$. I remark that the usual Hadamard composition is a special case of the Hadamard composition of the genus $m=2$. The Hadamard composition of genus $m\ge 1$ of functions $f_j$ I denote by ${(f_1\ast \dots \ast f_p)}_m$, i.e.,

$${(f_1\ast \dots \ast f_p)}_m(z)=\sum _{n=0}^{\infty }{a}_n{z}^n=\sum _{n=0}^{\infty }\left(\sum _{k_1+\cdots +k_p=m}{c}_{k_1\dots k_p}{a}_{n,1}^{k_1}·\dots ·a_{n,p}^{k_p}\right)z^n.$$

For a power series (1) with the convergence radius $R[f]\in [0,+\infty ]$ and a power series $l(z)={\displaystyle \sum _{n=0}^{\infty }}{l}_n{z}^n$ with the convergence radius $R[l]\in [0,+\infty ]$ and coefficients $l_n>0$ for all $n\ge 0$ the power series

$$D_l^{(k)}f(z)=\sum _{n=0}^{\infty }\frac{l_n}{l_{n+k}}{a}_{n+k}{z}^n$$

(3)

is called [5] Gelfond–Leont’ev derivative of the n-th order. If $l(z)=e^z$, then $D_l^{(k)}f(z)=f^{(k)}(z)$ is the usual derivative of the n-th order. The Gelfond–Leont’ev derivative is a very interesting object of investigations (see [6,7,8]). These derivatives found applications in the theory of univalent analytic functions. They allows researchers to describe the growth of these functions in other terms [7].

There are many papers on the Hadamard composition of analytic functions and the Dirichlet series [9,10,11]. For example, A. Gaisin and T. Belous [10] studied the maximal term of the Hadamard composition of the Dirichlet series with real exponents. Alower estimate for the sum of a Dirichlet series over a curve arbitrarily approaching the convergence line was obtained. Moreover, in [11] they established a criterion for the logarithm of the maximal term of a Dirichlet series whose absolute convergence domain is a half-plane to be equivalent to the logarithm of the maximal term of its Hadamard composition with another Dirichlet series of some class on the asymptotic set. S. Vakarchuk [9] investigated an interpolation problem for classes of analytic functions generated of the Hadamard compositions and obtained upper and lower bounds for various n-widths for these classes.

If $R[f]>0$, then, for $0\le r\le R[f]$, let $M(r,f)=max\{|f(z)|:|z|=r\}$ and $\mu (r,f)=max\{|a_n|r^n:n\ge 0\}$ be the maximal term of series (1). M. K. Sen [12,13] researched a connection between the growth of the maximal term of the derivative ${(f_1\ast f_2)}^{(k)}$ of the usual Hadamard composition $f_1\ast f_2$ of entire functions f and g and the growth of the maximal term of derivative $f_1^{(k)}\ast f_2^{(k)}$. In particular, he proved [13] that if the function has the order $\varrho$ and the lower order $\lambda$ then, for every $\epsilon >0$ and all $r\ge r_0(\epsilon )$,

$$r^{(k+2)\lambda -1-\epsilon }\le \frac{\mu (r,f^{(k+1)}\ast g^{(k+1)})}{\mu (r,{(f\ast g)}^{(k)})}\le r^{(k+2)\varrho -1+\epsilon }.$$

The research of M.K. Sen was continued in [14], where, instead of ordinary derivatives, the Gelfond–Leont’ev derivatives are considered. In particular, in [14] (see also [15] p. 128), the following lemma is proved.

Lemma 1

([14]). In order for an arbitrary series (1) the equalities $R[f]=+\infty$ and $R[D_l^{(k)}f]=+\infty$ to be equivalent, it is necessary and sufficient that

$$0<q=\underset{n\to \infty }{\underline{lim}}\sqrt[n]{l_n/l_{n+1}}\le \underset{n\to \infty }{\overline{lim}}\sqrt[n]{l_n/l_{n+1}}=Q<+\infty ,$$

(4)

and, for the equivalence of the equalities $R[f]=1$ and $R[D_l^{(k)}f]=1$, it is necessary and sufficient that

$$\underset{n\to \infty }{lim}\sqrt[n]{l_n/l_{n+1}}=1.$$

(5)

The generalization of the results from [14] to the case of Hadamard compositions of genus $m\ge 1$ has become an actual problem. It allows researchers to study the growth properties of these function classes and consider their applications in geometric function theory as it is achieved for usual the Gelfond–Leont’ev derivatives of univalent analytic functions in [6,7,8].

2. Convergence of Hadamard Compositions of Genus m

Clearly, if function (1) is a Hadamard composition of genus m of functions (2) then

$$|a_n|\le \sum _{k_1+\cdots +k_p=m}|c_{k_1\dots k_p}||a_{n,1}{|}^{k_1}·\dots ·{|a_{n,p}|}^{k_p}.$$

(6)

From $R[f]$, I denote the radius of the convergence of series (1) and suppose that $R[f_j]=R>0$ for all $1\le j\le p$. Then from the Cauchy–Hadamard formula, I have $\underset{n\to \infty }{\overline{lim}}\sqrt[n]{|a_{n,j}|}=1/R[f_j]=1/R$ and, thus, $|a_{n,j}{|\le (1/R+\epsilon )}^n$ for every $\epsilon >0$ and all $n\ge n_0(\epsilon )$. Therefore, (6) implies

$$|a_n|\le \sum _{k_1+\cdots +k_p=m}|c_{k_1\dots k_p}|{\left(\frac{1}{R}+\epsilon \right)}^{n{k}_1}·\dots ·{\left(\frac{1}{R}+\epsilon \right)}^{n{k}_p}={\left(\frac{1}{R}+\epsilon \right)}^{nm}\sum _{k_1+\cdots +k_p=m}|c_{k_1\dots k_p}|,$$

whence, $\sqrt[n]{|a_n|}\le (1+o(1)){(1/R+\epsilon )}^m$ as $n\to \infty$, i.e., $1/R[f]\le {(1/R+\epsilon )}^m$. In view of the arbitrariness of $\epsilon$, I obtain the inequality $R[F]\ge R^m$.

