Irregular finite order solutions of complex LDE's in unit disc

It is shown that the order and the lower order of growth are equal for all non-trivial solutions of $f^{(k)}+A f=0$ if and only if the coefficient $A$ is analytic in the unit disc and $\log^+ M(r,A)/\log(1-r)$ tends to a finite limit as $r\to 1^-$. A family of concrete examples is constructed, where the order of solutions remain the same while the lower order may vary on a certain interval depending on the irregular growth of the coefficient. These coefficients emerge as the logarithm of their modulus approximates smooth radial subharmonic functions of prescribed irregular growth on a sufficiently large subset of the unit disc. A result describing the phenomenon behind these highly non-trivial examples is also established. En route to results of general nature, a new sharp logarithmic derivative estimate involving the lower order of growth is discovered. In addition to these estimates, arguments used are based, in particular, on the Wiman-Valiron theory adapted for the lower order, and on a good understanding of the right-derivative of the logarithm of the maximum modulus.


Introduction and main results
The balance between the growth of coefficients and the growth and oscillation of solutions has been a central theme of research concerning linear differential equations in a complex domain for over a half of century. In the case of the complex plane C, the classical result of Wittich [23,Satz 1] states that the analytic coefficients A 0 , . . . , A k−1 are polynomials if and only if all solutions of f (k) + A k−1 f (k−1) + · · · + A 1 f ′ + A 0 f = 0 (1.1) are entire functions of finite order of growth. Further, each order belongs to a finite set of rational numbers induced by the degrees of the polynomial coefficients [10].
In the particular case when the coefficient A ≡ 0 of is a polynomial, all non-trivial solutions f of (1. The oscillation of solutions is equally intimately connected to the growth of coefficients, see [1,7]. The connection between the growth of coefficients and the growth of solutions is also well understood in the case of the unit disc D = {z ∈ C : |z| < 1}. In this setting, the analogue of polynomial coefficients in C are coefficients belonging to the Korenblum space A −∞ [7,15]. Indeed, the analytic coefficients belong to A −∞ if and only if all solutions of (1.1) are of finite order of growth. In the case of (1.2), all non-trivial solutions are of maximal growth and have order of growth uniquely determined by the coefficient A ∈ A −∞ [4,5,7]. However, almost nothing is known about the lower order of growth of solutions. Our principal objective is to show that the degree and the lower degree of growth of the coefficient A ∈ A −∞ control the lower order of growth of non-trivial solutions. In particular, we demonstrate that solutions having different order and lower order are possible in the case of the unit disc, contrasting sharply with the analogous situation in the complex plane.
To reach concrete statements, some notation is needed. For a function f analytic in D, the order and the lower order of growth (with respect to the maximum modulus) are defined by Our first result relates the irregular growth of the coefficient to the irregular growth of non-trivial solutions of (1.2). This shows that the unit disc case is, in a certain sense, similar to the plane case. The discussion on the principal ingredients of the proof of Theorem 1 is postponed to the end of the present section.
Theorem 2 below unfolds a family of examples, and shows that the difference between σ M,deg (A) and λ M,deg (A) does not uniquely determine the lower order of solutions of (1.2). The reason why this result differs radically from the corresponding situation in the plane is that the Korenblum space is a much richer family of functions in D than the set of polynomials is in C.
The proof of Theorem 2 is rather involved and constitutes the bulk of the paper. The first part of the proof is a laborious construction of a smooth radial subharmonic function ϕ of irregular growth. Then, we show that there exists an analytic function A such that log |A| approximates ϕ with sufficient precision in a large subset of D. The upper bound for the lower order of growth of any solution of (1.2) follows from the irregular growth of the coefficient by a growth estimate for solutions of linear differential equations. Meanwhile, the lower bound for the lower order is established by using a recent integrated logarithmic derivative estimate. Theorem 2 is proved in Section 2.
