Cmjv02i06p0846 by Journal of Computer and Mathematical Sciences

J. Comp. & Math. Sci. Vol.2 (6), 846-854 (2011)

A Certain Subclass of Regular Functions with Negative Coefficients Associated with Conic Regions for Contraction Operators on Hilbert Space C. SELVARAJ1, A. ALMA JULIET PAMELA2 and M.THIRUCHERAN1 1

Presidency College (Autonomous), Chennai − 600 005, Tamil Nadu, India 2 Stella Matutina College of Education (Autonomous), Chennai − 600 083, Tamil Nadu, India ABSTRACT In this paper the authors introduce a new subclass Tµ,β(α) of analytic functions with negative coefficients associated with conic regions for operators on Hilbert space and obtain various results including characterization theorem. Keywords: Hilbert space, negative coefficients, regular function, conic regions.

We denote by S*(α) the class of all functions satisfying (1.2).

1. INTRODUCTION Let A denote the class of normalised functions of the form f(z) = z +

∞

∑a z , n

(1.1)

n=2

which are regular in the unit disk ∆ = {z : |z| < 1}. Also let S denote the class of all functions in A, which are univalent in the disk ∆. A function f(z) ∈ S is said to be starlike of order α if  zf ′(z)  Re   > α, (z ∈ ∆, 0 ≤ α < 1).  f(z) 

(1.2)

A function f(z) ∈ S is said to be convex of order α in ∆ if and only if  zf ′′(z)  Re1 +  > α, (z ∈ ∆, 0 ≤ α < 1). f ′(z)  

(1.3)

We denote by K(α) the class of all functions satisfying (1.3). Definition5. The function f ∈ S is said to be uniformly convex (UCV) if and only if

 zf ′′(z)  zf ′′(z) Re1 + , (z ∈ ∆) ≥ f ′(z)  f ′(z) 

Journal of Computer and Mathematical Sciences Vol. 2, Issue 6, 31 December, 2011 Pages (780-898)

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C. Selvaraj, et al., J. Comp. & Math. Sci. Vol.2 (6), 846-854 (2011)

Definition12. The function f ∈ S is said to be β-uniformly convex of order α if and only if

D n f(z) :=

 zf ′′(z)  zf ′′(z) Re 1 + + α, (z ∈ ∆) >β f ′(z)  f ′(z) 

function

F (z)≡ 2F1 (a,b,c;z)

2 1

∞

(a)n (b)n z , where c ≠ 0,−1,−2,... (1.4) (c)n n!

∑ n=0

and (x)n is the Pochhammer symbol defined in terms of the Gamma function by (x)n :=

Γ(x + n) Γ(x)

1 (n = 0) = x(x + 1)...(x+ n −1) (n ∈ N := {1,2,3,...}). (1.5) 

The Hadamard product of two functions ∞

∞

f(z) = ∑ a n z n

and

n =0

g(z) = ∑ b n z n n =0

is defined by ∞

f(z) ∗ g(z) = ∑ a n b n z n .

z(z n -1f(z)) (n) n!

(1.9)

and is called the Ruscheweyh derivative of f(z).

0 ≤ α < 1 and 0 ≤ β ≤ 1. The Gaussian hypergeometric 2F1(z) is defined by

In fact, for µ = n ∈ N0 = {0, 1, 2, …} (1.8) may be expressed as

(1.6)

n =0

Ruscheweyh15 introduced an operator Dµ : A → A, defined by the convolution

Let T denote the subclass of S consisting of functions of the form f(z) = z −

∞

∑a z , n

(a n ≥ 0).

(1.10)

n=2

We also denote by T*(α) , K*(α) and β − UCV*(α) the subclasses of T that are respectively starlike of order α, convex of order α and β-uniformly convex of order α . Let Tµ(α) 19 denote the class of functions of the form (1.10) satisfying the condition  (D µ f(z))′  Re z µ  > α, (µ > −1, α < 1, z ∈ ∆).  D f(z) 

Note that D0(z) = f(z), D′f(z) = zf ′(z), so that T0(α) = T*(α), T1(α) = K(α). Let Tµ,β(α) denote the subclass of T satisfying  (D µ f(z) ) ′  (D µ f(z) ) ′ Re  z − 1 + α, >βz µ D f(z) D µ f(z)  

0 ≤ α < 1, 0 ≤ β ≤ 1.

