IJMMSInternational Journal of Mathematics and Mathematical Sciences1687-04250161-1712Hindawi Publishing Corporation36907810.1155/2010/369078369078Research ArticleDifferential Subordination Defined by New Generalised Derivative Operator for Analytic FunctionsHarayzeh Al-AbbadiMa'moun1DarusMaslina1WallaceDorothySchool of Mathematical SciencesFaculty of Science and TechnologyUniversiti Kebangsaan Malaysia43600 Bangi Selangor (Darul Ehsan)Malaysiaukm.my201074201020100906200920122009090320102010Copyright © 2010This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

A new generalised derivative operator μλ1,λ2n,m is introduced. This operator generalised many well-known operators studied earlier by many authors. Using the technique of differential subordination, we will study some of the properties of differential subordination. In addition we investigate several interesting properties of the new generalised derivative operator.

1. Introduction and Preliminaries

Let 𝒜 denote the class of functions of the form

f(z)=  z  +  k=2akzk,where  ak  is  complex  number, which are analytic in the open unit disc U={z  :  |z|  <  1} on the complex plane . Let S,S*(α),C(α)  (0α<1) denote the subclasses of 𝒜 consisting of functions that are univalent, starlike of order α, and convex of order α in U, respectively. In particular, the classes S*(0)=S*  and  C(0)=C are the familiar classes of starlike and convex functions in U, respectively. A function fC(α) if Re(1+zf/f)>α. Furthermore a function f analytic in U is said to be convex if it is univalent and f(U) is convex.

Let (U) be the class of holomorphic function in unit disc U={z  :  |z|<1}.

Let

𝒜n={f(U):  f(z)=z+an+1zn+1+,  (zU)}, with 𝒜1=𝒜.

For a and n={1,2,3,} we let

[a,n]={f(U):  f(z)=z+anzn+an+1zn+1+,  (zU)}. Let f(z)=z+k=2akzk  and  g(z)=z+k=2bkzk  be analytic in the open unit disc U={z:  |z|<1}. Then the Hadamard product (or convolution) f*g of the two functions f, g is defined by

f(z)*g(z)=(f*g)(z)=z+k=2akbkzk. Next, we state basic ideas on subordination. If f and g are analytic in U, then the function f is said to be subordinate to g, and can be written as

fg,f(z)g(z),(zU), if and only if there exists the Schwarz function w, analytic in U, with w(0)=0  and  |w(z)|<1  such that  f(z)=g(w(z)), (zU)  .

Furthermore if g is univalent in U, then fg if and only if f(0)=g(0) and f(U)g(U) (see [1, page 36]).

Let ψ:  3×U and let h be univalent in U. If p is analytic in U and satisfies the (second-order) differential subordination

ψ(p(z),zp(z),z2p′′(z);z)h(z),(zU), then p is called a solution of the differential subordination.

The univalent function q is called a dominant of the solutions of the differential subordination, or more simply a dominant, if pq for all p satisfying (1.6). A dominant q̃ that satisfies q̃q for all dominants q of (1.6) is said to be the best dominant of (1.6). (Note that the best dominant is unique up to a rotation of U.)

Now, (x)k denotes the Pochhammer symbol (or the shifted factorial) defined by

(x)k={1for  k=0,x{0},x(x+1)(x+2)(x+k-1)for  k={1,2,3,},  x. To prove our results, we need the following equation throughout the paper:

μλ1,λ2n,m+1f(z)=(1-λ1)[μλ1,λ2n,mf(z)*ϕλ2(z)]+λ1z[μλ1,λ2n,mf(z)*ϕλ2(z)],(zU), where n,m0={0,1,2,}, λ2λ10, and ϕλ2(z) is analytic function given by

ϕλ2(z)=z+k=2zk1+λ2(k-1). Here μλ1,λ2n,m   is the generalized derivative operator which we shall introduce later in the paper. Moreover, we need the following lemmas in proving our main results.

Lemma 1.1 (see [<xref ref-type="bibr" rid="B5">2</xref>, page 71]).

