Unless otherwise mentioned, we assume throughout this paper that \(\gamma>2,\,\sigma >0\) and \(-1\le B<A\le 1.\)
Theorem 1
Let
$$\begin{aligned} \frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_\sigma ^{\gamma -2}f(z)}{\mathcal {P}_\sigma ^{\gamma -1} f(z)}- \frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma }f(z)} \right) \prec \frac{1+A z}{1+B z}\qquad (\sigma>0,\,\gamma >2) \end{aligned}$$
(15)
then we have
$$\begin{aligned} \frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma } f(z)}\prec \hbar (z) \prec \frac{1+A z}{1+B z}\qquad (z\in \Delta ) \end{aligned}$$
where
$$\begin{aligned} \hbar (z)=(1+Bz)^{-1}\left[ \,{_2}F_1\bigg (1,1;1+\sigma ;Bz/(1+Bz)\bigg ) +\frac{\sigma A z}{\sigma +1 } \,{_2}F_1\bigg (1,1;2+\sigma ;Bz/(1+Bz)\bigg ) \right] \end{aligned}$$
(16)
and \(\hbar (z)\) is the best dominant. Furthermore,
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma } f(z)} \right) >\eta \end{aligned}$$
where
$$\begin{aligned} \eta =(1-B)^{-1}\left[ \,{_2}F_1\bigg (1,1;1+\sigma ;B/(B-1)\bigg ) -\frac{\sigma A}{\sigma +1}\, {_2}F_1\bigg (1,1;2+\sigma ;B/(B-1)\bigg ) \right] . \end{aligned}$$
(17)
Proof
Suppose that
$$\begin{aligned} q(z)= \frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma } f(z)}, \end{aligned}$$
(18)
then \(q(z)=1+a_1 z+a_2 z^2+...\) is analytic in \(\Delta\) with \(q(0)=1\). Using the logarithmic differentiation of the both sides of (18) with respect to z and with the aid of the identity (5), we get
$$\begin{aligned} \left( \frac{1}{\sigma } \right) \frac{ zq'(z)}{ q(z)}= \frac{\mathcal {P}_\sigma ^{\gamma -2}f(z)}{\mathcal {P}_\sigma ^{\gamma -1} f(z)}-\frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma } f(z)} \end{aligned}$$
(19)
By using (18) and (19), we obtain
$$\begin{aligned} \frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_\sigma ^{\gamma -2}f(z)}{\mathcal {P}_\sigma ^{\gamma -1} f(z)}- \frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma }f(z)} \right) =q(z)+\left( \frac{1}{\sigma }\right) zq'(z). \end{aligned}$$
Thus, by using Lemma 1, for \(\delta =\sigma\), we have
$$\begin{aligned} \frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma } f(z)}\prec \sigma z^{-\sigma } \int _0^z s^{\sigma -1}\frac{1+A s}{1+Bs}ds =\hbar (z). \end{aligned}$$
Using the identities (7) and (8), we can written \(\hbar (z)\) as following:
$$\begin{aligned} \hbar (z)= & {} \sigma \int _0^1 t^{\sigma -1}\frac{1+A tz}{1+Btz}dt \\= & {} \sigma \bigg [ \int _0^1 t^{\sigma -1}(1+Btz)^{-1}dt+A z \int _0^1 t^{\sigma }(1+Btz)^{-1}dt\bigg ]\\= & {} (1+Bz)^{-1}\bigg [\, _2F_1\bigg (1,1;1+\sigma , Bz/(1+Bz) \bigg )+\frac{\sigma A z}{\sigma +1}\, _2F_1\bigg (1,1;2+\sigma , Bz/(1+Bz)\bigg ) \bigg ]. \end{aligned}$$
This completes the proof of (16) of Theorem 1. Now, to prove the assertion (17) of Theorem 1, it suffices to show that
$$\begin{aligned} \inf _{|z|<1}\{\hbar (z)\}=\hbar (-1). \end{aligned}$$
(20)
Indeed, for \(|z|\le r<1,\)
$$\begin{aligned} \text {Re}\bigg (\frac{1+Az}{1+Bz}\bigg )\ge \frac{1-A r}{1-Br}. \end{aligned}$$
Upon setting
$$\begin{aligned} K(t,z)=\frac{1+Atz}{1+Btz} \text { and } d\nu (t)=\sigma t^{\sigma -1} dt\qquad \qquad (0\le t\le 1,\, z\in \Delta ), \end{aligned}$$
which is a positive measure on [0, 1], we get
$$\begin{aligned} \hbar (z)=\int _0^1 K(t,z) d\nu (t), \end{aligned}$$
so that
$$\begin{aligned} \text {Re}(\hbar (z))\ge \int _0^1 \bigg (\frac{1-Atr}{1-Btr}\bigg ) d\nu (t)=\hbar (-r)\qquad \qquad (|z|\le r<1). \end{aligned}$$
