Sufficient Conditions for Starlike Functions Associated with the Lemniscate of Bernoulli

Let -1\leq B<A\leq 1. Condition on \beta, is determined so that 1+\beta zp'(z)/p^k(z)\prec(1+Az)/(1+Bz)\;(-1<k\leq3) implies p(z)\prec \sqrt{1+z}. Similarly, condition on \beta is determined so that 1+\beta zp'(z)/p^n(z) or p(z)+\beta zp'(z)/p^n(z)\prec\sqrt{1+z}\;(n=0, 1, 2) implies p(z)\prec(1+Az)/(1+Bz) or \sqrt{1+z}. In addition to that condition on \beta is derived so that p(z)\prec(1+Az)/(1+Bz) when p(z)+\beta zp'(z)/p(z)\prec\sqrt{1+z}. Few more problems of the similar flavor are also considered.


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 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 zf ′ (z)/f (z) ≺ ϕ(z). For special choices of ϕ, S * (ϕ) reduce to well-known subclasses of starlike functions. For example, when −1 ≤ B < A ≤ 1, S * [A, B] := S * ((1 + Az)/(1 + Bz)) is the Janowski starlike functions [5] (see [9]) 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 Sokó l and Stankiewicz [12] and studied recently by Ali et al. [2,4]. A function f ∈ A is in the class SL if zf ′ (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 : (zf ′ (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 |(zf ′ (z)/f (z)) a − b| < b was considered by Paprocki and Sokó l [10]. Clearly S * [2, 1] =: SL. For some radius problems related with lemniscate of Bernoulli see [3,4,11,12]. Estimates for the initial coefficients of functions in the class SL is available in [11]. Let p be an analytic function defined on D with p(0) = 1. Recently Ali et al. [2] determined conditions for p(z) Motivated by the works in [2,3,4,10,11], in Section 2 condition on β is determined so that Further condition on β is obtained in each case so that p(z) ≺ √ 1 + z when p(z) + βzp ′ (z)/p n (z), n = 0, 1, 2. At the end of this section problem p(z) + βzp ′ (z)/p(z) ≺ √ 1 + z implies p(z) ≺ (1 + Az)/(1 + Bz) is also considered.
Silveramn [13] introduced the class G b by Further this result was improved by Obradovič and Tuneski [8] Tuneski [14] further obtained the . Inspired by the work of Silverman [13], Nunokawa et al. [7], obtained the, sufficient conditions for function in the class G b to be strongly starlike, strongly convex, or starlike in D.
can be written as Recently Ali et al. [1], obtained condition on the constants A, B, D, E ∈ [−1, 1] and β so that p(z) ≺ (1 + Az)/(1 + Bz) when 1 + βzp ′ (z)/p n (z) ≺ (1 + Dz)/(1 + Ez), n = 0, 1. In Section 3, alternate and easy proof of results [1, Lemma 2.1, 2.10] are discussed. Further this section has been concluded with condition on A, B, D, E ∈ [−1, 1] and β such that The following results are required in order to prove our main results: Corollary 3.4h, p.135] Let q be univalent in D, and let ϕ be analytic in a domain D containing q(D). Let zq ′ (z)ϕ(q(z)) be starlike. If p is analytic in D, p(0) = q(0) and satisfies then p ≺ q and q is the best dominant.
The following is a more general form of the above lemma: Suppose that > 0 for z ∈ D.

Results Associated with Lemniscate of Bernoulli
In the first result condition on β is obtained so that the subordination Let p be an analytic function defined on D with p(0) = 1 satisfies is starlike in the unit disk D. Consider the subordination Thus in view of Lemma 1.1, it follows that p(z) ≺ q(z). In order to prove our result, we need to prove .
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 and the proof is complete now.
Proof. Define the function q : D → C by Since 1 − B 2 r 2 > 0 (|B| ≤ 1, 0 < r < 1) and so Re(zQ ′ (z)/Q(z)) > 0, this shows that Q is starlike in D. It follows from Lemma 1.1, that the subordination implies p(z) ≺ q(z). Now we need to prove the following in order to prove lemma: and this completes the proof.
Let p be an analytic function defined on D with p(0) = 1 satisfies Proof. Let the function q : D → C be defined by A computation shows that .
Let p be an analytic function defined on D with p(0) = 1 satisfies Proof. Let the function q : D → C be defined by with q(0) = 1. Then Let z = re it , −π ≤ t ≤ π, 0 < r < 1. Then Since 1 − A 2 r 2 > 0 (|A| ≤ 1, 0 < r < 1). Hence Re(zQ ′ (z))/Q(z) > 0, this shows that Q is starlike in D. An application of Lemma 1.1 reveals that the subordination implies p(z) ≺ q(z). Now our result established if we prove Rest part of the proof is similar to that of Lemma 2.2 and therefore it is skipped here.
Consider the function Q defined by 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 satisfies 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 and so Consider the function Q defined by .

Further
Re 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 satisfies 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 Consider the function Q defined by .

Further
Re Thus the function Q is starlike and the result now follows by an application of Lemma 1.3.
Let p be an analytic function defined on D with p(0) = 1 satisfies Then p(z) ≺ 1+Az 1+Bz .
A computation shows that Q(z) is starlike univalent in D. Further Thus by Lemma 1.2, it follows that p(z) ≺ q(z). In order to prove our result, we need to prove that The subordination Φ(z) ≺ h(z) is equivalent to the subordination z ≺ Φ −1 (h(z)). Now in order to prove result it is enough to show |Φ −1 (h(e it ))| ≥ 1, −π ≤ t ≤ π. Now This completes the proof.

Sufficient condition for Janowski Starlikeness
The However, an alternate proof of the same is presented below which is much easier than that of given by Ali et al. [1]: Let p be an analytic function defined on D with p(0) = 1 satisfies Then p(z) ≺ 1+Az 1+Bz .
Proof. Define the function q : D → C by Then q is convex in D with q(0) = 1. Further computation shows that and Q is starlike in D. It follows from Lemma 1.1, that the subordination implies p(z) ≺ q(z). In view of the above result it is sufficient to prove Let z = e it , π ≤ t ≤ π. Thus It should be noted that Ali et al. [1] made the assumption AB > 0 in order to prove the result [1, Lemma 2.10], whereas in the following lemma this condition has been dropped: Then p(z) ≺ 1+Az 1+Bz .
Proof. As above define the function q : D → C by Then q is convex in D with q(0) = 1. A computation shows that and Q is starlike in D. It follows from Lemma 1.1, that the subordination implies p(z) ≺ q(z). Now we need to prove  Then p(z) ≺ 1+Az 1+Bz .
Remaining part of the proof is similar to that of Lemma 3.1 and therefore it is skipped here.