### MATHEMATICA SCANDINAVICA 109(2)(2011

```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 . 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 .
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  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 . Some applications
involving these results are also considered.
Let us consider the Dziok differintegral operator Ωαβ  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  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. )
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  or the Libera operator  for special
choices of β and ν.
By using the definitions of fractional calculus, Srivastava and Owa  (see also ,,) 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  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)
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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 , 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 .
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 . After some adaptation Theorem 3.2j (, p. 97) becomes the following lemma.
Lemma 1. 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
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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),
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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 , 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 . In the famous paper  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  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
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´ w, ul. Rejtana 16A,
Department of Mathematics, Institute of Mathematics, University of Rzeszo
´ w, Poland
35-310 Rzeszo