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eISSN 0454-8124
pISSN 1225-6951

### Articles

Kyungpook Mathematical Journal 2018; 58(3): 507-517

Published online September 30, 2018

### Certain Geometric Properties of an Integral Operator Involving Bessel Functions

Kuppathai Appasamy Selvakumaran*
Department of Mathematics, R. M. K College of Engineering and Technology, Puduvoyal-601206, Tamil Nadu, India
e-mail : selvaa1826@gmail.com

Róbert Szász
Sapientia Hungarian University of Transylvania, Tg. Mures, Romania
e-mail : szasz_robert2001@yahoo.com

Received: April 17, 2017; Revised: August 19, 2018; Accepted: August 27, 2018

### Abstract

In this article, we introduce a new integral operator involving normalized Bessel functions of the first kind and we obtain a set of sufficient conditions for univalence. Our results contain some interesting corollaries as special cases. Further, as particular cases, we improve some of the univalence conditions proved in [2].

Keywords: analytic functions, Bessel functions, integral operator, univalent functions.

### 1. Introduction and Preliminaries

Let be an open unit disk and let be the class of functions f of the form

f(z)=z+n=2anzn,

which are analytic in and satisfying the normalization condition f(0) = f′ (0) − 1 =0. Let be the subclass of consisting of all univalent functions f in . Further, by ℘ we denote the class of analytic functions p in with p(0) = 1 and Re p(z) > 0 for all .

The Bessel function of the first kind of order κ (κ ∈ ℛ) is given by

w(z)=Jκ(z)=n0(-1)nn!Γ(κ+n+1)(z2)2n+κ,         z.

We consider the normalized Bessel function of the first kind defined by

ϕκ(z)=2κΓ(κ+1)z1-κ/2Jκ(z1/2)=z+n=1(-1)nzn+14nn!(κ+1)(κ+n),

where κ ≠ −1, −2, …. It is easy to verify that ϕ1/2(z)=zsinz and ϕ-1/2(z)=zcosz. Recently, R. Szász and P. A. Kupán [16] and Á. Baricz [1] have studied the univalence of the normalized Bessel function of the first kind ϕκ(z).

Very recently several authors studied the problem of integral operators which preserve the class . For example, H. M. Srivastava et al. [10] and D. Breaz et al. [4] extended univalence conditions for a family of integral operators. L. F. Stanciu et al. [14] obtained some sufficient conditions for certain families of integral operators. H. M. Srivastava et al. [12] gave a set of sufficient conditions for the univalence, starlikeness and convexity of a certain newly-defined general family of integral operators in the open unit disk. E. Deniz et al. [5] gave sufficient conditions for certain families of integral operators, which are defined by means of the normalized Bessel functions, to be univalent in the open unit disk. For recent investigations on normalized Bessel functions of first kind and on the univalence of integral operators we refer to [3, 6, 11, 13, 17].

We now introduce an integral operator , involving normalized Bessel functions of the first kind, defined by

F(z)=Fκ1,,κn,α1,,αn,β,p1,,pm(z)={β0ztβ-1i=1n(ϕκi(t)t)1/αij=1mpj(t)dt}1/β,

where αi, β are nonzero complex numbers, κi ∈ ℛ for all i = 1, 2, …, n, pj(z) ∈ ℘ for all j = 1, 2, …, m and n, m are positive integers. We remark here that if p1(z) = p2(z) = · · · = pm(z) = 1, then we obtain the integral operator Fκ1, …, κn, α1, …, αn, β(z) defined in [2].

As remarked, the integral operator F(z) in (1.4) generalizes the integral operator defined in [2]. The objective of defining this new integral operator is to find out whether the univalence is preserved by the operator if the integrand is multiplied by a finite product of analytic functions p(z) ∈ ℘. As a consequence, we obtain some sufficient conditions for the integral operator F(z), defined by the equation (1.4), to be in the class . In this article, our aim is to improve the results of Á. Baricz and B.A. Frasin from [2] by giving sufficient conditions for the parameters of the integral operator F(z) to be univalent in the open unit disk. To prove our results we need the following lemmas.

### Lemma 1.1

([8]) Let υ be a complex number such that Re υ> 0 and let. If

1-|z|2ReυReυ|zh(z)h(z)|1

for all, then for any complex number ω with Re ωRe υ the function

Fω(z)={ω0ztω-1h(t)dt}1/ω

is in the class.

