Volume 9 (2008), Issue 3, Article 81, 7 pp.
ON STARLIKENESS AND CONVEXITY OF ANALYTIC FUNCTIONS
SATISFYING A DIFFERENTIAL INEQUALITY
SUKHWINDER SINGH, SUSHMA GUPTA, AND SUKHJIT SINGH
D EPARTMENT OF A PPLIED S CIENCES
BABA BANDA S INGH BAHADUR E NGINEERING C OLLEGE
FATEHGARH S AHIB -140407 (P UNJAB )
INDIA.
ss_billing@yahoo.co.in
D EPARTMENT OF M ATHEMATICS
S ANT L ONGOWAL I NSTITUTE OF E NGINEERING & T ECHNOLOGY
L ONGOWAL -148106 (P UNJAB )
INDIA.
sushmagupta1@yahoo.com
sukhjit_d@yahoo.com
Received 03 March, 2008; accepted 10 July, 2008
Communicated by S.S. Dragomir
A BSTRACT. In the present paper, the authors investigate a differential inequality defined by
multiplier transformation in the open unit disk E = {z : |z| < 1}. As consequences, sufficient
conditions for starlikeness and convexity of analytic functions are obtained.
Key words and phrases: Multivalent function, Starlike function, Convex function, Multiplier transformation.
2000 Mathematics Subject Classification. 30C45.
1. I NTRODUCTION
P
k
Let Ap denote the class of functions of the form f (z) = z p + ∞
k=p+1 ak z , p ∈ N =
{1, 2, . . . }, which are analytic in the open unit disc E = {z : |z| < 1}. We write A1 = A. A
function f ∈ Ap is said to be p-valent starlike of order α (0 ≤ α < p) in E if
0 zf (z)
> α, z ∈ E.
<
f (z)
We denote by Sp∗ (α), the class of all such functions. A function f ∈ Ap is said to be p-valent
convex of order α (0 ≤ α < p) in E if
zf 00 (z)
> α, z ∈ E.
< 1+ 0
f (z)
064-08
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S UKHWINDER S INGH , S USHMA G UPTA ,
AND
S UKHJIT S INGH
Let Kp (α) denote the class of all those functions f ∈ Ap which are multivalently convex of
order α in E. Note that S1∗ (α) and K1 (α) are, respectively, the usual classes of univalent
starlike functions of order α and univalent convex functions of order α, 0 ≤ α < 1, and will
be denoted here by S ∗ (α) and K(α), respectively. We shall use S ∗ and K to denote S ∗ (0) and
K(0), respectively which are the classes of univalent starlike (w.r.t. the origin) and univalent
convex functions.
For f ∈ Ap , we define the multiplier transformation Ip (n, λ) as
n
∞ X
k+λ
p
ak z k , (λ ≥ 0, n ∈ Z).
(1.1)
Ip (n, λ)f (z) = z +
p
+
λ
k=p+1
The operator Ip (n, λ) has recently been studied by Aghalary et.al. [1]. Earlier, the operator
I1 (n, λ) was investigated by Cho and Srivastava [3] and Cho and Kim [2], whereas the operator
I1 (n, 1) was studied by Uralegaddi and Somanatha [11].PI1 (n, 0) is the well-known Sălăgean
n
k
[10] derivative operator Dn , defined as: Dn f (z) = z + ∞
k=2 k ak z , n ∈ N0 = N ∪ {0} and
f ∈ A.
A function f ∈ Ap is said to be in the class Sn (p, λ, α) for all z in E if it satisfies
α
Ip (n + 1, λ)f (z)
(1.2)
<
> ,
Ip (n, λ)f (z)
p
for some α (0 ≤ α < p, p ∈ N). We note that S0 (1, 0, α) and S1 (1, 0, α) are the usual classes
S ∗ (α) and K(α) of starlike functions of order α and convex functions of order α, respectively.
In 1989, Owa, Shen and Obradovic̆ [8] obtained a sufficient condition for a function f ∈ A
to belong to the class Sn (1, 0, α) = Sn (α).
Recently, Li and Owa [4] studied the operator I1 (n, 0).
