Integral Transforms and Special Functions Vol. 18, No. 9, September 2007, 639–651
Some inclusion properties for certain subclasses of strongly starlike and strongly convex functions involving a family of fractional integral operators J. K. PRAJAPAT†, R. K. RAINA‡ and H. M. SRIVASTAVA*§ †Department of Mathematics, Sobhasaria Engineering College NH-11 Gokulpura, Sikar 332001, Rajasthan, India ‡10/11 Ganpati Vihar, Opposite Sector 5, Udaipur 313002, Rajasthan, India §Department of Mathematics and Statistics, University of Victoria, Victoria, British Columbia V8W 3P4, Canada A known family of fractional integral operators (with the Gauss hypergeometric function in the kernel) is used here to define some new subclasses of strongly starlike and strongly convex functions of order β and type α in the open unit disk U. For each of these new function classes, several inclusion relationships associated with the fractional integral operators are established. Some interesting corollaries and consequences of the main inclusion relationships are also considered. Keywords: Analytic functions; Starlike functions; Convex functions; Close-to-convex functions; Strongly starlike functions; Strongly convex functions; Gauss hypergeometric function; Fractional integral operators; Inclusion properties 2000 Mathematics Subject Classification: Primary: 26A33, 30C45, 33C20; Secondary: 30C75
1.
Introduction, definitions and preliminaries
Let A denote the class of functions f (z) normalized by f (z) = z +
∞
an z n ,
(1)
n=2
which are analytic in the open unit disk U = {z : z ∈ C
and
|z| < 1}.
Suppose also that S ∗ (α), K(α) and C(α) denote, respectively, the subclasses of A consisting of functions, which are starlike of order α (0 α < 1) in U, convex of order α (0 α < 1) in U and close-to-convex of order α (0 α < 1) in U. If f (z) ∈ A satisfies the following *Corresponding author. Email:
[email protected]
Integral Transforms and Special Functions ISSN 1065-2469 print/ISSN 1476-8291 online © 2007 Taylor & Francis http://www.tandf.co.uk/journals DOI: 10.1080/10652460701449668
640
inequality:
J. K. Prajapat et al.
arg zf (z) − α < π β (z ∈ U; 0 α < 1; 0 < β 1), f (z) 2
(2)
then the function f (z) is said to be strongly starlike of order β and type α in U. We denote this subclass of A by Ss∗ (α, β). If, on the other hand, f (z) ∈ A satisfies the following inequality: arg 1 + zf (z) − α < π β (z ∈ U; 0 α < 1; 0 < β 1), (3) f (z) 2 then the function f (z) is said to be strongly convex of order β and type α in U. We denote this subclass of A by Kc (α, β). It is obvious that f (z) ∈ Kc (α, β) ⇐⇒ zf (z) ∈ Ss∗ (α, β). We also observe that Ss∗ (0, β) = Ss∗ (β) and
Kc (0, β) = Kc (β),
where Ss∗ (β) and Kc (β) are, respectively, the classes of strongly starlike functions of order β (0 < β 1) in U and strongly convex functions of order β (0 < β 1) in U. Furthermore, we have the following relationships: Ss∗ (α, 1) = S ∗ (α) and
Kc (α, 1) = K(α)
(see, for details, [1] and [2]; see also [3]). λ,μ,η due We recall here the following family of generalized fractional integral operators I0,z to Srivastava et al. [4] (see also [5]). λ,μ,η
DEFINITION 1 Let λ > 0 and μ, η ∈ R. Then the fractional integral operator I0,z is defined by t z−λ−μ z λ,μ,η (z − t)λ−1 2 F1 λ + μ, −η; λ; 1 − f (t)dt, (4) I0,z f (z) = (λ) 0 z where the function f (z) is analytic in a simply-connected region of the complex z-plane containing the origin, with the order f (z) = O(|z|ε ) (z −→ 0; ε > max{0, μ − η} − 1), it being understood that (z − t)λ−1 denotes the principal value satisfying the following inequality 0 arg(z − t) < 2π. The function 2 F1 occurring in the kernel of equation (4) is the familiar Gauss hypergeometric function (see, for example, [2]). DEFINITION 2 Under the hypothesis of Definition 1, let min{λ + η, −μ + η, −μ} > −2.
