Logarithmic Coefficients and a Coefficient Conjecture for Univalent Functions

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by  M. Obradović

LOGARITHMIC COEFFICIENTS AND A COEFFICIENT
CONJECTURE FOR UNIVALENT FUNCTIONS
MILUTIN OBRADOVIC´, SAMINATHAN PONNUSAMY, AND KARL-JOACHIM WIRTHS
Abstract. Let U(λ) denote the family of analytic functions f(z), f(0) = 0 =
f(cid:48)(0)−1, in the unit disk D, which satisfy the condition (cid:12)(cid:12)(cid:0)z/f(z)(cid:1)2f(cid:48)(z)−1(cid:12)(cid:12)<λ
7 forsome0<λ≤1. Thelogarithmiccoefficientsγn off aredefinedbytheformula
1 log(f(z)/z) = 2(cid:80)∞ γ zn. In a recent paper, the present authors proposed a
n=1 n
0 conjecture that if f ∈ U(λ) for some 0 < λ ≤ 1, then |a | ≤ (cid:80)n−1λk for n ≥ 2
2 n k=0
and provided a new proof for the case n = 2. One of the aims of this article

r is to present a proof of this conjecture for n = 3,4 and an elegant proof of the
p
A inequality for n=2, with equality for f(z)=z/[(1+z)(1+λz)]. In addition, the
authors prove the following sharp inequality for f ∈U(λ):

6
(cid:88)∞ 1(cid:18)π2 (cid:19)
|γ |2 ≤ +2Li (λ)+Li (λ2) ,
] n 4 6 2 2
V n=1
C where Li2 denotes the dilogarithm function. Furthermore, the authors prove two
. such new inequalities satisfied by the corresponding logarithmic coefficients of
h
some other subfamilies of S.
t
a
m
[
1. Introduction
2
v
Let A be the class of functions f analytic in the unit disk D = {z ∈ C : |z| < 1}
3
1 with the normalization f(0) = 0 = f(cid:48)(0)−1. Let S denote the class of functions f
4 from A that are univalent in D. Then the logarithmic coefficients γ of f ∈ S are
5 n
0 defined by the formula
.
1 (cid:18) (cid:19) ∞
1 f(z) (cid:88)
0 (1) log = γ zn, z ∈ D.
n
7 2 z
1 n=1
: These coefficients play an important role for various estimates in the theory of
v
i univalent functions. When we require a distinction, we use the notation γ (f)
X n
instead of γ . For example, the Koebe function k(z) = z(1−eiθz)−2 for each θ has
r n
a logarithmic coefficients γ (k) = einθ/n, n ≥ 1. If f ∈ S and f(z) = z +(cid:80)∞ a zn,
n n=2 n
thenby(1)itfollowsthat2γ = a andhence,bytheBieberbachinequality,|γ | ≤ 1.
1 2 1
Let S(cid:63) denote the class of functions f ∈ S such that f(D) is starlike with respect to
theorigin. Functionsf ∈ S(cid:63) arecharacterizedbytheconditionRe(zf(cid:48)(z)/f(z)) > 0
in D. The inequality |γ | ≤ 1/n holds for starlike functions f ∈ S, but is false for
n
the full class S, even in order of magnitude. See [4, Theorem 8.4 on page 242]. In [6],
2010 Mathematics Subject Classification. 30C45.
Key words and phrases. Univalent, starlike, convex and close-to-convex functions, subordina-
tion, logarithmic coefficients and coefficient estimates .
File: Ob-S-Wirths4˙LogCoef2016˙1˙˙arXiv.tex, printed: 7-4-2017, 0.57.
1
2 M. Obradovi´c, S. Ponnusamy and K.-J. Wirths
Girela pointed out that this bound is actually false for the class of close-to-convex
functionsinDwhichisdefinedasfollows: Afunctionf ∈ Aiscalledclose-to-convex,
denoted by f ∈ K, if there exists a real α and a g ∈ S(cid:63) such that
(cid:18) zf(cid:48)(z)(cid:19)
Re eiα > 0, z ∈ D.
