JFS Journal of Function Spaces 2314-8888 2314-8896 Hindawi 10.1155/2019/6157394 6157394 Research Article A New Class of Analytic Functions Associated with Sălăgean Operator http://orcid.org/0000-0003-1484-7643 Arif Muhammad 1 Ahmad Khurshid 1 http://orcid.org/0000-0001-7963-2470 Liu Jin-Lin 2 Sokół Janusz 3 Sugimoto Mitsuru 1 Department of Mathematics Abdul Wali Khan University Mardan 23200 Mardan Pakistan awkum.edu.pk 2 Department of Mathematics Yangzhou University Yangzhou 225002 China yzu.edu.cn 3 University of Rzeszów Faculty of Mathematics and Natural Sciences ul. Prof. Pigonia 1 35-310 Rzeszów Poland ur.edu.pl 2019 322019 2019 29 06 2018 15 01 2019 322019 2019 Copyright © 2019 Muhammad Arif et al. This 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.

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.

National Natural Science Foundation of China 11571299 Natural Science Foundation of Jiangsu Province BK20151304
1. Introduction and Definitions

Let A denote the set of all functions of the form(1)fz=z+k=2akzk,which are analytic in the open unit disk U=zC:z<1. Suppose that f and g are analytic in U; then we say that f is subordinate to g, written as fg 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=gwzzU.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)fzgzzUf0=g0and  fUgU.Furthermore, let PA,B, -1B<A1, denote the family of all functions q that are analytic in the open unit disk U with q0=1 and satisfy (4)qz1+Az1+Bz.Geometrically, by (3), a function q is in the class P[A,B], -1B<A1, 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 class P[A,B], we now define the classes S[A,B] and C[A,B] of Janowski starlike and Janowski convex functions as below: (6)SA,B=fA:zfzfzPA,B,zU,CA,B=fA:zfzfzPA,B,zU.These classes were introduced by Janowski . The extension of Janowski function was discussed by Kuroki, Owa, and Srivastava  by choosing the complex parameters A and B with the following conditions: (7)iAB,B<1,A1and  Re1-AB¯A-BiiAB,B=1,A1,and  Re1-AB¯>0.Later on, Kuroki and Owa  discussed the fact that the condition A1 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 , Arif et al. , Polatog˘lu , Cho , Cho et al. [8, 9], Liu and Noor , Liu and Patel , Liu and Srivastava [12, 13], etc.

For a function f of the form (1) and n=1,2,3, the Sa˘la˘gean operator  is defined by (8)Dnfz=z+k=2knakzk.It is easy to see that the series Dnf(z) is convergent in the unit disc for each nN. Further, we have D1f(z)=zf(z). Also, we consider the following differential operator: (9)D-1fz=0zfξξdξ=z+k=2k-1akzk,D-nfz=D-1(D-n-1fz=z+k=2k-nakzk,for any integers. Then, for fA given by (1), we know that(10)Dnfz=z+k=2knakzkn=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.

If fA, then fQsb(A,B,j) if and only if(11)1+1b2Dj+1fzDjfz-Djf-z-11+Az1+Bz,where -1B<A1, j=1,2,3,, and bC0.

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 set j=0 in (11), we get the class QsbA,B,0Ssb,A,B defined as (12)Ssb,A,B=fA:1+1b2zfzfz-f-z-11+Az1+Bz,zU,and further, by making b=2, A=1, B=-1 in Ssb,A,B, we obtain the familiar class Ss of starlike functions with respect to symmetrical pints studied in . Also, by putting j=1 in (11), we obtain the family Csb,A,B defined by (13)Csb,A,B=fA:1+1b2zfzfz-f-z-11+Az1+Bz,zU,and further, taking b=2, A=1, and B=-1 in Csb,A,B, we get the set Cs intoduced by Das and Singh . For more work, see .

We will assume throughout our discussion, unless otherwise stated, that (14)-1B<A1and  bC0.

2. A Set of Lemmas Lemma 2 (see [<xref ref-type="bibr" rid="B19">19</xref>]).

If p(z)=1+p1z+p2z+ belongs to the class P[A,B], then (15)pnA-Bn1.

Lemma 3 (see [<xref ref-type="bibr" rid="B19">19</xref>]).

Let N be analytic and M starlike functions in U with N(0)=M(0)=0. Then, for -1B<A1, (16)Nz/Mz-1A-BNz/Mz<1implies (17)Nz/Mz-1A-BNz/Mz<1,zU.

3. The Main Results and Their Consequences Theorem 4.

If fQsb(A,B,j), then the odd function(18)Ψz=12fz-f-zsatisfies(19)1+1bDj+1ΨzDjΨz-11+Az1+Bz.

Proof.

If fQsb(A,B,j), then there exists pP[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-z21+Az1+Bz.It follows (19).

Theorem 5.

A function fA is in the class Qsb(A,B,j), if and only if there exists pP[A,B] such that(23)Dj+1fz=z21+b(pz-1expb20zpt+p-t-2tdt.

Proof.

If fQsb(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+1bzDjΨzDjΨz-1=pz+p-z2.Equation (25) gives (26)DjΨz=zexpb20zpt+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 fA satisfies (23), then fQsb(A,B,j).