Letting \(r\rightarrow 1^-\) in the above inequality, we obtain the assertion (20). The result in (17) is the best possible as the function \(\hbar (z)\) is the best dominant of (16). The proof of Theorem 1 is completed. \(\square\)
Letting \(\sigma =1\) in Theorem 1 and using the identities (11) and (12), we have
Corollary 1
Let
$$\begin{aligned} \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}^{\gamma -2}f(z)}{\mathcal {P}^{\gamma -1} f(z)}- \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma }f(z)} \right) \prec \frac{1+A z}{1+B z}\qquad (\gamma >2) \end{aligned}$$
then we have
$$\begin{aligned} \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)}\prec \hbar (z) \prec \frac{1+A z}{1+B z}\qquad (z\in \Delta ) \end{aligned}$$
where
$$\begin{aligned} \hbar (z)=\left\{ \begin{array}{lll} \bigg (1-\frac{A}{B} \bigg )\frac{\ln (1+Bz)}{Bz}+\frac{A}{B} &{} \quad &{} \quad B\ne 0 \\ &{} \quad &{} \\ 1+\frac{A}{2}z &{} \quad &{} \quad B=0, \end{array} \right. \end{aligned}$$
(21)
and \(\hbar (z)\) is the best dominant. Furthermore,
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)} \right) >\eta \end{aligned}$$
where
$$\begin{aligned} \eta =\left\{ \begin{array}{lll} \bigg (\frac{A}{B}-1 \bigg )\frac{\ln (1-B)}{B}+\frac{A}{B} &{} \quad &{} \quad B\ne 0 \\ &{} \quad &{} \\ 1- \frac{A}{2} &{} \quad &{} B=0. \end{array} \right. \end{aligned}$$
(22)
Letting \(B\ne 0\) in Corollary 1, we have
Corollary 2
If
$$\begin{aligned} \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}^{\gamma -2}f(z)}{\mathcal {P}^{\gamma -1} f(z)}- \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma }f(z)} \right) \prec \frac{1+\frac{B\ln (1-B)}{B+\ln (1-B)} z}{1+B z}\qquad (\gamma >2) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)} \right) >0\qquad \qquad (z\in \Delta ). \end{aligned}$$
Letting \(B=-1\) in Corollary 2, we obtain the following special case
Example 1
If
$$\begin{aligned} \text {Re}\bigg ( \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}^{\gamma -2}f(z)}{\mathcal {P}^{\gamma -1} f(z)}- \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma }f(z)} \right) \bigg )> \frac{2\ln 2-1}{2\ln 2-2}\thickapprox -0.61\qquad (\gamma >2) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)} \right) >0\qquad \qquad (z\in \Delta ). \end{aligned}$$
Letting \(A=1-2\lambda ,\,(0\le \lambda <1)\) and \(B=-1\) in Corollary 1, we have
Corollary 3
If
$$\begin{aligned} \text {Re}\bigg (\frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}^{\gamma -2}f(z)}{\mathcal {P}^{\gamma -1} f(z)}- \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma }f(z)} \right) \bigg )>\lambda \qquad (\gamma >2) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)} \right) >(2\lambda -1)+2(1-\lambda )\ln 2. \end{aligned}$$
Letting \(\sigma =2\) in Theorem 1 and using the identities (12) and (13), we have
Corollary 4
Let
$$\begin{aligned} \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_2^{\gamma -2}f(z)}{\mathcal {P}_2^{\gamma -1} f(z)}- \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma }f(z)} \right) \prec \frac{1+A z}{1+B z}\qquad (\gamma >2) \end{aligned}$$
then we have
$$\begin{aligned} \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)}\prec \hbar (z) \prec \frac{1+A z}{1+B z}\qquad (z\in \Delta ) \end{aligned}$$
where
