# Inclusion of the Generalized Bessel Functions in the Janowski Class

Authors: Saiful R. Mondal; Mohammed Al Dhuain
Journal: International Journal of Analysis
 (2016)

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## Abstract

Sufficient conditions onA, B, p, b, and c are determined that will ensure the generalized Bessel function up,b,c satisfies the subordination up,b,c(z)≺1+Az/(1+Bz). In particular this gives conditions for (-4κ/c)(up,b,c(z)-1), c≠0, to be close-to-convex. Also, conditions for up,b,c(z) to be Janowski convex and zup,b,c(z) to be Janowski starlike in the unit disk D=z∈C:z<1 are obtained.


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## Body

## 1. Introduction

LetA denote the class of analytic functions f defined in the open unit disk D={z:|z|<1} normalized by the conditions f(0)=0=f′(0)-1. If f and g are analytic in D, then f is subordinate to g, written f(z)≺g(z), if there is an analytic self-map w of D satisfying w(0)=0 and f=g∘w. For -1≤B<A≤1, let P[A,B] be the class consisting of normalized analytic functions p(z)=1+c1z+⋯ in D satisfying(1)pz≺1+Az1+Bz.For instance, if 0≤β<1, then P[1-2β,-1] is the class of functions p(z)=1+c1z+⋯ satisfying Rep(z)>β in D.The classS∗[A,B] of Janowski starlike functions [1] consists of f∈A satisfying(2)zf′zfz∈PA,B.For 0≤β<1, S∗[1-2β,-1]≔S∗(β) is the usual class of starlike functions of order β; S∗[1-β,0]≔Sβ∗=f∈A:|zf′(z)/f(z)-1|<1-β; and S∗[β,-β]≔S∗[β]={f∈A:|zf′(z)/f(z)-1|<β|zf′(z)/f(z)+1|}. These classes have been studied, for example, in [2, 3]. A function f∈A is said to be close-to-convex of order β [4, 5] if Rezf′(z)/g(z)>β for some g∈S∗≔S∗(0).This article studies the generalized Bessel functionup(z)=up,b,c(z) given by the power series(3)upz=F10κ,-c4z=∑k=0∞-1kck4kκkzkk!,where κ=p+b+1/2≠0,-1,-2,-3,…. The function up(z) is analytic in D and solution of the differential equation(4)4z2u′′z+4κzu′z+czuz=0,if b, p, c in C, such that κ=p+(b+1)/2≠0,-1,-2,-3,…, and z∈D. This normalized and generalized Bessel function of the first kind of order p also satisfies the following recurrence relation:(5)4κup′z=-cup+1z,which is useful tool to study several geometric properties of up. There have been several works [6–11] studying geometric properties of the function up(z), for example, on its close-to-convexity, starlikeness, and convexity, and radius of starlikeness and convexity.In Section2 of this paper, sufficient conditions on A, B, c, and κ are determined that will ensure up satisfies the subordination up(z)≺(1+Az)/(1+Bz). It is to be understood that a computationally intensive methodology with shrewd manipulations is required to obtain the results in this general framework. The benefits of such general results are that, by judicious choices of the parameters A and B, they give rise to several interesting applications, which include extending the results of previous works. Using this subordination result, sufficient conditions are obtained for -4κ/cu′(z)∈P[A,B], which next readily gives conditions for (-4κ/c)(up(z)-1) to be close-to-convex. Section 3 gives emphasis to the investigation of up(z) to be Janowski convex as well as of zup(z) to be Janowski starlike.The following lemma is needed in sequel.Lemma 1 (see [5, 12]).
LetΩ⊂C and Ψ:C2×D→C satisfy(6)Ψiρ,σ;z∉Ωwhenever z∈D, ρ is real, and σ≤-(1+ρ2)/2. If p is analytic in D, with p(0)=1, and Ψ(p(z),zp′(z);z)∈Ω for z∈D, then Rep(z)>0 in D.In the caseΨ:C3×D→C, then the condition in Lemma 1 is generalized to(7)Ψiρ,σ,μ+iν;z∉Ω.ρ is real, σ+μ≤0, and σ≤-(1+ρ2)/2.


