A STUDY ON STARLIKE AND CONVEX
PROPERTIES FOR HYPERGEOMETRIC FUNCTIONS
Study on Starlike and
Convex Properties
A. O. MOSTAFA
Department of Mathematics
Faculty of Science
Mansoura University
Mansoura 35516, Egypt
A. O. Mostafa
vol. 10, iss. 3, art. 87, 2009
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EMail: adelaeg254@yahoo.com
Received:
07 May, 2008
Accepted:
21 August, 2009
Communicated by:
S.S. Dragomir
2000 AMS Sub. Class.:
30C45.
Key words:
Starlike, Convex, Hypergeometric function, Integral operator.
Abstract:
The objective of the present paper is to give some characterizations for a (Gaussian) hypergeometric function to be in various subclasses of starlike and convex
functions. We also consider an integral operator related to the hypergeometric
function.
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Contents
1
Introduction
3
2
Main Results
5
3
An Integral Operator
13
Study on Starlike and
Convex Properties
A. O. Mostafa
vol. 10, iss. 3, art. 87, 2009
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1.
Introduction
Let T be the class consisting of functions of the form:
(1.1)
f (z) = z −
∞
X
an z n
(an > 0),
n=2
which are analytic and univalent in the open unit disc U = {z : |z| < 1}. Let T (λ, α)
be the subclass of T consisting of functions which satisfy the condition:
zf 0 (z)
(1.2)
Re
> α,
λzf 0 (z) + (1 − λ)f (z)
for some α (0 ≤ α < 1), λ (0 ≤ λ < 1) and for all z ∈ U.
Also, let C(λ, α) denote the subclass of T consisting of functions which satisfy
the condition:
0
f (z) + zf 00 (z)
(1.3)
Re
> α,
f 0 (z) + λzf 00 (z)
for some α (0 ≤ α < 1), λ (0 ≤ λ < 1) and for all z ∈ U.
From (1.2) and (1.3), we have
(1.4)
f (z) ∈ C(λ, α) ⇔ zf 0 (z) ∈ T (λ, α).
We note that T (0, α) = T ∗ (α), the class of starlike functions of order α (0 ≤ α < 1)
and C(0, α) = C(α), the class of convex functions of order α (0 ≤ α < 1) (see
Silverman [6]).
Let F (a, b; c; z) be the (Gaussian ) hypergeometric function defined by
(1.5)
F (a, b; c; z) =
∞
X
(a)n (b)n
n=0
(c)n (1)n
zn,
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Convex Properties
A. O. Mostafa
vol. 10, iss. 3, art. 87, 2009
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where c 6= 0, −1, −2, ..., and (θ)n is the Pochhammer symbol defined by
(
1,
n=0
(θ)n =
θ(θ + 1) · · · (θ + n − 1) n ∈ N = {1, 2, ...}.
We note that F (a, b; c; 1) converges for Re(c − a − b) > 0 and is related to the
Gamma function by
(1.6)
F (a, b; c; 1) =
Γ(c)Γ(c − a − b)
.
Γ(c − a)Γ(c − b)
Silverman [7] gave necessary and sufficient conditions for zF (a, b; c; z) to be in
T (α) and C(α), also examining a linear operator acting on hypergeometric functions. For other interesting developments on zF (a, b; c; z) in connection with various
subclasses of univalent functions, the reader can refer the to works of Carlson and
Shaffer [2], Merkes and Scott [4], Ruscheweyh and Singh [5] and Cho et al. [3].
∗
Study on Starlike and
Convex Properties
A. O. Mostafa
vol. 10, iss. 3, art. 87, 2009
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2.
Main Results
To establish our main results, we need the following lemma due to Altintas and Owa
[1].
Lemma 2.1.
(i) A function f (z) defined by (1.1) is in the class T (λ, α) if and only if
(2.1)
∞
X
Study on Starlike and
Convex Properties
A. O. Mostafa
(n − λαn − α + λα)an ≤ 1 − α.
vol. 10, iss. 3, art. 87, 2009
n=2
(ii) A function f (z) defined by (1.1) is in the class C(λ, α) if and only if
(2.2)
∞
X
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n(n − λαn − α + λα)an ≤ 1 − α.
n=2
Theorem A.
