Starlikeness associated with lemniscate of Bernoulli

Theorem 2.2. [8, Theorem 2.3b, p. 28] Let ψ ∈Ψn[Ω, q] with q(0) = a. Thus for p ∈H[a, n] such that (2.1) ψ(p(z), zp′(z), z2p′′(z); z) ∈Ω ⇒ p(z) ≺q(z). If Ωis a simply connected region which is not the whole complex plane, then there is a conformal mapping h from D onto Ωsatisfying h(0) = ψ(a, 0, 0; 0). Thus, for p ∈H[a, n], (2.1) can be written as (2.2) ψ(p(z), zp′(z), z2p′′(z); z) ≺h(z) ⇒ p(z) ≺q(z). The univalent function q is said to be the dominant of the solutions of the second order differe

Theorem 2.3. Let p ∈H[1, n] with p(z) ̸≡1 and n ≥1. Let Ω⊂C and ψ : C3 × D →C with domain D satisfy ψ(r, s, t; z) ̸∈Ωwhenever z ∈D, for r = √ 2 cos 2θeiθ, s = me3iθ/(2 √ 2 cos2θ) and Re(t/s + 1) ≥3m/4 where m ≥ n ≥1 and −π/4 < θ < π/4. For z ∈D, if (p(z), zp′(z), z2p′′(z); z) ∈D, and ψ(p(z), zp′(z), z2p′′(z); z) ∈Ω, then p(z) ≺√1 + z.

Theorem 3.1. Let p ∈H[1, n] with p(z) ̸≡1 and n ≥1. Let Ω⊂C and ψ : C2 × D →C with domain D satisfy ψ(r, s; z) ̸∈Ωwhenever z ∈D, for r = √ 2 cos2θeiθ and s = me3iθ/(2 √ 2 cos2θ) where m ≥n ≥1 and −π/4 < θ < π/4. For z ∈D, if (p(z), zp′(z); z) ∈D and ψ(p(z), zp′(z); z) ∈Ω, then p(z) ≺ √1 + z. Likewise for an analytic function h, if Ω= h(D), then the above theorem becomes ψ(p(z), zp′(z); z) ≺h(z) ⇒p(z) ≺ √ 1 + z. Using the above theorem, now some sufficient conditions are determined for

Lemma 3.2. Let p be analytic in D and p(0) = 1 and β0 = 1.1874. Let p(z) + βzp′(z) p3(z) ≺ √ 1 + z (β > β0), then p(z) ≺ √ 1 + z.

Lemma 3.3. Let p be analytic in D and p(0) = 1 and β0 = 3.58095. Let p(z) + βzp′(z) p4(z) ≺ √ 1 + z (β > β0), then p(z) ≺ √ 1 + z.

Lemma 3.4. Let β, γ > 0 and p be analytic in D such that p(0) = 1. If p2(z) + zp′(z) βp(z) + γ ≺1 + z, then p(z) ≺ √ 1 + z.

Lemma 3.5. Let p be analytic in D with p(0) = 1. Let β be a complex number such that Re β > 0. If p2(z) + βzp′(z)p(z) ≺1 + z, then p(z) ≺ √ 1 + z.

Lemma 3.6. Let β > 0 and p be analytic in D with p(0) = 1. If p2(z) + βzp′(z) ≺1 + z, then p(z) ≺ √ 1 + z.

Lemma 3.7. Let β > 0 and p be analytic in D with p(0) = 1. If p2(z) + βzp′(z) p(z) ≺1 + z, then p(z) ≺ √ 1 + z.

Lemma 3.8. Let β0 = 2 √ 2. Let p be analytic in D with p(0) = 1. If p2(z) + βzp′(z) p2(z) ≺1 + z (β > β0), then p(z) ≺ √ 1 + z.

Lemma 3.9. Let β0 = 2 and p be analytic in D with p(0) = 1. If p2(z) + βzp′(z)p(z) ≺2 + z 2 −z (β ≥β0), then p(z) ≺ √ 1 + z. The lower bound β0 is best possible.

Lemma 4.1. Let p be analytic in D such that p(0) = 1. If zp′(z) + z2p′′(z) ≺ 3z 8 √ 2, then p(z) ≺ √ 1 + z.

Theorem 4.2. Let f be a function in A. If f satisfies the subordination zf ′(z) f(z)  1 + zf ′′(z) f ′(z) −zf ′(z) f(z)  + zf ′(z) f(z)

Lemma 4.3. Let p be analytic in D such that p(0) = 1 and let p ∈H[1, 2]. If p2(z) + zp′(z) + z2p′′(z) ≺1 +  1 + 3 2 √ 2  z, then p(z) ≺ √ 1 + z.