g(z)
For 0 ≤ β < 1, a function f ∈ S is said to belong to the class of starlike functions
of order β, denoted by f ∈ S(cid:63)(β), if Re(zf(cid:48)(z)/f(z)) > β for z ∈ D. Note that
S(0) =: S(cid:63). The class of all convex functions of order β, denoted by C(β), is then
defined by C(β) = {f ∈ S : zf(cid:48) ∈ S(cid:63)(β)}. The class C(0) =: C is usually referred
to as the class of convex functions in D. With the class S being of the first priority,
its subclasses such as S(cid:63), K, and C, respectively, have been extensively studied in
the literature and they appear in different contexts. We refer to [4, 7, 10, 12] for a
general reference related to the present study. In [5, Theorem 4], it was shown that
the logarithmic coefficients γ of every function f ∈ S satisfy
n
(cid:88)∞ π2
(2) |γ |2 ≤
n
6
n=1
and the equality is attained for the Koebe function. The proof uses ideas from
the work of Baernstein [3] on integral means. However, this result is easy to prove
(see Theorem 1) in the case of functions in the class U := U(1) which is defined as
follows:
(cid:40) (cid:12) (cid:12) (cid:41)
(cid:12)(cid:18) z (cid:19)2 (cid:12)
U(λ) = f ∈ A : (cid:12) f(cid:48)(z)−1(cid:12) < λ, z ∈ D ,
(cid:12) f(z) (cid:12)
(cid:12) (cid:12)
where λ ∈ (0,1]. It is known that [1, 2, 11] every f ∈ U is univalent in D and hence,
U(λ) ⊂ U ⊂ S for λ ∈ (0,1]. The present authors have established many interesting
properties of the family U(λ). See [10] and the references therein. For example, if
f ∈ U(λ) for some 0 < λ ≤ 1 and a = f(cid:48)(cid:48)(0)/2, then we have the subordination
2
relations
f(z) 1 1
(3) ≺ = , z ∈ D,
z 1+(1+λ)z +λz2 (1+z)(1+λz)
and
z
+a z ≺ 1+2λz +λz2, z ∈ D.
2
f(z)
Here ≺ denotes the usual subordination [4, 7, 12]. In addition, the following conjec-
ture was proposed in [10].
Conjecture 1. Suppose that f ∈ U(λ) for some 0 < λ ≤ 1. Then |a | ≤ (cid:80)n−1λk
n k=0
for n ≥ 2.
In Theorem 1, we present a direct proof of an inequality analogous to (2) for
functions in U(λ) and in Corollary 1, we obtain the inequality (2) as a special case
for U. At the end of Section 2, we also consider estimates of the type (2) for some
interesting subclasses of univalent functions. However, Conjecture 1 remains open
for n ≥ 5. On the other hand, the proof for the case n = 2 of this conjecture is due
to [17] and an alternate proof was obtained recently by the present authors in [10,
Logarithmic Coefficients for Univalent Functions 3
Theorem 1]. In this paper, we show that Conjecture 1 is true for n = 3,4. and our
proof includes an elegant proof of the case n = 2. The main results and their proofs
are presented in Sections 2 and 3.
2. Logarithmic coefficients of functions in U(λ)
Theorem 1. For 0 < λ ≤ 1, the logarithmic coefficients of f ∈ U(λ) satisfy the
inequality
(cid:88)∞ 1 (cid:18)π2 (cid:19)
(4) |γ |2 ≤ +2Li (λ)+Li (λ2) ,
n 2 2
4 6
n=1
where Li denotes the dilogarithm function given by
2
(cid:88)∞ zn (cid:90) 1 log(1/t)
Li (z) = = z dt.
2 n2 1−tz
n=1 0
The inequality (4) is sharp. Further, there exists a function f ∈ U such that |γ | >
n
(1+λn)/(2n) for some n.
Proof. Let f ∈ U(λ). Then, by (3), we have
z
≺ (1−z)(1−λz)
f(z)
which clearly gives
∞ (cid:114) ∞
(cid:88) f(z) −log(1−z)−log(1−λz) (cid:88) 1
(5) γ zn = log ≺ = (1+λn)zn.
n
z 2 2n
n=1 n=1
Again, by Rogosinski’s theorem (see [4, 6.2]), we obtain
(cid:32) (cid:33)
(cid:88)∞ (cid:88)∞ 1 1 (cid:88)∞ 1 (cid:88)∞ λn (cid:88)∞ λ2n
|γ |2 ≤ (1+λn)2 = +2 +
n 4n2 4 n2 n2 n2
n=1 n=1 n=1 n=1 n=1
and the desired inequality (4) follows. For the function
z
g (z) = ,
λ
(1−z)(1−λz)
we find that γ (g ) = (1+λn)/(2n) for n ≥ 1 and therefore, we have the equality
n λ
in (4). Note that g (z) is the Koebe function z/(1−z)2.
