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Abstract

In this paper, we derive several subordination results and integral means result for certain class of analytic functions defined by means of q- differential operator. Some interesting corollaries and consequences of our results are also considered.

Results & Lemmas (10)

Each result is auto-extracted from the paper via OCR - the math is shown verbatim and may carry occasional transcription artifacts. Click to expand the full statement.

Lemma 1.3 Lemma 1.3 ([28]) The sequence bn ∞ n=1 is a subordinating factor sequence if and only if ℜ  1 + 2 ∞ X n=1 bnzn > 0, z ∈∆.
Lemma 1.3 ([28]) The sequence {bn}∞ n=1 is a subordinating factor sequence if and only if ℜ  1 + 2 ∞ X n=1 bnzn > 0, z ∈∆.
Lemma 1.4 Lemma 1.4 Assume that ∞ X n=2 [n]m q [1 + (n −1)λ]ζ[n(β + 1) −(α + β)]|an| ≤1 −α, (4) then f ∈Sζ,m λ,q (α, β), where −1 ≤α < 1, β ≥0, λ > 0…
Lemma 1.4 Assume that ∞ X n=2 [n]m q [1 + (n −1)λ]ζ[n(β + 1) −(α + β)]|an| ≤1 −α, (4) then f ∈Sζ,m λ,q (α, β), where −1 ≤α < 1, β ≥0, λ > 0 and m, ζ ∈N0. The result is sharp for the function fn(z) = z − 1 −α [n]m q [1 + (n −1)λ]ζ[n(β + 1) −(α + β)]zn.
Theorem 2.1 Theorem 2.1 Let the function f be defined by (1) be in the class S∗,ζ,m λ,q (α, β), where −1 ≤α < 1, β ≥0, λ > 0, ζ ∈N0. Also let K denote…
Theorem 2.1 Let the function f be defined by (1) be in the class S∗,ζ,m λ,q (α, β), where −1 ≤α < 1, β ≥0, λ > 0, ζ ∈N0. Also let K denote the familiar class of functions f ∈A which are also univalent and convex in ∆. Then (1 + q)m(1 + λ)ζ(β + 2 −α) 2[1 −α + (1 + q)m(1 + λ)ζ(β + 2 −α)](f ∗g)(z) ≺g(z), z ∈∆, g ∈K, (5) and ℜ(f(z)) > −1 −α + (1 + q)m(1 + λ)ζ(β + 2 −α) (1 + q)m(1 + λ)ζ(β + 2 −α) , z ∈∆. (6)
Corollary 2.2 Corollary 2.2 Let f, defined by (1), be in the class M∗ λ(ζ, α, β), where −1 ≤α < 1, β ≥0, λ > 0, ζ ∈N0. Then (1 + λ)ζ(β + 2 −α) 2[1 −α + (1…
Corollary 2.2 Let f, defined by (1), be in the class M∗ λ(ζ, α, β), where −1 ≤α < 1, β ≥0, λ > 0, ζ ∈N0. Then (1 + λ)ζ(β + 2 −α) 2[1 −α + (1 + λ)ζ(β + 2 −α)](f ∗g)(z) ≺g(z) z ∈∆, g ∈K and ℜ(f(z)) > −1 −α + (1 + λ)ζ(β + 2 −α) (1 + λ)ζ(β + 2 −α) , z ∈∆. The constant (1+λ)ζ(β+2−α) 2[1−α+(1+λ)ζ(β+2−α)] is the best estimate.
Corollary 2.3 Corollary 2.3 Let f, defined by (1), be in the class β −UST (α). Then β + 2 −α 2(β + 3 −2α)(f ∗g)(z) ≺g(z), −1 ≤α < 1, β ≥0, z ∈∆, g ∈K and…
Corollary 2.3 Let f, defined by (1), be in the class β −UST (α). Then β + 2 −α 2(β + 3 −2α)(f ∗g)(z) ≺g(z), −1 ≤α < 1, β ≥0, z ∈∆, g ∈K and ℜ(f(z)) > −β + 3 −2α β + 2 −α , z ∈∆. The constant β+2−α 2(β+3−2α) is the best estimate.
