Abstract
In this paper our aim is to establish some geometric properties (like
univalence, starlikeness, convexity and close-to-convexity) for the generalized Bessel
functions of the first kind. In order to prove our main results, we use the technique
of differential subordinations developed by Miller and Mocanu, and some classical
results of Ozaki and Fej´er.
Results & Lemmas (17)
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Lemma 1.1
Lemma 1.1 ([15]). Let E be a set in the complex plane C and ψ: C3×D 7→C a function, that satisfies the admissibility condition ψ(ρi, σ, µ +…
Lemma 1.1 ([15]). Let E be a set in the complex plane C and ψ : C3×D 7→C a function, that satisfies the admissibility condition ψ(ρi, σ, µ + νi; z) /∈E, where z ∈D, ρ, σ, µ, ν ∈R with µ+σ ≤0 and σ ≤−(1+ρ2)/2. If h : D →C, which satis- fies h(0) = 1, is analytic and for all z ∈D we have ψ ¡ h(z), zh′(z), z2h′′(z); z ¢ ∈E, then Re h(z) > 0 for all z ∈D. In particular, if we only have ψ : C2 × D 7→C, the admissibility condition reduces to ψ(ρi, σ; z) /∈E for all z ∈D and ρ, σ ∈R with σ ≤−(1 + ρ2)/2.
Lemma 1.2
Lemma 1.2 ([18]). If the function f(z) = z + a2z2 +... + anzn +... is analytic in D and in addition 1 ≥2a2 ≥... ≥nan ≥... ≥0 or 1 ≤2a2 ≤...…
Lemma 1.2 ([18]). If the function f(z) = z + a2z2 + . . . + anzn + . . . is analytic in D and in addition 1 ≥2a2 ≥. . . ≥nan ≥. . . ≥0 or 1 ≤2a2 ≤. . . ≤ nan ≤. . . ≤2, then f is close-to-convex with respect to the convex function z 7→ −log(1−z). Moreover, if the odd function g(z) = z+b3z3 +. . .+b2n−1z2n−1 +. . . is analytic in D and if 1 ≥3b3 ≥. . . ≥(2n + 1)b2n+1 ≥. . . ≥0 or 1 ≤3b3 ≤. . . ≤ (2n + 1)b2n+1 ≤. . . ≤2, then g is univalent in D. We note that, as Ponnusamy and Vuorinen [20] pointe
Lemma 1.3
Lemma 1.3 ([11]). If the function f(z) = a1z+a2z2+...+anzn+..., where a1 = 1 and an ≥0 for all n ≥2, is analytic in D and if the sequences…
Lemma 1.3 ([11]). If the function f(z) = a1z+a2z2+. . .+anzn+. . ., where a1 = 1 and an ≥0 for all n ≥2, is analytic in D and if the sequences {nan}n≥1, {nan −(n + 1)an+1}n≥1 both are decreasing, then f is starlike in D. Moreover, if for the analytic function g(z) = b1 + b2z + . . . + bn+1zn + . . ., where b1 = 1 and bn ≥0 for all n ≥2, we have that {bn}n≥1 is a convex decreasing sequence, i.e., bn −2bn+1 + bn+2 ≥0 and bn −bn+1 ≥0 for all n ≥1, then Re[g(z)] > 1/2 for all z ∈D. It is important t
Proposition 2.14.
Proposition 2.14. If b, p, c ∈C such that 2p + b + 1 ̸= 0, −2, −4,..., and z ∈C, then for the generalized Bessel function of the first kind…
Proposition 2.14. If b, p, c ∈C such that 2p + b + 1 ̸= 0, −2, −4, . . . , and z ∈C, then for the generalized Bessel function of the first kind of order p the following recursive relations hold: (i) zwp−1(z) + czwp+1(z) = (2p + b −1)wp(z); (ii) zw′ p(z) + (p + b −1)wp(z) = zwp−1(z); (iii) zw′ p(z) + czwp+1(z) = pwp(z); (iv) [z−pwp(z)]′ = −cz−pwp+1(z); (v) 2(2p + b + 1)u′ p(z) = −cup+1(z).
Theorem 2.16.
