# Possible Intervals forT- and M-Orders of Solutions of Linear Differential Equations in the Unit Disc

Authors: Martin Chuaqui; Janne Gröhn; Janne Heittokangas; Jouni Rättyä
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

In the case of the complex plane, it is known that there exists a finite set
of rational numbers containing all possible growth orders of solutions off(k)+ak-1(z)f(k-1)+⋯+a1(z)f′+a0(z)f=0 with polynomial coefficients. In the present paper, it is shown by an example that a unit disc counterpart of such finite set does not contain all possible T- and M-orders of solutions, with respect to Nevanlinna characteristic and maximum modulus, if the coefficients are analytic functions belonging either to weighted Bergman spaces or to weighted Hardy spaces. In contrast to a finite set, possible intervals for T- and M-orders are introduced to give detailed information about the growth of solutions. Finally, these findings yield sharp lower bounds for the sums of T- and M-orders of functions in the solution bases.


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## Body

## 1. Introduction

This research is a continuation of recent activity in the field of complex differential equations. In particular, the present paper concerns linear differential equations of the type(1.1)f(k)+ak-1(z)f(k-1)+⋯+a1(z)f′+a0(z)f=0,
where the coefficients a0(z),…,ak-1(z) are analytic functions in the unit disc 𝔻:={z:|z|<1} of the complex plane ℂ. A variety of publications in the existing literature illustrate that the connection between the growth of coefficient functions and the growth of solutions is relatively well understood. On the one hand, the growth estimates in [1] have been proven to be instrumental tools in estimating the growth of solutions when the growth of coefficients is known. On the other hand, proofs of the converse direction have taken advantage of the method of order reduction as well as different types of logarithmic derivative estimates.For an analytic function in𝔻, it is known that T- and M-orders of growth, with respect to Nevanlinna characteristic and maximum modulus, are not equal in general. This is in contrast to the corresponding case in ℂ. Hence, there are two distinct cases in 𝔻 to work with. First, if the growth of solutions is measured by using the T-order, then it is natural to express the other growth aspects by means of integration as well. In particular, it is reasonable to consider coefficient functions belonging to some weighted Bergman spaces and use integrated estimates for logarithmic derivatives [2]. Second, if the growth of solutions is measured by using the M-order, then it is natural to express the other growth aspects by means of the maximum modulus function. In particular, it is sensible to restrict the growth of the maximum modulus of coefficient functions, which leads to weighted Hardy spaces, and work with estimates for the maximum modulus of logarithmic derivatives involving exceptional sets [3].The main focus of this paper is in improving the lower bounds for the growth of solutions of (1.1) given in [2, 3] and explore some consequences, which are motivated by the following observations.By the classical results inℂ making use of Newton-Puiseux diagram, there is a finite set containing the possible growth orders of solutions of (1.1) assuming that coefficients are polynomials. In particular, Gundersen-Steinbart-Wang showed that this finite set consists of rational numbers obtained from simple arithmetic with the degrees of the polynomial coefficients in (1.1) [4, Theorem  1]. Their proof relies on classical Wiman-Valiron theory in ℂ. Even though a recent unit disc counterpart of Wiman-Valiron theory [5] has been successfully applied to differential equations, the possible orders of solutions of (1.1) in 𝔻 have been obtained only by assuming that coefficients are α-polynomial regular. These α-polynomial regular functions have similar growth properties than polynomials in the sense that maximal growth is attained in every direction. However, they appear to be only a relatively small subset of the Korenblum space, which characterizes finite-order solutions of (1.1) in 𝔻 [6, Theorem  6.1]. Note that in the case of ℂ, all solutions of (1.1) are of finite order if and only if coefficients are polynomials [7, Satz  1].In the present paper, it is shown by an example that a unit disc counterpart of the finite set constructed by Gundersen-Steinbart-Wang does not contain all possible orders of solutions of (1.1), provided that the coefficients belong either to weighted Bergman spaces or to weighted Hardy spaces. In contrast to a finite set, we introduce possible intervals for T-orders and M-orders, giving detailed information about the growth of solutions. Finally, these findings are applied to estimate the sums of T- and M-orders of functions in the solution bases of (1.1) from below.


## 2. Results and Motivation

The results concerningT- and M-orders of solutions of (1.1) are given, respectively, in Sections 2.1-2.2 and 2.3-2.4. Due to the similarities of the assertions, we omit the proofs of results regarding M-orders of solutions of (1.1), excluding the sketched proof of Theorem 2.5 in Section 7.Letℳ(𝔻) and ℋ(𝔻) denote the sets of all meromorphic and analytic functions in 𝔻. For simplicity, we write α+:=max{α,0} for any α∈ℝ, |f(z)|≲|g(z)| if there exists a constant C>0 independent of z such that |f(z)|≤C|g(z)|, and f(z)~g(z) if there exist constants C1>0 and C2>0 independent of z such that C1|g(z)|≤|f(z)|≤C2|g(z)|.


### 2.1. Growth of Solutions with Respect to Nevanlinna Characteristic

TheT-order of growth of f∈ℳ(𝔻) is defined as (2.1)σT(f):=limsup⁡r→1-log⁡+T(r,f)-log⁡(1-r),
where T(r,f) is the Nevanlinna characteristic of f. For p>0 and α>-1, the weighted Bergman spaceAαp consists of those f∈ℋ(𝔻) for which (2.2)‖f‖Aαp:=(∫D|f(z)|p(1-|z|2)αdm(z))1/p<∞.
Functions of maximal growth in ⋂q<α<∞Aαp are distinguished by denoting f∈𝔸qp, if q=inf{α>-1:f∈Aαp}.If the growth of the coefficients is expressed by means of integration, then it is natural to consider the growth of solutions of (1.1) with respect to T-order.Theorem A (see [2, Theorems  1 and  2]).
Suppose thataj∈𝔸αj1/(k-j), where αj≥0 for j=0,…,k-1, and denote αk:=0.  (i)
Let0≤α<∞. Then all solutions f of (1.1) satisfy σT(f)≤α if and only if maxj=0,…,k-1{αj}≤α.(ii)
All nontrivial solutionsf of (1.1) satisfy(2.3)min⁡j=1,…,k{k(α0-αj)j+αj}≤σT(f)≤max⁡j=0,…,k-1{αj}.(iii)
Ifq∈{0,…,k-1} is the smallest index for which αq=maxj=0,…,k-1{αj}, then each solution base of (1.1) contains at least k-q linearly independent solutions f such that σT(f)=αq.The assumptionaj∈𝔸αj1/(k-j) in Theorem A(i) cannot be replaced by aj∈Aαj1/(k-j); see [8]. We refine Theorem A and then further underscore its consequences.Theorem 2.1.
Suppose thataj∈𝔸αj1/(k-j), where αj≥-1 for j=0,…,k-1, and let q∈{0,…,k-1} be the smallest index for which αq=maxj=0,…,k-1{αj}. If s∈{0,…,q}, then each solution base of (1.1) contains at least k-s linearly independent solutions f such that
(2.4)min⁡j=s+1,…,k{(k-s)(αs-αj)j-s+αj}≤σT(f)≤αq+,
where αk:=-1.The cases=0 clearly reduces to Theorem A(ii). If s=q, then the condition αq=maxj=0,…,k-1{αj} implies that (2.5)min⁡j=q+1,…,k{(k-q)(αq-αj)j-q+αj}=min⁡j=q+1,…,k{(k-j)(αq-αj)j-q+αq}=αq,
where the minimum is attained for j=k. Hence the assertion of Theorem 2.1 for s=q is contained in Theorem A(iii). Our contribution is to extend the first inequality in (2.4) for s∈{1,…,q-1}. Theorem 2.1 is proved in Section 4, and the sharpness and the special cases k=2 and k=3 are further discussed in Section 3.1.Letq∈{0,…,k-1} be the smallest index for which αq=maxj=0,…,k-1{αj}. If αq≤0, then all solutions in each solution base of (1.1) are of zero T-order by Theorem A(ii). Suppose that αq>0. In order to state the following corollaries of Theorem 2.1, we denote
(2.6)βT(s):=min⁡j=s+1,…,k{(k-s)(αs-αj)j-s+αj},s=0,…,q,
where αk:=-1. Moreover, we define
(2.7)s⋆:=min⁡{s∈{0,…,q}:βT(s)>0}.
Remark that βT(q)>0, since (2.5) implies αq=βT(q).Corollary 2.2.
Suppose thataj∈𝔸αj1/(k-j), where αj≥-1  for j=0,…,k-1, and let q∈{0,…,k-1} be the smallest index for which αq=maxj=0,…,k-1{αj}>0. Then each solution base of (1.1) admits at most s⋆≤q solutions f satisfying σT(f)<βT(s⋆). In particular, there are at most s⋆≤q solutions f satisfying σT(f)=0.To estimate the quantity∑j=1kσT(fj) by using Theorem 2.1, we set (2.8)γT(j):=max⁡{βT(0),…,βT(j)},j=0,…,q.
Evidently γT(j)>0 for j∈{s⋆,…,q}, and γT(j)≤0 for j∈{0,…,s⋆-1}.Corollary 2.3.
Suppose thataj∈𝔸αj1/(k-j), where αj≥-1 for j=0,…,k-1, and let q∈{0,…,k-1} be the smallest index for which αq=maxj=0,…,k-1{αj}>0. Let {f1,…,fk} be a solution base of (1.1). If q=0, then ∑j=1kσT(fj)=kα0, while if q≥1, then
(2.9)(k-q)αq+∑j=s⋆q-1γT(j)≤∑j=1kσT(fj)≤kαq.Note that the sum in (2.9) is considered to be empty, if s⋆=q.


### 2.2. Gundersen-Steinbart-Wang Method forT-Order

We proceed to give an alternative statement of Theorem2.1 and its corollaries by modifying the key steps in [4]. This yields a natural way to define possible intervals for T-orders of solutions of (1.1). As a consequence, we get a useful estimate following from Corollary 2.3.Setδj:=(αj+1)(k-j) for all j=0,…,k-1. Let s1∈{0,…,k-1} be the smallest index satisfying αs1=maxj=0,…,k-1{αj}>0, which is equivalent to (2.10)δs1k-s1=max⁡j=0,…,k-1{δjk-j}>1.
If s1 cannot be found, then all solutions of (1.1) are of zero T-order by Theorem A(ii). Otherwise, for a given sm, m∈ℕ, let sm+1∈{0,…,sm-1} be the smallest index satisfying (2.11)δsm+1-δsmsm-sm+1=max⁡j=0,…,sm-1{δj-δsmsm-j}>1.
Eventually this process will stop, yielding a finite list of indices s1,…,sp such that p≤k and s1>s2>⋯>sp≥0. Further, set (2.12)BT(t):=δst-δst-1st-1-st-1,t=1,…,p,
where s0:=k and δk:=0. Due to resemblance between (2.12) and [4, Equation (2.4)], it seems plausible that the possible nonzero T-orders of solutions of (1.1) in the unit disc case could be found among the numbers ℬT(t), where t=1,…,p. However, Example 3.1 shows that this is not the case.The following lemma allows us to view the results in Section2.1 in a new perspective. In particular, Lemma 2.4 emphasizes the connection between ℬT and γT.Lemma 2.4.
One has the following:(i)
ℬT(1)>ℬT(2)>⋯>ℬT(p)>0;(ii)
βT(st)=ℬT(t) for all t∈{1,…,p};(iii)
γT(q)=ℬT(1), γT(j)=ℬT(t) for all st≤j<st-1 and t∈{2,…,p}, and γT(j)≤0 for all j<sp. In particular, sp=s⋆.By relying on Lemma2.4, Theorem 2.1, and Corollary 2.2, we proceed to state possible intervals forT-orders of functions in solution bases of (1.1) in the case aj∈𝔸αj1/(k-j), where αj≥-1 for j=0,…,k-1. In fact, each solution base of (1.1) contains the following: (i)
at leastk-s1 solutions f satisfying σT(f)=ℬT(1);(ii)
at leastk-st solutions f satisfying σT(f)∈[ℬT(t),ℬT(1)] for t=2,…,p;(iii)
at mostsp solutions f satisfying σT(f)∈[0,ℬT(sp)). For the following application, let {f1,…,fk} be a solution base of (1.1). Knowing the possible intervals for T-orders, we get
(2.13)∑j=1kσT(fj)≥(k-s1)BT(1)+⋯+(sp-1-sp)BT(p)+sp⋅0=δsp+sp-k.
In view of Lemma 2.4, the lower estimates in (2.9) and (2.13) are equal.Finally, we point out a useful consequence of (2.13). If sp=0, then δsp+sp=δ0. If sp>0, then (δ0-δsp)/sp≤1 by (2.11), and δsp+sp≥δ0. Hence, in both cases we can state that
(2.14)∑j=1kσT(fj)≥δsp+sp-k≥δ0-k≥α0k,
where the equalities hold, if α0=maxj=0,…,k-1{αj}>0.


