MATHEMATICA SCANDINAVICA 109(2)(2011), 253-268 ——————————————————— CERTAIN RESULTS ON SUBORDINATIONS ASSOCIATED WITH DIFFERINTEGRAL OPERATOR ´L JACEK DZIOK, RAVINDER KRISHNA RAINA, JANUSZ SOKO Abstract. We investigate several properties of the linear differintegral Dziok–operator and a associated class of multivalent analytic functions which generalize that one introduced and studied by J. Dziok, Applications of the Jack Lemma, Acta Math. Hungar. 105(1-2)(2004) 93–102. Several theorems are an extension of earlier results of the above paper. 1. Introduction Let A(p), p ∈ N, denote the class of functions of the form ∞ X p (1) f (z) = z + an+p z n+p n=1 which are analytic and p-valent in the open unit disc U = {z : |z| < 1} on the complex plane C. For two functions f, g ∈ A(p), we say that f is subordinate to g, written as f ≺ g, if and only if there exists an analytic Schwarz function ω, ω(0) = 0 and |ω(z)| < 1 in U such that f (z) = g(ω(z)). In particular, if g is univalent in U, then we have the following equivalence (2) f (z) ≺ g(z) ⇐⇒ f (0) = g(0) and f (|z| < 1) ⊂ g(|z| < 1). The idea of subordination was used for defining many classes of functions studied in geometric function theory. Let us recall zf 0 (z) ∗ (3) Sp [ϕ] := f ∈ A(p) : ≺ ϕ(z), z ∈ U , f (z) zf 00 (z) (4) Kp [ϕ] := f ∈ A(p) : 1 + 0 ≺ ϕ(z), z ∈ U , f (z) 1+z where ϕ is analytic in U with ϕ(0) = p. For p = 1 and ϕ(z) = 1−z we obtain the well known classes S ∗ , K of starlike and convex functions, respectively. For other functions ϕ we can obtain classes investigated many times earlier. If we restrict our attention to the functions ϕ which map U onto a disc or a half-plane then we obtain the classes 1 + Az ∗ (5) Sp p , −1 ≤ B < A ≤ 1, 1 + Bz and 1 + Az Kp p , −1 ≤ B < A ≤ 1, 1 + Bz introduced and investigated for p = h 1 by W.i Janowski [5]. For h p = 1, iB = −1 and A = 1 − 2α ∗ 1+(1−2α)z ∗ these classes become the classes S1 = S (α) and K1 1+(1−2α)z = K(α) of α-starlike and 1−z 1−z α-convex functions what were introduced by Robertson [10]. 2000 Mathematics Subject Classification. Primary 30C45, secondary 30C80. Key words and phrases. Hadamard product (or convolution); Differential subordination; Linear operator; Convolution operator; Analytic multivalent functions; Starlike function; Convex function. 1 ´L JACEK DZIOK, RAVINDER KRISHNA RAINA, JANUSZ SOKO 2 One can consider the conditions joining zf 0 (z)/f (z) with zf 00 (z)/f 0 (z). In this way Mocanu introduced in [6] the class zf 00 (z) zf 0 (z) 1+z (6) M(α) := f ∈ A : Re α 1 + 0 + (1 − α) ≺ for z ∈ U f (z) f (z) 1−z of convex functions of order α, α ∈ R , which give a continuous passage from convex to starlike functions. In this paper we give a generalization of main theorems contained in [3]. Some applications involving these results