In the present paper, we investigate the differential inequality
(1 − α)Ip (n + 1, λ)f (z) + αIp (n + 2, λ)f (z)
<
> M (α, β, γ, λ, p)
(1 − β)Ip (n, λ)f (z) + βIp (n + 1, λ)f (z)
where α and β are real numbers and M (α, β, γ, λ, p) is a certain real number given in Section
2, for starlikeness and convexity of f ∈ Ap . We obtain sufficient conditions for f ∈ Ap to be
a member of Sn (p, λ, γ), for some γ (0 ≤ γ < p, p ∈ N). Many known results for starlikeness
appear as corollaries to our main result and some new results regarding convexity of analytic
functions are obtained.
2. M AIN R ESULT
We shall make use of the following lemma of Miller and Mocanu to prove our result.
Lemma 2.1 ([6, 7]). Let Ω be a set in the complex plane C and let ψ : C2 × E → C. For u =
u1 +iu2 , v = v1 +iv2 , assume that ψ satisfies the condition ψ(iu2 , v1 ; z) ∈
/ Ω, for all u2 , v1 ∈ R,
with v1 ≤ −(1 + u22 )/2 and for all z ∈ E. If the function p, p(z) = 1 + p1 z + p2 z 2 + · · · , is
analytic in E and if ψ(p(z), zp0 (z); z) ∈ Ω, then < p(z) > 0 in E.
We, now, state and prove our main theorem.
Theorem 2.2. Let α ≥ 0, β ≤ 1, λ ≥ 0 and 0 ≤ γ < p be real numbers such that β(1 − γp ) <
and β ≤ α. If f ∈ Ap satisfies the condition
(1 − α)Ip (n + 1, λ)f (z) + αIp (n + 2, λ)f (z)
(2.1)
<
> M (α, β, γ, λ, p),
(1 − β)Ip (n, λ)f (z) + βIp (n + 1, λ)f (z)
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 81, 7 pp.
1
2
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S TARLIKENESS AND C ONVEXITY OF A NALYTIC F UNCTIONS
3
then
<
Ip (n + 1, λ)f (z)
Ip (n, λ)f (z)
>
γ
p
i.e., f (z) ∈ Sn (p, λ, γ) where,
M (α, β, γ, λ, p) =
(1−α)γ
p
+
αγ 2
p2
−
1−β 1−
α(1− γp )
2(p+λ)
γ
p
.
Proof. Since 0 ≤ γ < p, let us write µ = γp . Thus, we have 0 ≤ µ < 1.
Now we define,
Ip (n + 1, λ)f (z)
= µ + (1 − µ)r(z),
Ip (n, λ)f (z)
(2.2)
z ∈ E.
Therefore r(z) is analytic in E and r(0) = 1.
Differentiating (2.2) logarithmically, we obtain
0
(2.3)
0
0
zIp (n + 1, λ)f (z) zIp (n, λ)f (z)
(1 − µ)zr (z)
−
=
,
Ip (n + 1, λ)f (z)
Ip (n, λ)f (z)
µ + (1 − µ)r(z)
z ∈ E.
Using the fact that
0
zIp (n, λ)f (z) = (p + λ)Ip (n + 1, λ)f (z) − λIp (n, λ)f (z).
Thus (2.3) reduces to
0
Ip (n + 2, λ)f (z)
(1 − µ)zr (z)
= µ + (1 − µ)r(z) +
.
Ip (n + 1, λ)f (z)
(λ + p)[µ + (1 − µ)r(z)]
Now, a simple calculation yields
(1 − α)Ip (n + 1, λ)f (z) + αIp (n + 2, λ)f (z)
(1 − β)Ip (n, λ)f (z) + βIp (n + 1, λ)f (z)
0
(1−µ)zr (z)
(1 − α) + α µ + (1 − µ)r(z) + (λ+p)[µ+(1−µ)r(z)]
=
[µ + (1 − µ)r(z)]
(1 − β) + β[µ + (1 − µ)r(z)]
0
(z)
(1 − α)[µ + (1 − µ)r(z)] + α [µ + (1 − µ)r(z)]2 + (1−µ)zr
(λ+p)
=
(1 − β) + β[µ + (1 − µ)r(z)]
(2.4)
0
= ψ(r(z), zr (z); z)
where,
2
(1 − α)[µ + (1 − µ)u] + α (µ + (1 − µ)u) +
ψ(u, v; z) =
(1 − β) + β[µ + (1 − µ)u]
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 81, 7 pp.
