Starlike Functions Related to the Bell Numbers

Citation for this paper:

Cho, N.E., Kumar, S., Kumar, V., Ravichandran, V. & Srivastava, H.M. (2019).

Starlike Functions Related to the Bell Numbers. Symmetry, 11(2), 219.

https://doi.org/10.3390/sym11020219

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Starlike Functions Related to the Bell Numbers

Nak Eun Cho, Sushil Kumar, Virendra Kumar, V. Ravichandran and H. M. Srivastava

February 2019

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This article was originally published at:

https://doi.org/10.3390/sym11020219

symmetry

S S

Article

Starlike Functions Related to the Bell Numbers

Nak Eun Cho1,*, Sushil Kumar2, Virendra Kumar3 , V. Ravichandran4 and H. M. Srivastava5,6,*

1 _{Department of Applied Mathematics, Pukyong National University, Busan 48513, Korea} 2 _{Bharati Vidyapeeth’s College of Engineering, Delhi 110063, India; sushilkumar16n@gmail.com}

3 _{Department of Mathematics, Ramanujan College, University of Delhi, Kalkaji, New Delhi 110019, India;}

vktmaths@yahoo.in

4 _{Department of Mathematics, National Institute of Technology, Tiruchirappalli, Tamil Nadu 620015, India;}

vravi68@gmail.com or ravic@nitt.edu

5 _{Department of Mathematics and Statistics, University of Victoria, Victoria, BC V8W 3R4, Canada} 6 _{Department of Medical Research, China Medical University Hospital, China Medical University,}

Taichung 40402, Taiwan

* Correspondence: necho@pknu.ac.kr (N.E.C.); harimsri@math.uvic.ca (H.M.S.)

Received:19 January 2019; Accepted: 12 February 2019; Published: 13 February 2019




 Abstract: The present paper aims to establish the first order differential subordination relations between functions with a positive real part and starlike functions related to the Bell numbers. In addition, several sharp radii estimates for functions in the class of starlike functions associated with the Bell numbers are determined.

Keywords:differential subordination; starlike functions; Bell numbers; radius estimate

MSC:30C45; 30C55; 30C80

1. Introduction

LetAbe a class of analytic functions f in the open unit diskD:= {z∈ C:|z| <1}and normalized

by the conditions f(0) = 0 and f0(0) = 1. SupposeS is a subclass ofA consisting of univalent functions. An analytic function f is subordinate to g, written as f ≺ g, if there exists an analytic function w :D → Dwith|w(z)| ≤ |z|such that f(z) =g(w(z)) (z∈ D). Moreover, if g is univalent in D, then the equivalent conditions for subordination can be written as f(0) =g(0) and f(D) ⊆g(D).

By imposing some geometric and analytic conditions over the functions in the classS, many authors considered several subclasses ofS. Various subclasses of starlike and convex functions were studied in the literature, and they can be unified by considering an analytic univalent function ϕ with a positive real part inD, symmetric about the real axis and starlike with respect to ϕ(0) = 1, and ϕ0(0) > 0.

Ma and Minda [1] studied the class

S∗(ϕ):=  f ∈ A: z f 0_(z) f(z) ≺ ϕ(z)  .

The classS∗(ϕ)for various choice of the domain ϕ(D)was considered in recent years. The class

S∗[A, B]:= S∗((1+Az)/(1+Bz))(−1≤B<A≤1)was introduced by Janowski [2]. For 0≤α≤1,

the classS∗₍

α) := S∗[1−2α,−1] is the class of starlike functions of order α. Uralegaddi et al. [3]

defined the class

M(β):=  f ∈ A: Re z f 0_(z) f(z)  <β (β>1)  = S∗ 1+ (1−2β)z 1−z  .

Several authors considered various special cases of the class of Janowski starlike functions by considering some specific functions, namely ϕq(z) := z +

√

1+z2_{, ϕ}

0(z) := 1+ (z/k)((k+z)/(k−z)) (k = √2+1) , ϕs(z) := 1+sin z, and Gα(z) := 1+z/(1−αz2). Some of

those classes are: S_L∗ := S∗(√1+z) [4], S_q∗ := S∗(ϕq(z)) [5], Se∗ = S∗(ez) [6], SR∗ = S∗(ϕ0) [7], S∗

s = S∗(ϕs)[8]) ,BS∗(α):= S∗(Gα(z)), 0≤α<1 [9,10]. For a brief survey on these classes, readers

may refer to [11,12].

It should be noted that the special cases of ϕ, mentioned above, are univalent in the unit disk. In 2011, Dziok et al. [13,14] considered ϕ to be a non-univalent function associated with the Fibonacci numbers, defined by ˜p(z):= ϕ(z) = 1+τ 2_z2 1−τz−τ2z2, τ:=  1−√5/2

which maps the unit diskDon to a shell-like domain in the right-half plane. Further, they defined the

classS∗

F :={f ∈ A: z f0(z)/ f(z) ≺ ˜p(z)}. The functions f ∈ SF∗are starlike of order √

5/10.

Motivated by the above defined classes, we consider a function associated with the Bell Numbers. For a fixed non-negative integer n, the Bell numbers Bncount the possible disjoint partitions of a set

with n elements into non-empty subsets or, equivalently, the number of equivalence relations on it. The Bell numbers Bnsatisfy a recurrence relation involving binomial coefficients Bn+1=∑nk=0(

n k)Bk.