are also considered. Let us consider the Dziok differintegral operator Ωαβ [3] defined for f ∈ A(p) : Z z 1−α−β z α Ωβ f (z) = (z − ξ)α−1 ξ β−1 f (ξ)dξ (Reα > 0, Reβ > −p), Γ(α) 0 Z z z 1−α−β d α Ωβ f (z) = (z − ξ)α ξ β−1 f (ξ)dξ (−1 < Reα ≤ 0, Reβ > −p), Γ(1 + α) dz 0 γ dm Ωβ f (z) =z (Reα ≤ −1, α = γ − m, −1 < Reγ ≤ 0, Reβ > −p, m ∈ N). dz m z 1−α−β−m The multiplicities of (z − ξ)α−1 , (z − ξ)γ are removed by requiring log(z − ξ) ∈ R for z − ξ > 0. If f (z) = z p + ap+1 z p+1 + · · · , then in terms the gamma function Γ the operator Ωαβ becomes Ωαβ f (z) (7) 1−α−β Ωαβ f (z) ∞ X Γ(n + β)an n = z Γ(n + β + α) n=p ( Γ(β) P∞ (β) a n n n z f or α + β 6= 0 Γ(β+α) n P∞n=p(β)(β+α) = n an n Γ(β) n=p (n−1)! z f or α + β = 0 hP i P∞ ∞ (β)n n n Γ(β) z ∗ an z f or α + β = 6 0 Γ(β+α) hP n=p (β+α)n P n=p i = ∞ ∞ (β) n n n Γ(β) f or α + β = 0 n=p (n−1)! z ∗ n=p an z (z ∈ U), where (x)n is the Pochhammer symbol Γ(x + n) 1 f or n = 0, x 6= 0 (x)n = = x(x + 1)...(x + n − 1) f or k ∈ N = {1, 2, 3, ...}, Γ(x) and ∗ denotes the Hadamard product "∞ # "∞ # "∞ # X X X an z n ∗ bn z n = an bn z n . n=p n=p n=p Ruscheweyh [11] introduced an operator Dλ : A(1) → A(1), λ ≥ −1 such that (8) ∞ X z (λ + 1)n n+1 D f (z) = ∗ f (z) = z + z , λ+1 (1 − z) n! n=1 λ which implies that for n ∈ N ∪ {0} z(z n−1 f (z))(n) n! n so D f (z) is called the Ruscheweyh differential operator. Thus we observe from (8) and (7) that for f ∈ A(p) 1 z p−1 Dλ [z 1−p f (z)] = Ω−λ f (z). Γ(λ + 1) 1+λ Dn f (z) = MATHEMATICA SCANDINAVICA 109(2)(2011), 253-268———————————————————CERTAIN RESU Next we recall the generalized Bernardi integral operator Jβ,ν : A(1) → A(1), defined by (cf. [1]) Zz ν+β (9) Jβ,ν [f (z)] = tν−1 f β (t)dt (β, ν ∈ C ), zν 0 for β = 1 it becomes the Bernardi operator and with f (z) = z + a2 z 2 + a3 z 3 + . . . we have ∞ X ν+1 J1,ν [f (z)] = an z n ν + n n=1 while for f ∈ A(p) we have 1+ν (ν)n Ω1ν f (z). ν The Bernardi operator becomes the Livingston operator [9] or the Libera operator [8] for special choices of β and ν. By using the definitions of fractional calculus, Srivastava and Owa [15] (see also [2],[13],[14]) have b λ : A(1) → A(1), by defined the linear operator Ω b λ f (z) = Γ(2 − λ)z λ Dzλ f (z) Ω (λ 6= 2, 3, 4, . . .), z p−1 J1,ν [z 1−p f (z)] = where Dzλ f (z) = 1 d Γ(1−λ) dz dn Rz 0 f (ζ) dζ (z−ζ)λ (0 ≤ λ < 1) Dzλ−n f (z) (n ≤ λ < n + 1; n = 0, 1, 2, . . .) and (z − ζ)λ is removed by requiring log(z − ζ) to be real when z − ζ > 0. Then it is easily observed that for f ∈ A(p) b λ [z 1−p f (z)] = (1 − λ)Γ(1 − λ)Ω1 f (z). z p−1 Ω 1−λ Following the idea of Mocanu class M(α) let us consider for σ ∈ [0, 1] and h(z) = 1+a1 z+a2 z 2 +· · · the class of functions ( " # ) α 00 α 0 z(Ω f (z)) z(Ω f (z)) β β (10) J (σ, Ωαβ ; h) := f ∈ A(p) : σ 1 + + (1 − σ) ≺ ph(z), z ∈ U (Ωαβ f (z))0 Ωαβ f (z) dz n In this paper we shall focus particulary on the class α α 1 + Az (11) J (σ, Ωβ ; A, B) := J σ, Ωβ ; , 1 + Bz where A, B ∈ C, |B| ≤ 1. The class J (σ, Ωαβ ; A, B) give a continuous passage from J (0, Ωαβ ; A, B) to J (1, Ωαβ : A, B). Notice that for A = −B = 1 the class J (0, Ωαβ ; 1, −1) becomes the class Gp (α, β) considered in [3] and J (σ, Ω0β ; 1, −1) = M(σ), where M(σ) is the Mocanu class (6). After some calculation we can obtain from (7) (12) Ωα−1 f (z) z(Ωαβ f (z))0 β = + 1 − (α + β) Ωαβ f (z) Ωαβ f (z) (13) (Ωα−1 f (z))0 z(Ωαβ f (z))00 β 1+ = + 1 − (α + β). (Ωαβ f (z))0 (Ωαβ f (z))0 Therefore the class (10) becomes ( (14) J (σ, Ωαβ ; h) = 2. Main results " # ) α−1 0 z(Ωα−1 f (z)) Ω f (z) β β f ∈ A(p) : σ + (1 − σ) α + 1 − (α + β) ≺ ph(z) . (Ωαβ f (z))0 Ωβ f (z) ´L JACEK DZIOK, RAVINDER KRISHNA RAINA, JANUSZ SOKO 4 It is easy to verify that for the functions h(z) = 1+a1 z +a2 z 2 +· · · and F (z) = cp z p +p+1 z p+1 +· · · which are analytic in U the subordination zF 00 (z) zF 0 (z) σ 1+ 0 + (1 − σ) ≺ ph(z) (z ∈ U) F (z) F (z) can be rewritten as the Briot–Bouquet type differential subordination (15) s(z) + σzs0 (z) ≺ h(z) (z ∈ U), ps(z) where (16) s(z) = zF 0 (z) = 1 + b1 z + · · · pF (z) (z ∈ U). Therefore, the subordination in (10) is of the Briot–Bouquet type differential subordination and we can write σzs0 (z) α (17) J (σ, Ωβ ; h) = f ∈ A(p) : s(z) + ≺ h(z), z ∈ U , ps(z) where 0 z Ωαβ f (z) s(z) = = 1 + b1 z + · · · pΩαβ f (z) (18) (z ∈ U). One of the basic results in the theory of Briot–Bouquet differential subordinations is a theorem from [4], which says (in its particular case) that if h is convex univalent with positive real part in U and the functions s, h satisfy (15) with real σ and p, then s(z) ≺ h(z). Thus by (17) we directly get the following theorem. Theorem 1. If h is convex univalent with positive real part in U and f ∈ J (σ, Ωαβ ; h), then 0 z Ωαβ f (z) (19) ≺ h(z) (z ∈ U). pΩαβ f (z) Corollary 1. Under the assumptions of Theorem 1 we have J (σ, Ωαβ ; h) ⊂ J (0, Ωαβ ; h) for all σ ∈ (0, 1]. Recall that J (0, Ωαβ ; 1, −1) = Gp (α, β), where the class Gp (α, β) was considered in [3]. Let f be given by (1). If we denote (20) h(z) = pz 1−p Ωαβ f (z) α 0 Ωβ f (0) ∞ Γ(β + α + p) X Γ(k + β + p − 1) = ak+p−1 z k Γ(β + p) k=1 Γ(k + β + α + p − 1) ∞ β + α + p − 1 X (β + p − 1)k = ak+p−1 z k β + p − 1 k=1 (β + α + p − 1)k then we can rewrite (19) in the form (21) zh0 (z) + (p − 1)h(z) ≺ h(z) (z ∈ U). ph(z) (z ∈ U), MATHEMATICA SCANDINAVICA 109(2)(2011), 253-268———————————————————CERTAIN RESU Thus for h(z) = (22) 1+Az , 1+Bz |B| ≤ 1, (21) gives zh0 (z) 1 + Az 1 + z(p(A − B) + B) ≺p +1−p= h(z) 1 + Bz 1 + Bz (z ∈ U). Therefore we have thus proved the following corollary. Corollary 2. Suppose that |B| ≤ 1. If a function f has the form (1) and belongs to the class J (σ, Ωαβ ; A, B), then the function (20) satisfies (22). This means that h belongs to the class ∗ 1 + z(p(A − B) + B) S1 1 + Bz defined in (5). Theorem 1 can sometimes be improved whenever we know more about the function h. If h is a special bilinear transformation a result better than (19) will be derived by employing the results from the book [7]. After some adaptation Theorem 3.2j ([7], p. 97) becomes the following lemma. Lemma 1.[7] Suppose that β > 0 and n is a positive integer. Let us denote Rβ,n (z) = β 1+z 2nz + 1 − z 1 − z2 (z ∈ U) and let h be a convex univalent function in U, with h(0) = 1 such that (23) βh(z) ≺ Rβ,n (z) (z ∈ U), then the Briot–Bouquet differential equation q(z) + nzq 0 (z) = h(z) (z ∈ U) βq(z) has a univalent solution qn analytic in U. Furthermore if the functions h and s(z) = 1 + an z n + an+1 z n+1 + · · · satisfy the Briot–Bouquet differential subordination s(z) + zs0 (z) ≺ h(z) (z ∈ U) , βs(z) then (24) s(z) ≺ qn (z) ≺ h(z) (z ∈ U) , and the function qn is the best dominant of the subordination (24) in the sense that if there exists a function p such that s(z) ≺ p(z), then qn (z) ≺ p(z). Notice that the function Rβ,n is called the open door function and it is univalent in U, Rβ,n (0) = β. Thus, by (2) in order to verify (23) it is sufficient to show that h(0) = 1 and βh(U) ⊂ Rβ,n (U). p The set Rβ,n (U) is the complex plane with slits along the half-lines Rew = 0 and |Imw| ≥ γ = n 1 + 2β/n. See Fig.1 below. Theorem 2. Suppose that A, B ∈ C, |B| ≤ 1, σ ∈ (0, 1], p ∈ N and one of the following conditions is satisfied iϕ (i) B = eiϕ , A = re f or some r ∈ (−1, p ∞) \ {1}, ϕ ∈ R, 2 (ii) |B| < 1 and Im[AB] ≥ (1 − |B| ) 1 + 2p/σ and Re[1 − AB] ≥ |A − B|, p (iii) |B| < 1 and Im[AB] < (1 − |B|2 ) 1 + 2p/σ 2 p 2 and Re[1 − AB] + Im[AB] − σ (1 − |B|2 ) 1 + 2p/σ ≥ |A − B|. p ´L JACEK DZIOK, RAVINDER KRISHNA RAINA, JANUSZ SOKO 6 If f (z) = ap z p + ap+1 z p+1 + · · · belongs to the class J (σ, Ωαβ ; A, B), then the function (18) satisfies 0 z Ωαβ f (z) σ 1 + Az s(z) = ≺ ≺ α pΩβ f (z) pgp (z) 1 + Bz (25) (z ∈ U), where (26) gp (z) = 1 R 1+Btz σp ( BA −1) p −1 t σ dt, if B 6= 0 1+Bz 0 R1 pA(t−1)z p −1 e σ t σ dt, if B = 0. 0 and the function σ pgp (z) is the best dominant of the subordination (25). Proof. We shall to use Lemma 1 with β = p/σ, n = 1 and h(z) = (1 + Az)/(1 + Bz), |B| ≤ 1. Then (23) becomes (27) p 1 + Az p1+z 2z ≺ Rp/σ,n (z) = + (z ∈ U) . σ 1 + Bz σ 1 − z 1 − z2 We will try to find A and B such that (27) is satisfied for given p and σ. The right-hand side of (27) is the open door function which is univalent in U. By (2) in order to verify (27) it is sufficient to show that p h(U) ⊂ Rp/σ,n (U). σ (28) For |B| = 1, A 6= B the set h(U) is the half-plane bounded by 1 + A/B 1 − A/B + it, t ∈ R . (29) ∂h(U) = w ∈ C : w = 2 2 So in this case (28) is satisfied whenever ∂h(U) is parallel to the imaginary axis and on the right of its, what gives 1 − A/B 1 + A/B (30) |B| = 1, A 6= B, Im = 0 and Re ≥ 0. 