Introduction and preliminaries Let A be the class of functions of the from f (z

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https://doi.org/10.5831/HMJ.2019.41.1.101

GEOMETRIC PROPERTIES OF GENERALIZED DINI FUNCTIONS

Erhan Deniz^∗ and S¸eyma G¨oren

Abstract. In this paper our aim is to establish some geometric properties (like starlikeness, convexity and close-to-convexity) for the generalized and normalized Dini functions. In order to prove our main results, we use some inequalities for ratio of these functions in normalized form and classical result of Fejer.

1. Introduction and preliminaries Let A be the class of functions of the from

f (z) = z +

∞

X

n=2

anzⁿ, which are analytic in the open unit disk

U = {z : |z| < 1, z ∈ C}.

Bessel functions of the first kind play an important role in applied mathematics and other branches of engineering. Many authors who study on geometric funtions theory are interested in some geometric properties such as univalency, starlikeness, convexity and close-to-convexity of Bessel functions (see, [1, 2, 3, 4, 5, 6, 7], [10]). We denote by S the subclass of A consisting of functions f (z) ∈ A which are univalent in U . A function f ∈ S is said to be starlike of order β(0 ≤ β < 1) if and only if

< zf⁰(z) f (z)

> β.

Received May 22, 2018. Accepted January 14, 2019.

2010 Mathematics Subject Classification. 33C10, 30C45.

Key words and phrases. Bessel functions of the fist kind, convex functions, starlike functions, close-to-convex functions, Dini function.

This research has been supported by Kafkas University Scientific Research Projects Coordination Unit. Project Number 2018-FEF-15.

*Corresponding author

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We detone by S^∗(β) the class of all such functions. Also, we note that S^∗(0) = S^∗ is the usual class of starlike funtions in U . A function f ∈ S is said to be convex of order β(0 ≤ β < 1) if and only if

<

1 +zf⁰⁰(z) f⁰(z)

> β.

We detone by C(β) the class of all such functions. Also, we note that C(0) = C is the usual class of convex functions in U . A function f ∈ S , g ∈ S^∗ is said to be close-to-convex of order β(0 ≤ β < 1) if and only if

< zf⁰(z) g (z)

> β.

We detone by K (β) the class of all such functions. Also, we note that K (0) = K is the usual close-to- convex functions in U .

Let us consider the following second-order linear homogenaus differential equation (see, for details, [11]):

(1) z²ω⁰⁰(z) + bzω⁰(z) +cz²− ν²+ (1 − b)ν ω(z) = 0 (b, c, ν ∈ C) . The function ω_ν,b,c(z), which is called the generalized Bessel function of the first kind of order ν, is defined as a particular solution of (1). We have the following familiar series representation for the function ω_ν,b,c(z) (2) ω_ν,b,c(z) =

∞

X

n=0

(−c)ⁿ

n!Γ (ν + n + (b + 1) /2)

z 2

2n+ν

(z ∈ C) where Γ stands for the Euler gamma function. The series (2) permits the study of Bessel, modifiye Bessel and spherical Bessel functions in a unified manner. Each of these particular cases of the function ω_ν,b,c(z) is worthy of mention here.

For b = c = 1 in (2) , we obtain the familiar Bessel function of the first kind of order ν defined by

Jν(z) =

∞

X

n=0

(−1)ⁿ n!Γ (ν + n + 1)

z 2

2n+ν

(z ∈ C) .

For b = 1 and c = −1 in (2) , we obtain the modified Bessel function of the first kind of order ν defined by

I_ν(z) =

∞

X

n=0

1 n!Γ (ν + n + 1)

z 2

2n+ν

(z ∈ C) . For b = 2 and c = 1 in (2), the function ω_ν,b,c(z) reduces to√

2j_ν(z)√ π, where j_ν is the spherical Bessel function of the first kind of order ν

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defined by

j_ν(z) =r π 2

∞

X

n=0

(−1)ⁿ n!Γ (ν + n + 32)

z 2

2n+ν

(z ∈ C) .

Deniz et al. [6] considered the function ψ_ν,b,c(z) defined, in terms of the generalized and normalized Bessel function ων,b,c(z), by the transforma- tion

(3) ψν,b,c(z) = 2^νΓ

ν + b + 1 2

z^1−ν/2ων,b,c(√ z).