In Theorem 2 the order of non-trivial solutions is always p 2 /k − 1, while the lower order can be any pregiven number on the interval [p 1 /k − 1, p 1 /k − p 1 /p 2 ]. In this case, the lower order of growth is strictly smaller than the order of growth. The following theorem reveals the general phenomenon induced by the irregular growth of the coefficient. (1.5) (b) if 2k < k 2 + p 2 −2k p 2 < p 1 ( ≤ ) p 2 , then all non-trivial solutions f of (1.2) satisfy The following result shows, analogously to the plane situation in (1.3), that if A has regular growth in D, then the solutions of (1.2) have regular growth as well.
We may re-write (1.6) in the form To see that the upper bound in (1.7) is sharp, choose α = p 1 /p 2 in Theorem 2. It would be desirable to show that the lower bound in (1.7) can be replaced by the value p 1 /k − 1, which corresponds to α = 1 in Theorem 2. However, it is not known whether this is true, unless p 2 = p 1 . In this case, (1.7) reduces to the equality σ M (f ) = λ M (f ) by (1.4). In fact, we already know this, even under weaker hypothesis, by Theorem 1. The proof of Theorem 3 is given in Section 5, and it depends on the Wiman-Valiron theory adapted for the lower order of growth. Therefore, we need a good understanding of the quantities Here the plus sign refers to the right derivative. In Section 3, we prove that each function f analytic in D satisfies Both estimates are shown to be sharp.
In addition to the Wiman-Valiron theory, the proof of Theorem 3 strongly relies on a new logarithmic derivative estimate involving the lower order of growth, which is stated as Theorem 5 below. This result complements a known logarithmic derivative estimate for functions of finite maximum modulus order given in terms of a proximate order [5]. To the best of our knowledge, logarithmic derivative estimates involving the lower order of growth do not appear in the existing literature. The upper density of a measurable set E ⊂ [0, 1) is defined as where m 1 (F ) denotes the one-dimensional Lebesgue measure of the set F .
The proof of Theorem 5 is given in Section 4, and it is based on an approach similar to that in [5,Theorem 1.2]. In the present paper, we take advantage of the full strength of Linden's result, which is stated as Theorem D below. This result provides a local estimate for the behavior of an analytic function in terms of a quantity depending on the maximum modulus, and it turns out that this quantity can be further estimated in terms of the lower order of growth.
The proof of Theorem 1 depends on Theorem 3 and on an auxiliary result on the lower order of solutions of (1.2). The proof and the auxiliary result are presented in Section 6. In the final Section 7 we discuss an example, which addresses the case when the condition λ M,deg (A) > 2k in Theorem 3(b) is not satisfied. Then the correlations between the growth indicators of the coefficient and of solutions become even more complicated.

Proof of Theorem 2
The laborious proof is divided into three parts.
We still need to prove that (2.3) admits a unique solution (r, ε) where r ∈ (r ⋆ n , 1). By combining the equations, we eliminate ε and obtain (2.5) Consider the interval [r ⋆ n , 1). Let g L = g L (r) be the function in (2.4) and let g R = g R (r) be the function in (2.5). By a straight-forward differentiation, provided C > e p 1 p 2 −p 1 , and hence the function g R is strictly increasing. Another differentiation gives p 2 −p 1 , and hence the function g L is strictly decreasing. Therefore there exists at most one point r ∈ [r ⋆ n , 1) for which g L (r) = g R (r). Let a, b be positive constants such that (log C) 1/2 > a > b (further restrictions apply later), and define {α n } and {β n } by Then r ⋆ n < α n < β n < 1. Recall the inequalities 1 − x < log(1/x) < (1/x)(1 − x), valid for all x ∈ (0, 1). On one hand, we estimate Now that we have the factor log C 1− rn 3 2 in the lower estimate for g L (α n ), and the factor log C 1− rn in the upper estimate for g L (α n ), it is easy to see that g L (α n ) > g R (α n ) if a (and hence C also) is sufficiently large. On the other hand, we estimate and therefore g L (β n ) < g R (β n ) assuming that b > 0 is sufficiently small (This reasoning can be easily modified such that in the definitions of α n and β n , the powers 1 2 and 2 are replaced by 1). As both functions g L and g R are continuous, this ensures the existence of a unique value r ∈ (α n , β n ) for which g L (r) = g R (r). This value is denoted by r ′′ n . Define r n+1 by (viii) which guarantees r n → 1 − , as n → ∞, since {η n } is unbounded. Finally, define ε n+1 by (vi). The reasoning above then proves (vii).