2 2 After some calculation (30) becomes the condition (i). The problem of finding A and B such that (27) is satisfied is somewhat more complicated when |B| < 1. Notice that in this case the function h(z) = (1 + Az)/(1 + Bz) maps the open unit disc U onto the disc D(C, R) = {w ∈ C : |w − C| < R}, where (31) C= 1 − AB , 1 − |B|2 R= |A − B| . 1 − |B|2 To verify (28) we should to find C and R such that the disc σp D(C, R) contains p/σ and lies on the right half-plane or such that the disc σp D(C, R) contains p/σ and contains a little of imaginary axis p outside the slits Rew = 0 and |Imw| ≥ γ = 1 + 2p/σ. This means that σp D(C, R) lies partially on the open door, see Fig 1. MATHEMATICA SCANDINAVICA 109(2)(2011), 253-268———————————————————CERTAIN RESU 6 Im n = 1, p = 2, σ = 1/2 γ = 3, A= B= p/σ = 4 −13 √ , 4 10 −3 √ , 4 10 p C σ p R σ = 1, = = √ p σ γi '$ s =4 q qq 1 ε 10, ε = 1 + √ s &% - Re −γi 10 Fig.1. Rβ,n (U) It all depends on the situation the center C and on the length of the radius R given in (31). If h p i p (32) Im C ≥ γ = 1 + 2p/σ, σ then the center pC/σ lies outside of the horizontal strip between ±γi and (28) is satisfied whenever hp i p (33) Re C ≥ R. σ σ The requirements |B| < 1, (32) and (33) grow (ii) after some calculations. If h p i p (34) Im C < γ = 1 + 2p/σ, σ then the center pC/σ lies on of the horizontal strip between ±γi and (28) is satisfied under a slightly weaker conditions than (ii). In this case the disc σp D(C, R) can lie a little on the line segment from −γi to γi. Thus the requirements p p p p (35) C − γi ≥ R and C + γi ≥ R σ σ σ σ are sufficient for (28) in this case. Joining conditions |B| < 1, (34) and (35) we get (iii). If we consider γ ∈ [0, 1) and f ∈ J (σ, Ωαβ ; A, B) with A = 1 − 2γ , B = −1, then the condition (i) p is satisfied and Theorem 2 gives α 0 2γ 1 + 1 − z z Ωβ f (z) p σ (36) ≺ ≺ (z ∈ U), pΩαβ f (z) pgp (z) 1−z where Z1 (37) gp (z) = 1 − tz 1−z 2(γ−1) σ p t σ −1 dt. 0 From the subordination (36) we find that 0 z Ωαβ f (z) Re >γ Ωαβ f (z) so we obtain a result below. (z ∈ U), ´L JACEK DZIOK, RAVINDER KRISHNA RAINA, JANUSZ SOKO 8 Corollary 3. If γ ∈ [0, 1) and f ∈ J (σ, Ωαβ ; A, B) with A = 1 − h(z) = pΩαβ f (z) [Ωαβ f ]0 (0) 2γ , p B = −1, then the function (z ∈ U) is a p−valently starlike function of order γ. Using the known identity Z2 tb−1 (1 − t)c−b−1 (1 − zt)−a dt = Γ(b)Γ(c − b) 2 F1 (a, b, c; z) Rec > Reb > 0 Γ(c) 0 with a = 2(1−γ) , b = σp , c = σ+p−2(1−γ) we can express the best dominant (37) of the subordination σ σ (36) in terms of the hypergeometric function 2 F1 (a, b, c; z) σ−2(1−γ) p 2(γ−1) Z1 Γ Γ σ p σ σ 1 − tz 2(1 − γ) p σ + p − 2(1 − γ) −1 2 F1 gp (z) = t σ dt = , , ;z . 