Differential subordination related to conic sections

J. Math. Anal. Appl. 317 006 650 658 www.elsevier.com/locate/jmaa Differential subordination related to conic sections Stanisława Kanas Department of Mathematics, Rzeszów University of Technology, W. Pola, PL-35-959 Rzeszów, Poland Received 9 September 00 Available online 1 October 005 Submitted by J. Noguchi Abstract In this paper some problems of the theory of differential subordination are investigated in connection with conic domains. In particular, fundamental conditions for functions mapped the unit disk onto domains bounded by parabolas and hyperbolas are deduced. 005 Elsevier Inc. All rights reserved. Keywords: Differential subordination; Domains bounded by conic sections 1. Introduction and definitions Following Miller and Mocanu cf. [8, p. 1], denote by Q the set of functions q analytic and injective on Ū \ Eq, where { } Eq = ζ U: lim =, z ζ and are such that q ζ 0forζ U \ Eq. Miller and Mocanu [8] formulated for functions in Q the fundamental lemma in the theory of differential subordinations which is the key lemma for numerous problems of analytic and univalent functions, see below. Lemma 1.1. [8, p. ] Let pz = a + a n z n + be analytic in U with pz a and n 1, and let q Q with q0 = a. If there eist points z 0 U and ζ 0 U \ Eq such that pz 0 = qζ 0 and pu r0 qu, where r 0 = z 0, then there eists m n 1 such that E-mail addresses: skanas@prz.rzeszow.pl, skanas@prz.edu.pl. 00-7X/$ see front matter 005 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.005.09.03

S. Kanas / J. Math. Anal. Appl. 317 006 650 658 651 1 z 0 p z 0 = mζ 0 q ζ 0 and R z 0p z 0 ζ0 q ζ 0 p + 1 mr z 0 q ζ 0 + 1. Lemma 1.1, known also as the etension of the Jack s lemma, have been used by Miller and Mocanu and other mathematicians in order to prove directly some results in the theory of univalent functions. For instance, Miller and Mocanu proved that if pz + zp z + z p z n + 1Mz then pz Mz. In this case we obviously have p0 = 0, and the obtained result is sharp. However Miller and Mocanu constructed a special tool which makes proofs very short and easy. Such a tool is called the admissibility condition. Definition 1.1. [8, p. 7] Let Ω beasetinc, q Q and n be a positive integer. The class of admissible functions, Ψ n [Ω,q], consists of those functions ψ : C 3 U C that satisfy the admissibility condition ψr,s,w; z / Ω 1.1 whenever r = qζ, s = mζ q ζ, and R w ζq s + 1 mr ζ q ζ + 1, z U, ζ U \ Eq and m n. Making use of Lemma 1.1 and the above admissibility condition, Miller and Mocanu formulated and proved the following: Theorem 1.1. [8, p. 8] Let ψ Ψ [Ω,q] with q0 = a. Ifpz = a + a n z n + satisfies then p q. ψ pz,zp z, z p z; z Ω, 1. Applying Theorem 1.1 it suffices to check the admissibility condition in order to prove p q. Miller and Mocanu in their monograph [8] considered in details two special cases of Lemma 1.1; when the function q maps the unit disk onto a disk and onto a half-plane. Their results have been used by many authors and found many applications in the geometric theory of univalent functions. For k [0,, set Ω k = { u + iv: u >k u 1 + k v,u>0 }. 1.3 Note that Ω k is the domain bounded by a conic section: line for k = 0, a right branch of a hyperbola when 0 <k<1, a parabola when k = 1, and finally an ellipse when k>1. Moreover, 1 Ω k for all k and each Ω k is conve and symmetric about the real ais. The author and Wiśniowska [1,,5] considered the family Ω k in their study of k-uniformly conve and k-starlike functions and gave the eplicit formulas for conformal mappings q k : U Ω k so that q k 0 = 1 and q k 0>0 as follows:

65 S. Kanas / J. Math. Anal. Appl. 317 006 650 658 Theorem 1. [1,]. The conformal map q k : U Ω k with q k 0 = 1 and q k 0>0 is given by 1+z 1 z for k = 0, 1 + sinh [Ak tanh 1 z] for k 0, 1, 1 k q k z = 1 + log 1+ z 1. 1 for k = 1, z 1 + k 1 sin π Kt F z/ t,t for k>1, where Ak = /π arccos k, Fw, t is the Jacobi F-function: w d Fw, t = 1 1 t, 0 Kt = F1,tis the complete elliptic integral of the first kind, and t 0, 1 is such that πk t k = cosh = cosh μt, Kt K t = K 1 t. The quantity μt is known as the modulus of the Grötzsch ring U \[0,t] for t 0, 1. We will abbreviate q := q 1 and Ω := Ω 1. The domain Ω is enclosed by the parabola u = v + 1 with focus at 1. Such a domain was also treated in the study of uniformly conve functions due to Rønning [7], and Ma and Minda [6], independently.. Main results The motivation of that paper is the supplement the subordination theory connected with domains Ω k. We will discuss variations of Lemma 1.1 involving the subordination to functions those map the unit disk onto conic regions. The results given here may be used in solving etremal problems for families of univalent functions. Some ideas of differential subordinations related to domains bounded by conic sections has been developed by the author [1], by the author and Lecko [], and by Kim and Lecko [3]. First we consider the case when qu is a domain bounded by the parabola u = v + 1. Then Eq ={1} and we have: Theorem.1. Let pz = 1 + p n z n + be analytic in U with pz 1 and n 1. If there eist points z 0 = r 0 e iθ 0 U and ζ 0 U \ Eq, such that pu r0 qu and pz 0 = qζ 0 = 1 + log + i log, π > 0,.1 then there eist m n 1 such that 1 z 0 p z 0 = m = m π pz0 1 sinh + 1 + i m log + 1 pz0 1,