Clearly B0= B1= 1, B2 =2, B3 =5, B4 =15, B5 =52, and B6= 203. For more details, see [15–21].

Kumar et al. [22] considered the function Q(z):=eez−1= ∞

∑

n=0 Bnz n n! =1+z+z 2₊5 6z 3₊5 8z 4_{+ · · · (z∈ D)}

which is starlike with respect to 1 and it’s coefficients generate the Bell numbers. Kumar et al. [22] defined the classS_B∗byS_B∗:= S∗(Q). From [1], note that the function f ∈ S_B∗if and only if there exists an analytic function q, satisfying q(z) ≺Q(z) (z∈ D), such that

f(z) =I(q(z)) =z exp Z z 0 q(t) −1 t dt  . The above representation shows that the functions in the classS∗

B can be seen as an integral

transform I(q(z))of the function q with f(0) =0 and f0(0) =1. The reader may refer to the paper [23] and the references cited therein for integral transform related works. The authors in [22] determined sharp coefficient bounds on the six initial coefficients, Hankel determinant, and on the first three consecutive higher order Schwarzian derivatives for functions in the classS∗

B.

LetPbe the class of analytic functions p :D → Cwith p(0) =1 and Re p(z) >0 (z∈ D). In 1989,

Nunokawa et al. [24] showed that if 1+zp0(z) ≺ 1+z, then p(z) ≺ 1+z. In 2007, Ali et al. [25] computed the condition on β, in each case, for which

1+βz p 0_(z) pj_(z) ≺ 1+Dz 1+Ez (j=0, 1, 2) implies p(z) ≺ 1+Az 1+Bz,

A, B, C, D, E, F∈ [−1, 1]. Further, Ali et al. [26] determined some sufficient conditions for normalized analytic functions to lemniscate starlike functions. Recently, Kumar and Ravichandran [27] obtained sufficient conditions for first order differential subordinations so that the corresponding analytic function belongs to the classP. In 2016, Tuneski [28] gave a criteria for analytic functions to be Janowski starlike. For more details, see [11,29–33].

Motivated by above works, in Section2, using the theory of differential subordination developed by Miller and Mocanu, a sharp bound on parameter β is determined in each case so that p(z) ≺Q(z), whenever 1+βz p0(z)/pj(z)(j = 0, 1, 2)is subordinate to the function ϕ0(z)or

√

1+z or Gα(z)or

in the classS_B∗as an application of these subordination results. In Section3,S∗_B-radius for the class of Janowski starlike functions and some other well-known classes of analytic functions are investigated.

2. Differential Subordinations

Theorem1provides estimate on β so that p(z) ≺Q(z)holds, whenever 1+βz p0(z) ≺ϕ0(z)or

ϕs(z)or √

1+z or Gα(z)or(1+Az)/(1+Bz)or ϕs(z)or ϕq(z)or ez.

To prove our main results, we need the following lemma due to Miller and Mocanu:

Lemma 1. ([32] Theorem 3.4h, p. 132) Let q be analytic inDand let ψ and ν be analytic in a domain U

containing q(D)with ψ(w) 6=0 when w∈q(D). Set

Q(z):=zq0(z)ψ(q(z)) and h(z):=ν(q(z)) + Q(z).

Suppose that

(i) either h is convex, orQis starlike univalent inDand

(ii) Re zh

0_(z) Q(z)



>0 for z∈ D.

If p is analytic inD, with p(0) =q(0), p(D) ⊆U and

ν(p(z)) +zp0(z)ψ(p(z)) ≺ν(q(z)) +zq0(z)ψ(q(z)),

then p≺q, and q is most dominant.

Theorem 1. Let l(e) = (1−e(1−e)/e₎−1_{, 0<α<1, 0<B<A<1, and p be an analytic function defined}

inDwith p(0) =1.

Set

Υβ(z, p(z)) =1+βz p

0_(z).

Then, the following are sufficient for p(z) ≺Q(z). (a) Υβ(z, p(z)) ≺ ϕ0(z)for β≥l(e)(1− √ 2+log 2) ≈0.59533. (b) Υβ(z, p(z)) ≺ √ 1+z for β≥l(e)(2(1−log2)) ≈1.30984. (c) Υβ(z, p(z)) ≺Gα(z)for β≥l(e)₂√1_αlog 1+√α 1−√α.

(d) Υβ(z, p(z)) ≺ 1+Az1+Bz for β≥l(e)A−BB log(1−B)−1.

(e) Υβ(z, p(z)) ≺ ϕs(z)for β≥l(e)∑∞n=0 (−1) n (2n+1)!(2n+1) ≈2.01905. (f) Υ_β(z, p(z)) ≺ ϕq(z)for β≥l(e)(2− √ 2−log 2+log(1+√2)) ≈1.65198. (g) Υβ(z, p(z)) ≺ezfor β≥l(e)∑∞n=0(−1) n−1 n!n ≈0.785166.

The lower bound on β in each case is sharp.

Proof. Let the functions ν and ψ be defined by ν(w) =1 and ψ(w) =β.

