Deficient Values of Solutions of Linear Differential Equations

Differential equations of the form f′′+A(z)f′+B(z)f=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f'' + A(z)f' + B(z)f = 0$$\end{document} (*) are considered, where A(z) and B(z)≢0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$B(z) \not \equiv 0$$\end{document} are entire functions. The Lindelöf function is used to show that for any ρ∈(1/2,∞)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho \in (1/2, \infty )$$\end{document}, there exists an equation of the form (*) which possesses a solution f with a Nevanlinna deficient value at 0 satisfying ρ=ρ(f)≥ρ(A)≥ρ(B)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho =\rho (f)\ge \rho (A)\ge \rho (B)$$\end{document}, where ρ(h)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho (h)$$\end{document} denotes the order of an entire function h. It is known that such an example cannot exist when ρ≤1/2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho \le 1/2$$\end{document}. For smaller growth functions, a geometrical modification of an example of Anderson and Clunie is used to show that for any ρ∈(2,∞)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho \in (2, \infty )$$\end{document}, there exists an equation of the form (*) which possesses a solution f with a Valiron deficient value at 0 satisfying ρ=ρlog(f)≥ρlog(A)≥ρlog(B)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho =\rho _{\log }(f)\ge \rho _{\log }(A)\ge \rho _{\log }(B)$$\end{document}, where ρlog(h)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho _{\log }(h)$$\end{document} denotes the logarithmic order of an entire function h. This result is essentially sharp. In both proofs, the separation of the zeros of the indicated solution plays a key role. Observations on the deficient values of solutions of linear differential equations are also given, which include a discussion of Wittich’s theorem on Nevanlinna deficient values, a modified Wittich theorem for Valiron deficient values, consequences of Gol’dberg’s theorem, and examples to illustrate possibilities that can occur.


Introduction
The solutions of the linear differential equation with entire coefficients A 0 (z), . . . , A n−1 (z), A 0 (z) ≡ 0, are entire, and it is well known that the zeros of any solution f ≡ 0 of (1.1) are of multiplicity ≤ n − 1.
The main focus of this paper is on solutions of (1.1) which have less than the usual frequency of zeros. A standard measurement of the frequency of a c-point (c ∈ C) of an entire function f is the Nevanlinna deficiency δ N (c, f ) defined by If δ N (c, f ) > 0, then c is said to be a Nevanlinna deficient value of f .
Let ρ( f ) denote the order of an entire function f . It is known [6, p. 207] that an entire function f cannot possess a finite Nevanlinna deficient value when ρ( f ) ≤ 1/2. Thus it can be asked: Theorem 1 is sharp with respect to ρ ∈ (1/2, ∞) because, as noted above, the result does not hold for ρ ≤ 1/2. The solution f in the proof of Theorem 1 is the classical Lindelöf function L ρ in (6.1). Thus the crux of the proof is to find entire coefficients A(z) and B(z) satisfying (1.2) and (1.3) below show. In addition, Theorem 1 addresses all the orders in the infinite interval (1/2, ∞), not just the integer orders.

