Volume 10 (2009), Issue 3, Article 87, 8 pp.
A STUDY ON STARLIKE AND CONVEX PROPERTIES FOR
HYPERGEOMETRIC FUNCTIONS
A. O. MOSTAFA
D EPARTMENT OF M ATHEMATICS
FACULTY OF S CIENCE
M ANSOURA U NIVERSITY
M ANSOURA 35516, E GYPT
adelaeg254@yahoo.com
Received 07 May, 2008; accepted 21 August, 2009
Communicated by S.S. Dragomir
A BSTRACT. 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.
Key words and phrases: Starlike, Convex, Hypergeometric function, Integral operator.
2000 Mathematics Subject Classification. 30C45.
1. I NTRODUCTION
Let T be the class consisting of functions of the form:
∞
X
(1.1)
f (z) = z −
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:
(1.3)
Re
f 0 (z) + zf 00 (z)
f 0 (z) + λzf 00 (z)
> α,
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)
134-08
f (z) ∈ C(λ, α) ⇔ zf 0 (z) ∈ T (λ, α).
2
A. O. M OSTAFA
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
F (a, b; c; z) =
(1.5)
∞
X
(a)n (b)n
n=0
(c)n (1)n
zn,
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].
2. M AIN R ESULTS
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
(n − λαn − α + λα)an ≤ 1 − α.
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
n(n − λαn − α + λα)an ≤ 1 − α.
n=2
Theorem A.
(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
(2.3)
c>a+b+1−
(1 − λα)ab
.
1−α
(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
Γ(c)Γ(c − a − b)
(1 − λα)ab
(2.4)
1+
≤ 2.
Γ(c − a)Γ(c − b)
(1 − α)(c − a − b − 1)
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 87, 8 pp.
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S TUDY ON S TARLIKE AND C ONVEX P ROPERTIES
3
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
∞
ab X (a + 1)n−2 (b + 1)n−2 n
= z − z ,
c n=2 (c + 1)n−2 (1)n−1
(2.5)
according to (i) of Lemma 2.1, we must show that
∞
X
(a + 1)n−2 (b + 1)n−2 c (n − λαn − α + λα)
≤ (1 − α) .
(c + 1)n−2 (1)n−1
ab
n=2
(2.6)
Note that the left side of (2.6) diverges if c < a + b + 1. Now
∞
X
(n − λαn − α + λα)
n=2
(a + 1)n−2 (b + 1)n−2
(c + 1)n−2 (1)n−1
∞
X
(a + 1)n (b + 1)n
(a + 1)n (b + 1)n
= (1 − λα)
(n + 1)
+ (1 − α)
(c + 1)n (1)n+1
(c + 1)n (1)n+1
n=0
n=0
∞
X
∞
X
(a + 1)n (b + 1)n
∞
(1 − α)c X (a)n (b)n
(c + 1)n (1)n
ab
(c)n (1)n
n=0
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)
= (1 − λα)
+
Hence, (2.6) is equivalent to
(2.7)
(1 − α)(c − a − b − 1)
Γ(c + 1)Γ(c − a − b − 1)
(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
or equivalently,
c>a+b+1−
(1 − λα)ab
.
1−α
(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
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 87, 8 pp.
(a)n−1 (b)n−1
≤ 1 − α.
(c)n−1 (1)n−1
http://jipam.vu.edu.au/
4
A. O. M OSTAFA
Now,
∞
X
(2.8)
(n − λαn − α + λα)
n=2
=
∞
X
(a)n−1 (b)n−1
(c)n−1 (1)n−1
[(n − 1)(1 − λα) + (1 − α)]
n=2
(a)n−1 (b)n−1
(c)n−1 (1)n−1
∞
∞
X
X
(a)n (b)n
(a)n (b)n
= (1 − λα)
n
+ (1 − α)
(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 − α)
−1
= (1 − λα)
c n=0 (c + 1)n (1)n
(c)n (1)n
n=0
ab Γ(c + 1)Γ(c − a − b − 1)
Γ(c)Γ(c − a − b)
= (1 − λα)
+ (1 − α)
−1
c
Γ(c − a)Γ(c − b)
Γ(c − a)Γ(c − b)
Γ(c)Γ(c − a − b)
ab(1 − λα)
=
(1 − α) +
− (1 − α).
Γ(c − a)Γ(c − b)
(c − a − b − 1)
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.
(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 (2.11)
n(n − λαn − α + λα)
≤ (1 − α) .
(c
+
1)
(1)
ab
n−2
n−1
n=2
Note that for c > a + b + 2 , the left side of (2.11) converges. Writing
(n + 2)[(n + 2)(1 − λα) − α(1 − λ)]
= (n + 1)2 (1 − λα) + (n + 1)(2 − α − λα) + (1 − α),
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 87, 8 pp.
