# Generalized Lommel–Wright function and its geometric properties – Journal of Inequalities and Applications

#### ByHanaa M. Zayed and Khaled Mehrez

Sep 7, 2022

Our first two theorems of this section contain some interesting and applicable results involving the order of starlikeness and the order of convexity inside

$\mathbb{U}$

using some technical manipulations of the gamma and digamma functions which improve slightly the results given in [33].

### Theorem 1

Let (kappa _{1},kappa _{2}geq 0) such that (kappa _{1}+kappa _{2}geq frac{1}{2}). Also, assume that

${}_{}$
κ
3
,mN

and

$$0leq alpha leq 1- frac{3(kappa _{1}+1)^{-m}Gamma (kappa _{1}+kappa _{2}+1)e}{4Gamma (kappa _{1}+kappa _{2}+kappa _{3}+1)-3(e-1)(kappa _{1}+1)^{-m}Gamma (kappa _{1}+kappa _{2}+1)}=:widetilde{alpha}_{max},$$

(2.1)

then (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)in mathcal{S}^{ast}(alpha )).

### Proof

To prove that (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)in mathcal{S}^{ast}(alpha )) for all

$z\in \mathbb{U}$

, it is sufficient to show that

$$bigglvert frac{z(mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z))^{prime}}{mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)}-1 biggrvert < 1-alpha,$$

for

$z\in \mathbb{U}$

. Using the maximum modulus theorem of an analytic function as well as the well-known inequality (vert z_{1}+z_{2} vert leq vert z_{1} vert + vert z_{2} vert ), we get

begin{aligned} bigglvert bigl(mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z) bigr)^{ prime }- frac{mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)}{z} biggrvert & = Bigglvert sum _{n=1}^{infty} frac{(-1)^{n}n ( 2n+1 ) }{4^{n}(kappa _{1}+kappa _{2}+1)_{nkappa _{3}}[ ( kappa _{1}+1 ) _{n}]^{m}}z^{n} Biggrvert \ & = Bigglvert sum_{n=1}^{infty} frac{(-1)^{n} ( 2n^{2}+n ) }{4^{n}(kappa _{1}+kappa _{2}+1)_{nkappa _{3}}[ ( kappa _{1}+1 ) _{n}]^{m}}e^{in theta} Biggrvert \ & < sum_{n=1}^{infty} frac{n ( 2n+1 ) }{4^{n}(kappa _{1}+kappa _{2}+1)_{nkappa _{3}}[ ( kappa _{1}+1 ) _{n}]^{m}} \ & =sum_{n=1}^{infty} frac{Gamma ^{m} ( kappa _{1}+1 ) Gamma ( kappa _{1}+kappa _{2}+1 ) Gamma (2n+2)Gamma ( n+1 ) }{4^{n}Gamma ( kappa _{1}+kappa _{2}+nkappa _{3}+1 ) Gamma ^{m} ( kappa _{1}+n+1 ) Gamma (2n+1)Gamma ( n ) }, end{aligned}

for

$\theta \in \mathbb{R}$

and

$z\in \mathbb{U}$

. Using the fact that the gamma function satisfies (Gamma (z+1)=zGamma (z)), we get

$$Gamma biggl( z+frac{1}{2} biggr) = frac{1cdot 3cdot cdot cdot (2n-1)}{sqrt{pi}},$$

and so

${}_{}$

(
2
)

2
n

=
4
n

(
1
)

n

(

3

2

)

n
,nN.

Now,

begin{aligned} &bigglvert bigl(mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z) bigr)^{ prime }- frac{mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)}{z} biggrvert \ &quad < frac{Gamma ^{m} ( kappa _{1}+1 ) Gamma ( kappa _{1}+kappa _{2}+1 ) }{Gamma ( 3/2 ) } \ & qquad {}times sum_{n=1}^{infty} frac{ [ Gamma ( n+1 ) ] ^{2}Gamma (n+3/2)}{Gamma ( kappa _{1}+kappa _{2}+1+nkappa _{3} ) Gamma ^{m} ( kappa _{1}+n+1 ) Gamma (2n+1)Gamma ( n ) }. end{aligned}

(2.2)

Suppose that

$\mathrm{F}\left(n\right):=\frac{}{}$

[
Γ
(
n
+
1
)
]

2

Γ
(
n
+
3
/
2
)

Γ
(

κ
1

+

κ
2

+
n

κ
3

+
1
)

Γ
m

(

κ
1

+
n
+
1
)
Γ
(
2
n
+
1
)

,nN.

(2.3)

Differentiating (2.3) logarithmically with respect to n, we find

begin{aligned} mathrm{F}^{prime}(n) ={}&bigl[2psi (n+1)+psi (n+3/2)-kappa _{3}psi ( kappa _{1}+kappa _{2}+1+kappa _{3}n) \ &{}-mpsi (kappa _{1}+n+1)-2 psi (2n+1)bigr] mathrm{F}(n) \ ={}&bigl[2psi (n+1)-2psi (2n+1)+psi (n+3/2)-kappa _{3}psi ( kappa _{1}+ kappa _{2}+1+kappa _{3}n) \ &{}-mpsi ( kappa _{1}+n+1)bigr]mathrm{F}(n), end{aligned}

(2.4)

here, ψ stands for the digamma function defined by

$$psi (z)=frac{partial}{partial z} bigl[ log Gamma (z) bigr] = frac {Gamma ^{prime}(z)}{Gamma (z)}.$$

By using the fact that the digamma function is increasing on ((0,infty )) and (psi (z)geq 0) for all (zgeq x^{ast}), where (x^{ast}simeq 1.461632144ldots) is the abscissa of the minimum of the gamma function, and with the help of (2.4), we deduce that the sequence ({ mathrm{F}(n) } _{ngeq 1}) is decreasing. Then we get

begin{aligned} bigglvert bigl(mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z) bigr)^{ prime }- frac{mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)}{z} biggrvert & < frac{3(kappa _{1}+1)^{-m}Gamma ( kappa _{1}+kappa _{2}+1 ) }{4Gamma ( kappa _{1}+kappa _{2}+kappa _{3}+1 ) } sum_{n=1}^{infty}frac{1}{Gamma (n)} \ & = frac{3e(kappa _{1}+1)^{-m}Gamma ( kappa _{1}+kappa _{2}+1 ) }{4Gamma ( kappa _{1}+kappa _{2}+kappa _{3}+1 ) }. end{aligned}

On the other hand,

begin{aligned} bigglvert frac{mathfrak{J}_{kappa _{1},kappa _{2}}^{lambda _{3},m}(z)}{z} biggrvert ={}& Bigglvert 1+sum _{n=1}^{infty} frac {(-1)^{n}(2n+1)}{4^{n}(kappa _{1}+kappa _{2}+1)_{nkappa _{3}}[ ( kappa _{1}+1 ) _{n}]^{m}}z^{n} Biggrvert \ geq{}& 1- Bigglvert sum_{n=1}^{infty} frac{(-1)^{n}(2n+1)}{4^{n}(kappa _{1}+kappa _{2}+1)_{nkappa _{3}}[ ( kappa _{1}+1 ) _{n}]^{m}}z^{n} Biggrvert \ geq{}& 1- Bigglvert sum_{n=1}^{infty} frac{(-1)^{n}(2n+1)}{4^{n}(kappa _{1}+kappa _{2}+1)_{nkappa _{3}}[ ( kappa _{1}+1 ) _{n}]^{m}}e^{intheta} Biggrvert \ >{}&1- frac{Gamma ^{m} ( kappa _{1}+1 ) Gamma ( kappa _{1}+kappa _{2}+1 ) }{Gamma ( 3/2 ) } \ &{} times sum_{n=1}^{infty} frac{Gamma (n+3/2) [ Gamma ( n+1 ) ] ^{2}}{Gamma ( kappa _{1}+kappa _{2}+1+nkappa _{3} ) Gamma ^{m} ( kappa _{1}+n+1 ) Gamma (2n+1)Gamma ( n+1 ) } \ geq{}& 1- frac{3 ( kappa _{1}+1 ) ^{-m}Gamma ( kappa _{1}+kappa _{2}+1 ) }{4Gamma ( kappa _{1}+kappa _{2}+kappa _{3}+1 ) } sum_{n=1}^{infty} frac{1}{Gamma ( n+1 ) } \ ={}& frac{4Gamma ( kappa _{1}+kappa _{2}+kappa _{3}+1 ) -3(e-1) ( kappa _{1}+1 ) ^{-m}Gamma ( kappa _{1}+kappa _{2}+1 ) }{4Gamma ( kappa _{1}+kappa _{2}+kappa _{3}+1 ) } , end{aligned}

for

$\theta \in \mathbb{R}$

and

$z\in \mathbb{U}$

. Putting everything together, we see that

$|\frac{}{}$

z

(

J

κ
1

,

κ
2

κ
3

,
m

(
z
)
)

