On the Fourth Coefficient of the Inverse of a Starlike Function of Positive Order

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On the Fourth Coeﬃcient of the Inverse of a Starlike Function

of Positive Order

Toshiyuki SUGAWA1 _{and Li-Mei WANG}2;

1_{Graduate School of Information Sciences, Tohoku University, Sendai 980-8579, Japan} 2_{School of Statistics, University of International Business and Economics, Beijing 100029, China}

We consider the inverse function z ¼ gðwÞ of a (normalized) starlike function w ¼ f ðzÞ of order on the unit disk of the complex plane with 0 < < 1. Krzyz˙, Libera and Złotkiewicz obtained sharp estimates of the second and the third coefficients of gðwÞ in their paper (1979). Prokhorov and Szynal gave sharp estimates of the fourth coefficient of gðwÞ as a consequence of the solution to an extremal problem in 1981. We give a straightforward proof of the estimate of the fourth coefficient of gðwÞ together with explicit forms of the extremal functions. KEYWORDS: starlike function of order , inverse function, coefficient estimates

1. Introduction

We denote by A the set of analytic functions f ðzÞ on the unit disk D ¼ fz 2 C : jzj < 1g normalized by f ð0Þ ¼ f0_{ð0Þ 1 ¼ 0. Let}_{S denote the subset of A consisting of univalent ones. Each function f 2 S is expanded in the form}

f ðzÞ ¼ z þX

1

n¼2

anzn: ð1:1Þ

The Bieberbach conjecture asserts that janj n for n 2 and it is ﬁnally proved completely by de Branges in 1984

(see, for instance, [4]). We note that equality holds for the Koebe function KðzÞ ¼ z=ð1 zÞ2 ¼z þ 2z2_þ_3z3_þ_.

The Koebe one-quarter theorem implies that f ðDÞ contains the disk jwj < 1=4 so that the inverse function z ¼ gðwÞ of a univalent function w ¼ f ðzÞ inS is analytic on jwj < 1=4. We expand g ¼ f1 in the form

gðwÞ ¼ w þX

1

n¼2

bnwn: ð1:2Þ

Lo¨wner [10] showed the sharp inequalities

jbnj ð2nÞ! n!ðn þ 1Þ!¼ 1 n þ 1 2n n ; n ¼ 2; 3; . . .

(see also [4, Sect. 7.9]). We note that equalities hold for K1_{ðwÞ ¼ ½1 2w}pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi_{1 4w}_{=ð2wÞ. A function f 2}_{A is}

called starlike of order if

Re z f

0_ðzÞ

f ðzÞ

> ; z 2 D;

for a constant 2 ½0; 1Þ. We denote bySðÞ the set of those functions f . A function inSð0Þ ¼:S is simply called starlike and known to map D univalently onto a starlike domain in C with respect to the origin. HenceSðÞ (S) is contained inS for 0 < 1. For a function f in SðÞ, Robertson [12] proved the sharp inequality

janj

ð2 2 þ n 1Þ ð2 2ÞðnÞ ¼

ð2 2Þn1

ðn 1Þ! ;

for each n 2. Here, ðaÞn is the Pochhammer symbol and means aða þ 1Þ ða þ n 1Þ for n 1 and ðaÞ0¼1.

Equalities hold above for the function KðzÞ ¼ z ð1 zÞ2ð1Þ ¼ X1 n¼1 ð2 2Þn1 ðn 1Þ! z n_:

2010 Mathematics Subject Classiﬁcation: Primary 30C45; Secondary 30C50

The ﬁrst author is supported in part by JSPS KAKENHI Grant Number JP17H02847. The second author is supported by ‘‘the Fundamental Research Funds for the Central Universities’’ in UIBE (No. 18YB02) and National Natural Science Foundation of China (No. 11901086).

_{Corresponding author. E-mail: [email protected]} Received November 28, 2020; Accepted February 2, 2021

#Graduate School of Information Sciences, Tohoku University ISSN 1340-9050 print/1347-6157 online

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It is also known that f ðDÞ contains the disk jwj < 22ð1Þ_{for every f 2}_S_{ðÞ. However, sharp inequalities are less}

known for the inverse functions of starlike functions of order > 0. Krzyz˙, Libera and Złotkiewicz [6] gave the sharp estimates jb2j 2ð1 Þ and jb3j ð1 Þð5 6Þ for 0 2₃, 1 for 2₃ < 1. ( Equality is attained by g ¼ K1

for jb2j and for jb3j with 0 2=3 and by g ¼ K;21 for jb2j with 2=3 < 1,

where

K;nðzÞ ¼ fKðznÞg1=n¼

z

ð1 zn_Þ2ð1Þ=n; 0 < 1; n ¼ 1; 2; 3; . . . :

Campschroer [2, Theorem 2.IX] proved the following theorem, except for the explicit expression of the bound in the ﬁrst case and for the second case, both of which were shown by Kapoor and Mishra [5].

Theorem A. Let f1ðwÞ ¼ w þ b2w2þb3w3þ for an f 2SðÞ. Then

jbnj ðn 2n þ 2Þn1 n! for 0 2 n, ðn 2n þ k þ 1Þnk nðn 1Þðn k 1Þ! for k n < k þ 1 n , k ¼ 2; . . . ; n 2, 2ð1 Þ n 1 ; for n 1 n < 1. 8 > > > > > > < > > > > > > :

Moreover, equality holds when f ¼ Kfor the ﬁrst case, and equality holds when f ¼ K;n1 for the last case.

We remark that the above result was extended to the so-called Janowski starlike functions by Ali and Vasudevarao [1].

Though the sharp bound of jbnj is not known for n 5 and 2=n < < ðn 1Þ=n, Prokhorov and Szynal [11]

obtained the sharp bounds of jb4jfor ð; Þ-convex functions, which contains the case of starlike functions of order .

