Asymptotic functions of entire functions
aa r X i v : . [ m a t h . C V ] F e b ASYMPTOTIC FUNCTIONS OF ENTIRE FUNCTIONS
AIMO HINKKANEN , JOSEPH MILES , JOHN ROSSI Dedicated to the memory of Walter K. Hayman, FRS
Abstract. If f is an entire function and a is a complex number, a is said to be an asymptotic value of f if there exists a path γ from 0 to infinity such that f ( z ) − a tends to 0 as z tends to infinityalong γ . The Denjoy–Carleman–Ahlfors Theorem asserts that if f has n distinct asymptotic values, then the rate of growth of f is atleast order n/
2, mean type. A long-standing problem asks whetherthis conclusion holds for entire functions having n distinct asymp-totic (entire) functions, each of growth at most order 1 /
2, minimaltype. In this paper conditions on the function f and associated as-ymptotic paths are obtained that are sufficient to guarantee that f satisfies the conclusion of the Denjoy–Carleman–Ahlfors Theorem.In addition, for each positive integer n , an example is given of anentire function of order n having n distinct, prescribed asymptoticfunctions, each of order less than 1 / Introduction
Suppose that f ( z ) is an entire function in the complex plane C . Anentire function a ( z ) is said to be an asymptotic function for f if thereexists a path γ in C from 0 to infinity such that f ( z ) − a ( z ) tends to0 as z tends to infinity along γ ; we then say that f is asymptotic to a on γ . In the case that a ( z ) is a constant function, it is said to be anasymptotic value of f . The Denjoy–Carleman–Ahlfors Theorem ([2];in a very general form for subharmonic functions this can be found in[17], Theorem 8.9, p. 562) states that if f has n distinct asymptoticvalues then the rate of growth of f is at least order n/
2, mean type.This bound is known to be sharp ([1], p. 1). For consider f ( z ) = R z ( w − n/ sin w n/ ) dw . Then f has order n/
2, and if 0 ≤ ν ≤ n − z = re νiπ/n , where r >
0, we have f ( z ) → e νiπ/n Z ∞ x − n/ sin( x n/ ) dx as r = | z | → ∞ . Mathematics Subject Classification.
Primary 30D20; Secondary 31A05.
It is an open question [16], Problem 2.3, whether the analogue ofthe Denjoy–Carleman–Ahlfors Theorem holds for asymptotic functions a ( z ) with growth at most order 1 /
2, minimal type. For such functionsit is known that the minimum modulus is unbounded on a sequenceof circles with radii tending to infinity. Absent this minimum mod-ulus condition on the asymptotic functions a ( z ), there are easy ex-amples of entire functions of finite order with infinitely many asymp-totic functions. Such an example is f ( z ) = sin √ z √ z or f ( z ) = e − z , and a j ( z ) = jf ( z ) for each positive integer j . Note that both f ( z ) and a j ( z ) tend to 0 as z → ∞ along the positive real axis.Throughout this paper we let B (0 , r ) denote the disk { z ∈ C : | z |
Theorem A. If f is entire with n distinct asymptotic functions oforder less than / , then the growth of f is at least order n/ , meantype. In [17], Theorem 8.13, p. 577, Hayman showed that Fenton’s methodscan be used to obtain this result for asymptotic functions with growthno faster than order 1 /
4, minimal type. For an earlier result of Somorjaiwith asymptotic functions of order < /
30, see [21].Stronger results are known if the asymptotic paths are rays. In thiscase Denjoy ([4], [5]) showed that if f has order µ then the numberof distinct asymptotic functions with order less than 1 / (2 + µ − ) is atmost 2 µ .Dudley Ward and Fenton [6] obtained the following result. SYMPTOTIC FUNCTIONS 3
Theorem B.
Suppose that f is analytic in the sector D = { z ∈ C : | arg z | < η } for some η ∈ (0 , π ) and is continuous on ∂D . Supposethat a ( z ) and b ( z ) are entire, each with order less than / (2 + 2 ηπ − ) ,are not identically zero and satisfy (1) f ( te iη ) − a ( te iη ) → as t → + ∞ and (2) f ( te − iη ) − b ( te − iη ) → as t → + ∞ . Then lim inf r →∞ log M ( r, D, f ) r π/ (2 η ) > . Elementary arguments lead from Theorem B to the following corol-lary.
Corollary A.
