DEMONSTRATIO MATHEMATICA
Vol. XLIII
No 4
2010
J. K. Prajapat, R. K. Raina
NEW SUFFICIENT CONDITIONS FOR STARLIKENESS
OF ANALYTIC FUNCTIONS INVOLVING A FRACTIONAL
DIFFERINTEGRAL OPERATOR
Abstract. In this paper we apply a fractional differintegral operator to a class of
analytic functions and derive certain new sufficient conditions for the starlikeness of this
class of functions. The usefulness of the main results are depicted by deducing several
interesting corollaries and relevances with some of the earlier results are also pointed out.
1. Introduction and definitions
Let H(U) represent a space of analytic functions in the open unit disk
U = {z ∈ C : |z| < 1}, then for a ∈ C and n ∈ N, we let An and H[a, n]
denote, respectively, the subclasses of the class H(U) defined by
(1.1)
An = f ∈ H(U); f (z) = z + an+1 z n+1 + . . . , ; z ∈ U
and
(1.2)
H[a, n] = {f ∈ H(U); f (z) = a + an z n + an+1 z n+1 + ... ; z ∈ U},
with A1 = A. A function f ∈ An is said to be in the class of starlike functions
of order α (0 ≤ α < 1) in U, denoted by Sn∗ (α), if
′ zf (z)
(1.3)
ℜ
>α
(z ∈ U).
f (z)
For α = 0, the class Sn∗ (0) = Sn∗ is the class of starlike functions in U. Let
P(η) denote the class of functions f ∈ An of the form (1.1) which satisfies
the inequality
f (z)
(1.4)
(z ∈ U).
z − 1 < η
2000 Mathematics Subject Classification: 30C45.
Key words and phrases: analytic functions, starlike function, fractional differintegral
operators, differential subordination.
806
J. K. Prajapat, R. K. Raina
If f and g are two analytic in U, then we say that the function f (z) is
subordinate to g(z) in U and write f (z) ≺ g(z) (z ∈ U), if there exists a
Schwarz function w(z) (analytic in U with w(0) = 0, and |w(z)| < 1), such
that f (z) = g(w(z)), z ∈ U. In particular, if the function g(z) is univalent
in U, the above subordination is equivalent to f (0) = g(0) and f (U) ⊂ g(U).
We recall here the following definition of fractional calculus (that is fractional integral and fractional derivative of an arbitrary order) considered by
Owa [8] (see also, Dziok [1] and Owa and Srivastava [11]).
Definition 1. The fractional integral of order λ (λ > 0) is defined for
a function f (z) analytic in a simply-connected region of the complex plane
containing the origin by
\
z
(1.5)
Dz−λ f (z)
1
f (ζ)
=
dζ,
Γ(λ) 0 (z − ζ)1−λ
where the multiplicity (z − ζ)λ−1 is removed by requiring log(z − ζ) to be
real when z − ζ > 0.
Definition 2. Under the hypothesis of Definition 1, the fractional derivative operator of order λ (λ ≥ 0) is defined by

z
1
d
f (ζ)


dζ (0 ≤ λ < 1),

Γ(1 − λ) dz 0 (z − ζ)λ
λ
(1.6) Dz f (z) =
n


 d Dλ−n f (z) (n ≤ λ < n + 1; n ∈ N = N ∪ {0}),
0
dz n z
\
where the multiplicity (z − ζ)−λ is removed as in Definition 1.
For the purpose of this paper, we define here a fractional differintegral
operator
Ωλz : An → An (−∞ < λ < 2; n ∈ N)
for a function f (z) of the form (1.1) by
(1.7)
Ωλz f (z) = z +
∞
X
Γ(2 − λ) Γ(k + 1)
ak z k
Γ(k + 1 − λ)
k=n+1
= Γ(2 − λ) z λ Dzλ f (z)
(−∞ < λ < 2; z ∈ U),
where Dzλ f (z) in (1.7) represents, respectively, a fractional integral of f (z)
of order λ when −∞ < λ < 0 and a fractional derivative of f (z) of order λ
when 0 ≤ λ < 2. It is easily seen from (1.7) that
Ω0z f (z) = f (z),
Ω1z f (z) = zf ′ (z)
New sufficient conditions of starlikeness of analytic functions
807
\
and
z
2
f (ζ) dζ := L(f )(z),
z0
Ω−1
z f (z) =
where L(f )(z) denotes the Libera integral operator [3]. We also observe that
the fractional differintegral operator Ωλz f (z) satisfies the following three-term
recurrence relation:
′
(1.8) z Ωλz f (z) = (1 − λ)Ωλ+1
f (z) + λ Ωλz f (z) (−∞ < λ < 1; z ∈ U).
z
The fractional operator Ωλz (0 ≤ λ < 1) was ealier studied by Srivastava
and Aouf [10] (see also, Patel and Mishra [9]).