By using the well-known Pochhammer symbol (or the shifted factorial) (λ)_µdefined, for λ, µ ∈ C and in terms of the Euler Γ- function, by

(λ)_µ= Γ (λ + µ) Γ (λ) =

1

λ (λ + 1) ... (λ + n − 1)

(µ = 0; λ ∈ C − {0}) (µ = n ∈ N; λ ∈ C) . It is being understood conventionally that (0)₀ := 1, and we obtain the following series representation for the function ψν,b,c(z) given by (3)

ψν,b,c(z) = z +

∞

X

n=1

(−c)ⁿ 4ⁿ(κ)_n

zⁿ⁺¹ n!

where N := {1, 2, 3, ...}, Z⁻₀ := {0, −1, −2, ...} and κ := ν + ^b+1₂ ∈ Z/ ⁻₀ In [7] authors defined and studied the generalized Dini functions Dν,a,b,c: C → C by

D_ν,a,b,c(z) = (a − ν) ω_ν,b,c(z) + zω⁰_ν,b,c(z) (a, b, c, ν ∈ C) and its normalized form

ϕ_ν,a,b,c(z) = 2 a

ν

Γ

ν + b + 1 2

z¹⁻^ν²D_ν,a,b,c √ z

where ω_ν,b,c is the generalized Bessel function of first kind. From the definition of ϕ_ν,a,b,c(z) we obtain series representation as

ϕ_ν,a,b,c(z) = z +

∞

X

n=1

(−c)ⁿ(2n + a) a4ⁿn! (κ)_n zⁿ⁺¹. To prove our main results we need the following Lemmas.

Lemma 1.1. [8] If the function f (z) = P∞

n=1Anzⁿ, where A1 = 1 and A_n ≥ 0 for all n ≥ 2 is analytic in U , and the sequences {nA_n− (n + 1)A_n+1}_n≥1 and {nAn}_n≥1 both are decreasing, then f is starlike in U .

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Lemma 1.2. [9] If f ∈ A satisfy

f (z) z − 1

< 1 for each z ∈ U , then f is starlike in U_1/2=z : |z| < ¹₂ .

Lemma 1.3. [9] If f ∈ A satisfy |f⁰(z) − 1| < 1 for each z ∈ U , then f is convex in U_1/2 =z : |z| < ¹₂ .

Lemma 1.4. [9] If f ∈ A satisfy |f⁰(z) − 1| < ^√²

5 for each z ∈ U , then f is starlike in U = {z : |z| < 1}.

Lemma 1.5. [10] If 0 ≤ β < 1, f ∈ A satisfy |zf⁰⁰(z)| < ^1−β₄ for each z ∈ U , then <(f⁰(z)) > ^1+β₄ in U = {z : |z| < 1}.

Recently Baricz et al [4] studied the close-to-convexity of Dini functions. Some monotonicity properties and functional inequalities for the modified Dini function are discussed in [5]. Further some geometric properties of Dini functions are studied in [7]. In this paper we determine the conditions on parameters that ensure the generalized and normalized Dini function ϕ_ν,a,b,c(z) to be starlike of order β, convex of order β and close-to-convex of the given order in U and U_1/2. We also study the starlikeness by using the Fejer lemma (Lemma 1.1) which completely different method.

2. Geometric Properties of generalized Dini Functions Throughout our present investigation, we tacitly assume that κ :=

ν + ^b+1₂ .

Theorem 2.1. Let a ∈ R⁺, b ∈ R, c ∈ C, κ ∈ R − {Z⁻∪ {0}} with β ∈ [0, 1) and z ∈ U . If

κ >

∆a,c,β−q

∆²_a,c,β− 8a(β −1) |c| (−8(2 + a)(β −2) + (16 + a − 2β) |c|) 16a(β − 1)

then ϕ_v,a,b,c(z) ∈ S^∗(β), where

∆a,c,β = 4(β − 2) |c| + a(8 − 5 |c| + β(3 |c| − 8)).