Theorem 2.16. If b, c, p ∈R and κ = p + (b + 1)/2, then the functions wp and up satisfy the following properties: (i) If κ ≥|c|/4 + 1, then…
Theorem 2.16. If b, c, p ∈R and κ = p + (b + 1)/2, then the functions wp and up satisfy the following properties: (i) If κ ≥|c|/4 + 1, then Re up(z) > 0 for all z ∈D; (ii) If κ ≥|c|/4 and c ̸= 0, then up is univalent in D; (iii) If κ ≥|c|/4 + 1/2 and c ̸= 0, then up is convex in D; (iv) If κ ≥|c|/4 + 3/2 and c ̸= 0, then z 7→zup(z) is starlike in D; (v) If κ ≥|c|/2 + 1 and c ̸= 0, then z 7→zup(z) is starlike of order 1/2 in D; (vi) If κ ≥|c|/2 + 1 and c ̸= 0, then z 7→z1−pwp(z) is starlike in D.
Lemma 1.1.
Lemma 1.1. Hence by Lemma 1.1 we conclude Re h(z) = Re up(z) > 0 for all z ∈D. (ii) When κ ≥|c|/4 and c ̸= 0, then the above result implies…
Lemma 1.1. Hence by Lemma 1.1 we conclude Re h(z) = Re up(z) > 0 for all z ∈D. (ii) When κ ≥|c|/4 and c ̸= 0, then the above result implies Re up+1(z) > 0 for all z ∈D. Using part (v) of Proposition 2.14 we conclude that Re · −4κ c u′ p(z) ¸ = Re up+1(z) > 0 for all z ∈D. This in turn implies that up is close-to-convex with respect to the function ϕ(z) = −(cz)/(4κ). Now, since every close to convex function is univalent, it follows that up is univalent in D. We note that the univalence of the fu
Corollary 2.22.
Corollary 2.22. Let Jp: D →C be defined by Jp(z) = 2pΓ(p+1)z−pJp(z). Then the following assertions are true: (i) If Re p ≥1/4, then Re…
Corollary 2.22. Let Jp : D →C be defined by Jp(z) = 2pΓ(p+1)z−pJp(z). Then the following assertions are true: (i) If Re p ≥1/4, then Re Jp(z1/2) > 0 for all z ∈D; (ii) If Re p ≥−3/4, then z 7→Jp(z1/2) is univalent in D; (iii) If Re p ≥−1/4 + (Im p)2/6, then z 7→Jp(z1/2) is convex in D; (iv) If Re p ≥3/4 + (Im p)2/6, then z 7→zJp(z1/2) is starlike in D; (v) If Re p ≥1/2 + (Im p)2/4, then z 7→zJp(z1/2) is starlike of order 1/2 in D;
Corollary 2.23.
Corollary 2.23. The following assertions are true: (i) If Re p ≥1/4, then Re Ip(z1/2) > 0 for all z ∈D; (ii) If Re p ≥−3/4, then z…
Corollary 2.23. The following assertions are true: (i) If Re p ≥1/4, then Re Ip(z1/2) > 0 for all z ∈D; (ii) If Re p ≥−3/4, then z 7→Ip(z1/2) is univalent in D; (iii) If Re p ≥−1/4 + (Im p)2/6, then z 7→Ip(z1/2) is convex in D; (iv) If Re p ≥3/4 + (Im p)2/6, then z 7→zIp(z1/2) is starlike in D; (v) If Re p ≥1/2 + (Im p)2/4, then z 7→zIp(z1/2) is starlike of order 1/2 in D; (vi) If Re p ≥1/2 + (Im p)2/4, then z 7→z1−pIp(z) is starlike in D.
Corollary 2.24.
Corollary 2.24. Let Sp: D →C be defined by Sp(z) =2pΓ(p + 3/2)z−pSp(z). Then the following assertions are true: (i) If Re p ≥−1/4, then Re…
Corollary 2.24. Let Sp : D →C be defined by Sp(z) =2pΓ(p + 3/2)z−pSp(z). Then the following assertions are true: (i) If Re p ≥−1/4, then Re Sp(z1/2) > 0 for all z ∈D; (ii) If Re p ≥−5/4, then z 7→Sp(z1/2) is univalent in D; (iii) If Re p ≥−3/4 + (Im p)2/6, then z 7→Sp(z1/2) is convex in D; (iv) If Re p ≥1/4 + (Im p)2/6, then z 7→zSp(z1/2) is starlike in D; (v) If Re p ≥(Im p)2/4, then z 7→zSp(z1/2) is starlike of order 1/2 in D; (vi) If Re p ≥(Im p)2/4, then z 7→z1−pSp(z) is starlike in D. 3. Con
Theorem 3.1.