Suppose that f is entire and that γ , γ , . . . , γ n aredistinct rays emanating from . Suppose for ≤ j ≤ n that a j ( z ) aredistinct entire functions of order less than / (2 + 2 n − ) and that f ( z ) is asymptotic to a j ( z ) on γ j . Then lim inf r →∞ log M ( r, f ) r n/ > . Hinkkanen and Rossi [18] obtained the conclusion of Theorem Bunder the relaxed assumption that the open set D is bounded by paths γ and γ , not necessarily disjoint away from 0, with (1) holding on γ and (2) holding on γ and that the angular measure m ( D ∩ S (0 , r )) of D ∩ S (0 , r ) satisfies m ( D ∩ S (0 , r )) ≤ η for all large r .For further results in this direction, see [8], [9], [10], [11], and [12].The principal result of this paper is the following theorem. Theorem 1.1.
Suppose that γ , γ , . . . , γ n are n simple segmental pathsfrom to infinity, disjoint except at the origin, arranged in coun-terclockwise fashion. Let D j be the Jordan domain between γ j and γ j +1 , where γ n +1 = γ . Suppose that f is entire and that a j ( z ) , for ≤ j ≤ n , are distinct entire functions with order ρ ( a j ) < / suchthat f is asymptotic to a j on γ j . Suppose that there exists a number κ > ρ := max { ρ ( a j ) : 1 ≤ j ≤ n } such that (3) lim sup z →∞ z ∈ Dj log | f ( z ) || z | κ > AIMO HINKKANEN, JOSEPH MILES, JOHN ROSSI for each j . Then (4) lim inf r →∞ log M ( r, f ) r n/ > . We note in particular that if (3) holds with κ = 1 /
2, then (4) holds.From Theorem 1.1 we obtain the following corollary.
Corollary 1.2.
Suppose for ≤ j ≤ n that a j ( z ) are distinct entirefunctions with order ρ ( a j ) < / . For ≤ j ≤ n , let γ j be the ray with arg z = 2 πj/n . Suppose that f is entire and that f is asymptotic to a j on γ j . Then lim inf r →∞ log M ( r, f ) r n/ > . In comparing Corollary 1.2 with Corollary A, we note that Corol-lary 1.2 treats asymptotic functions a ( z ) of all orders less than 1 /
2, butrequires that the rays be equally spaced.We also show that Fenton’s result in the special case that the a j ( z )are polynomials follows quite easily from Theorem 1.1.Finally, in Section 3 we give an example of an entire function of order n having n distinct, prescribed asymptotic functions of order < / f of order n with 2 n such prescribed asymptotic functions.2. Proofs
A lemma.
Fundamental to our approach is the following lemma.
Lemma 2.1.
Suppose that γ and γ are simple segmental paths from to infinity, disjoint except for the origin. Let D be a Jordan domainwith ∂D = γ ∪ γ . For t > , let Φ( t ) be the angular measure of D ∩ S (0 , t ) . Suppose that a and a are distinct entire functions withorders ρ ( a ) and ρ ( a ) satisfying ρ = max { ρ ( a ) , ρ ( a ) } < / , suchthat f is asymptotic to a j on γ j for j = 1 , . Suppose that there existsa number κ > ρ such that lim sup z →∞ z ∈ D log | f ( z ) || z | κ > . Then there exists R > such that for all R > R , we have log M ( R, D, f ) ≥ π (cid:26) π Z RR dtt Φ( t ) (cid:27) ≥ π (cid:18) RR (cid:19) / . SYMPTOTIC FUNCTIONS 5
Proof.
Without loss of generality, we may assume that κ < / κ and κ such that ρ < κ < κ < κ . There exists anumber C ≥ z ∈ ∂D we havemax { log | a ( z ) | , log | a ( z ) |} < C + | z | κ . There exists M ′ ≥ j = 1 , | f ( z ) − a j ( z ) | ≤ M ′ for all z ∈ γ j and hencelog | f ( z ) | ≤ log + | a j ( z ) | + M ′ + log 2 < C + | z | κ + M ′ + log 2 . With M = C + M ′ + log 2, we have(5) log | f ( z ) | < M + | z | κ for all z ∈ ∂D .Set A = 20 (cid:0) − κ (cid:1) − κ > . We choose R so large that for all r ≥ R we have1 + M + A r κ < r κ . The following lemma is from [17], Lemma 8.13, p. 583.
