First Order Differential Subordinations for Carathéodory Functions

Shweta Gandhi; Sushil Kumar; V. Ravichandran

doi:10.5666/KMJ.2018.58.2.257

Article

Kyungpook Mathematical Journal 2018; 58(2): 257-270

Published online June 23, 2018

First Order Differential Subordinations for Carathéodory Functions

Shweta Gandhi, Sushil Kumar and V. Ravichandran^*

Department of Mathematics, University of Delhi, Delhi–110007, India, e-mail : gandhishwetagandhi@gmail.com, Bharati Vidyapeeth’s College of Engineering, Delhi–110063, India, e-mail : sushilkumar16n@gmail.com, Department of Mathematics, University of Delhi, Delhi–110007, India, e-mail : vravi@maths.du.ac.in; vravi68@gmail.com

Received: April 19, 2017; Accepted: September 9, 2017

Abstract

Go to

Abstract
1. Introduction
2. Main Results
Acknowledgements
References

The well-known theory of differential subordination developed by Miller and Mocanu is applied to obtain several inclusions between Carathéodory functions and starlike functions. These inclusions provide sufficient conditions for normalized analytic functions to belong to certain class of Ma-Minda starlike functions.

Keywords: Differential subordination, starlike function, Carathé,odory functions, Janowski function

1. Introduction

Go to

Abstract
1. Introduction
2. Main Results
Acknowledgements
References

Let be the class of analytic functions f with f(0) = 0 and f′(0) = 1, defined in the open unit disk := {z ∈ ℂ : |z| < 1} and let be the subset of containing one-to-one functions. Let f and g be analytic in . The function f is subordinate to g, written as f ≺ g, if there is a Schwarz function w analytic in with w(0) = 0 and |w(z)| < 1 such that f(z) = g(w(z)) for all . In particular, if the function g is univalent in then f ≺ g if and only if f(0) = g(0) and . Let ℘ be the class of Carathéodory functions of the form p(z) = 1+c₁z + c₂z² + · · · for all z in such that Re(p(z)) > 0. These functions are analytic in . A Carathéodory function f maps into the right-half plane. The function p(z) = (1+z)/(1 − z) is a prominent example of Carathéodory function which maps conformally onto the right-half plane. The first two important results involving first-order differential implications were given in 1935 by Goluzin and in 1947 by Robinson. Goluzin [7] proved that if zp′(z) is subordinate to a convex function h then p(z)≺∫0zh(t)t-1dt. After this basic result, many authors investigated several aspects of first order differential subordination. This first order differential subordination has many applications in the theory of univalent functions. Miller and Mocanu [15] discussed the general theory of differential subordination.

Several classes of starlike and convex functions are characterized by the quantities zf′(z)/f(z) or 1+zf″(z)/f′(z) and unified classes of starlike and convex functions using concept of Hadamard product and subordination. Further, Shanmugam [22] studied the class Sg*(ω) of all satisfying z(f * g) ′/(f * g) ≺ ω, where ω is a convex function, g is a fixed function in . For g(z) = z/(1 − z)^α the class was investigated by Padmanabhan and Parvatham [19]. When g is z/(1 − z) and z/(1−z)², the subclass Sg*(ω) reduces to and respectively. In 1992, Ma and Minda [13] considered the class Ω consisting of analytic univalent functions ω with the positive real part mapping onto domains symmetric with respect to real axis and starlike with respect to ω(0) = 1 and ω′(0) > 0. For such an ω ∈ Ω, Ma and Minda [13] proved distortion, growth, and covering theorems. The class generalizes many subclasses of , for example, (−1 ≤ B < A ≤ 1) [10], SL*:=S*(1+z) [25], Se*:=S*(ez) [14], SS*:=S*(ϕS(z)) [4], SC*:=S*(ϕC(z)) [23], SR*:=S*(ϕk(z)) [11], and S⦅*:=S*(ϕ⦅(z)) [20] where ϕC(z):=1+43z+23z2, ϕ_S(z) := 1+sinz, ϕk(z):=1+zk(k+z)(k-z)(k=2+1) and ϕ⦅(z):=z+1+z2 respectively.

