Kyungpook Mathematical Journal 2018; 58(2): 257-270
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
*Corresponding Author.
Received: April 19, 2017; Accepted: September 9, 2017; Published online: June 23, 2018.

This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract

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, Carathéodory functions, Janowski function
1. Introduction

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 fg, 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 fg 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 $ϕC(z):=1+43z+23z2$, ϕS(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)/pj(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)/pj(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 $1+z$, ez, (1+Az)/(1 + Bz), whenever 1 + βzp′(z)/pj(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

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)/pj(z) ≺ φ ((j = 0, 1, 2). Then

(a) p(z) ≺ ez holds for β ≥ βj (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 ≥ βj (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) ≺ ϕS(z) holds for β ≥ βj (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) ≺ ϕk(z) holds for β ≥ βj (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) ≺ ϕC(z) holds for β ≥ βj (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(qβ(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) ≺ ez holds if the subordination qβ(z) ≺ ez holds. Since qβ is an increasing function on (−1, 1), qβ(z) ≺ ez holds if

$e-1≤qβ(-1)≤qβ(1)≤e.$

This gives a necessary condition for p(z) ≺ ez to hold. Surprisingly, this necessary condition is also sufficient. This can be seen by looking at the graph of the exponential function. The inequalities qβ(−1) ≥ e−1 and qβ(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 qβ(z) ≺ ez 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) ≺ ez 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(qβ(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) ≺ ez 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 qβ(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) ≺ qβ(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 qβ(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)//qβ(z) implies p(z) ≺ qβ(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 qβ(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) ≺ qβ(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) $1+βzp′(z)pj(z)≺ez$for β ≥ βj (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 ≥ βj (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 ≥ βj (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 ≥ βj (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) = ez − 1 is starlike and the function h(z) = ν(qβ(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) ≺ qβ(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 qβ as:

$qβ(z)=exp (1β∑n=1∞znn!n),$

which is analytic in . The function qβ(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(qβ(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) ≺ qβ(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 qβ 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) ≺ qβ(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) ≺ ϕS(z):

(a) $1+βzp′(z)≺1+Az1+Bz$for 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) $1+βzp′(z)p(z)≺1+Az1+Bz$for 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) $1+βzp′(z)p2(z)≺1+Az1+Bz$for 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(qβ(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) ≺ qβ(z). As in Theorem 2.1(a), the subordination p(z) ≺ ϕS(z) holds if β ≥ max{β1, β2}.

(b) Let the function qβ(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) ≺ qβ(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) ≺ ϕS(z) holds if β ≥ max{β3, β4}.

(c) Let the function qβ(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(qβ(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) ≺ qβ(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) ≺ ϕS(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) ≺ ϕC(z), φ(z) and $1+Cz1+Dz$, ez. 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) ≺ ϕC(z):

(a) $1+βzp′(z)≺1+Az1+Bz$for β ≥ max{β1, β2}, where

$β1=3(A-B) log(1-B)-12B and β2=(A-B) log(1+B)2B.$

(b) $1+βzp′(z)p(z)≺1+Az1+Bz$for 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) $1+βzp′(z)p2(z)≺1+Az1+Bz$for 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) $1+βzp′(z)≺1+Az1+Bz$for β ≥ max{β1, β2}, where

$β1=(A-B) log(1-B)-1B(2-2) and β2=(A-B) log(1+B)2B.$

(b) $1+βzp′(z)p(z)≺1+Az1+Bz$for 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) $1+βzp′(z)p2(z)≺1+Az1+Bz$for 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) $1+βzp′(z)≺1+Az1+Bz$for 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) $1+βzp′(z)p(z)≺1+Az1+Bz$for 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) $1+βzp′(z)p2(z)≺1+Az1+Bz$for 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) ≺ ez.

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

The authors are thankful to the referee for his useful comments.

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