### 2.3. Growth of Solutions with Respect to Maximum Modulus

Alongside of theT-order, we may also define the M-order of growth of f∈ℋ(𝔻) by (2.15)σM(f):=limsup⁡r→1-log⁡+log⁡+M(r,f)-log⁡(1-r),
where M(r,f):=max|z|=r|f(z)| is the maximum modulus of f. It is well known that the inequalities (2.16)σT(f)≤σM(f)≤σT(f)+1
are satisfied for all f∈ℋ(𝔻), and all possibilities allowed by (2.16) can be assumed [9, Theorems  3.5–3.7]. A function f∈ℋ(𝔻) is said to belong to the weighted Hardy spaceHα∞, if there exists α≥0 such that (2.17)sup⁡z∈D(1-|z|2)α|f(z)|<∞.
Functions of maximal growth in ⋂α>pHα∞ are distinguished by denoting f∈ℍp∞, if p=inf{α≥0:f∈Hα∞}. Remark that H0∞=H∞ is the space of all bounded analytic functions in 𝔻. The union ⋃α>0Hα∞ is also known as the Korenblum space𝒜-∞ [10], and since [11] ℍp∞ is also known as Gp.If the growth of coefficients is measured by means of maximum modulus estimates, then it is natural to consider the growth of solutions with respect toM-order.Theorem B (see [3, Theorem  1.4]).
Suppose thataj∈ℍ(pj+1)(k-j)∞, where pj≥-1 for j=0,…,k-1, and denote pk:=-1.  (i)
Suppose that(2.18)min⁡j=1,…,k{k(p0-pj)j+pj}>1,
and let 1≤α<∞. Then all solutions f of (1.1) satisfyσM(f)≤α if and only if maxj=0,…,k-1{pj}≤α.(ii)
All nontrivial solutionsf of (1.1) satisfy σM(f)≤maxj=0,…,k-1{pj+}, and(2.19)min⁡j=1,…,k{k(p0-pj)j+pj}≤max⁡{σM(f),1}.(iii)
Suppose that (2.18) holds. If q∈{0,…,k-1} is the smallest index for which pq=maxj=0,…,k-1{pj}, then each solution base of (1.1) contains at least k-q linearly independent solutions f such that σM(f)=pq.To conclude [3, Equation  (4.17)] in the proof of Theorem B, the inequality [3, Equation  (1.9)], corresponding to (2.18), must be strict. By a simple modification of the proof of Theorem B, the assumption (2.18) can be relaxed to (2.20)max⁡j=0,…,k-1{pj}>1,
which allows us to apply Theorem B(iii) also in the case that there are solutions f satisfying σM(f)≤1. To see that (2.20) is in fact a weaker assumption than (2.18), we refer to [2, Example  10], which is further considered in Section 3.2. In this case (2.21)min⁡j=1,…,k{k(p0-pj)j+pj}=-4,max⁡j=0,…,k-1{pj}>1.
Note that by taking j=k in (2.18), we obtain p0>1. Hence (2.18) implies (2.20).Theorem2.5 corresponds to Theorem 2.1.Theorem 2.5.
Suppose thataj∈ℍ(pj+1)(k-j)∞, where pj≥-1 for j=0,…,k-1, and let q∈{0,…,k-1} be the smallest index for which pq=maxj=0,…,k-1{pj}. If s∈{0,…,q}, then each solution base of (1.1) contains at least k-s linearly independent solutions f such that
(2.22)min⁡j=s+1,…,k{(k-s)(ps-pj)j-s+pj}≤max⁡{σM(f),1}.Note that (2.22) gives information on σM(f) only in the case when the minimum in (2.22) is strictly greater than 1. The case s=0 in Theorem 2.5 reduces to Theorem B(ii), and the case s=q reduces to Theorem B(iii) with the assumption (2.20), since now (2.23)min⁡j=q+1,…,k{(k-q)(pq-pj)j-q+pj}=pq=max⁡j=0,…,k-1{pj},
where the minimum is attained for j=k. For a similar argumentation, see (2.5). Our contribution is to extend (2.22) for s∈{1,…,q-1}. The proof of Theorem 2.5 is sketched in Section 7, and the sharpness and the the special cases k=2 and k=3 are further discussed in Section 3.2.Letq∈{0,…,k-1} be the smallest index for which pq=maxj=0,…,k-1{pj}. If pq≤1, then all solutions f in each solution base of (1.1) satisfy σM(f)≤1 by Theorem B(ii). Suppose that pq>1. In order to state the following corollaries of Theorem 2.5, we denote (2.24)βM(s):=min⁡j=s+1,…,k{(k-s)(ps-pj)j-s+pj},s=0,…,q,
where pk:=-1. Moreover, we define (2.25)s⋆:=min⁡{s∈{0,…,q}:βM(s)>1}.
Remark that βM(q)>1, since (2.23) implies pq=βM(q).Corollary 2.6.
Suppose thataj∈ℍ(pj+1)(k-j)∞, where pj≥-1 for j=0,…,k-1, and let q∈{0,…,k-1} be the smallest index for which pq=maxj=0,…,k-1{pj}>1. Then each solution base of (1.1) admits at most s⋆≤q solutions f satisfying σM(f)<βM(s⋆). In particular, there are at most s⋆≤q solutions f satisfying σM(f)≤1.To estimate the quantity∑j=1kσM(fj) by using Theorem 2.5, we set (2.26)γM(j):=max⁡{βM(0),…,βM(j)},j=0,…,q.
Evidently γM(j)>1 for j∈{s⋆,…,q}, and γM(j)≤1 for j∈{0,…,s⋆-1}.Corollary 2.7.
Suppose thataj∈ℍ(pj+1)(k-j)∞, where pj≥-1 for j=0,…,k-1, and let q∈{0,…,k-1} be the smallest index for which pq=maxj=0,…,k-1{pj}>1. Let {f1,…,fk} be a solution base of (1.1). If q=0, then ∑j=1kσM(fj)=kp0, while if q≥1, then
(2.27)(k-q)pq+∑j=s⋆q-1γM(j)≤∑j=1kσM(fj)≤kpq.Note that the sum in (2.27) is considered to be empty, if s⋆=q.


### 2.4. Gundersen-Steinbart-Wang Method forM-Order

We proceed to give an alternative statement of Theorem2.5 and its corollaries by modifying the key steps in [4]. This yields a natural way to define the possible intervals for M-orders of solutions of (1.1). As a consequence, we get a useful estimate following from Corollary 2.7.Setδj:=(pj+1)(k-j) for all j=0,…,k-1. Let s1∈{0,…,k-1} be the smallest index satisfying ps1=maxj=0,…,k-1{pj}>1, which is equivalent to (2.28)δs1k-s1=max⁡j=0,…,k-1{δjk-j}>2.
If s1 cannot be found, then all solutions f of (1.1) satisfy σM(f)≤1 by Theorem B(ii). Otherwise, for a given sm, m∈ℕ, let sm+1∈{0,…,sm-1} be the smallest index satisfying (2.29)δsm+1-δsmsm-sm+1=max⁡j=0,…,sm-1{δj-δsmsm-j}>2.
Eventually this process will stop, yielding a finite list of indices s1,…,sp such that p≤k and s1>s2>⋯>sp≥0. Further, set (2.30)BM(t):=δst-δst-1st-1-st-1,t=1,…,p,
where s0:=k and δk:=0. By Example 3.1, it is possible that (1.1) possesses a solution f of M-order strictly greater than one such that σM(f)≠ℬM(t) for all t=1,…,p.The following lemma, which can be proved similarly than Lemma2.4, allows us to view the results in Section 2.3 in a new perspective.Lemma 2.8.
One has the following:(i)
ℬM(1)>ℬM(2)>⋯>ℬM(p)>1;(ii)
βM(st)=ℬM(t) for all t∈{1,…,p};(iii)
γM(q)=ℬM(1), γM(j)=ℬM(t) for all st≤j<st-1 and t∈{2,…,p}, and γM(j)≤1 for all j<sp. In particular, sp=s⋆.By relying on Lemma2.8, Theorem 2.5, and Corollary 2.6, we proceed to state possible intervals forM-orders of functions in solution bases of (1.1) in the case aj∈ℍ(pj+1)(k-j)∞, where pj≥-1 for j=0,…,k-1. In fact, each solution base of (1.1) contains the following: (i)
at leastk-s1 solutions f satisfying σM(f)=ℬM(1);(ii)
at leastk-st solutions f satisfying σM(f)∈[ℬM(t),ℬM(1)] for t=2,…,p;(iii)
at mostsp solutions f satisfying σM(f)∈[0,ℬM(sp)).For results of the same type, we refer to [12, Theorem  1] and [13,  Corollary   1]. To compare (i) and (ii) to the estimates given in [13, Corollary  1], note that there is −1 in (2.30) instead of −2 in [13, Equation  (1.3)]. Evidently, assertions (i) and (ii) improve the estimates given for the M-orders of solutions in [13, Corollary    1]. Moreover, by means of (2.16) we see that (i) and (ii) reduce to [13, Corollary    1], if we consider the growth of solutions of (1.1) with respect to T-order.For the following application, let{f1,…,fk} be a solution base of (1.1). Knowing the possible intervals for M-orders, we get (2.31)∑j=1kσM(fj)≥(k-s1)BM(1)+⋯+(sp-1-sp)BM(p)+sp⋅0=δsp+sp-k.
Corresponding to the case in Section 2.2, by means of Lemma 2.8 we see that the lower estimates in (2.27) and (2.31) are equal.Finally, we point out a practical estimate, which is a consequence of (2.31). If sp=0, then δsp+sp=δ0-sp. If sp>0, then (δ0-δsp)/sp≤2 by (2.29), and δsp+sp≥δ0-sp. Hence, in both cases we can state that (2.32)∑j=1kσM(fj)≥δ0-sp-k≥p0k-sp.
We conclude that if s1=0, then the equalities hold in (2.32), since in this case sp=s1=0. Note that if (2.18) holds, then we can conclude that sp=0.


## 2.1. Growth of Solutions with Respect to Nevanlinna Characteristic

TheT-order of growth of f∈ℳ(𝔻) is defined as (2.1)σT(f):=limsup⁡r→1-log⁡+T(r,f)-log⁡(1-r),
where T(r,f) is the Nevanlinna characteristic of f. For p>0 and α>-1, the weighted Bergman spaceAαp consists of those f∈ℋ(𝔻) for which (2.2)‖f‖Aαp:=(∫D|f(z)|p(1-|z|2)αdm(z))1/p<∞.
Functions of maximal growth in ⋂q<α<∞Aαp are distinguished by denoting f∈𝔸qp, if q=inf{α>-1:f∈Aαp}.If the growth of the coefficients is expressed by means of integration, then it is natural to consider the growth of solutions of (1.1) with respect to T-order.Theorem A (see [2, Theorems  1 and  2]).
Suppose thataj∈𝔸αj1/(k-j), where αj≥0 for j=0,…,k-1, and denote αk:=0.  (i)
Let0≤α<∞. Then all solutions f of (1.1) satisfy σT(f)≤α if and only if maxj=0,…,k-1{αj}≤α.(ii)
All nontrivial solutionsf of (1.1) satisfy(2.3)min⁡j=1,…,k{k(α0-αj)j+αj}≤σT(f)≤max⁡j=0,…,k-1{αj}.(iii)
Ifq∈{0,…,k-1} is the smallest index for which αq=maxj=0,…,k-1{αj}, then each solution base of (1.1) contains at least k-q linearly independent solutions f such that σT(f)=αq.The assumptionaj∈𝔸αj1/(k-j) in Theorem A(i) cannot be replaced by aj∈Aαj1/(k-j); see [8]. We refine Theorem A and then further underscore its consequences.Theorem 2.1.
Suppose thataj∈𝔸αj1/(k-j), where αj≥-1 for j=0,…,k-1, and let q∈{0,…,k-1} be the smallest index for which αq=maxj=0,…,k-1{αj}. If s∈{0,…,q}, then each solution base of (1.1) contains at least k-s linearly independent solutions f such that
(2.4)min⁡j=s+1,…,k{(k-s)(αs-αj)j-s+αj}≤σT(f)≤αq+,
where αk:=-1.The cases=0 clearly reduces to Theorem A(ii). If s=q, then the condition αq=maxj=0,…,k-1{αj} implies that (2.5)min⁡j=q+1,…,k{(k-q)(αq-αj)j-q+αj}=min⁡j=q+1,…,k{(k-j)(αq-αj)j-q+αq}=αq,
where the minimum is attained for j=k. Hence the assertion of Theorem 2.1 for s=q is contained in Theorem A(iii). Our contribution is to extend the first inequality in (2.4) for s∈{1,…,q-1}. Theorem 2.1 is proved in Section 4, and the sharpness and the special cases k=2 and k=3 are further discussed in Section 3.1.Letq∈{0,…,k-1} be the smallest index for which αq=maxj=0,…,k-1{αj}. If αq≤0, then all solutions in each solution base of (1.1) are of zero T-order by Theorem A(ii). Suppose that αq>0. In order to state the following corollaries of Theorem 2.1, we denote
(2.6)βT(s):=min⁡j=s+1,…,k{(k-s)(αs-αj)j-s+αj},s=0,…,q,
where αk:=-1. Moreover, we define
(2.7)s⋆:=min⁡{s∈{0,…,q}:βT(s)>0}.
Remark that βT(q)>0, since (2.5) implies αq=βT(q).Corollary 2.2.
Suppose thataj∈𝔸αj1/(k-j), where αj≥-1  for j=0,…,k-1, and let q∈{0,…,k-1} be the smallest index for which αq=maxj=0,…,k-1{αj}>0. Then each solution base of (1.1) admits at most s⋆≤q solutions f satisfying σT(f)<βT(s⋆). In particular, there are at most s⋆≤q solutions f satisfying σT(f)=0.To estimate the quantity∑j=1kσT(fj) by using Theorem 2.1, we set (2.8)γT(j):=max⁡{βT(0),…,βT(j)},j=0,…,q.
Evidently γT(j)>0 for j∈{s⋆,…,q}, and γT(j)≤0 for j∈{0,…,s⋆-1}.Corollary 2.3.
Suppose thataj∈𝔸αj1/(k-j), where αj≥-1 for j=0,…,k-1, and let q∈{0,…,k-1} be the smallest index for which αq=maxj=0,…,k-1{αj}>0. Let {f1,…,fk} be a solution base of (1.1). If q=0, then ∑j=1kσT(fj)=kα0, while if q≥1, then
(2.9)(k-q)αq+∑j=s⋆q-1γT(j)≤∑j=1kσT(fj)≤kαq.Note that the sum in (2.9) is considered to be empty, if s⋆=q.