(1−µ)v
(λ+p)
.
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S UKHWINDER S INGH , S USHMA G UPTA ,
AND
S UKHJIT S INGH
Let u = u1 + iu2 and v = v1 + iv2 , where u1 , u2 , v1 , v2 are reals with v1 ≤ −
have
1+u22
.
2
Then, we
< ψ(iu2 , v1 ; z)
=
[(1 − α)µ + αµ2 ][1 − β(1 − µ)]
[1 − β(1 − µ)]2 + β 2 (1 − µ)2 u22
+
h
≤
(1 − µ)2 [(1 − α)β − α(1 − β(1 − µ)) + 2αβµ]u22 +
α(1−µ)[1−β(1−µ)]v1
p+λ
[1 − β(1 − µ)]2 + β 2 (1 − µ)2 u22
i
(1 − α)µ + αµ2 − α(1−µ)
[1 − β(1 − µ)]
2(λ+p)
[1 − β(1 − µ)]2 + β 2 (1 − µ)2 u22
h
(1 − µ)2 [(1 − α)β − α(1 − β(1 − µ)) + 2αβµ] −
+
α(1−µ)[1−β(1−µ)]
2(p+λ)
i
u22
[1 − β(1 − µ)]2 + β 2 (1 − µ)2 u22
A + Bu22
[1 − β(1 − µ)]2 + β 2 (1 − µ)2 u22
= φ(u2 ), say
=
≤ max φ(u2 )
(2.5)
where,
α(1 − µ)
2
[1 − β(1 − µ)]
A = (1 − α)µ + αµ −
2(λ + p)
and
B = (1 − µ)2 [(1 − α)β − α(1 − β(1 − µ)) + 2αβµ] −
α(1 − µ)[1 − β(1 − µ)]
.
2(p + λ)
0
It can be easily verified that φ (u2 ) = 0 implies that u2 = 0. Under the given conditions, we
00
observe that φ (0) < 0. Therefore,
(2.6)
max φ(u2 ) = φ(0) = M (α, β, γ, λ, p).
Let
Ω = {w : < w > M (α, β, γ, λ, p)}.
Then from (2.1) and (2.4), we have ψ(r(z), zr0 (z); z) ∈ Ω for all z ∈ E, but ψ(iu2 , v1 ; z) ∈
/ Ω,
in view of (2.5) and (2.6). Therefore, by Lemma 2.1 and (2.2), we conclude that
Ip (n + 1, λ)f (z)
γ
<
> .
Ip (n, λ)f (z)
p
3. C OROLLARIES
By taking p = 1 and λ = 0 in Theorem 2.2. We have the following corollary.
Corollary 3.1. Let α ≥ 0, β ≤ 1 and 0 ≤ γ < 1 be real numbers such that β(1 − γ) <
β ≤ α. If f ∈ A satisfies the condition
(1 − α)Dn+1 f (z) + αDn+2 f (z)
<
> M (α, β, γ, 0, 1),
(1 − β)Dn f (z) + βDn+1 f (z)
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 81, 7 pp.
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2
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S TARLIKENESS AND C ONVEXITY OF A NALYTIC F UNCTIONS
then
5
Dn+1 f (z)
> γ,
Dn f (z)
<
i.e. f (z) ∈ Sn (γ), where,
(1 − α)γ + αγ 2 − α(1−γ)
2
M (α, β, γ, 0, 1) =
.
1 − β(1 − γ)
By taking p = 1, n = 0 and λ = 0 in Theorem 2.2. We have the following corollary.
Corollary 3.2. Let α ≥ 0, β ≤ 1 and 0 ≤ γ < 1 be real numbers such that β(1 − γ) <
β ≤ α. If f ∈ A satisfies the condition
zf 0 (z) + αz 2 f 00 (z)
<
> M (α, β, γ, 0, 1),
(1 − β)f (z) + βzf 0 (z)
then
<
1
2
and
zf 0 (z)
> γ,
f (z)
i.e. f (z) ∈ S ∗ (γ), where,
(1 − α)γ + αγ 2 − α(1−γ)
2
M (α, β, γ, 0, 1) =
.