(a) Define the function qβ:D → Cby

qβ(z) =1− 1 βk  z+2k log1−z k 

is a solution of the differential equation βzq0(z) = ϕ0(z) −1 and is analytic in D. Now consider

the function

Q(z) =zq0_β(z)ψ(qβ(z)) =ϕ0(z) −1=

k+z−2k2 k−z . It can be easily seen thatQis starlike inDand the function h is defined by

satisfies the following inequality Re zh 0_(z) Q(z)  =Re zQ 0_(z) Q(z)  >0(z∈ D). Therefore, from Lemma1, we conclude that

1+βz p0(z) ≺1+βzq0_β(z)implies p≺qβ. (1)

Now the subordination p≺Q holds if subordination q_β ≺Q. Thus, the subordination q_β≺Q holds if the inequalities

Q(−1) ≤qβ(−1) ≤qβ(1) ≤Q(1)

hold and these yield a necessary condition for subordination p ≺ Q to hold. In view of the graph of the respective function, the necessary condition is also sufficient condition. The inequalities qβ(−1) ≥Q(−1)and qβ(1) ≤Q(1)yield β≥β1and β≥β2, where

β1= 1

−√2+log 2

1−e(1−e)/e and β2=

1−√2−2 log(2−√2)

e(e−1)/e₋₁ .

Now the subordination qβ≺Q holds if β≥max{β1, β2} =β1.

(b) The function qβ(z) = β+2( √ 1+z−log(1+√1+z) +log 2−1) β

is an analytic solution of the first order differential equation βzq0(z) =√1+z−1 inD. The function Q defined by Q(z) = zq0_β(z)ψ(q_β(z)) =

√

1+z−1 is starlike in D and the function h(z) :=

ν(q(z)) + Q(z)satisfies Re(zh0(z)/Q(z)) = Re(zQ0(z)/Q(z)) > 0, z ∈ D. Therefore, in view of

the subordination relation1, the required subordination p≺Q holds if subordination qβ ≺Q holds.

Thus, the subordination qβ≺Q holds if the inequalities

Q(−1) ≤qβ(−1) ≤qβ(1) ≤Q(1)

hold which in-turn yield a necessary condition for subordination p≺Q. The inequalities qβ(−1) ≥

Q(−1) and qβ(1) ≤ Q(1) yield β ≥ β1 = 2(1−log 2)/1−e(1−e)/e and β ≥ β2 = 2(

√

2−1+

log 2−log(1+√2))/(e(1−e)/e−1), respectively. Therefore, the subordination qβ ≺ Q holds if β≥

max{β1, β2} = β1.

(c) The analytic function

qβ(z) = 2√αβ+log1+ √ αz 1−√αz 2√αβ

is a solution of the differential equation βzq0_β(z) =Gα(z) −1 inD. Now computation shows that

Q(z) =zq0_β(z)ψ(q_β(z)) = z

1−αz2

is starlike inD. Note that the function h(z):=ν(q(z)) + Q(z) =1+ Q(z)satisfies Re(zh0(z)/Q(z)) =

Re(zQ0(z)/Q(z)) > 0 in D. Therefore, in view of the subordination relation 1, the required

subordination p≺Q holds if subordination qβ ≺Q. Similar to as in part (a), the desired subordination

p≺Q holds if β≥max{β1, β2} =β1, where β1=l(e)g(α)and β2= −l(e)g(α)such that

g(α) = 1

2√αlog

1+√α

(d) Consider the analytic function

q_β(z) = Bβ+ (A−B)log(1+Bz)

Bβ which is a solution of differential equation

βzq0(z) = (A−B)z

1+Bz .

Since the function(A−B)z/(1+Bz)is starlike inD, it follows thatQ(z) =zq0_β(z)ψ(q_β(z))is

starlike inD. The function h : D → Cdefined by h(z) := ν(q_β(z)) +Q(z) = 1+Q(z)satisfies

Re(zh0(z)/Q(z)) > 0 (z ∈ D). Thus, as in previous case, the subordination p ≺ Q holds if

β≥max{β1, β2} =β1, where β1= (A−B)log(1−B)−1 B(1−e(1−e)/e₎ and β2= (A−B)log(1+B) B(e(1−e)/e₋₁₎ .

(e) The differential equation

dq dz =

sin z

βz

has an analytic solution

qβ(z) =1+ 1 β ∞

∑

n=0 (−1)nz2n+1 (2n+1)!(2n+1)

inD. Now the functionQ(z) = zq0_β(z)ψ(q_β(z)) = sin z is starlike in Dand the function h(z) :=

ν(q(z)) + Q(z) =1+ Q(z), satisfies Re(zh0(z)/Q(z)) =Re(zQ0(z)/Q(z)) >0 holds. As in part (a),

the desired subordination p(z) ≺Q(z)holds if β≥max{β1, β2} = β1, where

β1= 1 (1−e(1−e)/e₎ ∞

∑

n=0 (−1)n (2n+1)!(2n+1) ≈2.01905 and β2= 1 (e(e−1)₋₁₎ ∞

∑

n=0 (−1)n (2n+1)!(2n+1) ≈0.206779.

(f) The differential equation

dq dz =

z+√1+z2₋₁

βz

has an analytic solution qβ(z) =

β+ (z+

√

1+z2_−log(1+√_1+z2_{) −1+log 2)}

β .

Computation shows that the function

Q(z) =zq0_β(z)ψ(qβ(z)) =z+

p

1+z2₋₁

is starlike inD. As before, the function h(z):=ν(q(z)) + Q(z)satisfies Re(zh0(z)/Q(z)) >0, z∈ D.

