Application of subordination and superordination for multivalent analytic functions associated with differintegral operator

1.   Introduction

Let $H(U)$ be the class of analytic functions in the open unit disc $U = \{z\in \mathbb{C}:\left\vert z\right\vert < 1\}$ and let $H[a, \upsilon ]$ be the subclass of $H(U)$ including form-specific functions

we denote by $H = H[1, 1].$

Also, $A(p)$ should denote the class of multivalent analytic functions in $U$, with the power series expansion of the type:

Upon differentiating $j$-times for each one of the (1.1) we obtain:

Numerous mathematicians, for instance, have looked at higher order derivatives of multivalent functions (see ^{[1,3,6,9,16,27,28,31]}).

For $f, \hslash \in H$, the function $f$ is subordinate to $\hslash$ or the function $\hslash$ is said to be superordinate to $f$ in $U$ and we write $f(z)\prec \hslash (z)$, if there exists a Schwarz function $\omega$ in $U$ with $\omega (0) = 0$ and $\left\vert \omega (z)\right\vert < 1$, such that $f(z) = \hslash (\omega (z)),$ $z\in U.$ If $\hslash$ is univalent in $U$, then $f(z)\prec \hslash (z)$ iff $f(0) = \hslash (0)$ and$\ f(U)\subset \hslash (U).$ (see ^[7,21]).

In the concepts and common uses of fractional calculus (see, for example, ^[14,15] see also ^[2]; the Riemann-Liouville fractional integral operator of order $\alpha \in \mathbb{C}$ $(\Re (\alpha) > 0)$ is one of the most widely used operators (see ^[29]) given by:

applying the well-known (Euler's) Gamma function $\Gamma (\alpha).$ The Erd élyi-Kober fractional integral operator of order $\alpha \in \mathbb{C} \; (\Re (\alpha) > 0)$ is an interesting alternative to the Riemann-Liouville operator $I_{0+}^{\alpha },$ defined by:

which corresponds essentially to (1.3) when $\sigma -1 = \eta = 0,$ since

Mainly motivated by the special case of the definition (1.4) when $x = \sigma = 1$, $\eta = \nu -1$ and $\alpha = \rho -\nu$, here, we take a look at the integral operator $\Im _{p}(\nu, \rho, \mu)$ with $f\in A(p)$ by (see ^[11])

Evaluating (Euler's) Gamma function by using the Eulerian Beta-function integral as following:

we readily find that

It is readily to obtain from (1.5) that

The integral operator $\Im _{p}(\nu, \rho; \ell)f(z)$ should be noted as a generalization of several other integral operators previously discussed for example,

(ⅰ) If we set $p = 1,$ we get $\tilde{I}(\nu, \rho; \ell)f(z)$ defined by Ŕaina and Sharma (^[22] with $m = 0$);

(ⅱ) If we set $\nu = \beta, \rho = \beta +1$ and $\ \ell = 1,$ we obtain $\Im _{p}^{\beta }f(z)(\beta > -p)$ it was presented by Saitoh et al.^[24];

(ⅲ) If we set $\nu = \beta, \rho = \alpha +\beta -\delta +1, \ \ell = 1,$ we obtain $\Re _{\beta, p}^{\alpha, \delta }f(z)(\delta > 0;$ $\alpha \geq \delta -1;$ $\beta > -p)$ it was presented by Aouf et al. ^[4];

(ⅳ) If we put $\nu = \beta, \rho = \alpha +\beta, \ \ell = 1,$ we get $Q_{\beta, p}^{\alpha }f(z)(\alpha \geq 0;\beta > -p)$ it was investigated by Liu and Owa ^[18];

(ⅴ) If we put $p = 1,$ $\nu = \beta, \rho = \alpha +\beta, \ \ell = 1,$ we obtain $\Re _{\beta }^{\alpha }f(z)(\alpha \geq 0;\beta > -1)$ it was introduced by Jung et al. ^[13];

