ACTA UNIVERSITATIS APULENSIS

No 17/2009

CRITERIA FOR UNIVALENCE OF CERTAIN INTEGRAL OPERATORS

V. Ravichandran Abstract. In this paper, we determine conditions on β, αi and fi (z) so o1/β n R Q z is univalent in that the integral operator β 0 ξ β−1 ni=1 (fi (ξ)/ξ)1/αi dξ the open unit disk. We also obtain similar results for the integral operator  R z β−1 1 P β 0 ξ exp ( ni=1 αi fi (ξ)) dξ β . 2000 Mathematics Subject Classification: 30C45. 1. Introduction An analytic function that maps different points in the open unit disk U := {z ∈ C : |z| < 1} to different points in C is a univalent function. Such functions have been studied for long time. Let A be the class of all analytic functions f (z) defined in U and normalized by the conditions f (0) = 0 = f 0 (0) − 1. Let S be the subclass of A consisting of univalent functions. Pascu [2] has proved the following theorem: Theorem 1 ([1],[2]) Let β ∈ C, γ ∈ R and <β ≥ γ > 0. If f ∈ A satisfies 1 − |z|2γ zf 00 (z) f 0 (z) ≤ 1 (z ∈ U ), γ then the integral operator  Z Fβ (z) = β

z

ξ

0

is in S.

141

β−1 0

f (ξ)dξ

 β1

V. Ravichandran - Criteria for Univalence of Certain Integral Operators

We denote by Sβ the class of functions f ∈ A satisfying 2 0 z f (z) f 2 (z) − 1 < β (0 < β ≤ 1; z ∈ U ) while Sβ∗ denote the class of functions f ∈ A satisfying 0 zf (z) < β (0 < β ≤ 1; z ∈ U ). − 1 f (z) Using Theorem 1, Pescar [4] proved the following theorem. Theorem 2 ([4]) Let α, β ∈ C and <β ≥ <α ≥ 3/|α|. If f ∈ S1 satisfies the condition |f (z)| ≤ 1 (z ∈ U ), then the integral operator ( Z Hα,β (z) = β

z

ξ β−1



0

f (ξ) ξ

) β1

 α1 dξ

(1)

is in S. Using Theorem 1, Breaz and Breaz [1] extended Theorem 2 and obtained the following theorem. Theorem 3 (Theorem 1, p.260 [1]) Let α, β ∈ C and <β ≥ <α > 3n/|α|. If fi ∈ S1 (i = 1, 2, · · · , n) satisfies the conditions |fi (z)| ≤ 1

(z ∈ U,

i = 1, 2, · · · , n),

then the integral operator ( Z Fα,β (z) = β

) β1

1

z

ξ β−1

0

 n  Y fi (ξ) α i=1

ξ

dξ

is in S. Using Theorem 1, Pescar [5] obtained the following theorem. 142

(2)

V. Ravichandran - Criteria for Univalence of Certain Integral Operators

Theorem 4 (Theorem 1, p.452 [5]) Let f ∈ A and β ∈ C satisfies √ 3 3 1 ≤ <β ≤ |β| ≤ . 2 If |zf 0 (z)| ≤ 1

(z ∈ U ),

then the function  Z Tβ (z) = β

z

ξ β−1


β ef (ξ) dξ

 β1

0

is in S. By Schwarz’s Lemma (see below), the only function f ∈ A with |f (z)| ≤ 1 is f (z) = z and hence the hypothesis of Theorem 3–4 are satisfied only by a single function f (z) = z. In this paper, we extend Theorem 2 and Theorem 4 to obtain a sufficient conditions for univalence of a more general integral operator. We also prove some related results. The class of functions to which our theorems are applicable is non-trivial (when Mi 6= 1 for some i). To prove our main results, we need the following lemma: Lemma 1 (Schwarz’s Lemma) If the function w(z) is analytic in the unit disk U , w(0) = 0 and |w(z)| ≤ 1 for all z ∈ U , then |w(z)| ≤ |z| (z ∈ U ),

|w0 (0)| ≤ 1

and the either of these equalities holds if and only if w(z) = z, where || = 1.

2. Univalence Criteria For fi ∈ A (i = 1, 2, · · · , n) and α1 , α2 , . . . , αn , β ∈ C, we define an integral operator by ( Z Fα1 ,α2 ,··· ,αn ;β (z) := β

) β1

1

z

ξ β−1

0

 n  Y fi (ξ) αi i=1

ξ

dξ

.