Example 2 If w(z) is any entire function, then f (z) = e z satisfies
For smaller growth functions f , we obtain an analogous result to Theorem 1 by considering Valiron deficient values and logarithmic order. The Valiron deficiency δ V (c, f ) of a c-point of an entire function f is defined by Observe that finite logarithmic order implies zero order, non-constant polynomials have logarithmic order one, and that there do not exist any non-constant entire functions of logarithmic order < 1, see [3,4]. A classical result of Valiron [1,21] says that any entire function f satisfying T (r , f ) = O log 2 r has no finite Valiron deficient values. Entire functions f satisfying ρ log ( f ) < 2 have this growth rate. Hence we state our main result regarding Valiron deficient values as follows. Theorem 3 is essentially sharp with respect to ρ ∈ (2, ∞) because the result does not hold when ρ < 2, and the only unsettled case is when ρ log ( f ) = 2 and f has infinite logarithmic type.
The solution f in Theorem 3 is a laborious modification of a function due to Anderson and Clunie [1], and its symmetric geometric construction takes a substantial portion of this paper. More precisely, the proof of [1,Thm. 2] involves a canonical product with negative real zeros of unbounded multiplicity. Such a function is a real entire function (real on reals), but it cannot be a solution of (1.2) because it has zeros with unbounded multiplicities. Thus we modify the reasoning in [1] in such a way that the revised canonical product f has only simple zeros lying symmetrically in the left half-plane, where f has the pre-determined logarithmic order. The zeros of f will be pairs of complex conjugates so that f becomes real entire, and, in addition, f (r ) = M(r , f ) holds. These properties are crucial for proving that δ V (0, f ) = 1. A consequence of the proof of Theorem 3 is that the zeros of the modified Anderson-Clunie function f are uniformly logarithmically q-separated for any q > ρ log ( f ) − 1.
After preparations in Sect. 5, we prove Theorem 1 in Sect. 6, while after preparations in Sect. 7, we prove Theorem 3 in Sect. 8

Wittich's Theorem
On the topic of possible deficiencies of solutions of equations of the form (1.1), we recall the following well-known result of Wittich on Nevanlinna deficient values.
Wittich's theorem ([15,Thm. 4.3], [22]) Suppose that a solution f of (1.1) is admissible in the sense that as r → ∞ outside a possible exceptional set of finite linear measure. Then 0 is the only possible finite Nevanlinna deficient value for f. In particular, this is true for transcendental solutions of (1.1) with polynomial coefficients.
The assumption on admissibility of f cannot be removed in Wittich's theorem. The next example shows that if the growth of at least one of the coefficients in (1.1) is at least that of a solution f , then any c ∈ C can be a Picard value of f .

Example 4
For an arbitrary c ∈ C and an arbitrary entire function w(z), the function f (z) = e z + c with c as its Picard value solves the equation The following example shows that 0 may or may not be a deficient value for an admissible solution.
which has 0 as a deficient value. On the other hand, from [9, Ex. 5], the function g(z) = exp z 2 + e z is an admissible solution of We have T (r , g) ∼ N (r , 1/g) as r → ∞ by [19,, so that δ N (0, g) = 0. Observe that f 1 (z) = exp z 2 and f 2 (z) = e z also satisfy (2.2), where δ N (0, The next example gives an equation of the form (1.2) whose non-trivial solutions are all admissible solutions with 0 as a deficient value.

Example 6
The functions f (z) = exp z 2 /2 sin z and g(z) = exp z 2 /2 cos z discussed in [8, p. 416] are linearly independent admissible solutions of It follows that all the non-trivial solutions of (2.3) are admissible and have 0 as a deficient value. Moreover, if w(z) is any entire function, then f and g satisfy More examples of this kind can be generated by using [8,Ex. 4].  [18, p. 276] shows that E V ( f ) always has logarithmic capacity zero. Improvements of this result are due to Hyllengren [14] in the finite order case and to Hayman [12] in the infinite order case. See also [6,Ch. 4].
Wittich's theorem can easily be modified to deal with Valiron deficient values. To achieve this, we need to avoid all exceptional sets, and thus the reasoning works only for finite-order solutions. This modified result is particularly valuable when ρ( f ) ≤ 1/2, as we know that f does not have finite Nevanlinna deficient values in this case. Solutions of zero-order are also possible, provided that at least one of the coefficients is transcendental [9]. For the convenience of the reader, a proof is given.

