# Differential Subordination Defined by New Generalised Derivative Operator for Analytic Functions

Authors: Ma'moun Harayzeh Al-Abbadi; Maslina Darus
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

A new generalised derivative operatorμλ1,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.


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## Body

## 1. Introduction and Preliminaries

Let𝒜 denote the class of functions of the form(1.1)f(z)=z+∑k=2∞akzk,whereakiscomplexnumber,
which are analytic in the open unit disc U={z∈ℂ:|z|<1} on the complex plane ℂ. Let S,S*(α),C(α)(0≤α<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*andC(0)=C are the familiar classes of starlike and convex functions in U, respectively. A function f∈C(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.Letℋ(U) be the class of holomorphic function in unit disc U={z∈ℂ:|z|<1}.Let(1.2)𝒜n={f∈ℋ(U):f(z)=z+an+1zn+1+⋯,(z∈U)},
with 𝒜1=𝒜.Fora∈ℂ and n∈ℕ={1,2,3,…} we let(1.3)ℋ[a,n]={f∈ℋ(U):f(z)=z+anzn+an+1zn+1+⋯,(z∈U)}.
Let f(z)=z+∑k=2∞akzkandg(z)=z+∑k=2∞bkzkbe analytic in the open unit disc U={z∈ℂ:|z|<1}. Then the Hadamard product (or convolution) f*g of the two functions f, g is defined by(1.4)f(z)*g(z)=(f*g)(z)=z+∑k=2∞akbkzk.
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(1.5)f≺g,f(z)≺g(z),(z∈U),
if and only if there exists the Schwarz function w, analytic in U, with w(0)=0and|w(z)|<1such thatf(z)=g(w(z)), (z∈U).Furthermore ifg is univalent in U, then f≺g if and only if f(0)=g(0) and f(U)⊂g(U) (see [1, page 36]).Letψ:ℂ3×U→ℂ and let h be univalent in U. If p is analytic in U and satisfies the (second-order) differential subordination(1.6)ψ(p(z),zp′(z),z2p′′(z);z)≺h(z),(z∈U),
then p is called a solution of the differential subordination.The univalent functionq is called a dominant of the solutions of the differential subordination, or more simply a dominant, if p≺q for all p satisfying (1.6). A dominant q̃ that satisfies q̃≺q 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(1.7)(x)k={1fork=0,x∈ℂ∖{0},x(x+1)(x+2)⋯(x+k-1)fork∈ℕ={1,2,3,…},x∈ℂ.
To prove our results, we need the following equation throughout the paper:(1.8)μλ1,λ2n,m+1f(z)=(1-λ1)[μλ1,λ2n,mf(z)*ϕλ2(z)]+λ1z[μλ1,λ2n,mf(z)*ϕλ2(z)]′,(z∈U),
where n,m∈ℕ0={0,1,2,…}, λ2≥λ1≥0, and ϕλ2(z) is analytic function given by(1.9)ϕλ2(z)=z+∑k=2∞zk1+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 [2, page 71]).
Leth be analytic, univalent, and convex in U, with h(0)=a, γ≠0 and, Reγ≥0. If p∈ℋ[a,n]and
(1.10)p(z)+zp′(z)γ≺h(z),(z∈U),
then
(1.11)p(z)≺q(z)≺h(z),(z∈U),
where q(z)=(γ/nzγ/n)∫0zh(t)t(γ/n)-1dt,(z∈U).The functionq is convex and is the best (a,n)-dominant.Lemma 1.2 (see [3]).
Letg be a convex function in U and let
(1.12)h(z)=g(z)+nαzg'(z),
where α>0 and n is a positive integer.
If
(1.13)p(z)=g(0)+pnzn+pn+1zn+1+⋯,(z∈U),
is holomorphic in U and
(1.14)p(z)+αzp′(z)≺h(z),(z∈U),
then
(1.15)p(z)≺g(z),
and this result is sharp.Lemma 1.3 (see [4]).
Letf∈𝒜, if
(1.16)Re(1+zf′′(z)f′(z))>-12,
then
(1.17)2z∫0zf(t)dt,(z∈U,z≠0),
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 operatorμλ1,λ2n,mf(z). Note that differential subordination has been studied by various authors, and here we follow similar works done by Oros [5] and G. Oros and G. I. Oros [6].In order to derive our new generalised derivative operator, we define the analytic function(2.1)Fλ1,λ2m(z)=z+∑k=2∞(1+λ1(k-1))m(1+λ2(k-1))m-1zk,
where m∈ℕ0={0,1,2,…} and λ2≥λ1≥0. Now, we introduce the new generalised derivative operator μλ1,λ2n,m as follows.Definition 2.1.
