On Fuzzy Differential Subordination

Proposition 1.1 ([3]). If (M, FM) = (N, FN), then we have M = N, where M = supp(M, FM), N = supp(N, FN).

Proposition 1.2 ([3]). If (M, FM) ⊆(N, FN), then we have M ⊆N, where M = supp(M, FM), N = supp(N, FN). We also need the following notations and results from the classical complex analysis [5]. For D ⊂C, we denote by H(D) the class of holomorphic functions on D, and by Hn(D) the class of holomorphic and univalent functions on D. In this paper, we denote by H(U) the set of holomorphic functions in the unit disc U = {z ∈C : |z| < 1} with ∂U = {z ∈C : |z| = 1} the boundary of the unit disc. For a ∈C

Proposition 1.3 ([5]). Let Let D ⊂C, z0 ∈D be a fixed point, and let the functions f, g ∈H(D). If f(z) < Fg(z), z ∈D, then 1. f(z0) = g(z0), 2. f(D) ⊆g(D), where f(D) = supp(f(D), Ff(D)), g(D) = supp(g(D), Fg(D)). The equality occurs if and only if Ff(D)f(z) = Fg(D)g(z). Denoted by S∗= {f ∈A : Rezf ′(z) f(z) > 0, z ∈U} the class of normalized starlike functions in U, K = {f ∈A : Rezf ′′(z) f′(z) + 1 > 0, z ∈U} the class of normalized convex functions in U and by C = {f ∈A : ∃ϕ ∈K, Re f ′(z)

Theorem 1.1 ([5]). Let h be analytic in U, let φ be analytic in domain D containing h(U) and suppose a) Reφ[h(z)] > 0, z ∈U and b) h(z) is convex. If p is analytic in U , with p(0) = h(0), p(U) ⊂D and ψ(C2 ×U) →C, ψ(p(z), zp′(z)) = p(z)+zp′(z).φ[p(z)] is analytic in U, then Fψ(C2×U)ψ(p(z), zp′(z)) ≤Fh(U)h(z), implies Fp(U)p(z) ≤Fh(U)h(z), z ∈U, where ψ(C2 × U) = supp(C2 × U, Fψ(C2×U)ψ(p(z), zp′(z)) = {z ∈C : 0 < Fψ(C2×U)ψ(p(z), zp′(z)) ≤1}, h(U) = supp(U, Fh(U)h(z)) = {z ∈C : 0 < Fh(U)h(z) ≤1}.

Theorem 1.2 ([5]). Let h be convex in U and let P : U →C, with ReP(z) > 0. If p is analytic in U and ψ : C2 × U →C, ψ(p(z), zp′(z)) = p(z) + P(z)zp′(z) is analytic in U, then Fψ(C2×U)[p(z) + P(z)zp′(z)] ≤Fh(U)h(z), implies Fp(U)P(z) ≤Fh(U)h(z), z ∈U.

Theorem 1.3 ([5]). (Hallenbeck and Ruscheweyh) Let h be a convex func- tion with h(0) = a, and let γ ∈C∗be a complex number with Reγ ≥0. If p ∈ H[a, n] with p(0) = a and ψ : C2 ×U →C, ψ(p(z)+zp′(z)) = p(z)+ 1 γ zp′(z) is analytic in U, then Fψ(C2×U)[p(z) + 1 γ zp′(z)] ≤Fh(U)h(z), implies Fp(U)p(z) ≤Fq(U)q(z) ≤Fh(U)h(z), z ∈U,

Proposition 2.1. Let q be univalent in U and let θ and φ be analytic in a domain D containing q(U), with φ(w) ̸= 0, when w ∈q(U). Set Q(z) = zq′(z).φ[q(z)] and h(z) = θ[q(z)] + Q(z) and suppose that either (i) Q is starlike, or (ii) h is convex. In addition, assume that (iii) Re(zh′(z) Q(z ) = Re(θ′[q(z)] φ[q(z)] + zQ′(z) Q(z) ) > 0. If p is analytic in U, with p(0) = q(0), p(U) ⊂D and ψ : C2 × U →C, ψ(p(z), zp′(z)) = p(z) + zp′(z).φ(p(z)) is analytic in U, then Fψ(C2×U)[p(z) + zp′(z).φ(p(z)] ≤F