Lemma 2.2.
Let φ be a non-negative continuous convex function of log t for ≤ t < ∞ with φ (0) = 0 , and suppose that for some δ > ,we have Z ∞ δ φ ( t ) t / dt < ∞ . Let D be a domain in the plane such that every boundary point of D is regular for Dirichlet’s problem and such that for all r ∈ (0 , ∞ ) , thecircle S (0 , r ) intersects the complement of D . Then there is a function u , continuous and non-negative in D and harmonic in D , such that u ( z ) = φ ( | z | ) for all z ∈ ∂D and such that for all z ∈ D , we have (6) φ ( | z | ) ≤ u ( z ) ≤ | z | / Z ∞ | z | φ ( t ) t / dt. Hayman used an estimate for harmonic measure that is not as strongas is known. Hence the number 20 on the right hand side of (6) can bereduced but we do not try to do this as it is not important for us.We use φ ( t ) = t κ for our earlier choice of κ and calculate20 | z | / Z ∞ | z | φ ( t ) t / dt = A | z | κ . AIMO HINKKANEN, JOSEPH MILES, JOHN ROSSI
We apply Lemma 2.2 to obtain a harmonic function u on D such thatfor all z ∈ D we have | z | κ ≤ u ( z ) ≤ A | z | κ . By our assumptions, there exists z ∈ D with | z | > R such that(7) log | f ( z ) | > | z | κ > A | z | κ + M + 1 . We set R = | z | .Suppose that R > R . Let D ( R ) be the component of D ∩ B (0 , R )containing z . We note that ∂D ( R ) lies in ∂D except for a non-emptyunion of open arcs in S (0 , R ). Let U = U R be the harmonic functionin D ( R ) that satisfies U ( z ) = 0 if z ∈ ∂D ( R ) lies in the interior of anarc in S (0 , R ) lying in ∂D ( R ), while U ( z ) = | z | κ if z ∈ ∂D ( R ) and | z | < R (and thus z ∈ ∂D ). Let u be the harmonic function obtainedabove from Lemma 2.2 with the choice φ ( t ) = t κ . Evidently U − u is harmonic on D ( R ) and U ( z ) − u ( z ) ≤ z ∈ ∂D ( R ). Thus U ( z ) − u ( z ) ≤ z ∈ D ( R ).Let M ′ ( R, D, f ) = max {| f ( z ) | : | z | = R, z ∈ ∂D ( R ) } ≤ M ( R, D, f ) . Write H ( R ) = S (0 , R ) ∩ ∂D ( R ). Let ω ( R, z ) denote the harmonicmeasure of H ( R ) at z ∈ D ( R ). The function w ( z ) = U ( z ) + M + ω ( R, z ) log M ′ ( R, D, f )is harmonic on D ( R ) and log | f ( z ) | ≤ w ( z )for all z ∈ ∂D ( R ) by (5). Then for all z ∈ D ( R )(8) log | f ( z ) | ≤ w ( z ) ≤ u ( z ) + M + ω ( R, z ) log M ( R, D, f ) . Setting z = z we obtain from (7) A | z | κ + M +1 ≤ log | f ( z ) | ≤ A | z | κ + M + ω ( R, z ) log M ( R, D, f ) , implying that 1 ≤ ω ( R, z ) log M ( R, D, f ) . If we now let Φ ∗ ( t ) be the angular measure of D ( R ) ∩ S (0 , t ), wehave Φ ∗ ( t ) ≤ Φ( t ), and by [14], Theorem 6.2, (6.4), p. 149 and p. 158,we get(9) ω ( R, z ) ≤ π exp (cid:26) − π Z RR dtt Φ ∗ ( t ) (cid:27) ≤ π exp (cid:26) − π Z RR dtt Φ( t ) (cid:27) . SYMPTOTIC FUNCTIONS 7
We rearrange to obtainlog M ( R, D, f ) ≥ π (cid:26) π Z RR dtt Φ( t ) (cid:27) . The second inequality in our conclusion follows from the fact thatΦ( t ) ≤ π .2.2. Proof of Theorem 1.1.
