This is to certify that the thesis entitled COEFFICIENTS OF BLOCH FUNCTIONS \ I presented by John Joseph Neitzke has been accepted towards fulfillment of the requirements for Ph.D. d . Mathematics egree 1n Major professor Date J01; 30, #790 0-7639 LIBRARY Michigan Sum University ,, I I ' i Q . OVERDUE FINES: 4-5“ “3.; 25¢ per day per its ‘ ( “(Wklj narumm LIBRARY MATERIALS: Place in book return to remove charge fro- circulation records “5H3; (f ’1'}; 1 {auflhvflrm- ; b" COEFFICIENTS OF BLOCH FUNCTIONS By John Joseph Neitzke A DISSERTATION Submitted to Michigan State University, in partial fuifillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1980 ABSTRACT COEFFICIENTS 0F BLOCH FUNCTIONS By John Joseph Neitzke A Bloch function is a function analytic in the open unit disc D which satisfies the restrictive growth condition that (1-|z|2)|f'(z)| is bounded in D. The space of all Bloch functions is denoted by B. The subspace of 8 consisting of those functions for which (1-lzl2)|f'(z)| tends to zero as |z| + 1 is called 80. The Bloch norm is defined by Hf|h3= |f(0)| + 228(1- |z| 2)|f' (z)|. With this norm, the Bloch functions form a Banach space. .In this dissertation, the coefficients of the power series expansion E akzk of functions f in B and 30 are examined. Conditions on the coefficients which imply that f is in B or in B0 are given, as well as other conditions which are necessary for f to be in B or in 80. Certain facts concerning coefficients of Bloch functions are known. If f is Bloch, the coefficients are bounded but need not tend to zero. If f is in 80, the coefficients ndo converge to zero. It can be shown that for Bloch functions, kzl klakl =0(n) and Z Iakl2 is bounded. A natural place to search-for conditions shich characterize Bloch functions is among expressions similar to these. ’ n Mathews [1] proved that Z kJ |ak| = 0(nJ) for some integer k=1 3/2) n J 3_1 is sufficient, and Z klakl = 0(n is necessary for f to k=1 John Joseph Neitzke be a Bloch function, and that the first is also necessary if the coefficients have arguments lying in an interval of length n/Z. Mathews also gave an example of a Bloch function satisfying E klakl = 0(np), for P between 1 and 3/2. k-l In this dissertation, we extend one of Mathews' results, that E klakl2 being bounded is sufficient for f to be Bloch, to accom- Uate a wider class of functions. We then state and prove analogues of Mathew's theorems for functions in 80. We show that if for some J 3_1, E leakl = 0(nJ), then f is in 80, and if the arguments of the cgefficients lie in an interval of length n/Z, this condition is also necessary. We use this to prove that a Hadamard gap series with coefficients tending to zero is in 80. We also show that 3/2) n X klakl = o(n is necessary for f to be in 80. k=1 We then examine conditions involving sums from m to 2m. If 2m all the ak are real and nonnegative, then 2 lakl g_M < a is k=m necessary and sufficient for f to be Bloch, and the convergence of these partial sums to zero is necessary and sufficient for f to be in 80. Without the real nonnegative restriction, we obtain a series of results which parallel those for sums taken from 1 to n. We prove that f is Bloch if 2%“ kJ'llakl = 0(m‘J'1) for some integer J 3_1; with the arguments of the coefficients restricted to an interval of length n/Z, this is also necessary. Without this restriction, P-l/Z). 2m a necessary condition is that 2 kp'llakl = 0(m For functions k=m in 30, these results hold with o in place of 0. We give an example of a Bloch function which is bounded with coefficients tending to zero, with the property that if the part of the power series expansion of f from k = m to k = 2m is John Joseph Neitzke considered as a function, then for selected m, the sum of the coefficients and the Bloch norm of the function both diverge to infinity as m increases. 1. J. H. Mathews, Coefficients of uniformly normal-Bloch functions, Yokahama Math. J. 25 (19735,_§7—31. ACKNOWLEDGMENTS The author wishes to express appreciation to those who have assisted in the writing of this dissertation, especially Dr. Peter Lappan, his major professor. Thanks are also expressed to Dr. Stephen Dragosh and Dr. Thomas McCoy for their careful reading of the manuscript. ii II. III. IV. VI. TABLE OF CONTENTS PRELIMINARIES ..................... AN EXTENSION OF A THEOREM OF MATHEWS ......... CONDITIONS ON COEFFICIENTS FOR THE CLASS 80 ...... CONDITIONS ON BLOCKS OF COEFFICIENTS ......... A BOUNDED BLOCH FUNCTION OF WHICH SECTIONS ARE . . . . ARBITRARILY LARGE OPEN PROBLEMS ..................... BIBLIOGRAPHY ..................... 