Therefore, the desired subordination p≺Q holds if β≥max{β1, β2} =β1, where

β1=

2−√2−log 2+log(1+√2)

and

β2= √

2+log 2−log(1+√2)

e(1−e)/e₋₁ ≈0.267979.

(g) The differential equation

dq dz =

ez−1

βz

has an analytic solution

qβ(z) =1+ 1 β ∞

∑

n=0 zn n!n.

Note that the functionQ(z) =zq0_β(z)ψ(q_β(z)) =ezis starlike in the unit diskDand the function

h(z) := ν(q(z)) + Q(z) =1+ Q(z)satisfies Re(zh0(z)/Q(z)) = Re(zQ0(z)/Q(z)) > 0. Now the

subordination p≺Q holds if β≥max{β1, β2} =β1, where

β1= 1 (1−e(1−e)/e₎ ∞

∑

n=0 (−1)n−1 n!n ≈0.785166 and β2= 1 (e(e−1)₋₁₎ ∞

∑

n=0 1 n!n≈0.288069. This ends the proof.

Theorem1also provides the following various sufficient conditions for the normalized analytic functions f to be in the classS_B∗.

Let function f ∈ Aand set Υβ  z, z f 0_(z) f(z)  =1+βz f 0_(z) f(z)  1− z f 0_(z) f(z) + z f00(z) f0(z)  . If either of the following subordination holds

(a) Υβ  z, z f_{f (z)}0(z)≺ϕ0(z) (β≥0.59533), (b) Υβ  z, z f_{f (z)}0(z)≺√1+z(β≥1.30984), (c) Υβ  z, f_{f (z)}0(z)≺Gα(z) (β≥ _(1−e(11−e)/e₎ 1 2√αlog 1+√α 1−√α), (d) Υβ  z, f_{f (z)}0(z)≺ 1+Az_1+Bz (β≥ 1 (1−e(1−e)/e₎ A−B B log(1−B)−1), (e) Υβ  z, z f_{f (z)}0(z)≺ϕs(z) (β≥2.01905), (f) Υ_βz, z f_{f (z)}0(z)≺ϕq(z) (β≥1.65198), (g) Υβ  z, z f_{f (z)}0(z)≺ez(β≥0.785166), then f ∈ S∗ B.

The next result gives sharp lower bound on β such that subordination p≺Q holds, whenever 1+βz p0(z)/p(z) ≺ϕ0(z)or ϕs(z)or

√

1+z or Gα(z)or(1+Az)/(1+Bz)or ϕs(z)or ϕq(z)or ez. Theorem 2. Let0<α<1, 0<B<A<1, and p be an analytic function defined inDwith p(0) =1.

Set

Ωβ(z, p(z)) =1+β

zp0(z)

p(z) .

Then, the following conditions are sufficient for subordination p≺Q. (a) Ωβ(z, p(z)) ≺ϕ0(z)for β≥

e(2(1+√2) log√2−1

(e−1)(1+√2) ≈0.441266.

(b) Ωβ(z, p(z)) ≺

√

1+z for β≥ 2e(1−log 2)_e−1 ≈0.970868. (c) Ωβ(z, p(z)) ≺Gα(z)for β≥ _2(e−1)e √_αlog

1+√α

1−√α.

(e) Ωβ(z, p(z)) ≺ϕs(z)for β≥ e e−1∑∞n=0 (−1)n (2n+1)!(2n+1) ≈1.49655. (f) Ωβ(z, p(z)) ≺ϕq(z)for β≥ e e−1(2− √ 2+log(1+√2) −log 2) ≈1.22447. (g) Ω_β(z, p(z)) ≺ez_{for β≥} 1 e−1∑∞n=0n!n1 ≈0.766987.

The lower bound on β in each case is sharp.

Proof. Let us define ν(w) =1 and ψ(w) =β/w for all w∈ C.

(a) The function

q_β(z) =exp  − 1 βk  z+2k log1−z k 

satisfies the differential equation βzq0(z)/q(z) = ϕ0(z) −1. Clearly, the functionQ :D →defined

byQ(z) = zq0_β(z)ψ(q_β(z)) = (z−2k2+k)/(k−z) is starlike inD. Further, the function h(z) :=

ν(q_β(z)) + Q(z)satisfies Re(zh0(z)/Q(z)) >0(z∈ D). Thus, using Lemma1, it follows that

1+βzp

0_(z)

p(z) ≺1+β

zq0_β(z)

q_β(z) implies p≺qβ. (2)

Now using Theorem1(a), the subordination p≺Q holds if β≥max{β1, β2} = β1, where

β1= (−1+2(1+√2)log√2)e (e−1)(1+√2) and β2= − (1+2(1+√2)log(2−√2)) (e−1)(1+√2) . (b) The function qβ(z) =exp  2 β √ 1+z−log(1+√1+z) +log 2−1 

is a solution of the differential equation

βzq

0_(z)

q(z) = √

1+z−1. Moreover, the functionQ(z) =zq0_β(z)ψ(q_β(z)) =

√

1+z−1 is starlike inDand a computation

shows that the function h(z) := ν(q(z)) + Q(z)satisfies Re(zh0(z)/Q(z)) > 0 (z ∈ D). Now the

desired subordination p≺Q holds if β≥max{β1, β2} = β1, where β1=2e(1−log 2)/(e−1)and

β2=2(−1+ √

2+log 2−log(1+√2))/(e−1). (c) Consider the function qβdefined by

qβ(z) =exp  1 2√αβlog 1+√αz 1−√αz  .

