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The logarithmic derivative of areally mean p-valent functions
J. Math. Anal. Appl. 296 (2004) 650–657 www.elsevier.com/locate/jmaa
The logarithmic derivative of areally mean p-valent functions ✩ Xin-Han Dong Department of Mathematics, Hunan Normal University, Changsha 410081, China Received 23 November 2003 Available online 10 June 2004 Submitted by S. Ruscheweyh
Abstract Let U(p, k) denote the class of functions f which are areally mean p-valent and have k maximal growth directions in ∆ = {z: |z| < 1}. In this paper we give an asymptotic formula of the mean square of the logarithmic derivative f /f for f ∈ U(p, k) with some restricted condition. This generalizes the corresponding result in [J. Anal. Math. 36 (1979) 36–43]. Also we construct a function f ∈ U(p, k) which does not satisfy this asymptotic formula. 2004 Elsevier Inc. All rights reserved. Keywords: Areally mean p-valent function; Logarithmic derivative; Asymptotic behavior
1. Introduction Let f be analytic in the unit disk ∆ = {z: |z| < 1} and let n(f = w, D) be the number of roots of the equation f (z) = w in D (⊂ ∆). We say that f is areally mean p-valent in ∆ (we denote the class by U(p)) if for every R > 0, 1 n(f = w, ∆) du dv p, πR 2 |w|R
where w = u + vi, and p is a positive real number, not necessarily an integer. It is clear that if f is univalent and analytic in ∆ then n(f = w, ∆) 1 for each w ∈ C. Hence a ✩
The research is supported in part by the National Natural Science Foundation of China (No. 19871026). E-mail address: [email protected].
0022-247X/$ – see front matter 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2004.04.032
X.-H. Dong / J. Math. Anal. Appl. 296 (2004) 650–657
651
function f analytic and univalent in ∆ must be areally mean 1-valent. Let f ∈ U(p) and let z1 , z2 , . . . , zq be the zeros of f (z) in ∆, repeated according to multiplicity, so that q p [1, p. 38]. We choose an appropriate constant b such that ∞
log
f (z) = 2p cn z n h(z)
(1.1)
n=1
q is analytic in |z| < 1, where h(z) = b j =1 (z − zj ). For integer k, we say that f is in U(p, k) if f ∈ U(p) and if we can find k distinct points t1 = eiθ1 , . . . , tk = eiθk on |z| = 1, − and a constant d > 0 and a sequence {rn }∞ n=1 with rn → 1 as n → ∞ such that 1 (1.2) d(1 − rn )−2p/ k f (rn eiθs ) M(rn ) (1 − rn )−2p/ k d for s = 1, 2, . . . , k and all n 1, where M(r) = M(r, f ) = max|z|=r |f (z)|. For f ∈ U(p, k) we say that f attains maximal growth on the k rays ts , s = 1, 2, . . . , k. The class U(p, k) was introduced by Eke in [3, p. 148]. We proved in [4, p. 333] that (1.2) is an equivalent definition of U(p, k) which is more explicit than the original. In this paper, we consider the asymptotic growth of 2π iθ 2 f (re ) f 1 dθ = I2 r, f 2π f (reiθ )
(1.3)
0
for f ∈ U(p, k). This topic has been investigated by Duren and Leung [5], [2] Hayman 2 |c |2 , n and Elhosh [7]. A relative problem is to discuss the asymptotic behavior of N n n=1 where {cn } are defined by (1.1). In [5], Duren and Leung proved Theorem A. Suppose that f (z) = z + a2 z2 + · · · is univalent and analytic in ∆ with positive Hayman index, i.e., limr→1 (1 − r)2 M(r) = α > 0. If ∞
log
f (z) =2 cn z n , z
|z| < 1,
n=1
then there exist constants C1 > 0, C2 > 0 such that f (1 − r)I2 r, C1 , 0 < r < 1, f and N −1
N
n2 |cn |2 C2 ,
N = 1, 2, . . . .
