Thamer Khalil MS. Al-Khafaji* | Asmaa KH. Abdul-Rahman | Lieth A. Majed
© 2022 IIETA. This article is published by IIETA and is licensed under the CC BY 4.0 license (http://creativecommons.org/licenses/by/4.0/).
OPEN ACCESS
The current study is fully devoted to studying differential subordination and superordination theorems of analytic functions with some sandwich results involving linear operators $I_{s, a, \mu}^\lambda$. This operator was obtained by the Hadamard product with the family of integral operators and the Hurwitlz-Lerch Zeta function. The current results demonstrate the possibility and capability of extracting the Sandwich theorem, and the conclusion includes differential subordination and differential superordination.
integral operator, univalent function, convex function, linear operator, sandwich theorem
Through the integral operator, Cotirlă [1], obtained a sandwich theorem by different method. Recently, Shammugam et al. [2] and Goyal et al. [3] and Atshan [4], Atshan & HadiAbd [5], Ibrahim et al. [6], Atshan & Hussain [7], and Atshan & Jawad [8], studied sandwich theorems for another condition. Previous research established those differential subordinations and superordinations can be used to obtain adequate conditions to meet the sandwich implication of a large number of well-known sandwich theorems.
Let me use and called H=H(∇) is the class of functions (analytic) in the disk $\nabla=\{z \in \mathbb{C}:|z|<1\}$ . $\forall n \in(+i n t e g e r)$. In addition, $a \in \mathbb{C}$. Now, we define H [a, n] is the subclass of H. H [a, n] include the shape:
$\mathcal{F}(z)=a+a_n z^n+a_{n+1} z^{n+1}+(a$ belong to $\mathbb{C})$ (1)
Too, suppose φ subclass of H include of the functions in the shape:
$\mathcal{F}(z)=z+\sum_{n=2}^{\infty} a_n z^n$ (2)
Let $F, g$ belong to $\varphi$. The function $f$ is called subordinate to $g$, or $g$ is called superordinate to $\mathcal{F}$, if $\exists$ schwarz analytic function in $\nabla$ is $w$, with zero= $w($ zero $)$ and the absolute value of $w(z)$ less than $1,(z \in \nabla)$ s.t $\mathcal{F}(z)$ equal $g(w(z))$. we shall write $\mathcal{F} \prec g$.
If $g$ is univalent in $\nabla$, then $\mathcal{F} \prec g$ iff $F$ (zero) $=g($ zero $), \mathcal{F}(\nabla)$ is subset of $g(\nabla)$.
Now: Suppose $h, p \in H$ and $\psi(r, \lambda, t ; z): \mathbb{C}^3 \times \nabla \longrightarrow \mathbb{C}$. If the variable $p$ and the set of concepts $\psi\left(p(z)\right.$ and $z p^{\prime}(z)$ plus $\left.z^2 p^{\prime \prime}(z) ; z\right)$ are function univalent in the disk $\nabla$ in addition if $p$ in the shape the 2 nd-order superordinations.
$h(z)<\psi\left(p(z), z p^{\prime}(z)\right.$ and $\left.z^2 p^{\prime \prime}(z): z\right)$ (3)
So, p gives title a solution of the superordination of (3). (If f is subordinate to g, then g is superordinate to f). Now An analytic q is said to be a subordinant of (3), if $\mathfrak{q} \prec p \forall$ the functions p satisfies (3). Now $\tilde{q}$ univalent subordinant have the property $\mathrm{q} \prec \tilde{\mathrm{q}}$,∀ differential subordinants q of (3) is known the best subordinant. The two researchers Miller & Mocanu [9] they own set on the analytic h; q and ψ by the set of concepts:
$\begin{aligned} h(z) \prec \psi\left(p(z), z \frac{d p}{d z}, z^2 \frac{d^2 p}{d z^2}(z) ; z\right) & \Rightarrow \mathbb{q}(z) \prec p(z) \end{aligned}$ (4)
Komatu [10] gave the family of integral operator:
$J_\mu^\lambda: \Sigma \rightarrow \Sigma$.