Modified Wittich's theorem
Suppose that a finite-order solution f of (1.1) satisfies (2.1) as r → ∞ without an exceptional set. Then 0 is the only possible finite Valiron deficient value for f.
which in turn can be written as Using a version of the lemma on the logarithmic derivative valid for functions of finite order (where no exceptional set appears) together with the first main theorem and the assumption (2.1) without an exceptional set, we obtain that m(r , f , c) = O(log r ) + o(T (r , f )), as r → ∞ without an exceptional set. The property (2.1) guarantees that f is transcendental, even if the coefficients are polynomials. Thus, m(r , f , c) = o(T (r , f )) as r → ∞ without an exceptional set. This proves that c is not a Valiron deficient value of f . Theorems 1 and 3 address a different question than that in Wittich's theorem and the modified Wittich's theorem. That said, we mention for independent interest that the proofs of Theorems 1 and 3 do not reveal whether the indicated solution f is an admissible solution of (1.2) or not.

Gol'dberg's Theorem and Sets of Deficiencies
Any zero of a non-trivial solution of (1.1) must have multiplicity ≤ n − 1. Conversely, we have the following result.
Gol'dberg's theorem [10, p. 300] Let f ≡ 0 be an entire function whose zeros all have multiplicity at most n − 1, n ∈ N. Then f is a solution of some differential equation of the form (1.1).
If n = 1, then f has no zeros, and the proof of Gol'dberg's theorem is trivial. For the convenience of the reader, we prove the case n = 2.
Proof of Gol'dberg's theorem for n = 2. Let f ≡ 0 be any entire function whose zeros are all simple. We assume that f has at least one zero, since otherwise the proof is trivial.
We construct entire functions A(z) and B(z) ≡ 0 such that f solves (1.2). For A(z) to be entire, at the zeros z k of f , A(z) needs to solve the interpolation problem where σ k ∈ C. Note that (3.1) can always be solved: If {z k } is a finite sequence, then A(z) can be chosen to be the Lagrange interpolation polynomial, while if {z k } is an infinite sequence, then A(z) can be constructed by means of Mittag-Leffler series. Let ζ = z k be fixed. Along with (3.1), we may require that This guarantees that f (z) + A(z) f (z) ≡ 0. After an entire A(z) satisfying (3.1) and (3.2) has been found, we define B(z) by which is entire and ≡ 0. This completes the proof.

Remark 1
If f ≡ 0 is any entire function, then for a suitable constant c, the function g = f − c has all simple zeros. A consequence of this easy observation is that many properties of f , such as the number of deficient values, remain valid for g, and by Gol'dberg's theorem, g solves some equation of the form (1.2). Gol'dberg's theorem does not give information about the orders of the coefficients in (1.1). Thus, although Gol'dberg's theorem is useful in the above discussions, it cannot be used to prove the respective inequalities in Theorems 1 and 3.

Separation of Zeros
We recall a few concepts from [20, Ch. V]. To shorten the notation, for q > 0, let q denote the standard sequence space consisting of sequences {w n } in C satisfying n |w n | q < ∞. We say that an infinite sequence {z n } in C\{0} with no finite limit points has a finite exponent of convergence In all cases, p ≤ λ. The Weierstrass convergence factors are e 0 (z) = 1 and e k (z) = exp where k ∈ N. If {z n } has finite genus p ≥ 0, then the canonical product converges uniformly on compact subsets of C, and hence represents an entire function having zeros precisely at the points z n . We have p ≤ λ ≤ ρ( f ).
Our first observation in this section is that, even though the double inequalities in Theorem 1 occur for solutions of (1.2), they do not always hold, as many examples show, including Examples 2 and 8.