(1.4)
(1.5)
n=1
Our aim is to prove thatthe function (1 − r)I2 (r, f /f ) in (1.4) has a limit as r → 1− . 2 2 Hence the sequence N −1 N n=1 n |cn | in (1.5) also has a limit as N → ∞. In fact, we prove these results for f ∈ U(p, k) with some restricted condition (see (2.1)). It is obvious that if f satisfies the normalized condition, i.e., f (z) = zp (1 + a1 z + · · ·), then this restricted condition is trivial (see Corollary 2.2). Also, we construct a function f ∈ U(p, k) 2 |c |2 = o(log N). n such that (1 − r)I2 (r, f /f ) = o(log(1/(1 − r))) and N −1 N n n=1
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X.-H. Dong / J. Math. Anal. Appl. 296 (2004) 650–657
2. The theorem Theorem 2.1. Suppose that f ∈ U(p, k), and that 1
1 n f = w, r0 |z| < 1 du dv dR < ∞, R3
(2.1)
|w|R
0
where w = u + vi and r0 = (1/4)(3 + max1j q |zj |). Then 2p2 f → as r → 1 (1 − r)I2 r, f k
(2.2)
and N
−1
N
n2 |cn |2 →
n=1
1 k
as N → ∞,
(2.3)
where {cn } are defined as in (1.1). Proof. It is easy to check that 2π 2π f (z) 2 r 2 f r 2 h (z) 2 log = r I2 r, dθ − dθ f 2π h(z) 2π h(z) 2
0
+
r2 Re π
0
2π
f (z) h (z) dθ. f (z) h(z)
0
Let z = reiθ (r r0 ). Then computation shows r2 π
2π
q f (z) h (z) r2 dθ = f (z) h(z) πi j =1
0
=
|z|=r q q j =1 s=1
=
q q j =1 s=1
1 πi
1 f (z) dz f (z) r 2 − zj z
|z−zs |=ε
r2
r2 dz − zj z z − zs
2r 2 . r 2 − zj zs
We write r2 B1 (r) = − 2π
2π 2 q q h (z) r2 dθ + 2 . h(z) 2 r − zj zs 0
j =1 s=1
It is clear that B1 (r) is a bounded function on [r0 , 1]. Let t1 = eiθ1 , . . . , tk = eiθk be the maximal growth directions of f (z). Then we have by the above that
X.-H. Dong / J. Math. Anal. Appl. 296 (2004) 650–657
∞ f r I2 r, n2 |cn |2 r 2n + B1 (r) = 4p2 f n=1 2 k 2 k ∞ ∞ 1 n 2n 1 2 2 2 n 2n = 4p n c n − t j r − 4p tj r k nk n=1 j =1 n=1 j =1
k ∞ 4p + Re 2p ncn tjn r 2n + B1 (r). k
653
2
j =1
(2.4)
n=1
By Theorem 3.1 in [4, p. 334], (2.1) implies 2 ∞ k 1 n ncn − t j < ∞, nk j =1
n=1
hence ∞ n=1
2 k 1 o(1) , n2 cn − t nj r 2n = kn 1−r
r → 1.
(2.5)
j =1
Observe the representation 2p
∞
ncn tjn r 2n =
n=1
r 2 tj f (r 2 tj ) r 2 tj h (r 2 tj ) − , f (r 2 tj ) h(r 2 tj )
and note the asymptotic formula r 2 tj f (r 2 tj ) 2p + o(1) = , f (r 2 tj ) k(1 − r 2 )
r → 1 (see (14.2) in [3, p. 192]),
we get
k ∞ 8p2 + o(1) 4p n 2n , Re 2p ncn tj r = k k(1 − r 2 ) j =1
r → 1.
n=1
It follows by (2.4)–(2.6) that k 2 ∞ 2 1 4p 1 f o(1) r 2 I2 r, = − 4p2 tjn r 2n + f k 1−r k 1−r n=1
=
2p2
1 o(1) + k 1−r 1−r
j =1
as r → 1.