## 2.2. Gundersen-Steinbart-Wang Method forT-Order

We proceed to give an alternative statement of Theorem2.1 and its corollaries by modifying the key steps in [4]. This yields a natural way to define possible intervals for T-orders of solutions of (1.1). As a consequence, we get a useful estimate following from Corollary 2.3.Setδj:=(αj+1)(k-j) for all j=0,…,k-1. Let s1∈{0,…,k-1} be the smallest index satisfying αs1=maxj=0,…,k-1{αj}>0, which is equivalent to (2.10)δs1k-s1=max⁡j=0,…,k-1{δjk-j}>1.
If s1 cannot be found, then all solutions of (1.1) are of zero T-order by Theorem A(ii). Otherwise, for a given sm, m∈ℕ, let sm+1∈{0,…,sm-1} be the smallest index satisfying (2.11)δsm+1-δsmsm-sm+1=max⁡j=0,…,sm-1{δj-δsmsm-j}>1.
Eventually this process will stop, yielding a finite list of indices s1,…,sp such that p≤k and s1>s2>⋯>sp≥0. Further, set (2.12)BT(t):=δst-δst-1st-1-st-1,t=1,…,p,
where s0:=k and δk:=0. Due to resemblance between (2.12) and [4, Equation (2.4)], it seems plausible that the possible nonzero T-orders of solutions of (1.1) in the unit disc case could be found among the numbers ℬT(t), where t=1,…,p. However, Example 3.1 shows that this is not the case.The following lemma allows us to view the results in Section2.1 in a new perspective. In particular, Lemma 2.4 emphasizes the connection between ℬT and γT.Lemma 2.4.
One has the following:(i)
ℬT(1)>ℬT(2)>⋯>ℬT(p)>0;(ii)
βT(st)=ℬT(t) for all t∈{1,…,p};(iii)
γT(q)=ℬT(1), γT(j)=ℬT(t) for all st≤j<st-1 and t∈{2,…,p}, and γT(j)≤0 for all j<sp. In particular, sp=s⋆.By relying on Lemma2.4, Theorem 2.1, and Corollary 2.2, we proceed to state possible intervals forT-orders of functions in solution bases of (1.1) in the case aj∈𝔸αj1/(k-j), where αj≥-1 for j=0,…,k-1. In fact, each solution base of (1.1) contains the following: (i)
at leastk-s1 solutions f satisfying σT(f)=ℬT(1);(ii)
at leastk-st solutions f satisfying σT(f)∈[ℬT(t),ℬT(1)] for t=2,…,p;(iii)
at mostsp solutions f satisfying σT(f)∈[0,ℬT(sp)). For the following application, let {f1,…,fk} be a solution base of (1.1). Knowing the possible intervals for T-orders, we get
(2.13)∑j=1kσT(fj)≥(k-s1)BT(1)+⋯+(sp-1-sp)BT(p)+sp⋅0=δsp+sp-k.
In view of Lemma 2.4, the lower estimates in (2.9) and (2.13) are equal.Finally, we point out a useful consequence of (2.13). If sp=0, then δsp+sp=δ0. If sp>0, then (δ0-δsp)/sp≤1 by (2.11), and δsp+sp≥δ0. Hence, in both cases we can state that
(2.14)∑j=1kσT(fj)≥δsp+sp-k≥δ0-k≥α0k,
where the equalities hold, if α0=maxj=0,…,k-1{αj}>0.


## 2.3. Growth of Solutions with Respect to Maximum Modulus

Alongside of theT-order, we may also define the M-order of growth of f∈ℋ(𝔻) by (2.15)σM(f):=limsup⁡r→1-log⁡+log⁡+M(r,f)-log⁡(1-r),
where M(r,f):=max|z|=r|f(z)| is the maximum modulus of f. It is well known that the inequalities (2.16)σT(f)≤σM(f)≤σT(f)+1
are satisfied for all f∈ℋ(𝔻), and all possibilities allowed by (2.16) can be assumed [9, Theorems  3.5–3.7]. A function f∈ℋ(𝔻) is said to belong to the weighted Hardy spaceHα∞, if there exists α≥0 such that (2.17)sup⁡z∈D(1-|z|2)α|f(z)|<∞.
Functions of maximal growth in ⋂α>pHα∞ are distinguished by denoting f∈ℍp∞, if p=inf{α≥0:f∈Hα∞}. Remark that H0∞=H∞ is the space of all bounded analytic functions in 𝔻. The union ⋃α>0Hα∞ is also known as the Korenblum space𝒜-∞ [10], and since [11] ℍp∞ is also known as Gp.If the growth of coefficients is measured by means of maximum modulus estimates, then it is natural to consider the growth of solutions with respect toM-order.Theorem B (see [3, Theorem  1.4]).
Suppose thataj∈ℍ(pj+1)(k-j)∞, where pj≥-1 for j=0,…,k-1, and denote pk:=-1.  (i)
Suppose that(2.18)min⁡j=1,…,k{k(p0-pj)j+pj}>1,
and let 1≤α<∞. Then all solutions f of (1.1) satisfyσM(f)≤α if and only if maxj=0,…,k-1{pj}≤α.(ii)
All nontrivial solutionsf of (1.1) satisfy σM(f)≤maxj=0,…,k-1{pj+}, and(2.19)min⁡j=1,…,k{k(p0-pj)j+pj}≤max⁡{σM(f),1}.(iii)
Suppose that (2.18) holds. If q∈{0,…,k-1} is the smallest index for which pq=maxj=0,…,k-1{pj}, then each solution base of (1.1) contains at least k-q linearly independent solutions f such that σM(f)=pq.To conclude [3, Equation  (4.17)] in the proof of Theorem B, the inequality [3, Equation  (1.9)], corresponding to (2.18), must be strict. By a simple modification of the proof of Theorem B, the assumption (2.18) can be relaxed to (2.20)max⁡j=0,…,k-1{pj}>1,
which allows us to apply Theorem B(iii) also in the case that there are solutions f satisfying σM(f)≤1. To see that (2.20) is in fact a weaker assumption than (2.18), we refer to [2, Example  10], which is further considered in Section 3.2. In this case (2.21)min⁡j=1,…,k{k(p0-pj)j+pj}=-4,max⁡j=0,…,k-1{pj}>1.
Note that by taking j=k in (2.18), we obtain p0>1. Hence (2.18) implies (2.20).Theorem2.5 corresponds to Theorem 2.1.Theorem 2.5.
Suppose thataj∈ℍ(pj+1)(k-j)∞, where pj≥-1 for j=0,…,k-1, and let q∈{0,…,k-1} be the smallest index for which pq=maxj=0,…,k-1{pj}. If s∈{0,…,q}, then each solution base of (1.1) contains at least k-s linearly independent solutions f such that
(2.22)min⁡j=s+1,…,k{(k-s)(ps-pj)j-s+pj}≤max⁡{σM(f),1}.Note that (2.22) gives information on σM(f) only in the case when the minimum in (2.22) is strictly greater than 1. The case s=0 in Theorem 2.5 reduces to Theorem B(ii), and the case s=q reduces to Theorem B(iii) with the assumption (2.20), since now (2.23)min⁡j=q+1,…,k{(k-q)(pq-pj)j-q+pj}=pq=max⁡j=0,…,k-1{pj},
where the minimum is attained for j=k. For a similar argumentation, see (2.5). Our contribution is to extend (2.22) for s∈{1,…,q-1}. The proof of Theorem 2.5 is sketched in Section 7, and the sharpness and the the special cases k=2 and k=3 are further discussed in Section 3.2.Letq∈{0,…,k-1} be the smallest index for which pq=maxj=0,…,k-1{pj}. If pq≤1, then all solutions f in each solution base of (1.1) satisfy σM(f)≤1 by Theorem B(ii). Suppose that pq>1. In order to state the following corollaries of Theorem 2.5, we denote (2.24)βM(s):=min⁡j=s+1,…,k{(k-s)(ps-pj)j-s+pj},s=0,…,q,
where pk:=-1. Moreover, we define (2.25)s⋆:=min⁡{s∈{0,…,q}:βM(s)>1}.
Remark that βM(q)>1, since (2.23) implies pq=βM(q).Corollary 2.6.
Suppose thataj∈ℍ(pj+1)(k-j)∞, where pj≥-1 for j=0,…,k-1, and let q∈{0,…,k-1} be the smallest index for which pq=maxj=0,…,k-1{pj}>1. Then each solution base of (1.1) admits at most s⋆≤q solutions f satisfying σM(f)<βM(s⋆). In particular, there are at most s⋆≤q solutions f satisfying σM(f)≤1.To estimate the quantity∑j=1kσM(fj) by using Theorem 2.5, we set (2.26)γM(j):=max⁡{βM(0),…,βM(j)},j=0,…,q.
Evidently γM(j)>1 for j∈{s⋆,…,q}, and γM(j)≤1 for j∈{0,…,s⋆-1}.Corollary 2.7.
Suppose thataj∈ℍ(pj+1)(k-j)∞, where pj≥-1 for j=0,…,k-1, and let q∈{0,…,k-1} be the smallest index for which pq=maxj=0,…,k-1{pj}>1. Let {f1,…,fk} be a solution base of (1.1). If q=0, then ∑j=1kσM(fj)=kp0, while if q≥1, then
(2.27)(k-q)pq+∑j=s⋆q-1γM(j)≤∑j=1kσM(fj)≤kpq.Note that the sum in (2.27) is considered to be empty, if s⋆=q.


