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Article

A Criterion for Subfamilies of Multivalent Functions of Reciprocal Order with Respect to Symmetric Points

1
Department of Mechanical Engineering, Sarhad University of Science and Information Technology, Peshawar 25000, Pakistan
2
Department of Mathematics and Statistics, University of Victoria, Victoria, BC V8W 3R4, Canada
3
Department of Medical Research, China Medical University Hospital, China Medical University, Taichung 40402, Taiwan
4
Department of Mathematics, Abdul Wali Khan University Mardan, Mardan 23200, Pakistan
5
Department of Mathematics, Aligarh Muslim University, Aligarh 202002, India
*
Author to whom correspondence should be addressed.
Fractal Fract. 2019, 3(2), 35; https://doi.org/10.3390/fractalfract3020035
Submission received: 19 May 2019 / Revised: 19 June 2019 / Accepted: 20 June 2019 / Published: 25 June 2019

Abstract

:
In the present research paper, our aim is to introduce a new subfamily of p-valent (multivalent) functions of reciprocal order. We investigate sufficiency criterion for such defined family.
MSC:
Primary 30C45, 30C10; Secondary 47B38

1. Introduction

Let us suppose that A p represents the class of p-valent functions f z that are holomorphic (analytic) in the region E = z : z < 1 and has the following Taylor series representation:
f ( z ) = z p + k = 1 a p + k z p + k .
Two points p and p are said to be symmetrical with respect to o if o is the midpoint of the line segment p p .
If f ( z ) and g ( z ) are analytic in E , we say that f ( z ) is subordinate to g ( z ) , written as f ( z ) g ( z ) , if there exists a Schwarz function, w ( z ) , which is analytic in E with w 0 = 0 and w z < 1 such that f ( z ) = g ( w ( z ) ) . Furthermore, if the function g ( z ) is univalent in E , then we have the following equivalence, see [1].
f ( z ) g ( z ) ( z E ) f ( 0 ) = g ( 0 ) and f ( E ) g ( E ) .
Let N α denotes the class of starlike functions of reciprocal order α α > 1 and is given below
N α : = f z A : Re z f ( z ) f ( z ) < α , z E .
This class was introduced by Uralegaddi et al. [2] amd further studied by the Owa et al. [3]. After that Nunokawa and his coauthors [4] proved that f z N α , 0 < α < 1 2 , if and only if the following inequality holds
2 α z f z f z 1 < 1 , z E .
Later on, Owa and Srivastava [5] in 2002 generalized this idea for the classes of multivalent convex and starlike functions of reciprocal order α α > p , and further studied by Polatoglu et al. [6]. For more details of the related concepts, see the article of Dixit et al. [7], Uyanik et al. [8], and Arif et al. [9].
For 1 t < s 1 with s 0 t , 0 < α < 1 , and p N , we introduce a subclass of A p consisting of all analytic p-valent functions of reciprocal order α , denoted by N α p S s , t and is defined as
N α p S s , t = f z A p : Re s p t p z f ( z ) f ( s z ) f ( t z ) < p α , z E ,
or equivalently
s p t p z f ( z ) f ( s z ) f ( t z ) p 2 α p 2 α .
Many authors studied sufficiency conditions for various subclasses of analytic and multivalent functions, for details see [4,10,11,12,13,14,15,16,17].
We will need the following lemmas for our work.
Lemma 1
(Jack’s lemma [18]). Let Ψ be a non-constant holomorphic function in E and if the value of | Ψ | is maximum on the circle | z | = r < 1 at z , then z Ψ ( z ) = k Ψ ( z ) , where k 1 is a real number.
Lemma 2
(See [1]). Let H C and let Φ : C 2 × E * C be a mapping satisfying Φ i a , b , z H for a , b R such that b 1 + a 2 2 . If p z = 1 + c 1 z 1 + c 2 z 2 + is regular in E * and Φ p z , z p z , z H z E * , then Re p z > 0 .
Lemma 3
(See [15]). Let p ( z ) = 1 + c 1 z + c 2 z 2 + be analytic in E and η be analytic and starlike (with respect to the origin) univalent in E with η ( 0 ) = 0 . If z p ( z ) η ( z ) , then
p ( z ) 1 + 0 z η ( t ) t d t .
This result is the best possible.