Hence, it follows that, if $R[f_j]=+\infty$ for all j, then $R[{(f_1\ast \dots \ast f_p)}_m]=+\infty$, and, if $R[f_j]=1$ for all j, then $R[{(f_1\ast \dots \ast f_p)}_m]\ge 1$.

In order for $R[{(f_1\ast \dots \ast f_p)}_m]=1$, additional conditions on $a_{n,j}$ are required. For example, I say that the function $f_1$ is dominant if $|c_{m0\dots 0}||a_{n,1}{|}^m>0$ for all $n\ge 0$ and $|a_{n,j}|=o(|a_{n,1}|)$ as $n\to \infty$ for all $2\le j\le p$.

We put

$${\Sigma }_n^{\prime }=\sum _{k_1+\cdots +k_p=m,k_1\ne m}{c}_{k_1\dots k_p}{a}_{n,1}^{k_1}·\dots ·a_{n,p}^{k_p}=\sum _{k_1+\cdots +k_p=m}{c}_{k_1\dots k_p}{a}_{n,1}^{k_1}·\dots ·a_{n,p}^{k_p}-c_{m0\dots 0}{a}_{n,1}^m.$$

Since, for each monomial of the polynomial ${\Sigma }_n^{\prime }$, the sum of the exponents is equal to m, I have

$${\displaystyle \frac{|a_{n,1}{|}^{k_1}·\dots ·{|a_{n,p}|}^{k_p}}{|a_{n,1}{|}^m}}={\displaystyle \frac{|a_{n,2}{|}^{k_2}·\dots ·{|a_{n,p}|}^{k_p}}{|a_{n,1}{|}^{m-k_1}}}\to 0,n\to \infty ,$$

and, thus, ${\Sigma }_n^{\prime }=o(|a_{n,1}{|}^m)$ as $n\to +\infty$.

Since

$$|c_{m0\dots 0}||a_{n,1}{|}^m-|{\Sigma }_n^{\prime }|\le |a_n|\le |c_{m0\dots 0}||a_{n,1}{|}^m+|{\Sigma }_n^{\prime }|,$$

we have $|a_n|=(1+o(1))|c_{m0\dots 0}||a_{n,1}{|}^m$ as $n\to \infty$, whence

$$1=\frac{1}{R[f_j]}\le \frac{1}{R[f_1]}=\underset{n\to \infty }{\underline{lim}}\sqrt[nm]{\frac{|a_n|}{|c_{m0\dots 0}|}}=\underset{n\to \infty }{\underline{lim}}{\left(\sqrt[n]{|a_n|}\right)}^{1/m}={\left(\frac{1}{R[f]}\right)}^{1/m},$$

i.e., $R[{(f_1\ast \dots \ast f_p)}_m]\le 1$ and, thus, $R[{(f_1\ast \dots \ast f_p)}_m]=1$.

It is easy to check that

$$D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m(z)=\sum _{n=0}^{\infty }\frac{l_n}{l_{n+k}}\left(\sum _{k_1+\cdots +k_p=m}{c}_{k_1\dots k_p}{a}_{n+k,1}^{k_1}·\dots ·a_{n+k,p}^{k_p}\right)z^n$$

(7)

is the Gelfond–Leont’ev derivative of the Hadamard composition of genus m, and

$${(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m(z)=\sum _{n=0}^{\infty }{\left(\frac{l_n}{l_{n+k}}\right)}^m\left(\sum _{k_1+\cdots +k_p=m}{c}_{k_1\dots k_p}{a}_{n+k,1}^{k_1}·\dots ·a_{n+k,p}^{k_p}\right)z^n$$

(8)

is the Hadamard composition of genus m of the Gelfond–Leont’ev derivatives.

Lemma 2.

If condition (4) holds, then the equalities $R[{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m]=+\infty$ and $R[D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]=+\infty$ are equivalent, and, if condition (5) holds, then the equalities $R[{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m]=1$ and $R[D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]=1$ are equivalent.

Proof. 

Indeed, using the Cauchy–Hadamard formula from (7) and (8) with

$$a_{n+k}=\sum _{k_1+\cdots +k_p=m}{c}_{k_1\dots k_p}{a}_{n+k,1}^{k_1}·\dots ·a_{n+k,p}^{k_p},$$

(9)

we have

$$\frac{1}{R[{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m]}=\underset{n\to \infty }{\overline{lim}}\sqrt[n]{|a_{n+k}|\frac{l_n}{l_{n+k}}{\left(\frac{l_n}{l_{n+1}}\right)}^{m-1}}\ge$$

$$\ge \underset{n\to \infty }{\overline{lim}}\sqrt[n]{|a_{n+k}|\frac{l_n}{l_{n+k}}}\underset{n\to \infty }{\underline{lim}}\sqrt[n]{{\left(\frac{l_n}{l_{n+1}}\right)}^{m-1}}=\frac{1}{R[D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]}\underset{n\to \infty }{\underline{lim}}{\left(\frac{l_n}{l_{n+k}}\right)}^{(m-1)/n}$$

and, similarly,

$$\frac{1}{R[{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m]}\le \frac{1}{R[D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]}\underset{n\to \infty }{\overline{lim}}{\left(\frac{l_n}{l_{n+k}}\right)}^{(m-1)/n}.$$

The last inequality yields the validity of Lemma 2. □

3. Hadamard Compositions of Gelfond–Leont’ev Derivatives of Entire Functions

We will remind that the most widely-used descriptions of entire transcendental function f are the lower order $\lambda [f]=\underset{r\to +\infty }{\underline{lim}}{\displaystyle \frac{lnlnM(r,f)}{lnr}}$ and the order $\varrho [f]=\underset{r\to +\infty }{\overline{lim}}{\displaystyle \frac{lnlnM(r,f)}{lnr}}$. In view of the Cauchy inequality, I have

$$\mu (r,f)\le M(r,f)\le \sum _{n=0}^{\infty }|a_n|{(2r)}^n{2}^{-n}\le 2\mu (2r,f),$$

whence it follows that $\lambda [f]=\underset{r\to +\infty }{\underline{lim}}{\displaystyle \frac{lnln\mu (r,f)}{lnr}}$ and $\varrho [f]=\underset{r\to +\infty }{\overline{lim}}{\displaystyle \frac{lnln\mu (r,f)}{lnr}}$.

