We derive the q-analogue of the well-known Ruscheweyh differential operator using the concept of q-derivative. Here, we investigate several interesting properties of this q-operator by making use of the method of differential subordination.

1. Introduction

Recently, the area of q-analysis has attracted the serious attention of researchers. This great interest is due to its application in various branches of mathematics and physics. The application of q-calculus was initiated by Jackson [1, 2]. He was the first to develop q-integral and q-derivative in a systematic way. Later, from the 80s, geometrical interpretation of q-analysis has been recognized through studies on quantum groups. It also suggests a relation between integrable systems and q-analysis. In ([3–5]) the q-analogue of Baskakov Durrmeyer operator has been proposed, which is based on q-analogue of beta function. Another important q-generalization of complex operators is q-Picard and q-Gauss-Weierstrass singular integral operators discussed in [6–8]. The authors studied approximation and geometric properties of these q-operators in some subclasses of analytic functions in compact disk. Very recently, other q-analogues of differential operators have been introduced in [9]; see also ([10, 11]). These q-operators are defined by using convolution of normalized analytic functions and q-hypergeometric functions, where several interesting results are obtained. From this point, it is expected that deriving q-analogues of operators defined on the space of analytic functions would be important in future. A comprehensive study on applications of q-analysis in operator theory may be found in [12].

We provide some notations and concepts of q-calculus used in this paper. All the results can be found in [12–14]. For n∈ℕ,0<q<1, we define
(1)[n]q=1-qn1-q,[n]q!={[n]q[n-1]q⋯[1]q,n=1,2,…;;1,n=0.

As q→1,[n]q→n, and this is the bookmark of a q-analogue: the limit as q→1 recovers the classical object.

For complex parameters a,b,c,q(c∈ℂ∖{0,-1,-2,…},|q|<1), the q-analogue of Gauss’s hypergeometric function 2Φ1(a,b;c,q,z) is defined by
(2)2Φ1(a,b;c,q,z)=∑k=0∞(a,q)k(b,q)k(q,q)k(c,q)kzk,(z∈𝕌),
where (n,q)k is the q-analogue of Pochhammer symbol defined by
(3)(n,q)k={1,k=0;(1-n)(1-nq)(1-nq2)⋯(1-nqk-1),k∈ℕ.

The q-derivative of a function h(x) is defined by
(4)Dq(h(x))=h(qx)-h(x)(q-1)x,q≠1,x≠0,
and Dq(h(0))=f′(0). For a function h(z)=zk observe that
(5)Dq(h(z))=Dq(zk)=1-qk1-qzk-1=[k]qzk-1;;
then limq→1Dq(h(z))=limq→1[k]qzk-1=kzk-1=h′(z), where h′(z) is the ordinary derivative.

Next, we state the class 𝒜 of all functions of the following form:
(6)f(z)=z+∑k=2∞akzk,
which are analytic in the open unit disk 𝕌={z∈ℂ:|z|<1}. If f and g are analytic functions in 𝕌, we say that f is subordinate to g; written f≺g, if there is a function w analytic in 𝕌, with w(0)=0,|w(z)|<1, for all z∈𝕌, such that f(z)=g(w(z)) for all z∈𝕌. If g is univalent, then f≺g if and only if f(0)=g(0) and f(𝕌)⊆g(𝕌).

For each A and B such that -1≤B<A≤1, we define the function
(7)h(A,B;z)=1+Az1+Bz,(z∈𝕌).
It is well known that h(A,B;z) for -1≤B≤1 is the conformal map of the unit disk onto the disk symmetrical with respect to the real axis having the center (1-AB)/(1-B2) for B≠±1 and radius (A-B)/(1-B2). The boundary circle cuts the real axis at the points (1-A)/(1-B) and (1+A)/(1+B).

Definition 1.