Indeed, for given real numbers and , they solved the extremal problem of ﬁnding the maximum je3þe1e2þe31j

over all the analytic functions !ðzÞ ¼ e1z þ e2z2þe3z3þ on D with j!ðzÞj jzj, and then they derived their main

results from the solution to this extremal problem (see also [2, Lemma 2.VI]). They also provided the information of extremal functions in terms of the function ! determined by the relation z f0_{ðzÞ= f ðzÞ ¼ ð1 þ ð1 2Þ!ðzÞÞ=ð1 !ðzÞÞ.}

Therefore, the expression of the extremal functions is not very explicit. We will give a more detailed proof of it with explicit forms of extremal functions as follows. (See also [2, Theorem 2.VIII].)

Theorem 1.1 (cf. Prokhorov-Szynal [11, Theorem 200_{]). Let f}1_{ðwÞ ¼ w þ b}

2w2þb3w3þb4w4þ for a function

f 2SðÞ with 0 < 1. Then the following sharp inequality holds:

jb4j 2 3ð1 Þð3 4Þð7 8Þ if 0 1¼ 3 5, 4ð2 3Þ3=2ð3 4Þð7 8Þ 3pffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 8 if 1 2 ¼ 35 pffiffiffiffiffi33 48 0:609488, 2 3 11 12 6 3=2 if 2 3, 2 3ð1 Þ if 3 < 1, 8 > > > > > > > > > > > > < > > > > > > > > > > > > : where 3¼ 1 24 21 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 13 þ 2pffiffiffiffiffi11 3 q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 13 2pffiffiffiffiffi11 3 q 0:685369:

If 0 1, then equality holds for f ¼ K.

If 1 2, then equality holds for

f ðzÞ ¼ z

ð1 zÞ2ð1Þð1 ei0_zÞ2ð1Þð1Þ;

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0¼arccos 4ð722_{103 þ 36Þ} 5 8 ; ¼ ð5 3Þð7 8Þ 2ð1 Þ2_{ð72 43Þ ð1 Þ}pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi_{ð3 2Þð72 43Þð24}2_{33 þ 11Þ}:

If 2 3, then equality holds for

f ðzÞ ¼ z ð1 þ zÞð1Þð1sÞð1 zÞð1Þð1þsÞ; where s ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 11 12 4pffiffiffi6ð1 Þ s :

If 3 < 1, then equality holds for f ¼ K;3.

Let BðÞ be the sharp bound for jb4j given in the above theorem. The graph of the function BðÞ in the range

1=2 < 1 is exhibited in Fig. 1.

The organization of the present note is as follows. In the next section, a few preliminary results are given. Some of them may be useful in diﬀerent situations. Section 3 will be devoted to the proof of the main theorem except for the technical Lemma 3.1, which will be shown in Sect. 4.

As the reader will observe, our proof is quite involved with heavy computations. It is quite surprising that Prokhorov and Szynal were able to show it (even in a much general form) as early as in 1981. Finally, the authors would like to thank Professor Ponnusamy for conveying the paper [11] to their attention. Before that, the authors prepared the manuscript, being unaware of [11]. The authors believe that the present paper has a merit in the sense that it may give new techniques to the reader to attack similar problems.

2. Preliminaries

Recall that an analytic function p on D with pð0Þ ¼ 1 and Re p > 0 is called a Carathe´odory function. We will denote byP the set of Carathe´odory functions. Therefore, we can say that a function f 2 A is starlike of order if and only if pðzÞ ¼ ½z f0_{ðzÞ= f ðzÞ =ð1 Þ belongs to}_{P. In this case, we can write}

z f0ðzÞ

f ðzÞ ¼ þ ð1 ÞpðzÞ: ð2:1Þ

It is therefore fundamental to investigate the coeﬃcients of Carathe´odory functions in order to study starlike functions of order . We will introduce several necessary results in this section. The next result is essentially due to Libera and Złotkiewicz [9] (the present version is found in [8]).

Lemma 2.1. Let pðzÞ ¼ 1 þ 2c1z þ 2c2z2þ2c3z3þ be a Carathe´odory function. Then

c1 ¼u;

c2 ¼u2þvð1 juj2Þ;

c3 ¼u3þ ð1 juj2Þvð2u uvÞ þ ð1 juj2Þð1 jvj2Þw

for some u; v; w 2 D. Conversely, for complex numbers c1; c2; c3expressed as above, there is a Carathe´odory function

p satisfying pðzÞ ¼ 1 þ 2c1z þ 2c2z2þ2c3z3þOðz4Þ.

We will also make use of the next result derived from Carathe´odory–Toeplitz theorem (see [7, Lemma 2.6]).

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Lemma 2.2. Let pðzÞ ¼ 1 þ 2c1z þ 2c2z2þ be a Carathe´odory function with c1¼c 2 ½0; 1Þ. If c2¼c2þ

vð1 c2Þfor some v 2 C with jvj ¼ 1, then

pðzÞ ¼ 1 1 þ "1z 1 "1z þ2 1 þ "2z 1 "2z ; ð2:2Þ where "1¼c qei’ ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 c2 p and "2¼c þ q1ei’ ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 c2 p with ’ ¼1 2arg v 2 ð 2; 2, and 1 ¼1=ð1 þ q 2_Þ, 2¼

q2_{=ð1 þ q}2_Þ_{and q ¼ ðc cos ’ þ}pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi_{1 c}2_sin2_’_Þ=pffiffiffiffiffiffiffiffiffiffiffiffiffi_{1 c}2_.

We remark that the above condition on c1 ¼c and c2 means that the equality jc2c21j ¼1 jc1j2 holds.

Furthermore, the condition c ¼ 1 corresponds to the case when 1¼1.

