Applications of Subordination Chains and Fractional Integral in Fuzzy Differential Subordinations

Abstract

Fuzzy differential subordination theory represents a generalization of the classical concept of differential subordination which emerged in the recent years as a result of embedding the concept of fuzzy set into geometric function theory. The fractional integral of Gaussian hypergeometric function is defined in this paper and using properties of the subordination chains, new fuzzy differential subordinations are obtained. Dominants of the fuzzy differential subordinations are given and using particular functions as such dominants, interesting geometric properties interpreted as inclusion relations of certain subsets of the complex plane are presented in the corollaries of the original theorems stated. An example is constructed as an application of the newly proved results.

1. Introduction

In the theory of differential equations, there are often examples of differential inequalities which can be used for obtaining bounds on a function using its derivatives of a certain order. The ideas regarding differential inequalities for real functions were extended to complex functions and a new theory emerged called differential subordination theory. This theory was introduced by S.S. Miller and P.T. Mocanu in a paper published in 1978 [1] and another one published in 1981 [2]. This theory developed nicely during the next years and its main features can be found in the monograph published in 2000 [3] by the authors who have introduced it.

The present paper deals with a special form of differential subordination named fuzzy differential subordination. This concept was introduced in 2012 [4] after the notion of subordination in fuzzy sets theory was defined in 2011 [5]. This merge between fuzzy sets theory and geometric function theory was inspired by the constant concerns of mathematicians to embed the notion of fuzzy set into already established mathematical theories. The notion of fuzzy set was introduced by Lotfi A. Zadeh in 1965 [6] and had an extraordinary evolution being used nowadays in many branches of science and technique. Some applications of the notion in fuzzy linear fractional programming can be seen in [7]. Data analysis of the quadcopter control and movement involving PID (Proportional Integral Derivative) and Fuzzy-PID control is carried out in [8] and an evolution of the concept of fuzzy normed linear spaces is presented in [9]. Some steps in the development of the concept of fuzzy set can be followed in the nice review papers [10,11].

The word “fuzzy” associated to the concept of differential subordination is related to the definition of the fuzzy differential subordination where the codomain of the involved function is the interval $\left[0, 1\right]$. The relevance of the results from the perspective of the fuzzy set theory is made clear by the example contained in the paper but since the attempt to introduce the new theory of fuzzy differential subordination is rather new, so far, no applications in other fields of research or science have emerged. Hopefully, applications of this concept will appear after further investigations.

The theory of fuzzy differential subordination follows the general theory of differential subordination and develops by adapting most of the ideas of the classical theory for obtaining new results. The first paper concerned introducing the notions of fuzzy dominant and fuzzy best dominant [12] and then, the special form of Briot-Bouquet fuzzy differential subordination was considered in studies [13]. Different types of operators were added to the researches [14,15,16] and the dual concept of fuzzy differential superordination was also introduced in 2017 [17] following the classical theory of differential superordination introduced by S.S. Miller and P.T. Mocanu in 2003 [18].

In the last few years, the development of both theories of fuzzy differential subordination and superordination continued. Nice results were obtained using different types of operators such as an integral operator [19], a linear operator [20], a differential operator [21,22], or a Mittag-Leffler-type Borel distribution series [23].

A prolific study is conducted involving fractional calculus in fuzzy differential subordination and superordination. Interesting results were published using the Atangana–Baleanu fractional integral of Bessel functions [24] and the fractional integral of confluent hypergeometric function [25,26].

Keeping those studies in mind, in this paper the fractional integral is applied to Gaussian hypergeometric function for obtaining a new operator used in the study regarding fuzzy differential subordinations. It is worth mentioning that Gaussian hypergeometric function is used for studies in geometric function theory since hypergeometric functions were connected to the theory of univalent functions by L. de Branges in 1985 when he used the generalized hypergeometric function in the proof of Bieberbach’s conjecture [27]. Univalence conditions were obtained for Gaussian hypergeometric function in early studies [28] and in very recent ones [29]. In recent studies, differential inequalities were used and geometric properties were obtained by proving inclusion relations between subsets of the complex plane. Such an approach can also be seen in recent publications [30,31] and will be taken regarding the original results contained in this paper too. The subordination chains and differential inequalities were used for proving that Gaussian hypergeometric function belongs to Carathéodory class of analytic functions in [32].

2. Preliminaries

The notations, definitions and previously proved results used during the study are presented in this section.

Considering $U=\left\{z\in ℂ: \left|z\right|<1\right\}$ the unit disc of the complex plane, $H\left(U\right)$ denotes the class of holomorphic functions in $U$ and the following subclasses are well-known:

$$A_n=\left\{f\in H\left(U\right) :f\left(z\right)=z+a_{n+1}{z}^{n+1}+\cdots , z \in U\right\},$$

with $A_1=A$, and

$$H\left[a,n\right]=\left\{f\in H\left(U\right) :f\left(z\right)=a+a_n{z}^n+a_{n+1}{z}^{n+1}+\cdots , z \in U\right\},$$

with $H_0=H\left[0,1\right]$, when $a\in ℂ,$ and $n$ a natural number different from 0.

Denote by $S$ the class of univalent functions $f\in A.$

The class of starlike functions of order $\alpha$ is defined as:

$$S^{*}\left(\alpha \right)=\left\{f\in A:Re\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}>\alpha \right\},$$

when $0<\alpha <1$ and for $\alpha =0,$ the class of starlike functions is obtained, denoted by $S^{*}$.

The class of convex functions of order $\alpha$ is defined as:

$$K\left(\alpha \right)=\left\{f\in A:Re\left(\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}+1\right)>\alpha \right\},$$

when $0<\alpha <1$ and for $\alpha =0,$ the class of convex functions is obtained, denoted by $K$.

Carathéodory class of analytic functions is defined as:

$$\mathcal{P}=\left\{p\in H\left(U\right):p\left(0\right)=1, Re p\left(z\right)>0,z\in U\right\}.$$

Definition 1 ([33] (p. 81)).

A function$f\in H\left(U\right)$is close-to-convex if there exists a function$\phi$convex in$U$such that:

$$Re\frac{f\prime \left(z\right)}{\phi \prime \left(z\right)}>0, z\in U.$$

Remark 1.

(a) In particular, if$Re f^{\prime }\left(z\right)>0,z\in U,$we say that function$f$is close-to-convex with respect to the identical function $\phi \left(z\right)=z, z\in U.$

(b) Based on Alexander’s duality theorem we have that function$\phi$is convex if and only if function$g\left(z\right)=z{\phi }^{\prime }\in S^{*}.$Hence, we get that function$f$is close-to-convex if and only if$g\in S^{*}$such that

$$Re\frac{zf\prime \left(z\right)}{g\left(z\right)}>0, z\in U.$$

Definition 2 ([3]).

Let$f$and$F$be members of$H\left(U\right)$. The function$f$is said to be subordinate to$F$, written$f\prec F$or$f\left(z\right)\prec F\left(z\right)$, if there exists a function$w$, analytic in$U$, with conditions$w\left(0\right)=0$and$\left|w\left(z\right)\right|<1$and such that$f\left(z\right)=F\left(w\left(z\right)\right).$If$F$is univalent, then$f\prec F$if and only if$f\left(0\right)=F\left(0\right)$and$f\left(U\right) \subset F\left(U\right).$

Definition 3 ([3]).

A function$L\left(z,t\right), z\in U, t\ge 0,$is a subordination chain if$L\left(·,t\right)$is analytic and univalent in$U$for all$t\ge 0, L\left(·,t\right)$is continuously differentiable on${ℝ}^{+}$for all$z\in U$and$L\left(z,s\right)\prec L\left(z,t\right)$when$0\le s\le t.$

Lemma 1 ([3]).

The function$L\left(z,t\right)=a_1\left(t\right)z+a_2\left(t\right)z^2+\cdots$with$a_1\left(t\right)\ne 0$for$t\ge 0$and$\underset{t\to \infty }{\lim }\left|a_1\left(t\right)\right|=\infty$is a subordination chain if

$$Re\frac{z\frac{\partial L\left(z,t\right)}{\partial z}}{\frac{\partial L\left(z,t\right)}{\partial t}}>0, z\in U,t\ge 0.$$

Definition 4 ([3]).