In 1989, Nunokawa et al. [17] showed that zp′(z) ≺ z implies p(z) − 1 ≺ z and further authors [18] have shown some sufficient condition for Carathéodory functions. Further, Ali et al. [2] obtained the estimates on β for which 1 + βzp′(z)/p^j(z) ≺ (1 + Dz)/(1 + Ez) (j = 0, 1, 2) implies p(z) ≺ (1 + Az)/(1 + Bz), where A,B,D,E ∈ [−1, 1]. In 2013, Sivaprasad Kumar et al. [24] determined the bound on β with −1 < E < 1 and |D| ≤ 1 such that 1 + βzp′(z)/p^j(z) ≺ (1 + Dz)/(1 + Ez) (j = 0, 1, 2) implies p(z)≺1+z. Recently, Kumar and Ravichandran [12] determined certain sufficient conditions for first order differential subordinations to imply the corresponding analytic solution is subordinate to a function with positive real part. For related results, see [1, 3, 5, 6, 8, 9, 21, 26].

Motivated by these earlier works, in the next section, we obtain the sharp bound on β so that the Carathéodory function p is subordinate to a starlike function with positive real part like 1+z, e^z, (1+Az)/(1 + Bz), whenever 1 + βzp′(z)/p^j(z) ((j = 0, 1, 2) is subordinate to certain well known starlike functions. Our estimates on β are sharp. Several sufficient conditions for functions to belong to the above defined classes can be obtained as an application of the subordination results for Carathéodory functions.

2. Main Results

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Abstract
1. Introduction
2. Main Results
Acknowledgements
References

The first result of this section gives the sharp bound on β such that each of the first order differential subordination 1+βzp′(z) ≺ ϕ_⦅(z), 1+βzp′(z)/p(z) ≺ ϕ⦅(z), 1 + βzp′(z)/p²(z) ≺ ϕ⦅(z) implies p is subordinated to some well known starlike functions.

Theorem 2.1

Let p be analytic function defined inwith p(0) = 1 satisfying the subordination 1 + βzp′(z)/p^j(z) ≺ φ_⦅ ((j = 0, 1, 2). Then

(a) p(z) ≺ e^z holds for β ≥ β_j (j = 0, 1, 2), where
β0=e(2-2-log 2+log (1+2))e-1≈1.22447,β1=2+log(2)-log(1+2)≈1.22599
and
β2=e(2+log 2-log(1+2))e-1≈1.93948.
(b) p(z) ≺ ϕ⦅(z) holds for β ≥ β_j (j = 0, 1, 2), where
β0=2-2-log 2+log (1+2)2-2≈1.32132,β1=2+log 2-log(1+2)log(1+2)≈1.391
and
β2=2+2+(1+2)log 2-(1+2)log(1+2))2≈2.09289.
(c) p(z) ≺ ϕ_S(z) holds for β ≥ β_j (j = 0, 1, 2), where
β0=(2+log 2-log (1+2)) csc(1)≈1.45696,β1=2+log 2-log(1+2)log (1+sin(1))≈2.00796
and
β2=(2+log 2-log(1+2))(1+csc(1))≈2.68294.
(d) p(z) ≺ ϕ_k(z) holds for β ≥ β_j (j = 0, 1, 2) where
β0=2+2-(3+22)(log 2-log(1+2))≈4.51128,β1=2-2+log 2-log(1+2)log(2+2)-log(3+22)≈4.11214,
and
β2=2(2-(1+2)(log 2-log(1+2))≈3.73726.
(e) p(z) ≺ ϕ_C(z) holds for β ≥ β_j (j = 0, 1, 2) where
β0=32(2-2-log 2+log(1+2))≈1.16102,β1=2+log 2-log(1+2)log 3≈1.11594,
and
β2=32(2+log 2-log(1+2))≈1.83898.

The bounds on β are best possible.

The following lemma will be used in our investigation.

Lemma 2.2

([16, Theorem 3.4h, p. 132]) Let q be analytic inand let ψ and ν be analytic in a domain U containingwith ψ(w) ≠ 0 when. Set Q(z) := zq′(z)ψ(q(z)) and h(z) := ν(q(z)) + Q(z). Suppose that (i) either h is convex, or Q is starlike univalent inand (ii) Re (zh′(z)/Q(z)) > 0 for. If p is analytic in, with p(0) = q(0),and ν(p(z)) + zp′(z)ψ(p(z)) ≺ ν(q(z)) + zq′(z)ψ(q(z)), then p(z) ≺ q(z), and q is best dominant.

