Sufficient conditions for starlike functions associated with the lemniscate of Bernoulli

Abstract

Let $−1≤B<A≤1$. The condition on β is determined so that $1+βz p ′ (z)/ p k (z)≺(1+Az)/(1+Bz)$ ($−1<k≤3$) 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 $D:={z∈C:|z|<1}$ 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 $f≺g$, if there is a Schwarz function w with $|w(z)|≤|z|$ such that $f(z)=g(w(z))$. If g is univalent, then $f≺g$ 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 $f∈A$ 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 $−1≤B<A≤1$, $S ∗ [A,B]:= S ∗ ((1+Az)/(1+Bz))$ is the class of Janowski starlike functions [1] (see [2]) and $S ∗ [1−2α,−1]$ 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 $f∈A$ 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, $SL:={f∈A:| ( z f ′ ( z ) / f ( z ) ) 2 −1|<1}$. For $b≥1/2$ and $a≥1$, 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 $(1−β)p(z)+β p 2 (z)+βz p ′ (z)$ is subordinated to $1 + z$. Motivated by the works in [4–8], in Section 2 the condition on β is determined so that $p(z)≺ 1 + z$ when $1+βz p ′ (z)/ p k (z)≺(1+Az)/(1+Bz)$ ($−1<k≤3$). 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 $G b ⊂ S ∗ (2/(1+ 1 + 8 b ))$, $0<b≤1$. Further, this result was improved by Obradovič and Tuneski [10] by showing $G b ⊂ S ∗ [0,b]⊂ S ∗ (2/(1+ 1 + 8 b ))$, $0<b≤1$. 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

$$1+ z p ′ ( z ) p 2 ( z ) ≺1+bz⟹p(z)≺ 1 + A z 1 + B z .$$

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 $p≺q$ 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 $w∈q(D)$. Set

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

Suppose that

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

• (2) $Re( z h ′ ( z ) Q ( z ) )>0$ for $z∈D$.

If

$$ν ( p ( z ) ) +z p ′ (z)φ ( p ( z ) ) ≺ν ( q ( z ) ) +z q ′ (z)φ ( q ( z ) ) ,$$

(1.1)

then $p≺q$ 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) $Reϕ[q(z)]>0$ and either

• (2) q is convex, or

• (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

$$1+β z p ′ ( z ) p k ( z ) ≺ 1 + A z 1 + B z (−1<B<A≤1)$$

implies $p(z)≺ 1 + z$.

Lemma 2.1 Let $|β|≥ 2 ( k + 3 ) / 2 (A−B)+|Bβ|$, $−1<k≤3$. Let p be an analytic function defined on $D$ with $p(0)=1$ satisfying

$$1+β z p ′ ( z ) p k ( z ) ≺ 1 + A z 1 + B z (−1<B<A≤1),$$

then $p(z)≺ 1 + z$.

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

$$Q(z):=β z q ′ ( z ) q k ( z ) = β z 2 ( 1 + z ) ( k + 1 ) / 2 (−1<k≤3)$$

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 $z= e i t$, $−π≤t≤π$, 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 $(A−B)|β|≤0$ which is absurd. Thus

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

for $|β|≥ 2 ( k + 3 ) / 2 (A−B)+|Bβ|$. 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 $p(z)≺(1+Az)/(1+Bz)$ ($−1≤B<A≤1$). 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 $(A−B)β≥ 2 ( 1 + | B | ) 2 + ( 1 − B ) 2$ and $−1≤B<A≤1$. 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 $q:D→C$ by

$$q(z)= 1 + A z 1 + B z (−1≤B<A≤1)$$

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 $z=r e i t$, $r∈(0,1)$, $−π≤t≤π$. 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 $| Φ − 1 (h( e i t ))|≥1$, $z= e i t$, $−π≤t≤π$. 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 $(A−B)β≥ 2 ( 1 + | B | ) 2 + ( 1 − B ) 2$. Therefore $Φ(z)≺h(z)$ and this completes the proof. □

