Mathematics and Informatics ICTAMI 2005 - Alba Iulia, Romania
ON A CLASS OF α-CONVEX FUNCTIONS
Mugur Acu
Abstract. In this paper we define a general class of α-convex functions with respect to a convex domain D contained in the right half plane by using a generalized S˘al˘agean operator introduced by F.M. Al-Oboudi in [5] and we give some properties of this class.
2000 Mathematical Subject Classification: 30C45
Key words and phrases: α-convex functions, Libera-Pascu integral opera- tor, Briot-Bouquet differential subordination, generalized S˘al˘agean operator
1.Introduction
Let H(U) be the set of functions which are regular in the unit disc U, A ={f ∈ H(U) :f(0) =f0(0)−1 = 0}, Hu(U) ={f ∈ H(U) :f is univalent in U} and S ={f ∈A:f is univalent in U}.
Let Dn be the S˘al˘agean differential operator ([10]) defined as:
Dn :A→A, n∈N and
D0f(z) = f(z) D1f(z) =Df(z) = zf0(z)
Dnf(z) = D(Dn−1f(z)).
Remark 1.1Iff ∈S, f(z) =z+P∞
j=2
ajzj, j = 2,3, ...,z ∈U thenDnf(z) = z+
∞
P
j=2
jnajzj.
The aim of this paper is to define a general class of α-convex functions with respect to a convex domain D contained in the right half plane by using a generalized S˘al˘agean operator introduced by F.M. Al-Oboudi in[5] and to obtain some leftoperties of this class.
2.Preliminary results
We recall here the definitions of the well - known classes of starlike func- tions, convex functions and α-convex functions (see [6])
S∗ =
(
f ∈A: Rezf0(z)
f(z) >0, z∈U
)
,
Sc =CV =K =
(
f ∈H(U); f(0) =f0(0)−1 = 0, Re
(
1 + zf00(z) f0(z)
)
>0, z ∈U
)
,
Mα ={f ∈H(U), f(0) =f0(0)−1 = 0,ReJ(α, f;z)>0, z∈U , α∈R}
where
J(α, f;z) = (1−α)zf0(z)
f(z) +α 1 + zf00(z) f0(z)
!
We observe that M0 = S∗ and M1 =Sc where S∗ and Sc are the class of starlike functions, respectively the class of convex functions.
Remark 2.1. By using the subordination relation, we may define the class Mα thus if f(z) = z +a2z2 +..., z ∈ U, then f ∈ Mα if and only if J(α, f;z)≺ 1+z1−z, z ∈U, where by ”≺” we denote the subordination relation.
Let consider the Libera-Pascu integral operator Lα :A→A defined as:
f(z) = LαF(z) = 1 +a za
z
Z
0
F(t)·ta−1dt, α∈C ,Reα ≥0. (1) In the case a = 1,2,3, ... this operator was introduced by S. D. Bernardi and it was studied by many authors in different general cases.
Definition 2.1.[5]Letn ∈Nandλ≥0. We denote withDλnthe operator defined by
Dλn:A→A ,
D0λf(z) = f(z), D1λf(z) = (1−λ)f(z) +λzf0(z) =Dλf(z), Dnλf(z) = DλDn−1λ f(z).
Remark 2.2.[5] We observe that Dnλ is a linear operator and for f(z) = z+
∞
P
j=2
ajzj we have
Dnλf(z) = z+
∞
X
j=2
(1 + (j −1)λ)najzj.
Also, it is easy to observe that if we consider λ= 1 in the above definition we obtain the S˘al˘agean differential operator.
Definition 2.2 [3]Letq(z)∈ Hu(U), with q(0) = 1 andq(U) = D, where D is a convex domain contained in the right half plane, n ∈N and λ≥0. We say that a function f(z)∈A is in the class SL∗n(q) if D
n+1 λ f(z)
Dnλf(z) ≺q(z), z ∈U. Remark 2.3. Geometric interpretation: f(z) ∈ SL∗n(q) if and only if
Dλn+1f(z)
Dλnf(z) take all values in the convex domain D contained in the right half- plane.