1
From the relation (5), we cannot conclude that
1+λn
|γ (f)| ≤ |γ (g )| = for f ∈ U(λ).
n n λ
2n
Indeed for the function f defined by
λ
z
(6) f (z) =
λ
(1−z)(1−λz)(1+(λ/(1+λ))z)
we find that
z λ−(1+λ)2 λ2
= 1+ z + z3
f (z) 1+λ 1+λ
λ
4 M. Obradovi´c, S. Ponnusamy and K.-J. Wirths
1.0
1.0
0.5 0.5
0.0 0.0
-0.5 -0.5
-1.0 -1.0
-0.5 0.0 0.5 1.0 -1.0 -0.5 0.0 0.5
(a )λ = 0.25 (b) λ = 0.5
1.0
1.0
0.5
0.5
0.0 0.0
-0.5
-0.5
-1.0
-1.0
-1.0 -0.5 0.0 0.5 -1.0 -0.5 0.0 0.5
(c) λ = 0.75 (d) λ = 1
z
Figure 1. The image of f (z) =
λ
(1−z)(1−λz)(1+(λ/(1+λ))z)
under D for certain values of λ
and
(cid:18) z (cid:19)2 2λ2 (cid:18) (1+2λ)(1−λ)(cid:19)
f(cid:48)(z)−1 = − z3 = − 1− z3
f (z) λ 1+λ 1+λ
λ
which clearly shows that f ∈ U(λ). The images of f (z) under D for certain values
λ λ
of λ are shown in Figures 1(a)-(d). Moreover, for this function, we have
(cid:18) (cid:19) (cid:18) (cid:19)
f (z) λ
λ
log = −log(1−z)−log(1−λz)−log 1+ z
z 1+λ
∞
(cid:88)
= 2 γ (f )zn,
n λ
n=1
where
1 (cid:18)1+λn λn (cid:19)
γ (f ) = +(−1)n .
n λ 2 n n(1+λ)n
Logarithmic Coefficients for Univalent Functions 5
This contradicts the above inequality at least for even integer values of n ≥ 2.
Moreover, with these γ (f ) for n ≥ 1, we obtain
n λ
(cid:88)∞ 1 (cid:88)∞ (cid:26)(1+λn)2 (−1)n (cid:20)(cid:18) λ2 (cid:19)n
|γ (f )|2 = +2
n λ 4 n2 n2 1+λ
n=1 n=1
(cid:41)
(cid:18) λ (cid:19)n(cid:21) 1 (cid:18) λ (cid:19)2n
+ +
1+λ n2 1+λ
and by a computation, it follows easily that
(cid:88)∞ 1 (cid:18)π2 (cid:19)
|γ (f )|2 = +2Li (λ)+Li (λ2)
n λ 2 2
4 6
n=1
1 (cid:20) (cid:18) −λ2 (cid:19) (cid:18) −λ (cid:19)(cid:21) 1 (cid:18) λ2 (cid:19)
+ Li +Li + Li
2 2 1+λ 2 1+λ 4 2 (1+λ)2
1 (cid:18)π2 (cid:19) 1
= +2Li (λ)+Li (λ2) + A(λ)
2 2
4 6 4
1 (cid:18)π2 (cid:19)
< +2Li (λ)+Li (λ2) for 0 < λ ≤ 1,
2 2
4 6
and we complete the proof, provided A(λ) < 0 for 0 < λ ≤ 1. Now, we claim that
(cid:20) (cid:18) −λ2 (cid:19) (cid:18) −λ (cid:19)(cid:21) (cid:18) λ2 (cid:19)
A(λ) := 2 Li +Li +Li < 0.
2 1+λ 2 1+λ 2 (1+λ)2
Because Li (z2) = 2(Li (z)+Li (−z)), the last claim is equivalent to
2 2 2
A(λ) (cid:18) −λ (cid:19) (cid:20) (cid:18) λ (cid:19) (cid:18) −λ2 (cid:19)(cid:21)
= 2Li + Li +Li < 0
2 2 2
2 1+λ 1+λ 1+λ
for 0 < λ ≤ 1. According to the integral representation of Li (z) given in the
2
statement of Theorem 1, we can write
(cid:90) 1
A(λ) = −2λ B(λ,t)log(1/t)dt,
0
where
2 1 λ
B(λ,t) = − +
1+λ+tλ 1+λ−tλ 1+λ+tλ2
(1+λ)−3tλ λ
= +
(1+λ)2 −t2λ2 1+λ+tλ2
N(λ,t)
=
[(1+λ)2 −t2λ2][1+λ+tλ2]
with
N(λ,t) = (1+λ)3 −(3−λ)(1+λ)λt−4λ3t2.
