General Mathematics Vol. 18, No. 4 (2010), 53–60
On certain classes of p-valent functions defined by
Sălăgean operator 1
Hesam Mahzoon, S. Latha
Abstract
Let Ap be the class of analytic functions f which are of the form
∞
X
f (z) = z p +
am z m , (p ∈ N = {1, 2, 3, ...}), defined in the open
m=p+1
unit disk U = {z ∈ C : |z| < 1}. We introduce the class M(α, β, n, p)
and also the subclass M? (α, β, n, p). The aim of the present paper is to
derive some convolution properties for functions belonging to the class
M? (α, β, n, p).
2010 Mathematics Subject Classification: 30C45
Key words and phrases: Analytic functions, Hadamard product, p-valent
functions and Sălăgean operator.
1
Introduction
Let A be the class of functions f of the form
(1)
f (z) = z +
∞
X
am z m .
m=2
which are analytic in the open unit disk U = {z ∈ C : |z| < 1}.
Sălăgean [5] has introduced the following operator Dn : A −→ A defined by
1
Received 17 January, 2009
Accepted for publication (in revised form) 11 February, 2009
53
54
H. Mahzoon, S. Latha
0
D0 f (z) = f (z), D1 f (z) = D(f (z)) = zf (z) and Dn f (z) = D(Dn−1 f (z)), for
n ∈ N = {1, 2, 3, ...}. We note that if f is of the form ( 1), then
∞
X
Dn f (z) = z +
(2)
mn am z m
m=2
Let Ap be the class of functions f of the form
(3)
f (z) = z p +
∞
X
am z m ,
(p ∈ N = {1, 2, 3, ...})
m=p+1
which are analytic in the open unit disk U = {z ∈ C : |z| < 1}.
We define Dn : Ap −→ Ap by D0 f (z) = f (z), D1 f (z) = D(f (z)) =
and Dn f (z) = D(Dn−1 f (z)) for n ∈ N = {1, 2, 3, ...}.
We note that if f is of the form ( 3), then
µ ¶n
∞
X
m
n
p
am z m
(4)
D f (z) = z +
p
1 0
zf (z)
p
m=p+1
We introduce the subclass M(α, β, n, p) consisting of functions f ∈ Ap which
satisfy
¯
¯
(
)
0
¯ z(Dn f (z))0
¯
z(Dn f (z))
¯
¯
(5)
<
<
α
−
1
(z ∈ U)
¯
¯ + β,
n
n
¯ D f (z)
¯
D f (z)
for some α ≤ 0, β > 1 and n ∈ N0 = {0, 1, 2, 3, ...}.
We note that M(α, β, 0, 1) ≡ MD(α, β) and M(α, β, 1, 1) ≡ N D(α, β). The
classes MD(α, β) and N D(α, β) were studied by Owa [3] and M(α, β, n, 1) ≡
M(α, β, n), the class introduced by Mahzoon and Latha [2].
2
Coefficient inequalities
We derive sufficient conditions for f to be in the class M(α, β, n, p) in terms
of coefficient inequalities.
Theorem 1 If f ∈ Ap satisfies
µ ¶n
∞
X
m
{|m − β + 1| + |m − β − 1| − 2α(m − 1)} |am |
p
m=p+1
On certain classes of p-valent functions defined by Sălăgean operator
55
≤ {|(1 + β − p)| − |(p − β + 1)| + 2α(p − 1)}
for some α ≤ 0, β > 1 and n ∈ N0 , then f ∈ M(α, β, n, p).
Proof. We suppose that
µ ¶n
∞
X
m
{|m − β + 1| + |m − β − 1| − 2α(m − 1)}|am |
p
m=p+1
≤ {|(1 + β − p)| − |(p − β + 1)| + 2α(p − 1)}
for f ∈ Ap . It suffices to show that
¯
¯
!
¯Ã
¯
¯ z(Dn f (z))0
¯
¯ z(Dn f (z))0
¯
¯
¯
¯
¯
−
α
−
1
−
β
+
1
¯
¯
¯
¯
¯ Dn f (z)
¯
Dn f (z)
¯
¯
¯Ã
¯ < 1,
¯
¯
!
0
0
¯
¯
¯ z(Dn f (z))
¯
¯ z(Dn f (z))
¯
¯
¯
¯
−α¯
− 1¯ − β − 1 ¯
n
n
¯
¯
¯
¯
D f (z)
D f (z)
(z ∈ U).
We have
¯
¯
!
