IJBRARY Michigan State University This is to certify that the dissertation entitled A STUDY OF VALUE DISTRIBUTION, ZERO SETS, INEQUALITIES AND COEFFICIENT CONDITIONS FOR ANALYTIC FUNCTIONS WITH SLOW GROWTH AND LACUNARY SERIES WITH HADAMARD GAPS presented by PETER TIEN-YU CHERN has been accepted towards fulfillment of the requirements for PH.D. degmin MATHEMATICS flfié‘a Major professor Date flirt/4r} /S-/ /l7§f ' / MSU i: an Affirmative Action/Equal Opportunity Institution 0- 12771 ———— ~ __ MSU LIBRARIES .—;—. RETURNING MATERIALS: Place in book drop to remove this checkout from your record. FINES will be charged if book is returned after the date stamped below. A STUDY OF VALUE DISTRIBUTION, ZERO SETS, INEQUALITIES, AND COEFFICIENT CONDITIONS FOR ANALYTIC FUNCTIONS WITH SLOW GROWTH AND LACUNARY SERIES WITH HADAMARD GAPS By Peter Tien-Yu Chem A THESIS Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1988 ABSTRACT A STUDY OF VALUE DISTRIBUTION, ZERO SETS, INEQUALITIES, AND COEFFICIENT CONDITIONS FOR ANALYTIC FUNCTIONS WITH SLOW GROWTH AND LACUNARY SERIES WITH HADAMARD GAPS By Peter Tien-Yu Chern Let n(r) be the number of zeros of f e B in |z| ‘ r, where B is the space of Bloch functions. Anderson, Clunie, and Ponmerenke [J. Reine Angew Math. 279 (1974), p. 36] asked: is it true that n(r) = °[I%; log log I%;] as r -+ 1‘? The attempt to answer this problem led to the following main result concerning with the lower bound of the Nevanlinna counting function N(r,f,a) = N(r,a). There exists a Bloch function f in |z| < 1 such that for each a e C lim inf "913‘? t x , r -+ 1" log log I: where K is a positive constant. Let 80 be the little Bloch space. We show that % log+[o(log I%;) + 1] is an upper bound of the Nevanlinna characteristic function T(r,f) as r -9 1‘ for f c 80, and we show by example that this upper bound is sharp. We next establish inequalities for lacunary series with Hadamard gaps along small circular arcs in |z| < 1. Applications to the value distribution of B and B‘" are given. Peter Tien—Yu Chern Let B“ (a > 1), L:(D) (1 i p < a), and 5”“ (0 < n < o), be the a-Bloch space, the Bergman p-space in [z] < 1, .and the Shapiro- Shields class; respectively. We show that for each a e t, the least LE upper bound and the greatest lower bound of N(r,a) for 8“, and 5”“ all have the same order log I%;' We obtain some results about the zero sets of a—Bloch functions (a > 1). We show that these "at-Bloch zero sets" are quite different from Blaschke sequences. Some sufficient and some necessary conditions on the coefficients for B“ and 8: are shown. We also give conditions on blocks of coefficients for B“ and 8:. ACKNOWLEDGMENTS I wish to thank Professor Peter Lappan for suggesting the investigation of the tenth problem submitted by Anderson, Clunie, and Pommerenke and for his patient guidance throughout the preparation of this thesis. 11' II. III. IV. THUBLJEGJF'CKNWTEBVPS Title INTRODUCTION . . . . . . . . . . . . . . . . THE VALUE DISTRIBUTION OF BLOCH FUNCTIONS AND LITTLE BLOCH FUNCTIONS . . . 1.1 The value distribution of Bloch functions. 1.2 The value distribution of little Bloch functions . . . . . . . . . . . . . . SOME INEQUALITIES or LACUNARY SERIES WITH HADAMARD GAPS ALONG CIRCLES [2] = r < 1 2.0 Introduction . . . . . . . . . . . . . . 2.1 Inequalities of Dirichlet series with Hadamard gaps and bounded coefficients along small circular arcs in |z| = r < 1 2.2 Inequalities of Lacunary series with Hadamard gaps and bounded coefficients along circles lzl = r ( 1 . . . . . 2.3 Inequalities of Dirichlet series with Hadamard gaps and unbounded coefficients along small circular arcs |z| = r < 1 2.4 Inequalities of Lacunary series with Hadamard gaps and unbounded coefficients along circles |z| = r < 1 . . . . . 2.5 Applications . . . . COUNTING FUNCTIONS OF ANALYTIC FUNCTIONS WITH SLOW GROWTH . . . . . . . . . . . . . . . . 3.1 Introduction . . . . . . . . . . . . . 3.2 An upper bound . 3.3 A lower bound ZEROS 0F a—BLOCH FUNCTIONS . . . . . . 4.0 Introduction ..... . . . . . . 4.1 Zero sets of a-Bloch functions . . m Page 10 10 15 24 24 27 33 35 39 4O 43 43 45 49 49 50 Title Page COEFFICIENTS 0F a-BLOCH FUNCTIONS . . . . . . . . 55 5.1 Introduction . . . . . . . . . ....... 55 5.2 An extension of a theorem of Mathews . . . . 55 5.3 Conditions on coefficients for the little a-Bloch space H: ............. 57 5.4 Conditions on Blocks of coefficients in the class B“ . . . . . ......... 62 BIBLIOGRAPHY . . . . . . . . . . . . . ..... 67 iv INTRODUCTION Let D denote the unit disc {2: |z| < l}, C the complex plane, and i: = d: U {a}. A function f in D is called an a-Bloch function (a > 0) if f is analytic in D and satisfies (0.1) sup (1 — IzI)“ If'(z)| < . . zeD The set of all a-Bloch functions is denoted by B“. A Bloch function is precisely a l-Bloch function. Let B: (a > 0) be the set of functions f analytic in D such that (0.2) lim (1 - Izl)“ |f‘(z)| = o . IZI—fl’ B: is called the space of little a-Bloch functions. A lacunary series with Hadamard gaps means a function f analytic in D which can be expressed as a power series in the form "k (0.3) f(z) = Z a k=o kz , nk+1/nk i q > 1 . The purpose of this paper is to study B“ and) some lacunary series with Hadamard gaps. We obtain some results concerning the value distribution (Chapter 1, 3), the zero sets (Chapter 4), the coefficient conditions for a-Bloch functions in |z| < 1 (Chapter 5), and some useful inequalities for lacunary series with Hadamard gaps along circles in lzl = r < 1 (Chapter 2). 1 Let us begin with some background in value distribution theory. Picard (1880) proved that if f(z) is a non-rational function meromorphic in c, then f takes every a e I: infinitely often, with at most two exceptional values. Let f be a meromorphic function in c. If a c C, we denote by n(r,a) the number of solutions of f(z) = a in |z| i r. We set M(r,f) = max [f(z)]. The order of f is defined by lzl=r (0.4) p = lim sup 1° 10 M r f r—bw logr ’ and the order of the roots of f(z) = a is defined by (0.5) p(a) = lim sup mg)- . raw logr Hadamard (1893) showed that p(a) ‘ p for all a e C. Borel (1897) extended Picard’s result by proving that p(a) = p for all a, with at most two exceptions. Hadamard’s theory lacked precision and did not work well for functions of infinite order or meromorphic functions. In the case when f is meromorphic in C, the maximum modulus M(r,f) does not behave satisfactorily as an indicator of growth, since M(r,f) is infinite whenever f(z) has a pole on lzl = r. For a e 3:, the Nevanlinna counting function is defined by r (0.6) N(r,a) =10 n(t,a) 51:- . We put 2n (0.7) m(r,a) = 51;} Iog+ |f(reia)|d0 , and 0 2n 1 I (0.8) m(r,a) = - log+ . d9 , a e C . 2" J.0 f(rele) - a The Nevanlinna characteristic Of f is defined by (0 9) T(r,f) = m(r.°) + N(r,a) . Nevanlinna (1925) stated his First Fju‘ndamental Theorem as follows: If a e I: and f(O) 1 a, f is meromorphic in C, then (0.10) m(r,a) + N(r,a) = T(r,f) + 0(1) . The function T(r) gives an excellent description of the growth of any meromorphic function in a finite disk or in the whole plane. N(r,a) measures the number of roots of the equation f(z) = a in |z| ‘ r, while m(r,a) measures the average closeness of f(z) to a on |z| : r. We deduce at once Hadamard's inequality, namely p(a) 1‘ p, for every a. However Nevanlinna’s aim was to obtain a sharper version of Borel’s inequality. He did this by showing that in general it is the term N(r,a) which dominates in (0.10). He showed that if q i 3, and a,,...,aq are q distinct complex values then 9 (0.11) (q - 2)T(r.f) ‘ 2 N(r,av) - N.