J

κ
1

,

κ
2

κ
3

,
m

(
z
)

1|<

3

(

κ
1

+
1
)

m

Γ
(

κ
1

+

κ
2

+
1
)
e

4
Γ
(

κ
1

+

κ
2

+

κ
3

+
1
)

3
(
e

1
)

(

κ
1

+
1
)

m

Γ
(

κ
1

+

κ
2

+
1
)

,zU,

and then we conclude that (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)in mathcal{S}^{ast}(alpha )). □

### Theorem 2

Suppose that (kappa _{1}geq 0), (kappa _{2}geq 0, kappa _{3}),

$m\in \mathbb{N}$

, and

$$0leq alpha leq 1- frac{3(kappa _{1}+1)^{-m}Gamma (kappa _{1}+kappa _{2}+1)e}{2Gamma (kappa _{1}+kappa _{2}+kappa _{3}+1)-3(e-1)(kappa _{1}+1)^{-m}Gamma (kappa _{1}+kappa _{2}+1)}=:widehat{alpha}_{max},$$

(2.5)

then (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)in mathcal{K}(alpha )).

### Proof

To prove that (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)in mathcal{K}(alpha )) for all

$z\in \mathbb{U}$

, it is sufficient to show that

$$bigglvert frac{z(mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z))^{prime prime}}{(mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z))^{prime}}-1 biggrvert < 1-alpha,$$

for

$z\in \mathbb{U}$

. As in Theorem 1, we shall base the proof on the maximum modulus theorem of an analytic function to get

begin{aligned} biglvert zbigl(mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z) bigr)^{prime prime} bigrvert ={}& Bigglvert sum _{n=1}^{ infty}frac{(-1)^{n}n(n+1) ( 2n+1 ) }{4^{n}(kappa _{1}+kappa _{2}+1)_{nkappa _{3}}[ ( kappa _{1}+1 ) _{n}]^{m}}z^{n} Biggrvert \ < {}&sum_{n=1}^{infty} frac{n(n+1) ( 2n+1 ) }{4^{n}(kappa _{1}+kappa _{2}+1)_{nkappa _{3}}[ ( kappa _{1}+1 ) _{n}]^{m}} \ ={}&sum_{n=1}^{infty} frac{n(n+1) ( 2n+1 ) Gamma ( kappa _{1}+kappa _{2}+1 ) Gamma ^{m} ( kappa _{1}+1 ) }{4^{n}Gamma ( kappa _{1}+kappa _{2}+nkappa _{3}+1 ) Gamma ^{m} ( kappa _{1}+n+1 ) } frac{(2)_{n}(2)_{2n}(1)_{n}}{(2)_{n}(2)_{2n}(1)_{n}} \ ={}& frac{Gamma ( kappa _{1}+kappa _{2}+1 ) Gamma ^{m} ( kappa _{1}+1 ) }{Gamma ( 3/2 ) } \ &{} times sum_{n=1}^{infty} frac{Gamma (n+1)Gamma ( n+2 ) Gamma ( n+3/2 ) }{Gamma ( kappa _{1}+kappa _{2}+1+nkappa _{3} ) Gamma ^{m} ( kappa _{1}+n+1 ) Gamma (2n+1)Gamma ( n ) }, end{aligned}

for

$z\in \mathbb{U}$

. Using the increasing property of the digamma functions, it is easy to observe that

$\mathrm{G}\left(n\right)=\frac{}{}$

Γ
(
n
+
1
)
Γ
(
n
+
2
)
Γ
(
n
+
3
/
2
)

Γ
(

κ
1

+

κ
2

+
1
+
n

κ
3

)

Γ
m

(

κ
1

+
n
+
1
)
Γ
(
2
n
+
1
)

,nN,

is a strictly decreasing function of n. Thus, we get

$$biglvert zbigl(mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z) bigr)^{prime prime} bigrvert < frac{3(kappa _{1}+1)^{-m}eGamma ( kappa _{1}+kappa _{2}+1 ) }{2Gamma ( kappa _{1}+kappa _{2}+kappa _{3}+1 ) }.$$

Further computations yield

begin{aligned} biglvert bigl(mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z) bigr)^{ prime } bigrvert geq{}& 1- Bigglvert sum _{n=1}^{infty} frac {(-1)^{n}(n+1)(2n+1)}{4^{n}(kappa _{1}+kappa _{2}+1)_{nkappa _{3}}[ ( kappa _{1}+1 ) _{n}]^{m}}z^{n} Biggrvert \ >{}&1-sum_{n=1}^{infty} frac{(n+1)(2n+1)}{4^{n}(kappa _{1}+kappa _{2}+1)_{nkappa _{3}}[ ( kappa _{1}+1 ) _{n}]^{m}} \ ={}&1- frac{Gamma ( kappa _{1}+kappa _{2}+1 ) Gamma ^{m} ( kappa _{1}+1 ) }{Gamma ( 3/2 ) } \ &{} times sum_{n=1}^{infty} frac{Gamma ( n+2 ) Gamma (n+3/2)Gamma ( n+1 ) }{Gamma ( kappa _{1}+kappa _{2}+1+nkappa _{3} ) Gamma ^{m} ( kappa _{1}+n+1 ) Gamma (2n+1)Gamma ( n+1 ) } \ geq{}& 1- frac{3 ( kappa _{1}+2 ) ^{-m}Gamma ( kappa _{1}+kappa _{2}+1 ) }{2Gamma ( kappa _{1}+kappa _{2}+kappa _{3}+1 ) } sum_{n=1}^{infty} frac{1}{Gamma ( n+1 ) } \ ={}& frac{2Gamma ( kappa _{1}+kappa _{2}+kappa _{3}+1 ) -3(e-1) ( kappa _{1}+1 ) ^{-m}Gamma ( kappa _{1}+kappa _{2}+1 ) }{2Gamma ( kappa _{1}+kappa _{2}+kappa _{3}+1 ) }. end{aligned}

Combining everything together to get

$|\frac{}{}$

z

(

J

κ
1

,

κ
2

κ
3

,
m

(
z
)
)

(

J

κ
1

,

κ
2

κ
3

,
m

(
z
)
)

1|<

3

(

κ
1

+
1
)

m

Γ
(

κ
1

+

κ
2

+
1
)
e

2
Γ
(

κ
1

+

κ
2

+

κ
3

+
1
)

3
(
e

1
)

(

κ
1

+
1
)

m

Γ
(

κ
1

+

κ
2

+
1
)

,zU,

and from the above inequality, we conclude that (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)in mathcal{K}(alpha )). □

### Remark 1

It is worth noting that special cases will follow if we set (kappa _{1}=0), (kappa _{3}=m=1), and (kappa _{1}=1/2), (kappa _{3}=m=1), respectively, in Theorems 1 and 2.

In the following results, that is, Theorems 3 and 4, the starlikeness and convexity with its order have been evaluated where the leading concept of the proofs comes from Lemma 1.

### Theorem 3

Assume that (kappa _{1}), (kappa _{2}), (kappa _{3}) are positive numbers,

$m\in \mathbb{N}$

such that

$$mln ( kappa _{1}+2 ) +kappa _{3}ln (kappa _{1}+ kappa _{2}+1+kappa _{3})- frac{m}{kappa _{1}+2}- frac{kappa _{3}}{kappa _{1}+kappa _{2}+1+kappa _{3}}geq frac{5}{3},$$

and

$$0leq alpha leq 1- frac{Gamma (kappa _{1}+kappa _{2}+1)}{(kappa _{1}+1)^{m}Gamma (kappa _{1}+kappa _{2}+1+kappa _{3})-Gamma (kappa _{1}+kappa _{2}+1)}=: widetilde{delta}_{max},$$

(2.6)

then (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)in mathcal{S}^{ast}(alpha )).