The object of the present note is to obtain sufficient conditions for fractional differintegral operator Ωλz f (z) to be in the class of starlike functions
of order α. We also consider some special cases of our results which lead
to various interesting corollaries, and relevances of some of these with other
known results are also mentioned.
2. Main results
Before stating and proving our main results, we require the following
lemmas.
Lemma 1. (Hallenbeck and Ruscheweyh [2]) Let h(z) be a convex (univalent) function in U with h(0) = a, and let the function φ ∈ H[a, n] and
(2.1)
φ(z) +
zφ′ (z)
≺ h(z)
γ
then
(2.2)
φ(z) ≺ ψ(z) :=
\
z
γ
nz
(ℜ(γ) ≥ 0 (γ 6= 0); z ∈ U),
γ
n
γ
t n −1 h(t) dt ≺ h(z)
(z ∈ U),
0
and ψ(z) is convex and is the best dominant.
Lemma 2. ([5]) Let F (z) and G(z) be analytic functions in the unit disk
U and F (0) = G(0). If H(z) = zG′ (z) is a starlike function in U and
zF ′ (z) ≺ zG′ (z), then
\H(t)
z
F (z) ≺ G(z) = G(0) +
0
t
dt.
Lemma 3. ([6]) Suppose that the function ψ : C2 × U → C satisfies the
condition ℜ{ψ(ix, y; z)} ≤ δ for all x, y ≤ −(1 + x2 )/2 and z ∈ U. If p(z) =
1 + p1 z + . . . is analytic in U and ℜ{ψ(p(z), zp′ (z); z)} > δ for z ∈ U, then
ℜ(p(z)) > 0 in U.
The following result is a simple corollary of Lemma 1.
808
J. K. Prajapat, R. K. Raina
Lemma 4. (see also [4]). Let β < 1. If the function p(z) is analytic in U
with p(0) = 1 and
ℜ p(z) + zp′ (z) > β
(z ∈ U),
then
ℜ(p(z)) > (2β − 1) + 2(1 − β) ln 2
(z ∈ U),
and the result is sharp.
Our first main result is contained in the following:
Theorem 1. Let n be a positive integer, −∞ < µ < 1, 0 ≤ λ < 1,
0 ≤ α < 1 and
(1 − α)(β + n)
p
,
(2.3)
M = Mn (λ, µ, α) =
β{|β + α − 1| + β 2 + (β + n)2 }
where β = (1 − λ)(1 − µ). If
Ωµ+1
f (z) − λΩµz f (z)
z
∈ P(M ),
1−λ
then
Ωµz f (z) ∈ Sn∗ (α).
Proof. Let us assume that
Ωµz f (z)
(−∞ < µ < 1; z ∈ U),
(2.4)
P (z) =
z
and (for convenience sake), we put
1 − λ = l, 1 − µ = m and 1 − α = a,
then in view of (1.7), P (z) is of the form (1.2) and so is analytic in U with
P (0) = 1. Differenting (2.4) and using the identity (1.8), we obtain
(2.5)
Ωµ+1
f (z) − λΩµz f (z)
1
z
= P (z) +
zP ′ (z)
lz
lm
≺ 1 + Mz
(z ∈ U),
and using Lemma 1 (for γ = lm), we deduce that
lmM
z.
(2.6)
P (z) ≺ q(z) = 1 +
lm + n
By setting R = lmM/(lm + n), (2.6) can then be written as
(2.7)
|P (z) − 1| < R,
and if we put
(2.8)
z Ωµz f (z)
Ωµz f (z)
′
= a p(z) − a + 1,
New sufficient conditions of starlikeness of analytic functions
809
where p(z) is of the form (1.2), then in view of (1.7), it is easy to observe
µ
′
z f (z))
that the function z(Ω
is analytic (and has no poles) in the open unit
µ
Ωz f (z)
disk U. Using (1.8), (2.4) and (2.5), we get
P (z)[a p(z) + lm − a]
Ωµ+1 f (z) − λΩµ f (z)
z
z
− 1 = − 1
(2.9)
lz
lm
(lm + n)R
.
lm
We now establish the starlikeness of Ωµz f (z) by showing that ℜ(p(z)) > 0
for all z ∈ U. Suppose this is false, then since p(0) = 1, there exists a point
z0 ∈ U and a real number ρ such that p(z0 ) = iρ. We proceed to show now
that at such a point the opposite of the inequality (2.9) holds true which
means that
<
(2.10)
|P (z0 )(aiρ + lm − a) − lm| ≥ (lm + n)R
for all real ρ. Following [7], if we put P (z0 ) = u(z0 ) + iv(z0 ) = u + iv, then
E = |P (aiρ + lm − a) − lm|2
= (u2 + v 2 )[l2 m2 + a2 (1 + ρ2 ) − 2alm] − 2lm[u(lm − a) − vaρ] + l2 m2
= (u2 + v 2 )a2 ρ2 + 2amlvρ + |P (lm − a) − lm|2 .