(r) + S(r) . v=l where N,(r) measures the zeros of f ' and the multiple poles of f and S(r) is a term in general much smaller than T(r). This is t_h_e Second Fundamental Theorem. The Second Fundamental Theorem gives 4 an extension both of Picard’s result and Borel’s result. The deficiency of a is defined as follows: _ .. nra_ _ . N(r,a) (0.12) 6(a) - 1112f W - l Ilia—amp T(r) . Applying (0.12), Nevanlinna obtained the deficiency relation (0-13) 2 6(a) ‘ 2 . The concepts defined above are for meromorphic functions in c. For functions defined in the disc D, we can get similar results by replacing r -t w by r -> 1" and by making a few other minor changes. The first significant result concerning with the value distribution of Bloch functions was obtained by Anderson, Clunie, and Pommerenke [ACP]. They proved that if f c B, than (0.14) lim sup N(r,f,0) ‘ lim sup TQ‘Jf) ‘ 1/2 . r-91‘ log logr}; r->1‘ log 1031-}; They also asked ([ACP] tenth prob., p. 36) if it is true that n(r 0) = o(-l- log log -l-) as r -# 1‘ ’ l-r l-r for f c B? A non-constant meromorphic function f in D is called admissible [Hay] if lin sup 1&5? = +0 . r -% 1' log T:; Although the Second Fundamental Theorem provides a lower bound for N(r,f) in terms of T(r), N,(r), and S(r), it doesn’t provide us with a practical tool to evaluate lim inf N(r,f). The problem is r -9 1‘ that (0.15) S(r,f) = 0(log T(r,f) + log I%F) as r —» I- . (see Hayman [Hay]). By (0.14) T(r,f) = 0(log log 1%?) as r -+ 1‘, and hence (0.16) S(r,f) = 0(log 3%?) as r -+ 1- . Thus S(r,f) has order log I%; which is larger than that of T(r,f). Therefore the Second Fundamental Theorem does not work on this problem. The Second Fundamental Theorem works for the admissible functions only. Bloch functions are not admissible. In Chapter 1, we develop a method to evaluate the lower bound of N(r,f) as r -+ 1' for analytic functions in D with slow growth, especially for Bloch functions. The attempt to answer the tenth problem led to the main results in Chapter 1. Theorem 1.1.1. If f(z) is a lacunary series with Hadamard gaps in D, and if 211 (1 1.1) lim inf (I—r)’ I |f'(re19)|2 de a constant > 0 r -+ 1‘ 0 then lim inf T(r,f) b C > 0 r -+ I- log log 1%; 6 where Cq is a constant depending only on 9. o n Theorem 1.1.2. If f(z) = 2 a z k with ak ¢ 0, k=0 k n /n i q > 1, lim sup | l > 0, and lin T(r,f) = +6, then for k+1 k k __’ . BR [‘91- each a e C, (1.1.7) N(r,f,a) = T(r,f) + 0(1) as r -+ l” . Theorem 1.1.3. There exists a Bloch function f in D which is not little Bloch such that for each a e C (1.1.8) lim inf N(gféi i K r '9 1' log log 1:; where K is a positive constant. Theorem 1.2.1. If f e 90, then T(r,f) ‘ % log’[o(log 1%?) + 1] as r—tl‘. Theorem 1.2.2. If Mr) is a non-negative strictly decreasing continuous function in (0,1) with N1‘) = 0, then there exists a little Bloch function f in D such that T(r,f) i K log+ [T(r) log é] as r —t l" , where K is a positive constant. Mm 1.2.3. If T(r) is a non-negative strictly decreasing 0 continuous function in (0,1) with 9(1‘) =0 and 2 9[1--l-(-}_— = k=2 2 ‘ +0, then there exists a little Bloch function g in D such that N(r,g,a) h K log“+ [9(r) log YEP] as r -* 1", for each a e C, where K is a positive constant. More-1.2.4. There exists a little Bloch function g in D such that (1.1.8) lim inf M i K for each a e C , r -9 1" log log I: where K is a positive constant. In Chapter 2 we deal with some inequalities for gap series with Hadamard gaps along small circular arcs in lzl = r < 1. Zygmund [Zyl], Waterman [Wa], Binmore [Bil], and Gnuschke [Gn] established some inequalities between certain sums and integrals of series with Hadamard gaps along paths leading from inside of D to the boundary of D. By way of contrast, we obtain some results along circles lzl = r < 1. We begin with a Lemma of Binmore and then consider Dirichlet series with Hadamard gaps and bounded coefficients. We establish some inequalities along separated circular arcs in D (see Theorem 2.1.2). Then we use a kind of interpolation method for the sequences of exponents to establish the inequalities along small circular arcs close to [2] = 1 (see Theorem 2.2.1). We deal also with the case with unbounded coefficients. Some applications giving examples for exhibiting the value distribution for Bloch functions and a-Bloch functions (a > 1) are given. In Chapter 3, we deal with counting functions of analytic functions with slow growth. A function f in D is said to be in the Shapiro-Shields class 3 -k (0 < k < .) if r is analytic in D and satisfies sup (1 — 121)“ lf(2)l < a . zeD Let L:(D) (l i p < a) be the Bergman space of analytic function f such that |f|p is integrable on |z| < l. The main purpose of Chapter 3 is to show that the least upper bound and the greatest lower bound of N(r,a) for B“ (a > 1), L:(D) (l i p < 0), and S"k (0 < k < s) all have the same order log 3%; . In Chapter 4, we deal with the zero sets for a—Bloch functions (a > 1). It is well known that the zeros {zk} of nontrivial Hp (Hardy space 0 < p i 0) functions are completely characterized by the Blaschke condition I (1-|z|) 1) are quite different from the Blaschke sequences. By constructing a Horowitz—type infinite product [Ho], adopting Beller’s technique [Bel], and using a result on the coefficients of the a-Bloch functions [Ya] we obtain the following main results. Theorem 4.1. If a > 1, then there is an a—Bloch zero set which is not a Bloch zero set. Corollarj 4.1. For each a > 1, there is an a-Bloch zero set which does not satisfy the Blaschke condition. Theorem 4.2. If 1 < a < fl and if 40: - 3 < B then there exists a fl-Bloch zero set which is not an a-Bloch zero set. In Chapter 5, we study the coefficients of a-Bloch functions. Coefficients of Bloch functions have been studied by Anderson, Clunie, and Pommerenke [ACP], Mathews [Ma], Neitzke [Ne], and Fernandez [Fe], among others. In this chapter we investigate coefficients of a-Bloch functions. In section 2 following Bennett, Stegenga, and Timoney [BST] we give two sufficient conditions on coefficients for B“. In section 3 we obtain one sufficient condition and three necessary conditions on coefficients of little a—Bloch functions. In section 4, we establish some sufficient conditions and some necessary conditions on blocks of coefficients for the classes 8“ and 8:. The results we obtain in this chapter are related to results of Hardy [Ha], Titchmarch [Ti], Mathews [Ma] and Neitzke [Ne]. CHAPTER I THE VALUE DISTRIBUTION OF BLOCH FUNCTIONS AND LITTLE BLOCH FUNCTIONS Anderson, Clunie, and Pommerenke [ACP] have listed twelve open problems for Bloch functions. The tenth problem says the following. Let n(r) be the number of zeros of f e B in lzl ‘ r. Is it true that l-r l-r ' The attempt to answer this problem led to the results of this chapter. 51.1 The value distribution of Bloch functions o n _ k . Theorem 1.1.1. If f(z) — kilo 8k 2 , w1th ak 7f 0, nk+l/nk q > 1, and if 2 2" ‘ 2 (1.1.1) lim inf (l—r) [ |f‘(re19)| do n K* > 0 , r -) 1‘ 0 T(r,f) then lim inf r —9 1— log log -l-_r i Cq, where Cq is a positive constant depending only on q. To prove Theorem 1.1.1 we need the following two lemmas. 10 11 ELLI. ([Zy2, Vol. 1, p. 215]). Q If h(x) = 2 (a cos n x + b sin n x), n /n i q > 1, and if a k=1 k k k k k+l k 2 (a; + b;) < 0, then lhl, i Aqlhla, where A9 is a positive k=1 constant depending only on q. Lemma 1.1.2. ([Zy2, Vol. 1, p. 216]) If g(x) is a nonnegative measurable function defined on a set E in the real line 6? with positive linear measure IE] such that -1-Ig(x)dxiA>O and -1—Iga(x)dx£B, II?! E IEI B then for 0 < 6 < l, we have lEé' i |E|(l — 6)2 Aa/B, where E6 = {x c E | g(x) i 6A}. Proof of Theorem 1.1.1. For any 0 < r < 1, if fr(z) = f(rz), |z| < 1, then fr c L3 (lzl = 1). It follows from Lemma 1.1.1 that we have lfrl, 5 Ag If,” Putting EM = {0 e [0,2n] | lfr(ei9)| i Aq Ifrla/Z}, taking a = 1/2, A = A9 lfrla, B = lfrl: , g = lfrl, E = [0,2n] and applying Lemma 1.1.2, we have I= _ (1.1.2) lEr’al s 2n (1 — 2) Ag — A E 2 2 q ° According to the definition of Nevanlinna characteristic and (1.1.2) we obtain 12 1f T(r,f) = 5% log+ [f(re19)| d9 4. . h 3% I log |f(re19)|’ d0 E r,6 IE | A’ Ifrl’ 1'16 + z E 4n log [ 4 A’ A’ Ifrl’ 5 .3 log ._9__z__£ Thus + 2 (1.1.3) T(r,f) n cq log Ifrla + Bq, where Cq and Bq are positive constants depending only on q. Applying Parseval’s identity, we have 0 2 2 an-Z 1 2n . 