We denote by$Q$the set of functions$q$that are analytic and injective on$\overline{U}\backslash E\left(q\right)$where

$$E\left(q\right)=\left\{\zeta \in \partial U: \underset{z\to \zeta }{\lim }q\left(z\right)=\infty \right\},$$

and are such that$q\prime \left(\zeta \right)\ne 0$for$\zeta \in \partial U\backslash E\left(q\right), \partial U=\left\{z\in ℂ: \left|z\right|=1\right\}.$

Lemma 2 ([3]).

Let$q\in Q$with$q\left(0\right)=a$and let$p\left(z\right)=a+a_n{z}^n+\cdots$be analytic in$U$with$p\left(z\right)\ne a$and$n\ge 1.$If$p$is not subordinate to$q$, then there exist points$z_0=r_0{e}^{i{\theta }_0}\in U$and${\zeta }_0\in \partial U\backslash E\left(q\right)$and an$m\ge n\ge 1$for which$p\left(U_{r_0}\right)\subset q\left(U\right)$and

(i). 

$p\left(z_0\right)=q\left({\zeta }_0\right),$

(ii). 

$z_{0}p\prime \left(z_0\right)=m{\zeta }_{0}q\left({\zeta }_0\right)$

(iii). 

$Re\left(\frac{z_{0}p″\left(z_0\right)}{p\prime \left(z_0\right)}+1\right)\ge mRe\left(\frac{{\zeta }_{0}q″\left({\zeta }_0\right)}{q\prime \left({\zeta }_0\right)}+1\right).$

The definition of Gaussian hypergeometric function is reminded in [3] and presented below:

Definition 5 ([3]).

Let a, b, c$\in ℂ$,$c\ne 0,-1,-2,\dots$The function

$$\begin{gathered} F\left(a,b,c;z\right)=1+\frac{ab}{c}·\frac{z}{1!}+\frac{a\left(a+1\right)b\left(b+1\right)}{c\left(c+1\right)}·\frac{z^2}{2!}+\cdots ={\displaystyle \sum }_{k=0}^{\infty }\frac{{\left(a\right)}_k·{\left(b\right)}_k}{{\left(c\right)}_k}·\frac{z^k}{k!} \\ =\frac{\Gamma \left(c\right)}{\Gamma \left(a\right)\Gamma \left(b\right)}{\displaystyle \sum }_{k=0}^{\infty }\frac{\Gamma \left(a+k\right)\Gamma \left(b+k\right)}{\Gamma \left(c+k\right)}·\frac{z^k}{k!}, z\in U, \end{gathered}$$

(1)

is called Gaussian hypergeometric function, where

$${\left(d\right)}_k=\frac{\Gamma \left(d+k\right)}{\Gamma \left(d\right)}=d\left(d+1\right)\left(d+2\right)\dots \left(d+k-1\right) with {\left(d\right)}_0=1,$$

And

$$\Gamma \left(z\right)={{\displaystyle \int }}_0^{\infty }{e}^{-t}{t}^{z-1}dt with \Gamma \left(z+1\right)=z·\Gamma \left(z\right), \Gamma \left(1\right)=1, \Gamma \left(n+1\right)=n!.$$

The fractional integral of order $\lambda$ used by Owa [34] and Owa and Srivastava [35] is defined as:

Definition 6 ([34,35]).

The fractional integral of order$\lambda (\lambda >0)$is defined for a function$f$by the following expression:

$$D_z^{-\lambda }f\left(z\right)=\frac{1}{\Gamma \left(\lambda \right)}{{\displaystyle \int }}_0^z\frac{f\left(t\right)}{{\left(z-t\right)}^{1-\lambda }}dt,$$

(2)

where$f$is an analytic function in a simply-connected region of the z-plane containing the origin and the multiplicity of${\left(z-t\right)}^{1-\lambda }$is removed by requiring$log\left(z-t\right)$to be real when$z-t>0.$

Definition 7 ([4]).

A pair$\left(A, F_A\right)$, where$F_A : X \to \left[0, 1\right]$and$A=\left\{x\in X:0<F_A\left(x\right)<1\right\},$is called fuzzy subset of$X$. The set$A$is called the support of the fuzzy set$\left(A, F_A\right)$and$F_A$is called the membership function of the fuzzy set$\left(A, F_A\right).$One can also denote$A=supp\left(A, F_A\right)$.

Definition 8 ([4,5]).

Let$D\subset ℂ$and let$z_0\in D$be a fixed point and let the functions$f,g$be holomorphic in$U.$The function$f$is said to be fuzzy subordinate to function$g$and write$f{\prec }_{F}g$or$f\left(z\right){\prec }_{F}g\left(z\right)$if there exists a function$G:ℂ\to \left[0,1\right]$such that:

1. $f\left(z_0\right)=g\left(z_0\right);$

2. $G\left(f(z)\right)\le G\left(g(z)\right), z\in D.$

Remark 2.

Such a function$G:ℂ\to \left[0,1\right]$can be considered:

$$G\left(z\right)=\frac{\left|z\right|}{1+\left|z\right|} or G\left(z\right)=\frac{1}{1+\left|z\right|}.$$

Condition 2 from Definition 9 is then written as:

$$\frac{\left|f\left(z\right)\right|}{1+\left|f\left(z\right)\right|}\le \frac{\left|g\left(z\right)\right|}{1+\left|g\left(z\right)\right|} or \frac{1}{1+\left|f\left(z\right)\right|}\le \frac{1}{1+\left|g\left(z\right)\right|}, z\in D,$$

and it is equivalent to$f\left(D\right)\subset g\left(D\right).$

If $\mathrm{D}=\mathrm{U}=\left\{\mathrm{z}\in ℂ: \left|\mathrm{z}\right|<1\right\},$ and $\mathrm{g}$ is univalent in $\mathrm{U},$ then conditions 1 and 2 become $\mathrm{f}\left(0\right)=\mathrm{g}\left(0\right), \mathrm{f}\left(\mathrm{U}\right)\subset \mathrm{g}\left(\mathrm{U}\right),\mathrm{g}\left(\partial \mathrm{U}\right)\notin \mathrm{f}\left(\mathrm{U}\right),$ where $\partial \mathrm{U}=\left\{\mathrm{z}\in ℂ: \left|\mathrm{z}\right|=1\right\}.$

Definition 9.

Ref. [5] Let $\psi :{ℂ}^3\times U\to ℂ, a\in ℂ$ and let $h$ be univalent in $U,$ with $h\left(0\right)=\psi \left(a,0,0;0\right).$ Let $q$ be univalent in $U$ with $q\left(0\right)=a$ and let $p$ be analytic in $U$ with $p\left(0\right)=a.$ Function $\psi \left(p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right);z\right)$ is also analytic in $U$ and let $G:ℂ\to \left[0,1\right]$. If $p$ satisfies the (second-order) fuzzy differential subordination

$$\begin{gathered} G\left(\psi \left(p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right);z\right)\right)\le G\left(h\left(z\right)\right) i.e., \\ \psi \left(p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right);z\right){\prec }_{F}h\left(z\right), \end{gathered}$$

(3)

then $p$ is called a fuzzy solution of the fuzzy differential subordination. The univalent function $q$ is called a fuzzy dominant of the fuzzy solutions of the fuzzy differential subordination, or more simply a fuzzy dominant, if $p\left(z\right){\prec }_{F}q\left(z\right) or G\left(p\left(z\right)\right)\le G\left(q\left(z\right)\right), z\in U,$ for all $p$ satisfying (3). A fuzzy dominant $\tilde{q}$ that satisfies $\tilde{q}\left(z\right){\prec }_{F}q\left(z\right) or G\left(\tilde{q}\left(z\right)\right)\le G\left(q\left(z\right)\right), z\in U,$ for all fuzzy dominants $q$ of (3) is said to be fuzzy best dominant of (3).

