Bulletin of Mathematical Analysis and Applications
ISSN: 1821-1291, URL: http://www.bmathaa.org
Volume 3 Issue 4(2011), Pages 232-239.
APPLICATION OF AN INTEGRAL OPERATOR FOR P-VALENT
FUNCTIONS
(COMMUNICATED BY SHIGEYOSHI OWA)
MASLINA DARUS, IMRAN FAISAL AND ZAHID SHAREEF
Abstract. By making use of an integral operator defined in an open unit disk,
we introduce and study certain new subclasses of p-valent functions. Inclusion
relationships are established and integral preserving properties of functions in
these subclasses are discussed.
1. Introduction and preliminaries
Let Ap denote the class of functions of the form
f (z) = z p +
∞
∑
ak z k , p ∈ N = {1, 2, · · · }.
(1.1)
k=p+1
which are analytic in an open unit disk U = {z : |z| < 1}.
Next we define some well known subclasses of p-valent functions as follows:
{
}
( ′ )
zf (z)
∗
Sp (ξ) =
f ∈ Ap : ℜ
> ξ, 0 ≤ ξ < p, p ∈ N, z ∈ U ;
f (z)
{
}
(
)
zf ′′ (z)
Kp (ρ, ξ) =
f ∈ Ap : ℜ 1 + ′
> ξ, 0 ≤ ξ < p, p ∈ N, z ∈ U ;
f (z)
{
}
( ′ )
zf (z)
∗
Kp (ρ, ξ) =
f ∈ Ap : ∃g(z) ∈ Sp (ξ) ∧ ℜ
> ρ, 0 ≤ ρ, ξ < p, p ∈ N, z ∈ U ;
g(z)
{
}
(
)
(zf ′ (z))′
∗
Kp (ρ, ξ) =
f ∈ Ap : ∃g(z) ∈ Cp (ξ) ∧ ℜ
> ρ, 0 ≤ ρ, ξ < p, p ∈ N, z ∈ U .
g ′ (z)
2000 Mathematics Subject Classification. 30C45.
Key words and phrases. Analytic Functions; Integral operator; Inclusion properties.
c
⃝2011
Universiteti i Prishtinës, Prishtinë, Kosovë.
Submitted May 8, 2011. Published December 1, 2011.
M.Darus, I. Faisal, and Z. Shareef were supported by the grant MOHE: UKM-ST-06FRGS0244-2010. The authors also would like to thank the referee for the comments and suggestion
given to improve the article.
232
APPLICATION OF AN INTEGRAL OPERATOR
233
The classes Sp∗ (ξ), Sp∗ , Cp (ξ), Cp , Kp (ρ, ξ), K1 (ρ, ξ), Kp∗ (ρ, ξ) and K1∗ (ρ, ξ) were
introduced by Patil and Thakare [13], Goodman [14], Owa [15], Aouf [16], Libera
[17] and Noor [18, 19] respectively.
Also note that
f (z) ∈ Cp (ξ) if and only if
Similarly
zf ′ (z)
p
∈ Sp∗ (ξ), 0 ≤ ξ < p, p ∈ N, z ∈ U·
f (z) ∈ Kp (ξ) if and only if
zf ′ (z)
p
∈ Kp∗ (ξ), 0 ≤ ξ < p, p ∈ N, z ∈ U·
For a function f ∈ Ap , we define a differential operator as follow:
Υ0 f (z)
=
Υ1λ (p, α, β, µ)f (z)
=
Υ1λ (p, α, β, µ)f (z)
=
Dp,λ (α, β, µ)f (z);
Υ2λ (p, α, β, µ)f (z)
=
..
.
D(Υ1λ (p, α, β, µ)f (z));
Υnλ (p, α, β, µ)f (z)
=
D(Υλn−1 (p, α, β, µ)f (z))·
f (z);
pµ + pλ zf ′ (z)
α − pµ + β − pλ
)f (z) + (
)
;
(
(α + β)
(α + β)
p
(1.2)
If f is given by (1.1) then from (1.2) we have
Υnλ (p, α, β, µ)f (z)
=
zp +
∞ (
∑
α + (µ + λ)(k − p) + β )n
ak z k ,
α+β
(1.3)
k=p+1
where f ∈ Ap , α, β, µ, λ ≥ 0, α + β ̸= 0, n ∈ No = {0, 1, · · · }.
