STARLIKE AND UNIFORMLY CONVEX FUNCTIONS INVOLVING ...

International Journal of Pure and Applied Mathematics Volume 92 No. 5 2014, 691-701 ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version) url: http://www.ijpam.eu doi: http://dx.doi.org/10.12732/ijpam.v92i5.5

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STARLIKE AND UNIFORMLY CONVEX FUNCTIONS INVOLVING GENERALIZED HYPERGEOMETRIC FUNCTION C. Ramachandran1 , L. Vanitha2 , G. Murugusundaramoorthy3 § 1,2 Department

of Mathematics University College of Engineering Villupurm Anna University Villupuram, 605 602, Tamilnadu, INDIA 3 School of Advanced Sciences VIT University Vellore, 632014, INDIA Abstract: In this paper, we obtained some conditions on the parameters of generalized hypergeometric function and also deals with mapping properties of various subclasses of starlike and uniformly convex functions defined through a generalized hypergeometric function. AMS Subject Classification: 30C45 Key Words: generalized hypergeometric function, starlike function, uniformly convex function 1. Introduction Let A denotes the class of functions of the form : ∞ X an z n f (z) = z +

(1)

n=2

which are analytic in the unit disk U = {z : |z| < 1}.As usual, we denote by S the subclass of A consisting of functions which are normalized by f (0) = 0 = f ′ (0) − 1 and also univalent in U. Denote by T [13] the subclass of A consisting Received:

February 26, 2014

§ Correspondence

author

c 2014 Academic Publications, Ltd.

url: www.acadpubl.eu

692

C. Ramachandran, L. Vanitha, G. Murugusundaramoorthy

of functions of the form f (z) = z −

∞ X

an z n , an ≥ 0, n = 2, 3, . . . ..

(2)

n=2

Also, for functions f ∈ A given by (1) and g ∈ A given by g(z) = z+ we define the Hadamard product (or convolution) of f and g by (f ∗ g)(z) = z +

∞ X

an bn z n , z ∈ U.

P∞

n=2 bn z

n,

(3)

n=2

A f ∈ A is said to be starlike of order α (0 ≤ α < 1), if and only function zf ′ (z) if ℜ f (z) > α (z ∈ U). This function class is denoted by S ∗ (α). We also write S ∗ (0) ≡ S ∗ , where S ∗ denotes the class of functions f ∈ A that f (U) is starlike with respect to the origin. A function f ∈ A is said to be convex of ′′ (z) > α (z ∈ U). This class is order α (0 ≤ α < 1) if and only if ℜ 1 + zff ′ (z) denoted by K(α). Further, K(0) = K, the well-known standard class of convex functions. It is an established fact that f ∈ K(α) ⇐⇒ zf ′ ∈ S ∗ (α). We recall the definitions of the class of uniformly convex functions denoted by U CV, introduced by Goodman [5, 6], studied extensively by Rønning [11, 12], and independently by Ma and Minda [7]. Definition 1.1. A function f of the form (1) is said to be uniformly convex in U, if and only if ! ′′ ′′ zf (z) zf (z) , (z ∈ U). ℜ 1+ ′ ≥ ′ f (z) f (z)

Further the class α − SP related to the class α − U CV by means of the well-known Alexander equivalence between the usual classes of convex K and starlike S ∗ functions are defined and the analytic criterion for functions in these classes are given as below: Definition 1.2. Sp (α) if

A function f of the form (1)is said to be in the class ′

ℜ

zf (z) f (z)

!

and f ∈ U CV(α), α ≥ 0 if ′′

′′ zf (z) ≥ α ′ − 1 , f (z)

zf (z) ℜ 1+ ′ f (z)

!

′′ zf (z) , ≥ α ′ f (z)

(z ∈ U)

(z ∈ U)

STARLIKE AND UNIFORMLY CONVEX FUNCTIONS INVOLVING... 693 Also, U CT (α) = U CV(α) ∩ T and T S p (α) = Sp (α) ∩ T . It is of interest to note that U CV(1) ≡ U CV and ≡ U CV(0) ≡ K; Sp (0) ≡ S ∗ and Sp (1) ≡ Sp . (see[16]) we recall the following results due to Subramanian et al., [16]. Lemma 1.3. A function f (z) of the form (2) is in the class T S p (α) if and only if n=∞ X [(α + 1)n − α] an ≤ 1