Using the known results presented in this section, new fuzzy differential subordinations are studied in the next section. First, the fractional integral of Gaussian hypergeometric function is defined and certain useful properties are stated. Fuzzy differential subordinations involving fractional integral of Gaussian hypergeometric function are investigated in the two main theorems proved. A fuzzy dominant is given for the first fuzzy differential subordination considered and the fuzzy best dominant is obtained for the second fuzzy differential subordination of the study. Interesting corollaries are obtained when specific functions are used as fuzzy dominant and fuzzy best dominant in the two original theorems.

3. Results

The first original result presented in this paper is the definition of the fractional integral of Gaussian hypergeometric function obtained using the functions given by (1) and (2).

Definition 10.

Let $\lambda >0.$The fractional integral of Gaussian hypergeometric function is the analytic function in$U$given by:

$$\begin{gathered} D_z^{-\lambda }F\left(a,b,c;z\right)=\frac{1}{\Gamma \left(\lambda \right)}{{\displaystyle \int }}_0^z\frac{\frac{\Gamma \left(c\right)}{\Gamma \left(a\right)\Gamma \left(b\right)}{{\displaystyle \sum }}_{k=0}^{\infty }\frac{\Gamma \left(a+k\right)\Gamma \left(b+k\right)}{\Gamma \left(c+k\right)}·\frac{z^k}{k!}}{{\left(z-t\right)}^{1-\lambda }}dt \\ =\frac{\Gamma \left(c\right)}{\Gamma \left(a\right)\Gamma \left(b\right)}{\displaystyle \sum }_{k=0}^{\infty }\frac{\Gamma \left(a+k\right)\Gamma \left(b+k\right)}{\Gamma \left(c+k\right)\Gamma \left(\lambda +k+1\right)}·z^{k+\lambda }, z\in U. \end{gathered}$$

(4)

Remark 3.

Certain useful properties of the fractional integral of Gaussian hypergeometric function are next given:

(a)$D_z^{-\lambda }F\left(a,b,c;z\right)=\frac{1}{\Gamma \left(\lambda +1\right)}·z^{\lambda }+\frac{\Gamma \left(c\right)}{\Gamma \left(a\right)\Gamma \left(b\right)}·\frac{\Gamma \left(a+1\right)\Gamma \left(b+1\right)}{\Gamma \left(c+1\right)\Gamma \left(\lambda +2\right)}·z^{\lambda +1}+\cdots =\frac{1}{\Gamma \left(\lambda +1 \right)}·z^{\lambda }+\frac{ab}{c}·\frac{1}{\Gamma \left(\lambda +2 \right)}·z^{\lambda +1}+\cdots$We conclude that$D_z^{-\lambda }F\left(a,b,c;z\right)\in H\left[0,\lambda \right].$

(b)${\left[D_z^{-\lambda }F\left(a,b,c;z\right)\right]}^{\prime }=\frac{\lambda }{\Gamma \left(\lambda +1 \right)}·z^{\lambda -1}+\frac{ab}{c}·\frac{\lambda +1}{\Gamma \left(\lambda +2 \right)}·z^{\lambda }+\cdots$

(c) For$\lambda =1,$

$$D_z^{-1}F\left(a,b,c;z\right)=z+\frac{ab}{c}·\frac{z^2}{2}+\cdots ,$$

which shows that$D_z^{-1}F\left(a,b,c;z\right)\in A.$

We can also write:

$${\left[D_z^{-1}F\left(a,b,c;z\right)\right]}^{\prime }=1+\frac{ab}{c}·z+\cdots and {\left[D_z^{-1}F\left(a,b,c;0\right)\right]}^{\prime }=1\ne 0.$$

In the first theorem proved in this paper, a fuzzy dominant is given for the investigated fuzzy differential subordination involving the fractional integral of Gaussian hypergeometric function. Both Lemmas 1 and 2 are used for the proof of this result.

Theorem 1.

Let$h\in H\left(U\right)$and let$\varphi \in H\left(D\right),$where D is a domain which contains$h\left(U\right).$Let$G:ℂ\to \left[0,1\right], G\left(z\right)=\frac{1}{1+\left|z\right|}.$

Assume that:

(j) 

$Re \varphi \left(h\left(z\right)\right)>0,z\in U,$

(jj) 

$H\left(z\right)=zh\prime \left(z\right)·\varphi \left(h\left(z\right)\right)$is a starlike function in$U.$

If the fractional integral of Gaussian hypergeometric function given by (4) is an analytic function in$U$, with$D_z^{-\lambda }F\left(a,b,c;0\right)=h\left(0\right), D_z^{-\lambda }F\left(U\right)\subset D,$and satisfies the fuzzy differential subordination

$$D_z^{-\lambda }F\left(a,b,c;z\right)+z{\left[D_z^{-\lambda }F\left(a,b,c;z\right)\right]}^{\prime }·\varphi \left(D_z^{-\lambda }F\left(a,b,c;z\right)\right){\prec }_{F}h\left(z\right), z\in U,$$

(5)

equivalently written as

$$\frac{1}{1+\left|D_z^{-\lambda }F\left(a,b,c;z\right)+z{\left[D_z^{-\lambda }F\left(a,b,c;z\right)\right]}^{\prime }·\varphi \left(D_z^{-\lambda }F\left(a,b,c;z\right)\right)\right|}\le \frac{1}{1+\left|h\left(z\right)\right|}, z\in U,$$

or

$$G\left(D_z^{-\lambda }F\left(a,b,c;z\right)+z{\left[D_z^{-\lambda }F\left(a,b,c;z\right)\right]}^{\prime }·\varphi \left(D_z^{-\lambda }F\left(a,b,c;z\right)\right)\right)\le G\left(h\left(z\right)\right), z\in U,$$

then

$$D_z^{-\lambda }F\left(a,b,c;z\right){\prec }_{F}h\left(z\right), z\in U,$$

equivalently written as

$$\frac{1}{1+\left|D_z^{-\lambda }F\left(a,b,c;z\right)\right|}\le \frac{1}{1+\left|h\left(z\right)\right|}, z\in U,$$

or

$$G\left(D_z^{-\lambda }F\left(a,b,c;z\right)\right)\le G\left(h\left(z\right)\right), z\in U.$$

Proof. 

Assume that functions $D_z^{-\lambda }F\left(a,b,c;z\right), h\left(z\right)$ and $H\left(z\right)$ satisfy conditions (i)–(iii) from Lemma 2 on $\overline{U}=\left\{z\in ℂ: \left|z\right|\le 1\right\}.$ Otherwise, those functions can be replaced by $D_{rz}^{-\lambda }F\left(a,b,c;z\right), h_r\left(z\right)$ and $H_r\left(z\right),$ where $0<r<1,$ and the new functions satisfy conditions (i)–(iii) from Lemma 2 on $\overline{U}.$ When $r\to 1,$ we get that $D_{rz}^{-\lambda }F\left(a,b,c;z\right)\to D_z^{-\lambda }F\left(a,b,c;z\right), h_r\left(z\right)\to h\left(z\right)$ and $H_r\left(z\right)\to H\left(z\right)$ for $z\in \overline{U}.$

Next, we prove that function $h$ is close-to-convex, hence univalent.