We observe that
|P (lm − a) − lm| = |(lm − a)(P − 1) − a)| ≥ a − |lm − a|R
and this implies that
E ≥ (u2 + v 2 )a2 ρ2 + 2vρalm + (a − |lm − a|R)2 .
(2.11)
If we let
(2.12)
F (ρ) = E − l2 m2 M 2
≥ (u2 + v 2 )a2 ρ2 + 2vρalm + (a − |lm − a|R)2 − (lm + n)2 R2 ,
then (2.10) holds if F (ρ) ≥ 0 for any real ρ. The R.H.S of the inequality
(2.12) contains a quadratic expression in ρ whose first coefficient is always
positive, and hence the inequality F (ρ) ≥ 0 would hold if its discriminant
(∆) is negative which yields that
(2.13) ∆ = a2 l2 m2 v 2 − (u2 + v 2 )[(a − |lm − a|R)2 − (lm + n)2 R2 ] ≤ 0.
The inequality (2.13) can be expressed as
(2.14)
v 2 l2 m2 − (a − |lm − a|R)2 + (lm + n)2 R2
≤ u2 (a − |lm − a|R)2 − (lm + n)2 R2
810
J. K. Prajapat, R. K. Raina
and by using (2.7) and (2.14), we get
R2
(a − |lm − a|R)2 − (lm + n)2 R2
v2
≤
≤
,
u2
1 − R2
l2 m2 − (a − |lm − a|R)2 + (lm + n)2 R2
and noting that the denominator in the exteme right-hand side of the above
inequality is positive under the stated conditions with the theorem implies
that ∆ ≤ 0. Elementary calculations reveal that on using the condition
(2.13) in (2.12), the inequality F (ρ) ≥ 0 is not satisfied. This contradicts
our assumption about (2.9). Thus, it follows that ℜ(p(z)) > 0, which in
view of (2.8) is equivalent to
′ z Ωµz f (z)
ℜ
= (1 − α)ℜ(p(z)) + α ≥ α,
Ωµz f (z)
and hence Ωµz f ∈ Sn∗ (α).
Putting µ = α in Theorem 1, we get
Corollary 1. If 0 ≤ λ < 1, 0 ≤ α < 1 and
α+1
Ωz f (z) − λΩαz f (z)
− 1
(1 − λ)z
(1 − λ)(1 − α) + n
p
<
,
(1 − λ) λ(1 − α) + (1 − λ)2 (1 − α)2 + [(1 − λ)(1 − α) + n]2
then Ωαz f (z) ∈ Sn∗ (α).
For λ = 0, Theorem 1 gives
Corollary 2. If −∞ < µ < 1, 0 ≤ α < 1 and
µ+1
Ω
f (z)
(1 − α)(1 − µ + n)
p
− 1 <
,
z
(1 − µ) |α − µ| + (1 − µ)2 + (1 − µ + n)2
then Ωµz f (z) ∈ Sn∗ (α).
Next when µ = −1 in Theorem 1, we get
Corollary 3. If 0 ≤ λ < 1, 0 ≤ α < 1 and
f (z) − λLf (z)
− 1
(1 − λ)z
(1 − α)[2(1 − λ) + n]
p
<
,
2(1 − λ) |1 + α − 2λ| + 4(1 − λ)2 + [2(1 − λ) + n]2
then L(f )(z) ∈ Sn∗ (α).
It may be observed that for µ = 0, Theorem 1 corresponds to the result
of Mocanu and Oras [7, p. 558, Theorem 2.1].
Our next result is given by the following:
811
New sufficient conditions of starlikeness of analytic functions
Theorem 2. If f (z) ∈ An , −∞ < λ < 1, 0 < η ≤ 1, and
λ+1
Ωz f (z) − Ωλz f (z) < η
(z ∈ U),
(2.15)
1−λ
z
then Ωλz f (z) ∈ P(η).
Proof. We note that the condition (2.15) is equivalent to
η
Ωλ+1
f (z) − Ωλz f (z)
z
≺
z
z
1−λ
(2.16)
for z ∈ U. Suppose
Ωλz f (z)
and
G(z) = 1 + ηz,
z
then F (z) and G(z) are analytic functions in U and F (0) = G(0) = 1. Also,
zG′ (z) = ηz is a starlike function in U. Applying now Lemma 2 and in the
process using the recurrence relation (1.8), we arrive at the desired result.