Define as follows:
$J_\mu^\lambda \mathcal{F}(z)=z+\sum_{n=1}^{\infty}(\mu / n+\mu-1)^\lambda a_n z^n$
$(z$ belong to $\nabla, n>1, \lambda \geq 0)$
By function Hurwitlz-Lerch Zeta:
$\phi(z, s, a)=\sum_{n=0}^{\infty} \frac{z^k}{(k+a)^s}, a \in \mathbb{C} / z_0^{-}, s \in \mathbb{C}$ when $0<|z|$
greater than 1
Using convolution where $G_{s, a(z)}$ by:
Then a linear operator $\mathrm{I}_{\mathrm{s}, \mathrm{a}, \mu}^\lambda \mathcal{F}(z): \Sigma \rightarrow \Sigma[11]$ is defined:
$I_{s, a, \mu}^\lambda \mathcal{F}(z)=G_{s, a(z)} * J_\mu^\lambda f(z)$ (5)
$=z+\sum_{n=2}^{\infty}\left(\frac{1+a}{k+a}\right)^s\left(\frac{\mu}{\mu+n-1}\right)^\lambda a_n z^n$ (6)
From Eq. (6) get:
$z\left(I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)\right)^{\prime}=\mu I_{s, a, \mu}^\lambda \mathcal{F}(z)-(\mu-1) I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)$ (7)
AL-Ameedee et al. [12] got conditions for the concepts certain analytic.
$\mathbb{q}_1(z) \prec z \mathcal{F}^{\prime}(z) / \mathcal{F}(z)<\mathbb{q}_2(z)$
where, $\mathbb{q}_1$ and $\mathfrak{q}_2$ are univalent in $\nabla$ in addition $\mathbb{q}_1($ zero $)=$ $\mathfrak{q}_2($ zero $)=1$.
The fundamental objective our research is to find some properties of normalized analytic functions $\mathcal{F}$ like sufficient condition.
$\mathbb{q}_1(z)<\left(\frac{I_{s, a, \mu}^\lambda \mathcal{F}(z)}{z}\right)^\gamma<\mathbb{q}_2(z)$
And:
$\mathrm{q}_1(z) \prec\left(\frac{\mathrm{t} I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)+(1-\mathrm{t}) I_{s, a, \mu}^\lambda \mathcal{F}(z)}{z}\right)^\gamma \prec \mathrm{q}_2(z)$
We got univalent functions $q_1$ and $q_2$ in $\nabla$ with, $1=$ $\mathbb{q}_2(0)=\mathbb{q}_1(0)$
In the next step, we need some basic information [13-20].
Definition (1) [13]. The set functions $\mathcal{F}$ Denote by $\mathrm{Q}$, is analytic and one to one $\bar{\nabla} \backslash E(\mathcal{F})$, where $\bar{\nabla}=\nabla \cup\{z \in \partial \nabla\}$ and $\mathrm{E}(\mathcal{F})=$ $\left\{\zeta \in \partial \nabla: \operatorname{limit}_{z \rightarrow \zeta} \mathcal{F}(z)=\infty\right\}$ s.t
$\mathcal{F}^{\prime}(\zeta) \neq 0$ for $\zeta \in \partial \nabla \backslash E(\mathcal{F})$.
$\mathcal{F}(z)=a$, then: $Q$ (zero $)=Q_0$ and $Q($ one $)=Q_1=\{\mathcal{F} \in Q: \mathcal{F}(0)=1\}$.
Lemma (1)[13]. In $\nabla$ suppose be univalent and suppose $\theta$ in addition $\phi$ analytic function in $D$ consists of $q(D)$ and $0 \neq \phi(w)$ such that $w \in q(\nabla) . z \mathrm{q}^{\prime}(z) \phi(\mathrm{q}(z))=Q(z)$ and $h(z)=\theta(\mathrm{q}(z))+Q(z)$. let:
i. $\nabla$ contains the starlike univalent function $Q(z)$.
ii. $\operatorname{Re}\left\{h^{\prime}(z) z / Q(z)\right\}$ greater than $0, \forall z$ belong to $\nabla$.
If the analytic function $\mathrm{p}$ in $\nabla$, too $\mathrm{p}($ zero $)=\mathrm{q}$ (zero), $p(D)$ sub set $D$,
$\theta(p(z))$ plus $z p^{\prime}(z) \phi(p(z)) \prec \theta(\mathrm{q}(z))$ plus $z \mathrm{q}^{\prime}(z) \phi(\mathrm{q}(z))$
So $\mathfrak{q}$ is the best dominant in addition $q \prec$ qu."
