27. Application of an integral operator for p-valent Functions

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ISSN: 1821-1291, URL: http://www.bmathaa.org Volume 3 Issue 4(2011), Pages 232-239.

APPLICATION OF AN INTEGRAL OPERATOR FOR P-VALENT FUNCTIONS

(COMMUNICATED BY SHIGEYOSHI OWA)

MASLINA DARUS, IMRAN FAISAL AND ZAHID SHAREEF

Abstract. By making use of an integral operator defined in an open unit disk, we introduce and study certain new subclasses of p-valent functions. Inclusion relationships are established and integral preserving properties of functions in these subclasses are discussed.

1. Introduction and preliminaries

LetAp denote the class of functions of the form

f(z) =zp+ ∞

∑

k=p+1

akzk, p∈N={1,2,· · · }. (1.1)

which are analytic in an open unit diskU={z:|z|<1}.

Next we deﬁne some well known subclasses of p-valent functions as follows:

S_p∗(ξ) = {

f ∈Ap:ℜ

( zf′(z)

f(z) )

> ξ,0≤ξ < p, p∈N, z∈U }

;

Kp(ρ, ξ) =

{

f ∈Ap:ℜ

( 1 +zf

′′₍_z₎

f′(z) )

> ξ,0≤ξ < p, p∈N, z∈U }

;

Kp(ρ, ξ) =

{

f ∈Ap:∃g(z)∈Sp∗(ξ)∧ ℜ

( zf′(z)

g(z) )

> ρ,0≤ρ, ξ < p, p∈N, z∈U }

;

K_p∗(ρ, ξ) = {

f ∈Ap:∃g(z)∈Cp(ξ)∧ ℜ

(

(zf′(z))′ g′(z)

)

> ρ,0≤ρ, ξ < p, p∈N, z∈U }

.

2000Mathematics Subject Classiﬁcation. 30C45.

Key words and phrases. Analytic Functions; Integral operator; Inclusion properties. c

⃝2011 Universiteti i Prishtinës, Prishtinë, Kosovë. Submitted May 8, 2011. Published December 1, 2011.

M.Darus, I. Faisal, and Z. Shareef were supported by the grant MOHE: UKM-ST-06-FRGS0244-2010. The authors also would like to thank the referee for the comments and suggestion given to improve the article.

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The classes Sp∗(ξ), Sp∗, Cp(ξ), Cp, Kp(ρ, ξ), K1(ρ, ξ), Kp∗(ρ, ξ) and K1∗(ρ, ξ) were

introduced by Patil and Thakare [13], Goodman [14], Owa [15], Aouf [16], Libera [17] and Noor [18, 19] respectively.

Also note that

f(z)∈Cp(ξ) if and only if zf′(z)

p ∈Sp∗(ξ), 0≤ξ < p, p∈N, z∈U·

Similarly

f(z)∈Kp(ξ) if and only if zf′(z)

p ∈Kp∗(ξ), 0≤ξ < p, p∈N, z∈U·

For a functionf ∈Ap, we deﬁne a diﬀerential operator as follow:

Υ0f(z) = f(z);

Υ1_λ(p, α, β, µ)f(z) = (α−pµ+β−pλ

(α+β) )f(z) + (

pµ+pλ (α+β))

zf′(z) p ;

Υ1_λ(p, α, β, µ)f(z) = Dp,λ(α, β, µ)f(z);

Υ2_λ(p, α, β, µ)f(z) = D(Υ1_λ(p, α, β, µ)f(z)); ..

.

Υnλ(p, α, β, µ)f(z) = D(Υ n₋1

λ (p, α, β, µ)f(z))· (1.2)

Iff is given by (1.1) then from (1.2) we have

Υn_λ(p, α, β, µ)f(z) = zp+ ∞

∑

k=p+1

(_α_{+ (}_µ₊_λ₎₍_k₋_p_{) +}_β α+β

)n

akzk, (1.3)

wheref ∈Ap, α, β, µ, λ≥0, α+β= 0̸ , n∈No={0,1,· · · }.

This generalizes some well known diﬀerential operators available in literature (see for examples [1]-[11]).