Lemma 2.3 Let $(A−B)β≥( 2 −1)(1+|A|)(1+|B|)$ and $−1≤B<A≤1$. 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 $q:D→C$ be defined by

$$q(z)= 1 + A z 1 + B z (−1≤B<A≤1).$$

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 $z=r e i t$, $r∈(0,1)$, $−π≤t≤π$. 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 $1+AB r 2 +(A+B)rcost≥(1−Ar)(1−Br)>0$ for $A+B≥0$ and, similarly, $1+AB r 2 +(A+B)rcost≥(1+Ar)(1+Br)>0$ for $A+B≤0$, 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 $| Φ − 1 (h( e i t ))|≥1$, $−π≤t≤π$. 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 $(A−B)β≥( 2 −1)(1+|A|)(1+|B|)$. Therefore $Φ(z)≺h(z)$ and this completes the proof. □

Lemma 2.4 Let $(A−B)β≥( 2 −1) ( 1 + | A | ) 2 + ( 1 − A ) 2$ and $−1≤B<A≤1$. 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 $q:D→C$ be defined by

$$q(z)= 1 + A z 1 + B z (−1≤B<A≤1)$$

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 $z=r e i t$, $−π≤t≤π$, $0<r<1$. 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 $q:D→C$ 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,

$$Re ( z Q ′ ( z ) Q ( z ) ) =1−Re ( z 1 + z ) ≥ 1 2 >0.$$

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 $−1≤B<A≤1$, $(A−B)β≥ 2 (1+|A|)(1+|B|)+ | A | 2 −1$ 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 $q:D→C$ by $q(z)=(1+Az)/(1+Bz)$, $−1≤B<A≤1$. 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 $φ(w):=β/w$ ($β≠0$). Clearly, the functions ν and φ are analytic in ℂ 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 $z= e i t$, $−π≤t≤π$. 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 $| Φ − 1 (h( e i t ))|≥1$, $−π≤t≤π$. 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 $(A−B)β≥ 2 (1+|A|)(1+|B|)+ | A | 2 −1$. 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 $−1≤B<A≤1$, $−1≤E<D≤1$ and $β(A−B)≥(D−E)(1+ B 2 )+|2B(D−E)−Eβ(A−B)|$. Let p be an analytic function defined on $D$ with $p(0)=1$ satisfying

$$1+βz p ′ (z)≺ 1 + D z 1 + E z ,β≠0.$$

Then $p(z)≺ 1 + A z 1 + B z$.

Proof Define the function $q:D→C$ by

$$q(z)= 1 + A z 1 + B z ,−1≤B<A≤1.$$

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 $z= e i t$, $π≤t≤π$. 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 $|β|(A−B)≥(D−E)(1+ B 2 )+|(2B(D−E)−Eβ(A−B))|$. 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 $−1≤B<A≤1$, $−1≤E<D≤1$ and $β(A−B)≥(D−E)(1+|AB|)+|(A+B)(D−E)−Eβ(A−B)|$. Let p be an analytic function defined on $D$ with $p(0)=1$ satisfying

$$1+β z p ′ ( z ) p ( z ) ≺ 1 + D z 1 + E z ,β≠0.$$

Then $p(z)≺ 1 + A z 1 + B z$.

Proof As above, define the function $q:D→C$ by

$$q(z)= 1 + A z 1 + B z ,−1≤B<A≤1.$$

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 $z= e i t$, $π≤t≤π$. 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 $|β|(A−B)≥(D−E)(1+|AB|)+|(A+B)(D−E)−Eβ(A−B)|$. Hence $q(D)⊂h(D)$, that is, $q(z)≺h(z)$, this completes the proof. □

Lemma 3.3 Assume that $−1≤B<A≤1$, $−1≤E<D≤1$ and $|β|(A−B)≥(D−E)(1+ A 2 )+|2A(D−E)−Eβ(A−B)|$. 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 $q:D→C$ by

$$q(z)= 1 + A z 1 + B z ,−1≤B<A≤1.$$

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.