Proposition 1.

If condition (4) holds, then for every k

$$\lambda [{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m]=\lambda [D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]=\lambda [{(f_1\ast \dots \ast f_p)}_m]$$

and

$$\varrho [{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m]=\varrho [D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]=\varrho [{(f_1\ast \dots \ast f_p)}_m].$$

Proof. 

At first, let $k=1$. From condition (4), the existence of the numbers $0<q_1\le q_2<+\infty$ follows such that $q_1^n\le l_n/l_{n+1}\le q_2^n$ for all $n\ge 0$. Therefore, using (9), I obtain

$$r\mu (r,D_l^{(1)}{(f_1\ast \dots \ast f_p)}_m)=max\left\{\frac{l_n}{l_{n+1}}|a_{n+1}|r^{n+1}:n\ge 0\right\}\le$$

$$\le \frac{1}{q_2}max\left\{|a_{n+1|}{(q_{2}r)}^{n+1}:n\ge 0\right\}\le \frac{\mu (q_{2}r,{(f_1\ast \dots \ast f_p)}_m)}{q_2}$$

and, by analogy,

$$r\mu (r,D_l^{(1)}{(f_1\ast \dots \ast f_p)}_m)\ge \frac{\mu (q_{1}r,{(f_1\ast \dots \ast f_p)}_m)}{q_1}$$

for all large-enough r.

$lnr=o(ln\mu (r,f))$ as $r\to +\infty$ for each entire transcendental function, hence, I obtain

$$\begin{array}{c}(1+o(1))ln\mu (q_{1}r,{(f_1\ast \dots \ast f_p)}_m)\le ln\mu (r,D_l^{(1)}{(f_1\ast \dots \ast f_p)}_m)\le \\ \le (1+o(1))ln\mu (q_{2}r,{(f_1\ast \dots \ast f_p)}_m)\end{array}$$

as $r\to +\infty$. Hence, it follows that $\lambda [D_l^{(1)}{(f_1\ast \dots \ast f_p)}_m]=\lambda [{(f_1\ast \dots \ast f_p)}_m]$ and $\varrho [D_l^{(1)}{(f_1\ast \dots \ast f_p)}_m]=\varrho [{(f_1\ast \dots \ast f_p)}_m]$.

Since $D_l^{(k+1)}f(z)=D_l^{(1)}{D}_l^{(k)}f(z)$, the equalities $\lambda [D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]=\lambda [{(f_1\ast \dots \ast f_p)}_m]$ and $\varrho [D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]=\varrho [{(f_1\ast \dots \ast f_p)}_m]$ are proved.

On the other hand,

$$\mu (r,{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m)=max\left\{{\left(\frac{l_n}{l_{n+k}}\right)}^{m-1}\frac{l_n}{l_{n+k}}|a_{n+k}|r^n:n\ge 0\right\}\ge$$

$$\ge max\left\{q_1^{n(m-1)}\frac{l_n}{l_{n+k}}|a_{n+k}|r^n:n\ge 0\right\}=max\left\{\frac{l_n}{l_{n+k}}|a_{n+k}|{(q_1^{(m-1)}r)}^n:n\ge 0\right\}=$$

$$=\mu (q_1^{(m-1)}r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)$$

and, by analogy,

$$\mu (r,{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m)\le \mu (q_1^{(m-1)}r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m),$$

whence, as above, I obtain $\lambda [{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m]=\lambda [D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]$ and $\varrho [(D_l^{(k)}{f}_1$$\ast \dots \ast D_l^{(k)}{f}_p{)}_m]=\varrho [D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]$. □

Now let me establish a connection between the growth of a function $f={(f_1\ast \dots \ast f_p)}_m$ and the growth of functions $f_j$. Since $|a_n|\le {\displaystyle \sum _{k_1+\cdots +k_p=m}}|c_{k_1\dots k_p}||a_{n,1}{|}^{k_1}·\dots ·{|a_{n,p}|}^{k_p}$, I have

$$|a_n|r^{mn}\le \sum _{k_1+\cdots +k_p=m}|c_{k_1\dots k_p}|(|a_{n,1}|r^n{)}^{k_1}·\dots ·(|a_{n,p}{|r^n)}^{k_p},$$

i.e.,

$$\mu (r^m,f)\le \sum _{k_1+\cdots +k_p=m}|c_{k_1\dots k_p}|\mu {(r,f_1)}^{k_1}·\dots ·\mu {(r,f_p)}^{k_p},$$

whence, for all r large enough, I get

$$ln\mu (r^m,f)\le \sum _{k_1+\cdots +k_p=m}ln(|c_{k_1\dots k_p}|\mu {(r,f_1)}^{k_1}·\dots ·\mu {(r,f_p)}^{k_p})+K_1=$$

$$=\sum _{k_1+\cdots +k_p=m}(ln|c_{k_1\dots k_p}|+k_{1}ln\mu (r,f_1)+\dots +k_{p}ln\mu (r,f_p))+K_1\le$$

$$\le \sum _{k_1+\cdots +k_p=m}(k_{1}ln\mu (r,f_1)+\dots +k_{p}ln\mu (r,f_p))+K_2,(K_j\equiv \mathrm{const}>0).$$

(10)

Let $max\{\varrho [f_j]:1\le j\le p\}=\varrho <+\infty$. Then, $ln\mu (r,f_j)\le r^{\varrho +\epsilon }$ for every $\epsilon >0$ all $r\ge r_0(\epsilon )$ and all j. Therefore, in view of (10)

$$ln\mu (r^m,f)\le r^{\varrho +\epsilon }\sum _{k_1+\cdots +k_p=m}(k_1+\dots +k_p)+K_2,$$

whence

$$\varrho [f]=\underset{r\to +\infty }{\overline{lim}}\frac{lnln\mu (r^m,f)}{ln{r}^m}=\underset{r\to +\infty }{\overline{lim}}\frac{lnln\mu (r^m,f)}{mlnr}\le \frac{\varrho +\epsilon }{m},$$

and, in view of the arbitrariness of $\epsilon$, I obtain $m\varrho [f]\le \varrho$.

