We assume throughout this paper unless otherwise mentioned that\(\ p\in \mathbb {N},\ 0\leq \lambda <1,\;\mu < p,\;\eta >\max \{\lambda,\mu \}-p-1,\ -1\leq B<A\leq 1,\ 0\leq \zeta < p-q,\ \xi <1,\ \sigma >0,\ 0< c\leq 1\) and the powers are considered principal ones.
Theorem 1
Assume that \(1\leq q\leq p\ {and}\ f\left (z\right)\in \mathcal {A}\left (p\right) \) satisfy
$$(1-\sigma)\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}+\sigma \frac{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\prec \sqrt{1+cz}, $$
then
$$ \frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\prec Q(z)\prec \sqrt{1+cz}, $$
(15)
where
$$ Q(z)={\left(1+cz\right)^{\frac{1}{2}}}_{\,2}F_{1}\left(-\frac{1} {2},1;\frac{p-\mu}{\sigma}+1;\frac{cz}{1+cz}\right), $$
(16)
is the best dominant of (15). Furthermore,
$$ \mathfrak{R}\left\{ \frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\right\} >M, $$
(17)
where
$$M={\left(1-c\right)^{\frac{1}{2}}}_{\,2}F_{1}\left(-\frac{1}{2},1;\frac{p-\mu}{\sigma}+1;\frac{c}{c-1}\right). $$
The estimate in (17) is the best possible.
Proof
Putting
$$ \phi(z)=\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\ (z\in \mathbb{U}), $$
(18)
then ϕ(z) is analytic in \(\mathbb {U}\). After some computations, we get
$$(1-\sigma)\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}+\sigma \frac{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}} $$
$$=\phi(z)+\left(\frac{\sigma}{p-\mu}\right) z\phi^{\prime}(z)\prec \sqrt{1+cz}. $$
where the influence of \(h(z) = \sqrt {1+cz}\) under certain values of c is illustrated by Fig. 1. To apply Lemma 1, it suffies to show that h(z) is convex, therefore for z=reiθ, r∈(0,1), θ∈[−π,π], we have
$$1+\frac{zh^{\prime\prime}}{h^{\prime}}=1-\frac{cz}{2\left(1+cz\right) }=\frac{2+cz}{2\left(1+cz\right) }, $$
and
$$\begin{array}{*{20}l} \mathfrak{R}\left(1+\frac{zh^{\prime\prime}}{h^{\prime}}\right) & =\frac{2+3cr\cos\theta+c^{2}r^{2}}{\left\vert 1+cre^{i\theta}\right\vert^{2}}\geq\frac{2-3cr+c^{2}r^{2}}{\left\vert 1+cre^{i\theta}\right\vert^{2}}\\ & =\frac{\left(2-cr\right) \left(1-cr\right) }{\left\vert 1+cre^{i\theta}\right\vert^{2}}>0. \end{array} $$
This implies that h is convex in \(\mathbb {U}\).
Now, by using Lemma 1 (with n=1) and making a change of variables followed by the use of (4) and (5), we deduce that
$$\begin{array}{*{20}l} \frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}} & \prec Q(z)=\frac {p-\mu}{\sigma}z^{-\frac{p-\mu}{\sigma}}\int_{0}^{z}t^{\frac{p-\mu}{\sigma} -1}\left(1+ct\right)^{\frac{1}{2}}dt\\ & ={\left(1+cz\right)^{\frac{1}{2}}}_{\,2}F_{1}\left(-\frac{1}{2},1;\frac{p-\mu}{\sigma}+1;\frac{cz}{1+cz}\right), \end{array} $$
this proves (15). Next, it is enough to show that
$$\inf_{\left\vert z\right\vert <1}\left\{ \mathfrak{R}(Q(z)\right\} =Q(-1). $$
Indeed
$$\mathfrak{R}\left\{ \sqrt{1+cz}\right\} \geqslant\sqrt{1-cr}\ \left(\left\vert z\right\vert \leq r<1\right). $$
Setting
$$G(z,s)=\sqrt{1+czs}\;\text{and\ }d\nu(s)=\frac{p-\mu}{\sigma}s^{\frac{p-\mu }{\sigma}-1}ds\;(0\leq s\leq1), $$
which is a positive measure on the closed interval [0,1], we get
$$Q(z)=\int\limits_{0}^{1}G(z,s)d\nu(s), $$
so that
$$\mathfrak{R}\left\{ Q(z)\right\} \geqslant\int_{0}^{1}\sqrt{1-cr}d\nu(s)=Q(-r)\;(\left\vert z\right\vert \leq r<1). $$
Letting r→1− in the above inequality, we obtain (17). To show that the result in (17) is sharp, let us suppose that
$$\mathfrak{R}\left\{ \frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\right\} >M_{1}, $$
that is
$$\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\prec\frac{1+\left(1-2M_{1}\right) z}{1-z}. $$
From (15), we have
$$\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\prec\frac{1+\left(1-2M\right) z}{1-z}, $$