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(i) If a, b > −1, c > 0 and ab < 0, then zF (a, b; c; z) is in T (λ, α) if and only if
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(2.3)
(1 − λα)ab
c>a+b+1−
.
1−α
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(ii) If a, b > 0 and c > a + b + 1, then F1 (a, b; c; z) = z[2 − F (a, b; c; z)] is in
T (λ, α) if and only if
(1 − λα)ab
Γ(c)Γ(c − a − b)
1+
≤ 2.
(2.4)
Γ(c − a)Γ(c − b)
(1 − α)(c − a − b − 1)
Proof. (i) Since
∞
ab X (a + 1)n−2 (b + 1)n−2 n
zF (a, b; c; z) = z +
z
c n=2 (c + 1)n−2 (1)n−1
(2.5)
X
ab ∞ (a + 1)n−2 (b + 1)n−2 n
= z − z ,
c n=2 (c + 1)n−2 (1)n−1
Study on Starlike and
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A. O. Mostafa
according to (i) of Lemma 2.1, we must show that
(2.6)
∞
X
(n − λαn − α + λα)
n=2
vol. 10, iss. 3, art. 87, 2009
(a + 1)n−2 (b + 1)n−2 c ≤ (1 − α) .
(c + 1)n−2 (1)n−1
ab
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Note that the left side of (2.6) diverges if c < a + b + 1. Now
∞
X
(a + 1)n−2 (b + 1)n−2
(n − λαn − α + λα)
(c + 1)n−2 (1)n−1
n=2
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= (1 − λα)
∞
X
n=0
= (1 − λα)
(n + 1)
(a + 1)n (b + 1)n
(a + 1)n (b + 1)n
+ (1 − α)
(c + 1)n (1)n+1
(c + 1)n (1)n+1
n=0
∞
X
(a + 1)n (b + 1)n
n=0
∞
X
(c + 1)n (1)n
∞
(1 − α)c X (a)n (b)n
+
ab
(c)n (1)n
n=1
Γ(c + 1)Γ(c − a − b − 1) (1 − α)c Γ(c)Γ(c − a − b)
= (1 − λα)
+
−1 .
Γ(c − a)Γ(c − b)
ab
Γ(c − a)Γ(c − b)
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Hence, (2.6) is equivalent to
Γ(c + 1)Γ(c − a − b − 1)
(1 − α)(c − a − b − 1)
(2.7)
(1 − λα) +
Γ(c − a)Γ(c − b)
ab
h c ci
≤ (1 − α) +
= 0.
ab
ab
Thus, (2.7) is valid if and only if
(1 − λα) +
(1 − α)(c − a − b − 1)
≤ 0,
ab
Study on Starlike and
Convex Properties
A. O. Mostafa
vol. 10, iss. 3, art. 87, 2009
or equivalently,
c>a+b+1−
(1 − λα)ab
.
1−α
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(ii) Since
F1 (a, b; c; z) = z −
∞
X
(a)n−1 (b)n−1
n=2
(c)n−1 (1)n−1
zn,
by (i) of Lemma 2.1, we need only to show that
∞
X
(n − λαn − α + λα)
n=2
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(a)n−1 (b)n−1
≤ 1 − α.
(c)n−1 (1)n−1
Now,
(2.8)
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∞
X
(n − λαn − α + λα)
n=2
=
∞
X
n=2
(a)n−1 (b)n−1
(c)n−1 (1)n−1
[(n − 1)(1 − λα) + (1 − α)]
(a)n−1 (b)n−1
(c)n−1 (1)n−1
∞
∞
X
X
(a)n (b)n
(a)n (b)n
= (1 − λα)
+ (1 − α)
n
(c)n (1)n
(c)n (1)n
n=1
n=1
∞
∞
X
X
(a)n (b)n
(a)n (b)n
= (1 − λα)
+ (1 − α)
.