λ > 0 and Then the fractional integral operator
λ,μ,η
J0,z
f (z) : A −→ A
is defined by λ,μ,η
J0,z
f (z) =
(2 − μ)(2 + λ + η) μ λ,μ,η z I0,z f (z). (2 − μ + η)
(5)
Inclusion properties involving fractional integral operators
641
LEMMA 1 (see [4, p. 415, Lemma 3]). If λ > 0 and n > μ − η − 1, then λ,μ,η n
z =
I0,z
(n + 1) (n − μ + η + 1) n−μ z . (n − μ + 1) (n + λ + η + 1)
(6)
If f (z) ∈ A is of the form (1), it is easily seen from equations (5) and (6) that ∞
λ,μ,η J0,z f (z)
(2 − μ)(2 + λ + η) (n + 1)(n − μ + η + 1) an zn . (7) =z+ (2 − μ + η) (n − μ + 1)(n + λ + η + 1) n=2 λ,μ,η
The generalized fractional integral operator J0,z satisfies the following three-term recurrence relation: λ+1,μ,η λ,μ,η λ+1,μ,η z J0,z f (z) = (λ + η + 2)J0,z f (z) − (λ + η + 1)J0,z f (z). (8) We observe from equation (5) that λ,−λ,η
J0,z
f (z) = (2 + λ)z−λ Dz−λ f (z) (λ > 0),
(9)
where Dz−λ is the relatively more familiar fractional integral operator of order λ (λ > 0) (cf. [6], [7] and [1]; see also [2]). Upon setting μ = 0 and η = α − 1 in equations (5) and (7), we obtain the integral operator (or, equivalently, the mutiplier transformation) λα , which was introduced and studied by Jung et al. [8] (see also [9], [10] and [11]) as follows: λ,0,α−1 J0,z f (z) =: λα f (z) ∞ (λ + α + 1) (n + α) an z n (α + 1) n=2 (n + λ + α) λ + α λ z α−1 t λ−1 = 1 − t f (t) dt α zα 0 z
=z+
(10)
(λ > 0; α > −1; f ∈ A). We also deduce the following obvious relationship from equation (10): zλ+α−1 λα f (z) =
(λ + α + 1) λ Qα f (z), (α + 1)
(11)
where Qλα denotes the operator (which is essentially a Riemann–Liouville fractional integral operator) studied recently by Dziok [12]. On the other hand, if we set μ = 0,
λ=1
and
η =α−1
in equations (5) and (7), we obtain the well-known Bernardi-Libera-Livingston operator Iα (see [1] and [2]) defined by 1,0,α−1 J0,z f (z) =: Iα f (z)
=z+
∞ α+1 α+n
n=2
α+1 = zα
an z n
z
t α−1 f (t) dt 0
(α > −1; f ∈ A).
(12)
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Moreover, we have 1,0,−1 J0,z f (z) =: J [f ](z)
=z+ = 0
z
∞ 1 an z n n n=2
f (t) dt t
(f ∈ A).
(13)
The operator in J [f ](z) is the familar Alexander integral operator (cf., e.g., [1] and [2]; see also [13]). λ,μ,η Using the generalized fractional integral operator J0,z , we now introduce the following subclasses of A: Ss∗ (λ, μ, η, α, β) := f : f (z) ∈ A,
Ss∗ (α, β)
λ,μ,η z J0,z f (z)
λ,μ,η J0,z f (z)
∈
λ,μ,η J0,z f (z)
λ,μ,η z J0,z f (z) ∈ Kc (α, β) and λ,μ,η = α J0,z f (z)
and
λ,μ,η
J0,z
f (z)
= α
(z ∈ U)
(14)
and Kc (λ, μ, η, α, β) := f : f (z) ∈ A,
(z ∈ U) . (15)
It is obvious from the definitions (14) and (15) that f (z) ∈ Kc (λ, μ, η, α, β) ⇐⇒ z f (z) ∈ Ss∗ (λ, μ, η, α, β). In this paper, we establish some inclusion relationships for the above-mentioned function λ,μ,η classes involving the generalized integral operator J0,z defined by equation (5) above. Some corollaries and consequences of our main inclusion relationships are also mentioned.