Lemma (2)[14]. Suppose the convex univalent function in $\nabla$ is $q$ in addition let $\alpha$ belong to $c, \beta$ belong to $C \backslash\{z e r o\}$ and:
$\operatorname{Re}\left\{\left(z \mathrm{q}^{\prime \prime}(z) / \mathrm{q}^{\prime}(z)\right)+1\right\}$
greater than $\max \left\{\right.$ zero, $\left.-\operatorname{Re}\left(\frac{\alpha}{\beta}\right)\right\}$
then:
$\alpha p(z)+\beta z p^{\prime}(z) \prec \alpha \mathrm{q}^{(z)}+\beta z q^{\prime}(z)$,
So, the function is the best dominant in addition $p \prec q$.
Lemma (3) [14]. Let the convex univalent function in $\nabla$ is $\mathrm{q}$ in addition suppose $\beta$ belong to $C$. In addition, suppose that $\operatorname{Re}(\beta)>$ zero. If the function $p \in H[\mathrm{q}($ zero $), 1]$ intersection with $Q$ and the univalent in $\nabla$ is $p(z)$ plus $\beta z p^{\prime}(z)$ when:
$\mathrm{q}(z)+\beta z \mathrm{q}^{\prime}(z) \prec p(z)$ plus $\beta z p^{\prime}(z)$
i.e., the function $q$ is the best subordinant and $\mathrm{q} \prec p$.
Lemma (4) [8]. Suppose the convex univalent function in $\nabla$ is $q$ in addition suppose $\phi$ and $\theta$ are analytic function in $\mathrm{D}$ contains $q(\nabla)$. Then, suppose:
(1) Real $\left\{\theta^{\prime}(\mathrm{q}(z)) \div \phi(\mathrm{q}(z))\right\}$ greator than zero, $\forall$ complex variable $z$ belong to $\nabla$.
(2) the function $Q(z)$ equal $z \mathrm{q}^{\prime}(z)$ product $\phi(q(z))$ is function univalent star like in $\nabla$.
If $p$ belong to $Q \cap H[q(z e r o), 1]$, in addition $p(\nabla)$ subset $D, \theta(p(z))$ plus $\lambda p^{\prime}(z) \phi(p(z))$ is univalent function in unit disk.
$\theta(\mathrm{q}(z))$ plus $z \mathrm{q}^{\prime}(z) \phi(q(z)) \prec \theta(p(z))$ plus $z p^{\prime}(z)$ product $\phi(p(z))$
then, $\mathrm{q}<p$ and the function $q$ satisfy the best subordinate.
Theorem (1). Let the convex univalent in $\nabla$ is $q$ with $1=q($ zero $), 0 \neq \varepsilon \in c, \gamma$ greater than to zero and let $q$ satisfies
$\operatorname{Real}\left\{\left[z \mathrm{q}^{\prime \prime}(z) / \mathrm{q}^{\prime}(z)\right]+1\right\}>\operatorname{maximmum}\left\{\operatorname{Zero},-\operatorname{Real}\left(\frac{\gamma}{\varepsilon}\right)\right\}$ (8)
when, $\mathcal{F} \in \varphi$ satisfies the subordination,
$(1-\varepsilon \mu)\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma+\varepsilon \mu\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma\left(\frac{I_{s, a, \mu}^\lambda \mathcal{F}(z)}{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}\right) \prec \mathrm{q}(z) p l u s \frac{\varepsilon}{\gamma} z q^{\prime}(z)$ (9)
then,
$\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma \prec q(z)$ (10)
and the "best dominant of $(9)$ is q.
Proof. $p$ defined as follows:
$p(z)=\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma$ (11)
By purely mathematical operations for condition (11) with respect to z.
$z p^{\prime}(z) / p(z)=\gamma\left(-1+\frac{z\left(l_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)\right)^{\prime}}{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}\right)$ (12)
Then, obtain the following subordination in view of (7).