Forf∈𝒜 the operator μλ1,λ2n,m is defined by μλ1,λ2n,m:𝒜→𝒜(2.2)μλ1,λ2n,mf(z)=Fλ1,λ2m(z)*Rnf(z),(z∈U),
where n,m∈ℕ0=ℕ∪{0},λ2≥λ1≥0,andRnf(z) denotes the Ruscheweyh derivative operator [7], given by
(2.3)Rnf(z)=z+∑k=2∞c(n,k)akzk,(n∈ℕ0,z∈U),
where c(n,k)=(n+1)k-1/(1)k-1.
Iff is given by (1.1), then we easily find from equality (2.2) that
(2.4)μλ1,λ2n,mf(z)=z+∑k=2∞(1+λ1(k-1))m(1+λ2(k-1))m-1c(n,k)akzk,(z∈U),
where n,m∈ℕ0={0,1,2…},λ2≥λ1≥0,andc(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 casesμ0,λ2n,1≡Rn and μλ1,0n,0≡Rn [7], the Salagean derivative operator μ1,00,m≡Sn [8], the generalised Ruscheweyh derivative operator in two cases μλ1,λ2n,1≡Rλn and μλ1,λ2n,0≡Rλn [9], the generalised Salagean derivative operator μλ1,00,m≡Sβn introduced by Al-Oboudi [10], and the generalised Al-Shaqsi and Darus derivative operator μλ1,0n,m≡Dλ,βn that can be found in [11].Now, we remind the well-known Carlson-Shaffer operatorL(a,c) [12] associated with the incomplete beta function ϕ(a,c;z), defined by(2.5)L(a,c):𝒜→𝒜,L(a,c)f(z):=ϕ(a,c;z)*f(z),(z∈U),
where(2.6)ϕ(a,c;z)=z+∑k=2∞(a)k-1(c)k-1zk,a is any real number, and c∉z0-;z0-={0,-1,-2,…}.It is easily seen that(2.7)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(2.8)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.
Forn,m∈ℕ0,λ2≥λ1≥0, and 0≤α<1, let Rλ1,λ2n,m(α) denote the class of functions f∈𝒜 which satisfy the condition
(2.9)Re(μλ1,λ2n,mf(z))′>α,(z∈U).
Also let Kλ1,λ2n,m(δ) denote the class of functions f∈𝒜 which satisfy the condition
(2.10)Re(μλ1,λ2n,mf(z)*ϕλ2(z))′>δ,(z∈U).Remark 2.4.
It is clear thatRλ1,00,1(α)≡R(λ1,α), and the class of functions f∈𝒜 satisfying
(2.11)Re(λ1zf′′(z)+f′(z))>α,(z∈U)
is studied by Ponnusamy [13] and others.Now we begin with the first result as follows.Theorem 2.5.
Let(2.12)h(z)=1+(2α-1)z1+z,(z∈U),
be convex in U, with h(0)=1 and 0≤α<1. If n,m∈ℕ0,λ2≥λ1≥0, and the differential subordination
(2.13)(μλ1,λ2n,m+1f(z))′≺h(z),(z∈U),
holds, then
(2.14)(μλ1,λ2n,mf(z)*ϕλ2(z))′≺q(z)=2α-1+2(1-α)λ1z1/λ1σ(1λ1),
where σ is given by
(2.15)σ(x)=∫0ztx-11+tdt,(z∈U).
The function q is convex and is the best dominant.Proof.
By differentiating (1.8), with respect to z, we obtain
(2.16)(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
(2.17)[μλ1,λ2n,mf(z)*ϕλ2(z)]′+λ1z[μλ1,λ2n,mf(z)*ϕλ2(z)]′′≺h(z)=1+(2α-1)z1+z.
Let
(2.18)p(z)=[μλ1,λ2n,mf(z)*ϕλ2(z)]′=[z+∑k=2∞(1+λ1(k-1))m(1+λ2(k-1))mc(n,k)akzk]′=1+p1z+p2z2+⋯,(p∈ℋ[1,1],z∈U).
Using (2.18) in (2.17), the differential subordination becomes
(2.19)p(z)+λ1zp′(z)≺h(z)=1+(2α-1)z1+z.
By using Lemma 1.1, we have
(2.20)p(z)≺q(z)=1λ1z1/λ1∫0zh(t)t(1/λ1)-1dt=1λ1z1/λ1∫0z(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
(2.21)[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.