### Proof

From Theorem 1, we have

begin{aligned} bigglvert bigl(mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z) bigr)^{ prime }- frac{mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)}{z} biggrvert < {}& frac{Gamma ^{m} ( kappa _{1}+1 ) Gamma ( kappa _{1}+kappa _{2}+1 ) }{4} \ & {}times sum_{n=1}^{infty} frac{n ( 2n+1 ) }{4^{n-1}Gamma ^{m} ( kappa _{1}+n+1 ) Gamma ( kappa _{1}+kappa _{2}+1+nkappa _{3} ) }, end{aligned}

for

$z\in \mathbb{U}$

. Letting

$$mathrm{D}_{1}(x)= frac{x ( 2x+1 ) }{Gamma ^{m} ( kappa _{1}+x+1 ) Gamma ( kappa _{1}+kappa _{2}+1+kappa _{3}x ) },quad xgeq 1.$$

(2.7)

Hence,

$$frac{mathrm{D}_{1}^{prime}(x)}{mathrm{D}_{1}(x)}=frac{1}{x}+ frac {2}{2x+1}-mpsi (kappa _{1}+x+1)-kappa _{3}psi (kappa _{1}+ kappa _{2}+1+kappa _{3}x):=mathrm{D}_{2}(x)$$

and

$$mathrm{D}_{2}^{prime}(x)=-frac{1}{x^{2}}- frac{4}{ ( 2x+1 ) ^{2}}-mfrac{partial}{partial x}psi (kappa _{1}+x+1)-kappa _{3}^{2}frac{partial}{partial x}psi (kappa _{1}+kappa _{2}+1+kappa _{3}x).$$

From Lemma 1, we have

begin{aligned} mathrm{D}_{2}^{prime}(x) < {}&-frac{1}{x^{2}}- frac{4}{ ( 2x+1 ) ^{2}}-m biggl( frac{1}{kappa _{1}+x+1}+ frac{1}{2 ( kappa _{1}+x+1 ) ^{2}} biggr) \ &{} -kappa _{3}^{2} biggl( frac{1}{kappa _{1}+kappa _{2}+1+kappa _{3}x}+ frac{1}{2 ( kappa _{1}+kappa _{2}+1+kappa _{3}x ) ^{2}} biggr) < 0, end{aligned}

under the given hypotheses, which leads to (mathrm{D}_{2}(x)) is a strictly decreasing function on ([1,infty )) with (mathrm{D}_{2}(1)<0) to conclude that (mathrm{D}_{2}(x)<0) for all (xgeq 1). Consequently, (mathrm{D}_{1}^{prime}(x)<0) under the given hypotheses, that is, (mathrm{D}_{1}(x)) is a strictly decreasing function on ([1,infty )) and

begin{aligned} bigglvert bigl(mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z) bigr)^{ prime }- frac{mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)}{z} biggrvert & < frac{3Gamma ^{m} ( kappa _{1}+1 ) Gamma ( kappa _{1}+kappa _{2}+1 ) }{4Gamma ^{m} ( kappa _{1}+2 ) Gamma ( kappa _{1}+kappa _{2}+1+kappa _{3} ) } sum_{n=1}^{infty} frac{1}{4^{n-1}} \ & = frac{Gamma ( kappa _{1}+kappa _{2}+1 ) }{ ( kappa _{1}+1 ) ^{m}Gamma ( kappa _{1}+kappa _{2}+1+kappa _{3} ) }. end{aligned}

Similarly, we can show that

begin{aligned} bigglvert frac{mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)}{z} biggrvert & geq 1- frac{3Gamma ( kappa _{1}+kappa _{2}+1 ) }{4 ( kappa _{1}+1 ) ^{m}Gamma ( kappa _{1}+kappa _{2}+kappa _{3}+1 ) } sum_{n=1}^{infty}frac{1}{4^{n-1}} \ & = frac{ ( kappa _{1}+1 ) ^{m}Gamma ( kappa _{1}+kappa _{2}+kappa _{3}+1 ) -Gamma ( kappa _{1}+kappa _{2}+1 ) }{ ( kappa _{1}+1 ) ^{m}Gamma ( kappa _{1}+kappa _{2}+1+kappa _{3} ) }, end{aligned}

for

$z\in \mathbb{U}$

, which ultimates our proof. □

Using arguments similar to Theorem 3, we get the following result regarding the order of convexity by using (1.6) and (1.7).

### Theorem 4

Assume that (kappa _{1}), (kappa _{2}), (kappa _{3}) are positive numbers,

$m\in \mathbb{N}$

such that

$$mln ( kappa _{1}+2 ) +kappa _{3}ln (kappa _{1}+ kappa _{2}+1+kappa _{3})- frac{m}{kappa _{1}+2}- frac{kappa _{3}}{kappa _{1}+kappa _{2}+1+kappa _{3}}geq frac{13}{6},$$

and

$$0leq alpha leq 1- frac{2Gamma (kappa _{1}+kappa _{2}+1)}{(kappa _{1}+1)^{m}Gamma (kappa _{1}+kappa _{2}+1+kappa _{3})-2Gamma (kappa _{1}+kappa _{2}+1)}=:widehat{delta}_{max},$$

(2.8)

then (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)in mathcal{K}(alpha )).

In the next two theorems, we are going with other results including the order of starlikeness and the order of convexity that are evaluated using the sharp inequalities for the shifted factorial, which improve slightly the results given in [33].

### Theorem 5

Suppose that

$$0leq alpha leq 1-frac{mathrm{R}_{1}}{mathrm{R}_{2}}=: widetilde{zeta }_{max},$$

where

begin{aligned} mathrm{R}_{1} ={}&48(kappa _{1}+2)^{m}Gamma ( kappa _{1}+kappa _{2}+2)Gamma (kappa _{1}+kappa _{2}+1+2kappa _{3}) \ &{}times bigl[ ( kappa _{1}+kappa _{2}+1+gamma ) ^{kappa _{3}}(kappa _{1}+1+ gamma )^{m}-1 bigr] \ &{} +40Gamma (kappa _{1}+kappa _{2}+2)Gamma (kappa _{1}+kappa _{2}+1+ kappa _{3}) bigl[ ( kappa _{1}+kappa _{2}+1+gamma ) ^{ kappa _{3}}(kappa _{1}+1+gamma )^{m}-1 bigr] \ &{} +21(kappa _{1}+2)^{m} ( kappa _{1}+kappa _{2}+1+gamma ) ^{-kappa _{3}}(kappa _{1}+1+gamma )^{-m}prod_{ ell =1}^{2}Gamma ( kappa _{1}+kappa _{2}+1+ell kappa _{3}), end{aligned}

and

begin{aligned} mathrm{R}_{2} ={}&64(kappa _{1}+kappa _{2}+1) prod_{ell =1}^{2}Gamma (kappa _{1}+kappa _{2}+1+ell kappa _{3})prod _{ ell =1}^{2}(kappa _{1}+ell )^{m} \ & {}times bigl[ ( kappa _{1}+kappa _{2}+1+gamma ) ^{ kappa _{3}}(kappa _{1}+1+gamma )^{m}-1 bigr] -48( kappa _{1}+2)^{m}Gamma ( kappa _{1}+kappa _{2}+1+2kappa _{3}) \ &{} times Gamma (kappa _{1}+kappa _{2}+2) bigl[ ( kappa _{1}+ kappa _{2}+1+gamma ) ^{kappa _{3}}(kappa _{1}+1+gamma )^{m}-1 bigr] \ &{} -20Gamma (kappa _{1}+kappa _{2}+2)Gamma (kappa _{1}+kappa _{2}+1+ kappa _{3}) bigl[ ( kappa _{1}+kappa _{2}+1+gamma ) ^{ kappa _{3}}(kappa _{1}+1+gamma )^{m}-1 bigr] \ &{} -7(kappa _{1}+2)^{m} ( kappa _{1}+kappa _{2}+1+gamma ) ^{-kappa _{3}}(kappa _{1}+1+gamma )^{-m}prod_{ ell =1}^{2}Gamma ( kappa _{1}+kappa _{2}+1+ell kappa _{3}), end{aligned}

with (kappa _{1}>-1), (kappa _{2}geq 0), (kappa _{3}geq 1), (gamma geq max {gamma _{1},gamma _{2}}), where (gamma _{1}) and (gamma _{2}) are given in Lemma 3and (mathrm{R}_{2}>0), then (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}in mathcal{S}^{ast }(alpha )).