Now we deﬁne the integral operator forf(z)∈Ap as follows:

{0

p(α, β, µ, λ)f(z) =f(z);

{1

p(α, β, µ, λ)f(z) =

α+β µ+λz

p−(α_µ+₊_λβ)

∫ z

o

t(αµ++βλ)−p−1_f₍_t₎_dt_;

{2

p(α, β, µ, λ)f(z) =

α+β µ+λz

p−(α_µ₊+β_λ)

∫ z

o

t(αµ++βλ)−p−1{1

p(α, β, µ, λ)f(t)dt;

.. .

{m

p (α, β, µ, λ)f(z) =

α+β µ+λz

p−(α_µ+₊_λβ)

∫ z

o

t(αµ++βλ)−p−1{m−1

p (α, β, µ, λ)f(t)dt·

This implies

{m

p(α, β, µ, λ)f(z) =z p₊

∞

∑

k=2

(

α+β

α+ (µ+λ)(k−p) +β )m

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where α ≥ 0, β ≥ 0, µ ≥0, λ ≥ 0, p ∈ N, m ∈ N0 = N∪ {0}, f(z) ∈ Ap, z ∈ U).

From (1.4) we have (

µ+λ )

z (

{m+1

p (α, β, µ, λ)f(z)

)′ =

( α+β

)(

{m

p(α, β, µ, λ)f(z)

) −

(

α+β−p(µ+λ) )(

{m+1

p (α, β, µ, λ)f(z)

) ·(1.5)

Using the operator {m

p(α, β, µ, λ)f(z) deﬁned in (1.4), we introduce the following

subclasses of p-valent functions:

S∗_m(p, ξ, λ, α, β, µ) = {

f ∈Ap:{mp (α, β, µ, λ)f ∈Sp∗(ξ)

} ;

Cm(p, ξ, λ, α, β, µ) =

{

f ∈Ap:{mp (α, β, µ, λ)f ∈Cp(ξ)

} ;

Km(p, ρ, ξ, λ, α, β, µ) =

{

f ∈Ap:{mp (α, β, µ, λ)f ∈Kp(ρ, ξ)

} ;

K∗_m(p, ρ, ξ, λ, α, β, µ) = {

f ∈Ap:{mp (α, β, µ, λ)f ∈Kp∗(ρ, ξ)

} .

2. _{Inclusion relationships}

In this section, we establish various inclusion relationships for the functions belong-ing to the new subclasses of p-valent functions.

Lemma 2.1.[20, 21] Let φ(µ, ν) be a complex function, ϕ:D →C, D⊂C×C, and let µ = µ1+iµ1, ν = ν1+iν1. Suppose that φ(µ, ν) satisﬁes the following

conditions:

(1) φ(µ, ν) is continuous inD; (2) (1,0)∈D andℜ{φ(1,0)}>0;

(3) ℜ{φ(iµ2, ν1)} ≤0 for all (iµ2, ν1)∈D such thatν1≤ −12(1 +µ 2 2)·

Leth(z) = 1 +c1z+c2z2+· · · be analytic inU, such that (h(z), zh′(z))∈D for

allz∈U. Ifℜ{φ(h(z), zh′(z))}>0(z∈U) , thenℜ{h(z)}>0 forz∈U·

Theorem 2.2. Letf(z)∈Ap andα, β, µ, λ≥0, p∈N, m∈N0 =N∪ {0}, z∈U·

Then

S∗_m(p, ξ, λ, α, β, µ)⊆S∗_m+1(p, ξ, λ, α, β, µ)⊆S∗_m+2(p, ξ, λ, α, β, µ)·

Proof. To proveS∗m(p, ξ, λ, α, β, µ)⊆S∗m+1(p, ξ, λ, α, β, µ), letf(z)∈S∗m(p, ξ, λ, α, β, µ)

and assume that

z({m+1p (α, β, µ, λ)f(z))′ {m+1

p (α, β, µ, λ)f(z)

=ξ+ (p−ξ)h(z), 0≤ξ <1, z∈U· (2.1)

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Using (1.5) and (2.1), we have

z({mp(α, β, µ, λ)f(z))′ {m

p (α, β, µ, λ)f(z)

−ξ= (p−ξ)h(z) + (p−ξ)zh ′₍_z₎

(α+β_µ+λ−p) +ξ+ (p−ξ)h(z). (2.2)

Taking h(z) = µ = µ1+iµ1 and zh′(z) = ν = ν1+iν1, we deﬁne the function

φ(µ, ν) by:

φ(µ, ν) = (p−ξ)µ+ (p−ξ)ν

(α+β_µ+λ−p) +ξ+ (p−ξ)µ·

This implies

(i)φ(µ, ν) is continuous inD= (C−(

α+β µ+λ−p)+ξ

ξ−p )×C,

(ii) (1,0)∈D andℜ{φ(1,0)}>1−ξ,

(iii) For all (iµ2, ν1)∈D such thatν1≤ −1₂(1 +µ22).Therefore

ℜ{φ(iµ2, ν1)}=ℜ{

(p−ξ)ν

(α+β_µ+λ −p) +ξ+ (p−ξ)iµ2

}= [(

α+β

µ+λ −p) +ξ](p−ξ)ν

((α+β_µ+λ −p) +ξ)2_{+ (}_p−_ξ₎2_µ2 2

,

⇒

ℜ{φ(iµ2, ν1)}=

[(α+β_µ+λ −p) +ξ](p−ξ)ν

((α+β_µ+λ−p) +ξ)2_{+ (}_p−_ξ₎2_µ2 2

≤ −[(

α+β

µ+λ −p) +ξ](p−ξ)(1 +µ 2 2)

2((α+β_µ+λ−p) +ξ)2_{+ (}_p−_ξ₎2_µ2 2

<0·

Hence, the functionφ(µ, ν) satisﬁes the conditions of Lemma 2.1. This shows that ℜ{h(z)}>0(z∈U), that is,f(z)∈S∗_m+1(p, ξ, λ, α, β, µ)·

Theorem 2.3. If f(z)∈Ap and α, β, µ, λ≥0, p∈N, m∈N0 =N∪ {0}, z ∈U·

Then

Cm(p, ξ, λ, α, β, µ)⊆Cm+1(p, ξ, λ, α, β, µ)⊆Cm+2(p, ξ, λ, α, β, µ).

Proof. Letf ∈Cm(p, ξ, λ, α, β, µ)⇒{mp (α, β, µ, λ)f ∈Cp(ξ),⇔ z_p({mp(α, β, µ, λ)f)′ ∈

Sp∗(ξ)⇔{mp (α, β, µ, λ)f( zf′

p )∈Sp∗(ξ),⇔ zf′

p ∈Cm(p, ξ, λ, α, β, µ)⊆Cm+1(p, ξ, λ, α, β, µ),

⇒ zf′

p ∈Cm+1(p, ξ, λ, α, β, µ)⇔{ m+1

p (α, β, µ, λ)f( zf′

p )∈Sp∗(ξ),⇔ zp({ m+1

p (α, β, µ, λ)f)′∈

Sp∗(ξ)⇒{m+1p (α, β, µ, λ)f ∈Cp(ξ),⇒f ∈Cm+1(p, ξ, λ, α, β, µ)·

Then

Km(p, ρ, ξ, λ, α, β, µ)⊆Km+1(p, ρ, ξ, λ, α, β, µ)⊆Km+2(p, ρ, ξ, λ, α, β, µ)·

Proof. Letf(z)∈Km(p, ρ, ξ, λ, α, β, µ) implies

ℜ (

z({m

p(α, β, µ, λ)f(z))′ {m

p(α, β, µ, λ)g(z)

)

> ρ, g(z)∈S∗m(p, ξ, λ, α, β, µ),0≤ρ <1, z∈U·

Sinceg(z)∈S∗_m(p, ξ, λ, α, β, µ)⊆S∗_m+1(p, ξ, λ, α, β, µ), let

z({m+1p (α, β, µ, λ)g(z))′ {m+1

p (α, β, µ, λ)g(z)

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Suppose that (

z({m+1p (α, β, µ, λ)f(z))′ {m+1

p (α, β, µ, λ)g(z)

)

=ρ+ (p−ρ)h(z), 0≤ρ <1, z∈U· (2.4)

Whereh(z) = 1 +c1z+c2z2+· · · ·Using (1.5), (2.3) and (2.4) we get

z({m

p(α, β, µ, λ)f(z))′ {m

p(α, β, µ, λ)g(z)

=

z({m+1

p (α,β,µ,λ)zf′(z))′ {m+1

p (α,β,µ,λ)g(z)

+ (α+β_µ+λ−p)(ρ+ (p−ρ)h(z))

ξ+ (p−ξ)H(z) + (α+β_µ+λ −p) ·(2.5)