We compute which implies that: (I) for every δ > 0 there exists a constant C > 0 such that independently of n; (II) we also have for any fixed C. By (iii),(v) and (vi), (2.6) The right-hand side of (2.6) contains four terms. The first term is the constant p 1 − p 2 . Since |ε n | ≤ (p 2 − p 1 )/2 by the assumption, the second term can be made arbitrarily close to zero by choosing sufficiently large C. By choosing C sufficiently large, the fourth term can be made arbitrarily close to zero as well. To estimate the third term, we compute We conclude that the third term in the right-hand side of (2.6) can be made arbitrarily close to the value p 2 − p 1 by choosing r 1 ∈ (0, 1) sufficiently close to 1 and C sufficiently large. This means that, for any δ > 0, we obtain |ε n+1 | < δ by choosing r 1 ∈ (0, 1) sufficiently close to 1 and C sufficiently large. That said, we may assume |ε n+1 | < (p 2 − p 1 )/2. The same computation also shows that ε n → 0 as n → ∞. Therefore (vi) ensures the asymptotic equality which gives a relatively precise location of r ′′ n with respect to r n . (b) Let ϕ be the piecewise-defined function in the assertion. It is immediate that ϕ is continuous in each subinterval. At endpoints of subintervals, we conclude that continuity at r n is clear; continuity at r ′ n follows from (i); continuity at r n is clear and the same happens with r ⋆ n ; and continuity at r ′′ n follows from (vi). By straight-forward differentiation, Also the derivative ϕ ′ is continuous in each subinterval. At endpoints of subintervals, we conclude that continuity at r n follows by (iii); continuity at r ′ n , r n and r ⋆ n is clear; and continuity at r ′′ n follows from (vii). Another straight-forward differentiation gives (2.1).
We need to prove the properties in (2.2). Note that We only give details on cases where the upper and lower estimates are different. The remaining computations are similar and hence omitted. For r ′′ n−1 ≤ r < r n , we deduce the upper estimate and the lower estimate For r n ≤ r < r ′ n , we have the upper estimate and, by (i), the lower estimate For r ′ n ≤ r < r n , we have the upper estimate The following lower estimate is immediate For r n ≤ r < r ⋆ n , we have the upper estimate and the lower estimate For r ⋆ n ≤ r < r ′′ n , we have, by (ii), the upper estimate and the lower estimate ϕ(r) log 1 This completes the proof of (b).

2.2.
Approximation of subharmonic functions. We will find an analytic function A in D such that log |A| approximates the radial function ϕ, constructed in Lemma 6, with sufficient precision (logarithmic growth is approximated at a log log accuracy). In order to do this we use [6, Theorem 1], stated as Theorem A below. Some more notation is needed. We say that a differentiable function b and cb(r) < 3(1 − r)/4 for all 0 < c < 3/4. Observe that b is not required to be monotonic. Standard computations show that the decreasing functions b 1 (r) = (1 − r)/C and b 2 (r) = (1 − r)/(log(C/(1 − r))) for C > 1 are of regular variation. These functions play a role in our application.