2 (1.1.4) 2 nk lakl r = E; I lf'(re19)| d0 k=0 0 1 i (K*/2) (l-r)’ for r near 1 , where the last inequality follows from the hypothesis in Theorem 1.1.1. Integrating twice in (1.1.4) with respect to r, we have 2 “It In '2 r 2nk(2nk-l) k a 2n 1 i (Kt/2) log 1:; for r near 1 , 2 k k=0 and it follows that 13 a l 2" - 2 (1.1.5) Ifrla = E; |f(re19)| d9 0 2n = X lakla r k k=1 1 i (K*/2) logl_r for r near I . Combining (1.1.3) and (1.1.5), we obtain (1.1.6) lim inf ““3 r -+ 1' log log 1%; i C . q This completes the proof of Theorem 1.1.1. o n _ k - Theorem 1.1.2. If f(z) - kEo ak z with HR ¢ 0, nk+1/nk i q > 1, lim sup lakl > 0, and lim T(r,f) : +a, then for each a e C, k -9 w r -9 1‘ (1.1.7) N(r,f,a) = T(r,f) + 0(1) as r -9 l- . Proof of Theorem 1.1.2. Since f is a lacunary series with Hadamard gaps and lim sup Ink] > 0, by a result of Murai [Mu2], k -9 0 lim inf m(r,f,a) is finite, for each a e C. Since for each a e C, r -9 1' T(r,f) = N(r,f,a) + m(r,f,a) + 0(1) and N(r,f,a) is an increasing function with respect to r, hence N(r,f,a) = T(r,f) + 0(1) as r -+ I“. This completes the proof of Theorem 1.1.2. Theorem 1.1.3. There exists a Bloch function f in D which is not little Bloch such that 14 (1.1.8) lim inf "(’1f18% a K , for each a c c , r'-9 1" log log I:; where K is a positive constant . “ “k k Proof of Theorem 1.1.3. If f(z) = X z with nk = 2 , then f k=0 is Bloch but not little Bloch [Ya]. We want to show that (1.1.8) is true for this f. According to Theorem 1.1.1 and Theorem 1.1.2, it suffices to show (1.1.1). By Parseval’s formula, we have a 2n 0 2n -2 51—51- |f‘(re19)|2 do = z n’ r k (I—r)2 . 2n k 0 k=0 . 1 For any 0 < r < I, there ex1sts n. such that n‘_1 6 I:; < n" 1 i 1 - r > -1- and l - i r < l - - . If we fix r, “1-1 “1 "1—1 "1 we obtain hence w a 2n -2 a 2 an-Z 2 2 nk r (l—r) h n! r (l-r) k=0 2 a 2n3-2 [ _l l g n, r n, 2n gr' 1 4":-1 *[1-,. l c-l i e”4 . It follows that 15 Zn . lim inf (I-r)’ I lf'(re1°)l’ d6 u 3—1’ r -+ 1’ o e which completes the proof of Theorem 1.1.3. A non—constant meromorphic function f in D is called admissible [Hayl] if lim sup -I£EL§1 = +0. r -+ 1' log I:; Theorem 1.1.4. There exists a non-admissible analytic function f in D such that mgr,a) 6(a,f) = lim inf T(r f) r -+ 1’ = 0 for each complex number a. Proof of Theorem 1.1.4. Let f be as that in the proof of Theorem 1.1.3. Then f is Bloch, and by (0.14) we have that lim sup -I££L§l = 0. Since T(r,f) = +w as r -+ 1‘ and ak = 1, by r -+ 1' log 1:; Murai’s result [Mu2] we have 6(a,f) = 0, for each complex number a. This completes the proof of Theorem 1.1.4. 51.2 The value distribution of little Bloch functions It is natural to ask the following: Can we find a little Bloch function f such that for each a e C (1.1.8) holds. The answer is affirmative and will be given in Theorem 1.2.4. We first show that % log+[o(log'T%;) + 1] is an upper bound of T(r,f) as r -+ 1’, for f e Bo (Theorem 1.2.1). We next obtain a result about the lower bound of T(r,f) for some little Bloch functions in D (Theorem 1.2.2). The result of Theorem 1.2.1 is sharp in the sense of Theorem 1.2.2. Furthermore, in Theorem 1.2.3, we obtain a result about the lower 16 bound of the counting function of some little Bloch functions. By applying Theorem 1.2.3, we show that there exists a little Bloch function f in D such that for each a e C, (1.1.8) holds (Theorem 1.2.4). The main results in this section are in the following. Theorem 1.2.1. If f e 80, then T(r,f) ‘ % log+[o(log I—l-T) + l] as r -9 1’. Theorem 1.2.2. If Mr) is a non-negative strictly decreasing continuous function in (0,1) with N1") = 0, then there exists a little Bloch function f in D such that T(r,f) i K log+ [C(r) log I%;] as r -+ 1’ , where K is a positive constant. Theorem 1.2.3. If T(r) is a non~negative strictly decreasing 0 continuous function in (0,1) with C(l‘) = 0 and 2 CI] - -E%- = k=2 2 ‘ +w, then there exists a little Bloch function g in D such that N(r,g,a) i K log+ [C(r) log -l-] as r -+ 1’, for each a e C, l-r where K is a positive constant. Theorem 1.2.4. There exists a little Bloch function g in D such that (1.1.8) lim inf N(r,g,a) i K for each a c C , r -9 1’ log log I%; 17 where K is a positive constant. Theorem 1.2.1 provides an upper bound of T(r,f) for little Bloch functions. The proof is due in essence to Anderson, Clunie and Pommerenke [ACP]. Theorem 1.2.1. If f 6 80, then T(r,f) 5 % log+[o(log 1%?) + 1] as r -+ 1‘. Proof of Theorem 1.2.1. Let f(z) = 2 anzn be a little Bloch n=0 function in D. According to Parseval’s identity, we have on 27: (1.2.1) 2 nalazlrzn—z - 5% I lf'(re1°)|’ d9 n=1 0 = O[ 1 2I as r -+ 1’, since f e 80 . (l-r) Integrating twice in (1.2.1) with respect to r, we have 2 2 2n 0 n IanI r l - (1.2.2) “El 2n(2n-1) - o(log I:;) as r -+ 1 . Since f is in 80, we obtain 1 2" - (1.2.3) T(r,f) = 5— I log+|f(re19)|d9 " 0 2n 1 1 - = 5 - 5; I0 Iog+|f(re19)|’ d0 27! 1 l - 2 i 5 - 5; I0 log+[|f(re19)| + I]d0 2n 1 .1 '0 3 i 2 logI 2" I0 (|f(re1 )| + l)d9] log+[o(log I%;) + 1] as r -9 1‘ . NIH 18 This completes the proof of Theorem 1.2.1. Corollary 1.2.2. If f e 9,, then N(r,f,a) ‘ % 1og+[o(1og Flt) + 1] + 0(1) as r -’ 1". Proof of Corollar1 1.2.2. According to the First Fundamental Theorem of Nevanlinna, we have m(r,a) + N(r,a) = T(r,f) + 0(1) for each a e C . By Theorem 1.2.1, T(r,f) 6 log+[o(log I%;) + 1] as r -4 1’ . NIH This implies that N(r,f,a) 6 NIH log+[o(log 31;) + 1] + 0(1) as r —-) 1- . Theorem 1.2.2. If Mr) is a non-negative strictly decreasing continuous function in (0,1) with N1") = 0. Then there is a little Bloch function f in D such that T(r,f) a K log+[ 1 - r 3 2-‘, l - 2-(“1) < r 6 l - 2-.. If we fix r, (1.2.4) yields 1 r 2 1 2n 2 (1.2.5) T(r) 2" I0 If (re )l d9 9 2 2 23 2(2 —1) (l-r) i a! 2 r - C(r) a2 f i _§ 229 r_2(2 —I) (2-0)2 at v '—T:‘ I v m Combining (1.2.4) and (1.2.5), we obtain . 2" a (1.2.6) 5% I0 If'(rei9)|’ d6 : kzz 22k a; r2(2k-1) i T(r) -‘ . <1-r)’ Since T(t) is decreasing in (0,1), we have 20 1 r 3 1 (1-t) dt ds n 9(r) I dt ds 1' 8 (1.2.7) I I v(t) , 0 o 0 0 (l-t) = 9(r) [log 1%; + r], for 0 < r < 1. Integrating twice in (1.2.6) with respect to r, using (1.2.7), we have a 22k 82 2 k "k -4 .1. (1.2.8) 2 2nk(2nk-l) r h e 9(r) I log l-r + r] , k=2 where nk = 2k . It follows from (1.2.8) that 21! .1 i9 3 — 3 .1. (1.2.9) 2" I0 |f(re )I de - Irria a v(r) log l~r , as r -+ 1‘ . Since f is a lacunary series with Hadamard gaps, we have + 3 (1.1.3) T(r,f) I Cq log Ifrla + Bq . Combining (1.2.9) and (1.1.3), we obtain (1.2.10) T(r,f) i K log+(¢(r)log 1%?) , as r -9’l‘ , where K is a positive constant. This completes the proof of Theorem 1.2.2. Iheorem.l.2.3. If C(r) is a nonnegative strictly decreasing continuous function in (0,1) with lim V(r) = 0 and r+1‘ 21 Z CII — -kl— : +o, then there exists a little Bloch function g "1 k=2 2 such that N(r,g,a) i K log+[v(r)log I%;] as r -+ 1’ , for each a e C , where K is a positive constant. To prove Theorem 1.2.3 we need the following lemma. Lemma 1.2.1. [Mul] Let f be analytic in D and of unbounded type and let p(z) be a non-constant polynomial. For each 0 e [0,2n], let g9(z) = f(z) + e16 p(z). Then meas{0 e [0,2n] : 6(a,g9) = 0 for each a c C] = 2", where mess is the linear Lebesgue measure. a k Proof of Theorem 1.2.3. Let f(z) = 2 ak z2 be the function k=2 constructed in the proof of Theorem 1.2.2. According to Theorem 1.2.2, we have (1.2.10) T(r,f) a» x Iog+(v(r)1og 14;) as r —’ I- . Since f is a lacunary series with Hadamard gaps and I ah2 = +w, k=2 according to ([Zy2, vol 1, 8.12 Theorem, p. 214]), the set {e19 | lim f(reie) exists and finite] has linear Lebesgue measure zero, r91" and hence f is of unbounded type. Let ge(z) = f(z) + ei9 2. According to Lemma 1.2.1, there exists a 9 e [0.2"] such that 6(a,g9) = 0 for each a c C. If g = g9, then g 6 80 since f e 80. Also, 22 (1.2.11) T(r,f) T(r,g(z) - eiez) I\ T(r.E) + T(r. - e19 2) + 0(1) Thus) + 0(1) . Combining (1.2.10) and (1.2.11), we obtain T(r,g) s x Iog+(v(r)1og 1%?) as r -+ I- . Since 6(a,g) : 0, for each a e C, and g is of unbounded type, we obtahi N(r,g,a) h K log+(v(r)log I%;) as r -+ 1‘ . This completes the proof of Theorem 1.2.3. Theore!;1.2.4. There exists a little Bloch function g in D such that (1.1.8) lim inf __!I£1512%_ a x for each a e c , r -+ 1’ log log 1:; where K is a positive constant. Proof of Theore! 