Skip to main content

Sufficient conditions for starlike functions associated with the lemniscate of Bernoulli

Abstract

Let 1B<A1. The condition on β is determined so that 1+βz p (z)/ p k (z)(1+Az)/(1+Bz) (1<k3) implies p(z) 1 + z . Similarly, the condition on β is determined so that 1+βz p (z)/ p n (z) or p(z)+βz p (z)/ p n (z) 1 + z (n=0,1,2) implies p(z)(1+Az)/(1+Bz) or 1 + z . In addition to that, the condition on β is derived so that p(z)(1+Az)/(1+Bz) when p(z)+βz p (z)/p(z) 1 + z . A few more problems of the similar flavor are also considered.

MSC:30C80, 30C45.

1 Introduction

Let A be the class of analytic functions defined on the unit disk D:={zC:|z|<1} normalized by the condition f(0)=0= f (0)1. For two analytic functions f and g, we say that f is subordinate to g or g is superordinate to f, denoted by fg, if there is a Schwarz function w with |w(z)||z| such that f(z)=g(w(z)). If g is univalent, then fg if and only if f(0)=g(0) and f(D)g(D). For an analytic function φ whose range is starlike with respect to φ(0)=1 and is symmetric with respect to the real axis, let S (φ) denote the class of Ma-Minda starlike functions consisting of all fA satisfying z f (z)/f(z)φ(z). For special choices of φ, S (φ) reduces to well-known subclasses of starlike functions. For example, when 1B<A1, S [A,B]:= S ((1+Az)/(1+Bz)) is the class of Janowski starlike functions [1] (see [2]) and S [12α,1] is the class S (α) of starlike functions of order α and S := S (0) is the class of starlike functions. For φ(z):= 1 + z , the class S (φ) reduces to the class SL introduced by Sokół and Stankiewicz [3] and studied recently by Ali et al. [4, 5]. A function fA is in the class SL if z f (z)/f(z) lies in the region bounded by the right half-plane of the lemniscate of Bernoulli given by | w 2 1|<1. Analytically, SL:={fA:| ( z f ( z ) / f ( z ) ) 2 1|<1}. For b1/2 and a1, a more general class S [a,b] of the functions f satisfying | ( z f ( z ) / f ( z ) ) a b|<b was considered by Paprocki and Sokół [6]. Clearly, S [2,1]=:SL. For some radius problems related with the lemniscate of Bernoulli, see [3, 5, 7, 8]. Estimates for the initial coefficients of functions in the class SL are available in [8].

Let p be an analytic function defined on D with p(0)=1. Recently Ali et al. [4] determined conditions for p(z) 1 + z when 1+βz p (z)/ p k (z) with k=0,1,2 or (1β)p(z)+β p 2 (z)+βz p (z) is subordinated to 1 + z . Motivated by the works in [48], in Section 2 the condition on β is determined so that p(z) 1 + z when 1+βz p (z)/ p k (z)(1+Az)/(1+Bz) (1<k3). Similarly, the condition on β is determined so that p(z)(1+Az)/(1+Bz) when 1+βz p (z)/ p n (z) 1 + z , n=0,1,2. Further, the condition on β is obtained in each case so that p(z) 1 + z when p(z)+βz p (z)/ p n (z) 1 + z , n=0,1,2. At the end of this section, the problem p(z)+βz p (z)/p(z) 1 + z implies p(z)(1+Az)/(1+Bz) is also considered.

Silverman [9] introduced the class G b by

G b := { f A : | z f ( z ) / f ( z ) z f ( z ) / f ( z ) 1 | < b }

and proved G b S (2/(1+ 1 + 8 b )), 0<b1. Further, this result was improved by Obradovič and Tuneski [10] by showing G b S [0,b] S (2/(1+ 1 + 8 b )), 0<b1. Tuneski [11] further obtained the condition for G b S [A,B]. Inspired by the work of Silverman [9], Nunokawa et al. [12] obtained the sufficient conditions for a function in the class G b to be strongly starlike, strongly convex, or starlike in D. By setting p(z)=z f (z)/f(z), the inclusion G b S [A,B] can be written as

1+ z p ( z ) p 2 ( z ) 1+bzp(z) 1 + A z 1 + B z .

