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Research Articles

Inclusion and properties neighbourhood for certain p-valent functions associated with complex order and q–p-valent Cătaş operator

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Pages 1226-1232 | Received 01 Jun 2020, Accepted 10 Aug 2020, Published online: 01 Sep 2020

Abstract

Here in our paper, we present two new subclasses of multivalent analytic functions and complex order by using q–p-valent Cătaş operator. Also we obtain coefficient estimates and consequent inclusion relationships involving the Ni,δp,q(f)-neighbourhood of these classes.

1. Introduction

Let Ai(p) denote the class of functions in the next formula: (1) f(z)=zp+k=i+pakzk (i,pN={1,2,}),(1) which are analytic and p-valent in the open unit disk U={z:|z|<1}. We note that A1(p)=A(p) (see [Citation1,Citation2]) and A1(1)=A. Also let Ti(p) denote the subclass of Ai(p) which can express in the form: (2) f(z)=zpk=i+pakzk (ak0, i,pN).(2) In [Citation3,Citation4] (see also ([Citation5–11]) the q-derivative Dq,p of f is defined as follows: (3) Dq,pf(z)=f(z)f(qz)(1q)zfor z0,f(0)for z=0,(fA(p))(3) provided that f(0) exists. From (Equation1) (with i = 1) and (Equation3), we deduce that (4) Dq,pf(z)=pqzp1+k=p+1kqakzk1,(4) where [k]q is q-integer number k defined by (5) [k]q=1qk1q=1+k=1n1qk, 0<q<1, 0q=0.(5)

We observe that limq1Dq,pf(z)=limq1f(z)f(qz)(1q)z=f(z), for a function f which is differentiable in a given subset of C. For fTi(p), we introduce the q–p-valent Cătaş operator Iq,pn(ϰ,ϱ)f(z):Ti(p)Ti(p) (ϱ,ϰ0,nN0=N{0},0<q<1,i,pN) as follows: (6) Iq,p0(ϰ,ϱ)f(z)=f(z),Iq,p1(ϰ,ϱ)f(z)=(1ϰ)f(z)+ϰp+ϱqzϱ1Dq,p(zϱf(z))=zpk=i+pp+ϱq+ϰk+ϱqp+ϱqp+ϱqakzk,Iq,p2(ϰ,ϱ)f(z)=(1ϰ)Iq,p1(ϰ,ϱ)f(z)+ϰp+ϱqzϱ1Dq,p(zϱIq,p1(ϰ,ϱ)f(z))=zpk=i+pp+ϱq+ϰk+ϱqp+ϱqp+ϱq2akzk,Iq,pn(ϰ,ϱ)f(z)=(1ϰ)Iq,pn1(ϰ,ϱ)f(z)+ϰp+ϱqzϱ1Dq,p(zϱIq,pn1(ϰ,ϱ)f(z)) (nN).(6) From (Equation2) and (Equation6), we can obtain (7) Iq,pn(ϰ,ϱ)f(z)=zpk=i+pΨq,pn(k,ϰ,ϱ)akzk,(7) where (8) Ψq,pn(k,ϰ,ϱ)=p+ϱq+ϰk+ϱqp+ϱqp+ϱqn(nN0, i,pN,ϱ, ϰ0).(8) We note that

  1. limq1Iq,pn(ϰ,ϱ)f(z)=Ipn(ϰ,ϱ)f(z),Iq,1n(ϰ,ϱ)f(z)=Iqn(ϰ,ϱ)f(z), limq1Iqn(ϰ,ϱ)f(z)=In(ϰ,ϱ)f(z) (Cătaş [Citation12]);

  2. Iq,pn(1,ϱ)f(z)=Iq,pn(ϱ)f(z) =fTi(p):Iq,pn(ϱ)f(z)=zpk=i+pk+ϱqp+ϱqnakzk,nN0, i,pN,ϱ0, 0<q<1;

  3. Iq,pn(ϰ,0)f(z)=Dq,pn(ϰ)f(z) =fTi(p):Dq,pn(ϰ)f(z)=zpk=i+ppq+ϰkqpqpqnakzk,nN0, i,pN, ϰ0, 0<q<1;

  4. Iq,pn(1,0)f(z)=Dq,pnf(z) =fTi(p):Dq,pnf(z)=zpk=i+pkqpqnakzk,nN0, i,pN, 0<q<1; where limq1Dq,pnf(z)=Dpnf(z) (see [Citation13–15]);