$$\begin{aligned} \hbar (z)=\left\{ \begin{array}{lll} \bigg (\frac{A}{B}-1 \bigg )\bigg (\frac{2[\ln (1+Bz)-Bz]}{B^2z^2}\bigg )+\frac{A}{B} &{} \quad &{} \quad B\ne 0 \\ &{} \quad &{} \\ \frac{2}{3}Az+1 &{} \quad &{} B=0, \end{array} \right. \end{aligned}$$
(23)
and \(\hbar (z)\) is the best dominant. Furthermore,
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)} \right) >\eta \end{aligned}$$
where
$$\begin{aligned} \eta =\left\{ \begin{array}{lll} \bigg (\frac{A}{B}-1 \bigg )\frac{2[\ln (1-B)+B]}{B^2}+\frac{A}{B} &{} \quad &{} \quad B\ne 0 \\ &{} \quad &{} \\ 1-\frac{2}{3}A &{} \quad &{} B=0, \end{array} \right. \end{aligned}$$
(24)
Letting \(B\ne 0\) in Corollary 4, we have
Corollary 5
If
$$\begin{aligned} \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_2^{\gamma -2}f(z)}{\mathcal {P}_2^{\gamma -1} f(z)}- \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma }f(z)} \right) \prec \frac{1+\frac{B\ln (1-B)}{B+\ln (1-B)} z}{1+B z}\qquad (\gamma >2) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)} \right) >0 \qquad \qquad (z\in \Delta ). \end{aligned}$$
Letting \(B=-1\) in Corollary 5, we obtain the following special case
Example 2
If
$$\begin{aligned} \text {Re}\bigg ( \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_2^{\gamma -2}f(z)}{\mathcal {P}_2^{\gamma -1} f(z)}- \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma }f(z)} \right) \bigg )> \frac{4\ln 2-3}{4 \ln 2-2}\thickapprox -0.29 \qquad (\gamma >2) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)} \right) >0 \qquad \qquad (z\in \Delta ). \end{aligned}$$
Letting \(A=1-2\lambda ,\,(0\le \lambda <1)\) and \(B=-1\) in Corollary 4, we have
Corollary 6
If
$$\begin{aligned} \text {Re}\bigg ( \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_2^{\gamma -2}f(z)}{\mathcal {P}_2^{\gamma -1} f(z)}- \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma }f(z)} \right) \bigg )>\lambda \qquad (\gamma >2) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)} \right) >(2\lambda -1)-4(1-\lambda )(\ln 2-1). \end{aligned}$$
Letting \(\sigma =3\) in Theorem 1 and using the identities (13) and (14), we have
Corollary 7
Let
$$\begin{aligned} \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_3^{\gamma -2}f(z)}{\mathcal {P}_3^{\gamma -1} f(z)}- \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma }f(z)} \right) \prec \frac{1+A z}{1+B z}\qquad (\gamma >2) \end{aligned}$$
then we have
$$\begin{aligned} \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)}\prec \hbar (z) \prec \frac{1+A z}{1+B z}\qquad (z\in \Delta ) \end{aligned}$$
where
$$\begin{aligned} \hbar (z)=\left\{ \begin{array}{lll} \bigg (1-\frac{A}{B} \bigg )\frac{3}{B^3z^3}\bigg [\ln (1+Bz)-Bz+\frac{B^2 z^2}{2}\bigg ]+\frac{A}{B} &{} \quad &{} \quad B\ne 0 \\ &{} \quad &{} \\ 1+ \frac{3}{4}Az &{} \quad &{} B=0, \end{array} \right. \end{aligned}$$
(25)
and \(\hbar (z)\) is the best dominant. Furthermore,
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)} \right) >\eta \end{aligned}$$
where
$$\begin{aligned} \eta =\left\{ \begin{array}{lll} \bigg (\frac{A}{B}-1 \bigg )\frac{3}{B^3}\bigg [\ln (1-B)+B+\frac{B^2}{2}\bigg ]+\frac{A}{B} &{} \quad &{} \quad B\ne 0 \\ &{} \quad &{} \\ 1- \frac{3}{4}A &{} \quad &{} B=0. \end{array} \right. \end{aligned}$$
(26)
Letting \(B\ne 0\) in Corollary 7, we have
Corollary 8
If
$$\begin{aligned} \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_3^{\gamma -2}f(z)}{\mathcal {P}_3^{\gamma -1} f(z)}- \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma }f(z)} \right) \prec \frac{1+\frac{B\ln (1-B)}{B+\ln (1-B)} z}{1+B z}\qquad (\gamma >2) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)} \right) >0\qquad \qquad (z\in \Delta ). \end{aligned}$$