F (z) =
Putting λ = 0, n = 1 in Theorem 2, we get
Corollary 4. If f (z) ∈ A, 0 < η ≤ 1 and
′
f (z) − f (z) < η
(z ∈ U),
z then f (z) ∈ P(η).
On the other hand, if λ = −1, n = 1 in Theorem 2, then we get
Corollary 5. If f (z) ∈ A, 0 < η ≤ 1 and
f (z) − L(f )(z) η
<
2
z
(z ∈ U),
then L(f )(z) ∈ P(η).
Lastly, we prove the following result.
Theorem 3. Let f (z) ∈ An and
′
′ λ+1
λ
(2.17) ℜ (1 − λ) Ωz f (z) + λ Ωz f (z)
> 1 − 2β
(z ∈ U),
then
Ωλz f (z) ∈ Sn∗ ,
provided that
(2.18)
0<β<
3
.
4[2(ln 2 − 1)2 + 1]
Proof. Let f (z) ∈ An such that
ℜ{(1 − λ)(Ωλ+1
f (z))′ + λ(Ωλz f (z))′ } > 1 − 2β
z
(z ∈ U),
812
J. K. Prajapat, R. K. Raina
then by using (1.8) and Lemma 4, we get
(2.19)
ℜ{(Ωλz f (z))′ } > 1 + 4β(ln 2 − 1)
(z ∈ U).
Applying Lemma 4 once more, (2.19) would then yield
λ
Ωz f (z)
> 1 − 8 β(ln 2 − 1)2 > 0
(z ∈ U).
(2.20)
ℜ
z
If we set
Ωλz f (z)
z(Ωλz f (z))′
p(z) =
and
h(z)
=
,
Ωλz f (z)
z
then it is clear that p(z) is analytic in U with p(0) = 1, and (2.20) readily
gives
ℜ(h(z)) > 1 − 8β(ln2 − 1)2 > 0.
(2.21)
On performing simple calculations, we get
(2.22)
(1 − λ)(Ωλ+1
f (z))′ + λ(Ωλz f (z))′ = h(z)[p2 (z) + zp′ (z)]
z
= ψ(p(z), zp′ (z); z),
where ψ(u, v; z) = h(z)(u2 + v). Using (2.17) and (2.22), we infer that
ℜ{ψ(p(z), zp′ (z); z)} > 1 − 2β,
(2.23)
for z ∈ U, and for real x, y ≤ − 21 (1 + x2 ), we find that
1
1
ℜ{ψ(ix, y; z)} = (y − x2 )ℜ(h(z)) ≤ − (1 + 3x2 )ℜ(h(z)) ≤ − ℜ(h(z))
2
2
for z ∈ U. The above inequality in view of (2.21) gives
ℜ{ψ(ix, y; z)} ≤ 1 − 2β,
for z ∈ U, provided that β satisfies (2.18). Hence by Lemma 3, we conlude
that ℜ(p(z)) > 0 in U which implies that Ωλz f (z) ∈ S ∗ .
For λ = 0, Theorem 3 gives
Corollary 6. If f (z) ∈ An and
ℜ{f ′ (z) + zf ′′ (z)} > 1 − 2β
then f (z) ∈
Sn∗ ,
(z ∈ U),
provided that β satisfies (2.18).
For n = 1 Corollary 6 corresponds to the result of Ling and Ding [3].
On the other hand, if λ = −1, then Theorem 3 yields the following result.
Corollary 7. If f (z) ∈ An and
z
2
2
′
ℜ 2f (z) − f (z) + 2 f (t) dt > 1 − 2β
z
z 0
\
then L(f )(z) ∈ Sn∗ , provided that β satisfies (2.18).
(z ∈ U),
New sufficient conditions of starlikeness of analytic functions
813
Acknowledgement. Authors thank the referee for his useful suggestions.
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J. K. Prajapat
DEPARTMENT OF MATHEMATICS
CENTRAL UNIVERSITY OF RAJASTHAN
CITY ROAD, KISHANGARH - 305802
DISTT.-AJMER, RAJASTHAN, INDIA
E-mail: [email protected]
R. K. Raina
M. P. UNIVERSITY OF AGRICULTURE AND TECHNOLOGY
UDAIPUR 313001, RAJASTHAN, INDIA
Present address:
10/11 GANPATI VIHAR, OPPOSITE SECTOR 5,
UDAIPUR 313002, RAJASTHAN, INDIA
E-mail: [email protected]
Received May 7, 2009; revised version October 29, 2009.