(i)

n=2

and (ii) U CT (α) if and only if

n=∞ X

n [(α + 1)n − α] an ≤ 1.

n=2

For, a1 , a2 , ..., ap and b1 , b2 , ..., bq be complex numbers with bj = 6 0, −1, ...; j = 1, 2, ..., q, then the generalized hypergeometric functions p Fq (z) is defined by p Fq (z) =p Fq (a1 , a2 , ..., ap ; b1 , b2 , ..., bq , z) =

∞ X (a1 )n (a2 )n ...(ap )n z n , (b ) (b ) ...(b ) (1) 1 n 2 n p n n n=0

p ≤ q + 1 where (λ)n is the Pochhammer symbol defined by 1 if n = 0 (λ)n = λ(λ + 1) . . . (λ + n − 1) if n = 1, 2, . . . .

(4)

(5)

We note that the function in the series in (4) converges absolutely in the entire complex plane for p < q + 1, and for p = q + 1 in the unit disc, the condition p ≤ q + 1 stated with the definition (3) will be hold true throughout this!paq p P P per.We note that the function p Fq (1) converges whenever ℜ bj − aj > j=1

i=1

0.

Merkes and Scott [8] used continued fractions to find sufficient conditions for the function z2 F1 (z) to be in the class S ∗ (α), 0 ≤ α < 1. Silverman [14] and Dixit and Pathak [2] determined sufficient conditions for the function z2 F1 (z) to be in the class S ∗ (α)and U CV(α) respectively and various geometric properties of this function was discussed in [3, 4, 9, 14, 15]. In this paper,we we obtain sufficient condition for function h(z),given by ′

h(z) = (1 − µ)τ (z) + µzτ (z)

(6)

where µ ≥ 0 and τ (z) = zp Fq (z) belonging to the classes Sp (α)and U CT (α).

694

C. Ramachandran, L. Vanitha, G. Murugusundaramoorthy 2. Main Results Theorem 2.1.

If ai > 0, (i = 1, 2, ..., p) and

q P

j=1

bj >

p P

ai + 2, then a

i=1

sufficient condition for the function h(z) to be in the class T S p (α), 0 ≤ α < 1, is that

(a1 )2 (a2 )2 ...(ap )2 µ(α + 1) p Fq (a1 + 2; b1 + 2, 1) (b1 )2 (b2 )2 ...(bp )2 (a1 )(a2 )...(ap ) + (µ(α + 2) + α + 1) p Fq (a1 + 1; b1 + 1, 1) + (b1 )(b2 )...(bp )

P Fq (1)

≤ 2. (7)

The condition (7) is necessary and sufficient for the function h1 (z), defined by to be in the class T S p (α). h1 (z) = z 2 − h(z) z Proof. Since

h(z) = z +

∞ X

(1 − µ + µn)

n=2

(a1 )n−1 (a2 )n−1 ...(ap )n−1 z n . (b1 )n−1 (b2 )n−1 ...(bp )n−1 (1)n−1

According to Lemma 1.3 we need only to show that ∞ X

[n(α + 1) − α] (1 − µ + µn)

n=2

(a1 )n−1 (a2 )n−1 ...(ap )n−1 ≤ 1. (b1 )n−1 (b2 )n−1 ...(bp )n−1 (1)n−1

Now from (8) ∞ X

[(n + 1)(α + 1) − α] (1 − µ + µ(n + 1))

n=1

=

∞ X

n=1

µ(α + 1)n2 + (µ + α + 1)n + 1

(a1 )n (a2 )n ...(ap )n (b1 )n (b2 )n ...(bp )n (1)n

(a1 )n (a2 )n ...(ap )n . (b1 )n (b2 )n ...(bp )n (1)n

(8)

STARLIKE AND UNIFORMLY CONVEX FUNCTIONS INVOLVING... 695 Using the fact that (λ)n = λ(λ + 1)n−1 yields = µ(α + 1)

∞ X

n=1

(a1 )n (a2 )n ...(ap )n (b1 )n (b2 )n ...(bp )n (1)n−2

+ (µα + 2µ + α + 1)