1,2,3. Let T(r) = (log I%;)"/’, for o < r < 1. 0(r) is a non-negative strictly decreasing continuous function in (0’1) With ¢(1—) = 0 and 2 'Il ’ 'E%TI = 2 ((k—I)Iog 2)"/2 k=2 2 k- (log 2)"/’ 2 ———l-—- = +o. k=2 (k-I)‘/’ 23 Applying Theorem 1.2.3, there exists a little Bloch function g in D such that N(r,g,a) i K log+ V(r)log -l- as r -9 1’ . 1-r ”l wins we] MIMI-17f +0.1... 1.1 1 l _ 2 log log l-r as r -9 1 . This completes the proof of Theorem 1.2.4. CHAPTER II SOME INEQUALITIES or LACUNARY SERIES WITH HADAMARD GAPS ALONG CIRCLES lzl = r < 1 52.0 Introduction The motivation of this chapter is to find more Bloch functions f, in addition to the one found in Chapter 1, such that for each a e C (1.1.8) lim inf "(r’f'a) n K > 0 . r -) 1‘ log log1-_—r- It is natural to look at a lacunary series with Hadamard gaps co nk f(z) = Z ak z , |z| < l , nk+1/nk i A > 1 . k=0 According to Murai’s result [Mu2], if lim sup Iakl ) 0, then 6(a,f), k-)o the deficiency of f at a satisfies 6(a,f) = 0. Therefore, according to Theorem 1.1.1 and Theorem 1.1.2, in order to show (1.1.8), it suffices to show that (1.1.1) is true. Throughout this chapter, we consider f to be an analytic function in D with Hadamard gaps. We will explore an inequality concerning the radial variation of f due to Zygmund and some of its extensions. In 1944, Zygmund [Zyl] showed that 1 w _1 I |f(x)ldx < a => 2 lakl nk < o . 0 k=0 24 25 It is clear that, with the obvious modification, the radius 1 can be replaced by any radius and that the converse is valid for arbitrary power series. In 1950, Waterman [Wall showed that l m I (l - x)“ |f(x)|p dx < a => 2 laklp n;(a+1) < a 0 k=0 for D > 1, 0 6 a 6 fl - 1. In 1967, Halasz [Hal] gave a new and simpler proof for Zygmund’s result and also showed that 1 o I (1 — x)"1 |f(x)|dx < a => 2 lakl < s . 0 k=0 In 1972, Binmore [BiZ] proved Waterman’s result for a = fl = 2. We say that the curve C C D goes to the boundary of D if C = (z(t) : 0 6 t < 1} where |z(t)| -’ 1 as t -’ 1‘. In 1983, Zygmund’s result was generalized by Gnuschke and Pommenerke [GP], where the radius was replaced by an arbitrary curve which goes to the boundary of D. In 1984, Gnuschke proved that if C is a half-open curve C in D with OeC, COED1¢ then 5 _ z lakl nk (“*1) . M, I (1 — IzI)“ If(z)I” IdzI k=0 C for a: e R, p > 0, where M, depends only on a, p and A. In Gnuschke’s result, if we choose at : 2, fl = l and let f' replace f, we obtain 1 a -s (1 — IzI)’ lf'(z)l IdzI » - 2 In a I n IC It1 k=0 kit It lakl 2 k ZlI—I II M 8 kOn 26 This result resembles (1.1.1) except for the fact that C is not a circle in D. It would be desirable if we could replace C by circles of the form [2] = r < 1 in the above. Gnuschke’s result was based on the following lemma which was essentially proved by Binmore [Bil]. Binmore’sie-a. Let p e N and let w¢,,...,wp e C such that (i) min Re wk = q, and O‘k‘p (ii) 0 < so 6 le — wkl 6 1/2 for j # k . Then there exist co,...,cp e C such that p -w .t (2.0.1) h(t) = 2 c. e J .I=o 9 satisfies h(l) = 1 and both (2.0.2) |h(t)| ‘ th eq(1—t) for 0 s t < s , and (2.0.3) lcjl ‘ zpseq/af for j = O,...,p . In section 1, we establish inequalities for Dirichlet series with Hadamard gaps and bounded coefficients along separated small circular arcs in |z| = r < 1 (see Theorem 2.1.2). In section 2, we use a kind of interpolation method, when the radius r is sufficiently near to l, to establish inequalities along small circular arcs in |z| = r < 1 (see Theorem 2.2.1). We also extend the inequalities along circles [2| = r < 1. In section 3 and 4 we deal also with the case with 27 unbounded coefficients. In section 5 we present some applications, giving examples for exhibiting the value distribution for Bloch functions and a-BIOCh functions (a > 1). 52.1 Inequalities of Dirichlet series with Hadamard ggps and bounded coefficients along small circular arcs in 124:: r < 1. Theorem 2.1.1. Let {ak} be a sequence of complex numbers such that (2.1.1) [ah/82' < M < w, for any k, l, where M is a constant . Let {yk} be a sequence of positive numbers such that (2.1.2) 1 < A 6 y /y 6 A , JAk/e)‘k ’ 1 < constant < l, k+1 k 0 o for k = 1,2,..., where A0, A are constants. If m -yks e , Re 5 > 0, and p > 0 , then there exists a positive integer p which depends on A, 10, and M such that if "j (j = 0,...,p) satisfy the hypotheses of the Binmore Lemma with p + 1 6 q 6 p + 2, where q = Min Re w., then we have O‘j‘p p -w./y (2.1.3) lavlp‘K 2 |f(e J VII” i=0 for v = 1,2,..., where K is a constant depending on A, A0, 3, and NL 28 Proof of Theorem 2.1.1. For 3 = 0,...,p, let Cj be the coefficients of the function h in the Binmore Lemma. We have, for a fixed v, p -w./y p a -y w./y (2.1.4) 2 c. f(e J v) = 2 c. 2 ak e R J V jeo J j-O J k-O ° 9 -wj(yk/yv) l M to M O ('D I a h(y /y ) . k=0 k k v Hence it follows from (2.0.3), (2.1.4), h(l) : l, and triangle inequality that 2 .89 p -wJ-/yv p -wJ-/yv (2.1.5) -2;- 2 lf(e )l * Z ICJI lf(e )l «o i=0 0:0 fl 5 I Z a h(y /y )l k=0 k k v a k ~IaI1— 2 I—llh(Y/y)l. v { kiv av k v } where so ‘ le - wk] 6 1/2 for j f k. Because p + 1 6 q 6 p + 2, {yk} is increasing, and |h(t)| 6 2tp eq<1-t) for t i 0, we have Y Ir“ V 1 - yk/yv lp (2.1.6) |h(yk/yv)| e 2e“ e , if k < v, and y 1 — y /y 9 (2.1.7) lh(yk/Yv)l ‘ 28 I §E e k v I V if k > v . 29 Because of (2.1.5), (2.1.6), and (2.1.7), we obtain p -w./y (2.1.8) 2 We J V)I 5:0 YR “Elavl 2 Y 1 - T; P a I 1 - 2Me 2 -—-e I 2p!eq kv yv We shall show that the expression in curly brackets on the right of (2.1.8) exceeds 1/2. Since yk+1/yk i A, we have v-l y 1 - y /y p v-l _ 1 _ k-v p (,5. “I W 1° 1. < E I x-J e1 — Azj Ip . j=1 l " 7t—«j/AJ. Since e ° < constant < 1 for each j, we can choose p sufficiently large such that (2.1.10) v21 [ 11 e1 - yk/yv ]p < 1 k=1 yv 8Me Since yk+1/yk < A0, we have s y 1 - y /y p w . 1 _ J P (2.1.11) k=2+1 I ;5 e k V I o §=1[ A: e A I By (2.1.2) and (2.1.11), we can choose p sufficiently large such that w y 1 - y /y p (2.1.12) 2 I -5 e k V I < -l—-. k=v+1 yv 8Me 30 Combining (2.1.10) and (2.1.12), we can choose p sufficiently large (depending on A, A0, and M) such that the expression in curly brackets of (2.1.8) exceeds 1/2. Thus we obtain -wj/yv) p (2.1.13) IavI 4 Al 2 If(e I , i=0 where p and K, depends on A, A0 and M. It follows that -w./y J V)|fi p p (2.1.14) Ia I 6 K 2 If(e v 3 . J=0 where K, = X? if 0 < p s 1, and K, = (p + IN?"1 Rf if p > 1. This completes the proof of Theorem 1.2.1. Theorem_2,1.2. If f is the same as in Theorem 2.1.1 and if a e R, then there exists a positive integer v0 depending on {yk], A, A0, and M such that, for F > O, l +— IavI’ C 4V. a , , (2.1.15) «+1 4 K I (I - r) If(re19)I de y c v -211 for v 6 v0, c e [0,2n], where r = e yv, with p the number given by Theorem 2.1.1, and K is a constant depending on A, A0, M, and 5. Proof of Theorem 2.1.2. We first prove the case c = 0. Let p be the integer resulting from Theorem 2.1.1. For v = 1,2,..., .1 = 0'1!”'