and so
$$\frac{1+\left(1-2M\right) z}{1-z}\prec\frac{1+\left(1-2M_{1}\right) z}{1-z}, $$
which implies that M≤M1, that is, M cannot be decreased and the estimate in (17) is the best possible. □
For \(f\in \mathcal {A}(p)\) the generalized Bernardi-Libera-Livingston integeral operator Fp,υ is defined by (see [18]):
$$\begin{array}{*{20}l} F_{p,\upsilon}f(z) & =\frac{\upsilon+p}{z^{p}}\int\limits_{0}^{z} t^{\upsilon-1}f(t)dt\\ & =\left(z^{p}+\sum\limits_{n=1}^{\infty}\frac{\upsilon+p}{\upsilon +p+n}z^{p+n}\right) \ast f(z)\\ & ={z^{p}}_{\,3}F_{2}\left(1,1,\upsilon+p;1,\upsilon+p+1;z\right) \ast f(z)\ (\upsilon>-p). \end{array} $$
(19)
Lemma 4
If \(f\in \mathcal {A}\left (p\right),\ \)then (i) \(\mathcal {M}_{0,z}^{\lambda,\mu,\eta,p}\left (F_{p,\upsilon }f\right) =F_{p,\upsilon }\left (\mathcal {M}_{0,z}^{\lambda,\mu,\eta,p}f\right),\)
(ii)
$$ z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\right)^{\prime}=(p+\upsilon)\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)-\upsilon \mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z), $$
(20)
(iii)
$$ {}z\left(\! \mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\!\right)^{\left(q\right) }\!=(p+\upsilon)\left(\! \mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\!\right)^{\left(q-1\right) }-\left(\upsilon+q-1\right) \left(\! \mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\!\right)^{\left(q-1\right) }. $$
(21)
Proof
Since \(f(z)\in \mathcal {A}(p)\), then
$$\begin{array}{*{20}l} {}\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}\left(F_{p,\upsilon}f\right) &\,=\,\left[ {z^{p}}_{3}F_{2}\left(1,p+1,p+1-\mu+\eta;p+1-\mu,p+1-\lambda +\eta;z\right) \right] \ast\left(F_{p,\upsilon}f\right) \\ & \,=\,\left[ {z^{p}}_{3}F_{2}\left(1,p+1,p+1-\mu+\eta;p+1-\mu,p+1-\lambda +\eta;z\right) \right] \\ &\quad\ast \left[ {z^{p}}_{3}F_{2}\left(1,1,\upsilon+p;1,\upsilon+p+1;z\right) \ast f(z)\right], \end{array} $$
and
$$\begin{array}{*{20}l} {}F_{p,\upsilon}\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f\right) &={z^{p}}_{3}F_{2}\left(1,1,\upsilon+p;1,\upsilon+p+1;z\right) \ast\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f\right) \\ & ={z^{p}}_{3}F_{2}\left(1,1,\upsilon+p;1,\upsilon+p+1;z\right) \ast\\ & \left[ {z^{p}}_{3}F_{2}\left(1,p+1,p+1-\mu+\eta;p+1-\mu,p+1-\lambda+\eta;z\right) \ast f(z)\right]. \end{array} $$
Now, the first part of this lemma follows. Also, the recurrence relation of Fp,υ is given by
$$ z\left(F_{p,\upsilon}f(z)\right)^{\prime}=(p+\upsilon)f(z)-\upsilon F_{p,\upsilon}f(z). $$
(22)
If we replace f(z) by \(\mathcal {M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\ \)and using the first part of this lemma, we get (20). If we differentiate (20) q-times, we obtain (21). □
Theorem 2
Suppose that \(1\leq q\leq p\ {and}\ f\left (z\right)\in \mathcal {A}(p)\) satisfy
$$(1-\sigma)\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon }f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1} }+\sigma\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\prec \sqrt{1+cz}, $$
where Fp,υ defined by (19), then
$$ \frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\prec \varphi(z)\prec\sqrt{1+cz}, $$
(23)
where φ(z) given by
$$\varphi(z)={\left(1+cz\right)^{\frac{1}{2}}}_{\,2}F_{1}\left(-\frac{1} {2},1;\frac{\upsilon+p}{\sigma}+1;\frac{cz}{1+cz}\right), $$
is the best dominant of (23). Further,
$$ \mathfrak{R}\left\{ \frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon }f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1} }\right\} >L, $$
(24)
where
$$L={\left(1-c\right)^{\frac{1}{2}}}_{\,2}F_{1}\left(-\frac{1}{2},1;\frac{\upsilon+p}{\sigma}+1;\frac{c}{c-1}\right). $$
The result is the best possible.