It can be verified that the function qβis a solution of the differential equation

βzq

0_(z)

q(z) =

1 1−αz2.

Now the functionQ(z) =zq0_β(z)ψ(q_β(z)) =1/(1−αz2)is starlike inDand the function h(z):=

ν(q(z)) + Q(z)satisfies Re(zh0(z)/Q(z)) >0(z∈ D). Now, as in previous cases, p≺Q holds only if β≥max{β1, β2} =β1, where

β1= e 2(e−1)√αlog 1+√α 1−√α and β2= 1 2(e−1)√αlog 1+√α 1−√α.

(d) Let the function qβ(z) = exp((A−B)log(1+Bz)/βB) be an analytic solution of the

differential equation 1+βzq 0_(z) q(z) = 1+Az 1+Bz.

Now the desired subordination p ≺ Q holds if β ≥ max{β1, β2} = β1, where β1 = e(A−

B)log(1−B)−1/B(e−1)and β2=e(A−B)log(1+B)/B(e−1).

(e) The differential equation βzq0(z)/q(z) =sin z has an analytic solution given by

qβ(z) =exp 1 β ∞

∑

n=0 (−1)nz2n+1 (2n+1)!(2n+1) ! .

As in part Theorem2(a), the subordination p≺Q holds if β≥max{β1, β2} = β1where

β1= e e−1 ∞

∑

n=0 (−1)n (2n+1)!(2n+1) ≈1.49655 and β2= 1 e−1 ∞

∑

n=0 (−1)n (2n+1)!(2n+1) ≈0.55055.

(f) The solution of the differential equation dq dz = z+√1+z2₋₁ βz is given by qβ(z) =exp z+√1+z2_−log(1+√_1+z2_{) −1+log 2} β ! .

As in proof of Theorem2(a), the desired result holds if β ≥ max{β1, β2} = β1, where β1 =

e(2−√2+log(1+√2) −log 2)/(e−1)and β2= ( √

2−log(1+√2) +log 2)/(e−1). (g) The differential equation βzq0(z)/q(z) =ez_{−1 has a solution}

qβ(z) =exp 1 β ∞

∑

n=1 zn n!n !

analytic inD. Thus, as previous, the subordination p≺Q holds if β≥max{β1, β2} =β2, where

β1= e e−1 ∞

∑

n=0 (−1)n−1 n!n ≈0.581976 and β2= 1 e−1 ∞

∑

n=0 1 n!n ≈0.766987. This ends the proof.

Next, Theorem2also provides the following various sufficient conditions for the normalized analytic functions f to be in the classS∗

B. Let the function f ∈ Aand set

Ωβ  z, z f 0_(z) f(z)  =1+β  1−z f 0_(z) f(z) + z f00(z) f0(z)  . If either of the following subordination conditions are fulfilled:

(a) Ωβ  z, z f_{f (z)}0(z)≺ϕ0(z) (β≥0.441266), (b) Ωβ  z, z f_{f (z)}0(z)≺√1+z(β≥0.970868), (c) Ωβ  z, z f_{f (z)}0(z)≺Gα(z) (β≥ _2(e−1)e √_αlog 1+√α 1−√α), (d) Ωβ  z, z f_{f (z)}0(z)≺ 1+Az 1+Bz (β≥ B(e−1)e (A−B)log(1−B) −1_), (e) Ωβ  z, z f_{f (z)}0(z)≺ϕs(z) (β≥1.49655), (f) Ωβ  z, z f_{f (z)}0(z)≺ϕq(z) (β≥1.22447), (g) Ωβ  z, z f_{f (z)}0(z)≺ez(β≥0.766987), then f ∈ S_B∗.

In the following theorem, the sharp lower bound on β is obtained so that the subordination p≺Q holds, whenever 1+βz p0(z)/p2(z) ≺ϕ0(z)or ϕs(z)or

√

1+z or Gα(z)or(1+Az)/(1+Bz)or ϕs(z)

or ϕq(z)or ez. These results can be proved by defining the functions ν, ψ :D →defined by ν(w) =1

and ψ(w) =β/w2and proceeding in a similar fashion as in the proofs of Theorems1and2.

Theorem 3. Let0<α<1, 0<B<A<1, and p be an analytic function defined inDwith p(0) =1.

Set

Ξβ(z, p(z)) =1+β

zp0(z)

p2_(z).

Then, the following conditions are sufficient for p≺Q. (a) Ξβ(z, p(z)) ≺ ϕ0(z)for β≥ 1+2(√2+1) log(2−√2) (1+√2)(e(1−e)₋₁₎ ≈0.798642. (b) Ξ_β(z, p(z)) ≺√1+z for β≥ 2(−1+ √ 2+log 2−log(1+√2)) 1−e1−e ≈0.550768. (c) Ξβ(z, p(z)) ≺Gα(z)for β≥ e e−1 ee−1₋₁₂√1_αlog 1+√α 1−√α. (d) Ξβ(z, p(z)) ≺ 1+Az 1+Bz for β≥ e(1−e)/e 1−e(1−e)/e (A−B) log (1−B)−1 B . (e) Ξβ(z, p(z)) ≺ ϕs(z)for β≥ e e−1 ee−1₋₁∑∞n=0 (−1)n (2n+1)!(2n+1) ≈1.15278. (f) Ξβ(z, p(z)) ≺ ϕq(z)for β≥ e e−1 ee−1₋₁( √ 2−log(1+√2) +log 2) ≈1.49397. (g) Ξβ(z, p(z)) ≺ezfor β≥ e e−1 ee−1₋₁∑∞n=0n!n1 ≈1.60597.