11 14 24 32 37 39 I. PRELIMINARIES The beginnings of Bloch functions date back to 1925, when André Bloch [6] studied the class G of functions f holomorphic in the unit disc D = {z : [2] < 1} with normalization f'(0) = 1. A schlicht disc in the Riemann image surface W of f is an open disc P in W such that there is an open connected set A in D which f maps one-to-one onto P. Let df(z) denote the radius of the largest schlicht disc in W, centered at f(z). Set r(f) = sup df(z) 260 and b = inf {r(f) : f c G}. Bloch proved that b is positive. Since then, the work of Bloch has prompted results of various types. Bounds for b were calculated. Other mathematicians generalized Bloch's result to Rn and C". Another group considered the function-theoretic implications for the unit disc. More recently, a Banach space of functions related to those examined by Bloch, called the Bloch functions, has been studied [10]. The first activity following Bloch's paper was in the search for bounds for b. Landau first proved that b > .396 [15]. Grunsky and Ahlfors proved b<:.472 [2]. Ahlfors subsequently proved b >%/§> .433 [1]. Generalizations of Bloch's result to Rn and Cn are found in the papers of Bochner [7], Takahashi [25] and Sakaguchi [23]. Recently, applications have been found for Bloch functions 1 2 in the study of functions of bounded mean oscillation (BMOA) and vanishing mean oscillation (VMOA). Since much of the interest in Bloch functions has stemmed from their position as a subset of the space of normal functions, we will state the definition of normal function to show how it relates to the definition of Bloch function. A function f(z) meromorphic in the unit disc D is called normal if the family F = {f(S(z))}Ses is a normal family in the sense of Monte]; that is, every sequence of elements of F has a subsequence which converges uniformly on compact subsets of D, either to a function meromorphic in D or identically to infinity. Here, 3 is the set of all conformal self-mappings of D, S = {5(2) = u jail-: |a| < 1, [pl = 1}. 1+az The following characterization is equivalent and could be used as a definition: A function f(z) meromorphic in D is normal if and only if there is a finite constant M, depending only on f(z), such that (1) sup (1-|z|2) M7 5. M [18]. 1+|f(z)| There are several alternative characterizations of Bloch func- tions. We shall use only the second two below; we include the others for reference. 1. A function f analytic in the unit disc D is a Bloch function if and only if it is finitely normal; that is, if the family F = {f(S(z)) - f(S(0))}SCS forms a normal family where the constant infinity is not permitted as a limit [3]. 3 2. Equivalently, a function f analytic in D is a Bloch function if and only if there is a finite constant M such that (2) sup (1-|z|2)|f'(z)|.5_M [3]. One reason why (1-|z|2)|f'(z)| is used in (2) rather than, say, (1-|z|)|f'(z)|, is that it is an "invariant" form of the derivative: with S(,z) = 3::2- , where |a| < 1, the derivative 1+az of f(S(z)) is fl ( 3+2 ) 1'53 1+‘5‘z (152)? with z = 0, this reduces to (1-lal2)|f'(a)|. 3. The preceding condition (2) may be replaced by (3) lf'(z)| = 0((1-|z|)'1) as lzl + 1 [19]. 4. Defining df(z) as previously, f(z) is a Bloch function if and only if sup df(z) < w [3]. 5. f(z) is a Bloch function if and only if its indefinite integral satisfies a certain smoothness condition: Let ¢(z) = 2"1 J; f(w) dw = RED ak(k+1)'lzk. Then f(z) is a Bloch function if and only if ¢(z) is continuous in |z| §_1 and (4) lmax |¢(eitz) + ¢(e“'tz) - 2¢(z)| = om as t + o [3]. 2 <1 6. Let G denote the complex linear space of all functions 8 'bnz" analytic in D for which 0 HQIU;= [9(0ll +257J0 I0 Ig'(re )| dr d6 < m. 9(2) = n "M 4 Then the space of Bloch functions is the dual of G [3]. Denote the space of all Bloch functions by 3. With the Bloch norm, defined by (5) Ilf||3= |wa + sup(1-|2|2)If'(z)l. ZeD B is a Banach space. It is clear from a comparison of the defini- tions that the Bloch functions are contained in the space of normal functions; however, since the sum of two normal functions need not be normal [17], the normal functions are not even a linear space. In the paper of Pommerenke [21] in which the Bloch functions were defined, the