+

∞ X

n=1

∞ X

(a1 )n (a2 )n ...(ap )n (b1 )n (b2 )n ...(bp )n (1)n−1 n=1

(a1 )n (a2 )n ...(ap )n (b1 )n (b2 )n ...(bp )n (1)n

(a1 )2 (a2 )2 ...(ap )2 = µ(α + 1) p Fq (a1 + 2; b1 + 2, 1) (b1 )2 (b2 )2 ...(bp )2 (a1 )(a2 )...(ap ) + (µ(α + 2) + α + 1) p Fq (a1 + 1; b1 + 1, 1) +P Fq (1) − 1. (b1 )(b2 )...(bp ) But this last expression is bounded by 1 if and only if( 7) holds. Since h1 (z) = z −

∞ X

(1 − µ + µn)

n=2

(a1 )n−1 (a2 )n−1 ...(ap )n−1 z n . (b1 )n−1 (b2 )n−1 ...(bp )n−1 (1)n−1

(9)

The condition (7) is also necessary for h1 (z) to be in the class T S p (α), 0 ≤ α < 1 from the Lemma 1.3 n Q Theorem 2.2. If ai > −1, (i = 1, 2, ..., p), bj > 0, (j = 1, 2, ..., q), ai < 0 and

q P

j=1

bj >

p P

i=1

ai + 2, then a necessary and sufficient condition for the function

i=1

h(z) to be in the class T S p (α), 0 ≤ α < 1, is that (a1 + 1)(a2 + 1)...(ap + 1) µ(α + 1) p Fq (a1 + 2; b1 + 2, 1) (b1 + 1)(b2 + 1)...(bp + 1) b1 b2 ...bq + (µ(α + 2) + α + 1) p Fq (a1 + 1; b1 + 1, 1) + p Fq (1) ≥ 0. a1 a2 ...ap The condition

q P

j=1

bj >

p P

i=1

ai + 1 −

n Q

(10)

ai is necessary and sufficient for the

i=1

function h(z) to be in the class T S 0 = S ∗ , a class of starlike function with negative coefficients.

696


Proof. Since, a1 a2 ...ap h(z) = z − b1 b2 ...bq ∞ X (a1 + 1)n−2 (a2 + 1)n−2 ...(ap + 1)n−1 z n . (11) (1 − µ + µn) (b1 + 1)n−2 (b2 + 1)n−2 ...(bp + 1)n−2 (1)n−1 n=2

If we show the inequality ∞ X

[n(α + 1) − α] (1 − µ + µn)

n=2

(a1 + 1)n−2 (a2 + 1)n−2 ...(ap + 1)n−2 (b1 + 1)n−2 (b2 + 1)n−2 ...(bp + 1)n−2 (1)n−1 b1 b2 ...bq ≤ a1 a2 ...ap (12)

then by Lemma 1.3, Theorem 2.2 is obtained. The left hand side of the equation q p P P (12) converges if bj > ai + 2.From (12) j=1

∞ X

i=1

[(n + 2)(α + 1) − α] [1 − µ + µ(n + 2)]

n=0

= µ(α + 1)

(a1 + 1)n (a2 + 1)n ...(ap + 1)n (b1 + 1)n (b2 + 1)n ...(bp + 1)n (1)n+1

∞ (a1 + 1)(a2 + 1)...(ap + 1) X (a1 + 2)n (a2 + 2)n ...(ap + 2)n (b1 + 1)(b2 + 1)...(bp + 1) n=0 (b1 + 2)n (b2 + 2)n ...(bp + 2)n (1)n + (µ(α + 2) + α + 1)

∞ X

n=0

+

∞ X

n=1

(a1 + 1)n (a2 + 1)n ...(ap + 1)n (b1 + 1)n (b2 + 1)n ...(bp + 1)n (1)n

(a1 + 1)n−1 (a2 + 1)n−1 ...(ap + 1)n−1 (b1 + 1)n−1 (b2 + 1)n−1 ...(bp + 1)n−1 (1)n

(a1 + 1)(a2 + 1)...(ap + 1) = µ(α + 1) (b1 + 1)(b2 + 1)...(bp + 1)

p Fq (a1

+ (µ(α + 2) + α + 1) p Fq (a1 + 1; b1 + 1, 1) +

+ 2; b1 + 2, 1)

b1 b2 ...bq {p Fq (1) − 1}. a1 a2 ...ap

STARLIKE AND UNIFORMLY CONVEX FUNCTIONS INVOLVING... 697 Hence (12) is equivalent to (a1 + 1)(a2 + 1)...(ap + 1) µ(α + 1) p Fq (a1 + 2; b1 + 2, 1) (b1 + 1)(b2 + 1)...(bp + 1) + (µ(α + 2) + α + 1) p Fq (a1 + 1; b1 + 1, 1)