(ⅵ) If we put $p = 1,$ $\nu = \alpha -1, \ \rho = \beta -1, \ \ell = 1,$ we obtain $L(\alpha, \beta)f(z)(\alpha, \beta \in \mathbb{C} \backslash \mathbb{Z} _{0}, \mathbb{Z} _{0} = \{0, -1, -2, ...\})$ which was defined by Carlson and Shaffer ^[8];

(ⅶ) If we put $p = 1,$ $\nu = \nu -1, \ \rho = j, \ \ell = 1$ we obtain $I_{\nu, j}f(z)(\nu > 0;j\geq -1)$ it was investigated by Choi et al. ^[10];

(ⅷ) If we put $p = 1,$ $\nu = \alpha, \rho = 0, \ \ell = 1,$ we obtain $D^{\alpha }f(z)(\alpha > -1)$ which was defined by Ruscheweyh ^[23];

(ⅸ) If we put $p = 1,$ $\nu = 1, \ \rho = m, \ \ell = 1,$ we obtain $I_{m}f(z)(m\in \mathbb{N} _{0})$ which was introduced by Noor ^[21];

(ⅹ) If we set $p = 1,$ $\nu = \beta, \rho = \beta +1, \ \ell = 1$ we obtain $\Im _{\beta }f(z)$ which was studied by Bernadi ^[5];

(ⅹⅰ) If we set $p = 1,$ $\nu = 1, \ \rho = 2, \ \ell = 1$ we get $\Im f(z)$ which was defined by Libera ^[17].

2.   Key lemmas

We state various definition and lemmas which are essential to obtain our results.

Definition 1. (^[20], Definition 2, p.817) We denote by $Q$ the set of the functions $f\ $that are holomorphic and univalent on $\overline{ U}\setminus E(f)$, where

and satisfy $f^{\prime }(\zeta)\neq 0$ for $\zeta \in \partial U\setminus E(f)$.

Lemma 1. (^[12]; see also (^[19], Theorem 3.1.6, p.71)) Assume that $h(z)$ is convex (univalent) function in $U$ with $h(0) = 1$, and let $\varphi (z)\in H,$ is analytic in $U$. If

where $\gamma \neq 0$ and $\mathit{\text{Re}}(\gamma)\geq 0$. Then

and $\Psi (z)$ is the best dominant.

Lemma 2. (^[26]; Lemma 2.2, p.3) Suppose that $q$ is convex function in $U$ and let$\ \ \psi \in C$ with $\varkappa \in C^{\ast } = C\backslash \{0\}$ with

If $\lambda (z)$ is analytic in $U$, and

therefore $\lambda (z)\prec q(z)$, and $q$ is the best dominant.

Lemma 3. (^[20]; Theorem 8, p.822) Assume that $q$ is convex univalent in $U$ and suppose $\delta \in C$, with $\mathit{\text{Re}}\left(\delta \right) > 0$. If $\lambda \in H[q(0), 1]\cap Q$ and $\lambda (z)+\delta z\lambda ^{\prime }(z)$ is univalent in $U$, then

implies

and $q$ is the best subordinant.

For $a, \varrho, c$ and $c(c\notin \mathbb{Z} _{0}^{-})$ real or complex number the Gaussian hypergeometric function is given by

The previous series totally converges for $z\in U$ to a function analytical in $U$ (see, for details, (^[30], Chapter 14)) see also ^[19].

