Coefficients estimates for functions inBn(α)

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PII. S0161171203302339 http://ijmms.hindawi.com © Hindawi Publishing Corp.

COEFFICIENTS ESTIMATES FOR FUNCTIONS IN

B

n

(α)

S. ABDUL HALIM

Received 12 February 2003

We consider functionsf, analytic in the unit disc and of the normalised form

f (z)=z+∞_k=2akzk. For functionsf∈Bn(α), the class of functions involving

the S˘al˘agean diﬀerential operator, we give some coeﬃcient estimates, namely,|a2|,

|a3|, and|a4|.

2000 Mathematics Subject Classiﬁcation: 30C45.

1. Introduction. LetAbe the class of functionsfwhich are analytic in the unit discD= {z:|z|<1}and are of the form

f (z)=z+

∞

j=2

ajzj. (1.1)

For functionsf∈A, we introduce the subclassBn(α)given by the following deﬁnition.

Definition1.1. Forα >0 andn=0,1,2,...,a functionf normalised by

(1.1) belongs toBn(α)if and only if, forz∈D,

ReD

n_{f (z)}α

zα >0, (1.2)

where Dn _{denotes the diﬀerential operator with} _Dn_{f (z)}₌_D(Dn−1_{f (z))} ₌ z[Dn−1_{f (z)]}_and_D◦_{f (z)}₌_{f (z)}_.

Remark1.2. The diﬀerential operatorDnwas introduced by S˘al˘agean [5].

Forn=1,B1(α)denotes the class of Bazilevi´c functions with logarithmic growth studied [4,6,7], amongst others. In [2], the author established some properties of the classBn(α)including showing thatBn(α)forms a subclass ofS, the class of all analytic, normalized, and univalent functions inD. The classB0(α)was initiated by Yamaguchi [8].

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Lemma2.1. Letp∈Pand let it be of the formp(z)=1+∞i=1cizi. Then

(i) |ci| ≤2fori≥1,

(ii) |c2−µc_{1| ≤}2 _{2 max}_{₁_,_|₁₋₂_µ_|}_{for any}_µ_∈_C_.

Lemma2.2(see [3]). If the functions1+∞v=1bvzvand1+

∞

v=1cvzvbelong

toP, then the same is true for the function1+(1/2)∞_v=1bvcvzv.

Lemma 2.3 (see [3]). Let h(z)=1+h1z+h2z2_{+ ···} _{and let} ₁₊_g(z) ₌

1+g1z+g2z2_+···_{be functions in}_P_{. Set}_γ0₌₁_{and for}_v_≥₁_,

γv=2−v



1+1₂

v µ=1 v µ hµ 

. (2.1)

IfAkis deﬁned by

∞

v=1

(−1)v+1_γ

v−1

g(z)v=

∞

k=1

Akzk, (2.2)

then

Ak≤2. (2.3)

3. Results

Theorem3.1. Ifα >0,n=0,1,2,...,andf∈Bn(α)(nis ﬁxed) withf (z)=

z+∞j=2ajzj, then the following inequalities hold:

a2≤₍2₁α₊n−_α)1_n, (3.1)

a3≤

            

2αn−1 (2+α)n



1−

α−1

α

α2₊₂_α α2₊₂_α₊₁

n

, for0< α <1,

2αn−1

(2+α)n, forα≥1,

(3.2)

_a4_≤

                    

2αn−1 (3+α)n

+₍4(1−α)α2n−2

1+α)n₍₂₊_α)n

1+(1−2α)(2+α)nαn−1

3(1+α)2n

, for0< α <1,

2αn−1

(3+α)n, forα≥1.