Recently Ali et al. [13], obtained the condition on the constants A,B,D,E[1,1] and β so that p(z)(1+Az)/(1+Bz) when 1+βz p (z)/ p n (z)(1+Dz)/(1+Ez), n=0,1. In Section 3, alternate and easy proofs of results [[13], Lemmas 2.1, 2.10] are discussed. Further, this section is concluded with the condition on A,B,D,E[1,1] and β such that 1+βz p (z)/ p 2 (z)(1+Dz)/(1+Ez) implies p(z)(1+Az)/(1+Bz).

The following results are required in order to prove our main results.

Lemma 1.1 [[14], Corollary 3.4h, p.135]

Let q be univalent in D, and let φ be analytic in a domain D containing q(D). Let z q (z)φ(q(z)) be starlike. If p is analytic in D, p(0)=q(0) and satisfies

z p (z)φ ( p ( z ) ) z q (z)φ ( q ( z ) ) ,

then pq and q is the best dominant.

The following is a more general form of the above lemma.

Lemma 1.2 [[14], Corollary 3.4i, p.134]

Let q be univalent in D, and let φ and ν be analytic in a domain D containing q(D) with φ(w)0 when wq(D). Set

Q(z):=z q (z)φ ( q ( z ) ) ,h(z):=ν ( q ( z ) ) +Q(z).

Suppose that

  1. (1)

    h is convex or Q(z) is starlike univalent in D and

  2. (2)

    Re( z h ( z ) Q ( z ) )>0 for zD.

If

ν ( p ( z ) ) +z p (z)φ ( p ( z ) ) ν ( q ( z ) ) +z q (z)φ ( q ( z ) ) ,
(1.1)

then pq and q is the best dominant.

Lemma 1.3 [[14], Corollary 3.4a, p.120]

Let q be analytic in D, let ϕ be analytic in a domain D containing q(D) and suppose

  1. (1)

    Reϕ[q(z)]>0 and either

  2. (2)

    q is convex, or

  3. (3)

    Q(z)=z q (z)ϕ[q(z)] is starlike.

If p is analytic in D, with p(0)=q(0), p(D)D and

p(z)+z p (z)ϕ [ p ( z ) ] q(z),

then p(z)q(z).

2 Results associated with the lemniscate of Bernoulli

In the first result, condition on β is obtained so that the subordination

1+β z p ( z ) p k ( z ) 1 + A z 1 + B z (1<B<A1)

implies p(z) 1 + z .

Lemma 2.1 Let |β| 2 ( k + 3 ) / 2 (AB)+|Bβ|, 1<k3. Let p be an analytic function defined on D with p(0)=1 satisfying

1+β z p ( z ) p k ( z ) 1 + A z 1 + B z (1<B<A1),

then p(z) 1 + z .

Proof Let q(z)= 1 + z . A computation shows that the function

Q(z):=β z q ( z ) q k ( z ) = β z 2 ( 1 + z ) ( k + 1 ) / 2 (1<k3)

is starlike in the unit disk D. Consider the subordination

1+β z p ( z ) p k ( z ) 1+β z q ( z ) q k ( z ) .

Thus in view of Lemma 1.1, it follows that p(z)q(z). In order to prove our result, we need to prove

1 + A z 1 + B z 1+ β z q ( z ) q k ( z ) =1+ β z 2 ( 1 + z ) ( k + 1 ) / 2 :=h(z).

Let w=Φ(z)= 1 + A z 1 + B z . Then Φ 1 (w)= w 1 A B w . The subordination Φ(z)h(z) is equivalent to z Φ 1 (h(z)). Thus in order to prove the result, we need only to show | Φ 1 (h( e i t ))|1. For z= e i t , πtπ, we have

| Φ 1 ( h ( e i t ) ) | | β | 2 ( A B ) ( 2 cos ( t / 2 ) ) ( k + 1 ) / 2 + | B β | =:g(t).

A calculation shows that g(t) attains its minimum at t=0. Further, the value of g(t) at π or −π comes out to be 1/|B| which is naturally greater than the value at the extreme point t=0 because if g(0)g(π), then (AB)|β|0 which is absurd. Thus

g(0)= | β | 2 ( k + 3 ) / 2 ( A B ) + | B β | 1

for |β| 2 ( k + 3 ) / 2 (AB)+|Bβ|. Hence Φ(z)h(z), and the proof is complete now. □

Next result depicts the condition on β such that 1+βz p (z) 1 + z implies p(z)(1+Az)/(1+Bz) (1B<A1). On subsequent lemmas, similar results are obtained by considering the expressions 1+βz p (z)/p(z) and 1+βz p (z)/ p 2 (z).

