VFAST Transactions on Mathematics

Lemma 1. If p ∈P and has the form (8) , then |cm+n – λcmcn| ≤ ( 2 for 0 ≤λ ≤1, 2 |2λ – 1| otherwise. (9) |cn| ≤ 2 (10) and for complex number µ, we have

Lemma 2. [52]Let α, β, γ and a satisfies that a, α ∈(0, 1) and 8a (1 – α) h (αβ – 2γ)2 + (α (a + α) – β)2i + α (1 – α) (β – 2aα)2 ≤4α2 (1 – α)2 α (1 – a) . If p ∈P, then 2γc2 1 + 2ac2 2 + 4αc1c3 – 3βc2 1c2 – 2c4 ≤4.

Lemma 3. [19]Let p 2 P and has the form (2:1) ; then 2c2 = c2 1 + 4 c2 1
x; 4c3 = c3 1 + 4 c2 1

Lemma 4. [43]If p 2 P and has the form (2:1) ; then Jc3 1 Kc1c2 + Lc3  2 jJj + 2 jK 2Jj + 2 jK J + Lj : 40

Theorem 1. Let f 2 RLp and has the form (1:1) ; then jap+1j  p 2 (p + 1); jap+2j  p 2 (p + 2); jap+3j  p 2 (p + 3); jap+4j 

Theorem 2. Let f 2 RLp and of the form (1:1) : Then ap+2 a2 p+1  p 2 (p + 2) max  1;

Corollary 1. Let f 2 RLp and given the form (1:1) : Then ap+2 a2 p+1  p 2 (p + 2): (3.10)

Theorem 3. Let f 2 RLp be the series form (1:1) : Then jap+1ap+2 ap+3j  p 2 (p + 3):

Theorem 4. Let f 2 RLp be the series form (1:1) : Then ap+1ap+3 a2 p+2  p2 4 (p + 2)2 :

Theorem 5. Let f 2 RLp be given the form (1:1). Then jH3;p (f)j  p2 5p4 + 50p3 + 179p2 + 268p + 136
8 (p + 2)3 (p + 3)2 (p + 4) ; where H3;p (f) =

Corollary 2. Let f 2 RL be given by (1:1). Then jH3;1 (f)j  319 8640 = 0:0369: VFAST Transactions on Mathematics 44

Theorem 6. Let f 2 SLp and has the form (1:1) : Then jap+1j  p 2; jap+2j   p 4 for p  2; p 16 (2p 1) otherwise: ; jap+3j

Corollary 3. Let f 2 S L and be given by (1:1) : Then ja2j  1 2; ja3j  1 4; ja4j  1 6; ja5j  1 8: These inequalities are best possible and for these see [48, 57].

Theorem 7. Let f 2 SLp and has given by (1:1) : Then for  2 C ap+2 a2 p+1  p 4 max  1; p p 2 + 1 4

Corollary 4. Let f 2 SLp and be given by (1:1) : Then ap+2 a2 p+1  p 4 : (4.9)

Theorem 8. Let f 2 SLp and has the form (1:1). Then jap+1ap+2 ap+3j  2p3 + 37p 384 :

Theorem 9. Let f 2 SLp and be given by (1:1) : Then ap+1ap+3 a2 p+2  p2 16:

Theorem 10. Let f 2 SLp and has the form (1:1) : Then jH3;p (f)j  8 > < > : p2 147456 2p2 (p) + 37 (p) + 2304p + 4608
for p  2; p2 147456

Corollary 5. Let f 2 S L be given by (1:1). Then jH3;1 (f)j  49 768 = 0:0638: This result proves that this bound improves the estimate which was proved in [57]. 5 BOUNDS OF jH3;p (f)j FOR THE SET CLp

Theorem 11. Let f 2 CLp be given the form (1:1). Then jap+1j  p2 2 (p + 1); jap+2j  ( p2 4(p+2) for p  2; p2(2p1) 16(p+2) otherwise: ; jap+3j 

Corollary 6. Let f 2 CLp and be given by (1:1) : Then ja2j  1 6; ja3j  1 12; ja4j  1 24; ja4j  1 40: The result is best possible.

Theorem 12. Let f 2 CLp and be given by (1:1) : Then for  2 C ap+2 a2 p+1  p2 4 (p + 2) max  1;

Corollary 7. Let f 2 CLp and has the form. Then ap+2 a2 p+1  p2 4 (p + 2): (5.10)

Theorem 13. Let f 2 CLp and given by (1:1) : Then jap+1ap+2 ap+3j   2p2 348 (p + 1) (p + 2) (p + 3) (p) ; where  (p) = 2p4 + 6p3 15p2 24p 26 + 2 2p4 + 6p3 + 5p2 + 24p + 14

Corollary 8. Let f 2 CLp and be given by (1:1) : Then ja2a3 a4j  1 24: VFAST Transactions on Mathematics 51

Theorem 14. Let f 2 CLp and has the form (1:1) : Then ap+1ap+3 a2 p+2  8 < : p4 16(p+2)2 for p  2; p4(p2+8p+2016p34p4) 768(p+1)(p+2)2(p+3) otherwise: :

Theorem 15. If f 2 CLp and has the form (1:1) : Then jH3;p (f)j  8 > > > > > > > > > > > >

Corollary 9. Let f 2 CL be given the form (1:1) : Then jH3;1 (f)j  19 4320 = 0:00439: Conclusions: In this article, we studied Hankel determinant H3;p (f) for the fam- ilies RLp; CLp and SLp of multivalent functions subordinated with Bernoulli lem- niscate. Further, by putting "p" is equal to "1" in one of the obtained result, we have deduce an improved bound of third Hankel determinant for the family S L. Data Availability: No data were used to support this study. Con‡icts of Interest: The au