Condition (jj) from the hypothesis of the theorem states that $H$ is starlike in $U.$ We evaluate:

$$\frac{zh\prime \left(z\right)}{H\left(z\right)}=\frac{zh\prime \left(z\right)}{zh\prime \left(z\right)·\varphi \left(h\left(z\right)\right)}=\frac{1}{\varphi \left(h\left(z\right)\right)}, z\in U.$$

Using condition (j) from hypothesis of the theorem, we can write:

$$Re\frac{zh\prime \left(z\right)}{H\left(z\right)}=Re\frac{1}{\varphi \left(h\left(z\right)\right)}>0, z\in U.$$

Since function $H$ is starlike in $U$, from Remark 1 we have that function $h$ is close-to-convex in $U$, hence univalent in $U.$

Let $L:\overline{U}\times \left[0,\infty \right)\to ℂ$ be given by:

$$\begin{gathered} L\left(z,t\right)=a_1\left(t\right)z+a_2\left(t\right)z^2+\cdots =h\left(z\right)+tH\left(z\right)=h\left(z\right)+tz{h}^{\prime }\left(z\right)·\varphi \left(h\left(z\right)\right), z \\ \in U, t\ge 0, \end{gathered}$$

(6)

which is an analytic function in $\overline{U}$ for any $t\ge 0,$ and continuously differentiable on $\left[0,\infty \right)$ for all $z\in \overline{U}.$

Differentiating (6) with respect to $z$, we obtain:

$$\begin{gathered} \frac{\partial L\left(z,t\right)}{\partial z}=a_1\left(t\right)+2a_2\left(t\right)z^2+\cdots =h\left(z\right)+tH\left(z\right) \\ =h^{\prime }\left(z\right)+t{h}^{\prime }\left(z\right)·\varphi \left(h\left(z\right)\right)+tz{h}^{″}\left(z\right)·\varphi \left(h\left(z\right)\right) \\ +tz{h}^{\prime }\left(z\right){\left[\varphi \left(h\left(z\right)\right)\right]}^{’}, \end{gathered}$$

(7)

for $z\in \overline{U}, t\ge 0,$

$$\frac{\partial L\left(0,t\right)}{\partial z}=a_1\left(t\right)=h^{\prime }\left(0\right)+t{h}^{\prime }\left(0\right)·\varphi \left(h\left(0\right)\right)=h^{\prime }\left(0\right)·\left[1+t\varphi \left(h\left(0\right)\right)\right].$$

Since $h$ is univalent, we have that $h^{\prime }\left(0\right)\ne 0$ and using condition (j) we have $Re \varphi \left(h\left(z\right)\right)>0$ which gives that $Re \varphi \left(h\left(0\right)\right)>0$ and $\varphi \left(h\left(0\right)\right)\ne 0.$

$$a_1\left(t\right)=\frac{\partial L\left(0,t\right)}{\partial z}=h\prime \left(0\right)·\left[1+t\varphi \left(h\left(0\right)\right)\right]\ne 0, t\ge 0.$$

Since $h^{\prime }\left(0\right)\ne 0$ and $\varphi \left(h\left(0\right)\right)\ne 0$ we conclude that $a_1\left(t\right)\ne 0$ and we have:

$$\underset{t\to \infty }{\lim }\left|a_1\left(t\right)\right|=\underset{t\to \infty }{\lim }\left|h\prime \left(0\right)·\left[1+t\varphi \left(h\left(0\right)\right)\right]\right|=\infty .$$

(8)

Using condition (jj) from the hypothesis of the theorem we can write:

$$Re\frac{z{H}^{\prime }\left(z\right)}{H\left(z\right)}>0,z\in U.$$

(9)

Differentiating (6) with respect to $t$, we obtain:

$$\frac{\partial L\left(z,t\right)}{\partial t}=z{h}^{\prime }\left(z\right)·\varphi \left(h\left(z\right)\right),z\in \overline{U}, t\ge 0.$$

Using conditions (j) and (jj) and relation (9) we write:

$$Re\frac{z\frac{\partial L\left(z,t\right)}{\partial z}}{\frac{\partial L\left(z,t\right)}{\partial t}}=Re \frac{1}{\varphi \left(h\left(z\right)\right)}+tRe\frac{z{H}^{\prime }\left(z\right)}{H\left(z\right)}>0,z\in \overline{U}, t\ge 0.$$

(10)

Using (8) and (10), Lemma 1 can be applied and we have that function $L\left(z,t\right)$ given by (6) is a subordination chain.

Considering the definition of the subordination chain, we have:

$$L\left(z,s\right){\prec }_{F}L\left(z,t\right), 0\le s\le t.$$

(11)

Using (6) we know that $L\left(z,0\right)=h\left(z\right), z\in U$ and relation (11) becomes:

$$h\left(z\right){\prec }_{F}L\left(z,t\right), z\in \overline{U}, t\ge 0,$$

which, using Remark 2, is equivalent to

$$h\left(U\right)\subset \left\{L\left(z,t\right): z\in \overline{U}, t\ge 0\right\} \mathrm{and} L\left(\zeta ,t\right)\notin h\left(U\right) \mathrm{for} \left|\zeta \right|=1, t\ge 0.$$

(12)

Since $D_z^{-\lambda }F\left(a,b,c;0\right)=h\left(0\right)$ and $h$ is a univalent function, using fuzzy subordination (5) we get:

$$D_z^{-\lambda }F\left(a,b,c;z\right)+z{\left[D_z^{-\lambda }F\left(a,b,c;z\right)\right]}^{\prime }·\varphi \left(D_z^{-\lambda }F\left(a,b,c;z\right)\right)\in h\left(U\right), z\in U.$$

(13)

For $z=z_0\in U$, relation (13) becomes:

$$D_z^{-\lambda }F\left(a,b,c;z_0\right)+z_0{\left[D_z^{-\lambda }F\left(a,b,c;z_0\right)\right]}^{\prime }·\varphi \left(D_z^{-\lambda }F\left(a,b,c;z_0\right)\right)\in h\left(U\right).$$

(14)

Lemma 2 will be applied in order to prove that fuzzy differential subordination (5) implies $D_z^{-\lambda }F\left(a,b,c;z\right){\prec }_{F}h\left(z\right), z\in U$.

Let’s assume that $D_z^{-\lambda }F\left(a,b,c;z\right){\nprec }_{F}h\left(z\right), z\in U$. Using Lemma 2, we have that there exist points $z_0\in U$ and ${\zeta }_0\in \partial U$ and an $m\ge 1$ such that $D_z^{-\lambda }F\left(a,b,c;z_0\right)=h\left({\zeta }_0\right)$ and $z_0{\left[D_z^{-\lambda }F\left(a,b,c;z_0\right)\right]}^{\prime }=m{\zeta }_0{h}^{\prime }\left({\zeta }_0\right).$ Then:

$$\begin{gathered} D_z^{-\lambda }F\left(a,b,c;z_0\right)+z_0{\left[D_z^{-\lambda }F\left(a,b,c;z_0\right)\right]}^{\prime }·\varphi \left(D_z^{-\lambda }F\left(a,b,c;z_0\right)\right) \\ =h\left({\zeta }_0\right)+m{\zeta }_0{h}^{\prime }\left({\zeta }_0\right)·\varphi \left(h\left({\zeta }_0\right)\right). \end{gathered}$$

(15)

For $z={\zeta }_0\in \partial U, \left|{\zeta }_0\right|=1$ relation (6) becomes:

$$L\left({\zeta }_0,t\right)=h\left({\zeta }_0\right)+t{\zeta }_0{h}^{\prime }\left({\zeta }_0\right)·\varphi \left(h\left({\zeta }_0\right)\right), t\ge 0.$$

(16)

For $t=m>1,$ relation (16) becomes:

$$L\left({\zeta }_0,m\right)=h\left({\zeta }_0\right)+m{\zeta }_0{h}^{\prime }\left({\zeta }_0\right)·\varphi \left(h\left({\zeta }_0\right)\right).$$

(17)

By using (17) in (15), we have:

$$D_z^{-\lambda }F\left(a,b,c;z_0\right)+z_0{\left[D_z^{-\lambda }F\left(a,b,c;z_0\right)\right]}^{\prime }·\varphi \left(D_z^{-\lambda }F\left(a,b,c;z_0\right)\right)=L\left({\zeta }_0,m\right), m\ge 1.$$

(18)

Using $\zeta ={\zeta }_0, \left|{\zeta }_0\right|=1, t=m$ in (12) we obtain:

$$L\left({\zeta }_0,m\right)\notin h\left(U\right) \mathrm{for} \left|{\zeta }_0\right|=1, m\ge 1.$$

(19)

Using (19) in (18) we get:

$$D_z^{-\lambda }F\left(a,b,c;z_0\right)+z_0{\left[D_z^{-\lambda }F\left(a,b,c;z_0\right)\right]}^{\prime }·\varphi \left(D_z^{-\lambda }F\left(a,b,c;z_0\right)\right)\notin h\left(U\right),z_0\in U.$$

Since this contradicts (14) we conclude that the assumption made is false, hence

$$D_z^{-\lambda }F\left(a,b,c;z\right){\prec }_{F}h\left(z\right), z\in U.$$

(20)

□

The following corollary of Theorem 1 is obtained by taking:

$$h\left(z\right)=\frac{z}{1-z}, h\left(0\right)=0, h\in H\left(U\right).$$

Corollary 1.

