MEROMORPHIC FUNCTIONS OF THE FORM f(z) =

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1 MEROMORPHIC FUNCTIONS OF THE FORM f(z) = /(z z n ) J.K. LANGLEY AND JOHN ROSSI Abstract. P We prove some results on the zeros of functions of the form f(z) =, with complex a z z n n, using quasiconformal surgery, Fourier series methods, and Baernstein s spread theorem. Our results have applications to fixpoints of entire functions. M.S.C classification: 30D35. Keywords: meromorphic functions, zeros, critical points, logarithmic potentials, quasiconformal surgery.. Introduction A number of recent papers [7,, 2] have concerned zeros of meromorphic functions represented as infinite sums () f(z) =, z n, C, z n, z z n <. z n 0 We assume throughout that 0 and that z n without repetition. By (), (2) n(r) = = o(r), r. z n r If the z n are all non-zero and the are all real and positive, then () gives T (r, u) (3) f = u x iu y, u(z) = log z/z n, lim = 0, r r in which u is subharmonic in the plane. We will need the following fundamental result [3, p.327] on functions of the form () with complex. Here we use standard notation from [6]. Theorem. ([3]). Let f be given by (), and let 0 < p <. Then (4) m(r, f) = o(), 2π 0 z n f(re iθ ) p dθ = o(), r, so that in particular δ(, f) = 0. If f has finite lower order then δ(a, f) = 0 for all a C \ {0}, and the same conclusion holds for a = 0 if in addition (5) <, 0. This research was carried out during a visit by the first author to Virginia Tech. He thanks the Department of Mathematics for support and hospitality. Both authors thank David Drasin for helpful discussions.

2 2 J.K. LANGLEY AND JOHN ROSSI We state next some of the main results from [7, ]. Theorem.2 ([7, ]). Assume that f is given by () with all the real, and let n(r) be defined by (2). (a) If all the are integers and for all but finitely many n then f has infinitely many zeros. (b) If all the are integers and n(r) = o( r) as r then f has infinitely many zeros. (c) If inf{ : n N} > 0 then f has infinitely many zeros. Theorem.2, (c) represents a substantial step in the direction of the following conjecture from [7]. Conjecture.. If all the are real and positive in () then f has infinitely many zeros. Equivalently, subharmonic functions u as in (3), with as in (), have infinitely many critical points. It was conjectured further in [7] that if the in () are real and n(r) = o( r) as r then f must have zeros, but we give counter-examples to this in Examples 2.2 and 2.3. The key fact used in [] to prove Theorem.2, (c) is that inf{ } > 0 implies that T (r, f) = O(r) as r. The first main result of the present paper refines Theorem.2, (c) to allow finitely many complex, and this turns out to require application of the Ahlfors spiral theorem [8, p.600]. Our theorem also establishes Conjecture. when f has finite order and all but finitely many of the z n lie close to the real axis. Theorem.3. Let f be given by (), with the real and positive. Assume that f has finite order, that (6) =, and that either (i) (7) lim inf r or (ii) there exists ε > 0 with T (r, f) r <, (8) z n z n z n ε for all large n. Let S(z) be a rational function. Then f(z) S(z) has infinitely many zeros. In Theorem.3 and some of our subsequent results we consider the zeros of f S, with S a rational function, rather than of f. The effect of this is to allow in particular finitely many residues which are not real and positive. In Example 2. we show that the hypothesis (6) is not redundant in Theorem.3. On the other hand, if f has finite lower order and is absolutely convergent then the proof of Theorem. from [3] goes through to give δ(0, f S) = 0 for some choices of rational S: see Proposition 3.. Next, we make two remarks about Theorem.3, (ii). The first applies a theorem of Miles [23]. Suppose that the exponent of convergence of the z n is infinite, but that of the non-real z n is finite. Then if f is given by () and S is meromorphic

3 MEROMORPHIC FUNCTIONS OF THE FORM f(z) = P an/(z zn) 3 of finite order, f S has zeros with infinite exponent of convergence. If this is not the case, we can write f S = F G in which F is an entire function of infinite order, with real zeros, and G is meromorphic of finite order. Miles result [23] gives N(r, /F ) = o(t (r, F )) on a set of logarithmic density. This implies the existence of a sequence r m with log T (r m, F ) log r m, m(r m, f S) m(r m, /F ) m(r m, G) ( o())t (r m, F ), which contradicts (4) and the fact that S has finite order. Second, Ostrowski proved in [27] that if f is given by () with (9) Im(z n ) zn 2 < z n 0 and δ(a, f) > 0 for some a, then T (r, f) = O(r) as r. Ostrowski s result may therefore, if the are positive, be combined with the method of Theorem.2, (c) to show that f has infinitely many zeros. However, (8) only implies (9) if z n 0 z n ε <. We turn our attention next to zeros of f as given by (), with the complex. The following conjecture, which obviously implies Conjecture., seems likely to be true. Conjecture.2. If (0) sup{ arg : n N} < π/2 then f as given by () has infinitely many zeros. Conjecture.2 is certainly true if f has finite lower order and is absolutely convergent, by Theorem.. For the case in which = and f has order at most 2, we have the following result in support of Conjecture.2. Theorem.4. Let f be given by (), of order σ 2, and write () = x + n x n + iy n, x + n 0, x n 0, y n R, and (2) n + (r) = Assume that z n r x + n, n (r) = z n r (3) lim n(r) = =, r and that there exist positive constants δ, d such that x n. (4) n (r) ( δ)n + (r), n(r) d n + (r) for all large r. Let S(z) be a rational function. Then: (5) δ(0, f S) cos πσ, σ < 2 ; δ(0, f S) <, σ = 2.

