On differential subordinations in the complex plane

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1 Nunokawa et al. Journal of Inequalities Applications 015) 015:7 DOI /s R E S E A R C H Open Access On differential subordinations in the complex plane Mamoru Nunokawa 1, Nak Eun Cho *,OhSangKwon 3 Janusz Sokół 4 * Correspondence: necho@pknu.ac.kr Department of Applied Mathematics, College of Natural Sciences, Pukyong National University, Pusan, , Korea Full list of author information is available at the end of the article Abstract The purpose of this work is to present a new approach to solve some problems in differential subordination theory. We also discuss the new results closely related to the generalized Briot-Bouquet differential subordination. MSC: Primary 30C45; secondary 30C80 Keywords: analytic functions; convex functions; convex of order alpha; univalent functions; differential subordination 1 Introduction Let H denote the class of all analytic functions in the unit disc D = z : z <1} in the complex plane C.RecallthatasetE C is said to be starlike with respect to a point w 0 E if only if the linear segment joining w 0 to every other point w E lies entirely in E,while asete is said to be convex if only if it is starlike with respect to each of its points, that is, if only if the linear segment joining any two points of E lies entirely in E.Aunivalent function f maps D onto a convex domain E if only if [1] Re 1+ zf } z) >0 forallz D, 1.1) f z) then f is said to be convex in D or briefly convex). Let A denote the subclass of H consisting of functions normalized by f 0) = 0, f 0) = 1. We denote by K the set of all functions f A that are convex univalent in D. Wesaythatf A is convex of order α, 0 α <1,when Re 1+ zf } z) > α for all z D. 1.) f z) Functions that are convex of order α were introduced by Robertson in []. For two analytic functions f, g,we say that f is subordinate to g,written as f g,if only if there exists an analytic function ω with property ωz) z in D such that f z)=gωz)). In particular, if g is univalent in D, then we have the following equivalence: f z) gz) f 0) = g0) f D) gd). 1.3) The idea of subordination was used for defining many of classes of functions studied in geometric function theory. For obtaining the main result, we shall use the methods of 015 Nunokawa et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License which permits unrestricted use, distribution, reproduction in any medium, provided the original work is properly credited.

2 Nunokawa et al. Journal of Inequalities Applications 015) 015:7 Page of 8 differential subordinations. The main results in the theory of differential subordinations were introduced by Miller Mocanu in [3] [4]. A function p, analytic in D, issaid to satisfy a first-order differential subordination if φ pz), zp z) ) hz), 1.4) pz), zp z)) D C for z D, φ : C C φpz), zp z)) is analytic in D, h is analytic univalent in D. The function q issaid tobe a dominant of the differential subordination 1.4) ifp q for all p satisfying 1.4). If q is a dominant of 1.4) q q for all dominants q of 1.4), then we say that q is the best dominant of the differential subordination 1.4). The purpose of the present paper is to investigate interesting new results in connection with differential subordination to improve some results obtained by Miller Mocanu [5]. Also we remark that the reader may refer to the recent results obtained by Sokół Nunokawa [6] as applications of differential subordination. The following lemma will be required in our present investigation. Lemma 1.1 [3], [4,p.4]) Assume that Q is the set of functions f H that are injective on D \ Ef ), Ef ):= ζ : ζ D lim f z)= z ζ }, are such that f ζ ) 0 ζ D) \ Ef ) ). Let ψ Q with ψ0) = a, let ϕz)=a + a m z m + be analytic in D with ϕz) a m N. If ϕ ψ in D, then there exist points z 0 = r 0 e iθ D ζ 0 D \ Eψ) for which ϕ z < r 0 ) ψd), ϕz 0 )=ψζ 0 ) z 0 ϕ z 0 )=sζ 0 ψ ζ 0 ) 1.5)

3 Nunokawa et al. Journal of Inequalities Applications 015) 015:7 Page 3 of 8 for some s m. Moreover, z0 ϕ } z 0 ) ζ0 ψ } ζ 0 ) Re +1 sre ) ϕ z 0 ) ψ ζ 0 ) To prove the main results, we also need the following lemma, which is a generalization of a result due to Nunokawa [7, 8]. Lemma 1. Let pz) be a function analytic in z D of the form pz)=1+ c n z n, c m 0, n=m with pz) 0in z <1.If there exists a point z 0 D such that arg pz) } < πϕ for z < z 0 arg pz0 ) } = πϕ for some ϕ >0,then we have z 0 p z 0 ) pz 0 ) = ilϕ, l m a + 1 ) m a when arg pz 0 ) } = πϕ 1.7) l m a + 1 ) m a when arg pz 0 ) } = πϕ, 1.8) pz0 ) } 1/ϕ = ±ia a >0. Main results Theorem.1 Let Bz) Cz) be analytic in D with Im Cz) } < Re Bz) }..1) If pz) is analytic in D with p0) = 1, if arg Bz)zp z)+cz)pz) } π < + tz),.)

