ON THE SECOND HANKEL DETERMINANT OF MEAN UNIVALENT FUNCTIONS

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1 ON THE SECOND HANKEL DETERMINANT OF MEAN UNIVALENT FUNCTIONS 1. Let S be the class of functions By W. K. HAYMAN [Received 3 May 1966] /(2) = <V n (LI) 0 areally mean univalent f in \z\ < 1, and normalized so that In a recent paper, Pommerenke (6) has considered the Hankel determinants of the coefficients of such functions, and in particular the second {2) Hankel determinant A n = a n a n+2 a 2 n+1. He proved in particular that if f(z) is starlike and univalent then 4^) = 0(1) as n-»ao, (1.2) and noted that from a previous result of the author (2) it follows that for /(z) in S we have A n = o{n) as n -> oo. (1.3) In this paper we show that (1.2) fails for general univalent functions, and that for/(z) in S the best possible result is More precisely, we have A n < 2) = o{n 1/2 ) as n -+ oo. OO THEOBEM 1. Suppose thatf(z) = *Za n z n e S. Then o \A {2) \ = \Vn + 2-a n+ i 2 \<An 1/2, n = 2, 3,..., (1.4) where A 1 is an absolute constant. Further, In the opposite direction we have A n = oin 1 ' 2 ) as n -+ oo. (1.5) THEOREM 2. Given any positive sequence e n, which tends to zero as n -> oo, there exists f(z) in S such that W 2) l>v* 1/2 > (1.6) for infinitely many n. t For a definition of this term see e.g. ((1) 94). Proc. London Math. Soe. (3) 18 (1968) 77-94

2 78 W. K. HAYMAN THEOREM 3. For every fixed n ^ 3, there exists f{z) in S such that (1.7) Theorems 2 and 3 show that (1.4) and (1.5) are essentially best possible. It is much harder to decide whether these results are still sharp for univalent as opposed to mean univalent functions. In this direction we can only prove THEOREM 4. Given e n satisfying the hypotheses of Theorem 2, there exists a univalent function f(z) in S such that for infinitely many n. (1.8) Thus at any rate (1.2) is false even for univalent/(z). We denote as usual by A an absolute constant, not necessarily the same each time it occurs, and by A v A 2,... particular absolute constants. Constants depending on parameters a, j8,..., etc. are denoted by 4(a,j8,...),4(a,j8 f...). 2. In the next four sections we prove Theorem 1. The technique uses previous ideas of Pommerenke (5) and the author ((1) (2)), and also an inequality for the moduli of a mean p-valent function at two points due to Lucas (4). Let n be a positive integer such that n ^ 6, and set J^ = 2 1 ~ V M 1. We chose 0 O = 6 0 {n) so that 0 ^ 0 O < 2TT and /(«), (2.2) lei-pi With this (or any other) choice of the value 9 0 we have identically In order to achieve our results we shall estimate the various terms on the right-hand side of (2.3). It follows from a previous result of the author proved for the case a 0 = 0 (2) that \a n+1 -e^oa n \<A. (2.4) In Lemmas 1 and 8 we reprove (2.4) for S, and estimate a n and a n+2 ~ Z er * e a n+i + e~ 2i0 a n ^ terms of M x. It turns out on the whole that the smaller M x is, the better is our estimate for \a n \ and the worse that

3 ON THE SECOND HANKEL DETERMINANT 79 for \a n+2 2e- ie a n+1 + e~ 2i0o a n \. However, by taking a product we obtain (1.4) in all cases. We start by estimating a n, which is relatively simple. LEMMA 1. We have, with the above notation, We choose p in the range 1-^psU-i, (2.5) and note that _ p x ~ n and hence, by Schwarz's inequality, \fia i n \<AF\f(p*')\d8 (2.6) By a known inequality ((1) Lemma 3.1, p. 46), we can choose p in the range (2.5) so that 2^J o (\f'(pe ie )\ 2 /\f(pe ie )\)dd < M x n. (2.7) At the same time we have ((1) (3.10), p. 45), for \ < p < 1, Ji " (2.8) where M(t,f) = max /(z). \z\=l It follows from the theorem of Cartwright and Spencer ((I) Theorem 2.5, p. 31) that M(t,f) <A 2 (l-t)-\ 0<t<l. (2.9) We define t 0 by the equation