Suppose now that the function $f_1$ is dominant. Then $|a_n|=(1+o(1))|c_{m0\dots 0}||a_{n,1}{|}^m$ as $n\to \infty$ and, thus, $|a_n|r^{mn}=(1+o(1))|c_{m0\dots 0}|(|a_{n,1}{|r^n)}^m$ as $n\to \infty$, whence it follows that

$$A_1\mu {(r,f_1)}^m\le \mu (r^m,f)\le A_2\mu {(r,f_1)}^m,(A_1,A_2=\mathrm{const}>0).$$

(11)

Using (11), as above, I obtain $m\varrho [f]=\varrho [f_1]$ and $m\lambda [f]=\lambda [f_1]$. Thus, the following statement is proved.

Proposition 2.

Let $f_1,\dots ,f_p$ be entire transcendental functions, and condition (4) holds. If $\varrho [f_1]=\dots =\varrho [f_p]=\varrho$, then $m\varrho [{(f_1\ast \dots \ast f_p)}_m]\le \varrho$, and, if among functions $f_1,\dots ,f_p$ there is a dominant function $f_1$, then $m\varrho [{(f_1\ast \dots \ast f_p)}_m]=\varrho$. Moreover, if among functions $f_1,\dots ,f_p$ there is a dominant function $f_1$ then $m\lambda [{(f_1\ast \dots \ast f_p)}_m]=\lambda [f_1]$.

Let us now examine the growth of the ratio ${\displaystyle \frac{\mu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}}$, $j\ge k$. Let $\nu (r,f)=max\{n:|a_n|r^n=\mu (r,f)\}$ be the central index of series (1). Then, $\mu (r,f)=|a_{\nu (r,f)}|r^{\nu (r,f)}$ and, therefore,

$$\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)=\frac{l_{\nu (r,D_l^{(k)}f)}}{l_{\nu (r,D_l^{(k)}f)+k}}|a_{\nu (r,D_l^{(k)}f)+k}|r^{\nu (r,D_l^{(k)}f)}=$$

$$={\left(\frac{l_{\nu (r,D_l^{(k)}f)+k}}{l_{\nu (r,D_l^{(k)}f+k-j)}}\right)}^{m-1}\frac{l_{\nu (r,D_l^{(k)}f)}}{l_{\nu (r,D_l^{(k)}f)+k-j}}\times$$

$$\times {\left(\frac{l_{\nu (r,D_l^{(k)}f)+k-j}}{l_{\nu (r,D_l^{(k)}f)+k-j+j}}\right)}^m|a_{\nu (r,D_l^{(k)})+k-j+j}|r^{\nu (r,D_l^{(k)}f)+k-j}{r}^{j-k}\le$$

$$\le {\left(\frac{l_{\nu (r,D_l^{(k)}f)+k}}{l_{\nu (r,D_l^{(k)}f)+k-j}}\right)}^{m-1}\frac{l_{\nu (r,D_l^{(k)}f)}}{l_{\nu (r,D_l^{(k)}f)+k-j}}{r}^{j-k}\mu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)$$

and

$$\mu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)=$$

$$={\left(\frac{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m}}{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m+j}}\right)}^m|a_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)+j}|r^{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m}=$$

$$={\left(\frac{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m}}{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m+j}}\right)}^m|a_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)+j-k+k}|r^{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m+j-k}{r}^{k-j}$$

$$={\left(\frac{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m}}{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m+j}}\right)}^{m-1}\frac{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m}}{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m+j-k}}\times$$

$$\times \frac{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m+j-k}}{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m+j}}|a_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)+j-k+k}|r^{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m+j-k}{r}^{k-j}\le$$

$$\le {\left(\frac{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m}}{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m+j}}\right)}^{m-1}\frac{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m}}{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m+j-k}}{r}^{k-j}\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m).$$

Hence, it follows that

$${\left(\frac{l_{\nu (r,D_l^{(k)}f)+k-j}}{l_{\nu (r,D_l^{(k)}f)+k}}\right)}^{m-1}\frac{l_{\nu (r,D_l^{(k)}f)+k-j}}{l_{\nu (r,D_l^{(k)}f)}}\le \frac{r^{j-k}\mu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}\le$$

$$\le {\left(\frac{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m}}{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m+j}}\right)}^{m-1}\frac{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m}}{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m+j-k}}.$$

(12)

Using (12), I prove such a theorem.

Theorem 1.

Let $f_j$ be entire transcendental functions, $1\le j\le p$, and (4) hold with $q>1$. Then, for each $j\ge k$

$$\underset{r\to +\infty }{\underline{lim}}\frac{1}{lnr}lnln\frac{r^{j-k}\mu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}=\lambda [{(f_1\ast \dots \ast f_p)}_m]$$

(13)

and

$$\underset{r\to +\infty }{\overline{lim}}\frac{1}{lnr}lnln\frac{r^{j-k}\mu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}=\varrho [{(f_1\ast \dots \ast f_p)}_m].$$

(14)

Proof. 