Let f∈𝒜. Denote by ℛqλ the q-analogue of Ruscheweyh operator defined by
(8)ℛqλf(z)=z+∑k=2∞[k+λ-1]q![λ]q![k-1]q!akzk,
where [a]q and [a]q! are defined in (1).

From the definition we observe that, if q→1, we have
(9)limq→1ℛqλf(z)=z+limq→1[∑k=2∞[k+λ-1]q![λ]q![k-1]q!akzk]=z+∑k=2∞(k+λ-1)!(λ)!(k-1)!akzk=ℛλf(z),
where ℛλ is Ruscheweyh differential operator which was defined in [15] and has been studied by many authors, for example [16–18].

It can also be shown that this q-operator is hypergeometric in nature as
(10)ℛqλf(z)=z2Φ1(qλ+1,q,q,q;z)*f(z),
where 2Φ1 is the q-analogue of Gauss hypergeometric function defined in (2), and the symbol (*) stands for the Hadamard product (or convolution).

The following identity is easily verified for the operator ℛqλ:
(11)qλz(Dq(ℛqλf(z)))=[λ+1]qℛqλ+1f(z)-[λ]qℛqλf(z).

2. Main Results

Before we obtain our results, we state some known lemmas.

Let P(β) be the class of functions of the form
(12)ϕ(z)=1+c1z+c2z2+⋯,
which are analytic in 𝕌 and satisfy the following inequality:
(13)Re(ϕ(z))>β,(0≤β<1;z∈𝕌).

Lemma 2 (see [<xref ref-type="bibr" rid="B16">19</xref>]).

Let ϕj∈P(βj) be given by (12), where (0≤βj<1;j=1,2); then
(14)(ϕ1*ϕ2)∈P(β3),(β3=1-2(1-β1)(1-β2)),
and the bound β3 is the best possible.

Lemma 3 (see [<xref ref-type="bibr" rid="B15">20</xref>]).

Let the function ϕ, given by (12), be in the class P(β). Then
(15)Reϕ(z)>2β-1+2(1-β)1+|z|,(0≤β<1).

Lemma 4 (see [<xref ref-type="bibr" rid="B19">21</xref>]).

The function (1-z)γ≡eγlog(1-z),γ≠0, is univalent in 𝕌 if and only if γ is either in the closed disk |γ-1|≤1 or in the closed disk |γ+1|≤1.

We now generalize the lemmas introduced in [22] and [23], respectively, using q-derivative.

Lemma 5.

Let h(z) be analytic and convex univalent in 𝕌 and h(0)=1 and let g(z)=1+b1z+b2z2+⋯ be analytic in 𝕌. If
(16)g(z)+zDq(g(z))c≺h(z),(z∈𝕌;c≠0),
Then, for Re(c)≥0,
(17)g(z)≺czc∫0ztc-1h(t)dt.

Proof.

Suppose that h is analytic and convex univalent in 𝕌 and g is analytic in 𝕌. Letting q→1 in (16), we have
(18)g(z)+zg′(z)c≺h(z),(z∈𝕌;c≠0).

Then, from Lemma in [22], we obtain
(19)g(z)≺czc∫0ztc-1h(t)dt.

Lemma 6.

Let q(z) be univalent in 𝕌 and let θ(w) and ϕ(w) be analytic in a domain D containing q(𝕌) with ϕ(w)≠0 when w∈q(𝕌). Set Q(z)=zDq(q(z))ϕ(q(z)),h(z)=θ(q(z)+Q(z)) and suppose that

If p(z) is analytic in 𝕌, with p(0)=q(0),p(𝕌)⊂D, and
(20)θ(p(z))+zDq(p(z))ϕ(p(z))≺θ(q(z))+zDq(q(z))ϕ(q(z))=h(z),
then p(z)≺q(z) and q(z) is the best dominant.

The proof is similar to the proof of Lemma 5.

Theorem 7.