2. Main Results

Theorem 1.
Let f ( z ) A p and satisfies
n = 1 α p + n + p s p + n t p + n s p t p a n + p p 2 1 2 α 1 .
Then f ( z ) N α p S s , t .
Proof. 
Let us assume that the inequality (5) holds. It suffices to show that
2 α s p t p z f ( z ) f ( s z ) f ( t z ) p p .
Consider
2 α s p t p z f ( z ) f ( s z ) f ( t z ) p = p 2 α 1 s p t p z p + n = 1 ( 2 α p + n s p t p p s n + p t n + p ) a n + p z n + p s p t p z p + n = 1 s n + p t n + p a n + p z n + p p 2 α 1 s p t p + n = 1 ( 2 α p + n s p t p + p s n + p t n + p ) a n + p s p t p n = 1 s n + p t n + p a n + p
The last expression is bounded above by p if
p 2 α 1 s p t p + n = 1 ( 2 α p + n s p t p + p s n + p t n + p ) a n + p < p s p t p n = 1 s n + p t n + p a n + p .
Hence
n = 1 α p + n + p s p + n t p + n s p t p a n + p p 2 1 2 α 1 .
This shows that f ( z ) NS p s , t , α . This completes the proof. □
Theorem 2.
If f ( z ) A p satisfies the condition
1 + z f ( z ) f ( z ) z f ( s z ) f ( t z ) f ( s z ) f ( t z ) < 1 α , 1 2 α < 1 ,
then f ( z ) N α p S s , t .
Proof. 
Let us set
q z = 1 α s p t p z f ( z ) p f ( s z ) f ( t z ) 1 α 1 .
Then clearly q ( z ) is analytic in E with q 0 = 0 . Differentiating logarithmically, we have
1 + z f ( z ) f ( z ) z f ( s z ) f ( t z ) f ( s z ) f ( t z ) = 1 α z q z α 1 α q z .
So
1 + z f ( z ) f ( z ) z f ( s z ) f ( t z ) f ( s z ) f ( t z ) = 1 α z q z α 1 α q z .
From (7), we have
1 α z q z α 1 α q z < 1 α .
Next, we claim that q z < 1 . Indeed, if not, then for some z E , we have
max z z 0 q z = q z 0 = 1 .
Applying Jack’s lemma to q z at the point z 0 , we have
q z 0 = e i θ , z 0 q z 0 q z 0 = k , k 1 .
Then
1 + z 0 f ( z 0 ) f ( z 0 ) z f ( s z 0 ) f ( t z 0 ) f ( s z 0 ) f ( t z 0 ) = 1 α z 0 q z 0 α 1 α q z 0 = 1 α z 0 q z 0 q z 0 1 1 α α e i θ = 1 α k α e i θ 1 α 1 α 1 1 α α e i θ .
Therefore
1 + z 0 f ( z 0 ) f ( z 0 ) z f ( s z 0 ) f ( t z 0 ) f ( s z 0 ) f ( t z 0 ) 2 1 α 2 1 α 2 + α 2 2 α 1 α cos θ .
Now the right hand side has minimum value at cos θ = 1 , therefore we have
1 + z 0 f ( z 0 ) f ( z 0 ) z f ( s z 0 ) f ( t z 0 ) f ( s z 0 ) f ( t z 0 ) 2 1 α 2 .
But this contradicts (7). Hence we conclude that q z < 1 for all z E , which shows that
1 α s p t p z f ( z ) p f ( s z ) f ( t z ) 1 α 1 < 1 .
This implies that
s p t p z f ( z ) p f ( s z ) f ( t z ) 1 < 1 α 1 .
Now we have
s p t p z f ( z ) p f ( s z ) f ( t z ) 1 2 α s p t p z f ( z ) p f ( s z ) f ( t z ) 1 + 1 1 2 α < 1 α 1 + 1 1 2 α = 1 2 α .
This implies that f ( z ) N α p S s , t .  □
Theorem 3.
If f ( z ) A p satisfies the condition
Re 1 z f ( z ) f ( z ) + z f ( s z ) f ( t z ) f ( s z ) f ( t z ) > α 2 α 1 , 0 α 1 2 α 1 2 α , 1 2 α < 1 ,
then f ( z ) N α p S s , t for 0 α < 1 .
Proof. 
Let
q z = p f ( s z ) f ( t z ) s p t p z f ( z ) α 1 α .
Then clearly q z is analytic in E . Applying logarithmic differentiation, we have
1 z f ( z ) f ( z ) + z f ( s z ) f ( t z ) f ( s z ) f ( t z ) = 1 α z q z α + 1 α q z = Ψ q z , z q z , z ,
where
Ψ u , v ; t = 1 α v α + 1 α u .
Now for all x , y R satisfying the inequality y 1 + x 2 2 , we have
Ψ i x , y , z = 1 α y α + 1 α i x .
Therefore
Re Ψ i x , y , z α 1 α 1 + x 2 2 α 2 + 1 α 2 x 2 , α 2 α 1 , 0 α 1 2 , α 1 2 α , 1 2 α < 1 .
We set