Example 8
Let {z n } be the sequence defined by z 2n−1 = 2 n and z 2n = 2 n + ε n , where {ε n } is any fixed sequence satisfying Thus the sequence {z n } has non-zero distinct points, and its exponent of convergence is equal to zero. Then the canonical product is an entire function of order zero. Moreover, Then the reasoning in the proof of [2, Cor. 1] shows that Thus no finite order A(z) can satisfy (3.1), even though ρ( f ) = 0. As before, we set B(z) to be the function in (3.3), and then ρ(B) = ∞.
Obviously, many zeros of the function f in (4.2) are close together. As stated in Sect. 1, the separation of zeros of the indicated solutions play a key role in the proofs of Theorems 1 and 3. For illustrative purposes, we discuss some examples regarding uniformly q-separated sequences (defined below).
Following an analogous definition in the unit disc [7], we say that a sequence {z n } of finite genus p is uniformly q-separated for q ≥ 0 provided that there exists a constant C > 0 such that An elementary differentiation of (4.1) followed by a substitution z = z k yields so that we may write (4.5) equivalently as The definition of a q-separated sequence in [13] assumes that C = 1 in (4.7).

Example 9
We prove that the zeros of the function f in (4.2) are not uniformly qseparated for any q ≥ 0. Let q ≥ 0, and set k = 2n − 1. We have for some constant K > 0 independent of k. Thus, for every C > 0, as n → ∞ (or equivalently as k → ∞).

Remark 2
In Example 8, we showed that the function f in (4.2) could not satisfy the conclusion in Theorem 1, and above we showed that the zeros of this function are not uniformly q-separated for any q ≥ 0. In contrast, the zeros of the indicated solution f in Theorem 1 (the Lindelöf function L ρ ) are uniformly q-separated for every q > ρ( f ), see Sect. 6.
For completeness, we construct an example of a uniformly q-separated sequence for q > 0 and λ = 0.

Example 10
The sequence {z n } given by z n = 2 n is uniformly 0-separated and has zero exponent of convergence [13, p. 299]. For a fixed q > 0, choose γ n ∈ min {1/2, 2 n exp (−2 nq )} , 1), and define w n = z n + γ n . Then {w n } is also uniformly 0-separated and has zero exponent of convergence. Let {ζ n } denote the union sequence {z n } ∪ {w n }. Construct canonical products P 1 (z) and P 2 (z) with zero sequences {z n } and {w n }, respectively, and define P(z) = P 1 (z)P 2 (z). Then a calculation in the spirit of Example 9, with P in place of f , shows that {ζ n } is uniformly q-separated. The details are omitted.

Preparations for the Proof of Theorem 1
The following auxiliary result is a modification of [13,Cor. 3.3] that is needed to find an entire A(z) satisfying the interpolation problem (3.1) in our proof of Theorem 1.
As we see, the growth of such an A(z) depends heavily on the uniform q-separation of the zeros of f . This needs to be taken into account when proving the inequality ρ( f ) ≥ ρ(A).