## 2. Close-To-Convexity of Generalized Bessel Functions

In this section, one main result on the close-to-convexity of the generalized Bessel function with several consequences is discussed in detail.Theorem 2.
Let-1≤B≤3-22≈0.171573. Suppose B<A≤1 and c,κ∈R satisfy(8)κ-1≥1+B1+A4A-Bc.Further let A, B, κ, and c satisfy either the inequality(9)κ-12+κ-11+B1-B-κ-1A+B2A-Bc+1+B21+A41-BA-Bc≥1-A21-B216A-B2c2whenever(10)2κ-11-BA+Bc+1+B21+Ac≥12A-B1-Bc2or the inequality(11)κ-1A+B2A-Bc+1+B21+A41-BA-Bc2≤c24κ-12+κ-11+B1-B-1-AB216A-B2c2whenever(12)2κ-11-BA+Bc+1+B21+Ac<12A-B1-Bc2.If (1+B)up(z)≠(1+A) for all z∈D, then up(z)∈P[A,B].Proof.
Define the analytic functionp:D→C by (13)pz=-1-A-1-Bupz1+A-1+Bupz.Then, a computation yields(14)upz=1-A+1+Apz1-B+1+Bpz,up′z=2A-Bp′z1-B+1+Bpz2,up′′z=2A-B1-B+1+Bpzp′′z-41+BA-Bp′z21-B+1+Bpz3.Thus, using identities (14), the Bessel differential equation (4) can be rewritten as (15)z2p′′z-21+B1-B+1+Bpzzp′z2+κzp′z+1-B+1+Bpz1-A+1+Apz8A-Bcz=0.
AssumeΩ={0}, and define Ψ(r,s,t;z) by(16)Ψr,s,t;z≔t-21+B1-B+1+Brs2+κs+1-B+1+Br1-A+1+Ar8A-Bcz.It follows from (15) that Ψ(p(z),zp′(z),z2p′′(z);z)∈Ω. To ensure Rep(z)>0 for z∈D, from Lemma 1, it is enough to establish ReΨ(iρ,σ,μ+iν;z)<0 in D for any real ρ, σ≤-(1+ρ2)/2, and σ+μ≤0.
Withz=x+iy∈D in (16), a computation yields(17)ReΨiρ,σ,μ+iν;z=μ-21-B21-B2+1+B2ρ2σ2+κσ-ρ1-AB4A-Bcy+1-B1-A-1+B1+Aρ28A-Bcx.Since σ≤-(1+ρ2)/2 and B∈[-1,3-22], (18)21-B21-B2+1+B2ρ2σ2≥21-B21-B2+1+B2ρ21+ρ224≥1+B21-B.Thus (19)ReΨiρ,σ,μ+iν;z≤κ-1σ-1+B21-B-ρ1-AB4A-Bcy+1-B1-A-1+B1+Aρ28A-Bcx≤-12κ-11+ρ2-1+B21-B-ρ1-AB4A-Bcy+1-B1-A-1+B1+Aρ28A-Bcx=p1ρ2+q1ρ+r1≔Qρ,where (20)p1=-12κ-1-1+B1+Acx8A-B,q1=-1-AB4A-Bcy,r1=-12κ-1+1-B1-A8A-Bcx-1+B21-B.
Condition (8) shows that (21)p1=-12κ-1-1+B1+Acx8A-B<-12κ-1-1+B1+A4A-Bc≤0.Since maxρ∈Rp1ρ2+q1ρ+r1=(4p1r1-q12)/(4p1) for p1<0, it is clear that Q(ρ)<0 when (22)1-AB216A-B2c2y2<4-12κ-1-1+B1+A8A-Bcx-12κ-1+1-B1-A8A-Bcx-1+B21-B,with |x|,|y|<1. As y2<1-x2, the above condition holds whenever (23)1-AB2c216A-B21-x2≤κ-1+1+B1+Acx4A-Bκ-1-1-B1-A4A-Bcx+1+B1-B,that is, when(24)c216x2+κ-1A+B2A-Bc+1+B21+A41-BA-Bcx+κ-12+κ-11+B1-B-1-AB216A-B2c2≥0.To