Definition 2.3 [4]Letq(z)∈ Hu(U), with q(0) = 1 andq(U) = D, where D is a convex domain contained in the right half plane, n ∈N and λ≥0. We say that a function f(z)∈A is in the class SLcn(q) if D
n+2 λ f(z)
Dn+1λ f(z) ≺q(z), z ∈U. Remark 2.4. Geometric interpretation: f(z) ∈ SLcn(q) if and only if
Dλn+2f(z)
Dλn+1f(z) take all values in the convex domain D contained in the right half- plane.
The next theorem is result of the so called ”admissible functions method”
introduced by P.T. Mocanu and S.S. Miller (see [7], [8], [9]).
Theorem 2.1. Let h convex in U and Re[βh(z) + γ] > 0, z ∈ U. If p ∈ H(U) with p(0) = h(0) and p satisfied the Briot-Bouquet differential subordination
p(z) + zp0(z)
βp(z) +γ ≺h(z), then p(z)≺h(z).
3. Main results
Definition 3.1 Letq(z)∈ Hu(U), withq(0) = 1, q(U) =D, where Dis a convex domain contained in the right half plane, n ∈N, λ ≥0 and α ∈[0,1].
We say that a function f(z)∈A is in the class M Ln,α(q) if Jn,λ(α, f;z) = (1−α)Dn+1λ f(z)
Dnλf(z) +αDλn+2f(z)
Dλn+1f(z) ≺q(z), z ∈U.
Remark 3.1 Geometric interpretation: f(z) ∈ M Ln,α(q) if and only if Jn,λ(α, f;z) take all values in the convex domain D contained in the right half-plane.
Remark 3.2 It is easy to observe that if we choose different function q(z) we obtain variously classes of α-convex functions, such as (for example), for λ = 1 and n = 0, the class of α-convex functions, the class of α-uniform convex functions with respect to a convex domain (see [2]), and, for λ = 1, the class U Dn,α(β, γ), β ≥ 0, γ ∈ [−1,1), β+γ ≥ 0 (see [1]), the class of α-n-uniformly convex functions with respect to a convex domain (see [2]).
Remark 3.3 We have M Ln,0(q) =SL∗n(q) and M Ln,1(q) =SLcn(q).
Remark 3.4 For q1(z) ≺ q2(z) we have M Ln,α(q1) ⊂ M Ln,α(q2). From the above we obtain M Ln,α(q)⊂M Ln,α
1+z 1−z
Theorem 3.1 For all α, α0 ∈ [0,1], with α < α0, we have M Ln,α0(q) ⊂ M Ln,α(q).
Proof. From f(z)∈M Ln,α0(q) we have Jn,λ(α, f;z) = (1−α)Dn+1λ f(z)
Dnλf(z) +αDλn+2f(z)
Dλn+1f(z) ≺q(z), (2) where q(z) is univalent in U with q(0) = 1 and maps the unit discU into the convex domain D contained in the right half-plane.
With notation
p(z) = Dn+1λ f(z) Dnλf(z)
where
p(z) = 1 +p1z+. . . andf(z) = z+
∞
X
j=2
ajzj
we have
p(z) +α0λ· zp0(z) p(z) =
= Dλn+1f(z)
Dλnf(z) +α0λ Dλnf(z) Dn+1λ f(z) ·z
Dλn+1f(z)0Dλnf(z)−Dn+1λ f(z) (Dnλf(z)0
(Dλnf(z))2 =
= Dλn+1f(z)
Dλnf(z) +α0λ Dλnf(z) Dλn+1f(z)
zDn+1λ f(z)0
Dλnf(z) − Dλn+1f(z)
Dλnf(z) · z(Dλnf(z)0 Dnλf(z)
=
= Dλn+1f(z)
Dλnf(z) +α0λ· Dλnf(z) Dλn+1f(z)
z z+
∞
P
j=2
(1 + (j−1)λ)n+1ajzj
!0
Dλnf(z) −
− Dλn+1f(z) Dnλf(z) ·
z z+ P∞
j=2
(1 + (j−1)λ)najzj
!0
Dλnf(z)
=
= Dn+1λ f(z)
Dnλf(z) +α0λ· Dnλf(z) Dn+1λ f(z)
z 1 +
∞
P
j=2
j(1 + (j−1)λ)n+1ajzj−1
!