4 4 J.K. LANGLEY AND JOHN ROSSI Conditions () and (4) are obviously satisfied if (0) holds. Theorem.4 allows infinitely many non-real, but the proof is totally dependent on minimum modulus results for functions of order at most 2, and in particular on results for the extremal case of the cos πρ theorem [8]. For f of order between 2 and and with arg sufficiently small, the following rather weaker result is applicable, based on the method of quasiconformal surgery [5, 6, 28]. The statement of the theorem is somewhat technical, but both it and Theorem.4 have subsequent applications to fixpoints of entire functions. Theorem.5. Let 0 < σ < and let f be given by () with order ρ < and (6) Re( ) > 2 + σ, < r n = < /σ, t n = arg ( π/2, π/2), and assume that, for all n, ( ) tan sin tn r n cos t n ρ (7) ( ( )) 2 ( ( )) < k 0 < 2 rn log 2 +r + n 2 2rn cos tan sin tn + ρ. tn r n cos t n Let S(z) be a rational function. Then f(z) S(z) has infinitely many zeros. Note that (6) automatically gives (3). The hypotheses (6) and (7) are required in order to facilitate quasiconformal surgery and to control the dilatation arising therefrom, and these conditions seem very unlikely to be sharp. Obviously if is real and positive and satisfies (6) then (7) holds, and Theorem.5 provides a result applying when the are sufficiently close to the positive real axis. Next, for f as in () but of possibly larger growth than in Theorems.4 and.5, we have the following result, based on Baernstein s spread theorem [2]. Theorem.6. Let < α < and 0 < σ. Let f be given by (), of finite lower order µ, such that (3) holds and (8) arg z n < b < C(µ, σ) = 2 µ sin σ 2, arg + arg z n < c < π 2, for all n. Let S(z) be a rational function. Then δ(0, f S) < σ. Corollary.. Let f be transcendental entire, of at most order, convergence class, and with zero sequence (z n ). If lim n arg z n = 0, then δ(0, f /f) = 0. Corollary. follows at once from Theorem.6, since f /f has a representation (), and establishes a conjecture of Fuchs [, 25] in the case where the zeros of f accumulate at a single ray. We observe next that the hypotheses on arg in Theorems.4,.5 and.6 are not redundant. In Examples 2.2 and 2.3 we construct functions f of the form (), with real, such that f(z) has no zeros. Thus results for complex require, in general, some condition on the lines of (5) or some hypothesis on arg. Our methods have an application to the fixpoints of entire functions of order less than. Whittington [3] proved that if F is a transcendental entire function with T (r, F ) = o( r) as r then F has infinitely many fixpoints z with either F (z) = or F (z) >, the proof based on applying the cos πρ theorem [3, 8] to F and the residue theorem to /(z F (z)). In the same paper Whittington gave an example of an entire function F of order 2 with only attracting fixpoints i.e. F (z) = z implies F (z) < (see also [4]). We prove here the following theorem.

5 MEROMORPHIC FUNCTIONS OF THE FORM f(z) = P an/(z zn) 5 Theorem.7. Let F be transcendental and meromorphic in the plane, with finitely many poles and of order at most 2. Let 0 < c <. Then F has infinitely many fixpoints z with F (z) = z, F (z) c. Theorem.7 is proved by writing /(z F (z)) in the form () and applying Theorem.4. Using again the method of quasiconformal surgery [5, 6, 28] we establish the following result on the multipliers at fixpoints of functions of order between 2 and. Theorem.8. Let F be transcendental and meromorphic with finitely many poles in the plane, and with order ρ ( 2, ). Let 0 < d <, π 6(log /d)2 + π < ρ 2 + ρ. Then F has infinitely many fixpoints u n with F (u n ) = u n, F (u n ) > d. The dependence of d on ρ in Theorem.8 seems unlikely to be sharp, in particular as ρ 2. However, the function z + ez has order and only superattracting fixpoints. We conclude the paper by proving some results when the denominators in () are replaced by a larger power. Theorem.9. Let k 2 be an integer, and let (9) F (z) = (z z n ) k+, z n, z n 0 zn k+ <. Then, as r outside a set of finite measure, ( ) k + (20) ( o())t (r, F ) < N(r, /F ). k Example 2.4 shows that the constant in (20) is sharp, and that Theorem.9 fails for k =, even if (2) < d > 0. z n 0 z +d n However, if we assume that (z n ) has finite exponent of convergence, and that z n 0 /z n <, then we do get a result for k =. Theorem.0. Let F be as in (9), with k =, and assume that (22) z n L <, < z n 0 for some L > 0. Then δ(0, F ) <. z n 0 The authors thank the referee for some very helpful comments and suggestions. z n

6 6 J.K. LANGLEY AND JOHN ROSSI Example Examples The following example shows that (6) is not redundant in Theorem.3. Let g(z) = i z 2 (e iz ), and let S be the principal part of g at 0, so that Let f(z) = S( ) = 0, g(z) + S(z) = O(), z 0. n Z,n 0 4π 2 n 2 (z 2πn), h(z) = f(z) g(z) S(z). Then f satisfies (), (7) and (8), using (4), and h is entire. Since g(z) = o() as z in the union of the circles z = π(2m + ), m N, we have h 0 by (4). Thus f S has no zeros. Example 2.2. Let z n be real and positive with z large and z n+ > 4z n for each n, and let g(z) = ( z/z n ), r n = 2 z n. We estimate g (z m ) in the following standard way. For z = r m with m large we have g (z) g(z) = z z n 2 <, log g(z) > log M(r m, g) O(). r n Let h(z) = g(z)/(z z m ). Applying the maximum principle to /h(z) in r m z r m shows that and it follows that log g (z m ) = log h(z m ) > 2 log M(r m, g) <, = /g (z n ) R. Let f be defined by (). Then f(z) /g(z) is entire. But on the circle z = r n, for large n, the function g(z) is large, while f(z) 2rn a m = o(). m= It follows that f(z) /g(z) 0, and so f has no zeros. Thus a function f(z) as in () may have real residues and arbitrarily small growth, but fail to have zeros. Example 2.3. Let H(z) = z cos z, J(z) = z + k Z ( ) k+ w k (z w k ), w k = (2k + )π. 2