A polar rectangle is a set of the form Q = {z = re iθ : r 1 < r < r 2 , θ 1 < θ < θ 2 } ⊂ D. We write ℓ r (Q) = r 2 − r 1 and ℓ θ (Q) = θ 2 − θ 1 . We say that a positive Borel measure µ in D admits a regular partition, with respect to a function b of regular variation, if there exist a sequence {Q (l) } of polar rectangles in D and a decomposition µ = µ (l) such that Further, µ is locally regular with respect to a function b of regular variation if there exists r 0 ∈ (0, 1) such that Note that (i) the behavior of µ in compact subsets of D is not relevant, since b(r) → 0 + as r → 1 − ; (ii) µ is not necessarily finite, as the hyperbolic measure dµ(z) = dm 2 (z)/(1 − |z| 2 ) 2 is locally regular with respect to b 1 (here and later m 2 is the two-dimensional Lebesgue measure); (iii) by Fubini's theorem, the condition (2.7) is equivalent to Finally, let Z f stand for the zero set of an analytic function f . Recall that each subharmonic function u : D → R induces a unique positive measure, namely the Riesz measure µ = µ u , such that dµ = 1 2π ∆u dm 2 . Theorem A ([6, Theorem 1]). Let u : D → R be a subharmonic function such that µ u admits a regular partition and µ u is locally regular, with respect to some b : [0, 1) → (0, 1) of regular variation. Then there exists an analytic function f in D such that sup and for each ε > 0 there exist r 1 = r 1 (ε) ∈ (0, 1) and C = C(ε) > 0 such that Lemma 7. Let ϕ be as in Lemma 6. Then there exists an analytic function A in D with the following properties: Proof. Recall that r ′′ 0 = 0 and define r n,0 = r ′′ n−1 for all n ∈ N. Define the annuli as in (2.1): n , and note that the restriction of the Riesz measure 1 2π ∆ϕ of the function ϕ on the boundary of any annulus is the zero measure. Define inductively r n,k , k ∈ {0, . . . , m n }, where m n will be specified later. Given r n,k we choose the numbers , where the square brackets denote the integer part of a real number. In order to divide the annulus {ζ : r n,k < |ζ| < 1} into the polar rectangles, define Q j n,k = ζ : r n,k < |ζ| < r n,k+1 , ψ j−1 n,k ≤ arg ζ < ψ j n,k , j = 1, . . . , where r n,k+1 is chosen such that µ ϕ (Q j n,k ) = 2 for all j, that is, Such r n,k+1 exists and is unique, because (x − r n,k )/(1 − x) increases to infinity as x → 1 − . In particular, We continue the process until the interval (r n,m , r n,m+1 ] contains r n , that is, when either r n,m+1 = r n or r n,m+1 > r ′ n , because the Laplacian of ϕ vanishes on A ′ n . Set m n = m. We redefine Q j n,mn−1 as Q j n,mn−1 = ζ : r n,mn−1 < |ζ| < r n , ψ j−1 n,mn−1 ≤ arg ζ < ψ j n,mn−1 , where ψ n,mn−1 is defined in the following way. We set ψ 0 n,mn−1 = 0 and define ψ j n,mn−1 for 1 ≤ j ≤ s n − 1 by the equation µ ϕ (Q j n,mn−1 ) = 2, where the value s n is uniquely defined by the condition s n = min{j : ψ j n,mn−1 > 2π}. Redefine Q sn−1 n,mn−1 so that the collection  covers the whole annulus A n . Let Q n1 denote this modified rectangle Q sn−1 n,mn−1 and note that its side lengths are comparable to 1 − r n while 2 ≤ µ ϕ (Q n1 ) < 4.
Since on A n and A ′′ n we have we can similarly define partitions of the Riesz measures of the restrictions ϕ An and ϕ A ′′ n , respectively. Then covers the whole annulus A ′′ n , which implies 2 ≤ µ ϕ (Q n3 ) < 4. By construction, Q n2 and Q n3 are polar rectangles with side lengths comparable to 1 − r n and 1 − r ′′ n , respectively. Then consider the annulus A * n . We choose ψ * 0 n = 0, and ψ * j n such that for Q * j n = {ζ : r n < |ζ| < r * n , ψ j−1 n ≤ arg ζ < ψ j n }, we have The value s * n is uniquely defined by the condition s * n = min{j : ψ * j n,m−1 > 2π}. Then we obtain a partition {Q * j is modified, and denoted by Q n4 , such that s * n −1 j=0 Q * j n covers the whole annulus A * n , which implies 2 ≤ µ ϕ (Q n4 ) < 4. By construction, Q n4 is a polar rectangle with side lengths comparable to 1− rn The fact that these integrals converge depends on {η n } in Lemma 6. Careful inspection shows that the condition η n → ∞, as n → ∞, is sufficient for our purposes.
Although the subharmonicity of u 2 follows by standard arguments [8,11], we discuss it in detail for the convenience of the reader. More precisely, we show that u 2 is well-defined, it is subharmonic in D and its Riesz measure coincides with ν/(2π).