Let h be analytic, univalent, and convex in U, with h(0)=a, γ0 and, Re  γ0. If p[a,n]  and p(z)+zp(z)γh(z),(zU), then p(z)q(z)h(z),(zU), where q(z)=(γ/nzγ/n)0zh(t)  t(γ/n)-1dt,(zU).

The function q is convex and is the best (a,n)-dominant.

Lemma 1.2 (see [<xref ref-type="bibr" rid="B6">3</xref>]).

Let g be a convex function in U and let h(z)=g(z)+nαzg'(z), where α>0 and n is a positive integer.

If

p(z)=g(0)+pnzn+pn+1zn+1+,(zU), is holomorphic in U and p(z)+αzp(z)h(z),(zU), then p(z)g(z), and this result is sharp.

Lemma 1.3 (see [<xref ref-type="bibr" rid="B7">4</xref>]).

Let f𝒜, if Re(1+zf′′(z)f(z))>-12, then 2z0zf(t)dt,(zU,z0), belongs to the class of convex functions.

2. Main Results

In the present paper, we will use the method of differential subordination to derive certain properties of generalised derivative operator μλ1,λ2n,mf(z). Note that differential subordination has been studied by various authors, and here we follow similar works done by Oros  and G. Oros and G. I. Oros .

In order to derive our new generalised derivative operator, we define the analytic function

Fλ1,λ2m(z)=z+k=2(1+λ1(k-1))m(1+λ2(k-1))m-1zk, where m0={0,1,2,} and λ2λ10. Now, we introduce the new generalised derivative operator μλ1,λ2n,m as follows.

Definition 2.1.

For f𝒜 the operator μλ1,λ2n,m is defined by μλ1,λ2n,m:𝒜𝒜μλ1,λ2n,mf(z)=Fλ1,λ2m(z)*Rnf(z),(zU), where n,m0=  {0},λ2λ10,and  Rnf(z) denotes the Ruscheweyh derivative operator , given by Rnf(z)=z+k=2c(n,k)akzk,(n0,  zU), where c(n,k)=(n+1)k-1/(1)k-1.

If f is given by (1.1), then we easily find from equality (2.2) that

μλ1,λ2n,mf(z)=z+k=2(1+λ1(k-1))m(1+λ2(k-1))m-1  c(n,k)akzk,(zU), where n,m0={0,1,2},λ2λ10,and  c(n,k)=(nn+k-1)=(n+1)k-1/(1)k-1.

Remark 2.2.

Special cases of this operator include the Ruscheweyh derivative operator in two cases μ0,λ2n,1Rn and μλ1,0n,0Rn , the Salagean derivative operator μ1,00,mSn , the generalised Ruscheweyh derivative operator in two cases μλ1,λ2n,1Rλn and μλ1,λ2n,0Rλn , the generalised Salagean derivative operator μλ1,00,mSβn introduced by Al-Oboudi , and the generalised Al-Shaqsi and Darus derivative operator μλ1,0n,mDλ,βn that can be found in .

Now, we remind the well-known Carlson-Shaffer operator L(a,c)  associated with the incomplete beta function ϕ(a,c;z), defined by

L(a,c):𝒜𝒜,L(a,c)f(z):=ϕ(a,c;z)*f(z),(zU), where

ϕ(a,c;z)=z+k=2(a)k-1(c)k-1zk,a is any real number, and cz0-;  z0-={0,-1,-2,}.

It is easily seen that

μλ1,00,0f(z)=μ0,00,mf(z)=μ0,λ20,1f(z)=μ0,11,2f(z)=L(a,a)f(z)=f(z),μλ1,01,0f(z)=μ0,01,mf(z)=μ0,λ21,1f(z)=μλ1,10,0f(z)=L(2,1)f(z)=zf(z), and also

μλ1,0a-1,0f(z)=μ0,λ2a-1,1f(z)=μ0,0a-1,mf(z)=L(a,1)f(z), where a=1,2,3,.

Next, we give the following.

Definition 2.3.