7 MEROMORPHIC FUNCTIONS OF THE FORM f(z) = P an/(z zn) 7 Then it is evident that J(z) satisfies (), and there is an entire function G such that H J = G. However, since there exist arbitrarily large r such that H(z) is small on the whole circle z = r, estimate (4) shows that G 0 and so J has no zeros. Again J has real residues, this time with = O(log r), r. Example 2.4. z n r We show here that the constant in (20) is sharp, and that Theorem.9 fails for k =. Let g(z) = /(e z ). A simple induction argument shows that, for each k N, g (k) (z) = ez P k (e z ) (e z ) k+, in which P k is a polynomial of degree at most k. Expanding the numerator in powers of e z we see that there exists d k > 0 such that ( ) m(r, g (k) ) d k m r, e z = o(r), and so T (r, g (k) ) (k + )T (r, e z ), N(r, /g (k) ) (k )T (r, e z ) + O(). In particular, g has no zeros. By periodicity there exists a constant c k such that g (k) c k (z) = + O(), z 2πin, n Z. (z 2πin) k+ Set G k (z) = c k n Z (z z n ) k+, z n = 2πin. Then each G k, k N, is of the form (9). Also g (k) G k is entire of order at most. Let δ be small and positive and let z be large with arg z π/2 δ or π arg z π/2 δ. Then g (k) (z) is small and, with the d j positive constants, z z n d max{ z, z n }, G k (z) d 2 z 2 z n k 2, so that G k (z) is also small. Applying the Phragmén-Lindelöf principle now gives g (k) G k. Clearly G, which has no zeros, satisfies (2). n Z We need the following lemmas. 3. Preliminaries Lemma 3. ([2]). Suppose that d > and that G is transcendental and meromorphic in the plane of order less than d. Let R 0 > 0. Then there exist uncountably many R > R 0 such that the length L(r, R, G) of the level curves G(z) = R lying in z r satisfies (23) L(r, R, G) r (3+d)/2, r log R.

8 8 J.K. LANGLEY AND JOHN ROSSI Lemma 3.2 ([9]). Let < r < R < and let g be meromorphic in z R. Let I(r) be a subset of [0, 2π] of Lebesgue measure µ(r). Then log + g(re iθ ) dθ Rµ(r) ( ) + log + T (R, g). 2π R r µ(r) I(r) Lemma 3.3 ([7]). Let S(r) be an unbounded positive non-decreasing function on [r 0, ), continuous from the right, of order ρ and lower order µ. Let A >, B >. Then G = {r r 0 : S(Ar) BS(r)} satisfies ( ) ( ) log A log A logdensg ρ, logdensg µ. log B log B The next lemma is a standard application of Tsuji s estimate for harmonic measure [30]. Lemma 3.4. Let u be subharmonic and non-constant in the plane, and let U be a domain such that u 0 on U and sup{u(z) : z U} > 0. For t > 0 let θ U (t) be the angular measure of the intersection of U with the circle z = t, except that θ U (t) = if the whole circle z = t lies in U. Then there exists R 0 with r R 0 tθu in which B(2r, u) = sup{u(z) : z = 2r}. πdt log B(2r, u) + O(), r, (t) Lemma 3.5. Let f be as in (), of finite lower order, and suppose that δ(0, f S) > 0, for some rational function S. Then S( ) = 0. Proof. Since f S has finite lower order Lemmas 3.2 and 3.3 give c 0 > 0 and arbitrarily large r such that f S is small on a subset E r of the circle z = r of angular measure at least c 0. If S( ) 0, this contradicts (4). Since several of our results (Theorems.3,.4,.5 and.6) rely on the assumption that diverges, we include for completeness the following immediate extension of the method of Theorem. from [3]. Proposition 3.. Suppose that f is given by (), with <, and that f has finite lower order. If S is a rational function with (24) lim zs(z) λ =, z then δ(0, f S) = 0. Examples 2.2, with S = λ = 0, and 2. show that (24) is not redundant in Proposition 3.. Proof. Assume that f and S are as in the hypotheses, but that δ(0, f S) > 0. Then S( ) = 0, by Lemma 3.5. Following [3, p.333], write in which z(f(z) S(z)) = g(z) + λ zs(z), g(z) = z n z z n

9 MEROMORPHIC FUNCTIONS OF THE FORM f(z) = P an/(z zn) 9 satisfies the requirements of (). The assumption (24) gives lim z (λ zs(z)) 0. Now (4), applied to g, shows that g(z) = o(), f(z) S(z) = O(r) for all z on z = r, apart from a set I(r) of angular measure o() as r. Lemma 3.2 gives (δ(0, f S) o())t (r, f S) o(t (2r, f S)) which contradicts Lemma 3.3, since f has finite lower order. 4. Proof of Theorem.3 To prove Theorem.3 assume that f,, z n are as in () and (6) with the real and positive. Assume further that f satisfies at least one of (7) and (8), but that f S has finitely many zeros, for some rational function S. We may assume that all z n are non-zero. Let G and the positive integer N satisfy (25) N > 3 + ρ(f), f(z) S(z) = z N G(z). Then G is transcendental and meromorphic in the plane, of finite order and with finitely many poles. By Lemma 3.5, we have S( ) = 0. Let r 0 be large and positive, so large that neither G nor S has poles in r 0 z <. Thus log G(z) is subharmonic in z > r 0. Using Lemma 3. and (25), choose a large positive R, in particular with (26) R > M(2r 0, G), L(r, R, G) r N r log R. We may also assume that G has no multiple points with G(z) = R, and that the level curves G(z) = R are never tangent to any circle z = t > 0 or ray arg z = t R. Lemma 4.. The set {z : G(z) > R} has finitely many unbounded components V j. If U is one of the V j then U lies in z > 2r 0. Further, the finite boundary U of U is the union of countably many pairwise disjoint level curves of G, each either simple and going to infinity in both directions, or simple closed. Finally, we have (27) I = f(z) S(z) dz <, and B(0, 2r 0 ) lies in an unbounded component of C \ U. U Proof. There exist finitely many V j, since G has finite order, and each lies in z > 2r 0, by the choice of R. The assertion concerning the components of U holds since the level curves G(z) = R do not intersect. To prove (27) let T = log R, and partition U into its intersections with the disc z T and the annuli 2 m T < z 2 m+ T, m Z, m 0. Since (25) gives (28) f(z) S(z) R z N R, z U U, we get I R (T N + ) 2 (N )(m+) T N 2 Nm T N <, m=0 using (26), since r 0 and R are large.