Let >0, α>0, and -1≤B<A≤1. If f∈𝒜 satisfies
(21)(1-α)ℛqλf(z)z+αℛqλ+1f(z)z≺h(A,B;z),
then
(22)Re((ℛqλf(z)z)1/n)>([λ+1]qqλα∫01u([λ+1]q/qλα)-1(1-Au1-Bu)du)1/nu([λ+1]q/qλα)-1GRKRRRRQQQIQZM(n≥1).

The result is sharp.

Proof.

Let
(23)g(z)=ℛqλf(z)z,
for f∈𝒜. Then the function g(z)=1+b1z+⋯ is analytic in 𝕌. By using logarithmic q-differentiation on both sides of (23) and multiplying by z, we have
(24)zDq(g(z))g(z)=zRqλf(z)Rqλf(z)-1;;
by making use of identity (11), we obtain
(25)zDq(g(z))g(z)=[λ+1]qqλRqλ+1f(z)Rqλf(z)-[λ]qqλ-1.

Taking into account that [λ+1]q=[λ]q+qλ, we obtain
(26)qλ[λ+1]qzDq(g(z))+g(z)=ℛλ+1f(z)z.

From (11), (23), and (26), we get
(27)g(z)+qλα[λ+1]qzDq(g(z))≺h(A,B;z).

Now, applying Lemma 5, we have
(28)g(z)≺[λ+1]qqλαz-[λ+1]q/qλα∫01t([λ+1]q/qλα)-1(1+At1+Bt)dt,
or by the concept of subordination
(29)ℛqλf(z)z=[λ+1]qqλα∫01u([λ+1]q/qλα)-1(1+Auw(z)1+Buw(z))du.

In view of -1≤B<A≤1 and λ>0, it follows from (29) that
(30)Re(ℛqλf(z)z)>[λ+1]qqλα∫01u([λ+1]q/qλα)-1(1-Au1-Bu)du,
with the aid of the elementary inequality Re(w1/n)≥(Rew)1/n for Rew>0 and n≥1. Hence, inequality (22) follows directly from (30). To show the sharpness of (22), we define f∈𝒜 by
(31)ℛqλf(z)z=[λ+1]qqλα∫01u([λ+1]q/qλα)-1(1+Auz1+Buz)du.

For this function, we find that
(32)(1-α)ℛqλf(z)z+αℛqλ+1f(z)z=1+Az1-Bz,ℛqλf(z)z⟶[λ+1]qqλα∫01u([λ+1]q/qλα)-1(1-Au1-Bu)duℛqλf(z)z⟶[λ+1]qqλαRSQRRRRRRKasz⟶-1.

This completes the proof.

Corollary 8.

Let A=2β-1andB=-1, where 0≤β<1 and α,λ>1. If f satisfies
(33)(1-α)ℛqλf(z)z+αℛqλ+1f(z)z≺h(2β-1,-1;z),
then
(34)Re((ℛqλf(z)z)1/n)>(2(1-β)[λ+1]qqλα∫01u([λ+1]q/qλα)-11+udu(2β-1)u[λ+1]q/qλα+2(1-β)[λ+1]qqλα∫01u([λ+1]q/qλα)-11+udu)1/nu([λ+1]q/qλα)-1GRKRRRRRRQIQZM(n≥1).

Proof.

Following the same steps as in the proof of Theorem 7 and considering g(z)=ℛqλf(z)/z, the differential subordination (27) becomes
(35)g(z)+qλα[λ+1]qzDq(g(z))≺1+(2β-1)z1+z.

Let λ>0 and 0≤ρ<1. Let γ be a complex number with γ≠0 and satisfy either |2γ(1-ρ)([λ+1]q/qλ)-1|≤1 or |2γ(1-ρ)([λ+1]q/qλ)+1|≤1. If f∈𝒜 satisfies the condition
(37)Re(ℛqλ+1f(z)ℛqλf(z))>ρ,(z∈𝕌),
then
(38)(ℛqλf(z)z)γ≺1(1-z)2γ(1-ρ)([λ+1]q/qλ),(z∈𝕌),
where q(z) is the best dominant.

