Ma-Minda φ-classes studied in this paper:
Results & Lemmas (18)
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Lemma 3.1
Lemma 3.1 If K is smooth (analytic) in ∪, ⋎∈C( 1+ξ 1–ξ ) is convex and g ∈S∗( 1+ξ 1–ξ ) is starlike then ⋎∗(Kg) ⋎∗g (∪) ⊆co K(∪) , (3.8)…
Lemma 3.1 If K is smooth (analytic) in ∪, ⋎∈C( 1+ξ 1–ξ ) is convex and g ∈S∗( 1+ξ 1–ξ ) is starlike then ⋎∗(Kg) ⋎∗g (∪) ⊆co K(∪) , (3.8) where co(K(∪)) is the closed convex hull of K(∪).
Lemma 3.2
Lemma 3.2 For analytic functions h,ℏ∈∪, the subordination h ≺ℏimplies that 2π 0 h(ξ) p dθ ≤ 2π 0 ℏ(ξ) p dθ, (3.9) where ξ =…
Lemma 3.2 For analytic functions h,ℏ∈∪, the subordination h ≺ℏimplies that 2π 0 h(ξ) p dθ ≤ 2π 0 ℏ(ξ) p dθ, (3.9) where ξ = reiθ, 0 < r < 1, and p is a positive number. Some of the few studies in q-calculus are realized by comparison between two differ- ent values of calculus. Class Ξ ν,k q1,q2(σ) shows the relation between the q1- and q2-calculus depending on the operator (2.7).
Theorem 4.1
Theorem 4.1 Let ⋎∈ and let the function g:= Ψ k q2 ∗⋎∈S∗( 1+ξ 1–ξ ), ξ ∈∪. If ⋎∈Ξ 0,k q1,q2(σ), q1 ̸= q2 and the function Φk(ξ) ∈C( 1+ξ…
Theorem 4.1 Let ⋎∈ and let the function g := Ψ k q2 ∗⋎∈S∗( 1+ξ 1–ξ ), ξ ∈∪. If ⋎∈Ξ 0,k q1,q2(σ), q1 ̸= q2 and the function Φk(ξ) ∈C( 1+ξ 1–ξ ) then ⋎∈Ξ ν,k q1,q2(σ), σ(0) = 1.
Corollary 4.2
Corollary 4.2 Let ⋎be a function from and σ(ξ) be a convex univalent function in ∪ such that σ(0) = 1. Then Ξ 0,k q1,q2(σ) ⊂Ξ ν,k…
Corollary 4.2 Let ⋎be a function from and σ(ξ) be a convex univalent function in ∪ such that σ(0) = 1. Then Ξ 0,k q1,q2(σ) ⊂Ξ ν,k q1,q2(σ). In general, we have the following result:
Theorem 4.3
Theorem 4.3 Let ⋎∈ and let the function G:= Ψ k q2 ∗Φk ν ∗⋎∈S∗( 1+ξ 1–ξ ), ξ ∈∪. If ρ1:= Ψq1 ∗Φν ≺r ρ2:= Ψq2 ∗Φν for some r < 1 and the…
Theorem 4.3 Let ⋎∈ and let the function G := Ψ k q2 ∗Φk ν ∗⋎∈S∗( 1+ξ 1–ξ ), ξ ∈∪. If ρ1 := Ψq1 ∗Φν ≺r ρ2 := Ψq2 ∗Φν for some r < 1 and the function ρ2 ∈C( 1+ξ 1–ξ ) then Ξ ν,k q1,q2(σ) ⊂Ξ ν,k+1 q1,q2 (σ). (4.6)
Theorem 4.4
Theorem 4.4 Let ⋎∈ and let the function H:= Ψ k q2 ∗Φk ν1 ∗⋎∈S∗( 1+ξ 1–ξ ), ξ ∈∪. If Φk ν1 ≺r Φk ν2 for some r < 1 then Ξ ν1,k q1,q2(σ) ⊂Ξ…