Note that j"1j ¼ j"2j ¼1. Thus, they can be expressed by "1¼ei1 and "2 ¼ei2for some 1; 22 R. Note also that

1¼

1 c2

2 c2_þ_c2_{cos 2’ þ c}pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi_{2ð1 þ cos 2’Þ c}2_sin2_2’ and 2¼1 1: ð2:3Þ

It is convenient to notice the following fact. The function f 2A determined by the equation ð2.1Þ for the function p in the form of ð2.2Þ should satisfy the equation

f0_ðzÞ f ðzÞ 1 z¼ 1 z½ þ ð1 ÞpðzÞ 1 ¼ ð1 Þ 21"1 1 "1z þ 22"2 1 "2z :

Integrating both sides, we obtain

logf ðzÞ

z ¼ 2ð1 Þ1logð1 "1zÞ 2ð1 Þ2logð1 "2zÞ; and hence, the expression

f ðzÞ ¼ z

ð1 "1zÞ2ð1Þ1ð1 "2zÞ2ð1Þ2

:

By taking the rotation "1f ð"1zÞ ¼ ei1f ðei1zÞ, we can replace "1; "2 by 1; " ¼ "1"2¼ei0, respectively, where 0¼

21. Moreover, we may assume that Im " 0 by considering f ðzÞ if necessary. We now compute

cos 0¼Reð"1"2Þ ¼2c21 c ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 c2 p ðq q1Þcos ’ ¼2c2sin2’ 1 ¼ c2ð1 cos 2’Þ 1:

Thus an extremal function can be given in the form

f ðzÞ ¼ z

ð1 zÞ2ð1Þ1_{ð1 e}i0_zÞ2ð1Þ2; ð2:4Þ

where 0 ¼arccos½c2ð1 cos 2’Þ 1.

For real constants A; B; C we consider the quantity jA þ 2BeiþCe2ijfor 2 R. Since this is invariant under the transformation 7! , it is a function of x ¼ cos . Thus we can deﬁne a function : ½1; 1 ! ½0; þ1Þ satisfying

ðcos Þ ¼ jA þ 2BeiþCe2ij: ð2:5Þ

We define Sgn x ¼ x=jxj for x 2 R with x 6¼ 0. The next elementary lemma (cf. [3, Lemma 3.3]) is useful in the sequel. Lemma 2.3. Let A; B; C be real constants with AC < 0 and set ¼ BðA þ CÞ=ð2ACÞ. Then the function satisfies the inequality ðxÞ ð1Þ ¼ jA 2B þ Cj ¼ Sgn BðA þ 2B CÞ if 1, ð Þ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 B 2 AC s jA Cj if 1 1, ð1Þ ¼ jA þ 2B þ Cj ¼ Sgn BðA þ 2B þ CÞ if 1 8 > > > < > > > : for 1 x 1.

3. Proof of the Main Theorem

We are ready to show Theorem 1.1.

Let f 2SðÞ be of the form ð1.1Þ. Then we may write z f0_ðzÞ

f ðzÞ ¼ þ ð1 ÞpðzÞ ¼ 1 þ 2 X1 n¼1

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for a Carathe´odory function pðzÞ ¼ 1 þ 2ðc1z þ c2z2þ Þ, where we put ¼ 1 for the sake of simplicity.

By a comparison of the coeﬃcients, we have the relations a2¼2c1; a3¼22c21þc2; a4¼ 4 3 3_c3 1þ2 2_c 1c2þ 2 3c3: Let g ¼ f1 _{be of the form ð1.2Þ. Then simple computations yield}

b2¼ a2;

b3¼ a3þ2a2;

b4¼ a4þ5a2a35a32: ð3:1Þ

By using these relations, we have

b4¼ 2 3ðc312c1c2þ32 2_c3 1Þ ¼ 2 3Q; where Q ¼ c312c1c2þ322c31. Let M ¼ MðÞ denote the quantity

supfjc312c1c2þ322c31j: pðzÞ ¼ 1 þ 2c1z þ 2c2z2þ2c3z3þOðz4Þfor some p 2Pg:

Thus we need to show

MðÞ ¼ ð4 1Þð8 1Þ if 1 > 1¼ 2 5, 2ð3 1Þ3=2ð4 1Þð8 1Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi8 3 if 1 2 ¼ 13 þpffiffiffiffiffi33 48 , 1 12 1 6 3=2 if 2 3 ¼1 3, 1 if 3 > 0 8 > > > > > > > > > > < > > > > > > > > > > : ð3:2Þ

with enough information about the extremal functions, in addition. We now use Lemma 2.1 to parametrize the quantity Q as

Q ¼ ð8 1Þð4 1Þu32ð6 1Þð1 juj2Þuv ð1 juj2Þuv 2þ ð1 juj2Þð1 jvj2Þw ¼Ae3i þ2Beið þÞþCeið2 Þþ ð1 s2Þð1 t2Þw;

where u ¼ sei , v ¼ tei and

A ¼ ð8 1Þð4 1Þs3; B ¼ ð6 1Þð1 s2Þst; C ¼ ð1 s2Þst2:

Note here that the condition jc2c21j ¼1 jc1j2 is equivalent to t ¼ jvj ¼ 1. Since the argument of w can be taken

arbitrarily, we have the sharp inequality

jQj jA þ 2Beið2 ÞþCe2ið2 Þj þ ð1 s2Þð1 t2Þ ¼ðcosð 2 ÞÞ þ ð1 s2Þð1 t2Þ;

where is deﬁned in ð2.5Þ. We next divide the proof into two cases. Case I: When 0 < 3=4, in this case 1=4 < 1.

We note that 8 1 > 6 1 > 4 1 > 0. Hence A > 0, C < 0 and B < 0 as long as st > 0. In order to apply Lemma 2.3, we compute

¼ BðA þ CÞ

2AC ¼

ð6 1Þfð8 1Þð4 1Þs2_t2_{ð1 s}2_Þg

2s2_{tð4 1Þð8 1Þ} : ð3:3Þ

Therefore, by Lemma 2.3, MðÞ is the supremum of the following quantity over s; t 2 ð0; 1:

Fðs; tÞ ¼ ð1Þ þ ð1 s2Þð1 t2Þ ¼F1ðs; tÞ if 1, ð Þ þ ð1 s2Þð1 t2Þ ¼F2ðs; tÞ if 1 1, ð1Þ þ ð1 s2Þð1 t2Þ ¼F3ðs; tÞ if 1 . 8 > < > : Here,

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F1ðs; tÞ ¼ ð8 1Þð4 1Þs3þ2ð6 1Þð1 s2Þst ð1 s2Þst2þ ð1 s2Þð1 t2Þ; F2ðs; tÞ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð6 1Þ242_s2 ð8 1Þð4 1Þ s fð8 1Þð4 1Þs2þ ð1 s2Þt2g þ ð1 s2Þð1 t2Þ; F3ðs; tÞ ¼ ð8 1Þð4 1Þs3þ2ð6 1Þð1 s2Þst þ ð1 s2Þst2þ ð1 s2Þð1 t2Þ:

By its form, the function Fðs; tÞ is continuous on ½0; 12¼ ½0; 1 ½0; 1. We observe that 1 if and only if

s2 ð6 1Þt

2

ð6 1Þt2_{2ð8 1Þð4 1Þt þ ð8 1Þð6 1Þð4 1Þ}¼2ðtÞ 2_;

and that 1 if and only if

s2 ð6 1Þt

2

ð6 1Þt2_þ_{2ð8 1Þð4 1Þt þ ð8 1Þð6 1Þð4 1Þ}¼1ðtÞ 2

:

Here, 0 1ðtÞ 2ðtÞ and 1ðtÞ < 1 for 0 t 1. Note that

s1:¼ 1ð1Þ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6 1 82_{ð24 5Þ} s 2 ð0; 1Þ; ð3:4Þ

which is a decreasing function in > 0. In view of the form of 1=jðtÞ2, we observe that 1ðtÞ and 2ðtÞ are monotone

increasing and continuous on 0 t 1 and 2ðtÞ ¼ 1 when

t ¼ t1:¼

6 1

2 :

For a ﬁxed , we divide the set ½0; 12 into three closed regions 1, 2, 3 according as 1, 1 1, 1 ,

respectively. More precisely, we can deﬁne them by 1 ¼ fðs; tÞ 2 ½0; 12: 2ðtÞ sg, 2¼ fðs; tÞ 2 ½0; 12: 1ðtÞ

s 2ðtÞg, 3¼ fðs; tÞ 2 ½0; 12: s 1ðtÞg. Then M ¼ MðÞ ¼ max s;t2½0;1Fðs; tÞ ¼ maxfM1; M2; M3g; where Mj¼ max ðs;tÞ2j Fjðs; tÞ; j ¼ 1; 2; 3:

According to the conditions t12 ð0; 1Þ and t11 of t1, we consider two subcases to ﬁnd the value MðÞ.

Subcase Ia:When 1=4 < < 1=2, we have t12 ð0; 1Þ. See Fig. 2 for the domains 1; 2; 3 in this subcase.

At this point, it is worth noting that M Fjð0; 0Þ ¼ 1 for any 2 ð1=4; 1=2Þ. Our strategy to compute M is as

follows. In order to ﬁnd the value of Mj, we ﬁrst look for critical points of the function Fjin the interior Int jof j. If

there is no critical point, the maximum must be taken on the boundary @j. Thus we have only to look at the boundary

values of Fj.

On the value M1: By the identity

@F1

@t ðs; tÞ ¼ 2ð1 s

2_{Þfð6 1 þ tÞs tg;}

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critical points of F1 in Int 1 lie on the curve s ¼ t=ð6 1 þ tÞ. Since 2ðtÞ2 t 6 1 þ t 2 ¼ 4 2_t2_{ð6 1 2tÞ} ð6 1 þ tÞ2fð6 1Þt2_{2ð8 1Þð4 1Þt þ ð8 1Þð6 1Þð4 1Þg}> 0

for 0 < t < t1, there is no critical point of F1ðs; tÞ in Int 1. Thus the maximum M1 of F1 is taken on @1. Simple

computations yield

F1ð1; tÞ ¼ ð8 1Þð4 1Þ;

F1ðs; 0Þ ¼ ð8 1Þð4 1Þs3þ1 s2¼: ’ðsÞ:

Since ’0_{ðsÞ ¼ f3ð8 1Þð4 1Þs 2gs, the function ’ðsÞ is decreasing in 0 < s < 2=½3ð8 1Þð4 1Þ and}

increasing in 2=½3ð8 1Þð4 1Þ < s. In particular, the maximum of F1ðs; 0Þ is taken at s ¼ 0 or s ¼ 1 on ½0; 1.

Noting that the function F1ð2ðtÞ; tÞ takes the value F1ð0; 0Þ ¼ 1 for t ¼ 0 and F1ð1; t1Þ ¼F1ð1; 0Þ for t ¼ t1, the

maximum of F1on 1is attained on the common boundary curve s ¼ 2ðtÞ (0 t t1) with 2. Hence, we conclude

that M1M2 so that M ¼ maxfM2; M3g.

On the value M2: Since

@F2 @t ¼2tð1 s 2_Þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 4 2_{ð1 s}2_Þ ð8 1Þð4 1Þ s 1 ! > 0

for s; t 2 ð0; 1Þ, the function F2ðs; tÞ is increasing in 0 t 1 for a ﬁxed s 2 ð0; 1Þ. Therefore, the maximum M2 is

taken on either the segment ½s1; 1 f1g or the common boundary s ¼ 1ðtÞ (0 t 1) with 3. If we put

M4¼ max s1s1

F2ðs; 1Þ;

then we have M ¼ maxfM3; M4g.

We now look for the maximum of the function ðxÞ ¼ ð8 1Þð4 1ÞF2ð ffiffiffi x p ; 1Þ2¼ fð6 1Þ242xgð322x 12x þ 1Þ2 on s2

1x 1. Here we remark that max s2

1x1

ðxÞ ¼ ð8 1Þð4 1ÞM2 4. Since

0ðxÞ ¼ 12ð322x 12x þ 1Þf19231362þ29 2 42ð8 3Þxg; the function ðxÞ has possibly a critical point at

x ¼ ðÞ :¼192

3₁₃₆2_þ_{29 2}

42_{ð8 3Þ} :

In view of the formulae

ðÞ 1 ¼ð8 1Þð4 1Þð5 2Þ 42_{ð8 3Þ} and ðÞ s2₁¼ð8 1Þð4 1Þð288 2_{156 þ 17Þ} 82_{ð8 3Þð24 5Þ} ; we see that s2 1< ðÞ < 1 precisely when ð13 þ ffiffiffiffiffi 33 p

Þ=48 < < 2=5 for under consideration. Note here that ð13 þ ffiffiffiffiffi

33 p

Þ=48 0:390512 > 3=8. Also, we note that 0_{ð1Þ ¼ 12ð8 1Þ}2_{ð4 1Þ}2_{ð5 2Þ. According to the above}

observation, we consider the following three cases.