Lemma 2.2 Let (AB)β 2 ( 1 + | B | ) 2 + ( 1 B ) 2 and 1B<A1. Let p be an analytic function defined on D with p(0)=1 satisfying

1+βz p (z) 1 + z ,

then p(z) 1 + A z 1 + B z .

Proof Define the function q:DC by

q(z)= 1 + A z 1 + B z (1B<A1)

with q(0)=1. A computation shows that

Q(z)=βz q (z)= β ( A B ) z ( 1 + B z ) 2

and

z Q ( z ) Q ( z ) = 1 B z 1 + B z .

Let z=r e i t , r(0,1), πtπ. Then

Re ( 1 B z 1 + B z ) = Re ( 1 B r e i t 1 + B r e i t ) = 1 B 2 r 2 | 1 + B r e i t | 2 .

Since 1 B 2 r 2 >0 (|B|1, 0<r<1) and so Re(z Q (z)/Q(z))>0, this shows that Q is starlike in D. It follows from Lemma 1.1 that the subordination

1+βz p (z)1+βz q (z)

implies p(z)q(z). Now we need to prove the following in order to prove the lemma:

1 + z 1+βz q (z)=1+β ( A B ) z ( 1 + B z ) 2 =:h(z).

Let w=Φ(z)= 1 + z . Then Φ 1 (w)= w 2 1. The subordination Φ(z)h(z) is equivalent to the subordination z Φ 1 (h(z)). Now in order to prove the result, it is enough to show | Φ 1 (h( e i t ))|1, z= e i t , πtπ. Now

| Φ 1 ( h ( e i t ) ) | =| ( 1 + β ( A B ) e i t ( 1 + B e i t ) 2 ) 2 1|1implies that|1+β ( A B ) e i t ( 1 + B e i t ) 2 | 2 .

Further,

| 1 + β ( A B ) e i t ( 1 + B e i t ) 2 | = | 1 + ( 2 B + β ( A B ) ) e i t + B 2 e 2 i t | | 1 + 2 B e i t + B 2 e 2 i t | Re ( 2 B + β ( A B ) + B 2 e i t + e i t ) 1 + 2 | B | + B 2 = 2 B + β ( A B ) + ( 1 + B 2 ) cos t ( 1 + | B | ) 2 2 B + β ( A B ) ( 1 + B 2 ) ( 1 + | B | ) 2 2

for (AB)β 2 ( 1 + | B | ) 2 + ( 1 B ) 2 . Therefore Φ(z)h(z) and this completes the proof. □

Lemma 2.3 Let (AB)β( 2 1)(1+|A|)(1+|B|) and 1B<A1. Let p be an analytic function defined on D with p(0)=1 satisfying

1+β z p ( z ) p ( z ) 1 + z ,

then p(z) 1 + A z 1 + B z .

Proof Let the function q:DC be defined by

q(z)= 1 + A z 1 + B z (1B<A1).

A computation shows that

Q(z):= β z q ( z ) q ( z ) = β ( A B ) z ( 1 + A z ) ( 1 + B z )

and

z Q ( z ) Q ( z ) = 1 A B z 2 ( 1 + A z ) ( 1 + B z ) .

Let z=r e i t , r(0,1), πtπ. Then

Re ( 1 A B z 2 ( 1 + A z ) ( 1 + B z ) ) = Re ( 1 A B r 2 e 2 i t ( 1 + A r e i t ) ( 1 + B r e i t ) ) = ( 1 A B r 2 ) ( 1 + ( A + B ) r cos t + A B r 2 ) | 1 + A r e i t | 2 | 1 + B r e i t | 2 .

Since 1+AB r 2 +(A+B)rcost(1Ar)(1Br)>0 for A+B0 and, similarly, 1+AB r 2 +(A+B)rcost(1+Ar)(1+Br)>0 for A+B0, it follows that Q is starlike in D. Lemma 1.1 suggests that the subordination

1+β z p ( z ) p ( z ) 1+β z q ( z ) q ( z )

implies p(z)q(z). Now we have to prove

1 + z 1+β z q ( z ) q ( z ) =1+ β ( A B ) z ( 1 + A z ) ( 1 + B z ) =:h(z).