Theorem 4.4 Let ⋎∈ and let the function H := Ψ k q2 ∗Φk ν1 ∗⋎∈S∗( 1+ξ 1–ξ ), ξ ∈∪. If Φk ν1 ≺r Φk ν2 for some r < 1 then Ξ ν1,k q1,q2(σ) ⊂Ξ ν2,k q1,q2(σ). (4.12)
Theorem 5.1
Theorem 5.1 Consider the operator [Sk ν ]q ⋎(ξ), ⋎∈. If the coefficients of ⋎satisfy the inequality | ⋎n | ≤( 1 nν )k, ν ∈(0,1) then 2π 0…
Theorem 5.1 Consider the operator [Sk ν ]q ⋎(ξ), ⋎∈. If the coefficients of ⋎satisfy the inequality | ⋎n | ≤( 1 nν )k, ν ∈(0,1) then 2π 0 [Sk ν ]q ⋎(ξ) ξ p dθ ≤ 2π 0
Theorem 5.2
Theorem 5.2 Consider the operator [Sk ν ]q ⋎(ξ), ⋎∈. If the coefficients of ⋎satisfy the inequality | ⋎n | ≤( 1 nν )k, ν ∈(0,1) then 2π 0…
Theorem 5.2 Consider the operator [Sk ν ]q ⋎(ξ), ⋎∈. If the coefficients of ⋎satisfy the inequality | ⋎n | ≤( 1 nν )k, ν ∈(0,1) then 2π 0 Sk ν
Theorem 5.3
Theorem 5.3 Consider the operator Sκ,k q ψ(z), ψ ∈Λ. If the coefficients of ψ satisfy the inequality |ϑn| ≤( 1 nκ )k, κ ∈(0,∞) then there is…
Theorem 5.3 Consider the operator Sκ,k q ψ(z), ψ ∈Λ. If the coefficients of ψ satisfy the inequality |ϑn| ≤( 1 nκ )k, κ ∈(0,∞) then there is a probability measure μ on (∂U)2, for all δ > 1.
Theorem 6.1
Theorem 6.1 If the function ⋎∈Vq(ψ) is given by (2.1), then | ⋎2 | ≤ 1 [Q2]kq, | ⋎3 | ≤ 1 [Q3]kq. (6.6)
Theorem 6.1 If the function ⋎∈Vq(ψ) is given by (2.1), then | ⋎2 | ≤ 1 [Q2]kq , | ⋎3 | ≤ 1 [Q3]kq . (6.6)
Lemma 6.2
Lemma 6.2 Consider functions f1,f2,f3: ∪→C such that ℜ(f1) ≥a ≥0. If f ∈H[1,n] (the set of analytic functions having the expansion f (ξ) =…
Lemma 6.2 Consider functions f1,f2,f3 : ∪→C such that ℜ(f1) ≥a ≥0. If f ∈H[1,n] (the set of analytic functions having the expansion f (ξ) = 1 + ϕ1ξ + ···) and ℜ aξ 2f ′′(ξ) + f1(ξ)ξf ′(ξ) + f2(ξ)f (ξ) + f3(ξ) > 0, a ≥0,ξ ∈∪, then ℜ(f (ξ)) > 0.
Lemma 6.3
Lemma 6.3 Let ♭be convex in ∪and suppose f1,f2,f3: ∪→C are analytic functions such that ℜ(f1) ≥a ≥0. If g ∈H[0,m] (the set of analytic…
Lemma 6.3 Let ♭be convex in ∪and suppose f1,f2,f3 : ∪→C are analytic functions such that ℜ(f1) ≥a ≥0. If g ∈H[0,m] (the set of analytic functions with the expansion g(ξ) = g1ξ m + ···), m ≥1 and aξ 2g′′(ξ) + f1(ξ)ξg′(ξ) + f2(ξ)g(ξ) + f3(ξ) ≺♭(ξ), a ≥0,ξ ∈∪, then g(ξ) ≺♭(ξ).
