Bank-Laine functions, the Liouville transformation and the Eremenko-Lyubich class
aa r X i v : . [ m a t h . C V ] D ec Bank-Laine functions, the Liouville transformationand the Eremenko-Lyubich class
J.K. LangleyDecember 24, 2018
For Larry Zalcman, on the occasion of his retirement from Journal d’Analyse Math´ematique
Abstract
The Bank-Laine conjecture concerning the oscillation of solutions of second order ho-mogeneous linear differential equations has recently been disproved by Bergweiler and Ere-menko. It is shown here, however, that the conjecture is true if the set of finite criticaland asymptotic values of the coefficient function is bounded. It is also shown that if E isa Bank-Laine function of finite order with infinitely many zeros, all real and positive, thenits zeros must have exponent of convergence at least / , and an example is constructedvia quasiconformal surgery to demonstrate that this result is sharp. MSC 2000: 30D35. If f is a non-constant entire function, let ρ ( f ) = lim sup r → + ∞ log + T ( r, f )log r , λ ( f ) = lim sup r → + ∞ log + N ( r, /f )log r ≤ ρ ( f ) , denote its order of growth and the exponent of convergence of its zeros [11]. In their landmarkpaper [1], Bank and Laine proved the following results on the oscillation of solutions of y ′′ + A ( z ) y = 0 . (1) Theorem 1.1 ([1])
Let A be an entire function, let f , f be linearly independent solutions of(1) and let E = f f , so that λ ( E ) = max { λ ( f ) , λ ( f ) } .(i) If A is a polynomial of degree n > then λ ( E ) = ( n + 2) / .(ii) If λ ( E ) < ρ ( A ) < + ∞ then ρ ( A ) ∈ N = { , , . . . } .(iii) If A is transcendental and ρ ( A ) < / then λ ( E ) = + ∞ . The case where / ≤ ρ ( A ) < was considered by Rossi [22] and Shen [23]. heorem 1.2 ([22, 23]) Let A be an entire function of order ρ ( A ) and let E = f f , where f , f are linearly independent solutions of (1). If ρ ( A ) = 1 / then λ ( E ) = + ∞ , while ρ ( A ) + 1 λ ( E ) ≤ if / < ρ ( A ) < . (2) In particular, if / ≤ ρ ( A ) < then ρ ( E ) > . The methods of [1] focused on the product E = f f of linearly independent solutions f j of (1),and in particular on the equation A = (cid:18) E ′ E (cid:19) − E ′′ E − c E , c = W ( f , f ) , (3) linking E and A , in which the Wronskian W ( f , f ) = f f ′ − f ′ f is constant by Abel’s identity.The paper [1] inspired much subsequent activity concerning the zeros of solutions of (1) and, moregenerally, linear differential equations with entire coefficients [16], and gave rise to the Bank-Laineconjecture – let A be a transcendental entire function of finite order ρ ( A ) and let f , f be linearlyindependent solutions of (1) : if λ ( f f ) is finite then ρ ( A ) ∈ N . However, two remarkable recentpapers of Bergweiler and Eremenko [5, 6] show via quasiconformal constructions not only thatthe Bank-Laine conjecture is false, but also that the inequality (2) is sharp.When A is a non-constant polynomial in (1), satisfying A ( z ) = a n z n (1 + o (1)) as z → ∞ ,there are n + 2 critical rays given by arg z = θ ∗ , where a n e i ( n +2) θ ∗ is real and positive, and theLiouville transformation Y ( Z ) = A ( z ) / y ( z ) , Z = Z zz A ( t ) / dt, (4) may be applied in sectors symmetric about these rays. This reduces (1) to a sine-type equation d YdZ + (cid:18) O (1) Z (cid:19) Y = 0 , for which solutions asymptotic to e ± iZ on a sectorial region in the Z plane are delivered byHille’s method [14, 15]. On one side of the critical ray, one of the corresponding solutions A ( z ) − / e ± iZ (1 + o (1)) of (1) is large while the other is small, and these roles are reversed asthe critical ray is crossed.In contrast, for transcendental entire A , although a local analogue of Hille’s method wasdeveloped in [17], applying on small neighbourhoods of maximum modulus points of A , theanalytic continuation and estimation of Z in (4) present substantial difficulties. However, it turnsout that for a certain class of entire functions A the transformation (4) may be adapted so as tobe readily applicable on components where | A ( z ) | is large.The Eremenko-Lyubich class B plays a key role in complex dynamics [3, 9, 25] and consists ofthose transcendental meromorphic functions A with the following property: there exists a positivereal number M = M ( A ) such that all finite critical and asymptotic values of A have modulusless than M . Now suppose that A ∈ B is entire. Then, by standard results from [21, p.287](see also [4]), all components U M of the set { z ∈ C : | A ( z ) | > M } correspond to logarithmic ingularities of A − over ∞ ; in particular, v = log A ( z ) maps each such U M conformally ontothe half-plane H given by Re v > log M . Under the change of variables A ( z ) = e v , z = φ ( v ) , A ′ ( z ) A ( z ) = dvdz = 1 φ ′ ( v ) , (5) in which z = φ ( v ) is the inverse mapping from H to U M , a solution y ( z ) of (1) on U M transformsto a solution w ( v ) = y ( z ) on H of w ′′ ( v ) − φ ′′ ( v ) φ ′ ( v ) w ′ ( v ) + e v φ ′ ( v ) w ( v ) = 0 , (6) and the second formula in (4) becomes, for a suitable choice of z = φ ( v ) , Z = Z vv e u/ φ ′ ( u ) du. (7) The fact that φ ′ varies relatively slowly on H , by classical theorems on conformal mappings [13],makes it possible to prove the following theorem. Theorem 1.3