(c)n (1)n−1
(c)n (1)n
n=1
n=1
Noting that (θ)n = θ((θ + 1)n−1 then, (2.8) may be expressed as
∞
∞
X
ab X (a + 1)n−1 (b + 1)n−1
(a)n (b)n
(1 − λα)
+ (1 − α)
c n=1 (c + 1)n−1 (1)n−1
(c)n (1)n
n=1
"∞
#
∞
X (a)n (b)n
ab X (a + 1)n (b + 1)n
= (1 − λα)
+ (1 − α)
−1
c n=0 (c + 1)n (1)n
(c)n (1)n
n=0
Γ(c)Γ(c − a − b)
ab Γ(c + 1)Γ(c − a − b − 1)
= (1 − λα)
+ (1 − α)
−1
c
Γ(c − a)Γ(c − b)
Γ(c − a)Γ(c − b)
ab(1 − λα)
Γ(c)Γ(c − a − b)
=
(1 − α) +
− (1 − α).
Γ(c − a)Γ(c − b)
(c − a − b − 1)
Study on Starlike and
Convex Properties
A. O. Mostafa
vol. 10, iss. 3, art. 87, 2009
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But this last expression is bounded above by 1 − α if and only if (2.4) holds.
Theorem B.
(i) If a, b > −1, ab < 0, and c > a + b + 2, then zF (a, b; c; z) is in C(λ, α) if and
only if
(2.9) (1 − λα)(a)2 (b)2 + (3 − 2λα − α)ab(c − a − b − 2)
+ (1 − α)(c − a − b − 2)2 > 0.
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(ii) If a, b > 0 and c > a + b + 2, then F1 (a, b; c; z) = z[2 − F (a, b; c; z)] is in
C(λ, α) if and only if
Γ(c)Γ(c − a − b)
(1 − λα)(a)2 (b)2
(2.10)
1+
Γ(c − a)Γ(c − b)
(1 − α)(c − a − b − 2)2
3 − 2λα − α
ab
+
≤ 2.
1−α
c−a−b−1
Proof. (i) Since zF (a, b; c; z) has the form (2.5), we see from (ii) of Lemma 2.1,
that our conclusion is equivalent to
∞
X
(a + 1)n−2 (b + 1)n−2 c ≤ (1 − α) .
n(n − λαn − α + λα)
(c + 1)n−2 (1)n−1
ab
n=2
(2.11)
Note that for c > a + b + 2 , the left side of (2.11) converges. Writing
(n + 2)[(n + 2)(1 − λα) − α(1 − λ)]
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A. O. Mostafa
vol. 10, iss. 3, art. 87, 2009
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2
= (n + 1) (1 − λα) + (n + 1)(2 − α − λα) + (1 − α),
we see that
∞
X
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(n + 2)[(n + 2)(1 − λα) − α(1 − λ)]
n=0
= (1 − λα)
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∞
X
n=0
(n + 1)2
(a + 1)n (b + 1)n
(c + 1)n (1)n+1
(a + 1)n (b + 1)n
(c + 1)n (1)n+1
∞
X
∞
X
(a + 1)n (b + 1)n
(a + 1)n (b + 1)n
+ (2 − α − λα)
(n + 1)
+ (1 − α)
(c + 1)n (1)n+1
(c + 1)n (1)n+1
n=0
n=0
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= (1 − λα)
∞
X
(n + 1)
n=0
+ (2 − α − λα)
(a + 1)n (b + 1)n
(c + 1)n (1)n
∞
X
(a + 1)n (b + 1)n
n=0
= (1 − λα)
(c + 1)n (1)n
+ (1 − α)
∞
X
(a + 1)n (b + 1)n
n=0
∞
X
(c + 1)n (1)n+1
∞
X
(a + 1)n (b + 1)n
(a + 1)n (b + 1)n
n
+ (3 − α − 2λα)
(c + 1)n (1)n
(c + 1)n (1)n
n=0
n=0
+ (1 − α)
A. O. Mostafa
∞
X
(a + 1)n−1 (b + 1)n−1
n=1
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Convex Properties
vol. 10, iss. 3, art. 87, 2009
(c + 1)n−1 (1)n
∞
(1 − λα)(a + 1)(b + 1) X (a + 2)n (b + 2)n
=
(c + 1)
(c + 2)n (1)n
n=0
+ (3 − α − 2λα)
∞
X
(a + 1)n (b + 1)n
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∞
c X (a)n (b)n
+ (1 − α)
ab n=1 (c)n (1)n
(c + 1)n (1)n
n=0
Γ(c + 1)Γ(c − a − b − 2)
=
− (1 − λα)(a + 1)(b + 1)
Γ(c − a)Γ(c − b)
(1 − α)c
(1 − α)
(c − a − b − 2)2 −
.