Proof of Theorem 2.1

(a) Case (i)(j = 0) The function defined by

qβ(z)=1+1β(z+1+z2-log(1+1+z2)-1+log 2)

is analytic in and satisfies the differential equation 1+βzqβ′(z)=ϕ⦅(z) for . Consider the functions ν(w) = 1 and ψ(w) = β. The function defined by Q(z)=zqβ′(z)ψ(qβ(z))=βzqβ′(z)=ϕ⦅(z)-1 is starlike in . If the function defined by h(z) := ν(q_β(z)) + Q(z) = 1+Q(z), then zh′(z)/Q(z) = zQ′(z)/Q(z) is a function with positive real part. By making use of Lemma 2.2, we see that the subordination

1+βzp′(z)≺1+βzqβ′(z) implies p(z)≺qβ(z).

The desired subordination p(z) ≺ e^z holds if the subordination q_β(z) ≺ e^z holds. Since q_β is an increasing function on (−1, 1), q_β(z) ≺ e^z holds if

e-1≤qβ(-1)≤qβ(1)≤e.

This gives a necessary condition for p(z) ≺ e^z to hold. Surprisingly, this necessary condition is also sufficient. This can be seen by looking at the graph of the exponential function. The inequalities q_β(−1) ≥ e⁻¹ and q_β(1) ≤ e reduces to β ≥ β₀ and β≥β0*, where

β0=2e-2e-e log (2)+e log (1+2)e-1 and β0*=2+log(2)-log(1+2)e-1

respectively. Thus, the subordination q_β(z) ≺ e^z holds if β≥max{β0,β0*}=β0.

Case (ii)(j = 1) The analytic function

qβ(z)=exp(1β(z+1+z2-log(1+1+z2)-1+log 2))

satisfies 1+βzqβ′(z)/qβ(z)=ϕ⦅(z). Consider the functions ν(w) = 1 and ψ(w) = β/w. The function defined by Q(z)=zqβ′(z)ψ(qβ(z))=ϕ⦅(z)-1 is starlike in . The function h(z) := ν(q(z)) + Q(z) = 1 + Q(z) satisfies Re(zh′(z)/Q(z)) > 0 in . By Lemma 2.2, the subordination

1+βzp′(z)p(z)≺1+βzqβ′(z)qβ(z) implies p(z)≺qβ(z).

Thus as in the proof of case(i), the subordination p(z) ≺ e^z holds if β≥2+log 2-log(1+2).

Case (iii)(j = 2). The analytic function

qβ(z)=(1-1β(z+1+z2-log(1+1+z2)-1+log 2))-1

satisfies the differential equation 1+βzqβ′(z)/qβ2(z)=ϕ⦅(z). Consider the functions ν(w) = 1 and ψ(w) = β/w². The function defined by Q(z) = zq′(z)ψ(q(z)) = zq′(z)/q²(z) = ϕ_⦅(z) − 1 is starlike in . The function h(z) := ν(q_β(z)) + Q(z) = 1+Q(z) satisfies Re(zh′(z)/Q(z)) > 0 in . Therefore, by use of Lemma 2.2, it follows that the subordination

1+βzp′(z)p2(z)≺1+βzqβ′(z)qβ2(z) implies p(z)≺qβ(z).

As in proof of case(i), the subordination p(z) ≺ e^z holds if

β≥e(2+log 2-log(1+2))(e-1).

The proofs of part (b)–(e) of this theorem are similar to the proof of part (a).

By applying Theorem 2.1(a) to the function p(z) = zf′(z)/f(z) we see that any one of the following is a sufficient condition for f to be in Se*.

Let β0=e(2-2-log 2+log(1+2))e-1≈1.22447,β1=2+log(2)-log(1+2)≈1.22599, and β2=e(2+log 2-log(1+2))e-1≈1.93948. If satisfying the following subordinations

1+β(zf′(z)f(z))i(1-zf′(z)f(z)+zf″(z)f′(z))≺ϕ⦅(z) for β≥βi+1, (i=-1,0,1),

then f∈Se*.