First observe that the integrand in (2.8), as the logarithm of modulus of the primary factor of genus 1, admits the estimate By Lemma 6(viii) we get 1 − r n+1 = (1 − r ′′ n )/η n ≤ (1 − r n )/η n , and hence To prove that u 2 is subharmonic in D, it is sufficient to verify it locally. First, we need an upper estimate of ν on compact subsets of D. For any r * Second, let K be an arbitrary compact subset of D. Note that for any z ∈ D, where h(z, ζ) is harmonic in z throughout the unit disc. In K, the function u 2 can be represented as a sum of the logarithmic potential of the finite measure ν K , which is subharmonic by [18,Theorem 3.1.2], and a harmonic function. Therefore u 2 is subharmonic in K, and consequently in the whole unit disc. By [18,Theorem 3.7.4] the Riesz measure of u 2 coincides with ν/(2π). Similar arguments show that functions u * j are subharmonic in D with the Riesz measures respectively. The convergence of the integrals in (2.9) follow from the convergence of the sums It then follows from the definition of u 1 that ∆u 1 = ∆ϕ − ∆u 2 − 4 j=1 ∆u * j , so that supp ∆u 1 is contained in the closure of the set where the equality ∆u 1 = ∆ϕ holds. We conclude that the Riesz measure of u 1 is ∆ϕ dm 2 /(2π) B 1 . We proceed to approximate the subharmonic functions u 1 and u 2 using Theorem A, and show that |u * 1 (z)|, |u * 2 (z)|, |u * 3 (z)| and |u * 4 (z)| are uniformly small. We estimate u * 1 and omit the considerations of u * 2 and u * 3 , which are similar. We consider three cases. First, let |ϕ ζ (z)| = |(ζ − z)/(1 − ζz)| ≤ 1/2, that is, ζ belongs to the pseudo-hyperbolic disc D ph (z, 1/2). Then We begin with S 1 . Since uniformly for all n ∈ N, we deduce To estimate S 2 , suppose that |ϕ z (ζ)| ≥ 1/2. Then | log |ϕ z (ζ)|| ≤ log 2 and Q n1 \D ph (z,1/2) for some constant M > 1 independent of n, we conclude 1 − r n+1 (1 − r 1 ) M n for all n ∈ N. Therefore and therefore In conclusion, we have proved Similar arguments show |u * 2 (z)| + |u * 3 (z)| log log We then consider u 4 . Now We begin with T 1 . There exists a constant c > 0 such that Since there exists a constant κ > 1 such that uniformly for all n ∈ N. We deduce To estimate T 2 , suppose that |ϕ ζ (z)| ≥ 1/2. Then Therefore, as in the case of S 2 , we deduce Therefore, In conclusion, |u * 4 (z)| log log We then approximate u 1 and u 2 separately. The Riesz measure µ 1 = µ u 1 admits a regular partition with respect to the function b 1 (z): (i) µ 1 (Q j n,k ) = µ 1 ( Q j n,k ) = µ 1 (Q ′′ j n,k ) = 2 for all n, k, j; (ii) the interiors of all polar rectangles Q j n,k , Q j n,k and Q ′′ j n,k are pairwise disjoint for all n, k, j. (iii) the sides of each rectangle are comparable; (iv) diam Q j n,k ≍ b 1 (r n,k ), diam Q j n,k ≍ b 1 ( r n,k ) and diam Q ′′ j n,k ≍ b 1 (r ′′ n,k ); (v) µ 1 D \ n ( k,j Q j n,k ∪ k,j Q j n,k ∪ k,j Q ′′ j n,k ) = 0; (vi) for appropriate values n, k, j, the measure µ 1 admits the representation It is easy to check that µ 1 is locally regular with respect to b 1 , that is, for some constant ρ 0 ∈ (0, 1). In fact, we have µ 1 (D(z, t)) t 2 /(1 − |z|) 2 for all 0 < t ≤ 1−|z| 2 , which together with b 1 (r) = (1 − r)/2 gives local regularity of µ 1 with respect to b 1 .