This proves (2.2). (2.2) and the first equality in (2.4) give (1 − r)
∞ n=1
n2 |cn |2 r n =
√ f 1 1 (1 − r)I r, + o(1) = + o(1) 2 2 4p f k
(2.6)
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X.-H. Dong / J. Math. Anal. Appl. 296 (2004) 650–657
as r → 1. By virtue of Tauberian theorem [8], we obtain N −1
N n=1
1 n2 |cn |2 → , k
N → ∞.
2
Corollary 2.2. Suppose that f (z) = zp (1 + a2 z + · · ·) ∈ U(p, k), where f (z) may be defined in D = ∆ \ (−1, 0], if necessary, and that ∞
log
f (z) = 2p cn z n , p z
|z| < 1.
n=1
Then f 2p2 (1 − r)I2 r, → f k
as r → 1−
and N −1
N
n2 |cn |2 →
n=1
1 k
as N → +∞.
Proof. From the definition of areally mean p-valent function, it is easy to see that there exist ρ0 ∈ (0, 1) and R0 > 0 such that
n f = w, ρ0 |z| < 1 = 0, |w| < R0 . Hence (2.1) holds. 2 Corollary 2.3. Suppose that f (z) = z + a2 z2 + · · · ∈ U(1, 1) and ∞
log
f (z) =2 cn z n , z
|z| < 1.
n=1
Then f (1 − r)I2 r, →2 f
as r → 1−
and N −1
N
n2 |cn |2 → 1 as N → ∞.
n=1
If f is univalent and analytic in ∆ with positive Hayman index, then f ∈ U(1, 1). Therefore, Theorem A follows directly by Corollary 2.3.
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655
3. An example Observing Theorem 2.1, naturally one can ask whether the condition (2.1) is necessary, i.e., whether the limit limr→1 (1 − r)I2 (r, f /f ) exists, or whether the function (1 − r)I2 (r, f /f ) is bounded on [r0 , 1) for f ∈ U(p, k). We now construct a function f (z) ∈ U(p, k) by revising Hayman’s example in [2, p. 166], which satisfies 1 f (1 − r)I2 r, = o log as r → 1. f 1−r 2 2 Similar to the proof of Lemma 1 in [5], we have N −1 N n=1 n |cn | = o(log N). Theorem 3.1. There exists a circumferentially mean p-valent function f (z) in ∆, attaining maximal growth on k rays (naturally, f ∈ U(p, k)), such that f 1−r > 0. (3.1) I2 r, lim sup f r→1 log(1/(1 − r)) Proof. Our goal is to construct a circumferentially mean p-valent function f (z) in ∆, attaining maximal growth on k rays, such that for any ε > 0 there are a positive constant C(ε) (depending only on ε) and sequences tj , tj (both → 1− as n → ∞) with (1 − tj ) × (1 − tj )−1 kC(ε), while tj |z|tj
2 2 f (z) 1 dx dy (1 − ε)πp log f (z) k k(1 − tj )
(3.2)
for all large j . Thus (3.1) follows easily from (3.2). Firstly, we consider the case p = 1 and k = 1. For positive integer n, take positive integer j satisfying 2(j −1) n < 2j , 2
2
(3.3) j2
and define d−n by −2 log d−n = 2 − n + j . Set S = s = σ + ti: σ ∈ (−∞, ∞), |t| < π , ∞ π 1 |t| < π , T= s = −n + + ti: 2 2 n=3 ∞ π Γ = s = σ ± i: −n − d−n σ −n + d−n , 2 n=3
and Ω = S \ {T ∪ Γ }. Then Ω is a simply connected domain in the s plane, and is a symmetrical about the real axis. From the Riemann mapping theorem, there exists an analytic function ψ1 (z) which
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X.-H. Dong / J. Math. Anal. Appl. 296 (2004) 650–657