Let w=Φ(z)= 1 + z . Then Φ 1 (w)= w 2 1. The subordination Φ(z)h(z) is equivalent to the subordination z Φ 1 (h(z)). Now in order to prove the result, it is enough to show | Φ 1 (h( e i t ))|1, πtπ. Now

| Φ 1 ( h ( e i t ) ) | = | ( 1 + β ( A B ) e i t ( 1 + A e i t ) ( 1 + B e i t ) ) 2 1 | 1 implies that | 1 + β ( A B ) e i t ( 1 + A e i t ) ( 1 + B e i t ) | 2 .

Further,

| 1 + β ( A B ) e i t ( 1 + A e i t ) ( 1 + B e i t ) | Re ( 1 + β ( A B ) e i t ( 1 + A e i t ) ( 1 + B e i t ) ) 1 + ( A B ) β ( 1 + | A | ) ( 1 + | B | ) 2

for (AB)β( 2 1)(1+|A|)(1+|B|). Therefore Φ(z)h(z) and this completes the proof. □

Lemma 2.4 Let (AB)β( 2 1) ( 1 + | A | ) 2 + ( 1 A ) 2 and 1B<A1. Let p be an analytic function defined on D with p(0)=1 satisfying

1+β z p ( z ) p 2 ( z ) 1 + z ,

then p(z) 1 + A z 1 + B z .

Proof Let the function q:DC be defined by

q(z)= 1 + A z 1 + B z (1B<A1)

with q(0)=1. Then

Q(z)= β z q ( z ) q 2 ( z ) = β ( A B ) z ( 1 + A z ) 2

and

z Q ( z ) Q ( z ) = 1 A z 1 + A z .

Let z=r e i t , πtπ, 0<r<1. Then

Re ( 1 A z 1 + A z ) = 1 A 2 r 2 | 1 + A r e i t | 2 .

Since 1 A 2 r 2 >0 (|A|1, 0<r<1). Hence Re(z Q (z))/Q(z)>0, this shows that Q is starlike in D. An application of Lemma 1.1 reveals that the subordination

1+β z p ( z ) p 2 ( z ) 1+β z q ( z ) q 2 ( z )

implies p(z)q(z). Now our result is established if we prove

1 + z 1+β z q ( z ) q 2 ( z ) =1+β ( A B ) z ( 1 + A z ) 2 =:h(z).

The rest of the proof is similar to that of Lemma 2.2, and therefore it is skipped here. □

In the next result, the condition on β is obtained so that p(z)+βz p (z) 1 + z implies p(z) 1 + z . On subsequent lemmas, similar results are discussed by considering the expressions p(z)+βz p (z)/p(z) and p(z)+βz p (z)/ p 2 (z).

Lemma 2.5 Let p be an analytic function defined on D with p(0)=1 satisfying p(z)+βz p (z) 1 + z , β>0. Then p(z) 1 + z .

Proof Define the function q:DC by q(z)= 1 + z with q(0)=1. Since q(D)={w:| w 2 1|<1} is the right half of the lemniscate of Bernoulli, q(D) is a convex set, and hence q is a convex function. Let us define ϕ(w)=β, then

Reϕ [ q ( z ) ] =β>0.

Consider the function Q defined by

Q(z):=z q (z)ϕ ( q ( z ) ) =β z 2 1 + z .

Further,

Re ( z Q ( z ) Q ( z ) ) = 1 Re ( z 2 ( 1 + z ) ) 3 4 > 0 .

Thus the function Q is starlike, and the result now follows by an application of Lemma 1.3. □

Lemma 2.6 Let p be an analytic function defined on D with p(0)=1 satisfying

p(z)+β z p ( z ) p ( z ) 1 + z ,β>0.

Then p(z) 1 + z .

Proof As before, let q be given by q(z)= 1 + z with q(0)=1. Then q is a convex function. Let us define ϕ(w)=β/w. Since q(D)={w:| w 2 1|<1} is the right half of the lemniscate of Bernoulli, so

Reϕ [ q ( z ) ] = β | 1 + z | 2 Re( 1 + z )>0.

Consider the function Q defined by

Q(z):=β z q ( z ) q ( z ) =β z 2 ( 1 + z ) .

Further,

Re ( z Q ( z ) Q ( z ) ) =1Re ( z 1 + z ) 1 2 >0.

Thus the function Q is starlike, and the result now follows by an application of Lemma 1.3. □

Lemma 2.7 Let p be an analytic function defined on D with p(0)=1 satisfying

p(z)+β z p ( z ) p 2 ( z ) 1 + z ,β>0.

Then p(z) 1 + z .