1 − β(1 − γ)
By taking p = 1, n = 0, λ = 0 and β = 1 in Theorem 2.2. We have the following corollary.
Corollary 3.3. Let α ≥ 1 and
1
2
< γ < 1 be real numbers. If f ∈ A satisfies the condition
zf 00 (z)
< 1+α 0
> M (α, 1, γ, 0, 1),
f (z)
then
<
zf 0 (z)
> γ,
f (z)
i.e. f (z) ∈ S ∗ (γ), where
1
M (α, 1, γ, 0, 1) = 1 − α(1 − γ) 1 +
2γ
By taking p = 1, n = 0, λ = 0 and β = 0 in Theorem 2.2, we have the following result of
Ravichandran et. al. [9].
Corollary 3.4. Let α ≥ 0 and 0 ≤ γ < 1 be real numbers. If f ∈ A satisfies the condition
zf 0 (z)
zf 00 (z)
<
1+α 0
> M (α, 0, γ, 0, 1),
f (z)
f (z)
then
<
zf 0 (z)
> γ,
f (z)
i.e. f (z) ∈ S ∗ (γ), where,
α(1 − γ)
.
2
Remark 1. In the case when γ = α2 , Corollary 3.4 reduces to the result of Li and Owa [5].
M (α, 0, γ, 0, 1) = (1 − α)γ + αγ 2 −
By taking p = 1, n = 0 and λ = 1 in Theorem 2.2, we have the following corollary.
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 81, 7 pp.
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S UKHWINDER S INGH , S USHMA G UPTA ,
AND
S UKHJIT S INGH
Corollary 3.5. Let α ≥ 0, β ≤ 1 and 0 ≤ γ < 1 be real numbers such that β(1 − γ) <
β ≤ α. If f ∈ A satisfies the condition
1 (2 − α)f (z) + (2 + α)zf 0 (z) + αz 2 f 00 (z)
<
> M (α, β, γ, 1, 1),
2
(2 − β)f (z) + βzf 0 (z)
then
1
zf 0 (z)
<
1+
> γ,
2
f (z)
where,
(1 − α)γ + αγ 2 − α(1−γ)
4
M (α, β, γ, 1, 1) =
.
1 − β(1 − γ)
1
2
and
1
2
and
By taking p = 1, n = 1 and λ = 0 in Theorem 2.2, we have the following corollary.
Corollary 3.6. Let α ≥ 0, β ≤ 1 and 0 ≤ γ < 1 be real numbers such that β(1 − γ) <
β ≤ α. If f ∈ A satisfies the condition
0
000
zf (z) + (2α + 1)z 2 f 00 (z) + αz 3 f (z)
<
> M (α, β, γ, 0, 1),
zf 0 (z) + βz 2 f 00 (z)
then
zf 00 (z)
> γ,
< 1+ 0
f (z)
i.e. f (z) ∈ K(γ), where,
(1 − α)γ + αγ 2 − α(1−γ)
2
.
M (α, β, γ, 0, 1) =
1 − β(1 − γ)
By taking p = 1, n = 1, λ = 0 and β = 0 in Theorem 2.2, we have the following corollary.
Corollary 3.7. Let α ≥ 0 and 0 ≤ γ < 1 be real numbers. If f ∈ A satisfies the condition
000
zf 00 (z)
z 2 f (z)
< 1 + (2α + 1) 0
+α 0
> M (α, 0, γ, 0, 1),
f (z)
f (z)
then
zf 00 (z)
< 1+ 0
> γ,
f (z)
i.e., f (z) ∈ K(γ), where,
α(1 − γ)
.
2
Remark 2. In the main result, the real number M (α, β, γ, λ, p) may not be the best possible
as authors have not obtained the extremal function for it. The problem is still open for the best
possible real number M (α, β, γ, λ, p).
M (α, 0, γ, 0, 1) = (1 − α)γ + αγ 2 −
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J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 81, 7 pp.
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