10 0 J.K. LANGLEY AND JOHN ROSSI We prove the last assertion of the lemma by contradiction, and thus assume that B(0, 2r 0 ) lies in a bounded component of the complement of U. Then there exists a simple closed curve γ, a component of U, such that B(0, 2r 0 ) lies in the interior U of γ. Since z n, for large n, is a pole of f S with residue > 0 we get, using (27), f(z) S(z) dz 2π O() 2π O(), γ z n U z n <r 0 which contradicts (6) provided r 0 was chosen large enough. Lemma 4.2. Let U be one of the V j, and let C be the unbounded component of U which separates U from B(0, 2r 0 ). Let D be the component of C \ C which contains U. Fix w 0 U and define a single valued branch of log z, continuous on the closure of D, with arg w 0 π. Then we have log z = O(log z ) as z in the closure of D. Proof. Define a subharmonic function v(z) on C as follows: (29) v(z) = log G(z) log R, z U ; v(z) = 0, z U. Then v has finite order, since G has. The boundary of D is the curve C, and v(z) = 0 there. Thus the Ahlfors spiral theorem [8, pp ] gives arg z = O(log z ) as z tends to infinity on D. Since arg z is monotone on arcs of circles centred at the origin, the result follows. Lemma 4.3. Let u be defined by (3), and let U be one of the V j. Then u(z) = O(log z ) as z in the closure of U. Proof. Using U = V j, w 0, D and the same branch of log z as in Lemma 4.2, there exist a constant c and a function S, analytic and bounded in z > r 0, such that (30) S(z) = d dz (c log z + S (z)) for z in the closure of D. By (3) and (25) we have (3) u x iu y = f(z) = S(z) + z N G(z). For each w in U, join w 0 to w by a path σ w consisting of part of the ray arg z = arg w 0, part of the circle z = w, and part of the boundary of U. Using Lemma 4. and (28) we get a constant c 2 such that σ w f(z) S(z) dz < c 2 for all w in U. This gives, using (30) and (3), (u x iu y )(dx + idy) c log w + O() σ w and the result follows using Lemma 4.2. Lemma 4.4. There exist positive constants k, k 2 with the following property. With u as in (3), define: u (z) = max{u(z) k log z k 2, 0}, z > 2r 0 ; u (z) = 0, z 2r 0.

11 MEROMORPHIC FUNCTIONS OF THE FORM f(z) = P an/(z zn) Then u is non-constant and subharmonic in the plane, with u (z) = 0 on the union of the closures of the V j, and T (r, u ) = o(r) as r. Proof. Choose k and k 2 using Lemma 4.3, so that u(z) k log z +k 2 on z = 2r 0 and on the union of the closures of the V j. Thus u is subharmonic, with T (r, u ) = o(r) by (3). Finally, u is non-constant by (6). We may now complete the proof of Theorem.3 in case (i), in which (7) holds. Let U be one of the V j, and let W be a component of the set {z : u (z) > 0}. Then (7), (25) and (29) give lim inf r T (r, v)/r <. Since v vanishes off U, while u vanishes on the closure of U, a standard application of the Cauchy-Schwarz inequality as in [30] gives θu (t) + θw (t) 2 π for large t, and a contradiction arises on applying Lemma 3.4 to v and u. To finish the proof of Theorem.3, it remains only to dispose of case (ii), and we assume henceforth that (8) holds. Lemma 4.5. For r > 0 and 0 < t < π/2 let V + (r, t) = {z : z = r, t < arg z < π t}, V (r, t) = {z : z = r, π + t < arg z < 2π t}. Let δ > 0. Then for all r in a set E δ of lower logarithmic density at least δ, we have u (z) 0 on at least one of the sets V + (r, δ), V (r, δ). Proof. We may assume that δ is small. Then we have (32) f(z) S(z) = O(), z, δ/4 arg z π δ/4, since S( ) = 0 and (8) gives z n = O( z z n ) for large z as in (32). By Lemma 3.3, there exist c 3 > 0 and a set E of lower logarithmic density at least δ such that T (4r, G) < c 3 T (2r, G), r E. For r E, since δ is assumed small, Lemma 3.2 and (25) now give (33) 2 log G(z) > T (2r, G), 4 log f(z) S(z) < T (2r, G), z I r, in which I r is a subset of the circle z = 2r, of angular measure at least 8δ > 0. Let δ 2 = min{δ, δ } and let r be large, with r E. Without loss of generality I r V + (2r, δ 2 /2) has angular measure at least δ 2, and we apply the two-constants theorem [26] to log f(z) S(z) in the interior Ω of the region r/2 z 2r, δ 2 /4 arg z π δ 2 /4. This gives positive c 4, c 5 independent of r such that, using (32) and (33), 4 log f(z) S(z) T (2r, G)ω(z, I r, Ω)+O() c 4 T (2r, G)+c 5, z V + (r, δ 2 ), where ω(z, I r, Ω) denotes harmonic measure, and so V + (r, δ 2 ) V j, for some j, using (25) again. Lemma 4.5 now follows from Lemma 4.4. Lemma 4.6. u has lower order at least.