## 2.4. Gundersen-Steinbart-Wang Method forM-Order

We proceed to give an alternative statement of Theorem2.5 and its corollaries by modifying the key steps in [4]. This yields a natural way to define the possible intervals for M-orders of solutions of (1.1). As a consequence, we get a useful estimate following from Corollary 2.7.Setδj:=(pj+1)(k-j) for all j=0,…,k-1. Let s1∈{0,…,k-1} be the smallest index satisfying ps1=maxj=0,…,k-1{pj}>1, which is equivalent to (2.28)δs1k-s1=max⁡j=0,…,k-1{δjk-j}>2.
If s1 cannot be found, then all solutions f of (1.1) satisfy σM(f)≤1 by Theorem B(ii). Otherwise, for a given sm, m∈ℕ, let sm+1∈{0,…,sm-1} be the smallest index satisfying (2.29)δsm+1-δsmsm-sm+1=max⁡j=0,…,sm-1{δj-δsmsm-j}>2.
Eventually this process will stop, yielding a finite list of indices s1,…,sp such that p≤k and s1>s2>⋯>sp≥0. Further, set (2.30)BM(t):=δst-δst-1st-1-st-1,t=1,…,p,
where s0:=k and δk:=0. By Example 3.1, it is possible that (1.1) possesses a solution f of M-order strictly greater than one such that σM(f)≠ℬM(t) for all t=1,…,p.The following lemma, which can be proved similarly than Lemma2.4, allows us to view the results in Section 2.3 in a new perspective.Lemma 2.8.
One has the following:(i)
ℬM(1)>ℬM(2)>⋯>ℬM(p)>1;(ii)
βM(st)=ℬM(t) for all t∈{1,…,p};(iii)
γM(q)=ℬM(1), γM(j)=ℬM(t) for all st≤j<st-1 and t∈{2,…,p}, and γM(j)≤1 for all j<sp. In particular, sp=s⋆.By relying on Lemma2.8, Theorem 2.5, and Corollary 2.6, we proceed to state possible intervals forM-orders of functions in solution bases of (1.1) in the case aj∈ℍ(pj+1)(k-j)∞, where pj≥-1 for j=0,…,k-1. In fact, each solution base of (1.1) contains the following: (i)
at leastk-s1 solutions f satisfying σM(f)=ℬM(1);(ii)
at leastk-st solutions f satisfying σM(f)∈[ℬM(t),ℬM(1)] for t=2,…,p;(iii)
at mostsp solutions f satisfying σM(f)∈[0,ℬM(sp)).For results of the same type, we refer to [12, Theorem  1] and [13,  Corollary   1]. To compare (i) and (ii) to the estimates given in [13, Corollary  1], note that there is −1 in (2.30) instead of −2 in [13, Equation  (1.3)]. Evidently, assertions (i) and (ii) improve the estimates given for the M-orders of solutions in [13, Corollary    1]. Moreover, by means of (2.16) we see that (i) and (ii) reduce to [13, Corollary    1], if we consider the growth of solutions of (1.1) with respect to T-order.For the following application, let{f1,…,fk} be a solution base of (1.1). Knowing the possible intervals for M-orders, we get (2.31)∑j=1kσM(fj)≥(k-s1)BM(1)+⋯+(sp-1-sp)BM(p)+sp⋅0=δsp+sp-k.
Corresponding to the case in Section 2.2, by means of Lemma 2.8 we see that the lower estimates in (2.27) and (2.31) are equal.Finally, we point out a practical estimate, which is a consequence of (2.31). If sp=0, then δsp+sp=δ0-sp. If sp>0, then (δ0-δsp)/sp≤2 by (2.29), and δsp+sp≥δ0-sp. Hence, in both cases we can state that (2.32)∑j=1kσM(fj)≥δ0-sp-k≥p0k-sp.
We conclude that if s1=0, then the equalities hold in (2.32), since in this case sp=s1=0. Note that if (2.18) holds, then we can conclude that sp=0.


## 3. Sharpness Discussion

### 3.1. Sharpness of Theorem2.1

We may assume thatmaxj=0,…,k-1{αj}>0, for otherwise all solutions are of zero T-order. If k=2, then the statement of Theorem 2.1 is contained in Theorem A, and all the assertions are sharp [2, Examples  3 and 6].Ifk=3, then we have three different cases to consider.  (A1)
Ifα1,α2≤α0, then all nontrivial solutions f of (1.1) satisfy σT(f)=α0. In this case s=0=q.(A2)
Ifα0<α1 and α2≤α1, then in every solution base {f1,f2,f3} of (1.1) there are at least two solutions f1 and f2 such that σT(fj)=α1 for both j=1,2, and all solutions fj satisfy
(3.1)σT(fj)≥min⁡{3α0-2α1,32α0-12α2,α0},j=1,2,3.
In this case s=0 or s=1=q.(A3)
Ifα0,α1<α2, then in every solution base {f1,f2,f3} of (1.1) there is at least one solution f1 such that σT(f1)=α2, two solutions f1 and f2 such that
(3.2)σT(fj)≥min⁡{2α1-α2,α1},j=1,2,
and all solutions fj satisfy (3.1). In this case s=0, s=1, or s=2=q.It is clear that the assertion in (A1) is sharp, and so are the ones in (A2) by [2, Example   10]. Moreover, [2,  Example   9] shows that the assertions in (A3) are sharp for s=0 and s=2. Example 3.2 shows the sharpness of the assertions in (A3) for s=0,1,2. That is, in all cases there exists a solution for which the lower bound for the T-order of growth is attained.


### 3.2. Sharpness of Theorem2.5

We may assume thatmaxj=0,…,k-1{pj}>1, for otherwise all solutions f of (1.1) satisfy max{σM(f),1}=1, and we cannot conclude anything from (2.22). If k=2, then the statement of Theorem 2.5 is contained in Theorem B, and all the assertions are sharp by [2, Examples  3 and 6]. In the case of [2, Examples  3 and 6], for β>1, linearly independent solutions f1 and f2 satisfy σM(f1)=β and σM(f2)=β+2. Moreover, aj∈ℍ(pj+1)(2-j)∞, where p0=β+1 and p1=β+2. Note that max{p0,p1}=p1=β+2>1, and hence q=1. An easy computation shows the sharpness for s=0 and for s=q=1. In the case of [2, Example  6], for β>1, linearly independent solutions f1 and f2 satisfy σM(f1)=β and σM(f2)=β. Moreover, aj∈ℍ(pj+1)(2-j)∞, where p0=β and p1=-1/2. Note now that max{p0,p1}=p0=β>1, and hence q=0. This example shows the sharpness for s=q=0. For another example, see [3, Example  2].Ifk=3, then we have three different cases to consider.  (B1)
Ifp1,p2≤p0, then s=0=q, and all nontrivial solutions f of (1.1) satisfy σM(f)=p0 by (2.23).(B2)
Ifp0<p1, and p2≤p1, then in every solution base {f1,f2,f3} of (1.1) there are at least two solutions f1 and f2 such that σM(fj)=p1 for both j=1,2, and all solutions fj satisfy
(3.3)max⁡{σM(fj),1}≥min⁡{3p0-2p1,32p0-12p2,p0},j=1,2,3.
In this case s=0 or s=1=q.(B3)
Ifp0,p1<p2, then in every solution base {f1,f2,f3} of (1.1) there is at least one solution f1 such that σM(f1)=p2, two solutions f1 and f2 such that
(3.4)max⁡{σM(fj),1}≥min⁡{2p1-p2,p1},j=1,2,
and all solutions fj satisfy (3.3). In this case s=0, s=1, or s=2=q.It is clear that the assertion in (B1) is sharp. By [2, Example  10], we see that the assertion in (B2) corresponding to s=1 is sharp. In this case, for β>1, linearly independent solutions f1,f2, and f3 satisfy σM(f1)=σM(f2)=β and σM(f3)=0. Now aj∈ℍ(pj+1)(3-j)∞, where p0=(2/3)β, p1=β+2, and p2=0. Moreover, by [2, Example   9], we see that the assertions in (B3) are sharp for s=0 and s=2. In this case for β>1, linearly independent solutions f1, f2, and f3 satisfy σM(f1)=σM(f2)=β and σM(f3)=2β. Moreover, aj∈ℍ(pj+1)(3-j)∞, where p0=(4/3)β, p1=β, and p2=2β. Example 3.2 shows the sharpness of the assertions in (B3) for s=0,1,2. That is, in all cases there exists a solution for which equality holds in (2.22).


### 3.3. Examples

Example3.1 shows that a unit disc counterpart of the finite set constructed by Gundersen-Steinbart-Wang does not contain growth orders of solutions of (3.5)f′′+a1(z)f′+a0(z)f=0,
if coefficients belong either to weighted Bergman spaces or to weighted Hardy spaces.Example 3.1.
Letα,β∈ℝ be constants satisfying 1<β<α<2β-1. Then the functions
(3.6)f1(z)=(1-z)α+βexp⁡((11-z)α+(11+z)β),f2(z)=(1-z)α+βexp⁡(11+z)β
are linearly independent analytic solutions of (3.5), where
(3.7)a0(z)=β2(1+z)2+2β+β(α+3β)(γ+z)(1-z)(1+z)2+β-α(α+β)(1-z)2+α-αβ(1-z)1+α(1+z)1+β+β(α+β)(1-z)2,a1(z)=2β(1+z)1+β-α(1-z)1+α+α+2β-11-z
belong to ℋ(𝔻), and γ=(α+β-2)/(α+3β)∈(0,1/2).
It is clear thataj∈𝔸αj1/(2-j), where α0=β-1 and α1=α-1. We calculate that s1=1, s2=0, ℬT(1)=α-1, and ℬT(2)=2β-α-1. Hence [2β-α-1,α-1] is the only possible interval for T-orders of solutions of (3.5). Since σT(f2)=β-1, we conclude that the T-order of a solution does not have to be one of the endpoints.
On the other hand, it is also clear thataj∈ℍ(pj+1)(2-j)∞, where p0=β and p1=α. We calculate that s1=1, s2=0, ℬM(1)=α, and ℬM(2)=2β-α. Hence [2β-α,α] is the only possible interval for M-orders of solutions of (3.5). Since σM(f2)=β, we conclude that the M-order of a solution does not have to be one of the endpoints.
The following example demonstrates the sharpness of Theorems2.1 and 2.5 in the case that they do not reduce to known results.Example 3.2   2.
Letβ>1, and denote g(z)=(5/(1-z))β. Then the functions
(3.8)fj(z)=(1-z)βexp⁡(g(z)j),j=1,2,3,
are linearly independent solutions of f′′′+a2(z)f′′+a1(z)f′+a0(z)f=0, where
(3.9)a2(z)=P2(g(z))(1-z)Q(g(z)),a1(z)=P1(g(z))(1-z)2Q(g(z)),a0(z)=β3P0(g(z))(1-z)3Q(g(z))
are such that
(3.10)P2(ζ)=54βζ8-27βζ7-24βζ6+(108β+54)ζ5-(82β+63)ζ4+3βζ3+(22β+39)ζ2-(6β+24)ζ+6,P1(ζ)=-108β2ζ10+72β2ζ9+(27β2-54β)ζ8+(27β-135β2)ζ7+(24β+15β2)ζ6+(64β2-108β-18)ζ5+(21+82β-51β2)ζ4-(3β+15β2)ζ3+(31β2-22β-13)ζ2+(8+6β-14β2)ζ+2β2-2,P0(ζ)=108ζ11-234ζ10+126ζ9+123ζ8-276ζ7+183ζ6-104ζ5+40ζ4-6ζ3-4ζ2,Q(ζ)=-18ζ5+21ζ4-13ζ2+8ζ-2.
The zeros of Q(ζ) lie in the open disc of radius 1+21/18 centered at the origin by [14, Lemma  1.3.2]. Since |g(z)|>|5/(1-z)|>5/2>39/18 for all z∈𝔻, we conclude that a0,a1,a2∈ℋ(𝔻). In fact, the coefficients a0,a1,anda2 satisfy
(3.11)a2(z)∼(11-z)3β+1,a1(z)∼(11-z)5β+2,a0(z)∼(11-z)6β+3,
in a neighborhood of z=1, while they are bounded in a neighborhood of any boundary point in ∂𝔻∖{1}.
Note thataj∈𝔸αj1/(3-j), where α2=3β-1, α1=(5/2)β-1, and α0=2β-1. Evidently σT(fj)=βj-1 for j=1,2,3. We deduce that there is one solution f3 such that σT(f3)=α2=3β-1, two solutions f2 and f3 such that
(3.12)σT(f3)>σT(f2)=min⁡{2α1-α2,α1}=2β-1,
and three solutions f1, f2, and f3 such that
(3.13)σT(f3)>σT(f2)>σT(f1)=min⁡{3α0-2α1,32α0-12α2,α0}=β-1.
That is, in all cases s=0,1,2 there exists a solution for which the lower bound in (2.4) is attained. Further, this example is in line with Corollary 2.2, since all solutions f1, f2, and f3 are of strictly positive T-order, and in this case s*=0.
NowγT(0)=βT(0)=β-1, γT(1)=βT(1)=2β-1, and γT(2)=βT(2)=3β-1. It follows that for the solution base {f1,f2,f3} equality holds in the first inequality in (2.9), and for the solution base {f1+f3,f2+f3,f3} equality holds in the last inequality in (2.9). This shows the sharpness of Corollary 2.3.
On the other hand,aj∈ℍ(pj+1)(3-j)∞, where p2=3β, p1=5β/2, and p0=2β. Evidently σM(fj)=βj for j=1,2,3. We deduce that there is one solution f3 such that σM(f3)=p2=3β>1, two solutions f2 and f3 such that
(3.14)σM(f3)>σM(f2)=min⁡{2p1-p2,p1}=2β>1,
and three solutions f1, f2, and f3 such that
(3.15)σM(f3)>σM(f2)>σM(f1)=min⁡{3p0-2p1,32p0-12p2,p0}=β.
That is, in all cases s=0,1,2 there exists a solution for which the lower bound in (2.22) is attained. Further, this example is in line with Corollary 2.6, since all solutions f1, f2, and f3 are of M-order strictly greater than 1, and in this case s⋆=0.
NowγM(0)=βM(0)=β, γM(1)=βM(1)=2β, and γM(2)=βM(2)=3β. It follows that for the solution base {f1,f2,f3} equality holds in (2.27), and for the solution base {f1+f3,f2+f3,f3} upper bound for the sum of M-orders is attained. This shows the sharpness of Corollary 2.7.