Λ = ζ : Re ζ > α 2 α 1 , 0 α 1 2 , α 1 2 α , 1 2 α < 1 .
Then Ψ i x , y ; z Λ for all real x, y such that y 1 + x 2 2 . Moreover, in view of (10), we know that Ψ q z , z q z , z Λ . So applying Lemma 2, we have
Re q z > 0 ,
which shows that the desired assertion of Theorem 6 holds.  □
Theorem 4.
If f ( z ) A p satisfies
Re f ( s z ) f ( t z ) s p t p z f ( z ) 1 β z f ( z ) f ( z ) + β z f ( s z ) f ( t z ) f ( s z ) f ( t z ) > 2 α + β 3 α 1 2 p ,
then f ( z ) N α p S s , t for 0 < α < 1 and β 0 .
Proof. 
Let
h z = p f ( s z ) f ( t z ) s p t p z f ( z ) α 1 α .
where h z is clearly analytic in E such that h 0 = 1 . We can write
p f ( s z ) f ( t z ) s p t p z f ( z ) = α + 1 α h z .
After some simple computation, we have
β z f ( z ) f ( z ) + β z f ( s z ) f ( t z ) f ( s z ) f ( t z ) = β α + 1 α h z + z h z α + 1 α h z
It follows from (12) that
p f ( s z ) f ( t z ) s p t p z f ( z ) 1 β z f ( z ) f ( z ) + β z f ( s z ) f ( t z ) f ( s z ) f ( t z ) = β 1 α z h z + 1 α 1 + β h z + α 1 + β = Ψ h z , z h z , z
where
Ψ u , v , t = β 1 α v + 1 α 1 + β u + α 1 + β .
Now for some real numbers x and y satisfying y 1 + x 2 2 , we have
Re Ψ i x , y , z β 1 α 1 + x 2 2 + α 1 + β = 1 2 2 α + β 3 α 1 .
If we set
Λ = ζ : Re ζ > 1 2 2 α + β 3 α 1 ,
then Ψ i x , y , z Λ Furthermore, by virtue of (11), we know that Ψ h z , z h z , z Λ . Thus by using Lemma 2, we have
Re h z > 0 ,
which implies that the assertion of Theorem 7 holds true.  □
Theorem 5.
If f ( z ) A p satisfies the condition
p 2 α s p t p z f ( z ) f ( s z ) f ( t z ) p β z γ ,
then f ( z ) N α p S s , t with 0 < α < 1 , 0 < β γ + 1 and γ 0 .
Proof. 
Let we define
ϝ z = z p 2 α s p t p z f ( z ) f ( s z ) f ( t z ) .
Then ϝ z is regular in E and ϝ 0 = 0 . The condition (14) gives
p 2 α s p t p z f ( z ) f ( s z ) f ( t z ) = ϝ z z
It follows from (13) that
ϝ z z p β z γ .
This implies that
ϝ z z = 0 z ϝ t t d t 0 z ϝ t t d t p β z γ + 1 γ + 1 ,
and therefore
ϝ z z < p ,
which further gives
s p t p z f ( z ) p f ( s z ) f ( t z ) 1 2 α < 1 2 α .
Hence f ( z ) N α p S s , t .  □
Theorem 6.
If f ( z ) A p satisfies
s p t p z f ( z ) f ( s z ) f ( t z ) 1 + z f ( z ) f ( z ) z f ( s z ) f ( t z ) f ( s z ) f ( t z ) < p 1 α α ,
then f ( z ) N α p S s , t , where p p + 1 < α < 1 .
Proof. 
Let
q z = p f ( s z ) f ( t z ) s p t p z f ( z ) .
Then q z is clearly analytic in E such that q 0 = 1 . After logarithmic differentiation and some simple computation, we have
z 1 q z q z = 1 + z f ( z ) f ( z ) z f ( s z ) f ( t z ) f ( s z ) f ( t z ) .
From (16) and (17), we find that
z 1 q z = s p t p z f ( z ) p f ( s z ) f ( t z ) 1 + z f ( z ) f ( z ) z f ( s z ) f ( t z ) f ( s z ) f ( t z ) .
Now by condition (15), we have
z 1 q z p 1 α α z = Θ z ,
where Θ 0 = 0 . Applying Lemma 3, we have
1 q z 1 + 0 z Θ t t d t = α + p 1 α z α ,
which implies that
q z α α + p 1 α z = H z .
We can write
Re 1 + z H z H z = Re α p 1 α z α + p 1 α z α p 1 α α + p 1 α .
Now since p 1 + p < α < 1 , therefore we have
Re 1 + z H z H z > 0 .
This shows that H is convex univalent in E and H E is symmetric about the real axis, therefore
Re H z H 1 0 .
Combining (16), (18), and (19), we deduce that
Re q z > α ,
which implies that f ( z ) N α p S s , t .  □