!p! and ‘r s [0,1] define functions 31 (2.1.16) 2 .(r) - 9+1 J + 7 -2V'J(T). V.J ’ yv + 4(p+1)yv i and cVJ“) = e For fixed v and 7, let w‘j = yv zv J(r), j = 0,...,p. Then we have all of the following: + H Re wj = p for j = 0,...,p, for j i k, fill-J , — ‘ IwJ wkI _ j-k I '"3 "k' 4p+4 ‘ 4p+4 if j > k. If we put q = p + 1, then "j (j = 0,...,p) satisfies the hypotheses of the Binmore Lemma with “o = l/(4p + 4). By applying Theorem 2.1.1, we obtain 5 p -wj/y (2.1.17) Ia I 4 x 2 If(e V 1 J20 ”II” = K: 2 If(cmom”. In > o. where K, is a constant depending on A, A0, M, and fl. Let Fv be the arc described by the union of the images of the functions (v j’ j = 0,...,p. Because of (2.1.16) we have -Re 2 .(r) - Bil ' - ' via] 1 yv (2.1.18) ICV’J(Z)I - Izv’j(T)Ie - _ZTP:T7;V e and -(p+l)/yv 1 - Icmml = 1 -e By (2.1.17) and (2.1.18), we have 32 p a 2 If(tv’J(z))lp It; j(r) I <1 — ltv’j(r)l) i=0 IaVI” e’(P*1’/’v —(p+1)/yv , K. 4(p+1>yv ‘1 ‘ e lavlp e-(P+1)/Yv _(p+1)/yv a = a+, [yv (1 - e )l 4K: (p+1)yv -(p+1)/yv -(p+1)/yv Since e -+ 1 and yv(1 — e ) -+ p + 1 as v -9 °. there exists v0, depending on {yk}, A, A0, and M, such that -(p+1)/y -(p+1)/y a e V 5 % and [yv (1 — e v)]a i [ 2%; ] for V ) V0, and hence p p . , IaVI” (2.1.19) )10 If(tv’j(r))l lfv,j(r)l (1 - lfv’j(r)l) i K: “+1 v for v 6 v0, 1' e [0,1] where K; is a constant depending only on A0, A, M, and 43. If we integrate (2.1.19) with respect to 'r from O to 1, then we obtain Ia I” 1 V . l— p I r ) p ' < ) 1 - ( > I“ d (2.1.20) “+1 K: '10 0 I (tv’j(r )I Irv j r l ( Irv“j T I 7 TV J- 1 p (vtj(1) p a z i: ~10 It 'f(cv.J)l H”mil (1 _ 'tV.JI) dCVIJ J v.5(0) _1_ 4y . e l. I V (1 - r)“ If(re19)I” d0 K, 0 -(p+1)/yv where r = e for v 3 v0, with p the number given by Theorem 2.1.1. Thus the proof of Theorem 2.1.2 is Complete for c = 0. 33 For the case c i 0, replace (2.1.16) by -z .(T) . I V,J e 211 - ._.L.t.1_ . e zv,J(T) yv + Cl + 4(p+1)yv 1 . Cv’j(7) e . and repeat the previous procedure. This gives us c + —l— I lfl 4y . :11 K I V (1 — r)“ If(re19)I5 do , v c -(p+1)/yv where r = e for v 3 v0, with p the number given by Theorem 2.1.1 and K is a constant depending on A, A0, M, and 5. This completes the proof of Theorem 2.1.2. 52.2 Inequalitieg of La_9_1_1nary series with Hadamard gaps and bounded coefficients along circles lat] = r < 1. Theorem 2.2.1. Let {akl be a sequence of complex numbers satisfying (2.1.1). Let {nk} be a sequence of positive integers satisfying (2.1.2) with A = A0. If G nk f(z) = 2 ak z , IzI < 1 , k=0 then there exists a positive integer v0, depending on {nk}, A, and M, such that, for any 3 i n , if n 6 s < n , then vo v v+1 laI In . (2.2.1) 2 I: 43 (I-r)“ If(re1°)|” for any 0 «2 [0,27!], fl ) 0, and a c R, where r = e‘(P+1)/9 with p given by Theorem 2.1.1, and K is a constant depending on A, p, and M. 34 Proof of TheoregL2.2.l. Let {yk} be a sequence of positive numbers satisfying (2.1.2) with A = A0. Because yk+1/yk = nk+1/nk = A, we can put nk = dyk for any k = 1,2,..., where d is a positive constant. For Re t > 0, if (2.2.2) f(e- then f is a Dirichlet series with exponents {-ykdt} and co- efficients {8 Because {yk} and {ak} satisfy (2.1.2) and (2.1.1) k}' respectively, it follows from Theorem 2.1.2 there exists v0 depending on {yk}, A, and M such that l + ———- . IaVI” ° 4w . , (2.2.3) 2+1 ‘ K I (1-r)“ If(re19)| de y c v for v 6 v0, c e [0,2n], a c R, p > 0, where r = e-(p+1)/yv and p is a positive integer depending only on A and M, where K is a constant depending on A, F, and M. Note that the inequality (2.2.3) holds for v i v0, where Va is dependent only on {nk}, A, and M, and Va is independent of the particular sequence {yk} satisfying (2.1.2) with A = A0. If s i nyo, there exists a DV such that nv 6 s < nv+1' We can construct a sequence {yk} of positive numbers satisfying (2.1.2), with A = A0, and yv = s. From our previous argUment, we obtain I IavIfi ° + '4? a . 5 (2.2.4) s“*1 o 2 IC (l-r) If(re19)I do Since 3 is arbitrary, we have completed the proof of Theorem 2.2.1. 35 Theorem 2.2.2. If f, {ak}, {n are the same in Theorem 2.2.1 k} then there exists a positive integer v0, depending on {nk}, A, and M, such that, for any 5 i "Vo’ if nv 6 s < nv+1, then 6 lavl 27! a . 5 (2.2.5) 4 K I (l-r) If(re19)l do 0 1"v where a e R, p > 0, r = e‘IPHVB with p a constant depending on A and M, and K is a constant depending on A, n, a, and M. Proof of Theorem 2.2.2. Since 5 fl 3 fl 2,, - lavl lavl _ lavl lavl 1/48 A«so: Aana no: so: v v+l the desired result (2.2.5) follows from Theorem 2.2.1. 52.3 Inegzalities of Dirichlet series with Hadamard ggps and ugbojunded coefficients along small circular arcs 121 = r < 1. Theorem_2.3.l. Let {ak} be a sequence of complex nmnbers such that (2.3.1) 1 < Is 1' < M < s for k = 0,1,..., k+1/a where M is a constant . Let {yk} be a sequence of positive numbers satisfying 36 (2.3.2) 1 < A 6 yk+l/yk 6 A0 and (MAW/elk“1 < constant < 1, for k = 0,1,..., where A and A0 are constants. If —s a -yks f(e ) = Z a e , Re 3 > 0 and D > 0 , k=0 k then there exists a positive integer p, which depends on A, A0, and M, such that if wj(j = 0,...,p) satisfy the hypotheses of the Binmore Lemma with p + l 6 q 6 p + 2, where q = Min Re w., then we have O‘J‘p p -W-/y (2.3.3) IavIp 4 K E If(e J v)'fl i=0 for v = 1,2,..., where K is a constant depending on A, A0, 3, and Proof of Theorem 2.3.1. According to (2.0.3), (2.1.4), h(l) = l, and triangle inequality, for a fixed v, we have , q p -w /y (2.3.4) 39$ 2 We J v>I . Iavl {1— 1: I331 wok/nu}. a j=0 ksiv v where so and h are given in the Binmore Lemma. By (2.3.4), (2.1.6) and (2.1.7) we obtain 2 -w./y (2.3.5) 2 If(e J ">I i=0 P P a l8 l v-1 y 1 - y /y i -2-!- [l - 2e2 2 -£ e k v] 2p!eq k=0 yv Y m a y l-SI-BP I: e —"e “I I k=v+l v yv 37 We shall show that the expression in curly brackets on the right of (2.3.5) exceeds 1/2. Since yk+1/yk i A, we have k-v v-l yk 1 - y /y p v-l _ 1 - Ao p (2.3.6) kEO[S’:e k ”I skEOIxJ‘Ve I . $3 In .1 ‘9}? 1 — A;J Because e /AJ ( constant < l for each. j, we can choose p sufficiently large such that (— v-1 y 1 - y /y p (2.3.7) 2 I —"- e k " I 12 . k=0 1"; 8e Since yk+1/yk < A0, we have a y 1 - y /y p (2.3.8) 2 ,Jsl IJ-‘e k vI k=v+l v yv ( .2 M4 [13.1 - “I". 3;" l (I. .014 .1 - “I". J 1 '=1 . j_ Since (MA)J/eA 1 < constant < l for any j = 1,2,..., we can choose p sufficiently large such that a 3k [ yk 1 - yk/yv 1P 1 _ e < _— (2.3.9) I: I—l k=v+1 av yv 8e Combining (2.3.7) and (2.3.9), we can choose p sufficiently large (depending on A, A0, and M) such that the expression in curly 38 brackets of (2.3.5) exceeds 1/2. Thus we obtain ‘W . J/y p v (2.3.10) IavI 6 Ki 2 If(e )I i=0 where p and K, depend on A, A0, and M. It follows that p -w./y (2.3.3) In I” . x I If(e J “II” V 3 J30 where R2 = K? if 0 6 R < l and K2 = (p + Up"1 Kp if D > 1. 1 This completes the proof of Theorem 2.3.1. Theorem 2.3.2. If f is the same as in Theorem 2.3.1 then there exists a positive integer v0 (depending on {yk}, A, A0, and M) such that, for F > 0, Ia I” C " ”‘1 '0 (2 3.10) “:1 s x I V (I—r)“ If(re1 )I” do V C -(p+1)/yv for any v i v0, c 6 [0,26], where r = e , with p a constant integer given in Theorem 2.3.1, and K a constant depending on A, A0, M and fl. Proof of Theorem 2.3.2. Inequality (2.3.3) follows from Theorem 2.3.1. Repeating the same procedure as in the proof of Theorem 2.1.2, the result is obtained. 