(3)

When α1 = α2 = · · · = αn = α, Fα1 ,α2 ,··· ,αn ;β (z) becomes the integral operator Fα,β (z) considered in Theorem 3. 143

V. Ravichandran - Criteria for Univalence of Certain Integral Operators

Theorem 5 Let 0 < βi ≤ 1. Let fi ∈ Sβi (i = 1, 2, · · · , n) satisfy the conditions |fi (z)| ≤ Mi (Mi ≥ 1; z ∈ U, i = 1, 2, · · · , n). If α1 , α2 , . . . , αn , β ∈ C, <β ≥ γ and γ :=

n X 1 + Mi (1 + βi )

|αi |

i=1

,

(4)

then the function Fα1 ,α2 ,··· ,αn ;β (z) defined by (3) is in S. Proof.

Define the function h(z) by n  zY

Z h(z) := 0

i=1

fi (ξ) ξ

 α1

i

dξ.

Then we have h(0) = h0 (0) − 1 = 0. Also a simple computation yields 1

0

h (z) =

 n  Y fi (z) αi z

i=1

and

n

zh00 (z) X 1 = h0 (z) αi i=1



 zfi0 (z) −1 . fi (z)

(5)

From equation (5), we have 00   n X zh (z) 1 zfi0 (z) ≤ +1 h0 (z) |αi | fi (z) i=1   n X 1 z 2 fi0 (z) fi (z) = +1 . |αi | fi2 (z) z i=1 From the hypothesis, we have |fi (z)| ≤ Mi by Schwartz Lemma, yields |fi (z)| ≤ Mi |z| (z ∈ U,

144

(z ∈ U,

(6)

i = 1, 2, · · · , n) which,

i = 1, 2, · · · , n).

V. Ravichandran - Criteria for Univalence of Certain Integral Operators

Using this inequality in the inequality (6), we obtain 00  2 0  n X zh (z) z fi (z) 1 ≤ Mi 2 +1 h0 (z) |αi | fi (z) i=1  2 0  n X z fi (z) 1 Mi 2 − 1 + 1 + Mi . ≤ |α | f (z) i i i=1 Since fi ∈ Sβi , in view of (4), the equation (7) yields 00 n X zh (z) 1 + Mi (1 + βi ) < = γ. h0 (z) |αi | i=1

(7)

(8)

Multiply (8) by 1 − |z|2γ , γ we obtain 1 − |z|2γ γ

00 zh (z) 2γ h0 (z) ≤ 1 − |z| < 1 (z ∈ U ).

Since <β ≥ γ > 0, it follows from Theorem 1 that  β1  Z z β−1 0 ξ h (ξ)dξ ∈ S. β 0

Since  Z β 0

z

ξ β−1 h0 (ξ)dξ

 β1

( Z = β

ξ β−1

0

) β1

1

z

 n  Y fi (ξ) αi i=1

ξ

dξ

= Fα1 ,α2 ,··· ,αn ;β (z), the function Fα1 ,α2 ,··· ,αn ;β (z) defined by (3) is in S. Theorem 6 Let α1 , α2 , . . . , αn , β ∈ C, 0 < βi ≤ 1 (i = 1, 2, . . . , n) and <β ≥ γ :=

n X βi . |α | i i=1

(9)

If fi ∈ Sβ∗i (i = 1, 2, · · · , n), then the function Fα1 ,α2 ,··· ,αn ;β (z) defined by (3) is in S. 145

V. Ravichandran - Criteria for Univalence of Certain Integral Operators

Proof.

From (5), we get 00 X 0 n zh (z) zfi (z) 1 ≤ − 1 h0 (z) fi (z) |α | i i=1

and by using (9) and fi ∈ Sβ∗i , we obtain 00 n zh (z) X βi ≤ = γ. h0 (z) |α | i i=1 This, as in the proof of Theorem 5, shows that 1 − |z|2γ zh00 (z) h0 (z) < 1 (z ∈ U ) γ and therefore, by Theorem 1, z

 Z β

ξ β−1 h0 (ξ)dξ

 β1 ∈ S.

0

Therefore the function Fα1 ,α2 ,··· ,αn ;β (z) defined by (3) is in S. Example 1 Let α, β ∈ C, 0 < βi ≤ 1 (i = 1, 2, . . . , n) satisfy n

1 X <β ≥ βi . |α| i=1 If fi ∈ Sβ∗i (i = 1, 2, · · · , n), then the function Fα,β (z) defined by (2) is in S. β1 In particular, if α, β ∈ C, 0 < β1 ≤ 1 satisfy <β ≥ |α| and f ∈ Sβ∗1 , then the function Hα,β (z) defined by (1) is in S. For i = 1, 2, . . . , n, let αi , β ∈ C and fi (z) ∈ A. Define the integral operator Tα1 ,...,αn ;β (z) by ( Z Tα1 ,...,αn ;β (z) = β

z

ξ β−1 exp

0

n X i=1

146

! αi fi (ξ) dξ

) β1 .

(10)

V. Ravichandran - Criteria for Univalence of Certain Integral Operators

Theorem 7 Let fi ∈ A (i = 1, 2, · · · , n) satisfy |zfi0 (z)| ≤ Mi

(Mi ≥ 1;

z ∈ U ).