By Theorem A there exists an analytic function A 1 in D such that and for each ε > 0 there exist ρ 1 = ρ 1 (ε) ∈ (0, 1), where The Riesz measure µ 2 = µ u 2 admits a regular partition with respect to the function b 2 (z): (i) µ 2 (Q * j n ) = 2 for all n, j; (ii) the interiors of all polar rectangles Q * j n are pairwise disjoint for all n, j; (iii) the sides of each rectangle are comparable; (iv) diam Q * j n ≍ b 2 (r * n ); (v) µ 2 D \ n ( j Q * j n ) = 0; (vi) for all appropriate values n, j, the measure µ 2 admits the representation Corresponding to the above, the measure µ 2 is locally regular with respect to b 2 . We have which yields the required property. By Theorem A there exists an analytic function A 2 in D such that and for each ε > 0 there exist ρ 2 = ρ 2 (ε) ∈ (0, 1), Since ϕ(z) = ϕ(|z|), there exists a constant Y 1 > 0 such that Combining (2.10), (2.11), (2.12), (2.14), and (2.16) we obtain as |z| → 1 − , where the comparison constant depends on ε. This proves Part (a) in Lemma 7. It follows from the proof of Theorem A that for each rectangle Q j n,k , Q j n,k , Q ′′ j n,k there corresponds exactly two zeros of A 1 lying in a neighborhood of the rectangle of radius equal to the diameter of the rectangle. Hence the number of zeros of A 1 in each such neighborhood is uniformly bounded by some constant P 1 . Suppose that ∂D(0, r) ∩ Q j n,k = ∅ for some n, k, j. Then the linear measure of ∂D(0, r) ∩ E 1 ε is at most a constant multiple of P 1 ε. The same is also true for Q j n,k , Q ′′ j n,k , in addition to Q j n,k . Similarly, Q * j n corresponds exactly two zeros of A 2 lying in a neighborhood of the rectangle of radius equal to the diameter of the rectangle. Thus, the number of zeros of A 2 in each such neighborhood is uniformly bounded by some constant P 2 . Consequently, the linear measure of ∂D(0, r) ∩ E 2 ε is at most a constant multiple of P 2 ε, provided that ∂D(0, r) ∩ Q * j n = ∅. By fixing a sufficiently small ε > 0, the linear measure of E ε ∩∂D(0, r) is less than 2πr, where r ∈ (1/2, 1). So, if r is sufficiently close to 1, then there exists z r ∈ ∂D(0, r) \ E ε with |z r | = r. Therefore, Theorem B ([7, Corollary 6]). Let 0 < R < ∞ and f meromorphic in a domain containing D(0, R). Suppose that j, k are integers with k > j ≥ 0, and f (j) ≡ 0. Then Let k ∈ N be fixed and k ≤ p 1 < p 2 ≤ p < ∞ be constants such that p 2 ≥ 2k. By Our argument is based on Lemmas 6 and 7, and we take advantage of the same notation. Choose η n = n + 1 for all n ∈ N in Lemma 6. As a general property we note that, if 0 < r < 1 and 1 − ρ = C(1 − r) q for some C > 0 and q > 1, then ρ r

(2.19)
If f is a solution of f (k) + Af = 0, where log |A| is approximated with ϕ as in Lemma 7, then the growth estimate [ for all θ ∈ R as n → ∞, which gives the desired upper bound for the lower order of growth.
Clearly, it suffices to estimate the integral of the coefficient A. Note also that lim inf We continue in two separate parts.
(i) Let r ≥ r ′ N . Then r ′ N ≤ r ≤ r N +1 . Lemma 7 gives a lower bound for the modulus of the coefficient only outside the exceptional set. Since the exceptional set lies (almost completely) outside the annulus A ′ n = {z ∈ D : r n < |z| < r ′ n }, we have a good control of the coefficient there. Note that ∆ϕ vanishes on these annuli. Then, , and therefore By the discussion preceding the statement of Lemma 6, lim n→∞ log η n log 1 1−r ′ n = 0, and we obtain log (ii) Let r < r ′ N . Then r N < r < r ′ N . Choose ε ∈ (0, π C 4 ). It follows from Lemma 7 that m 1 (∂D(0, r) \ E ε ) ≥ 2π − C 4 ε ≥ π. The, applying Lemma 6, we deduce . Hence, By choosing different values for the parameter p ∈ [p 2 , ∞), we obtain the assertion. Note that the lower order of growth for solutions (1.2) is as large as possible if p = p 2 , which corresponds to the value α = p 1 /p 2 . However, α = 1 corresponds to the case when both terms in (2.20) are of similar growth.

