# A New Class of Analytic Functions Associated with Sălăgean Operator

Authors: Muhammad Arif; Khurshid Ahmad; Jin-Lin Liu; Janusz Sokół
Journal: Journal of Function Spaces
 (2019)

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## Abstract

The main object of the present paper is to investigate a number of useful properties such as inclusion relation, distortion bounds, coefficient estimates, and subordination results for a new subclass of analytic functions which are defined here by means of a linear operator. Several known consequences of the results are also pointed out.


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## Body

## 1. Introduction and Definitions

LetA denote the set of all functions of the form(1)fz=z+∑k=2∞akzk,which are analytic in the open unit disk U=z∈C:z<1. Suppose that f and g are analytic in U; then we say that f is subordinate to g, written as f≺g or f(z)≺g(z), if there exists a function w, which is analytic in U with w0=0 and w(z)<1 such that (2)fz=gwzz∈U.Therefore, f(z)≺g(z) implies f(U)⊂g(U). Moreover, if the function g is univalent in U, then the following equivalence holds:(3)fz≺gzz∈U⇔f0=g0andfU⊂gU.Furthermore, let PA,B, -1≤B<A≤1, denote the family of all functions q that are analytic in the open unit disk U with q0=1 and satisfy (4)qz≺1+Az1+Bz.Geometrically, by (3), a function q is in the class P[A,B], -1≤B<A≤1, if and only if q(0)=1 and q(U)⊂Ω[A,B], where circular domain Ω[A,B] is defined by(5)ΩA,B=ω:ω-1-AB1-B2<A-B1-B2forB≠-1,ω:Reω>1-A2forB=-1.The domain Ω[A,B], B≠-1, represents an open circular disk centered on the real axis with diameter end points D1=1-A/1-B and D2=1+A/1+B with 0<D1<1<D2.With the help of the classP[A,B], we now define the classes S∗[A,B] and C[A,B] of Janowski starlike and Janowski convex functions as below: (6)S∗A,B=f∈A:zf′zfz∈PA,B,z∈U,CA,B=f∈A:zf′z′f′z∈PA,B,z∈U.These classes were introduced by Janowski [1]. The extension of Janowski function was discussed by Kuroki, Owa, and Srivastava [2] by choosing the complex parameters A and B with the following conditions: (7)iA≠B,B<1,A≤1andRe1-AB¯≥A-BiiA≠B,B=1,A≤1,andRe1-AB¯>0.Later on, Kuroki and Owa [3] discussed the fact that the condition A≤1 can be omitted from the conditions in part (i) of (7). Janowski functions are being studied and extended in different directions by several renowned mathematicians like Noor and Arif [4], Arif et al. [5], Polatog˘lu [6], Cho [7], Cho et al. [8, 9], Liu and Noor [10], Liu and Patel [11], Liu and Srivastava [12, 13], etc.For a functionf of the form (1) and n=1,2,3…, the Sa˘la˘gean operator [14] is defined by (8)Dnfz=z+∑k=2∞knakzk.It is easy to see that the series Dnf(z) is convergent in the unit disc for each n∈N. Further, we have D1f(z)=zf′(z). Also, we consider the following differential operator: (9)D-1fz=∫0zfξξdξ=z+∑k=2∞k-1akzk,D-nfz=D-1(D-n-1fz=z+∑k=2∞k-nakzk,for any integers. Then, for f∈A given by (1), we know that(10)Dnfz=z+∑k=2∞knakzkn=0,±1,±2,…,.Using this Sa˘la˘gean operator Dnalong with the concepts of Janowski functions, we now define a subclass of A as follows.Definition 1.
Iff∈A, then f∈Qsb(A,B,j) if and only if(11)1+1b2Dj+1fzDjfz-Djf-z-1≺1+Az1+Bz,where -1≤B<A≤1, j=1,2,3,⋯, and b∈C∖0.Special Cases. In literature, various interesting subfamilies of analytic and univalent functions associated with circular domain have been studied from a number of different view points which are closely related with the class Qsb(A,B,j).For example, if we setj=0 in (11), we get the class QsbA,B,0≡Ss∗b,A,B defined as (12)Ss∗b,A,B=f∈A:1+1b2zf′zfz-f-z-1≺1+Az1+Bz,z∈U,and further, by making b=2, A=1, B=-1 in Ss∗b,A,B, we obtain the familiar class Ss∗ of starlike functions with respect to symmetrical pints studied in [15]. Also, by putting j=1 in (11), we obtain the family Csb,A,B defined by (13)Csb,A,B=f∈A:1+1b2zf′z′fz-f-z′-1≺1+Az1+Bz,z∈U,and further, taking b=2, A=1, and B=-1 in Csb,A,B, we get the set Cs intoduced by Das and Singh [16]. For more work, see [17–21].We will assume throughout our discussion, unless otherwise stated, that(14)-1≤B<A≤1andb∈C∖0.