(3.3)

Remark3.2. Whenn=1, the above results reduce to those obtained by

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COEFFICIENTS ESTIMATES FOR FUNCTIONS IN n 3763

Proof. Forf∈Bn(α),Deﬁnition 1.1gives

ReD n_{f (z)}α

zα >0. (3.4)

Inequality (3.4) suggests that there existsp∈P such that forz∈D,

Dn_{f (z)}α

zα =αnp(z). (3.5)

Next, writingDn_{f (z)}α _as_z[Dn−1_{f (z)}α_]_and _p(z)₌₁₊∞

i=1ciziin (3.5), it

follows that

Dn−1_{f (z)}α₌_αn



zα−1₊∞

i=1

cizi+α−1



 (3.6)

and integration gives

Dn−1_{f (z)}α

zα =αn−

1



1+

∞

i=1 α cizi

(i+α)



. (3.7)

Now, repeating the process, we are able to establish the following relation which holds in general for anyk=0,1,2,...,n

Dn−k_{f (z)}α

zα =αn−k



1+

∞

i=1

αk cizi

(i+α)k



. (3.8)

In particular, whenn=k, we have

D0_{f (z)}α

zα =

_{f (z)}

z

α

=1+

∞

i=1

αn cizi

(i+α)n. (3.9)

On comparing coeﬃcients in (3.9) withf (z)=z+∞j=2ajzj, we obtain

αa2=₍αnc1

1+α)n, (3.10)

αa3=₍αnc2

2+α)n+

α(1−α)a2 2

2 , (3.11)

αa4=₍αnc3

3+α)n+

α(1−α)(α−2)a3 2

6 +α(1−α)a3a2. (3.12)

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Eliminatinga2in (3.11), we have

a3=₍αn−1c2

2+α)n+

(1−α)

2

_α_n−1_c1

(1+α)n

2

=₍₂α₊n−_α)1_n

c2−(α−₂1)₍(₁2₊+_α)α)2nnαn−

1_c2 1

=₍αn−1

2+α)n

_c2

−µc2 1

≤₍2αn−1

2+α)nmax

1,|1−2µ|,

(3.13)

where we usedLemma 2.1(ii) with

2µ=(α₍−₁₊1)α_α)n−_n 1

₂₊_α

1+α

n

. (3.14)

Sinceµ≥0 forα≥1, both inequalities in (3.2) are easily obtained. We now prove (3.3). Using (3.10) and (3.11) in (3.12) gives

a4=₍₃α₊n−_α)1_n

c3+(1−α)(₍₁3₊+_α)α)_nnαn−1

c1c2 (2+α)n+

(1−2α)αn−1_c3 1

6(1+α)2n

.

(3.15)

First, we consider the case 0< α <1/2. Applying the triangle inequality with

Lemma 2.1(i) in (3.15) results in the inequality

a4≤₍2αn−1

3+α)n

1+2(1−α)(3+α) n_αn−1

(1+α)n

1

(2+α)n+

(1−2α)αn−1

3(1+α)2n

(3.16)

which is the ﬁrst inequality in (3.3).

For the case 1/2≤α <1, we use Carathéodory-Toeplitz result which states that for someεwith|ε|<1,

c2=c 2 1

2 +ε

2−|c1|₂2

. (3.17)

Thus, (3.15) becomes

a4=₍₃α₊n−_α)1_n

c3+(1−α)(₍3₁+₊α)_α)_nnαn−1c1

×

c2 1

2(2+α)n+

(1−2α)αn−1_c2 1

6(1+α)2n +

ε (2+α)n

2−|c1|2

2

.

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We then have

_a4_≤ αn−1

(3+α)n

_c3₊(1−α)(3+α)n_αn−1_c1

(1+α)n₍₂₊_α)n

c2 1

2w−

_c12

2 ε+2ε

,

(3.19)

where

w=1+(1−2α)αn−1(2+α)n

3(1+α)2n . (3.20)

Since 0< w≤1 and|ε|<1, it is easily shown that

a4≤

αn−1 (3+α)n

c3+

(1−α)(3+α)nαn−1_c1 (1+α)n₍₂₊_α)n

c12

2 (w−1)+2

(3.21)

and the result follows trivially when using|c1| ≤2 and|c3| ≤2.

Finally, we consider (3.3) for the caseα≥1. Here, we use a method intro-duced by Nehari and Netanyahu [3] which was also used by Singh [6] and the author in [1].

First, lethandgbe deﬁned as inLemma 2.3, and sincep∈P,Lemma 2.2

indicates that

1+G(z)=1+1₂

∞

k=1

gkckzk (3.22)

also belongs toP.