(1.12) Then

D µ f(z) :=

z ∗ f(z), (µ ≥ −1; z ∈ ∆). (1.7) (1 − z)µ +1

D f(z) = z 2 F1 (1, µ + 1;1; z) ∗ f(z). µ

T0,0 (α) = T ∗ (α), T1,0 (α) = K ∗ (α), T0,β (α) = S0 (1, β, α) [16],

It is easy to verify that (1.8)

(1.11)

T1,β (α) = β − UCV ∗ (α). [12]

Journal of Computer and Mathematical Sciences Vol. 2, Issue 6, 31 December, 2011 Pages (780-898)

C. Selvaraj, et al., J. Comp. & Math. Sci. Vol.2 (6), 846-854 (2011)

Finally, let H be a complex Hilbert space. Let A be a bounded linear operator on H. For a complex valued function f analytic on a domain D of the complex plane containing the spectrum σ(A) of A, f(A) will denote the operator on H define by the usual RieszDunford integral10. 1 f(z)(zI − A) −1 dz. f(A) := 2π Γ

∫

(1.13)

Proof. If f ∈ Tµ,β(α) , then  (Dµ f(z))′  (D µ f(z))′ Rez µ −1 + α  > βz µ D f(z)  D f(z) 

Now βz

 (D µ f(z))′  (D µ f(z))′ − 1 − Re z µ − 1 µ D f(z)  D f(z)   (D µ f(z))′  ≤ (β + 1)z µ − 1  D ∞ f(z) 

Here I stands for the identity operator on H, Γ is a positively oriented simple closed rectifiable contour, such that the inside domain Ω of Γ contains σ(A) and Γ ∪ Ω ⊂ D. Also f(A) can be defined by the series f(A) :=

∞

∑ n =0

f (n) (0) n A , n!

(n − 1)(µ + 1) n a nzn (n − 1)! = (β + 1) n = 2 ∞ (µ + 1) n z− an n = 2 (n − 1)! ∞

∑

(n − 1)(µ + 1) n an (n − 1)! ≤ 1 − α. ∞ (µ + 1) n 1− an n = 2 (n − 1)!

(β + 1)

(1.14)

≤

∑ n=2

(µ + 1) n a nzn . (n − 1)!

Therefore,

(1.15)

f ∈ T µ, B (α) ⇔

∞

⇔

∞

∑ n=2

⇔

∞

∑a z , n

n =2

(an ≥ 0) belongs to the class Tµ,β(α) if and only if

∑ n=2

[n(β + 1) − (α + β)] b n (µ)a n ≤ 1, (1 − α)

(1.16)

where b n (µ) =

 n(β + 1) − (α + β)   b n (µ)a n ≤ 1, 1− α 

∑  n=2

∞

 (β + 1)(n − 1)( µ + 1) n (1 − α)(µ + 1) n  + a n ≤ 1 − α (n − 1)! (n − 1)!  (µ + 1) n [(β + 1)(n − 1) + (1 − α)]a n ≤ 1 − α (n − 1)!

∑  n=2

In our investigation, A* always denote the conjugate operator of A. Lemma 1.1. The function f(z) = z −

∑

which converges in the norm topology . If f is defined by (1.10), then ∞

∞

n =2

D µ f(z) = z −

848

where b n (µ) =

(µ + 1) n −1 . (n − 1)!

Note that, the condition  (D µ f(z))′  (D µ f(z))′ Rez µ +α >βz µ D f(z)  D f(z) 

is equivalent to (µ + 1) n −1 (µ + 1)(µ + 2)...(µ + n − 1) = . (n − 1)! (n − 1)!