Proof Let q be given by q(z)= 1 + z with q(0)=1. Then q is a convex function. Let us define ϕ(w)=β/ w 2 and

Reϕ [ q ( z ) ] =Re ( β 1 + z ) > β 2 >0.

Consider the function Q defined by

Q(z):=β z q ( z ) q 2 ( z ) =β z 2 ( 1 + z ) 3 2 .

Further,

Re ( z Q ( z ) Q ( z ) ) =1 3 2 Re ( z 1 + z ) 1 4 >0.

Thus the function Q is starlike, and the result now follows by an application of Lemma 1.3. □

In the next result, the condition on β is obtained such that p(z)+βz p (z)/p(z) 1 + z implies that p(z)(1+Az)/(1+Bz).

Lemma 2.8 Let 1B<A1, (AB)β 2 (1+|A|)(1+|B|)+ | A | 2 1 and

1 β max { 0 , A B ( 1 + | A | ) ( 1 + | B | ) 1 | B | 1 + | B | } .

Let p be an analytic function defined on D with p(0)=1 satisfying

p(z)+β z p ( z ) p ( z ) 1 + z .

Then p(z) 1 + A z 1 + B z .

Proof Define the function q:DC by q(z)=(1+Az)/(1+Bz), 1B<A1. Consider the subordination

p(z)+β z p ( z ) p ( z ) q(z)+β z q ( z ) q ( z ) .

Thus, in view of Lemma 1.2, the above subordination can be written as (1.1) by defining the functions ν and φ as ν(w):=w and φ(w):=β/w (β0). Clearly, the functions ν and φ are analytic in and φ(w)0. Let the functions Q(z) and h(z) be defined by

Q(z):=z q (z)φ ( q ( z ) ) =β z q ( z ) q ( z )

and

h(z):=ν ( q ( z ) ) +Q(z)=q(z)+β z q ( z ) q ( z ) .

A computation shows that Q(z) is starlike univalent in D. Further,

z h ( z ) Q ( z ) = 1 β +1+ z q ( z ) q ( z ) z q ( z ) q ( z ) .

Let z= e i t , πtπ. Then

Re ( e i t h ( e i t ) Q ( e i t ) ) = 1 β + Re ( 1 B e i t 1 + B e i t ( A B ) e i t ( 1 + A e i t ) ( 1 + B e i t ) ) 1 β + 1 | B | 1 + | B | A B ( 1 + | A | ) ( 1 + | B | ) > 0 .

Thus by Lemma 1.2, it follows that p(z)q(z). In order to prove our result, we need to prove that

Φ(z):= 1 + z q(z)+β z q ( z ) q ( z ) = 1 + A z 1 + B z + β ( A B ) z ( 1 + A z ) ( 1 + B z ) :=h(z).

The subordination Φ(z)h(z) is equivalent to the subordination z Φ 1 (h(z)). Now in order to prove the result, it is enough to show | Φ 1 (h( e i t ))|1, πtπ. Now

| Φ 1 ( h ( e i t ) ) |=| ( 1 + A e i t 1 + B e i t + β ( A B ) e i t ( 1 + A e i t ) ( 1 + B e i t ) ) 2 1|1

implies

| 1 + A e i t 1 + B e i t + β ( A B ) e i t ( 1 + A e i t ) ( 1 + B e i t ) | 2 .

Further,

| 1 + A e i t 1 + B e i t + β ( A B ) e i t ( 1 + A e i t ) ( 1 + B e i t ) | Re ( 1 + A e i t 1 + B e i t + β ( A B ) e i t ( 1 + A e i t ) ( 1 + B e i t ) ) 1 | A | 1 + | B | + ( A B ) β ( 1 + | A | ) ( 1 + | B | ) 2

for (AB)β 2 (1+|A|)(1+|B|)+ | A | 2 1. This completes the proof. □

3 Sufficient condition for Janowski starlikeness

The following first two results (Lemmas 3.1, 3.2) are essentially due to Ali et al. [[13], Lemmas 2.1, 2.10]. However, an alternate proof of the same result, which is much easier than that given by Ali et al. [13], is presented below.

Lemma 3.1 Assume that 1B<A1, 1E<D1 and β(AB)(DE)(1+ B 2 )+|2B(DE)Eβ(AB)|. Let p be an analytic function defined on D with p(0)=1 satisfying

1+βz p (z) 1 + D z 1 + E z ,β0.

Then p(z) 1 + A z 1 + B z .