(i) When 1=4 < ð13 þpffiffiffiffiffi33Þ=48, ðxÞ is decreasing on ½s2

1; 1 and therefore F2ðs; 1Þ is decreasing on ½s1; 1.

Hence, we have M4 ¼F2ðs1; 1Þ M3 because ðs1; 1Þ 2 @3.

(ii) When ð13 þpffiffiffiffiffi33Þ=48 < < 2=5, ðxÞ is increasing on ½s2₁; ðÞ and decreasing on ½ðÞ; 1. Hence ðxÞ takes the maximum at x ¼ ðÞ. A straightforward calculation produces

ððÞÞ ¼4ð8 1Þ 3_{ð4 1Þ}3_{ð3 1Þ}3 2_{ð8 3Þ} and therefore M4¼ 2ð8 1Þð4 1Þð3 1Þ3=2 ð8 3Þ1=2 :

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(iii) When 2=5 < 1=2, ðxÞ is increasing on ½s2

1; 1 and so is F2ðs; 1Þ on ½s1; 1. Thus M4¼F2ð1; 1Þ ¼

ð8 1Þð4 1Þ.

On the value M3: The last step is to solve the extremal problem on 3. To ﬁnd a critical point of F3, we solve the

system of equations @F3 @t ¼2ð1 s 2_{Þfsð6 1 þ tÞ tg ¼ 0;} @F3 @s ¼ 3ð8 1 þ tÞð4 1 þ tÞs 2 2ð1 t2Þs þ tð12 2 þ tÞ ¼ 0:

Solving the ﬁrst equation, we have the relation s ¼ t=ð6 1 þ tÞ. We substitute it into the second equation and simplify to

12tfð7 2Þt þ 2ð6 1Þð3 1Þg

ð6 1 þ tÞ2 ¼0:

Hence a candidate of critical points in Int 3 is only ðs0; t0Þ, where

s0¼

2ð1 3Þ

; t0 ¼

2ð6 1Þð3 1Þ

7 2 :

By the requirement 0 < s0< 1, the parameter should satisfy 2=7 < < 1=3. After some computations, we have the

expression

F3ðs0; t0Þ ¼145

16

432 þ 432

2 _{¼: hðÞ:}

Since h0ðÞ ¼ 16ð6 þ 1Þð3 1Þ2=2 > 0 in this range, we have F3ðs0; t0Þ hð1=3Þ ¼ 1 ¼ F3ð0; 0Þ. Therefore, this

critical point does not contribute to the maximum of F3. At any event, we conclude that the maximum M3is attained on

the boundary of 3.

Next we shall look at the values of F3on @3. First, F3ð0; tÞ ¼ 1 t21 ¼ Fð0; 0Þ for t 2 ½0; 1. Second, we will

prove the following result in the next section.

Lemma 3.1. Let 1=4 < < 1=2. Then F3ð1ðtÞ; tÞ maxf1; F3ðs1; 1Þg for 0 t 1.

Therefore, we see that the maximum of F3on 3is taken either on the segment ½0; s1 f1g or at the point ð0; 0Þ. We

now have F3ðs; 1Þ ¼ ð12 1Þs 322s3¼: GðsÞ for s 2 ½0; s1. Let s2¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 12 1 p 4pffiffiffi6 ; which is the positive critical point of GðsÞ. Since

s2₁s2₂¼ 288 2_{156 þ 17} 962_{ð24 5Þ} ; s22 ½0; s1if and only if ð13 þ ffiffiffiffiffi 33 p

Þ=48 for 2 ð1=4; 1=2Þ. So we divide the problem into two cases. (1) When 1=4 < < ð13 þpffiffiffiffiffi33Þ=48, we have

GðsÞ Gðs2Þ ¼

ð12 1Þ3=2 6pffiffiffi6 for 0 s s1. Hence M3¼maxf1; Gðs2Þgin this case.

(2) When ð13 þpffiffiffiffiffi33Þ=48 < 1=2, GðsÞ is increasing in ½0; s1so that

GðsÞ Gðs1Þ ¼

9ð8 1Þð4 1Þpffiffiffiffiffiffiffiffiffiffiffiffiffiffi6 1 2pffiffiffi2ð24 5Þ3=2 for s 2 ½0; s1. Hence M3¼maxf1; Gðs1Þgin this case.

We now summarize what we have seen. So far, we had MðÞ ¼ M ¼ maxfM3; M4gand M3is computed in (1) and (2)

above. On the other hand M4is computed in (i), (ii) and (iii) on the value M2. Hence, we have the following conclusion.

(a) When 1=4 < ð13 þpffiffiffiffiffi33Þ=48 ¼ 2, we have M ¼ M3¼maxf1; Gðs2Þg. It is easy to see that 1 Gðs2Þ ¼

ð12 1Þ3=2=6pffiffiffi6 precisely if 3, where 3 0:31463 is the unique real solution to the equation

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3¼ 1 24 3 þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 13 þ 2pffiffiffiffiffi11 3 q þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 13 2pffiffiffiffiffi11 3 q ;

which is the same as in ð3.2Þ.

Next we shall give extremal functions. First, for 1=4 < 3, M ¼ 1 is attained when s ¼ 0, t ¼ 0 and

jwj ¼ 1, which corresponds to pðzÞ ¼ ð1 þ wz3_{Þ=ð1 wz}3_{Þ. Taking w ¼ 1, we get f ¼ K}

;3. Second, for

3 2, M ¼ Gðs2Þis attained when t ¼ 1, s ¼ s2and cos ¼ 1, where ¼ 2 . We may assume ¼ 0

so that ¼ ¼ 0. Now we will apply Lemma 2.2 with c ¼ s2, ’ ¼ =2 ¼ 0, where s2¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1112 4pffiffi6ð1Þ q . Then q ¼ ð1 þ s2Þ= ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 s2 2 p

so that 1¼ ð1 s2Þ=2, 2¼ ð1 þ s2Þ=2 and "1¼ 1, "2 ¼ 1. Therefore, with the help of

ð2.4Þ, we obtain the extremal function as in the third case of Theorem 1.1.