(D µ f(z))′ (D µ f(z))′ −1 < β z µ + (1 − α). µ D f(z) D f(z)

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C. Selvaraj, et al., J. Comp. & Math. Sci. Vol.2 (6), 846-854 (2011)

Corollary 1.2. For β = 0, we get ∞

∑ n=2

n −α b n (µ)a n ≤ 1, the equivalent condition 1− α

obtained for Tµ(α) in19. Corollary 1.3. If the function f(z) of the form (1.13) belongs to Tµ,β(α), then for |z| ≤ r < 1, an ≤

where b n (µ) =

Proof. Assume that (2.1) holds. Then we get A(D µ f)′(A) − D µ f(A) − β A(D µ f)′(A) − D µ f(A) − (1 − α) D µ f(A) ∞

∞

= A − ∑ nb n (µ)a n A n − A + ∑ b n (µ)a n A n n =2

n =2

∞

− β A − ∑ nb n (µ)a n A − A + ∑ b n (µ)a n A n

(1 − α) , [n(β + 1) − (α + β)] b n (µ)

n=2

n =2

∞

− (1 − α) A − ∑ b n (µ)a n A n n =2

(µ + 1) n −1 . (n − 1)!

where b n (µ) =

(µ + 1) n −1 . (n − 1)!

∞

∑ (n − 1)b n=2

From Lemma 1.1, we define a new class Tµ,β(α, A) as follows:

∞

(µ)a n A n − β ∑ (n − 1)b n (µ)a n A n n =2

∞

− (1 − α) A − ∑ b n (µ)a n A n n=2

Definition 1.4. Let Tµ,β(α, A) denote the class of functions of the form f(z) = z −

∞

≤ ∑ [(n − 1) + (n − 1)β + (1 − α)]b n (µ)a n − (1 − α) n =2 ∞

∞

∑a z n

= ∑ [n(β + 1) − (α + β)]b n (µ)a n − (1 − α) ≤ 0.

(a n ≥ 0)

n =2

satisfying the condition: A(D µ f ) ′(A) − D µ f(A)

≤ β A(D µ f ) ′ (A) − D µ f(A) + (1 − α) D µ f(A) ,

µ > −1, 0 ≤ β < 1, 0 ≤ α < 1, and all operators A with ||A|| < 1, A ≠ θ (θ denote the zero operator on H ). In this paper we propose to investigate some results for the class Tµ,β(α, A) and also consider applications of fractional calculus for operators on Hilbert Space. Such type of work has been carried out by various authors12,17,18,19. 2. MAIN RESULTS

∑ n =2

[n(β + 1) − (α + β)] b n (µ)a n ≤ 1, (1 − α)

A(Dµ f)′(A) − D µ f(A) ≤ β A(Dµ f)′(A) − D µ f(A) + (1 − α) D µ f(A) .

Then ∞

∑ (n − 1)b (µ)a A n

n=2

≤β

∞

∑ (n − 1)b (µ)a A n

+ (1 − α)A −

n =2

∞

∑ b (µ)a A n

n =2

Choose A = eI (0 < e < 1) . We have ∞

Theorem 2.1. A function f(z) ∈ Tµ,β(α, A) for all operators A on H with ||A|| < 1, A ≠ θ, µ > −1, 0 ≤ β < 1, 0 ≤ α < 1, if and only if ∞

Hence f(z) ∈ Tµ,β(α) . Conversely, assume that

(2.1)

∑[(n − 1) + (n − 1)β]b (µ)a e n

n=2

e−

∞

∑

≤ 1 − α.

(2.2)

b n (µ)a n e n

n =2

Upon clearing the denominator in (2.2) and letting e → 1, we get

Journal of Computer and Mathematical Sciences Vol. 2, Issue 6, 31 December, 2011 Pages (780-898)

850

C. Selvaraj, et al., J. Comp. & Math. Sci. Vol.2 (6), 846-854 (2011) ∞

∑[(n − 1) + (n −1)β]b (µ)a n

n =2

∞

Thus,

∑  n =2

 ≤ (1 − α) 1 − 

∞



∑ b (µ)a . n

n =2

n(β + 1) − (α + β)   b n (µ)a n ≤ 1. 1− α 

where µn ≥ 0 and

∑ µ f (z), n n

∑µ

= 1. ∞

∑ µ f (z). n n

Then

n =1

f n (z) = z −

∞

∑µ

(1 − α) zn . [n(β + 1) − (α + β)]b n (µ)

Thus we have,

Corollary 2.1. For β = 0, we get

∑a z

∞

Proof. Assume that f(z) =

(1 − α) z n , (n ≥ 2). [n(β + 1) − (α + β)]