maps ∆(1, 1) conformally onto Ω and has the properties ψ1 (0) = 0 and ψ1 (0) > 0. Define f1 (z) = exp{ψ1 (z)} and ψ2 (z) = log(1/(1 − z)2 ) (with ψ2 (0) = 0) for z ∈ ∆. Clearly, f1 (z) and ψ2 (z) both are univalent and analytic in ∆. Since Re ψ2 (z) −2 log 2 and | Im ψ2 (z)| < π for z ∈ ∆, we have ψ2 (∆) ⊂ Ω and ψ2 (0) = ψ1 (0). It follows from the subordination principle [6, p. 190] that there exists a function ϕ(z) analytic in ∆ such that |ϕ(z)| |z| and (1 − z)−2 = f1 (ϕ(z)) for |z| < 1. This shows that M(r) = M(r, f1 ) (1 − r)−2 , 0 < r < 1, and hence f1 has a positive Hayman index α > 0. So f ∈ U(1, 1). + − For given ε > 0, we define E−n and E−n (n = 3, 4, . . .) as in [2, p. 167]. From the proof of Theorem 3 in [2, p. 167–177], we know that (5.14) in [2, p. 167] holds also for + − domain Ω, that is, if s ∈ E−n or E−n , then 2
2
2j + j − C1 (ε) 2d(0, s, Ω) < 2j + j + C1 (ε),
(3.4)
where d(s1 , s2 , Ω) stands for the hyperbolic distance of two points s1 , s2 in Ω, j is defined by (3.3), and C1 (ε) is a suitable positive constant depending only on ε. We define rj , rj by 2
2j + j + C1 (ε) = log
1 , 1 − rj
2
2j + j − C1 (ε) = log
2 . 1 − rj
Clearly 0 < rj < rj < 1, rj → 1 (j → ∞) and (1 − rj )(1 − rj )−1 = 2e2C1 (ε) = C(ε). Noting
1 1 + |z| , d 0, ψ1 (z), Ω = log 2 1 + |z| + − we obtain by (3.4) that if s = ψ1 (z) ∈ E−n or E−n then
log
1 2 1 + |z| log log . 1 − rj 1 − |z| 1 − rj
Thus +
− ∪ ψ1−1 E−n ⊂ z: rj |z| rj , ψ1−1 E−n and hence rj |z|rj
2 f1 (z) f (z) dx dy = 1
2 ψ (z) dx dy 2
2
j −1 2
+ area E−n
1
rj |z|rj
n=2(j−1)
2
j2 ε π (j −1)2 2 2 −2 1− 2 2 j2
π(1 − ε) 2 + j − C1 (ε) 1 = π(1 − ε) log for all large j. 1 − rj
(3.5)
In the general case, we define f (z) = f1 (zk )p/ k , |z| < 1. Since f1 is univalent in ∆, we get by Lemma 5.1 in [1, p. 145] that f is circumferentially mean p-valent in ∆. The
X.-H. Dong / J. Math. Anal. Appl. 296 (2004) 650–657
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fact that f1 has positive Hayman index gives f ∈ U(p, k). Set tj = rj , tj = r j , then (1 − tj )(1 − tj )−1 k(1 − rj )(1 − rj )−1 = kC(ε). By (3.5), 2 2 2 2 f (z) f1 (z) dx dy = p dx dy π(1 − ε)p log 1 f (z) k f1 (z) k 1 − rj 1/ k
tj |z|tj
1/ k
rj |z|rj
1 (1 − ε)πp2 log k k(1 − tj )
for all large j . Thus the proof of Theorem 3.1 is now complete. 2
References [1] [2] [3] [4] [5] [6] [7] [8]
W.K. Hayman, Multivalent Functions, second ed., Cambridge Univ. Press, Cambridge, 1994. W.K. Hayman, The logarithmic derivative of multivalent functions, Michigan Math. J. 27 (1980) 149–179. B.G. Eke, The asymptotic behaviour of areally mean valent functions, J. Anal. Math. 20 (1967) 147–212. X.H. Dong, On a theorem of Bazilevic for areally mean p-valent functions attaining maximal growth on k rays, Israel J. Math. 100 (1997) 327–337. P.L. Duren, Y.L. Leung, Logarithmic coefficients of univalent functions, J. Anal. Math. 36 (1979) 36–43. P.L. Duren, Univalent Functions, Springer-Verlag, Berlin, 1983. M.M. Elhosh, On mean p-valent functions, Rev. Romaine Math. Pures Appl. 34 (1989) 11–15. E.C. Titchmarsh, The Theory of Functions, Oxford Univ. Press, 1952.