Theorem 2.3

Let −1 < B < A < 1 and p be analytic inwith p(0) = 1. Anyone of the following subordination conditions is sufficient for p(z) ≺ (1 + Az)/(1 + Bz):

(a) 1 + βzp′(z) ≺ ϕ_⦅(z) for β ≥ max{β₁, β₂}, where
β1=(1+B)(2+log 2-log(1+2))A-B
and
β2=(1-B)(2-2-log 2+log(1+2))A-B;
(b) 1+βzp′(z)p(z)≺ϕ⦅(z)for β ≥ max{β₃, β₄}, where
β3=2-2+log 2-log(1+2)log(1-A)-log(1-B)
and
β4=2+log 2-log(1+2)log(1+A)-log(1+B);
(c) 1+βzp′(z)p2(z)≺ϕ⦅(z)for β ≥ max{β₅, β₆}, where
β5=(1+A) (2+log 2-log(1+2))A-B
and
β6=(1-A) (2-2-log 2+log(1+2))A-B.

Proof

(a) We take the functions q_β(z), ν, ψ, Q(z), and h(z) as in the proof of case (i) of the Theorem 2.1(a). By an application of Lemma 2.2, the subordination 1+βzp′(z)≺1+βzqβ′(z) implies p(z) ≺ q_β(z). Similar reasoning as in the proof of Theorem 2.1(a), the necessary subordination p(z) ≺ (1 + Az)/(1 + Bz) holds if β ≥ max{β₁, β₂}.

Let C0=1-2-log 2/(1+2). A simple calculation gives that if B ≥ C₀, then β ≥ β₁ or B ≤ C₀, then β ≥ β₂.

(b) By defining the functions q_β(z), ν, ψ, Q(z), and h(z) as in case (ii) of the Theorem 2.1(a), Lemma 2.2 is applied. Therefore, the subordination 1 + βzp′(z)/p(z) ≺ 1 + βzq′(z)//q_β(z) implies p(z) ≺ q_β(z). By the similar reasoning as in the proof of Theorem 2.1(b), the necessary subordination p(z) ≺ (1 + Az)/(1 + Bz) holds if β ≥ max{β₃, β₄}.

(c) Consider the functions q_β(z), ν, ψ, Q(z), and h(z) as in case (iii) of the Theorem 2.1(a). By the application of Lemma 2.2, the subordination 1+βzp′(z)/p2(z)≺1+βzq′(z)//qβ2(z) implies p(z) ≺ q_β(z). Similar reasoning as in the proof of Theorem 2.1(c), the necessary subordination p(z) ≺ (1 + Az)/(1 + Bz) holds if β ≥ max{β₅, β₆}.

By taking C1=(1+log(1+2))/((1+2)+log 2), it can be seen that if B ≥ C₁, then β ≥ β₅ or B ≤ C₁, then β ≥ β₆.

For a function , by applying Theorem 2.3 to p(z) = zf′(z)/f(z) to, we see that any one of the following is a sufficient condition for :

1+β(zf′(z)f(z)) (1-zf′(z)f(z)+zf″(z)f′(z))≺ϕ⦅(z) for β≥max{β1,β2},1+β(1-zf′(z)f(z)+zf″(z)f′(z))≺ϕ⦅(z) for β≥max{β2,β4},1+β(zf′(z)f(z))-1 (1-zf′(z)f(z)+zf″(z)f′(z))≺ϕ⦅(z) for β≥max{β5,β6}.

Next result provides the bound on β such that the subordination p(z) ≺ ϕ⦅(z) holds whenever 1 + βzp′(z), 1 + βzp′(z)/p(z), or 1 + βzp′(z)/p²(z) is subordinate to some well known starlike functions.