From condition (4), with $q>1$ the existence of numbers $1<q_1\le q_2<+\infty$, it follows such that $q_1^{nk}\le l_n/l_{n+k}\le q_2^{nk}$ for all $n\ge n_0$. Therefore,

$${\left(\frac{l_{\nu (r,D_l^{(k)}f)+k-j}}{l_{\nu (r,D_l^{(k)}f)+k}}\right)}^{m-1}\frac{l_{\nu (r,D_l^{(k)}f)+k-j}}{l_{\nu (r,D_l^{(k)}f)}}\ge q_1^{(m-1)(\nu (r,D_l^{(k)}f)+k-j)j}{q}_1^{(\nu (r,D_l^{(k)}f)+k-j)(j-k)}=$$

$$=q_1^{(mj-k)(\nu (r,D_l^{(k)}f)+k-j)},$$

$${\left(\frac{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}}{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)+j}}\right)}^{m-1}\frac{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}}{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)+j-k}}\le$$

$$q_2^{(m-1)\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}{q}_2^{(j-k)\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}=q_2^{(jm-k)\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)},$$

and, thus, (12) implies

$$(mj-k)(\nu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)+k-j)ln{q}_1\le ln\frac{r^{j-k}\mu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}\le$$

$$\le (mj-k)\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)ln{q}_2.$$

(15)

It is well known ([16], p. 13) that

$$ln\mu (r,f)=ln\mu (1,f)+\underset{1}{\overset{r}{\int }}\frac{\nu (t,f)}{t}dt.$$

(16)

Hence, I get $ln\mu (r,f)\le ln\mu (1,f)+\nu (r,f)lnr$, $ln\mu (r,f)\ge ln\mu (1,f)+\nu (r/2,f)ln2$ and, thus, $\lambda [f]=\underset{r\to +\infty }{\underline{lim}}{\displaystyle \frac{ln\nu (r,f)}{lnr}}$, $\varrho [f]=\underset{r\to +\infty }{\overline{lim}}{\displaystyle \frac{ln\nu (r,f)}{lnr}}$. Therefore, by Proposition 1

$$\underset{r\to +\infty }{\underline{lim}}{\displaystyle \frac{ln\nu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}{lnr}}=\lambda [D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]=\lambda [{(f_1\ast \dots \ast f_p)}_m],$$

$$\underset{r\to +\infty }{\overline{lim}}{\displaystyle \frac{ln\nu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}{lnr}}=\varrho [D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]=\varrho [{(f_1\ast \dots \ast f_p)}_m],$$

and these equalities hold good if, instead of $D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m$, it is possible to put ${(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m$. From this and (15), I obtain (13) and (14). Theorem 1 is proved. □

For $j=k$, Theorem 1 implies the following statement.

Corollary 1.

Let $f_j$ be entire functions, $1\le j\le p$, and (4) holds with $q>1$. Then, for each $k\ge 1$

$$\underset{r\to +\infty }{\underline{lim}}\frac{1}{lnr}lnln\frac{\mu (r,{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}=\lambda [{(f_1\ast \dots \ast f_p)}_m]$$

and

$$\underset{r\to +\infty }{\overline{lim}}\frac{1}{lnr}lnln\frac{\mu (r,{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}=\varrho [{(f_1\ast \dots \ast f_p)}_m].$$

From Corollary 1 and Proposition 2, I obtain the following corollary.

Corollary 2.

Suppose that, among entire functions $f_1,\dots ,f_p$, there is a dominant function $f_1$. If condition (4) holds with $q>1$ then, for each $k\ge 1$,

$$\underset{r\to +\infty }{\underline{lim}}\frac{1}{lnr}lnln\frac{\mu (r,{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}=\frac{\lambda [f_1]}{m}$$

and

$$\underset{r\to +\infty }{\overline{lim}}\frac{1}{lnr}lnln\frac{\mu (r,{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}=\frac{\varrho [f_1]}{m}.$$

Let us now consider the case when, instead of condition (5), the stronger condition

$$0<\underset{n\to \infty }{\overline{lim}}\frac{l_n}{(n+1)l_{n+1}}\le \underset{n\to \infty }{\overline{lim}}\frac{l_n}{(n+1)l_{n+1}}<+\infty$$

(17)

is fulfilled.

Theorem 2.

Let $f_j$ be entire functions, $1\le j\le p$, and (17) hold, then, for each $j\ge k$,

$$\underset{r\to +\infty }{\underline{lim}}\frac{1}{lnr}ln\frac{\mu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}=(mj-k)\lambda [{(f_1\ast \dots \ast f_p)}_m]+k-j$$

(18)

and

$$\underset{r\to +\infty }{\overline{lim}}\frac{1}{lnr}ln\frac{\mu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}=(mj-k)\varrho [{(f_1\ast \dots \ast f_p)}_m]+k-j.$$

(19)

Proof. 

From condition (17), the existence of numbers $0<h_1\le h_2<+\infty$ follows such that ${(h_{1}n)}^k\le l_n/l_{n+k}\le {(h_{2}n)}^k$. Therefore,

$${\left(\frac{l_{\nu (r,D_l^{(k)}f)+k-j}}{l_{\nu (r,D_l^{(k)}f)+k}}\right)}^{m-1}\frac{l_{\nu (r,D_l^{(k)}f)+k-j}}{l_{\nu (r,D_l^{(k)}f)}}\ge$$

$$\ge h_1^{(m-1)j}{(\nu (r,D_l^{(k)}f)+k-j)}^{(m-1)j}{h}_1^{j-k}{(\nu (r,D_l^{(k)}f)+k-j)}^{j-k}=$$

$$=h_1^{mj-k}{(\nu (r,D_l^{(k)}f)+k-j)}^{mj-k},$$

$${\left(\frac{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}}{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)+j}}\right)}^{m-1}\frac{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}}{l_{\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)+j-k}}\le$$

$${(h_2\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m))}^{j(m-1)}{(h_2\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m))}^{j-k}=$$

$$={(h_2\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m))}^{mj-k},$$

and, thus, (12) implies

$$(mj-k)ln(\nu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m))+k-j)+(mj-k)ln{h}_1\le$$

$$\le (j-k)lnr+ln\frac{r^{j-k}\mu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}\le$$

$$\le (mj-k)ln(\nu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m+(mj-k)ln{h}_2.$$

Hence, as above, I obtain

$$j-k+\underset{r\to +\infty }{\underline{lim}}\frac{1}{lnr}ln\frac{\mu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}=(mj-k)\lambda [{(f_1\ast \dots \ast f_p)}_m]$$

and

$$j-k+\underset{r\to +\infty }{\overline{lim}}\frac{1}{lnr}ln\frac{\mu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}=(mj-k)\varrho [{(f_1\ast \dots \ast f_p)}_m],$$

i.e., (18) and (19) hold. Theorem 2 is proved. □

For $j=k+1$, Theorem 2 implies the following statement.