4 Nunokawa et al. Journal of Inequalities Applications 015) 015:7 Page 4 of 8 tz)= argbz)i + Cz)} argbz)i + Cz)} π/ when argbz)i + Cz)} [0, π/], when argbz)i + Cz)} π/, π], then we have Re pz) } >0, z D..3) Proof By Lemma 1., ifrepz)} > 0 does not hold for all z D, then there exists a point z 0, z 0 <1,suchthat arg pz) } < π for z < z 0 arg pz0 ) } = π, z 0 p z 0 ) pz 0 ) = il, l 1 a + 1 ) 1 when arg pz 0 ) } = π a.4) l 1 a + 1 ) 1 when arg pz 0 ) } = π a,.5) pz 0 )=±ia a >0. For the case pz 0 )=ia, a > 0, we are going to show that We have arg Bz0 )z 0 p z 0 )+Cz 0 )pz 0 ) } π + tz 0)..6) arg Bz0 )z 0 p z 0 )+Cz 0 )pz 0 ) } [ = pz arg 0 ) Bz 0 ) z 0p z 0 ) + Cz 0 )]} pz 0 ) = arg pz 0 ) [ Bz 0 )il + Cz 0 ) ]}..7)

5 Nunokawa et al. Journal of Inequalities Applications 015) 015:7 Page 5 of 8 By.1), we have ImBz 0 )il + Cz 0 )} > 0. Therefore, from.7)weobtain arg Bz0 )z 0 p z 0 )+Cz 0 )pz 0 ) } = π + argbz)il + Cz)} argbz)il + Cz)} π/ π + argbz)i + Cz)} when argbz)il + Cz)} [0, π/], when argbz)il + Cz)} π/,π] when argbz)i + Cz)} [0, π/], argbz)i + Cz)} π/ when argbz)i + Cz)} π/, π] = π + tz 0)..8) This contradicts.). For the case pz 0 )= ia, a > 0, the proof runs as in the first case. Remark.1 Theorem.1improvesaresultobtainedbyMillerMocanu [5,p.08]. Corollary. Let gz) be analytic in D with g0) = 1 Imzg z)/gz)} <1.If f z) = z + is analytic in D arg gz)f z) } < π + vz), z D, vz)= argi +1 zg z)/gz)} argi +1 zg z)/gz)} π/ when argi +1 zg z)/gz)} [0, π/], when argi +1 zg z)/gz)} π/, π], then we have } gz)f z) Re >0, z z D..9) Proof We put Bz)=1,Cz)=1 zg z)/gz), pz)=gz)f z)/z. Thenpz) is analytic in D, p0) = 1 Im Cz) } < Re Bz) } =1. Moreover,.)becomes arg gz)f z) } π < + vz), z D. Hence, applying Theorem.1, we obtain.9)immediately. Theorem.3 Let Bz) Cz) be analytic in D with } Cz) Re 1, z D..10) Bz) If pz) is analytic in D with p0) = 0, if Bz)zp z)+cz)pz) < Bz)+Cz), z D,.11)

6 Nunokawa et al. Journal of Inequalities Applications 015) 015:7 Page 6 of 8 then we have pz) <1, z D..1) Proof By Lemma 1.1,ifpz) z in D, then there exist points z 0 = r 0 e iθ D ζ 0, ζ 0 =1 for which p z < r 0 ) D, pz 0 )=ζ 0 z 0 p z 0 )=sζ 0.13) for some s 1. Then, by.10), we have Bz 0 )z 0 p z 0 )+Cz 0 )pz 0 ) = sbz 0 )+Cz 0 ) = Bz 0 ) s + Cz 0 )/Bz 0 ) = Bz0 ) s + Re Cz0 )/Bz 0 ) } + iim Cz 0 )/Bz 0 ) } Bz0 ) 1+Re Cz0 )/Bz 0 ) } + iim Cz 0 )/Bz 0 ) } = Bz 0 )+Cz 0 ), which contradicts.11). Therefore, pz) <1 in D. Theorem.4 Let Bz) Cz) be analytic in D with } Cz) Im 1, z D..14) Bz) Bz) If pz) is analytic in D with p0) = 0, if Bz)zp z)+cz)pz) < 1+ }) Bz) zp z) Cz) pz) + Re, z D,.15) Bz) then we have pz) <1, z D..16) Proof Applying the same method as in the proof of Theorem.3,wehave Bz 0 )z 0 p z 0 )+Cz 0 )pz 0 ) = pz 0 ) Bz 0) z 0p z 0 ) + Cz 0 ) pz 0 ) = sbz0 )+Cz 0 )