39 52.4 Inequalities of Lacunary series with Hadamard ggps and unbounded coefficients along circles IgL: r < l. Theorem 2.4.1. Let {ak} be a sequence of complex numbers satisfying (2.3.1) and let {nk} be a sequence of positive integers satisfying (2.3.2) with A = A0. If f(z) = ”Z ak :k then there exists a positive integer v0, depending on {nkI:0A, and M, such that,if s 6 my with nv 6 s < n then 0 v+l’ c + l— 45 (I-r)“ If(re1°)|” do Ie-VI‘9 (2.4.1) s“+1 ‘ 2 IC for c e [0,2n], a e R, B > 0, where r = e-(p+l)/s’ p depends on A, and M and K are constants depending on A, p, and M. Proof of Theorem 2.4.1. Since {8k} satisfies (2.3.1), and {nk] satisfies (2.1.2) and (2.3.2) with A = A0, we can apply Theorem 2.3.2 to the procedure of the proof of Theorem 2.2.1 to obtain the desired result. Theorem 2.4.2. If f(z), {ak}, and {DR} are the same in Theorem 2.4.1, then there exists a positive integer v0 such that, for any 2 ' 6 s nyo, 1f nv s < nv+1 then p a I 2" . (2.4.2) Z s x I (I—r)“ If(re10)Ip do 3 0 for a e R, F > 0 where r = e-(p+l)/s’ p depends on A, and M and K are constants depending on A, p, and M. 40 Proof of Theorem 2.4.2. Since 2, IaVI’ IavI’ IaVI” IaVI” -- = Bus and 6 = < , 1/48 a: O: at a at a A s A n n s v v+1 (2.4.2) follows from Theorem 2.4.1 by repeating the augments used in the previous Section 52.3. 52.5 Applications Theorem 2.5.1. Let {ak} be a sequence of complex numbers and {nk} be a sequence of positive integers satisfying n a (2.5.1) lim sup IakI > 0 , I-Eil-§:ll < M < w , and k -+ w k k n - l k (2.5.2) 39-1—— = A > 1 , Jfll. < constant < 1 , n - l k k A - 1 e for k = 0,1,2,..., where M, A are positive constants. If m n f(z) = 2 ak z k k=0 then (2.5.3) lim inf N(r,f,ai 6 KA for each a e C , rI-é 1’ log log 1:; where K; is a positive constant depending on A. Proof of Theorem 2.5.1. Since f is a lacunary series with Hadamard gaps and lim sup Iakl > 0, according to Theorem 1.1.1 and k-fio 41 Theorem 1.1.2 we need only show that 2n . (1.1) lim inf (1 — r)’ I If'(re1°)|’ do s positive constant. r -+ l- 0 a nk-l Because f‘(z) = 2 nk ak z with {nk ak} satisfying (2.3.1) k=0 and {11k - 1} satisfying (2.3.2), we can apply Theorem 2.4.2 for a = fl = 2 to obtain the desired result. Theorem 2.5.2. If a is a positive rational number and if o nk 6/2 f(z) = 2 ak z , where {nk} satisfies (2.5.2) and ak = nk , k=0 then (2.5.4) lim inf Eiitiffll s K. for each a e c , r -+ 1’ log 1:; where K; is a positive constant depending on A. Proof of Theorem 2.5.2. Since f is a lacunary series with Hadamard gaps and unbounded coefficients, we have lim T(r,f) = +w, r -+ l- and it follows from Theorem 1.1.2 that (2.5.5) N(r,f,a) = T(r,f) + o(l) as r -+ 1‘ for each a c C . Thus, to prove (2.5.4) it suffices to show that (2.5.6) lim inf -21£L§l i K as r -9 l“ , where r -9 1' log 1:; K is a positive constant. 0 nk-l Now f'(z) = 2 11k ak z , and since {nk ak} satisfies (2.3.1), k=0 the sequence {nk - 1} satisfies (2.3.2), thus, we can apply Theorem 42 2.4.2 for a = 6 + 2, fl = 2, to obtain Inwal (2.5.7) I = -——;:§—- 4 x I: (1 - r)2 If (re “)1 nv -(1>+1)/nv for n is sufficiently large, r = e , and p and K v constants depending on A and M. Thus we have (2.5.8) I If'(rei°)|2 do a'———ll§-— as r -+ 1- . 0 (1_ r)2+t: Inequality (1.1.3) says that T(r,f) 6 CA log+ lfrl: , where C; are is a positive constant depending on A. Applying Parseval’s identity, we have m 2 2 2nk-2 1 2n (2.5.9) REG nk IakI r = 2; ID If (re 0)I’ 1 F as r -+ 1' 2n 3(1-r)2+8 where the last inequality follows from (2.5.7). Integrating (2.5.9) twice with respect to r, we have 112 Zn 2 2n (2nk - 1) “311'2 k ‘ K1 1 c k=0 k k (l-r) where K1 is a positive constant, and it follows that 2 1 2" '9 2 (2.5.10) Ifrna = 5; o If(rc1 )I do a 2n = 2 Iakl2 r k 6 K2 a as rI-é 1‘ . k=0 (l-r) Combining (1.1.3), (2.5.10) we Obtain the desired result (2.5.6). as r -9 1’ , CHAPTER III COUNTING FUNCTIONS OF ANALYTIC FUNCTIONS WITH SLOW GROWTH 53.1 IntroduLction Value distribution theory of analytic functions with slow growth have been a subject of recent interest, in particular with respect to bounds on the counting function. A function f in D is said to be in the Shapiro-Shields class S 'k (0 < k < a) if f is analytic in D and satisfies k sup (1 - lzl) |f(z)| < w . zeD Let L:(D) (l i p < w) be the Bergman space of analytic functions that are p-th power integrable in [z] < 1 and let 8“ (a > 0) be the «— Bloch space. Let f be a meromorphic function in D. If a e i, we denote by n(r,a) the number of solutions of f(z) = a in lzl ‘ r. For a e 3:, the Nevanlinna counting function is defined by r N(r,a) = I n(r,a) gi . The main purpose of this chapter is to show that the least upper bound and the greatest lower bound of N(r,a) for B“ (a > 1), L:(D) l l i p < w, and S’n (O < n < a) all have the same order log 17;. 43 44 53.2 An upmr bound. Theorg! 3.1. If f e s'-k, k > 0, then T(r,f) = 0(log T%F) as r —9 1’ and for each a e C, N(r,f,a) = 0(log TE?) as r -’ 1". Proof of Theore- 3.1. Let f(z) = 2 an 2". Since f is analytic n=0 in D, we obtain 1 2" - T(r,f) = 5; I log+ If(re19)Ide 0 2n _ _l + '9 3 — 4" I0 log If(re1 )| dB 211 . El I 1og[If(reiG)’ + 1]d9 " 0 2n .1. .1 'a 3 ‘ 2 log { 2" I0 (If(re1 )I + 1)de } ‘11 l -k -5 og0———;i +0(1). (feS) (l-r) _ _l_ _ - 0(1og 1-r) as r -9 1 , and hence T(r,f) = 0(log'I%;) as r -9 1’. Therefore for each a e c, N(r,f,a) = 0(log T§;). Corollary 3.1.1. If f e B“ (a > 1), then T(r,f) = 0(108 TE?) and for each a e C, N(r,f,a) = 0(log 1%?) as r -’ 1'. Proof. Since 8“ C S "(“"l [BeZ], the results follow from Theorem 3.1. 45 Corollary 3.1.2. If f e L:(D), then T(r,f) = 0(log 1%?) and for each a e C, N(r,f,a) = 0(log 1%?) as r -9 1’. Proof. Since L:(D) C S"3/P [Be2], the results follow from Theorem 3.1. 63. A lower bound It is natural to ask the following: For each a > 1, can we find a function f e B“ such that for each a e C, lim inf ELELELEl i K r—’l- log-1_—r > 0, where K is a constant. The answer is affirmative and will be given in Theorem 3.2. We need the following lemma. o n Lemma 3.3.1. [Ya] Let f(z) = Z ak z k be analytic in k=0 lzl < 1, with nk+1/nk i q > 1. Then f e 8“ (a > 0) iff . l-a 11m sup laklnk < w. k—)m Theorem 3.2. If a > 1, then there exists a function f e B“ such that for each a e C, lim inf ELELifEl h K > 0, where K is a r—)l" log-1:? constant. Proof. Since a > 1, there exists a positive integer m such Q 1. - “k - .. _ that 1 + 2m < a. Let f(z) — k§1 ak z with nk+1/nk — 2, n, — 1, 1 1 ~a+— "m 1 —-a 2m 0 = 2 = = and ak nk -+ a. We have ak nk nk < 13k 1, and thus lim sup 3 n l_a < 1. According to Lemma 3.3.1, we have f e B“. k _+ a k k We first show that lim inf I££L£%- h K > 0 , where K is a constant . r41- 1031:; 46 Applying Parseval’s identity, we have -bmz'" . (3.3.1) .11_£1§_ I If'(re19)I’ de 770 0 2n ~2 = 2 a: n: r k (l-r)bm k=1 where bm - 2 + 2 (m ‘) For each 0 < r < 1, there exists a n; such that ‘ r < I - -l . Fix r, and let 1 be given by the last in- "2-1 "1 equality. Then 2n —2 (3.3.2) 2 hm r k (1-r)bm 'v 3 II N V’ H V [1 ’ 1r1,11]4n"—1 * e—‘ . Combining (3.3.1) and (3.3.2), we obtain a 2n “2 2 2 k e" (3.3.3) 2 a n r i ‘---- . = k (l—r)bm Integrating twice in (3.3.3) with respect to r, we have 0 n2 82 rznk k k L 47 where cm = 2-(m-1)’ and it follows that z 1 2" - 2 (3.3.5) Ifrla = 2; I0 If(re19)| d9 0 2n : 2 laklz r k k=1 5 K --l-- , where K, is a positive constant. ’ (l-r)Cm According to (1.1.3) and (3.3.5), we obtain (3.3.6) lim inf 115L§%— e K , r -9 1' log 1:; where K is a positive constant. Since f is a lacunary series with Hadamard gaps, lim sup lanl > O, n-—)o and lim T(r,f) = +w, by Theorem 1.1.2 (Chapter 1), we obtain (1.1.7) r91“ for each a e C, T(r,f) = N(r,f,a) + 0(1) as r -+ 1‘. Combining (3.3.6) and (1.1.7), we have, for each a e C lim inf ELELELEZ i K as r -+ l‘ . r -+ 1’ log -- l-r This completes the proof. Corollary 3.2.1. If k > 0, then there exists a f e S"k such that for each a e C, lim inf ELELETSl h L, where L is a r -+ 1’ log -- l-r positive constant. 