  5. Iq,1n(1,ϱ)f(z)=Iqn(ϱ)f(z), limq1Iqn(ϱ)f(z)=In(ϱ)f(z) (see [Citation16,Citation17]);

  6. Iq,1n(ϰ,0)f(z)=Dq,ϰnf(z), limq1Dq,ϰnf(z)=Dϰnf(z) (see [Citation18–20]);

  7. Iq,1n(1,0)f(z)=Dqnf(z) (see [Citation21]), limq1Dqnf(z)=Dnf(z) (see Sălăgean [Citation22]) see also [Citation23,Citation24].

Now by using the operator Iq,pn(ϰ,ϱ)f(z) we defined the classes Sqn(i,p,ϰ,ϱ,σ,b,β) and Kqn(i,p,ϰ,ϱ,σ,b,β) as follows:

Definition 1

Let f(z)Ti(p). Then fSqn(i,p,ϰ,ϱ,σ,b,β) if it satisfies the following inequality: (9) 1b(1σ)zDq,p(Iq,pn(ϰ,ϱ)f(z))+σzDq,p(zDq,p(Iq,pn(ϰ,ϱ)f(z)))(1σ)Iq,pn(ϰ,ϱ)f(z)+σzDq,p(Iq,pn(ϰ,ϱ)f(z))pq<β(bC=C{0},nN0,p,iN,0<q<1,ϱ,ϰ0,0σ1,0<β1).(9)

We note that:

  1. limq1Sqn(i,p,ϰ,ϱ,σ,b,β)=Sn(i,p,ϰ,ϱ,σ,b,β) fTi(p):1bz(Ipn(ϰ,ϱ)f(z))+σz2(Ipn(ϰ,ϱ)f(z))(1σ)Ipn(ϰ,ϱ)f(z)+σz(Ipn(ϰ,ϱ)f(z))p<β(bC,nN0,p,iN,ϱ,ϰ0,0σ1,0<β1)1bz(Ipn(ϰ,ϱ)f(z))+σz2(Ipn(ϰ,ϱ)f(z))(1σ)Ipn(ϰ,ϱ)f(z)+σz(Ipn(ϰ,ϱ)f(z));

  2. Sqn(i,p,1,ϱ,σ,b,β)=Sqn(i,p,ϱ,σ,b,β) fTi(p):1b(1σ)zDq,p(Iq,pn(ϱ)f(z))+σzDq,p(zDq,p(Iq,pn(ϱ)f(z)))(1σ)Iq,pn(ϱ)f(z)+σzDq,p(Iq,pn(ϱ)f(z))(1σ)zDq,p(Iq,pn(ϱ)f(z))+σzDq,p(zDq,p(Iq,pn(ϱ)f(z)))(1σ)Iq,pn(ϱ)f(z)+σzDq,p(Iq,pn(ϱ)f(z))pq<β(bC,nN0,p,iN,0<q<1,ϱ0,0σ1,0<β1)z(Ipn(ϰ,ϱ)f(z))+σz2(Ipn(ϰ,ϱ)f(z))(1σ)Ipn(ϰ,ϱ)f(z)+σz(Ipn(ϰ,ϱ)f(z));

  3.  Sqn(i,1,ϰ,ϱ,σ,b,β)=Sqn(i,ϰ,ϱ,σ,b,β) fTi(1):1b(1σ)zDq(Iqn(ϰ,ϱ)f(z))+σzDq(zDq(Iqn(ϰ,ϱ)f(z)))(1σ)Iqn(ϰ,ϱ)f(z)+σzDq(Iqn(ϰ,ϱ)f(z))1(1σ)zDq(Iqn(ϰ,ϱ)f(z))+σzDq(zDq(Iqn(ϰ,ϱ)f(z)))(1σ)Iqn(ϰ,ϱ)f(z)+σzDq(Iqn(ϰ,ϱ)f(z))<β(bC,iN,nN0,0<q<1,ϱ,ϰ0,0σ1,0<β1)z(Ipn(ϰ,ϱ)f(z))+σz2(Ipn(ϰ,ϱ)f(z))(1σ)Ipn(ϰ,ϱ)f(z)+σz(Ipn(ϰ,ϱ)f(z));