For n,m0,λ2λ10, and 0α<1, let Rλ1,λ2n,m(α) denote the class of functions f𝒜 which satisfy the condition Re(μλ1,λ2n,mf(z))>α,(zU). Also let Kλ1,λ2n,m(δ) denote the class of functions f𝒜 which satisfy the condition Re(μλ1,λ2n,mf(z)*ϕλ2(z))>δ,(zU).

Remark 2.4.

It is clear that Rλ1,00,1(α)R(λ1,α), and the class of functions f𝒜 satisfying Re(λ1zf′′(z)+f(z))>α,(zU) is studied by Ponnusamy  and others.

Now we begin with the first result as follows.

Theorem 2.5.

Let h(z)=1+(2α-1)z1+z,(zU), be convex in U, with h(0)=1 and 0α<1. If n,m0,λ2λ10, and the differential subordination (μλ1,λ2n,m+1f(z))h(z),(zU), holds, then (μλ1,λ2n,mf(z)*ϕλ2(z))q(z)=2α-1+2(1-α)λ1z1/λ1σ(1λ1), where σ is given by σ(x)=0ztx-11+tdt,(zU). The function q is convex and is the best dominant.

Proof.

By differentiating (1.8), with respect to z, we obtain (μλ1,λ2n,m+1f(z))=[μλ1,λ2n,mf(z)*ϕλ2(z)]+λ1z[μλ1,λ2n,mf(z)*ϕλ2(z)]′′. Using (2.16) in (2.13), differential subordination (2.13) becomes [μλ1,λ2n,mf(z)*ϕλ2(z)]+λ1z[μλ1,λ2n,mf(z)*ϕλ2(z)]′′h(z)=1+(2α-1)z1+z. Let p(z)=[μλ1,λ2n,mf(z)*ϕλ2(z)]=[z+k=2(1+λ1(k-1))m(1+λ2(k-1))m  c(n,k)akzk]=1+p1z+p2z2+,(p[1,1],  zU). Using (2.18) in (2.17), the differential subordination becomes p(z)+λ1zp(z)h(z)=1+(2α-1)z1+z. By using Lemma 1.1, we have p(z)q(z)=1λ1z1/λ10zh(t)  t(1/λ1)-1dt=1λ1z1/λ10z(1+(2α-1)t1+t)  t(1/λ1)-1dt=2α-1+2(1-α)λ1z1/λ1σ(1λ1), where σ is given by (2.15), so we get [μλ1,λ2n,mf(z)*ϕλ2(z)]q(z)=2α-1+2(1-α)λ1z1/λ1σ(1λ1). The function q is convex and is the best dominant. The proof is complete.

Theorem 2.6.

If n,m0,λ2λ10, and 0α<1, then one has Rλ1,λ2n,m+1(α)Kλ1,λ2n,m(δ), where δ=2α-1+2(1-α)λ1σ(1λ1), and σ is given by (2.15).

Proof.

Let f  Rλ1,λ2n,m+1(α), then from (2.9) we have Re(μλ1,λ2n,m+1f(z))>α,(zU), which is equivalent to (μλ1,λ2n,m+1f(z))h(z)=1+(2α-1)z1+z. Using Theorem 2.5, we have [μλ1,λ2n,mf(z)*ϕλ2(z)]  q(z)=2α-1+2(1-α)λ1z1/λ1σ(1λ1). Since q is convex and q(U) is symmetric with respect to the real axis, we deduce that Re[μλ1,λ2n,mf(z)*ϕλ2(z)]>Re  q(1)=δ=δ(α,λ1)=2α-1+2(1-α)λ1σ(1λ1), from which we deduce Rλ1,λ2n,m+1(α)Kλ1,λ2n,m(δ). This completes the proof of Theorem 2.6.

Remark 2.7.

Special case of Theorem 2.6 with λ2=0 was given earlier in .

Theorem 2.8.

Let q be a convex function in U, with q(0)=1, and let h(z)=q(z)+λ1zq(z),(zU). If n,m0,λ2λ10,and f𝒜 and satisfies the differential subordination (μλ1,λ2n,m+1f(z))h(z),(zU), then [μλ1,λ2n,mf(z)*ϕλ2(z)]q(z),(zU), and this result is sharp.

