On the exceptional set in Nevanlinna's second fundamental theorem in the unit disc

Publicacions Matem`atiques, Vol 40 (1996), 135–156. ON THE EXCEPTIONAL SET IN NEVANLINNA’S SECOND FUNDAMENTAL THEOREM IN THE UNIT DISC Arturo Fern´ andez Arias and Francisco Rodr´ ıguez Mateos Abstract A general example of an analytic function in the unit disc possessing an exceptional set in Nevanlinna’s second fundamental theorem is built. It is used to show that some conditions on the size of the exceptional set are sharp, extending analogous results for meromorphic functions in the plane. 1. Introduction We shall use the standard terminology of Nevanlinna theory (see, for example, [4], [6], [7]). In [3] we prove the following theorems, sharpening a result of R. Nevanlinna [7, p. 247] and extending to meromorphic functions in the unit disc analogous results in the plane [1]. Theorem A. Let Fbe a meromorphic function in the unit disc D= {z||z|<1}and λ≥0a positive real number. Then the error term S(r, F )in Nevanlinna’s second fundamental theorem satisfies (1.1) S(r, F )=O(log+T(r, F )) + Olog 1 1−r as r→1outside a set Eλsuch that (1.2) Eλ dr (1 −r)λ<∞. Theorem B. Let Fbe meromorphic in the unit disc and such that (1.3) log 1 1−r=o(T(r, F )),as r→1. 136 A. Fern´ andez Arias, F. Rodr´ ıguez Mateos Then the error term S(r, F )satisfies (1.4) S(r, F )=o(T(r, F )) as r→1outside a set E, independent of λ, such that (1.5) E1 1−rλ dr < ∞ for every λ>0. Theorem C. Let Fbe a meromorphic function in the unit disc satisfying (1.3). Then the error term S(r, F )satisfies (1.4) outside a set E which can be contained in a sequence of intervals [rn,r n+δn]such that rnis increasing and tends to 1and (1.6) δn<1−rn eΨ(n)2,where Ψ(1) = 1,Ψ(n)=eΨ(n−1). In this paper we show that these results are sharp. To do this, we shall give an example of a meromorphic function in the unit disc possessing a suitable exceptional set. The basic ideas of the example go back to W. K. Hayman [5] and A. Fern´andez [2], although to obtain the desired properties a more complicated construction is needed. 2. Statement of the results Theorem 1. For every function φ(r),0≤r<1, such that (2.1) φ(r)(1 −r)λ→∞,as r→1,for all positive λ, we can construct a meromorphic function Fsatisfying (1.3) such that (2.2) S(r, F ) log T(r, F ) 1−r →∞ and (2.3) S(r, F )>2T(r, F ) in a set E⊂[0,1] satisfying (2.4) E φ(r)dr =∞. This result shows that Theorems A and B are sharp. Next theorem gives a converse result to Theorem C. The exceptional set in Nevanlinna theory in the disc 137 Theorem 2. Let Ψ(n)be defined as in (1.6). Then for every increasing sequence L(n)such that there is no Nin Nfor which (2.5) Ψ(n+N)≥L(n),n∈N, there is a meromorphic function in the unit disc F, satisfying (1.3), for which we cannot find a sequence of disjoint intervals [rn,r n+δn]such that rnis increasing and tends to one and δnsatisfies (2.6) δn<1−rn eL(n)2 and such that S(r, F )=o(T(r, F )) outside the union n[rn,r n+δn]. The proofs of Theorems 1 and 2 rely on the following construction, which gives an example of a