establish inequality (24), consider the polynomial R given by (25)Rx≔mx2+nx+r,x<1,where (26)m≔c216,n≔κ-1A+B2A-Bc+1+B21+A41-BA-Bc,r≔κ-12+κ-11+B1-B-1-AB216A-B2c2.Constraint (10) yields |n|≥2|m|, and thus R(x)≥m+r-|n|. Now inequality (9) readily implies that (27)Rx≥m+r-n=c216+κ-12+κ-11+B1-B-1-AB216A-B2c2-κ-1A+B2A-Bc+1+B21+A41-BA-Bc=κ-12+κ-11+B1-B-κ-1A+B2A-Bc+1+B21+A41-BA-Bc-1-A21-B216A-B2c2≥0.Now consider the case of constraint (12), which is equivalent to |n|<2m. Then the minimum of R occurs at x=-n/(2m), and (11) yields (28)Rx≥4mr-n24m≥0.
EvidentlyΨ satisfies the hypothesis of Lemma 1, and thus Rep(z)>0; that is,(29)-1-A-1-Bupz1+A-1+Bupz≺1+z1-z.Hence there exists an analytic self-map w of D with w(0)=0 such that(30)-1-A-1-Bupz1+A-1+Bupz=1+wz1-wz,which implies that up(z)≺(1+Az)/(1+Bz).Theorem2 gives rise to simple conditions on c and κ to ensure up(z) maps D into a half plane.Corollary 3.
Letc≤0 and κ≥max⁡{5/4,1+c2/2}. Then Reup(z)>c/(c-1).Proof.
ChooseA=-(c+1)/(c-1) and B=-1 in Theorem 2. Then condition (8) is equivalent to κ≥1 and (10) reduces to κ≥5/4, and clearly both hold for κ≥max⁡{5/4,1+c2/2}. The proof will complete if hypothesis (9) holds; that is,(31)κ-12≥12κ-1c2.Since κ≥1+c2/2, it follows that (32)κ-12-12κ-1c2=κ-1κ-1-c22≥0,which establishes (31).Corollary 4.
Letc∈R. Then if κ≥|c|/2, then Reup(z)>1/2.Proof.
PutA=0 and B=-1 in Theorem 2. Condition (8) reduces to κ≥1, which holds in all cases. It is sufficient to establish conditions (10) and (9) or, equivalently,(33)4κ-1-c≥0,κ-12-12κ-1c≥0.The hypothesis κ-1≥|c|/2 implies that 4(κ-1)-|c|≥|c|≥0 and (κ-1)2-1/2(κ-1)|c|≥(κ-1)(κ-1-c/2)≥0.Next theorem gives the sufficient condition for close-to-convexity whenB≥3-22.Theorem 5.
Let3-22≤B<A≤1 and c,κ∈R satisfy(34)κ-1≥1+B1+A4A-Bc.Suppose A, B, κ, and c satisfy either the inequality(35)κ-12-16κ-1B1-B1+B3-κ-1A+B2A-Bc+4B1-B21+A1+B3A-Bc≥1-A21-B216A-B2c2whenever(36)κ-11+B3A+Bc+8B1-B21+Ac≥c24A-B1+B3or the inequality(37)κ-1A+B2A-Bc+4B1-B21+A1+B3A-Bc2≤c24κ-12+16κ-1B1-B1+B3-1-AB216A-B2c2whenever(38)κ-1A+B1+B3+8B1-B21+Ac<c24A-B1+B3.If (1+B)up(z)≠(1+A) for all z∈D, then up(z)∈P[A,B].Proof.
