Our first two theorems of this section contain some interesting and applicable results involving the order of starlikeness and the order of convexity inside
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
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
, 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
. 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
and
. 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
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
(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
and
. Putting everything together, we see that
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}),
, 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
, 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
. 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
. Using the increasing property of the digamma functions, it is easy to observe that
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
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,
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
. 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
, 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,
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
, then (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z) ast z/(1+z)) is starlike in
.
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
(2.10)
where
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
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
, 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
, 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
. 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
, 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
, 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
It remains to show that (nA_{n}+(n+2)A_{n+2}geq 2(n+1)A_{n+1}) for all
. Since (A_{n+2}>0) for all
, 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
, 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
, we have
which ends the proof. □
Theorem 8
Suppose that (kappa _{1}geq 0) and
. Then (mathfrak{J}_{kappa _{1},kappa _{2}}^{kappa _{3},m}(z)ast z/(1+z)) is starlike in
.
Proof
Under the hypotheses (kappa _{1}geq 0) and
, 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
. 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
, 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
, 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
, then, (mathfrak{J}_{kappa _{1},kappa _{2}}^{ kappa _{3},m}(z)ast z/(1+z)) is convex in
.
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
. We then have
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
We shall show that ((n+1)B_{n+1}leq nB_{n}) for all
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
, 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
. Bearing in mind that
, 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
. 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
, 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),
. 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
, and according to Lemma 2, we end the proof of the theorem. □
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