definitions above included the additional re- striction f(0) = 0. Anderson, Clunie and Pommerenke dropped this assumption; the changes this entails are minor [3]. Let B0 denote the subspace of 8 consisting of all f(z) for which (5) (1-|z|2)|f'(z)| + o as |z| + 1. As a subspace of B, B0 is separable, (strongly) closed, nowhere dense, and is the closure of the polynomials in the Bloch norm [3]. We have the following alternative characterizations of 80: f(z) is in 30 if and only if |f'(z)| = o((1-|z|)'1) as |z| + 1. Equivalently, f(z) is in 30 if and only if df(z) has limit 0 as |z| + 1. Another characterization is that f(z)e:B0 if and only if the maximum in (4) is equal to o(t) as t + 1 [3]. More recently, Bloch functions have been studied in connection with BMOA and VMOA functions. BMOA, or "bounded mean oscillation" denotes the space of functions f e H2 for which 5 2 2 1 2 1- w ”in =—[ |f(z)| —L—1— ld2| W 2 2‘" 30 W IZ'WIZ is bounded for w in D, where we set = .EEIL. - fw(s) f(1+fig) f(w) (s e o, w e o). VMOA, or vanishing mean oscillation, denotes the subspace of functions with llfwllz-r 0 as |w| + 1. It can be shown that BMOAC B and VMOAC'B0 [22]. Campbell, Cima and Stephenson [8] gave an example of a Bloch function which is in all HP classes, but not in BMOA. As an analytic function in D, a Bloch function f(z) has a no power series expansion f(z) = Z anzn . The coefficients {an} n=0 are known to satisfy several conditions. If f(z) e B, then the coefficients are bounded but need not tend to zero. Hayman [14] raised the question of finding further conditions which imply an + 0. This was answered in part by Pommerenke [21]: If the Bloch function f(z) has radial limits almost everywhere, then an-+ 0 as n +~mu Pommerenke [21] gave an example of a Bloch function with radial limits almost nowhere. Timoney [26] gave an example of a Bloch function which has radial limits almost everywhere, but which is not in any Hp space for 0 < p §_m. Anderson, Clunie and Pommerenke [3] gave two results on power series where the modului a of the coefficients are given but the n arguments a are independent random variables with the uniform n distribution on [0, 2n]. Set f(z,a) = n20 anei(“n) 2". As a first result, suppose that 6 1/2 { E k2(a )2} = o(ntiog(n)>'1’2). k=1 k Then f(z,a) is a Bloch function with probability 1. In particular, if 2 an log(n) < m, then f(z,a) is in 80 with probability 1. n=1 In the other direction, given a decreasing sequence {pn} of positive numbers, where pn +~0 as n +»w, there exists a sequence {an} with a > O and Z an2 pn log(n) < m such that with proba- n=1 n bility 1, f(z,a) is not Bloch. There is also the following characterization of the Bloch func- tions by a quadratic form: Let f(z) = X anzn be a Bloch function. n=1 Then 2 on 00 a °° IW u+v+1 v (6) ugo vgo UTV;T_- w" WV 'E' K v20 V+1 , where the wv are complex numbers for v = 1, 2, ... , and K = 2|[fH3., The double sum converges if the right-hand sum does. Conversely, (6) implies f is a Bloch function and IIflIB 12K [3]. Let f(z) = Z anzn be a Bloch function. By (2), there is a n=0 number M < m such that |f'(z)|§_M/(1 - lzlz) for all z in D. Then [3] p 2n . p 2n (7) I I |f'(re1e)|2 do r dr '1_I J M2 de r dr. oo 00—7(1_r)2 Examining first the right side of (7), we find that it is equal to 2 JP ———7—”2 d TI r T' =- 0 (1 - r )2 7 p n MZ/(l - r2) I o «an/u- ph. For the left side of (7), Parseval's formula gives 2n 2n k; k2 laklzr 2“ 2 =( |f'(reie)|2 de. 0 Thus, the left side of (7) is p w _ J 2n 2 kzlakl2 er 2 r dr 0 k=1 00 p - 2n 2 kzlakl2 I r2k 1 dr k=1 0 2nk21k2|ak|2p2kl(2k) , z klak [2p 2“ [3]. k=1 - Thus, (7) reduces toco 2 p2k kil klakiP '< M2 9 2/(1- -p 2) or equivalently I klakl2 p2(k'1)(1-p2) :.M2- k=1 Setting p2 = 1 - 1/n, we obtain kEI klaki2(1 - 1/n)k-1 n-1 < M2 Since the partial sums of this series are increasing for fixed n, and (1 - 1/n)"'1 5_(1 - 1/h)"'1 for k = 1, 2, ... , n, 8 n n (1 - 1/n)"'1 n"1 A klaklz 5 kzl klaklzfl - 1/n)"’1 n'l. We have n (a) (1 - 1/n)"‘1 n'1 z klaklz 3M2. k=1 Since (1 - 1/n)""'1 approaches e'1 from above as n +~w, (8) yields n e"1 n"1 X klakl2 §_M2 k=1 or (9) E kla |2 < neM2 k=1 k “ Since it is also true that 2m 2 2m 2 kgm klakl 5_ 2m kgmlakl , and all of the terms in the sum in (9) are positive, we also have (10) 22'" laklz 3 2eM2. k=m We note that the last inequality is independent of m. The two inequalities (9) and (10) are conditions on the coefficients