Corollary 3.

Let $f_j$ be entire functions, $1\le j\le p$, and (17) hold, then, for each $k\ge 1$,

$$\underset{r\to +\infty }{\underline{lim}}\frac{1}{lnr}ln\frac{\mu (r,{(D_l^{(k+1)}{f}_1\ast \dots \ast D_l^{(k+1)}{f}_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}=((m-1)k+m)\lambda [{(f_1\ast \dots \ast f_p)}_m]-1$$

and

$$\underset{r\to +\infty }{\overline{lim}}\frac{1}{lnr}ln\frac{\mu (r,{(D_l^{(k+1)}{f}_1\ast \dots \ast D_l^{(k+1)}{f}_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}=((m-1)k+m)\varrho [{(f_1\ast \dots \ast f_p)}_m]-1.$$

Choosing $l_n=1/n!$ and $m=2$ from Corollary 3, I get the above result of M.K. Sen [12], i.e., Lemma 1.

4. Hadamard Compositions of Gelfond–Leont’ev Derivatives for Functions Analytic in a Disk

For the functions analytic in the disk $U=\{z:|z|<1\}$, the lower order ${\lambda }_U[f]$ and the order ${\varrho }_U[f]$ are defined as

$${\lambda }_U[f]=\underset{r\uparrow 1}{\underline{lim}}\frac{{ln}^{+}lnM(r,f)}{-ln(1-r)},{\varrho }_U[f]=\underset{r\uparrow 1}{\overline{lim}}\frac{{ln}^{+}lnM(r,f)}{-ln(1-r)}.$$

Since

$$\mu (r,f)\le M(r,f)\le \sum _{n=0}^{\infty }|a_n|{\left(\frac{1+r}{2}\right)}^n{\left(\frac{2r}{1+r}\right)}^n\le \frac{1+r}{1-r}\mu \left(\frac{1+r}{2},f\right),$$

and in view of (16) for $r\ge 1/e$

$$\nu \left(\frac{1+r}{2},f\right)\ge \underset{1/e}{\overset{(1+r)/2}{\int }}\frac{\nu (t,f)}{t}dt=ln\mu \left(\frac{1+r}{2},f\right)-ln\mu \left(\frac{1}{2},f\right)\ge$$

$$\ge \underset{r}{\overset{(1+r)/2}{\int }}\frac{\nu (t,f)}{t}dt\ge \nu (r,f)ln\frac{1+r}{2r}=(1+o(1)\frac{1-r}{2}\nu (r,f),r\uparrow 1,$$

I obtain

$$\underset{r\uparrow 1}{\underline{lim}}\frac{{ln}^{+}ln\mu (r,f)}{-ln(1-r)}={\lambda }_U[f],\underset{r\uparrow 1}{\overline{lim}}\frac{{ln}^{+}ln\mu (r,f)}{-ln(1-r)}={\varrho }_U[f]$$

and

$${\lambda }_U[f]\le \underset{r\uparrow 1}{\underline{lim}}\frac{{ln}^{+}\nu (r,f)}{-ln(1-r)}\le {\lambda }_U[f]+1,{\varrho }_U[f]\le \underset{r\uparrow 1}{\overline{lim}}\frac{{ln}^{+}\nu (r,f)}{-ln(1-r)}\le {\varrho }_U[f]+1.$$

(20)

Proposition 3.

If condition (17) holds, then, for every k,

$${\lambda }_U[{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m]={\lambda }_U[D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]={\lambda }_U[{(f_1\ast \dots \ast f_p)}_m]$$

and

$${\varrho }_U[{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m]={\varrho }_U[D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]={\varrho }_U[{(f_1\ast \dots \ast f_p)}_m].$$

Proof. 

At first, I remark that, for each function f and analytic in U, the equalities ${\lambda }_U[f^{\prime }]={\lambda }_U[f]$ and ${\varrho }_U[f^{\prime }]={\varrho }_U[f]$ are true.

Indeed, from Cauchy formula $f^{\prime }(z)={\displaystyle \frac{1}{2\pi i}}\underset{|\tau -z|=(1-|z|)/2}{\int }{\displaystyle \frac{f(\tau )d\tau }{{(\tau -z)}^2}}$ I have $M(r,f^{\prime })\le 2M((1+r)/2,f)/(1-r)$, and since $f(z)-f(0)=\underset{0}{\overset{z}{\int }}{f}^{\prime }(\tau )d\tau$, the inequality $M(r,f)\le M(r,f^{\prime })+|f(0)|$ holds, from which the necessary equalities follow.

Now, in view of (17) $h_1(n+1)\le l_n/l_{n+1}\le h_2(n+1)$, I have, for $k=1$,

$$\mu (r,D_l^{(1)}({(f_1\ast \dots \ast f_p)}_m))=max\left\{\frac{l_n}{l_{n+1}}|a_{n+1}|r^n:n\ge 0\right\}\le$$

$$\le h_{2}max\{(n+1)|a_{n+1}|r^n:n\ge 0\}=h_2\mu (r,{({(f_1\ast \dots \ast f_p)}_m)}^{\prime })$$

and by analogy $\mu (r,D_l^{(1)}{(f_1\ast \dots \ast f_p)}_m)\ge h_1\mu (r,{({(f_1\ast \dots \ast f_p)}_m)}^{\prime })$. Hence, it follows that ${\lambda }_U[D_l^{(1)}{(f_1\ast \dots \ast f_p)}_m]={\lambda }_U[{(f_1\ast \dots \ast f_p)}_m]$ and ${\varrho }_U[D_l^{(1)}{(f_1\ast \dots \ast f_p)}_m]={\varrho }_U[{(f_1\ast \dots \ast f_p)}_m]$ and, thus, ${\lambda }_U[D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]={\lambda }_U[{(f_1\ast \dots \ast f_p)}_m]$ and ${\varrho }_U[D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]={\varrho }_U[{(f_1\ast \dots \ast f_p)}_m]$.