Suppose that A is a transcendental entire function in the Eremenko-Lyubich class B , and let E = f f , where f , f are linearly independent solutions of (1). Then exactly one ofthe following holds.(A) The functions A and E satisfy ρ ( A ) = ρ ( E ) = 1 and T ( r, A ) + T ( r, E ) = O ( r ) as r → + ∞ . (8) (B) There exists d > such that the zeros of E satisfy n ( r, /E ) > exp (cid:0) dr / (cid:1) as r → + ∞ , (9) and in particular ρ ( E ) = λ ( E ) = + ∞ . It follows from Theorem 1.3 that the Bank-Laine conjecture, despite being false in general [5],is true when the coefficient A is entire and in the class B . An example going back to [1] showsthat each of conclusions (A) and (B) can occur: if A ( z ) = − e z − / then (1) has solutions f ( z ) = e − z/ exp ( − e z ) , f ( z ) = e − z/ exp ( e z ) , f ( z ) f ( z ) = e − z , ρ ( f f ) = 1 , as well as solutions g ( z ) = e − z/ sinh ( e z ) , g ( z ) = e − z/ cosh ( e z ) , λ ( g g ) = + ∞ . An example will be given in Section 4 to show that the exponent / in (9) is sharp.The second main result of this paper concerns the location of zeros of Bank-Laine functions,that is, entire functions E such that E ( z ) = 0 implies E ′ ( z ) = ± . By [2, Lemma C], an entirefunction E is a Bank-Laine function if and only if E = f f , where f , f are linearly independentsolutions of (1) with A entire and W ( f , f ) = 1 . Although a Bank-Laine function with norestriction on its growth may have an arbitrary sequence ( a n ) of zeros, subject only to a n → ∞ without repetition [24], the following result was proved in [7] concerning Bank-Laine functionswith real zeros. heorem 1.4 ([7]) Let E be a Bank-Laine function of finite order, with infinitely many zeros,all real, and denote by n ( r ) the number of zeros of E lying in [ − r, r ] . Then n ( r ) = o ( r ) as r → + ∞ . If, in addition, all zeros of E are positive, then n ( r ) = O ( r ) as r → + ∞ . The first assertion of Theorem 1.4 is evidently sharp, because of sin z . The next theoremwill establish a sharp lower bound for λ ( E ) when E is a Bank-Laine function of finite order withinfinitely many zeros, all real and positive. Here it is sufficient to consider the case where E is real entire, because otherwise it is possible to write E = Π e P + iQ , where Π is the canonicalproduct over the zeros of E , while P and Q are real polynomials; thus e iQ ( z ) = ± at every zeroof E and F = Π e P is also a Bank-Laine function. Theorem 1.5
Let E be a real Bank-Laine function of finite order, with infinitely many zeros,all real and positive. Then the exponent of convergence λ ( E ) of the zeros of E is at least / . Moreover, if λ ( E ) = 3 / then E and the associated coefficient function A have order ρ ( E ) = ρ ( A ) = 3 / . To demonstrate the sharpness of Theorem 1.5, quasiconformal techniques will be used inSection 6 to construct a real Bank-Laine function E , with only positive zeros, such that E andits associated coefficient function A satisfy λ ( E ) = ρ ( E ) = ρ ( A ) = 3 / , so that A provides afurther counter-example to the Bank-Laine conjecture.The author thanks the referee for an extremely careful reading of the manuscript and fornumerous helpful suggestions. The following lemma is an extension of a method from [17], and provides bounds for the errorterms in Hille’s method [14, 15].
Lemma 2.1
Let c > and < ε < π . Then there exists d > , depending only on c and ε ,with the following properties. Suppose that the function A is analytic, with | − A ( z ) | ≤ c | z | − ,on a domain containing Ω = Ω
R,S = { z ∈ C : 1 ≤ R ≤ | z | ≤ S < + ∞ , | arg z | ≤ π − ε } . Then the equation (1) has linearly independent solutions U ( z ) , V ( z ) satisfying U ( z ) = e − iz (1 + δ ( z )) , U ′ ( z ) = − ie − iz (1 + δ ( z )) ,V ( z ) = e iz (1 + δ ( z )) , V ′ ( z ) = ie iz (1 + δ ( z )) , (10) in which | δ j ( z ) | ≤ d | z | for z ∈ Ω ∗ R,S = Ω
R,S \ { z ∈ C : Re( z ) < , | Im( z ) | < R } . (11) Proof.
Let X = Se iσ , where σ = min { π/ , π − ε } . Choose an analytic solution v on Ω of v ′′ + 2 iv ′ − F v = 0 , F = 1 − A, (12)4 uch that v ( X ) = 1 , v ′ ( X ) = 0 , and write L ( z ) = v ( z ) −