48 Proof. Setting a = g + l, we have R = 2(a - 1). Since a“ c s-=(¢-1) = s-k [Be2], the result follows from Theore- 3.2.1. Corollary 3.2.2. If p > 0, then there exists a function f c L:(D) such that fer each a e C, lim inf ELE‘ELEI i K, r-Ol‘ log—- where K is a positive constant. 1 - +2 Proof. Since .«rkc 1.p (n), for e > o, the result follows from Corollary 3.2.1. Relark. Combining results in 53.2 and 93.3, we obtain that the least upper bound and the greatest lower bound of N(r,a), a e C, for a“ (a > 1), 12(1)) (1 i p < a), 54! (o < k < .) all have the same order log 1%; . CHAPTER IV ZEROS OF FUNCTIONS IN THE a-BLOCH SPACES 54.0 Introduction A function f(z) analytic in the unit disc D is called an a-Bloch function (a ) 0) if f satisfies sup (1 — Izl)“ lf'(z)l < . zeD The set of all a—Bloch functions is denoted by B“. A Bloch function is precisely a l-Bloch function. It is known that the zeros {zk} of nontrivial Hp (Hardy space 0 < p i 0) functions are completely characterized by the Blaschke condition I: (1“IZ|)<° k=1 k In particular, all of the Hp spaces admit the same zero sets [Du]. In this chapter, we obtain some results on the zero sets of a-Bloch functions. We show that these "ax-Bloch zero sets" (a > 1) are quite different from the Blaschke sequences. Our main results are as follows. Theorem 4.1. If a > I, then there is an a-Bloch zero set which is not a Bloch zero set. 49 50 Corollary 4.1. For each a > 1, there is an a-Bloch zero zet which does not satisfy the Blaschke condition. Theorem 4.2. If 1 < a < p and if e < 21; (p + 3) then there is a fl-Bloch zero set which is not an a-Bloch zero set. The first and the second theorems are proved by constructing a Horowitz-type infinite product [Bel], [Ho], adopting Beller’s technique [Bel], and using a result on the coefficients of the a-Bloch functions (see Theorem 5.1 of this dissertation). Corollary 4.1 is an immediate consequence of Theorem 4.1. 54.] Zero sets of a-Bloch functions Beller’s Le-a. [Bel] If f e L:(D), (0 < p < a) and if {Zn} 1 - z is the ordered zero set of f, then lim sup ini—gn—L 6 i . n-dm Corollary 4.1. If f e 8, {Zn} is the ordered zero set of f, then Proof: Since B C L: for each 0 < p < a, the required result follows from Beller’s Lei-a. Theorem 4.1. If a > 1, then there is an a-Bloch zero set which is not a Bloch zero set. Proof of Theorem 4.1. Since a: > 1, there exists a positive number q such that l + g < a. Let 51 0 (n -n_) (4.1.1) f(z) = n 1 + bk 2 k k 1 k=0 k l where nk = 22 , n-, = 0, h1" = nkq. Since the radius of convergence w (nk - nk_1) of 2 bk 2 is 1, it follows that f is analytic in k=0 |z| < l, and its zeros are precisely the zeros of the factors of the right side of (4.1.1). Let {zn};:1 be the ordered zero set of f. Clearly, for nk_1 ‘ n 6 nk, we have l ul=b-mk—%fl) =n“%—"”” n k k In particular, -log n 1 - [2D | : 1 - exp {g(n _ nk )} k k k-i log nk log n > g(nk _ nk-1) > 9(nk) for suff1c1ently large k . Thus we obtain 1 - 2 (4.1.2) lim sup —:,—'i-l-Ll l . n a n 0g n q It follows from Corollary 4.1 that f l B, and hence {Zn} is not a Bloch zero set. We next shall show that f e B“, for each a > 1. Put f(z) = a 2 8k zk, ak i 0. According to [Theorem 5.1], it suffices to show that k=0 n 2 n n 2 k ak = 0(n“). Since a > 1 + - and 2 k a i n 2 k=1 9 k=1 k=l suffices to show that 52 n 2/ (4.1.3) 3 sh = 0(n q) . k=l n m Let Sn = 2 [a I, it follows from [Bel] that S = (l + b ). _ k “m _ k k—l k-O w _1 a 1 a 1 l/q Since 2 bk = Z 1/ = 2 __i < w , we obtain k=1 k=l n q k=1 2 k 2 w _1 m m (4.1.4) S < H (l + b ) H b < K n b , n” k=1 k k=0 k k=0 k for some positive constant K . In addition “ k l l. 2 2 1 2M+‘ - 1 m m 21‘ q q k=0 6 2 - 1 (4.1.5) fl bk = n 2 = 2 = 2 = k=0 k=0 1 111+: _ l n+1 2/q 2 _(2 1) —(2 ) m _ = 2g < 2q = 22 = n q m (4.1.4) and (4.1.5) implies 2/q (4.1.6) Snm < K(nm) . If nm-, < n < nm - nm_,, we have _ 2/q 2/q (4.1.7) sn — Sum—1 < K nm__i < K nm . If nm - nm_, ‘ n < nm, we have (.1. S. = s... < Km} < x Hag)” < K 22/q 112/q . 53 Combining (4.1.6), (4.1.7) and (4.1.8), we obtain (4.1.3). Thus f e B“. This completes the proof. Corollary 4.2. For each a > 1, there exists an a—Bloch function f such that f does not satisfy the Blaschke condition. This is an immediate consequence of Theorem 4.1. Theorem; 1.3, If 1 < a < B and if 4a - 3 < )3 then there exists a fl-Bloch ordered zero set which is not an a-Bloch ordered zero set. Proof of Theorem 4.2. Let s > 0, p = 2-(-a%l-) + a, and w (n - n _ ) f(z) = n [1 + bk 2 k k ‘ ] k=0 k _ 2 _ 1/p where nk - 2 , bk — nk Let {Zn}n=x be the ordered zero set of f. For nk_1 < n ‘ nk, nl=b-mk~%fl) =n N%_n“9 n k k Do the same procedure to reaching (4.1.2), we obtain 1 - 2 (4.1.9) lim sup -:T-l-fll l = 1 > 1 . nae " ”g" P —J—+. ——L-+m 2(a-l) 2(a-1) —2(a11) + 28 It follows from Beller’s Lemma and (4.1.9) that f t La 1 . -2(a-1) 2(3—1) + " . Since 8“ C S’ 5 L a for each n ) 0 [Be2], it follows that f t B“. 54 a we next shall show that f c 85. Put f(z) = X ak 2k, a i 0. k=0 k According to [Thm. 5.1], to show f e 8p it is only necessary to show n n that 2 k a = 0(np). Let SD = 2 ak. As in the proof of Theorem k=1 k .. k=0 4.1, we obtain sn ‘ x 22/9 nZ/p for n =1,2,3,... . Also, n n 1 + g 2 k 8k 5 n 2 8k = n SD = 0(n P) k=0 k=0 and g _ 1 _ 4(a - 1) _ = 1 + P - 1 + 2 [ 1 + c] — 1 + 1 + 28(a - 1) < 1 + 4a 4 2(a-1) = 4a - 3 < 5 . n Thus 2 k ak = 0(nfl). This completes the proof of Theorem 4.2. k=0 CHAPTER V COEFFICIENTS OF u—BLOCH FUNCTIONS 95. 1 Introduction Coefficients of Bloch functions have been studied by Anderson, Clunie, and Pommerenke [ACP], Mathews [Ma], Neitzke [Ne], and Fernandez [Fe], among others. In this chapter we investigate coefficients of a-Bloch functions. In section 2 following Bennett, Stegenga, and Timoney [BST] we give two sufficient conditions on coefficients for B“. In section 3 we obtain one sufficient condition and three necessary conditions on coefficients of little a-Bloch functions. In section 4, we establish some sufficient conditions and some necessary conditions on blocks of coefficients for the classes B“ a: and 80. The results we obtain in this chapter are related to results of Hardy [Ha], Titchmarch [Ti], Mathews [Ma] and Neitzke [Ne]. 55.2 _A_3n extension of a theore- of Mathews Theorem 5.0 [BST] If f(z) = 2 an zn is holomorphic in D n=0 n and an a 0. Then f e a“ iff 2 k ak = 0(n“), and then f e a: n k=1 iff X k ak = o(n“). k=1 ” n J J+ —1 Theore- 5.1. Let f(z) = )3 an z". If x k lakl =0(n °‘ ) n=0 R J for some J = 1,2,..., then f e B“. 55 56 0 Proof of Theorem 5.1. For J = 1, put g(z) = 2 lanlzn, since n=0 n a 2 klakl = 0(n ), it follows from Theorem 5.0, that g e 8“. Now k=l “ k 1 IZIIf'(Z)l ‘ Z klakl IZI = g'(l2|) ‘ constant —-----1; . k=1 (l-Izl) and hence f c B“ . . . . (J) C For J > I, it is easy to verify that If (z)| £ “+J_1 , (1-IZI) and the result follows by successive integration. Corollary 5.]. [Ma] Let f(z) = 2 an zn be holomorphic n=0 n in D, if there is a fixed integer J i 0 such that 2 leakl ‘ k=J CnJ for each n, then f e 8. Proof. Setting a = l, the result follows from Theorem 5.1. Q Theorem 5.2. Let f(z) = 2 an 2“ be holomorphic in D. If a a 1 n=0 n and 2 kq—‘laqu = 0(na_1) for some q, 1 < q < m, then f c B“. k=1 Proof of Theorem 5.2. Applying Theorem 5.1, it suffices to show = 0(n“). n that 2 klakl k=1 57 u use 1 zklal=2k 1...; k:]_ k k=1 k n 1 l/p n % l/q ( n d p- 9 H01 er ‘ [kgl k ] [ REI k lakl ] inequalitY) :1. = 0(n) 0[n q ] 1 1+“ : 0[n q ] = 0(n“) - Q Corollary 5.2. [Ma] Let f(z) = 2 an 2“, if 2 nlanl2 < 0 n=0 n=1 then f e 3. Proof. Setting a = l and q = 2, the result follows from Theorem 5.2. Corollary 5.3. ([Ne, Theorem 1, p. 11]). Q G Let f(z) = 2 an 2 , if 2 kq-ilaqu ( w, for some q > 1, then n=0 n=l f e 8. Proof. Putting a = 1, the result follows from Theorem 5.2. ' a 95.3 Conditions on coefficients for the little a-Bloch space 80 II M :5” N r Theorem 5.3. Let f(z) be a function in B: where f(z) l—a a then R k = 0(1) as k -9 a. -Ji 58 Proof of Theorem 5.3. From Cauchy’s formula _ l f'gz) lakl ‘ I 2nik I _ k dz lzl-r z 2 . - . ‘ 2%E I W I 2.152121 rie19 I d9 0 k ike r e °[ 1 J 2"“ (1-r)“ "1 o(k r1—k(l - r)—a) as r -+ l‘ . The minimum value of the last term occurs for r = l - a/(k - 1). Evaluating for this r, we obtain l-k -a -i 1—k -a _ -1 a a _ a—i k r (1 ‘ r) ‘ k [1 ’ k~1] [k—I] ‘ o(k ) as k -9 w Thus kl—a ak = 0(1) as k -+ 0. Corollary 5.4. [Ne] Let f(z) = 2 a zn. If f 6 Bo then an -9 0 as n -+ w. Proof. The result follows from Theorem 5.3, by setting a = l. 0 Theorem 5.4. Let f(z) = X a 2“. If n k' [ n 2 ”'2‘“? la l = o 2 k sz (R J). k k for some fixed J = 1,2,..., then f e 8:. To prove Theorem 5.4. We need the following Lemma 5.1. 