Since $l_n/l_{n+1}\ge h_1(n+1)\ge h_1$, I have ${(l_n/l_{n+k})}^{m-1}\ge h_1^{k(m-1)}$, i. e.

$$\mu (r,{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m)=max\left\{{\left(\frac{l_n}{l_{n+k}}\right)}^{m-1}\frac{l_n}{l_{n+k}}|a_{n+k}|r^n:n\ge 0\right\}\ge$$

$$\ge h_1^{k(m-1)}\mu (r,D_l^{(1)}({(f_1\ast \dots \ast f_p)}_m))$$

and, thus, ${\lambda }_U[D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]\le {\lambda }_U[{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m]$ and ${\varrho }_U[D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]\le {\varrho }_U[{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m]$.

On the other hand,

$${\left(\frac{l_n}{l_{n+k}}\right)}^{m-1}\le {\left(h_2^k(n+1)\dots (n+k)\right)}^{m-1}\le$$

$$\le h_2^{k(m-1)}(n+1)\dots (n+k)(n+k+1)\dots (n+2k)\dots (n+(m-2)k+1)\dots (n+(m-1)k).$$

Therefore,

$$\mu (r,{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m)\le h_2^{k(m-1)}max\left\{(n+1)\dots (n+(m-1)k)\frac{l_n}{l_{n+k}}|a_{n+k}|r^n:n\ge 0\right\}=$$

$$=h_2^{k(m-1)}max\left\{{\left(\frac{l_n}{l_{n+k}}|a_{n+k}|r^{n+(m-1)k}\right)}^{((m-1)k)}:n\ge 0\right\}=$$

$$=h_2^{k(m-1)}max\left\{{\left(r^{k(m-1)}\frac{l_n}{l_{n+k}}|a_{n+k}|r^n\right)}^{(k(m-1))}:n\ge 0\right\}=h_2^{k(m-1)}\mu (r,F^{(k(m-1))}),$$

where $F(z)=z^{k(m-1)}{D}_l^{(k)}{(f_1\ast \dots \ast f_p)}_m(z)$. Hence it follows that ${\lambda }_U[{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m]\le {\lambda }_U[F^{(k(m-1))}]={\lambda }_U[F]={\lambda }_U[D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]$ and ${\varrho }_U[{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m]\le {\varrho }_U[D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m]$. Proposition 3 is proved. □

The following statement is an analog of Proposition 2.

Proposition 4.

Let $f_1,\dots ,f_p$ be functions analytic in U and condition (17) holds. If ${\varrho }_U[f_1]=\dots ={\varrho }_U[f_p]=\varrho$, then ${\varrho }_U[{(f_1\ast \dots \ast f_p)}_m]\le m\varrho$, and if. among functions $f_1,\dots ,f_p$. there is a dominant function $f_1$, then ${\varrho }_U[{(f_1\ast \dots \ast f_p)}_m]=m\varrho$. Moreover, if, among functions $f_1,\dots ,f_p$, there is a dominant function $f_1$, then ${\lambda }_U[{(f_1\ast \dots \ast f_p)}_m]=m{\lambda }_U[f_1]$.

Proof. 

Indeed, since ${\varrho }_U[f_1]=\dots ={\varrho }_U[f_p]=\varrho$, for every $\epsilon >0$ and all $r\in [r_0(\epsilon ,1)$ I have $ln\mu (r,f_j)\le {(1/(1-r))}^{\varrho +\epsilon }$, and (10) implies $ln\mu (r^m,f)\le C{(1/(1-r))}^{\varrho +\epsilon }$ for $r\in [r_0(\epsilon ,1)$, where $C=\mathrm{const}$. Therefore,

$${\varrho }_U[f]=\underset{r\uparrow 1}{\overline{lim}}\frac{{ln}^{+}ln\mu (r^m,f)}{-ln(1-r^m)}\le (\varrho +\epsilon )\underset{r\uparrow 1}{\overline{lim}}\frac{-ln(1-r)}{-ln(1-r^m)}\le (\varrho +\epsilon )\underset{r\uparrow 1}{\overline{lim}}\frac{1-r^m}{1-r}=(\varrho +\epsilon )m,$$

and, in view of the arbitrariness of $\epsilon$, I obtain ${\varrho }_U[{(f_1\ast \dots \ast f_p)}_m]\le m\varrho$.

If the function $f_1$ is dominant, then, from (11), I obtain ${\varrho }_U[f]=m{\varrho }_U[f_1]$ and ${\lambda }_U[f]=m{\lambda }_U[f_1]$. □

Using (12) and (20), I can investigate the growth of the ratio ${\displaystyle \frac{\mu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}}$ in the case of analytic functions in U. I will not dwell on this, but rather study the growth of the ratio ${\displaystyle \frac{\mu (r,D_l^{(j)}{(f_1\ast \dots \ast f_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}}$ for $j>k$ and the ratio ${\displaystyle \frac{\mu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}{\mu (r,{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m)}}$ for $j>k$.