We now turn to the proof of Theo-rem 1.1. For each j , let Φ j ( t ) be the angular measure of D j ∩ S (0 , t ).We apply Lemma 2.1 on each D j to conclude that there exists R ( j ) > R > R ( j ),log M ( R, D j , f ) ≥ π (cid:26) π Z RR ( j ) dtt Φ j ( t ) (cid:27) . Let R = max { R ( j ) : 1 ≤ j ≤ n } . Then for all R > R we havelog M ( R, f ) ≥ π (cid:26) π Z RR ( j ) dtt Φ j ( t ) (cid:27) for each j with 1 ≤ j ≤ n .We have n ≤ n X j =1 j ( t ) ! n X j =1 Φ j ( t ) ! ≤ π n X j =1 j ( t ) , implying that n Z RR dtt ≤ π n X j =1 Z RR dtt Φ j ( t ) . Thus for all
R > R , there exists j = j ( R ) such that n Z RR dtt ≤ π Z RR dtt Φ j ( t ) . So for
R > R , with j = j ( R ) we havelog M ( R, f ) ≥ π (cid:26) π Z RR ( j ) dtt Φ j ( t ) (cid:27) ≥ π (cid:18) RR (cid:19) n/ , establishing (4). AIMO HINKKANEN, JOSEPH MILES, JOHN ROSSI
Proof of Corollary 1.2.
We now prove Corollary 1.2. Let D j be the sector between γ j and γ j +1 (with γ n +1 = γ ). From elementaryconsiderations we have lim sup r →∞ log M ( r, f ) r / > . Trivially for each R there exists j = j ( R ) such that log M ( R, f ) =log M ( R, D j , f ). Thus there exists j such thatlim sup z →∞ z ∈ Dj log | f ( z ) || z | / > . We may thus apply Lemma 2.1 on this D j to conclude that there exists R = R ( j ) > R > R log M ( R, f ) ≥ log M ( R, D j , f ) ≥ π (cid:26) π Z RR ( j ) dt πtn − (cid:27) = π (cid:18) RR (cid:19) n/ . (cid:3) Remark.
Our proofs in fact yield somewhat stronger versions ofTheorem 1.1 and Corollary 1.2. The assumption that all of the asymp-totic functions are distinct can be weakened. Both Theorem 1.1 andCorollary 1.2 depend on applications of Lemma 2.1, each applicationrequiring only that the asymptotic functions associated with adjacentpaths are distinct. Thus both results hold even if f has fewer than n distinct asymptotic functions on the n disjoint paths, provided onlythat asymptotic functions associated with adjacent paths are distinct.Similar considerations apply to the Denjoy–Carleman–Ahlfors The-orem and Fenton’s theorem, as an examination of the arguments in[2] and [7] shows. In fact, more can be said in the case of asymptoticvalues. A path to infinity on which f tends to a constant a , finite orinfinite, corresponds to a singularity of f − lying over a . More thanone singularity can lie over a given a . Certain singularities are classi-fied as direct singularities. If f is meromorphic of order ρ , Ahlfors [3]showed that f − has no more than max { ρ, } direct singularities. If f is entire, all singularities lying over infinity are direct. From this itfollows that if f is entire of order ρ , f has no more than 2 ρ singular-ities lying over finite points, implying the Denjoy–Carleman–AhlforsTheorem. For details, including a necessary modification of Ahlfors’sarguments, see [19], pp. 309–313 or [20], pp. 303–307.2.4. Fenton’s theorem for polynomials a j . We next turn to aproof, based on Theorem 1.1, of Fenton’s theorem in the case that
SYMPTOTIC FUNCTIONS 9 all a j are polynomials. We treat the situation where not all a j areassumed to be distinct.Suppose that f is entire and that F = { γ , γ , . . . , γ m } is a collectionof paths from 0 to infinity, each with an associated polynomial a j suchthat f is asymptotic to a j on γ j . Suppose that two of these paths, say γ j and γ k , intersect on a sequence of points tending to infinity. Then a j − a k tends to 0 on this sequence, implying that a j ≡ a k ; in thiscircumstance, we delete one of γ j and γ k , obtaining a subcollection of F . If two paths in this subcollection intersect on a sequence tendingto infinity, we again delete one of them. We continue until we arriveat a subcollection F ′ with the property that there is a disk D centeredat the origin containing all points of intersection of any two membersof F ′ . After renumbering, we write F ′ = { γ , γ , . . . , γ q } . After an ob-vious modification of the paths on D , we