Consider

$$h\left(z\right)=\frac{z}{1-z}, h\left(0\right)=0, h\in H\left(U\right),$$

and let$\varphi \in H\left(D\right),$where D is a domain which contains$h\left(U\right).$

Let

$$G:ℂ\to \left[0,1\right], G\left(z\right)=\frac{1}{1+\left|z\right|}.$$

assume that:

(j)$Re \varphi \left(\frac{z}{1-z}\right)>0,z\in U, and$

(jj)$H\left(z\right)=\frac{z}{1-z^2}·\varphi \left(\frac{z}{1-z}\right)$is a starlike function in U.

If the fractional integral of Gaussian hypergeometric function given by (4) is an analytic function in$U$, with$D_z^{-\lambda }F\left(a,b,c;0\right)=h\left(0\right), D_z^{-\lambda }F\left(U\right)\subset D,$and satisfies the fuzzy differential subordination

$$D_z^{-\lambda }F\left(a,b,c;z\right)+z{\left[D_z^{-\lambda }F\left(a,b,c;z\right)\right]}^{\prime }·\varphi \left(D_z^{-\lambda }F\left(a,b,c;z\right)\right){\prec }_F\frac{z}{1-z}, z\in U,$$

(21)

equivalently written as

$$\begin{gathered} \frac{1}{1+\left|D_z^{-\lambda }F\left(a,b,c;z\right)+z{\left[D_z^{-\lambda }F\left(a,b,c;z\right)\right]}^{\prime }·\varphi \left(D_z^{-\lambda }F\left(a,b,c;z\right)\right)\right|} \\ \le \frac{1}{1+\left|\frac{z}{1-z}\right|}, z\in U, \end{gathered}$$

or

$$G\left(D_z^{-\lambda }F\left(a,b,c;z\right)+z{\left[D_z^{-\lambda }F\left(a,b,c;z\right)\right]}^{\prime }·\varphi \left(D_z^{-\lambda }F\left(a,b,c;z\right)\right)\right)\le G\left(\frac{z}{1-z}\right), z\in U,$$

then

$$D_z^{-\lambda }F\left(a,b,c;z\right){\prec }_F\frac{z}{1-z}, z\in U,$$

equivalently written as

$$\frac{1}{1+\left|D_z^{-\lambda }F\left(a,b,c;z\right)\right|}\le \frac{1}{1+\left|\frac{z}{1-z}\right|}, z\in U,$$

or

$$G\left(D_z^{-\lambda }F\left(a,b,c;z\right)\right)\le G\left(\frac{z}{1-z}\right), z\in U,$$

which implies

$$Re D_z^{-\lambda }F\left(a,b,c;z\right)>-\frac{1}{2}, z\in U.$$

Proof. 

Using relation (20) obtained in the proof of Theorem 1, by putting $h\left(z\right)=\frac{z}{1-z},$ we obtain:

$$D_z^{-\lambda }F\left(a,b,c;z\right){\prec }_{F}h\left(z\right)=\frac{z}{1-z}, z\in U.$$

(22)

The convexity of function $h$ must be proved in order to obtain the conclusion of this corollary.

We evaluate:

$$h^{\prime }\left(z\right)=\frac{1}{{\left(1-z\right)}^2}, h^{″}\left(z\right)=\frac{2}{{\left(1-z\right)}^3}.$$

We can write

$$\frac{zh″\left(z\right)}{h\prime \left(z\right)}+1=\frac{1+z}{1-z}, z\in U.$$

(23)

We analyze the function

$$f\left(z\right)=\frac{1+z}{1-z}, f^{\prime }\left(z\right)=\frac{2}{{\left(1-z\right)}^2}, f^{\prime }\left(0\right)=2\ne 0, z\in U.$$

Since from $f\left(z_1\right)=f\left(z_2\right)$ we obtain that $z_1=z_2,$ we conclude that function $f$ is univalent in $U$ with $f^{\prime }\left(0\right)\ne 0.$ This means that $f$ is conformal mapping of the unit disk into the half-plane $S=\left\{w\in ℂ:Re w>0\right\}$ from which we can write:

$$Re f\left(z\right)=Re\frac{1+z}{1-z}>0, z\in U.$$

(24)

Using (24) in (23) we obtain:

$$Re \left[\frac{zh″\left(z\right)}{h\prime \left(z\right)}+1\right]>0, z\in U,$$

hence $h\in K$ which gives that (22) is equivalent to

$$Re D_z^{-\lambda }F\left(a,b,c;z\right)>Re\frac{z}{1-z}, z\in U.$$

(25)

Since $h\left(z\right)=\frac{z}{1-z}$ is a convex function, hence univalent, and since $h^{\prime }\left(0\right)=1\ne 0,$ we conclude that this function is a conformal mapping of the unit disc into the half-plane $S\prime =\left\{w\in ℂ:Re w>-\frac{1}{2}\right\}$ from which we can write:

$$Re \frac{z}{1-z}>-\frac{1}{2}, z\in U.$$

(26)

Using (26) in (25) we obtain:

$$Re D_z^{-\lambda }F\left(a,b,c;z\right)>-\frac{1}{2}, z\in U.$$

□

The best fuzzy dominant is obtained for the fuzzy differential subordination involving the fractional integral of Gaussian hypergeometric function studied in the next theorem.

Theorem 2.

Let q be a univalent function in$U$with$q\left(0\right)=0, \alpha , \beta \in H\left(D\right),$where D is a domain which contains$q\left(U\right).$Let

$$G:ℂ\to \left[0,1\right], G\left(z\right)=\frac{1}{1+\left|z\right|}.$$

assume that:

(l)$H\left(z\right)=zq\prime \left(z\right)·\beta \left(q\left(z\right)\right)$is a starlike function in$U;$

(ll)$Re\frac{{\alpha }^{\prime }\left(q\left(z\right)\right)}{\beta \left(q\left(z\right)\right)}>0, z\in U;$

(lll)$h\left(z\right)=\alpha \left(q\left(z\right)\right)+H\left(z\right)=\alpha \left(q\left(z\right)\right)+z{q}^{\prime }\left(z\right)·\beta \left(q\left(z\right)\right), z\in U,$is analytic in$U.$

If the fractional integral of Gaussian hypergeometric function given by (4) is an analytic function in$U$, with$D_z^{-1}F\left(a,b,c;0\right)=q\left(0\right)=0, D_z^{-1}F\left(U\right)\subset D,$and satisfies the fuzzy differential subordination

$$\begin{gathered} \alpha \left(D_z^{-1}F\left(a,b,c;z\right)\right)+z{\left[D_z^{-1}F\left(a,b,c;z\right)\right]}^{\prime }·\beta \left(D_z^{-1}F\left(a,b,c;z\right)\right){\prec }_F \\ \alpha \left(q\left(z\right)\right)+z{q}^{\prime }\left(z\right)·\beta \left(q\left(z\right)\right), z\in U, \end{gathered}$$