Theorem 4.4. Let f be a function in A such that zf ′(z)/f(z) has Taylor series expansion of the form 1 + a2z2 + a3z3 + . . . . If f satisfies the subordination zf ′(z) f(z) 2 + zf ′(z) f(z)  1 + zf ′′(z) f ′(z) −zf ′(z) f(z)  + zf ′(z) f(z)

Lemma 4.5. Let γ ≥β > 0 be such that 4γ −β ≥1. Let p be analytic in D such that p(0) = 1 and γzp′(z) + βz2p′′(z) ≺ z 8 √ 2 for γ ≥β > 0 and 4γ −β ≥1, then p(z) ≺ √ 1 + z.

Theorem 4.6. Let f be a function in A. Let γ, β be as stated in Lemma 4.5. If f satisfies the subordination γ zf ′(z) f(z)  1 + zf ′′(z) f ′(z) −zf ′(z) f(z)  + β zf ′(z) f(z)

Lemma 5.1. Let p be analytic function on D and p(0) = 1. Let β0 = 2 √ 2( √ 2−1) ≈ 1.17. If 1 + βzp′(z) ≺ √ 1 + z (β ≥β0), then p(z) ≺ √ 1 + z.

Theorem 5.2. Let β0 = 2 √ 2( √ 2 −1) ≈1.17 and f ∈A. (1) If f satisfies the subordination 1 + β zf ′(z) f(z)  1 + zf ′′(z) f ′(z) −zf ′(z) f(z)  ≺ √

Lemma 5.3. Let p be analytic function on D and p(0) = 1. Let β0 = 4( √ 2 −1) ≈ 1.65. If 1 + β zp′(z) p(z) ≺ √ 1 + z (β ≥β0), then p(z) ≺ √ 1 + z.

Theorem 5.4. Let β0 = 4( √ 2 −1) ≈1.65 and f ∈A. (1) If f satisfies the subordination 1 + β  1 + zf ′′(z) f ′(z) −zf ′(z) f(z)  ≺ √ 1 + z (β ≥β0), then f ∈SL. (2) If 1 + βzf ′′(z)/f ′(z) ≺√1 + z (β ≥β0), then f ′(z) ≺√1 + z.

Lemma 5.5. Let p be analytic function on D and p(0) = 1. Let β0 = 4 √ 2( √ 2−1) ≈ 2.34. If 1 + β zp′(z) p2(z) ≺ √ 1 + z (β ≥β0), then p(z) ≺ √ 1 + z.

Theorem 5.6. Let β0 = 4 √ 2( √ 2 −1) ≈2.34 and f ∈A. If f satisfies the subor- dination 1 −β + β 1 + zf ′′(z)/f ′(z) zf ′(z)/f(z)  ≺ √ 1 + z (β ≥β0), then f ∈SL. Kumar et al. introduced that for every β > 0, p(z) ≺√1 + z whenever p(z) +

Lemma 5.7. Let β > 0 and p be analytic in D and p(0) = 1 such that p(z) + βzp′(z) ≺ √ 1 + z, then p(z) ≺ √ 1 + z.

Theorem 5.8. Let β > 0 and f be a function in A. (1) If f satisfies the subordination zf ′(z) f(z) + β zf ′(z) f(z)  1 + zf ′′(z) f ′(z) −zf ′(z) f(z)  ≺ √ 1 + z, then f ∈SL. (2) If f ′(z) + βzf ′′(z) ≺√1 + z, then f ′(z) ≺√1 + z.

Lemma 5.9. Let β > 0 and p be analytic in D and p(0) = 1 such that p(z) + βzp′(z) p(z) ≺ √ 1 + z, then p(z) ≺ √ 1 + z.

Theorem 5.10. Let β > 0 and f be a function in A. (1) If f satisfies the subordination zf ′(z) f(z) + β  1 + zf ′′(z) f ′(z) −zf ′(z) f(z)  ≺ √ 1 + z, then f ∈SL. (2) If f satisfies the subordination z2f ′(z)

Lemma 5.11. Let β > 0 and p be analytic in D and p(0) = 1 such that p(z) + βzp′(z) p2(z) ≺ √ 1 + z, then p(z) ≺ √ 1 + z.

Theorem 5.12. Let β > 0 and f be a function in A. If f satisfies the subordination zf ′(z) f(z) −β + β 1 + zf ′′(z)/f ′(z) zf ′(z)/f(z)  ≺ √ 1 + z, then f ∈SL. Acknowledgements The first author is supported by University Grants Commission(UGC), UGC- Ref. No.:1069/(CSIR-UGC NET DEC, 2016). References [1] R. M. Ali, N. E. Cho, V. Ravichandran, and S. S. Kumar, Differential subordination for