of a function which are necessary for that function to be a Bloch function. A natural place to search for sufficient condi- tions is among inequalities which are similar to but weaker than these two. It is also of interest to see how far these inequalities can be weakened and still remain necessary. Mathews [19] gave conditions for which, as in (9), the summa- tion is from 1 to n. He proved that 9 n (11) Z klakl= 0(n) as n +-m k=1 is a sufficient condition, and " 3/2 (12) X kla l = 0(n ) as n +'m k=1 k is a necessary condition for f(z) to be a Bloch function, and if the coefficients are restricted to lie in one sector of the plane with central angle n/Z, (11) is also necessary. Theorem 1 of Chapter II is a consequence of Mathews result (11). Using the Holder inequality, we establish that for 1 < q < m, 2 kq-llaqu < m is a sufficient condition for f(z) to be a Bloch function. Two examples follow, illustrating the use of the theorem. In Chapter III, we prove a series of theorems which are the analogues of Mathews' results for functions in 80. The first theorem of the chapter shows that if for some integer J 3_1, n (13) z kJIakI = o(nJ), k=1 then f(z) is in 80. Two consequences of this are proved. One is the analogue of the theorem of Chapter II, the other is that a Hadamard gap series with coefficients approaching zero as n +.m. is in 80. Theorem 5 then shows that if the coefficients of f(z) are constrained to lie in one sector of the plane with central angle n/Z, then (13) is also necessary. An example demonstrates that there are functions in 30 for which 10 n X klakl = o(np) k=1 for 1 < p < 3/2. The last theorem of the chapter shows n z klakl = ocn3’2) k=1 is a necessary condition, where the arguments of the ak's are not restricted. In Chapter IV, motivated by (10), we examine conditions on "blocks" of coefficients, looking at sums taken from m to 2m. First we prove that if all of the coefficients are real and nonnegative, 2m then 2 ak 5_ K < m is a necessary and sufficient condition for k=m 2m f(z) to be Bloch, and Z ak -+ 0 is a necessary and sufficient k=m condition for f(z) to be in 80. without the real nonnegative restriction on the coefficients, we obtain a series of theorems which parallel those of Mathews and of Chapter III. The change made from those theorems is, in general, a substitution of the limits m to 2m for the limits 1 to n on the finite sums, and removal of a factor of n from the bounds on those sums. In Chapter V, we give a single example of a bounded Bloch func- tion, with coefficients summing to 1, for which the behavior of sections of the power series from m to 2m is bad for selected values of m: Considering this section of the power series as deter- mining a function, as m increases, the sum of the coefficients and the Bloch norm both diverge to infinity. The last chapter contains a short list of open problems. II. AN EXTENSION OF A THEOREM 0F MATHEWS Mathews [19] proved that a sufficient condition for f(z) = akz to be a Bloch function is: If there exists a fixed constant C, C < m, such that Z klakl2 5_ C. k=1 The method of proof in Mathews' paper relies on the Cauchy - Schwarz inequality. By using the Holder inequality, we can establish the following: Theorem 1: If X kq-llaqu i C , where 1< q < co and k=1 C is a positive constant, then f(z) is a Bloch function. Proof: Set p = q/(q - 1). Then 1 < p < m and p + q = pq. By using the Holder inequality, we can show that "M: H klakl , E k(p-1)/p k1/p lakl k=1 1/p n 1/q (k(P'1”P))p} - {kgl (kl/P)q Iaqu} E1 kp'1}1/p - { E kq'1 laqu}1/q [A A P’hfi 7V p—I IA u on O O A 3 3 v O O 11 12 for a suitable constant B. By [19, Theorem 2(i)], f(z) is a Bloch function. we remark that if q = 2, the preceding theorem is exactly the theorem of Mathews quoted earlier. Next, we will give two examples and use the preceding theorem to show that each is indeed a Bloch function. The first uses a value of q near 1, the second employs a large q. Example 1: Let 8 > 0. Then kE1 k-e/(1+e)(k1/k _ 1)k/(1+e) 2k is a Bloch function. Proof: Set q = 1 + e and ak = k(1/q)"1(k1/k - 1)k/q. Then “ -1 kzl kq laqu kE1 kq-1(k(1/q)-1(k1/k _ 1)k/q)q kxl kQ‘l kl'q (kl/k _ 1)k oo 2 (kl/k _ 1)k. The last sum is bounded [20, p.219], so by the preceding theorem, 2 k(1/Q)-1(k1/k _ 1)k/q zk k=1 is a Bloch function. Since 1/q - 1 = -e/(1 + a), this is exactly what we wished to prove. 