Theorem 2.4

Let p be analytic inwith p(0) = 1. Anyone of following subordination conditions is sufficient for p(z) ≺ ϕ_⦅(z):

(a) 1+βzp′(z)pj(z)≺ezfor β ≥ β_j (j = 0, 1, 2) where
β0=12-2∑n=1∞(-1)n-1n!n≈1.35988, β1=1log (2+1)∑n=1∞1n!n≈1.49528
and
β2=2+12∑n=1∞1n!n≈2.24979,
(b) 1+βzp′(z)pj(z)≺ϕS(z)for β ≥ β_j (j = 0, 1, 2), where
β0=12-2∑n=0∞(-1)n(2n+1)!(2n+1)≈1.61506,β1=1log(2+1)∑n=0∞(-1)n(2n+1)!(2n+1)≈1.07342
and
β2=2+12∑n=0∞(-1)n(2n+1)!(2n+1)≈1.61506;
(c) 1+βzp′(z)pj(z)≺ϕk(z)for β ≥ β_j (j = 0, 1, 2), where
β0=(-1-(22+2)log22+12(2+1)≈0.327693,β1=(-1-(22+2)log22+1(2+1) log(2+1)≈0.743597
and
β2=(-1-(22+2)log22+12 ≈1.11881;
(d) 1+βzp′(z)pj(z)≺ϕC(z)for β ≥ β_j (j = 0, 1, 2) where
β0=12-2≈1.70711, β1=53(log(2+1))≈1.89099
and
β2≥5(2+1)32≈2.84518.

Proof

(a) Case (i) (j = 0) The function

qβ(z)=1+1β∑n=1∞znn!n

is analytic and satisfies the differential equation 1+βzqβ′(z)=ez for . Consider the functions ν, ψ as in Theorem 2.1(a). The function Q(z) = βzq′(z) = e^z − 1 is starlike and the function h(z) = ν(q_β(z)) + Q(z) satisfies an inequality Re(zh′(z)/Q(z))) > 0 in . By making use of Lemma 2.2, the subordination 1+βzp′(z)≺1+βzqβ′(z) implies p(z) ≺ q_β(z). As in the proof of Theorem 2.1(a), the necessary subordination p(z) ≺ ϕ_⦅(z) holds if β ≥ β₀.

Case (ii) (j = 1) Define the function q_β as:

qβ(z)=exp (1β∑n=1∞znn!n),

which is analytic in . The function q_β(z) satisfies the differential equation 1+βzqβ′(z)/qβ(z)=ez. Consider the functions ν and ψ as in Theorem 2.1(a). Note that the function Q(z)=βzqβ′(z)/qβ(z)=ez-1 is starlike and the function h(z) = ν(q_β(z))+Q(z) satisfies an inequality Re(zh′(z)/Q(z))) > 0 in . From the view of Lemma 2.2, the subordination 1+βzp′(z)/p(z)≺1+βzqβ′(z)/qβ(z) implies p(z) ≺ q_β(z). As in the proof of Theorem 2.1(a), the necessary subordination p(z) ≺ ϕ_⦅(z) holds if β ≥ β₁.

Case (iii) (j = 2) Define the function q_β as:

qβ(z)=(1-1β∑n=1∞znn!n)-1

which is analytic in and satisfies the differential equation 1+βzqβ′(z)/qβ2(z)=ez for . We take the functions ν and ψ as in Theorem 2.1(a). The function Q(z)=βzqβ′(z)/qβ2(z)=ez-1 is starlike and the function h(z) = 1+Q(z) satisfies Re(zh′(z)/Q(z))) > 0 in . From Lemma 2.2, the subordination 1+βzp′(z)/p2(z)≺1+βzqβ′(z)/qβ2(z) implies p(z) ≺ q_β(z). The desired subordination p(z) ≺ ϕ_⦅(z) holds if β ≥ β₂ as in the proof of Theorem 2.1(a).

(b) The analytic functions

q1β(z)=1+1β∑n=0∞(-1)nz2n+1(2n+1)!(2n+1), q2β(z)=exp(1β∑n=0∞(-1)nz2n+1(2n+1)!(2n+1))

and

q3β(z)=(1-1β∑n=0∞(-1)nz2n+1(2n+1)!(2n+1))-1.

satisfies the differential equations 1+βzq′(z)/qiβj(z)=ϕS(z) for where i, j = 0, 1, 2. The required subordination p(z) ≺ ϕ_⦅(z) holds for all three cases if β ≥ β₀, β₁, and β₂ as in Theorem 2.4(a) respectively.