may suppose that the pathsin F ′ are simple, segmental, disjoint except at the origin, and num-bered in counterclockwise order. If any two adjacent paths in F ′ , say γ j and γ j +1 (with γ q +1 = γ ), are associated with the same asymptoticpolynomial, we delete one of these paths. We continue this processuntil we obtain a subcollection F ′′ = { γ , γ , . . . , γ n } of disjoint sim-ple segmental paths numbered in counterclockwise order such that thepolynomials associated with any two adjacent paths are distinct.Every collection F has such a subcollection F ′′ . Our task is to showthat f has growth at least order n/
2, mean type, where n is the numberof paths in such a subcollection F ′′ . We note that if all the asymptoticpolynomials associated with the paths in F are distinct, then F = F ′′ ,after obvious modifications of the paths on a disk D . Similarly, F = F ′′ (after obvious modifications) if all the paths in F are disjoint outsidesome disk and the asymptotic polynomials associated with adjacentpaths are distinct.We adopt the notation of Theorem 1.1. We consider such a collection F ′′ and for 1 ≤ j ≤ n write a j ( z ) = d j X p =0 b pj z p where d j is the degree of a j .Fix j and assume, without loss of generality, that d j ≥ d j +1 . Replac-ing f by f − a j +1 and a j by a j − a j +1 without changing notation, wemay assume a j +1 ≡ a j
0. Our goal is to show that in D j , themodified function log | f ( z ) | grows at least like | z | / , clearly implyingthat our original log | f ( z ) | does as well. If a j is constant, it follows from familiar arguments in the case ofasymptotic values that f is unbounded in D j . On ∂D j , we havelog | f ( z ) | ≤ M for some M >
1. For every z ∈ ∂D j with | z | < R ,we have (8) with u ( z ) = 0. We note that ω ( R, z ) = O (cid:18) | z | R (cid:19) / from (9). Letting M j ( R ) = M ( R, D j , f ), we find from (8) that if thereis a sequence of R → ∞ on which log M j ( R ) = o ( R / ), then f isbounded in D j , which is a contradiction. Thus we must have(10) lim inf R →∞ log M j ( R ) R / > . Suppose now that d j ≥
1. Let α be a value that f takes infinitelyoften, say (at least) at distinct values c p for 1 ≤ p ≤ d j . Set P ( z ) = Q d j p =1 ( z − c p ) and g ( z ) = f ( z ) − αP ( z ) . Then g is entire, and g ( z ) → z → ∞ along γ j +1 (since f ( z ) → z → ∞ along γ j , we have f ( z ) − a j ( z ) →
0, so that g ( z ) = o (1) + a j ( z ) − αP ( z ) = o (1) + a j ( z ) P ( z ) = o (1) + b d j j since deg a j = deg P = d j . Recall that b d j j = 0. Applying the previousarguments to g instead of f , we see that (10) holds if M j ( R ) refers to g , and hence also holds if M j ( R ) refers to f . The above works in alldomains D j , so that we may apply Theorem 1.1 to conclude thatlim inf R →∞ log M ( R, f ) R n/ > . An example
Theorem 3.1.
Let n be a positive integer. For ≤ j ≤ n , supposethat a j is an entire function of growth no faster than order n , minimaltype. There exists an entire function f of order n , mean type, such thateach a j , for ≤ j ≤ n , is an asymptotic function of f . Our proof is an adaptation of a technique that Fuchs and Hayman([13], see also [15], pp. 80–83) used to assign deficiencies of an entirefunction arbitrarily subject only to the condition that the sum of thedeficiencies does not exceed 1. We prove the following lemma.
SYMPTOTIC FUNCTIONS 11
Lemma 3.2.
Let n be a positive integer. Let c n = Z ∞ e − t n dt. Let the path Γ be the boundary of a sector of opening πn given by Γ( t ) = (cid:26) − te − iπ/n , −∞ < t < ,te iπ/n , ≤ t < ∞ . Let Ω = { re iθ : r > and | θ | < π/n } be the inside of Γ and Ω = { re iθ : r > and π/n < θ < π − π/n } be the outside. Then thereexists an entire function ϕ such that (11) ϕ ( z ) = (cid:26) e z n − (cid:0) c n π sin πn (cid:1) z − + O ( | z | − ) , z ∈ Ω , − (cid:0) c n π sin πn (cid:1) z − + O ( | z | − ) , z ∈ Ω . uniformly as z → ∞ . Proof of Lemma 3.2.
