Second Hankel determinant for bi-starlike and bi-convex functions of order \b{eta}

0.15 MB·English

by  Erhan Deniz

SECOND HANKEL DETERMINANT FOR BI-STARLIKE AND BI-CONVEX
FUNCTIONS OF ORDER β
ERHANDENIZ,MURATC¸AG˘LAR,ANDHALITORHAN
Abstract. In the present investigation the authors obtain upper bounds for the second Han-
5 kel determinant H2(2) of theclasses bi-starlikeandbi-convex functions oforder β, represented
1 by Sσ∗(β)and Kσ(β), respectively. In particular, the estimates for the second Hankel determi-
0 nantH2(2)ofbi-starlikeandbi-convexfunctionswhichareimportantsubclassesofbi-univalent
2 functionsarepointedout.

r
p
A
1. Introduction and definitions
7

Let denote the family of functions f analytic in the open unit disk = z C: z <1 of
] A U { ∈ | | }
V the form
C ∞
(1.1) f(z)=z+ a zn.
. n
h n=2
X
t
a Let denote the class of all functions in which are univalent in . The Koebe one-quarter
S A U
m theorem (see [7]) ensures that the image of under every f contain a disk of radius 1(cid:30)4. So,
U ∈S
[ every f has an inverse function f−1 satisfying f−1(f(z))=z (z ) and
∈S ∈U
3 f(f−1(w))=w (w <r0(f); r0(f) 1(cid:30)4)
| | ≥
v where f 1(w)=w a w2+(2a2 a )w3 (5a3 5a a +a )w4+....
2 − − 2 2− 3 − 2− 2 3 4
A function f is saidto be bi-univalentin if both f(z)and f 1(z)are univalentin . Let
8 −
∈A U U
6 σ denote the class of bi-univalent functions in given by (1.1).
U
1 Twoofthemostfamoussubclassesofunivalentfunctionsaretheclass (β)ofstarlikefunctions
∗
0 of order β and the class (β)of convex functions of order β. By definitiSon, we have
. K
1 zf (z)
′
0 ∗(β)= f : >β; z ; 0 β <1
S ∈S ℜ f(z) ∈U ≤
5 (cid:26) (cid:18) (cid:19) (cid:27)
1 and
: zf (z)
v (β)= f : 1+ ′′ >β; z ; 0 β <1 .
Xi K (cid:26) ∈S ℜ(cid:18) f′(z) (cid:19) ∈U ≤ (cid:27)
The classes consisting of starlike and convex functions are usually denoted by = (0) and
∗ ∗
r S S
= (0), respectively.
a
K K
For 0 β < 1, a function f σ is in the class (β) of bi-starlike functions of order β, or
≤ ∈ Sσ∗
(β)ofbi-convex functions of order β ifbothf anditsinversemapf 1 are,respectively,starlike
σ −
K
or convex of order β. These classes were introduced by Brannan and Taha [2] in 1985. Especially
the classes (0) = and (0) = are bi-starlike and bi-convex functions, respectively. In
Sσ∗ Sσ∗ Kσ Kσ
1967, Lewin [17] showed that for every functions f σ of the form (1.1), the second coefficient of
∈
f satisfy the inequality a < 1.51. In 1967, Brannan and Clunie [1] conjectured that a √2
2 2
| | | | ≤
for f σ. Later, Netanyahu [18] proved that max a = 4/3. In 1985, Kedzierawski [13]
f σ 2
∈ ∈ | |
proved Brannan and Clunie’s conjecture for f . In 1985, Tan [25] obtained the bound for a
∈ Sσ∗ 2
namely a < 1.485 which is the best known estimate for functions in the class σ. Brannan and
2
| |
Taha[2]obtainedestimatesontheinitialcoefficients a and a forfunctionsintheclasses (β)
| 2| | 3| Sσ∗
and (β). Recently, Deniz [6] and Kumar et al. [15] both extended and improved the results of
σ
K
BrannanandTaha[2]bygeneralizingtheirclassesusingsubordination. Theproblemofestimating
coefficients a , n 2 is still open. However,a lot of results for a , a and a were provedfor
n 2 3 4
| | ≥ | | | | | |
2000 Mathematics Subject Classification. Primary30C45,30C50;Secondary30C80.
Key words and phrases. Bi-univalentfunctions, bi-starlikefunctions of order β,bi-convex functions of order β,
secondHankeldeterminant.