meromorphic function in the unit disc possessing an exceptional set in Nevanlinna’s second fundamental theorem. Theorem 3. Suppose that 0<α<1and that {rn}is a sequence of positive numbers satisfying (2.7) rn<1,1−rn+1 <α 2(1 −rn)n=1,2,... Let {λn}be the sequence of integers defined by (2.8) λn=     1n=1 1 1−rnλ2 n−1+1 n=2,3,... and {mn}the sequence of integers (2.9) mn=   1,n=1 √λn(1 −rn−1) 1−rn,n=2,3,... Write (2.10) F(z)= ∞  n=1 mn−1  k=0     1−rn exp 2kπi mn−z     λn . 138 A. Fern´ andez Arias, F. Rodr´ ıguez Mateos Then F(z)is analytic in the unit disc Dand satisfies (1.3). Further, if {sn}is the sequence of positive numbers defined by (2.11) 1 1−sn =1 1−rnλn 2(n−1) λn−11 λn−λn−1 and (2.12) s n=sn+δn,δ n=o1−rn λn, then (2.13) T(r, F ) T(r, F )→∞ and (2.14) T(r, F ) log T(r, F ) 1−r →∞ as r→1in E=n[sn,s  n]. 3. Proof of Theorem 3 We note that, by (2.7), rn→1asn→∞. We consider z∈D,z=re iϑ such that (3.1) α(1 −rN)<1−r<1−rN and note that, from (2.7), this implies 1−rN+1 <α(1 −r). Next we write ϑk,n =2kπ mn . For each n>1 we can find k0such that |eiϑ k0,n −z|≤|eiϑ k,n −z|,k=0,1,... ,m n−1. The exceptional set in Nevanlinna theory in the disc 139 Then, if we write (3.2) αk,n =|ϑ−ϑk,n|if |ϑ−ϑk,n|≤π 2π−|ϑ−ϑk,n|if |ϑ−ϑk,n|≥πk=0,1,... ,m n−1 we have αk0,n ≤π mn αk,n ≥(2 |k−k0|−1) π mn ,if 0 <|k−k0|≤mn−1 2 αk,n ≥(2 (mn−|k−k0|)−1) π mn ,if |k−k0|≥mn+1 2 and, in any case, αk,n ≤π. Now we consider ngiven large and, for the sake of simplicity and without loss of generality, suppose that k0=0. Then we have, for 1 ≤k≤(mn−1)/2, αk,n ≥(2 k−1) π mn≥√kπ mn∼√kπ(1 −rn) √λn(1 −rn−1), so that |eiϑk,n −re iϑ|2=1+r2−2rcos(ϑk,n −ϑ) =1+r2−2rcos αk,n >1+r2−2r1−α2 k,n 2+α4 k,n 4!  =(1−r)2+rα2 k,n 1−α2 k,n 12  (3.3) ≥(1 −r)2+rkπ 2(1 −rn)2 λn(1 −rn−1)21−π2 12 =(1−r)2+crk(1 −rn)2 λn(1 −rn−1)2,c=π21−π2 12. We note that for (mn+1)/2≤k≤mn−1 a similar result holds in terms of k=mn−k, and that for ko= 0 the only change needed in (3.3) is writing |k−k0|instead of k. We have  1−rn eiϑ k,n −z λn =1−rn 1−rλn(1 −r)2 |eiϑk,n −z|2 λn 2 140 A. Fern´ andez Arias, F. Rodr´ ıguez Mateos and from (3.3) (1 −r)2 |eiϑk,n −z|2 λn 2 <    (1 −r)2 (1 −r)2+crk(1 −rn)2 λn(1 −rn−1)2     λn 2 =1+ crk(1 −rn)2 λn(1 −rn−1)2(1 −r)2−λn 2 . Next we show that in the sum mn−1  k=0  1−rn eiϑk,n −z λn = 1−rn eiϑk0,n −z λn + mn−1  k=0 k=k0  1−rn eiϑk,n −z λn for nlarge the main term is that one corresponding to k0. We recall that, from Euler’s summation formula, we have for any real function fpossessing continuous derivative fin the interval [y,x], 0 < y<x,  y<k≤x k∈N f(k)=x y f(t)dt +x y (t−[t])f(t)dt +f(x)([x]−x)−f(y)([y]−y), so that taking in this formula f(t)=1+ crt(1 −rn)2 λn(1 −rn−1)2(1 −r)2−λn 2 y=1,x=mn−1 2 and noting that for this election [x]−x=[y]−y=0 f(t)<0 for t>0 The exceptional set in Nevanlinna theory in the disc 141 we have x  k=2 f(k)≤2λn(1 −r)2(1 −rn−1)2 (λn−2) cr(1 −rn)21+ cr(1 −rn)2 λn(1 −rn−1)2(1 −r)2−λn 2+1 , so that x  k=1 f(k)<1+ cr(1 −rn)2 λn(1 −rn−1)2(1 −r)2−λn 2 1+2λn(1 −r)2(1 −rn−1)2 (λn−2) cr(1 −rn)21+ cr(1 −rn)2 λn(1 −rn−1)2(1 −r)2. From (3.1) we see that