## 3.1. Sharpness of Theorem2.1

We may assume thatmaxj=0,…,k-1{αj}>0, for otherwise all solutions are of zero T-order. If k=2, then the statement of Theorem 2.1 is contained in Theorem A, and all the assertions are sharp [2, Examples  3 and 6].Ifk=3, then we have three different cases to consider.  (A1)
Ifα1,α2≤α0, then all nontrivial solutions f of (1.1) satisfy σT(f)=α0. In this case s=0=q.(A2)
Ifα0<α1 and α2≤α1, then in every solution base {f1,f2,f3} of (1.1) there are at least two solutions f1 and f2 such that σT(fj)=α1 for both j=1,2, and all solutions fj satisfy
(3.1)σT(fj)≥min⁡{3α0-2α1,32α0-12α2,α0},j=1,2,3.
In this case s=0 or s=1=q.(A3)
Ifα0,α1<α2, then in every solution base {f1,f2,f3} of (1.1) there is at least one solution f1 such that σT(f1)=α2, two solutions f1 and f2 such that
(3.2)σT(fj)≥min⁡{2α1-α2,α1},j=1,2,
and all solutions fj satisfy (3.1). In this case s=0, s=1, or s=2=q.It is clear that the assertion in (A1) is sharp, and so are the ones in (A2) by [2, Example   10]. Moreover, [2,  Example   9] shows that the assertions in (A3) are sharp for s=0 and s=2. Example 3.2 shows the sharpness of the assertions in (A3) for s=0,1,2. That is, in all cases there exists a solution for which the lower bound for the T-order of growth is attained.


## 3.2. Sharpness of Theorem2.5

We may assume thatmaxj=0,…,k-1{pj}>1, for otherwise all solutions f of (1.1) satisfy max{σM(f),1}=1, and we cannot conclude anything from (2.22). If k=2, then the statement of Theorem 2.5 is contained in Theorem B, and all the assertions are sharp by [2, Examples  3 and 6]. In the case of [2, Examples  3 and 6], for β>1, linearly independent solutions f1 and f2 satisfy σM(f1)=β and σM(f2)=β+2. Moreover, aj∈ℍ(pj+1)(2-j)∞, where p0=β+1 and p1=β+2. Note that max{p0,p1}=p1=β+2>1, and hence q=1. An easy computation shows the sharpness for s=0 and for s=q=1. In the case of [2, Example  6], for β>1, linearly independent solutions f1 and f2 satisfy σM(f1)=β and σM(f2)=β. Moreover, aj∈ℍ(pj+1)(2-j)∞, where p0=β and p1=-1/2. Note now that max{p0,p1}=p0=β>1, and hence q=0. This example shows the sharpness for s=q=0. For another example, see [3, Example  2].Ifk=3, then we have three different cases to consider.  (B1)
Ifp1,p2≤p0, then s=0=q, and all nontrivial solutions f of (1.1) satisfy σM(f)=p0 by (2.23).(B2)
Ifp0<p1, and p2≤p1, then in every solution base {f1,f2,f3} of (1.1) there are at least two solutions f1 and f2 such that σM(fj)=p1 for both j=1,2, and all solutions fj satisfy
(3.3)max⁡{σM(fj),1}≥min⁡{3p0-2p1,32p0-12p2,p0},j=1,2,3.
In this case s=0 or s=1=q.(B3)
Ifp0,p1<p2, then in every solution base {f1,f2,f3} of (1.1) there is at least one solution f1 such that σM(f1)=p2, two solutions f1 and f2 such that
(3.4)max⁡{σM(fj),1}≥min⁡{2p1-p2,p1},j=1,2,
and all solutions fj satisfy (3.3). In this case s=0, s=1, or s=2=q.It is clear that the assertion in (B1) is sharp. By [2, Example  10], we see that the assertion in (B2) corresponding to s=1 is sharp. In this case, for β>1, linearly independent solutions f1,f2, and f3 satisfy σM(f1)=σM(f2)=β and σM(f3)=0. Now aj∈ℍ(pj+1)(3-j)∞, where p0=(2/3)β, p1=β+2, and p2=0. Moreover, by [2, Example   9], we see that the assertions in (B3) are sharp for s=0 and s=2. In this case for β>1, linearly independent solutions f1, f2, and f3 satisfy σM(f1)=σM(f2)=β and σM(f3)=2β. Moreover, aj∈ℍ(pj+1)(3-j)∞, where p0=(4/3)β, p1=β, and p2=2β. Example 3.2 shows the sharpness of the assertions in (B3) for s=0,1,2. That is, in all cases there exists a solution for which equality holds in (2.22).


## 3.3. Examples

Example3.1 shows that a unit disc counterpart of the finite set constructed by Gundersen-Steinbart-Wang does not contain growth orders of solutions of (3.5)f′′+a1(z)f′+a0(z)f=0,
if coefficients belong either to weighted Bergman spaces or to weighted Hardy spaces.Example 3.1.
Letα,β∈ℝ be constants satisfying 1<β<α<2β-1. Then the functions
(3.6)f1(z)=(1-z)α+βexp⁡((11-z)α+(11+z)β),f2(z)=(1-z)α+βexp⁡(11+z)β
are linearly independent analytic solutions of (3.5), where
(3.7)a0(z)=β2(1+z)2+2β+β(α+3β)(γ+z)(1-z)(1+z)2+β-α(α+β)(1-z)2+α-αβ(1-z)1+α(1+z)1+β+β(α+β)(1-z)2,a1(z)=2β(1+z)1+β-α(1-z)1+α+α+2β-11-z
belong to ℋ(𝔻), and γ=(α+β-2)/(α+3β)∈(0,1/2).
It is clear thataj∈𝔸αj1/(2-j), where α0=β-1 and α1=α-1. We calculate that s1=1, s2=0, ℬT(1)=α-1, and ℬT(2)=2β-α-1. Hence [2β-α-1,α-1] is the only possible interval for T-orders of solutions of (3.5). Since σT(f2)=β-1, we conclude that the T-order of a solution does not have to be one of the endpoints.
On the other hand, it is also clear thataj∈ℍ(pj+1)(2-j)∞, where p0=β and p1=α. We calculate that s1=1, s2=0, ℬM(1)=α, and ℬM(2)=2β-α. Hence [2β-α,α] is the only possible interval for M-orders of solutions of (3.5). Since σM(f2)=β, we conclude that the M-order of a solution does not have to be one of the endpoints.
The following example demonstrates the sharpness of Theorems2.1 and 2.5 in the case that they do not reduce to known results.Example 3.2   2.
Letβ>1, and denote g(z)=(5/(1-z))β. Then the functions
(3.8)fj(z)=(1-z)βexp⁡(g(z)j),j=1,2,3,
are linearly independent solutions of f′′′+a2(z)f′′+a1(z)f′+a0(z)f=0, where
(3.9)a2(z)=P2(g(z))(1-z)Q(g(z)),a1(z)=P1(g(z))(1-z)2Q(g(z)),a0(z)=β3P0(g(z))(1-z)3Q(g(z))
are such that
(3.10)P2(ζ)=54βζ8-27βζ7-24βζ6+(108β+54)ζ5-(82β+63)ζ4+3βζ3+(22β+39)ζ2-(6β+24)ζ+6,P1(ζ)=-108β2ζ10+72β2ζ9+(27β2-54β)ζ8+(27β-135β2)ζ7+(24β+15β2)ζ6+(64β2-108β-18)ζ5+(21+82β-51β2)ζ4-(3β+15β2)ζ3+(31β2-22β-13)ζ2+(8+6β-14β2)ζ+2β2-2,P0(ζ)=108ζ11-234ζ10+126ζ9+123ζ8-276ζ7+183ζ6-104ζ5+40ζ4-6ζ3-4ζ2,Q(ζ)=-18ζ5+21ζ4-13ζ2+8ζ-2.
The zeros of Q(ζ) lie in the open disc of radius 1+21/18 centered at the origin by [14, Lemma  1.3.2]. Since |g(z)|>|5/(1-z)|>5/2>39/18 for all z∈𝔻, we conclude that a0,a1,a2∈ℋ(𝔻). In fact, the coefficients a0,a1,anda2 satisfy
(3.11)a2(z)∼(11-z)3β+1,a1(z)∼(11-z)5β+2,a0(z)∼(11-z)6β+3,
in a neighborhood of z=1, while they are bounded in a neighborhood of any boundary point in ∂𝔻∖{1}.
Note thataj∈𝔸αj1/(3-j), where α2=3β-1, α1=(5/2)β-1, and α0=2β-1. Evidently σT(fj)=βj-1 for j=1,2,3. We deduce that there is one solution f3 such that σT(f3)=α2=3β-1, two solutions f2 and f3 such that
(3.12)σT(f3)>σT(f2)=min⁡{2α1-α2,α1}=2β-1,
and three solutions f1, f2, and f3 such that
(3.13)σT(f3)>σT(f2)>σT(f1)=min⁡{3α0-2α1,32α0-12α2,α0}=β-1.
That is, in all cases s=0,1,2 there exists a solution for which the lower bound in (2.4) is attained. Further, this example is in line with Corollary 2.2, since all solutions f1, f2, and f3 are of strictly positive T-order, and in this case s*=0.
NowγT(0)=βT(0)=β-1, γT(1)=βT(1)=2β-1, and γT(2)=βT(2)=3β-1. It follows that for the solution base {f1,f2,f3} equality holds in the first inequality in (2.9), and for the solution base {f1+f3,f2+f3,f3} equality holds in the last inequality in (2.9). This shows the sharpness of Corollary 2.3.
On the other hand,aj∈ℍ(pj+1)(3-j)∞, where p2=3β, p1=5β/2, and p0=2β. Evidently σM(fj)=βj for j=1,2,3. We deduce that there is one solution f3 such that σM(f3)=p2=3β>1, two solutions f2 and f3 such that
(3.14)σM(f3)>σM(f2)=min⁡{2p1-p2,p1}=2β>1,
and three solutions f1, f2, and f3 such that
(3.15)σM(f3)>σM(f2)>σM(f1)=min⁡{3p0-2p1,32p0-12p2,p0}=β.
That is, in all cases s=0,1,2 there exists a solution for which the lower bound in (2.22) is attained. Further, this example is in line with Corollary 2.6, since all solutions f1, f2, and f3 are of M-order strictly greater than 1, and in this case s⋆=0.
NowγM(0)=βM(0)=β, γM(1)=βM(1)=2β, and γM(2)=βM(2)=3β. It follows that for the solution base {f1,f2,f3} equality holds in (2.27), and for the solution base {f1+f3,f2+f3,f3} upper bound for the sum of M-orders is attained. This shows the sharpness of Corollary 2.7.


## 4. Proof of Theorem2.1

The following lemma on the order reduction procedure originates fromℂ.Lemma C (see [4, Lemma  6.4]).
Letf0,1,f0,2,…,f0,m be  m≥2 linearly independent meromorphic solutions of
(4.1)y(k)+a0,k-1(z)y(k-1)+⋯+a0,0(z)y=0,k≥m,
where a0,0(z),…,a0,k-1(z) are meromorphic functions in 𝔻. For 1≤p≤m-1, set
(4.2)fp,j=(fp-1,j+1fp-1,1)′,j=1,…,m-p.
Then fp,1,fp,2,…,fp,m-p are linearly independent meromorphic solutions of
(4.3)y(k-p)+ap,k-p-1(z)y(k-p-1)+⋯+ap,0(z)y=0,
where
(4.4)ap,j(z)=∑n=j+1k-p+1(nj+1)ap-1,n(z)fp-1,1(n-j-1)(z)fp-1,1(z)
for j=0,…,k-p-1. Here an,k-n(z)≡1 for all n=0,…,p.Lemma D (see [15, Theorem  E(b)]).
Letk and j be integers satisfying k>j≥0, and let ε>0. If f is meromorphic in 𝔻 such that σT(f)<∞, and f(j)≢0, then
(4.5)∫D|f(k)(z)f(j)(z)|1/(k-j)(1-|z|)σT(f)+εdm(z)<∞.


