RELATIVE GROWTH OF ENTIRE DIRICHLET SERIES WITH DIFFERENT GENERALIZED ORDERS

For entire functions F and G deﬁned by Dirichlet series with exponents increasing to + 1 formulas are found for the ﬁnding the generalized order ϱ (cid:11);(cid:12) [ F G = 1 ) and the generalized lower order (cid:21) (cid:11);(cid:12) [ F ] G = lim (cid:27) ! + 1 (cid:11) ( M (cid:0) 1 G ( M F ( (cid:27) ))) (cid:12) ( (cid:27) ) of F with respect to G , where M F ( (cid:27) ) = sup fj F ( (cid:27) + it ) j : t 2 R g and (cid:11) and (cid:12) are positive increasing to + 1 functions.


Introduction
Let f and g be entire transcendental functions and M f (r) = max{|f (z)| : |z| = r}. For the study of relative growth of the functions f and g Ch. Roy [1] used the order ϱ g [f ] = lim r→+∞ ln M −1 g (M f (r))/ ln r and the lower order λ g [f ] = lim r→+∞ ln M −1 g (M f (r))/ ln r of the function f with respect to the function g. Researches of relative growth of entire functions was continued by S.K. Data, T. Biswas and other mathematicians (see, for example, [2,3,4,5]) in terms of maximal terms, Nevanlinna characteristic function and k-logarithmic orders. In [6] it is considered a relative growth of entire functions of two complex variables and in [7] the relative growth of entire Dirichlet series is studied in terms of R-orders.
Suppose that Λ = (λ n ) is an increasing to +∞ sequence of non-negative numbers, and by S(Λ) we denote a class of entire Dirichlet series For σ < +∞ we put M F (σ) = sup{|F (σ + it)| : t ∈ R}. We remark that the function M F (σ) is continuous and increasing to +∞ on (−∞, +∞) and, therefore, there exists the function M −1 F (x) inverse to M F (σ), which increase to +∞ on (x 0 , +∞).
Theorem (A). Let β ∈ L and γ ∈ L. Except for the cases, when ϱ γ, is true and subject to the condition of the generalized regular (γ, β)-growth of G this inequality converts into an equality. Except for the cases, when λ γ, is true and subject to the condition of the generalized regular (γ, β)-growth of G this inequality converts into an equality.
Theorem (B). Let 0 < p < +∞ and one of conditions is executed: for each c ∈ (0, +∞) and If the function G has generalized regular (γ, β)-growth and κ n [G] : If, moreover, κ n [F ] ↗ +∞ as n 0 ≤ n → ∞ then Similar results in terms of R-types are obtained in [10]. Here we consider the general case when α ̸ = β.

Analogues of Theorem (A)
We begin from the following general theorem.
are true for each function γ ∈ L except for the cases ϱ γ, are true for each function γ ∈ L except for the cases λ γ, Proof. Indeed, and i. e. inequalities (2) are proved. The proof of (3) is similar. Indeed, and  [11], and if we choose α(x) = β(x) = ln x for x ≥ 3 then we obtain the definition of the logarithmic order ϱ l [G] and the lower logarithmic order λ l [G].
For the characteristic of the relative growth of the function F with respect to a function G in Ritt's scale we use in the logarithmic scale we use Then Theorem 1 implies the following statement.
For a more detailed description of the growth of Dirichlet series of finite nonzero order use the type. If 0 < ϱ R [F ] < +∞ then the quantities are called the R-type and the lower R-type of function F . Similarly, the quantities are called the logarithmic type and the lower logarithmic type of function F . Therefore, by analogy, if 0 < ϱ α,β [F ] < +∞ then we define the generalized (α, β)-type and the lower generalized (α, β)-type of F as follows Similarly, if 0 < ϱ α,β [F ] G < +∞ then we define the generalized (α, β)-type and the lower generalized (α, β)-type of the function F with respect to the function G as follows Theorem 2. Let α ∈ L, β ∈ L and γ ∈ L. If the function G has the regular generalized and Proof. Since G has the regular generalized (γ, α)-growth, by Theorem 1 (see Remark 1) we . Therefore, and Estimates (4) are proved. The proof of (5) is similar and we will omit it.
Theorem 2 implies the following statement.
Corollary 2. If the function G has the regular growth and If the function G has the regular logarithmic growth and If the function G has the regular logarithmic growth and