13 Example 2: For any a > 0, kzl (k(Iog(k))€)'1 2k is a Bloch function. Proof: We have kEI kq‘l ()€)'1>q = E (k<1ogck>>q8>'l. k=1 This converges for q > 1/e [20, p.210], and by Theorem 1, we are done. III. CONDITIONS ON COEFFICIENTS FOR THE CLASS 30 In this chapter, we will adapt several results from Mathews' paper, stated for Bloch functions, to functions in 80. In the theorem following, we will give a proof, using Cauchy's formula, of the known fact [3] that the coefficients of a function in 80 tend to zero. Theorem 2: Let f(z) be a function in 80, f(z) = kio akzk. Then ak + 0. Proof: From Cauchy's formula, f' z 1 lakL = §n1E I dz l |z|=r 2 2n ie . . 1 f' re . TB 5. EEE'IO ‘;£;iial"‘e d9 1 ( -1 1-k 2"| -i(k-1)el ggfi' 0 (l-r) )r 0 e de k'1 o((l-r)'1) r1"k = o(k'l rl'k (1-r)'1). The minimum value of the last term occurs for r 1 - k-l. Evaluating for this r, we obtain 1?; <1 - %)1"‘ <-,1;>'1 which tends to e'1 as k +-m. Thus ak + 0. In Theorem 3, we will establish conditions sufficient for a holo- morphic function to be in the class 30. We will then investigate necessary conditions. 14 15 Theorem 3: We have the following sufficient conditions: (i). Let f(z) = kE akzk. If there exists a fixed integer J, =0 J 3_1, such that n X leakl = 0(nJ)a k=1 then f(z) is in Bo. (ii)- Let f(Z) = Z akzk. If there exists a p, 1 < p < m, k=0 and a w, w < p - 1, such that (14) E k(p-W)/(p-1) la lp/(p-l) = o(n(p-w-1)/(p-l)), k=1 k then f(z) is in 80. (iii). Let a > 1, and let {kj} be a sequence for which kj > akj_1. If akj + 0, then on k3. .2 akj (z) e 30 J=0 We shall need the following lemmas, which are extensions of a theorem of Titchmarsh [28, 9.224]. Lemma 1: Let h(x) = Z akxk and g(x) = X bkxk, where k=O k=O ak 3_0, bk 3_0, and both series converge for |x| < 1 and diverge for |x|_3 1. If ak = o(bk) as k +1w, then h(x) = o(g(x)) as x + 1. Proof: Given a > 0, there exists an integer N such that 8 an < 2 bn_ for n 3_N. Then ” k h(x) = Z a x = k=0 k II M O! X’ X ' + x ll 2M + H n! 3' X IA N k e X a x + —- X b x k=0 k 2 k=N+1 k N k e kzo akx + 2- g(x). IA Let N be fixed. Since g(x) +-w as xi+ 1, we may choose 6 > 0 k N so that kXO akx < §-9(x) for x > 1 - 6. Thus h(x) < eg(x) for x > 1 - 6, or h(x)/g(x) < e for x > 1 -6 . Since 6 is arbitrary, h(x) = o(g(x)) as x + 1. Lemma 2: Let h(x) = Z a xk and g(x) = X b xk both '---- _ k _ k k-O k-O n converge for |x| < 1 and diverge for -|X|.Z 1, and set 5n = Z ak k=O n and tn = Z bk' Suppose that 5n and tn are positive, is and k=0 . " {tn are divergent, and 5n = o(tn). Then h(x) = o(g(x)). k n Proof: We have h(x) = (1 - x) X skx and k=O n g(x) = (1 - x) Z thk. By the preceding lemma, k=0 "£0 snxn = o(ngo tnxn), and the result is immediate. We remark that if, in particular, 5n = o(n), then h(x) = o((1 - x)‘1). We will now prove Theorem 3. n Proof’of (i1: Set 5 = Z k J " k=1 J ak and tn = n . For J = 1, 5n = o(n), so by the remark above, with h(x) = kzl klaklxk and x = [z], |f'(z)| = o((1 - Iz|)'1). For J > 1, we can show 17 f‘d’ = o((l - IZIJ)'1). Successive integration then completes the proof of part (i) Proof of (ii): Suppose 1< p < co, p + q = pq, and w < p - 1. Applying the Holder inequality, 5 I I k a k=1 k E kw/p k(p-wllp Ia l k=1 k < ( E (kW/P)p)1/P .( E (k(P'W)/P)q la IQ)1/q '_ k=1 k=1 k . (0(nw+1))1/p .( E k((p-w)/p)-(p/(p-1))Iaklp//)(P-1)/P 1 = 0(n(W+1)/P) . (o(n(P'W’1)/(P'1)))(P‘1)/P = 0(n(w+1)/p) . o("(p-w-lflp) = o("(w+1)/p + (p-w-1)/p) = o(n) An application of the first part of this theorem now completes the proof of part (ii). We note that if p = 2 and w = 0, then f(z) e 80 if n 2 k2 lakl2 = o(n). k=1 1 Proof of(i11): Since ikj'l <:kj, we have kJ._1 ('3 kj, so k3. - kj_1> kj - E k). = (1 - ;)kj. We also note that, writing 1((J) instead of kj to make the formulas more readable, 18 (1%) I<(1)Izl"‘(“')'1 < (k(j) - k(j-1)>IZIk(j)’1 < lzlk(j'1) + |z|k(j'1)+1 + ... + lzlk(j)-1- Let s > 0, and let J be such that for j 3_J, ak(j) < g-(l - gi- Then J-l jgl k(j) ak(j) < zk(j)-1 km-I; + 1 km la -l|2I 3-:0 H1) The second sum is bounded by 1 °° . k ' -1 gm '3) 3-1, km |z| (3’ < E-(1 - l) 1 |z|t ‘2- .“ t=kIJ-1) 1-51; : g. X IZIt t=k(J-I) <§ 1 , l-IZI Since the first sum is a polynomial of degree k(J-1)—1, it is bounded, and hence is less than 5- 1 21-121 if |z| is sufficiently near 1. Thus, 1 |f'(z)| < e l-lzl for |z| near 1. Since a is arbitrary, |f'(z)| = o< 1| I) ' 1 - 2 as |z| + 1. We note that the functions in (iii) are called Hadamard gap series. If the coefficients converge to zero, the series is in 80. 