## 2. A Set of Lemmas

Lemma 2 (see [19]).
Ifp(z)=1+p1z+p2z+⋯ belongs to the class P[A,B], then (15)pn≤A-B∀n≥1.Lemma 3 (see [19]).
LetN be analytic and M starlike functions in U with N(0)=M(0)=0. Then, for -1≤B<A≤1, (16)N′z/M′z-1A-BN′z/M′z<1implies (17)Nz/Mz-1A-BNz/Mz<1,z∈U.


## 3. The Main Results and Their Consequences

Theorem 4.
Iff∈Qsb(A,B,j), then the odd function(18)Ψz=12fz-f-zsatisfies(19)1+1bDj+1ΨzDjΨz-1≺1+Az1+Bz.Proof.
Iff∈Qsb(A,B,j), then there exists p∈P[A,B] such that (20)bpz-1=2Dj+1fzDjfz-Djf-z-1,bp-z-1=-2Dj+1f-zDjfz-Djf-z-1.This gives (21)1+1bDj+1ΨzDjΨz-1=pz+p-z2.Because p(z)≺(1+Az)/(1+Bz) and (1+Az)/(1+Bz) is univalent, then by (3) we have (22)pz+p-z2≺1+Az1+Bz.It follows (19).Theorem 5.
A functionf∈A is in the class Qsb(A,B,j), if and only if there exists p∈P[A,B] such that(23)Dj+1fz=z21+b(pz-1exp⁡b2∫0zpt+p-t-2tdt.Proof.
Iff∈Qsb(A,B,j), then it is equivalent to(24)1+1b2Dj+1fzDjfz-Djfz-1=1+1bDj+1fzDjΨz-1=pzfor some p belonging to the class P[A,B]. From Theorem 4, we also have(25)1+1bDj+1ΨzDjΨz-1=1+1bzDjΨz′DjΨz-1=pz+p-z2.Equation (25) gives (26)DjΨz=zexp⁡b2∫0zpt+p-t-2tdtand using this in (24) provides (23). It is easy to verify that if p belongs to the class P[A,B] and f∈A satisfies (23), then f∈Qsb(A,B,j).Theorem 6.
Letf∈Qsb(A,B,j). Then(27)a2≤bA-B2j+1,a3≤bA-B2·3j+1,And, for n≥2,(28)a2n≤bA-B2j+nnjn!∏k=1n-1bA-B+2kand(29)a2n+1≤bA-B2nn!2n+1j∏k=1n-1bA-B+2k.Proof.
Iff∈Qsb(A,B,j), then by Definition 1 we have(30)1+1b2Dj+1fzDjfz-Djf-z-1=1+Awz1+Bwz.Put (31)pz=1+∑k=1∞pkzk=1+Awz1+Bwz.From (30), we can write(32)1+1b2Dj+1fzDjfz-Djf-z-1=1+∑k=1∞pkzk.Also, from (32) and (10) we have (33)z+2j+1a2z2+3j+1a3z3+…+2nj+1a2nz2n+2n+1j+1a2n+1z2n+1+…=z+3ja3z3+5ja5z5+…+2n-1ja2n-1z2n-1+2n+1ja2n+1z2n+1+…×1+bp1z+p2z2+p3z3+…+p2nz2n+p2n+1z2n+1+….Equating the coefficients of like powers of z, we have(34)2j+1a2=bp1,2·3ja3=bp2,(35)4j+1a4=bp3+3bp1a3,5j+1a5=bp4+3jbp2a3+5a5z5,(36)2nj+1a2n=bp2n-1+3ja3bp2n-3+5ja5bp2n-5+…+b2n-1ja2n-1p1,(37)2n+1j+1a2n+1=bp2n+3ja3bp2n-2+5ja5bp2n-4+…+b2n-1ja2n-1p2.Using Lemma 2 and (34), we easily get (38)a2≤bA-B2j+1,a3≤bA-B2·3j+1,which gives (27). Using Lemma 2, (27), and (35) we get(39)a4≤bA-B2·4j+1bA-B+2,a5≤bA-B8·5jbA-B+2.It follows that (28) and (29) hold for n=2. We now prove (28) and (29) by using induction. Equation (36) in conjunction with Lemma 2 yields(40)a2n≤bA-B2nj+11+∑r=1n-12r+1ja2r+1.We assume that (28) and (29) hold for 3,4,…,(n-1). Then from (40) we obtain(41)a2n≤bA-B2nj+11+∑r=1n-12r+1jbA-B2rr!2r+1j∏k=1r-1bA-B+2k=bA-B2nj+11+∑r=1n-1bA-B2rr!