$p^{\prime}(z) /_{p(z)}=\gamma\left(\mu\left(-1+\frac{I_{s, a, \mu}^\lambda \mathcal{F}(z)}{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}\right)+\left(-1+\frac{I_{s, a, \mu}^\lambda \mathcal{F}(z)}{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}\right)\right)$
Therefore,
$\frac{z p^{\prime}(z)}{\gamma}=\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma\left(\mu\left(-1+\frac{I_{s, a, \mu}^\lambda \mathcal{F}(z)}{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}\right)+\left(-1+\frac{I_{s, a, \mu}^\lambda \mathcal{F}(z)}{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}\right)\right)$
from the hypothesis the subordination (9) becomes:
$p(z) p$ lus $\frac{\varepsilon}{\gamma} z p^{\prime}(z)<\mathrm{q}(z) p \operatorname{lus} \frac{\varepsilon}{\gamma} z \mathrm{q}^{\prime}(z)$
we obtain (10) by application of (Lemma (2)) with $\beta=\frac{\varepsilon}{\gamma}$ and $\alpha=1$,
Corollary (1). Let $0 \neq \varepsilon \in \mathbb{C}, \gamma>0$ and:
$\operatorname{Re}\left\{\frac{2 z}{-z+1}+1\right\}>\max \left\{\right.$ zero, $\left.-\operatorname{Real}\left(\frac{\gamma}{\varepsilon}\right)\right\}$
If $\mathcal{F} \in \varphi$ got the subordination:
$(1-\varepsilon \mu)\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma+\varepsilon \mu\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma\left(\frac{I_{s, a, \mu}^\lambda \mathcal{F}(z)}{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}\right) \prec\left(\frac{1-z^2+2 \frac{\varepsilon}{\gamma} z}{\left(1-z^2\right)}\right)$
So, $\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma$ subordinat to (z plus $1 / \mathrm{z}$ minus 1 ).
In addition, $q(\mathrm{z})$ equal $(\mathrm{z}+1 / \mathrm{z}-1)$ is the perfect dominant.
Theorem (2). Suppose the convex univalent in $\nabla$ is $q$ in addition $1=q($ zero $)$, zero not equal $q(\mathrm{z})$, $(\mathrm{z}$ belong to $\nabla$) and suppose:
$\operatorname{Re}\left\{1-\frac{\gamma}{\varepsilon}+\frac{z q^{\prime \prime}(z)}{q^{\prime}(z)}\right\}>0$ (13)
where, $\varepsilon>0, \varepsilon \in C \backslash\{$ zero $\}$ and $z \in \nabla$.
Let the starlike univalent in $\nabla$ is $\left(-\varepsilon z \mathrm{q}^{\prime}(z)\right)$. If $\mathcal{F} \in \varphi$ then:
$\emptyset(\gamma, \mathrm{s}, \lambda, \mathrm{a}, \varepsilon ; z) \prec \gamma q(z)-\varepsilon z \mathrm{q}^{\prime}(z)$ (14)
where,
$\begin{aligned} \emptyset(\gamma, \mathrm{s}, \lambda, \mathrm{a}, \varepsilon ; z)=& \gamma\left(\frac{\mathrm{t}_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)+(1-\mathrm{t}) I_{s, a, \mu}^\lambda \mathcal{F}(z)}{z}\right)^\gamma-\\ & \gamma \varepsilon\left(\frac{\left.\mathrm{I}_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)+(1-\mathrm{t}) I_{s, a, \mu}^\lambda \mathcal{F}(z)\right)}{z}\right)^\gamma\left(-1+\frac{\left.\mathrm{t} I_{s, a, \mu}^\lambda \mathcal{F}(z)\right)+(1-\mathrm{t}) I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{\left.\mathrm{t} I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)+I_{s, a, \mu}^\lambda \mathcal{F}(z)\right)}\right) \end{aligned}$ (15)
then,
$\left(\frac{\mathrm{t} I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)+(1-\mathrm{t}) I_{s, a, \mu}^\lambda \mathcal{F}(z)}{z}\right)^\gamma \prec \mathrm{q}(z)$ (16)
and the best dominant of $(14)$ is $q(z)$.
Proof: Now we will start defined $p$ :
$p(z)=\left(\frac{\mathrm{t} I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)+(1-\mathrm{t}) I_{s, a, \mu}^\lambda \mathcal{F}(z)}{z}\right)^\gamma$ (17)
Setting: $\emptyset(w)=-\varepsilon$ in addition $\theta(w)=\gamma w, w$ not equal to zero. Then, $\theta(w)$ and $\emptyset(w)$ is analytic in $\mathrm{C}$ with $\mathrm{C} /\{$ zero $\}$ respectively. So, $\emptyset(w)$ not equal to zero, $w$ belong to $\mathrm{C} /\{0\}$, then;
$z \mathrm{q}^{\prime}(z) \cdot \emptyset \mathrm{q}(z)=-\varepsilon z \mathrm{q}^{\prime}(z)=\mathrm{Q}(z)$
and
$\theta \mathrm{q}(z)$ plus $Q(z)$ equal $\gamma \mathrm{q}(z)-\varepsilon z \mathrm{q}^{\prime}(z)=\mathrm{h}(z)$
$Q(z)$ is a starlike univalent in $\nabla$
$\operatorname{Real}\left\{\frac{z \mathrm{~h}^{\prime}(z)}{\mathrm{Q}(z)}\right\}=\operatorname{Real}\left\{1-\frac{\gamma}{\varepsilon}+\frac{z \mathrm{q}^{\prime \prime}(z)}{\mathrm{q}^{\prime}(z)}\right\}>0$.