2. A set of main inclusion relationships In order to derive our main results, we recall here the following lemma. LEMMA 2 (see [14]). Let a function p(z) be analytic in U with p(0) = 1 and p(z) = 0 (z ∈ U). If there exists a point z0 ∈ U such that | arg(p(z))| < then
π π β (|z| < |z0 |) and | arg(p(z0 ))| = β (0 < β 1), 2 2 z0 p (z0 ) = ikβ, p(z0 )
(16)
(17)
Inclusion properties involving fractional integral operators
where
π 1 1 a+ when arg p(z0 ) = β, 2 a 2 1 π 1 k− a+ when arg p(z0 ) = − β, 2 a 2
k
and
1/β
p(z0 )
643
(18) (19)
= ±ia (a > 0).
Our first main inclusion relationship is given by Theorem 1 below. THEOREM 1 Let f ∈ A. Suppose also that λ > 0,
min{λ + η, −μ + η, −μ} > −2, 0 α < 1 and 0 < β 1.
Then Ss∗ (λ, μ, η, α, β) ⊂ Ss∗ (λ + 1, μ, η, α, β). Proof Let f ∈ Ss∗ (λ, μ, η, α, β). Then, upon setting λ+1,μ,η z J0,z f (z) 1 p(z) = −α λ+1,μ,η 1−α J0,z f (z)
(z ∈ U),
(20)
we see that the function p(z) = 1 + c1 z + c2 z2 + · · · is analytic in U, with p(0) = 1
and p(z) = 0
(z ∈ U).
Using the identity (8) in equation (20), we find that λ,μ,η
J0,z
f (z)
λ+1,μ,η J0,z f (z)
=
α + λ + η + 1 + (1 − α)p(z) . λ+η+2
(21)
Differentiating both sides of equation (21) with respect to z, we get λ,μ,η z J0,z f (z) (1 − α)zp (z) − α = (1 − α)p(z) + . λ,μ,η α + λ + η + 1 + (1 − α)p(z) J0,z f (z) Suppose now that there exists a point z0 ∈ U such that the conditions (16) to (19) of Lemma 2 are satisfied. Thus, if π arg p(z0 ) = β (z0 ∈ U), 2 then λ,μ,η z0 J0,z f (z0 ) z0 p (z0 )/p(z0 ) − α = (1 − α)p(z0 ) 1 + λ,μ,η α + λ + η + 1 + (1 − α)p(z0 ) J0,z f (z0 ) ikβ . = (1 − α) a β e(iπβ/2) 1 + α + λ + η + 1 + (1 − α)a β e(iπβ/2)
644
J. K. Prajapat et al.
This implies that arg
λ,μ,η z0 J0,z f (z0 ) λ,μ,η J0,z f (z0 )
−α
=
πβ λ,η + k;α,β , 2
where
λ,η
k;α,β := arg 1 + ⎛
ikβ α + λ + η + 1 + (1 − α)a β e(iπβ/2)
(22)
(α + λ + η + 1)2 + (1 − α)2 a 2β + 2(α + λ + η + 1)(1 − α)a β cos(πβ/2) ⎜ +kβ(1 − α)a β sin(πβ/2) = arg ⎜ ⎝ (α + λ + η + 1)2 + (1 − α)2 a 2β + 2(α + λ + η + 1)(1 − α)a β cos(πβ/2) ⎞ +i
⎟ kβ(α + λ + η + 1 + (1 − α)a cos(πβ/2)) ⎟ 2 2 2β β (α + λ + η + 1) + (1 − α) a + 2(α + λ + η + 1)(1 − α)a cos(πβ/2) ⎠ β
⎛
⎞
⎜ = tan−1 ⎜ ⎝
⎟ kβ(α + λ + η + 1 + (1 − α)a β cos(πβ/2)) ⎟ (α + λ + η + 1)2 + (1 − α)2 a 2β + 2(α + λ + η + 1)(1 − α)a β cos(πβ/2) ⎠ +kβ(1 − α)a β sin(πβ/2)
0, since
1 1 k a+ 1 2 a
and
z0 ∈ U.
