Results & Lemmas (12)
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Lemma 1.
Lemma 1. Let ϕi(z) ∈MS∗(αi) where 0 ≤αi < 1 (i = 0, 1, 2,..., k −1). Then for k −1 ≤ k−1 i=0 αi < k, we have zk−1 k−1 i=0 ϕi(z) ∈MS∗…
Lemma 1. Let ϕi(z) ∈MS∗(αi) where 0 ≤αi < 1 (i = 0, 1, 2, ..., k −1). Then for k −1 ≤ k−1 i=0 αi < k, we have zk−1 k−1 i=0 ϕi(z) ∈MS∗ k−1 i=0
Lemma 2
Lemma 2 (see [2]). Suppose that h(z) = 1 z + ∞ n=1 cnzn ∈MS∗ (2.6) Then |cn| ≤ 2 n + 1 (n ∈N) (2.7) Each of these inequality is sharp,…
Lemma 2 (see [2]). Suppose that h(z) = 1 z + ∞ n=1 cnzn ∈MS∗ (2.6) Then |cn| ≤ 2 n + 1 (n ∈N) (2.7) Each of these inequality is sharp, with the extremal function given by h(z) = z−1
Lemma 3
Lemma 3 (see [1]). Let p ∈P[A, B] and p(z) = 1 + ∞ n=1 cnzn Then |cn| ≤A −B This result is sharp.
Lemma 3 (see [1]). Let p ∈P[A, B] and p(z) = 1 + ∞ n=1 cnzn Then |cn| ≤A −B This result is sharp.
Lemma 4
Lemma 4 (see [4]). Let p ∈P[A, B], then for |z| = r < 1 1 −Ar 1 −Br ≤Rep(z) ≤|p(z)| ≤1 + Ar 1 + Br (2.9) These bounds are sharp.
Lemma 4 (see [4]). Let p ∈P[A, B], then for |z| = r < 1 1 −Ar 1 −Br ≤Rep(z) ≤|p(z)| ≤1 + Ar 1 + Br (2.9) These bounds are sharp.
Lemma 5
Lemma 5 (see [7]) Suppose that g ∈MS∗, then (1 −r)2 r ≤|g(z)| ≤(1 + r)2 r (|z| = r; 0 < r < 1) (2.10)
Lemma 5 (see [7]) Suppose that g ∈MS∗, then (1 −r)2 r ≤|g(z)| ≤(1 + r)2 r (|z| = r; 0 < r < 1) (2.10)
Lemma 6
Lemma 6(see [6]). Let −1 ≤B2 ≤B1 < A1 ≤A2 ≤1. Then 1 + A1z 1 + B1z ≺1 + A2z 1 + B2z (2.11)
Lemma 6(see [6]). Let −1 ≤B2 ≤B1 < A1 ≤A2 ≤1. Then 1 + A1z 1 + B1z ≺1 + A2z 1 + B2z (2.11)
Theorem 1
Theorem 1 Let g(z) = 1 z + ∞ n=1 bnzn ∈MS∗ k−1 k
Theorem 1 Let g(z) = 1 z + ∞ n=1 bnzn ∈MS∗ k−1 k
Theorem 2.
Theorem 2. Let f(z) given by (1.1) and −1 ≤B < A ≤1. if ∞ n=1
Theorem 2. Let f(z) given by (1.1) and −1 ≤B < A ≤1. if ∞ n=1
Theorem 3.
Theorem 3. Let f ∈MK(k)[A, B] (−1 ≤B < A ≤1) and gk(z) is given by (1.1) and (1.14) respectively. Then for k ≥1, we have n k=1 |kak +…
Theorem 3. Let f ∈MK(k)[A, B] (−1 ≤B < A ≤1) and gk(z) is given by (1.1) and (1.14) respectively. Then for k ≥1, we have n k=1 |kak + Bk|2 − n−1 k=1 |A.Bk + kBak|2 < (A −B)2 (3.4)
Theorem 4.
Theorem 4. Suppose that f(z) = 1 z + ∞ n=1 anzn ∈MK(k) [A, B] Then |an| ≤(A −B) n −1 + 2 n m=1
Theorem 4. Suppose that f(z) = 1 z + ∞ n=1 anzn ∈MK(k) [A, B] Then |an| ≤(A −B) n −1 + 2 n m=1
Theorem 5.
Theorem 5. Let f ∈MK(k)[A, B]. Then (1 −r)2 r2 1 −Ar 1 −Br ≤|f ′(z)| ≤(1 + r)2 r2 1 + Ar 1 + Br (|z| = r, 0 < r < 1) (3.12)
Theorem 5. Let f ∈MK(k)[A, B]. Then (1 −r)2 r2 1 −Ar 1 −Br ≤|f ′(z)| ≤(1 + r)2 r2 1 + Ar 1 + Br (|z| = r, 0 < r < 1) (3.12)
Theorem 6.
Theorem 6. Let −1 ≤B2 ≤B1 < A1 ≤A2 ≤1. Then MK(k) (A1, B1) ⊂MK(k) (A2, B2) (3.13)
Theorem 6. Let −1 ≤B2 ≤B1 < A1 ≤A2 ≤1. Then MK(k) (A1, B1) ⊂MK(k) (A2, B2) (3.13)
Function classes studied:
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