Corresponding author. [email protected] (ErhanDeniz).
1
2 ERHANDENIZ,MURATC¸AG˘LAR,ANDHALITORHAN
some subclasses of σ (see [3], [5], [9], [11], [21], [23], [24], [26], [27]). Unfortunatelly, none of them
are not sharp.
One of the important tools in the theory of univalent functions is Hankel Determinants which
are utility, for example, in showing that a function of bounded characteristic in , i.e., a function
U
whichisaratiooftwoboundedanalyticfunctions,withitsLaurentseriesaroundtheoriginhaving
integral coefficients, is rational [4]. The Hankel determinants [19] H (n)(n = 1,2,..., q = 1,2,...)
q
of the function f are defined by
a a ... a
n n+1 n+q 1
−
a a ... a
n+1 n+2 n+q
Hq(n)= .. .. ..  (a1 =1).
. . .
 
 a a ... a 
 n+q−1 n+q n+2q−2 
 
This determinant was discussed by several authors with q =2. For example, we can know that
the functional H (1) = a a2 is known as the Fekete-Szego¨ functional and they consider the
2 3 − 2
further generalized functional a µa2 where µ is some real number (see, [8]). In 1969, Keogh
3 − 2
and Merkes [14] proved the Fekete-Szego¨ problem for the classes and . Someone can see the
∗
S K
Fekete-Szego¨ problem for the classes (β) and (β) at special cases in the paper of Orhan et.al.
∗
S K
[20]. On the other hand, very recently Zaprawa [28], [29] have studied on Fekete-Szego¨ problem
for some classes of bi-univalent functions. In special cases, he gave Fekete-Szego¨ problem for the
classes (β) and (β). In 2014, Zaprawa[28] proved the following resuts for µ R,
Sσ∗ Kσ ∈
1 β; 1 µ 3
f ∈Sσ∗(β)⇒ a3−µa22 ≤ 2(1 β−) µ 1 ; µ 23≤and≤µ2 1
(cid:26) − | − | ≥ 2 ≤ 2
(cid:12) (cid:12)
and (cid:12) (cid:12)
f ∈Kσ(β)⇒ a3−µa22 ≤ (1 β1)−3βµ; 1 ; µ 234≤anµd≤µ34 2 .
(cid:26) − | − | ≥ 3 ≤ 3
(cid:12) (cid:12)
The second Hankel determ(cid:12)inant H2(cid:12)(2) is given by H2(2) = a2a4 − a23. The bounds for the
second Hankel determinant H (2) obtained for the classes and in [12]. Recently, Lee et al.
2 ∗
S K
[16] established the sharp bound to H (2) by generalizing their classes using subordination. In
2
| |
their paper, one can find the sharp bound to H (2) for the functions in the classes (β) and
2 ∗
| | S
(β).
K
Inthispaper,weseekupperboundforthefunctionalH (2)=a a a2 forfunctionsf belonging
2 2 4− 3
to the classes (β)and (β).
Sσ∗ Kσ
Let be the class of functions with positive real part consisting of all analytic functions :
P P
C satisfying p(0)=1 and p(z)>0.
U → ℜ
To establish our main results, we shall require the following lemmas.
Lemma 1.1. [22] If the function p is given by the series
∈P
(1.2) p(z)=1+c z+c z2+...