Thus, in view of equations (2) and (14), this last inequality in conjunction with equation (22) would contradict our assumption that f (z) ∈ Ss∗ (λ, μ, η, α, β). On the other hand, if we set
π arg p(z0 ) = − β, 2
then it can similarly be shown that λ,μ,η z0 J0,z f (z0 ) π arg − α − β, λ,μ,η 2 J0,z f (z0 ) since k−
1 1 a+ −1 2 a
and
z0 ∈ U,
which again contradicts the assumption that f (z) ∈ Ss∗ (λ, μ, η, α, β). Hence, the function p(z) defined by equation (20) satisfies the following inequality: π | arg p(z) | < β 2
(z ∈ U),
Inclusion properties involving fractional integral operators
645
which implies that λ+1,μ,η π z J0,z f (z) −α < β arg λ+1,μ,η 2 J0,z f (z)
(z ∈ U).
This completes the proof of Theorem 1. We next prove the following inclusion relationships. THEOREM 2 Let f ∈ A. Suppose also that λ > 0,
min{λ + η, −μ + η, −μ} > −2, 0 α < 1
and
0 < β 1.
Then Kc (λ, μ, η, α, β) ⊂ Kc (λ + 1, μ, η, α, β). Proof To prove the inclusion relationship asserted by Theorem 2, we observe from Theorem 1 that λ,μ,η
f (z) ∈ Kc (λ, μ, η, α, β) ⇐⇒ J0,z f (z) ∈ Kc (α, β) λ,μ,η ⇐⇒ z J0,z f (z) ∈ Ss∗ (α, β) λ,μ,η ⇐⇒ J0,z zf (z) ∈ Ss∗ (α, β) ⇐⇒ zf (z) ∈ Ss∗ (λ, μ, η, α, β) =⇒ zf (z) ∈ Ss∗ (λ + 1, μ, η, α, β) λ,μ,η ⇐⇒ J0,z zf (z) ∈ Ss∗ (α, β) λ+1,μ,η f (z) ∈ Ss∗ (α, β) ⇐⇒ z J0,z λ+1,μ,η
⇐⇒ J0,z
f (z) ∈ Kc (α, β)
⇐⇒ f (z) ∈ Kc (λ + 1, μ, η, α, β),
which establishes Theorem 2. THEOREM 3 Let f ∈ A. Suppose also that λ > 0, min{λ + η, −μ + η, −μ} > −2, 0 α < 1, 0 < β 1 and that
λ,μ,η z J0,z Ic f (z) λ,μ,η
J0,z
Ic f (z)
and
c > −α,
= α (z ∈ U).
Then f (z) ∈ Ss∗ (λ, μ, η, α, β) =⇒ Ic f (z) ∈ Ss∗ (λ, μ, η, α, β).
(23)
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Proof We begin by assuming that f (z) ∈ Ss∗ (λ, μ, η, α, β) and defining a function q(z) by 1 q(z) = 1−α
λ,μ,η z J0,z Ic f (z) λ,μ,η
J0,z
Ic f (z)
−α
(z ∈ U),
(24)
where q(z) = 1 + d1 z + d2 z2 + · · · is analytic in U, with q(0) = 1
and q(z) = 0
(z ∈ U).
It can easily be verified from equations (7) and (12) that λ,μ,η λ,μ,η λ,μ,η z J0,z Ic f (z) = (c + 1)J0,z f (z) − c J0,z Ic f (z).