Ifn,m∈ℕ0,λ2≥λ1≥0, and 0≤α<1, then one has
(2.22)Rλ1,λ2n,m+1(α)⊂Kλ1,λ2n,m(δ),
where
(2.23)δ=2α-1+2(1-α)λ1σ(1λ1),
and σ is given by (2.15).Proof.
Letf∈Rλ1,λ2n,m+1(α), then from (2.9) we have
(2.24)Re(μλ1,λ2n,m+1f(z))′>α,(z∈U),
which is equivalent to
(2.25)(μλ1,λ2n,m+1f(z))′≺h(z)=1+(2α-1)z1+z.
Using Theorem 2.5, we have
(2.26)[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
(2.27)Re[μλ1,λ2n,mf(z)*ϕλ2(z)]′>Req(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 Theorem2.6 with λ2=0 was given earlier in [11].Theorem 2.8.
Letq be a convex function in U, with q(0)=1, and let
(2.28)h(z)=q(z)+λ1zq′(z),(z∈U).
If n,m∈ℕ0,λ2≥λ1≥0,and f∈𝒜 and satisfies the differential subordination
(2.29)(μλ1,λ2n,m+1f(z))′≺h(z),(z∈U),
then
(2.30)[μλ1,λ2n,mf(z)*ϕλ2(z)]′≺q(z),(z∈U),
and this result is sharp.Proof.
Using (2.18) in (2.16), differential subordination (2.29) becomes
(2.31)p(z)+λ1zp′(z)≺h(z)=q(z)+λ1zq′(z),(z∈U).
Using Lemma 1.2, we obtain
(2.32)p(z)≺q(z),(z∈U).
Hence
(2.33)[μλ1,λ2n,mf(z)*ϕλ2(z)]′≺q(z),(z∈U).
And the result is sharp. This completes the proof of the theorem.We give a simple application for Theorem2.8.Example 2.9.
Forn=0,m=1,λ2≥λ1≥0,q(z)=(1+z)/(1-z),f∈𝒜, and z∈U and applying Theorem 2.8, we have
(2.34)h(z)=1+z1-z+λ1z(1+z1-z)′=1+2λ1z-z2(1-z)2.
By using (1.8) we find that
(2.35)μλ1,λ20,1f(z)=(1-λ1)f(z)+λ1zf′(z).
Now,
(2.36)μλ1,λ20,1f(z)*ϕλ2(z)=(1-λ1)[z+∑k=2∞akzk1+λ2(k-1)]+λ1[z+∑k=2∞akkzk1+λ2(k-1)].
A straightforward calculation gives the following:
(2.37)[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=2∞k(1+λ1(k-1))zk]z.
Similarly, using (1.8), we find that
(2.38)μλ1,λ20,2f(z)=(1-λ1)[μλ1,λ20,1f(z)*ϕλ2(z)]+λ1z[μλ1,λ20,1f(z)*ϕλ2(z)]′,
then
(2.39)(μλ1,λ20,2f(z))′=(μλ1,λ20,1f(z)*ϕλ2(z))′+λ1z(μλ1,λ20,1f(z)*ϕλ2(z))′′.
By using (2.37) we obtain
(2.40)(μλ1,λ20,1f(z)*ϕλ2(z))′′=∑k=2∞k(k-1)ak(1+λ1(k-1))1+λ2(k-1)zk-2.
We get
(2.41)(μλ1,λ20,2f(z))′=1+∑k=2∞kak(1+λ1(k-1))1+λ2(k-1)zk-1+λ1∑k=2∞k(k-1)ak(1+λ1(k-1))1+λ2(k-1)zk-1=1+∑k=2∞kak(1+λ1(k-1))21+λ2(k-1)zk-1=[f(z)*ϕλ2(z)]*[z+∑k=2∞k(1+λ1(k-1))2zk]z.
From Theorem 2.8 we deduce that
(2.42)[f(z)*ϕλ2(z)]*[z+∑k=2∞k(1+λ1(k-1))2zk]z≺1+2λ1z-z2(1-z)2,
implies that
(2.43)[f(z)*ϕλ2(z)]*[z+∑k=2∞k(1+λ1(k-1))zk]z≺1+z1-z,(z∈U).Theorem 2.10.
Letq be a convex function in U, with q(0)=1 and let
(2.44)h(z)=q(z)+zq′(z),(z∈U).
If n,m∈ℕ0,λ2≥λ1≥0, and f∈𝒜 and satisfies the differential subordination
(2.45)(μλ1,λ2n,mf(z))′≺h(z),
then
(2.46)μλ1,λ2n,mf(z)z≺q(z),(z∈U).