12 2 J.K. LANGLEY AND JOHN ROSSI Proof. This is a standard application of Lemma 3.4. Let Ω be a component of the set {z : u (z) > 0}. Let δ > 0. Then by Lemma 4.5 we have θω (t) π + 2δ for all t in a set E δ of lower logarithmic density at least δ. Lemma 3.4 gives r πdt log B(2r, u ) O() tθω (t) π dt ( δ o())π log r π + 2δ t π + 2δ R 0 E δ [R 0,r] as r, and since δ is arbitrary the result follows. Lemma 4.7. Let δ > 0. Then there exists c(δ) > 0 such that for all large r we have (34) u (w) c(δ) w ε, w = r, δ arg w π δ. Proof. By (8) there exist c 6 = c 6 (δ) > 0 and r > 0 such that (35) z n w c 6 max{ z n, w }, δ arg w π δ, w = r r. For w as in (35) we get, without loss of generality, 0 log w/z n log w/z n = log w/z n w/z n = log + w(z n z n ) z n (z n w), and so (8) gives for large n, since log + t log( + t ) t, log w/z n log w/z n This gives, for w as in (35), using (2), (36) u(w) u(w) O(log r)+c 6 r w z n ε c 6 max{ z n, w }. t ε dn(t)+c 6 r t ε dn(t) = O(r ε ). But, by Lemma 4.5, u (w) vanishes and so u(w) O(log r) on at least one of the arcs V + (r, δ), V (r, δ), and (34) now follows from (36). Let η be small and positive. Since u has finite order Lemma 3.3 gives a positive constant c 7 such that (37) T (2r, u ) c 7 T (r, u ), r F, in which F has lower logarithmic density at least η/2. Let δ be small and positive, in particular so small that (38) δ < η/2, 24δc 7 <. Using Lemma 4.7 we find that, for all r in a set F 2 of lower logarithmic density at least δ, we have T (r, u ) 4δB(r, u ) + O(r ε ) 2δT (2r, u ) + O(r ε ). This gives, by (37) and (38), T (r, u ) = O(r ε ), r F 3 = F F 2, and F 3 has lower logarithmic density at least η. This contradicts Lemma 4.6, and the proof of Theorem.3 is complete. r

13 MEROMORPHIC FUNCTIONS OF THE FORM f(z) = P an/(z zn) 3 5. Proof of Theorem.4 Suppose that f and S are as in the statement of Theorem.4, but that (5) fails. Then we have S( ) = 0, by Lemma 3.5. We may assume that all the z n are non-zero. Throughout the proof we use c to denote a positive constant, not necessarily the same at each occurrence. We also write (39) z = re iθ, r = z, θ = arg z [0, 2π]. Lemma 5.. If r is large and positive and none of the z n lie on z = r then (40) cn(r) Further, (4) lim sm(s, H) =, s 2π Proof. We have, by the residue theorem, z n <r I = 2π 0 0 zf(z) dθ. H(w) = f(w) S(w). zf(z)dθ = 2π z n <r and so, using (4), I 2π Re( ) = 2π(n+ (r) n (r)) cn + (r) cn(r), which proves (40). Now (4) follows from (3), (40) and Lemma 3.5. Since (5) fails by assumption, it follows immediately from (4) and an application to /H of the cos πλ theorem for functions with deficient poles [3, p.262] (see also [5]) that σ = 2. We now write H = f /f 2, with f, f 2 entire functions of order at most 2 with no common zeros. By (4), (4) and Lemma 3.5 we have T (r, H) N(r, H) + O() N(r, /f 2 ) + O(). Also, since (5) fails by assumption, we get δ(0, H) =, N(r, /f ) = N(r, /H) = o(t (r, H)) = o(n(r, /f 2 )). Now (4) gives m 0 (r, f 2 ) = min{ f 2 (z) : z = r} = o(rm(r, f )), and exactly as in [, pp ] we obtain (42) log m 0 (r, f 2 ) o(log M(r, f 2 )), log M(r, f ) o(log M(r, f 2 )). Lemma 5.2. There exist a set E of logarithmic density and, for each r E, a subset U r of [0, 2π] such that (43) m(u r ) = o(), zf(z) = O(), r = z, θ = arg z V r = [0, 2π] \ U r, in which m(u r ) denotes Lebesgue measure.

14 4 J.K. LANGLEY AND JOHN ROSSI Proof. The relations (42) imply that f 2 is extremal for the cos πρ theorem. Let the positive function ψ(r) tend to 0 slowly as r. Results of Drasin and Shea [8] (see also [9, section II]) give log f 2 (z) ψ(r) log M(r, f 2 ), z = r E, arg z V r. Using (42) we get zh(z) = o() for z V r and (43) follows, using (4) and the fact that S( ) = 0. We now apply to zf(z) the method of [24, Theorem ]. Since n(r) has order at most, we may apply [24, p.98] with M = 6, ρ = and R 0 = min{ z n } to obtain (2.) and (2.2) of [24] for all r in a set E = E M of positive lower logarithmic density. Note that the lemma of [24, p.98] is stated only for integer-valued functions, but the proof goes through for n(r) as in (2). We may assume that E E, with E as in Lemma 5.2, and that E { z n } =. Assume henceforth that r is large and in E, and that z satisfies (39). Write and (44) z n <r z n >r z = z z n z n <r m=0 z = z z n z n >r z z n (z n /z) m = k=0 m Z,m 0 (z/z n ) k = m= z m z m z n <r z n >r z m n z m n, the change of order of summation justified since the first double series in (44) is plainly absolutely convergent. Thus (45) zf(z) = m Z b m (r)e imθ, in which (46) b m (r) = and z n >r (47) b m (r) = z n <r (r/z n ) m, m > 0, (r/z n ) m, m 0. Fix a large positive integer q, in particular with q > 2M(ρ+) = 24. Then (2.) and (2.2) of [24] may be applied to (46) and (47) exactly as in [24, p.202], to give (2.) and (2.2) of [24], with b m (r, F j ) and n j (r) replaced by b m (r) and n(r) respectively. This leads to (48) b m (r) 2 cn(r) 2. Write m Z,m {0,,...,q} (49) zf(z) = P (z) + s(r, θ), P (z) = Then (48) gives (50) s = s 2 = ( 2π 2π 0 q b m (r)e imθ. m=0 ) s(r, θ) 2 2 dθ cn(r).