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S TUDY ON S TARLIKE AND C ONVEX P ROPERTIES
5
we see that
∞
X
(n + 2)[(n + 2)(1 − λα) − α(1 − λ)]
n=0
= (1 − λα)
∞
X
(n + 1)2
n=0
(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
(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
∞
X
= (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)
∞
X
n
n=0
+ (1 − α)
(c + 1)n (1)n
+ (1 − α)
∞
X
(a + 1)n (b + 1)n
n=0
∞
X
(c + 1)n (1)n+1
(a + 1)n (b + 1)n
(a + 1)n (b + 1)n
+ (3 − α − 2λα)
(c + 1)n (1)n
(c + 1)n (1)n
n=0
∞
X
(a + 1)n−1 (b + 1)n−1
n=1
(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
∞
+ (1 − α)
c X (a)n (b)n
ab n=1 (c)n (1)n
(c + 1)n (1)n
Γ(c + 1)Γ(c − a − b − 2)
=
− (1 − λα)(a + 1)(b + 1)
Γ(c − a)Γ(c − b)
(1 − α)
(1 − α)c
+(3 − α − 2λα)(c − a − b − 2) +
(c − a − b − 2)2 −
.
ab
ab
This last expression is bounded above by abc (1 − α) if and only if
n=0
(1 − λα)(a + 1)(b + 1) + (3 − α − 2λα)(c − a − b − 2) +
(1 − α)
(c − a − b − 2)2 ≤ 0,
ab
which is equivalent to (2.9).
(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
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 87, 8 pp.
(a)n+1 (b)n+1
(c)n+1 (1)n+1
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6
A. O. M OSTAFA
= (1 − λα)
∞
X
(n + 2)
2 (a)n+1 (b)n+1
(c)n+1 (1)n+1
n=0
− α(1 − λ)
∞
X
(n + 2)
n=0
(a)n+1 (b)n+1
.
(c)n+1 (1)n+1
Writing (n + 2) = (n + 1) + 1, we have
(2.13)
∞
∞
X
∞
(a)n+1 (b)n+1 X (a)n+1 (b)n+1
(a)n+1 (b)n+1 X
(n + 2)
=
(n + 1)
+
(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
(a)n+1 (b)n+1
n=0
(c)n+1 (1)n
+
∞
X
(a)n+1 (b)n+1
n=0
(c)n+1 (1)n+1
and
(2.14)
∞
X
(n + 2)2
n=0
=
∞
X
(a)n+1 (b)n+1
(c)n+1 (1)n+1
(n + 1)
n=0
=
∞
∞
X
(a)n+1 (b)n+1
(a)n+1 (b)n+1 X (a)n+1 (b)n+1
+2
+
(c)n+1 (1)n
(c)n+1 (1)n
(c)n+1 (1)n+1
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
(c)n+1 (1)n
n=0
+ (1 − α)
∞
X
(a)n (b)n
n=1
(c)n (1)n
.
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.
Putting λ = 0 in (i) of Theorem B, we have:
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].
3. A N I NTEGRAL O PERATOR
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
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 87, 8 pp.
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S TUDY ON S TARLIKE AND C ONVEX P ROPERTIES
7
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
α(1 − λ)(c − 1)2
Γ(c + 1)Γ(c − a − b) (1 − λα) α(1 − λ)(c − a − b)
(3.2)
−
+
≤ 0.
Γ(c − a)Γ(c − b)
ab
(a − 1)2 (b − 1)2
(a − 1)2 (b − 1)2
Proof. Since
∞
ab X (a + 1)n−2 (b + 1)n−2 n
G(a, b; c; z) = 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 (n − λαn − α + λα)
≤ (1 − α) .
(c + 1)n−2 (1)n
ab
n=2
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
− α(1 − λ)
= (1 − λα)
(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
− 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
= (1 − λα)
ab n=0 (c)n (1)n
= (1 − λα)
∞
X (a − 1)n (b − 1)n
(c − 1)2
(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.
− α(1 − λ)
ab
Now, we observe that G(a, b; c; z) ∈ C(λ, α) if and only if zF (a, b; c; z) ∈ 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
(1 − λα)ab
c>a+b+1−
.
1−α
Remark 2. Putting λ = 0 in the above results, we obtain the results of Silverman [7].
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 87, 8 pp.
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8
A. O. M OSTAFA
R EFERENCES
[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 functions, J. Math.
Anal. Appl., 15 (1984), 737–745.
[3] N.E. CHO, S.Y. WOO AND S. OWA, Uniform convexity properties for hypergeometric functions,
Fract. Calculus Appl. Anal., 5(3) (2002), 303–313.
[4] E. MERKES AND B.T. SCOTT, Starlike hypergeometric functions, Proc. Amer. Math. Soc., 12
(1961), 885–888.
[5] St. RUSCHEWEYH AND V. SINGH, On the order of starlikeness of hypergeometric functions, J.
Math. Anal. Appl., 113 (1986), 1–11.
[6] H. SILVERMAN, Univalent functions with negative coefficients, Proc. Amer. Math. Soc., 51 (1975),
109–116.
[7] H. SILVERMAN, Starlike and convexity properties for hypergeometric functions, J. Math. Anal.
Appl., 172 (1993), 574–581.
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 87, 8 pp.
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