Lemma 6.4
Lemma 6.4 Let a,b,c ∈R be such that a ≥0, b ≥–a, c ≥–b. If q ∈H[0,1], where q(ξ) = q1ξ + ··· and aξ 2q′′(ξ) + bξq′(ξ) + cq(ξ) ≺ξ, a ≥0,ξ…
Lemma 6.4 Let a,b,c ∈R be such that a ≥0, b ≥–a, c ≥–b. If q ∈H[0,1], where q(ξ) = q1ξ + ··· and aξ 2q′′(ξ) + bξq′(ξ) + cq(ξ) ≺ξ, a ≥0,ξ ∈∪, then q(ξ) ≺ ξ b+c, which is the best dominant.
Theorem 6.5
Theorem 6.5 Let ⋎∈Vq(ξ) and F(ξ) = ξ([Skν ]q⋎(ξ))′ ([Skν ]q⋎(ξ)). If ℜ(ξF(ξ)) > –1, ξ ∈∪, then [Sk ν ]q⋎∈ S∗(starlike with respect to the…
Theorem 6.5 Let ⋎∈Vq(ξ) and F(ξ) = ξ([Skν ]q⋎(ξ))′ ([Skν ]q⋎(ξ)) . If ℜ(ξF(ξ)) > –1, ξ ∈∪, then [Sk ν ]q⋎∈ S∗(starlike with respect to the origin).
Theorem 6.6
Theorem 6.6 Let ⋎∈ and F(ξ) = ξ([Skν ]q⋎(ξ))′ [Skν ]q⋎(ξ). If ξF′(ξ) F(ξ)
Theorem 6.6 Let ⋎∈ and F(ξ) = ξ([Skν ]q⋎(ξ))′ [Skν ]q⋎(ξ) . If ξF′(ξ) F(ξ)
Lemma 6.3
Lemma 6.3, we have P(ξ) = 1 + ([Sk ν ]q ⋎(ξ))′′ ([Skν ]q ⋎(ξ))′ – ξ([Sk ν ]q ⋎(ξ))′ [Skν ]q ⋎(ξ) ≺ψ(ξ). Consequently, we get ⋎∈Vq(ψ). □
Lemma 6.3, we have P(ξ) = 1 + ([Sk ν ]q ⋎(ξ))′′ ([Skν ]q ⋎(ξ))′ – ξ([Sk ν ]q ⋎(ξ))′ [Skν ]q ⋎(ξ) ≺ψ(ξ). Consequently, we get ⋎∈Vq(ψ). □
Theorem 6.7
Theorem 6.7 Let ⋎∈ and F(ξ) = ξ([Skν ]q⋎(ξ))′ [Skν ]q⋎(ξ). If ξF′(ξ) F(ξ)
Theorem 6.7 Let ⋎∈ and F(ξ) = ξ([Skν ]q⋎(ξ))′ [Skν ]q⋎(ξ) . If ξF′(ξ) F(ξ)
Lemma 6.4
Lemma 6.4, we have P(ξ) = 1 + ([Sk ν ]q ⋎(ξ))′′ ([Skν ]q ⋎(ξ))′ – ξ([Sk ν ]q ⋎(ξ))′ [Skν ]q ⋎(ξ) ≺ξ. Consequently, we obtain ⋎∈Vq(ξ). □ 7…
Lemma 6.4, we have P(ξ) = 1 + ([Sk ν ]q ⋎(ξ))′′ ([Skν ]q ⋎(ξ))′ – ξ([Sk ν ]q ⋎(ξ))′ [Skν ]q ⋎(ξ) ≺ξ. Consequently, we obtain ⋎∈Vq(ξ). □ 7 Conclusion In this paper, we presented different types of integral inequalities based on q-calculus and conformable differential operator. These inequalities described the relations between the quantum conformable differential operators for different orders. Acknowledgements The authors would like to thank the Associate Editor for her/his advice in preparing the a
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