∞

n =1

n=2

f(z) = z −

f(z) =

expressed in the form

n =1

The result is sharp for the function f(z) = z −

Then f(z) ∈ Tµ,β(α, A) if and only if it can be

∈ Tµ (0, α) = Tµ (α)

∞

∑

n =2

n=2

 n −α  if and only if  b n (µ)a n ≤ 1, the n =2  1 − α  ∞

[n(β + 1) − (α + β)]b n (µ) 1− α µn 1− α [n(β + 1) − (α + β)]b n (µ)

∑

∞

∑µ

= 1 − µ 1 ≤ 1.

n =2

characterisation obtained for Tµ(α) in . Corollary 2.2. Let f(z) be in Tµ,β(α, A). Then an ≤

(1 − α) , [n(β + 1) − (α + β)]b n (µ)

where b n (µ) =

(2.3)

(µ + 1) n −1 , µ > −1, n = 2, 3, … (n − 1)!

and 0 ≤ β ≤ 1, 0 ≤ α < 1.

Hence f(z) ∈ Tµ,β(α, A). Conversely, let f(z) given by (1.10) be in the class Tµ,β(α, A). From Corollary 2.2, we obtain an ≤

1− α , n = 2,3,... [n(β + 1) − (α + β)]b n (µ)

We may set µn =

[n(β + 1) − (α + β)]b n (µ) a n , n = 2,3,... 1− α

and µ 1 = 1 −

∞

∑µ . n

Thus we obtain

n =2

The result is best possible for the function f n (z) = z −

(1 − α)z , (n ≥ 2). [n(β + 1) − (α + β)]b n (µ)

f(z) =

(1 − α) f n (z) = z − zn . [n(β + 1) − (α + β)]b n (µ)

n =1

Theorem 2.3. Let f1(z) = z and

∞

∑ µ g (z).

Theorem 2.4. If f(z) ∈ Tµ,β(α, A), µ ≥ 0, 0 ≤ α < 1 and 0 ≤ β < 1 and ||A|| < 1, A ≠ θ, then f(A) ≥ A −

(1 − α) A [2(β + 1) − (α + β)](µ + 1)

and

Journal of Computer and Mathematical Sciences Vol. 2, Issue 6, 31 December, 2011 Pages (780-898)

(2.4)

851

C. Selvaraj, et al., J. Comp. & Math. Sci. Vol.2 (6), 846-854 (2011)

f(A) ≤ A +

2 (1 − α) A . (2.5) [2(β + 1) − (α + β)](µ + 1)

Proof. By Theorem 2.1, we have ∞

∑ n =2

[n(β + 1) − (α + β)]b n (µ) a n ≤ 1. 1− α

∞ [2(β + 1) − (α + β)] (µ + 1) a n ≤ 1− α n=2

∑

∞

∑ n =2

[n(β + 1) − (α + β)] b n (µ)a n ≤ 1 1− α

for µ ≥ 0, which gives ∞

∑

an ≤

n =2

(1 − α) [2(β + 1) − (α + β)](µ + 1)

(2.6)

Therefore we have f(A) ≥ A − A

n =2

≥ A −

(1 − α) A [2(β + 1) − (α + β)](µ + 1) 2

where a > 0 and b, c ∈ R.

(3.1)

1 g ′(A), Γ(1 − a)

(3.2)

where

(1 − α) ≤ A + A [2(β + 1) − (α + β)](µ + 1)

∫

g(z) = z −b 2 F1 (b − a + 1,−c;1 − a;1 − t)f(tz)(1 − t) −a dt.