Lemma 3
Suppose that {z n } is an infinite sequence of non-zero points in C with finite exponent of convergence λ, and that {z n } is uniformly q-separated for some q ≥ 0 (and C > 0). Let {σ n } be an infinite sequence of points in C, not necessarily distinct, and let h : [0, ∞) → [1, ∞) be a continuous and non-decreasing function such that |σ n | ≤ h(|z n |) for n ∈ N.
Then there exists an entire function A(z) ≡ 0 such that and, for any given α > 1, Proof Without loss of generality, we may suppose that {z n } is ordered according to increasing moduli. Moreover, I (x) is well defined and non-decreasing by continuity. Let P(z) be the canonical product of genus p ≤ λ having {z n } as its zero sequence. By uniform q-separation, the points z n are simple, and hence 1/P (z n ) ∈ C for any n. where c n = σ n /P (z n ) and {q n } is a sequence such that each q n ≥ 0 is the smallest integer satisfying for |z n | > e, and set q n = 0 otherwise. Then H (z) is meromorphic in C and A = P H is an entire function that satisfies (5.1). Hence it suffices to prove (5.2).
By the assumptions, there exists a constant C 1 > 0 such that where C 2 > 0 is a constant. Set so that the inequality log |c n | log |z n | ≤ g(|z n |) holds for every n such that |z n | ≥ e. As in the proof of [13,Cor. 3.3], we would like to apply [13, Thm. 3.1] next, but the monotonicity of g(t) is not known, in particular when 0 ≤ q < 1. Thus we replace g(t) with which is a non-decreasing function by continuity. Now [13, Thm. 3.1] gives us Indeed, it is apparent from the proof of [13,Thm. 3.1] that the finitely many indices n for which |z n | < e have no affect on this conclusion. The assertion (5.2) then follows from ρ(A) ≤ max{ρ(P), ρ(H )}. (b) Suppose that σ n = 0 for at least one n. If σ n = 0 for all n, we may choose A(z) = P(z), in which case (5.1) and (5.2) clearly hold. Hence we may suppose that σ n = 0 for at least one index n but not for all n. Let {s n } denote the subsequence of {σ n } consisting of all the non-zero points. Let {z n } = {ζ n } ∪ {ξ n } be a partition of the sequence {z n } such that each ζ n corresponds to s n . In other words, the interpolation problem (5.1) is transformed into finding an entire function A(z) such that A(ζ n ) = s n and A(ξ n ) = 0. The possibility that σ n = 0 for some n is now included, and the growth of the majorant function g(t) has been considered in more detail. (b) Lemma 3 will be applied to the target sequence where f is a canonical product with genus p ≥ 0. A very similar target sequence appears in the proof of [2, Thm. 1]. We have the representation (c) Let f be an entire function having simple zeros at the points a n , and let {b n } be a target sequence. Then the Lagrange interpolation series is an entire solution to the interpolation problem L(a n ) = b n , provided that L(z) converges uniformly on compact subsets of C, see [16, p. 195]. In our case, however, the target sequence is unbounded, and the zero-sequence {a n } is uniformly q-separated. The latter means in general that | f (a n )| can tend to zero exponentially as n → ∞. Thus it seems unlikely that the Lagrange interpolation series could be used in proving Theorem 1, and the use of Lemma 3 instead seems to be justified.

Proof of Theorem 1
The solution f will be the Lindelöf function L ρ of order ρ > 1/2, which is a canonical product with simple zeros precisely at the points z n = −n 1/ρ , n ≥ 1, on the negative real axis. We may write where α = 1/ρ and p = ρ (= integer part of ρ) is the genus of L ρ . It is known (see [5, p. 294] or [17, p. 54]) that where q ≥ 0 is an integer. It follows that δ N (0, L ρ ) > 0 for every ρ > 1/2. Since entire functions of order ≤ 1/2 have no finite Nevanlinna deficient values, the Lindelöf functions illustrate the sharpness of this inequality.
The crux of the proof of Lemma 5 is a simple identity where the integrand x α−1 is decreasing for 0 < α < 1 and non-decreasing for α ≥ 1. We omit the details.

Lemma 6
Let L ρ be the Lindelöf function of order ρ > 1/2 with zeros z k = −k 1/ρ of genus p = ρ . Then the (target) sequence σ k = −L ρ (z k )/L ρ (z k ) can be written as
The estimates in the previous proof seem to have some flexibility. Hence the result is unlikely to be sharp, but it is nevertheless more than enough for our use.

Lemma 7
Let L ρ be the Lindelöf function of order ρ ∈ (1/2, ∞) with zeros z k = −k 1/ρ of genus p = ρ . Then there exists a constant C > 0 such that In particular, the zero sequence of L ρ is uniformly q-separated for every q > ρ.
Proof Set α = 1/ρ for brevity. As the final conclusion is trivial, it suffices to prove (6.6). We do this in two steps.
We estimate P 1 (k) from below by where k ≥ 2. When estimating P 2 (k) from below, we apply the right endpoint rule to the decreasing function x → log (1 + k α /(x α − k α )) on [k + 1, ∞), and make use of Part (1) in the proof of Lemma 6. We conclude that for some C 1 = C 1 (α) > 0 and for all k ≥ 2.
A trivial elimination of the Weierstrass convergence factor allows us to argue analogously as in estimating P 1 (k) in Part (1) of the proof. In addition, we make use of the inequality n n ≤ e n n!, which is valid for all n ∈ N. We have where k ≥ 2 and D 1 = log(1/α) + 1 − α ≥ 0 for 0 < α ≤ 1.
Before estimating Q 2 (k), we make the preliminary manipulations A simple reasoning based on geometric series yields The estimate for T 2 (k) in the proof of Lemma 6 then yields Q 2 (k) −1 ≤ exp (D 2 k log k) for some D 2 = D 2 (α) > 0 and for all k ≥ 2.
Finally, we combine the estimates for Q 1 (k) and Q 2 (k), and obtain where C is any constant satisfying C > D 1 + D 2 and k ≥ 2 is large enough. This completes the proof of (6.6) in the case ρ ∈ [1, ∞).