Using (2.3) and (2.4) we have

z(({m+1

p (α, β, µ, λ)zf′(z)))′

({m+1p (α, β, µ, λ)g(z))

= [ρ+ (p−ρ)h(z)][ξ+ (p−ξ)H(z)] + (p−ρ)zh′(z)·(2.6)

Simplifying simultaneously (2.5) and (2.6), we get

z({m

p(α, β, µ, λ)f(z))′ {m

p(α, β, µ, λ)g(z)

−ρ= (p−ρ)h(z) + (p−ρ)zh ′₍_z₎

(α+β_µ+λ−p+ξ) + (p−ξ)H(z)· (2.7)

φ(µ, ν) by

φ(µ, ν) = (p−ρ)µ+ (p−ρ)ν

(α+β_µ+λ −p+ξ) + (p−ξ)H(z)·

Clearly conditions (i) and (ii) of Lemma 2.1 inD =C×Care satisﬁed. For (iii), we proceed as follows.

ℜ{φ(iµ2, ν1)}=

ν1(p−ρ)[(α+β_µ+λ −p+ξ) + (p−ξ)h1(x1, y1)]

[(α+β_µ+λ−p+ξ) + (p−ξ)h1(x1, y1)]2+ [(p−ξ)h2(x2, y2)]2

·

WhereH(z) =h1(x1, y1) +ih2(x2, y2),h1(x1, y1) andh2(x2, y2) being functions of

xandyandℜ(h1(x1, y1))>0. Sinceν1≤ −1₂(1 +µ22),implies

ℜ{φ(iµ2, ν1)}=−

1 2

(1 +µ2

2)(p−ρ)[( α+β

µ+λ−p+ξ) + (p−ξ)h1(x1, y1)]

[(α+β_µ+λ−p+ξ) + (p−ξ)h1(x1, y1)]2+ [(p−ξ)h2(x2, y2)]2

<0·

Applying Lemma 2.1. on φ(µ, ν), gives ℜ{h(z)} > 0(z ∈ U). This shows that f(z)∈Km+1(p, ρ, ξ, λ, α, β, µ)·

The following theorem can be proved in a similar manner.

Then

K∗_m(p, ρ, ξ, λ, α, β, µ)⊆K∗_m+1(p, ρ, ξ, λ, α, β, µ)⊆K∗_m+2(p, ρ, ξ, λ, α, β, µ)·

3. _{Integral operator}

Forc >−pandf(z)∈Ap, the integral operatorLc,p:Ap→Ap is deﬁned by

Lc,p(f) =

c+p zc

∫ z

0

tc−1f(t)dt· (3.1)

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Theorem 3.1. Letc >−p, 0≤ξ < p·If f(z)∈S∗m(p, ξ, λ, α, β, µ), thenLc(f)∈

S∗m(p, ξ, λ, α, β, µ))·

Proof. Using (3.1) we get

z({m_p(α, β, µ, λ)Lc,pf(z))′ = (c+p)({mp(α, β, µ, λ)f(z))−c({ m

p(α, β, µ, λ)Lc,pf(z))·(3.2)

Let

z({m

p(α, β, µ, λ)Lcf(z))′ {m

p(α, β, µ, λ)Lcf(z)

=ξ+ (p−ξ)h(z)· (3.3)

Whereh(z) = 1 +c1z+c2z2+· · · ·By using (3.1) and (3.2) we get

z({m

p (α, β, µ, λ)f(z))′ {m

p(α, β, µ, λ)f(z)

−ξ= (p−ξ)h(z) + (p−ξ)zh ′₍_z₎

ξ+ (p−ξ)h(z) +c·

φ(µ, ν) by

φ(µ, ν) = (p−ξ)µ+ (p−ξ)ν ξ+c+ (p−ξ)µ·

Clearly conditions (i) and (ii) of Lemma 2.1 inD= (

C− {ξ+c ξ₋p}

)

×Care satisﬁed. We proceed for (iii) as follows;

ℜ{φ(iµ2, ν1)}=

(ξ+c)(p−ξ)ν1

[ξ+c]2_{+ [(}_p₋_ξ₎_µ 2]2

≤ −(ξ+c)(p−ξ)(1 +µ22)

2[ξ+c]2_{+ 2[(}_p₋_ξ₎_µ 2]2

<0·

Applying Lemma 2.1. we haveℜ{h(z)}>0(z∈U), that is,Lc(f)∈S∗m(p, ξ, λ, α, β, µ)·

Theorem 3.2. Let c >−p, 0≤ξ < p· Iff(z)∈Cm(p, ξ, λ, α, β, µ), then Lc(f)∈

Cm(p, ξ, λ, α, β, µ)·

Proof. For the proof, use Theorem 3.1 and the fact thatf(z)∈Cp(ξ)⇔ zf

′_(z)

p ∈

S_p∗(ξ).