(13)

b1 b2 ...bq b1 b2 ...bq b1 b2 ...bq + + = 0. p Fq (1) ≤ a1 a2 ...ap a1 a2 ...ap a1 a2 ...ap

Thus (13) is valid if and only if (a1 + 1)(a2 + 1)...(ap + 1) µ(α + 1) p Fq (a1 + 2; b1 + 2, 1) (b1 + 1)(b2 + 1)...(bp + 1) + (µ(α + 2) + α + 1) p Fq (a1 + 1; b1 + 1, 1) +

b1 b2 ...bq p Fq (1) ≥ 0. a1 a2 ...ap

This is equivalent to (10). This completes the proof of the Theorem 2.2. Theorem 2.3.

If ai > 0, i = 1, 2, ..., p, and

q P

j=1

bj >

p P

ai + 3, then a

i=1

sufficient condition for the function h(z) to be in the class U CT (α), 0 ≤ α < 1, is that (a1 )3 (a2 )3 ...(ap )3 µ(α + 1) p Fq (a1 + 3; b1 + 3, 1) (b1 )3 (b2 )3 ...(bq )3 (a1 )2 (a2 )2 ...(ap )2 (14) + (4µα + 5µ + α + 1) p Fq (a1 + 2; b1 + 2, 1) (b1 )2 (b2 )2 ...(bq )2 a1 a2 ...ap + (2µα + 4µ + α + 3) p Fq (a1 + 1; b1 + 1, 1) +p Fq (1) ≤ 2. b1 b2 ...bq The condition (14) is a necessary and sufficient condition for the function h1 (z) defined by (9) to be in the class U CT (α). Proof. Taking h(z) defined by (6), in view of (ii) of Lemma 1.3, we need only to show that ∞ X

n=2

n [n(α + 1) − α] (1 − µ + µn)

(a1 )n−1 (a2 )n−1 ...(ap )n−1 ≤ 1. (b1 )n−1 (b2 )n−1 ...(bp )n−1 (1)n−1

(15)

698


The left hand side of (15) converges if

q P

bj >

j=1

∞ X

p P

ai + 3. From (15)

i=1

(a1 )n+1 (a2 )n+1 ...(ap )n+1 (b1 )n+1 (b2 )n+1 ...(bp )n+1 (1)n+1

(n + 2) [(n + 2)(α + 1) − α] (1 − µ + µ(n + 2))

n=0

= µ(α + 1)

∞ X

n=0

(a1 )n+3 (a2 )n+3 ...(ap )n+3 (b1 )n+3 (b2 )n+3 ...(bp )n+3 (1)n

+ (4µα + 5µ + α + 1) + (2µα + 4µ + α + 3)

∞ X

(a1 )n+2 (a2 )n+2 ...(ap )n+2 (b ) (b ) ...(bp )n+2 (1)n n=0 1 n+2 2 n+2 ∞ X

n=0

∞

X (a1 )n (a2 )n ...(ap )n (a1 )n+1 (a2 )n+1 ...(ap )n+1 + . (b1 )n+1 (b2 )n+1 ...(bp )n+1 (1)n (b1 )n (b2 )n ...(bp )n (1)n n=1

(16)

Since (a)n+k = (a)k (a + k)n , we may write (16) as ∞ (a1 )3 (a2 )3 ...(ap )3 X (a1 + 3)n (a2 + 3)n ...(ap + 3)n = µ(α + 1) (b1 )3 (b2 )3 ...(bq )3 (b + 3)n (b2 + 3)n ...(bp + 3)n (1)n n=0 1 ∞ (a1 )2 (a2 )2 ...(ap )2 X (a1 + 2)n (a2 + 2)n ...(ap + 2)n + (4µα + 5µ + α + 1) (b1 )2 (b2 )2 ...(bq )2 (b1 + 2)n (b2 + 2)n ...(bp + 2)n (1)n

n=0

+ (2µα + 4µ + α + 3)

a1 a2 ...ap b1 b2 ...bq

∞ X

n=0

+

(a1 + 1)n (a2 + 1)n ...(ap + 1)n (b1 + 1)n (b2 + 1)n ...(bp + 1)n (1)n

∞ X

n=0

(a1 )n (a2 )n ...(ap )n − 1. (b1 )n (b2 )n ...(bp )n (1)n

The last expression is bounded by 1 if and only if (14) holds,hence the proof of the Theorem 2.3.