1 2
then the sharp estimate c 2 (k =1,2,...) holds.
k
| |≤
Lemma 1.2. [10] If the function p is given by the series (1.2), then
∈P
(1.3) 2c = c2+x(4 c2)
2 1 − 1
(1.4) 4c = c3+2(4 c2)c x c (4 c2)x2+2(4 c2) 1 x2 z,
3 1 − 1 1 − 1 − 1 − 1 −| |
(cid:16) (cid:17)
for some x, z with x 1 and z 1.
| |≤ | |≤
2. Main results
Our first main result for the class (β) as follows:
Sσ∗
Theorem 2.1. Let f(z) given by (1.1) be in the class (β), 0 β <1. Then
Sσ∗ ≤
4(1 β)2 4β2 8β+5 , β 0,29 √137
(2.1) a a a2 3 − − ∈ −32
(cid:12) 2 4− 3(cid:12)≤ (1−β)2 (cid:0)1163ββ22−−1246ββ+−57 ,(cid:1) β ∈ h29−3√2137,1i.
(cid:12) (cid:12) (cid:16) (cid:17) (cid:16) (cid:17)

SECOND HANKEL DETERMINANT FOR BI-STARLIKE AND BI-CONVEX FUNCTIONS OF ORDER β 3
Proof. Let f (β)and g =f 1. Then
∈Sσ∗ −
zf (z) wg (w)
′ ′
(2.2) =β+(1 β)p(z)and =β+(1 β)q(w)
f(z) − g(w) −
where p(z)=1+c z+c z2+... and q(w)=1+d w+d w2+... in .
1 2 1 2
P
Comparing coefficients in (2.2), we have
(2.3) a = (1 β)c ,
2 1
−
(2.4) 2a a2 = (1 β)c ,
3− 2 − 2
(2.5) 3a 3a a +a3 = (1 β)c
4− 3 2 2 − 3
and
(2.6) a = (1 β)d ,
2 1
− −
(2.7) 3a2 2a = (1 β)d ,
2− 3 − 2
(2.8) 10a3+12a a 3a = (1 β)d .
− 2 3 2− 4 − 3
From (2.3) and (2.6), we arrive at
(2.9) c = d
1 1
−
and
(2.10) a =(1 β)c .
2 1
−
Now, from (2.4), (2.7) and (2.10), we get that
(1 β)
(2.11) a =(1 β)2c2+ − (c d ).
3 − 1 4 2− 2
Also, from (2.5) and (2.8), we find that
2 5 1
(2.12) a = (1 β)3c3+ (1 β)2c (c d )+ (1 β)(c d ).
4 3 − 1 8 − 1 2− 2 6 − 3− 3
Thus, we can easily establish that
1 1
a a a2 = (1 β)4c4+ (1 β)3c2(c d )
2 4− 3 −3 − 1 8 − 1 2− 2
(cid:12)
(cid:12) (cid:12) (cid:12)1 1
(2.13) (cid:12) (cid:12) +(cid:12)(cid:12) (1 β)2c1(c3 d3) (1 β)2(c2 d2)2 .
6 − − − 16 − −
(cid:12)
(cid:12)
(cid:12)
According to Lemma 1.2 and (2.9), we write (cid:12)
2c =c2+x(4 c2) 4 c2
(2.14) 2d22=d121+x(4−−d121) (cid:27)=⇒c2−d2 = −2 1(x−y)
and
4c = c3+2(4 c2)c x c (4 c2)x2+2(4 c2) 1 x2 z,
3 1 − 1 1 − 1 − 1 − 1 −| |
(cid:16) (cid:17)
4d = d3+2(4 d2)d y d (4 d2)y2+2(4 d2) 1 y 2 w,
3 1 − 1 1 − 1 − 1 − 1 −| |
(cid:16) (cid:17)
(2.15)
c3 c 4 c2 c 4 c2 4 c2
c d = 1 + 1 − 1 (x+y) 1 − 1 (x2+y2)+ − 1 1 x2 z 1 y 2 w .