Proof
Taking
$$ \Theta(z)=\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\ (z\in\mathbb{U)}, $$
(25)
then Θ is analytic in \(\mathbb {U}\). After some calculations, we have
$$\begin{array}{*{20}l} & (1-\sigma)\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}+\sigma\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\\ & =\Theta(z)+\left(\frac{\sigma}{p+\upsilon}\right) z\Theta^{\prime }\left(z\right) \prec\sqrt{1+cz}. \end{array} $$
By employing the same technique that was used in proving Theorem 1, the remaining part of the theorem can be proved. □
Theorem 3
Let\(\ q\in \mathbb {N}_{0}\ {and}\ p>q+\zeta.\ \)If \(f\left (z\right) \in \mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta,\xi \right),\ \)then \(f\left (z\right) \in \mathcal {S}_{p,q}^{\lambda +1,\mu +1,\eta +1}\left (\zeta,\xi \right) \ \)for |z|<R(p,q,μ,ζ,ξ) where
$$ R\left(p,q,\mu,\zeta,\xi\right) =\min\left\{ r>0:t\left(r\right) =0\right\}, $$
(26)
and
$$t\left(r\right) =1-\frac{2r}{\left(p-q-\zeta\right) \left\vert \left(1-\xi\right) \left(1-r\right)^{2}-\left\vert \xi+\frac{q+\zeta-\mu }{p-q-\zeta}\right\vert \left(1-r^{2}\right) \right\vert }. $$
Proof
Assume that \(f\left (z\right) \in \mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta,\xi \right) \) and
$$ u\left(z\right) =\frac{1}{p-q-\zeta}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q\right) }}-\zeta\right), $$
(27)
then, u(z) is analytic in \(\mathbb {U}\) with \(u(0)=1,\ \mathfrak {R}\left \{ u\left (z\right) \right \} >\xi.\) After some computations, we have
$$ {}u\left(z\right) +\frac{zu^{\prime}\left(z\right) }{\left(p-q-\zeta\right) u(z)+\left(q+\zeta-\mu\right) }\,=\,\frac{1}{p-q-\zeta}\!\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q\right) }}-\zeta\right). $$
(28)
Letting \(v(z)=\frac {u(z)-\xi }{1-\xi },\ \)then, v(0)=1 with \(\mathfrak {R}\left \{ v\left (z\right) \right \} >0.\ \)Substituting in (28), we obtain
$$\begin{array}{*{20}l} & \frac{1}{p-q-\zeta}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q\right) }}-\zeta\right) -\xi\\ & =\left(1-\xi\right) \left[ v\left(z\right) +\frac{zv^{\prime}\left(z\right) }{\left(p-q-\zeta\right) \left[ \left(1-\xi\right) v(z)+\xi\right] +\left(q+\zeta-\mu\right) }\right], \end{array} $$
and so
$$\begin{array}{*{20}l} & \mathfrak{R}\left\{ \frac{1}{p-q-\zeta}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q\right)}}-\zeta\right) -\xi\right\} \\ & \geq\left(1-\xi\right) \ \left[ \mathfrak{R}\left\{ v\left(z\right)\right\} -\frac{\left\vert zv^{\prime}\left(z\right) \right\vert }{\left(p-q-\zeta\right) \left\vert \left(1-\xi\right) \left\vert v\left(z\right) \right\vert -\left\vert \xi+\frac{q+\zeta-\mu}{p-q-\zeta}\right\vert\, \,\right\vert }\right] \\ & \geq\left(1-\xi\right) \ \left[ \mathfrak{R}\left\{ v\left(z\right)\right\} -\frac{\left\vert zv^{\prime}\left(z\right) \right\vert }{\left(p-q-\zeta\right) \left\vert \left(1-\xi\right) \ \mathfrak{R}\left\{ v\left(z\right) \right\} -\left\vert \xi+\frac{q+\zeta-\mu}{p-q-\zeta}\right\vert\, \,\right\vert }\right]. \end{array} $$
Applying the following well-known estimate [19]:
$$\mathfrak{R}\left\{ v(z)\right\} \geq\frac{1-r}{1+r}\ \text{and\ }\frac{\left\vert zv^{\prime}(z)\right\vert }{\mathfrak{R}\left\{ v(z)\right\} }\leq\frac{2nr^{n} }{1-r^{2n}}\ (\left\vert z\right\vert =r<1), $$
for n=1, we get