19 Next we consider necessary conditions on the coefficients. The first theorem in this direction shows when the sufficient condition of Theorem 3, with an additional condition on the arguments of the coefficients, is also necessary. Theorem 4: Let f(z) = Z akzk be in 80. If there is some k=0 a such that for each k, a §_arg(ak) §_a + n/Z, then n 2 klakl = o(n). k=1 Proof: There is no loss of generality in assuming that a = 0. Since f(z) is in 30’ we have “4 mu-Inrh |f'(2)l Z k a z k=0 k and IPGH aux-uni» Therefore, N ll kEI k Re(ak) k'l o((l - |2|)'1). where 0 §_Re(ak). Similarly, N ll gIkap “1 uu-Idrh. where 0 §_Im(ak). Thus, =2uu Inrh=ou1-un4x z klaklzk'l k=1 k By [12, Theorem 96], if ck 3_0, g(x) = kZO ckx and g(x) = o((l - x)'1) as x + 1, then kZO ck = 0(n). Here, we set 20 Ck-l = klakl and note that "MS W 9’ W H °° k s. 2 kla I I2! . k=1 " Hence Z klakl = o(n). Corollary: If F(z) = Z |ak|zk is in 80, then so is k=0 f(z) = Z akzk. k=O Example 3: Let 1 < p < 3/2. There exists a function f(z) n in 80 for which 2 k ak = o(np). Let 6 > 0. Then 1 k= m . 2p-2 ' f(2) = z kp-2-6 elk Zk k=1 is the required function. 00 - . a Proof: Hardy [13] has shown that if F(z) = Z k b e1k zk ' k=1 where O < 0, and 1 . IF(Z)I '1 0((1_ |z|)1-b-a/2) . Set a = 2p - 2, and note that 1 < p < 3/2, so 0 < a < 1. Then with b = 1 - p + 6, m . 2p-2 F(z) = Z kp'l'6 e1k zk k=1 and 1 ‘ 0((1 - IzI)1‘5') _ 21 = o((1 - IzI)‘1). Setting F(z) = zf'(z), it follows that f(z) is in 30. The coefficients of this function are . 2p-2 = kp-2-6 e1 k ak SD n n kzl kIakl = kzl kp 1'6 = 0(np'5) = o(np). In the proofs of the following lemma and of Theorem 5, we will use the following fact: It follows from an exercise in Titchmarsh [28, D-242] that if a > 1, and the ak are real, ak 3_O, then for _ m k _ -a f(z) - Z ak z - (1 - z)- , k=1 a is asymptotic to k0"1 F(a) and f(z) is asymptotic to k Na.) 2 ka'lzk k=0 as |Z| + 1. For the proof of the next theorem, we will need the following lemma, which is adapted from [11, p.45]. 8 a 2n is analytic in D and Lemma 3: If g(z) = n "M O n g(z) = o((1 - lzl)'a), where a > 0, then for 3» < y . 22 N _ k Proof: Assume that y < 1/2. Set 5 = Z k Yla Ir . By the —————- N k=1 k Cauchy - Schwarz inequality, N _ 1/2 N 1/2 sNg{zckh§ -{z unnhfi k=1 k=1 _ 2 . 1/2 (0(n 21*1))1’2' {ii' [0",g(rele)|2 d9} IA 1/2 IA 0(N'Yfl/2) - {-2-}; [ZNIOHI - r)'°‘)|2 d9} = o<~‘Y+1/2) - o((1 - r)'a). Thus (1 - r) 2 5k rk k=1 G(r) = (1 - r) o((1 - r)'a)ok§ 0(k'YI’1/2) rk =1 o((1 - r>1'“) . k2 k'**1’2 r“ =1 )1'a) . 2 k3/2 'Y-l rk k=1 = o((1 r = 0111 - r)1'“) 0((1 - r)"3/2'Y’) = o((1 - r>1'“) o<<1 - r>Y'3/2) = o((1 - rlY'“'1/2>. For v.3 1/2, the result follows from the case when y < 1/2 by successive_integration. We now consider conditions on the coefficients which are necessary for a function f(z) to be in 80. 23 Theorem 5: Suppose that f(z) = Z ak zk is in 80. Then k=1 for each fixed integer P 3_1, we have n P P+1/2 kzl k lakl = o(n ). 3399:: Applying Lemma 3 with zf'(z) = 9(2), a = 1, y = -P + 1, we obtain ” - + 2 kPIakI 121“ = o((1-IZI)(P1’2)). k=1 Since the coefficients are all real and nonnegative, n n n P k k IZI" ZklaI: 2 la! I2! 1 ZklaVIIZI k=1 k k=1 k k=1 k for all |z| and n. Taking |z| = e'lln, and noting that 1.- e'l/n is asymptotic to l/n as n increases without bound, n , e'l kzl kplakl = 0((1 - e'l/n)'(P+1/2)) = o(nP+1/2). IV. CONDITIONS ON BLOCKS OF COEFFICIENTS Let f(z) = Z ak zk. In this chapter we shall give several k=1 results concerning "blocks" of coefficients for Bloch and 80 functions. We will consider the portion of the power series for f(z) between k = m and k = 2m. The integer "2" of the upper bound is not critical to the following analysis; simple revisions will accom- odate any number exceeding 1. f Define Bm = ii; lakl. We will first show that for functions whose power series have real nonnegative coefficients, we can charac- terize B and 80 by conditions on em. 00 Theorem 6: Suppose that f(z) = Z. ak 2k, where ak is real k=1 and nonnegative for all k. Then f(z) is a Bloch function if and only if there is a constant M < m. such that 8m'§_M for all m. Erggjg_ Suppose that Bm‘§_M < m» for all M. By [19, Theo- rem 2(i)], it is sufficient to show that there is a constant C such n that Z klakl 5_ Cn for all n. Fix n, and suppose that p is an k=1 integer such that 2p 5_ n < 2p+1. Then n n kgl klakl = kgl kak 1 3 n = kZI kak + kéz kak + ... + kgép kak .i 24 21 22 zp+1 g Zokak+ {lkak+ + Z kak k 2 =2 k=2p 21 2 22 -1 2p+1 < 2 Z ak+2 Zakw‘...