3 3
− 2 2 − 2 2 −| | − −| |
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
(cid:16)(cid:16) (cid:17) (cid:16) (cid:17) (cid:17)
4 ERHANDENIZ,MURATC¸AG˘LAR,ANDHALITORHAN
for some x,y, z,w with x 1, y 1, z 1 and w 1.Using (2.14) and (2.15) in (2.13), and
| |≤ | |≤ | |≤ | |≤
applying the triangle inequality we have
1 1
a a a2 = (1 β)4c4+ (1 β)3c2(4 c2)(x y)
2 4− 3 −3 − 1 16 − 1 − 1 −
(cid:12)
(cid:12)(cid:12)+1(1 β)(cid:12)(cid:12)2c1(cid:12)(cid:12)(cid:12) c31 + (4−c21)c1(x+y) (4−c21)c1(x2+y2)+ (4−c21) (1 x2)z (1 y 2)w
6 − 2 2 − 4 2 −| | − −| |
(cid:20) (cid:16) (cid:17)(cid:21)
1
(1 β)2(4 c2)2(x y)2
−64 − − 1 −
(cid:12)
1 1 (cid:12) 1
(1 β)4c4+ (1 β)2c4+(cid:12) (1 β)2c (4 c2)
≤ 3 − 1 12 − 1 (cid:12) 6 − 1 − 1
1 1
+ (1 β)3c2(4 c2)+ (1 β)2c2(4 c2) (x + y )
16 − 1 − 1 12 − 1 − 1 | | | |
(cid:20) (cid:21)
1 1 1
+ (1 β)2c2(4 c2) (1 β)2c (4 c2) (x2+ y 2)+ (1 β)2(4 c2)2(x + y )2.
24 − 1 − 1 − 12 − 1 − 1 | | | | 64 − − 1 | | | |
(cid:20) (cid:21)
Sincep ,so c 2.Lettingc =c,wemayassumewithoutrestrictionthatc [0,2].Thus,
1 1
∈P | |≤ ∈
for λ= x 1 and µ= y 1 we obtain
| |≤ | |≤
a a a2 T +T (λ+µ)+T (λ2+µ2)+T (λ+µ)2 =F(λ,µ)
2 4− 3 ≤ 1 2 3 4
where (cid:12) (cid:12)
(cid:12) (cid:12)
(1 β)2
T = T (c)= − 1+4(1 β)2 c4 2c3+8c 0,
1 1
12 − − ≥
1 h(cid:16) (cid:17) i
T = T (c)= (1 β)2c2(4 c2)(7 3β) 0,
2 2
48 − − − ≥
1
T = T (c)= (1 β)2c(4 c2)(c 2) 0,
3 3
24 − − − ≤
1
T = T (c)= (1 β)2(4 c2)2 0.
4 4 64 − − 1 ≥
Now we need to maximize F(λ,µ) in the closed square S = (λ,µ):0 λ 1,0 µ 1 .
{ ≤ ≤ ≤ ≤ }
Since T <0 and T +2T >0 for c [0,2), we conclude that
3 3 4
∈
F F (F )2 <0.
λλ µµ λµ
· −
Thus the function F cannot have a local maximum in the interior of the square S. Now, we
investigate the maximum of F on the boundary of the square S.
For λ=0 and 0 µ 1 (similarly µ=0 and 0 λ 1), we obtain
≤ ≤ ≤ ≤
F(0,µ)=G(µ)=(T +T )µ2+T µ+T .
3 4 2 1
i. The case T +T 0 : In this case for 0 < µ < 1 and any fixed c with 0 c < 2, it is
3 4
≥ ≤
clear that G(µ)=2(T +T )µ+T >0, that is, G(µ) is an increasing function. Hence, for fixed
′ 3 4 2
c [0,2), the maximum of G(µ) occurs at µ=1, and
∈
maxG(µ)=G(1)=T +T +T +T .
1 2 3 4
ii. The case T +T < 0 : Since T +2(T +T ) 0 for 0 < µ < 1 and any fixed c with
3 4 2 3 4
≥
0 c <2, it is clear that T +2(T +T )< 2(T +T )µ+T < T and so G(µ)> 0. Hence for
2 3 4 3 4 2 2 ′
≤
fixed c [0,2), the maximum of G(µ) occurs at µ=1.