(25)
Thus, by using equations (24) and (25), we find that λ,μ,η z J0,z f (z)
(1 − α)zq (z) . c + α + (1 − α)q(z)
− α = (1 − α)q(z) +
λ,μ,η J0,z f (z)
Suppose now that there exists a point z0 ∈ U such that π | arg q(z) | < β 2
(|z| < |z0 |)
π | arg q(z0 ) | = β. 2
and
Then, by applying Lemma 2, we can write z0 q (z0 ) = ikβ q(z0 )
and
1/β
q(z0 )
= ±ia
(a > 0).
If π arg q(z0 ) = β 2
(z0 ∈ U),
then λ,μ,η z0 J0,z f (z0 ) λ,μ,η
J0,z
f (z0 )
z0 q (z0 )/q(z0 ) − α = (1 − α)q(z0 ) 1 + c + α + (1 − α)q(z0 ) ikβ , = (1 − α)a β e(iπβ/2) 1 + c + α + (1 − α)a β e(iπβ/2)
which yields arg
λ,μ,η z0 J0,z f (z0 ) λ,μ,η
J0,z
f (z0 )
−α
=
π β + ck;α,β , 2
(26)
Inclusion properties involving fractional integral operators
where
647
ikβ := arg 1 + c + α + (1 − α)a β e(iπβ/2) ⎛ (c + α)2 + (1 − α)2 a 2β + 2(c + α)(1 − α)a β cos(πβ/2) ⎜ +kβ(1 − α)a β sin(πβ/2) = arg ⎜ ⎝ (c + α)2 + (1 − α)2 a 2β + 2(c + α)(1 − α)a β cos(πβ/2)
ck;α,β
⎞ +i
⎟ kβ[c + α + (1 − α)a β cos(πβ/2)] ⎟ (c + α)2 + (1 − α)2 a 2β + 2(c + α)(1 − α)a β cos(πβ/2) ⎠ ⎛
⎞
⎜ = tan−1 ⎜ ⎝
⎟ kβ(c + α + (1 − α)a cos(πβ/2)) ⎟ 2 2β β (c + α) + (1 − α) a + 2(c + α)(1 − α)a cos(πβ/2) ⎠ +kβ(1 − α)a β sin(πβ/2) β
2
0, since
1 1 1 a+ k a 2
z0 ∈ U.
and
This last inequality in conjunction with equation (26) would obviously contradict our assumption that f (z) ∈ Ss∗ (λ, μ, η, α, β). Similarly, in the case when π arg q(z0 ) = − β 2 we can prove that arg since
λ,μ,η z0 J0,z f (z0 ) λ,μ,η J0,z f (z0 )
1 1 a+ −1 k− 2 a
(z0 ∈ U),
−α
π − β, 2
and
z0 ∈ U,
contradicting once again the assumption that f (z) ∈ Ss∗ (λ, μ, η, α, β). We thus conclude that q(z) must satisfy the following argument inequality: π | arg q(z) | < β (z ∈ U). 2 This shows that
λ,μ,η π z J0,z Ic f (z) − α arg < β λ,μ,η 2 J0,z Ic f (z)
which completes the proof of Theorem 3.
(z ∈ U),
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J. K. Prajapat et al.
The proof of Theorem 4 below is analogous to that of Theorem 3. We, therefore, choose to omit its proof. THEOREM 4 Under the parametric constraints stated with Theorem 3, let f (z) ∈ A
λ,μ,η z J0,z Ic f (z) = α (z ∈ U). λ,μ,η J0,z Ic f (z)
and
Then f (z) ∈ Kc (λ, μ, η, α, β) =⇒ Ic f (z) ∈ Kc (λ, μ, η, α, β).
(27)
Remark Upon setting μ=0
and η = α − 1,
Theorems 1 to 4 would yield the corresponding known results due to Liu [15]. Furthermore, in view of the relationship (11), we can easily apply Theorems 1 to 4 in order to deduce the corresponding inclusion relationships involving the aforementioned operator Qλα , which was studied recently by Dziok (see [12]).