¯ Ã
¯
0
¯ z(Dn f (z))0
¯
¯
¯
z(Dn f (z))
¯
¯
¯
¯
−
α
−
1
−
β
+
1
¯
¯
¯
¯
n f (z)
n f (z)
¯
¯
D
D
¯
¯
¯Ã
¯
¯
¯
!
0
0
¯
¯
¯
¯
n
n
¯ z(D f (z))
¯
¯ z(D f (z))
¯
¯
¯
−
α
−
1
−
β
−
1
¯
¯
¯
¯
¯ Dn f (z)
¯
Dn f (z)
¯
¯
¯ z(Dn f (z))0 − αeiθ |z(Dn f (z))0 − (Dn f (z))| − βDn f (z) + Dn f (z) ¯
¯
¯
=¯
¯=
0
0
n
iθ
n
n
n
n
¯ z(D f (z)) − αe |z(D f (z)) − (D f (z))| − βD f (z) − D f (z) ¯
¯
¯
∞
∞
X
X
¯
¯ n
n
m−p
iθ n
n
m−p
¯ p (p−β +1)+
m (m−β +1)am z
−αe |p (p−1)+
m (m−1)am z
| ¯¯
¯
¯
¯
m=p+1
m=p+1
¯
¯
∞
∞
¯
¯
X
X
¯ n
n
m−p
iθ n
n
m−p ¯
m (m−β −1)am z
+αe |p (p−1)+
m (m−1)am z
|} ¯
¯−{p (1+β −p)−
¯
¯
m=p+1
∞
X
|pn (p−β +1|+
<
m=p+1
∞
X
mn |m−β +1||am |−αpn (p−1)−α
m=p+1
∞
X
|pn (1+β −p)|−
n
n
mn (m−1)|am |
m=p+1
∞
X
m |m−β −1||am |+αp (p−1)+α
m=p+1
.
n
m (m−1)|am |
m=p+1
The last expression is bounded above by 1 if
|pn (p − β + 1| +
∞
X
m=p+1
mn |m − β + 1||am | − αpn (p − 1) − α
∞
X
m=p+1
mn (m − 1)|am |
56
H. Mahzoon, S. Latha
≤ |pn (1+β−p)|−
∞
X
∞
X
mn |m−β+1||am |+αpn (p−1)+α
m=p+1
mn (m−1)|am |
m=p+1
which is equivalent to our condition
µ ¶n
∞
X
m
{|m − β + 1| + |m − β − 1| − 2α(m − 1)}|am |
p
m=p+1
≤ {|(1 + β − p)| − |(p − β + 1)| + 2α(p − 1)}
of the Theorem.
3
Relation for M? (α, β, n, p)
By Theorem 1, the class M? (α, β, n, p) is considered as the subclass of
M(α, β, n, p) consisting of f satisfying
µ ¶n
∞
X
m
{|m − β + 1| + |m − β − 1| − 2α(m − 1)}|am |
p
m=p+1
(6)
≤ {|(1 + β − p)| − |(p − β + 1)| + 2α(p − 1)}
for some α ≤ 0, β > 1 and n ∈ N0 .
We note that M? (α, β, 0, 1) ≡ MD? (α, β) and M? (α, β, 1, 1) ≡ N D? (α, β).
The classes MD? (α, β) and N D? (α, β) were studied by Owa [3] and
M? (α, β, n, 1) ≡ M? (α, β, n), the class studied by Mahzoon and Latha [2] .
Theorem 2 If f ∈ Ap , then M? (α1 , β, n, p) ⊆ M? (α2 , β, n, p) for some
α1 , α2 (α1 ≤ α2 ≤ 0).
Proof. For α1 ≤ α2 ≤ 0, we have
µ ¶n
∞
X
m
{|m − β + 1| + |m − β − 1| − 2α2 (m − 1)}|am |
p
m=p+1
µ ¶n
∞
X
m
≤
{|m − β + 1| + |m − β − 1| − 2α1 (m − 1)}|am |
p
m=p+1
and also,
(7)
{|(1 + β − p)| − |(p − β + 1)| + 2α1 (p − 1)}
(8)
≤ {|(1 + β − p)| − |(p − β + 1)| + 2α2 (p − 1)}
Therefore, if f ∈ M(α1 , β, n, p), then f ∈ M? (α2 , β, n, p).