Dλnf(z) −
− Dλn+1f(z) Dλnf(z) ·
z 1 + P∞
j=2
j(1 + (j−1)λ)najzj−1
!
Dλnf(z)
or
p(z)+α0·λ·zp(z)
p(z) = Dn+1λ f(z)
Dnλf(z) +α0λ· Dnλf(z) Dn+1λ f(z)
z+ P∞
j=2
j(1 + (j−1)λ)n+1ajzj Dλnf(z) −
− Dn+1λ f(z) Dnλf(z)
z+ P∞
j=2
j(1 + (j −1)λ)najzj Dnλf(z)
(3)
We have z+
∞
X
j=2
j(1 + (j −1)λ)n+1ajzj =z+
∞
X
j=2
((j−1) + 1) (1 + (j−1)λ)n+1ajzj =
=z+
∞
X
j=2
(1 + (j−1)λ)n+1ajzj +
∞
X
j=2
(j −1) (1 + (j−1)λ)n+1ajzj =
=z+Dn+1λ f(z)−z+
∞
X
j=2
(j−1) (1 + (j−1)λ)n+1ajzj =
=Dλn+1f(z) + 1 λ
∞
X
j=2
((j−1)λ) (1 + (j−1)λ)n+1ajzj =
=Dn+1λ f(z) + 1 λ
∞
X
j=2
(1 + (j−1)λ−1) (1 + (j −1)λ)n+1ajzj =
=Dλn+1f(z)− 1 λ
∞
X
j=2
(1 + (j −1)λ)n+1ajzj+ 1 λ
∞
X
j=2
(1 + (j−1)λ)n+2ajzj =
=Dn+1λ f(z)− 1 λ
Dλn+1f(z)−z+ 1 λ
Dλn+2f(z)−z=
=Dn+1λ f(z)− 1
λDλn+1f(z) + z λ + 1
λDλn+2f(z)− z λ =
= λ−1
λ Dn+1λ f(z) + 1
λDλn+2f(z) =
= 1 λ
λ−1)Dλn+1f(z) +Dn+2λ f(z). Similarly we have
z+
∞
X
j=2
j(1 + (j−1)λ)najzj = 1 λ
λ−1)Dλnf(z) +Dλn+1f(z).
From (3) we obtain
p(z) +α0·λ·zp0(z) p(z) =
= Dλn+1f(z)
Dλnf(z) +α0λ Dλnf(z) Dn+1λ f(z)
1
λ · (λ−1)Dn+1λ f(z) Dnλf(z) + +Dn+2λ f(z)
Dnλf(z) −Dn+1λ f(z)
Dλnf(z) (λ−1)− Dn+1λ f(z) Dnλf(z)
!2
=
= Dn+1λ f(z)
Dλnf(z) +α0Dn+2λ f(z)
Dn+1λ f(z) −α0Dn+1λ f(z) Dnλf(z) =
= Dn+1λ f(z)
Dnλf(z) (1−α0) +α0Dλn+2f(z)
Dλn+1f(z) =Jn,λ(α0, f;z) From (2) we have
p(z) + zp0(z)
1
α0λ ·p(z) ≺q(z)
with p(0) = q(0), Req(z) > 0, z ∈ U, α0 > 0 and λ ≥ 0. In this conditions from Theorem 2.1 we obtainp(z)≺q(z) or p(z) take all values in D.