(b) When ð13 þpffiffiffiffiffi33Þ=48 < < 2=5, we have M ¼ maxfM3; M4g. Noting M41, we obtain, more precisely,

M ¼ max 9ð8 1Þð4 1Þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6 1 p 2pffiffiffi2ð24 5Þ3=2 ; 2ð8 1Þð4 1Þð3 1Þ3=2 ð8 3Þ1=2 :

Letting m3 be the ﬁrst term in this maximum, we have

M4 m3 2 1 ¼ð144 2_{78 þ 13Þð288}2₁₅₆2_þ_17Þ2 81ð6 1Þð8 3Þ > 0 for in the current range. Hence, we obtain

M ¼ M4¼

2ð8 1Þð4 1Þð3 1Þ3=2 ð8 3Þ1=2 :

Next let us see forms of extremal functions. Recall that M4 was evaluated according as (i), (ii), (iii) occurs,

respectively, in the analysis of the value M_{ffiffiffiffiffiffiffiffiffi} 2. First, for 2< 1< 2=5, M ¼ M4 is attained when s ¼

ðÞ p

¼: s3, t ¼ 1 and cosð 2 Þ ¼ , where in ð3.3Þ is now computed as

¼ ð6 1Þfð8 1Þð4 1ÞðÞ ð1 ðÞÞg

2ðÞð4 1Þð8 1Þ :

We may take ¼ 0 so that Lemma 2.2 is applicable with ’ ¼ =2 2 ð=2; =2Þ and c ¼ s3. Now we obtain the

form of an extremal function in the second case of Theorem 1.1 by ð2.3Þ and ð2.4Þ, where we observe

cos 0¼s23ð1 cos Þ 1 ¼

4ð72241 þ 5Þ

8 3 ¼

4ð722103 þ 36Þ 5 8 and, after some simpliﬁcations,

1¼

ð5 2Þð8 1Þ

1443₅₈2₂pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi_{ð3 1Þð72 29Þð24}2_{15 þ 2Þ}:

(c) When 2=5 < 1=2, similarly, we have

M ¼ max 9ð8 1Þð4 1Þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6 1 p 2pffiffiffi2ð24 5Þ3=2 ; ð4 1Þð8 1Þ : Let HðÞ ¼ 9 ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6 1 p 2pffiffiffi2ð24 5Þ3=2 2 ¼ 81ð6 1Þ 82_{ð24 5Þ}3: Then H0ðÞ ¼ 81ð288 2_{75 þ 5Þ} 43_{ð24 5Þ}4 ¼ 81f2ð12 2Þ2þ3ð7 1Þg 43_{ð24 5Þ}4 < 0

for 2=5 < 1=2. Therefore, the function HðÞ is decreasing there so that

HðÞ Hð2=5Þ ¼354375 389344< 1 for 2=5 < 1=2. We thus get M ¼ ð4 1Þð8 1Þ.

Since this case corresponds to the condition s ¼ juj ¼ 1, we have the simplest extremal function f ¼ K.

Subcase Ib:When 1=2 1, t11. We focus only on the special case ¼ 1 since 1=2 < 1 can be proved by

the same technique showing in Case II. See Fig. 3 for the domains 1; 2; 3 when ¼ 1. Our aim is to show that

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If ¼ 1, F1; F2; F3 have the following forms. F1ðs; tÞ ¼ 21s3þ10ð1 s2Þst ð1 s2Þst2þ ð1 s2Þð1 t2Þ; F2ðs; tÞ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 25 4s2 21 r ½21s2þ ð1 s2Þt2 þ ð1 s2Þð1 t2Þ; F3ðs; tÞ ¼ 21s3þ10ð1 s2Þst þ ð1 s2Þst2þ ð1 s2Þð1 t2Þ:

A computation shows that @F1 @s ¼3s 2_{ð21 10t þ t}2_Þ_{2sð1 t}2_{Þ þ ð10 tÞt > 0;} _{for ðs; tÞ 2} 1 @F2 @t ¼2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 25 4s2 21 r ð1 s2Þt 2ð1 s2Þt > 0; for ðs; tÞ 2 2 F3ðs; tÞ 21s3þ12 < 21; for ðs; tÞ 2 3: Therefore Mð1Þ ¼ max s2½0;1fF1ð1; tÞ; F2ðs; 1Þ; 21g ¼ maxs2½0;1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 25 4s2 21 r ð1 þ 20s2Þ; 21 ( ) ¼21

since F1ð1; tÞ 21 and F2ð1ðtÞ; tÞ ¼ F3ð1ðtÞ; tÞ < 21 for t 2 ½0; 1. The extremal function is the Koebe function K0.

Case II:When 3=4 < 1, for any function f ðzÞ ¼ z þP

1 n¼2

anzn2SðÞ, we have f ðzÞ ¼1 f ð zÞ 2SðÞ Sð0Þ

where 0_{is any constant in ð}

3; 3=4Þ and 3 ¼ ð1 0Þ=ð1 Þ > 1. In view of ð3.1Þ and the result proved in Case I for

f , we have j 3ða4þ5a2a35a32Þj 2 3ð1 0_Þ which is ja4þ5a2a35a32j 2 3ð1 Þ: On the other hand the function K;3 makes the equality hold.

In this way, we have proven ð3.2Þ as required and ﬁnish the proof of the main theorem, up to Lemma 3.1.

4. Proof of Lemma 3.1

The remaining task is to show that F3ð1ðtÞ; tÞ maxf1; F3ðs1; 1Þg for 0 t 1, where s1¼s1ðÞ is given in ð3.4Þ.

The standard method using Langrange multipliers seems to involve messy calculations. Thus we will take another approach. It is convenient in the sequel to consider the inverse function 1: ½0; s1 ! ½0; 1 of 1: ½0; 1 ! ½0; s1.