15 Set MEROMORPHIC FUNCTIONS OF THE FORM f(z) = P an/(z zn) 5 (5) P (z) 2 = so that q k= q h k (r)e ikθ, h k (r) = 2π (52) h k (r) h 0 (r) = P 2 = By (5), since [0, 2π] = U r V r, P (z) 2 dθ = h 0 (r)m(v r ) V r 2π 0 q b m (r) 2. m=0 U r k 0 P (z) 2 e ikθ dθ, h k (r)e ikθ dθ and so (43) and (52) give (53) P (z) 2 dθ h 0 (r)(m(v r ) 2qm(U r )) (2π o())h 0 (r). V r But on V r we have, using (43) and (49), zf(z) = O(), and so (53) gives h 0 (r) c ( s + O()) 2 dθ c V r Thus, using (50), P (z) = s(r, θ) + O(), 2π (54) h 0 (r) c ( s + O()) 2 cn(r) 2. 0 ( s + O()) 2 dθ. Now (3), (40) and (43) give ( ) 2 n(r) 2 c zf(z) dθ cm(u r ) zf(z) 2 dθ U r U r and so, using (43) again and (45), (48), (52) and (54), we get n(r) 2 cm(u r ) zf(z) 2 cm(u r ) b m (r) 2 o(n(r) 2 ), m Z which is obviously a contradiction. Theorem.4 is proved. 6. A theorem on zeros of entire functions Theorem.3 leads to a result on the zeros of entire functions of small growth, which will be required for the proof of Theorem.5. We begin with: Lemma 6.. Let h be transcendental and meromorphic of finite order in the plane, with finitely many poles and with T (r, h) (55) lim inf = 0. r r Assume that (z n ) is on-zero sequence tending to infinity without repetition such that all but finitely many zeros of h lie in the set {z n }. Assume further that / = h (z n ) 0 for each n, and that (56) <. z n

16 6 J.K. LANGLEY AND JOHN ROSSI Then we may write (57) = f(z) S(z), h(z) where f is given by () and S is a rational function with S( ) = 0. Proof. Assume that h is as in the statement. Using (56), define f(z) by (). Then there exists a rational function S, with S( ) = 0, such that (58) g(z) = f(z) + S(z) h(z) is entire. By (4) we have (59) T (r, f) N(r, /h) + o(), T (r, g) 2T (r, h) + O(log r). By (58) we need only show that g 0. Choose ρ > ρ(h), and suppose that r is large and positive and the circle z = r does not meet the union U 0 of the discs B(z n, z n ρ ). Then for z = r we have f(z) (2r)ρ+. z n r/2 z n z n 2r z n r/2< z n <2r Hence using (56) there exist a set E 0 of finite measure and a positive integer ρ 2 such that (60) f(z) r ρ2, z = r E 0. Suppose first that g is transcendental, and set g (z) = g(z)z ρ2. Let r 0 > be so large that h has no poles in z r 0 and (6) M(r 0, g ) > 4, M(r 0, h) > 4 and S(z) < for z r 0. Choose K > M(r 0, g ) + M(r 0, h) and let U be an unbounded component of the set {z : g (z) > K}, and U 2 an unbounded component of the set {z : h(z) > K}. Then both U j lie in z > r 0. For t > 0 and j =, 2, let θ j (t) be the angular measure of the intersection of U j with the circle z = t, and let θj (t) = θ U j (t), as defined in Lemma 3.4. Then by (58), (60) and (6) we have θ (t) + θ 2 (t) 2π for r E 0 and so π (62) θ (t) + π θ2 (t) 2, r E 0. Using (55) and (59) and the fact that E 0 has finite measure, a standard application of Lemma 3.4 and (62) now gives a contradiction. Suppose finally that g is a polynomial. Lemmas 3.2 and 3.3 give arbitrarily large r such that h is large on a subset E r of the circle z = r of angular measure at least c 3 > 0. Using (4), (58) and the fact that S( ) = 0 we get g(z) = o() for at least one z E r. Thus g 0. This proves Lemma 6.. Theorem 6.. Let h be as in Lemma 6., and assume in addition that h (z n ) is real and positive for each n. Then /h (z n ) <. To prove Theorem 6., assume that h is as in the statement, but that =, in which = /h (z n ). Then we have (57) and, by (55) and (59), f satisfies the hypotheses of Theorem.3. Thus f S has infinitely many zeros and this contradicts (57). z n