Analogues of Theorem (B)
We need the following lemma.
Now we prove the following theorem.
Proof. Since G has generalized regular (γ, α)-growth, by Theorem 1 and by Lemma 1 . Therefore, On the other hand, let P α,β > 0. Then for every ε ∈ (0, P α,β ) there exists an increasing to +∞ sequence (n k ) of integers such that and, thus, whence in view of the arbitrariness of ε we get ϱ α,β [F ] G ≥ P α,β . For P α,β = 0 the last inequality is obvious. Equality (8) is proved. For the proof of (9) we remark that since G has generalized regular (α, β)-growth, by Theorem 1 and Lemma 1 On the other hand, let p α,β < +∞. Then for every ε > 0 there exists an increasing to +∞ sequence (n k ) of integers such that and, as above, whence in view of the arbitrariness of ε we get λ α,β [F ] G ≤ p α,β . For p α,β = +∞ the last inequality is obvious. Equality (9) is proved, and the proof of Theorem 3 is complete.
For the study of the relative growth in classical scales we need the following lemmas.
This result can be directly obtained using Lemma 2. It is easy to see also that the functions α(x) = β(x) = γ(x) = ln x do not satisfy the conditions of Theorem 3. However, the following statement is correct.
If, moreover, κ n [F ] ↗ +∞ as n 0 ≤ n → ∞ then Proof. Since G has generalized regular logarithmic growth, by Corollary 2 ϱ l, and by Lemma 3 On the other hand, if P l > 0 then for every ε ∈ (0, P l ) there exists an increasing to +∞ sequence (n k ) of integers such that i. e. by Lemma 3 whence in view of the arbitrariness of ε we get ϱ l [F ] G ≥ P l . For P l = 0 the last inequality is obvious. Equality (10) is proved. The proof of (11) is similar.
The condition ln n = o(λ n ln λ n ) as n → ∞ implies the condition lim n→∞ ln ln n/ ln λ n ≤ 1. Therefore, using Lemmas 2 and 3 it is easy to prove the following statement. Let us turn to the results about the relative growth of functions in terms of their types. For classic growth scales, we can use Lemmas 2 and 3, and for generalized orders we need such lemma.
Using Lemma 4, we prove the following theorem.
If the function G has strongly regular generalized (γ, α)-growth (i. e. 0 < t γ, . Proof. Since the function G has strongly regular generalized (γ, α)-growth, by Theorem On the other hand, let Q > 0. Then for every Q 1 ∈ (0, Q) there exists an increasing to ∞ sequence (n k ) of integers such that exp and, thus, Next three statements are proved in general a way and we will drop their proofs. Using Corollary 2 and Lemma 2, we get the following statement. Proposition 4. If the function G has the strongly regular growth (i. e. 0 < t R [G] = = T R [G] < +∞), ln n = o(λ n ), λ n+1 ∼ λ n and κ n [G] ↗ +∞ as n 0 ≤ n → ∞ then If, moreover, κ n [F ] ↗ +∞ as n 0 ≤ n → ∞ then Since the condition ln ln n = o(ln λ n ) as n → ∞ implies the condition ln n = = o(λ ϱ/(ϱ−1) n ) as n → ∞ for every ϱ > 1, using Corollary 2 and Lemma 3, we get the next statement. Finally, since the condition ln ln n = o(ln λ n ) as n → ∞ implies the condition ln n = o(λ n ) as n → ∞, using Corollary 2 and Lemmas 2 and 3, we get the next statement.