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FullText JMA 34_08.pdf - Journal of Mathematics and Applications

FullText JMA 34_08.pdf - Journal of Mathematics and Applications

92 Inclusion

92 Inclusion relationship and Fekete-Szegö like inequalities . . . and We have z(Dα,β m f(z))′ Dα,β m f(z) = −1 + (1 + 2αβ + α − β) m a 0 z + [ 2(1 + 6αβ + 2α − 2β) m a 1 − (1 + 2αβ + α − β) 2m a 2 ] 0 z 2 (20) + . . . z(Dm α,β f(z))′′ (D m α,β f(z))′ = −2 − 2(1 + 6αβ + 2α − 2β) m a 1 z 2 + . . . (21) (1 − 2γ)z(D m α,β f(z))′ + (1 − γ)z 2 (D m α,β f(z))′′ (1 − γ)(D m α,β f(z))′ − γD m α,β f(z) = −1 + γ(1 + 2αβ + α − β) m a 0 z − [ 2(1 − 2γ)(1 + 6αβ + 2α − 2β) m a 1 + γ 2 (1 + 2αβ + α − β) 2m a 2 0] z 2 + . . . (22) Using (20), (21) and (22) in (19) we find and c 1 = −(1 − γ)(1 + 2αβ + α − β) m a 0 c 2 = −2(1 − 2γ)(1 + 6αβ + 2α − 2β) m a 1 − (1 − γ 2 )(1 + 2αβ + α − β) 2m a 2 0 which give and if γ /∈ where { 1 2 , 1 } . a 1 = Therefore, we have c 1 a 0 = − , if γ ≠ 1 (23) (1 − γ)(1 + 2αβ + α − β) m [ −1 2(1 − 2γ)(1 + 6αβ + 2α − 2β) m c 2 − 1 + γ ] 1 − γ c2 1 a 1 − µa 2 0 = −1 2(1 − 2γ)(1 + 6αβ + 2α − 2β) m (c 2 − vc 2 1) v = (1 − γ2 )(1 + 2αβ + α − β) 2m − 2(1 − 2γ)(1 + 6αβ + 2α − 2β) m µ (1 − γ) 2 (1 + 2αβ + α − β) 2m . Now, the result (16) follows by an application of Lemma 3.1. −c 2 If γ = 1, then a 0 = 0 and a 1 = 2(1 + 6αβ + 2α − 2β) m . Since |c 2| ≤ 2 it follows 1 that |a 1 | ≤ (1 + 6αβ + 2α − 2β) m which proves (17). Also, if γ = 1 2 , then a 1 = 0 and c 1 = − 1 2 (1 + 2αβ + α − β)m a 0 (24)

Dorina Răducanu and Halit Orhan and Erhan Deniz 93 and c 2 = 3 4 (1 + 2αβ + α − β)2m a 2 0. 2 √ 6 Since |c 1 | ≤ 2 and |c 2 | ≤ 2 it follows that |a 0 | ≤ and thus, (18) 3(1 + 2αβ + α − β) m is proved. The bounds are sharp for the functions f 1 (z) and f 2 (z) defined by 1 − γ Dα,β m f 1(z) − γ z(Dα,β m f 1(z)) ′ = 1 F 1 (z) , where − zF ′ 1(z) F 1 (z) = 1 + z 1 − z respectively, 1 − γ Dα,β m f 2(z) − γ z(Dα,β m f 2(z)) ′ = 1 F 2 (z) , where − zF ′ 2(z) F 2 (z) = 1 + z2 1 − z 2 . Obviously, the functions F 1 , F 2 ∈ Σ ∗ and f 1 , f 2 ∈ HΣ ∗ m(α, β, γ). If we consider first m = 0, then γ = 0 and then m = γ = 0 , respectively in Theorem 3.1, we obtain the following consequences. Corollary 3.1 Let f(z) given by (1) be in the class HΣ ∗ (γ). Then, for any complex number µ { } |a 1 − µa 2 1 0| ≤ |1 − 2γ| max 1; |3γ2 − 2γ − 1 + 4(1 − 2γ)µ| |1 − γ| 2 , { } 1 if γ /∈ 2 , 1 |a 1 − µa 2 0| ≤ 1 , if γ = 1 |a 1 − µa 2 0| ≤ 2√ 6 |µ| , if γ = 0. 3 The bounds are sharp. Corollary 3.2 If f(z) given by (1) belongs to the class Σ ∗ m(α, β), then for any complex number µ where |a 1 − µa 2 0| ≤ 1 max {1; Φ(α, β, µ, m)} (1 + 6αβ + 2α − 2β) m Φ(α, β, µ, m) = |4(1 + 6αβ + 2α − 2β)m µ − (1 + 2αβ + α − β) 2m | (1 + 2αβ + α − β) 2m . The bound is sharp.

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