(27)

equivalently written as

$$\begin{gathered} \frac{\left|\alpha \left(D_z^{-1}F\left(a,b,c;z\right)\right)+z{\left[D_z^{-1}F\left(a,b,c;z\right)\right]}^{\prime }·\beta \left(D_z^{-1}F\left(a,b,c;z\right)\right)\right|}{1+\left|\alpha ·D_z^{-1}F\left(a,b,c;z\right)+z{\left[D_z^{-1}F\left(a,b,c;z\right)\right]}^{\prime }·\beta \left(D_z^{-1}F\left(a,b,c;z\right)\right)\right|} \\ \le \frac{\left|\alpha \left(q\left(z\right)\right)+z{q}^{\prime }\left(z\right)\beta \left(q\left(z\right)\right)\right|}{1+\left|\alpha \left(q\left(z\right)\right)+z{q}^{\prime }\left(z\right)\beta \left(q\left(z\right)\right)\right|}, z\in U, \end{gathered}$$

or

$$\begin{gathered} G\left(\alpha \left(D_z^{-1}F\left(a,b,c;z\right)\right)+z{\left[D_z^{-1}F\left(a,b,c;z\right)\right]}^{\prime }·\beta \left(D_z^{-1}F\left(a,b,c;z\right)\right)\right) \\ \le G\left(\alpha \left(q\left(z\right)\right)+z{q}^{\prime }\left(z\right)\beta \left(q\left(z\right)\right)\right), z\in U, \end{gathered}$$

then

$$D_z^{-1}F\left(a,b,c;z\right){\prec }_{F}q\left(z\right), z\in U,$$

equivalently written as

$$\frac{\left|D_z^{-1}F\left(a,b,c;z\right)\right|}{1+\left|D_z^{-1}F\left(a,b,c;z\right)\right|}\le \frac{\left|q\left(z\right)\right|}{1+\left|q\left(z\right)\right|}, z\in U,$$

or

$$G\left(D_z^{-1}F\left(a,b,c;z\right)\right)\le G\left(q\left(z\right)\right), z\in U,$$

and q is the best fuzzy dominant.

Proof. 

Just like in the proof of Theorem 1, assume that functions $D_z^{-1}F\left(a,b,c;z\right), h\left(z\right)$ and $H\left(z\right)$ satisfy conditions (i)–(iii) from Lemma 2 on $\overline{U}=\left\{z\in ℂ: \left|z\right|\le 1\right\}.$

Next, we prove that function $h$ given by (lll) is univalent.

We differentiate $h$ and after a short calculation we obtain:

$$Re\frac{zh\prime \left(z\right)}{H\left(z\right)}=Re\frac{{\alpha }^{\prime }\left(q\left(z\right)\right)}{\beta \left(q\left(z\right)\right)}+Re\frac{H\prime \left(z\right)}{H\left(z\right)}, z\in U.$$

Using conditions (l) and (ll) from hypothesis of the theorem, we can write:

$$Re\frac{zh\prime \left(z\right)}{H\left(z\right)}>0, z\in U.$$

Since function $H$ is starlike in $U$, from Remark 1 we have that function $h$ is close-to-convex in $U$, hence univalent in $U.$

Let $L:\overline{U}\times \left[0,\infty \right)\to ℂ$ be given by:

$$L\left(z,t\right)=a_1\left(t\right)z+a_2\left(t\right)z^2+\cdots =\alpha \left(q\left(z\right)\right)+\left(t+1\right)z{q}^{\prime }\left(z\right)·\beta \left(q\left(z\right)\right), z\in \overline{U}, t\ge 0,$$

(28)

which is an analytic function in $\overline{U}$ for any $t\ge 0,$ and continuously differentiable on $\left[0,\infty \right)$ for all $z\in \overline{U}.$

Differentiating (28) with respect to $z$, we obtain:

$$\begin{gathered} \frac{\partial L\left(z,t\right)}{\partial z}=a_1\left(t\right)+2a_2\left(t\right)z^2+\cdots = \\ =q^{\prime }\left(z\right)·{\alpha }^{\prime }\left(q\left(z\right)\right)+\left(t+1\right)q^{\prime }\left(z\right)·\beta \left(q\left(z\right)\right)+\left(t+1\right)z{q}^{″}\left(z\right) \\ ·\beta \left(q\left(z\right)\right)+\left(t+1\right)z{\left[q^{\prime }\left(z\right)\right]}^2·\beta \prime \left(q\left(z\right)\right), \end{gathered}$$

(29)

for $z\in \overline{U}, t\ge 0.$

For $z=0$, relation (29) becomes:

$$\frac{\partial L\left(0,t\right)}{\partial z}=a_1\left(t\right)=q^{\prime }\left(0\right){\alpha }^{\prime }\left(q\left(0\right)\right)+\left(t+1\right)q^{\prime }\left(0\right)·\beta \left(q\left(0\right)\right).$$

(30)

Using condition (l) we have:

$$H^{\prime }\left(z\right)=q^{\prime }\left(z\right)·\beta \left(q\left(z\right)\right)+z{q}^{″}\left(z\right)·\beta \left(q\left(z\right)\right)+z{\left[q^{\prime }\left(z\right)\right]}^2·{\beta }^{\prime }\left(q\left(z\right)\right), z\in \overline{U}.$$

(31)

For $z=0$, relation (31) becomes:

$$H^{\prime }\left(0\right)=q^{\prime }\left(0\right)·\beta \left(q\left(0\right)\right).$$

(32)

Since function $H$ is starlike, hence univalent in $U,$ we have that $H^{\prime }\left(0\right)\ne 0.$ From (32) we obtain that $q^{\prime }\left(0\right)·\beta \left(q\left(0\right)\right)\ne 0$. Since $q$ is univalent in $U$, we have $q^{\prime }\left(0\right)\ne 0$ and we conclude that $\beta \left(q\left(0\right)\right)\ne 0.$

Using relation (32) and the fact that $\beta \left(q\left(0\right)\right)\ne 0$, from relation (30) we write:

$$a_1\left(t\right)=\frac{\partial L\left(0,t\right)}{\partial z}=q\prime \left(0\right)·\beta \left(q\left(0\right)\right)·\left(\frac{{\alpha }^{\prime }\left(q\left(0\right)\right)}{\beta \left(q\left(0\right)\right)}+t+1\right)\ne 0, t\ge 0,$$

and

$$\underset{t\to \infty }{\lim }\left|a_1\left(t\right)\right|=\underset{t\to \infty }{\lim }\left|q^{\prime }\left(0\right)·\beta \left(q\left(0\right)\right)·\left(\frac{{\alpha }^{\prime }\left(q\left(0\right)\right)}{\beta \left(q\left(0\right)\right)}+t+1\right)\right|=\infty .$$

(33)

Differentiating (28) with respect to $t$, we obtain:

$$\frac{\partial L\left(z,t\right)}{\partial t}=z{q}^{\prime }\left(z\right)·\beta \left(q\left(z\right)\right),z\in \overline{U}, t\ge 0.$$

(34)

Using relations (29) and (34) we write:

$$Re\frac{z\frac{\partial L\left(z,t\right)}{\partial z}}{\frac{\partial L\left(z,t\right)}{\partial t}}=Re \frac{{\alpha }^{\prime }\left(q\left(z\right)\right)}{\beta \left(q\left(z\right)\right)}+\left(t+1\right)Re\frac{z{H}^{\prime }\left(z\right)}{H\left(z\right)}>0,z\in \overline{U}, t\ge 0.$$

(35)

Considering (35), conditions (l) and (ll) and $t\ge 0$, we deduce:

$$Re\frac{z\frac{\partial L\left(z,t\right)}{\partial z}}{\frac{\partial L\left(z,t\right)}{\partial t}}>0,z\in \overline{U}, t\ge 0.$$

(36)

Lemma 1 can be applied using (33) and (36), and we have that function $L\left(z,t\right)$ given by (28) is a subordination chain.