On certain classes of p-valent functions defined by Sălăgean operator
57
Convolution of the class M? (α, β, n, p)
4
For analytic functions f1 and f2 given by
(9)
p
f1 (z) = z +
∞
X
am,1 z
m
p
and f2 (z) = z +
m=p+1
∞
X
am,2 z m
m=p+1
the Hadamard product ( f1 ∗ f2 )(z) of f1 and f2 is defined by
( f1 ∗ f2 )(z) = z p +
(10)
∞
X
am,1 am,2 z m
m=p+1
?
Theorem 3 µIf f1 ∈ M? (α, β1 , n, p) and f2 ∈ M
¶ (α, β2 , n, p) for some α
√
β1 −p
β2 − p
(α ≤ 2− 5), p =
6 1 : α>
and α >
, β1 , β2 (p < β1 , β2 ≤ p + 1)
1−p
1−p
Q+R
and n ∈ N0 , then ( f1 ∗ f2 ) ∈ M? (α, β, n, p), where β = 1 +
with
S
Q = pn [(β1 − p) + α(p − 1)][(β2 − p) + α(p − 1)][p(1 − α) + 1]
R = (p + 1)n (p − 1)(1 − α)[p(1 − α) + (1 − β1 )][p(1 − α) + (1 − β2 )]
S = pn [(β1 − p) + α(p − 1)][(β2 − p) + α(p − 1)]+
+(p + 1)n [p(1 − α) + (1 − β1 )][p(1 − α) + (1 − β2 )].
Proof. From ( 6), for f ∈ M? (α, β, n, p) with p < β ≤ p + 1 and α >
(p 6= 1) , we have
µ ¶n
∞
X
m
(11)
{(m − β + 1) + (m − β − 1) − 2α(m − 1)}|am |
p
β−p
,
1−p
m=p+1
µ ¶n
∞
X
m
{(m−β+1)+|m−β−1|−2α(m−1)}|am | ≤ 2pn [(β−p)+α(p−1)].
≤
p
m=p+1
That is, if f ∈ M? (α, β, n, p), then
µ ¶n
∞
X
m
[m(1 − α) + (α − β)]
(12)
|am | ≤ 1.
p
[(β − p) + α(p − 1)]
m=p+1
Conversely, if f satisfies
µ ¶n
∞
X
m
[m(1 − α) + (1 − β + α)]
|am | ≤ 1,
(13)
p
[(β − p) + α(p − 1)]
m=p+1
58
H. Mahzoon, S. Latha
then f ∈ M? (α, β, n, p). From ( 12), if f1 ∈ M(α, β1 , n, p), then
µ ¶n
∞
X
m
[m(1 − α) + (α − β1 )]
(14)
|am,1 | ≤ 1
p
[(β1 − p) + α(p − 1)]
m=p+1
and also if f2 ∈ M? (α, β2 , n, p), then
µ ¶n
∞
X
m
[m(1 − α) + (α − β2 )]
|am,2 | ≤ 1.
p
[(β2 − p) + α(p − 1)]
(15)
m=p+1
Applying the Schwarz’s inequality, we have the following inequality
(16) s
µ ¶2n
∞
q
X
m
[m(1 − α) + (α − β1 )][m(1 − α) + (α − β2 )]
|am,1 ||am,2 | ≤ 1
p
[(β1 − p) + α(p − 1)][(β2 − p) + α(p − 1)]
m=p+1
by ( 14) and ( 15). From ( 13) and ( 16), if the following inequality
µ ¶n
∞
X
[m(1 − α) + (1 − β + α)]
m
|am,1 ||am,2 |
p
[(β − p) + α(p − 1)]
m=p+1
(17)
∞
X
≤
sµ
m=p+1
m
p
¶2n
[m(1 − α) + (α − β1 )][m(1 − α) + (α − β2 )]
[β1 − p) + α(p − 1)][(β2 − p) + α(p − 1)]
q
|am,1 ||am,2 |
is satisfied, then we say that (f1 ∗ f2 ) ∈ M? (α, β, n, p) this inequality holds
true if
µ ¶n
q
m
[m(1 − α) + (1 − β + α)]
|am,1 ||am,2 |
p
[(β − p) + α(p − 1)]
sµ
(18)
≤
m
p
¶2n
[m(1 − α) + (α − β1 )][m(1 − α) + (α − β2 )]
[(β1 − p) + α(p − 1)][(β2 − p) + α(p − 1)]
for all m ≥ p + 1. Therefore, we have
µ ¶n
m
[m(1 − α) + (1 − β + α)]
p
[(β − p) + α(p − 1)]
µ
(19)
≤
m
p
¶2n
[m(1 − α) + (α − β1 )][m(1 − α) + (α − β2 )]
[β1 − p) + α(p − 1)][(β2 − p) + α(p − 1)]
On certain classes of p-valent functions defined by Sălăgean operator
59
Q+R
where
S
n
Q = p [(β1 − p) + α(p − 1)][(β2 − p) + α(p − 1)][p(1 − α) + 1]
R = (p + 1)n (p − 1)(1 − α)[p(1 − α) + (1 − β1 )][p(1 − α) + (1 − β2 )]
S = pn [(β1 − p) + α(p − 1)][(β2 − p) + α(p − 1)]+
+(p + 1)n [p(1 − α) + (1 − β1 )][p(1 − α) + (1 − β2 )].