17 MEROMORPHIC FUNCTIONS OF THE FORM f(z) = P an/(z zn) 7 The hypothesis (55) is not redundant in Theorem 6.. In [20] an entire function E is constructed with zero sequence (z n ) such that z n+ /z n > 2 and E (z n ) = for all n, while T (r, E) = O(r) as r. Corollary 6.. Let 0 < c < c 2 < and let h be transcendental and meromorphic with finitely many poles in the plane. Assume that all but finitely many zeros z of h have h (z) R, c < h (z) < c 2. Then h has order ρ(h). To prove Corollary 6., assume that h has order less than. Then we may choose a sequence (z n ) such that all but finitely many zeros of h lie in {z n }, and /h (z n ) (c, c 2 ), and such that z n <. Since /h (z n ) obviously diverges, this contradicts Theorem 6.. The obvious example h(z) = e z shows that Corollary 6. is sharp. 7. Application of quasiconformal surgery Lemma 7.. Let 0 ρ <, 0 < κ <, and let g be quasimeromorphic with finitely many poles in the plane and with the following properties: (i) there exist Jordan curves σ n, τ n in C such that σ n lies in the interior domain V n of τ n, for each n N, and V n τ n lies in the exterior domain of τ n, for n n; (ii) g has no poles in V n τ n and maps V n into the interior domain U n of σ n ; (iii) g is conformal on U n, for each n, and meromorphic on Y = C\ (V n τ n ); (iv) we have (63) log M(r, g) < r ρ+o(), r ; (v) we have (64) g z κ g z a.e. Set K = +κ κ. Then there exist a K-quasiconformal homeomorphism φ of the extended plane, fixing 0,,, and a function h meromorphic in the plane of order at most ρk, with finitely many poles, such that g φ h φ. Further, φ is conformal on W = U n and on the interior of C \ m= g m (W ). Proof. This is basically Shishikura s lemma on quasiconformal surgery [5, 28]. Let W 0 = W, W m+ = g m (W ) \ g m (W ), m 0, H = C \ g m (W ). Define a Beltrami coefficient µ(z) on C as follows. For z W 0 H set µ = 0. Assuming that µ has been defined on W m, define µ for w W m+ by (65) µ(w) = µ g(w) + µ(g(w))a(w) + µ(g(w))µ g (w)a(w), A = g w g w. Thus µ is defined inductively a.e. in C. We assert next that (64) and (65) give (66) µ(w) κ a.e. This is obviously true for w W 0 H, and we have (66) for w V n using (64), since µ(g(w)) = 0 for such w, by (ii). Now suppose that w W m, w E = (V n τ n ). Then µ g (w) = 0, by (iii), and so (65) gives µ(w) µ(g(w)). Thus induction gives (66) on the W m. m=0

18 8 J.K. LANGLEY AND JOHN ROSSI We then define φ to be a quasiconformal homeomorphism of the extended plane fixing 0,,, and with complex dilation µ, and (65) gives µ φ g = µ φ a.e., so that φ g = h φ, with h meromorphic, with finitely many poles. It remains only to prove that h has order at most ρk. But this follows from (63) and the fact that h = φ g φ, using the standard estimate [, 22] φ(z) c z K, φ (z) c 2 z K, z. Next, for 0 < R < S <, α C \ {0} and β [ π/2, π/2], define ψ(z) = ψ(α, β, R, S, z) by (67) ψ(z) = ψ(α, β, R, S, z) = αz ( z R), ψ(z) = αze iβ ( z S), and (68) ψ(z) = ψ(α, β, R, S, z) = αz exp ( ) iβ log z /R log S/R In the domain {z C : R < z < S, 0 < arg z < 2π} write (R z S). ζ = log z = σ + iθ, φ(z) = log ψ(z)/α = σ + i(θ + β(σ log R)(log S/R) ) with σ, θ real. Then 2φ ζ = φ σ + iφ θ = iβ(log S/R), 2φ ζ = φ σ iφ θ = 2 + iβ(log S/R). Thus ψ is quasiconformal in the plane, with (69) µ ψ 2 β 2 4(log S/R) 2 + β 2 Lemma 7.2. Let 0 < c < and let F be meromorphic with finitely many poles in the plane, and with order ρ <. Suppose that (u n ) tends to infinity without repetition, and that all but finitely many fixpoints of F lie in the set {u n : n N}. Suppose further that, for each n N, (70) F (u n ) = u n, F (u n ) = λ n e iθn, 0 λ n c <, θ n R, and that θ n satisfies a.e. (7) min{ θ n, θ n π } δ n π/2. Then (72) lim sup q n ρ n + ρ, q n = In (72) we set q n = 0 whenever λ n = 0. δ n. 4(log /λn ) 2 + δn 2 Proof. Assume that F, u n are as in the statement, but that (72) fails. Then we may assume that there exists κ 0 (0, ) such that, for all n, δ n (73) q n = κ 0, ρk 0 <, K 0 = + κ 0. 4(log /λn ) 2 + δn 2 κ 0 We define a quasimeromorphic function g by modifying F as follows. Let u n be such that λ n 0. Then (70) and Schröder s functional equation [29, p.66] give a neighbourhood B(u n, ρ n ), with ρ n small and positive, and a function φ n defined and conformal on B(0, ρ n ) such that (74) φ n (0) = 0, φ n(0) =, F (z) u n = φ n (λ n e iθn φ n (z u n )),

19 MEROMORPHIC FUNCTIONS OF THE FORM f(z) = P an/(z zn) 9 for z B(u n, ρ n ). Take a small positive r n, with B(0, r n ) φ n (B(0, 2 ρ n)), and define ψ n by ψ n (w) = ψ(λ n e ipn, θ n p n, λ n r n, r n, w), with ψ as in (67) and (68). Here p n is 0 or π and is chosen according to (7) so that θ n p n δ n. Then (69) gives µ ψn κ 0, with κ 0 as in (73). Set U n = {u n + v : φ n (v) B(0, λ n r n )}, V n = {u n + v : φ n (v) B(0, r n )}, so that U n V n B(u n, 2 ρ n), and set Then g is conformal on U n with g(z) u n = φ n (ψ n (φ n (z u n ))), z V n. g(z) u n = φ n (±λ n (φ n (z u n ))), g (u n ) = ±λ n [ c, c]. Also g maps V n into U n. For z not in the union of the V n, we set g(z) = F (z). Thus g is quasimeromorphic and has the same poles and fixpoints as F. By Lemma 7. there exist a quasiconformal homeomorphism φ of the extended plane, fixing 0,,, and a function H meromorphic with finitely many poles in the plane, and with order at most ρk 0 <, with K 0 as in (73), such that g φ H φ. Also φ is conformal on the union W of the U n, and on the interior of C \ m= g m (W ). Obviously z is a fixpoint of g if and only if φ(z) is a fixpoint of H. If u n has λ n 0 then g and φ are conformal on U n and we get H(w n ) = w n, H (w n ) = g (u n ) = ±λ n [ c, c], w n = φ(u n ). Suppose next that u n has λ n = 0. Then u n lies in an open disc Y n disjoint from the closures of the V m, with F (Y n ) Y n, and g = F on Y n. Hence g and φ are analytic on Y n and in this case we get H (w n ) = 0. The function h(z) = z H(z) thus satisfies the hypotheses of Corollary 6., since h(w n ) = 0, h (w n ) [ c, + c]. But h has order less than, and this contradiction proves Lemma Proof of Theorem.5 Assume that f and S are as in the statement, but that f S has finitely many zeros. Define F by = f(z) S(z). z F (z) Then F is transcendental and meromorphic in the plane, with finitely many poles, and with order ρ <. Further, each pole z n of f is, for large n, a fixpoint of F with F (z n ) = b n, b n = = rn e itn. By (6) we have λ 2 n = b n 2 = + r2 n 2r n cos t n r 2 n < M 0 <, Re(b n ) = r n cos t n > 0 and sin t n arg b n = θ n ( π/2, π/2), tan(θ n ) =. r n cos t n Applying Lemma 7.2 gives (72), which contradicts (7).