1−z σ σ σ Γ σ+p−2(1−γ) 0 σ The sharp order of p−valently starlikeness is min{Regp (z) : |z| < 1}. The proof that it is gp (−1) one can find in [7], pp. 112–114. Theorem 3. Let h be a convex function. If 0 ≤ σ1 ≤ σ2 ≤ 1, then (38) J (σ2 , Ωαβ ; h) ⊂ J (σ1 , Ωαβ ; h). Proof. If σ1 = 0, then Corollary 1 follows (38). Suppose that σ1 > 0, f ∈ J (σ2 , Ωαβ ; h) and s is given by (18). Then σσ12 ∈ (0, 1] and σ1 zs0 (z) σ1 σ2 zs0 (z) σ1 (39) s(z) + = s(z) + + 1− s(z) ≺ h(z) ps(z) σ2 ps(z) σ2 because by (17) and by (19) in (39) there is a convex combination of functions subordinated to the convex function h so this combination is subordinated to h too. Therefore f ∈ J (σ1 , Ωαβ ; h). Lemma 2. Suppose that f ∈ J (σ, Ωαβ ; A, B) and one of the following conditions is satisfied (j) B = eiϕ , A = reiϕ f or some r ∈ (1 − p2 , ∞) \ {1}, ϕ ∈ R, (jj) |B| < 1 and Re 1 − [p(A − B) + B]B ≥ p|A − B|. 0 Then the function h(z) = pz 1−p Ωαβ f (z)/ Ωαβ f (0) given in (20) belongs to the class S ∗ of starlike functions. Proof. By Corollary 2 the function h given in (20) satisfies (22). Thus h is a starlike function whenever the function 1 + z(p(A − B) + B) g(z) = (z ∈ U) 1 + Bz satisfies Re[g(z)] > 0 for z ∈ U. If |B| = 1, then the set g(U) is a half-plane. Repeating the consideration from the first part of the proof of Theorem 2 we obtain that the conditions 1 − [p(A − B) + B]/B 1 + [p(A − B) + B]/B |B| = 1, A 6= B, Im = 0 and Re ≥0 2 2 are equivalent to Re[g(z)] > 0. After some calculation those requirements become (j). For |B| < 1 the set g(U) is is a disc. In this case Re[g(z)] > 0 if and only if the real part of its center is greater MATHEMATICA SCANDINAVICA 109(2)(2011), 253-268———————————————————CERTAIN RESU than or equal to the length of its radius. Using (31), after suitable adaptation, we get p(A − B) 1 − [p(A − B) + B]B Re ≥ 1 − |B|2 1 − |B|2 and then we obtain (jj). Theorem 4. Let p ∈ N. Suppose that A and B satisfy (j) or (jj) given in Lemma 2. If α and β are such the function ∞ (k + p − 1)z k (p + β + α)(1 + p + β + α) X e H(z) = p (k + p − 1 + β + α)(k + p + β + α) k=1 (40) (z ∈ U) belongs to the class K of convex functions, then ; A, B). J (σ, Ωαβ ; A, B) ⊂ J (σ, Ωα+2 β (41) Proof. The main idea of the proof is to take advantage of several convolution results due mainly to St. Ruscheweyh. Let f ∈ J (σ, Ωαβ ; A, B). It is easy to check that Ωα+2 f (z) = g(z) ∗ Ωαβ f (z), β (42) where (43) g(z) = ∞ X n=p zn (n + β + α)(n + β + α + 1) (z ∈ U). From (42) and (43) it follows that " # 00 z(Ωα+2 f (z)) z(Ωα+2 f (z))0 β β (44) σ 1+ + (1 − σ) (Ωα+2 f (z))0 Ωα+2 f (z) β β # " z(Ωαβ f ∗ g)0 (z) z(Ωαβ f ∗ g)00 (z) = σ 1+ + (1 − σ) (Ωαβ f ∗ g)0 (z) (Ωαβ f ∗ g)(z) zg 0 (z) ∗ z(Ωαβ f )0 (z) g(z) ∗ z(Ωαβ f )0 (z) + (1 − σ) zg 0 (z) ∗ Ωαβ f (z) g(z) ∗ Ωαβ f (z) " # " # z 2−p g 0 (z) ∗ z(z 1−p Ωαβ f )0 (z) z 1−p g(z) ∗ z(z 1−p Ωαβ f )0 (z) = σ − (1 − p) + (1 − σ) − (1 − p) z 2−p g 0 (z) ∗ z 1−p Ωαβ f (z) z 1−p g(z) ∗ z 1−p Ωαβ f (z) = σ = σ z 2−p g 0 (z) ∗ z(z 1−p Ωαβ f )0 (z) z 1−p g(z) ∗ z(z 1−p Ωαβ f )0 (z) + (1 − σ) − (1 − p) z 2−p g 0 (z) ∗ z 1−p Ωαβ f (z) z 1−p g(z) ∗ z 1−p Ωαβ f (z) 1−p pz g(z) e ∗ zh0 (z) H(z) ∗ zh0 (z) g 0 (0) = σ + (1 − σ) pz1−p g(z) − (1 − p), e H(z) ∗ h(z) ∗ h(z) 0 g (0) where h is given in (20). For Rex ≥ 0 or x = 0 the function (45) ˜ z) = h(x; ∞ X (1 + x)z k k=1 (k + x) is convex univalent [11]. In the famous paper [12] it was proved the P`olya–Schoenberg conjecture that the class of convex univalent functions is preserved under convolution. Under our assumptions ´L JACEK DZIOK, RAVINDER KRISHNA RAINA, JANUSZ SOKO 10 on α, β, p the function (46) ge(z) = pz 1−p g(z) g 0 (0) ∞ X (p + β + α)(1 + p + β + α) zk (k + p − 1 + β + α)(k + p + β + α) k=1 # "∞ # "∞ X 1+p+β+α X p+β+α = zk ∗ zk k + p − 1 + β + α k + p + β + α k=1 k=1 = (z ∈ U) is in K as the convolution of two functions from K. From (22) we have 1 + Aω(z) 0 (47) zh (z) = h(z) p + 1 − p := h(z) [qA,B (ω(z)) + 1 − p] , 1 + Bω(z) where ω is an analytic function with ω(0) = 0 and |ω(z)| < 1 for z ∈ U. Using (46) and (47) in (44) we obtain " # 00 z(Ωα+2 f (z)) z(Ωα+2 f (z))0 β β (48) σ 1+ + (1 − σ) (Ωα+2 f (z))0 Ωα+2 f (z) β β = σ e H(z) ∗ h(z)qA,B (ω(z)) ge(z) ∗ h(z)qA,B (ω(z)) , + (1 − σ) e ge(z) ∗ h(z) H(z) ∗ h(z) In the paper [12] it was also proved that if q(z) = 1 + b1 z + · · · , f ∈ K and g ∈ S ∗ , then (49) f (z) ∗ q(z)g(z) ∈ coq(U) (z ∈ U), f (z) ∗ g(z) e ∈ K, by (46) where coq(U) denotes the closed convex hull of the set q(U). By our assumptions H ∗ ge ∈ K and by Lemma 2 we have h ∈ S . Thus from (49) we can see that in (48) we have a convex combination of values of coqA,B (ω(U)). Because coqA,B (ω(U)) ⊂ coqA,B (U), where qA,B (z) is convex univalent function, we deduce from (2) that # " 00 z(Ωα+2 f (z)) z(Ωα+2 f (z))0 1 + Az β β , 1+ + (1 − σ) ≺p α+2 α+2 0 1 + Bz (Ωβ f (z)) Ωβ f (z) what means f ∈ J (σ, Ωα+2 ; A, B). β On setting α + β = −1, p ≥ 2 in Theorem 4 then the function (40) becomes ∞ X (1 + p − 2)z k e H(z) = (z ∈ U) k+p−2 k=1 and in view of (45) it belongs to the class of convex function K, so we get the following result. Corollary 4. Let β + α = −1 and p ≥ 2. Then ; A, B). J (σ, Ωαβ ; A, B) ⊂ J (σ, Ωα+2 β Next, if we set α + β = 0, p≥1 MATHEMATICA SCANDINAVICA 109(2)(2011), 253-268———————————————————CERTAIN RESU in Theorem 4 then the function (40) becomes e H(z) = (1 + p) ∞ X zk k+p k=1 (z ∈ U) and in view of (45) it belongs to the class of convex function K, so we get the result below. Corollary 5. Let β + α = 0 and p ≥ 1. Then J (σ, Ωαβ ; A, B) ⊂ J (σ, Ωα+2 ; A, B). β References [1] S. D. Bernardi, Convex and starlike univalent functions, Trans. Amer. Math. Soc. 135(1969) 429-446. [2] J. Dziok, Classes of analytic functions involving some integral operator, Folia Sci. Univ. Tech. Resoviensis 20(1995) 21–39. [3] J. Dziok, Applications of the Jack Lemma, Acta Math. Hungar. 