Clearly, B(1,t) > 0 for t ∈ [0,1) and it follows that, A(1) < 0. On the other hand,
since N(λ,t) is a decreasing function of t for t ∈ [0,1], we obtain that
N(λ,t) ≥ N(λ,1) = (1+λ)3 −(3−λ)(1+λ)−4λ3 = 1−λ3 +λ2(1−λ) > 0
6 M. Obradovi´c, S. Ponnusamy and K.-J. Wirths
for 0 < λ < 1. Consequently, B(λ,t) > 0 for all t ∈ [0,1] and for 0 < λ < 1. This
observation shows that A(λ) < 0 for 0 < λ ≤ 1. This proves the claim and thus, the
proof is complete. (cid:3)
Corollary 1. The logarithmic coefficients of f ∈ U satisfy the inequality
(cid:88)∞ (cid:88)∞ 1 π2
(7) |γ |2 ≤ = .
n n2 6
n=1 n=1
We have equality in the last inequality for the Koebe function k(z) = z(1−eiθz)−2.
Further there exists a function f ∈ U such that |γ | > 1/n for some n.
n
Remark 1. From the analytic characterization of starlike functions, it is easy to
see that for f ∈ S(cid:63),
zf(cid:48)(z) (cid:18) (cid:18)f(z)(cid:19)(cid:19)(cid:48) (cid:88)∞ 2z
−1 = z log = 2 nγ zn ≺
n
f(z) z 1−z
n=1
and thus, by Rogosinski’s result, we obtain that |γ | ≤ 1/n for n ≥ 1. In fact for
n
starlike functions of order α, α ∈ [0,1), the corresponding logarithmic coefficients
satisfy the inequality |γ | ≤ (1−α)/n for n ≥ 1. Moreover, one can quickly obtain
n
that
(cid:88)∞ π2
|γ |2 ≤ (1−α)2
n
6
n=1
if f ∈ S(cid:63)(α), α ∈ [0,1) (See also the proof of Theorem 2 and Remark 3). As
remarked in the proof of Theorem 1, from the relation (7), we cannot conclude the
same fact, namely, |γ | ≤ 1/n for n ≥ 1, for the class U although the Koebe function
n
k(z) = z/(1−z)2 belongs to U ∩S(cid:63). For example, if we set λ = 1 in (6), then we
have
z (cid:16) z(cid:17) 3 z3
= (1−z)2 1+ = 1− z + ,
f (z) 2 2 2
1
where f ∈ U and for this function, we obtain
1
(cid:88)∞ (cid:88)∞ (cid:18)1 1 (cid:19)2
|γ (f )|2 = +(−1)n
n 1 n n2n+1
n=1 n=1
π2 1 (cid:18)1(cid:19) (cid:18) 1(cid:19)
= + Li +Li −
2 2
6 4 4 2
π2 1 (cid:20) (cid:18)1(cid:19) (cid:18)−1(cid:19)(cid:21)
= + Li +3Li ,
2 2
6 2 2 2
where we have used the fact that Li (z2) = 2(Li (z)+Li (−z)). From the proof of
2 2 2
Theorem 1, we conclude that
(cid:88)∞ π2
|γ (f )|2 < ,
n 1
6
n=1
Logarithmic Coefficients for Univalent Functions 7
because
(cid:18) (cid:19) (cid:18) (cid:19)
1 −1
Li +3Li < 0.
2 2
2 2
As a direct approach, it is easy to see that
(cid:88)∞ 1 (cid:88)∞ z2k (cid:88)∞ z2k−1
Li (z)+3Li (−z) = (1+3(−1)n)zn = −2
2 2 n2 k2 (2k −1)2
n=1 k=1 k=1
and thus, we obtain that
(cid:88)∞ π2 1 (cid:88)∞ 1 (cid:18) 1 1 (cid:19) π2 (cid:88)∞ 1 (cid:18) 4k −1 (cid:19)
|γ (f )|2 = + − = −
n 1 6 2 4k k2 (k −1/2)2 6 4k k2(2k −1)2
n=1 k=1 k=1
and thus,
(cid:88)∞ π2
|γ (f )|2 < .
n 1
6
n=1
On the other hand, it is a simple exercise to verify that f ∈/ S(cid:63). The graph of this
1
function is shown in Figure 1(d).