Corollary 2.4 Corollary 2.4 Let f, defined by (1), be in the class β −UKV(α). Then β + 2 −α 2β + 5 −3α(f ∗g)(z) ≺g(z), −1 ≤α < 1, β ≥0, z ∈∆, g ∈K and…
Corollary 2.4 Let f, defined by (1), be in the class β −UKV(α). Then β + 2 −α 2β + 5 −3α(f ∗g)(z) ≺g(z), −1 ≤α < 1, β ≥0, z ∈∆, g ∈K and ℜ(f(z)) > −2β + 5 −3α 2(β + 2 −α), z ∈∆. The constant β+2−α 2β+5−3α is the best estimate. Putting m = 0, ζ = 0 and β = 0 in Theorem 2.1, we obtain the next two results obtained by Frasin [6].
Corollary 2.5 Corollary 2.5 Let f, defined by (1), be in the class S∗(α). Then 2 −α 6 −4α(f ∗g)(z) ≺g(z), z ∈∆, g ∈K and ℜ(f(z)) > −3 −2α 2 −α, z ∈∆. The…
Corollary 2.5 Let f, defined by (1), be in the class S∗(α). Then 2 −α 6 −4α(f ∗g)(z) ≺g(z), z ∈∆, g ∈K and ℜ(f(z)) > −3 −2α 2 −α , z ∈∆. The constant 2−α 6−4α is the best estimate.
Corollary 2.6 Corollary 2.6 Let f, defined by (1), be in the class K(α). Then 2 −α 5 −3α(f ∗g)(z) ≺g(z, ) z ∈∆, g ∈K and ℜ(f(z)) > −5 −3α 2(2 −α), z ∈∆.…
Corollary 2.6 Let f, defined by (1), be in the class K(α). Then 2 −α 5 −3α(f ∗g)(z) ≺g(z, ) z ∈∆, g ∈K and ℜ(f(z)) > −5 −3α 2(2 −α), z ∈∆. The constant 2−α 5−3α is the best estimate.
Lemma 3.1 Lemma 3.1 ([17]) If the functions f and g are analytic in ∆with g ≺f, then for η > 0, and 0 < r < 1, 2π Z 0 |g(reiθ)|ηdθ ≤ 2π Z 0…
Lemma 3.1 ([17]) If the functions f and g are analytic in ∆with g ≺f, then for η > 0, and 0 < r < 1, 2π Z 0 |g(reiθ)|ηdθ ≤ 2π Z 0 |f(reiθ)|ηdθ. In [22], Silverman found that the function f2(z) = z −z2 2 is often extremal over the family T and applied this function to resolve his integral means inequality, conjectured in [23] and settled in [24], that
Theorem 3.2 Theorem 3.2 Suppose f ∈Sζ,m λ,q (α, β), η > 0, and f2 is defined by f2(z) = z − 1 −α (1 + q)m[1 + λ]ζ[β + 2 −α]z2. Then for z = reiθ, 0 < r…
Theorem 3.2 Suppose f ∈Sζ,m λ,q (α, β), η > 0, and f2 is defined by f2(z) = z − 1 −α (1 + q)m[1 + λ]ζ[β + 2 −α]z2. Then for z = reiθ, 0 < r < 1 we have 2π Z 0 |f(z)|η dθ ≤ 2π Z 0 |f2(z)|η dθ.

Definitions (1)

Def 1.1 Definition 1.1 (Subordination Principle) Let g be analytic and univalent in ∆. If f is analytic in ∆, f(0) = g(0) and f(∆) ⊂g(∆), then the…
Definition 1.1 (Subordination Principle) Let g be analytic and univalent in ∆. If f is analytic in ∆, f(0) = g(0) and f(∆) ⊂g(∆), then the function f is subordinate to g in ∆and we write f ≺g. Definition 1.2 (Subordinating Factor Sequence) A sequence {bn}∞ n=1 of complex numbers is called a subordinating factor sequence if, whenever f is analytic , univalent and convex in ∆, we have the subordination given by ∞

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