N→∞is equivalent to r→1 and then, since (2.7) and (2.8) imply that λn(1 −rn−1)2→∞as n→∞,wehave cr(1 −rN)2 λN(1 −rN−1)2(1 −r)2→0,as r→1. We also have 2λN(1 −r)2(1 −rN−1)2 (λN−2) cr(1 −rN)2→0 and 1+ cr(1 −rN)2 λN(1 −rN−1)2(1 −r)2−λN 2 →0, so that we obtain, recalling that we are supposing that k0=0, (3.4) mN−1  k=1  1−rN eiϑk,N −z λN ∼2mN−1 2  k=1 1−rN 1−rλN (1 −r)2 |eiϑk,N −z|2 λN 2 <21−rN 1−rλNx  k=1 f(k)=o1−rN 1−rλN. We also have  1−rN 1−z λN <1−rN 1−rλN , 142 A. Fern´ andez Arias, F. Rodr´ ıguez Mateos so that (3.5) mN−1  k=0  1−rN eiϑk,N −z λN <1−rN 1−rλN (1 + o(1)). Next we consider rsufficiently close to 1 so that mN−1  k=0  1−rN eiϑk,N −z λN <21−rN 1−rλN . We write F(z)= 1 + 2 + 3 where (3.6)  1 = N−1  n=1 mn−1  k=0 1−rn eiϑk,n −zλn ≤ N−1  n=1 mn−1  k=0 1−rn 1−rλn = N−1  n=1 mn1−rn 1−rλn < N−1  n=1 mn1 1−rλn <(N−1) mN−11 1−rλN−1 ∼(N−1) λN−1(1 −rN−2) (1 −rN−1)1 1−rλN−1 <(N−1) λN−11 1−rλN−1+1 , since, for nlarge, mnis increasing by (2.8) and (2.9), and by (2.7) and (3.1) we have (1 −rN−1)>(1 −r). (3.7)  2 = mN−1  k=0 1−rN eiϑk,N −zλN ≤1−rN 1−rλN (1 + o(1)) ≤21−rN 1−rλN . The exceptional set in Nevanlinna theory in the disc 143 (3.8)  3 = ∞  n=N+1 mn−1  k=0 1−rn eiϑk,n −zλn ≤ ∞  n=N+1 mn1−rn 1−rλn ∼ ∞  n=N+1 √λn(1 −rn−1) (1 −rn)1−rn 1−rλn <1 1−r ∞  n=N+1 λn1−rn 1−rλn−1 <1 1−r ∞  n=N+1 (λn−1) αλn−1 <1 1−r ∞  n=1 nα n=c1 1−r,c 1=α (1 −α)2, since from (2.7) and (3.1) we have 1−rn 1−r<α, n=N+1,N +2,... From (3.6), (3.7) and (3.8) we conclude that F(z) converges in any compact subset of D, so that it is analytic in D. Writing f(z)=F(z)= 1 + 2 + 3  we have, as in (3.6) and (3.8), (3.9)  1  = N−1  n=1 λn mn−1  k=0 1−rn eiϑk,n −zλn1 eiϑk,n −z ≤ N−1  n=1 λnmn1−rn 1−rλn1 1−r <(N−1) λN−1mN−11 1−rλN−11 1−r <(N−1) λ3/2 N−11 1−rλN−1+2 (1 + o(1)). 150 A. Fern´ andez Arias, F. Rodr´ ıguez Mateos and we also obtain (3.22). Finally, for rN<r<s Nwe have T(r, F )≥T(rN,F)≥(1 −rN−2)λN−2log 1 1−rN ≥(1 −rN−2)λN−2log 1 1−r−(1 −rN−2)λN−2log 1−rN 1−sN , but (1 −rN−2)λN−2log 1−rN 1−sN =(1 −rN−2)λN−2 λN−λN−1λN−1log 1 1−rN + log (2(N−1)λN−1), so that this term tends to zero as Ntends to infinity. Thus we conclude that for any rclose enough to 1 we have T(r, F )≥(1 −rN−2)λN−2log 1 1−r(1 + o(1)), and therefore lim r→1 T(r, F ) log 1 1−r ≥lim r→1(1 −rN−2)λN−2=∞. This completes the proof of Theorem 3. 4. Proof of Theorem 1 Since to prove Theorem 1 we shall make use of Theorem 3, we first show that the set Eis in fact exceptional for the function F(z). We make use of the following two inequalities (4.1) S(r, F )≥m(r, F F) (4.2) m(r, F )≤m(r, F F)+m(r, F ), so that by (2.13) we deduce (4.3) S(r, F )≥m(r, F )−m(r, F ) =T(r, F )−T(r, F )=T(r, F )(1 + o(1)) The exceptional set in Nevanlinna theory in the disc 151 for r∈[sn,s n+δn] and nlarge, and we deduce from (2.14) and (4.3) S(r, F ) log T(r, F ) 1−r ≥T(r, F )(1 + o(1)) log T(r, F ) 1−r →∞ as n→∞, which is (2.2). From (2.13) we also have, for r∈[sn,s n+δn] and nlarge, m(r, F )=T(r, F )≥3T(r, F )=3m(r, F ) and, from (4.1) and (4.2), we conclude S(r, F )≥m(r, F F)≥2m(r, F )=2T(r, F ) which is (2.3). We