Let G(α) denote the class of locally univalent normalized analytic functions f in
the unit disk |z| < 1 satisfying the condition
(cid:18) zf(cid:48)(cid:48)(z)(cid:19) α
Re 1+ < 1+ for |z| < 1,
f(cid:48)(z) 2
and for some 0 < α ≤ 1. Set G(1) =: G. It is known (see [13, Equation (16)])
that G ⊂ S(cid:63) and thus, functions in G(α) are starlike. This class has been studied
extensively in the recent past, see for instance [9] and the references therein. We
now consider the estimate of the type (2) for the subclass G(α).
Theorem 2. Let 0 < α ≤ 1 and G(α) be defined as above. Then the logarithmic
coefficients γ of f ∈ G(α) satisfy the inequalities
n
∞
(cid:88) α
(8) n2|γ |2 ≤
n
4(α+2)
n=1
and
(cid:88)∞ α2 (cid:18) 1 (cid:19)
(9) |γ |2 ≤ Li .
n 4 2 (1+α)2
n=1
Also we have
α
(10) |γ | ≤ for n ≥ 1.
n
2(α+1)n
Proof. If f ∈ G(α), then we have (see eg. [8, Theorem 1] and [13])
zf(cid:48)(z) (1+α)(1−z) (cid:18) z/(1+α) (cid:19)
(11) −1 ≺ −1 = −α , z ∈ D,
f(z) 1+α−z 1−(z/(1+α))
8 M. Obradovi´c, S. Ponnusamy and K.-J. Wirths
which, in terms of the logarithmic coefficients γ of f defined by (1), is equivalent
n
to
(cid:88)∞ (cid:88)∞ zn
(12) (−2nγ )zn ≺ α .
n (1+α)n
n=1 n=1
Again, by Rogosinski’s result, we obtain that
∞ ∞
(cid:88) (cid:88) 1 α
4n2|γ |2 ≤ α2 =
n (1+α)2n α+2
n=1 n=1
which is (8).
Now, since the sequence A = 1 is convex decreasing, we obtain from (12)
n (1+α)n
and [15, Theorem VII, p.64] that
α
|−2nγ | ≤ αA = ,
n 1
1+α
which implies the desired inequality (10). As an alternate approach to prove this
inequality, we may rewrite (11) as
(cid:88)∞ (cid:18) (cid:18)f(z)(cid:19)(cid:19)(cid:48) (cid:18) z/(1+α) (cid:19)
(2nγ )zn = z log ≺ φ(z) = −α
n
z 1−(z/(1+α))
n=1
and, since φ(z) is convex in D with φ(cid:48)(0) = −α/(1+α), it follows from Rogosinski’s
result (see also [4, Theorem 6.4(i), p. 195]) that |2nγ | ≤ α/(1 + α). Again, this
n
proves the inequality (10).
Finally, we prove the inequality (9). From the formula (12) and the result of
Rogosinski (see also [12, Theorem 2.2] and [4, Theorem 6.2]), it follows that for
k ∈ N the inequalities
(cid:88)k α2 (cid:88)k 1
n2|γ |2 ≤
n 4 (1+α)2n
n=1 n=1
arevalid. Clearly, thisimpliestheinequality(8)aswell. Ontheotherhand, consider
these inequalities for k = 1,...,N, and multiply the k-th inequality by the factor
1 − 1 , if k = 1,...,N −1 and by 1 for k = N. Then the summation of the
k2 (k+1)2 N2
multiplied inequalities yields
(cid:88)N α2 (cid:88)N 1
|γ |2 ≤
k 4 k2(1+α)2k
k=1 k=1
α2 (cid:88)∞ 1
≤
4 k2(1+α)2k
k=1
α2 (cid:18) 1 (cid:19)
= Li for N = 1,2,...,
4 2 (1+α)2
which proves the desired assertion (9) if we allow N → ∞. (cid:3)
Logarithmic Coefficients for Univalent Functions 9
Corollary 2. The logarithmic coefficients γ of f ∈ G := G(1) satisfy the inequali-
n
ties
∞ ∞ (cid:18) (cid:19)
(cid:88) 1 (cid:88) 1 1
n2|γ |2 ≤ and |γ |2 ≤ Li .
n n 2
12 4 4
n=1 n=1
The results are the best possible as the function f (z) = z−1z2 shows. Also we have
0 2
|γ | ≤ 1/(4n) for n ≥ 1.
