Research Article | Open Access

# On a New Subclass of -Valent Close-to-Convex Mappings Defined by Two-Sided Inequality

**Academic Editor:**Henryk Hudzik

#### Abstract

A new subclass of -valent close-to-convex mappings defined by two-sided inequality is introduced. Some sufficient conditions for functions to be in are given.

#### 1. Introduction

Let be the class of functions of the formwhich are -valent analytic in the open unit disk . There and in the following, let , and be the sets of positive integers, complex numbers, and real numbers, respectively. A function analytic in is said to be close-to-convex if there is a convex function such thatfor all . The concept of close-to-convex was introduced by Kaplan [1] in 1952. A number of results for close-to-convex functions in have been obtained by several authors (see, e.g., [2–14]).

A function is said to be in the class if it satisfies the following two-sided inequality:for and . Note that if , then is -valent close-to-convex in . Furthermore, if , then (3) becomes A function is called -valent close-to-convex of order and type in .

Given two functions and , which are analytic in , we say that the function is subordinate to and write or , if there exists a Schwarz function , analytic in with and such that . In particular, if is univalent in , we have the following equivalence:

Throughout this paper, we letIn order to prove our main results, we need the following lemmas.

Lemma 1. *The function defined byis analytic and univalent convex in and*

*Proof. *In view of (6), it is easy to see that and the transformationmaps the convex region conformally onto the right-half -plane so that corresponding to . Sincemaps the right-half -plane onto , from (7), (9), and (11) we find that maps conformally onto with . The proof of Lemma 1 is completed.

Lemma 2 (see [15]). *Let the function be analytic and univalent in and let the functions and be analytic in a domain containing , with . Setand suppose that*(i)* is univalent starlike in ,*(ii)* If is analytic in with , , andthen . The function is the best dominant of (14).*

Lemma 3 (see [16]). *Let be analytic function in of the form with in . If there exists a point such that for and thenwherewhere and .*

In this paper we shall derive some criteria for a function to be in the class .

#### 2. Main Results

Our first result is the following theorem.

Theorem 4. *Let and . Also let and . If the function is analytic in with and and satisfieswhereis close-to-convex in , thenThe bounds and in (22) are sharp for the function defined by (7).*

*Proof. *We choosein Lemma 2. By Lemma 1, the function is analytic and univalent convex in andClearly, and are analytic in a domain containing and , with when . For the function given by is univalent starlike in becauseFurther, we have where is given by (21) and soFor , it follows from (27) and (29) that The other conditions of Lemma 2 are also satisfied. Therefore, we conclude that and the function is the best dominant of (20). The proof of the theorem is completed.

Theorem 5. *If satisfies andfor all , wherethen . The bounds and in (32) are the largest numbers such that (3) holds true.*

*Proof. *For satisfying , we define the function byThen is analytic in with and for all . Taking the logarithmic differentiations in both sides of (34), we haveandfor all . Putting and in Theorem 4 and using (36), we find that ifwhereis (close-to-convex) univalent in , then that is, .

For , , and , we haveandNow we consider the following two cases.

(i) If then we deduce from (38), (40), and (41) that which yieldsLet . Thenand it follows from (44) and (45) that(ii) If , then it follows from (38) to (41) that and soLet . Thenand from (48) and (49) we haveNoting that , we deduce from (46) and (50) that properly contains the region in the complex -plane. Therefore, if a function satisfies (32), then the subordination relation (37) holds true. This shows that .

Furthermore, for the function defined by (7), we haveHence, by using (46) and (50), we conclude that the bounds and in (32) are the best possible ones. The proof of the theorem is completed.

Theorem 6. *Let . Also let and . If satisfiesfor , thenIn particular, if , then or is -valent close-to-convex of order .*

*Proof. *We denote Then condition (52) becomesWe want to prove that If there exists a point such that then, from Lemma 3, we have where , and For the case , we havewhere The functionhas a negative derivativeHence Therefore, (60) becomes which contradicts (55). Thus For the case , we have . Applying the same method as the above, we get This contradicts (55). The proof of the theorem is completed.

Applying the same method as the above we can prove the following theorem.

Theorem 7. *Let . Also let and . If satisfies for , then In particular, if , then or is -valent close-to-convex of order .*

#### Data Availability

For every result in this paper, detailed proof has been given. Readers can understand this paper by reading these detailed proofs carefully. All results in this paper can be released.

#### Conflicts of Interest

There are no conflicts of interest among the authors.

#### Acknowledgments

This work is supported by National Natural Science Foundation of China (Grant no. 11571299), Natural Science Foundation of Jiangsu Province (Grant no. BK20151304), and Natural Science Foundation of Jiangsu Gaoxiao (Grant no. 17KJB110019).

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#### Copyright

Copyright © 2018 Yi-Hui Xu and Jin-Lin Liu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.