3.
Corollaries and consequences
In this concluding section, we consider some corollaries and consequences of our main results (Theorems 1 to 4) established in Section 2. First of all, on setting λ = 1,
μ=0
η = −1,
and
Theorem 1 would yield the following result. COROLLARY 1 Let f (z) ∈ A
and
f (z) = α 0
z
f (t) dt (z ∈ U). t
If f (z) satisfies the following inequality: π f (z) − α < β (0 α < 1; 0 < β 1), arg z 2 (f (t)/t) dt 0 then
z 2 0
1 1 − f (t) dt ∈ Ss∗ (α, β). t z
In its further special case when β = 1, Corollary 1 reduces to Corollary 2 below.
(28)
(29)
Inclusion properties involving fractional integral operators
COROLLARY 2 Let
f (z) ∈ A
f (z) = α
and
z
f (t) dt (z ∈ U). t
0
If f (z) satisfies the following inequality: (f (z)) > α
z
0
then
z 2 0
f (t) dt t
649
(0 α < 1),
1 1 − f (t) dt ∈ S ∗ (α). t z
(30)
(31)
Next, if we set λ=1
μ=η=0
and
in Theorem 1, we get Corollary 3 below. COROLLARY 3 Let
f (z) ∈ A
z
zf (z) = (α + 1)
and
f (t) dt (z ∈ U).
0
If f (z) satisfies the following inequality: π zf(z) − (α + 1) < β (0 α < 1; 0 < β 1), arg z 2 f (t) dt 0 then 6 z
z
1−
0
t f (t)dt ∈ Ss∗ (α, β). z
(32)
(33)
A further special case of Corollary 3 when β = 1 would immediately yield the following result. COROLLARY 4 Let
f (z) ∈ A
and
z
zf(z) = (α + 1)
f (t)dt (z ∈ U).
0
If f (z) satisfies the following inequality:
z
(zf (z)) > (α + 1)
f (t)dt
(0 α < 1),
(34)
0
then 6 z
z 0
1−
t f (t)dt ∈ S ∗ (α). z
By setting λ = 1,
μ=0
in Theorem 2, we arrive at Corollary 5 below.
and
η = −1
(35)
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J. K. Prajapat et al.
COROLLARY 5 Let f (z) ∈ A and zf (z) = αf (z) (z ∈ U). If f (z) ∈ Ss∗ (α, β), then
z 2 0
1 1 − f (t)dt ∈ Kc (α, β). t z
(36)
Upon setting λ=1
and μ = η = 0,
Theorem 2 would yield the following result. COROLLARY 6 Let f (z) ∈ A
z f (t) dt z2 f (z) = (α + 1) zf(z) −
and
(z ∈ U).
0
If f (z) satisfies the following inequality: π z2 f (z) z − (α + 1) < β (0 α < 1; 0 < β 1), arg 2 zf(z) − 0 f (t) dt then 6 z
(37)
z
t 1− f (t) dt ∈ Kc (α, β). z
0
(38)
If we set β = 1 in Corollary 6, we immediately get Corollary 7 below. COROLLARY 7 Let f (z) ∈ A
and
z
z f (z) = (α + 1) zf (z) − 2
f (t) dt
(z ∈ U).
0
If f (z) satisfies the following inequality: z 2 f (t) dt z f (z) > (α + 1) zf(z) −
(0 α < 1),
(39)
0
then 6 z
z 0
t 1− f (t) dt ∈ K(α). z
(40)
Finally, by setting λ = 1,
μ = 0,
η = −1
and
c=0
in Theorem 3, we get the following result. COROLLARY 8 If
then
f (z) ∈ A and z J J [f ](z) = α J J [f ](z) (z ∈ U),
(41)
J [f ](z) ∈ Ss∗ (α, β) =⇒ J J [f ](z) ∈ Ss∗ (α, β).
(42)
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