### Proof

To begin with, we note that if (fin mathcal{A }) satisfies (sum_{n=2}^{infty} ( n-alpha ) vert A_{n} vert leq 1-alpha ), then (fin mathcal{S}^{ast} ( alpha ) ) (see [30, Theorem 1]). Therefore, according to (1.5), it is sufficient to show that

$$mathrm{H}_{1}:=sum_{n=2}^{infty} ( n-alpha ) bigglvert frac{ ( -1 ) ^{n-1} ( 2n-1 ) }{4^{n-1}(kappa _{1}+kappa _{2}+1)_{(n-1)kappa _{3}} [ (kappa _{1}+1)_{n-1} ] ^{m}} biggrvert leq 1-alpha .$$

Since (kappa _{1}>-1), (kappa _{2}geq 0), and (kappa _{3}geq 1), we have

begin{aligned} mathrm{H}_{1} ={}&sum_{n=1}^{infty} frac{ ( 2n+1 ) ( n+1-alpha ) }{4^{n} (kappa _{1}+kappa _{2}+1)_{nkappa _{3}} [ (kappa _{1}+1)_{n} ] ^{m}} \ ={}& frac{3 ( 1-alpha ) }{4(kappa _{1}+kappa _{2}+1)_{kappa _{3}}(kappa _{1}+1)^{m}}+ frac{5 ( 3-alpha ) }{16(kappa _{1}+kappa _{2}+1)_{2kappa _{3}}(kappa _{1}+1)^{m}(kappa _{1}+2)^{m}} \ &{} +sum_{n=3}^{infty} frac{n ( 2n+1 ) }{4^{n}(kappa _{1}+kappa _{2}+1)_{nkappa _{3}} [ (kappa _{1}+1)_{n} ] ^{m}} \ &{}+ ( 1-alpha ) sum_{n=3}^{infty} frac{2n+1}{4^{n}(kappa _{1}+kappa _{2}+1)_{nkappa _{3}} [ (kappa _{1}+1)_{n} ] ^{m}}. end{aligned}

Using the fact that

$$n ( 2n+1 ) leq (21/64)cdot 4^{n},qquad 2n+1leq (7/64)cdot 4^{n}, quad ngeq 3,$$

which can be verified using the concept of mathematical induction and

$$(kappa _{1}+kappa _{2}+1) (kappa _{1}+kappa _{2}+1+gamma )^{(n-1) kappa _{3}}leq (kappa _{1}+kappa _{2}+1)_{nkappa _{3}},$$

for all (kappa _{1}>-1), (kappa _{2}geq 0), (kappa _{3}geq 1), and (gamma geq max {gamma _{1},gamma _{2}}) that follows from Lemma 3, we obtain

begin{aligned} mathrm{H}_{1} leq{}& frac{3 ( 1-alpha ) }{4(kappa _{1}+kappa _{2}+1)_{kappa _{3}}(kappa _{1}+1)^{m}}+ frac{5 ( 3-alpha ) }{16(kappa _{1}+kappa _{2}+1)_{2kappa _{3}}(kappa _{1}+1)^{m}(kappa _{1}+2)^{m}} \ &{} +frac{21}{64(kappa _{1}+kappa _{2}+1)(kappa _{1}+1)^{m}}sum_{n=3}^{infty} frac{1}{(kappa _{1}+kappa _{2}+1+gamma )^{kappa _{3}(n-1)}(kappa _{1}+1+gamma )^{m(n-1)}} \ &{} + frac{7 ( 1-alpha ) }{64(kappa _{1}+kappa _{2}+1)(kappa _{1}+1)^{m}} sum_{n=3}^{infty} frac{1}{(kappa _{1}+kappa _{2}+1+gamma )^{kappa _{3}(n-1)}(kappa _{1}+1+gamma )^{m(n-1)}} \ ={}& frac{3 ( 1-alpha ) }{4(kappa _{1}+kappa _{2}+1)_{kappa _{3}}(kappa _{1}+1)^{m}}+ frac{5 ( 3-alpha ) }{16(kappa _{1}+kappa _{2}+1)_{2kappa _{3}}(kappa _{1}+1)^{m}(kappa _{1}+2)^{m}} \ & {}+frac{21}{64(kappa _{1}+kappa _{2}+1)(kappa _{1}+1)^{m}}cdot frac {(kappa _{1}+kappa _{2}+1+gamma )^{-kappa _{3}}(kappa _{1}+1+gamma )^{-m}}{(kappa _{1}+kappa _{2}+1+gamma )^{kappa _{3}}(kappa _{1}+1+gamma )^{m}-1} \ &{} + frac{7 ( 1-alpha ) }{64(kappa _{1}+kappa _{2}+1)(kappa _{1}+1)^{m}} cdot frac{(kappa _{1}+kappa _{2}+1+gamma )^{-kappa _{3}}(kappa _{1}+1+gamma )^{-m}}{(kappa _{1}+kappa _{2}+1+gamma )^{kappa _{3}}(kappa _{1}+1+gamma )^{m}-1} \ leq{}& 1-alpha . end{aligned}

Thus, we conclude that (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}in mathcal{S}^{ast}(alpha )), as required. □

### Theorem 6

Suppose that

$$0leq alpha leq 1-frac{mathrm{T}_{1}}{mathrm{T}_{2}}=: widehat{zeta}_{max },$$

where

begin{aligned} mathrm{T}_{1} ={}&96(kappa _{1}+2)^{m}Gamma ( kappa _{1}+kappa _{2}+2)Gamma (kappa _{1}+kappa _{2}+1+2kappa _{3}) \ &{}times prod _{ ell =1}^{2} bigl[ ell ( kappa _{1}+ kappa _{2}+1+gamma ) ^{kappa _{3}}(kappa _{1}+1+gamma )^{m}-1 bigr] \ & {}+120Gamma (kappa _{1}+kappa _{2}+2)Gamma (kappa _{1}+kappa _{2}+1+kappa _{3})prod _{ell =1}^{2} bigl[ ell ( kappa _{1}+kappa _{2}+1+gamma ) ^{kappa _{3}}( kappa _{1}+1+gamma )^{m}-1 bigr] \ &{} +84(kappa _{1}+2)^{m} ( kappa _{1}+kappa _{2}+1+gamma ) ^{-kappa _{3}}(kappa _{1}+1+gamma )^{-m}prod_{ ell =1}^{2}Gamma ( kappa _{1}+kappa _{2}+1+ell kappa _{3}) \ &{} times bigl[ 2 ( kappa _{1}+kappa _{2}+1+gamma ) ^{ kappa _{3}}(kappa _{1}+1+gamma )^{m}-1 bigr], end{aligned}

and

begin{aligned} mathrm{T}_{2} ={}&64(kappa _{1}+kappa _{2}+1) prod_{ell =1}^{2}Gamma (kappa _{1}+kappa _{2}+1+ell kappa _{3})prod _{ ell =1}^{2}(kappa _{1}+ell )^{m} \ &{}times prod_{ell =1}^{2} bigl[ ell ( kappa _{1}+kappa _{2}+1+gamma ) ^{kappa _{3}}( kappa _{1}+1+gamma )^{m}-1 bigr] \ &{} -96(kappa _{1}+2)^{m}Gamma (kappa _{1}+ kappa _{2}+1+2kappa _{3})Gamma (kappa _{1}+kappa _{2}+2) \ &{}times prod_{ell =1}^{2} bigl[ ell ( kappa _{1}+kappa _{2}+1+gamma ) ^{kappa _{3}}( kappa _{1}+1+gamma )^{m}-1 bigr] \ & {}-60Gamma (kappa _{1}+kappa _{2}+1+kappa _{3})Gamma (kappa _{1}+ kappa _{2}+2)prod _{ell =1}^{2} bigl[ ell ( kappa _{1}+kappa _{2}+1+gamma ) ^{kappa _{3}}(kappa _{1}+1+ gamma )^{m}-1 bigr] \ &{} -21(kappa _{1}+2)^{m} ( kappa _{1}+kappa _{2}+1+gamma ) ^{-kappa _{3}}(kappa _{1}+1+gamma )^{-m}prod_{ ell =1}^{2}Gamma ( kappa _{1}+kappa _{2}+1+ell kappa _{3}) \ &{} times bigl[ 2 ( kappa _{1}+kappa _{2}+1+gamma ) ^{ kappa _{3}}(kappa _{1}+1+gamma )^{m}-1 bigr] \ &{} -14( kappa _{1}+2)^{m} ( kappa _{1}+kappa _{2}+1+gamma ) ^{-kappa _{3}}(kappa _{1}+1+ gamma )^{-m} \ &{} times bigl[ ( kappa _{1}+kappa _{2}+1+gamma ) ^{ kappa _{3}}(kappa _{1}+1+gamma )^{m}-1 bigr] prod _{ell =1}^{2} Gamma (kappa _{1}+ kappa _{2}+1+ell kappa _{3}), end{aligned}

with (kappa _{1}>-1), (kappa _{2}geq 0), (kappa _{3}geq 1), (gamma geq max {gamma _{1},gamma _{2}}), where (gamma _{1}) and (gamma _{2}) are given in Lemma 3and (mathrm{T}_{2}>0), then (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}in mathcal{K}(alpha )).