Theorem 6.

Let fQsb(A,B,j). Then(27)a2bA-B2j+1,a3bA-B2·3j+1,And, for n2,(28)a2nbA-B2j+nnjn!k=1n-1bA-B+2kand(29)a2n+1bA-B2nn!2n+1jk=1n-1bA-B+2k.

Proof.

If fQsb(A,B,j), then by Definition 1 we have(30)1+1b2Dj+1fzDjfz-Djf-z-1=1+Awz1+Bwz.Put (31)pz=1+k=1pkzk=1+Awz1+Bwz.From (30), we can write(32)1+1b2Dj+1fzDjfz-Djf-z-1=1+k=1pkzk.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)a2bA-B2j+1,a3bA-B2·3j+1,which gives (27). Using Lemma 2, (27), and (35) we get(39)a4bA-B2·4j+1bA-B+2,a5bA-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)a2nbA-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)a2nbA-B2nj+11+r=1n-12r+1jbA-B2rr!2r+1jk=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-B2sk=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.

If D>-1, then (45)QsbA,B,jQsbC,D,jif and only if (46)1-CD1-D2-1-AB1-B21-CD1-D2-1-AB1-B2.If D=-1, then (47)QsbA,B,jQsbC,D,jif and only if (48)C1-21-A1-B.

Proof.

For fQsb(A,B,j), we let (49)Hf,z=1+1b2Dj+1fzDjfz-Djf-z-1,zU.The values of the function H lie in Ω[A,B] (see (5)) since (50)fQsbA,B,jHf,UΩA,B.Similarly, (51)hQsbC,D,jHf,UΩC,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,BrC,Dand (54)sC,D-sA,BrC,D-rA,B.This is equivalent to (55)1-CD1-D2-1-AB1-B21-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,BΩC,-11-C21-A1-Band hence the result follows.

Theorem 8.

If fQsb(A,B,j), then FQsb(A,B,j), where(58)Fz=2z0zftdt.

Proof.

From (58), we can easily write (59)1+1b2Dj+1FzDjFz-DjF-z-1=2zDjfz+b-30zDjftdt+b-10zDjf-tdtb0zDjftdt+0zDjf-tdt.Let N and M be the numerator and denominator functions, respectively. Now we show that (60)Mz=b0zDjftdt+0zDjf-tdtis starlike. Since M(z)=bDjf(z)-Djf(-z), therefore (61)zMzMz=zDjfz-zDjf-z0zDjftdt+0zDjf-tdt.Let (62)0zDjftdt=hz.Then (61) becomes(63)zMzMz=122zhzhz-h-z+2-zh-zh-z-hz.Since fzQsb(A,B,j), it follows that (64)1+1b2zhzhz-h-z-11+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)NzMz=1+1b2Dj+1fzDjfz-Djf-z-1with  fQsbA,B,j.Thus (66)NzMz=1+Awz1+Bwz.This implies that (67)Nz/Mz-1A-BNz/Mz<1.By Lemma 3, we have (68)Nz/Mz-1A-BNz/Mz<1,zU,and this gives that FQsb(A,B,j).

Theorem 9.

If FQsb(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.

Since FQsb(1,-1,j), it follows that(70)Re1+1b2Dj+1FzDjFz-DjF-z-1>0,zU.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-30zDjftdt+b-10zDjf-tdtb0zDjftdt+0zDjf-tdt.Using (71) and (72), we can show that (73)1+1b2Dj+1fzDjfz-Djf-z-1=1-wz1+wz-2zwz1+w-z2-b1+wz21+w-z+b1-wzw-z1+wz.Thus fQsb(1,-1,j) if(74)Re1+1b2Dj+1fzDjfz-Djf-z-1=Re1-wz1+wz-2zwz1+w-z2-b1+wz21+w-z+b1-wzw-z1+wz>0in  z<r0.Now (75)Re1-wz1+wz=1-wz21+wz2and (76)Re2zwz1+w-z2-b1+wz21+w-z+b1-wzw-z1+wz2z1+w-z2-b1+wz21+w-z+b1-wzw-z1+wz×1-wz21-z2,where we have used the well-known estimate (77)wz1-wz21-z2for  z<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+wzand  h2z=11+w-z.Since (80)1-r1+wz1+r,this implies that (81)11+rhiz11-r,for  i=1,2.Therefore, we have (82)h1z+h2z21+r,and from (78) it follows that fQsb(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.

Let fQsb(A,B,j). Then for z=r, 0<r<1,(85)1-bArr+1-bBr21+r21-BrDj+1fz1+bArr+1-bBr21-r21+Br.

Proof.

Let us put h(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+r2hzr1-r2.Furthermore, for wA, one can easily obtain that (88)1-bAr+1-bBr1-Br(1+Bwz+bA-Bwz1+Bwz1+bAr+1-bBr1+Br.Applying the last inequalities along with (87) in (86) we easily obtain (85).

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors agree with the contents of the manuscript and there are no conflicts of interest among the authors.

Acknowledgments

This work is supported by National Natural Science Foundation of China (Grant no. 11571299) and Natural Science Foundation of Jiangsu Province of China (Grant no. BK20151304).

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