$$\mathfrak{R}\left\{ \frac{1}{p-q-\zeta}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q\right)}}-\zeta\right) -\xi\right\} \geq\left(1-\xi\right) t\left(r\right)\mathfrak{R}\left\{ v(z)\right\}. $$
It is easily seen that t(r) is positive, if |z|<R(p,q,μ,ζ,ξ), where R is given by (26). □
Theorem 4
Let \(f(z)\in \mathcal {A}(p),\ p>\mu,\ \gamma >0\ \)and
$$ \mathfrak{R}\left(\frac{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\prime}}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime}}-\frac{\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta +1,p}f(z)}{\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)}\right) <\frac{\gamma}{p-\mu}, $$
(29)
then
$$\mathfrak{R}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime}}{\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)}\right)^{-\frac{1}{2\gamma}}>\frac{1}{2}. $$
The result is sharp.
Proof
From (8), (29) may be written as
$$\mathfrak{R}\left(1+\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime\prime}}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime}}-\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime}}{\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)}\right) <\gamma, $$
or equivalently,
$$ 1+\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime\prime}}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime}}-\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime}}{\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)}\prec-\frac{2\gamma z}{1-z}. $$
(30)
Letting
$$\digamma(z)=\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime}}{\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)}\right)^{-\frac{1}{2\gamma}}, $$
then, we can express (30) as
$$ z\left(\log\digamma(z)\right)^{\prime}\prec z\left(\log\frac{1} {1-z}\right)^{\prime}. $$
(31)
Fom [20], (31) implies to
$$\digamma(z)\prec\frac{1}{1-z}, $$
or equivalently,
$$\mathfrak{R}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime}}{\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)}\right)^{-\frac {1}{2\gamma}}>\frac{1}{2}\ \left(z\in\mathbb{U}\right). $$
To show that the result is sharp, let
$$K(z)=z^{p}+\sum\limits_{n=1}^{\infty}\frac{(p+1)_{n}(p+1-\mu+\eta)_{n} }{(p+1-\mu)_{n}(p+1-\lambda+\eta)_{n}}\frac{2\gamma\left(2\gamma-1\right)...\left(2\gamma-n+1\right) }{n!}z^{p+n}, $$
and so
$$\begin{array}{*{20}l} \mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}K(z) & =z^{p}+\sum\limits_{n=1} ^{\infty}\frac{2\gamma\left(2\gamma-1\right)...\left(2\gamma-n+1\right) }{n!}z^{p+n}\\ & =z^{p}\left(1+z\right)^{2\gamma}. \end{array} $$
It is easy to check that K(z) satisfies (29) and
$$\mathfrak{R}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}K(z)\right)^{\prime}}{\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}K(z)}\right)^{-\frac {1}{2\gamma}}\rightarrow\frac{1}{2} $$
as z→1−. This ends our proof. □
Theorem 5
Consider that\(\ q\in \mathbb {N}_{0},\ p>q+\zeta \ {and}\ \)
$$ \left(p-q-\zeta\right) \left(1-A\right) +\left(q+\zeta-\mu\right) \left(1-B\right) \geq0. $$
(32)
(i) Suppose that \(\left (\mathcal {M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left (q\right) }\neq 0\ \)for all \(z\in \mathbb {U}^{\ast }:=\mathbb {U}\backslash \{0\},\ \)then
$$\mathcal{S}_{p,q}^{\lambda+1,\mu+1,\eta+1}\left(\zeta;A,B\right) \subset\mathcal{S}_{p,q}^{\lambda,\mu,\eta}\left(\zeta;A,B\right). $$