### Proof

Using the Alexander duality relation and according to [30, Corollary on p. 110], it suffices to show that

$$mathrm{H}_{2}:=sum_{n=2}^{infty}n ( n-alpha ) bigglvert frac{ ( -1 ) ^{n-1} ( 2n-1 ) }{4^{n-1}(kappa _{1}+kappa _{2}+1)_{(n-1)kappa _{3}} [ (kappa _{1}+1)_{n-1} ] ^{m}} biggrvert leq 1-alpha .$$

(2.9)

Since (kappa _{1}>-1), (kappa _{2}geq 0), and (kappa _{3}geq 1), we have

begin{aligned} mathrm{H}_{2} ={}& frac{6 ( 2-alpha ) }{4(kappa _{1}+kappa _{2}+1)_{kappa _{3}}(kappa _{1}+1)^{m}}+ frac{15 ( 3-alpha ) }{16(kappa _{1}+kappa _{2}+1)_{2kappa _{3}}(kappa _{1}+1)^{m}(kappa _{1}+2)^{m}} \ &{} +sum_{n=3}^{infty} frac{n^{2} ( 2n+1 ) }{4^{n}(kappa _{1}+kappa _{2}+1)_{nkappa _{3}} [ (kappa _{1}+1)_{n} ] ^{m}}+ ( 2-alpha ) sum_{n=3}^{infty} frac{n ( 2n+1 ) }{4^{n}(kappa _{1}+kappa _{2}+1)_{nkappa _{3}} [ (kappa _{1}+1)_{n} ] ^{m}} \ & {}+ ( 1-alpha ) sum_{n=3}^{infty} frac{ ( 2n+1 ) }{4^{n}(kappa _{1}+kappa _{2}+1)_{nkappa _{3}} [ (kappa _{1}+1)_{n} ] ^{m}}. end{aligned}

Recalling the fact that

begin{aligned}& n^{2} ( 2n+1 ) leq (63/64)cdot 4^{n}, qquad n(2n+1)leq (21/64) cdot 4^{n}, \& 2n+1leq (7/8)cdot 2^{n}, quad ngeq 3, end{aligned}

and Lemma 3, it follows that

begin{aligned} mathrm{H}_{2} leq{}& frac{6 ( 2-alpha ) }{4(kappa _{1}+kappa _{2}+1)_{kappa _{3}}(kappa _{1}+1)^{m}}+ frac{15 ( 3-alpha ) }{16(kappa _{1}+kappa _{2}+1)_{2kappa _{3}}(kappa _{1}+1)^{m}(kappa _{1}+2)^{m}} \ &{} +frac{63}{64(kappa _{1}+kappa _{2}+1)(kappa _{1}+1)^{m}}sum_{n=3}^{infty} frac{1}{(kappa _{1}+kappa _{2}+1+gamma )^{kappa _{3}(n-1)}(kappa _{1}+1+gamma )^{m(n-1)}} \ & {}+ frac{21 ( 2-alpha ) }{64(kappa _{1}+kappa _{2}+1)(kappa _{1}+1)^{m}} sum_{n=3}^{infty} frac{1}{(kappa _{1}+kappa _{2}+1+gamma )^{kappa _{3}(n-1)}(kappa _{1}+1+gamma )^{m(n-1)}} \ & {}+ frac{7 ( 1-alpha ) }{16(kappa _{1}+kappa _{2}+1)(kappa _{1}+1)^{m}} sum_{n=3}^{infty} frac{1}{2^{n-1}(kappa _{1}+kappa _{2}+1+gamma )^{kappa _{3}(n-1)}(kappa _{1}+1+gamma )^{m(n-1)}} \ ={}& frac{6 ( 2-alpha ) }{4(kappa _{1}+kappa _{2}+1)_{kappa _{3}}(kappa _{1}+1)^{m}}+ frac{15 ( 3-alpha ) }{16(kappa _{1}+kappa _{2}+1)_{2kappa _{3}}(kappa _{1}+1)^{m}(kappa _{1}+2)^{m}} \ & {}+frac{63}{64(kappa _{1}+kappa _{2}+1)(kappa _{1}+1)^{m}}cdot frac {(kappa _{1}+kappa _{2}+1+gamma )^{-kappa _{3}}(kappa _{1}+1+gamma )^{-m}}{(kappa _{1}+kappa _{2}+1+gamma )^{kappa _{3}}(kappa _{1}+1+gamma )^{m}-1} \ & {}+ frac{21 ( 2-alpha ) }{64(kappa _{1}+kappa _{2}+1)(kappa _{1}+1)^{m}} cdot frac{(kappa _{1}+kappa _{2}+1+gamma )^{-kappa _{3}}(kappa _{1}+1+gamma )^{-m}}{(kappa _{1}+kappa _{2}+1+gamma )^{kappa _{3}}(kappa _{1}+1+gamma )^{m}-1} \ & {}+ frac{7 ( 1-alpha ) }{32(kappa _{1}+kappa _{2}+1)(kappa _{1}+1)^{m}} cdot frac{(kappa _{1}+kappa _{2}+1+gamma )^{-kappa _{3}}(kappa _{1}+1+gamma )^{-m}}{2(kappa _{1}+kappa _{2}+1+gamma )^{kappa _{3}}(kappa _{1}+1+gamma )^{m}-1} \ leq{}& 1-alpha . end{aligned}

This proves the claim that (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}in mathcal{K}(alpha )). □

### Remark 2

1. Theorem 1, Theorem 3, and Theorem 5 assign sufficient conditions for starlikeness of (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}). As it appears in Table 1, the first one gives better result than the second and the third ones for suitable choices of the parameters.

2. Due to Theorem 2, Theorem 4, and Theorem 6, which assign sufficient conditions for convexity of (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}), it is important to observe as that Theorem 2 sometimes gives a better estimation than the others, while occasionally Theorem 4 is the best. See Table 2.

In the remainder of this section, we shall use Lemma 2 to prove (mathcal{I}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)= mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)ast z/(1+z)) is in the class of starlike and convex functions, respectively.

### Theorem 7

If (kappa _{1}geq (sqrt{13}-3)/2simeq 0.302776ldots) and

${}_{}$
κ
2
,
κ
3
,mN

, then (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z) ast z/(1+z)) is starlike in

$\mathbb{U}$

.

### Proof

From (1.5) and bearing in mind that (z/(1+z)) can be expressed as

$$frac{z}{1+z}=z+sum_{n=1}^{infty}(-1)^{n}z^{n+1},$$

we have

${}_{}^{}$
J

κ
1

,

κ
2

κ
3

,
m

(z)
z

1
+
z

=

n
=
1

A
n

z
n
,zU,

(2.10)

where

${}_{}$
A
n
=

2
n

1

4

n

1

[

(

κ
1

+
1
)

n

1

]

m

(

κ
1

+

κ
2

+
1
)

(
n

1
)

κ
3

for all nN,

Thanks to Lemma 2, we shall prove that (nA_{n}geq (n+1)A_{n+1}) and (nA_{n}-2(n+1)A_{n+1}+(n+2)A_{n+2}geq 0) for all

$n\in$
N
,

where (A_{1}=1), (A_{n}>0) for all (ngeq 2). Bearing in mind that

$$bigl(kappa _{1}+kappa _{2}+1+(n-1)kappa _{3} bigr) (kappa _{1}+ kappa _{2}+1)_{(n-1)kappa _{3}} leq (kappa _{1}+kappa _{2}+1)_{n kappa _{3}},$$