Considering the definition of the subordination chain, we have:

$$L\left(z,s\right){\prec }_{F}L\left(z,t\right), 0\le s\le t.$$

(37)

By taking $t=0$ in (28) we have $L\left(z,0\right)=h\left(z\right), z\in U$ and relation (37) becomes:

$$L\left(z,0\right)=h\left(z\right){\prec }_{F}L\left(z,t\right), z\in U, t\ge 0,$$

which, using Remark 2, is equivalent to

$$h\left(U\right)\subset \left\{L\left(z,t\right): z\in U, t\ge 0\right\} \mathrm{and} L\left(\zeta ,t\right)\notin h\left(U\right) \mathrm{for} \left|\zeta \right|=1, t\ge 0.$$

(38)

For $z=\zeta , \left|\zeta \right|=1,$ using relations (28) and (38) we obtain:

$$L\left(\zeta ,t\right)=\alpha \left(q\left(\zeta \right)\right)+\left(t+1\right)\zeta q^{\prime }\left(\zeta \right)·\beta \left(q\left(\zeta \right)\right)\notin h\left(U\right), t\ge 0.$$

(39)

For $\zeta ={\zeta }_0, \left|{\zeta }_0\right|=1,$ (39) ca be written as:

$$L\left({\zeta }_0,t\right)=\alpha \left(q\left({\zeta }_0\right)\right)+\left(t+1\right){\zeta }_0{q}^{\prime }\left({\zeta }_0\right)·\beta \left(q\left({\zeta }_0\right)\right)\notin h\left(U\right), t\ge 0.$$

(40)

Since $D_z^{-1}F\left(a,b,c;0\right)=h\left(0\right)$ and $h$ is a univalent function, using fuzzy subordination (27) we get:

$$\alpha (D_z^{-1}F\left(a,b,c;z\right))+z{\left[D_z^{-1}F\left(a,b,c;z\right)\right]}^{\prime }·\beta \left(D_z^{-1}F\left(a,b,c;z\right)\right)\in h\left(U\right), z\in U.$$

(41)

For $z=z_0\in U$, relation (41) becomes:

$$\alpha (D_z^{-1}F\left(a,b,c;z_0\right))+z_0{\left[D_z^{-1}F\left(a,b,c;z_0\right)\right]}^{\prime }·\beta \left(D_z^{-1}F\left(a,b,c;z_0\right)\right)\in h\left(U\right).$$

(42)

Lemma 2 will be applied in order to prove that fuzzy differential subordination (27) implies $D_z^{-1}F\left(a,b,c;z\right){\prec }_{F}q\left(z\right), z\in U$.

Let’s assume that $D_z^{-1}F\left(a,b,c;z\right){\nprec }_{F}q\left(z\right), z\in U$. Using Lemma 2, we have that there exist points $z_0\in U$ and ${\zeta }_0\in \partial U$ and an $m\ge 1$ such that $D_z^{-1}F\left(a,b,c;z_0\right)=q\left({\zeta }_0\right)$ and $z_0{\left[D_z^{-1}F\left(a,b,c;z_0\right)\right]}^{\prime }=m{\zeta }_0{q}^{\prime }\left({\zeta }_0\right).$ Then, using (42) we obtain:

$$\begin{gathered} \alpha \left(D_z^{-1}F\left(a,b,c;z_0\right)\right)+z_0{\left[D_z^{-1}F\left(a,b,c;z_0\right)\right]}^{\prime }·\beta \left(D_z^{-1}F\left(a,b,c;z_0\right)\right) \\ =\alpha \left(q\left({\zeta }_0\right)\right)+m{\zeta }_0{q}^{\prime }\left({\zeta }_0\right)·\beta \left(q\left({\zeta }_0\right)\right). \end{gathered}$$

(43)

If in (40) we put $t=m-1\ge 0,$ we ca write:

$$L\left({\zeta }_0,m-1\right)=\alpha \left(q\left({\zeta }_0\right)\right)+m{\zeta }_0{q}^{\prime }\left({\zeta }_0\right)·\beta \left(q\left({\zeta }_0\right)\right)\notin h\left(U\right).$$

(44)

Using (44) in (43) we get:

$$\alpha \left(D_z^{-1}F\left(a,b,c;z_0\right)\right)+z_0{\left[D_z^{-1}F\left(a,b,c;z_0\right)\right]}^{\prime }·\beta \left(D_z^{-1}F\left(a,b,c;z_0\right)\right)\notin h\left(U\right).$$

(45)

Since (45) contradicts (41), we conclude that the assumption made is false, hence

$$D_z^{-1}F\left(a,b,c;z\right){\prec }_{F}q\left(z\right), z\in U.$$

(46)

Since function $q$ is a univalent solution of the equation from (lll), the conclusion is that function $q$ is the best fuzzy dominant of the fuzzy differential subordination (27). $\square$

If we use function $q\left(z\right)=\frac{z}{1+z}, q\left(0\right)=0,$ in Theorem 2, the following corollary is obtained:

Corollary 2.

Let the convex function$q\left(z\right)=\frac{z}{1+z}, z\in U, q\left(0\right)=0,$and let$\alpha , \beta \in H\left(D\right),$where D is a domain which contains$q\left(U\right).$Let

$$G:ℂ\to \left[0,1\right], G\left(z\right)=\frac{\left|z\right|}{1+\left|z\right|}.$$

assume that:

(l)$H\left(z\right)=\frac{z}{{\left(1+z\right)}^2}·\beta \left(\frac{z}{1+z}\right),$is a starlike function in$U;$

(ll)$Re\frac{{\alpha }^{\prime }\left(\frac{z}{1+z}\right)}{\beta \left(\frac{z}{1+z}\right)}>0, z\in U;$

(lll)$h\left(z\right)=\alpha \left(\frac{z}{1+z}\right)+H\left(z\right)=\alpha \left(\frac{z}{1+z}\right)+\frac{z}{{\left(1+z\right)}^2}·\beta \left(\frac{z}{1+z}\right), z\in U,$is analytic in$U.$

If the fractional integral of Gaussian hypergeometric function given by (4) is an analytic function in$U$, with$D_z^{-1}F\left(a,b,c;0\right)=q\left(0\right)=0, D_z^{-1}F\left(U\right)\subset D,$and satisfies the fuzzy differential subordination

$$\begin{gathered} \alpha \left(D_z^{-1}F\left(a,b,c;z\right)\right)+z{\left[D_z^{-1}F\left(a,b,c;z\right)\right]}^{\prime }·\beta \left(D_z^{-1}F\left(a,b,c;z\right)\right){\prec }_F \\ \alpha \left(\frac{z}{1+z}\right)+\frac{z}{{\left(1+z\right)}^2}·\beta \left(\frac{z}{1+z}\right), z\in U, \end{gathered}$$

(47)

equivalently written as

$$\begin{gathered} \frac{\left|\alpha \left(D_z^{-1}F\left(a,b,c;z\right)\right)+z{\left[D_z^{-1}F\left(a,b,c;z\right)\right]}^{\text{'}}·\beta \left(D_z^{-1}F\left(a,b,c;z\right)\right)\right|}{1+\left|\alpha ·D_z^{-1}F\left(a,b,c;z\right)+z{\left[D_z^{-1}F\left(a,b,c;z\right)\right]}^{\text{'}}·\beta \left(D_z^{-1}F\left(a,b,c;z\right)\right)\right|} \\ \le \frac{\left|\alpha \left(\frac{z}{1+z}\right)+\frac{z}{{\left(1+z\right)}^2}·\beta \left(\frac{z}{1+z}\right)\right|}{1+\left|\alpha \left(\frac{z}{1+z}\right)+\frac{z}{{\left(1+z\right)}^2}·\beta \left(\frac{z}{1+z}\right)\right|}, z\in U, \end{gathered}$$

or

$$\begin{gathered} G\left[\alpha \left(D_z^{-1}F\left(a,b,c;z\right)\right)+z{\left[D_z^{-1}F\left(a,b,c;z\right)\right]}^{\prime }·\beta \left(D_z^{-1}F\left(a,b,c;z\right)\right)\right] \\ \le G\left[\alpha \left(\frac{z}{1+z}\right)+\frac{z}{{\left(1+z\right)}^2}·\beta \left(\frac{z}{1+z}\right)\right], z\in U, \end{gathered}$$

then

$$D_z^{-1}F\left(a,b,c;z\right){\prec }_F\frac{z}{1+z}, z\in U,$$

(48)

equivalently written as

$$\frac{\left|D_z^{-1}F\left(a,b,c;z\right)\right|}{1+\left|D_z^{-1}F\left(a,b,c;z\right)\right|}\le \frac{\left|\frac{z}{1+z}\right|}{1+\left|\frac{z}{1+z}\right|}, z\in U,$$

or

$$G\left(D_z^{-1}F\left(a,b,c;z\right)\right)\le G\left(\frac{z}{1+z}\right), z\in U,$$

and$q\left(z\right)=\frac{z}{1+z}$is the best fuzzy dominant. Differential subordination (48) implies:

$$Re D_z^{-1}F\left(a,b,c;z\right)>\frac{1}{2}.$$

Proof. 