1+ 12 i Z zX ( e i ( t − z ) − F ( t ) v ( t ) dt, L ′ ( z ) = v ′ ( z ) − Z zX e i ( t − z ) F ( t ) v ( t ) dt, (13) so that L ′′ ( z ) = v ′′ ( z ) + 2 i Z zX e i ( t − z ) F ( t ) v ( t ) dt − F ( z ) v ( z ) = − iL ′ ( z ) . Since L ( X ) = L ′ ( X ) = 0 , the existence-uniqueness theorem implies that L ( z ) ≡ on Ω .Now let z ∈ Ω ∗ R,S and let γ z describe the clockwise arc of the circle | t | = S from X to thefirst point x of intersection with the line Im( t ) = Im( z ) , followed by the straight line segmentfrom x to z ; then | e i ( t − z ) | ≤ on γ z ⊆ Ω . Since L ( z ) = 0 , (13) gives | v ( z ) − | ≤ Z zX | F ( t ) v ( t ) | | dt | , | v ( z ) | ≤ Z zX | F ( t ) v ( t ) | | dt | . (14) Now parametrize γ z by t = ζ ( s ) , where s denotes arc length on γ z . Using (14), write H ( s ) = 1 + Z s | F ( ζ ( σ )) v ( ζ ( σ )) | dσ, H ′ ( s ) = | F ( ζ ( s )) v ( ζ ( s )) | ≤ | F ( ζ ( s )) | H ( s ) , and | v ( ζ ( s )) − | ≤ H ( s ) − (cid:18)Z s H ′ ( σ ) H ( σ ) dσ (cid:19) − ≤ exp (cid:18)Z s | F ( ζ ( σ )) | dσ (cid:19) − , which leads to | v ( z ) − | ≤ exp ( I z ) − , I z = Z zX | F ( t ) | | dt | . (15) Let d , d , . . . denote positive constants which depend only on c and ε . The circle | t | = S contributes at most d S − ≤ d | z | − to I z in (15), while the contribution J z from the horizontalpart of γ z satisfies: J z ≤ Z + ∞ Re z ct dt = c Re z ≤ d | z | if | arg z | ≤ π/ ; J z ≤ Z R cx + (Im z ) dx ≤ d | Im z | ≤ d | z | if π/ ≤ | arg z | ≤ π − ε .Since R ≥ , (13) and (15) now deliver | v ( z ) − | ≤ exp (cid:18) d | z | (cid:19) − ≤ d | z | ≤ d , | v ′ ( z ) | ≤ Z zX | F ( t ) | (1 + d ) | dt | ≤ d | z | . Now set V ( z ) = v ( z ) e iz ; then (12) implies that V solves (1), and the estimates (10) and (11) for V follow at once. To obtain U it is only necessary to apply the above argument to the equationsolved by y (¯ z ) for every solution y ( z ) of (1). ✷ Unbounded sectorial regions may be handled as follows. emma 2.2 Suppose that c > and < ε < π , and that the function A is analytic, with | − A ( z ) | ≤ c | z | − , on Ω ′ = { z ∈ C : 1 ≤ R ≤ | z | < + ∞ , | arg z | ≤ π − ε } . Then there exist d > , depending only on c and ε , and solutions U, V of (1) on Ω ′′ = { z ∈ C : R < | z | < + ∞ , | arg z | < π − ε } \ { z : Re( z ) ≤ , | Im( z ) | ≤ R } , such that U and V satisfy W ( U, V ) = 2 i and (10), with | δ j ( z ) | ≤ d/ | z | , on Ω ′′ .Proof. Taking a sequence S n → + ∞ yields solutions U n , V n of (1) on Ω ∗ R,S n , with correspondingerror terms δ j,n ( z ) , j = 1 , , , . Here the functions zδ j,n ( z ) are uniformly bounded, sincethe constant d is independent of S in (11). Thus, by normal families, it may be assumedthat the U n , V n , δ j,n converge locally uniformly on Ω ′′ . The limit functions U, V satisfy (10),with | δ j ( z ) | ≤ d/ | z | on Ω ′′ . Since W ( U, V ) is constant, by Abel’s identity, it follows that W ( U, V ) = 2 i . ✷ Finally, a change of variables z → − z shows that Lemmas 2.1 and 2.2 hold if Ω R,S and Ω ∗ R,S ,and correspondingly Ω ′ and Ω ′′ , are replaced by their reflections across the imaginary axis. Throughout this section let H = { v ∈ C : Re v > } and let φ : H → C \ { } be analytic andunivalent. For v, v ∈ H , define Z = Z ( v, v ) as in (7) by Z ( v, v ) = Z vv e u/ φ ′ ( u ) du = 2 e v/ φ ′ ( v ) − e v / φ ′ ( v ) − Z vv e u/ φ ′′ ( u ) du. (16) Since φ ( H ) the image of H under log φ contains no disc of radius greater than π ; thusapplying Bieberbach’s theorem and Koebe’s one quarter theorem [13, Theorems 1.1 and 1.2] to φ and log φ respectively gives, for u ∈ H , (cid:12)(cid:12)(cid:12)(cid:12) φ ′′ ( u ) φ ′ ( u ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ u , (cid:12)(cid:12)(cid:12)(cid:12) φ ′ ( u ) φ ( u ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ π Re u . (17) The fact that the estimates (17) are independent of φ is the key to the results of this sectionand the proof of Theorem 1.3. Lemma 3.1
Let ε be a small positive real number. Then there exists a large positive real number N , depending on ε but not on φ , with the following property.Let v ∈ H be such that S = Re v ≥ N , and define v , v , v , K and K by v j = 2 j S
128 + iT , T = Im v , K j = n v j + re iθ : r ≥ , − π j ≤ θ ≤ π j o . (18) Then the following three conclusions all hold:(i) Z = Z ( v, v ) satisfies, for v ∈ K , Z = Z ( v, v ) = Z vv e u/ φ ′ ( u ) du = 2 e v/ φ ′ ( v )(1 + δ ( v )) , | δ ( v ) | < ε. (19)6 ii) ψ = ψ ( v, v ) = log Z ( v, v ) is univalent on a domain containing K .(iii) There exists a domain D , with v ∈ D ⊆ K , mapped univalently by Z onto a sectorialregion M satisfying Z = Z ( v , v ) ∈ M = { Z ∈ C : | Z | / < | Z | < + ∞ , | arg( ηZ ) | < π/ } , (20) where η = 1 if Re Z ≥ and η = − if Re Z < .Proof. To prove (i) assume that S = Re v is large and let v ∈ K , so that S = Re v ≥ S
32 = 2 Re v . (21) Now v may be joined to v by a straight line segment L v which is parametrised with respect to s = Re u , and an elementary arc length estimate | du | ≤ (sec π/ ds ≤ ds holds on L v . Thus(17) delivers, for u ∈ L v , | φ ′ ( u ) | ≤ (cid:18) Ss (cid:19) | φ ′ ( v ) | , | φ ′′ ( u ) | ≤ s | φ ′ ( u ) | ≤ s (cid:18) Ss (cid:19) | φ ′ ( v ) | , (22) which implies by (21) that (cid:12)(cid:12)(cid:12)(cid:12) e v / φ ′ ( v ) e v/ φ ′ ( v ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18) S Re v (cid:19) exp (cid:18)