+(3 − α − 2λα)(c − a − b − 2) +
ab
ab
This last expression is bounded above by abc (1 − α) if and only if
(1−λα)(a+1)(b+1)+(3−α−2λα)(c−a−b−2)+
which is equivalent to (2.9).
(1 − α)
(c−a−b−2)2 ≤ 0,
ab
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(ii) In view of (ii) of Lemma 2.1, we need to show that
∞
X
n(n − λαn − α + λα)
n=2
(a)n−1 (b)n−1
≤ (1 − α) .
(c)n−1 (1)n−1
Now
(2.12)
∞
X
n(n − λαn − α + λα)
n=2
=
∞
X
(a)n−1 (b)n−1
(c)n−1 (1)n−1
(n + 2)[(n + 2)(1 − λα) − α(1 − λ)]
n=0
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A. O. Mostafa
(a)n+1 (b)n+1
(c)n+1 (1)n+1
∞
∞
X
X
(a)n+1 (b)n+1
2 (a)n+1 (b)n+1
= (1−λα) (n+2)
−α(1−λ)
(n+2)
.
(c)n+1 (1)n+1
(c)n+1 (1)n+1
n=0
n=0
Writing (n + 2) = (n + 1) + 1, we have
∞
X
∞
∞
(a)n+1 (b)n+1 X
(a)n+1 (b)n+1 X (a)n+1 (b)n+1
(2.13)
=
(n + 1)
+
(n + 2)
(c)n+1 (1)n+1
(c)n+1 (1)n+1 n=0 (c)n+1 (1)n+1
n=0
n=0
=
∞
X
n=0
∞
X
(a)n+1 (b)n+1
(a)n+1 (b)n+1
+
(c)n+1 (1)n
(c)n+1 (1)n+1
n=0
and
(2.14)
vol. 10, iss. 3, art. 87, 2009
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∞
X
(n + 2)2
n=0
∞
X
(a)n+1 (b)n+1
(c)n+1 (1)n+1
∞
∞
X
(a)n+1 (b)n+1
(a)n+1 (b)n+1 X (a)n+1 (b)n+1
=
(n + 1)
+2
+
(c)n+1 (1)n
(c)n+1 (1)n
(c)n+1 (1)n+1
n=0
n=0
n=0
=
∞
X
(a)n+1 (b)n+1
n=1
(c)n+1 (1)n−1
+3
∞
X
(a)n+1 (b)n+1
n=0
(c)n+1 (1)n
+
∞
X
(a)n (b)n
n=1
(c)n (1)n
.
Substituting (2.13) and (2.14) into the right side of (2.12), yields
(2.15) (1 − λα)
∞
X
(a)n+2 (b)n+2
n=0
(c)n+2 (1)n
+ (3 − 2λα − α)
∞
X
(a)n+1 (b)n+1
n=0
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(c)n+1 (1)n
+ (1 − α)
∞
X
(a)n (b)n
n=1
(c)n (1)n
A. O. Mostafa
.
Since (a)n+k = (a)k (a + k)n , we may write (2.15) as
Γ(c)Γ(c − a − b) (1 − λα)(a)2 (b)2 (3 − 2λα − α)ab
+
+ (1 − α) − (1 − α).
Γ(c − a)Γ(c − b) (c − a − b − 2)2
(c − a − b − 1)
By a simplification, we see that the last expression is bounded above by (1 − α) if
and only if (2.10) holds.
vol. 10, iss. 3, art. 87, 2009
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Putting λ = 0 in (i) of Theorem B, we have:
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Corollary 2.2. If a, b > −1, ab < 0, and c > a + b + 2, then zF (a, b; c; z) is in
C(α) if and only if
(a)2 (b)2 + (3 − α)ab(c − a − b − 2) + (1 − α)(c − a − b − 2)2 > 0.
Remark 1. Corollary 2.2, corrects the result obtained by Silverman [7, Theorem 4].
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3.