The lower bound on β in each case is sharp. Let f ∈ Aand set

Ξβ  z, z f 0_(z) f(z)  =1+β z f 0_(z) f(z) −1 1−z f 0_(z) f(z) + z f00(z) f0_(z)  . If either of the following subordination holds

(a) Ξβ  z, z f_{f (z)}0(z)≺ϕ0(z) (β≥0.798642), (b) Ξβ  z, z f_{f (z)}0(z)≺√1+z(β≥0.550768), (c) Ξβ  z, z f_{f (z)}0(z)≺Gα(z) (β≥ e e−1 ee−1₋₁₂√1_αlog 1+√α 1−√α), (d) Ξβ  z, z f_{f (z)}0(z)≺ 1+Az 1+Bz (β≥ e (1−e)/e 1−e(1−e)/e (A−B) log (1−B)−1 B ), (e) Ξβ  z, z f_{f (z)}0(z)≺ϕs(z) (β≥1.15278), (f) Ξβ  z, z f_{f (z)}0(z)≺ϕq(z) (β≥1.49397), (g) Ξβ  z, z f_{f (z)}0(z)≺ez_{(β≥1.60597),} then f ∈ S_B∗.

3. Radius Estimates

Let θ1and θ2be two sub-families ofA. The θ1radius of θ2is the largest number ρ∈ (0, 1)such

that r−1f(rz) ∈ θ1, 0 < r ≤ ρ for all f ∈ θ2. Grunsky [34] obtained the radius of starlikeness for

functions in the classS. Sokół [35] computed the radius of α-convexity and α-starlikeness for a class

S_L∗. In 2016, authors [7] determined theS_R∗-radius for various subclasses of starlike functions. For more results on radius problems, see [36–41].

The main technique involved in tackling theS_B∗-radius estimates for classes of functions f is the determination of the disk that contains the values of z f0(z)/ f(z). The associated technical lemma is achieved as:

Lemma 2. Let Q(z):=eez−1, z∈ D. Define the function r :[e1/e−1, ee−1] → R+by

r(a):= ( _ea−e1/e e , e 1 e−1≤a≤ e1/e+ee 2e ; ee−ea e , e 1/e_+ee 2e ≤a≤ee−1.

Then, the following holds:

{w∈ C:|w−a| <r(a)} ⊂ΩB⊂  w∈ C:|w−1| < e e_−e e  .

Proof. To prove the assertion, we let z=eit_{, t∈ (−π, π]. Therefore,}

Q(eit) =eeeit−1=u(t) +iv(t)

with

u(t):=cos sin(sin t)ecos t
exp ecos t_{cos(sin t) −1}
and

v(t):=sin sin(sin t)ecos t
expecos(t)cos(sin t) −1.

Now, consider the square of the distance of an arbitrary point(u(t), v(t))on the boundary of

∂Q(D)from(a, 0)and is given by

h(t) =d2(t) =a2−2aeecos tcos(sin t)−1cos sin(sin t)ecos t

+e2ecos tcos(sin t)−2.

Now we need to prove|w−a| <r(a)is the largest disk contained in Q(D). For this, we need to show that min−π≤t≤πd(t) = r(a). Since h is an even function, i.e., h(t) = h(−t), we need to

only consider the case when t∈ [0, π]. Now h0(t) =0 has three roots viz. 0, π and t0(a) ∈ (0, π).

Among these roots, the root t0(a)depends on a and graphics reveals that h is increasing in the interval [0, t0(a)]and decreasing in[t0(a), π], and therefore, h attains its minimum either at 0 or π. Further

computations give h(π) =



ea−e1/e2/e2 and h(0) = (ee−ea)2/e2. Hence, we have

min −π≤t≤πh(t) =min{h(0), h(π)} = ( h(π), e 1 e−1≤a≤ e1/e+ee 2e ; h(0), e1/e_2e+ee ≤a≤ee−1. Therefore, we can write

min −π≤t≤πd(t) = ( _ea−e1/e e , e 1 e−1≤a≤ e1/e+ee 2e ; ee−ea e , e 1/e_+ee 2e ≤a≤ee−1.

To find the circle of minimum radius with center at(1, 0)containing the domain Q(D), we need to find the maximum distance from(1, 0)to an arbitrary point on the boundary of the domain Q(D). The square of this distance function is given by

φ(t) = −2ee

cos t_{cos(sin t)−1}

cos sin(sin t)ecos t

+e2ecos tcos(sin t)−2+1.

The equation φ0(t) = 0 has two roots in [0, π], namely 0 and π. It is easy to see that

φ(0) = (e−ee)2/e2and φ(π/2) =  e−e1/e2/e2. Therefore, max{φ(0), φ(π)} =φ(0) =(e−e e₎2 e2 .