This generalizes some well known differential operators available in literature (see
for examples [1]-[11]).
Now we define the integral operator for f (z) ∈ Ap as follows:
{0p (α, β, µ, λ)f (z) = f (z);
{1p (α, β, µ, λ)f (z) =
{2p (α, β, µ, λ)f (z) =
α + β p−( α+β
µ+λ )
z
µ+λ
α + β p−( α+β
µ+λ )
z
µ+λ
∫
z
∫
z
α+β
t( µ+λ )−p−1 f (t)dt;
o
α+β
t( µ+λ )−p−1 {1p (α, β, µ, λ)f (t)dt;
o
..
.
{m
p (α, β, µ, λ)f (z)
α + β p−( α+β
µ+λ )
z
=
µ+λ
This implies
p
{m
p (α, β, µ, λ)f (z) = z +
∞
∑
k=2
(
∫
z
α+β
t( µ+λ )−p−1 {m−1
(α, β, µ, λ)f (t)dt·
p
o
α+β
α + (µ + λ)(k − p) + β
)m
ak z k ,
(1.4)
234
M. DARUS, I. FAISAL AND Z. SHAREEF
where α ≥ 0, β ≥ 0, µ ≥ 0, λ ≥ 0, p ∈ N, m ∈ N0 = N ∪ {0}, f (z) ∈ Ap , z ∈ U).
From (1.4) we have
(
) (
)′
µ + λ z {m+1
(α, β, µ, λ)f (z)
p
(
)
)(
α+β
{m
p (α, β, µ, λ)f (z)
(
−
=
)(
α + β − p(µ + λ)
)
{m+1
(α, β, µ, λ)f (z)
p
·(1.5)
Using the operator {m
p (α, β, µ, λ)f (z) defined in (1.4), we introduce the following
subclasses of p-valent functions:
{
S∗m (p, ξ, λ, α, β, µ) =
}
∗
f ∈ Ap : {m
p (α, β, µ, λ)f ∈ Sp (ξ) ;
{
f ∈ Ap :
Cm (p, ξ, λ, α, β, µ) =
}
{m
p (α, β, µ, λ)f
∈ Cp (ξ) ;
}
{
f ∈ Ap : {m
p (α, β, µ, λ)f ∈ Kp (ρ, ξ) ;
Km (p, ρ, ξ, λ, α, β, µ) =
{
K∗m (p, ρ, ξ, λ, α, β, µ)
=
f ∈ Ap :
}
{m
p (α, β, µ, λ)f
∈
Kp∗ (ρ, ξ)
.
2. Inclusion relationships
In this section, we establish various inclusion relationships for the functions belonging to the new subclasses of p-valent functions.
Lemma 2.1.[20, 21] Let φ(µ, ν) be a complex function, ϕ : D → C, D ⊂ C × C,
and let µ = µ1 + iµ1 , ν = ν1 + iν1 . Suppose that φ(µ, ν) satisfies the following
conditions:
(1) φ(µ, ν) is continuous in D;
(2) (1, 0) ∈ D and ℜ{φ(1, 0)} > 0;
(3) ℜ{φ(iµ2 , ν1 )} ≤ 0 for all (iµ2 , ν1 ) ∈ D such that ν1 ≤ − 12 (1 + µ22 )·
Let h(z) = 1 + c1 z + c2 z 2 + · · · be analytic in U, such that (h(z), zh′ (z)) ∈ D for
all z ∈ U. If ℜ{φ(h(z), zh′ (z))} > 0(z ∈ U) , then ℜ{h(z)} > 0 for z ∈ U·
Theorem 2.2. Let f (z) ∈ Ap and α, β, µ, λ ≥ 0, p ∈ N, m ∈ N0 = N ∪ {0}, z ∈ U·
Then
S∗m (p, ξ, λ, α, β, µ) ⊆ S∗m+1 (p, ξ, λ, α, β, µ) ⊆ S∗m+2 (p, ξ, λ, α, β, µ)·
Proof. To prove S∗m (p, ξ, λ, α, β, µ) ⊆ S∗m+1 (p, ξ, λ, α, β, µ), let f (z) ∈ S∗m (p, ξ, λ, α, β, µ)
and assume that
z({m+1
(α, β, µ, λ)f (z))′
p
= ξ + (p − ξ)h(z), 0 ≤ ξ < 1, z ∈ U·
(2.1)
{m+1
(α, β, µ, λ)f (z)
p
Where h(z) = 1 + c1 z + c2 z 2 + · · · .