Proof. We use the inequality

zϕ⁰_ν,a,b,c(z) ϕν,a,b,c(z) − 1

< 1 − β to prove the starlikeness of order β for the function ϕv,a,b,c(z). So by using the well- known triangle inequality

|z₁+ z₂| ≤ |z₁| + |z₂|

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and following inequalities

(4) 8(n − 1)!(κ + 1)_n−1≥ n [2(κ + 1)]ⁿ⁻¹,

2(n − 1)!(κ + 1)n−1≥ [2(κ + 1)]ⁿ⁻¹, n ∈ N we obtain

ϕ⁰_ν,a,b,c(z)−ϕ_ν,a,b,c(z) z

=

∞

X

n=1

n(2n + a) (−c)ⁿ a4ⁿn! (κ)_n zⁿ

≤

∞

X

n=1

n(2n + a) |c|ⁿ a4ⁿn! (κ)_n

= |c|

2κa

∞

X

n=1

n

4ⁿ⁻¹(n − 1)! (κ + 1)_n−1|c|ⁿ⁻¹ + |c|

4κ

∞

X

n=1

1

4ⁿ⁻¹(n − 1)! (κ + 1)_n−1|c|ⁿ⁻¹

≤ |c|

2κa+|c|

4κ+ 4 |c|

κa + |c|

2κ

^∞ X

n=2

|c|

8 (κ + 1)

n−1

=|c|²[14 + a] + |c| 8 (κ + 1) (2 + a) 4aκ (8 (κ + 1) − |c|) . Furthermore, if we use reverse triangle inequality

|z₁+ z₂| ≥ ||z₁| − |z₂||

and the inequalities (4) then we get

ϕ_ν,a,b,c(z) z

≥ 1 −

∞

X

n=1

(2n + a) (−c)ⁿ a4ⁿn! (κ)_n zⁿ

= 1 −

"

2 |c|

4κa +2 |c|

4κa

∞

X

n=2

|c|ⁿ⁻¹

4ⁿ⁻¹(n − 1)! (κ + 1)_n−1

#

−

"

|c|

4κ+ |c|

4κ

∞

X

n=2

|c|ⁿ⁻¹ 4ⁿ⁻¹n! (κ + 1)_n−1

#

≥ 1 −

"

|c|

2κa + |c|

4κ+ |c|

κa + |c|

4κ

^∞ X

n=2

|c|

8 (κ + 1)

n−1#

= −2 |c|²− |c| 4 (aκ + 2 (a + 2) (κ + 1)) + 32aκ (κ + 1) 4aκ (8 (κ + 1) − |c|) .

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By combining the above inequalities, we obtain

zϕ⁰_ν,a,b,c(z) ϕ_ν,a,b,c(z) − 1

≤ |c|²[14 + a] + |c| 8 (κ + 1) (2 + a)

−2 |c|²− |c| 4 (aκ + 2 (a+2) (κ+1)) + 32aκ (κ + 1). Consequently, by the hypothesis of theorem we have

|c|²[14 + a] + |c| 8 (κ + 1) (2 + a)

−2 |c|²− |c| 4 (aκ + 2 (a + 2) (κ + 1)) + 32aκ (κ + 1) < 1 − β that is,

zϕ⁰_ν,a,b,c(z) ϕν,a,b,c(z) − 1

< 1 − β.

So, ϕ_v,a,b,c(z) is starlike function of order β.

κ > 20 |c| + a−8√

5 + 10 +√

5 |c| + p5Ψ_a,c,β 16√

5a then ϕv,a,b,c(z) ∈ S^∗, where

Ψ_a,c,β = 80 |c|²+ 8a |c|

h 8√

5 + 10 |c| + 9√ 5 |c|

i +a²

h

64 + |c|

−16 + 32√

5 + 21 |c| + 8

√ 5 |c|

i . Proof. To prove this teorem, we use Lemma 1.4. From the well-known triangle inequality and the inequalities (4) we have

ϕ⁰_ν,a,b,c(z) − 1 ≤

∞

X

n=1

2n²+ n (a + 2) + a a4ⁿn! (κ)_n |c|ⁿ

≤

∞

X

n=1

2n

a4ⁿ(n − 1)! (κ)_n|c|ⁿ +a + 2

a

∞

X

n=1

1

4ⁿ(n − 1)! (κ)_n|c|ⁿ+

∞

X

n=1

1

4ⁿn! (κ)_n|c|ⁿ

≤ |c|

2κa+(a + 2) |c|

4κa + |c|

4κ + 4 |c|

κa +(a + 2) |c|

2κa + |c|

4κ

^∞ X

n=2

|c|

8 (κ + 1)

n−1

= (16 + a) |c|²+ 16 (a + 2) (κ + 1) |c|

4aκ (8 (κ + 1) − |c|) .