(c) The functions

q1β(z)=1-1βk(z+2k log(1-zk)), q2β(z)=exp(-1βk(z+2k log(1-zk)))

and

q3β(z)=(1+1βk(z+2k log(1-zk)))-1

satisfies the differential equations 1+βzq′(z)/qiβj(z)=ϕk(z) for , where i, j = 0, 1, 2. For all three cases, the required subordination p(z) ≺ ϕ_⦅(z) holds if β ≥ β₀, β₁, and β₂ as in Theorem 2.4(a) respectively.

(d) The analytic functions

q1β(z)=1+1β(43z+13z2) q2β(z)=exp (1β(43z+13z2))

and

q3β(z)=(1-1β(43z+13z2))-1

satisfy the differential equations 1+βzq′(z)/qiβj(z)=ϕC(z) for , where i, j = 0, 1, 2. Proceeding as in part (a) of this theorem, we get the subordination p(z) ≺ ϕ_⦅(z) if β ≥ β₀, β₁, and β₂ respectively.

Theorem 2.5

Let −1 < B < A < 1, B ≠ 0 and p(z) be the analytic function with p(0) = 1. Then each of the following subordination is sufficient for p(z) ≺ ϕ_S(z):

(a) 1+βzp′(z)≺1+Az1+Bzfor β ≥ max{β₁, β₂} where
β1=(A-B) log(1-B)-1B sin (1) and β2=(A-B) log(1+B)B sin (1).
(b) 1+βzp′(z)p(z)≺1+Az1+Bzfor β ≥ max{β₃, β₄} where
β3=(A-B) log(1-B)-1B log(1-sin (1))-1 and β4=(A-B) log(1+B)B log(1+sin (1).
(c) 1+βzp′(z)p2(z)≺1+Az1+Bzfor β ≥ max{β₅, β₆} where
β5=(A-B) (1-sin(1)) log(1-B)-1B sin (1)
and
β6=(A-B) (1+sin(1)) log(1+B)B sin (1).

The estimates on β are sharp.

Proof

(a) Define the analytic function

qβ(z)=1+(A-B) log(1+Bz)Bβ.

Consider the functions ν and ψ as in previous theorem. The function defined by Q(z)=zqβ′(z)ψ(qβ(z))=(A-B)z/(1+Bz). Since 1+βzqβ′(z)=(1+Az)/(1+Bz) and (1+Az)/(1 + Bz) − 1 is starlike in , then Q is starlike in . The function defined by h(z) := ν(q_β(z)) + Q(z) = 1+Q(z) satisfies Re(zh′(z)/Q(z)) > 0 in . Therefore, by making use of Lemma 2.2, it is easy to see that the subordination 1+βzp′(z)≺1+βzqβ′(z) implies p(z) ≺ q_β(z). As in Theorem 2.1(a), the subordination p(z) ≺ ϕ_S(z) holds if β ≥ max{β₁, β₂}.

(b) Let the function q_β(z) be defined by

qβ(z)=exp((A-B)log(1+Bz)Bβ).

Consider the functions ν, ψ as in Theorem 2.1. The function defined by Q(z)=zqβ′(z)ψ(qβ(z))=βzqβ′(z)/qβ(z)=(A-B)z/(1+Bz) is starlike in and the function h(z) := ν(q(z)) + Q(z) = 1+Q(z) satisfies Re(zh′(z)/Q(z)) > 0 in . Hence, from Lemma 2.2, the subordination p(z) ≺ q_β(z) holds, if 1+βzp′(z)/p(z)≺1+βzqβ′(z)/qβ(z) holds. As in Theorem 2.1(a), the required subordination p(z) ≺ ϕ_S(z) holds if β ≥ max{β₃, β₄}.

(c) Let the function q_β(z) be defined by

qβ(z)=(1-(A-B) log(1+Bz)Bβ)-1.

Consider the functions ν, ψ as in Theorem 2.1. The function defined by Q(z)=zqβ′(z)ψ(qβ(z))=βzqβ′(z)/qβ2(z)=(A-B)z/(1+Bz) is starlike in and the function h(z) := ν(q_β(z)) + Q(z) = 1+Q(z) satisfies an inequality Re(zh′(z)/Q(z)) > 0 in . Hence, from Lemma 2.2, the subordination p(z) ≺ q_β(z) holds, if subordination 1+βzp′(z)/p2(z)≺1+βzqβ′(z)/qβ2(z) holds. As in Theorem 2.1(a), the required subordination p(z) ≺ ϕ_S(z) holds if β ≥ max{β₅, β₆}.