The logarithmic derivative of areally mean p-valent functions ✩ Xin-Han Dong Department of Mathematics, Hunan Normal University, Changsha 410081, China Received 23 November 2003 Available online 10 June 2004 Submitted by S. Ruscheweyh
Abstract Let U(p, k) denote the class of functions f which are areally mean p-valent and have k maximal growth directions in ∆ = {z: |z| < 1}. In this paper we give an asymptotic formula of the mean square of the logarithmic derivative f /f for f ∈ U(p, k) with some restricted condition. This generalizes the corresponding result in [J. Anal. Math. 36 (1979) 36–43]. Also we construct a function f ∈ U(p, k) which does not satisfy this asymptotic formula. 2004 Elsevier Inc. All rights reserved. Keywords: Areally mean p-valent function; Logarithmic derivative; Asymptotic behavior
1. Introduction Let f be analytic in the unit disk ∆ = {z: |z| < 1} and let n(f = w, D) be the number of roots of the equation f (z) = w in D (⊂ ∆). We say that f is areally mean p-valent in ∆ (we denote the class by U(p)) if for every R > 0, 1 n(f = w, ∆) du dv p, πR 2 |w|R
where w = u + vi, and p is a positive real number, not necessarily an integer. It is clear that if f is univalent and analytic in ∆ then n(f = w, ∆) 1 for each w ∈ C. Hence a ✩
The research is supported in part by the National Natural Science Foundation of China (No. 19871026). E-mail address: [email protected].
0022-247X/$ – see front matter 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2004.04.032
X.-H. Dong / J. Math. Anal. Appl. 296 (2004) 650–657
651
function f analytic and univalent in ∆ must be areally mean 1-valent. Let f ∈ U(p) and let z1 , z2 , . . . , zq be the zeros of f (z) in ∆, repeated according to multiplicity, so that q p [1, p. 38]. We choose an appropriate constant b such that ∞
log
f (z) = 2p cn z n h(z)
(1.1)
n=1
q is analytic in |z| < 1, where h(z) = b j =1 (z − zj ). For integer k, we say that f is in U(p, k) if f ∈ U(p) and if we can find k distinct points t1 = eiθ1 , . . . , tk = eiθk on |z| = 1, − and a constant d > 0 and a sequence {rn }∞ n=1 with rn → 1 as n → ∞ such that 1 (1.2) d(1 − rn )−2p/ k f (rn eiθs ) M(rn ) (1 − rn )−2p/ k d for s = 1, 2, . . . , k and all n 1, where M(r) = M(r, f ) = max|z|=r |f (z)|. For f ∈ U(p, k) we say that f attains maximal growth on the k rays ts , s = 1, 2, . . . , k. The class U(p, k) was introduced by Eke in [3, p. 148]. We proved in [4, p. 333] that (1.2) is an equivalent definition of U(p, k) which is more explicit than the original. In this paper, we consider the asymptotic growth of 2π iθ 2 f (re ) f 1 dθ = I2 r, f 2π f (reiθ )
(1.3)
0
for f ∈ U(p, k). This topic has been investigated by Duren and Leung [5], [2] Hayman 2 |c |2 , n and Elhosh [7]. A relative problem is to discuss the asymptotic behavior of N n n=1 where {cn } are defined by (1.1). In [5], Duren and Leung proved Theorem A. Suppose that f (z) = z + a2 z2 + · · · is univalent and analytic in ∆ with positive Hayman index, i.e., limr→1 (1 − r)2 M(r) = α > 0. If ∞
log
f (z) =2 cn z n , z
|z| < 1,
n=1
then there exist constants C1 > 0, C2 > 0 such that f (1 − r)I2 r, C1 , 0 < r < 1, f and N −1
N
n2 |cn |2 C2 ,
N = 1, 2, . . . .