and, obtain:
$\emptyset(\gamma, \mathrm{s}, \lambda, \mathrm{a}, \mu, \varepsilon ; z)=\gamma \mathrm{p}(z)-\varepsilon \mathrm{zp}^{\prime}(z)$ (18)
So, $\emptyset(\gamma, \mathrm{s}, \lambda, \mathrm{a}, \mu, \varepsilon ; z)$ is given by (15).
By (14) with (18), we own:
$\gamma \mathrm{p}(z)-\varepsilon z \mathrm{p}^{\prime}(z) \prec \gamma q(z)-\varepsilon z q^{\prime}(z)$ (19)
Therefore, got $\mathrm{p}(z) \prec \mathrm{q}(z)$ by (Lemma (1)), and by using (17).
Corollary (2):
Let, Real $\left\{1-\frac{\gamma}{\varepsilon}+\frac{z 2 \mathrm{~B}}{(1+\mathrm{B} z)}\right\}>0$, and $-1 \leq \mathrm{B}<\mathrm{A} \leq 1$. where $\varepsilon$ belong to $\mathrm{C} /\{$ zero $\}$ in addition $z$ be long to $\nabla$, if $\mathcal{F} \in \varphi$ such that:
$\emptyset(\gamma, \mathrm{s}, \lambda, \mathrm{a}, \mu, \varepsilon ; z) \prec\left(\gamma\left(\frac{1+\mathrm{A} z}{1+\mathrm{B} z}\right)-\varepsilon z \frac{\mathrm{A}-\mathrm{B}}{(1+\mathrm{B} z)^2}\right)$
and $\emptyset(\gamma, \mathrm{s}, \lambda, \mathrm{a}, \mu, \varepsilon ; z)$ is given by condition $(8)$,
$\left(\frac{\mathrm{t} I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)+(1-\mathrm{t}) I_{s, a, \mu}^\lambda \mathcal{F}(z)}{z}\right)^\gamma \prec \frac{\mathrm{A} z+1}{\mathrm{Bz}+1}$
and the perfect dominant is $\mathrm{q}(z)=\frac{1+\mathrm{A} z}{1+\mathrm{B} z}$
Theorem (3): suppose the convex univalent in $\nabla$ is $q$ in addition $\gamma$ greator than zero, $q(0)=1$ and $\operatorname{Re}\{\varepsilon\}>0$.
Suppose that $\mathcal{F} \in \varphi$ satisfies: $\left(\frac{\left.I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)\right)}{z}\right)^\gamma \in \mathrm{Q}$ intersect $H[q(0), 1]$, so,
$(1-\varepsilon \mu)\left(\frac{I_{S, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma+\varepsilon \mu\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma\left(\frac{I_{S, a, \mu}^\lambda \mathcal{F}(z)}{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}\right)$
is univalent function in $\nabla$
if $\mathrm{q}(z)+\frac{\varepsilon}{\gamma} z \mathrm{q}^{\prime}(z) \prec(1-\varepsilon \mu)\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma+\varepsilon \mu\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma\left(\frac{I_{s, a, \mu}^\lambda \mathcal{F}(z)}{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}\right)$ (20)
then,
$\mathrm{q}(z)<\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma$ （21）
the best subordinant of (20) is q
Proof: p defined as follows:
$\mathrm{p}(z)=\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma$ (22)
Now we get (22) by differentiating:
$z p^{\prime}(z) / p(z)=\delta\left(\frac{z\left(I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)\right)^{\prime}}{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}-1\right)$ (23)
Using (7) in (23), we obtain:
$(1-\varepsilon \mu)\left(\frac{I_{S, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma+\varepsilon \mu\left(\frac{I_{S, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma\left(\frac{I_{S, a, \mu}^\lambda \mathcal{F}(z)}{I_{S, a, \mu}^{\lambda+1} \mathcal{F}(z)}\right)=\mathrm{q}(z)+\frac{\varepsilon}{\gamma} z \mathrm{q}^{\prime}(z)$
we get the result by using (Lemma (3)).