APPLICATION OF AN INTEGRAL OPERATOR
235
Using (1.5) and (2.1), we have
′
z({m
(p − ξ)zh′ (z)
p (α, β, µ, λ)f (z))
− ξ = (p − ξ)h(z) + α+β
.
m
{p (α, β, µ, λ)f (z)
( µ+λ − p) + ξ + (p − ξ)h(z)
(2.2)
Taking h(z) = µ = µ1 + iµ1 and zh′ (z) = ν = ν1 + iν1 , we define the function
φ(µ, ν) by:
φ(µ, ν) = (p − ξ)µ +
( α+β
µ+λ
(p − ξ)ν
·
− p) + ξ + (p − ξ)µ
This implies
( α+β −p)+ξ
(i) φ(µ, ν) is continuous in D = (C − µ+λξ−p
) × C,
(ii) (1, 0) ∈ D and ℜ{φ(1, 0)} > 1 − ξ,
(iii) For all (iµ2 , ν1 ) ∈ D such that ν1 ≤ − 12 (1 + µ22 ). Therefore
ℜ{φ(iµ2 , ν1 )} = ℜ{
( α+β
µ+λ
[( α+β
(p − ξ)ν
µ+λ − p) + ξ](p − ξ)ν
,
} = α+β
− p) + ξ + (p − ξ)iµ2
(( µ+λ − p) + ξ)2 + (p − ξ)2 µ22
⇒
ℜ{φ(iµ2 , ν1 )} =
[( α+β
µ+λ − p) + ξ](p − ξ)ν
2
2 2
(( α+β
µ+λ − p) + ξ) + (p − ξ) µ2
≤−
2
[( α+β
µ+λ − p) + ξ](p − ξ)(1 + µ2 )
2
2 2
2(( α+β
µ+λ − p) + ξ) + (p − ξ) µ2
< 0·
Hence, the function φ(µ, ν) satisfies the conditions of Lemma 2.1. This shows that
ℜ{h(z)} > 0(z ∈ U), that is, f (z) ∈ S∗m+1 (p, ξ, λ, α, β, µ)·
Theorem 2.3. If f (z) ∈ Ap and α, β, µ, λ ≥ 0, p ∈ N, m ∈ N0 = N ∪ {0}, z ∈ U·
Then
Cm (p, ξ, λ, α, β, µ) ⊆ Cm+1 (p, ξ, λ, α, β, µ) ⊆ Cm+2 (p, ξ, λ, α, β, µ).