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Therefore from hypothesis of theorem we obtain

ϕ⁰_ν,a,b,c(z) − 1

≤ (16 + a) |c|²+ 16 (a + 2) (κ + 1) |c|

4aκ (8 (κ + 1) − |c|) < 2

√ 5. So, from Lemma 1.4 ϕv,a,b,c(z) is starlike function.

κ >

Θ_a,c,β−q

Θ²_a,c,β+ Π_a,c,β+ Γ_a,c,β 16a(β − 1)

then ϕ_v,a,b,c(z) ∈ C(β), where

Θ_a,c,β = 8(β − 2) |c| + a(6 − 7 |c| + β(5 |c| − 8)), Π_a,c,β = 8a(β − 1)(4 |c| (−16 − 15 |c| + 4β(|c| + 2)),

Γa,c,β = a(−8 + |c| (−23 + 6 |c| + β(16 + |c|))).

To prove the convexity of order β of function ϕv,a,b,c(z), we have to show that

zϕ⁰⁰_ν,a,b,c(z) ϕ⁰_ν,a,b,c(z)

< 1 − β. By using the well-known triangle inequality

|z₁+ z2| ≤ |z₁| + |z₂|

and the inequalities (4) and 32(n − 1)!(κ + 1)_n−1≥ n²[2(κ + 1)]ⁿ⁻¹ for n ∈ N, we have

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z²ϕ⁰⁰_ν,a,b,c(z) ≤

∞

X

n=1

(2n + a) (n + 1) a4ⁿ(n − 1)! (κ)_n|c|ⁿ

≤ |c|

2aκ

∞

X

n=1

n²

4ⁿ⁻¹(n − 1)! (κ + 1)_n−1 |c|ⁿ⁻¹ +(a + 2) |c|

4aκ

∞

X

n=1

n

4ⁿ⁻¹(n − 1)! (κ + 1)_n−1 |c|ⁿ⁻¹ +|c|

4κ

∞

X

n=1

1

4ⁿ⁻¹(n − 1)! (κ + 1)_n−1 |c|ⁿ⁻¹

(8)

≤ |c|

2κa +16 |c|

κa

∞

X

n=2

|c|

8 (κ + 1)

n−1

+ a + 2 a

|c|

4κ+ a + 2 a

2 |c|

κ

∞

X

n=2

|c|

8 (κ + 1)

n−1

+|c|

4κ+ |c|

2κ

∞

X

n=2

|c|

8 (κ + 1)

n−1

= (44 − 7a) |c|²+ (8 (κ + 1) (a + 4) − a) |c| + 8a (κ + 1)

4aκ (8 (κ + 1) − |c|) .

Moreover, if we use reverse triangle inequality

|z₁+ z2| ≥ ||z₁| − |z₂||

and the inequalities (4), we have

zϕ⁰_ν,a,b,c(z) (6)

≥ 1 −

∞

X

n=1

2n²+ n (a + 2) + a a4ⁿn! (κ)_n |c|ⁿ

≥ 1 −

"_∞ X

n=1

2n

a4ⁿ(n − 1)! (κ)_n|c|ⁿ+

∞

X

n=1

a + 2

a4ⁿ(n − 1)! (κ)_n|c|ⁿ

#

−

∞

X

n=1

a

a4ⁿn! (κ)_n|c|ⁿ

≥ 1 −

"

|c|

2κa +4 |c|

κa

∞

X

n=2

|c|

8 (κ + 1)

n−1#

−

"

a + 2 a

|c|

4κ+ a + 2 a

|c|

2κ

∞

X

n=2

|c|

8 (κ + 1)

n−1#

−|c|

4κ− |c|

4κ

∞

X

n=2

|c|

8 (κ + 1)

n−1

= − (16 + a) |c|²− (4aκ + 8 (κ + 1) (2a + 4)) |c| + 32aκ (κ + 1)

4aκ (8 (κ + 1) − |c|) .