Corollary 2.5. If f(z) ∈ T0,β(α, A),0 ≤ α < 1, And ||A|| < 1, A ≠ θ, then

1− α 2 A ≤ f(A) [2(β + 1) − (α + β)] 1− α 2 ≤ A+ A . [2(β + 1) − (α + β)] A−

(2.7)

Corollary 2.6. If f(z)∈Tµ,0(α, A) = Tµ(α, A), 0 ≤ α < 1 and ||A|| < 1, A ≠ θ, then 1− α A (2 − α)(µ + 1)

∫

n =2

A −

1 A − b 2 F1 (a + b,−c; a;1 − t)f(tA)(1 − t) a −1 dt, Γ(a) 0

D a,0,b,Ac f(A) =

∞

∑a

I a,0,b,Ac f(A) =

and f(A) ≤ A + A

Definition 3.1.17 The fractional integral operator I 0,a,b,Ac of an invertible operator A is defined by

Definition 3.2.17 The fractional derivative operator of an invertible operator A is denoted by D 0,a,b,Ac f(A) and is defined as

∞

∑a

3. APPLICATIONS OF FRACTIONAL CALCULUS

2 1− α ≤ f(A) ≤ A + A , (2 − α)(µ + 1)

(2.8)

0 < a < 1 and b, c ∈ R. In both Definition 3.1 and 3.2 f(z) is an analytic function in a simply connected region of the z -plane containing the origin with the order f(z) = O(|z|ε), z → 0, where ε > max{0, b−c}−1 and the multiplicity of (1−t)a−1 is removed by requiring log (1−t) to be real, when (1−t) >0. Theorem 3.3. Let 2 > max{b − c, b, −c −a} and 2a > b(a + c). If f(z) ∈ Tµ,β(α, A), (µ ≥ 0), then I a,0,b,Ac f(A) ≤

1− b Γ(2 − b + c) A Γ(2 − b)Γ(a + 2 + c) (1 − α)Γ(2 − b + c) + A [2(β + 1) − (α + β)] Γ(2 − b)Γ(a + 2 + c)

the results obtained in19. Journal of Computer and Mathematical Sciences Vol. 2, Issue 6, 31 December, 2011 Pages (780-898)

2− b

(3.3)

852

C. Selvaraj, et al., J. Comp. & Math. Sci. Vol.2 (6), 846-854 (2011)

and 1− b Γ(2 − b + c) I 0,a, b,Ac f(A) ≥ A Γ(2 − b)Γ(a + 2 + c) (1 − α) Γ(2 − b + c) − A [2(β + 1) − (α + β)](µ + 1) Γ(2 − b)Γ(a + 2 + c)

I a,0,b,Ac f(A) ≥

(3.4) for a > 0, b, c ∈ R and all invertible operators A with (A ) ∗ (A ) = (A )(A ) ∗ (q ∈ N), ||A|| < 1 and ρ(A)ρ(A−1) ≤ 1, where ρ(A) is the spectral radius of A. 1 q

1 q

Γ(2 − b)Γ(a + 2 + c) b a, b, c A I 0, A f(A) Γ(2 − b + c)

=A−

∞

∑ n=2

=A−

Γ(n + 1 − b + c)Γ(n + 1)Γ(2 − b)Γ(a + 2 + c) anAn Γ(n + 1 − b)Γ(a + n + 1 + c)Γ(2 − b + c)

∞

∑b A , n

n=2

where Γ(n + 1 − b + c)Γ(n + 1)Γ(2 − b)Γ(a + 2 + c) bn = an. Γ(n + 1 − b)Γ(a + n + 1 + c)Γ(2 − b + c)

Put, for convenience, Φ(n) =

Γ(n + 1 − b + c)Γ(n + 1)Γ(2 − b)Γ(a + 2 + c) , n ∈ N − {1}. Γ(n + 1 − b)Γ(a + n + 1 + c)Γ(2 − b + c)

Then, it is easy to see Φ(n) is a nonincreasing function for integers n ≥ 2 and we have 0 < Φ(n) < 1. ∞

∞

∑b ≤ ∑ n

n =2

≤

∞

∑ n=2

[n(β + 1) − (α + β)] b n (µ)b n 1− α

[n(β + 1) − (α + β)] b n (µ)a n ≤ 1, 1− α

which gives ∞

(1 − α)

∑ [2(β + 1) − (α + β)](µ + 1) n =2

A). Therefore, we have

A −b

(3.5) Γ(2 − b + c) A A −b Γ(2 − b)Γ(a + 2 + c) (1 − α) − A [2(β + 1) − (α + β)] Γ(2 − b)Γ(a + 2 + c)

I a,0,b,Ac f(A) ≤

A −b .