∈
Also for c=2 we obtain
4
(2.16) F(λ,µ)= (1 β)2(4β2 8β+5).
3 − −
Taking into account the value (2.16), and the cases i and ii, for 0 µ 1 and any fixed c with
≤ ≤
0 c 2,
≤ ≤
maxG(µ)=G(1)=T +T +T +T .
1 2 3 4
For λ=1 and 0 µ 1 (similarly µ=1 and 0 λ 1), we obtain
≤ ≤ ≤ ≤
F(1,µ)=H(µ)=(T +T )µ2+(T +2T )µ+T +T +T +T .
3 4 2 4 1 2 3 4
Similarly to the above cases of T +T , we get that
3 4
maxH(µ)=H(1)=T +2T +2T +4T .
1 2 3 4
SECOND HANKEL DETERMINANT FOR BI-STARLIKE AND BI-CONVEX FUNCTIONS OF ORDER β 5
Since G(1) H(1) for c [0,2], maxF(λ,µ)=F(1,1) on the boundary of the square S. Thus
≤ ∈
the maximum of F occurs at λ=1 and µ=1 in the closed square S.
Let K :[0,2] R
→
(2.17) K(c)=maxF(λ,µ)=F(1,1)=T +2T +2T +4T .
1 2 3 4
Substituting the values of T ,T ,T and T in the function K defined by (2.17), yield
1 2 3 4
(1 β)2
K(c)= − 16β2 26β+5 c4+24(2 β)c2+48 .
48 − −
Assume that K(c) has a maximum(cid:2)(cid:0)value in an int(cid:1)erior of c [0,2], by el(cid:3)ementary calculation
∈
we find
(1 β)2
(2.18) K′(c)= − 16β2 26β+5 c3+12(2 β)c .
12 − −
As a result of some calculations we ca(cid:2)(cid:0)n do the followin(cid:1)g examine: (cid:3)
Case 1: Let 16β2 26β+5 0, that is, β 0,13 √89 . Therefore K (c) > 0 for c (0,2).
− ≥ ∈ −16 ′ ∈
Since K is an increasing function in the intervalh(0,2), maiximum point of K must be on the
boundary of c [0,2], that is, c=2. Thus, we have
∈
4
maxK(c)=K(2)= (1 β)2 4β2 8β+5 .
0 c 2 3 − −
≤ ≤
(cid:0) (cid:1)
Case 2: Let 16β2 26β + 5 < 0, that is, β 13 √89,1 . Then K (c) = 0 implies the
− ∈ −16 ′
real critical point c01 = 0 or c02 = 16−β122(22−6ββ+)5. Whe(cid:16)n β ∈ 13−(cid:17)1√689,29−3√2137 , we observe that
−
c02 ≥ 2, that is, c02 is out of the inqterval (0,2). Therefore th(cid:16)e maximum valueiof K(c) occurs at
c =0 or c=c which contradicts our assumption of having the maximum value at the interior
01 02
point of c [0,2]. Since K is an increasing function in the interval (0,2), maximum point of K
∈
must be on the boundary of c [0,2], that is, c=2. Thus, we have
∈
4
maxK(c)=K(2)= (1 β)2 4β2 8β+5 .
0 c 2 3 − −
≤ ≤
(cid:0) (cid:1)
When β 29 √137,1 we observe that c < 2, that is, c is interior of the interval [0,2].
∈ −32 02 02
Since K′′(c02)(cid:16)<0, the ma(cid:17)ximum value of K(c) occurs at c=c02. Thus, we have
12(2 β) 13β2 14β 7
maxK(c)=K(c )=K − − =(1 β)2 − − .
0≤c≤2 02 s16β2−26β+5! − (cid:18)16β2−26β+5(cid:19)
This completes the proof of the Theorem 2.1. (cid:3)
Forβ =0,Theorem2.1readilyyieldsthefollowingcoefficientestimatesforbi-starlikefunctions.
Corollary 2.2. Let f(z) given by (1.1) be in the class . Then
Sσ∗
20
a a a2 .