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Consequently, from (5), (6) and by the hypothesis of teorem we get

zϕ⁰⁰_ν,a,b,c(z) ϕ⁰_ν,a,b,c(z)

≤ (44 − 7a) |c|²+ (8 (κ + 1) (a + 4) − a) |c| + 8a (κ + 1)

−(16+a) |c|²−(4aκ+8 (κ+1)(2a+4)) |c| + 32aκ (κ+1)

< 1 − β

κ > 4 |c| + a(3 |c| − 8) + q

64a²+ 16a(a + 4) |c| + (a + 4)(9a + 4) |c|² 16a

then ϕv,a,b,c(z) is starlike in U_1/2.

Proof. By using the well-known triangle inequality and the second inequality in (4) we get

ϕ_ν,a,b,c(z)

z − 1

≤

∞

X

n=1

(2n + a) (−c)ⁿ a4ⁿn! (κ)_n zⁿ

≤ |c|

2κa + |c|

2κa

∞

X

n=2

|c|ⁿ⁻¹

4ⁿ⁻¹(n − 1)! (κ + 1)_n−1 +|c|

4κ+ |c|

4κ

∞

X

n=2

|c|ⁿ⁻¹ 4ⁿ⁻¹n! (κ + 1)_n−1

≤ |c|

2κa + |c|

4κ+ |c|

κa + |c|

4κ

∞

X

n=2

|c|

8 (κ + 1)

n−1

= 2 |c|²+ 8(κ + 1)(2 + a) |c|

4aκ(8(κ + 1) − |c|) . Since the hypothesis of theorem we obtain

2 |c|²+ 8(κ + 1)(2 + a) |c|

4aκ(8(κ + 1) − |c|) < 1.

Consequently, from Lemma 1.2 ϕ_v,a,b,c(z) is starlike in U_1/2.

κ > a(5 |c|−8)+8 |c|+

q

64a²+16a(3a+8) |c| + [64+a(33a+208)] |c|² 16a

then ϕ_v,a,b,c(z) is convex in U_1/2.

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Proof. By using the well-known triangle inequality and the inequalities (4) for n ∈ N, we have

ϕ⁰_ν,a,b,c(z) − 1 ≤

∞

X

n=1

2n²+ n (a + 2) + a a4ⁿn! (κ)_n |c|ⁿ

= 2 a

∞

X

n=1

n

4ⁿ(n − 1)! (κ)_n|c|ⁿ +a + 2

a

∞

X

n=1

1

4ⁿ(n − 1)! (κ)_n|c|ⁿ+

∞

X

n=1

1

4ⁿn! (κ)_n|c|ⁿ

≤ |c|

2κa +4 |c|

κa

∞

X

n=2

|c|

8 (κ + 1)

n−1

+ a + 2 a

|c|

4κ+ a + 2 a

|c|

2κ

∞

X

n=2

|c|

8 (κ + 1)

n−1

+|c|

4κ+ |c|

4κ

∞

X

n=2

|c|

8 (κ + 1)

n−1

= (16 + a) |c|²+ 16 (a + 2) (κ + 1) |c|

4aκ (8 (κ + 1) − |c|) . Thus, from the hypothesis of theorem we obtain

(16 + a) |c|²+ 16 (a + 2) (κ + 1) |c|

4aκ (8 (κ + 1) − |c|) < 1 and so we have required result by using Lemma 1.3.

κ >

$_a,c,β −q

$²_a,c,β+ ξ_a,c,β+ κa,c,β

16a(β − 1)

then <ϕ⁰_v,a,b,c(z) > ^1+β₄ or ϕv,a,b,c(z) is close-to-convexity of order ^1+β₄ where,

$_a,c,β = 8(β − 5) |c| + a [β(5 |c| − 8) − 13 |c|] , ξ_a,c,β = 8a(β − 1)(16 |c| [β(2 + |c| − 2(5 + 6 |c|)]), κa,c,β = a(−32 + |c| [−44 + 27 |c| + β(16 + |c|)] .