can assume without loss of generality that φ(r) is increasing for r close to 1. In fact, by (2.1), for every positive integer Nthere exists tN such that φ(r)≥1 1−rN ,r≥tN, where we can clearly assume that 1 −tN+1 ≤α(1 −tN), for a certain 0<α<1. Then, if we define φ1(r)=1 1−rN ,t N≤r<t N+1, for r≥t1,φ1(r) is increasing and satisfies (2.1) and also φ1(r)≤φ(r). Hence it is enough to prove (2.4) with φ1(r) instead of φ(r). We have shown that the function F(z) defined in (2.10) satisfies (2.2) and (2.3) as rtends to 1 through the sequence of intervals [sn,s n+δn] defined by (2.11) and (2.12). Thus it is enough to prove (2.4) for the set E=n[sn,s n+δn] and for this it is enough to show that (4.4)  n φ(sn)δn=∞, as this implies (2.4) since φis increasing. Next we show that we can choose {rn}and {δn}in Theorem 3 so that (4.5) φ(sn)δn≥1 152 A. Fern´ andez Arias, F. Rodr´ ıguez Mateos and so (4.4) holds. We can take {λn}as in (2.8) and (4.6) δn=(1 −rn)2 λn . Then, for nlarge, (4.7) φ(sn)δn∼φ(sn)(1 −rn)2 1 1−rnλ2 n−1 >φ(sn) 1 1−rn2λ2 n−1 . Let us assume that r1,... ,r n−1,λ 1,... ,λ n−1and therefore δ1,... ,δ n−1have already been defined. Then we define rnsuch that (2.7) is satisfied and such that (4.8) φ(r) 1 1−r2λ2 n−1≥1,r≥rn, which is possible by (2.1). Once rnhas been defined, we obtain λn,snand δnby (2.8), (2.11) and (4.6). So we have a well-defined function F(z), analytic in the unit disc, satisfying (2.2) and (2.3) in the set E=n[rn,r n+δn]. Since (4.5) follows from (4.7) and (4.8) and (4.5) implies (2.4), the proof of Theorem 1 is complete. 5. A preliminary result to Theorem 2 The following result will be used in the proof of Theorem 2. Theorem 4. We define the sequence {rn}by (5.1) 1 1−rn =φ(n)=Ψ(n)1/3, where Ψ(1) = 1,Ψ(n) = exp(Ψ(n−1)) as in (1.6). Then the function defined by (2.10) satisfies S(r, F )≥2T(r, F ) for rin a sequence of intervals (tn,t n+βn)such that (5.2) βn≥1 Ψ(n)for large n. The exceptional set in Nevanlinna theory in the disc 153 Proof: The sequence rngiven by (5.1) satisfies (2.7) for a certain α; hence F(z) is a well-defined function. In the proof of Theorem 1 we have shown that S(r, F )≥2T(r, F ) for rsufficiently close to 1 in a sequence of intervals [sn,s n+δn], where sn and δnare given by (2.11) and (2.12). We choose in particular δn=(1 −rn)2 λn . It remains to verify (5.2). We shall show by induction that there is N∈Nsuch that (5.3) λn (1 −rn)2≤φ(n+N)≤Ψ(n+N) for every n∈N. We assume nso large that φ(n) is much bigger than n and log φ(n); for such an nwe find Nso large that (5.3) holds. Then we prove that this inequality is true for all the following terms. In fact by (2.8) we have λn+1 (1 −rn+1)2=1 1−rn+1 λ2 n+1 1 (1 −rn+1)2 ≤1 1−rn+1 2λ2 n = exp 2λ2 nlog 1 1−rn+1  ≤exp 1 3φ(n+N)3=φ(n+N+1) ≤Ψ(n+N+1), since by our hypothesis on n 2λ2 nlog 1 1−rn+1 =2λ2 nlog φ(n+1) ≤1 3φ(n+N)3. This proves the inductive step, so that (5.3) holds for all large n.By increasing Nif necessary, we ensure that (5.3) is true for all n.Thuswe obtain δn=(1 −rn)2 λn≥1 Ψ(n+N) for some positive integer Nand all n. We conclude