12 Re ( v − v ) (cid:19) ≤ S exp ( − S/ < ε provided S is large enough. Moreover, (22) leads to (cid:12)(cid:12)(cid:12)(cid:12) e v/ φ ′ ( v ) Z vv e u/ φ ′′ ( u ) du (cid:12)(cid:12)(cid:12)(cid:12) ≤ Ψ( S ) = 8 S e S/ Z S e s/ s − ds. (24) Since lim S → + ∞ Ψ( S ) = 0 by L’Hˆopital’s rule, (21) implies that Ψ( S ) < ε/ if S is large enough.Thus (19) follows from (16), (23) and (24), which proves (i).Next, (19) gives, on K , ψ ( v ) = ψ ( v, v ) = log Z ( v, v ) = v φ ′ ( v ) + δ ( v ) , | δ ( v ) | ≤ | δ ( v ) | < ε. Since ε is small and S is large, (17), (18) and Cauchy’s estimate for derivatives now deliver (cid:12)(cid:12)(cid:12)(cid:12) ψ ′ ( v ) − (cid:12)(cid:12)(cid:12)(cid:12) ≤ v ≤ , (25) and hence Re ψ ′ ( v ) > , on a convex domain containing K , which proves (ii).Now let L = { v ∈ K : Re v ≥ S / } . Then, for v ∈ L , integration along the line segment from v to Re v + iT followed by that from Re v + iT to v yields, in view of (25), ψ ( v ) − ψ ( v ) = v − v η ( v ) , | η ( v ) | ≤ (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) log Re vS (cid:12)(cid:12)(cid:12)(cid:12) + tan π (cid:19) . (26)7 ince S is large this implies that, for v ∈ ∂L with Re v = S / , Re ( ψ ( v ) − ψ ( v )) ≤ − S
16 + 8 (cid:16) log 8 + tan π (cid:17) ≤ log 116 . On the other hand, all other v ∈ ∂L satisfy, by (18) and (26), | Im ( v − v ) | ≥ (cid:18) Re v − S (cid:19) tan π ≥ Re v π , | Im ( ψ ( v ) − ψ ( v )) | ≥ Re v π − (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) log Re vS (cid:12)(cid:12)(cid:12)(cid:12) + tan π (cid:19) ≥ π. Moreover,
Re ( ψ ( v ) − ψ ( v )) → + ∞ as v → ∞ in K , again by (26). Thus the strip (cid:26) ψ ( v ) + σ + iτ : σ ≥ log 18 , − π ≤ τ ≤ π (cid:27) lies in the interior of ψ ( L ) , which completes the proof of (iii) and the lemma. ✷ Proposition 3.1
There exists a positive real number N , independent of φ , with the followingproperty. If v ∈ H satisfies min { S , | e v / φ ′ ( v ) |} > N , S = Re v , and if w , w are linearly independent solutions of (6) with W ( w , w ) = ± φ ′ , | w ( v ) w ( v ) | ≥ , (27) then w w has a sequence of distinct zeros ζ m → ∞ in H which satisfy | φ ( ζ m ) | = O (log m ) as m → + ∞ . (28) Proof.
Observe first that, by Abel’s identity, the Wronskian of any two local solutions of (6)is a constant multiple of φ ′ . Fix a small positive ε and assume that v ∈ H , that w , w arelinearly independent solutions of (6) which satisfy (27), and finally that S and | e v / φ ′ ( v ) | areboth large. Let v , v , v , K and K be as in (18), and define Z by (16). By Lemma 3.1, Z = Z ( v , v ) is large and there exist η ∈ {− , } and a domain D ⊆ K , both as in conclusion(iii), so that M = Z ( D ) satisfies (20). The change of variables w ( v ) = e − v/ W ( Z ) , w j ( v ) = e − v/ W j ( Z ) , (29) transforms (6) on D to the equation on M given by W ′′ ( Z ) + (1 + G ( Z )) W ( Z ) = 0 , G ( Z ) = 116 e v φ ′ ( v ) (cid:18) φ ′′ ( v ) φ ′ ( v ) (cid:19) . (30) Here the derivatives in the first equation are with respect to Z , and | G ( Z ) | ≤ | Z | (31)8 n M = Z ( D ) , by (17), (19) and the fact that S = Re v is large. Now apply Lemma 2.2 with Ω ′ = { Z ∈ C : | Z | / ≤ | Z | < + ∞ , | arg( ηZ ) | ≤ π/ } ⊆ M , and let M = Ω ′′ , so that Z = Z ( v , v ) ∈ M ⊆ Ω ′ ⊆ M , by the choice of η . Since | Z | islarge, there exist solutions U ( Z ) , U ( Z ) of (30) on M , which satisfy W ( U , U ) = 2 i and | U ( Z ) e iZ − | + | U ( Z ) e − iZ − | ≤ d | Z | , (32) in which the positive constant d is independent of v and Z , by (31).Suppose first that, on M , W ( Z ) = σ U ( Z ) , W ( Z ) = σ U ( Z ) , σ j ∈ C \ { } . Then (19), (27) and (29) give ± φ ′ = W ( w , w ) = e − v/ W ( W , W ) dZdv = W ( W , W ) φ ′ = 2 iσ σ φ ′ , so that | σ σ | = 1 / . But Re v and | Z | are large, which implies, in view of (29) and (32), that w ( v ) w ( v ) = e − v / W ( Z ) W ( Z ) = e − v / σ σ U ( Z ) U ( Z ) is small, a contradiction.Because w and w are interchangeable, it now follows that at least one of W and W ,without loss of generality W , is a non-trivial linear combination W ( Z ) = A U ( Z ) − A U ( Z ) , A , A ∈ C \ { } , (33) of U , U on M . Fix a small positive κ and suppose that Z ∗ = 12 i log A A + πn, where n is an integer of large modulus and appropriate sign, depending on η . Then Z ∗ ∈ M and (32) implies that, on | Z − Z ∗ | = κ , i log A U ( Z ) A U ( Z ) − πn = Z − Z ∗ + J ( Z ) , | J ( Z ) | < κ. Hence W has a zero Z ∗∗ with | Z ∗∗ − Z ∗ | < κ , by Rouch´e’s theorem and (33).It follows that W ( Z ) has distinct zeros X , X , . . . , which tend to infinity in M and satisfy | X m | ≤ c + c m , where c , c , . . . denote positive constants which may depend on v and φ , butnot on m . By (19), these zeros X m satisfy, with ζ m ∈ K and ε small, X m = Z ( ζ m ) = 2 e ζ m / φ ′ ( ζ m )(1 + δ ( ζ m )) , | δ ( ζ m )) | < ε. Using (17) to estimate | log | φ ′ ( ζ m ) || then gives, in view of (18), | ζ m | ≤ c + c Re ζ m ≤ c + c log + | X m | + c log | ζ m | , | ζ m | ≤ c + c log + | X m | ≤ c + c log m. ow (28) is obtained by applying the Koebe distortion theorem [13, Theorem 1.3] with