As above, I have

$$\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)=\frac{l_{\nu (r,D_l^{(k)}f)}}{l_{\nu (r,D_l^{(k)}f)+k}}|a_{\nu (r,D_l^{(k)}f)+k}|r^{\nu (r,D_l^{(k)}f)}=$$

$$=\frac{l_{\nu (r,D_l^{(k)}f)}}{l_{\nu (r,D_l^{(k)}f)+k-j}}\frac{l_{\nu (r,D_l^{(k)}f)+k-j}}{l_{\nu (r,D_l^{(k)}f)+k-j+j}}|a_{\nu (r,D_l^{(k)}f)+k-j+j}|r^{\nu (r,D_l^{(k)}f)+k-j}{r}^{j-k}\le$$

$$\le \frac{l_{\nu (r,D_l^{(k)}f)}}{l_{\nu (r,D_l^{(k)}f)+k-j}}{r}^{j-k}\mu (r,D_l^{(j)}{(f_1\ast \dots \ast f_p)}_m)$$

and, similarly,

$$\mu (r,D_l^{(j)}{(f_1\ast \dots \ast f_p)}_m)\le \frac{l_{\nu (r,D_l^{(j)}f)}}{l_{\nu (r,D_l^{(j)}f)+j-k}}{r}^{k-j}\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m).$$

Therefore,

$$\frac{l_{\nu (r,D_l^{(k)}f)+k-j}}{l_{\nu (r,D_l^{(k)}f)}}\le \frac{r^{j-k}\mu (r,D_l^{(j)}{(f_1\ast \dots \ast f_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}\le \frac{l_{\nu (r,D_l^{(j)}f)}}{l_{\nu (r,D_l^{(j)}f)+j-k}}.$$

(21)

Using (21), I prove following theorem.

Theorem 3.

Let $f_j$ be analytic functions in U, $1\le j\le p$, and let (17) hold, then, for each $j\ge k$,

$$(j-k){\lambda }_U[{(f_1\ast \dots \ast f_p)}_m]\le \underset{r\uparrow 1}{\underline{lim}}\frac{1}{-ln(1-r)}ln\frac{\mu (r,D_l^{(j)}{(f_1\ast \dots \ast f_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}\le (j-k)({\lambda }_U[{(f_1\ast \dots \ast f_p)}_m]+1)$$

and

$$(j-k){\varrho }_U[{(f_1\ast \dots \ast f_p)}_m]\le \underset{r\uparrow 1}{\overline{lim}}\frac{1}{-ln(1-r)}ln\frac{\mu (r,D_l^{(j)}{(f_1\ast \dots \ast f_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}\le (j-k)({\varrho }_U[{(f_1\ast \dots \ast f_p)}_m]+1).$$

Proof. 

Since ${(h_{1}n)}^k\le l_n/l_{n+k}\le {(h_{2}n)}^k$, I have

$$\frac{l_{\nu (r,D_l^{(k)}f)+k-j}}{l_{\nu (r,D_l^{(k)}f)}}=\frac{l_{\nu (r,D_l^{(k)}f)+k-j}}{l_{\nu (r,D_l^{(k)}f)+k-j+j-k}}\ge {(h_1\nu (r,D_l^{(k)}f)+k-j)}^{j-k}$$

and

$$\frac{l_{\nu (r,D_l^{(j)}f)}}{l_{\nu (r,D_l^{(j)}f)+j-k}}\le {(h_2\nu (r,D_l^{(j)}f))}^{j-k}.$$

Therefore, (21) implies

$$(1+o(1))(j-k)ln\nu (r,D_l^{(k)}f)\le ln\frac{\mu (r,D_l^{(j)}{(f_1\ast \dots \ast f_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}\le (1+o(1))(j-k)ln\nu (r,D_l^{(j)}f)$$

as $r\uparrow 1$, i.e., in view of (20) and Proposition 3, I obtain

$$(j-k){\lambda }_U[f]\le \underset{r\uparrow 1}{\underline{lim}}\frac{1}{-ln(1-r)}ln\frac{\mu (r,D_l^{(j)}{(f_1\ast \dots \ast f_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}\le (j-k)({\lambda }_U[f]+1)$$

and

$$(j-k){\varrho }_U[f]\le \underset{r\uparrow 1}{\overline{lim}}\frac{1}{-ln(1-r)}ln\frac{\mu (r,D_l^{(j)}{(f_1\ast \dots \ast f_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}\le (j-k)({\varrho }_U[f]+1).$$

Q.E.D. □

Since series (8) differs from series (7), only that instead $l_k/l_{k+1}$, it contains ${(l_k/l_{k+1})}^m$, I will easily prove the inequalities

$${\left(\frac{l_{\nu (r,D_l^{(k)}f)+k-j}}{l_{\nu (r,D_l^{(k)}f)}})\right)}^m\le \frac{r^{j-k}\mu (r,D_l^{(j)}{(f_1\ast \dots \ast f_p)}_m)}{\mu (r,D_l^{(k)}{(f_1\ast \dots \ast f_p)}_m)}\le {\left(\frac{l_{\nu (r,D_l^{(j)}f)}}{l_{\nu (r,D_l^{(j)}f)+j-k}}\right)}^m,$$

whence, as in the proof of Theorem 3, I will come to the next theorem.

Theorem 4.

Let $f_j$ be analytic functions in U, $1\le j\le p$, and let (17) hold, then, for each $j\ge k$,

$$m(j-k){\lambda }_U[{(f_1\ast \dots \ast f_p)}_m]\le \underset{r\uparrow 1}{\underline{lim}}\frac{1}{-ln(1-r)}ln\frac{\mu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}{\mu (r,{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m)}\le$$

$$\le m(j-k)({\lambda }_U[{(f_1\ast \dots \ast f_p)}_m]+1)$$

and

$$m(j-k){\varrho }_U[{(f_1\ast \dots \ast f_p)}_m]\le \underset{r\uparrow 1}{\overline{lim}}\frac{1}{-ln(1-r)}ln\frac{\mu (r,{(D_l^{(j)}{f}_1\ast \dots \ast D_l^{(j)}{f}_p)}_m)}{\mu (r,{(D_l^{(k)}{f}_1\ast \dots \ast D_l^{(k)}{f}_p)}_m)}\le$$

$$\le m(j-k)({\varrho }_U[{(f_1\ast \dots \ast f_p)}_m]+1).$$

5. Discussion

In conclusion, I note that, in addition to the analytic continuation of functions, the usual Hadamard composition was used in other aspects of complex analysis (in particular, in the geometric function theory). One can naturally hope that the Hadamard composition of the genus m will also find applications in similar areas of mathematics.