that (5.2) holds, writing tn=sn−N,β n=δn−N, n>N. 154 A. Fern´ andez Arias, F. Rodr´ ıguez Mateos 6. Proof of Theorem 2 For the function Fof Theorem 4 there is a sequence of intervals (tn,t n+βn) such that for large nwe have (6.1) βn≥1 Ψ(n) and S(r, F )≥2T(r, F ) for r>r 0in n(tn,t n+βn). If Theorem 2 were false, we could find a sequence of disjoint intervals [rn,r n+δn] satisfying (6.2) δn<1−rn eL(n)2 and such that S(r, F )=o(T(r, F )) outside the union n[rn,r n+δn]. Then we should have for a certain N1∈N % n>N1 (tn,t n+βn)⊂% k [rk,r k+δk], and, since we are considering sequences of disjoint intervals, we should conclude that (6.3) (tn,t n+βn)⊂[rkn,r kn+δkn]n>N 1. By the way we constructed the intervals (tn,t n+βn) we can assume (6.4) 1 −tn+1 <α(1 −tn) for a certain 0 <α<1. We note that since tn→1asntends to infinity we also have rkn→1, so that knalso tends to infinity as n→∞. From (6.2), (6.3) and (6.4) we have tn+1 −tn>(1 −α)(1−tn)>(1 −α)(1−rkn−δkn) >(1 −α)&eL(kn)2−1'δkn. Since kn→∞as n→∞and, from (2.5), L(kn) also tends to infinity, we can find N2∈Nsuch that (1 −α)&eL(kn)2−1'>1,n>N 2, The exceptional set in Nevanlinna theory in the disc 155 so that tn+1 −tn>δ kn,n>N= max(N1,N 2). Therefore, for n>N, each interval [rkn,r kn+δkn] cannot meet more than one of the intervals (tn,t n+βn). Hence the sequence knis strictly increasing for n>N. Now using (6.1), (6.2) and (6.3) we must have that, for nlarge, 1 Ψ(n)≤βn≤δkn≤1−rkn eL(kn)2≤1 L(kn)≤1 L(n−N) since kn≥n−N. Hence there is n0∈Nsuch that Ψ(n)≥L(n−N),n>max(n0,N), that is Ψ(n+N)≥L(n), n>n 0. Then we should have Ψ(n+N+n0)≥L(n+n0)≥L(n),n∈N, which contradicts (2.5). This completes the proof of Theorem 2. References 1. A. Fern´ andez Arias, Some results about the size of the exceptional set in Nevanlinna’s second fundamental theorem, Collect. Math. 37 (1986), 229–238. 2. A. Fern´ andez Arias, On the size of the exceptional set in Nevanlinna theory, J. London Math. Soc. 34(2) (1986), 449–456. 3. A. Fern´ andez Arias and F. Rodr´ ıguez Mateos, On the size of the exceptional set in Nevanlinna’s second fundamental theorem for meromorphic functions in the unit disc, Rev. R. Acad. Ci. Madrid (to appear). 4. W. K. Hayman,“Meromorphic functions,” Clarendon Press, Oxford, 1964. 5. W. K. Hayman, Die Nevanlinna-Characteristic von meromorphen Funktionen und ihrer Integralen, in “Festband zum 70: Geburtstag von Rolf Nevanlinna,” Springer, Berlin, 1966, pp. 16–20. 6. R. Nevanlinna,“Le th´eor`eme de Picard-Borel et la th´eorie des fonctions m´eromorphes,” Gauthier-Villards, Paris, 1920. 156 A. Fern´ andez Arias, F. Rodr´ ıguez Mateos 7. R. Nevanlinna,“Analytic functions,” Springer-Verlag, Berlin, 1970. Keywords. Characteristic function, distribution, growth, exceptional set. 1991 Mathematics subject classifications: 30D35. Departamento de Matem´aticas Fundamentales Facultad de Ciencias Universidad Nacional de Educaci´on a Distancia Senda del Rey s/n 28040 Madrid SPAIN e-mail: [email protected] Primera versi´o rebuda el 30 de Maig de 1995, darrera versi´o rebuda el 30 d’Octubre de 1995