  4. Sqn(i,p,ϰ,0,σ,b,β)=Sqn(i,p,ϰ,σ,b,β) fTi(p):1b(1σ)zDq,p(Dq,pn(ϰ)f(z))+σzDq,p(zDq,p(Dq,pn(ϰ)f(z)))(1σ)Dq,pn(ϰ)f(z)+σzDq,p(Dq,pn(ϰ)f(z))pq<β(bC,nN0,p,iN,0<q<1,ϰ0,0σ1,0<β1)z(Ipn(ϰ,ϱ)f(z))+σz2(Ipn(ϰ,ϱ)f(z))(1σ)Ipn(ϰ,ϱ)f(z)+σz(Ipn(ϰ,ϱ)f(z));

  5. Sqn(i,p,1,0,σ,b,β)=Sqn(i,p,σ,b,β) fTi(p):1b(1σ)zDq,p(Dq,pnf(z))+σzDq,p(zDq,p(Dq,pnf(z)))(1σ)Dq,pnf(z)+σzDq,p(Dq,pnf(z))pq<β(bC,nN0,p,iN,0<q<1,0σ1,0<β1)z(Ipn(ϰ,ϱ)f(z))+σz2(Ipn(ϰ,ϱ)f(z))(1σ)Ipn(ϰ,ϱ)f(z)+σz(Ipn(ϰ,ϱ)f(z)).

Definition 2

Let fTi(p). Then fKqn(i,p,ϰ,ϱ,σ,b,β) if it satisfies the next inequality (10) 1b(1σ)Iq,pn(ϰ,ϱ)f(z)zp+σDq,p(Iq,pn(ϰ,ϱ)f(z))pqzp11<β(bC,nN0,p,iN,0<q<1,ϱ,ϰ0,0σ1,0<β1).(10)

We note that:

  1. limq1Kqn(i,p,ϰ,ϱ,σ,b,β)=Kn(i,p,ϰ,ϱ,σ,b,β) fTi(p):1b(1σ)Ipn(ϰ,ϱ)f(z)zp+σ(Ipn(ϰ,ϱ)f(z))pzp11<β(bC,nN0,p,iN,ϱ,ϰ0,0σ1,0<β1)Ipn(ϰ,ϱ)f(z)zp;

  2. Kqn(i,p,1,ϱ,σ,b,β)=Kqn(i,p,ϱ,σ,b,β) fTi(p):1b(1σ)Iq,pn(ϱ)f(z)zp+σDq,p(Iq,pn(ϱ)f(z))pqzp11<β(bC,nN0,p,iN,0<q<1,ϱ0,0σ1,0<β1)Ipn(ϰ,ϱ)f(z)zp;

  3. Kqn(i,p,1,0,σ,b,β)=Kqn(i,p,σ,b,β) fTi(p):1b(1σ)Dq,pnf(z)zp+σDq,p(Dq,pnf(z))pqzp11<β(bC,nN0,p,iN,0<q<1,0σ1,0<β1)Ipn(ϰ,ϱ)f(z)zp;

  4. Kqn(i,p,ϰ,0,σ,b,β)=Kqn(i,p,ϰ,σ,b,β) fTi(p):1b(1σ)Dq,pn(ϰ)f(z)zp+σDq,p(Dq,pn(ϰ)f(z))pqzp11<β(bC,nN0,p,iN,ϰ0,0<q<1,0σ1,0<β1)Dq,p(Dq,pn(ϰ)f(z))pqzp1.