Using relation (46) obtained in the proof of Theorem 2, by putting $h\left(z\right)=\frac{z}{1+z},$ we obtain:

$$D_z^{-\lambda }F\left(a,b,c;z\right){\prec }_{F}h\left(z\right)=\frac{z}{1+z},=q\left(z\right), z\in U.$$

(49)

The convexity of function $q\left(z\right)$ must be proved in order to obtain the conclusion of this corollary.

We evaluate:

$$q^{\prime }\left(z\right)=\frac{1}{{\left(1+z\right)}^2}, q^{″}\left(z\right)=\frac{-2}{{\left(1+z\right)}^3}.$$

We can write

$$\frac{zq″\left(z\right)}{q\prime \left(z\right)}+1=\frac{1-z}{1+z}, z\in U.$$

(50)

We analyze the function

$$r\left(z\right)=\frac{1-z}{1+z}, r^{\prime }\left(z\right)=\frac{-2}{{\left(1+z\right)}^2}, r^{\prime }\left(0\right)=-2\ne 0, z\in U.$$

Since from $r\left(z_1\right)=r\left(z_2\right)$ we obtain that $z_1=z_2,$ we conclude that function $r$ is univalent in $U$ with $r^{\prime }\left(0\right)\ne 0$ which means that $r$ is conformal mapping of the unit disk into the half-plane $S=\left\{w\in ℂ:Re w>0\right\}$ and we can write:

$$Re r\left(z\right)=Re\frac{1-z}{1+z}>0, z\in U.$$

(51)

Using (51) in (50) we obtain:

$$Re \left[\frac{zq″\left(z\right)}{q\prime \left(z\right)}+1\right]>0, z\in U,$$

hence $q\in K$ which gives that (48) is equivalent to

$$Re D_z^{-1}F\left(a,b,c;z\right)>Re\frac{z}{1+z}, z\in U.$$

(52)

Since $q\left(z\right)=\frac{z}{1+z}$ is a convex function, hence univalent, and since we have $q^{\prime }\left(z\right)=\frac{1}{{\left(1+z\right)}^2}\ne 0,$ we conclude that function $q$ is a conformal mapping of the unit disc into the half-plane $S\prime =\left\{w\in ℂ:Re w>\frac{1}{2}\right\}$ from which we can write:

$$Re \frac{z}{1+z}>\frac{1}{2}, z\in U.$$

(53)

Using (53) in (52) we obtain:

$$Re>\frac{1}{2}, z\in U.$$

□

An example is next given in order to show an application of the results proved in the first theorem.

Example 1.

Let$h$the convex function in$U,$

$$h\left(z\right)=\frac{z}{1-z},\varphi \in H\left(D\right), h\left(U\right)\subset D, \varphi \left(w\right)=1+h\left(z\right)=1+\frac{z}{1-z}=\frac{1}{1-z}.$$

Let

$$G:ℂ\to \left[0,1\right], G\left(z\right)=\frac{1}{1+\left|z\right|}.$$

Then we have:

$$Re \varphi \left(1+h\left(z\right)\right)=Re \frac{1}{1-z}=\frac{1-\rho \cos \alpha }{{\left(1-\rho \cos \alpha \right)}^2+{\rho }^2{\sin }^2\alpha }>0,$$

$$D_z^{-1}F\left(-1,1+i,1-i;z\right)=z-\frac{i}{2}{z}^2.$$

We calculate

$$\begin{gathered} D_z^{-1}F\left(-1,1+i,1-i;z\right)+z{\left[D_z^{-1}F\left(-1,1+i,1-i;z\right)\right]}^{\prime } \\ ·\varphi \left(D_z^{-1}F\left(-1,1+i,1-i;z\right)\right) \\ =-\frac{1}{2}{z}^4-\frac{3i}{2}{z}^3+z^2\left(i-\frac{3i}{2}\right)+2z. \end{gathered}$$

Using Theorem 1 we can write:

If

(j)$Re \varphi \left(h\left(z\right)\right)=Re \frac{1}{1-z}>0,z\in U,$

(jj)$H\left(z\right)=z·\frac{1}{{\left(1-z\right)}^2}·\varphi \left(\frac{1}{1-z}\right)$,

is a starlike function in$U$and the fuzzy differential subordination is satisfied

$$-\frac{1}{2}{z}^4-\frac{3i}{2}{z}^3+z^2\left(i-\frac{3i}{2}\right)+2z{\prec }_F\frac{z}{1-z}, z\in U,$$

equivalently written

$$\frac{1}{1+\left|-\frac{1}{2}{z}^4-\frac{3i}{2}{z}^3+z^2\left(i-\frac{3i}{2}\right)+2z\right|}\le \frac{1}{1+\left|\frac{z}{1-z}\right|}, z\in U.$$

then

$$z-\frac{i}{2}{z}^2{\prec }_F\frac{z}{1-z}, z\in U,$$

(54)

equivalently written

$$\frac{1}{1+\left|z-\frac{i}{2}{z}^2\right|}\le \frac{1}{1+\left|\frac{z}{1-z}\right|}, z\in U.$$

Indeed, if we let

$$f\left(z\right)=z-\frac{i}{2}{z}^2, f\left(0\right)=0, h\left(z\right)=\frac{z}{1-z}, h\left(0\right)=0 \Rightarrow f\left(0\right)=h\left(0\right).$$

Since $h\left(z\right)$ is a convex function as seen in Corollary 1, the fuzzy differential subordination (54) is equivalent to

$$Re \left(z-\frac{i}{2}{z}^2\right)>Re \left(\frac{z}{1-z}\right)>-\frac{1}{2}, z\in U.$$

We deduce that the image of the unit disk through function $f$ is:

$$f\left(U\right)\subset h\left(U\right)=\left\{w\in ℂ:Re w>-\frac{1}{2}\right\}.$$

Using conditions $f\left(0\right)=h\left(0\right)$ and $f\left(U\right)\subset h\left(U\right)$ and Remark 2, considering Definition 9 we conclude that

$$z-\frac{i}{2}{z}^2{\prec }_F\frac{z}{1-z}, z\in U.$$

4. Conclusions

The original results presented in Section 3 of this paper are part of the research concerning fuzzy differential subordination theory. Gaussian hypergeometric function for which univalence conditions have been previously obtained, is involved in the investigation combined with the fractional integral, another function which gives interesting outcome when considered in studies, as it is presented in the introduction of the paper. Fuzzy differential subordinations are obtained using the fractional integral of Gaussian hypergeometric function introduced in Definition 10. A fuzzy dominant is obtained for the fuzzy differential subordination investigated in Theorem 1 while a better result is given for the fuzzy differential subordination considered in Theorem 2, the best fuzzy dominant being found. The fuzzy differential subordinations are also interpreted in terms of inclusion relations between subsets of ℂ, an approach seen in recent studies cited in the Introduction. Interesting corollaries are written when particular functions with nice geometric properties are used as fuzzy dominant in Theorem 1 and as fuzzy best dominant in Theorem 2. An example is constructed in order to show the applicability of the results obtained in the study.

As future directions of study, fractional integral of Gaussian hypergeometric function can be used in similar studies involving the dual theory of fuzzy differential superordination. Sandwich-type results are likely to be obtained when combining the findings of the present study and the ones obtained using the dual theory as it can be seen for fractional integral of confluent hypergeometric function [25,26]. Then, the fractional integral of Gaussian hypergeometric function could be applied for introducing fuzzy classes of analytic functions as it has been carried out in earlier published papers such as [36]. A fractional derivative could also be used combined with Gaussian hypergeometric function in order to obtain new fuzzy differential subordinations as seen in [37].