z m
′
Proof. Let f ∈ Cm (p, ξ, λ, α, β, µ) ⇒ {m
p (α, β, µ, λ)f ∈ Cp (ξ), ⇔ p ({p (α, β, µ, λ)f ) ∈
′
zf
∗
Sp∗ (ξ) ⇔ {m
p (α, β, µ, λ)f ( p ) ∈ Sp (ξ), ⇔
zf ′
p
zf ′
p
∈ Cm (p, ξ, λ, α, β, µ) ⊆ Cm+1 (p, ξ, λ, α, β, µ),
′
⇒
∈ Cm+1 (p, ξ, λ, α, β, µ) ⇔
) ∈ Sp∗ (ξ), ⇔ pz ({m+1
(α, β, µ, λ)f )′ ∈
p
Sp∗ (ξ) ⇒ {m+1
(α, β, µ, λ)f ∈ Cp (ξ), ⇒ f ∈ Cm+1 (p, ξ, λ, α, β, µ)·
p
{m+1
(α, β, µ, λ)f ( zfp
p
Theorem 2.4. If f (z) ∈ Ap and α, β, µ, λ ≥ 0, p ∈ N, m ∈ N0 = N ∪ {0}, z ∈ U·
Then
Km (p, ρ, ξ, λ, α, β, µ) ⊆ Km+1 (p, ρ, ξ, λ, α, β, µ) ⊆ Km+2 (p, ρ, ξ, λ, α, β, µ)·
Proof. Let f (z) ∈ Km (p, ρ, ξ, λ, α, β, µ) implies
)
(
′
z({m
p (α, β, µ, λ)f (z))
> ρ, g(z) ∈ S∗m (p, ξ, λ, α, β, µ), 0 ≤ ρ < 1, z ∈ U·
ℜ
{m
p (α, β, µ, λ)g(z)
Since g(z) ∈ S∗m (p, ξ, λ, α, β, µ) ⊆ S∗m+1 (p, ξ, λ, α, β, µ), let
z({m+1
(α, β, µ, λ)g(z))′
p
{m+1
(α, β, µ, λ)g(z)
p
= ξ + (p − ξ)H(z)·
(2.3)
236
M. DARUS, I. FAISAL AND Z. SHAREEF
Suppose that
(
)
z({m+1
(α, β, µ, λ)f (z))′
p
{m+1
(α, β, µ, λ)g(z)
p
= ρ + (p − ρ)h(z), 0 ≤ ρ < 1, z ∈ U·
(2.4)
Where h(z) = 1 + c1 z + c2 z 2 + · · · · Using (1.5), (2.3) and (2.4) we get
′
z({m
p (α, β, µ, λ)f (z))
=
{m
p (α, β, µ, λ)g(z)
z({m+1
(α,β,µ,λ)zf ′ (z))′
p
{m+1
(α,β,µ,λ)g(z)
p
+ ( α+β
µ+λ − p)(ρ + (p − ρ)h(z))
ξ + (p − ξ)H(z) + ( α+β
µ+λ − p)
· (2.5)
Using (2.3) and (2.4) we have
z(({m+1
(α, β, µ, λ)zf ′ (z)))′
p
({m+1
(α, β, µ, λ)g(z))
p
= [ρ + (p − ρ)h(z)][ξ + (p − ξ)H(z)] + (p − ρ)zh′ (z)·(2.6)
Simplifying simultaneously (2.5) and (2.6), we get
′
z({m
(p − ρ)zh′ (z)
p (α, β, µ, λ)f (z))
−
ρ
=
(p
−
ρ)h(z)
+
·
{m
( α+β
p (α, β, µ, λ)g(z)
µ+λ − p + ξ) + (p − ξ)H(z)
(2.7)
Taking h(z) = µ = µ1 + iµ1 and zh′ (z) = ν = ν1 + iν1 , we define the function
φ(µ, ν) by
φ(µ, ν) = (p − ρ)µ +
(p − ρ)ν
·
( α+β
−
p
+
ξ) + (p − ξ)H(z)
µ+λ
Clearly conditions (i) and (ii) of Lemma 2.1 in D = C × C are satisfied. For (iii),
we proceed as follows.
ℜ{φ(iµ2 , ν1 )} =
ν1 (p − ρ)[( α+β
µ+λ − p + ξ) + (p − ξ)h1 (x1 , y1 )]
2
2
[( α+β
µ+λ − p + ξ) + (p − ξ)h1 (x1 , y1 )] + [(p − ξ)h2 (x2 , y2 )]
·
Where H(z) = h1 (x1 , y1 ) + ih2 (x2 , y2 ), h1 (x1 , y1 ) and h2 (x2 , y2 ) being functions of
x and y and ℜ(h1 (x1 , y1 )) > 0. Since ν1 ≤ − 21 (1 + µ22 ), implies
ℜ{φ(iµ2 , ν1 )} = −
(1 + µ22 )(p − ρ)[( α+β
1
µ+λ − p + ξ) + (p − ξ)h1 (x1 , y1 )]
< 0·
α+β
2 [( µ+λ − p + ξ) + (p − ξ)h1 (x1 , y1 )]2 + [(p − ξ)h2 (x2 , y2 )]2
Applying Lemma 2.1. on φ(µ, ν), gives ℜ{h(z)} > 0(z ∈ U). This shows that
f (z) ∈ Km+1 (p, ρ, ξ, λ, α, β, µ)·
The following theorem can be proved in a similar manner.