µ ( λ ) = φ ( v ) , v = 1 + λ − λ , | λ | < , ζ m = 1 + λ m − λ m . This yields, for large m , since ζ m tends to infinity in K , | φ ( ζ m ) | = | µ ( λ m ) | ≤ c (1 − | λ m | ) = c (cid:12)(cid:12)(cid:12)(cid:12) | ζ m | + 2Re ζ m + 14Re ζ m (cid:12)(cid:12)(cid:12)(cid:12) ≤ c | ζ m | ≤ c (log m ) . ✷ Let A , f and f be as in the hypotheses, without loss of generality satisfying W ( f , f ) = ± ,and set E = f f . Choose M > such that | A (0) | and all finite critical and asymptotic valuesof A have modulus at most M/ . It may be assumed that M ≤ , because otherwise the f j may be replaced by the functions g j ( z ) = M / f j ( z/M ) , which solve y ′′ + B ( z ) y = 0 , B ( z ) = M − A ( z/M ) . If T ( r, E ) = O ( r ) as r → + ∞ , then (3) delivers (8), and Theorems 1.1 and 1.2 force ρ ( A ) = ρ ( E ) = 1 .Assume henceforth that T ( r, E ) = O ( r ) as r → + ∞ and, following standard notation of theWiman-Valiron theory [12], denote by µ ( r, E ) the maximum term of the Maclaurin series of E ,and by ν ( r, E ) the central index. Then inequalities from [12] give T ( r, E ) ≤ log + M ( r, E ) ≤ log + µ (2 r, E ) + log 2 ≤ Z r ν ( t, E ) t dt + O (1) , (34) and so it may be assumed that ν ( r ) = ν ( r, E ) = O ( r ) as r → + ∞ .Let / < τ < . It follows from the Wiman-Valiron theory [12] that there exists a sequence ( z n ) satisfying | z n | = r n → + ∞ , | E ( z n ) | = M ( r n , E ) , lim n → + ∞ ν ( r n ) r n = + ∞ , (35) such that, if z = z n e σ , | σ | < ν ( r n ) − τ , then E ( z ) ∼ (cid:18) zz n (cid:19) ν ( r n ) E ( z n ) , E ′ ( z ) E ( z ) ∼ ν ( r n ) z , E ′′ ( z ) E ( z ) ∼ ν ( r n ) z , as well as, in view of (3), A ( z ) ∼ − ν ( r n ) z , A ( z ) − / ∼ ± izν ( r n ) . hus (35) delivers min {| E ( z n ) | , | A ( z n ) |} → + ∞ as n → + ∞ , while applying Cauchy’s estimatefor derivatives to A − / yields A ( z n ) − / A ′ ( z n ) = O (cid:18) r n ν ( r n ) · ν ( r n ) τ r n (cid:19) → . Take n so large that | E ( z n ) | > , log | A ( z n ) | > N , (cid:12)(cid:12)(cid:12)(cid:12) A ( z n ) / A ′ ( z n ) (cid:12)(cid:12)(cid:12)(cid:12) > N , (36) where N is the positive constant from Proposition 3.1. Then z n lies in a component C of { z ∈ C : | A ( z ) | > } , and C since | A (0) | < . Because all finite critical and asymptoticvalues of A have modulus at most / , a change of variables (5) gives a conformal equivalencebetween C and the right half-plane Re v > . Choose σ n , with Re σ n > , such that z n = φ ( σ n ) .Then e σ n = A ( z n ) and (5) and (36) imply that Re σ n > N , (cid:12)(cid:12) e σ n / φ ′ ( σ n ) (cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) A ( z n ) / A ′ ( z n ) (cid:12)(cid:12)(cid:12)(cid:12) > N . (37) A solution y ( z ) of (1) transforms under (5) to a solution w ( v ) = y ( z ) of (6), and { f , f } toa pair of solutions { w , w } of (6) with W ( w , w ) = ± φ ′ . Use (37) to apply Proposition 3.1,with v = σ n = φ − ( z n ) . Since | w ( v ) w ( v ) | = | E ( z n ) | > , by (36), the function E has asequence of distinct zeros φ ( ζ m ) satisfying (28). But this gives c > such that, for all large m ∈ N , and for r satisfying c (log m ) ≤ r < c (log( m + 1)) , n ( c (log m ) , /E ) ≥ m , n ( r, /E ) ≥ m + 13 >
13 exp (cid:0) ( r/c ) / (cid:1) , which establishes (9) and completes the proof of Theorem 1.3. ✷ The following example shows that the exponent / in (9) is sharp. Let A ( z ) = cos √ z ,which belongs to the Eremenko-Lyubich class B , and let f be a non-trivial solution of (1). Let ν ( r, f ) be the central index of f and apply to f the same results from the Wiman-Valiron theory[12] as used in (34) and subsequently. If r is large and lies outside an exceptional set of finitelogarithmic measure, and if | z | = r and | f ( z ) | = M ( r, f ) , then ν ( r, f ) z ∼ f ′′ ( z ) f ( z ) = − A ( z ) , ν ( r, f ) ≤ exp( c √ r ) , for some positive constant c . This upper bound for the non-decreasing function ν ( r, f ) thenholds for all large r , possibly with a larger c , and so applying to f the inequalities of (34) gives d > with n ( r, /f ) ≤ N (3 r, /f ) ≤ T (3 r, f ) + O (1) ≤ exp( d √ r ) as r → + ∞ . Because ρ ( A ) = 1 / , conclusion (A) of Theorem 1.3 cannot hold in this case, andso the exponent / in (9) is sharp. ✷ Proof of Theorem 1.5
Let E be as in the hypotheses, and assume that the zeros of E have exponent of convergence λ ≤ / . Then the canonical product Π over these zeros has order λ , and E = Π exp( P ) ,with P a real polynomial. If P has degree at least , then the zeros of E have Nevanlinnadeficiency δ (0 , E ) = 1 , which contradicts [7, Theorem 4.1] (see also [18, Theorem 2.1]). It maytherefore be assumed that E has order λ ≤ / .There exist an entire function A and solutions f , f of (1) such that W ( f , f ) = 1 and E = f f . Then f j ( z ) = 0 gives E ′ ( z ) = ( − j and each f j has infinitely many zeros, as maybe seen by considering the graph of E on the real axis. Define U by U = f f , U ′ U = W ( f , f ) f f = 1 E .