2 4− 3 ≤ 3
Our second main result for the class (cid:12) (β) is fo(cid:12)llowing:
(cid:12)σ (cid:12)
K
Theorem 2.3. Let f(z) given by (1.1) be in the class (β), 0 β <1. Then
σ
K ≤
(1 β)2 5β2+8β 32
(2.19) a a a2 − −
2 4− 3 ≤ 24 3β2 3β 4
(cid:18) − − (cid:19)
(cid:12) (cid:12)
Proof. Let f (β)and g (cid:12)=f 1. Th(cid:12)en
σ −
∈K
zf (z) wg (w)
′′ ′′
(2.20) 1+ =β+(1 β)p(z)and 1+ =β+(1 β)q(w)
f (z) − g (w) −
′ ′
where p(z)=1+c z+c z2+... and q(w)=1+d w+d w2+... in .
1 2 1 2
P
Now, equating the coefficients in (2.20), we have
(2.21) 2a = (1 β)c ,
2 1
−
(2.22) 6a 4a2 = (1 β)c ,
3− 2 − 2
(2.23) 12a 18a a +8a3 = (1 β)c
4− 3 2 2 − 3
6 ERHANDENIZ,MURATC¸AG˘LAR,ANDHALITORHAN
and
(2.24) 2a = (1 β)d ,
2 1
− −
(2.25) 8a2 6a = (1 β)d ,
2− 3 − 2
(2.26) 32a3+42a a 12a = (1 β)d .
− 2 3 2− 4 − 3
From (2.21) and (2.24), we arrive at
(2.27) c = d
1 1
−
and
1
(2.28) a = (1 β)c .
2 1
2 −
Now, from (2.22), (2.25) and (2.28), we get that
1 1
(2.29) a = (1 β)2c2+ (1 β)(c d ).
3 4 − 1 12 − 2− 2
Also, from (2.23) and (2.26), we find that
5 5 1
(2.30) a = (1 β)3c3+ (1 β)2c (c d )+ (1 β)(c d ).
4 48 − 1 48 − 1 2− 2 24 − 3− 3
Thus, we can easily establish that
1 1
a a a2 = (1 β)4c4+ (1 β)3c2(c d )
2 4− 3 −96 − 1 96 − 1 2− 2
(cid:12)
(cid:12) (cid:12) (cid:12)1 1
(2.31) (cid:12) (cid:12) +(cid:12)(cid:12) (1 β)2c1(c3 d3) (1 β)2(c2 d2)2 .
48 − − − 144 − −
(cid:12)
(cid:12)
Using (2.14) and (2.15) in (2.31), we have (cid:12)
(cid:12)
1 1
a a a2 = (1 β)4c4+ (1 β)3c2(4 c2)(x y)
2 4− 3 −96 − 1 192 − 1 − 1 −
(cid:12)
(cid:12)(cid:12)+ 1 (1 β(cid:12)(cid:12))2c(cid:12)(cid:12)(cid:12)1 c31 + (4−c21)c1(x+y) (4−c21)c1(x2+y2)+ (4−c21) (1 x2)z (1 y 2)w
48 − 2 2 − 4 2 −| | − −| |
(cid:20) (cid:16) (cid:17)(cid:21)
1
(1 β)2(4 c2)2(x y)2
−288 − − 1 −
(cid:12)
1 1 (cid:12) 1
(1 β)4c4+ (1 β)2c4+(cid:12) (1 β)2c (4 c2)
≤ 96 − 1 96 − 1 (cid:12) 48 − 1 − 1
1 1
+ (1 β)3c2(4 c2)+ (1 β)2c2(4 c2) (x + y )
192 − 1 − 1 96 − 1 − 1 | | | |
(cid:20) (cid:21)
1 1 1
+ (1 β)2c2(4 c2) (1 β)2c (4 c2) (x2+ y 2)+ (1 β)2(4 c2)2(x + y )2.