Following the investigations by Ruscheweyh [Citation25], Goodman [Citation26], Altintaş et al. [Citation27–29], Altintaş and Owa [Citation30] and see also ([Citation31–35]), we define the neighbourhood (i,δ) forfTi(p) by (11) Ni,δp(f)=g:gTi(p), g(z)=zpk=i+pbk zkandk=i+pkakbk δ.(11) In particular, if (12) S(z)=zp (pN),(12) we obtain (13) Ni,δp(S)=g:gTi(p), g(z)=zpk=i+pbk zkandk=i+pkbk δ.(13) Now we define the (q,i,δ) neighbourhood for fTi(p) (see [Citation36]) by (14) Ni,δp,q(f)=g:gTi(p), g(z)=zpk=i+pbk zkandk=i+pkqakbk δ.(14) In particular, if S(z) given by (Equation12), we immediately have (15) Ni,δp,q(S)=g:gTi(p), g(z)=zpk=i+pbk zkandk=i+pkqbk δ.(15) We note that limq1Ni,δp,q(f)=Ni,δp(f) and limq1Ni,δp,q(S)=Ni,δp(S) (see [Citation36]).

2. Coefficient estimates

In this research, we shall assume that bC,nN0,p,iN,ϱ,ϰ0,0<q<1,0σ1,0<β1 and Ψq,pn(k,ϰ,ϱ) is given by (Equation8).

In our present investigation of the inclusion relations, we shall require Lemmas 1 and 2.

Lemma 1

If fTi(p) is given by (Equation2), then fSqn(i,p,ϰ,ϱ,σ,b,β),

if and only if (16) k=i+p(kq+βbpq)[1+σ(kq1)]Ψq,pn×(k,ϰ,ϱ)akβb[1+σ(pq1)].(16)

Proof.

Let fSqn(i,p,ϰ,ϱ,σ,b,β). Then we have (17) (1σ)zDq,p(Iq,pn(ϰ,ϱ)f(z))+σzDq,p(zDq,p(Iq,pn(ϰ,ϱ)f(z)))(1σ)Iq,pn(ϰ,ϱ)f(z)+σzDq,p(Iq,pn(ϰ,ϱ)f(z))pq(1σ)zDq,p(Iq,pn(ϰ,ϱ)f(z))+σzDq,p(zDq,p(Iq,pn(ϰ,ϱ)f(z)))(1σ)Iq,pn(ϰ,ϱ)f(z)+σzDq,p(Iq,pn(ϰ,ϱ)f(z))>βb (zU),(17) or, equivalently, (18) k=i+p(kqpq)[1+σ(kq1)]Ψq,pn(k,ϰ,ϱ)ak zkp[1+σ(pq1)]k=i+p[1+σ(kq1)]Ψq,pn(k,ϰ,ϱ)ak zkp>βb.(18)

Setting z=r (0r<1) in (Equation18), so in the left side the denominator will be non negative for r = 0 and also for (0r<1). Now, by Assuming r1 through real values, (Equation18) leads up to the proof of Lemma 1.

Conversely, by applying (Equation16) and letting |z|=1, we observe from (Equation18) that (1σ)zDq,p(Iq,pn(ϰ,ϱ)f(z))+σzDq,p(zDq,p(Iq,pn(ϰ,ϱ)f(z)))(1σ)Iq,pn(ϰ,ϱ)f(z)+σzDq,p(Iq,pn(ϰ,ϱ)f(z))pq=k=i+p(kqpq)[1+σ(kq1)]Ψq,pn(k,ϰ,ϱ)ak zkp[1+σ(pq1)]k=i+p[1+σ(kq1)]Ψq,pn(k,ϰ,ϱ)ak zkpk=i+p(kqpq)[1+σ(kq1)]Ψq,pn(k,ϰ,ϱ)ak zkp[1+σ(pq1)]k=i+p[1+σ(kq1)]Ψq,pn(k,ϰ,ϱ)ak zkpk=i+p(kqpq)[1+σ(kq1)]Ψq,pn(k,ϰ,ϱ)ak [1+σ(pq1)]k=i+p[1+σ(kq1)]Ψq,pn(k,ϰ,ϱ)ak=βb. Thus we have fSqn(i,p,ϰ,ϱ,σ,b,β), by applying the maximum modulus theorem, which evidently completes the proof.