(3.6) But ||Ab|| = ||A||b (b > 0)12 (3.7) since A*A = AA*, ||A|| = ρ(A). So 1 = ||AA−1|| ≤ ||A|| ||A−1|| = ρ(A)ρ(A−1) ≤ 1. (3.8) This implies ||A−1|| = ||A||−1. (3.9) Therefore ||A−b|| = ||A||−b for all real b. Using (3.7) and (3.9) in (3.5) and (3.6), we get the desired inequality. Corollary 3.4. When β = 0, (3.5) and (3.6) become I 0,a,b,Ac f(A) ≥

1− b Γ(2 − b + c) A Γ(2 − b)Γ(a + 2 + c) (1 − α) Γ(2 − b + c) − A (2 − α)(µ + 1) Γ(2 − b)Γ(a + 2 + c)

2− b

and

By Theorem 2.1 we have 2(β + 1) − (α + β) (µ + 1) n 1− α

(1 − α) A [2(β + 1) − (α + β)] Γ(2 − b)Γ(a + 2 + c)

and

1 q

Proof. Let F(A) =

−

2−b

Γ(2 − b + c) A A −b Γ(2 − b)Γ(a + 2 + c)

I 0,a,b,Ac f(A) ≤

Γ(2 − b + c) A Γ(2 − b)Γ(a + 2 + c) +

1− b

(1 − α) Γ(2 − b + c) A (2 − α)(µ + 1) Γ(2 − b)Γ(a + 2 + c)

2−b

the result obtained in18. and f(z) ∈ Tµ,β(α,

Theorem 3.5. Let 1 > max{b−c−1, b, −2−c+a}, c+1 < (1−b)(2−a+c), and b(2−a+c) ≤ 2(1−a). If f(z) ∈ Tµ,β(α, A), (µ ≥ 0), then

Journal of Computer and Mathematical Sciences Vol. 2, Issue 6, 31 December, 2011 Pages (780-898)

853

C. Selvaraj, et al., J. Comp. & Math. Sci. Vol.2 (6), 846-854 (2011)

Γ(2 − b + c) A Γ(1 − b)Γ(3 − a + c)

D a,0,b,Ac f(A) ≤

0 < Ψ(n) < 1. Now proceeding as in Theorem 3.2, we get the desired estimates.

−b

1− b 2(1 − α)Γ(2 − b + c) A , (µ + 1)Γ(1 − b)Γ(3 − a + c)

(3.10) and Γ(2 − b + c) A Γ(1 − b)Γ(3 − a + c)

D a,0,b,Ac f(A) ≥

−b

1− b 2(1 − α)Γ(2 − b + c) A , (µ + 1)Γ(1 − b)Γ(3 − a + c)

−

(3.11) for 0 < a < 1, b, c ∈ R and all invertible operators A with (A ) ∗ (A ) = (A )(A ) ∗ , (q ∈ N), ||A|| < 1 and ρ(A)ρ(A−1) ≤ 1, where ρ(A) is the spectral radius of A. 1 q

1 q

Proof. Consider the function G(A) =

Γ(1 − b)Γ(3 − a + c) b +1 a,b,c A D 0,A f(A) Γ(2 − b + c) =A−

∞

∑ n =2

=A−

Γ(n + 1 − b + c)Γ(n + 1)Γ(1 − b)Γ(3 − a + c) a nAn Γ(n − b)Γ(n + 2 − a + c)Γ(2 − b + c)

∞

∑c A , n

n =2

where cn =

Γ(n + 1 − b + c)Γ(n + 1)Γ(1 − b)Γ(3 − a + c) an , Γ(n − b)Γ(n + 2 − a + c)Γ(2 − b + c)

0 < cn < n. Put, for convenience, Ψ(n) =

Γ(n + 1 − b + c)Γ(n + 1)Γ(1 − b)Γ(3 − a + c) , n ∈ N − {1}, Γ(n − b)Γ(n + 2 − a + c)Γ(2 − b + c)

And Ψ(n) =

cn . n

By the constraints of the theorem, note that Ψ(n) is non-increasing for n ≥ 2 and

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Journal of Computer and Mathematical Sciences Vol. 2, Issue 6, 31 December, 2011 Pages (780-898)