∏k=1r-1bA-B+2k,where we assumed (42)∏k=10bA-B+2k=1.In order to complete the proof, it is sufficient to show that(43)bA-B2sj+11+∑r=1s-1bA-B2rr!∏k=1r-1bA-B+2k=bA-B2j+ssjs!∏k=1s-1bA-B+2k,s=3,4,….Expression (43) is valid for s=3.
Let us suppose that (43) is true for 3,…,(s-1). Then from (41) (44)bA-B2sj+11+∑r=1s-1bA-B2rr!∏k=1r-1bA-B+2k=s-1j+1sj+1bA-B2s-1j+11+∑r=1s-2bA-B2rr!∏k=1r-1bA-B+2k+bA-B2sj+1bA-B2s-1s-1!∏k=1s-2bA-B+2k=s-1sbA-B2j+s-1sjs-1!∏k=1s-2bA-B+2k+bA-B2j+s-1sjs-1!bA-B2s∏k=1s-2bA-B+2k=bA-B2j+s-1sjs-1!∏k=1s-2bA-B+2kbA-B+2s-12s=bA-B2j+ssjs!∏k=1s-1bA-B+2k.Thus (43) holds and hence (41) with (43) implies (28). Similarly we can prove the coefficient estimates given in (29).Theorem 7.
IfD>-1, then (45)QsbA,B,j⊂QsbC,D,jif and only if (46)1-CD1-D2-1-AB1-B2≤1-CD1-D2-1-AB1-B2.If D=-1, then (47)QsbA,B,j⊂QsbC,D,jif and only if (48)C≥1-21-A1-B.Proof.
Forf∈Qsb(A,B,j), we let (49)Hf,z=1+1b2Dj+1fzDjfz-Djf-z-1,z∈U.The values of the function H lie in Ω[A,B] (see (5)) since (50)f∈QsbA,B,j⇔Hf,U⊂ΩA,B.Similarly, (51)h∈QsbC,D,j⇔Hf,U⊂ΩC,D.In the case B≠-1, Ω[A,B] is a disc D(A,B) with center s(A,B) and radius r(A,B)(52)sA,B=1-AB1-B2,rA,B=A-B1-B2,while it is a half plane for B=-1. Therefore for the case B≠-1, D≠-1, the inclusion relation Qsb(A,B,j)⊂Qsb(C,D,j) holds when (53)rA,B≤rC,Dand (54)sC,D-sA,B≤rC,D-rA,B.This is equivalent to (55)1-CD1-D2-1-AB1-B2≤1-CD1-D2-A-B1-B2.Furthermore, we have (56)ΩC,-1=ω:Reω>1-C2.The domain Ω[A,B] represents an open circular disk or a half plane on the right site of the point 1-A/1-B. Therefore, (57)ΩA,B⊂ΩC,-1⇔1-C2≤1-A1-Band hence the result follows.Theorem 8.
Iff∈Qsb(A,B,j), then F∈Qsb(A,B,j), where(58)Fz=2z∫0zftdt.Proof.