Hence, the radius of the smallest disk containing Q(D)is(e−ee_{)/e. This ends the proof.}

We now recall some classes and results related to them which are to be used for further development of this section. For−1≤B< A≤1, let

Pn[A, B]:= ( p(z) =1+ ∞

∑

k=n cnzn: p(z) ≺ 1 +Az 1+Bz ) .

Let us denotePn(α):= Pn[1−2α,−1]andP1(0) =:P. For f ∈ A, if we set p(z) =z f0(z)/ f(z)

and p(z) =1+z f00(z)/ f0(z), then the classP [A, B]is denoted byS∗[A, B]andK[A, B], respectively. These classes were introduced and studied by [2]. Further, letS∗₍

α):= S∗[1−2α,−1].

The following results will be needed:

Lemma 3. [42] If p∈ Pn[A, B], then, for|z| =r,

    p(z) −1−ABr 2n 1−B2_r2n     ≤ (A−B)r n 1−B2_r2n .

In particular, if p∈ Pn(α), then, for|z| =r,

    p(z) −(1+ (1−2α))r 2n 1−r2n     ≤ 2(1−α)r n 1−r2n . Lemma 4. [43] If p∈ Pn(α), then, for|z| =r,

    zp0(z) p(z)     ≤ 2(1−α)nr n (1−rn_{)(1+ (1−2α)r}n₎.

The main objective of this section is to determine theS∗

B-radii constants for functions belonging to

certain well-known subclasses ofA. LetGdenote the class of functions f ∈ S for which f(z)/z∈ P. The following theorem gives the sharpS_B∗-radius for the classG.

Theorem 4. Let f ∈ G. Then, the sharpS_B∗-radius is RS∗ B(G):= e−e1/e q 2e2_−2e1+1_{e +e}2/e_+e ≈0.222654.

Proof. Since f ∈ G, therefore, f(z)/z∈ P. Then, from Lemma2, we must have     z f0(z) f(z) −1     ≤ 2r 1−r2

Therefore, f ∈ S_B∗if 2r/(1−r2) ≤ (e−e1/e)/e, or equivalently if

(e−e1/e)r2+2er+e1/e−e≤0 which holds for all

r≤ e−e 1/e q 2e2_−2e1+1_{e +e}2/e_+e =: RS∗ B(G) ≈0.222654.

For verification of sharpness, consider the function f(z) =z(1+z)/(1−z). Then, f(z)/z∈ P

and at z=RS∗ B(G), we have RS∗ B(G)f 0_(R S∗ B(G)) f(RS∗ B(G)) −1= RS ∗ B(G) 1−RS∗ B(G) =1−e1e−1_.

Hence the result is sharp.

In the following theorem, we shall investigate sharpS_B∗-radius for the classS∗[A, B].

Theorem 5. Let f ∈ S∗[A, B]. Then, 1. for 0≤B<A≤1, the sharpS∗

B-radius for the classS∗[A, B]is

RS∗ B(S ∗ [A, B]) =min ( 1; √ e−e1/e √ eAB−e1/e_B2; e1/e_−e e1/e_B−eA ) . 2. for−1≤B<0≤A≤1, the sharpS_B∗-radius for the classS∗[A, B]is

RS∗ B(S ∗_{[A, B]) =} min    1; s −2e+e1/e_+ee −2eAB+e1/e_B2_+ee_B2; e1/e_−e e1/e_B−eA    .

Proof. Let f ∈ S∗[A, B]. Then using Lemma4, we see that f maps the disk|z| ≤r onto the disk     z f0(z) f(z) − 1−ABr2 1−B2_r2     ≤ (A−B)r 1−B2_r2.

The center of the above disk is at(c, 0)and the radius is R, where c := 1−ABr

2

1−B2_r2 and R :=

(A−B)r 1−B2_r2.

(1) We see that c≤ (e1/e+ee)/(2e)holds for all 0 ≤ B < A ≤ 1 and 0 < r< 1. Further, the condition 1−e1/e_{≤c is equivalent to}

−eABr2+e1/eB2r2−e1/e+e≥0 which holds for all

r≤

s

e−e1/e

eAB−e1/e_B2 =: r1.

Further computation shows that the condition R≤ (eea−e1/e)/e is equivalent to eAr−e1/eBr+

e1/e−e≤0 which holds for all

r≤ e 1/e_−e

e1/e_B−eA =: r2.

Now from Lemma2, f ∈ S_B∗for all|z| ≤RS∗

B(S

∗_{[A, B]) =min{1; r} 1; r2}.

(2) Let−1 ≤ B < 0 ≤ A ≤ 1. Then we see that e1/e−1 ≤ c holds for all 0 < r < 1. Further, c≤ (ee+e1/e)/2e is equivalent to

−2eABr2+e1/eB2r2+eeB2r2−e1/e−ee+2e≤0 which holds for

r≤

s

−2e+e1/e_+ee

−2eAB+e1/e_B2_+ee_B2 =: r3.

Now, as in the previous case R< (ec−e1/e)/e holds if r≤r2. Therefore,SB∗-radius for the class S∗_{[A, B]is R}

S∗

B(S

∗_{[A, B]) =min{1; r} 2; r3}.

The equality holds in case of the function f0defined by

f0(z) =

(

z(1+Bz)AB−1_{, B}6=_0;

zeAz, B=0. This ends the proof.