Proof.

Using (2.18) in (2.16), differential subordination (2.29) becomes p(z)+λ1zp(z)h(z)=q(z)+λ1zq(z),(zU). Using Lemma 1.2, we obtain p(z)q(z),(zU). Hence [μλ1,λ2n,mf(z)*ϕλ2(z)]q(z),(zU). And the result is sharp. This completes the proof of the theorem.

We give a simple application for Theorem 2.8.

Example 2.9.

For n=0,m=1,λ2λ10,q(z)=(1+z)/(1-z),f𝒜, and zU and applying Theorem 2.8, we have h(z)=1+z1-z+λ1z(1+z1-z)=1+2λ1z-z2(1-z)2. By using (1.8) we find that μλ1,λ20,1f(z)=(1-λ1)f(z)+λ1zf(z). Now, μλ1,λ20,1f(z)*ϕλ2(z)=(1-λ1)[z+k=2akzk1+λ2(k-1)]+λ1[z+k=2akkzk1+λ2(k-1)]. A straightforward calculation gives the following:   [μλ1,λ20,1f(z)*ϕλ2(z)]=1+k=2(1+λ1(k-1))kak1+λ2(k-1)  zk-1=z+k=2(kak(1+λ1(k-1))/(1+λ2(k-1)))zkz=[f(z)*ϕλ2(z)]*[z+k=2k(1+λ1(k-1))zk]z. Similarly, using (1.8), we find that μλ1,λ20,2f(z)=(1-λ1)[μλ1,λ20,1f(z)*ϕλ2(z)]+λ1z[μλ1,λ20,1f(z)*ϕλ2(z)], then (μλ1,λ20,2f(z))=(μλ1,λ20,1f(z)*ϕλ2(z))+λ1z(μλ1,λ20,1f(z)*ϕλ2(z))′′.   By using (2.37) we obtain (μλ1,λ20,1f(z)*ϕλ2(z))′′=k=2k(k-1)ak(1+λ1(k-1))1+λ2(k-1)zk-2. We get (μλ1,λ20,2f(z))=1+k=2kak(1+λ1(k-1))1+λ2(k-1)zk-1+λ1k=2k(k-1)ak(1+λ1(k-1))1+λ2(k-1)zk-1  =1+k=2kak(1+λ1(k-1))21+λ2(k-1)zk-1=[f(z)*ϕλ2(z)]*[z+k=2k(1+λ1(k-1))2zk]z. From Theorem 2.8 we deduce that [f(z)*ϕλ2(z)]*[z+k=2k(1+λ1(k-1))2zk]z  1+2λ1z-z2(1-z)2, implies that [f(z)*ϕλ2(z)]*[z+k=2k(1+λ1(k-1))zk]z  1+z1-z,(zU).

Theorem 2.10.

Let q be a convex function in U, with q(0)=1 and let h(z)=q(z)+zq(z),(zU). If n,m0,λ2λ10, and f𝒜 and satisfies the differential subordination (μλ1,λ2n,mf(z))h(z), then μλ1,λ2n,mf(z)zq(z),(zU). And the result is sharp.

Proof.

Let p(z)=μλ1,λ2n,mf(z)z=z+k=2((1+λ1(k-1))m/(1+λ2(k-1))m-1)c(n,k)akzkz=1+p1z+p2z2+,(p[1,1],  zU). Differentiating (2.47), with respect to z, we obtain (μλ1,λ2n,mf(z))=p(z)+zp(z),(zU). Using (2.48), (2.45) becomes p(z)+zp(z)h(z)=q(z)+zq(z). Using Lemma 1.2, we deduce that p(z)q(z),(zU), and using (2.47), we have μλ1,λ2n,mf(z)zq(z),(zU). This proves Theorem 2.10.

We give a simple application for Theorem 2.10.

Example 2.11.