### Lemma 1.2

([9]) Let ω be a complex number with Reω > 0 and c be a complex number such that |c| ≤ 1, c ≠ −1. Ifsatisfies the inequality

|c|z|2ω+(1-|z|2ω)zh(z)ωh(z)|1

for all, then the function

Fω(z)={ω0ztω-1h(z)dt}1/ω

is in the class.

### Lemma 1.3

(Miller-Mocanu, [7]) Let w(z) = 1 + anzn + … be analytic in U with w(z) ≢ 1. If Re w(z) ≯ 0, zU, then there is a point z0U, and there are two real numbers x, y ∈ ℛ such that

• w(z0) = ix,

• z0ω(z0)=y-(x2+1)2,

• Rez02w(z0)+z0w(z0)0.

The following results are crucial facts in the proofs of our main results.

### Lemma 1.4

([15]) Let κκ* ≃ −0.7745 …, and let ϕk(z) be the normalized Bessel function of the first kind defined by (1.3). If, then we have

|zϕκ(z)ϕκ(z)-1|1-ϕκ(1)ϕκ(1)

and the result is sharp.

### Lemma 1.5

([15]) The mapping u : (−1, ∞) → ℛ defined byu(κ)=1-ϕκ(1)/ϕκ(1)is strictly decreasing.

### Remark 1.6

In [2] Á. Baricz and B.A. Frasin proved the following result: Let κ(-5+5)/4 and consider the normalized Bessel function of the first kind defined by (1.3). If then we have

|zϕκ(z)ϕκ(z)-1|κ+24κ2+10κ+5,         zU.

This result and Lemma 1.4 imply that

1-ϕκ(1)ϕκ(1)κ+24κ2+10κ+5,         provided         κ>κ*-0.7745.

### Theorem 2.1

Let δ ∈ ℛ with δ > 0, let κ1, …, κn > κ* ≃ −0.7745 … and κ = min {κ1, …, κn}. Also, let αi ∈ ℂ {0} for i = 1, …, n and let Mj ≥ 0 and pj ∈ ℘ for j = 1, …, m. If

|zpj(z)pj(z)|Mj,         j=1,,m         (zU)and         (1-ϕκ(1)ϕκ(1))i=1n1|αi|+j=1mMjδ,

then for every complex number β with Re βδ > 0, the function F(z) defined by (1.4) is univalent in.

Proof

Let us consider the function

f(z)=0zi=1n(ϕκi(t)t)1/αij=1mpj(t)dt.

Clearly , since for all i = 1, …, n and hj ∈ ℘ for all j = 1, …, m. Then we have

f(z)=i=1n(ϕκi(z)z)1/αij=1mpj(z)         andzf(z)f(z)=i=1n1αi(zϕκi(z)ϕκi(z)-1)+j=1mzpj(z)pj(z).

Using (2.1) and (2.3) along with Lemma 1.4 and Lemma 1.5 we obtain

1-|z|2δδ|zf(z)f(z)|1-|z|2δδ[i=1n1|αi||zϕκi(z)ϕκi(z)-1|+j=1m|zpj(z)pj(z)|]1-|z|2δδ[i=1n1|αi|(1-ϕκi(1)ϕκi(1))+j=1mMj]1-|z|2δδ[(1-ϕκ(1)ϕκ(1))i=1n1|αi|+j=1mMj]

and by (2.2) we have

1-|z|2δδ|zf(z)f(z)|1,         zU.

Consequently, in view of Lemma 1.1, we obtain that F(z) is in the class .

By taking m = 1 in Theorem 2.1 we obtain the following result.

### Corollary 2.2

Let δ ∈ ℛ with δ > 0, M ≥ 0, p ∈ ℘, κ1, …, κn > κ* and κ = min {κ1, …, κn}. Also, let αi ∈ ℂ {0} for i = 1, …, n. If

|zp(z)p(z)|M,         (zU),and         (1-ϕκ(1)ϕκ(1))i=1n1|αi|+Mδ,

then for every complex number β with Re βδ, the integral operator

Fκ1,,κn,α1,,αn,β,p(z)={β0ztβ-1i=1n(ϕκi(t)t)1/αip(t)dt}1/β,         (zU)

is in the class.

