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Abstract
The article discusses criteria for univalence of analytic functions in the unit disc. Various families of analytic functions depending on real parameters are considered. A unified method for creating new sets of conditions ensuring univalence is presented. Applying this method we are able to find several families of new sharp criteria for univalence.
Results & Lemmas (11)
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Lemma 1
Lemma 1 Let u(x) = (1 −x2)λG(x). Then p(x, λ) = −u′′/u = 4λ(1 −λ)(1 −x2)−2 + 2λ(1 −x2)−1 + 4λx(1 −x2)−1 G′/G − G′/G 2 − G′/G ′.…
Lemma 1 Let u(x) = (1 −x2)λG(x). Then p(x, λ) = −u′′/u = 4λ(1 −λ)(1 −x2)−2 + 2λ(1 −x2)−1 + 4λx(1 −x2)−1 G′/G − G′/G 2 − G′/G ′ . (3.4) The proof is elementary and details are omitted. Some special cases of Lemma 1 are (a) u(x) =(1 −x2)a exp(λx2), p(x, a, λ) =4a(1 −a)x2(1 −x2)−2 + 2a(1 −x2)−1 + 8aλx(1 −x2)−1 −4λx2 −2λ =(2a + 2ax2 −4a2x2)(1 −x2)−2 + 8aλx2(1 −x2)−1 −(4λ2x2 + 2λ).
Lemma 2
Lemma 2 Let u(x, λ) = cos(πx/2)G(x, λ). Then p(x, λ) = −u′′(x)/u(x) =π2/4 + π G′/G tan(πx/2) − G′/G 2 − G′/G ′. (3.5) Again, the…
Lemma 2 Let u(x, λ) = cos(πx/2)G(x, λ). Then p(x, λ) = −u′′(x)/u(x) =π2/4 + π G′/G tan(πx/2) − G′/G 2 − G′/G ′ . (3.5) Again, the elementary calculation is omitted. Special cases of Lemma 2 are 7
Lemma 1
Repeated OCR label — verify against source
Lemma 1 We start with the first condition arising from (a), generated by u = (1 −x2)a exp(λx2).
Lemma 1 We start with the first condition arising from (a), generated by u = (1 −x2)a exp(λx2).
Theorem 1
Theorem 1 (A) Let p(x) = (2a + 2ax2 −4a2x2)(1 −x2)−2 + 8aλx2(1 −x2)−1 −(4λ2x2 + 2λ) (4.1) and let λ, a satisfy 1 −2a 4 ≤λ ≤1 2 h (1 + 2a)…
Theorem 1 (A) Let p(x) = (2a + 2ax2 −4a2x2)(1 −x2)−2 + 8aλx2(1 −x2)−1 −(4λ2x2 + 2λ) (4.1) and let λ, a satisfy 1 −2a 4 ≤λ ≤1 2 h (1 + 2a) −(1 + 6a)1/2i , 1 2 ≤a ≤1. (4.2) Then if f(z) is an analytic function in ∆satisfying
Theorem 2
Theorem 2 Let p(x) =4λ(1 −λ)x2(1 −x2)−2 + 2λ(1 −x2)−1 + µπ2/4 + µ(1 −µ)π2 tan2(πx/2)/4 −2µλπx tan(πx/2)(1 −x2)−1 (4.19) 11
Theorem 2 Let p(x) =4λ(1 −λ)x2(1 −x2)−2 + 2λ(1 −x2)−1 + µπ2/4 + µ(1 −µ)π2 tan2(πx/2)/4 −2µλπx tan(πx/2)(1 −x2)−1 (4.19) 11
Lemma 3
Lemma 3 For the interval (0,1) we have: (i) The function G(x):= (1 −x2) tan(πx/2) of (4.23) is convex. (ii) The function (1 −x2) tan(πx/2)…
Lemma 3 For the interval (0,1) we have: (i) The function G(x) := (1 −x2) tan(πx/2) of (4.23) is convex. (ii) The function (1 −x2) tan(πx/2) πx/2 decreases for 0 ≤x ≤1 and its maximal value at x = 0 is 1. (iii) G(x) ≤2 π(1 + x). (iv) π 2 G′(x) ≤1 + π2 4 (1 −x2). 14
Theorem 3
Theorem 3 (A) Let p(x) = (2a −4ab) + x2(2a + 4ab −4a2) (1 −x2)2 + (2b + 4ab) + x2(−2b −4b2 + 4ab) (1 + x2)2 (4.36) and let a, b satisfy 1 2…
Theorem 3 (A) Let p(x) = (2a −4ab) + x2(2a + 4ab −4a2) (1 −x2)2 + (2b + 4ab) + x2(−2b −4b2 + 4ab) (1 + x2)2 (4.36) and let a, b satisfy 1 2 ≤a ≤1, 1 4 −(5 + 4a) + (25 + 48a)1/2 ≤b ≤a −1 2.
Theorem 4
Theorem 4 Let p(x) = 2λπx tan(πx/2) + π2/4 −(2λ + 4λ2x2) (5.1) and let λ satisfy 0 ≤λ ≤λ0 = 1 8 (4 + π2) −(16 + π4)1/2. (5.2) Then if…
Theorem 4 Let p(x) = 2λπx tan(πx/2) + π2/4 −(2λ + 4λ2x2) (5.1) and let λ satisfy 0 ≤λ ≤λ0 = 1 8 (4 + π2) −(16 + π4)1/2 . (5.2) Then if f(z) is an analytic function in ∆satisfying |Sf(z)| ≤2p(|z|), z ∈∆, (5.3) then f(z) is univalent in ∆and condition (5.3) is sharp.
Theorem 5
Theorem 5 Let p(x) = π2 4 h 1 −λ2 sin2(πx/2) + 2λsin2(πx/2) cos(πx/2) −λ cos(πx/2) i. (5.10) 22
Theorem 5 Let p(x) = π2 4 h 1 −λ2 sin2(πx/2) + 2λsin2(πx/2) cos(πx/2) −λ cos(πx/2) i . (5.10) 22
Theorem 1
Repeated OCR label — verify against source
Theorem 1 we have τ = 4a(1 −a), which means by (4.7) that τ < 1 unless a = 1/2. 27
Theorem 1 we have τ = 4a(1 −a), which means by (4.7) that τ < 1 unless a = 1/2. 27
Theorem 2
Repeated OCR label — verify against source
Theorem 2 the following additional information: Suppose τ = 1. Then µ < 0 can not occur, i.e., the function arising for the values µ < 0…
Theorem 2 the following additional information: Suppose τ = 1. Then µ < 0 can not occur, i.e., the function arising for the values µ < 0 and λ + µ = 1/2 is not univalent. This means that ϕ(x) = p(x)(1 −x2)2 is not monotone in the interval (0,1). Indeed, monotonicity for ϕ would supply a function u(x) u = (1 −x2)1/2+|µ| cos−|µ|(πx/2) ̸= (1 −x2)1/2, contradicting the uniqueness argument. 28