Growth estimates for logarithm of maximum modulus
This section consists of preparations for the proof of Theorem 3. We use the Wiman-Valiron theory adapted for the lower order of growth.
Proposition 8. Let f be an analytic function in D.
The second part of the statement is similar to a result proved in [19,Corrigendum] for the maximum term and the central index. But since we have not found a proof in the existing literature, we offer a proof. Proposition 8 is a direct consequence of a growth result for convex functions that will be discussed next. Let Ω denote the class of convex functions h : (−∞, 0) → R satisfying the property lim x→0 − h(x) = ∞. It is easy to see that the right derivative h ′ + exists at every point and In particular, if α(h) = ∞, then α(h ′ + ) = ∞. Proposition 9. Let h ∈ Ω. Then the following statements are valid: Proof. Since h is nondecreasing and positive close to zero, there exists x 1 ∈ (−∞, 0) such that that and the inequalities α(h ′ + ) ≤ α(h) + 1 and β(h ′ + ) ≤ β(h) + 1 follow. Thus, in particular, (a) is proved.

Proof of Theorem 5
The proof of Theorem 5 is based on an approach from [5]. We only prove the difficult case λ M (f ) < σ M (f ) as the proof of the case λ M (f ) = σ M (f ) follows from [5,Corollary 1.3], which gives a logarithmic derivative estimate that can be extended to a radial set of upper density one by similar considerations as below.

Preparations.
We begin with a growth lemma, which originates from [17] and associates the order of growth with the lower order of growth.
By applying [17,Theorem 2] to the function f (Rz) at ζ/R, for |ζ| < R, we obtain the following result.

4.2.
Proof of the case k = 1 and j = 0. After the preparations in Section 4.1 we are finally ready to prove the special case k = 1, j = 0 and f (0) = 1. Denote z = re iϕ , where 0 < r < R < 1. By the differentiated Poisson-Jensen formula we have By the definition of u(z, h) we have Let R 0 ∈ (0, 1) be as in Theorem D. Denote p = 1 + (λ M (f ) − λ M (f )/σ M (f )) + for short. Let ε > 0, and let {R n }, {R ⋆ n } ⊂ (R 0 , 1) be the sequences in Lemma 11. Let ν ∈ N such that [r ν+1 , r ν+2 ] ⊂ [R ⋆ n , R n ] for some n ∈ N. Such ν and n exist, and the number of acceptable ν for given n increases to infinity as n → ∞ because of the identity n , R n ) between the points R ⋆ n < R n increases to infinity, as n → ∞, while the hyperbolic distance ̺ h (r ν+1 , r ν+m ), m ≥ 2, tends to the constant value (m − 1) log(2)/2, as ν → ∞.
The radial set (4.23) is of upper density one since D( E k,j ) = 1 and the set excluded in (4.23) is of upper density zero by (4.22).

Proof of Theorem 3
(a) Let f be a non-trivial solution of (1. Therefore , which is equivalent to (a). (b) We consider the case p 1 < p 2 only. The case p 1 = p 2 follows from Theorem 1, which in turn will be proved in Section 6 by Theorem 3(a). Assume 2k < k 2 + p 2 −2k p 2 < p 1 < p 2 < ∞, and let f be a non-trivial solution of (1.2). Then σ M (f ) = p 2 k − 1 > 1 by [  To proceed, we need the following result due to Strelitz.
With these preparations, we are finally ready to present the proof of Theorem 1. Denote σ M,deg (A) = p 2 and λ M,deg (A) = p 1 , for short.

Example
The following example addresses the case when the condition λ M,deg (A) > 2k is not satisfied. It seems that then the correlations between the growth indicators of the coefficient and the growth indicators of solutions of (1.2) become even more complicated. The reasoning below illustrates this situation for nonvanishing solutions.