Corollary (3): If $\mathcal{F} \in \varphi$ satisfies: $\left(\frac{\left.I_{S, a, \mu}^{\lambda+1} \mathcal{F}(z)\right)}{z}\right)^\gamma \in Q \cap H[\mathbb{Q}(0), 1]$ and let $\gamma>0$ and $\operatorname{Re}\{\varepsilon\}>0$.
and,
$(1-\varepsilon \mu)\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma+\varepsilon \mu\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma\left(\frac{I_{s, a, \mu}^\lambda \mathcal{F}(z)}{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}\right)$
be univalent in $\nabla$. If
$\frac{1-z^2+2 \frac{\varepsilon}{\gamma} z}{\left(1-z^2\right)} \prec(1-\varepsilon \mu)\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma+\varepsilon \mu\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma\left(\frac{I_{s, a, \mu}^\lambda \mathcal{F}(z)}{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}\right)$
then,
$\left(\frac{1+z}{1-z}\right) \prec\left(\frac{\left.I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)\right)}{z}\right)^\gamma$
and the best subordinant is $q(\mathrm{z})$ equal $\frac{1+z}{1-z}$.
Theorem (4): Let the convex univalent in $\nabla$ is $q$ in addition $1=q($ zero $)$, and let $q$ satisfies:
$\operatorname{Re}\left\{\frac{-\gamma \mathrm{q}^{\prime}(z)}{\varepsilon}\right\}>0$ (24)
"Let- $\gamma \mathrm{zq}$ ' $(\mathrm{z})$ " is a starlike univalent function "in $\nabla$ and let $\mathcal{F} \in \varphi$ satisfies: " $\left(\frac{\left.\mathrm{t} \mathrm{I}_{s, a, \mu}^{\hat{H},+} \mathcal{F}(z)+(1-\mathrm{t})\right)_{I_{s, a, \mu}^{\hat{H}}} \mathcal{F}(z)}{z}\right) \in \mathrm{Q} \cap$ $\mathrm{H}$ [q (zero), 1], "and $\emptyset(\gamma, \mathrm{s}, \lambda, \mathrm{a}, \mu, \varepsilon ; \mathrm{z})$ is a univalent function in $\nabla$, where $\emptyset(\gamma, \mathrm{s}, \lambda, \mathrm{a}, \mu, \varepsilon ; \mathrm{z})$ is given by (15). If:
$\gamma \mathrm{q}(z)-\varepsilon z \mathrm{q}^{\prime}(z) \prec \emptyset(\gamma, \mathrm{n}, \lambda, \mathrm{m}, \varepsilon ; z)$ (25)
Then,
$\mathrm{q}(z) \prec\left(\frac{\mathrm{t} I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)+(1-\mathrm{t}) I_{s, a, \mu}^\lambda \mathcal{F}(z)}{z}\right)^\gamma$ (26)
and the best subordinant of (25) that is q.
Proof: Defined p in the shape,
$\mathrm{p}(z)=\left(\frac{\mathrm{t} I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)+(1-\mathrm{t}) I_{s, a, \mu}^\lambda \mathcal{F}(z)}{z}\right)^\gamma$ (27)
by $\emptyset(w)=-\varepsilon$ and $\theta(w)=\gamma w, 0 \neq w$.
So, $\theta(w)$ and $\emptyset(w)$ are analytic in $C$ in addition $C \backslash\{$ zero $\}$ respectively and $\emptyset(w)$ does not equal zero, $w \in C \backslash\{z e r o\}$. Then:
$z \mathrm{q}^{\prime}(z) \emptyset \mathrm{q}(z)=-\varepsilon z \mathrm{q}^{\prime}(z)=\mathrm{Q}(z)$
We got starlike univalent" in $\nabla$ is $Q(z)$.
$\operatorname{Real}\left\{\frac{\theta^{\prime}(\mathrm{q}(z))}{\emptyset(\mathrm{q}(z))}\right\}=\operatorname{Re}\left\{\frac{-\gamma \mathrm{q}^{\prime}(z)}{\varepsilon}\right\}>0$
Now, obtain:
$\gamma \mathrm{p}(z)-\varepsilon z \mathrm{p}^{\prime}(z)=\emptyset(\gamma, \mathrm{s}, \lambda, \mathrm{a}, \mu, \varepsilon ; z)$ (28)
where, $\emptyset(\gamma, \mathrm{s}, \lambda, \mathrm{a}, \mu, \varepsilon ; z)$ is get by (15).