n
Remark 2. For the function f (z) = z − 1z2, we have that γ (f ) = − 1 for
0 2 n 0 n2n+1
n = 1,2,... and thus, it is reasonable to expect that the inequality |γ | ≤ 1 is
n n2n+1
valid for the logarithmic coefficients γ of each f ∈ G. But that is not the case as
n
the function f defined by f(cid:48)(z) = (1−zn)1 shows. Indeed for this function we have
n n n
zf(cid:48)(cid:48)(z) 1−2zn
1+ n =
f(cid:48)(z) 1−zn
n
showing that f ∈ G. Moreover,
n
f (z) 1
log n = − zn +··· ,
z n(n+1)
which implies that |γ (f )| = 1 for n = 1,2,..., and observe that 1 >
n n 2n(n+1) 2n(n+1)
1 for n = 2,3,.... Thus, we conjecture that the logarithmic coefficients γ of
n2n+1 n
each f ∈ G satisfy the inequality |γ | ≤ 1 for n = 1,2,.... Clearly, Corollary
n 2n(n+1)
2 shows that the conjecture is true for n = 1.
Remark 3. Let f ∈ C(α), where 0 ≤ α < 1. Then we have [18]
zf(cid:48)(z) (cid:88)∞
(13) −1 ≺ G (z)−1 = δ zn,
α n
f(z)
n=1
where δ is real for each n,
n

(2α−1)z
 if α (cid:54)= 1/2,
 (1−z)[(1−z)1−2α −1]
G (z) =
α −z
 if α = 1/2,

(1−z)log(1−z)
and
 1−2α
 if 0 ≤ α (cid:54)= 1/2 < 1,

 2[21−2α −1]
β(α) = G (−1) = inf G (z) =
α α 1
|z|<1  if α = 1/2

2log2
so that f ∈ S(cid:63)(β(α)). Also, we have [16]

(1−z)2α−1 −1
 if 0 ≤ α (cid:54)= 1/2 < 1,
f(z) K (z)  (1−2α)z
α
≺ =
z z  log(1−z)
 − if α = 1/2,
z
and K (z)/z is univalent and convex (not normalized in the usual sense) in D.
α
10 M. Obradovi´c, S. Ponnusamy and K.-J. Wirths
Now, the subordination relation (13), in terms of the logarithmic coefficients γ
n
of f defined by (1), is equivalent to
∞ ∞
(cid:88) (cid:88)
2 nγ zn ≺ G (z)−1 = δ zn, z ∈ D,
n α n
n=1 n=1
and thus,
k k
(cid:88) 1 (cid:88)
(14) n2|γ |2 ≤ δ2 for each k ∈ N.
n 4 n
n=1 n=1
Since f is starlike of order β(α), it follows that
zK(cid:48)(z) z
α −1 = G (z)−1 ≺ 2(1−β(α))
α
K (z) 1−z
α
and therefore, |δ | ≤ 2(1 − β(α)) for each n ≥ 1. Again, the relation (14) by the
n
previous approach gives
(cid:88)N 1 (cid:88)N δ2 (cid:88)N 1
|γ |2 ≤ k ≤ (1−β(α))2
k 4 k2 k2
k=1 k=1 k=1
for N = 1,2,..., and hence, we have
(cid:88)∞ 1 (cid:88)∞ δ2 (cid:88)∞ 1 π2
|γ |2 ≤ n ≤ (1−β(α))2 = (1−β(α))2
n 4 n2 n2 6
n=1 n=1 n=1
and equality holds in the first inequality for K (z). In particular, if f is convex then
α
β(0) = 1/2 and hence, the last inequality reduces to
(cid:88)∞ π2
|γ |2 ≤
n
24
n=1
which is sharp as the convex function z/(1−z) shows.
3. Proof of Conjecture 1 for n = 2,3,4
Theorem 3. Let f ∈ U(λ) for 0 < λ ≤ 1 and let f(z) = z+a z2+a z3+···. Then
2 3
1−λn
(15) |a | ≤ for 0 < λ < 1 and n = 2,3,4,
n
1−λ
and |a | ≤ n for λ = 1 and n ≥ 2. The results are the best possible.
n
Proof. The case λ = 1 is well-known because U = U(1) ⊂ S and hence, by the
de Branges theorem, we have |a | ≤ n for f ∈ U and n ≥ 2. Here is an alternate
n
proof without using the de Branges theorem. From the subordination result (3) with
λ = 1, one has
∞
f(z) 1 (cid:88)
≺ = nzn−1
z (1−z)2
n=1
and thus, by Rogosinski’s theorem [4, Theorem 6.4(ii), p. 195], it follows that |a | ≤
n
n for n ≥ 2.