Proof.

Let
(39)p(z)=(ℛqλf(z)z)γ,(z∈𝕌).Then, by making use of (11), (37), and (39), we obtain
(40)1+qλzDq(p(z))γ[λ+1]qp(z)≺1+(1-2ρ)z1-z,(z∈𝕌).

We now assume that
(41)q(z)=1(1-z)2γ(1-ρ)[λ+1]q/qλ,θ(w)=1,ϕ(w)=qλγ[λ+1]qw;
then q(z) is univalent by condition of the theorem and Lemma 4. Further, it is easy to show that q(z),θ(w), and ϕ(w) satisfy the conditions of Lemma 6. Note that the function
(42)Q(z)=zDq(q(z))ϕ(q(z))=2(1-ρ)z1-z
is univalent starlike in 𝕌 and
(43)h(z)=θ(q(z))+Q(z)=1+(1-2ρ)z1-z.

Combining (40) and Lemma 6 we get the assertion of Theorem 9.

Theorem 10.

Let α<1,λ>0 and -1≤Bi<Ai≤1. If each of the functions fi∈𝒜 satisfies the following subordination condition,
(44)(1-α)ℛqλfi(z)z+αℛqλ+1fi(z)z≺h(Ai,Bi;z),
then,
(45)(1-α)ℛqλΘ(z)z+αℛqλ+1Θ(z)z≺h(1-2γ,-1;z),
where
(46)Θ(z)=ℛqλ(f1*f2)(z),(47)γ=1-4(A1-B1)(A2-B2)(1-B1)(1-B2)×(1-[λ+1]qqλα∫01u([λ+1]q/qλα)-11+udu).

Proof.

we define the function hi by
(48)hi(z)=(1-α)ℛqλfi(z)z+αℛqλ+1fi(z)z(1-α)ℛqλfiz(fi∈𝒜,i=1,2);
we have hi(z)∈P(βi), where βi=(1-Ai)/(1-Bi)(i=1,2). By making use of (11) and (48), we obtain
(49)ℛqλfi(z)=[λ+1]qqλα∫01t([λ+1]q/qλα)-1hi(t)dt,(i=1,2),
which, in the light of (46), can show that
(50)ℛqλΘ(z)=[λ+1]qqλαz1-([λ+1]q/qλα)∫01t([λ+1]q/qλα)-1h0(t)dt,
where, for convenience,
(51)h0(z)=(1-α)ℛqλΘ(z)z+αℛqλ+1Θ(z)z=[λ+1]qqλαz1-([λ+1]q/qλα)×∫01t([λ+1]q/qλα)-1(h1*h2)(t)dt.

Note that, by using Lemma 2, we have (h1*h2)∈P(β3), where β3=1-2(1-β1)(1-β2).

Now, with an application of Lemma 3, we have
(52)Re(h0(z))=[λ+1]qqλα∫01u([λ+1]q/qλα)-1Re((h1*h2)(uz))du≥[λ+1]qqλα∫01u([λ+1]q/qλα)-1(2β3-1+2(1-β3)1+u|z|)du>[λ+1]qqλα∫01u([λ+1]q/qλα)-1(2β3-1+2(1-β3)1+u)du=1-4(A1-B1)(A2-B2)(1-B1)(1-B2)×(1-[λ+1]qqλα∫01u([λ+1]q/qλα)-11+udu)=γ,
which shows that the desired assertion of Theorem 10 holds.

Conflict of Interests

The authors declare that they have no competing interests regarding the publication of this paper.

Authors’ Contribution

Huda Aldweby and Maslina Darus read and approved the final manuscript.

Acknowledgments

The work presented here was partially supported by AP-2013-009 and DIP-2013-001. The authors also would like to thank the referees for the comments made to improve this paper.

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