By choosing n = 1 in Theorem 2.1 we have the following.

### Corollary 2.3

Let δ ∈ ℛ with δ > 0, κ > κ* and let α be a nonzero complex number. Also, let Mj ≥ 0 and pj ∈ ℘ for j = 1, …, m. If

|zpj(z)pj(z)|Mj,         j=1,,m         (zU)and         (1-ϕκ(1)ϕκ(1))1α+j=1mMjδ,

then for every complex number β with Re βδ, the integral operator

Fκ,α,β,p1,,pm(z)={β0ztβ-1(ϕκ(t)t)1/αj=1mpj(t)dt}1/β,         (zU)

is in the class.

Putting pj(z) = 1, Mj = 0 for j = 1, …, m and α1=α2==αn=1α in Theorem 2.1 we have

### Corollary 2.4

Let δ ∈ ℛ with δ > 0, let κ1, …, κn > κ* and κ = min {κ1, …, κn}. If α ∈ ℂ with Re α > 0 and

(1-ϕκ(1)ϕκ(1))αδn,

then for every complex number β with Re βδ, the function

Fκ1,,κn,α,β(z)={β0ztβ-1i=1n(ϕκi(t)t)αdt}1/β,

is univalent in.

The following corollary gives a better result than Theorem 2 of [2]. Replacing β by 1 + in Corollary 2.4 we have

### Corollary 2.5

Let n ∈ ℕ κ1, …, κn > κ* and κ = min {κ1, …, κn}. If α ∈ ℕ with Re α > 0 and

(1-ϕκ(1)ϕκ(1))α1nReα,

then the function

Fκ1,,κn,α,n(z)={(nα+1)0zi=1n(ϕκi(t))αdt}1/(nα+1),

is univalent in.

We now present some simple sufficient conditions for univalence of integral operators which involve sine and cosine functions. By choosing n = 1 in Corollary 2.5 we have the following particular cases.

### Corollary 2.6

Let α ≠ 0 be complex number and let κ > κ*. If 0<(1-ϕκ(1)ϕκ(1))αRe α, then the integral operator

Fκ,α(z)={(α+1)0z(ϕκ(t))αdt}1/(α+1),         (zU)

is in the class. In particular, ifsin 1-cos12sin 1αRe α, then the integral operator

F1/2,α(z)={(α+1)0z(tsin t)αdt}1/(α+1),         (zU)

is in the class. Moreover, ifsin 12cos 1αRe α, then the integral operator

F-1/2,α(z)={(α+1)0z(tcos t)αdt}1/(α+1),         (zU)

is in the class.

By using Lemma 1.2, we now present another set of sufficient conditions for the integral operator F(z) to be in the class . This theorem improves the results of Theorem 1 given in [2].

### Theorem 2.7

Let κ1, …, κn > κ* ≃ −0.7745 …, and κ = min {κ1, …, κn}. Let αi ∈ ℂ {0} for i = 1, …, n, β ∈ ℂ with Re β > 0. If

|zpj(z)pj(z)|Mj,         Mj0,pjP,j=1,,m      (zU)and         1|β|[(1-ϕκ(1)ϕκ(1))i=1n1|αi|+j=1mMj]1

then the function F(z) defined by (1.4) is univalent in.

Proof

Consider

f(z)=0zi=1n(ϕκi(t)t)1/αij=1mpj(t)dt.

Then it follows from (2.3) that

|zf(z)f(z)|i=1n1|αi||zϕκi(z)ϕκi(z)-1|+j=1m|zpj(z)pj(z)|.

Now, using (2.7), (2.8) and (2.9) along with Lemma 1.4 and Lemma 1.5, we obtain

||z|2β+(1-|z|2β)zf(z)βf(z)||z|2β+(1-|z|2β)1|β|(i=1n1|αi||zϕκi(z)ϕκi(z)-1|+j=1m|zpj(z)pj(z)|)|z|2β+(1-|z|2β)1|β|[i=1n1|αi|(1-ϕκi(1)ϕκi(1))+j=1mMj]|z|2β+(1-|z|2β)1|β|[(1-ϕκ(1)ϕκ(1))i=1n1|αi|+j=1mMj]|z|2β+(1-|z|2β)=1.

Finally by applying Lemma 1.2 with c = 1, we conclude that F(z) ∈ .