Proof Define the function q:DC by

q(z)= 1 + A z 1 + B z ,1B<A1.

Then q is convex in D with q(0)=1. Further computation shows that

Q(z)=βz q (z)= β ( A B ) z ( 1 + B z ) 2

and Q is starlike in D. It follows from Lemma 1.1 that the subordination

1+βz p (z)1+βz q (z)

implies p(z)q(z). In view of the above result, it is sufficient to prove

1 + D z 1 + E z 1+βz q (z)=1+β ( A B ) z ( 1 + B z ) 2 =h(z).

Let w=Φ(z)= 1 + D z 1 + E z . Then Φ 1 (w)= w 1 D E w and

Φ 1 ( h ( z ) ) = β ( A B ) z D ( 1 + B z ) 2 E ( 1 + B z ) 2 β E ( A B ) z = β ( A B ) z ( D E ) ( 1 + B 2 z 2 ) + ( 2 B ( D E ) β E ( A B ) ) z .

Let z= e i t , πtπ. Thus

| Φ 1 ( h ( e i t ) ) | | β | ( A B ) ( D E ) ( 1 + B 2 ) + | ( 2 B ( D E ) β E ( A B ) ) | 1

for |β|(AB)(DE)(1+ B 2 )+|(2B(DE)Eβ(AB))|. Hence q(D)h(D), that is, q(z)h(z), this completes the proof. □

It should be noted that Ali et al. [13] made the assumption AB>0 in order to prove the result [[13], Lemma 2.10], whereas in the following lemma this condition has been dropped.

Lemma 3.2 Assume that 1B<A1, 1E<D1 and β(AB)(DE)(1+|AB|)+|(A+B)(DE)Eβ(AB)|. Let p be an analytic function defined on D with p(0)=1 satisfying

1+β z p ( z ) p ( z ) 1 + D z 1 + E z ,β0.

Then p(z) 1 + A z 1 + B z .

Proof As above, define the function q:DC by

q(z)= 1 + A z 1 + B z ,1B<A1.

Then q is convex in D with q(0)=1. A computation shows that

Q(z)= β z q ( z ) q ( z ) = β ( A B ) z ( 1 + A z ) ( 1 + B z )

and Q is starlike in D. It follows from Lemma 1.1 that the subordination

1+β z p ( z ) p ( z ) 1+β z q ( z ) q ( z )

implies p(z)q(z). Now we need to prove

1 + D z 1 + E z 1+β z q ( z ) q ( z ) =1+β ( A B ) z ( 1 + B z ) 2 =h(z).

Let w=Φ(z)= 1 + D z 1 + E z . Then Φ 1 (w)= w 1 D E w and

Φ 1 ( h ( z ) ) = β ( A B ) z ( D E ) ( 1 + A z ) ( 1 + B z ) β E ( A B ) z = β ( A B ) z ( D E ) ( 1 + A B z 2 ) + ( ( A + B ) ( D E ) β E ( A B ) ) z .

Let z= e i t , πtπ. Thus

| Φ 1 ( h ( e i t ) ) | | β | ( A B ) ( D E ) ( 1 + | A B | ) + | ( A + B ) ( D E ) β E ( A B ) | 1

for |β|(AB)(DE)(1+|AB|)+|(A+B)(DE)Eβ(AB)|. Hence q(D)h(D), that is, q(z)h(z), this completes the proof. □

Lemma 3.3 Assume that 1B<A1, 1E<D1 and |β|(AB)(DE)(1+ A 2 )+|2A(DE)Eβ(AB)|. Let p be an analytic function defined on D with p(0)=1 satisfying

1+β z p ( z ) p 2 ( z ) 1 + D z 1 + E z .

Then p(z) 1 + A z 1 + B z .

Proof Define the function q:DC by

q(z)= 1 + A z 1 + B z ,1B<A1.

Then q is convex in D with q(0)=1. A computation shows that

Q(z)= β z q ( z ) q 2 ( z ) = β ( A B ) z ( 1 + A z ) 2

and

z Q ( z ) Q ( z ) = 1 A z 1 + A z .

As before, a computation shows Q is starlike in D. It follows from Lemma 1.1 that the subordination

1+β z p ( z ) p 2 ( z ) 1+β z q ( z ) q 2 ( z )

implies p(z)q(z). To prove result, it is enough to show that

1 + D z 1 + E z 1+β z q ( z ) q 2 ( z ) =1+β ( A B ) z ( 1 + A z ) 2 =h(z).