From (58), we can easily write (59)1+1b2Dj+1FzDjFz-DjF-z-1=2zDjfz+b-3∫0zDjftdt+b-1∫0zDjf-tdtb∫0zDjftdt+∫0zDjf-tdt.Let N and M be the numerator and denominator functions, respectively. Now we show that (60)Mz=b∫0zDjftdt+∫0zDjf-tdtis starlike. Since M′(z)=bDjf(z)-Djf(-z), therefore (61)zM′zMz=zDjfz-zDjf-z∫0zDjftdt+∫0zDjf-tdt.Let (62)∫0zDjftdt=hz.Then (61) becomes(63)zM′zMz=122zh′zhz-h-z+2-zh′-zh-z-hz.Since fz∈Qsb(A,B,j), it follows that (64)1+1b2zh′′zh′z-h′-z-1≺1+Az1+Bz,and h(z)∈Cs(b,A,B)⊂Ss∗(b,A,B)⊂Ss∗. Now, from (63), it follows that M(z) is starlike functions. Furthermore, (65)N′zM′z=1+1b2Dj+1fzDjfz-Djf-z-1withf∈QsbA,B,j.Thus (66)N′zM′z=1+Awz1+Bwz.This implies that (67)N′z/M′z-1A-BN′z/M′z<1.By Lemma 3, we have (68)Nz/Mz-1A-BNz/Mz<1,z∈U,and this gives that F∈Qsb(A,B,j).Theorem 9.
IfF∈Qsb(1,-1,j) in z<1, then f(z)=1/2(zF(z))′ belongs to the class Qsb(1,-1,j) in z<r0, where (69)r0=21+b+b2-2b+5.Proof.
SinceF∈Qsb(1,-1,j), it follows that(70)Re1+1b2Dj+1FzDjFz-DjF-z-1>0,z∈U.Equivalently, we have(71)1+1b2Dj+1FzDjFz-DjF-z-1=1-wz1+wz,where w is analytic in U with w(z)<1 and w(0)=0.
After simple computation we obtain(72)1+1b2Dj+1FzDjFz-DjF-z-1=2zDjfz+b-3∫0zDjftdt+b-1∫0zDjf-tdtb∫0zDjftdt+∫0zDjf-tdt.Using (71) and (72), we can show that (73)1+1b2Dj+1fzDjfz-Djf-z-1=1-wz1+wz-2zw′z1+w-z2-b1+wz21+w-z+b1-wzw-z1+wz.Thus f∈Qsb(1,-1,j) if(74)Re1+1b2Dj+1fzDjfz-Djf-z-1=Re1-wz1+wz-2zw′z1+w-z2-b1+wz21+w-z+b1-wzw-z1+wz>0inz<r0.Now (75)Re1-wz1+wz=1-wz21+wz2and (76)Re2zw′z1+w-z2-b1+wz21+w-z+b1-wzw-z1+wz≤2z1+w-z2-b1+wz21+w-z+b1-wzw-z1+wz×1-wz21-z2,where we have used the well-known estimate (77)w′z≤1-wz21-z2forz<1.Therefore (74) is true if(78)1b2z1-z2-2-b+1<2+wz+w-z1+wz1+w-z=h1z+h2z,where (79)h1z=11+wzandh2z=11+w-z.Since (80)1-r≤1+wz≤1+r,this implies that (81)11+r≤hiz≤11-r,fori=1,2.Therefore, we have (82)h1z+h2z≥21+r,and from (78) it follows that f∈Qsb(A,B,j) if (83)r1-r2-1-b<b1+r,or (84)r<21+b+b2-2b+5.Thus (74) is true if the last inequality holds.Theorem 10.
Letf∈Qsb(A,B,j). Then for z=r, 0<r<1,(85)1-bArr+1-bBr21+r21-Br≤Dj+1fz≤1+bArr+1-bBr21-r21+Br.Proof.
Let us puth(z)=Djf(z)-Djf(-z)/2. Then we obtain(86)Dj+1fz=hz1+bA-Bwz1+Bwz.Since h is odd starlike, it follows that(87)r1+r2≤hz≤r1-r2.Furthermore, for w∈A, one can easily obtain that (88)1-bAr+1-bBr1-Br≤(1+Bwz+bA-Bwz1+Bwz≤1+bAr+1-bBr1+Br.Applying the last inequalities along with (87) in (86) we easily obtain (85).



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