If we consider the function g : [0, α0]→C, g(u) =p(z) +u· λzp0(z)
p(z) ,
with g(0) =p(z)∈D and g(α0) =Jn,λ(α0, f;z)∈D, it easy to see that g(α) =p(z) +α·λzp0(z)
p(z) ∈D,0≤α < α0. Thus we have
Jn,λ(α, f;z)≺q(z)
or
f(z)∈M Ln,α(q).
From the above theorem we have
Corollary 3.1 For every n∈N and α∈[0,1], we have M Ln,α(q)⊂M Ln,0(q) =SL∗n(q) .
Remark 3.5 If we consider λ = 1 and n = 0 we obtain the Theorem 3.1 from [2]. Also, for λ= 1 and n∈N, we obtain the Theorem 3.3 from [2].
Remark 3.6 If we consider λ = 1 and D = Dβ,γ (see [1] or [2]) in the above theorem we obtain the Theorem 3.1 from [1].
Theorem 3.2 Let n ∈ N, α ∈ [0,1] and λ ≥ 1. If F(z) ∈ M Ln,α(q) then f(z) =LaF(z) ∈SL∗n(q), where La is the Libera-Pascu integral operator defined by (1).
Proof.
From (1) we have
(1 +a)F(z) =af(z) +zf0(z)
and, by using the linear operator Dn+1λ and if we consider f(z) = P∞j=2ajzj, we obtain
(1 +a)Dn+1λ F(z) = aDn+1λ f(z) +Dλn+1
z+
∞
X
j=2
jajzj
=
=aDn+1λ f(z) +z+
∞
X
j=2
(1 + (j−1)λ)n+1jajzj We have (see the proof of the above theorem)
z+
∞
X
j=2
j(1 + (j−1)λ)n+1ajzj = 1 λ
λ−1)Dn+1λ f(z) +Dλn+2f(z) (4)
Thus
(1 +a)Dn+1λ F(z) =aDn+1λ f(z) + 1 λ
λ−1)Dn+1λ f(z) +Dλn+2f(z)=
= a+ λ−1 λ
!
Dn+1λ f(z) + 1
λDλn+2f(z) or
λ(1 +a)Dn+1λ F(z) = (a+ 1)λ−1)Dλn+1f(z) +Dn+2λ f(z).
Similarly, we obtain
λ(1 +a)DnλF(z) = (a+ 1)λ−1)Dλnf(z) +Dn+1λ f(z).
Then
Dn+1λ F(z) DλnF(z) =
Dn+2λ f(z)
Dn+1λ f(z)· DDn+1λn f(z)
λf(z) + ((a+ 1)λ−1)· DDn+1λn f(z) λf(z)
((a+ 1)λ−1) + D
n+1 λ f(z) Dnλf(z)
With notation
Dλn+1f(z)
Dnλf(z) =p(z), p(0) = 1 we obtain
Dn+1λ F(z) DnλF(z) =
Dn+2λ f(z)
Dn+1λ f(z) ·p(z) + (a+ 1)λ−1 ·p(z)
p(z) + (a+ 1)λ−1 (5) Also, we obtain
Dn+2λ f(z)
Dn+1λ f(z) = Dλn+2f(z)
Dλnf(z) · Dnλf(z)
Dn+1λ f(z) = 1
p(z)· Dn+2λ f(z)
Dnλf(z) (6) We have
Dλn+2f(z) Dnλf(z) =
z+ P∞
j=2
(1 + (j−1)λ)n+2ajzj z+ P∞
j=2
(1 + (j−1)λ)najzj and
zp0(z) = zDn+1λ f(z)0
Dnλf(z) − Dλn+1f(z)
Dλnf(z) · z(Dλnf(z))0 Dnλf(z) =
=
z 1 +
∞
P
j=2
(1 + (j−1)λ)n+1jajzj−1
!