Letting \(B=-1\) in Corollary 8, we obtain the following special case
Example 3
If
$$\begin{aligned} \text {Re}\bigg (\frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_3^{\gamma -2}f(z)}{\mathcal {P}_3^{\gamma -1} f(z)}- \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma }f(z)} \right) \bigg )> \frac{12 \ln 2-19}{12\ln 2 -20}\thickapprox 0.91\qquad (\gamma >2) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)} \right) >0\qquad \qquad (z\in \Delta ). \end{aligned}$$
Letting \(A=1-2\lambda ,\,(0\le \lambda <1)\) and \(B=-1\) in Corollary 7, we have
Corollary 9
If
$$\begin{aligned} \text {Re}\bigg ( \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_3^{\gamma -2}f(z)}{\mathcal {P}_3^{\gamma -1} f(z)}- \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma }f(z)} \right) \bigg )>\lambda \qquad (\gamma >2) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)} \right) > (2\lambda -1)+3(1-\lambda )(2\ln 2-3). \end{aligned}$$
Theorem 2
Let
$$\begin{aligned} \frac{\mathcal {Q}_\sigma ^{\gamma -1}f(z)}{\mathcal {Q}_\sigma ^{\gamma } f(z)}\left( \frac{\mathcal {Q}_\sigma ^{\gamma -2}f(z)}{\mathcal {Q}_\sigma ^{\gamma -1} f(z)}-\frac{\sigma +\gamma -1}{\sigma +\gamma -2} \frac{\mathcal {Q}_\sigma ^{\gamma -1}f(z)}{\mathcal {Q}_\sigma ^{\gamma }f(z)}+\frac{\sigma +\gamma -1}{\sigma +\gamma -2} \right) \prec \frac{1+A z}{1+B z}\qquad (\sigma>0,\,\gamma >2) \end{aligned}$$
then we have
$$\begin{aligned} \frac{\mathcal {Q}_\sigma ^{\gamma -1}f(z)}{\mathcal {Q}_\sigma ^{\gamma } f(z)}\prec \hbar (z) \prec \frac{1+A z}{1+B z}\qquad (z\in \Delta ) \end{aligned}$$
where
$$\begin{aligned} \hbar (z)=(1+Bz)^{-1}\left[ \,_2F_1\bigg (1,1;\sigma +\gamma -1;Bz/(1+Bz) \bigg )+\frac{(\sigma +\gamma -2) A z}{\sigma +\gamma -1} \,_2F_1\bigg (1,1;\sigma +\gamma ;Bz/(1+Bz)\bigg ) \right] \end{aligned}$$
(27)
and \(\hbar (z)\) is the best dominant. Furthermore,
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{\gamma -1}f(z)}{\mathcal {Q}_\sigma ^{\gamma } f(z)} \right) >\eta \end{aligned}$$
where
$$\begin{aligned} \eta =(1-B)^{-1}\left[ \,_2F_1\bigg (1,1;\sigma +\gamma -1;B/(B-1)\bigg ) -\frac{(\sigma +\gamma -2) A }{\sigma +\gamma -1} \,_2F_1\bigg (1,1;\sigma +\gamma ;B/(B-1)\bigg ) \right] . \end{aligned}$$
(28)
Proof
Suppose that
$$\begin{aligned} q(z)= \frac{\mathcal {Q}_\sigma ^{\gamma -1}f(z)}{\mathcal {Q}_\sigma ^{\gamma } f(z)}, \end{aligned}$$
(29)
then \(q(z)=1+a_1 z+a_2 z^2+...\) is analytic in \(\Delta\) with \(q(0)=1\). Using the logarithmic differentiation of the both sides of (29) with respect to z, and with the aid of the identities (6), we get
$$\begin{aligned} \left( \frac{1}{\sigma +\gamma -2} \right) \frac{z q'(z)}{ q(z)}= \frac{\mathcal {Q}_\sigma ^{\gamma -2}f(z)}{\mathcal {Q}_\sigma ^{\gamma -1} f(z)}-\frac{\sigma +\gamma -1}{\sigma +\gamma -2}\frac{\mathcal {Q}_\sigma ^{\gamma -1}f(z)}{\mathcal {Q}_\sigma ^{\gamma } f(z)}+\frac{1}{\sigma +\gamma -2} \end{aligned}$$
(30)
By using (29) and (30), we obtain
$$\begin{aligned} \frac{\mathcal {Q}_\sigma ^{\gamma -1}f(z)}{\mathcal {Q}_\sigma ^{\gamma } f(z)}\left( \frac{\mathcal {Q}_\sigma ^{\gamma -2}f(z)}{\mathcal {Q}_\sigma ^{\gamma -1} f(z)}-\frac{\sigma +\gamma -1}{\sigma +\gamma -2} \frac{\mathcal {Q}_\sigma ^{\gamma -1}f(z)}{\mathcal {Q}_\sigma ^{\gamma }f(z)}+\frac{\sigma +\gamma -1}{\sigma +\gamma -2} \right) =q(z)+\left( \frac{1}{\sigma +\gamma -2}\right) zq'(z). \end{aligned}$$