4 80 W. K. HAYMAN and suppose first that 0 < t 0 < p. Then we deduce from (2.8) that Since t 0 > 0, we deduce that M x > A 2 so that I^pJ) < AM"\ (2.10) If M x ^ A 2, so that t 0 ^ 0, this inequality remains valid trivially since in this case UpJ) < M( P J) ^M,< AV*MU\ Also, if t 0 ^ p we have M x ^ A 2 (l p)~ 2, so that (2.8) yields < 4A 2 + 2A 2 /(l-p) < 6A 2 /(l-p) Thus (2.10) holds in all cases, and on combining this with (2.6) and (2.7) we deduce Lemma The estimates for a n+2-2e- i6o a n+1 + e~ 2i0o a n \ and a n+1 - e~ i0o a n \ lie considerably deeper. Our technique is very similar to that employed in (2), and we use the notation of that paper as well as that of 2 of the present paper. We set and note that if then (2-Zx) 2 /'(2) = S o K-i = {m-2)a m _ 2-2(m-l)z 1 a m _ 1 +mz 1 \ In order to prove Theorem 1 we need to estimate b n+1. Our technique is similar to that of ((2) Theorem 1). We assume that n ^ 6, and have, for 1 - (S/n) ^ p ^ 1 - (2/n), 1 f 2 V*- 2l ) 2 /'( '0

5 ON THE SECOND HANKEL DETERMINANT 81 We multiply both sides by p and integrate with respect to p. This gives I u n+li n A Cl 1 JO Ji (3/TI) {2/n) \pe^-z^\r(pe i9 )\pdpdd. We choose A > 2, and apply Schwarz's inequality. This gives, with z == pc^f n U/2, (3.2) where.f is the annulus 1-(3/TI) < \z\ < l-(2/n). Let F o be the part of \z\ < 1 in which /(z) < 1, and for v ^ 1 let JJ, be the part of \z\ < 1 where 2"" 1 ^ \f{z) < 2". Then if C o = 1, C, = 2<"- 1 >\ v > 1, we have since/(z) is mean univalent in \z\ < 1. Thus if/(z) is mean univalent in \z\ < 1, and A > 2, we see that rr \f'(z\\ 2 I JJ «l<ll+l/(2) v=0 \ Hence we deduce from (3.2) that IV 1 < 4<A)I At the same time we recall that J j ^ ] } (3.4) where J (3.5) r r 7lf 2P2 \l/2 G { R ) and Jlfj is given by (2.2). In order to estimate the right-hand sides of (3.4) and (3.5) we need a number of lemmas. f This is (3), p. 230, of (2). The inequality is actually stated with Ha,,.! a n on the left-hand side, but the proof gives (3.5) as shown.

6 82 W. K. HAYMAN LEMMA 2. Suppose that v{z) = v(x, y) is twice continuously differentiable in\z\ < p. Then if * ' ' we have, for 0 < p < 1, v(pe^)d<p = 2TTv{0)+ \\ dt\ \ ^dd. J o t This is Lemma 1 of (2). LEMMA 3. Iff(z) is regular in \z\ < 1 we write \f(z)\ = R and set M 2 B 2 G{R) = where M is a constant. Then we have, for \z\ < 1, ( 4Jf«(Jf»-2P) ' = (Jf2 + jr2)3 1/ (*) I > ( 3 ' 6 ) Also, ifx>2, ^ ^ ^ ' ( z ) l (3.7) The relations (3.6) and (3.7) are (4) and (5) of Lemma 2 of (2). To prove (3.8) we set <p(z) = (z z 1 ) 2 f(z) A/2 and use the limiting case M -> oo of (3.6). This shows that Also, so that 2p'( Z ) = 4( Z < 321 z - Zl» /( 2 ) A + 2A21 z - Zl «\f(z) 1^-21/' {2 ) 2. This gives (3.8). We apply Lemmas 2 and 3 to the integrals on the right-hand sides of (3.4) and (3.5). We need first LEMMA 4. Suppose that f(z)es and that G(R) = M 2 R 2 /{M 2 + R 2 ), where M is fixed. Then and \ 2n G{\f{re w )\)dd <, Amm{M^2,{\-r)-% 0 < r < 1, (3.9) / 2TT rp rp o G{\f(re ie )\)r\og dd ^ ^min{if 1/2,(l-p)- 1 }, 0 < p< I. (3.10) r