By (25) and (28);
$\gamma q(z)-\varepsilon z q^{\prime}(z)<\gamma \mathrm{p}(z)-\varepsilon \mathrm{p}^{\prime}(z)$ (29)
where, $\mathrm{q}(z) \prec \mathrm{p}(z)$ by (Lemma (3)). By using (27), will get to the desired result.
The concept of Sandwich represented by (Theorems (5) and (6)).
Theorem (5): Let the convex univalent in $\nabla$ is $\mathrm{q}_1$ (zero) $=1$, Real $\{\varepsilon\}>0$ and let the univalent in $\nabla$ is $\mathrm{q}_2($ zero $)=1$ and realize (8), let $\mathcal{F} \in \varphi$ such that.
$\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma \in \mathrm{Q} \cap \mathrm{H}[1,1]$
and,
$(1-\varepsilon \mu)\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma+\varepsilon \mu\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma\left(\frac{I_{s, a, \mu}^\lambda \mathcal{F}(z)}{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}\right)$
is univalent function in $\nabla$.
If,
$\mathrm{q}_1(z)+\frac{\varepsilon}{\gamma} \mathrm{zq}^{\prime}{ }_1(z)$ subordinant to $(1-\varepsilon \mu)\left(\frac{\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right.}{z}\right)^\gamma+\varepsilon \mu\left(\frac{I_{S, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma\left(\frac{I_{S,,,, \mu}^\lambda \mathcal{F}(z)}{I_{s, a, \mu}^{\lambda \lambda+1} \mathcal{F}(z)}\right) \prec \mathrm{q}_2(z)+\frac{\varepsilon}{\gamma} \mathrm{zq}_2^{\prime}(z)$
$\mathrm{q}_1(z)<\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma<\mathrm{q}_2(z)$
We got concepts best subordinant in addition best dominant $\mathrm{q}_1$ and $\mathrm{q}_2$ are respectively.
Theorem (6): Suppose the univalent convex in $\nabla$ is $\mathrm{q}_1, \mathrm{q}_1$ (zero) $=1$ and satisfies (24), let the univalent function in $\nabla$ is $\mathrm{q}_2$, $\mathrm{q}_2$ (zero) $=1$, realize (13), let $\mathcal{F} \in \varphi$ satisfies:
$\left(\frac{\mathrm{t} I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)+(1-\mathrm{t}) I_{s, a, \mu}^\lambda \mathcal{F}(z)}{z}\right)^\gamma \in \mathrm{Q} \cap \mathrm{H}[1,1]$,
and $\emptyset(\gamma, \mathrm{s}, \lambda, \mathrm{a}, \mu, \varepsilon ; \mathrm{z})$ is univalent in $\nabla$, where $\emptyset(\gamma, \mathrm{s}, \lambda, \mathrm{a}, \mu, \varepsilon ; \mathrm{z})$ we got by (15).
If $\gamma \mathrm{q}_1(z)-\varepsilon z \mathrm{q}_1^{\prime}(z) \emptyset(\gamma, \mathrm{s}, \lambda, \mathrm{a}, \mu, \varepsilon ; z)<\gamma \mathrm{q}_2(z)-\varepsilon z \mathrm{q}_2^{\prime}(z)$,
Then,
$\mathrm{q}_1(z) \prec\left(\frac{\mathrm{t} I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)+(1-\mathrm{t}) I_{s, a, \mu}^\lambda \mathcal{F}(z)}{z}\right) \prec \mathrm{q}_2(z)$
We got the concepts of best subordinant and best dominant $\mathbb{q}_1$ and $\mathbb{q}_2$, respectively.
The conclusion of this research gained subordination and superordination results by using the linear operator $I_{S, a, \mu}^{\lambda+1}$ for example for these results:
$1-\mathrm{q}_1(z) \prec\left(\frac{I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)}{z}\right)^\gamma \prec \mathrm{q}_2(z)$
$2-\mathrm{q}_1(z) \prec\left(\frac{\mathrm{t} I_{s, a, \mu}^{\lambda+1} \mathcal{F}(z)+(1-\mathrm{t}) I_{s, a, \mu}^\lambda \mathcal{F}(z)}{z}\right) \prec \mathrm{q}_2(z)$
[1] Cotîrlă, L. (2009). A differential sandwich theorem for analytic functions defined by the integral operator. Studia Universitatis Babes-Bolyai, Mathematica, (2).