(2.11)

for (kappa _{1}geq (sqrt{13}-3)/2) and

${}_{}$
κ
2
,
κ
3
,mN

, it is easy to observe that

begin{aligned} & nA_{n}-(n+1)A_{n+1} \ &quad = frac{n(2n-1)}{4^{n-1}(kappa _{1}+kappa _{2}+1)_{(n-1)kappa _{3}}[ ( kappa _{1}+1 ) _{n-1}]^{m}}- frac{(n+1)(2n+1)}{4^{n}(kappa _{1}+kappa _{2}+1)_{nkappa _{3}}[ ( kappa _{1}+1 ) _{n}]^{m}} \ & quad geq frac{1}{4^{n}(kappa _{1}+kappa _{2}+1)_{(n-1)kappa _{3}}[ ( kappa _{1}+1 ) _{n-1}]^{m}} \ &qquad {} times biggl[ 4n(2n-1)- frac{(n+1)(2n+1) ( kappa _{1}+n ) ^{-m}}{kappa _{1}+kappa _{2}+1+(n-1)kappa _{3}} biggr] , end{aligned}

(2.12)

that is,

$$nA_{n}-(n+1)A_{n+1}geq frac{mathrm{U}(n)}{4^{n}(kappa _{1}+kappa _{2}+1)_{(n-1)kappa _{3}}[ ( kappa _{1}+1 ) _{n}]^{m} (kappa _{1}+kappa _{2}+1+(n-1)kappa _{3} )},$$

where

$$mathrm{U}(n):=4n(2n-1) bigl(kappa _{1}+kappa _{2}+1+(n-1) kappa _{3} bigr) ( kappa _{1}+n ) ^{m}-(n+1) (2n+1).$$

Since (kappa _{1}geq (sqrt{13}-3)/2) and

${}_{}$
κ
2
,
κ
3
,mN

, we have

begin{aligned} mathrm{U}(n) :={}&4n(2n-1) bigl(kappa _{1}+kappa _{2}+1+(n-1)kappa _{3} bigr) ( kappa _{1}+n ) ^{m}-(n+1) (2n+1) \ geq{}& 4n(2n-1) bigl(kappa _{1}+kappa _{2}+1+(n-1) kappa _{3} bigr) ( kappa _{1}+n ) -(n+1) (2n+1) \ ={}&8kappa _{3}n^{4}+(8kappa _{1}kappa _{3}+8kappa _{1}+8kappa _{2}+8-12 kappa _{3})n^{3} \ &{}+bigl(8kappa _{1}^{2}+4 kappa _{1}+8kappa _{1} kappa _{2}-12kappa _{1}kappa _{3}-4kappa _{2}+4kappa _{3}-6bigr)n^{2} \ &{} +bigl(-4kappa _{1}^{2}-4kappa _{1}kappa _{2}-4kappa _{1}+4kappa _{1} kappa _{3}-3bigr)n-1=mathrm{:}widetilde{mathrm{U}}(n). end{aligned}

It is worth mentioning that ((n+1)A_{n+1}leq nA_{n}) if (widetilde {mathrm{U}}(n)geq 0) for all

$n\in \mathbb{N}$

. Noting that

$$1-4kappa _{1}^{2}-8kappa _{1}-4kappa _{2}-4kappa _{1}kappa _{2} leq -1,$$

which holds for (kappa _{1}geq (sqrt{13}-3)/2) and

${}_{}$
κ
2
,
κ
3
,mN

, it follows that

begin{aligned} widetilde{mathrm{U}}(n) geq{}& 8kappa _{3}n^{4}+(8 kappa _{1} kappa _{3}+8kappa _{1}+8kappa _{2}+8-12kappa _{3})n^{3} \ &{}+bigl(8kappa _{1}^{2}+4 kappa _{1}+8kappa _{1} kappa _{2}-12kappa _{1}kappa _{3}-4kappa _{2}+4 kappa _{3}-6bigr)n^{2} \ &{} +bigl(-4kappa _{1}^{2}-4kappa _{1}kappa _{2}-4kappa _{1}+4kappa _{1} kappa _{3}-3bigr)n+1-4kappa _{1}^{2}-8kappa _{1}-4kappa _{2}-4kappa _{1} kappa _{2} \ = {}&( n-1 ) bigl[8kappa _{3}n^{3}+(8+8kappa _{1}-4 kappa _{3}+8kappa _{1}kappa _{3}+8kappa _{2})n^{2} \ &{}+bigl(2+12kappa _{1}+8 kappa _{1}^{2}-4kappa _{1}kappa _{3}+4kappa _{2}+8kappa _{1}kappa _{2}bigr)n \ &{} -1+8kappa _{1}+4kappa _{1}^{2}+4kappa _{2}+4kappa _{1}kappa _{2} bigr]. end{aligned}

Furthermore, since (kappa _{1}geq (sqrt{13}-3)/2) and

${}_{}$
κ
2
,
κ
3
,mN

, we have

$$-1+8kappa _{1}+4kappa _{1}^{2}+4kappa _{2}+4kappa _{1}kappa _{2}geq -10-20kappa _{1}-8kappa _{1}^{2}-12kappa _{2}-8kappa _{1} kappa _{2}-4kappa _{3}-4kappa _{1}kappa _{3},$$

and so

begin{aligned} widetilde{mathrm{U}}(n) geq{}& ( n-1 ) bigl[8kappa _{3}n^{3}+(8+8 kappa _{1}-4kappa _{3}+8kappa _{3}+8kappa _{2})n^{2} \ &{}+bigl(2+12 kappa _{1} +8kappa _{1}^{2}-4kappa _{1}kappa _{3}+4 kappa _{2}+8 kappa _{1}kappa _{2}bigr)n \ &{} -10-20kappa _{1}-8kappa _{1}^{2}-12kappa _{2}-8kappa _{1} kappa _{2}-4kappa _{3}-4kappa _{1}kappa _{3} bigr] \ = {}&( n-1 ) ^{2} bigl[8kappa _{3}n^{2}+(8+4 kappa _{3}+8 kappa _{1}+8kappa _{1}kappa _{3}+8kappa _{2})n \ &{}+10+20kappa _{1}+8 kappa _{1}^{2}+4kappa _{3}+4kappa _{1}kappa _{3}+12kappa _{2}+8kappa _{1} kappa _{2} bigr]. end{aligned}

Continuing in this manner we get

$\begin{array}{}\end{array}$

U
˜

(
n
)

(
n

1
)

3

[
8

κ
3

n
+
8
+
8

κ
1

+
12

κ
3

+
8

κ
1

κ
3

+
8

κ
2

]

8

κ
3

(
n

1
)

3

0
,

κ
3

,
n

N
.

It remains to show that (nA_{n}+(n+2)A_{n+2}geq 2(n+1)A_{n+1}) for all

$n\in \mathbb{N}$

. Since (A_{n+2}>0) for all

$n\in \mathbb{N}$

, we easily get

begin{aligned} & nA_{n}-2(n+1)A_{n+1}+(n+2)A_{n+2} \ &quad >nA_{n}-2(n+1)A_{n+1} \ &quad =frac{1}{4^{n-1} } biggl[ frac{n(2n-1)}{(kappa _{1}+kappa _{2}+1)_{(n-1)kappa _{3}}[ ( kappa _{1}+1 ) _{n-1}]^{m}}- frac{2(n+1)(2n+1)}{4(kappa _{1}+kappa _{2}+1)_{nkappa _{3}}[ ( kappa _{1}+1 ) _{n}]^{m}} biggr] . end{aligned}

Using (2.11), we have

begin{aligned} & nA_{n}-2(n+1)A_{n+1}+(n+2)A_{n+2} \ &quad geq frac{1}{4^{n-1}(kappa _{1}+kappa _{2}+1)_{(n-1)kappa _{3}}[ ( kappa _{1}+1 ) _{n-1}]^{m}} \ & qquad {}times biggl[ 4n(2n-1)- frac{2(n+1)(2n+1) ( kappa _{1}+n ) ^{-m}}{kappa _{1}+kappa _{2}+1+(n-1)kappa _{3}} biggr] , end{aligned}

(2.13)

and so,

begin{aligned}&nA_{n}-2(n+1)A_{n+1}+(n+2)A_{n+2} \ &quad geq frac{mathrm{V}(n)}{4^{n}(kappa _{1}+kappa _{2}+1)_{(n-1)kappa _{3}}[ ( kappa _{1}+1 ) _{n}]^{m} (kappa _{1}+kappa _{2}+1+(n-1)kappa _{3} )}, end{aligned}

where

$$mathrm{V}(n):=4n(2n-1) bigl(kappa _{1}+kappa _{2}+1+(n-1) kappa _{3} bigr) ( kappa _{1}+n ) ^{m}-2(n+1) (2n+1).$$