### 4.1. Cases=1

Letk≥3, q≥2, s=1, and βT(1)>0, since otherwise there is nothing to prove. In particular, if α1≤0, then (2.4) is trivial, since by taking j=k in (2.6), we obtain βT(1)≤α1≤0. Let {f0,1,f0,2,…,f0,k} be a solution base of (1.1), and assume on the contrary to the assertion that there exist s+1=2 linearly independent solutions f0,1 and f0,2 such that max{σT(f0,1),σT(f0,2)}=:σ<βT(1). Then the meromorphic function g:=(f0,1/f0,2)′ satisfies σT(g)≤σ. Moreover, Lemma C implies that g satisfies (4.6)g(k-1)+a1,k-2(z)g(k-2)+⋯+a1,0(z)g=0,
where (4.7)a1,j(z)=a0,j+1(z)+∑n=j+2k(nj+1)a0,n(z)f0,1(n-j-1)(z)f0,1(z)
for j=0,1,…,k-2, and a0,k(z)≡1. Therefore (4.8)|a0,1(z)|≤|a1,0(z)|+∑n=2k(n1)|a0,n(z)||f0,1(n-1)(z)f0,1(z)|,
where (4.9)|a1,0(z)|≤|g(k-1)(z)g(z)|+|a1,k-2(z)||g(k-2)(z)g(z)|+⋯+|a1,1(z)||g′(z)g(z)|,
since g satisfies (4.6). Putting the last two inequalities together, we obtain (4.10)|a0,1(z)|≲∑j=1k-1|a1,j(z)||g(j)(z)g(z)|+∑n=2k|a0,n(z)||f0,1(n-1)(z)f0,1(z)|.
Let ε>0. Raising both sides to the power 1/(k-1) and integrating over the disc D(0,r) of radius r∈(0,1) with respect to (1-|z|2)α1-εdm(z), we obtain (4.11)∫D(0,r)|a0,1(z)|1/(k-1)(1-|z|2)α1-εdm(z)≲∑j=1k-1∫D|a1,j(z)|1/(k-1)|g(j)(z)g(z)|1/(k-1)(1-|z|2)α1-εdm(z)+∑n=2k∫D|a0,n(z)|1/(k-1)|f0,1(n-1)(z)f0,1(z)|1/(k-1)(1-|z|2)α1-εdm(z).To deal with the second sum in (4.11), consider (4.12)In:=∫D|a0,n(z)|1/(k-1)|f0,1(n-1)(z)f0,1(z)|1/(k-1)(1-|z|2)α1-εdm(z),n=2,…,k.
By Lemma D, (4.13)Ik=∫D|f0,1(k-1)(z)f0,1(z)|1/(k-1)(1-|z|2)α1-εdm(z)<∞
for ε>0 being small enough since σT(f0,1)≤σ<βT(1)≤α1. Moreover, by Hölder’s inequality, with indices (k-1)/(k-n) and (k-1)/(n-1), we have (4.14)In≤(∫D|a0,n(z)|1/(k-n)(1-|z|2)αn+εdm(z))(k-n)/(k-1)⋅(∫D|f0,1(n-1)(z)f0,1(z)|1/(n-1)(1-|z|2)ω1(n)dm(z))(n-1)/(k-1)
for all n=2,…,k-1, where (4.15)ωs(n):=(k-s)(αs-αn)n-s+αn-2k-n-sn-sε.
The first member in the product is finite since a0,n∈𝔸αn1/(k-n) for all n=2,…,k-1 by the assumption, and so is the second one for ε>0 small enough since (4.16)σT(f0,1)≤σ<βT(1)≤(k-1)(α1-αn)n-1+αn,n=2,…,k-1,
by the antithesis. Thus ∑n=2kIn is finite for ε>0 being small enough.To deal with the first sum in (4.11), denote (4.17)Jj:=∫D|a1,j(z)|1/(k-1)|g(j)(z)g(z)|1/(k-1)(1-|z|2)α1-εdm(z),j=1,…,k-1.
Lemma D implies that (4.18)Jk-1=∫D|g(k-1)(z)g(z)|1/(k-1)(1-|z|2)α1-εdm(z)<∞
for ε>0 being small enough since σT(g)≤σ<βT(1)≤α1. Moreover, by (4.7) we have (4.19)Jj≲∫D|a0,j+1(z)|1/(k-1)|g(j)(z)g(z)|1/(k-1)(1-|z|2)α1-εdm(z)+∑n=j+2k∫D|a0,n(z)|1/(k-1)|f0,1(n-j-1)(z)f0,1(z)|1/(k-1)|g(j)(z)g(z)|1/(k-1)(1-|z|2)α1-εdm(z)=:Kj+Lj,k+∑n=j+2k-1Lj,n
for all j=1,…,k-2. Since max{σT(g),σT(f0,1)}≤σ<βT(1), we deduce that Kj behaves like Ij+1 and hence ∑j=1k-2Kj<∞ for ε>0 being small enough. Moreover, by Hölder’s inequality, with indices (k-1)/(k-j-1) and (k-1)/j, and Lemma D we have (4.20)Lj,k≤(∫D|f0,1(k-j-1)(z)f0,1(z)|1/(k-j-1)(1-|z|2)α1-εdm(z))(k-j-1)/(k-1)⋅(∫D|g(j)(z)g(z)|1/j(1-|z|2)α1-εdm(z))j/(k-1)<∞
for all j=1,…,k-2 when ε>0 is sufficiently small. It remains to consider the double sum ∑j=1k-3∑n=j+2k-1Lj,n. By Hölder’s inequality, with indices (k-1)/(k-n) and (k-1)/(n-1), we have (4.21)Lj,n≤(∫D|a0,n(z)|1/(k-n)(1-|z|2)αn+εdm(z))(k-n)/(k-1)⋅(∫D|f0,1(n-j-1)(z)f0,1(z)|1/(n-1)|g(j)(z)g(z)|1/(n-1)(1-|z|2)ω1(n)dm(z))(n-1)/(k-1),
where ω1(n) is defined in (4.15). The first term in the product is bounded for all ε>0 since a0,n∈𝔸αn1/(k-n) for all n=3,…,k-1 by the assumption. One more application of Hölder's inequality, with indices (n-1)/(n-j-1) and (n-1)/j, together with Lemma D and the antithesis shows that also the second term in the product is bounded for ε>0 being small enough, and thus ∑j=1k-3∑n=j+2k-1Lj,n<∞ for ε>0 being small enough.We have proved that the right-hand side of (4.11) is uniformly bounded for all r∈(0,1), if ε>0 is small enough. However, a0,1∈𝔸α11/(k-1) by the assumption, and hence the left-hand side of (4.11) diverges as r→1-. Contradiction follows.


### 4.2. Cases>1

Letk≥3, q≥2, s>1, and βT(s)>0, since otherwise there is nothing to prove. In particular, it follows that αs>0. Let {f0,1,f0,2,…,f0,k} be a solution base of (1.1), and assume on the contrary to the assertion that there exist s+1 linearly independent solutions f0,1,…,f0,s+1 such that (4.22)σ:=max⁡{σT(f0,1),…,σT(f0,s+1)}<βT(s).
Then the meromorphic functions f1,j=(f0,j+1/f0,1)′ satisfy σT(f1,j)≤σ for all j=1,…,s. This in turn implies that f2,j=(f1,j+1/f1,1)′ satisfy σT(f2,j)≤σ for all j=1,…,s-1. In general, σT(fn,j)≤σ for all j=1,…,s-n+1 and n=1,…,s. Moreover, as in the case s=1, Lemma C implies (4.23)|a0,s(z)|≤|a1,s-1(z)|+∑n=s+1k(ns)|a0,n(z)||f0,1(n-s)(z)f0,1(z)|≤|a2,s-2(z)|+∑n=sk-1(ns-1)|a1,n(z)||f1,1(n-s+1)(z)f1,1(z)|+∑n=s+1k(ns)|a0,n(z)||f0,1(n-s)(z)f0,1(z)|≤⋯≤|as,0(z)|+∑m=0s-1∑n=s+1-mk-m(ns-m)|am,n(z)||fm,1(n-s+m)(z)fm,1(z)|,
where (4.24)|as,0(z)|≤|fs,1(k-s)(z)fs,1(z)|+∑m=1k-s-1|as,m(z)||fs,1(m)(z)fs,1(z)|.
Putting these inequalities together, we obtain (4.25)|a0,s(z)|≲∑m=0s∑n=s+1-mk-m-1|am,n(z)||fm,1(n-s+m)(z)fm,1(z)|+∑m=0s|fm,1(k-s)(z)fm,1(z)|.
Let ε>0. Raising both sides to the power 1/(k-s) and integrating over the disc D(0,r) with respect to (1-|z|2)αs-εdm(z), we obtain (4.26)∫D(0,r)|a0,s(z)|1/(k-s)(1-|z|2)αs-εdm(z)≲∑m=0s∑n=s+1-mk-m-1∫D|am,n(z)|1/(k-s)|fm,1(n-s+m)(z)fm,1(z)|1/(k-s)(1-|z|2)αs-εdm(z)+∑m=0s∫D|fm,1(k-s)(z)fm,1(z)|1/(k-s)(1-|z|2)αs-εdm(z)=:∑m=0s∑n=s+1-mk-m-1Im,n+∑m=0sJm.
Lemma D and the antithesis imply that ∑m=0sJm<∞ for ε>0 being small enough. Hence, in order to obtain a contradiction with (4.26) and the assumption a0,s∈𝔸αs1/(k-s), it suffices to show that Im,n<∞ for all m=0,…,s and n=s+1-m,…,k-m-1 when ε>0 is sufficiently small.By Hölder's inequality, with indices(k-s)/(k-n) and (k-s)/(n-s), we have (4.27)I0,n≤(∫D|a0,n(z)|1/(k-n)(1-|z|2)αn+εdm(z))(k-n)/(k-s)⋅(∫D|f0,1(n-s)(z)f0,1(z)|1/(n-s)(1-|z|2)ωs(n)dm(z))(n-s)/(k-s)
for all n=s+1,…,k-1, where ωs(n) is defined in (4.15). The first member in the product is finite since a0,n∈𝔸αn1/(k-n) for all n=s+1,…,k-1 by the assumption, and so is the second one for ε>0 being small enough, since (4.28)σT(f0,1)≤σ<βT(s)≤(k-s)(αs-αn)n-s+αn,n=s+1,…,k-1,
by the antithesis. In the general case Lemma C gives (4.29)Im,n=∫D|am,n(z)|1/(k-s)|fm,1(n-s+m)(z)fm,1(z)|1/(k-s)(1-|z|2)αs-εdm(z)≲∑n1=n+1k-m+1∫D|am-1,n1(z)|1/(k-s)|fm-1,1(n1-n-1)(z)fm-1,1(z)|1/(k-s)⋅|fm,1(n-s+m)(z)fm,1(z)|1/(k-s)(1-|z|2)αs-εdm(z)≲∑n1=n+1k-m+1∑n2=n1+1k-m+2∫D|am-2,n2(z)|1/(k-s)|fm-2,1(n2-n1-1)(z)fm-2,1(z)|1/(k-s)⋅|fm-1,1(n1-n-1)(z)fm-1,1(z)|1/(k-s)|fm,1(n-s+m)(z)fm,1(z)|1/(k-s)(1-|z|2)αs-εdm(z),
and finally (4.30)Im,n≲∑n1=n+1k-m+1∑n2=n1+1k-m+2⋯∑nm=nm-1+1kK(n,n1,…,nm),
where (4.31)K(n,n1,…,nm):=∫D|a0,nm(z)|1/(k-s)|f0,1(nm-nm-1-1)(z)f0,1(z)|1/(k-s)⋯|fm-2,1(n2-n1-1)(z)fm-2,1(z)|1/(k-s)⋅|fm-1,1(n1-n-1)(z)fm-1,1(z)|1/(k-s)|fm,1(n-s+m)(z)fm,1(z)|1/(k-s)(1-|z|2)αs-εdm(z).Ifnm=k, then a0,nm(z)≡1, and general form of Hölder's inequality with indices (4.32)nm-snm-nm-1-1,nm-snm-1-nm-2-1,…,nm-sn1-n-1,nm-sn-s+m
together with Lemma D shows that K(n,n1,…,nm)<∞ for ε>0 being small enough. If nm<k, then an appropriate application of Hölder’s inequality with indices (k-s)/(k-nm) and (k-s)/(nm-s) separates the coefficient from the solutions. The first term is finite by the assumption, and the second term can seen to be finite by another application of general form of Hölder's inequality with indices (4.32). This gives the desired contradiction, since the left-hand side of (4.26) diverges as r→1- and the right-hand side of (4.26) is uniformly bounded for all r∈(0,1).