Theorem 2.4. If ai > −1, (i = 1, 2, ..., p),

n Q

i=1

ai < 0 and

q P

j=1

bj >

p P

ai +3,

i=1

then a necessary and sufficient condition for the function h(z) to be in the class

STARLIKE AND UNIFORMLY CONVEX FUNCTIONS INVOLVING... 699 U CT (α), 0 ≤ α < 1, is that (a1 + 1)2 (a2 + 1)2 ...(ap + 1)2 µ(α + 1) p Fq (a1 + 3; b1 + 3, 1) (b1 + 1)2 (b2 + 1)2 ...(bp + 1)2 (a1 + 1)(a2 + 1)...(ap + 1) + (4µα + 5µ + α + 1) p Fq (a1 + 2; b1 + 2, 1) (b1 + 1)(b2 + 1)...(bp + 1) b1 b2 ...bq + (2µα + 4µ + α + 3) p Fq (a1 + 1; b1 + 1, 1) + p Fq (1) ≥ 0. a1 a2 ...ap (17)

Proof. Taking h(z) defined by (11), in view of (ii) of Lemma 1.3, we must show that ∞ X

n [n(α + 1) − α] (1 − µ + µn)

n=2

The left side of (18) converges if

(a1 + 1)n−2 (a2 + 1)n−2 ...(ap + 1)n−2 (b1 + 1)n−2 (b2 + 1)n−2 ...(bp + 1)n−2 (1)n−1 b1 b2 ...bq . ≤ a1 a2 ...ap (18)

q P

j=1

∞ X

bj >

p P

ai + 3. Now from (18)

i=1

(n + 2) [(n + 2)(α + 1) − α] (1 − µ + µ(n + 2))

n=0

= µ(α + 1)

∞ X

(a1 + 1)n+2 (a2 + 1)n+2 ...(ap + 1)n+2 (b1 + 1)n+2 (b2 + 1)n+2 ...(bp + 1)n+2 (1)n n=0

+ (4µα + 5µ + α + 1) + (2µα + 4µ + α + 3)

∞ X

n=0 ∞ X

n=0

+

(a1 + 1)n (a2 + 1)n ...(ap + 1)n (b1 + 1)n (b2 + 1)n ...(bp + 1)n (1)n+1

b1 b2 ...bq a1 a2 ...ap

X ∞

(a1 + 1)n+1 (a2 + 1)n+1 ...(ap + 1)n+1 (b1 + 1)n+1 (b2 + 1)n+1 ...(bp + 1)n+1 (1)n (a1 + 1)n (a2 + 1)n ...(ap + 1)n (b1 + 1)n (b2 + 1)n ...(bp + 1)n (1)n

(a1 )n+1 (a2 )n+1 ...(ap )n+1 (b ) (b ) ...(bp )n+1 (1)n+1 n=0 1 n+1 2 n+1

700


∞ (a1 + 1)2 (a2 + 1)2 ...(ap + 1)2 X (a1 + 3)n (a2 + 3)n ...(ap + 3)n = µ(α + 1) (b1 + 1)2 (b2 + 1)2 ...(bp + 1)2 (b + 3)n (b2 + 3)n ...(bp + 3)n (1)n n=0 1 ∞ (a1 + 1)(a2 + 1)...(ap + 1) X (a1 + 2)n (a2 + 2)n ...(ap + 2)n + (4µα + 5µ + α + 1) (b1 + 1)(b2 + 1)...(bp + 1) (b1 + 2)n (b2 + 2)n ...(bp + 2)n (1)n

n=0

∞ X

(a1 + 1)n (a2 + 1)n ...(ap + 1)n (b + 1)n (b2 + 1)n ...(bp + 1)n (1)n n=0 1 X ∞ b1 b2 ...bq (a1 )n (a2 )n ...(ap )n b1 b2 ...bq − . + a1 a2 ...ap (b ) (b ) ...(bp )n (1)n a1 a2 ...ap n=0 1 n 2 n b1 b2 ...bq This last expression is bounded by a1 a2 ...ap if and only if (17) holds. Thus the + (2µα + 4µ + α + 3)

proof of the Theorem 2.4 is completed.

Concluding Remak: If p = 2, q = 1, the results reduces to Dixit and Pathak[4] result.

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