(1.4)
(1.5)
n=1
Our aim is to prove thatthe function (1 − r)I2 (r, f /f ) in (1.4) has a limit as r → 1− . 2 2 Hence the sequence N −1 N n=1 n |cn | in (1.5) also has a limit as N → ∞. In fact, we prove these results for f ∈ U(p, k) with some restricted condition (see (2.1)). It is obvious that if f satisfies the normalized condition, i.e., f (z) = zp (1 + a1 z + · · ·), then this restricted condition is trivial (see Corollary 2.2). Also, we construct a function f ∈ U(p, k) 2 |c |2 = o(log N). n such that (1 − r)I2 (r, f /f ) = o(log(1/(1 − r))) and N −1 N n n=1
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X.-H. Dong / J. Math. Anal. Appl. 296 (2004) 650–657
2. The theorem Theorem 2.1. Suppose that f ∈ U(p, k), and that 1
1 n f = w, r0 |z| < 1 du dv dR < ∞, R3
(2.1)
|w|R
0
where w = u + vi and r0 = (1/4)(3 + max1j q |zj |). Then 2p2 f → as r → 1 (1 − r)I2 r, f k
(2.2)
and N
−1
N
n2 |cn |2 →
n=1
1 k
as N → ∞,
(2.3)
where {cn } are defined as in (1.1). Proof. It is easy to check that 2π 2π f (z) 2 r 2 f r 2 h (z) 2 log = r I2 r, dθ − dθ f 2π h(z) 2π h(z) 2
0
+
r2 Re π
0
2π
f (z) h (z) dθ. f (z) h(z)
0
Let z = reiθ (r r0 ). Then computation shows r2 π
2π
q f (z) h (z) r2 dθ = f (z) h(z) πi j =1
0
=
|z|=r q q j =1 s=1
=
q q j =1 s=1
1 πi
1 f (z) dz f (z) r 2 − zj z
|z−zs |=ε
r2
r2 dz − zj z z − zs
2r 2 . r 2 − zj zs
We write r2 B1 (r) = − 2π
2π 2 q q h (z) r2 dθ + 2 . h(z) 2 r − zj zs 0
j =1 s=1
It is clear that B1 (r) is a bounded function on [r0 , 1]. Let t1 = eiθ1 , . . . , tk = eiθk be the maximal growth directions of f (z). Then we have by the above that
X.-H. Dong / J. Math. Anal. Appl. 296 (2004) 650–657
∞ f r I2 r, n2 |cn |2 r 2n + B1 (r) = 4p2 f n=1 2 k 2 k ∞ ∞ 1 n 2n 1 2 2 2 n 2n = 4p n c n − t j r − 4p tj r k nk n=1 j =1 n=1 j =1
k ∞ 4p + Re 2p ncn tjn r 2n + B1 (r). k
653
2
j =1
(2.4)
n=1
By Theorem 3.1 in [4, p. 334], (2.1) implies 2 ∞ k 1 n ncn − t j < ∞, nk j =1
n=1
hence ∞ n=1
2 k 1 o(1) , n2 cn − t nj r 2n = kn 1−r
r → 1.