And the result is sharp.Proof.
Let(2.47)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∈ℋ[1,1],z∈U).
Differentiating (2.47), with respect to z, we obtain
(2.48)(μλ1,λ2n,mf(z))′=p(z)+zp′(z),(z∈U).
Using (2.48), (2.45) becomes
(2.49)p(z)+zp′(z)≺h(z)=q(z)+zq′(z).
Using Lemma 1.2, we deduce that
(2.50)p(z)≺q(z),(z∈U),
and using (2.47), we have
(2.51)μλ1,λ2n,mf(z)z≺q(z),(z∈U).
This proves Theorem 2.10.We give a simple application for Theorem2.10.Example 2.11.
Forn=0,m=1,λ2≥λ1≥0,q(z)=1/(1-z),f∈𝒜, and z∈U and applying Theorem 2.10, we have
(2.52)h(z)=11-z+z(11-z)′=1(1-z)2.
From Example 2.9, we have
(2.53)μλ1,λ20,1f(z)=(1-λ1)f(z)+λ1zf'(z),
so
(2.54)(μλ1,λ20,1f(z))′=f'(z)+λ1zf′′(z).
Now, from Theorem 2.10 we deduce that
(2.55)f'(z)+λ1zf′′(z)≺1(1-z)2
implies that
(2.56)(1-λ1)f(z)+λ1zf'(z)z≺11-z.Theorem 2.12.
Let(2.57)h(z)=1+(2α-1)z1+z,(z∈U),
be convex in U, with h(0)=1 and 0≤α<1. If n,m∈ℕ0,λ2≥λ1≥0,f∈𝒜, and the differential subordination holds as
(2.58)(μλ1,λ2n,mf(z))′≺h(z),
then
(2.59)μλ1,λ2n,mf(z)z≺q(z)=2α-1+2(1-α)ln(1+z)z.
The function q is convex and is the best dominant.Proof.
Let(2.60)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∈ℋ[1,1],z∈U).
Differentiating (2.60), with respect to z, we obtain
(2.61)(μλ1,λ2n,mf(z))′=p(z)+zp′(z),(z∈U).
Using (2.61), the differential subordination (2.58) becomes
(2.62)p(z)+zp′(z)≺h(z)=1+(2α-1)z1+z,(z∈U).
From Lemma 1.1, we deduce that
(2.63)p(z)≺q(z)=1z∫0zh(t)dt=1z∫0z(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
(2.64)μλ1,λ2n,mf(z)z≺q(z)=2α-1+2(1-α)ln(1+z)z.
The proof is complete.From Theorem2.12, we deduce the following corollary.Corollary 2.13.
Iff∈Rλ1,λ2n,m(α), then
(2.65)Re(μλ1,λ2n,mf(z)z)>(2α-1)+2(1-α)ln2,(z∈U).Proof.
Sincef∈Rλ1,λ2n,m(α), from Definition 2.3 we have
(2.66)Re(μλ1,λ2n,mf(z))′>α,(z∈U),
which is equivalent to
(2.67)(μλ1,λ2n,mf(z))′≺h(z)=1+(2α-1)z1+z.
Using Theorem 2.12, we have
(2.68)μλ1,λ2n,mf(z)z≺q(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
(2.69)Re(μλ1,λ2n,mf(z)z)>Req(1)=(2α-1)+2(1-α)ln2,(z∈U).Theorem 2.14.
Leth∈ℋ(U), with h(0)=1, h'(0)≠0 which satisfy the inequality
(2.70)Re(1+zh′′′(z)h′(z))>-12,(z∈U).
If n,m∈ℕ0,λ2≥λ1≥0,and f∈𝒜 and satisfies the differential subordination
(2.71)(μλ1,λ2n,mf(z))′≺h(z),(z∈U),
then
(2.72)μλ1,λ2n,mf(z)z≺q(z)=1z∫0zh(t)dt.Proof.
Let(2.73)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∈ℋ[1,1],z∈U).
Differentiating (2.73), with respect to z, we have
(2.74)(μλ1,λ2n,mf(z))′=p(z)+zp′(z),(z∈U).
Using (2.74), the differential subordination (2.71) becomes
(2.75)p(z)+zp′(z)≺h(z),(z∈U).
From Lemma 1.1, we deduce that
(2.76)p(z)≺q(z)=1z∫0zh(t)dt,
and using (2.73), we obtain
(2.77)μλ1,λ2n,mf(z)z≺q(z)=1z∫0zh(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 operatorμλ1,λ2n,m.



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