Let C₁ = log(1 − B₂)⁻¹ + sin(1) log((1 − B)(1 + B)⁻¹). A simple calculation gives that if B ≥ 0, then β ≥ β₁ or B ≤ 0, then β ≥ β₂. Further, if B ≥ C₁, then β ≥ β₅ or B ≤ C₁, then β ≥ β₆.

Next, Theorem 2.6–2.9 provides the sharp estimates on β so that the subordination 1+βzp′(z)pj(z)≺1+Az1+Bz implies the subordination p(z) ≺ ϕ_C(z), φ_⦅(z) and 1+Cz1+Dz, e^z. Proofs of these theorems are omitted as it is similar to Theorem 2.5.

Theorem 2.6

Let −1 < B < A < 1, B ≠ 0 and p(z) be an analytic function with p(0) = 1. Then each of the following subordination is sufficient for p(z) ≺ ϕ_C(z):

(a) 1+βzp′(z)≺1+Az1+Bzfor β ≥ max{β₁, β₂}, where
β1=3(A-B) log(1-B)-12B and β2=(A-B) log(1+B)2B.
(b) 1+βzp′(z)p(z)≺1+Az1+Bzfor β ≥ max{β₃, β₄}, where
β3=(A-B) log(1-B)-1B log 3 and β4=(A-B) log(1+B)B log 3.
(c) 1+βzp′(z)p2(z)≺1+Az1+Bzfor β ≥ max{β₅, β₆}, where
β5=(A-B) log(1-B)-12B and β6=3(A-B) log(1+B)2B.

The estimates on β are sharp.

Theorem 2.7

Let −1 < B < A < 1, B ≠ 0 and p(z) be an analytic function with p(0) = 1. Then each of the following subordination is sufficient for p(z) ≺ ϕ_⦅(z):

(a) 1+βzp′(z)≺1+Az1+Bzfor β ≥ max{β₁, β₂}, where
β1=(A-B) log(1-B)-1B(2-2) and β2=(A-B) log(1+B)2B.
(b) 1+βzp′(z)p(z)≺1+Az1+Bzfor β ≥ max{β₃, β₄}, where
β3=(A-B) log(1-B)-1B log (2-1)-1 and β4=(A-B) log(1+B)B log(2+1).
(c) 1+βzp′(z)p2(z)≺1+Az1+Bzfor β ≥ max{β₅, β₆}, where
β5=(A-B) log(1-B)-12B and β6=(2+1) (A-B) log(1+B)2B.

The estimates on β are sharp.

Theorem 2.8

Let −1 < B < A < 1, −1 < C < D < 1, B ≠ 0 and p(z) be an analytic function with p(0) = 1. Then each of the following subordinations is sufficient for p(z) ≺ (1 + Cz)/(1 + Dz):

(a) 1+βzp′(z)≺1+Az1+Bzfor β ≥ max{β₁, β₂}, where
β1=(A-B) (1-D) log(1-B)-1B(C-D) and β2=(A-B) (1+D) log(1+B)B(C-D).
(b) 1+βzp′(z)p(z)≺1+Az1+Bzfor β ≥ max{β₃, β₄} where
β3=(A-B) log(1-B)-1B(log(1-D)-log(1-C)) and β4=(A-B) log(1+B)B(log(1+C)-log(1+D)).
(c) 1+βzp′(z)p2(z)≺1+Az1+Bzfor β ≥ max{β₅, β₆} where
β5=(A-B) (C-D) log(1-B)-1B(1-C) and β6=(A-B) (1+C) log(1+B)B(C-D).

The estimates on β are sharp.

Theorem 2.9

Let −1 < B < A < 1, B ≠ 0. If the subordination

1+βzp′(z)p(z)≺1+Az1+Bz for β≥(A-B) log(1-B)-1B holds,

then p(z) ≺ e^z.

The subordination result in Theorem 2.9 was also investigated by the authors in [14, Theorem 2.16, p. 1019], which was not sharp.

Acknowledgements

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Abstract
1. Introduction
2. Main Results
Acknowledgements
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The authors are thankful to the referee for his useful comments.

References

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Abstract
1. Introduction
2. Main Results
Acknowledgements
References

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