Since (kappa _{1}geq (sqrt{13}-3)/2) and

${}_{}$
κ
2
,
κ
3
,mN

, we have

begin{aligned} mathrm{V}(n) geq{}& 4n(2n-1) ( kappa _{1}+n ) bigl( kappa _{1}+kappa _{2}+1+(n-1)kappa _{3} bigr)-2(n+1) (2n+1) \ ={}&8kappa _{3}n^{4}+(8kappa _{1}kappa _{3}+8kappa _{1}+8kappa _{2}+8-12 kappa _{3})n^{3} \ &{}+bigl(8kappa _{1}^{2}+4 kappa _{1}+8kappa _{1} kappa _{2}-12kappa _{1}kappa _{3}-4kappa _{2}+4kappa _{3}-8bigr)n^{2} \ &{} +bigl(-4kappa _{1}^{2}-4kappa _{1}kappa _{2}-4kappa _{1}+4kappa _{1} kappa _{3}-6bigr)n-2=:widetilde{mathrm{V}}(n). end{aligned}

Again, since (kappa _{1}geq (sqrt{13}-3)/2) and

${}_{}$
κ
2
,
κ
3
,mN

, we have

$\begin{array}{}\end{array}$

V
˜

(
n
)

(
n

1
)
[
8

κ
3

n
3

+
(
8
+
8

κ
1

4

κ
3

+
8

κ
1

κ
3

+
8

κ
2

)

n
2

+

(
12

κ
1

+
8

κ

1

2

4

κ
1

κ
3

+
4

κ
2

+
8

κ
1

κ
2

)

n

6
+
8

κ
1

+
4

κ

1

2

+
4

κ
2

+
4

κ
1

κ
2

]

(
n

1
)

2

[
8

κ
3

n
2

+
(
8
+
8

κ
1

+
4

κ
3

+
8

κ
1

κ
3

+
8

κ
2

)
n
+
8
+
20

κ
1

+
8

κ

1

2

+
4

κ
3

+
4

κ
1

κ
3

+
12

κ
2

+
8

κ
1

κ
2

]

(
n

1
)

3

[
8

κ
3

n
+
8
+
8

κ
1

+
12

κ
3

+
8

κ
1

κ
3

+
8

κ
2

]

8

κ
3

(
n

1
)

4

0

for

κ
3

,
n

N
,

which ends the proof. □

### Theorem 8

Suppose that (kappa _{1}geq 0) and

${}_{}$
κ
2
,
κ
3
,mN

. Then (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)ast z/(1+z)) is starlike in

$\mathbb{U}$

.

### Proof

Under the hypotheses (kappa _{1}geq 0) and

${}_{}$
κ
2
,
κ
3
,mN

, Lemma 2 and Theorem 7, we proceed to showing that ((n+1)A_{n+1}leq nA_{n}) and (nA_{n}+(n+2)A_{n+2} geq 2(n+1)A_{n+1}) for all

$n\in \mathbb{N}$

. At first, for (n=1), (mathrm{U}(1)=4(kappa _{1}+1)^{m}(kappa _{1}+kappa _{2}+1)-6geq 4( kappa _{1}+1)(kappa _{1}+2)-6>0) for (kappa _{1}geq 0 ), whilst for (ngeq 2), we find

$$frac{(n+1)(2n+1)}{4}< n(2n-1),quad ngeq 2.$$

(2.14)

Further, it can be shown that

$$Phi (n)= frac{Gamma (kappa _{1}+kappa _{2}+1)Gamma ^{m}(kappa _{1}+1)}{Gamma (kappa _{1}+kappa _{2}+1+(n-1)kappa _{3})Gamma ^{m}(kappa _{1}+n)},$$

is a decreasing function with respect to n as follows:

$$frac{Phi ^{prime}(n)}{Phi (n)}=-mpsi (kappa _{1}+n)-kappa _{3} psi bigl(kappa _{1}+kappa _{2}+1+(n-1)kappa _{3}bigr).$$

Since (kappa _{1}geq 0) and

${}_{}$
κ
2
,
κ
3
,mN

, we get (psi (kappa _{1}+n)geq 0) and (psi (kappa _{1}+kappa _{2}+1+(n-1)kappa _{3} )geq 0), these together with (2.12) lead to (nA_{n}geq (n+1)A_{n+1}), (ngeq 2). A similar argument may be used to prove that (nA_{n}+(n+2)A_{n+2}geq 2(n+1)A_{n+1}). For (n=1,2), it is easy to prove (nA_{n}+(n+2)A_{n+2}geq 2(n+1)A_{n+1}), whereas for (ngeq 3), we have

$$frac{(n+1)(2n+1)}{2}< n(2n-1), quad ngeq 3,$$

(2.15)

and since (mathrm{V}(n)) is a decreasing function with respect to (n, ngeq 3), this would lead to (nA_{n}+(n+2)A_{n+2}geq 2(n+1)A_{n+1}) for

$n\in \mathbb{N}$

, which asserts our claim. □

### Remark 3

It is important to note that Theorem 8 extends the range of validity for parameter (kappa _{1}) to (kappa _{1}geq 0).

### Theorem 9

If

${}_{}$
κ
1
,
κ
2
,
κ
3
,mN

, then, (mathfrak{J}_{kappa _{1},kappa _{2}}^{ kappa _{3},m}(z)ast z/(1+z)) is convex in

$\mathbb{U}$

.

### Proof

Using the classical Alexander theorem between the classes of starlike and convex functions, which asserts that (f(z)in mathcal{K }) if and only if (zf^{prime}(z)in mathcal{S}^{ast}), it is sufficient to prove that (z(mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z))^{prime} ast z/(1+z)) is starlike in

$\mathbb{U}$

. We then have

$z{}^{}$

(

J

κ
1

,

κ
2

κ
3

,
m

(
z
)
)

z

1
+
z

=

n
=
1

B
n

z
n
,zU,

where

$$B_{n}= frac{n(2n-1)}{4^{n-1} [ ( kappa _{1}+1 ) _{n-1} ] ^{m}(kappa _{1}+kappa _{2}+1)_{(n-1)kappa _{3}}} quad text{for all } ngeq 1.$$

To obtain the required result, we will use Lemma 2. It suffices to show that

$\left(n+1\right){}_{}$
B

n
+
1

n
B
n
andn
B
n
+(n+2)
B

n
+
2

2(n+1)
B

n
+
1

for all nN.

We shall show that ((n+1)B_{n+1}leq nB_{n}) for all

$n\in \mathbb{N}$

as follows:

begin{aligned} & nB_{n}-(n+1)B_{n+1} \ & quad = frac{n^{2}(2n-1)}{4^{n-1}(kappa _{1}+kappa _{2}+1)_{(n-1)kappa _{3}}[ ( kappa _{1}+1 ) _{n-1}]^{m}}- frac{(n+1)^{2}(2n+1)}{4^{n}(kappa _{1}+kappa _{2}+1)_{nkappa _{3}}[ ( kappa _{1}+1 ) _{n}]^{m}} \ & quad geq frac{mathrm{W}(n)}{4^{n}(kappa _{1}+kappa _{2}+1)_{(n-1)kappa _{3}}[ ( kappa _{1}+1 ) _{n}]^{m} (kappa _{1}+kappa _{2}+1+(n-1)kappa _{3} )}, end{aligned}

where

$$mathrm{W}(n):=4n^{2}(2n-1) bigl(kappa _{1}+kappa _{2}+1+(n-1) kappa _{3} bigr) ( kappa _{1}+n ) ^{m}-(n+1)^{2}(2n+1).$$

Since

${}_{}$
κ
1
,
κ
2
,
κ
3
,mN

, we have

begin{aligned} mathrm{W}(n) geq{}& 4n^{2}(2n-1) ( kappa _{1}+n ) bigl( kappa _{1}+kappa _{2}+1+(n-1)kappa _{3} bigr)-(n+1)^{2}(2n+1) \ ={}&8kappa _{3}n^{5}+ ( 8+8kappa _{1}-12kappa _{3}+8kappa _{1} kappa _{3}+8kappa _{2} ) n^{4} \ &{}+bigl(-6+4kappa _{1}+8kappa _{1}^{2}+4 kappa _{3}-12kappa _{1} kappa _{3}-4kappa _{2}+8kappa _{1}kappa _{2}bigr)n^{3} \ &{} +bigl(-5-4kappa _{1}-4kappa _{1}^{2}+4 kappa _{1}kappa _{3}-4kappa _{1}kappa _{2}bigr)n^{2}-4n-1=widetilde{mathrm{W}}(n). end{aligned}