(ii) Also, assuming that
$$ \frac{1-A}{1-B}\geq\frac{1}{p-q-\zeta}\max\left\{ \frac{p-2q-2\zeta+\mu-1} {2},-\left(q+\zeta-\mu\right) \right\}, $$
(33)
then
$$\mathcal{S}_{p,q}^{\lambda+1,\mu+1,\eta+1}\left(\zeta;A,B\right) \subset\mathcal{S}_{p,q}^{\lambda,\mu,\eta}\left(\zeta,\xi\right). $$
where the bound
$$ \xi\left(A,B\right) =\frac{1}{p-q-\zeta}\left[ \frac{p-\mu}{{\!~\!}_{2} F_{1}\left(1,\frac{2\left(p-q-\zeta\right) \left(A-B\right) } {1-B},p-\mu+1;\frac{1}{2}\right) }-\left(q+\zeta-\mu\right) \right], $$
(34)
is the best possible
Proof
Let \(f\left (z\right) \in \mathcal {S}_{p,q}^{\lambda +1,\mu +1,\eta +1}\left (\zeta ;A,B\right) \ \)and
$$ G\left(z\right) =z\left(\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q\right) }}{\delta\left(p,q\right) z^{p-q} }\right)^{\frac{1}{p-q-\zeta}}. $$
(35)
Since \(\left (\mathcal {M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left (q\right) }\neq 0\ \)for all \(z\in \mathbb {U}^{\ast },\ \) then G(z) is analytic in \(\mathbb {U}\) with G(0)=0 and G′(0)=1. Differentiating both sides of (35) logarithmically, we get
$$ \Psi\left(z\right) =\frac{zG^{\prime}\left(z\right) }{G\left(z\right)}=\frac{1}{p-q-\zeta}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q\right) }}-\zeta\right). $$
(36)
Using (10) in (36), we have
$$ \left(p-\mu\right) \frac{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q\right) }}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q\right) }}=\left(p-q-\zeta\right) \Psi\left(z\right) +\left(q+\zeta-\mu\right). $$
(37)
Differentiating both sides of (37) logarithmically, we get
$${}\frac{1}{p-q-\zeta}\!\left(\!\frac{z\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q\right) }}\,-\,\zeta\!\right)\! \!=\Psi\left(z\right) +\frac{z\Psi^{\prime}\left(z\right)}{\left(p-q-\zeta\right) \Psi\left(z\right) +\left(q+\zeta-\mu\right)}. $$
Combining this identity together with \(f\left (z\right)\in \mathcal {S}_{p,q}^{\lambda +1,\mu +1,\eta +1}\left (\zeta ;A,B\right),\ \)we obtain
$$\Psi\left(z\right) +\frac{z\Psi^{\prime}\left(z\right) }{\left(p-q-\zeta\right) \Psi\left(z\right) +\left(q+\zeta-\mu\right) } \prec\frac{1+Az}{1+Bz}\equiv h(z). $$
We will use Lemma 2 for \(\widetilde {\beta }=\left (p-q-\zeta \right),\ \widetilde {\gamma }=\left (q+\zeta -\mu \right).\ \)Since h(z) is a convex function in \(\mathbb {U\ }\)and
$$\mathfrak{R}\left[ \left(p-q-\zeta\right) \frac{1+Az}{1+Bz}+\left(q+\zeta -\mu\right) \right] >0, $$
whenever (32) holds. Then \(f(z)\in \mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta ;A,B\right) \ \)from Lemma 2. This completes the proof of (i). To prove (ii), we assume that (33) holds, then all the assumptions of Lemma 3 are satisfied for the above values of \(\widetilde {\beta },\ \widetilde {\gamma }\ \)and \(\widetilde {\alpha }=\frac {1-A}{1-B}.\ \)It follows that \(\mathcal {S}_{p,q}^{\lambda +1,\mu +1,\eta +1}\left (\zeta ;A,B\right) \subset \mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta,\xi \right) \) where ξ(A,B) given by (34) is the best possible. □
Theorem 6
Assume that\(\ q\in \mathbb {N}_{0},\ p>q+\zeta \ {and}\ \)
$$ \left(p-q-\zeta\right) \left(1-A\right) +\left(q+\zeta+\upsilon\right) \left(1-B\right) \geq0. $$
(38)
(i) Suppose that \(\left (\mathcal {M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon }f(z)\right)^{\left (q\right) }\neq 0\ \)for all \(z\in \mathbb {U}^{\ast },\ \)then