20 20 J.K. LANGLEY AND JOHN ROSSI 9. Proof of Theorem.6 Assume that f and σ are as in the statement, but that δ(0, f S) σ, for some rational function S. By Lemma 3.5 we have S( ) = 0. The second condition of (8) allows us to assume that b < π/2. Choose b 0, b, b 2 with b < b 0 < b < b 2 < min{π/2, C(µ, σ)}. Lemma 9.. There exists M > 0 such that for all large z lying outside the region arg z < b 0 we have f(z) M. Proof. This follows from (), since there exists M 2 > 0 such that for such z we have z z n M 2 z n for all n. Since δ(0, f S) σ, Baernstein s spread theorem [2] gives a sequence r m and, for each m, a subset I m of the circle z = r m, of angular measure at least min{2π, 2C(µ, σ)} o() 2b 2, such that log f(z) S(z) < T (r m, f) 2, z Im. For large m, we consider the subharmonic function v(z) = log f(z) S(z) on the domain Ω = {z : r m /4 < z < r m, b 0 < arg z < 2π b 0 }, and v is bounded above on Ω, by Lemmas 3.5 and 9.. Since the intersection J m of I m with the arc {z : z = r m, b < arg z < 2π b } has angular measure at least 2(b 2 b ), the two-constants theorem [26] and a standard estimate via conformal mapping for the harmonic measure of J m at r m /2 now give r m (f( r m /2) S( r m /2)) 0, m. Since S( ) = 0, applying the next lemma gives a contradiction. Lemma 9.2. We have lim r R,r + r f( r) =. Proof. If r > 0 then (8) gives arg(r + z n ) arg z n and so there exists M 3 > 0 such that ( ) an Re > M 3 r + z n r + z n. This gives, as r, using (3), r f( r) M 3 r r + z n M 3. 2 z n r z n r 0. Proof of Theorem.7 Assume that F is transcendental and meromorphic with finitely many poles in the plane, of order at most 2, and that all but finitely many fixpoints w of F have F (w) < c <. Set h(z) = z F (z). Then h has infinitely many zeros w, of which all but finitely many have h (w) < c <, c < h (w) < + c. By Lemma 6. we have (57), in which S is a rational function and = /h (z n ), /( + c), sup{ arg : n N} < π/2,

21 MEROMORPHIC FUNCTIONS OF THE FORM f(z) = P an/(z zn) 2 and each z n is a zero of h and so a fixpoint of F. The function f thus satisfies the hypotheses of Theorem.4. But this implies that f S has infinitely many zeros, contradicting (57). This proves Theorem.7.. Proof of Theorem.8 Assume that F, ρ and d are as in the hypotheses. Since ρ < and F is transcendental with finitely many poles, F has infinitely many fixpoints u. If we assume that all but finitely many of these have F (u) d, then applying Lemma 7.2 with δ n = π/2 gives a contradiction. 2. Proof of Theorem.9 We need the following special case of a result of Frank and Weissenborn. Theorem 2. ([2]). Let f be transcendental and meromorphic in the plane, with only simple poles, and let k 2 be an integer. Then (k )N(r, f) < N(r, /f (k) ) + o(t (r, f)) as r outside a set of finite measure. Assume now that F is as in (9). Then we have F = f (k) ( ) k z k, f(z) = (z z n )k!zn k. Applying (4) to f(z)z k, with replaced by ( ) k (k!) zn k, and using (9), we get m(r, f) o() + k log r, T (r, f) N(r, f) + O(log r). Thus (4), (9) and Theorem 2. give T (r, F ) N(r, F ) + S(r, f) (k + + o())n(r, f) outside a set of finite measure. This proves Theorem Proof of Theorem.0 ( ) k + k + o() N(r, /F ) Assume that F is as in the statement of Theorem.0, but with δ(0, F ) =. The assumptions (22) give F (z) = f (z), f(z) =, z z n in which f is meromorphic of finite order, using (4). Then N(r, /f ) = N(r, /F ) = o(t (r, F )) = o(t (r, f)) and so the counting function of the multiple points of f is o(t (r, f)). By a theorem of Eremenko [0], the function f has positive order ρ and sum of deficiencies 2, and at least one deficient value, a say, of f is non-zero. But then the results of [0] show that there exist arbitrarily large r such that f(z) is close to a on a subset of the circle z = r of angular measure close to π/ρ, and this contradicts (4).

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23 MEROMORPHIC FUNCTIONS OF THE FORM f(z) = P an/(z zn) 23 School of Mathematical Sciences, University of Nottingham, NG7 2RD, UK address: jkl@maths.nott.ac.uk Department of Mathematics, Virginia Polytechnic Institute and State University, Blacksburg VA address: rossi@math.vt.edu