(11)

Proof. By using the well-known triangle inequality and the inequalities (4) and 32(n − 1)!(κ + 1)n−1≥ n²[2(κ + 1)]ⁿ⁻¹, we have

zϕ⁰⁰_ν,a,b,c(z) ≤

∞

X

n=1

(2n + a) (n + 1) a4ⁿ(n − 1)! (κ)_n|c|ⁿ

= |c|

2aκ

∞

X

n=1

n²

4ⁿ⁻¹(n − 1)! (κ + 1)_n−1|c|ⁿ⁻¹ +(a + 2) |c|

4aκ

∞

X

n=1

n

4ⁿ⁻¹(n − 1)! (κ + 1)_n−1|c|ⁿ⁻¹ +|c|

4κ

∞

X

n=1

1

4ⁿ⁻¹(n − 1)! (κ + 1)_n−1|c|ⁿ⁻¹

≤ |c|

2κa +16 |c|

κa

∞

X

n=2

|c|

8 (κ + 1)

n−1

+ a + 2 a

|c|

4κ+ a + 2 a

2 |c|

κ

∞

X

n=2

|c|

8 (κ + 1)

n−1

+|c|

4κ+ |c|

2κ

∞

X

n=2

|c|

8 (κ + 1)

n−1

= (44 − 7a) |c|²+ (8 (κ + 1) (a + 4) − a) |c| + 8a (κ + 1)

4aκ (8 (κ + 1) − |c|) .

Thus, from the hypothesis of theorem we obtain

(44 − 7a) |c|²+ (8 (κ + 1) (a + 4) − a) |c| + 8a (κ + 1)

4aκ (8 (κ + 1) − |c|) < 1 − β 4 . Consequently by using Lemma 1.5, we have required result.

Theorem 2.7. Let a ∈ R⁺, b ∈ R and c < −0, 117874. If κ > N (c), then ϕν,a,b,c∈ S^∗, where N (c) =√

143c²+ 152c + 64 − 12c − 8.

From the definition of the function ϕ_ν,a,b,c we write ϕ_ν,a,b,c(z) =

∞

X

n=1

A_nzⁿ where

A_n= [2 (n − 1) + a] (−c)ⁿ⁻¹

a4ⁿ⁻¹(n − 1)! (κ)_n−1 , A₁ = 1.

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Since c < −0, 117874 we have An> 0 for all n ≥ 1. Let An= Fn+ Gn, where

Fn= 2 (−c)ⁿ⁻¹

a4ⁿ⁻¹(n − 2)! (κ)_n−1 > 0 and Gn= (−c)ⁿ⁻¹

4ⁿ⁻¹(n − 1)! (κ)_n−1 > 0.

To prove the theorem we use Lemma 1.1. Namely we must prove that the sequences {nA_n}_n≥1 and {nA_n− (n + 1)A_n+1}_n≥1 are decreasing.

First we prove that the sequence {nA_n}_n≥1 is decreasing. For this we must show that the sequences {nFn}_n≥2 and {nGn}_n≥1 is also decreasing.

Firstly, we show that the sequence {nFn}_n≥2 is decreasing. From the definition of F_n, we have

nF_n−(n + 1)F_n+1 = 2n (−c)ⁿ⁻¹

a4ⁿ⁻¹(n − 2)! (κ)_n−1 − 2 (n + 1) (−c)ⁿ a4ⁿ(n − 1)! (κ)_n

= 2 (−c)ⁿ⁻¹ a4ⁿ⁻¹(n − 2)! (κ)_n−1

n − (n + 1) (−c) 4 (n − 1) (κ + n − 1)

= Fn

4n (n − 1) (κ + n − 1) + (n + 1) c 4 (n − 1) (κ + n − 1)

.

Let U1(n) = 4n (n − 1) (κ + n − 1) + (n + 1) c. Since c is negative, κ >

N (c) > ^−3c₈ − 1 > ^−c₄ − 1 and (n − 1)² ≥ 2(n − 1) − 1 for n ≥ 2, we obtain

U₁(n) = nh

4 (n − 1)²+ 4 (n − 1) κ + ci + c

≥ n [8 (n − 1) − 4 + 4 (n − 1) κ + c] + c

≥ n [4 (κ + 1) + c] + c ≥ 8 (κ + 1) + 3c > 0.

Consequently we have shown that nFn− (n + 1)F_n+1 > 0. This shows that the sequence {nF_n}_n≥2 is strictly decreasing.