The remaining part of the proof is similar to that of Lemma 3.1, and therefore it is skipped here. □

Remark 3.4 When β=1, Lemma 3.3 reduces to [[13], Lemma 2.6] due to Ali et al.

References

  1. Janowski W: Extremal problems for a family of functions with positive real part and for some related families. Ann. Pol. Math. 1970/1971, 23: 159–177.

    MathSciNet  MATH  Google Scholar 

  2. Polatoğlu Y, Bolcal M: Some radius problem for certain families of analytic functions. Turk. J. Math. 2000, 24(4):401–412.

    MathSciNet  MATH  Google Scholar 

  3. Sokół J, Stankiewicz J: Radius of convexity of some subclasses of strongly starlike functions. Zesz. Nauk. Politech. Rzesz. Mat. 1996, 19: 101–105.

    MathSciNet  MATH  Google Scholar 

  4. Ali RM, Cho NE, Ravichandran V, Kumar SS: First order differential subordinations for functions associated with the lemniscate of Bernoulli. Taiwan. J. Math. 2012, 16(3):1017–1026.

    MATH  Google Scholar 

  5. Ali RM, Jain NK, Ravichandran V: Radii of starlikeness associated with the lemniscate of Bernoulli and the left-half plane. Appl. Math. Comput. 2012, 218(11):6557–6565. 10.1016/j.amc.2011.12.033

    MathSciNet  Article  MATH  Google Scholar 

  6. Paprocki E, Sokół J: The extremal problems in some subclass of strongly starlike functions. Zesz. Nauk. Politech. Rzesz. Mat. 1996, 20: 89–94.

    MathSciNet  MATH  Google Scholar 

  7. Ali RM, Cho NE, Jain NK, Ravichandran V: Radii of starlikeness and convexity of functions defined by subordination with fixed second coefficient. Filomat 2012, 26(3):553–561. 10.2298/FIL1203553A

    MathSciNet  Article  MATH  Google Scholar 

  8. Sokół J:Radius problems in the class SL . Appl. Math. Comput. 2009, 214(2):569–573. 10.1016/j.amc.2009.04.031

    MathSciNet  Article  MATH  Google Scholar 

  9. Silverman H: Convex and starlike criteria. Int. J. Math. Math. Sci. 1999, 22(1):75–79. 10.1155/S0161171299220753

    MathSciNet  Article  MATH  Google Scholar 

  10. Obradowič M, Tuneski N: On the starlike criteria defined by Silverman. Zesz. Nauk. Politech. Rzesz. Mat. 2001, 24: 59–64.

    MathSciNet  MATH  Google Scholar 

  11. Tuneski N: On the quotient of the representations of convexity and starlikeness. Math. Nachr. 2003, 248/249: 200–203. 10.1002/mana.200310015

    MathSciNet  Article  MATH  Google Scholar 

  12. Nunokawa M, Owa S, Saitoh H, Ahuja OP: On the quotient of the analytic representations of convex and starlike functions. Sūrikaisekikenkyūsho Kōkyūroku 1999, 1112: 63–69.

    MathSciNet  MATH  Google Scholar 

  13. Ali RM, Ravichandran V, Seenivasagan N: Sufficient conditions for Janowski starlikeness. Int. J. Math. Math. Sci. 2007., 2007: Article ID 62925

    Google Scholar 

  14. Miller SS, Mocanu PT Pure and Applied Mathematics 225. In Differential Subordinations: Theory and Applications. Dekker, New York; 2000.

    Google Scholar 

Download references

Acknowledgements

Dedicated to Professor Hari M Srivastava.

The research work is supported by a grant from University of Delhi and also by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (No. 2012-0002619).

Author information

Authors and Affiliations

Authors

Corresponding author

Correspondence to Nak Eun Cho.

Additional information

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors jointly worked on the results and they read and approved the final manuscript.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License ( https://creativecommons.org/licenses/by/2.0 ), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Reprints and Permissions

About this article

Cite this article

Kumar, S.S., Kumar, V., Ravichandran, V. et al. Sufficient conditions for starlike functions associated with the lemniscate of Bernoulli. J Inequal Appl 2013, 176 (2013). https://doi.org/10.1186/1029-242X-2013-176

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: https://doi.org/10.1186/1029-242X-2013-176

Keywords

  • starlike function
  • lemniscate of Bernoulli
  • subordination