First, proceed similarly to the proof of Theorem2 and derive the expression of ReΨ(iρ,σ,μ+iν;z) as given in (17). Now, for σ≤-(1+ρ2)/2, ρ∈R, and B≥3-22, (39)21-B21-B2+1+B2ρ2σ2≥21-B21-B2+1+B2ρ21+ρ224≥8B1-B1+B3,and then, with z=x+iy∈D and μ+σ<0, it follows that (40)ReΨiρ,σ,μ+iν;z≤-12κ-11+ρ2-1+B1+Aρ28A-Bcx-ρ1-AB4A-Bcy+1-B1-A8A-Bcx-8B1-B1+B3=p2ρ2+q2ρ+r2≔Q1ρ,where (41)p2=-12κ-1-1+B1+A8A-Bcx,q2=-1-ABcy4A-B,r2=-12κ-1+1-B1-A8A-Bcx-8B1-B1+B3.
Observe that inequality (34) implies that p2<0. Thus Q1(ρ)<0 for all ρ∈R provided q22≤4p2r2; that is, for |x|,|y|<1, (42)1-AB216A-B2c2y2≤κ-1+1+B1+A4A-Bcxκ-1-1-B1-A4A-Bcx+16B1-B1+B3.With y2<1-x2, it is enough to show, for |x|<1, (43)1-AB216A-B2c21-x2≤κ-1+1+B1+A4A-Bcxκ-1-1-B1-A4A-Bcx+16B1-B1+B3,which is equivalent to(44)R1x≔m1x2+n1x+r1≥0,where (45)m1≔c216,n1≔κ-1cA+B2A-B+4B1-B21+AA-B1+B3c,r1≔κ-12-κ-116B1-B1+B3-c21-AB216A-B2.
If (36) holds, then |n1|≥2|m1| and R1(x)≥m1+r1-|n1|, which is nonnegative from (35). On the other hand, if (38) holds, then |n1|<2|m1|, R1(x)≥(4m1r1-n12)/4m1, and (37) implies R1(x)≥0. Either case establishes (44).Theorem 6.
Let-1≤B≤3-22≈0.171573. Suppose B<A≤1 and c,κ∈R with c≠0 satisfying(46)κ≥1+B1+A4A-Bc.Further let A, B, κ, and c satisfy either (47)κ2+κ1+B1-B-κA+B2A-Bc+1+B21+A41-BA-Bc≥1-A21-B216A-B2c2whenever (48)2κ1-BA+Bc+1+B21+Ac≥12A-B1-Bc2or the inequality (49)κA+B2A-B+1+B21+A41-BA-Bc2≤c24κ2+κ1+B1-B-1-AB216A-B2c2when (50)2κ1-BA+Bc+1+B21+Ac<12A-B1-Bc2.If (1+B)up(z)≠(1+A) for all z∈D, then -4κ/cup′(z)∈P[A,B].Theorem 7.
Let3-22<B<A≤1. Suppose c,κ∈R and a≠0, such that (51)κ≥1+B1+A4A-Bc.Suppose A, B, κ, and c satisfy either (52)κ2+16κB1-B1+B3-κA+B2A-Bc+4B1-B21+A1+B3A-Bc≥1-A21-B216A-B2c2whenever (53)κ1+B3A+Bc+8B1-B21+Ac≥c24A-B1+B3or the inequality (54)κA+B2A-Bc+4B1-B21+A1+B3A-B2≤c24κ2+16κB1-B1+B3-1-AB216A-B2c2when (55)2κ1+B3A+Bc+8B1-B21+Ac<c24A-B1+B3.If (1+B)up(z)≠(1+A) for all z∈D, then -4κ/cup′(z)∈P[A,B].Corollary 8.
Letc≤-1 and(56)κ≥max⁡cc+12,c2c+1.Then (-4κ/c)(up(z)-1) is close-to-convex of order (c+1)/c with respect to the identity function.Corollary 9.
Letc be a nonzero real number and κ≥|c|/2. Then(57)Re-4κcup′z>12.