Thus, by using Lemma 1, for \(\delta =\sigma +\gamma -2\), the estimates (27) and (28) can be proved on the same lines as that of (16) and (17). This completes the proof of Theorem 2. \(\square\)
Letting \(\gamma =3-\sigma\) in Theorem 2 and using the identities (11) and (12), we have
Corollary 10
Let
$$\begin{aligned} \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}\left( 2+ \frac{\mathcal {Q}_\sigma ^{1-\sigma }f(z)}{\mathcal {Q}_\sigma ^{2-\sigma } f(z)}-2 \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma }f(z)} \right) \prec \frac{1+A z}{1+B z}\qquad (0<\sigma <1) \end{aligned}$$
then we have
$$\begin{aligned} \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}\prec \hbar (z) \prec \frac{1+A z}{1+B z}\qquad (z\in \Delta ) \end{aligned}$$
where \(\hbar (z)\) given as (21) and \(\hbar (z)\) is the best dominant. Furthermore,
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)} \right) >\eta \end{aligned}$$
where \(\eta\) given as (22).
Letting \(B\ne 0\) in Corollary 10, we have
Corollary 11
If
$$\begin{aligned} \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}\left( 2+ \frac{\mathcal {Q}_\sigma ^{1-\sigma }f(z)}{\mathcal {Q}_\sigma ^{2-\sigma } f(z)}-2 \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma }f(z)} \right) \prec \frac{1+\frac{B\ln (1-B)}{B+\ln (1-B)} z}{1+B z}\qquad (0<\sigma <1) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)} \right) >0\qquad \qquad (z\in \Delta ). \end{aligned}$$
Letting \(B=-1\) in Corollary 11, we obtain the following special case
Example 4
If
$$\begin{aligned} \text {Re}\bigg ( \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}\left( 2+ \frac{\mathcal {Q}_\sigma ^{1-\sigma }f(z)}{\mathcal {Q}_\sigma ^{2-\sigma } f(z)}-2 \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma }f(z)} \right) \bigg )> \frac{2\ln 2-1}{2\ln 2-2}\thickapprox -0.61\qquad (0<\sigma <1) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)} \right) >0\qquad \qquad (z\in \Delta ). \end{aligned}$$
Letting \(A=1-2\lambda ,\,(0\le \lambda <1)\) and \(B=-1\) in Corollary 10, we have
Corollary 12
If
$$\begin{aligned} \text {Re}\bigg ( \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}\left( 2+ \frac{\mathcal {Q}_\sigma ^{1-\sigma }f(z)}{\mathcal {Q}_\sigma ^{2-\sigma } f(z)}-2 \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma }f(z)} \right) \bigg )>\lambda \qquad (0<\sigma <1) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)} \right) >(2\lambda -1)+2(1-\lambda )\ln 2. \end{aligned}$$
Letting \(\gamma =4-\sigma\) in Theorem 2 and using the identities (12) and (13), we have
Corollary 13
Let
$$\begin{aligned} \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}\left( \frac{3}{2}+ \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}-\frac{3}{2} \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma }f(z)} \right) \prec \frac{1+A z}{1+B z}\qquad (0<\sigma <2) \end{aligned}$$
then we have
$$\begin{aligned} \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}\prec \hbar (z) \prec \frac{1+A z}{1+B z}\qquad (z\in \Delta ) \end{aligned}$$
where \(\hbar (z)\) given as (23) and \(\hbar (z)\) is the best dominant. Furthermore,
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)} \right) >\eta \end{aligned}$$
where \(\eta\) given as (24).