Obviously, (nB_{n}geq (n+1)B_{n+1}) if (widetilde{mathrm{W}}(n)geq 0) for all

$n\in \mathbb{N}$

. Bearing in mind that

${}_{}$
κ
1
,
κ
3
,
κ
2
N

, it follows that

begin{aligned} widetilde{mathrm{W}}(n) geq{}& 8kappa _{3}n^{5}+ ( 8+8 kappa _{1}-12 kappa _{3}+8kappa _{1}kappa _{3}+8kappa _{2} ) n^{4}\ &{}+bigl(-6+4 kappa _{1}+8kappa _{1}^{2}+4kappa _{3}-12kappa _{1}kappa _{3}-4kappa _{2}+8kappa _{1}kappa _{2} bigr)n^{3} \ &{} +bigl(-5-4kappa _{1}+8kappa _{1}^{2}+4 kappa _{3}-12kappa _{1} kappa _{3}-4 kappa _{2}+8kappa _{1}kappa _{2} bigr)n^{2} \ &{}-4n+7-8kappa _{1}-4 kappa _{1}^{2}-4 kappa _{2}-4kappa _{1}kappa _{2} \ geq{}& ( n-1 ) bigl[8kappa _{3}n^{4}+(8+8kappa _{1}-4 kappa _{3}+8kappa _{1}kappa _{3}+8kappa _{2})n^{3} \ &{}+bigl(2+12kappa _{1}+8 kappa _{1}^{2}-4kappa _{1}kappa _{3}+4kappa _{2}+8kappa _{1}kappa _{2}bigr)n^{2} \ & {}+ bigl( -3+8kappa _{1}+4kappa _{1}^{2}+4 kappa _{2}+4kappa _{1} kappa _{2} bigr) n-7+8 kappa _{1}+4kappa _{1}^{2}+4kappa _{2}+4 kappa _{1}kappa _{2} bigr] \ geq{}& ( n-1 ) ^{2} bigl[8kappa _{3}n^{3}+(8+8 kappa _{1}+4kappa _{3}+8kappa _{1} kappa _{3}+8kappa _{2})n^{2} \ &{}+bigl(2+12kappa _{1}+8kappa _{1}^{2}-4kappa _{1}kappa _{3}+4kappa _{2}+8kappa _{1} kappa _{2}bigr)n \ &{} +7+28kappa _{1}+12kappa _{1}^{2}+4kappa _{3}+4kappa _{1}kappa _{3}+16kappa _{2}+12kappa _{1}kappa _{2} bigr] \ geq{}& ( n-1 ) ^{3} bigl[8kappa _{3}n^{2}+(8+8 kappa _{1}+12kappa _{3}+8kappa _{1} kappa _{3}+8kappa _{2})n \ &{}+18+28kappa _{1}+8 kappa _{1}^{2}+16kappa _{3}+12kappa _{1}kappa _{3}+20kappa _{2}+8 kappa _{1}kappa _{2} bigr] \ geq{}& ( n-1 ) ^{4} [8kappa _{3}n+8+8kappa _{1}+20 kappa _{3}+8kappa _{1}kappa _{3}+8kappa _{2} ] \ geq{}& 8kappa _{3} ( n-1 ) ^{5}geq 0. end{aligned}

On the other hand, we prove that (nB_{n}+(n+2)B_{n+2}geq 2(n+1)B_{n+1}) for all

$n\in \mathbb{N}$

. We have

begin{aligned}&nB_{n}-2(n+1)B_{n+1}+(n+2)B_{n+2} \ &quad geq frac{mathrm{Y}(n)}{4^{n} [ ( kappa _{1}+1 ) _{n} ] ^{m}(kappa _{1}+kappa _{2}+1)_{(n-1)kappa _{3}}(kappa _{1}+kappa _{2}+1+(n-1)kappa _{3})}, end{aligned}

where

begin{aligned} mathrm{Y}(n) :={}&4n^{2}(2n-1) ( kappa _{1}+n ) ^{m} bigl(kappa _{1}+kappa _{2}+1+(n-1)kappa _{3} bigr)-2(n+1)^{2}(2n+1) \ geq{}& 4n^{2}(2n-1) ( kappa _{1}+n ) bigl(kappa _{1}+ kappa _{2}+1+(n-1)lambda _{3} bigr)-2(n+1)^{2}(2n+1) \ ={}&8kappa _{3}n^{5}+ ( 8+8kappa _{1}-12kappa _{3}+8kappa _{1} kappa _{3}+8kappa _{2} ) n^{4}+bigl(-8+4kappa _{1} \ &{}+8kappa _{1}^{2}+4 kappa _{3}-12kappa _{1} kappa _{3}-4kappa _{2}+8kappa _{1}kappa _{2}bigr)n^{3} \ &{} +bigl(-10-4kappa _{1}-4kappa _{1}^{2}+4 kappa _{1}kappa _{3}-4 kappa _{1}kappa _{2}bigr)n^{2}-8n-2=widetilde{mathrm{Y}}(n). end{aligned}

For

${}_{}$
κ
1
,
κ
3
,
κ
2
N

, we find

begin{aligned} widetilde{mathrm{Y}}(n) geq {}& ( n-1 ) bigl[8kappa _{3}n^{4}+(8+8 kappa _{1}-4kappa _{3}+8kappa _{1}kappa _{3}+8kappa _{2})n^{3} \ &{}+bigl(12 kappa _{1}+8kappa _{1}^{2}-4kappa _{1}kappa _{3}+4 kappa _{2}+8kappa _{1}kappa _{2}bigr)n^{2} \ &{} + bigl( -10+8kappa _{1}+4kappa _{1}^{2}+4 kappa _{2}+4kappa _{1} kappa _{2} bigr) n-18+8kappa _{1}+4kappa _{1}^{2}+4kappa _{2}+4 kappa _{1}kappa _{2} bigr] \ geq{}& ( n-1 ) ^{2} bigl[8kappa _{3}n^{3}+(8+8 kappa _{1}+4kappa _{3}+8kappa _{1} kappa _{3}+8kappa _{2})n^{2} \ &{}+bigl(8+20kappa _{1}+8kappa _{1}^{2}+4kappa _{3}+4kappa _{1}kappa _{3}+12kappa _{2}+8 kappa _{1}kappa _{2}bigr)n \ & {}-2+28kappa _{1}+12kappa _{1}^{2}+4kappa _{3}+4kappa _{1}kappa _{3}+16kappa _{2}+12kappa _{1}kappa _{2} bigr] \ geq{}& ( n-1 ) ^{3} bigl[8kappa _{3}n^{2}+(8+8 kappa _{1}+12kappa _{3}+8kappa _{1} kappa _{3}+8kappa _{2})n \ &{}+16+28kappa _{1}+8 kappa _{1}^{2}+16kappa _{3}+12kappa _{1}kappa _{3}+20kappa _{2}+8 kappa _{1}kappa _{2} bigr] \ geq{}& ( n-1 ) ^{4} [8kappa _{3}n+8+8kappa _{1}+20 kappa _{3}+8kappa _{1}kappa _{3}+8kappa _{2} ] \ geq{}& 8kappa _{3} ( n-1 ) ^{5}geq 0, end{aligned}

which completes the proof. We can propose another proof for the same result as outlined below. For (n=1), (mathrm{W}(1)=4(kappa _{1}+1)^{m}(kappa _{1}+kappa _{2}+1)-12 geq 4(kappa _{1}+1)(kappa _{1}+2)-12>0) for (kappa _{1}geq 1), whilst for (ngeq 2), we find

$$frac{(n+1)^{2}(2n+1)}{4}< n^{2}(2n-1),quad ngeq 2,$$

and since (mathrm{W}(n)) is a decreasing function with respect to (n, ngeq 2), we get (nB_{n}-(n+1)B_{n+1}geq 0),

$n\in \mathbb{N}$

. Secondly, it is easy to prove that (nB_{n}+(n+2)B_{n+2}geq 2(n+1)B_{n+1}) for (n=1,2), whereas for (ngeq 3), we have

$$frac{(n+1)^{2}(2n+1)}{2}< n^{2}(2n-1),quad ngeq 3,$$

and since (mathrm{Y}(n)) is a decreasing function with respect to (n, ngeq 3), it follows that (nB_{n}+(n+2)B_{n+2}geq 2(n+1)B_{n+1}) for

$n\in \mathbb{N}$

, and according to Lemma 2, we end the proof of the theorem. □