An Integral Operator
In the theorems below, we obtain similar results in connection with a particular integral operator G(a, b; c; z) acting on F (a, b; c; z) as follows:
Z z
(3.1)
G(a, b; c; z) =
F (a, b; c; t)dt.
0
Theorem C. Let a, b > −1, ab < 0 and c > max{0, a + b}. Then G(a, b; c; z)
defined by (3.1) is in T (λ, α) if and only if
Γ(c + 1)Γ(c − a − b) (1 − λα) α(1 − λ)(c − a − b)
−
(3.2)
Γ(c − a)Γ(c − b)
ab
(a − 1)2 (b − 1)2
α(1 − λ)(c − 1)2
+
≤ 0.
(a − 1)2 (b − 1)2
Proof. Since
∞
ab X (a + 1)n−2 (b + 1)n−2 n
z ,
G(a, b; c; z) = z − c n=2
(c + 1)n−2 (1)n
by (i) of Lemma 2.1, we need only to show that
∞
X
(a + 1)n−2 (b + 1)n−2 c ≤ (1 − α) .
(n − λαn − α + λα)
(c + 1)n−2 (1)n
ab
n=2
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A. O. Mostafa
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Now
∞
X
[n(1 − λα) − α(1 − λ)]
n=2
(a + 1)n−2 (b + 1)n−2
(c + 1)n−2 (1)n
∞
X
∞
X
(a + 1)n (b + 1)n
(a + 1)n (b + 1)n
= (1 − λα)
− α(1 − λ)
(n + 2)
(c + 1)n (1)n+2
(c + 1)n (1)n+2
n=0
n=0
= (1 − λα)
= (1 − λα)
∞
X
(a + 1)n (b + 1)n
n=0
∞
X
n=1
(c + 1)n (1)n+1
− α(1 − λ)
∞
X
(a + 1)n−1 (b + 1)n−1
n=1
(c + 1)n−1 (1)n+1
∞
(a + 1)n−1 (b + 1)n−1
c X (a)n (b)n
− α(1 − λ)
(c + 1)n−1 (1)n
ab n=1 (c)n (1)n+1
∞
∞
c X (a)n (b)n
c X (a)n (b)n
= (1 − λα)
− α(1 − λ)
ab n=1 (c)n (1)n
ab n=1 (c)n (1)n+1
"∞
#
c X (a)n (b)n
= (1 − λα)
−1
ab n=0 (c)n (1)n
∞
X (a − 1)n (b − 1)n
(c − 1)2
− α(1 − λ)
(a − 1)2 (b − 1)2 n=2 (c − 1)n (1)n
Γ(c + 1)Γ(c − a − b) (1 − λα) α(1 − λ)(c − a − b)
=
−
Γ(c − a)Γ(c − b)
ab
(a − 1)2 (b − 1)2
α(1 − λ)(c − 1)2 (1 − α)c
+
−
,
(a − 1)2 (b − 1)2
ab
which is bounded above by (1 − α) c if and only if (3.2) holds.
ab
Now, we observe that G(a, b; c; z) ∈ C(λ, α) if and only if zF (a, b; c; z) ∈
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A. O. Mostafa
vol. 10, iss. 3, art. 87, 2009
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T (λ, α). Thus any result of functions belonging to the class T (λ, α) about zF (a, b; c; z)
leads to that of functions belonging to the class C(λ, α). Hence we obtain the following analogous result to Theorem A.
Theorem 3.1. Let a, b > −1, ab < 0 and c > a + b + 2. Then G(a, b; c; z) defined
by (3.1) is in C(λ, α) if and only if
c>a+b+1−
(1 − λα)ab
.
1−α
Remark 2. Putting λ = 0 in the above results, we obtain the results of Silverman [7].
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A. O. Mostafa
vol. 10, iss. 3, art. 87, 2009
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References
[1] O. ALTINTAS AND S. OWA, On subclasses of univalent functions with negative
coefficients, Pusan Kyŏngnam Math. J., 4 (1988), 41–56.
[2] B.C. CARLSON AND D.B. SHAFFER, Starlike and prestarlike hypergeometric
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Study on Starlike and
Convex Properties
A. O. Mostafa
vol. 10, iss. 3, art. 87, 2009
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Title Page
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