105(1-2)(2004) 93–102. [4] P. Eenigenburg, S. S. Miller, P. T. Mocanu, M. O. Reade, On a Briot–Bouquet differential subordination, Seminar of geometric function theory, 1-13, Preprint, 82-4, Univ. Babe¸s-Bolyai, Cluj-Napoca, 1983. [5] W. Janowski, Extremal problems for a family of functions with positive real part and for some related families, Ann. Polon. Math. 23(1970) 159-177. [6] P. T. Mocanu, Une propri´et´e de convexit´e g´en´eralis´ee dans la th´eorie de la repr´esentation conforme, Mathematica (Cluj) 11(34) (1969) 127-133. [7] S. S. Miller, P. T. Mocanu, Differential subordinations: theory and applications, Series of Monographs and Textbooks in Pure and Applied Mathematics, Vol. 225, Marcel Dekker Inc., New York / Basel 2000. [8] R. J. Libera, Some classes of regular univalent functions, Proc. Amer. Math. Soc. 16(1965) 755–758. [9] A. E. Livingston, On the radius of univalence of certain analytic functions, Proc. Amer. Math. Soc. 17(1966) 352–357. [10] M. S. Robertson, Certain classes of starlike functions, Michigan Math. J. 76, no.1, (1954) 755–758. [11] St. Rusheweyh, New criteria for univalent functions, Proc. Amer. Math. Soc. 49(1975) 109–115. [12] St. Ruscheweyh, T. Sheil–Small, Hadamard product of schlicht functions and the Poyla-Schoenberg conjecture, Comm. Math. Helv. 48(1973) 119–135. [13] H. M. Srivastava, M. K. Aouf, A certain fractional derivative operator and its applications to a new class of analytic and multivalent functions with negative coefficients. I and II, J. Math. Anal. Appl. 171(1992), 1–13; ibid. 192(1995) 673–688. [14] H. M. Srivastava and S. Owa (Eds.), Current Topics in Analytic Function Theory, World Scientific Publishing Company, Singapore, New Jersey, London, and Hong Kong, 1992. [15] H. M. Srivastava, S. Owa (Editors), Univalent Functions, Fractional Calculus, and Their Applications, Halsted Press (Ellis Horwood Limited, Chichester), John Wiley and Sons, New York, Chichester, Brisbane and Toronto (1989). ´ w, ul. Rejtana 16A, Department of Mathematics, Institute of Mathematics, University of Rzeszo ´ w, Poland 35-310 Rzeszo E-mail address: [email protected] Department of Mathematics, Sobhasaria Engineering College 10/11 Ganpati Vihar, Opposite Sector 5, Udaipur 313002, India E-mail address: [email protected] ´ w University of Technology, ul. W. Pola 2, 35-959 Rzeszo ´ w, Department of Mathematics, Rzeszo Poland E-mail address: [email protected]

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