192 − 1 − 1 − 96 − 1 − 1 | | | | 576 − − 1 | | | |
(cid:20) (cid:21)
Since p , so c 2. Taking c = c, we may assume without restriction that c [0,2]. Thus,
1 1
∈ P | | ≤ ∈
for λ= x 1 and µ= y 1 we obtain
| |≤ | |≤
a a a2 M +M (λ+µ)+M (λ2+µ2)+M (λ+µ)2 =Ψ(λ,µ)
2 4− 3 ≤ 1 2 3 4
where (cid:12) (cid:12)
(cid:12) (cid:12)
(1 β)2
M = M (c)= − 1+(1 β)2 c4 2c3+8c 0,
1 1
96 − − ≥
1 h(cid:16) (cid:17) i
M = M (c)= (1 β)2c2(4 c2)(3 β) 0,
2 2
192 − − − ≥
1
M = M (c)= (1 β)2c(4 c2)(c 2) 0,
3 3
192 − − − ≤
1
M = M (c)= (1 β)2(4 c2)2 0.
4 4 576 − − 1 ≥
ThereforeweneedtomaximizeΨ(λ,µ)intheclosedsquareS= (λ,µ):0 λ 1,0 µ 1 .
{ ≤ ≤ ≤ ≤ }
To show that the maximum of Ψ we can follow the maximum of F in the Theorem 2.1. Thus the
maximum of Ψ occurs at λ=1 and µ=1 in the closed square S. Let Φ:[0,2] R defined by
→
(2.32) Φ(c)=maxΨ(λ,µ)=Ψ(1,1)=M +2M +2M +4M .
1 2 3 4
SECOND HANKEL DETERMINANT FOR BI-STARLIKE AND BI-CONVEX FUNCTIONS OF ORDER β 7
Substituting the values of M ,M ,M and M in the function Φ given by (2.32), yield
1 2 3 4
(1 β)2
Φ(c)= − 3β2 3β 4 c4+4(8 3β)c2+32 .
288 − − −
Assume that Φ(c) has a maximum(cid:2)(cid:0)value in an in(cid:1)terior of c [0,2], by (cid:3)elementary calculation
∈
we find
(1 β)2
Φ′(c)= − 3β2 3β 4 c3+2(8 3β)c .
72 − − −
Setting Φ(c)=0, since 0<c<2, and(cid:2)(cid:0)3β2 3β 4<(cid:1) 0 and 8 3β >(cid:3)0 for every β [0,1) we
′
− − − ∈
have the real criticalpoin c03 = 3β2(23β3−β8)4. Since c03 ≤2 for every β ∈[0,1) and so Φ′′(c03)<0,
− −
the maximum value of Φ(c) corrqesponds to c=c03, that is,
2(3β 8) (1 β)2 5β2+8β 32
maxΦ(c)=Φ(c )=Φ − = − − .
0<c<2 03 s3β2−3β−4! 24 (cid:18) 3β2−3β−4 (cid:19)
On the other hand,
(1 β)2 (1 β)2
Φ(0)= − and Φ(2)= − β2 2β+2 .
9 6 −
Consequently, since Φ(0)<Φ(2) Φ(c ) we obtain max Φ(cid:0)(c)=Φ(c )(cid:1).
≤ 03 0 c< 2 03
This completes the proof of the Theorem 2.3. ≤ ≤ (cid:3)
Forβ =0,Theorem2.3readilyyieldsthefollowingcoefficientestimatesforbi-convexfunctions.
Corollary 2.4. Let f(z) given by (1.1) be in the class . Then
σ
K
1
a a a2 .
2 4− 3 ≤ 3
(cid:12) (cid:12)
(cid:12) Referen(cid:12)ces
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DepartmentofMathematics,Facultyof Science andLetters, KafkasUniversity,Kars, Turkey.
E-mail address: [email protected] (Erhan Deniz), [email protected] (Murat C¸a˘glar)
DepartmentofMathematics,Facultyof Science,AtaturkUniversity, Erzurum,25240,Turkey.
E-mail address: [email protected] (Halit Orhan)