59 Le-a 5.1. ([Ne, p. 16] and [T1, p. 224]) " and g(x): 2b both have radius of ° k Let h(x) = Z ak x k x k=0 k=0 n convergence 1, let sn = Z ak, tn = 2 bk’ sn i 0, tn i 0 for k=0 k=0 each n, and suppose 2 sn and 2 tn are both divergent, and SD = o(tn). Then h(x) = o(g(x)) as x -+ 1’. a Proof of Theorem 5.4. If J = 1, 1et h(x) = 2 klaklxk, w k ka-l n k=0 1) g(x) = 2 b x where b = ; s = 2 kla | and t = 2 b , k=0 k k T(cz) n k=1 k n k=1 k by hypothesis, SD = o(tn) as n -+ 0. Applying Lemma 5.1, we obtain h(x) = o(g(x)) as x -+ 1‘ . Since tn ~ n“ as n -9 w, it follows from Theorem 5.0 that g(lzl) = o[-—-l——a-] as [2) -) 1". This implies that (1-IZI) lf'(2)l . 11021) = o) = o[—-—1—;] as IZI -+ 1". (1-lzl) thus f e 8:. For J ) 1 we have f(J)(z) = o[ 1 a] as |z| -+ 1’, and (l-lzl) the results follow by repeating the above procedure. Corollary 5.5. ([Ne, Thm. 3(1), p. 15]) Let f(z) = k ”MB 9 r N H h” 0 there exists a fixed integer J, J i 1 such that n J J 2 k lakl = o(n ) then f 6 Bo . k=1 60 Proof: The result follows from Theorem 5.4, by setting a = 1. k3“' n We obtain f e B; = 80, since 2 ~ nJ as n -+ a. k-l O Theora 5.5. Let f(z) = 2 a 21‘ be in 8:. If there isa G k=0 n such that O ‘ arg ak ‘ 9 + "/2 for each R. Then 2‘ klakl = o(n“) k=l k as n-9w. We need a lemma for the proof of Theorem 5.5. This Lemma is an immediate consequence of ([BST, Thm 1.4 and Thm l.10a]). Lemma 5.2. [BST] If g(x) = 2 ck xk = o[ 1 a] as x -9 1’, k=0 (l-x) n as x -+ 1’, where ck i 0 then 2 ok = 0(n“) as n -+ a. k=0 Proof of Theorem 5.5. There is no loss of generality in assuming that 9 = 0. Since f(z) e B: , we have 2 k a lf‘(2)l = o((l-Izl>‘“> k=1 .k-1 l k and = (f'(Z)) = o((l—|z|)-a) as |z| -+ 1- . Therefore 2 k Re(ak)zk" = o((1-|z|)'“) as [2) —+ 1- where o ‘ Re ak. Similarly 2 k 1n(ok)z1H = o((1—|z|)’“) as |z| -+ 1- where 0 ‘ Im(ak). Thus 61 2 klaklzk“ = o((l—Izl>‘“> . k=1 In applying the lemma above, put ck_, = klakl. We obtain g(|z|) = n o((l-lzl)‘“) as |z| -+ 1’. By Lemma 5.2, we have 2 klakl = 0(n“) k=l as n -+ a. This completes the proof. Theorem 5.6. If f(z) = 2 ak 2k is in B: then for any fixed k=0 1 n p+a~§ integer p = 1,2,..., 2 kplakl = o n as n -9 0. k=1 0 Proof of Theorem 5.6. Let g(z) = zf'(z) = 2 n an 2“. By a n=l result which is an extension of Hardy’s result ([Harl p. 45]), we have 1 p k -(p+a—§) - 2 k lakl IZI = 0 (1"IZI) as IZI "’ 1 - But n n n k ° k lzl 2 kplakl e 2 kplakl |z| ‘ 2 kplakl lzl for all z, |z| < 1 k=1 k=1 k=1 1 - a for each n = 1,2,3,... . Taking |z| = e , -l4v+a-% p..-1 kplakl = o[[l — e "I I = oln 2] as n -+ w; n e”1 2 k=l 50‘ since -l i l — e 4 2n the proof of Theorem 5. for n sufficiently large. This completes 62 55.4 Conditions on blocks of coefficiegts in the class ,gf Let f(z) = E ak zk. In this section we shall give several results concernitg1"blocks" of coefficients for a-Bloch and little a— Bloch functions. We will consider the portion of the power series for f(z) between R = m and k = 6m, where 6 = the gauss integer of a + 1. The integer "6" of the upper bound is not critical to the following analysis; simple revisions will accomodate any number exceeding 1. 6m Define flm = kin lakl. We will first show that for functions whose power series have real nonnegative coefficients, we can characterize Ba and B: by conditions on flm. 0 Theorem 5.7. Suppose that f(z) = 2 ak 2k, with ak 6 0. Then k=l f e B“ if and only if 3m 6 constant m““ for each m = 1,2,3,... Proof of Theorem 5.7: If f c B“, applying Theorem 5.0, we have 6m 6m 6m a a m 3 = m X a 6 2 k a 6 2 k a 6 constant(6m) 6 constant m . m _ k _ k _ k k-m k-m k—l If pm 6 constant ° 111""'1 for each m = 1,2,3,... . Fix n, and suppose that p is an integer such that 6p 6 n < 6p+1. To show n f e B“, by Theorem 5.1, that it suffices to show that Z k ak = k=l 0(n“). 63 n o 5’ 5P+* 2 k < 2 k a + 2 k + -°° + 2 k a k=1 8* k=1 k a 8“ op k 6 2 62 p+16p+‘ <5 )3 a +6 2 +-~+o 2 k=1 k k=6 8“ op 8* ‘ 6 la-i + 62 . da-i + ... + 6p+1 . 6p(a—1) + .— ‘ {6 + 62 + ... + 6P 1}6p(a 1) + _ 6 constant 6p 2 . na 1 6 constant na . This completes the proof of Theorem 5.7. Theorem 5.8. Suppose f(z) = E ak zk with ak 6 O for each R. k=l Let a i 1. If f e B: then flm = o(m“") as m -+ o. Proof of Theorem 5.8. If f e a: 1 m 6m 1 6m 1 a a—i 3 flm = 3; 2 Elk 6 3; I k 8k = 3; 0((6m) ) = 0((6m) ). m k-l 6m J-i J+a~2 Theorem 5.9. If 2 k lakl s M* m , for each positive k=m integer J, m and for a positive constant M*, then f(z) = {D X 8k ZR 6 Ba Proof of Theorem 5.9. By Theorem 5.1, it suffices to show that n 2 leakl = 0(nJ+a-1) as n -9 w. If 65 6 n < 69+‘, then R l 64 n J 5 J ‘2 J ‘8“ J 2 k lakl ‘ 2 k lakl + 2 k lakl + --' + 2 k lakl k=l k=1 6 65 6 32 63+! ‘ 6 2 RJ 1Iakl + 62 2 RJ llakl + .0. + 63+: 2 kJ llakl 1 6 53 ‘ 6M* + 62 M*(6J+a-2) + ... + 68+1 M*(6S(J+“—2)) (by hypothesis) 6M,[1 + 6J+a—l + 62(J+a-l) + ... _ 6s(J+a—l)] J+a-l 3+1 _ 8+1 : 6M*[(o ) - 1] ‘ 6M*(6(J+a 1)) 6J+a~l _ 1 6J+a-l _ 1 6 26M*(65)J+a_1 = 0(nJ+“_1) . This completes the proof of Theorem 5.9. Theorem 5.10. Let f(z) = 2 ak zk be in B“. If there is a 8 k=0 n such that 9 < arg ak 6 0 + 3 then 2 kla k=l = 0(n“) as n -9 w. 2’ k' Proof of Theorem 5.10. The result follows from repeating the procedure of the proof of Theorem 5.5, and applying ([BST, Theorem l.lO(a)]) putting V(n) = n“. N Theorem 5.11. Let f(z) = 2 ak zk be in B“. If there is an k=0 9 such that O 6 arg ak 6 0 + n/2 for each R, then there is a (6+l)m positive constant 0* such that 2 [akl 6 C*(ma-‘). k=6m 65 Proof of Theorem 5.11. It follows from Theorem 5.10 that n 2 klakl = 0(n“) as n -9 a. Then k=l (6+1)!!! (5+1)!!! (1 a 6m 2 lakl 6 2 klakl 6 constant((6+1)m) 6 constant m 6m 6m (6+1)m and thus 2 [akl 6 0* 111""1 where C* is a positive constant. k=6m Theorem 5.12. If f(z) = 2 an zn is an a—Bloch function. Then “:0 n p+a-% for any integer p h 0, we have 2 kplakl 6 C n . k 1 Proof of Theorem 5.12. Hardy has shown [Hal] that if g(z) = E on zn is analytic in D and g(z) = 0[--l--;], c > 0 then for k=0 (l-lzl) —o ( 7 < a + % a —7 n l 4E=1n 'CDI '2' - oll(l-|z|7—7+1/2]J. Setting f'(z) = zg(z), 7 = -p+l. We obtain Q l 2nficlufl=0[ _ ]. n=1 “ <1—Iz|>P*“ 1/2 Estimating 1 +“_1/2 , we have (1-12l)p a a p+a~ - n 2 np lanl |z|n 6 C 2 n |z| n=l n=l Titchmarsh has shown [Ti, p. 224] that the above inequality implies n that 2 kplakl 6 C np+“—1/2, for some positive constant C. k=l . 66 Theorem 5.13. If f(z) = 2 ak 2k 6 a“ then k=0 (GEDmkp—i lnp+a¢—3/2 lakl 6 constant for each positive integer P k=6m Proof of Theorem 5.13. It follows from Theorem 5.12 that (6+l)m _ (6+l)m (6+1)m 6m 2 RP llakl ‘ 2 kplakl ‘ 2 kplakl ‘ 6m 6m 1 (6+l)m 6 constant ((¢S+l)m)p+°“”1/2 => 2 kp-llakl = 0(mp+a—3/2) . 6m BIBLIOGRAPHY [ACP] [Bel] [Be2] [Bil] [Bi2] [BST] [DRS] [DU] [Gn] [GP] [Hal] [Harl] BIBLIOGRPAHY J.M. Anderson, J. Clunie, Ch. Pommerenke, "On Bloch functions and Normal functions", J. Reine Angew Math. 270 (1974), 12-37. E. Beller, "Zeros of AP functions and related classes of Analytic functions", Israel J. Math. Vol. 22, Nos. 1, 1975, 68-80. E. 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