(2.5)
j =1
Observe the representation 2p
∞
ncn tjn r 2n =
n=1
r 2 tj f (r 2 tj ) r 2 tj h (r 2 tj ) − , f (r 2 tj ) h(r 2 tj )
and note the asymptotic formula r 2 tj f (r 2 tj ) 2p + o(1) = , f (r 2 tj ) k(1 − r 2 )
r → 1 (see (14.2) in [3, p. 192]),
we get
k ∞ 8p2 + o(1) 4p n 2n , Re 2p ncn tj r = k k(1 − r 2 ) j =1
r → 1.
n=1
It follows by (2.4)–(2.6) that k 2 ∞ 2 1 4p 1 f o(1) r 2 I2 r, = − 4p2 tjn r 2n + f k 1−r k 1−r n=1
=
2p2
1 o(1) + k 1−r 1−r
j =1
as r → 1.
This proves (2.2). (2.2) and the first equality in (2.4) give (1 − r)
∞ n=1
n2 |cn |2 r n =
√ f 1 1 (1 − r)I r, + o(1) = + o(1) 2 2 4p f k
(2.6)
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X.-H. Dong / J. Math. Anal. Appl. 296 (2004) 650–657
as r → 1. By virtue of Tauberian theorem [8], we obtain N −1
N n=1
1 n2 |cn |2 → , k
N → ∞.
2
Corollary 2.2. Suppose that f (z) = zp (1 + a2 z + · · ·) ∈ U(p, k), where f (z) may be defined in D = ∆ \ (−1, 0], if necessary, and that ∞
log
f (z) = 2p cn z n , p z
|z| < 1.
n=1
Then f 2p2 (1 − r)I2 r, → f k
as r → 1−
and N −1
N
n2 |cn |2 →
n=1
1 k
as N → +∞.
Proof. From the definition of areally mean p-valent function, it is easy to see that there exist ρ0 ∈ (0, 1) and R0 > 0 such that
n f = w, ρ0 |z| < 1 = 0, |w| < R0 . Hence (2.1) holds. 2 Corollary 2.3. Suppose that f (z) = z + a2 z2 + · · · ∈ U(1, 1) and ∞
log
f (z) =2 cn z n , z
|z| < 1.
n=1
Then f (1 − r)I2 r, →2 f
as r → 1−
and N −1
N
n2 |cn |2 → 1 as N → ∞.
n=1
If f is univalent and analytic in ∆ with positive Hayman index, then f ∈ U(1, 1). Therefore, Theorem A follows directly by Corollary 2.3.
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3. An example Observing Theorem 2.1, naturally one can ask whether the condition (2.1) is necessary, i.e., whether the limit limr→1 (1 − r)I2 (r, f /f ) exists, or whether the function (1 − r)I2 (r, f /f ) is bounded on [r0 , 1) for f ∈ U(p, k). We now construct a function f (z) ∈ U(p, k) by revising Hayman’s example in [2, p. 166], which satisfies 1 f (1 − r)I2 r, = o log as r → 1. f 1−r 2 2 Similar to the proof of Lemma 1 in [5], we have N −1 N n=1 n |cn | = o(log N). Theorem 3.1. There exists a circumferentially mean p-valent function f (z) in ∆, attaining maximal growth on k rays (naturally, f ∈ U(p, k)), such that f 1−r > 0. (3.1) I2 r, lim sup f r→1 log(1/(1 − r)) Proof. Our goal is to construct a circumferentially mean p-valent function f (z) in ∆, attaining maximal growth on k rays, such that for any ε > 0 there are a positive constant C(ε) (depending only on ε) and sequences tj , tj (both → 1− as n → ∞) with (1 − tj ) × (1 − tj )−1 kC(ε), while tj |z|tj
2 2 f (z) 1 dx dy (1 − ε)πp log f (z) k k(1 − tj )
(3.2)
for all large j . Thus (3.1) follows easily from (3.2). Firstly, we consider the case p = 1 and k = 1. For positive integer n, take positive integer j satisfying 2(j −1) n < 2j , 2