Theorem 3.3. Let c > −p, 0 ≤ ξ < p· If f(z) ∈ Km(p, ρ, ξ, λ, α, β, µ), then

Lc(f)∈Km(p, ρ, ξ, λ, α, β, µ)·

Proof. Asf(z)∈Km(p, ρ, ξ, λ, α, β, µ) gives

ℜ (_z₍_{m

p (α, β, µ, λ)f(z))′ {m

p(α, β, µ, λ)g(z)

) > ρ·

Sinceg(z)∈S∗m(p, ξ, λ, α, β, µ)) implies Lc(g(z))∈S∗m(p, ξ, λ, α, β, µ))·Let

z({m

p (α, β, µ, λ)Lc(g(z))′ {m

p(α, β, µ, λ)Lcg(z)

=ξ+ (p−ξ)H(z), ℜ(H(z))>0, z∈U·

Also let

(_z₍_{m

p(α, β, µ, λ)Lcf(z))′ {m

p(α, β, µ, λ)Lcg(z)

)

=ρ+ (p−ρ)h(z), z∈U· (3.4)

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After doing calculations, we get

(_z₍_{m

p(α, β, µ, λ)f(z))′ {m

p(α, β, µ, λ)g(z)

) =

z({m

p(α,β,µ,λ)Lc(zf′(z))′

{m

p(α,β,µ,λ)Lcg(z) +c

({m

p(α,β,µ,λ)Lc(zf′(z))

{m

p(α,β,µ,λ)Lcg(z)

z({m

p(α,β,µ,λ)Lc(g(z))′

{m

p(α,β,µ,λ)Lcg(z) +c

·

Also

z({m

p(α, β, µ, λ)Lc(zf′(z))′ {m

p (α, β, µ, λ)Lcg(z)

= [ρ+ (p−ρ)h(z)][ξ+ (p−ξ)H(z)] + [(p−ρ)zh′(z)]·

Hence (_z₍_{m

p (α, β, µ, λ)f(z))′ {m

p (α, β, µ, λ)g(z)

)

−ρ= (p−ρ)h(z) + (p−ρ)zh ′₍_z₎

ξ+ (p−ξ)H(z) +c. (3.5)

φ(µ, ν) by

φ(µ, ν) = (p−ρ)µ+ (p−ρ)ν

ξ+ (p−ξ)H(z) +c· (3.6)

It is easy to see that the function φ(µ, ν) satisﬁes the conditions (i) and (ii) of Lemma 2.1 inD=C×C. To verify the condition (iii), we proceed as follows.

ℜ{φ(iµ2, ν1)}=

ν1(p−ρ)[(ξ+c) + (p−ξ)h1(x1, y1)]

[(ξ+c) + (p−ξ)h1(x1, y1)]2+ [(p−ξ)h2(x2, y2)]2·

WhereH(z) =h1(x1, y1) +ih2(x2, y2),h1(x1, y1) andh2(x2, y2) being functions of

xandyandℜ(h1(x1, y1))>0. By puttingν1≤ −1₂(1 +µ22),we obtain

ℜ{φ(iµ2, ν1)}=−

1 2

(1 +µ22)(p−ρ)[(ξ+c) + (p−ξ)h1(x1, y1)]

[(ξ+c) + (p−ξ)h1(x1, y1)]2+ [(p−ξ)h2(x2, y2)]2

<0·

By applying Lemma 2.1 we getℜ{h(z)}>0(z∈U), that is,Lc(f)∈Cm(p, ξ, λ, α, β, µ)·

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Maslina Darus

School of Mathematical Sciences Faculty of Science and Technology Universiti Ke-bangsaan Malaysia, Bangi 43600, Selangor Darul Ehsan Malaysia

E-mail address:[email protected] (Corresponding author)

Imran Faisal

E-mail address:[email protected]

Zahid Shareef