7 ON THE SECOND HANKEL DETERMINANT 83 This is Lemma 3 of (2). The result was stated with the special hypothesis /(0) = 0, but no use is made of this hypothesis in the proof. 4. When we apply Lemma 2 with v(z) = \z z x \ 2 O(R) or z z x 4 \f(z) A in order to estimate the integrals occurring on the right-hand sides of (3.5) and (3.4) respectively, and use the estimates of Lemma 3, we obtain in each case two or three terms arising from the two or three terms on the right-hand sides of (3.7) and (3.8). To deal with these terms new inequalities are required. The method of (2) which dealt with the case of (3.5) (at least when/(0) = 0) is not sufficiently powerful for (3.4), and we need oo LEMMA 5. Suppose that f{z) = a n z n is mean p-valent in o \z\ < 1, fj, p = max aj, = 1, and b > a > 0. Suppose that \f{z v )\ R v, v = 1, 2, where \z v \ < 1, v = 1, 2, and R x > R 2. Then 1 {2V(o6)-a}/2p This result is due to Lucas (4). Using it we now prove LEMMA 6. We have, with the notation of section 3, for /x ^ 1, 0 < p < 1-2/n, \\\z-z 1? \nz)\no^rdrde < ^jl + ^nnn^, (j^^^f"]}, (4.1) (4.2) where the integrals are taken over all points z = re i0 for which \f{z)\ < M^ and r < p. Let F v be the set of points in z \ < p for which M v+x^\f{z)\<m v, and suppose that K+i > M(hf), (4-3) so that F v lies in 121 ^ \. Then ^j 2(p-r) < 2(l~r).

8 84 W. K. HAYMAN Also, \f(z x ) = M 1 and 1/(2) ^ M v+1 = M X 2~ V in F v. Thus by Lemma 5, applied with p = 1, a = 1,6 = 4, and z 2 instead of z we have, in F v, or, n. (4.4) v x I»i I M x ~-l z m z ~ z ir Thus J J F» r (1 - p 1 )il4 1 i/^+1 J/ ' 5 J J Fv It follows from (4.4) that if F v is not void then A M V 2 A (MX 1 * (4.5) since z x - p > ^(1 + p) - p = (1 - p), and Zj_ = p x = 1 {l/(n + 2)}. Next we consider (4.2). We apply Lemma 5 with p = 1, a = 1, and yjb = 8/3. This gives, for/(z) in F v) Hence we have, in F v, This gives 25/18 < AiZr\ M-l*'«. ( n% \ 25/18 / / / TO 2 \ 25/18 < A since /(z) is mean univalent and \f(z) \ < M v in i^, or I 7Lf-l/12 (A 7\

9 ON THE SECOND HANKEL DETERMINANT 85 Suppose now that v 0 is the largest integer such that (4.3) holds for v = v 0. It follows from Cauchy's inequality that and hence, since /(z) e 8, we deduce that M(hf) > i At the same time we have, by (2.9), M(hf) ^ A. Thus \ < M H+1 ^ A. If F' denotes the disk \z\ < \, then clearly the integrals over F' contribute at most an absolute constant to the left-hand sides of (4.1) and (4.2). Again, if F" consists of all points in \ < \z\ < p where \f(z)\ < M Vo+1 then, since f(z) is mean-univalent, we have \f'(z) \hdrdd < A M * < A, (4.8) F" so that F" also contributes at most an absolute constant. On summing (4.5) from v = v x to v 0, where v x is the least integer which satisfies (4.6) and v ^ /A, and using (4.8), we obtain (4.1). Similarly, by summing (4.7) from v = 1 to v 0 and using (4.8) and the fact that M x = 0(n 2 ) by (2.9), we obtain (4.2). LEMMA 7. We have, with the notation of (3.4), (3.5), for 0 < p < l 2/n, for v ^ 1, and f n \pe^-z 1 \ 2 G y (\f(pe i^)\)d<p < A(l+n2->"*) (4.10) J o V-uzi/ 25/18 (4.11) It follows from Lemmas 2 and 3 that if 0 < r < 1 {2/n), z = re ie, (z) = R, then»2»r <^ff G v {R)r\og?-drdd JJ\z\<p r \z\<p \f(z)\<m v (4.12)