Theorem 3.1. If 0 ≤α < 1/2 and b, p, c ∈R, then the following assertions are true: (i) If 4κ ≥(1 −α)(1 −2α)−1/2|c| + 1, then Re up(z) > α…
Theorem 3.1. If 0 ≤α < 1/2 and b, p, c ∈R, then the following assertions are true: (i) If 4κ ≥(1 −α)(1 −2α)−1/2|c| + 1, then Re up(z) > α for all z ∈D; (ii) If 4κ ≥(1 −α)(1 −2α)−1/2|c| and c ̸= 0, then up is close-to-convex of order α in D.
Proposition 2.14
Proposition 2.14 we conclude that Re ·µ −4κ c ¶ u′ p(z) ¸ = Re up+1(z) > α for all z ∈D, i.e. up is close-to-convex of order α in D with…
Proposition 2.14 we conclude that Re ·µ −4κ c ¶ u′ p(z) ¸ = Re up+1(z) > α for all z ∈D, i.e. up is close-to-convex of order α in D with respect to the function ϕ(z) = −(cz)/(4κ). □
Theorem 3.3.
Theorem 3.3. If 0 ≤α < 1 and b, p, c ∈R such that c ̸= 0 and 4α2+ (|c| −6)α + 2 ≥0, then the functions wp and up have the following…
Theorem 3.3. If 0 ≤α < 1 and b, p, c ∈R such that c ̸= 0 and 4α2+ (|c| −6)α + 2 ≥0, then the functions wp and up have the following properties: (i) If 4(1 −α)κ ≥|c| + 2(1 −α)(1 −2α), then up is convex of order α in D;
Lemma 1.1
Lemma 1.1 to prove that Re q(z) > 0 for all z ∈D. For z = x + iy ∈D (with x, y ∈R) and ρ, σ ∈R satisfying σ ≤−(1 + ρ2)/2, we obtain Re…
Lemma 1.1 to prove that Re q(z) > 0 for all z ∈D. For z = x + iy ∈D (with x, y ∈R) and ρ, σ ∈R satisfying σ ≤−(1 + ρ2)/2, we obtain Re ψ(ρi, σ; x + iy) = 4(1 −α)σ −4(1 −α)2ρ2 + cx −2(1 −α)e2 ≤−2(1 −α)(3 −2α)ρ2 + cx −2(1 −α)(1 + e2) < |c| + 2(1 −α)(1 −2α) −4(1 −α)κ ≤0. By Lemma 1.1 we conclude that Re q(z) > 0 for all z ∈D. This result implies Re · 1 + zu′′ p(z) u′p(z) ¸ = (1 −α) Re q(z) + α > α for all z ∈D, which shows that up is convex of order α in D.
Theorem 4.1.
Theorem 4.1. If c < 0 and b, p ∈R, then up has the following properties:
Theorem 4.1. If c < 0 and b, p ∈R, then up has the following properties:
Lemma 1.2
Lemma 1.2 it follows that f is close-to-convex with respect to the convex function −log(1 −z). (ii) Set g(z) = zup(z2) = z+b3z3…
Lemma 1.2 it follows that f is close-to-convex with respect to the convex function −log(1 −z). (ii) Set g(z) = zup(z2) = z+b3z3 +. . .+b2n−1z2n−1 +. . ., where bn is defined by (2.11). Therefore we have 3b1 = −(3c)/(4κ) ≤1 and b2n−1 > 0 for all n ≥2. We want to show that {(2n −1)b2n−1}n≥2 is a decreasing sequence. Fix n ≥2. Then we have (2n −1)b2n−1 −(2n + 1)b2n+1 = b2n−1 · U2(n) 4n(κ + n −1), where U2(n) = 8n3 + 8(κ −3/2)n2 −4(κ −c/2 −1)n + c. Using the inequalities n3 ≥3n2 −3n + 1 and n2 ≥2n −1
Theorem 4.2.
Theorem 4.2. Let b, p be arbitrary real numbers and let c < 0. If 4κ ≥ −(c + 2) + p c2/2 −4c + 4, then the Alexander transform of the…
Theorem 4.2. Let b, p be arbitrary real numbers and let c < 0. If 4κ ≥ −(c + 2) + p c2/2 −4c + 4, then the Alexander transform of the function z 7→ zup(z) is close-to-convex with respect to the function −log(1−z) and it is starlike in D. Moreover, we have that Re up(z) > 1/2 holds for all z ∈D.
Corollary 4.5.
Corollary 4.5. If b, p are arbitrary real numbers and c < 0 such that κ ≥−c/4 −1 and κ ̸= 0, then the function up is univalent in D.
Corollary 4.5. If b, p are arbitrary real numbers and c < 0 such that κ ≥−c/4 −1 and κ ̸= 0, then the function up is univalent in D.
Function classes studied:
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