That is,

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t ¼ 1ðsÞ ¼ ð8 1Þð4 1Þs2_þ_spffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi_{ð8 1Þð4 1Þfð6 1Þ}2₄2_s2_g ð6 1Þð1 s2_Þ ¼s 2_{=ð6 1Þ þ s}pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi_{ð1 s}2_Þ 1 s2 for s 2 ½0; s1, where ¼ ð8 1Þð4 1Þ > 0 and ¼ 4 2 ð6 1Þ2 ¼ 42 þ 42 2 ð0; 1Þ:

Then it is enough to show that the function

wðsÞ ¼ F3ðs; 1ðsÞÞ

satisﬁes wðsÞ maxf1; F3ðs1; 1Þg for 0 s s1. A straightforward computation gives the expression

wðsÞ ¼ 1 þ s 2 1 þ svðsÞ; where vðsÞ ¼ 2s2þ ð 1Þs ð1 þ Þ þ 2ð1 þ sÞ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 s2 1 s :

We now analyze the behaviour of vðsÞ over ½0; s1. First we compute

v0ðsÞ ¼ 4s þ 1 þ2ð1 s 2s

2_Þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 Þð1 s2_Þ

p :

We note here that 1 s 2s2_{1 s}

12s21 for s 2 ½0; s1. By using the expression

1 s12s21¼1 1 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6 1 2ð24 5Þ s 1 ð6 1Þð24 5Þ¼: uðÞ;

we can check that u0_{ðÞ > 0 on 1=4 < < 1=2 and thus uðÞ is increasing there. We compute uð}17 52Þ ¼ 587 925 6 17pffiffiffiffi37 ¼0:0060 > 0. Hence, 1 s 2s 2_{0 over s 2 ½0; s}

1for 17=52 < 1=2. Note that 17=52 0:326923.

We divide the situation into four cases to show that wðsÞ maxf1; wðs1Þgover ½0; s1.

Case I: When 2 ½3=8; 1=2Þ, we have 1 and 1 s 2s21 s12s210 for s 2 ½0; s1. Then

v0ðsÞ 4s þ2ð1 s 2s

2_Þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 Þð1 s2_Þ

p > 0

for s 2 ½0; s1. Hence wðsÞ is increasing in ½0; s1and therefore

wðsÞ wðs1Þ; for s 2 ½0; s1:

Case II:When 2 ½17=52; 3=8, as we saw above, 1. Since ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 s2 1 p pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 s2 _{1 for 0 s s} 1, we have ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 s2 p v0ðsÞ ¼ ð4s þ 1Þpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 s2_þ2ð1 s 2s 2_Þ ffiffiffiffiffiffiffiffiffiffiffiffi 1 p 4s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 s2 1 q þ ð 1Þ þ2ð1 s 2s 2_Þ ffiffiffiffiffiffiffiffiffiffiffiffi 1 p ¼ 2ffiffiffiffiffiffiffiffiffiffiffiffi 1 p 2s2þ 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 Þð1 s2 1Þ q 1s þ 1 þð 1Þ ffiffiffiffiffiffiffiffiffiffiffiffi 1 p 2 ¼: 2ffiffiffiffiffiffiffiffiffiffiffiffi 1 p f ðsÞ

for s 2 ½0; s1. Since f ðsÞ is concave, f ðsÞ minf f ð0Þ; f ðs1Þgfor s 2 ½0; s1. Noting the estimate

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð4 1Þð8 1Þ p

2pffiffiffiffiffi21=13 ¼ 0:7050 > 7=10, we obtain

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f ð0Þ ¼ 1 þð 1Þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 Þ p 2 ¼1 ð3 8Þð6 1Þ 2pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið4 1Þð8 1Þ 1 5ð3 8Þð6 1Þ 7 ¼ 3ð16 5Þð5 1Þ 7 > 0: We also compute f ðs1Þ ¼1 2s21þ ð 1Þpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1 Þ 2 þ 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 Þð1 s2 1Þ q 1 s1 ¼1 þ3ð4 1Þð8 1Þ 2 2ð6 1Þð24 5Þ þ ð8 3Þð6 1Þ 2pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið4 1Þð8 1Þ 1 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6 1 48 10 s ¼: 1 þ q1ðÞ þ q2ðÞ q3ðÞ:

Simple calculations yield

q0₁ðÞ ¼ 3ð5 108 þ 956 2₃₆₄₈3_þ₄₆₀₈4_Þ 22_{ð24 5Þ}2_{ð6 1Þ}2 and q0₂ðÞ ¼3 þ 54 300 2_þ₅₄₄3 4½ð4 1Þð8 1Þ3=2 :

One can easily check that the numerators of q0₁ðÞ and q0₂ðÞ are concave on the interval ½17=52; 3=8, and that both of them take positive values at the end points 17/52 and 3/8. Therefore q1and q2are increasing on ½17=52; 3=8. Since q3

can be written in the form

q3ðÞ ¼ 1 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 8þ 1 4ð48 10Þ s ;

it is obvious that it is decreasing on ½17=52; 3=8. By virtue of the computation

1 þ ðq1þq2q3Þ 17 52 ¼23083 15725 125 34pffiffiffiffiffi21 65 17pffiffiffiffiffi37¼0:037 > 0; we conclude that v0_{ðsÞ is positive on ½0; s}

1. Therefore, no matter whether vðs1Þis positive or negative, we have

wðsÞ ¼ 1 þ s

2

1 þ svðsÞ maxfwð0Þ; wðs1Þg ¼maxf1; wðs1Þg; for s 2 ½0; s1:

Case III:When 2 ½8=25; 17=52, we observe thatpffiffiffiffiffiffiffiffiffiffiffiffi1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 ð8=25Þ¼pffiffiffiffiffiffiffiffi273=23 ¼ 0:71838 > 51=71 ¼ 1=A, because ¼ ðÞ is decreasing in 8=25 17=52, where A ¼ 71=51. Note also s1ð8=25Þ ¼ 25

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 23=134 p

=16. Then, for s 2 ½0; s1we have

vðsÞ ¼ 2s2þ ð 1Þs ð1 þ Þ þ 2ð1 þ sÞ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 s2 1 s 2s2þ ð 1Þs ð1 þ Þ þ 2Að1 þ sÞð1 s2=2Þ ¼: gðsÞ:

We will show that gðsÞ in increasing on ½0; s1in this case. To do that, we compute

g0ðsÞ ¼4s þ 1 þAð2 2s 3s 2 Þ ¼2ð2 AÞs þ 1 þAð2 3s 2_Þ;

which is a concave function of s. Thus g0_{ðsÞ minfg}0_{ð0Þ; g}0_ðs

1Þg for s 2 ½0; s1. Introducing the new variable x ¼

16 3, we compute 1 ¼ ð6 1Þ2ð8 3Þ ð4 1Þð8 1Þ¼ ðx 3Þð3x þ 1Þ2 ðx2_{1Þðx þ 3Þ} ¼: hðxÞ:

Note that the range of x is now ½53=25; 29=13. Since

h0ðxÞ ¼16ðx

3_þ_{3Þð3x þ 1Þ}

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for x 2 ½53=25; 29=13, hðxÞ is increasing and so that g0_{ð0Þ=ðÞ ¼ hðxÞ þ 2A hð53=25Þ þ 2A ¼ 46=}pffiffiffiffiffiffiffiffi₂₇₃

5819=2184 ¼ 0:1196 > 0. We also observe that hðxÞ is convex on 1 < x < þ1 by

h00ðxÞ ¼ 16ð6x

6_þ_3x5_þ_9x4_þ_82x3_þ_108x2_þ_{27x þ 21Þ}

ðx 1Þ3ðx þ 1Þ3ðx þ 3Þ3 :

In particular, ð 1Þ=ðÞ is a concave function of on ð1=4; 1=2Þ. Secondly, another computation yields g0_ðs 1Þ ¼2ð2 AÞs1ðÞ þ 1 þ A 2 5762_{216 þ 17} ð6 1Þð24 5Þ : ð4:1Þ

On the interval ½8=25; 17=52, we estimate, by convexity,

2ð2 AÞs0₁ðÞ > 2s0₁ð1=4Þ ¼ 20 and, by concavity, d d 1 ¼16h0ð16 3Þ 16h0ð7=3Þ ¼1431 50 > 20: Hence the sum of the ﬁrst two terms of g0_ðs

1Þ=ðÞ in ð4.1Þ is an increasing function of on ½8=25; 17=52. On the other

hand, by using the same change of variables as before, we have 5762216 þ 17 ð6 1Þð24 5Þ ¼ 2ð9x213Þ ð3x 1Þð3x þ 1Þ¼2 24 9x2₁;

which is obviously increasing in x > 1=3. Therefore, the third term is also increasing in . In summary, g0ðs1Þ=ðÞ is

increasing in so that its minimum over ½8=25; 17=52 is taken at ¼ 8=25 and its value is g0ðs1ð8=25ÞÞ

¼

34312pffiffiffiffiffiffiffiffi273þ13650pffiffiffiffiffiffiffiffiffiffi3082575pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi841386779746

292656 ¼0:0599 > 0:

Hence we have proved that gðsÞ is increasing on ½0; s1for each 2 ½8=25; 17=52. Thus we need only to show that the

value gðs1Þis negative. Now we write gðs1Þin the form

gðs1Þ ¼2s21þ ð 1Þs1 ð1 þ Þ þ 2Að1 þ s1Þð1 s21=2Þ

¼ ð2 AÞ½s2₁ þ ½ð 1Þs1 þ ½2A 1 þ 2A½s1 þA½2s31

¼ ð2 AÞT1þT2þT3þ2AT4þAT5: By virtue of T₁0ðÞ ¼ 16 3 ð24 5Þ2ð6 1Þ2; T₂0ðÞ ¼1152 2_{456 þ 49} ð24 5Þ2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 48 10 6 1 s ; T3ðÞ ¼ ð2A 1Þð4 1Þð8 1Þ 1; T4ðÞ ¼ ffiffiffi 2 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð4 1Þð8 1Þ ð6 1Þ2 s ffiffiffiffiffiffiffiffiffiffiffiffiffi 2T1ðÞ p ; T5ðÞ ¼ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð48 10Þð6 1Þ p ð4 1Þð8 1Þ ð6 1Þ3ð24 5Þ ¼: ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 ð48 10Þð6 1Þ p t5ðÞ and t0₅ðÞ ¼5 60 24 2_þ₁₉₂₀2₄₆₀₈4 ð24 5Þ2ð6 1Þ4 < 0; for 2 ½ 8 25; 17 52

by the concavity of the numerator of t0

5ðÞ, each Tj¼TjðÞ is increasing in 8=25 17=52. We thus obtain

gðs1Þ gðs1ð17=52ÞÞ ¼

7670451pffiffiffiffiffi3754764625

945465625 ¼ 0:00857 < 0: Hence, we see that vðsÞ is always negative on ½0; s1for 2 ½8=25; 17=52. Therefore in this case

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Case IV:When 2 ½1=4; 8=25, we have the estimates vðsÞ ¼ 2s2þ ð 1Þs ð1 þ Þ þ 2ð1 þ sÞ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 s2 1 s 2s2þ ð 1Þs ð1 þ Þ þ 2Að1 þ sÞ ¼: lðsÞ maxflð0Þ; lðs1Þg

for s 2 ½0; s1, where A ¼ 71=51 as before. Next we will show that both lð0Þ and lðs1Þare negative. First, since A < 7=5,

we have

lð0Þ ¼ ð1 þ Þ þ 2A < 1 þ14 5 ¼

4

5ð3 1Þð24 1Þ < 0: Second, a computation shows that

lðs1Þ ¼2s21þ ð 1Þs1 ð1 þ Þ þ 2Að1 þ s1Þ

¼2½s2₁ þ ½ð 1Þs1 þ ½2A 1 þ 2A½s1

¼2T1þT2þT3þ2AT4;

where T1, T2, T3and T4have the same form as in Case III. The same arguments as in Case III show that all the terms

Tj¼TjðÞ are increasing in as before. Hence, we obtain

lðs1Þ lðs1ð8=25ÞÞ ¼

37025pffiffiffiffiffiffiffiffiffiffi30824054049

110759375 ¼ 0:0180 < 0:

We ﬁnd that vðsÞ is always negative on ½0; s1for 2 ½1=4; 8=25. Therefore in this case too, we have

wðsÞ 1; for s 2 ½0; s1:

Considering the above four cases together, we ﬁnally proved that wðsÞ maxf1; wðs1Þgover ½0; s1.

Acknowledgments

The authors would like to thank the reviewers for their careful work and thoughtful suggestions which have helped improve this paper.

REFERENCES

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