10 86 W. K. HAYMAN say. It follows from (3.10) that I x < AminiMV^il-p)- 1 } < AM V 1/2 < An2-»'*, (4.13) by (2.9). Again, by (4.1) we have Finally, by Schwarz's inequality we have ) Jl [7 3 P< IT JJ\ ^ r \ g \ \ ^f g, \z\<p r JJ\z\< P {M v * + Mtf r (4.15) and On applying (4.14) with 1 instead of v, and (4.13), we deduce that 7 3 < and combining this with (4.12), (4.13), and (4.14), we have (4.10). Next we deduce from Lemma 2 and (3.8) that, with the previous notation, <A\»-2 1» /(«) a5/ia r]og^ r '0 JO ' (4.16) say. It follows at once from (4.2) that T «r 4(>n z /M^25/18 (A ]n\ To tackle J x we apply (4.12) with v = 1, so that i2 = \f(z) \ ^ ilf l5 2 Thus (4.12) to (4.15), applied with r instead of p, show that Also, by (2.9) we have on \z\ = r,

11 Thus ON THE SECOND HANKEL DETERMINANT 87 We multiply this inequality by r\og(p/r) and integrate from 0 to p. This gives 25/18 On combining (4.16), (4.17), and (4.18), we have (4.11). This completes the proof of Lemma 7. and 5. Proof of Theorem 1. We can deduce from the results obtained so far LEMMA 8. We have, with the notation of 2, W n+ x-^a n \<A (5.1) 2X25/36 j,» > 6. (5.2) To prove (5.1) we apply (3.5). Let v 0 be the largest integer for which n.2-"o/4 > x> Then (4.10) shows that v 0 I fl-{2/n) [>2n U/2 v 0 IA v=l(jl-(3/n) Also, if v ^ v 0 then, by (2.9), J v=l\^ Thus and 00 ( / l-(2/n) /«27r 2 Iz-Zi i'o+lwl-(3/n) JO f=i'o+l W/2 Iil^d/^JD < On combining these estimates with (3.5) we deduce (5.1). Similarly, we have from (3.4) and (4.11), taking A = 25/12, j so that In view of this, (3.1), and the fact that \a n \ < An, we obtain (5.2). This completes the proof of Lemma 8.

12 88 W. K. HAYMAN 5.1. We now apply (2.3) together with Lemmas 1 and 8. This gives 25/36-3/4 25/36 w) Since, by (2.9), M x < An 2, we deduce (1.4) at once. We also deduce (1.5) except when Suppose then, finally, that (5.3) holds. Then it follows from (4.11) that there exists a constant C independent of n and r such that p2ir \re id -z^\f{re e )\^i l Hd < C, 0 < r < 1, as n -> oo through a suitable sequence and with z x = (1 (1/n + 2))e ie {n), It then follows from ((1) Theorems 2.8 and 2.9) that 6 0 (n) approaches a limit 6 X as n -> oo. Thus we may replace z x by e idl and deduce that / 2w I I Qi6i_ r pi6 141 f(ifoio) 125/12 < Q 0 <?* < 1 J 0 We now prove, using a technique of Pommerenke (5), that r 1 I*(r) 2 rdr < +00, (5.4) where I*(r) = f "le^-re^l 2!/'^ In fact we have, by Schwarz's inequality, Thus by (3.3), which gives (5.4).