Theorem 2.5. If f (z) ∈ Ap and α, β, µ, λ ≥ 0, p ∈ N, m ∈ N0 = N ∪ {0}, z ∈ U·
Then
K∗m (p, ρ, ξ, λ, α, β, µ) ⊆ K∗m+1 (p, ρ, ξ, λ, α, β, µ) ⊆ K∗m+2 (p, ρ, ξ, λ, α, β, µ)·
3. Integral operator
For c > −p and f (z) ∈ Ap , the integral operator Lc,p : Ap → Ap is defined by
Lc,p (f ) =
c+p
zc
∫
z
tc−1 f (t)dt·
0
The operator Lc,p (f ) was introduced by Bernardi [12].
(3.1)
APPLICATION OF AN INTEGRAL OPERATOR
237
Theorem 3.1. Let c > −p, 0 ≤ ξ < p· If f (z) ∈ S∗m (p, ξ, λ, α, β, µ), then Lc (f ) ∈
S∗m (p, ξ, λ, α, β, µ))·
Proof. Using (3.1) we get
′
m
m
z({m
p (α, β, µ, λ)Lc,p f (z)) = (c + p)({p (α, β, µ, λ)f (z)) − c({p (α, β, µ, λ)Lc,p f (z))·(3.2)
Let
′
z({m
p (α, β, µ, λ)Lc f (z))
= ξ + (p − ξ)h(z)·
{m
p (α, β, µ, λ)Lc f (z)
(3.3)
Where h(z) = 1 + c1 z + c2 z 2 + · · · · By using (3.1) and (3.2) we get
′
z({m
(p − ξ)zh′ (z)
p (α, β, µ, λ)f (z))
−
ξ
=
(p
−
ξ)h(z)
+
·
ξ + (p − ξ)h(z) + c
{m
p (α, β, µ, λ)f (z)
Taking h(z) = µ = µ1 + iµ1 and zh′ (z) = ν = ν1 + iν1 , we define the function
φ(µ, ν) by
(p − ξ)ν
·
ξ + c + (p − ξ)µ
(
)
ξ+c
Clearly conditions (i) and (ii) of Lemma 2.1 in D = C − { ξ−p } × C are satisfied.
φ(µ, ν) = (p − ξ)µ +
We proceed for (iii) as follows;
ℜ{φ(iµ2 , ν1 )} =
(ξ + c)(p − ξ)ν1
−(ξ + c)(p − ξ)(1 + µ22 )
≤
< 0·
2
2
[ξ + c] + [(p − ξ)µ2 ]
2[ξ + c]2 + 2[(p − ξ)µ2 ]2
Applying Lemma 2.1. we have ℜ{h(z)} > 0(z ∈ U), that is, Lc (f ) ∈ S∗m (p, ξ, λ, α, β, µ)·
Theorem 3.2. Let c > −p, 0 ≤ ξ < p· If f (z) ∈ Cm (p, ξ, λ, α, β, µ), then Lc (f ) ∈
Cm (p, ξ, λ, α, β, µ)·
Proof. For the proof, use Theorem 3.1 and the fact that f (z) ∈ Cp (ξ) ⇔
Sp∗ (ξ).
zf ′ (z)
p
∈
Theorem 3.3. Let c > −p, 0 ≤ ξ < p· If f (z) ∈ Km (p, ρ, ξ, λ, α, β, µ), then
Lc (f ) ∈ Km (p, ρ, ξ, λ, α, β, µ)·
Proof. As f (z) ∈ Km (p, ρ, ξ, λ, α, β, µ) gives
(
′)
z({m
p (α, β, µ, λ)f (z))
ℜ
> ρ·
{m
p (α, β, µ, λ)g(z)
Since g(z) ∈ S∗m (p, ξ, λ, α, β, µ)) implies Lc (g(z)) ∈ S∗m (p, ξ, λ, α, β, µ))· Let
′
z({m
p (α, β, µ, λ)Lc (g(z))
= ξ + (p − ξ)H(z), ℜ(H(z)) > 0, z ∈ U·
{m
p (α, β, µ, λ)Lc g(z)
Also let
(
′
z({m
p (α, β, µ, λ)Lc f (z))
m
{p (α, β, µ, λ)Lc g(z)
Where h(z) = 1 + c1 z + c2 z 2 + · · · .