For n=0,m=1,λ2λ10,q(z)=1/(1-z),f𝒜, and zU and applying Theorem 2.10, we have h(z)=11-z+z(11-z)=1(1-z)2. From Example 2.9, we have μλ1,λ20,1f(z)=(1-λ1)f(z)+λ1zf'(z), so (μλ1,λ20,1f(z))=f'(z)+λ1zf(z). Now, from Theorem 2.10 we deduce that f'(z)+λ1zf(z)1(1-z)2 implies that (1-λ1)f(z)+λ1zf'(z)z11-z.

Theorem 2.12.

Let h(z)=1+(2α-1)z1+z,(zU), be convex in U, with h(0)=1 and 0α<1. If n,m0,λ2λ10,f𝒜, and the differential subordination holds as (μλ1,λ2n,mf(z))h(z), then μλ1,λ2n,mf(z)zq(z)=2α-1+2(1-α)ln(1+z)z. The function q is convex and is the best dominant.

Proof.

Let p(z)=μλ1,λ2n,mf(z)z=z+k=2((1+λ1(k-1))m/(1+λ2(k-1))m-1  )c(n,k)  akzkz=1+p1z+p2z2+,(p[1,1],  zU). Differentiating (2.60), with respect to z, we obtain (μλ1,λ2n,mf(z))=p(z)+zp(z),(zU). Using (2.61), the differential subordination (2.58) becomes p(z)+zp(z)h(z)=1+(2α-1)z1+z,(zU). From Lemma 1.1, we deduce that p(z)q(z)=1z0zh(t)dt=1z0z(1+(2α-1)t1+t)dt=1z[0z11+tdt+(2α-1)0zt1+tdt]=2α-1+2(1-α)ln(1+z)z. Using (2.60), we have μλ1,λ2n,mf(z)zq(z)=2α-1+2(1-α)ln(1+z)z. The proof is complete.

From Theorem 2.12, we deduce the following corollary.

Corollary 2.13.

If f  Rλ1,λ2n,m(α), then Re  (μλ1,λ2n,mf(z)z)>(2α-1)+2(1-α)ln2,(zU).

Proof.

Since f  Rλ1,λ2n,m(α), from Definition 2.3 we have Re  (μλ1,λ2n,mf(z))>α,(zU), which is equivalent to (μλ1,λ2n,mf(z))h(z)=1+(2α-1)z1+z. Using Theorem 2.12, we have μλ1,λ2n,mf(z)zq(z)=(2α-1)+2(1-α)ln(1+z)z. Since q is convex and q(U) is symmetric with respect to the real axis, we deduce that Re(μλ1,λ2n,mf(z)z)>Req(1)=(2α-1)+2(1-α)ln2,(zU).

Theorem 2.14.

Let h(U), with h(0)=1, h'(0)0 which satisfy the inequality Re(1+zh(z)h(z))>-12,(zU).   If n,m0,λ2λ10,and f𝒜 and satisfies the differential subordination (μλ1,λ2n,mf(z))h(z),(zU), then μλ1,λ2n,mf(z)zq(z)=1z0zh(t)dt.

Proof.

Let p(z)=μλ1,λ2n,mf(z)z=z+k=2((1+λ1(k-1))m/(1+λ2(k-1))m-1)c(n,k)akzkz=1+p1z+p2z2+,(p[1,1],  zU). Differentiating (2.73), with respect to z, we have (μλ1,λ2n,mf(z))=p(z)+zp(z),(zU). Using (2.74), the differential subordination (2.71) becomes p(z)+zp(z)h(z),(zU). From Lemma 1.1, we deduce that p(z)q(z)=1z0zh(t)dt, and using (2.73), we obtain μλ1,λ2n,mf(z)zq(z)=1z0zh(t)dt. From Lemma 1.3, we have that the function q is convex, and from Lemma 1.1, q is the best dominant for subordination (2.71). This completes the proof of Theorem 2.14.

3. Conclusion

We remark that several subclasses of analytic univalent functions can be derived and studied using the operator μλ1,λ2n,m.

Acknowledgment

This work is fully supported by UKM-ST-06-FRGS0107-2009, MOHE, Malaysia.

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