Proposition 2.2. Let q ∈H[p, p] be univalent, q(z) ̸= 0 and satisfies the following conditions. (i) zq′(z) q(z) is starlike, (ii) Re(q(z) α + 1 + zq ′′(z) q′(z) −zq′(z) q(z) ) > 0 for all α ̸= 0 and for all z ∈U. For p ∈H[p, p] with p(z) ̸= 0 in U and ψ : C2 × U →C, ψp(z), zp′(z)) = p(z) + α zp′(z) p(z) is analytic in U, then Fψ(C2×U)[p(z) + α zp′(z) p(z) ] ≤Fψ(C2×U)[q(z) + α zq′(z)

Proposition 2.3. Let q ∈H[p, p] be univalent, q(z) ̸= 0 and satisties the conditions: (i) zq′(z) q(z) is starlike, (ii) Re(q(z) α + 1 + zq ′′(z) q′(z) −zq′(z) q(z) ) > 0 for α ̸= 0 and for all z ∈U. For f ∈Ap with J(α, f; z) = (1 −α)zf′(z) f(z) + α(1 + zf ′′(z) f′(z) ), z ∈U and ψ : C2 × U →C,

Proposition 2.4. Let q ∈H[1, 1] be univalent and satisfies the following conditions: (i) q(z) is convex, (ii) Re[( 1 α + ρ) + zq ′′(z) q′(z) ] > 0 ρ ∈N = {1, 2, 3, ..}) for α ̸= 0 and for all z ∈U. For p ∈H[1, 1] in U and ψ : C2 × U →C, ψ(p(z), zp′(z)) = (1 −α + αρ)p(z) + αzp′(z) is analytic in U, then Fψ(C2×U)[(1 −α + αρ)(p(z) + αzp′(z)] ≤ Fψ(C2×U)[(1 −α + αρ)q(z) + αzq′(z)] = Fh(U)h(z), implies Fp(U)p(z) ≤Fq(U)q(z), and q is the best dominant.

Proposition 2.1, Fp(U)p(z) ≤Fq(U)q(z), z ∈U, and q is the best dominant. □

Theorem 2.1. Let q ∈H[1, 1] be univalent and satisfies the following con- ditions: (i) q(z) is convex, (ii) Re[( 1 α + ρ) + zq ′′(z) q′(z) ] > 0 (ρ ∈N = {1, 2, 3, ..}) for α ̸= 0 and for all z ∈U. For f ∈Ap with J(α, f; z) = (1 −α)zf′(z) f(z) + α(1 + zf ′′(z) f′(z) ), z ∈U and if ψ : C2 × U →C, ψ(q(z), zq′(z)) = (1 −α + αρ)q(z) + µzq′(z), then Fψ(C2×U)(f(z)

Corollary 2.1. Let q ∈H[1, 1] be univalent and satisfies the following con- ditions: (i) q(z) is convex, (ii) Re[( 1 α + 1) + zq ′′(z) q′(z) ] > 0 (ρ ∈N = {1, 2, 3, ..}) for α ̸= 0 and for all z ∈U. For p ∈H[1, 1] in U, if ψ : C2 × U →C ψ(p(z), zp′(z)) = p(z) + αzp′(z), then Fψ(C2×U)p(z) ≤Fq(U)q(z), z ∈U, and q is the best dominant.

Corollary 2.2. Let q ∈H[1, 1] be univalent, q(z) is convex for all z ∈U. For p ∈H[1, 1] in U if ψ : C2 × U →C, ψ(p(z), zp′(z)) = p(z) + zp′(z), then Fψ(C2×U)p(z) ≤Fψ(C2×U)q(z), z ∈U, and q is the best dominant.

Corollary 2.3. Let q ∈H[1, 1] be univalent, q(z) is convex for all z ∈U. For p ∈H[1, 1] in U if ψ : C2 × U →C, ψ(p(z), zp′(z)) = ρp(z) + zp′(z), (ρ ∈N = {1, 2, 3, ..}), then Fψ(C2×U)p(z) ≤Fψ(C2×U)q(z), z ∈U, and q is the best dominant. References [1] Ö.Ö. Kılıç, Sufficient Conditions for Subordination of Multivalent Functions, Journal of Inequalities and Applications, 2008 (2008), ArticleID 374756, 8 pages. [2] S.S. Miller and P.T. Mocanu, Differential subordinations, Theory and applications, Marce