For z ∈ C \ Γ, define E ( z ) = 12 πi Z Γ e w n w − z dw. We note that if w ∈ Γ, then w n = −| w | n and thus E is analytic for z ∈ C \ Γ. Let I = 12 πi Z Γ e w n dw = 12 πi Z ∞ e − t n (cid:0) e iπ/n − e − iπ/n (cid:1) dt = 1 π c n sin πn . Note that 1 w − z + 1 z = wz ( w − z ) . Thus(12) L := 12 πi Z Γ we w n z ( w − z ) dw = E ( z ) + (cid:16) c n π sin πn (cid:17) z . Consider z / ∈ Γ with | z | = R for large R >
0. Let Γ be the portionof Γ with modulus at most R/
2, Γ the portion of Γ with modulusbetween R/ R , and Γ the portion of Γ with modulus at least2 R . Let d n = Z ∞ te − t n dt. We have(13) (cid:12)(cid:12)(cid:12)(cid:12) πi Z Γ wz ( w − z ) e w n dw (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) πi Z Γ wz ( w − z ) e w n dw (cid:12)(cid:12)(cid:12)(cid:12) < (cid:18) πR + 1 πR (cid:19) d n . If the distance from z to Γ is at least 1, we set Γ ∗ = Γ . If thedistance from z to Γ is less than 1, we indent away from z the segment of Γ with distance less than 1 from z to the arc of circle with distanceat least 1 from z to obtain a modified Γ ∗ .By Cauchy’s Theorem we have Z Γ we w n z ( w − z ) dw = Z Γ ∗ we w n z ( w − z ) dw. We conclude that(14) (cid:12)(cid:12)(cid:12)(cid:12) πi Z Γ we w n z ( w − z ) dw (cid:12)(cid:12)(cid:12)(cid:12) = O (cid:0) Re − ( R/ n (cid:1) = O (cid:18) R (cid:19) . Combining (12), (13), and (14), we obtain(15) (cid:12)(cid:12)(cid:12)(cid:12) E ( z ) + (cid:16) c n π sin πn (cid:17) z (cid:12)(cid:12)(cid:12)(cid:12) = O (cid:18) R (cid:19) for z / ∈ Γ.Let E ( z ) be E ( z ) for z ∈ Ω and let E ( z ) be E ( z ) in Ω . Byreplacing a short segment of Γ by a circular arc lying in Ω , we seefrom Cauchy’s Theorem that E can be continued analytically acrossΓ onto Ω and, by the Cauchy Integral Formula, that E is entire and(16) E ( z ) = E ( z ) + e z n for all z ∈ Ω . We set ϕ ( z ) = E ( z ) and conclude from (15) and (16)that (11) holds. This proves Lemma 3.2.We now turn to the proof of Theorem 3.1. Letting ϕ be as inLemma 3.2 for 1 ≤ j ≤ n we set ϕ j ( z ) = ϕ (cid:0) e − πij/n z (cid:1) . Define(17) f ( z ) = n X j =1 ϕ j ( z ) a j ( z ) e z n . Clearly f is entire of order n mean type.Let γ j be the ray { re πij/n : r > } . Suppose that z ∈ γ j . Then e − πij /n z > . Thus ϕ j ( z ) = ϕ (cid:0) e − πij /n z (cid:1) = e z n + O ( | z | − ) = e | z | n + O ( | z | − ) . Hence ϕ j ( z ) a j ( z ) e z n − a j ( z ) = (cid:18) e z n + O ( | z | − ) e z n − (cid:19) a j ( z ) = O ( | z | − ) a j ( z ) e z n , implying that(18) lim z →∞ z ∈ γj (cid:18) ϕ j ( z ) a j ( z ) e z n − a j ( z ) (cid:19) = 0 . SYMPTOTIC FUNCTIONS 13
We now consider z ∈ γ j and j = j . We note that e − πij/n z ∈ Ω . By(11), | ϕ j ( z ) | = (cid:12)(cid:12) ϕ (cid:0) e − πij/n z (cid:1)(cid:12)(cid:12) = O ( | z | − ) . Consequently, for z ∈ γ j ,(19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X j =1 j = j ϕ j ( z ) a j ( z ) e z n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ n X j =1 j = j | a j ( z ) | O ( | z | − ) e | z | n = o (1) . The combination of (17), (18), and (19) completes the proof of The-orem 3.1.
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