13 ON THE SECOND HANKEL DETERMINANT 89 Also, e i(?1-2 2 /'(z) is subharmonic so that I*(r) is an increasing function of r. Thus in view of (5.4). If I372j/ ( r^;) as r->l, (5.5) then by (5.5). Also ^ J- -1L. I* l-^-\ = o{nv*), (5.6) v 2TT [^ +1J \n+l) ' K ' Thus the identity (2.3) shows that a n a n+2~ a n+l ~ ~ Z~ \ a n+2 e ^n+l) \ a n+l ^n) (5.7) It follows from ((1) Theorem 2.9, p. 38) that given e > 0 we have e for C(l r) ^ 0 0-J ^ TT, where C is a constant. Hence if 0 O = 0(r) is so chosen that then since by ((1) Theorem 2.8, p. 37) we have By (5.1) we have \ dl -d 0 \ = O(l-r), (5.8) ) > 0. where ^0 = 0 O (1 (1/^)). In view of (5.8) and \a n \ = O(n), we deduce that We substitute this result in (5.7), and deduce finally that 2\ ^ = oin 1? 2 ) n by (5.6). This completes the proof of Theorem 1.

14 90. W. K. HAYMAN 6. Proof of Theorems 2 and 3. We now consider Then g(z) maps z \ < 1 onto the half-plane P: u > 2 in the w = u + iv plane. Let E8(R) be the length of the arc of w = R which lies in P. We have 6(R) = 0, R < 2 so that, for R > 2, ^, R>2, 77" "... I 2i 2 Consider now W = G{z) = g{z) 2 = (3-z) 2 (l-z)" 2 = 9+ (4^ + 8)z' 1. (6.1) n=l Then (^(z) is clearly univalent in \z\ < 1. Let 2nRp(R) be the length of the arc of W\ = R, which is covered by the image of G(z). Then ( = 0, R < 4, Thus if R > 4, and p(r) = 0, R < 4. Also, if ^4(i2) denotes the area of that part of the image of z < 1 by G{z) which lies in W \ < R, then so that and if R ^ 4, \ h A{R) = 0, R < 4, A(R) ^ 2TT( E Yl- W < 27rf («- Theorems 2 and 3 are consequences of LEMMA 9. Suppose that b n (n = 1,2,...) are non-negative numbers such that b x = 0, S6 n ^ 1, I>6 n < 1. Then if a n = 4n + S + b n, n ^ l, the 1 2

15 ON THE SECOND HANKEL DETERMINANT 91 Suppose that the Lemma is proved. Then given the sequence e n of Theorem 2, we can choose b n to satisfy the conditions of Lemma 9 and such that b n+1 = I8e n /n 1/2 for infinitely many n, b n+1 = 0 otherwise. Also, we may assume that if 6 7l+1 ^ 0 then b n+2 = b n = 0. Then V = a 2 n+1 -a n+2 a n for infinitely many values of n. This gives Theorem 2. We also see, by choosing b n+1 = (n + l)~ 1/2, b k 0, k^n + l, that, for any fixed n ^ 1, we can have /wi = (a n+1 *-a n a n+2 ) = 8b n This gives Theorem 3. It remains to prove Lemma 9. Consider then a sequence b n satisfying the hypotheses of Lemma 9. Suppose that G(z) is given by (6.1), and set Let E{R) be the part of \z\ < 1 in which \F{z)\ < B. Clearly, so that in.27(12) we have Also, [ E(R) ff \ 9 '(z)\*dxdy < (( \ JJEiR) JJ\z\<\ Again, if R > 1, ff \F'{z)\*dxdy <$ \\ {\ 9 '{ and ff IO'(z)?P'( Z ) ^ ^ (f \G'{z)\*dxdyX' 2 JJE(R) \JE{R) 1 xfff \ \JJE(R)