Next, it follows from (2.2) that, withgreplaced byG,

A3=1₂g3c3−1₂γ1g1g2c1c2+1₈γ2g3 1c13

. (3.23)

Rewriting (3.15) as

α1−n₍₃₊_α)n_a4₌_c3₊(1−α)(3+α)nαn−1

(1+α)n₍₂₊_α)n c1c2

+(1−α)(1−₆₍2₁α)(₊_α)3₃+_nα)nα2n−2c3 1

(3.24)

and comparing it with (3.23), the required result is easily obtained since, by

Lemma 2.3,|A3| =((3+α)n_/(αn−1₎₎_|_{a4| ≤}_{2. This however is only true if we}

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Now sinceg1=g2=g3=2, it follows from (3.23) and (3.24) that

2γ1=(α₍−1)(3+α)nαn−1

1+α)n₍₂₊_α)n , (3.25)

γ2=(1−α)(1−2α)(3+α)nα2n−2

6(1+α)3n . (3.26)

However, from (2.1), we have

γ1=1

2

1+1

2h1

, (3.27)

γ2=1₄

1+h1+1₂h2

. (3.28)

Solving forh1by eliminatingγ1from (3.25) and (3.27), we obtain

h1=2(α−1)(3+α)

n_αn−1

(1+α)n(2+α)n −1

. (3.29)

Quite trivially, it can be seen that|h1| ≤2 forα≥1.

In a similar manner, eliminatingγ2from (3.26) and (3.28) and usingh1given by (3.29), we have

h2=2

1−2₃

1−_α1

_α2₊₃_α

α2₊₃_α₊₂

n₁₋₂_α

α

_α2₊₂_α

α2₊₂_α₊₁

n

+3

.

(3.30)

Forα≥1, elementary calculations show that|h2| ≤2. Next, we constructhby ﬁrst setting it to be of the form

h(z)=µ1(₁1₊−_zz)+µ2

1+λz2

1−λz2 (3.31)

with

µ1=1−(α₍₁−₊1)(_α)3_n+₍₂α)₊n_α)αn−_n 1,

µ2=(α₍−₁₊1)(_α)3_n+₍₂α)₊n_α)αn−_n 1,

λ=1−2₃

₁₋₂_α

α

_α2₊₂_α

α2₊₂_α₊₁

n

+3

.

(3.32)

It is readily seen that forα≥1, bothµ1andµ2are nonnegative andµ1+µ2=

1. Further, with a little bit of manipulation, it can be shown that|λ| ≤1 and the coeﬃcients ofzandz2_{in the expansion of}_h_{are respectively those given}

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References

[1] S. Abdul Halim,On the coeﬃcients of some Bazilevic functions of order, J.

Ramanu-jan Math. Soc.4(1989), no. 1, 53–64.

[2] ,On a class of analytic functions involving the S˘al˘agean diﬀerential operator,

Tamkang J. Math.23(1992), no. 1, 51–58.

[3] Z. Nehari and E. Netanyahu,On the coeﬃcients of meromorphic schlicht functions,

Proc. Amer. Math. Soc.8(1957), 15–23.

[4] M. Obradovi´c,Some results on Bazileviˇc functions, Mat. Vesnik37(1985), no. 1,

92–96.

[5] G. S. S˘al˘agean, Subclasses of univalent functions, Complex Analysis—Fifth

Romanian-Finnish Seminar, Part 1 (Bucharest, 1981), Lecture Notes in Math., vol. 1013, Springer, Berlin, 1983, pp. 362–372.

[6] R. Singh,On Bazileviˇc functions, Proc. Amer. Math. Soc.38(1973), 261–271.

[7] D. K. Thomas,On a subclass of Bazileviˇc functions, Int. J. Math. Math. Sci.8(1985),

no. 4, 779–783.

[8] K. Yamaguchi,On functions satisfyingR{f (z)/z}<0, Proc. Amer. Math. Soc.17

(1966), 588–591.

S. Abdul Halim: Institute of Mathematical Sciences, University of Malaya, 50603 Kuala Lumpur, Malaysia

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