Lemma 4. For $a, \varrho$ and $c\ (c\notin \mathbb{Z} _{0}^{-})$, real or complex parameters,

3.   Main results

Throughout the sequel, we assume unless otherwise indicated $-1\leq D < C\leq 1, \ \delta > 0,$ $\ell > 0, \ \nu, \rho \in \mathbb{R}, \ \nu > -\ell p, \ p\in \mathbb{N}$ and $(\rho -j)\geq 0.$ We shall now prove the subordination results stated below:

Theorem 1. Let $0\leq j < p,$ $0 < r\leq 1$ and for $f\in A(p)$ assume that

whenever $\delta \in (0, +\infty)\backslash \mathbb{N}.$ Let define the function $\Phi _{j}$ by

such that the powers are all the principal ones, i.e., log1 = 0. Whether

then

where

and $\left[ \frac{p!}{(p-j)!}\right] ^{\delta }p(z)$ is the best dominant of (3.3). Moreover, there are

where $\zeta$ is given by:

then (3.4) is the best possible.

Proof. Let

It is observed that the function $\phi (z)\in H, \ $which is analytic in $U$ and $\phi (0) = 1$. Differentiating (3.5) with respect to $z$, applying the given equation, the hypothesis (3.2), and the knowing that

we get

We can verify that the above equation $q(z)$ is analytic and convex in $U$ as following

Using Lemma 1, there will be

In order to calculate the integral, we define the integrand in the type

using Lemma 4 we obtain

On the other hand if $D = 0$ we have

where the identities (2.1)–(2.3), were used after changing the variable, respectively. This proof the inequality (3.3).

Now, we'll verify it

Indeed, we have

Setting

and

where $dv(s)$ is a positive measure on the closed interval [0, 1], we get that

so that

Now, taking $\sigma \rightarrow 1^{-}$ we get the result (3.7). The inequality (3.4) is the best possible since $\left[ \frac{p!}{(p-j)!} \right] ^{\delta }p(z)$ is the best dominant of (3.3).

If we choose $j = 1$ and $\alpha = \delta = 1$ in Theorem 1, we get:

Corollary 1. Let $0 < r\leq 1$. If

then

where $\zeta _{1}$ is given by:

then (3.8) is the best possible.

If we choose $\nu = \rho = 0$ and $\ \ell = 1$ in Theorem 1, we get:

Corollary 2. Let $0\leq j < p,$ $0 < r\leq 1$ and as $f\in A(p)$ assume that

whenever $\delta \in (0, +\infty)\backslash \mathbb{N}.$ Let define the function $\Phi _{j}$ by

such that the powers are all the principal ones, i.e., log1 = 0. If

then

where

and $\left[ \frac{p!}{(p-j)!}\right] ^{\delta }p_{1}(z)$ is the best dominant of (3.10). Morover, there are

where $\zeta _{2}$ is given by

then (3.11) is the best possible.

If we put $\delta = 1$ and $\ r = 1$ in Corollary 2, we get:

Corollary 3. Let $0\leq j < p,$ and for $f\in A(p)$ say it

Let define the function $\Phi _{j}$ by

If

then

where

and $\frac{p!}{(p-j)!}p_{2}(z)$ is the best dominant of (3.12). Morover there will be

where $\zeta _{3}$ is given by:

then (3.13) is the best possible.

For $C = 1, D = -1$ and $j = 1$ Corollary 3, leads to the next example:

Example 1. (i) For $f\in A(p)$ suppose that

Let define the function $\Phi _{j}$ by

then

and

where $\zeta _{4}$ is given by:

then (3.15) is the best possible.

(ii) For $p = \alpha = 1, \ (i)\ $leads to:

For $f\in A$ suppose that

Let define the function $\Phi _{j}$ by

then

So the estimate is best possible.

Theorem 2. Let $0\leq j < p,$ $0 < r\leq 1$ as for $f\in A(p).$ Assume that $\mathcal{F} _{\alpha }$ is defined by

If

then

where

and $\frac{p!}{(p-j)!}p(z)$ is the best dominant of (3.18). Moreover, there will be

where $\eta$ is given by:

then (3.19) is the best possible.

Proof. By using the definition (3.16) and the inequality (3.6), we have

for $0\leq j < p.$ Putting

we have that $\phi \in H$. Differentiating (3.21), and using (3.17), (3.20), we get

Following the techniques of Theorem 1, we can obtain the remaining part of the proof.