Let αi , β ∈ C (i = 1, 2, . . . , n) and γ > 0 be the smallest number such that 2

n X

Mi |αi | ≤ (2γ + 1)

2γ+1 2γ

.

(11)

i=1

Then, for β ∈ C, <β ≥ γ, the function Tα1 ,...,αn ;β (z) defined (10) is in S. Proof.

Define the function h(z) by Z h(z) :=

z

exp

n X

0

! αi fi (ξ) dξ.

i=1

Then we have h(0) = h0 (0) − 1 = 0. Also a simple computation yields ! n X 0 h (z) = exp αi fi (z) i=1

and

n

zh00 (z) X = αi zfi0 (z). 0 h (z) i=1

(12)

By Schwartz’s Lemma, we have |zfi0 (z)| ≤ |z|, (z ∈ U, i = 1, 2, · · · , n), and therefore we obtain, from (12), 00 n n X zh (z) X 0 ≤ |α | |zf (z)| ≤ ( Mi |αi |)|z|. i i h0 (z) i=1 i=1 Thus

n 1 − |z|2γ zh00 (z) |z|(1 − |z|2γ ) X Mi |αi | (z ∈ U ). h0 (z) ≤ γ γ i=1

For the function Q : [0, 1] → R defined by Q(t) = t(1 − t2γ ), γ > 0, the maximum is attained at the point t = 1/(2γ + 1)1/2γ and thus we have 2γ

Q(t) ≤

(2γ + 1) 147

2γ+1 2γ

.

V. Ravichandran - Criteria for Univalence of Certain Integral Operators

In view of this inequality and our assumption (11), we obtain 1 − |z|2γ zh00 (z) h0 (z) ≤ 1 (z ∈ U ) γ and, by Theorem 1, Tα1 ,...,αn ;β (z) is univalent in U for β ∈ C with <β ≥ γ. Also we have the following result. Theorem 8 Let αi , β ∈ C, 0 ≤ βi ≤ 1 (i = 1, 2, . . . , n). Let fi ∈ Sβ∗i (i = 1, 2, · · · , n) satisfy |fi (z)| ≤ Mi

(Mi ≥ 1;

z ∈ U;

i = 1, 2, . . . , n).

and γ > 0 be the smallest number such that 2

n X

Mi |αi |(1 + βi ) ≤ (2γ + 1)

2γ+1 2γ

.

(13)

i=1

Then, for β ∈ C, <β ≥ γ, the function Tα1 ,...,αn ;β (z) defined (10) is in S. Proof.

From (12), we have 00 X n zh (z) ≤ |αi | h0 (z) i=1

0 zfi (z) fi (z) |fi (z)| 0   n X zfi (z) − 1 |fi (z)| ≤ |αi | 1 + fi (z) i=1 ≤

n X

Mi |αi |(1 + βi )|z|.

i=1

The remaining part of the proof is similar to the proof of Theorem 7. Similarly we have the following result. Theorem 9 Let fi ∈ Sβi (i = 1, 2, · · · , n) satisfy |fi (z)| ≤ Mi

(Mi ≥ 1;

z ∈ U;

i = 1, 2, . . . , n).

Let αi , β ∈ C, 0 ≤ βi ≤ 1 (i = 1, 2, . . . , n), and γ > 0 be the smallest number such that n X 2γ+1 2 Mi2 |αi |(1 + βi ) ≤ (2γ + 1) 2γ . (14) i=1

Then, for β ∈ C, <β ≥ γ, the function Tα1 ,...,αn ;β (z) defined (10) is in S. 148

V. Ravichandran - Criteria for Univalence of Certain Integral Operators

References [1] D. Breaz and N. Breaz, An univalent condition for an integral operator, Nonlinear Funct. Anal. Appl. 11 (2006), no. 2, 259–263. [2] N. N. Pascu, On a univalence criterion II, in Itinerant seminar on functional equations, approximation and convexity (Cluj-Napoca, 1985), 153–154, Univ. “Babe¸s-Bolyai”, Cluj. [3] N. N. Pascu, An improvement of Becker’s univalence criterion, in Proceedings of the Commemorative Session: Simion Sto¨ılow (Bra¸sov, 1987), 43–48, Univ. Bra¸sov, Bra¸sov. [4] V. Pescar, New criteria for univalence of certain integral operators, Demonstratio Math. 33 (2000), no. 1, 51–54. [5] V. Pescar, Certain sufficient conditions for univalence, Hokkaido Math. J. 32(2) (2003), 451–455. [6] V. Pescar and S. Owa, Some criteria for univalence of certain integral operators, Int. J. Math. Math. Sci. 2004, no. 45-48, 2489–2494. Author: V. Ravichandran Department of Mathematics University of Delhi Delhi 110 007 India email: [email protected]

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