Remark 1. Let f ∈ S∗. Then, sinceS∗= S∗[0,−1], it follows from the above theorem, that theS_B∗-radius for starlike functions is r4:= (e−e1/e)/(e+e1/e) ≈0.30594. To see the sharpness, consider the Koebe function

k(z) =z/(1−z)2. Then, at z=r4, we have r4f0(r4) f(r4) = 1+r4 1−r4 =e1−1e_.

Because the function k is univalent too, it follows that theS∗

B-radius for the classSandS∗is r4. Therefore,

the radius r4can not be increased. Thus, we have the following:

Corollary 1. The sharpS_B∗-radius for the classesS andS∗is(e−e1/e_)/(e+e1/e_{) ≈0.30594.}

Let the classF1be defined by F₁:=  f ∈ A: Re f(z) g(z) >0 and Re g(z) z >0, g∈ A  . The following theorem gives the sharpS_B∗-radius for the classF1.

Theorem 6. Let f ∈ F1. Then, the sharpSB∗-radius is

RS∗ B(F1) = e−e1/e q 5e2_−2e1+1_{e +e}2/e_+2e ≈0.11557.

Proof. Since f ∈ F1, there is g∈ Asuch that Re(g(z)/z) >0. Define the functions p, h :D → Cby

p(z) = g(z)

z and h(z) = f(z)

g(z).

Then, through some assumptions, we have p, h∈ P. Now using Lemma4, we get     z f0(z) f(z) −1     ≤     zh0(z) h(z)     +     zp0(z) p(z)     ≤ 4r 1−r2 ≤ e−e1/e e ,

this holds if and only if(e−e1/e)r2+4er+e1/e−e≤0, that is if r≤ e−e 1/e q 5e2_−2e1+1_{e +e}2/e_+2e =: RS∗ B(F1) ≈0.11557.

Consider the functions f2and g2defined by

f2(z) =z 1 +z 1−z 2 and g2(z) =z 1 +z 1−z  .

Further, we have Re(f2(z)/g2(z)) > 0 and Re(g2(z)/z) > 0, and therefore f ∈ F1. Now a

computation shows that, for z=RS∗

B(F1), RS∗ B(F1)f 0 2(RSB∗(F1)) f2(RS∗ B(F1)) −1= 4RS ∗ B(F1) 1−RS∗ B(F1) 2 =1−e 1 e−1_.

Hence the result is sharp. Let us define the classF2by

F2:=  f ∈ A: Re f(z) g(z) >0 and Re g(z) z >1/2, g∈ A  . The following theorem gives the sharpS_B∗-radius for the classF2.

Theorem 7. Let f ∈ F2. Then, the sharpSB∗-radius is

S∗_B(F2) = 2e−e1/e q 17e2_−12e1+1 e +4e2/e+3e ≈0.145776.

Proof. Since f ∈ F2and g∈ Asatisfies Re(g(z)/z) > 1/2. Now define the functions p, h :D → C

by p(z) =g(z)/z and h(z) = f(z)/g(z). Then, it is clear that p∈ P (1/2)and h∈ P. Further, since f(z) =zp(z)h(z), it follows from Lemma4, get

    z f0(z) f(z) −1     ≤ 3r+r 2 1−r2 ≤ e−e1/e e provided−e1/er2+2er2+3er+e1/e−e≤0. This holds for

r≤ 2



e−e1/e q

17e2_−12e1+1_{e +4e}2/e_+3e

=:S_B∗(F₂) ≈0.145776.

Thus, f ∈ S∗

Bfor r≤ S∗B(F2).

For the sharpness of the result, consider the functions f3(z) = z

(1+z)

(1−z)2 and g3(z) =

z 1−z.

Then, we see that Re(f3(z)/g3(z)) >0 and Re(g3(z)/z) >1/2, and therefore, f ∈ F2. Now from

S∗ B(F2)f30(SB∗(F2)) f3(S_B∗(F2)) −1= 3S ∗ B(F2) + SB∗(F2)2 1− S∗ B(F2)2 =1−e1e−1_.

This confirms the sharpness of the result. Define the classF3by

F₃:=  f ∈ A:     f(z) g(z)−1     <1 and Reg(z) z >0, g∈ A  . The next result gives the sharpS_B∗-radius for the classF3.

Theorem 8. Let f ∈ F3. Then, the sharpS_B∗-radius is

S∗_B(F3) =

2e−e1/e q

17e2_−12e1+1_{e +4e}2/e_+3e

≈0.145776.

Proof. Since f ∈ F3, it follows that p ∈ P and h ∈ P (1/2), where the functions p, h :D → Care

defined by p(z) = g(z)/z and h(z) = g(z)/ f(z). Now since f(z) = zp(z)/h(z)from Lemma 4, we have     z f0(z) f(z) −1     ≤ 3r+r 2 1−r2 ≤ e−e1/e e which holds for all r≤ S_B∗(F3).

Consider the functions f4and g4defined by

f4(z) =

z(1+z)2

(1−z) and g4(z) =

z(1+z)

1−z . The results are sharp, since at z= S_B∗(F₃), we have

S∗_B(F3)f40(SB∗(F3))

f4(SB∗(F3))

=2−e1e−1_.

This completes the proof.

Author Contributions:All authors contributed equally.

Funding: This research was funded by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (No. 2016R1D1A1A09916450).

Conflicts of Interest:The authors declare no conflict of interest

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