16 92 W. K. HAYMAN Thus if 22 >1, ff \F'(z)\ 2 dxdy < TT{1 + 2(B-1) + (B-1) 2 } = TTM 2. (6.2) JJEiR) Also, (6.2) remains valid for R < 1, since in this case 2 (.R) is void, as Thus (6.2) is true in all cases, so that F(z) is mean univalent in \z\ < 1. Also, F(0) = 0(0) = 9, F'(Q) = O'(0) = 12, since b 1 = 0. Thus f(z) = F(z)/U e S, and the proof of Lemma 9, and so of Theorems 2 and 3, is complete. 7. Proof of Theorem 4. Let A(r) be a function of r, positive for 0 ^ r < 1 and such that A(r) -> 0 (7.1) as r -> 1. Then it was proved by Kennedy and the authorf (3) that there exists 00 real for real z and univalent in \z\ < 1, such that so that ((2) Theorem 3) 0 r) 2 f(r) = a > 0, and hence as n -> OD, and in addition a n- a n-l-*<x ( 7-2 ) 5 -* a (7 ' 3) /(-r) > exp(a(r)flog-j-) 1/2 } (7.4) for a sequence r = r n such that < r n < 1 and r n -> 1 as TI -^ oo. We now set so that, by (7.4), <p(z) = T l b n z* = (l-zmz), (7.5) o \? {-r)\ = (i +f )«/(-r) > 2exp A(r)(log I i 7 y /2 }, r = r tt. (7.6) f We also assumed that A(r) decreases with r. If this is not the case we can replace A(r) by A L (r) = sup A(p). If (7.4) holds with X x {r), then it also holds with A(r).

17 ON THE SECOND HANKEL DETERMINANT 93 Suppose now that e n is the sequence of Theorem 4, and set 9n = exp{ejlog n) 1 ' 2 and define X(r) by the equation ( 1 \ 1/2 - log n}, g{r) = g n r n t I l g^ = ^gg{r), 0 < r < 1. (7.7) 1 T) We proceed to show that (7.1) holds. Let e be a fixed positive number, and suppose that N o is chosen so large that e n < e, n^ N o, and that {log(w + 1)} 1/2 log(n + 1) < e(log n) 1/2 log n, n ^ 2V 0. We assume that 1 (l/n 0 ) < r < 1, and define N by the equation 1_JL< 1 Then oo rn N \ g(r) < 2exp( (logw) 1/2 ) t-o(l) < exp(e(log.af) 1/2 ) 2 oo Since «may be chosen as small as we please, we deduce that,1/2 / l V as r -> 1, log gr(r) = o! log j and hence (7.7) implies (7.1). Thus we can find/(z) univalent in \z\ < 1 and such that (7.4) and hence (7.6) holds. Suppose now that Then we deduce at once that, for 0 < r < 1, \?{-r)\ ^ hb n \r n I ^ ZQn I In view of (7.7) this contradicts (7.6). Thus we have, for infinitely many n,

18 94 ON THE SECOND HANKEL DETERMINANT and in view of (7.2), (7.3) we deduce that, for infinitely many n, iaexp{e n+2 (logw) 1/2 }. Since we may replace e n+2 by e n in this, we deduce Theorem 4. I would like to thank the referee for the correction of a large number of small errors in this manuscript. REFERENCES 1. W. K. HAYMAN, Multivalent functions, Cambridge Tracts in Mathematics and Mathematical Physics (1958), No 'On successive coefficients of univalent functions', J. London Math. Soc. 38 (1963), and P. B. KENNEDY, 'On the growth of multivalent functions, ibid. 33 (1958), K. W. LUCAS, 'A two-point modulus bound for areally mean ^-valent functions', J. London Math. Soc, to be published. 5. CH. POMMEEENKE, 'Uber die Mittelwerte und Koeffizienten multivalenter Funktionen', Math. Annalen 145 (1961/62), 'On the coefficients and Hankel determinants of univalent functions', J. London Math. Soc. 41 (1966), Imperial College London 8. WJ

1981] 209 Let A have property P 2 then [7(n) is the number of primes not exceeding r.] (1) Tr(n) + c l n 2/3 (log n) -2 < max k < ir(n) + c 2 n 2/3 (l

1981] 209 Let A have property P 2 then [7(n) is the number of primes not exceeding r.] (1) Tr(n) + c l n 2/3 (log n) -2 < max k < ir(n) + c 2 n 2/3 (l 208 ~ A ~ 9 ' Note that, by (4.4) and (4.5), (7.6) holds for all nonnegatíve p. Substituting from (7.6) in (6.1) and (6.2) and evaluating coefficients of xm, we obtain the following two identities. (p