Next, we show that the sequence {nG_n}_n≥1 is decreasing. From the definition of Gn, we have

nGn− (n + 1)G_n+1 = n (−c)ⁿ⁻¹

4ⁿ⁻¹(n − 1)! (κ)_n−1 −(n + 1) (−c)ⁿ 4ⁿ(n)! (κ)_n

= (−c)ⁿ⁻¹

4ⁿ⁻¹(n − 1)! (κ)_n−1

n + (n + 1) c 4n (κ + n − 1)

= Gn

4n³+ n²(4κ − 4) + nc + c 4n (κ + n − 1)

.

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Let V1(n) = 4n³ + n²(4κ − 4) + nc + c. Since n³ ≥ 3n² − 3n + 1, n² ≥ 2n − 1 for n ≥ 1 and κ > N (c) > −₂^c > −^c+4₈ , we have

V1(n) = 4n³+ n²(4κ − 4) + nc + c

≥ 4 3n²− 3n + 1 + n²(4κ − 4) + nc + c

= n²(4κ + 8) + n (c − 12) + c + 4

≥ n (c + 4 + 8κ) + c − 4κ − 4 > 0.

Consequently we have nG_n − (n + 1)G_n+1 > 0. This shows that the sequence {nGn}_n≥1 is strictly decreasing. Therefore the sequence {nA_n}_n≥1 is strictly decreasing.

Second, we prove that the sequence {nA_n−(n+1)A_n+1}_n≥1 is decreasing. For this we must show that the sequences {nFn−(n+1)F_n+1}_n≥2 and {nG_n− (n + 1)G_n+1}_n≥1 are also decreasing.

Now, we show that the sequence {nF_n− (n + 1)F_n+1}_n≥2is decreasing. For convenience we denote K_n= nF_n− (n + 1)F_n+1 for each n ≥ 1.

Thus we get

K_n− K_n+1= nF_n− 2(n + 1)F_n+1+ (n + 2)F_n+2

= 2 (−c)ⁿ⁻¹ a4ⁿ⁻¹(n − 2)! (κ)_n−1

×

n+ (n + 1) c

2 (n−1) (κ+n−1)+ (n + 2) c²

16n (n−1) (κ+n) (κ+n−1)

= F_n

U₂(n)

16n (n − 1) (κ + n) (κ + n − 1)

where

U2(n) = 16n⁵+ 16n⁴(2κ − 2) + 16n³ κ²− 3κ + 1 + c/2 +16n² κ − κ²+ (κ + 1) c/2 + n 8cκ + c² + 2c².

Our aim is to show that U₂(n) > 0. First we observe that n⁴ ≥ 4n³− 6n²+ 4n − 1 holds. By using this inequality we obtain

U2(n) ≥ n





16 4n³− 6n²+ 4n − 1 + 16n³(2κ − 2) +16n² κ²− 3κ + 1 + c/2

+16n κ − κ²+ (κ + 1) c/2 + 8cκ + c²



+ 2c²

= n

16n³(2κ + 2) + 16n² κ²− 3κ + 5 + c/2

+16n 4 + κ − κ²+ (κ + 1) c/2 + 8cκ + c²− 16

+2c² : = X₁(n).

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Clearly, the coefficient of n³in the above expression is nonnegative, since κ > N (c) > −1. Therefore using that n³ ≥ 3n²− 3n + 1, we obtain X₁(n) ≥ n





16 3n²− 3n + 1 (2κ + 2) +16n² κ²− 3κ + 5 + c/2

+16n 4 + κ − κ²+ (κ + 1) c/2 + 8cκ + c²− 16



+2c²

= n





16n² κ²+ 3κ + 1 + c/2

+16n −2 − 5κ − κ²+ (κ + 1) c/2 + 32κ + 8cκ + c²+ 16



+ 2c² : = X₂(n).

Similarly the coefficient κ²+ 3κ + 1 + c/2 of n² in the above expression is nonnegative, since κ > N (c). Therefore using that n² ≥ 2n − 1, we obtain

X₂(n) ≥ n





16 (2n − 1) κ²+ 3κ + 1 + c/2 +16n −2 − 5κ − κ²+ (κ + 1) c/2 + 32κ + 8cκ + c²+ 16



+ 2c²

= n

16n κ²+ κ + c + (κ + 1) c/2 + −16κ²− 16κ − 8c + 8cκ + c²

+ 2c² := X3(n).

Now, we observe that κ²+κ+c+(κ + 1) c/2 is also nonnegative, because κ > N (c). Therefore for n ≥ 2, we have

X₃(n) ≥ n16κ²+ 16κ + 40c + 24cκ + c² + 2c²:= X₄(n).