Author Contributions

Conceptualization, S.M., M.A. and H.M.S.; methodology, S.M. and M.A.; software, E.S.A.A.; validation, S.M., M.A. and H.M.S.; formal analysis, S.M.; investigation, S.M.; resources, F.G.; data curation, S.M. and M.A.; writing–original draft preparation, S.M.; writing–review and editing, E.S.A.A.; visualization, S.M. and H.M.S.; supervision, S.M. and M.A.; project administration, S.M.

Funding

This research received no external funding.

Acknowledgments

The authors would like to thank the reviewers of this paper for his/her valuable comments on the earlier version of the paper. They would also like to acknowledge Salim ur Rehman, Sarhad University of Science & Information Technology, for providing excellent research and academic environment.

Conflicts of Interest

The authors declare no conflict of interest.

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MDPI and ACS Style

Mahmood, S.; Srivastava, H.M.; Arif, M.; Ghani, F.; AbuJarad, E.S.A. A Criterion for Subfamilies of Multivalent Functions of Reciprocal Order with Respect to Symmetric Points. Fractal Fract. 2019, 3, 35. https://doi.org/10.3390/fractalfract3020035

AMA Style

Mahmood S, Srivastava HM, Arif M, Ghani F, AbuJarad ESA. A Criterion for Subfamilies of Multivalent Functions of Reciprocal Order with Respect to Symmetric Points. Fractal and Fractional. 2019; 3(2):35. https://doi.org/10.3390/fractalfract3020035

Chicago/Turabian Style

Mahmood, Shahid, Hari Mohan Srivastava, Muhammad Arif, Fazal Ghani, and Eman S. A. AbuJarad. 2019. "A Criterion for Subfamilies of Multivalent Functions of Reciprocal Order with Respect to Symmetric Points" Fractal and Fractional 3, no. 2: 35. https://doi.org/10.3390/fractalfract3020035

APA Style

Mahmood, S., Srivastava, H. M., Arif, M., Ghani, F., & AbuJarad, E. S. A. (2019). A Criterion for Subfamilies of Multivalent Functions of Reciprocal Order with Respect to Symmetric Points. Fractal and Fractional, 3(2), 35. https://doi.org/10.3390/fractalfract3020035

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