+2p Zak k=2 k=21 k=2p g(2+22+...+zp+1)M _<_2p+2M §_ 4Mn. Taking C = 4M, we see that f(z) is a Bloch function. Conversely, suppose that f(z) is a Bloch function. Since all of the ak's are real nonnegative, all of the arguments are zero. By a result of Mathews [19, Theorem 3], there is a constant C such that n kgl k ak 5_ Cn for all n. Then 2m 2m m kzm ak < kém k ak < 2mC so 2m k; ak 5_ 2C Theorem 7: Suppose f(z) = kEI ak zk with all ak real and nonnegative. Then 8m + 0 as m-rm if and only if f(z) is in 80. Proof: Suppose 8m + 0 as m +1w. To show that f(z) is in Bo, - n it will suffice to show that kf k ak = o(n), or that =1 -1 n . n 2 k ak + 0 as n + w. k=1 26 Let a > 0, and let N be large enough so that for p 3_N, 29*? 'Z ak’ < 6/8. k=2p We have N n 2 n n"1 2 kak = n"1 X kak+n"1 Z kak k=1 k=1 _ N k-2 +1 2" If n is large enough, say, n _>_ Q 2. 2N, then n'1 :21 kak < 8/2. Let us now concentrate on the second sum. If 2r g-n < 2r+1’ then n n-1 X kak =2" _1 péNTI-l 2N+2-1 n = n 2N kak + yN+1 kak + + Zrkak L.k=2 k 2 k=2 -1 2N+1 2r+1 5_ n X kak + 4- X kak Lk=2N =2r T 2N+1 2Y‘+1 : n-1 2n+1 2 ak + +2r+1 Z 3k _ k=2N k=2’ < e 2N+1 + 2N+2 + + 2r+1] _ ..- ..O r 8 2r 2r 2 5_ g- (21+N‘r + ... + 2'2 + 2'1 + 1 + 2) < §-. n Therefore, for n j? Q, n'1 X kak < 5. By Theorem 3(i), f(z) k=1 TS in 80. Conversely, if f is in 80, then by Theorem 4, 2m 2m 1 = m 1 1 2'Bm 2m kgm ak 5- 2m kém kak 5- 2m kgl kak +'o’ so Bm+ O. If we now consider functions with coefficients not restricted to be real nonnegative, we obtain several theorems which parallel the results of Mathews [19], with conditions on the coefficients between am and a2m’ Theorem 8: The following conditions are sufficient for f(z) = 2k to be a Bloch function: Z a k=0 k (i). If there is a constant M* such that 2m (15) 2 kJ‘lla | §_M*mJ'1, k=m k where J is an integer greater than or equal to 1. (ii). If there is a constant C such that for some q satisfying 1 < q < m, X kq'llaqu §_C. k=1 Proof of (i): Suppose that equation (15) is satisfied and 2r §_n < 2r+1. Then 22 2r+1 n 2 2 kdlakl: 2 leakl + z kJIak|+m+ z leakl k=1 k=1 k=2 _ r k-2 21 Zrfl 1 21 1 k‘]"1|ak|+...+2'”+1 2 kJ'llakl k=2° k=2” 5_ 2 M* (1 + 2J + 22d + ... + 2rd) < 2 M* 2(r+1)J :_ 2J+1 M* "J. By [19, Theorem 2(i)], f(z) is a Bloch function. 28 Proof of (ii): By part (i), with J = 2, it suffices to show that :Zm klakl_ < for some finite constant M*. The details of the proof, using the Holder inequality, are similar to those of Theorem 1; we omit them. The next theorem gives additional restrictions on the location of the coefficients, under which the sufficient condition of Theorem 8 for J = 1, that the 8m are bounded, is also necessary. Theorem 9: Let f(z) = X ak zk be a Bloch function. If there exists an a such that a g_arg(ak) g_a + n/Z for all k, then there is a constant C* such that 8m §_C* for all m. 3399:; By a theorem of Mathews [19], there exists a constant C such that E klakl §_Cn for all n. Then k=1 2m mBm g_ kzm klak|_ < 2Cm. Hence Bm §_2C. Set C* = 2C. We remark that an example used by Mathews [19] to illustrate n that a Bloch function f(z) may satisfy kzl klakl = 0(np) for 1 < p < 3/2 also satisfies 8m = 0(mp'1). For, 2m 2m p p m X laklf. 2 klakl = 0((2m) ) = 0(m ), k=m k=1 so 2m = o ”'1 . kgmlakl (m ) Next, we consider necessary conditions. 29 Theorem 10: Let f(z) = Z ak zk be a Bloch function. Then k=0 # for each fixed integer P 3_1 there is a constant C such that 2m 2 kP-llakl E. c# mP-l/Z. k=m 2599:: By a theorem of Mathews [19], there is a constant C such that n X kplakl iCflP+1/2. k=1 Then 2m 2m 2m m z kp’llakls z kplakl: z kplak|:C(2m)P+1/2. k=m k=m k=1 Thus 2m 2 kP-lla l < 2P+1/2CmP-1/2. k=m k '— Setting C# = 2P+1/ZC, we are done. We next turn our attention to functions in 30. These results are similar to the preceding Theorems 8, 9 and 10. The proofs of all but part of one of the following theorems are similar to those of earlier theorems and will be omitted. Theorem 11: The following are sufficient conditions for the k z to be in B : function f(z) = X a k=0 k 0 (i). If there is an integer J 3_1 such that 2m 2 kJ 1lakl = o(mJ 1). k=m (ii). If there is a number p, 1 < p < m, and a number w < p - 1 such that 30 @? k(p-w)/(p-1),ak,p/(p-1) g o(n(p-w-1)/(p-1)). k=m 2m Proof of (i): Let c > 0 be given, and set u(m) = 2 kJ'llakl. k=m . 