We can prove the lemma below like the proof of Lemma 1.

Lemma 2

Let fTi(p) is given by (Equation2). Then fKqn(i,p,ϰ,ϱ,σ,b,β),

if and only if (19) k=i+p[pq+σ(kqpq)]Ψq,pn(k,ϰ,ϱ)akβbpq.(19)

3. Neighbourhood properties for two new classes

Sqn(i,p,ϰ,ϱ,σ,b,β) and Kqn(i,p,ϰ,ϱ,σ,b,β)

In this part, we determine inclusion relations for each of the classes Sqn(i,p,ϰ,ϱ,σ,b,β) and Kqn(i,p,ϰ,ϱ,σ,b,β) involving (q,i,δ) neighbourhood defined by (Equation14) and (Equation15).

Theorem 1

Let fTi(p) be in the class Sqn(i,p,ϰ,ϱ,σ,b,β), then (20) Sqn(i,p,ϰ,ϱ,σ,b,β)Ni,θp,q(S),(20) since S(z) is defined by (Equation12) and θ is given by (21) θ=i+pqβb[1+σ(pq1)](i+pq+βbpq)[1+σ(i+pq1)]Ψq,pn(i+p,ϰ,ϱ) (pq>b).(21)

Proof.

If fSqn(i,p,ϰ,ϱ,σ,b,β), then by using (Equation16) we obtain (22) (i+pq+βbpq)[1+σ(i+pq1)]×Ψq,pn(i+p,ϰ,ϱ)k=i+pakk=i+p(kq+βbpq)×[1+σ(kq1)]Ψq,pn(k,ϰ,ϱ)akβb[1+σ(pq1)],(22) which readily yields (23) k=i+pakβb[1+σ(pq1)](i+pq+βbpq)[1+σ(i+pq1)]Ψq,pn(i+p,ϰ,ϱ).(23) Making use of (Equation16) with (Equation23), we obtain [1+σ(i+pq1)]Ψq,pn(i+p,ϰ,ϱ)k=i+pkqakβb[1+σ(pq1)]+(pqβb)×[1+σ(i+pq1)]Ψq,pn(i+p,ϰ,ϱ)k=i+pakβb[1+σ(pq1)]+(pqβb)βb[1+σ(pq1)](i+pq+βbpq)=i+pqβb[1+σ(pq1)]i+pq+βbpq. Hence (24) k=i+pkqaki+pqβb[1+σ(pq1)](i+pq+βbpq)[1+σ(i+pq1)]Ψq,pn(i+p,ϰ,ϱ)=θ (pq>b),(24) which, by means of the definition (Equation14), established the inclusion (Equation20) asserted by Theorem 1.

In analogous manner, by applying (Equation19) of Lemma 2 instead of (Equation16) of Lemma 1 to functions in the class Kqn(i,p,ϰ,ϱ,σ,b,β), we can prove the next inclusion relationship.

Theorem 2

Let fTi(p) belongs to Kqn(i,p,ϰ,ϱ,σ,b,β), then (25) Kqn(i,p,ϰ,ϱ,σ,b,β)Ni,δp,q(S),(25) where S(z) is defined by (Equation12) and δ is introduced by (26) δ=i+pqβbpq[pq+σ(i+pqpq)]Ψq,pn(i+p,ϰ,ϱ).(26)

4. Neighbourhood properties for two new classes Sqn(γ)(i,p,ϰ,ϱ,σ,b,β) and Kqn(γ)(i,p,ϰ,ϱ,σ,b,β)

In this part, we determine the neighbourhood for each of the classes Sqn(γ)(i,p,ϰ,ϱ,σ,b,β) and Kqn(γ)(i,p,ϰ,ϱ,σ,b,β), which we define as follows. A function fTi(p) belongs to the class Sqn(γ)(i,p,ϰ,ϱ,σ,b,β) if there exists a function ρ(z)Sqn(γ)(i,p,ϰ,ϱ,σ,b,β) such that (27) f(z)ρ(z)1<pqγ (zU; 0γ<pq).(27) Parallelly, f(z)Ti(p) belongs to the class Kqn(γ)(i,p,ϰ,ϱ,σ,b,β), if there exists a function ρ(z)Kqn(γ)(i,p,ϰ,ϱ,σ,b,β) such that (Equation27) holds true.