Letting \(B\ne 0\) in Corollary 13, we have
Corollary 14
If
$$\begin{aligned} \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}\left( \frac{3}{2}+ \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}-\frac{3}{2} \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma }f(z)} \right) \prec \frac{1+\frac{B\ln (1-B)}{B+\ln (1-B)} z}{1+B z}\qquad (0<\sigma <2) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)} \right) >0 \qquad \qquad (z\in \Delta ). \end{aligned}$$
Letting \(B=-1\) in Corollary 15, we obtain the following special case
Example 5
If
$$\begin{aligned} \text {Re}\bigg ( \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}\left( \frac{3}{2}+ \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}-\frac{3}{2} \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma }f(z)} \right) \bigg )> \frac{4\ln 2-3}{4 \ln 2-2}\thickapprox -0.29 \qquad (0<\sigma <2) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)} \right) >0 \qquad \qquad (z\in \Delta ). \end{aligned}$$
Letting \(A=1-2\lambda ,\,(0\le \lambda <1)\) and \(B=-1\) in Corollary 13, we have
Corollary 15
If
$$\begin{aligned} \text {Re}\bigg ( \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}\left( \frac{3}{2}+ \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}-\frac{3}{2} \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma }f(z)} \right) \bigg )>\lambda \qquad (0<\sigma <2) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)} \right) >(2\lambda -1)-4(1-\lambda )(\ln 2-1). \end{aligned}$$
Letting \(\gamma =5-\sigma\) in Theorem 2 and using the identities (13) and (14), we have
Corollary 16
Let
$$\begin{aligned} \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)}\left( \frac{4}{3}+ \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}-\frac{4}{3} \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma }f(z)} \right) \prec \frac{1+A z}{1+B z}\qquad (0<\sigma <3) \end{aligned}$$
then we have
$$\begin{aligned} \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)}\prec \hbar (z) \prec \frac{1+A z}{1+B z}\qquad (z\in \Delta ) \end{aligned}$$
where \(\hbar (z)\) given as (25) and \(\hbar (z)\) is the best dominant. Furthermore,
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)} \right) >\eta \end{aligned}$$
where given as (26).
Letting \(B\ne 0\) in Corollary 16, we have
Corollary 17
If
$$\begin{aligned} \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)}\left( \frac{4}{3}+ \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}-\frac{4}{3} \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma }f(z)} \right) \prec \frac{1+\frac{B\ln (1-B)}{B+\ln (1-B)} z}{1+B z}\qquad (0<\sigma <3) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)} \right) >0\qquad \qquad (z\in \Delta ). \end{aligned}$$
Letting \(B=-1\) in Corollary 17, we obtain the following special case
Example 6
If
$$\begin{aligned} \text {Re} \bigg (\frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)}\left( \frac{4}{3}+ \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}-\frac{4}{3} \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma }f(z)} \right) \bigg )> \frac{12 \ln 2-19}{12\ln 2 -20}\thickapprox 0.91\qquad (0<\sigma <3) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)} \right) >0\qquad \qquad (z\in \Delta ). \end{aligned}$$
Letting \(A=1-2\lambda ,\,(0\le \lambda <1)\) and \(B=-1\) in Corollary 16, we have
Corollary 18
If
$$\begin{aligned} \text {Re} \bigg (\frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)}\left( \frac{4}{3}+ \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}-\frac{4}{3} \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma }f(z)} \right) \bigg )>\lambda \qquad (0<\sigma <3) \end{aligned}$$
then
$$\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)} \right) > (2\lambda -1)+3(1-\lambda )(2\ln 2-3). \end{aligned}$$