Lemma 5.1
The coefficient function A in (1) has order at most λ but is transcendental.Proof. The first assertion is an immediate consequence of the Bank-Laine equation (3). Thesecond may be deduced from a theorem of Steinmetz [26], or from a combination of Theorem 1.4with the result of Edrei, Fuchs and Hellerstein [8] that if E is an entire function of finite orderand genus at least , all of whose zeros are positive, then is a Nevanlinna deficient value of E ,from which the transcendence of A follows using (3). It may also be proved using Hille’s methodas follows. Suppose that A is a polynomial. Since the f j have infinitely many positive zeros,the positive real axis must be one of the A ) critical rays for the equation, and each f j must be large in both adjacent sectors. Let L be the first other critical ray encountered whenmoving counter-clockwise from the positive real axis. Since the f j have only positive zeros, both f j must change from large to small as this critical ray L is crossed. A contradiction then arisesfrom the fact that linearly independent solutions cannot be small in the same sector, because theWronskian is a non-zero constant. ✷ Because U ′ /U has order at most / and is never , while all zeros and poles of U are simple, U has no critical values and finitely many asymptotic values [19]. Since U ′ /U is real, there exists θ ∈ R such that U = f /f = e iθ U , with U real meromorphic. But replacing f by f e iθ and f by f e − iθ leaves E unchanged, and so it may be assumed that θ = 0 and U is real meromorphic.Take zeros x , x , x ∈ R of f , with x < x < x , and let R be the supremum of all r > such that the branch of U − mapping to x admits unrestricted analytic continuation in theopen disc B (0 , r ) of centre and radius r . Then R is finite, and U maps a simply connecteddomain Ω , with x ∈ Ω , univalently onto B (0 , R ) . Moreover, U − has a singularity over some α with | α | = R , and Ω contains a path γ which tends to infinity and is mapped by U onto thehalf-open line segment [0 , α ) . If α is real then, because U is real meromorphic and univalent on Ω , the path γ must coincide with ( −∞ , x ] or [ x , + ∞ ) , contradicting the fact that x , x γ .Hence α R and, since U has finitely many critical and asymptotic values, U − has logarithmicsingularities over α and α . Lemma 5.2
Let F ( z ) = ( E ( z ) − E (0)) /z . Then there exist M > and disjoint non-emptycomponents Σ , Σ of the set { z ∈ C : | F ( z ) | > M } . roof. There exists
M > such that for each β ∈ { α, α } there is a component Ω β of the set { z ∈ C : | U ( z ) − β | < /M } mapped univalently by v = log 1 / ( U ( z ) − β ) onto the half-plane H given by Re v > log M . It may be assumed that M is so large that Ω β ∩ B (0 ,
1) = ∅ and Ω α ∩ Ω α = ∅ . Let φ : H → Ω β be the inverse function and write U ( z ) = β + e − v , z = φ ( v ) ∈ Ω β , v ∈ H . (38) Then E ( z ) = U ( z ) U ′ ( z ) = β + e − v − e − v · φ ′ ( v ) = − (1 + βe v ) φ ′ ( v ) , (39) and φ satisfies, on H , as in (17), (cid:12)(cid:12)(cid:12)(cid:12) φ ′′ ( v ) φ ′ ( v ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ v − log M , (cid:12)(cid:12)(cid:12)(cid:12) φ ′ ( v ) φ ( v ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ π Re v − log M . (40)
It follows from (38), (39) and (40) that there exists c > such that, as v → + ∞ on R , | z | = | φ ( v ) | = o ( v c ) = o ( e v | φ ′ ( v ) | ) = o ( | E ( z ) | ) , F ( z ) → ∞ , whereas if Re v = 1 + log M then | E ( z ) | ≤ | z | (1 + | β | M e ) (cid:12)(cid:12)(cid:12)(cid:12) φ ′ ( v ) φ ( v ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ | z | (1 + | β | M e )4 π, | F ( z ) | ≤ (1 + | β | M e )4 π + | E (0) | . Hence there exist M > such that the set { v ∈ H : | F ( φ ( v )) | > M } has a component whoseclosure with respect to the finite plane lies in H . ✷ The remainder of the proof follows lines fairly similar to [22, 23]. By (3) and well knownestimates for logarithmic derivatives [10], there exist positive integers M , M such that (cid:12)(cid:12)(cid:12)(cid:12) E ′ ( z ) E ( z ) (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) E ′′ ( z ) E ( z ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ | z | M , | A ( z ) | = 14 | E ( z ) | + O (cid:0) | z | M (cid:1) , (41) for all z outside a union U of countably many open discs, whose centres tend to infinity andwhose radii have finite sum. Choose a polynomial P , of degree at most M , such that B ( z ) = A ( z ) − P ( z ) z M +1 (42) is entire. For j = 1 , define a subharmonic function u j ( z ) on C by u j ( z ) = log | F ( z ) /M | on Σ j , with u j ( z ) = 0 on C \ Σ j , where M and Σ , Σ are as in Lemma 5.2. Similarly, let Σ be acomponent of the set { z ∈ C : | B ( z ) | > } , and set u ( z ) = log | B ( z ) | on Σ , with u ( z ) = 0 on C \ Σ . These u j have orders satisfying ρ ( u j ) ≤ ρ ( F ) = ρ ( E ) = λ , for j = 1 , , while ρ ( u ) ≤ ρ ( B ) = ρ ( A ) ≤ λ .For j = 1 , , and t > let θ j ( t ) be the angular measure of the intersection of Σ j with thecircle S (0 , t ) of centre and radius t . If j = 1 , and | z | is large and z ∈ Σ j ∩ Σ , then (42)implies that z lies in the exceptional set U of (41). Hence there exists a set F ⊆ [1 , + ∞ ) , offinite linear measure, such that if r ∈ [1 , + ∞ ) \ F then the following all hold: (a) S (0 , r ) doesnot meet U ; (b) Σ j ∩ Σ j ′ ∩ S (0 , r ) = ∅ for j = j ′ ; (c) no Σ j contains S (0 , r ) . et S be large and positive: then a well known consequence of Carleman’s estimate forharmonic measure [27, pp.116-7] gives, as r → + ∞ , rS ≤ Z [ S,r ] \ F X j =1 ! dtt + O (1) ≤ Z [ S,r ] \ F X j =1 θ j ( t ) ! X j =1 θ j ( t ) ! dtt + O (1) ≤ X j =1 Z [ S,r ] \ F πtθ j ( t ) dt + O (1) ≤ X j =1 log(max { u j ( z ) : | z | = 2 r } ) + O (1) ≤ X j =1 ( ρ ( u j ) + o (1)) log r ≤ (6 λ + o (1)) log r ≤ (9 + o (1)) log r. It follows at once that ρ ( u j ) = λ = 3 / for each j . ✷ The construction of an example demonstrating the sharpness of Theorem 1.5 will involve domains D , D , D and D defined by D = { u ∈ C : 0 < | u | < + ∞ , < arg u < π/ } ,D = E ∪ E ,E = { s + it : − π/ < s < , −∞ < t < + ∞} ,E = { s + it : − π/ < s < π/ , < t < + ∞} ,D = { v ∈ C : 0 < | v | < + ∞ , − π/ < arg v < } ,D = D ∪ { ζ ∈ C : | ζ | < , Re ζ > } . (43) Lemma 6.1
Let h : ( −∞ , → ( −∞ , be a continuous bijection, such that h (1) = 0 while h ′ is continuous and has positive upper and lower bounds for −∞ < y < (that is, there exists ε > such that ε < h ′ ( y ) < /ε for −∞ < y < ). Then there exists a homeomorphism ψ from the closure of D to that of D , such that: (A) ψ maps D quasiconformally onto D , with ψ ( z ) → ∞ and ψ ( z ) = O ( | z | ) as z → ∞ in D ; (B) ψ ( iy ) = ih ( y ) for −∞ < y ≤ ; (C) ψ ( z ) is real and strictly increasing as z describes the boundary of D clockwise from i to infinity.Proof. Let φ : D → D be a conformal bijection such that φ ( i ) = 0 and φ ( z ) → ∞ as z → ∞ in D . Then φ ( z ) is real and strictly increasing as z describes the boundary of D clockwisefrom i to infinity. Moreover, there exists a continuous bijection k : ( −∞ , → ( −∞ , with k (1) = 0 and φ ( iy ) = ik ( y ) . It is clear from the reflection principle that k ′ ( y ) is continuous andpositive for −∞ < y < , and it will be shown that k ′ ( y ) has positive upper and lower boundsfor −∞ < y < .Take the restriction of φ to { z ∈ D : | z | > r } , for some large positive r , and reflect twice,first across the imaginary axis and then across the real axis. This shows that L = lim y →−∞ k ′ ( y ) = lim z →∞ φ ′ ( z )14 xists and is finite and positive, and gives φ ( z ) = O ( | z | ) as z → ∞ in D .Next, extend φ : D → D by reflection across the imaginary axis to a conformal mappingonto the lower half-plane, and apply the reflection principle to φ ( u ) = φ ( e iu ) on the half-disc { u ∈ C : | u − π/ | < r , Im u > } , for some small positive r . This extended function has φ ′ ( π/ = 0 , which shows that L = lim y → − k ′ ( y ) exists and is finite and non-zero, and hencepositive by continuity.The function H = h ◦ k − is a continuous bijection from ( −∞ , to itself, and so there existsa homeomorphism η from the closure of D to itself given by η ( x + iy ) = x + iH ( y ) for x ≥ and y ≤ . Furthermore, the chain rule shows that H ′ ( y ) is continuous, with positive upper andlower bounds, for −∞ < y < . Hence η is C on D with ∂η∂z = η x + iη y = 1 − H ′ ( y ) , ∂η∂z = η x − iη y = 1 + H ′ ( y ) , which ensures that η is quasiconformal on D , and that η ( z ) = O ( | z | ) as z → ∞ in D .It now follows that ψ = η ◦ φ is a homeomorphism from the closure of D to that of D ,quasiconformal on D itself, and satisfies ψ ( iy ) = η ( φ ( iy )) = η ( ik ( y )) = iH ( k ( y )) = ih ( y ) for −∞ < y ≤ .Finally, ψ ( z ) = O ( | φ ( z ) | ) = O ( | z | ) as z → ∞ in D . ✷ Lemma 6.2
Let E be the closure of the domain D in (43), and define F on E \ D by F ( s + it ) = f ( s + it ) for −∞ < s ≤ − π/ , t ∈ R ,F ( s + it ) = f ( s + it ) for π/ ≤ s < + ∞ , ≤ t < + ∞ ,f ( u ) = − i exp(2 e iu ) ,f ( u ) = cot( u/
2) = − i (cid:18) e iu − e iu (cid:19) . (44) Then F extends to a mapping from E into the extended plane, continuous with respect to thespherical metric, with the following properties.(i) H = log F maps D quasiconformally onto the quadrant D , with H ( π/
2) = 0 .(ii) Let L be the path consisting of the line segment from π to followed by the negativeimaginary axis in the direction of − i ∞ . Then F ( u ) is real and strictly increasing as u describes L , and each u ∈ L has s > such that Im F ( u ) < on D ∩ B ( u , s ) .(iii) F is locally injective on E ;(iv) There exists c > such that | F ( u ) | ≤ exp exp( c | u | ) for u ∈ D lying on the circles | u | = (4 n + 1) π/ , n ∈ N .Proof. Using the principal argument and logarithm, set h ( y ) = − π y for −∞ < y ≤ ,h ( y ) = − π (cid:18) iy − iy (cid:19) = − π − i log (cid:18) iy − iy (cid:19) for < y ≤ . (45)15 or < y < this gives h ( y ) = − π y, h ′ ( y ) = 21 + y → as y → ,and so h ′ (0) = 2 . Thus h is a continuous bijection from ( −∞ , to ( −∞ , and h ′ exists andis continuous for −∞ < y < , with ≤ h ′ ( y ) ≤ there. Lemma 6.1 gives a homeomorphism ψ from the closure of D to that of D , such that ψ maps D quasiconformally onto D , with ψ ( z ) = O ( | z | ) as z → ∞ in D and ψ ( iy ) = ih ( y ) for −∞ < y ≤ . Furthermore, ψ ( z ) is realand strictly increasing as z describes the boundary of D clockwise from i to