Proof of Theorem 1
After these preparations, the actual proof of Theorem 1 is now easy. Let f = L ρ be the Lindelöf function of order ρ ∈ (1/2, ∞), which has simple zeros and the required growth. In addition, the zeros of f are uniformly q-separated for every q > ρ by Lemma 7. For K ≥ 1, define h : [0, ∞) → [1, ∞) by By Lemma 6, we may choose K ≥ 1 so that |σ k | ≤ h(|z k |) for all k. Moreover, for all p ≥ 0, we have log h(t) = O(log t). Thus and we conclude by Lemma 3 that there exists an entire function A(z) ≡ 0 satisfying (3.1) such that The zero sequence {z n } of f that determines A in Lemma 3 is fixed, but it is also uniformly q-separated for any q > ρ( f ) by Lemma 7. Thus the estimate in (6.8) provided by Lemma 3 holds for any q > ρ( f ), and, a fortiori, ρ(A) ≤ ρ( f ). Finally, we define the entire coefficient B(z) by (3.3), for which ρ(B) ≤ ρ(A) holds by the lemma on the logarithmic derivative. It remains to show that B(z) ≡ 0. Suppose on the contrary that B(z) ≡ 0, in which case (1.2) Since f (and hence f ) is of finite order, it follows that A(z) is a polynomial. This gives us ρ( f ) = deg(A) + 1 ≥ 1. Since ρ( f ) ∈ N, it is clear from the definition of the Lindelöf function that p = ρ( f ). However, according to the representation (4.6), the derivative f should have a p-fold zero at the origin, which is a contradiction. This completes the proof.

Preparations for the Proof of Theorem 3
The separation of the zeros of the constructed solution f of (1.2) plays a key role in the proof of Theorem 3. The zeros of f are uniformly logarithmically q-separated (defined below) for every q > ρ log ( f ) − 1.
For basic properties of entire (or, more generally, meromorphic) functions of finite logarithmic order, we refer to [3,4]. In particular, the logarithmic exponent of convergence of the zeros of an entire f is given by where n(r ) denotes the number of zeros of f in |z| < r , counting multiplicities. Similarly to the usual order, the functions T (r , f ) and log M(r , f ) have the same logarithmic order ρ log , and ρ log ( f ) = ρ log ( f ). Differing from the usual order, where This reflects the fact that polynomials are of logarithmic order one, and they have only finitely many zeros. We require a new concept on point separation. We say that a sequence {z n } of finite genus p is uniformly logarithmically q-separated for q ≥ 0 provided that there exists a constant C > 0 such that If f is given by (4.1), then using (4.6), we may write (7.1) equivalently as Certainly there exist sequences with zero exponent of convergence which are not uniformly logarithmically q-separated for any q ≥ 0. Moreover, uniformly logarithmically 0-separated sequences are uniformly 0-separated sequences, and vice versa. We will make use of this new separation concept for sequences of finite logarithmic exponent of convergence, in which case p = 0 and the corresponding f reduces to the form (4.2). The definition in (7.1) for general p is given for possible applications in the future. Moreover, one could replace the polynomial and logarithmic weights by some monotonic function ϕ : R + → R, and discuss uniform ϕ-separation.
In the next example we will discuss the case λ log = 1 and q > 0.
Then z n < w n < z n+1 , w n+1 ≥ z n+1 ≥ 3w n /2 and w n ≤ z n + 1/2 ≤ 4z n /3. A simple modification of the reasoning in Example 10 shows that the combined sequence {z n } ∪ {w n } has logarithmic exponent of convergence equal to one and the sequence is uniformly logarithmically q-separated.
To get an analogue of Lemma 3, we first modify [13,Thm. 3.1] to the case of finite logarithmic order. In the following statement, special attention has been paid to those points z n that are near the origin, as well as to minor monotonicity issues in [13, (3.2)].