By taking m = 1 in Theorem 2.7 we obtain the following result.

### Corollary 2.8

Let κ1, …, κn > κ* > and κ = min {κ1, …, κn}. Let αi ∈ ℂ {0} for i = 1, …, n, β ∈ ℂ with Re β > 0. If

|zp(z)p(z)|M,         M0,pP,zUand         1|β|[(1-ϕκ(1)ϕκ(1))i=1n1|αi|+M]1,

then the integral operator Fκ1, …, κn, α1, …, αn, β, p(z) defined by (2.4) is in the class.

Choosing n = 1 in Theorem 2.7 we have the following.

### Corollary 2.9

Let α be a nonzero complex number, β ∈ ℂ with Re β > 0. and let κ > κ*. If

|zpj(z)pj(z)|Mj,         Mj0,pjP,j=1,,m         (zU)and         1|β|[1α(1-ϕκ(1)ϕκ(1))+j=1nMj]1,

then the integral operator Fκ, α, β, p1,…,pm(z) defined by (2.5) is in the class.

Now by choosing n = 1 and pj(z) = 1, Mj = 0 for j = 1, …, m in Theorem 2.7 we have the following results. It is interesting to note that these results improve the conditions for univalence of the integral operators Fκ, α, β,(z), F1/2,α, β,(z) and F−1/2,α, β,(z) given in Corollary 2 of [2].

### Corollary 2.10

Let α be a nonzero complex number, β ∈ ℂ with Re β > 0. and let κ > κ*. If

(1-ϕκ(1)ϕκ(1))αβ,

then the integral operators Fκ, α, β,(z) defined by

Fκ,α,β(z)={β0ztβ-1(ϕκ(t)t)1/αdt}1/β,         (zU)

is in the class.

In particular, ifsin 1-cos 12sin 1αβ, then the integral operator

F1/2,α,β(z)={β0ztβ-1(sintt)1/αdt}1/β,         (zU)

is in the class.

Moreover, ifsin 12cos 1αβ, then the integral operator

F-1/2,α,β(z)={β0ztβ-1(cost)1/αdt}1/β,         (zU)

is in the class.

The following theorem shows that if the condition “β ∈ ℂ with Re β > 0” is replaced by a stronger one “β ∈ (0, ∞)” in Theorem 2.7, then we get a much better result.

### Theorem 2.11

Let κ1, …, κn > κ* ≃ −0.7745 …, and κ = min {κ1, …, κn}. Let αi ∈ ℂ {0} for i = 1, …, n, and β ∈ (0, ∞). If

|zpj(z)pj(z)|Mj,         Mj0,pjP,j=1,,m,         (zU),

and

1|β|[(1-ϕκ(1)ϕκ(1))i=1n1|αi|+j=1mMj]1

then the function F(z) defined by (1.4) is starlike-univalent in. Further, if β = 1, then the function F(z) is convex-univalent in.

Proof

Let φ be the function defined by ϕ(z)=zF(z)F(z). Then from (1.4) we get

ϕ(z)β+zϕ(z)ϕ(z)=β+i=1n1αi(zϕκi(z)ϕκi(z)-1)+j=1mzpj(z)pj(z).

It is easily seen that (2.10) implies

β+i=1n1αi(zϕκi(z)ϕκi(z)-1)+j=1mzpj(z)pj(z)0,   zU.

Thus (2.11), (2.12) and the minimum principle for harmonic functions imply that

ϕ(z)β+zϕ(z)ϕ(z)>0,         zU.

If the inequality Re φ(z) > 0, , does not hold, then according to the Miller-Mocanu lemma there is a point z0U and there are two real numbers x, y ∈ ℛ such that

• φ(z0) = ix

• z0ϕ(z0)=y-(x2+1)2.

Thus we get Re (ϕ(z0)β+z0ϕ(z0)ϕ(z0))=Re (ix+yix)=0 and this inequality contradicts (2.13) and consequently F is starlike-univalent in .

If β = 1, then we get

1+zF(z)F(z)=1+i=1n1αi(zϕκi(z)ϕκi(z)-1)+j=1mzpj(z)pj(z).

This equality along with (2.12) implies that 1+zF(z)F(z)>0, (z) and consequently F is convex when β = 1.

### Acknowledgement

The authors thank the referees for their valuable comments and helpful suggestions.

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