## 3. Janowski Starlikeness of Generalized Bessel Functions

This section contributes to finding conditions to ensure a normalized and generalized Bessel functionzup(z) in the class of Janowski starlike functions. For this purpose, first sufficient conditions for up(z) to be Janowski convex are determined, and then an application of relation (5) yields conditions for zup(z)∈S∗[A,B].Theorem 10.
Letc,κ∈R be such that A-Bup′(z)≠1+Bzup′′(z) for all z∈D and -1≤B<A≤1. Suppose (58)κ1+B≥1+B24A-Bc-1+A-B.Further let A, B, κ, and c satisfy(59)1+A-B+κ1+B1-A+B+κ1-B≥1-B2216A-B2c2+-AB2+1+B3+B2+1-B2Bκc2A-B.
If0∉up′(D) and 0∉up′′(D), then(60)1+zup′′zup′z≺1+Az1+Bz.Proof.
Define an analytic functionp:D→C by(61)pz≔A-Bup′z+1-Bzup′′zA-Bup′z-1+Bzup′′z.Then(62)zup′′zup′z=A-Bpz-1pz+1+Bpz-1,(63)z2up′′′z+zup′′zzup′′z-zup′′zup′z=zp′zpz-1-1+Bzp′zpz+1+Bpz-1=zp′zpz+1+Bpz-1-1+Bpz-1pz-1pz+1+Bpz-1.A rearrangement of (63) yields(64)zup′′′zup′′z=2zp′zpz-1pz+1+Bpz-1-1+zup′′zup′z.Thus,(65)zup′′′zup′′zzup′′zup′z=2A-Bpz-1zp′zpz-1pz+1+Bpz-12-A-Bpz-1pz+1+Bpz-1+A-B2pz-12pz+1+Bpz-12.Now a differentiation of (4) leads to (66)4z2up′′′z+4κ+1zup′′z+czup′z=0,which gives(67)zup′′′zup′′zzup′′zup′z+κ+1zup′′zup′z+c4z=0.Using (62) and (65), (67) yields (68)2A-Bzp′zpz+1+Bpz-12+A-B2pz-12pz+1+Bpz-12+A-Bpz-1κpz+1+Bpz-1+c4z=0,and equivalently(69)zp′z+A-B2+κ1+B2+cz1+B28A-Bpz2-A-B+κB-c1-B24A-Bzpz+A-B2-κ1-B2+cz1-B28A-B=0.
Define(70)Ψpz,zp′z,z≔zp′z+F1pz2+F2pz+F3,where (71)F1=A-B2+κ1+B2+cz1+B28A-B,F2=-A-B-κB+c1-B24A-Bz,F3=A-B2-κ1-B2+cz1-B28A-B.Thus, (69) yields Ψ(p(z),zp′(z),z)∈Ω={0}. Now, with z=x+iy∈D, let (72)G1≔ReF1=A-B2+κ1+B2+cx1+B28A-B=12A-B+κ1+B+cx1+B24A-B,G2≔ReiF2=-c1-B24A-By,G3≔ReF3=A-B2-κ1-B2+cx1-B28A-B=12A-B-κ1-B+c1-B24A-Bx.
Forσ≤-(1+ρ2)/2, ρ∈R, (73)ReΨiρ,σ,z=σ-G1ρ2+G2ρ+G3≤-1+2G12ρ2+G2ρ+2G3-12≔Qρ.Note that condition (58) implies (1+2G1)/2>0. In this case, Q has a maximum at ρ=G2/(1+2G1). Thus Q(ρ)<0 for all real ρ provided (74)G22≤1+2G11-2G3,x,y<1.Since y2<1-x2, it is left to show that (75)1-B2216A-B2c21-x2≤1+A-B+κ1+B+c1+B24A-Bx1-A+B-κ-1+B-c1-B24A-Bx,|x|<1. The above inequality is equivalent to(76)Hx≔h2A,Bx+h3A,B≥0,where (77)h2A,B=-AB2+1+B3+B2+1-B2Bκc2A-B,h3A,B=1+A-B+κ1+B1-A+B-κB-1-1-B2216A-B2c2.Since |x|<1, the left-hand side of inequality (76) satisfies (78)h2A,Bx+h3A,B≥-h2A,B+h3A,B.Now it is evident from (59) that H(x)≥0 which establishes inequality (76).
ThusΨ satisfies the hypothesis of Lemma 1, and hence Rep(z)>0, or equivalently (79)A-Bup′+1-Bzup′′A-Bup′-1+Bzup′′≺1+z1-z.By definition of subordination, there exists an analytic self-map w of D with w(0)=0 and (80)A-Bup′z+1-Bzup′′zA-Bup′z-1+Bzup′′z=1+wz1-wz.A simple computation shows that (81)1+zup′′zup′z=1+Awz1+Bwz,and hence(82)1+zup′′zup′z≺1+Az1+Bz.Relation (5) also shows that (83)zzupz′zupz=1+zup-1′′zup-1′z.Together with Theorem 10, relation (83) immediately yields the following result for zup(z)∈S∗[A,B].Theorem 11.
Letc and κ be real numbers such that A-Bup-1′(z)≠1+Bzup-1′′(z) for all z∈D and -1≤B<A≤1. Suppose(84)κ1+B≥1+B24A-Bc-A-2B.Further let A, B, κ, and c satisfy (85)A-2B+κ1+B2B-A+κ1-B≥1-B2216A-Bc2+B3-A-B1+B2+1-B2Bκ2A-Bc.Then zup(z)∈S∗[A,B].



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