Lemma 8 Suppose the following assumptions hold:
(a) {z n } is a sequence of distinct non-zero points in C with λ log < ∞. is meromorphic in C, and has simple poles exactly at the points z n with residue c n . Moreover, we have the growth estimates

4)
where G(x) = max e≤t≤x g(t) is a non-decreasing function for x ≥ e.
Proof Without loss of generality, we may suppose that {z n } is ordered according to increasing moduli. Moreover, G(x) is a well defined non-decreasing function by continuity. Let β = 3 √ α (> 1) and e < R < ∞. Suppose that |z| = r ≤ R, and write The expression S 1 (z) in (7.5) is a finite sum, and therefore it represents a rational meromorphic function in C. Hence, in order to prove that H (z) is meromorphic in C, it suffices to show that S 2 (z) converges uniformly. But this can be done analogously as in [13, p. 293] by making use of the fact that the genus of {z n } is p = 0.
Obviously, all poles z n of H (z) are simple and have residue c n . The inequality λ log +1 ≤ ρ log (H ) being clear from N (r , H ) ≤ T (r , H ), it remains to prove the second inequality in (7.4). This culminates in estimating S 1 (z). As q n ≥ 0 is the smallest integer satisfying (7.2), it follows that Proceeding as in [13, p. 294] (but replacing g with G to ensure monotonicity), we can find a constant C > 0 such that provided that |z| ≤ R and |z| / ∈ E ∪ [0, 1], where E ⊂ [1, ∞) has finite logarithmic measure. Let P be the canonical product associated with the sequence {z n }, and hence ρ log (P) = λ log + 1. Then P H is entire, and The second inequality in (7.4) follows from ρ log (H ) ≤ max{ρ log (P), ρ log (P H)}, which holds by using standard reasoning.

Lemma 9
Suppose that {z n } is an infinite sequence of non-zero points in C with λ log < ∞, and that {z n } is uniformly logarithmically q-separated for some q ≥ 0 (and C > 0). Let {σ n } be an infinite sequence of points in C, not necessarily distinct, and let h : [0, ∞) → [1, ∞) be a continuous and non-decreasing function such that |σ n | ≤ h(|z n |) for n ∈ N.
Then there exists an entire function A(z) ≡ 0 such that A(z n ) = σ n , n ∈ N, (7.6) and, for any given α > 1, is a non-decreasing function for x ≥ e.
Proof Without loss of generality, we may suppose that {z n } is ordered according to increasing moduli. Moreover, F(x) is well defined and non-decreasing by continuity. Let P(z) be the canonical product having {z n } as its zero sequence. By uniform logarithmic q-separation, the points z n are simple, and hence 1/P (z n ) ∈ C for any n.
(a) Suppose that σ n = 0 for all n. If H (z) denotes the Mittag-Leffler series in (7.3), where c n = σ n /P (z n ), then A = P H is entire and satisfies (7.6). Hence it suffices to prove (7.7). By the assumptions, there exists a constant C 1 > 0 such that |c n | ≤ C 1 |z n |h(|z n |)e C(log(1+|z n |)) q , n ∈ N.
Thus log |c n | log |z n | ≤ log h(|z n |)e C(log(1+|z n |)) q log |z n | + C 2 , |z n | ≥ e, (7.8) where C 2 > 0 is a constant. Choosing for t ≥ e, we deduce by Lemma 8 that where α > 1 and The assertion (7.7) then follows from ρ log (A) ≤ max{ρ log (P), ρ log (H )}. (b) Suppose that σ n = 0 for at least one n. If σ n = 0 for all n, we may choose A(z) = P(z), in which case (7.6) and (7.7) clearly hold. Hence we may suppose that σ n = 0 for at least one index n but not for all n. But this case is analogous to Part (b) in the proof of Lemma 3.