2
(3.3) j2
and define d−n by −2 log d−n = 2 − n + j . Set S = s = σ + ti: σ ∈ (−∞, ∞), |t| < π , ∞ π 1 |t| < π , T= s = −n + + ti: 2 2 n=3 ∞ π Γ = s = σ ± i: −n − d−n σ −n + d−n , 2 n=3
and Ω = S \ {T ∪ Γ }. Then Ω is a simply connected domain in the s plane, and is a symmetrical about the real axis. From the Riemann mapping theorem, there exists an analytic function ψ1 (z) which
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maps ∆(1, 1) conformally onto Ω and has the properties ψ1 (0) = 0 and ψ1 (0) > 0. Define f1 (z) = exp{ψ1 (z)} and ψ2 (z) = log(1/(1 − z)2 ) (with ψ2 (0) = 0) for z ∈ ∆. Clearly, f1 (z) and ψ2 (z) both are univalent and analytic in ∆. Since Re ψ2 (z) −2 log 2 and | Im ψ2 (z)| < π for z ∈ ∆, we have ψ2 (∆) ⊂ Ω and ψ2 (0) = ψ1 (0). It follows from the subordination principle [6, p. 190] that there exists a function ϕ(z) analytic in ∆ such that |ϕ(z)| |z| and (1 − z)−2 = f1 (ϕ(z)) for |z| < 1. This shows that M(r) = M(r, f1 ) (1 − r)−2 , 0 < r < 1, and hence f1 has a positive Hayman index α > 0. So f ∈ U(1, 1). + − For given ε > 0, we define E−n and E−n (n = 3, 4, . . .) as in [2, p. 167]. From the proof of Theorem 3 in [2, p. 167–177], we know that (5.14) in [2, p. 167] holds also for + − domain Ω, that is, if s ∈ E−n or E−n , then 2
2
2j + j − C1 (ε) 2d(0, s, Ω) < 2j + j + C1 (ε),
(3.4)
where d(s1 , s2 , Ω) stands for the hyperbolic distance of two points s1 , s2 in Ω, j is defined by (3.3), and C1 (ε) is a suitable positive constant depending only on ε. We define rj , rj by 2
2j + j + C1 (ε) = log
1 , 1 − rj
2
2j + j − C1 (ε) = log
2 . 1 − rj
Clearly 0 < rj < rj < 1, rj → 1 (j → ∞) and (1 − rj )(1 − rj )−1 = 2e2C1 (ε) = C(ε). Noting
1 1 + |z| , d 0, ψ1 (z), Ω = log 2 1 + |z| + − we obtain by (3.4) that if s = ψ1 (z) ∈ E−n or E−n then
log
1 2 1 + |z| log log . 1 − rj 1 − |z| 1 − rj
Thus +
− ∪ ψ1−1 E−n ⊂ z: rj |z| rj , ψ1−1 E−n and hence rj |z|rj
2 f1 (z) f (z) dx dy = 1
2 ψ (z) dx dy 2
2
j −1 2
+ area E−n
1
rj |z|rj
n=2(j−1)
2
j2 ε π (j −1)2 2 2 −2 1− 2 2 j2
π(1 − ε) 2 + j − C1 (ε) 1 = π(1 − ε) log for all large j. 1 − rj
(3.5)
In the general case, we define f (z) = f1 (zk )p/ k , |z| < 1. Since f1 is univalent in ∆, we get by Lemma 5.1 in [1, p. 145] that f is circumferentially mean p-valent in ∆. The
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fact that f1 has positive Hayman index gives f ∈ U(p, k). Set tj = rj , tj = r j , then (1 − tj )(1 − tj )−1 k(1 − rj )(1 − rj )−1 = kC(ε). By (3.5), 2 2 2 2 f (z) f1 (z) dx dy = p dx dy π(1 − ε)p log 1 f (z) k f1 (z) k 1 − rj 1/ k
tj |z|tj
1/ k
rj |z|rj
1 (1 − ε)πp2 log k k(1 − tj )
for all large j . Thus the proof of Theorem 3.1 is now complete. 2
References [1] [2] [3] [4] [5] [6] [7] [8]
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