If we choose $j = 1$ and $r = 1$ in Theorem 2, we get:

Corollary 4. For $f\in A(p)$ let the function $\mathcal{F} _{\alpha }$ define by 3.16. If

then

where $\eta _{1}$ is given by:

then (3.22) is the best possible.

Example 2. If we choose $p = C = \alpha = 1\ $and$\ D = -1\ $in Corollary 4, we obtain:

For

If

then

the result is the best possible.

Theorem 3. Let $0\leq j < p,$ $0 < r\leq 1$ as for $\theta > -p$ assume that $J_{p, \theta }:A(p)\rightarrow A(p)$ defined by

If

then

where

and $\frac{p!}{(p-j)!}p(z)$ is the best dominant of (3.25). Moreover, there will be

where $\beta$ is given by:

then (3.26) is the best possible.

Proof. Suppose

we have that $\phi \in H$. Differentiating the above definition, by using (3.24) and

we get

Now, we obtain (3.25) and the inequality (3.26) follow by using the same techniques in Theorem 1.

If we set $j = 1$ and $r = 1$ in Theorem 3, we get:

Corollary 5. For $\theta > -p,$ let the operator $J_{p, \theta }:A(p)\rightarrow A(p)$ defined by (3.25). If

then

where $\beta _{1}$ is given by:

then (3.27) is the best possible.

Example 3. If we choose $p = C = \theta = 1\ $and$\ D = -1\ $in Corollary 5, we get:

If

then

the result is the best possible.

Theorem 4. Let $q$ is univalent function in $U$, such that$\ q\ $satisfies

Let $0\leq j < p,$ $0 < r\leq 1$ and for $f\in A(p)$ assume that

whenever $\delta \in (0, +\infty)\backslash \mathbb{N}.$ Let the function $\Phi _{j}$ defined by (3.1), and assume that it satisfies:

Then,

and $q(z)$ is the best dominant of (3.30).

Proof. Let $\phi (z)$ is defined by (3.5), from Theorem 1 we get

Combining (3.29) and (3.31) we find that

The proof of Theorem 4 follows by using Lemma 2 and (3.32).

Taking $q(z) = \left(\frac{1+Cz}{1+Dz}\right) ^{r}$ in Theorem 4, we obtain:

Corollary 6. Suppose that

Let $0\leq j < p,$ $0 < r\leq 1$ and for $f\in A(p)$ satisfies

whenever $\delta \in (0, +\infty)\backslash \mathbb{N}.$ Let the function $\Phi _{j}$ defined by (3.1), satisfies:

Then,

so $\left(\frac{1+Cz}{1+Dz}\right) ^{r}$ is the best dominant of (3.33).

Taking $q(z) = \frac{1+Cz}{1+Dz}$ in Theorem 4, we get:

Corollary 7. Suppose that

Let $0\leq j < p,$ $0 < r\leq 1$ and for $f\in A(p)$ satisfies

whenever $\delta \in (0, +\infty)\backslash \mathbb{N}.$ Let the function $\Phi _{j}$ defined by (3.1), satisfies:

Then,

so $\frac{1+Cz}{1+Dz}$ is the best dominant of (3.34).

If we put $\nu = \rho = 0$ and $\ \ell = 1$ in Theorem 4, we get:

Corollary 8. Let $q$ is univalent function in $U$, such that$\ q\ $satisfies

For $f\in A(p)$ satisfies

Let the function $\Phi _{j}$ defined by (3.9), satisfies:

Then,

so $q(z)$ is the best dominant of (3.36).

Taking $C = 1$ and $D = -1$ in Corollaries 6 and 7 we get:

Example 4. (i) For $f\in A(p)$ assume that

Let the function $\Phi _{j}$ defined by (3.1), and assume that it satisfies:

Then,

so $\left(\frac{1+z}{1-z}\right) ^{r}$ is the best dominant of (3.37).