## 4.1. Cases=1

Letk≥3, q≥2, s=1, and βT(1)>0, since otherwise there is nothing to prove. In particular, if α1≤0, then (2.4) is trivial, since by taking j=k in (2.6), we obtain βT(1)≤α1≤0. Let {f0,1,f0,2,…,f0,k} be a solution base of (1.1), and assume on the contrary to the assertion that there exist s+1=2 linearly independent solutions f0,1 and f0,2 such that max{σT(f0,1),σT(f0,2)}=:σ<βT(1). Then the meromorphic function g:=(f0,1/f0,2)′ satisfies σT(g)≤σ. Moreover, Lemma C implies that g satisfies (4.6)g(k-1)+a1,k-2(z)g(k-2)+⋯+a1,0(z)g=0,
where (4.7)a1,j(z)=a0,j+1(z)+∑n=j+2k(nj+1)a0,n(z)f0,1(n-j-1)(z)f0,1(z)
for j=0,1,…,k-2, and a0,k(z)≡1. Therefore (4.8)|a0,1(z)|≤|a1,0(z)|+∑n=2k(n1)|a0,n(z)||f0,1(n-1)(z)f0,1(z)|,
where (4.9)|a1,0(z)|≤|g(k-1)(z)g(z)|+|a1,k-2(z)||g(k-2)(z)g(z)|+⋯+|a1,1(z)||g′(z)g(z)|,
since g satisfies (4.6). Putting the last two inequalities together, we obtain (4.10)|a0,1(z)|≲∑j=1k-1|a1,j(z)||g(j)(z)g(z)|+∑n=2k|a0,n(z)||f0,1(n-1)(z)f0,1(z)|.
Let ε>0. Raising both sides to the power 1/(k-1) and integrating over the disc D(0,r) of radius r∈(0,1) with respect to (1-|z|2)α1-εdm(z), we obtain (4.11)∫D(0,r)|a0,1(z)|1/(k-1)(1-|z|2)α1-εdm(z)≲∑j=1k-1∫D|a1,j(z)|1/(k-1)|g(j)(z)g(z)|1/(k-1)(1-|z|2)α1-εdm(z)+∑n=2k∫D|a0,n(z)|1/(k-1)|f0,1(n-1)(z)f0,1(z)|1/(k-1)(1-|z|2)α1-εdm(z).To deal with the second sum in (4.11), consider (4.12)In:=∫D|a0,n(z)|1/(k-1)|f0,1(n-1)(z)f0,1(z)|1/(k-1)(1-|z|2)α1-εdm(z),n=2,…,k.
By Lemma D, (4.13)Ik=∫D|f0,1(k-1)(z)f0,1(z)|1/(k-1)(1-|z|2)α1-εdm(z)<∞
for ε>0 being small enough since σT(f0,1)≤σ<βT(1)≤α1. Moreover, by Hölder’s inequality, with indices (k-1)/(k-n) and (k-1)/(n-1), we have (4.14)In≤(∫D|a0,n(z)|1/(k-n)(1-|z|2)αn+εdm(z))(k-n)/(k-1)⋅(∫D|f0,1(n-1)(z)f0,1(z)|1/(n-1)(1-|z|2)ω1(n)dm(z))(n-1)/(k-1)
for all n=2,…,k-1, where (4.15)ωs(n):=(k-s)(αs-αn)n-s+αn-2k-n-sn-sε.
The first member in the product is finite since a0,n∈𝔸αn1/(k-n) for all n=2,…,k-1 by the assumption, and so is the second one for ε>0 small enough since (4.16)σT(f0,1)≤σ<βT(1)≤(k-1)(α1-αn)n-1+αn,n=2,…,k-1,
by the antithesis. Thus ∑n=2kIn is finite for ε>0 being small enough.To deal with the first sum in (4.11), denote (4.17)Jj:=∫D|a1,j(z)|1/(k-1)|g(j)(z)g(z)|1/(k-1)(1-|z|2)α1-εdm(z),j=1,…,k-1.
Lemma D implies that (4.18)Jk-1=∫D|g(k-1)(z)g(z)|1/(k-1)(1-|z|2)α1-εdm(z)<∞
for ε>0 being small enough since σT(g)≤σ<βT(1)≤α1. Moreover, by (4.7) we have (4.19)Jj≲∫D|a0,j+1(z)|1/(k-1)|g(j)(z)g(z)|1/(k-1)(1-|z|2)α1-εdm(z)+∑n=j+2k∫D|a0,n(z)|1/(k-1)|f0,1(n-j-1)(z)f0,1(z)|1/(k-1)|g(j)(z)g(z)|1/(k-1)(1-|z|2)α1-εdm(z)=:Kj+Lj,k+∑n=j+2k-1Lj,n
for all j=1,…,k-2. Since max{σT(g),σT(f0,1)}≤σ<βT(1), we deduce that Kj behaves like Ij+1 and hence ∑j=1k-2Kj<∞ for ε>0 being small enough. Moreover, by Hölder’s inequality, with indices (k-1)/(k-j-1) and (k-1)/j, and Lemma D we have (4.20)Lj,k≤(∫D|f0,1(k-j-1)(z)f0,1(z)|1/(k-j-1)(1-|z|2)α1-εdm(z))(k-j-1)/(k-1)⋅(∫D|g(j)(z)g(z)|1/j(1-|z|2)α1-εdm(z))j/(k-1)<∞
for all j=1,…,k-2 when ε>0 is sufficiently small. It remains to consider the double sum ∑j=1k-3∑n=j+2k-1Lj,n. By Hölder’s inequality, with indices (k-1)/(k-n) and (k-1)/(n-1), we have (4.21)Lj,n≤(∫D|a0,n(z)|1/(k-n)(1-|z|2)αn+εdm(z))(k-n)/(k-1)⋅(∫D|f0,1(n-j-1)(z)f0,1(z)|1/(n-1)|g(j)(z)g(z)|1/(n-1)(1-|z|2)ω1(n)dm(z))(n-1)/(k-1),
where ω1(n) is defined in (4.15). The first term in the product is bounded for all ε>0 since a0,n∈𝔸αn1/(k-n) for all n=3,…,k-1 by the assumption. One more application of Hölder's inequality, with indices (n-1)/(n-j-1) and (n-1)/j, together with Lemma D and the antithesis shows that also the second term in the product is bounded for ε>0 being small enough, and thus ∑j=1k-3∑n=j+2k-1Lj,n<∞ for ε>0 being small enough.We have proved that the right-hand side of (4.11) is uniformly bounded for all r∈(0,1), if ε>0 is small enough. However, a0,1∈𝔸α11/(k-1) by the assumption, and hence the left-hand side of (4.11) diverges as r→1-. Contradiction follows.


## 4.2. Cases>1

Letk≥3, q≥2, s>1, and βT(s)>0, since otherwise there is nothing to prove. In particular, it follows that αs>0. Let {f0,1,f0,2,…,f0,k} be a solution base of (1.1), and assume on the contrary to the assertion that there exist s+1 linearly independent solutions f0,1,…,f0,s+1 such that (4.22)σ:=max⁡{σT(f0,1),…,σT(f0,s+1)}<βT(s).
Then the meromorphic functions f1,j=(f0,j+1/f0,1)′ satisfy σT(f1,j)≤σ for all j=1,…,s. This in turn implies that f2,j=(f1,j+1/f1,1)′ satisfy σT(f2,j)≤σ for all j=1,…,s-1. In general, σT(fn,j)≤σ for all j=1,…,s-n+1 and n=1,…,s. Moreover, as in the case s=1, Lemma C implies (4.23)|a0,s(z)|≤|a1,s-1(z)|+∑n=s+1k(ns)|a0,n(z)||f0,1(n-s)(z)f0,1(z)|≤|a2,s-2(z)|+∑n=sk-1(ns-1)|a1,n(z)||f1,1(n-s+1)(z)f1,1(z)|+∑n=s+1k(ns)|a0,n(z)||f0,1(n-s)(z)f0,1(z)|≤⋯≤|as,0(z)|+∑m=0s-1∑n=s+1-mk-m(ns-m)|am,n(z)||fm,1(n-s+m)(z)fm,1(z)|,
where (4.24)|as,0(z)|≤|fs,1(k-s)(z)fs,1(z)|+∑m=1k-s-1|as,m(z)||fs,1(m)(z)fs,1(z)|.
Putting these inequalities together, we obtain (4.25)|a0,s(z)|≲∑m=0s∑n=s+1-mk-m-1|am,n(z)||fm,1(n-s+m)(z)fm,1(z)|+∑m=0s|fm,1(k-s)(z)fm,1(z)|.
Let ε>0. Raising both sides to the power 1/(k-s) and integrating over the disc D(0,r) with respect to (1-|z|2)αs-εdm(z), we obtain (4.26)∫D(0,r)|a0,s(z)|1/(k-s)(1-|z|2)αs-εdm(z)≲∑m=0s∑n=s+1-mk-m-1∫D|am,n(z)|1/(k-s)|fm,1(n-s+m)(z)fm,1(z)|1/(k-s)(1-|z|2)αs-εdm(z)+∑m=0s∫D|fm,1(k-s)(z)fm,1(z)|1/(k-s)(1-|z|2)αs-εdm(z)=:∑m=0s∑n=s+1-mk-m-1Im,n+∑m=0sJm.
Lemma D and the antithesis imply that ∑m=0sJm<∞ for ε>0 being small enough. Hence, in order to obtain a contradiction with (4.26) and the assumption a0,s∈𝔸αs1/(k-s), it suffices to show that Im,n<∞ for all m=0,…,s and n=s+1-m,…,k-m-1 when ε>0 is sufficiently small.By Hölder's inequality, with indices(k-s)/(k-n) and (k-s)/(n-s), we have (4.27)I0,n≤(∫D|a0,n(z)|1/(k-n)(1-|z|2)αn+εdm(z))(k-n)/(k-s)⋅(∫D|f0,1(n-s)(z)f0,1(z)|1/(n-s)(1-|z|2)ωs(n)dm(z))(n-s)/(k-s)
for all n=s+1,…,k-1, where ωs(n) is defined in (4.15). The first member in the product is finite since a0,n∈𝔸αn1/(k-n) for all n=s+1,…,k-1 by the assumption, and so is the second one for ε>0 being small enough, since (4.28)σT(f0,1)≤σ<βT(s)≤(k-s)(αs-αn)n-s+αn,n=s+1,…,k-1,
by the antithesis. In the general case Lemma C gives (4.29)Im,n=∫D|am,n(z)|1/(k-s)|fm,1(n-s+m)(z)fm,1(z)|1/(k-s)(1-|z|2)αs-εdm(z)≲∑n1=n+1k-m+1∫D|am-1,n1(z)|1/(k-s)|fm-1,1(n1-n-1)(z)fm-1,1(z)|1/(k-s)⋅|fm,1(n-s+m)(z)fm,1(z)|1/(k-s)(1-|z|2)αs-εdm(z)≲∑n1=n+1k-m+1∑n2=n1+1k-m+2∫D|am-2,n2(z)|1/(k-s)|fm-2,1(n2-n1-1)(z)fm-2,1(z)|1/(k-s)⋅|fm-1,1(n1-n-1)(z)fm-1,1(z)|1/(k-s)|fm,1(n-s+m)(z)fm,1(z)|1/(k-s)(1-|z|2)αs-εdm(z),
and finally (4.30)Im,n≲∑n1=n+1k-m+1∑n2=n1+1k-m+2⋯∑nm=nm-1+1kK(n,n1,…,nm),
where (4.31)K(n,n1,…,nm):=∫D|a0,nm(z)|1/(k-s)|f0,1(nm-nm-1-1)(z)f0,1(z)|1/(k-s)⋯|fm-2,1(n2-n1-1)(z)fm-2,1(z)|1/(k-s)⋅|fm-1,1(n1-n-1)(z)fm-1,1(z)|1/(k-s)|fm,1(n-s+m)(z)fm,1(z)|1/(k-s)(1-|z|2)αs-εdm(z).Ifnm=k, then a0,nm(z)≡1, and general form of Hölder's inequality with indices (4.32)nm-snm-nm-1-1,nm-snm-1-nm-2-1,…,nm-sn1-n-1,nm-sn-s+m
together with Lemma D shows that K(n,n1,…,nm)<∞ for ε>0 being small enough. If nm<k, then an appropriate application of Hölder’s inequality with indices (k-s)/(k-nm) and (k-s)/(nm-s) separates the coefficient from the solutions. The first term is finite by the assumption, and the second term can seen to be finite by another application of general form of Hölder's inequality with indices (4.32). This gives the desired contradiction, since the left-hand side of (4.26) diverges as r→1- and the right-hand side of (4.26) is uniformly bounded for all r∈(0,1).