S. Kanas / J. Math. Anal. Appl. 317 006 650 658 653 Rz 0 p z 0 1 π, z0 p z 0 1 π, 3 R z 0p z 0 p + 1 mπ z 0 8 min + 1 >0 log 1 + / π. Proof. Making use of Lemma 1.1, it suffices to calculate ζ 0 q ζ 0 and ζ 0 q ζ 0 /q ζ 0, where ζ 0 = q 1 [pz 0 ]. For the function q given by 1., we have ζ 0 = q 1[ pz 0 ] = tanh pz0 1.. 8 Since q z = 1 log 1 + z z1 z 1 z and we obtain zq z + z q z = 1 + zq z q z = ζ 0 q ζ 0 = 1 = 1 + z 1 z + [ z 1 + z 1 z z log 1 + ] z 1 z + 1, z log 1 + z 1 1 z 1, z pz0 1 e pz 0 1 1 e pz 0 1/ pz0 1 sinh Setting ζ 0 = e it t 0, π,wehave.3 pz0 1.. log 1 + ζ 0 1 = log + i π ζ 0, >0, so that we immediately obtain.1. Combining this,.3,. and Lemma 1.1, we obtain the assertion 1. As a consequence of 1 and the inequalities: m n 1 and + 1 for >0, we conclude. Observe net, that 1 + ζ 0q ζ 0 q = π ζ 0 8 then, in view of Lemma 1.1 R z 0p z 0 p z 0 + 1 mπ 8 + 1 log + / + i + 1 mπ log =: + / 8 f [ 1 + 1 ] + log log, + /

65 S. Kanas / J. Math. Anal. Appl. 317 006 650 658 with m 1. Note that 1 1 f log + = 1 + 1 log log + / which equals 0 if and only if 1 1 log + 1 + 1 log = 0..5 By the substitution log = tt R, equality.5 becomes e t cosh t [t + ] tanh t t = 0. Denote gt = t + tanh t t. We will show gt = 0 if and only if t = 0. By the oddness of g we only need verify the case t 0. For t>0, we have tanh t<t, that implies gt < t t + =: ut. Moreover, it is easy to check that 8t/ for t 0, 0.8, tanh t st := t/5 + /5 for t 0.8,.5,.6 /t for t.5,. Then the function rt := t + st t 8t 3 / for t 0, 0.8, = t 3 /5 + t /5 + /0 t + /10 for t 0.8, 5., t/ fort 5.,, satisfies inequalities 0 rt gt ut, t 0. Both functions rt and ut attain its zeros at the only point t = 0 so does gt. That is equivalent to the fact that g attains its only zero at t = 0, or equivalently f = 0 if and only if = 1. Moreover f 1 = 8 8/π > 0, thus f f1 = 8, and the third assertion follows. Now, we concentrate on the case when qu is the region bounded by hyperbola. In this case we also have Eq ={1}. We first prove some lemma.

S. Kanas / J. Math. Anal. Appl. 317 006 650 658 655 Lemma.1. Let k 0, 1 and Ak = /π arccos k.alsoset ϕk = 1 k A k..7 Then ϕk > 0 for k 1/, 1 and ϕk < 0 in 0, 1/. Proof. Observe that ϕ k = k 1 k 8 arccos k, 1 k and set ψk:= k 1 k 8 arccos k. Then ϕ k > 0in0, 1 if ψk<0. We have ψ k = 8 + 1 k > 0 1 k if and only if k 0, + 1 and ψ k < 0 in + 1, 1. Since ψ0 = π <0 and ψ1 = 0 then there eists the only k 1 0, / + 1/ such that ψk 1 = 0, ψk<0in0,k 1 and ψk>0ink 1, 1. Equivalently, we have ϕ k > 0in0,k 1 and ϕ k < 0ink 1, 1 with ϕ k 1 = 0. It means that ϕ increases from ϕ0 = 1toϕk 1 and net ϕ decreases to ϕ1 = 0 with the only zero at some k 0 0,k 1. Since ϕ1/ = 0 thus k 0 = 1/, and the assertion follows. Theorem.. Let k 0, 1 and Ak = /π arccos k. Also, let pz = 1 + p n z n + be analytic in U with pz 1. If there eist points z 0 = r 0 e iθ 0 U and ζ 0 U \ Eq, such that pu r0 qu and cosh[ak log ] k 1 pz 0 = q k ζ 0 = 1 k + i sinh[ Ak log ],.8 1 k with >0, then there eists m n 1 such that 1 z 0 p m z 0 = Ak 1 k + i mkak 1 k + 1 cosh [ Ak log ] + 1 sinh [ Ak log ], Rz 0 p z 0 Ak 1 k, z0 p z 0 Ak 1 k, 3 R z 0p z 0 p z 0 Moreover for k 1/, 1, it holds R z 0p z 0 p z 0 + 1 m 1 Akk k + 1 min >0 cosh [Ak log ] k. + 1 Akk 1 k.