(ii) For $f\in A(p)$ say it

Let the function $\Phi _{j}$ defined by (3.1), and assume that it satisfies:

Then,

so $\frac{1+z}{1-z}$ is the best dominant of (3.38).

If we put $p = C = \alpha = \delta = 1, \ D = -1$ and $j = 0$ in Corollary 8 we get:

Example 5. For $f\in A$ suppose that

and

Then,

and $\left(\frac{1+z}{1-z}\right) ^{r}$ is the best dominant of (3.39).

Remark 1. For $\ \nu = \rho = 0, \ \ell = p = r = 1$ and $j = 0$ in Theorem 4, we get the results investigated by Shanmugam et al. (^[25], Theorem 3.1).

Theorem 5. Let $0\leq j < p,$ and for $f\in A(p)$ assume that

whenever $\delta \in (0, +\infty)\backslash \mathbb{N}.\ $Suppose that

such that $\left[ \frac{(p-j)!}{p!}\right] ^{\delta }\Phi _{j}(z)$ is univalent in $U$, where the function $\Phi _{j}$ is defined by (3.1). If $q$ is convex (univalent) function in $U$, and

then

so $q(z)$ is the best subordinate of (3.40).

Proof. Let $\phi$ is defined by (3.5), from (3.31) we get

The proof of Theorem 5 followes by an application of Lemma 3.

Taking $q(z) = \left(\frac{1+Cz}{1+Dz}\right) ^{r}$ in Theorem 5, we get:

Corollary 9. Let $0\leq j < p,$ $0 < r\leq 1$ and for $f\in A(p)$ assume that

Suppose that

such that $\left[ \frac{(p-j)!}{p!}\right] ^{\delta }\Phi _{j}(z)$ is univalent in $U$, where the function $\Phi _{j}$ is defined by (3.1)$.$ If

then

so $\left(\frac{1+Cz}{1+Dz}\right) ^{r}$ is the best dominant of (3.41).

Taking $q(z) = \frac{1+Cz}{1+Dz}\ $and$\ r = 1$ in Theorem 5, we get:

Corollary 10. Let $0\leq j < p,$ and for $f\in A(p)$ assume that

whenever $\delta \in (0, +\infty)\backslash \mathbb{N}.$ Assume that

such that $\left[ \frac{(p-j)!}{p!}\right] ^{\delta }\Phi _{j}(z)$ is univalent in $U$, where the function $\Phi _{j}$ is defined by (3.1). If

then

so $\frac{1+Cz}{1+Dz}$ is the best dominant of (3.42).

Combining results of Theorems 4 and 5, we have

Theorem 6. Let $0\leq j < p,$ and for $f\in A(p)$ assume that

Suppose that

such that $\left[ \frac{(p-j)!}{p!}\right] ^{\delta }\Phi _{j}(z)$ is univalent in $U$, where the function $\Phi _{j}$ is defined by (3.1). Let $q_{1}$ is convex (univalent) function in $U$, and assume that $q_{2}$ is convex in $U$, that $q_{2}$ satisfies (3.28). If

then

and $q_{1}(z)$ and $q_{2}(z)$ are respectively the best subordinate and best dominant of the above subordination.

4.   Conclusions

We used the application of higher order derivatives to obtained a number of interesting results concerning differential subordination and superordination relations for the operator $\Im _{p}(\nu, \rho; \ell)f(z)$ of multivalent functions analytic in $U, \ $the differential subordination outcomes are followed by some special cases and counters examples. Differential sandwich-type results have been obtained. Our results we obtained are new and could help the mathematicians in the field of Geometric Function Theory to solve other special results in this field.

Acknowledgments

This research has been funded by Deputy for Research & innovation, Ministry of Education through initiative of institutional funding at university of Ha'il, Saudi Arabia through project number IFP-22155.

Conflict of interest

The authors declare no conflict of interest.