infinity, and so is G = exp ◦ ψ , which is continuous on the closure of D and satisfies, by (45), G ( v ) = exp( ih ( y )) = − i exp(2 iy ) = − i exp(2 v ) for v = iy , −∞ < y ≤ ,G ( v ) = exp( ih ( y )) = − i (cid:18) iy − iy (cid:19) = − i (cid:18) v − v (cid:19) for v = iy , < y ≤ . (46) The next step is to set F ( u ) = G ( e iu ) on the closure of D . Now v = e iu maps D conformally onto D , with v → as Im u → + ∞ and v → ∞ as Im u → −∞ . Furthermore,the boundary of D is mapped by v = e iu as follows: the line Re u = − π/ to the negativeimaginary axis; the half-line Re u = π/ , ≤ Im u < + ∞ , to the segment v = iy , < y ≤ ;the real interval [0 , π/ to the arc of the unit circle from to i ; the negative imaginary axisto the real interval (1 , + ∞ ) . Hence (44) and (46) imply that F is well-defined and continuouson E , and that (i) and (ii) hold. Moreover, because ψ is injective on D and (44) implies thateach log | f j ( u ) | is positive for − π/ < Re u < π/ and negative for π/ < | Re u | < π/ , thefunction F is locally injective on E . Thus it remains only to prove (iv). By (44), it is enough tobound the growth of F ( u ) for u ∈ D , and hence it suffices to consider the continuous function G ( v ) on the closure of D . But, as v = e iu → ∞ in D , | F ( u ) | = | G ( v ) | ≤ exp( | ψ ( v ) | ) ≤ exp( O ( | v | )) = exp (cid:0) O ( | e iu | ) (cid:1) ≤ exp exp(2 | u | ) . ✷ Now define V ( z ) on the open upper half-plane H + by V ( z ) = F ( z / ) , in which z / is theprincipal branch and F is as in Lemma 6.2. Then V extends first to a (spherically) continuousfunction from the closed upper half-plane into the extended plane, mapping R into R ∪ {∞} ,and then to the whole plane via V ( z ) = V ( z ) . The extended function V is locally injective on C , in view of Lemma 6.2, and quasimeromorphic, by [20, Ch. I, Theorem 8.3]. Lemma 6.2(iv)delivers, as n → + ∞ , log + log + | V ( z ) | = O ( | z | / ) for | z | = r n = (cid:18) (4 n + 1) π (cid:19) / . (47) The remainder of the construction proceeds much as in [5]. Let D be the pre-image in H + of the domain D under u = z / , let E be its closure, and F the union of E and its reflectionacross the real axis. Then V is meromorphic off F and writing z = re iθ and u = se iη shows hat the complex dilatation µ V of V satisfies, for some C , C > , Z ≤| z | < + ∞ (cid:12)(cid:12)(cid:12)(cid:12) µ V ( z ) z (cid:12)(cid:12)(cid:12)(cid:12) dxdy ≤ Z ≤| z | < + ∞ ,z ∈ D r dr dθ = C Z ≤| u | < + ∞ ,u ∈ D s dη ds ≤ C Z + ∞ s ds = C . (48) Let φ be the (unique) quasiconformal homeomorphism of the extended plane which solves theBeltrami equation ∂φ∂z = µ V ( z ) ∂φ∂z a.e. and fixes each of , and ∞ [20]. In view of (48) andthe Teichm¨uller-Belinskii theorem [20, Ch. V, Theorem 6.1], there exists α ∈ C \ { } with φ ( z ) ∼ αz (49) as z → ∞ . Moreover, there exists a locally univalent meromorphic function U such that V = U ◦ φ on C . Let U ( z ) = U ( z ) : then U ( φ ( z )) = V ( z ) = V ( z ) = U ( φ ( z )) = U (cid:16) φ ( z ) (cid:17) . Hence φ ( z ) and φ ( z ) have the same complex dilation a.e. and both fix , and ∞ , which implies,by uniqueness, that φ is real on R and U is real meromorphic. Furthermore, φ ([0 , + ∞ )) =[0 , + ∞ ) and all zeros and poles of U are real and positive, while E = U/U ′ is a real Bank-Lainefunction with positive zeros.Now U satisfies, by Lemma 6.2 and (49), n ( r, /U ) + n ( r, U ) = O ( r / ) as r → + ∞ . (50) Let Π and Π be the canonical products over the zeros and poles of U respectively, which haveorder at most / , by (50), and write U = Π Π e h , E = U ′ U = Π ′ Π − Π ′ Π + h ′ , (51) where h is an entire function. For | z | = r n , where n is large, (47) and (49) yield log + log + | U ( φ ( z )) | = O ( | z | / ) = O ( | φ ( z ) | / ) . Thus the maximum principle delivers log + log + | Π ( ζ ) U ( ζ ) | = O ( | ζ | / ) as ζ → ∞ and hence log T ( r, Π U ) = O ( r / ) as r → + ∞ . (52) Combining (52) with (51) and the lemma of the logarithmic derivative leads to m ( r, h ′ ) ≤ m (cid:18) r, (Π U ) ′ Π U (cid:19) + m (cid:18) r, Π ′ Π (cid:19) + O (1) = O ( r / ) as r → + ∞ . Hence h ′ and E have order of growth at most / . Applying Theorem 1.5 then shows that E is areal Bank-Laine function whose zeros are all real and positive and have exponent of convergence / , and that E itself has order / , as has the associated coefficient function A . ✷ eferences [1] S. Bank and I. Laine, On the oscillation theory of f ′′ + Af = 0 where A is entire, Trans. Amer. Math. Soc.273 (1982), 351-363.[2] S. Bank and I. Laine, On the zeros of meromorphic solutions of second-order linear differential equations,Comment. Math. Helv. 58 (1983), 656-677.[3] W. Bergweiler, Iteration of meromorphic functions, Bull. Amer. Math. Soc. 29 (1993), 151-188.[4] W. Bergweiler and A. Eremenko, On the singularities of the inverse to a meromorphic function of finite order,Rev. Mat. Iberoamericana 11 (1995), 355-373.[5] W. Bergweiler and A. Eremenko, On the Bank-Laine conjecture, J. Eur. Math. Soc. 19 (2017), 1899-1909.[6] W. Bergweiler and A. Eremenko, Quasiconformal surgery and linear differential equations, to appear, J.Analyse Math.[7] D. 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