656 S. Kanas / J. Math. Anal. Appl. 317 006 650 658 Proof. Reasoning along the same line as in the proof of Theorem.1, we obtain and so that Setting we have and ζ 0 = q k 1 [ pz 0 ] = tanh [ sinh 1 1 k q k Ak sinh[ak tanh 1 z] z = 1 k, z1 z [pz 0 1] Ak { zq k z + z q k Ak z 1 + z z = 1 k 1 z z sinh[ Ak tanh 1 z ] 1 + zq k z q k z = 1 + z 1 z + log 1 + ζ 0 1 ζ 0 = logi, > 0, ζ 0 q k ζ 0 = Ak 1 + ζ 0q ζ 0 q ζ 0 + Ak cosh [ Ak tanh 1 z ]}, + 1 [ = Ak + i ], z 1 z Ak coth[ Ak tanh 1 z ]. + 1 1 cosh[ Ak log ] k + i 1 k 1 k sinh[ Ak log ] ], [ 1 + Ak k 1 k cosh [Ak log ] k + 1 sinh[ak log ] cosh [Ak log ] k Making use inequalities + 1/ and cosh[ak log ] 1for>0 we easily obtain the assertion, that holds for each k 0, 1. Now we prove the inequality 3. Observe that R ζ 0q k ζ 0 q k ζ 0 + 1 =: Akk 1 k L and 1 1 L {cosh [Ak log ] k } Ak + 1 = sinh[akt] {cosh. [Ak log ] k } Substituting t = log t R,wehavethatL = 0 if and only if wt = sinh t [ cosh [ Akt ] k ] Ak cosh t sinh [ Akt ] attains its zero for t R. Let us denote the function v by the condition wt = vtsinh t. Then w0 = 0, and similarly as for the function g in the proof of Theorem.1 we may prove that ].

S. Kanas / J. Math. Anal. Appl. 317 006 650 658 657 vt > 0fort>0 and k 1/, 1. The function w is odd, so that wt = 0 if and only if t = 0, or equivalently L = 0 if and only if = 1. Moreover, L 1 = 1 k A k 1 k is nonnegative by Lemma.1 for k 1/, 1. In conclusion the function L attains its only minimum at 0 = 1fork 1/, 1 and maimum if k 0, 1/, so that for k 1/, 1 R ζ 0q k ζ 0 q k ζ 0 + 1 Akk 1 k L1 = Akk 1 k. We now describe the class of admissible functions for this particular q as given in Theorems.1 and.. Since, for k = 1, qζ, ζ U is given by.1 that we can simplify as qv = v + 1 + iv, v R,.9 then the admissibility condition 1.1 becomes ψu+ iv,s,w; z / Ω, when v R, z U,.10 u = v + 1, Rs = m coshπv/, π Is = mv coshπv/, R w s + 1 m π, and m 1. In the case, when q = q k k 0, 1, in view of.8 the admissibility condition 1.1 has the form ψu+ iv,s,w; z / Ω, when t R, z U,.11 u = coshat sinhat, v= 1 k, cosh t coshat Rs = m 1 k A, 1 k Is = m Ak cosh t sinhat 1 k, R w s + 1 mak 1 k for k 1/, 1, and m 1, A = Ak = /π arccos k. In spite of simplifying conditions to be checked in proofs of differential subordinations for domains bounded by conic sections they remain incomparably more difficult than in the case when q maps the unit disk onto a disk or a half-plane. The author has proved [1] that the function ψpz,zp z = pz + zp z/pz satisfies the admissibility condition for q = q 1 and Ω = {w: Rw a > w 1 a } with a a 0 = 1/π. Also, under additional assumptions, the function ψpz,zp z = pz + zp z/[βpz + γ ] satisfies that condition with the same q and Ω [1]. References [1] S. Kanas, Techniques of the differential subordination for domains bounded by conic sections, Int. J. Math. Math. Sci. 003 003 389 00. [] S. Kanas, A. Lecko, Differential subordination for domains bounded by hyperbolas, Folia Sci. Tech. Resov. 175 1999 61 70. [3] Y.Ch. Kim, A. Lecko, On differential subordination for domains bounded by parabolas, Proc. Japan Acad. 75 1999 163 165.

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