7 Nunokawa et al. Journal of Inequalities Applications 015) 015:7 Page 7 of 8 By.14), we have = Bz0 ) s + Cz0 )/Bz 0 ) = Bz0 ) s + Re Cz0 )/Bz 0 ) } + iim Cz 0 )/Bz 0 ) } = Bz 0 ) }) Cz0 ) }) Cz0 ) s + Re + Im. Bz 0 ) Bz 0 ) Bz0 )z 0 p z 0 )+Cz 0 )pz 0 ) Bz0 ) Cz0 ) s + Re Bz 0 ) = }) + 1 Bz 0 ) 1+ Bz 0 ) s + Re Cz0 )/Bz 0 ) }) = 1+ Bz 0 ) }) z 0 p z 0 ) pz 0 ) + Re Cz0 ), Bz 0 ) which contradicts.15). Therefore, pz) <1 in D. Remark. Theorem.3 Theorem.4 improve a result obtained by Miller Mocanu [5,p.06]. Theorem.5 Let pz) be analytic in D with p0) = 1 Re pz) zp z)/p z) 1 } > α, z D..17) Then we have Re pz) } > α, z D,.18) α <1. Proof Putting qz) =pz) α)/1 α), q0) = 1, we have to prove that Reqz)} > 0 for z D.If.18) does not hold, then qz) ψz)=1+z)/1 z), z D. Hence by Lemma 1.1 there exist points z 0 = r 0 e iθ D ζ 0, ζ 0 =1 for which Re qz) } >0 for z < r 0 Re qz 0 ) } } 1+ζ0 = Re =0 1 ζ 0 z 0 q z 0 )=sζ 0 ψ ζ 0 )= sζ 0 1 ζ 0 ).19)

8 Nunokawa et al. Journal of Inequalities Applications 015) 015:7 Page 8 of 8 for some s 1. Then, by 1.6), we have z0 q } z 0 ) ζ0 ψ } ζ 0 ) Re +1 sre +1 = sre 1+ζ 0 =0..0) q z 0 ) ψ ζ 0 ) 1 ζ 0 Therefore, we have Re pz 0 ) z 0 p z 0 )/p z 0 ) 1 } = Re 1 α)qz 0 )+α ) z 0 q z 0 )/q z 0 ) 1 } Re 1 α)qz 0 )+α )} = α, which contradicts.18). This completesthe proof. Remark.3 Theorem.5improvesa resultobtained by MillerMocanu [5,p.07]. Competing interests The authors declare that they have no competing interests. Authors? contributions All authors jointly worked on the results they read approved the final manuscript. Author details 1 University of Gunma, Hoshikuki-cho 798-8, Chuou-Ward, Chiba, , Japan. Department of Applied Mathematics, College of Natural Sciences, Pukyong National University, Pusan, , Korea. 3 Department of Mathematics, Kyungsung University, Busan, , Korea. 4 Department of Mathematics, Rzeszów University of Technology, Al. Powstańców Warszawy 1,Rzeszów,35-959,Pol. Acknowledgements The authors would like to express their thanks to the referees for their valuable advice regarding a previous version of this paper. This research was supported by the Basic Science Research Program through the National Research Foundation of Korea NRF) funded by the Ministry of Education, Science Technology No ). Received: 3 November 014 Accepted: 17 December 014 References 1. Study, E: Konforme Abbildung einfach-zusammenhängender Bereiche. Teubner, Leipzig 1913). Robertson, MS: On the theory of univalent functions. Ann. Math. 37, ) 3. Miller, SS, Mocanu, PT: Differential subordinations univalent functions. Mich. Math. J. 8, ) 4. Miller, SS, Mocanu, PT: Differential Subordinations: Theory Applications. Series of Monographs Textbooks in Pure Applied Mathematics, vol. 5. Dekker, New York 000) 5. Miller, SS, Mocanu, PT: Differential subordination inequalities in the complex plane. J. Differ. Equ. 67), ) 6. Sokół, J, Nunokawa, M: On the subordination under Bernardi operator. Proc. Jpn. Acad., Ser. A, Math. Sci. 89, ) 7. Nunokawa, M: On properties of non-carathéodory functions. Proc. Jpn. Acad., Ser. A, Math. Sci. 686), ) 8. Nunokawa, M: On the order of strongly starlikeness of strongly convex functions. Proc. Jpn. Acad., Ser. A, Math. Sci. 697), )