Theorem 3

Let fTi(p) belongs to Sqn(γ)(i,p,ϰ,ϱ,σ,b,β) and (28) γ=pqθ(i+pq+βbpq)[1+σ(i+pq1)]Ψq,pn(i+p,ϰ,ϱ)i+pq(i+pq+βbpq)[1+σ(i+pq1)]Ψq,pn(i+p,ϰ,ϱ)βb[1+σ(pq1)],(28) then (29) Ni,θp,q(S)Sqn(γ)(i,p,ϰ,ϱ,σ,b,β),(29) where θ[p]qi+pq1βb[1+σ(pq1)](i+pq+βbpq)[1+σ(i+pq1)]Ψq,pn(i+p,ϰ,ϱ).

Proof.

Assume fNi,θp,q(S). From (Equation14) we find that (30) k=i+pkqakbkθ,(30) which readily implies that (31) k=i+pakbkθi+pq (p, iN).(31) Next, since ρ(z)Sqn(γ)(i,p,ϰ,ϱ,σ,b,β), by using (Equation23), we have (32) k=i+pbkβb[1+σ(pq1)](i+pq+βbpq)[1+σ(i+pq1)]Ψq,pn(i+p,ϰ,ϱ),(32) so f(z)ρ(z)1k=i+pakbk1k=i+pbkθ(i+pq+βbpq)[1+σ(i+pq1)]Ψq,pn(i+p,ϰ,ϱ)i+pq(i+pq+βbpq)[1+σ(i+pq1)]Ψq,pn(i+p,ϰ,ϱ)βb[1+σ(pq1)]=pqγ, provided that γ is given by (Equation28). Thus, with the above definition, fSqn(γ)(i,p,ϰ,ϱ,σ,b,β). This completes the proof.

We delete the details for proving Theorem 4 as they like the details for proving Theorem 3.

Theorem 4

Let fTi(p) belongs to Kqn(γ)(i,p,ϰ,ϱ,σ,b,β) and (33) γ=pqδ[pq+σ(i+pqpq)]Ψq,pn(i+p,ϰ,ϱ)i+pq[pq+σ(i+pqpq)]Ψq,pn(i+p,ϰ,ϱ)βpqb,(33) then (34) Ni,δp,q(S)Kqn(γ)(i,p,ϰ,ϱ,σ,b,β),(34) where (35) δpqi+pq1βpqb(pq+σ(i+pqpq)]Ψq,pn(i+p,ϰ,ϱ).(35)

Remark

  1. Letting q1 in Theorems 1–4, respectively, we obtain new results for the classes Sn(i,p,ϰ,ϱ,σ,b,β), Kn(i,p,ϰ,ϱ,σ,b,β), Sn(γ)(i,p,ϰ,ϱ,σ,b,β) and Kn(γ)(i,p,ϰ,ϱ,σ,b,β), respectively;

  2. Putting (i) ϰ=1, (ii) ϱ=0, (iii) ϰ=1 and ϱ=0 in Lemma 1, Theorems 1 and 3, respectively, we obtain new results for the classes Sqn(i,p,ϱ,σ,b,β),Sqn(i,p,ϰ,σ,b,β)

    and Sqn(i,p,σ,b,β), respectively;

  3. Putting (i) ϰ=1, (ii)ϱ=0, (iii) ϰ=1 and ϱ=0 in Lemma 2, Theorems 2 and 4, respectively, we obtain new results for the classes Kqn(i,p,ϱ,σ,b,β),Kqn(i,p,ϰ,σ,b,β)

    and Kqn(i,p,σ,b,β), respectively.

Acknowledgments

The authors are grateful to the referees for their valuable suggestions.

Availability of data and materials

During the current study, the data sets are derived arithmetically.

Disclosure statement

No potential conflict of interest was reported by the authors.

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