)
= ρ + (p − ρ)h(z), z ∈ U·
(3.4)
238
M. DARUS, I. FAISAL AND Z. SHAREEF
After doing calculations, we get
(
′
z({m
p (α, β, µ, λ)f (z))
{m
p (α, β, µ, λ)g(z)
)
=
′
′
z({m
p (α,β,µ,λ)Lc (zf (z))
{m
(α,β,µ,λ)L
g(z)
c
p
+c
′
({m
p (α,β,µ,λ)Lc (zf (z))
{m
(α,β,µ,λ)L
g(z)
c
p
′
z({m
p (α,β,µ,λ)Lc (g(z))
{m
p (α,β,µ,λ)Lc g(z)
·
+c
Also
′
′
z({m
p (α, β, µ, λ)Lc (zf (z))
= [ρ + (p − ρ)h(z)][ξ + (p − ξ)H(z)] + [(p − ρ)zh′ (z)]·
{m
p (α, β, µ, λ)Lc g(z)
Hence
(
′
z({m
p (α, β, µ, λ)f (z))
{m
p (α, β, µ, λ)g(z)
)
− ρ = (p − ρ)h(z) +
(p − ρ)zh′ (z)
.
ξ + (p − ξ)H(z) + c
(3.5)
Taking h(z) = µ = µ1 + iµ1 and zh′ (z) = ν = ν1 + iν1 , we define the function
φ(µ, ν) by
φ(µ, ν) = (p − ρ)µ +
(p − ρ)ν
·
ξ + (p − ξ)H(z) + c
(3.6)
It is easy to see that the function φ(µ, ν) satisfies the conditions (i) and (ii) of
Lemma 2.1 in D = C × C. To verify the condition (iii), we proceed as follows.
ℜ{φ(iµ2 , ν1 )} =
ν1 (p − ρ)[(ξ + c) + (p − ξ)h1 (x1 , y1 )]
·
[(ξ + c) + (p − ξ)h1 (x1 , y1 )]2 + [(p − ξ)h2 (x2 , y2 )]2
Where H(z) = h1 (x1 , y1 ) + ih2 (x2 , y2 ), h1 (x1 , y1 ) and h2 (x2 , y2 ) being functions of
x and y and ℜ(h1 (x1 , y1 )) > 0. By putting ν1 ≤ − 21 (1 + µ22 ), we obtain
ℜ{φ(iµ2 , ν1 )} = −
1
(1 + µ22 )(p − ρ)[(ξ + c) + (p − ξ)h1 (x1 , y1 )]
< 0·
2 [(ξ + c) + (p − ξ)h1 (x1 , y1 )]2 + [(p − ξ)h2 (x2 , y2 )]2
By applying Lemma 2.1 we get ℜ{h(z)} > 0(z ∈ U), that is, Lc (f ) ∈ Cm (p, ξ, λ, α, β, µ)·
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Maslina Darus
School of Mathematical Sciences Faculty of Science and Technology Universiti Kebangsaan Malaysia, Bangi 43600, Selangor Darul Ehsan Malaysia
E-mail address: maslina@ukm.my (Corresponding author)
Imran Faisal
School of Mathematical Sciences Faculty of Science and Technology Universiti Kebangsaan Malaysia, Bangi 43600, Selangor Darul Ehsan Malaysia
E-mail address: faisalmath@gmail.com
Zahid Shareef
School of Mathematical Sciences Faculty of Science and Technology Universiti Kebangsaan Malaysia, Bangi 43600, Selangor Darul Ehsan Malaysia
E-mail address: zahidmath@gmail.com