Proof of Theorem 3
Anderson The solution f to the problem above has real zeros with multiplicities tending to infinity. Moreover, the proof given in [1] is based upon a technical restriction ϕ(r ) = O(log r ), which yields ρ log ( f ) ∈ (2, 3). Next we present a modification of the Anderson-Clunie reasoning such that the solution f has only simple zeros, and hence it is suitable for being a solution of (1.2). In addition, f has arbitrary pre-determined logarithmic order on (2, ∞). and, in general, for any integer l ≥ 1, If m is even, then all zeros on |z| = b are pairwise complex conjugate numbers. If m is odd, then precisely one of the zeros on |z| = b lies on the negative real axis, while the rest are pairwise complex conjugates. Thus, if z ∈ R, we have P m,b (z) ∈ R due to the elementary identity ζ ζ = |ζ | 2 . In particular, P m,b is a real polynomial, and log P m,b (z) is analytic in −3π/4 < arg(z) < 3π/4. If P m,b has a real zero at ζ ∈ R − , then the modulus on any circle |z| = r is maximal when z = r . Suppose then that ζ, ζ is any pair of conjugate zeros of P m,b . Then the modulus on any circle |z| = r is maximal when z is simultaneously as far as possible from both points ζ, ζ that lie in 3π/4 ≤ arg(w) ≤ 5π/4, that is, when z = r . We repeat this reasoning for all zeros of P m,b , and conclude that Moreover, a simple geometric reasoning yields the growth estimates Next we define two non-decreasing sequences {b n } and {c n } of positive integers by setting b 1 = 1 = c 1 , and for n ≥ 2, where x denotes the integer part of x. This definition corresponds to the choice of points b n , c n in [1] where ϕ(r ) = (log r ) ρ−2 . We define a formal canonical product f in terms of the polynomial factors in (8.1) as (8.4) and prove that this function has the required properties. For r > 1 there exists an integer n such that b n < r ≤ b n+1 . If n(r ) denotes the number of zeros of f in |z| < r , counting multiplicities, then Thus the zeros of the formal product f have logarithmic exponent of convergence at most ≤ ρ − 1, so that f is entire, and, in fact, ρ log ( f ) ≤ ρ, see [4]. Let β = 2(ρ − 1)/(ρ − 2) > 2 for short. The auxiliary function is increasing on [0, 1] and decreasing on [1, ∞). Since g(0) = 1 and since g(x) → 1 as x → ∞, we have g(x) ≥ 1 for all x ≥ 0, that is, Therefore, we have c 1 = c 2 = 1, c 3 = 2 β ≥ 2 2 = 4, and it follows that the logarithmic exponent of convergence of the zero sequence of f is equal to ρ − 1, and thus ρ log ( f ) = ρ, see [4]. It remains to prove that δ V (0, f ) = 1, which is equivalent to lim inf r →∞ It suffices to show that For n ≥ 2, we have For a fixed r > 0 and all t > r , we deduce by (8.5) and L'Hospital's rule that as t → ∞. Therefore, Riemann-Stieltjes integration and integration by parts give us Using the well-known inequality [11, p. 18] log we then deduce that , (8.9) where n(b n ) = n j=1 c j ≥ c n = (log b n ) ρ−1 ≥ (log b n ) ρ−1 − 1.