## 5. Proof of Corollary2.3

The upper bound in (2.9) follows directly from Theorem 2.1. To conclude the lower bound in (2.9), assume that solutions f1,…,fk are given in increasing order with respect to T-order of growth; that is, σT(f1)≤⋯≤σT(fk). By applying Theorem 2.1 with s=0,…,q, we get the following sequence of successive statements. For all solutions f in the solution base, we have βT(0)≤σT(f); for k-1 solutions f in the solution base, we have βT(1)≤σT(f), ending up with the fact that k-q solutions f in the solution base satisfy αq=βT(q)=σT(f). Hence we have γT(0)=βT(0)≤σT(f1), γT(1)=max{βT(0),βT(1)}≤σT(f2) continuing to (5.1)γT(q-1)=max⁡{βT(0),…,βT(q-1)}≤σT(fq).Note thatαq=γT(q). To see this, note that βT(s)≤αs for every s=0,…,q, which follows by taking j=k in (2.6), and hence (5.2)αq=βT(q)≤γT(q)=max⁡{βT(0),…,βT(q)}≤max⁡{α0,…,αq}=αq.
The assertion follows by noting that if j∈{0,…,s⋆-1}, then γT(j)≤0, and we only have the trivial estimate σT(fj)≥0.


## 6. Proof of Lemma2.4

Letm∈{1,…,p}. By (2.11), we obtain (6.1)δj-δsm-1sm-1-j≤δsm-δsm-1sm-1-sm
for all 0≤j<sm-1, and the inequality (6.1) is strict for all 0≤j<sm. This estimate will be repeatedly needed later on.


### 6.1. Proof of Lemma2.4(i)

Let1≤t≤p-1. Note that ℬT(t)>0 by definition. Since st+1<st, (6.2)BT(t)-BT(t+1)=δst-δst-1st-1-st-δst+1-δstst-st+1=st-1-st+1st-st+1(δst-δst-1st-1-st-δst+1-δst-1st-1-st+1)>0
by (6.1), which proves the assertion of Lemma 2.4(i).


### 6.2. Proof of Lemma2.4(ii)

Sincemaxj=0,…,k-1{αj}=αs1>0, we get (6.3)βT(s1)=min⁡j=s1+1,…,k{k-jj-s1(αs1-αj)+αs1}=αs1=δs1k-s1-1=BT(1),
where the minimum is obtained with j=k. This proves the claim for t=1.Assume thatt∈{2,…,p}. To prove the claim, we need the following observations. If m∈{1,…,t-1}, then by (6.1) we get (6.4)δst-δsm-1sm-1-st-δst-δsmsm-st=sm-1-smsm-st(δsm-δsm-1sm-1-sm-δst-δsm-1sm-1-st)>0.
On the other hand, if m∈{1,…,t}, then (6.5)δsm-δjj-sm≥δsm-δsm-1sm-1-sm
for all sm<j≤sm-1. To verify (6.5), we consider the following two cases. If j=sm-1, then the equality in (6.5) holds. If j<sm-1, then by using (6.1), we obtain (6.6)δsm-δjj-sm=δsm-δsm-1j-sm-δj-δsm-1sm-1-jsm-1-jj-sm≥δsm-δsm-1j-sm-δsm-δsm-1sm-1-smsm-1-jj-sm=δsm-δsm-1sm-1-sm,
which proves (6.5).To complete the proof of(6.7)βT(st)=min⁡j=st+1,…,k{δst-δjj-st-1}=δst-δst-1st-1-st-1=BT(t),
we argue as follows. First, we show that βT(st)≥ℬT(t). If st<j≤st-1, then (6.5) holds for m=t. If j>st-1, then let m∈{1,…,t-1} be the smallest index such that sm<j. From (6.1), (6.4), and (6.5), we obtain (6.8)δst-δjj-st-δst-δsmsm-st=j-smj-st(δsm-δjj-sm-δst-δsmsm-st)>j-smj-st(δsm-δsm-1sm-1-sm-δst-δsm-1sm-1-st)>0,
which together with (6.4) shows that (6.9)δst-δjj-st>δst-δsmsm-st>⋯>δst-δst-1st-1-st.
Second, we note that equality in (6.7) follows by taking j=st-1.


### 6.3. Proof of Lemma2.4(iii)

Sinceαs1=maxj=0,…,k-1{αj}>0 and s1=q, we get by means of (5.2) that (6.10)BT(1)=δqk-q-1=αq=γT(q).
Let t∈{2,…,p}. We proceed to prove that γT(j)=ℬT(t) for all st≤j<st-1. Evidently, γT(j)=max{βT(0),…,βT(j)}≥βT(st). By Lemma 2.4(ii), we conclude that γT(j)≥ℬT(t). To prove that γT(j)≤ℬT(t), it is enough to show that βT(m)≤ℬT(t) for all m∈{0,…,j}. Since m+1≤st-1, we obtain by appealing to (6.1) that (6.11)βT(m)=min⁡j=m+1,…,k{δm-δjj-m-1}≤δm-δst-1st-1-m-1≤δst-δst-1st-1-st-1=BT(t).Ifsp>0, then for all m∈{0,…,sp-1} we have (6.12)βT(m)=min⁡j=m+1,…,k{δm-δjj-m-1}≤δm-δspsp-m-1≤0
by (2.11). Hence γT(j)=max{βT(0),…,βT(j)}≤0 for all j<sp. As a consequence we see that sp=s⋆.


## 6.1. Proof of Lemma2.4(i)

Let1≤t≤p-1. Note that ℬT(t)>0 by definition. Since st+1<st, (6.2)BT(t)-BT(t+1)=δst-δst-1st-1-st-δst+1-δstst-st+1=st-1-st+1st-st+1(δst-δst-1st-1-st-δst+1-δst-1st-1-st+1)>0
by (6.1), which proves the assertion of Lemma 2.4(i).


## 6.2. Proof of Lemma2.4(ii)

Sincemaxj=0,…,k-1{αj}=αs1>0, we get (6.3)βT(s1)=min⁡j=s1+1,…,k{k-jj-s1(αs1-αj)+αs1}=αs1=δs1k-s1-1=BT(1),
where the minimum is obtained with j=k. This proves the claim for t=1.Assume thatt∈{2,…,p}. To prove the claim, we need the following observations. If m∈{1,…,t-1}, then by (6.1) we get (6.4)δst-δsm-1sm-1-st-δst-δsmsm-st=sm-1-smsm-st(δsm-δsm-1sm-1-sm-δst-δsm-1sm-1-st)>0.
On the other hand, if m∈{1,…,t}, then (6.5)δsm-δjj-sm≥δsm-δsm-1sm-1-sm
for all sm<j≤sm-1. To verify (6.5), we consider the following two cases. If j=sm-1, then the equality in (6.5) holds. If j<sm-1, then by using (6.1), we obtain (6.6)δsm-δjj-sm=δsm-δsm-1j-sm-δj-δsm-1sm-1-jsm-1-jj-sm≥δsm-δsm-1j-sm-δsm-δsm-1sm-1-smsm-1-jj-sm=δsm-δsm-1sm-1-sm,
which proves (6.5).To complete the proof of(6.7)βT(st)=min⁡j=st+1,…,k{δst-δjj-st-1}=δst-δst-1st-1-st-1=BT(t),
we argue as follows. First, we show that βT(st)≥ℬT(t). If st<j≤st-1, then (6.5) holds for m=t. If j>st-1, then let m∈{1,…,t-1} be the smallest index such that sm<j. From (6.1), (6.4), and (6.5), we obtain (6.8)δst-δjj-st-δst-δsmsm-st=j-smj-st(δsm-δjj-sm-δst-δsmsm-st)>j-smj-st(δsm-δsm-1sm-1-sm-δst-δsm-1sm-1-st)>0,
which together with (6.4) shows that (6.9)δst-δjj-st>δst-δsmsm-st>⋯>δst-δst-1st-1-st.
Second, we note that equality in (6.7) follows by taking j=st-1.


## 6.3. Proof of Lemma2.4(iii)

Sinceαs1=maxj=0,…,k-1{αj}>0 and s1=q, we get by means of (5.2) that (6.10)BT(1)=δqk-q-1=αq=γT(q).
Let t∈{2,…,p}. We proceed to prove that γT(j)=ℬT(t) for all st≤j<st-1. Evidently, γT(j)=max{βT(0),…,βT(j)}≥βT(st). By Lemma 2.4(ii), we conclude that γT(j)≥ℬT(t). To prove that γT(j)≤ℬT(t), it is enough to show that βT(m)≤ℬT(t) for all m∈{0,…,j}. Since m+1≤st-1, we obtain by appealing to (6.1) that (6.11)βT(m)=min⁡j=m+1,…,k{δm-δjj-m-1}≤δm-δst-1st-1-m-1≤δst-δst-1st-1-st-1=BT(t).Ifsp>0, then for all m∈{0,…,sp-1} we have (6.12)βT(m)=min⁡j=m+1,…,k{δm-δjj-m-1}≤δm-δspsp-m-1≤0
by (2.11). Hence γT(j)=max{βT(0),…,βT(j)}≤0 for all j<sp. As a consequence we see that sp=s⋆.


## 7. Proof of Theorem2.5

Our proof of Theorem2.5 is parallel to the proof of [3,  Theorem  1.4], and hence we only outline the argumentation. We may assume that k≥3, q≥2, s∈{1,…,q-1}, and βM(s)>1 for otherwise there is nothing to prove by Theorem B; see the discussion after Theorem 2.5. In particular, if ps≤1, then (2.22) is trivial, since by taking j=k in (2.24), we obtain βM(s)≤ps≤1. On the contrary to the claim, assume that (1.1) admits s+1 linearly independent solutions f0,1,…,f0,s+1 such that (7.1)σM(f0,t)<α:=βM(s),t=1,…,s+1.
Remark that if α≤1, then there is nothing to prove in (2.22); so we may assume that α>1. Now max{β,1}<α, where β:=maxt=1,…,s+1{σM(f0,t)}<∞.Letε,δ∈(0,1). Now [3, Lemma  4.3] for m=s+1 implies that there exists a solution fs,1≢0 of (7.2)f(k-s)+as,k-s-1(z)f(k-s-1)+⋯+as,1(z)f′+as,0(z)f=0
of the from fs,1=gs,1/hs,1, where gs,1,hs,1∈ℋ(𝔻) and (7.3)max⁡{σM(gs,1),σM(hs,1)}≤max⁡{β,1}<α.
It is easy to see that α=βM(s) yields (k-l)pl≤(k-s)ps-(l-s)α for all l∈{s+1,…,k}. Hence, by [3, Lemma  4.3] and the assumption a0,j∈ℍ(pj+1)(k-j)∞, we get (7.4)M(r,as,j)≤(11-r)(ps+1)(k-s)-j(α+1)+ε,j=1,…,k-s-1,
for all r∈[0,1)∖E, where the set E satisfies the upper density condition: (7.5)D¯(E):=limsup⁡r→1-m(E∩[r,1))1-r≤δ<1.
Here m(Ω) is the Lebesque measure of the set Ω. We note that set E may not be the same at each occurrence; however, it always satisfies (7.5).Letη∈(δ,1). If we apply [3, Lemma  4.4] and use [3, Lemma  4.1] for the coefficient a0,s∈ℍ(ps+1)(k-s)∞, we conclude that for ε>0 being small enough, we have (7.6)M(r,as,0)≳(11-r)(ps+1)(k-s)-ε
for all r∈F∖E, where the set F⊂[0,1) satisfies D¯(F)≥η. In particular, we have D¯(F∖E)≥η-δ>0.On the other hand, by substitutingf=fs,1 in (7.2) and by applying [3,  Corollary  4.2] to fs,1, it follows that (7.7)|as,0(z)|≤|fs,1(k-s)(z)fs,1(z)|+∑j=1k-s-1|as,j(z)||fs,1(j)(z)fs,1(z)|≲(11-|z|)(ps+1)(k-s)-2ε
for all z∈𝔻, |z|∉E. By comparing (7.6) to (7.7), we get a contradictory inequality: (7.8)(ps+1)(k-s)-ε≤(ps+1)(k-s)-2ε.
This shows that each solution base of (1.1) contains at least k-s solutions f satisfying σM(f)≥βM(s).



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*Source: 928194-2011-09-22.xml*