Dλnf(z) −
−p(z)· z1 +P∞j=2(1 + (j−1)λ)njajzj−1 Dλnf(z)
or
zp0(z) = z+P∞j=2j(1 + (j −1)λ)n+1ajzj
Dnλf(z) −p(z)·
z+ P∞
j=2
j(1 + (j −1)λ)najzj Dnλf(z) .
(7) By using (4) and (7) we obtain
zp0(z) = 1 λ
(λ−1)Dn+1λ f(z) +Dλn+2f(z)
Dnλf(z) −p(z)(λ−1)Dnλf(z) +Dλn+1f(z) Dnλf(z)
!
=
= 1
λ (λ−1)p(z) + Dn+2λ f(z)
Dnλf(z) −p(z)((λ−1) +p(z))
!
=
= 1 λ
Dλn+2f(z)
Dλnf(z) −p(z)2
!
Thus
λzp0(z) = Dn+2λ f(z)
Dnλf(z) −p(z)2 or
Dλn+2f(z)
Dλnf(z) =p(z)2+λzp0(z).
From (6) we obtain
Dn+2λ f(z) Dn+1λ f(z) = 1
p(z)(p(z)2+λzp0(z)).
Then, from (5), we obtain Dλn+1F(z)
DnλF(z) = p(z)2+λzp0(z) + ((a+ 1)λ−1)p(z)
p(z) + ((a+ 1)λ−1) =p(z)+λ zp0(z)
p(z) + ((a+ 1)λ−1) where α∈C, Rea ≥0 and λ≥1.
If we denote D
n+1 λ F(z)
DλnF(z) =h(z), withh(0) = 1, we have fromF(z)∈M Ln,α(q) (see the proof of the above Theorem):
Jn,λ(α, F;z) = h(z) +α·λ· zh0(z)
h(z) ≺q(z) Using the hypothesis, from Theorem 2.1, we obtain
h(z)≺q(z) or
p(z) +λ zp0(z)
p(z) + ((a+ 1)λ−1) ≺q(z).
By using the Theorem 2.1 and the hypothesis we have p(z)≺q(z)
or
Dn+1λ f(z)
Dnλf(z) ≺q(z).
This means f(z) =LαF(z)∈SL∗n(q).
Remark 3.7 If we consider λ = 1 and n = 0 we obtain the Theorem 3.2 from [2]. Also, for λ= 1 and n∈N, we obtain the Theorem 3.4 from [2].
Remark 3.8 If we consider λ = 1 and D = Dβ,γ ( see [1] or [2]) in the above theorem we obtain the Theorem 3.2 from [1].
References
[1]M. Acu, On a subclass of α-uniform convex functions, Archivum Math- ematicum, no. 2, Tomus 41/2005, (to appear).
[2]M. Acu, Some subclasses of α-uniformly convex functions, Acta Mathe- matica Academiae Paedagogicae Nyiregyhaziensis, Vol. 21(2005), (to appear).
[3]M. Acu, On a subclass of n-starlike functions, (to appear).
[4]M. Acu, On a subclass of n-convex functions, (to appear).
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[6]P. Duren,Univalent functions, Springer Verlag, Berlin Heildelberg, 1984.
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[8]S. S. Miller and P. T. Mocanu, Univalent solution of Briot-Bouquet dif- ferential equation, J. Differential Equations 56(1985), 297-308.
[9]S. S. Miller and P. T. Mocanu, On some classes of first-order differential subordination, Mich. Math. 32(1985), 185-195.
[10]Gr. S˘al˘agean, On some classes of univalent functions, Seminar of geo- metric function theory, Cluj-Napoca, 1983.
Mugur Acu
University ”Lucian Blaga” of Sibiu Department of Mathematics
Str. Dr. I. Rat.iu, No. 5-7 550012 - Sibiu, Romania
E-mail address: acu [email protected]