2m J-1 J-1 Since k2 k lakl = o(m ), there is an increasing function s(j) =m .on the positive integers such that for m > 25(3), 2m . z kJ'llakl < 52-(J+3)mJ-1. k=m If n is sufficiently large, say, n 3_N1. then 1 -J 25( ) J-l n X k la I < 8/2. k=1 k Let n be a fixed integer greater that N1, and p an integer such that 25(9) §_n < 2500. Theorem 13: Let f(z) be a function in Bo, and p an integer, p 3_1. Then 2m -1 _ -1 2 kgm kp lakl - o(mp / ). V. A BOUNDED BLOCH FUNCTION OF WHICH SECTIONS ARE ARBITRARILY LARGE In this chapter we will give an example of a Bloch function with real coefficients for which the sum of the coefficients is finite, 8m is large for selected values of m, and if the the same portion of the power series from which the 3m are taken is considered as a function gm(z), the Bloch norm of gm(z) increases without bound as m +»m. Landau [15] gave a theorem which shows that a function f(z), analytic and bounded by 1 in D, satisfies sn g_Gn for all n, k where sn(z) = kgo ak z , = sn(1), and 5n " 1-3- ... -(2k-1)]2. G = z ( O O O n k=0 2 4 ... 2k I It is noted that Gn is asymptotic to %-log(n). Landau also gave the following example of a function for which 5n = Gn:- Define Kn(z) = E ('lk/z)(-z>k k-O _ 1 1.3 2 1-3- ... o(2n-1) n '1*22*212‘+~ * 21.”.n Z and set ” k f (z) = Z a z n k=0 n,k z" Kn(z'1) Kn(z) 32 33 1-3- ... -(2n-1) 1 n-1 n = 2-4- -2n + +22 I z 1+_1_z + +1'12‘3:4:.. '(ggal) zn 2 For this function, all the coefficients are real, a k > 0 for n 0 5_k 5_n, kgo an,k = Gn’ |fn(z)|= 1 for |z| = 1, fn(1) = 1, and |fn(z)| 5_1 for [z] 5_1. Since |fn(z)| is bounded by 1 on the closed unit disc, fn(z) is a Bloch function with Bloch norm ”anB i 2 [3]. As a bounded Bloch function, the coefficients of fn(z) satisfy a",k + O as k +»m» [21]. Ne inductively define a function n(i) on the positive integers as follows: Set n(1) = 2. If n(j) has been defined for j < i, let n(i) be an integer sufficiently large so that for k > n(i), . . 2i we have n(1) > 2n(1-1), Gn(i) 3_2 , and a"(3).k-n(j) S.(2n(i)(n(i)+1))'1 for j=1, 2, co. , i-lo Now, we define the function f(z) = Z dk zk by k=2 f(z) = igl 2'1 zn(1) fn(i)(z)' °° -i If [2| = 1, then |f(z)| 5_ 1:1 2 |fn(i)(z)l = ‘1. For |z| 5_1, |f(z)| 5_1. Thus f(z) is a Bloch function, with Bloch norm ||f||B _<_ 2 [21], [3]. Sumning the coefficients of f, 00 co -1 d = f(1) = 2 = 1. kgz k igl Let us now consider 8m for selected values of m, in particular for m = n(i). For 1 5_j < i, 34 2n(i) -j -1 -0 k=g(i) 2 Ia"(3').k-n(j)l S.("(1)+1)(ZN(i)(n(i)+1))) 2 J < Z'j/n(i) so 1-1 2n(i) -j 0 i=1 min) 2 lan(.1).k-n(j)| <1/no). We also have that 2n i) -i -1 n(i) kgéii) 2 lanci)’k'"(i)l = 2 kgo an(i).k = 2-1 Gn(i) > 2i. Thus, Bn(i) > 2‘ - 1/n(i), which is unbounded as i + m. 2n(i) k=n{i) right. We will show that the Bloch norm of gi(z) is unbounded Consider next dk zk as a function 91(2) in its own as i + m. We may write i-l -j 2n(i) k (16) 91(2) = jgl 2 k=§(,) an(i),k-n(j) z I . 2n(i) ., k 2 k3L(i) an(i),k—n(i) 2 ° Differentiating 2n i) 2'J a . . zk k=n i) "(J),k-D(J) for j < i, we get 2n(i) k-1 2'3 k3,“, kan(.i).k-n(.i)z 35 which in absolute value is less than -j . 2n(i) 2 2 "(1) k=§(1) lan(j).k-n(j)l : 21-3 n(i) (n(i)+1) <2n(i)(n(i)+1))'1 = 2'j. Thus the value of the derivative of the double sum in (16) is less than 1 in absolute value. The second, single sum in (16) may be rewritten as (17) Z'i z"(i) 13;: an(i),k 2k; we will denote the sum in (17) by ti(z). This function is a poly- nomial of degree n(i) with positive real coefficients, and hence the maximum absolute value of (17) occurs for z = 1. Differentiating 2'? z"(i) ti(z), we obtain (18) 2" z“(‘)'1 n(i) ti(z) + 2'1 z"(" ti'(z). All of the coefficients on ti'(z) are real nonnegative as well, so [ti‘(z)l has a maximum at 1. We have n(i) kgO ti'(1) k an(i),k [A "(1) (ti(1) ' an(i),O) §_n(i) ti(1)’ To find the Bloch norm of 91(2), we must find sup (1 - |2|2)|9i'(z)|. ch Because of the bounds found earlier on the double sum in (16), 36 Hal-(fills 1H2" 2"“) he)”, - 2. Since the second term in (18) is positive, it will suffice for us to show that (19) (1 - IZIZ) 2" z""‘°"1 n(i) t,(z) is bounded below on (0, 1) by a term which increases without bound as i +»m. The expression in (19) is greater than (1 . (2|) 2“ z"(‘)‘1 n(i) ti(z). Set 2 =1 - 1/(4n(i)). Then (1 - IzI) z"“"'1 n(i) 1/(4n(i))(1 - 1/(4n(i))>"“)'1 n(i) 1/4 (1 - 1/(4n(i)))"“"1. -1/4 which approaches the limit 1/4 e as i, and hence n(i), tends to infinity. We also have t,(1 - 1/(4n