for all m ≥ p + 1.
Let R(m) be the right hand side of the last inequality. Further, let us define
0
S(m) by numerator of R (m). Then, S(m) gives us that,
S(m) = pn [(β1 − p) + α(p − 1)][(β2 − p) + α(p − 1)]{pn [(β1 − p)
+α(p − 1)][(β2 − p) + α(p − 1)](1 − α) + 2mn (p − 1)(1 − α)2 [m(1 − α) + α]
+mn−1 (1 − α)[n(p − 1) − m][m(1 − α) + α]2 − nmn−1 [m(1 − α) + α]3
−nmn−1 (p−1)(1−α)[m(1−α)+α]β1 −nmn−1 (p−1)(1−α)[m(1−α)+α]β2
+nmn−1 [m(1−α)+α]2 β1 +nmn−1 [m(1−α)+α]2 β2 −nmn−1 [m(1−α)+α]β1 β2
+mn−1 (1−α)[m+n(p−1)]β1 β2 −mn (p−1)(1−α)2 β1 −mn (p−1)(1−α)2 β2 }
≤ pn [(β1 − p) + α(p − 1)][(β2 − p) + α(p − 1)]{pn (1 − α)
+2mn (p−1)(1−α)2 [m(1−α)+α]+mn−1 (1−α)[n(p−1)−m][m(1−α)+α]2
−nmn−1 [m(1 − α) + α]3 − 2nmn−1 (p − 1)(1 − α)[m(1 − α) + α]p
+2nmn−1 [m(1 − α) + α]2 (p + 1) − nmn−1 [m(1 − α) + α]p2
+mn−1 (1 − α)[m + n(p − 1)](p + 1)2 − 2mn (p − 1)(1 − α)2 p} ≤ 0
¶
µ
√
β2 − p
β1 − p
and α >
which shows that
for α ≤ 2 − 5 and p 6= 1 : α >
1−p µ
1−p
¶
√
β1 −p
β2 − p
R(m) is decreasing for m ≥ p+1, α ≤ 2− 5, p 6= 1 : α >
and α >
1−p
1−p
and for n ∈ N0 , p ∈ N.
Thus, R(p + 1) is the maximum of R(m) for α given by
µ
¶
√
β1 − p
β2 − p
α ≤ 2 − 5, p 6= 1 : α >
and α >
1−p
1−p
which is equivalent to β = 1 +
and for n ∈ N0 , p ∈ N.
References
[1] A.W. Goodman, Univalent functions, Volume I and Volume II, Mariner
Publ. Co. Inc. Tampa, Florida,1983.
[2] H. Mahzoon, S. Latha, A note on certain classes of analytic functions
defined by Sălăgean operator,(Communicated for publication).
60
H. Mahzoon, S. Latha
[3] J. Nishiwaki, S. Owa, Certain classes of analytic functions concerned with
uniformly starlike and convex functions, Applied Mathematics and Computation, 187, 2007,350 -355.
[4] J. Nishiwaki, S. Owa, Coefficient inequalities for analytic functions, Int. J.
Math. Math. Sci. 29 , 2002, 285-290.
[5] G.S. Sălăgean, Subclass of univalent functions, Lecture Notes in Math.,
Springer, Berlin, 1983, 362-372.
[6] B.A. Uralegaddi, M. D. Ganigi, S. M. Sarangi, Univalent functions with
positive Coefficients, Tamkang J.Math.25 1994, 225-230.
Hesam Mahzoon
Department of Studies in Mathematics
University of Mysore, Manasagangotri
Mysore-5700 06, Karnataka, India
e-mail: mahzoon hesam@yahoo.com
S. Latha
Department of Mathematics, Yuvaraja’s College
University of Mysore
Mysore - 570 005, India
e-mail: drlatha@gmail.com