Since κ > N (c), we have 16κ²+ 16κ + 40c + 24cκ + c²> 0 . Thus we obtain for n ≥ 2

X4(n) ≥ 32κ²+ 32κ + 80c + 48cκ + 4c² := X5(n) > 0.

Thus, we have proved a chain of inequalities

U₂(n) ≥ X₁(n) ≥ X₂(n) ≥ X₃(n) ≥ X₄(n) ≥ X₅(n) > 0,

which implies Kn−K_n+1> 0. Consequently this shows that the sequence {K_n}_n≥2 is strictly decreasing.

Next, we show that the sequence {nG_n− (n + 1)G_n+1}_n≥1is decreasing.

For convenience we denote P_n= nG_n− (n + 1)G_n+1for n ≥ 1. Therefore we find that

Pn− P_n+1 = nGn− 2 (n + 1) G_n+1+ (n + 2) Gn+2

= Gn

V2(n)

16n (n + 1) (κ + n − 1) (κ + n)

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where V₂(n) = n

16n⁴+ 32κn³+ n² 16κ²+ 16κ − 16 + 8c +n 16κ²− 16κ + 16c + 8κc + 16κc + 8c + c²

+8cκ+2c². Our aim is to show that V2(n) > 0. First we observe that n⁴ ≥ 4n³− 6n²+ 4n − 1 holds. By using this inequality we obtain

V₂(n) ≥ = n





32n³(κ + 2) + n² 16κ²+ 16κ − 112 + 8c +n 16κ²− 16κ + 16c + 8κc + 64

+16κc + 8c + c²− 16



+8cκ+2c² : = Y₁(n).

Clearly, the coefficient of n³in κ+2 the above expression is nonnegative, since κ > N (c). Therefore using that n³ ≥ 3n²− 3n + 1, we obtain

Y₁(n) ≥ n





n² 16κ²+ 112κ + 80 + 8c

+n 16κ²− 112κ + 16c + 8κc − 128 +16κc + 8c + 32κ + c²+ 48



+ 8cκ + 2c² : = Y₂(n).

Similarly the coefficient of n² in the above expression is nonnegative, since κ > N (c). Therefore using that n² ≥ 2n − 1, we obtain

Y2(n) ≥ n

n 48κ²+ 112κ + 32c + 8κc + 32

−16κ²− 80κ + 16κc + c²− 32

+ 8cκ + 2c² := Y3(n).

Now, we observe that 48κ²+ 112κ + 32c + 8κc + 32 is also nonnegative, because κ > N (c). Therefore for n ≥ 1 we can deduce easily that

Y3(n) ≥ n 32 κ²+ κc + c + 1 + c² + 8κc + c²:=Y4(n).

Since κ > N (c) we have 32 κ²+ κc + c + 1 + c²> 0 for n ≥ 1. Thus we get

Y₄(n) ≥ 32 κ²+ κc + c + 1 + 3c² := Y₅(n) > 0.

Consequently we have proved a chain of inequalities

V₂(n) ≥ Y₁(n) ≥ Y₂(n) ≥ Y₃(n) ≥ Y₄(n) ≥ Y₅(n) > 0.

This implies that the sequence {P_n}_n≥1 is strictly decreasing.

Since the sequences {Kn} and {P_n} are strictly decreasing, {nA_n− (n + 1)A_n+1}_n≥1

is also strictly decreasing.

Consequently using Lemma 1.1 ϕ_ν,a,b,c∈ S^∗.

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[3] ´A. Baricz, M. C¸ a˘glar and E. Deniz, Starlikeness of Bessel functions and their derivatives, Math. Ineq. Appl. 19(2) (2016), 439–449.

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[8] L. Fejer, Untersuchungen ¨uber Potenzreihen mit mehrfach monotoner Koeffizien- tenfolg e, Acta Litterarum ac Scientiarum 8 (1936), 89–115.

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Erhan Deniz

Department of Mathematics, Kafkas University, Kars 36100, Turkey.

E-mail: edeniz36@gmail.com S¸eyma G¨oren

Department of Mathematics, Kafkas University, Kars 36100, Turkey.

E-mail: seyma mat36@hotmail.com