$$\mathcal{S}_{p,q}^{\lambda,\mu,\eta}\left(\zeta;A,B\right) \subset F_{p,\upsilon}\left(\mathcal{S}_{p,q}^{\lambda,\mu,\eta}\left(\zeta;A,B\right) \right). $$
(ii) Also, assuming that
$$ \frac{1-A}{1-B}\geq\frac{1}{p-q-\zeta}\max\left\{ -\frac{q+2\zeta+\upsilon +1}{2},-\left(p+\zeta+\upsilon\right) \right\}, $$
(39)
then
$$\mathcal{S}_{p,q}^{\lambda,\mu,\eta}\left(\zeta;A,B\right) \subset \mathcal{S}_{p,q}^{\lambda,\mu,\eta}\left(\zeta;\tau\left(A,B\right) \right). $$
where the bound
$$ {}\tau\left(A,B\right) =\frac{1}{p-q-\zeta}\left[ \frac{2p-q+\upsilon}{{\!~\!}_{2}F_{1}\left(1,\frac{2\left(p-q-\zeta\right) \left(A-B\right)}{1-B},2p-q+\upsilon;\frac{1}{2}\right) }-\left(p+\zeta+\upsilon\right) \right], $$
(40)
is the best possible.
Proof
Let \(f\left (z\right) \in \mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta ;A,B\right) \ \)and
$$ H\left(z\right) =z\left(\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\right)^{\left(q\right) }}{\delta\left(p,q\right) z^{p-q}}\right)^{\frac{1}{p-q-\zeta}}. $$
(41)
Since \(\left (\mathcal {M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon }f(z)\right)^{\left (q\right) }\neq 0\ \)for all \(z\in \mathbb {U}^{\ast },\ \)then H(z) is analytic in \(\mathbb {U}\) with H(0)=0 and H′(0)=1. Differentiating both sides of (41) logarithmically, we get
$$ \Phi\left(z\right) =\frac{zH^{\prime}\left(z\right) }{H\left(z\right)}=\frac{1}{p-q-\zeta}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\right)^{\left(q\right) }}-\zeta\right). $$
(42)
Using (21) in (42), we have
$$ \left(p+\upsilon\right) \frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q\right) }}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\right)^{\left(q\right) }}=\left(p-q-\zeta\right) \Phi\left(z\right) +\left(q+\zeta+\upsilon\right). $$
(43)
Differentiating both sides of (43) logarithmically, we get
$${}\frac{1}{p-q-\zeta}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q\right) }}-\zeta\right) =\Phi\left(z\right) +\frac{z\Phi^{\prime}\left(z\right) }{\left(p-q-\zeta\right) \Phi\left(z\right) +\left(q+\zeta+\upsilon\right) }. $$
Combining this identity together with \(f\left (z\right) \in \mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta ;A,B\right),\ \)we obtain
$$\Phi\left(z\right) +\frac{z\Phi^{\prime}\left(z\right) }{\left(p-q-\zeta\right) \Phi\left(z\right) +\left(q+\zeta+\upsilon\right) }\prec\frac{1+Az}{1+Bz}\equiv h(z). $$
We will use Lemma 2 for \(\widetilde {\beta }=\left (p-q-\zeta \right),\ \overline {\gamma }=\left (q+\zeta +\upsilon \right).\ \)Since h(z) is a convex function in \(\mathbb {U\ }\)and
$$\mathfrak{R}\left[ \left(p-q-\zeta\right) \frac{1+Az}{1+Bz}+\left(q+\zeta+\upsilon\right) \right] >0, $$
whenever (38) holds. Then \(f(z)\in F_{p,\upsilon }\left (\mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta ;A,B\right) \right) \ \)from Lemma 2. This proves (i). To prove (ii), we assume that (39) holds, then all the assumptions of Lemma 3 are satisfied for \(\widetilde {\beta },\ \overline {\gamma }\ \) which stated above and \(\widetilde {\alpha }=\frac {1-A}{1-B}.\ \)It follows that
$$\mathcal{S}_{p,q}^{\lambda,\mu,\eta}\left(\zeta;A,B\right) \subset\mathcal{S}_{p,q}^{\lambda,\mu,\eta}\left(\zeta;\tau\left(A,B\right)\right), $$
where τ(A,B) given by (40) is the best possible □