[2] Shammugam, T.N., Ramachandran, C., Darus, M., Sivasubramanian, S. (2007). Differential sandwich theorems for some subclasses of analytic functions involving a linear operator. Acta Mathematica Universitatis Comenianae, 76(2): 287-294.
[3] Goyal, S.P., Goswami, P., Silverman, H. (2008). Subordination and superordination results for a class of analytic multivalent functions. International Journal of Mathematics and Mathematical Sciences, 2008: 561638. https://doi.org/10.1155/2008/561638
[4] Atshan, W.G. (2020). On some sandwich theorems of analytic functions involving Noor-Salagean operator. Advances in Mathematics Scientific Journal, 9(10): 1857-8438. https://doi.org/10.37418/amsj.9.10.74
[5] Atshan, W.G., HadiAbd, E. (2017). Differential subordination and superordinationof univalent functions by using generalized integral operator and Bernardin-libera-Livingston integral operator. American Journal of Scientific Research, (109): 13-18.
[6] Ibrahim, R.W., Elobaid, R.M., Obaiys, S.J. (2020). On the connection problem for Painleve differential equation in view of geometric function theory. Mathematics, 8(7): 1198. https://doi.org/10.3390/math8071198
[7] Atshan, W.G., Hussain, K.O. (2016). Subordination results for certain subclass of univalent functions. European Journal of Scientific Research, 142(1): 5-11.
[8] Atshan, W.G., Jawad, S.A.A.H. (2019). On differential sandwich results for analytic functions. Journal of Al-Qadisiyah for Computer Science and Mathematics, 11(1): 96.
[9] Miller, S.S., Mocanu, P.T. (2003). Subordinants of differential superordinations. Complex Variables, 48(10): 815-826. https://doi.org/10.1080/02781070310001599322
[10] Komatu, Y. (1990). On analytic prolongation of a family of operators. Mathematica (cluj), 32(55): 141-145.
[11] Atshan, W.G., Badawi, E.I. (2019). On sandwich theorems for certain univalent functions defined by a new operator. Journal of Al-Qadisiyah for Computer Science and Mathematics, 11(2): 72-80.
[12] AL-Ameedee, S.A., Atshan, W.G., Ali AL-Maamori, F. (2020). On sandwich results of univalent functions defined by a linear operator. Journal of Interdisciplinary Mathematics, 23(4): 803-809. https://doi.org/10.1080/09720502.2020.1727613
[13] Miller, S.S., Mocanu, P.T. (2000). Differential Subordinations: Theory and Applications (1st ed.). CRC Press. https://doi.org/10.1201/9781482289817
[14] Bulboacă, T. (2002). Classes of first-order differential superordinations. Demonstratio Mathematica, 35(2): 287-292.
[15] Bulboacă, T. (2005). Differential subordinations and superordinations: Recent results. Casa Cărt̨ii de S̨tiint̨ă, XXXV(2): 287-292. https://doi.org/10.1515/dema-2002-0209
[16] Murugusundaramoorthy, G., Magesh, N. (2006). Differential subordinations and superordinations for analytic functions defined by the Dziok-Srivastava linear operator. Journal of Inequalities in Pure andApplied Mathematics, 7(4): 1-9.
[17] Murugusundaramoorthy, G., Magesh, N. (2009). Differential subordinations and superordinations for analytic functions defined by convolution structure. Studia Universitatis Babes-Bolyai, Mathematica, 2.
[18] Murugusundaramoorthy, G., Magesh, N. (2006). Differential subordinations and superordinations for analytic functions defined by the Dziok-Srivastava linear operator. Journal of Inequalities in Pure andApplied Mathematics, 7(4): 1-9.
[19] Aouf, M.K., Bulboacă, T. (2010). Subordination and superordination properties of multivalent functions defined by certain integral operator. Journal of the Franklin Institute, 347(3): 641-653. https://doi.org/10.1016/j.jfranklin.2010.01.001
[20] Lupas, A.A. (2011). On special differential subordinations using a generalized Salagean operator and Ruscheweyh derivative. Journal of Computational Analysis and Applications, 61(4): 1048-1058. https://doi.org/10.1016/j.camwa.2010.12.055