����������� ������ 55 (2003), 83–96

UDK 517.546

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A CLASS OF UNIVALENT FUNCTIONS DEFINED BY USING HADAMARD PRODUCT

M. K. Aouf, H. M. Hossen and A. Y. Lashin

Abstract. In this paper we introduce the class L_{α}^{∗}(λ, β) of functions defined by f ∗ S_{α}(z) of f(z) and S_{α} = \frac{z}{(1 − z)^{2(1−α)}} . We determine coefficient estimates, closure theorems, distortion theorems and radii of close-to-convexity, starlikeness and convexity. Also we find integral operators and some results for Hadamard products of functions in the class L_{α}^{∗}(λ, β). Finally, in terms of the operators of fractional calculus, we derive several sharp results depicting the growth and distortion properties of functions belonging to the class L_{α}^{∗}(λ, β).

1. Introduction

Let A denote the class of functions of the form

f(z) = z + ∑_{n=2}^{∞} a_{n}z^{n}  (1.1)

which are analytic in the open unit disc U = { z : |z| < 1 }. And let S denote the subclass of A consisting of analytic and univalent functions f(z) in U.

A function f(z) from S is said to be starlike of order α if and only if

Re \{\frac{zf′(z)}{f(z)}\} > α (z ∈ U)

for some α, 0 ⩽ α < 1. We denote the class of all starlike functions of order α by S^{∗}(α). Further, a function f(z) from S is said to be convex of order α if and only if

Re \{1 + \frac{zf′′(z)}{f′(z)}\} > α (z ∈ U)

for some α, 0 ⩽ α < 1. And we denote the class of all convex functions of order α by K(α). We note that f(z) ∈ K(α) if and only if zf′(z) ∈ S^{∗}(α). The classes S^{∗}(α) and K(α) were first introduced by Robertson [7], and later were studied by Schild [9], MacGregor [2] and Pinchuk [6].

AMS Subject Classification: 30 C 45

Keywords and phrases: Univalent functions, Hadamard product, fractional calculus.

Now, the function

S_{α}(z) = \frac{z}{(1 − z)^{2(1−α)}} (0 ⩽ α < 1)

is the well-known extremal function for the class S^{∗}(α). Setting

C(α, n) = \frac{1}{(n − 1)!} ∏_{k=2}^{n}(k − 2α) (n ⩾ 2),

S_{α}(z) can be written in the form S_{α}(z) = z + ∑_{n=2}^{∞} C(α, n)z^{n}. Then we can see that C(α, n) is a decreasing function in α and satisfies

lim_{n→∞} C(α, n) = \{ \begin{matrix}∞, & α < 1/2, \\ 0, & α > 1/2, \\ 1, & α = 1/2. \end{matrix}

Let f ∗ g(z) denote the Hadamard product (convolution) of two functions f(z) and g(z), that is, if f(z) is given by (1.1) and g(z) is given by g(z) = z+∑_{n=2}^{∞} b_{n}z^{n}, then

f ∗ g(z) = z + ∑_{n=2}^{∞} a_{n}b_{n}z^{n}.

Let T denote the subclass of S consisting of functions of the form

f(z) = z − ∑_{n=2}^{∞} a_{n}z^{n}  (a_{n} ⩾ 0).  (1.2)

We say that a function f(z) defined by (1.1) belongs to the class L_{α}(λ, β) if f(z) satisfies the following condition

Re \{\frac{(f ∗ S_{α}(z))′}{λ(f ∗ S_{α}(z))′ + (1 − λ)}\} > β  (1.3)

for some α, 0 ⩽ α < 1, λ, 0 ⩽ λ < 1, β, 0 ⩽ β < 1 and for all z ∈ U.

Further we denote by L_{α}^{∗}(λ, β) the class obtained by taking intersection of the class L_{α}(λ, β) with T, that is L_{α}^{∗}(λ, β) = L_{α}(λ, β) ∩ T. We note that:

(i) L_{1/2}^{∗}(0, β) = T^{∗∗}(β) (Sarangi and Uralegaddi [8] and Al-Amiri [1]); (ii) L_{1/2}^{∗}(λ, β) represents the class of functions f(z) ∈ T satisfying the condition

Re \{\frac{f′(z)}{λf′(z) + (1 − λ)}\} > β,

where 0 ⩽ λ < 1 and 0 ⩽ β < 1; (iii) L_{α}(0, β) represents the class of functions f(z) ∈ T satisfying the condition

Re{(f ∗ S_{α}(z))′} > β.

2. Coefficient estimates

Theorem 1.Let the function f(z) be defined by (1.2). Then f(z) is in the class L_{α}^{∗}(λ, β) if and only if

∑_{n=2}^{∞} n(1 − λβ)C(α, n)a_{n} ⩽ 1 − β.  (2.1)

The result is sharp.

Proof. Assume that inequality (2.1) holds and let |z| < 1. Then we have

|\frac{(f ∗ S_{α}(z))′}{λ(f ∗ S_{α}(z))′ + (1 − λ)} − 1| = |\frac{−(1 − λ) ∑_{n=2}^{∞} nC(α, n)a_{n}z^{n−1}}{1 − λ ∑_{n=2}^{∞} nC(α, n)a_{n}z^{n−1}} |

< \frac{(1 − λ) ∑_{n=2}^{∞} nC(α, n)a_{n}}{1 − λ ∑_{n=2}^{∞} nC(α, n)a_{n}} ⩽ 1 − β.

This shows that the values of \frac{(f ∗ S_{α}(z))′}{λ(f ∗ S_{α}(z))′ + (1 − λ)} lie in the circle centered at

w = 1 whose radius is 1 − β. Hence f(z) satisfies condition (1.3).

Conversely, assume the function f(z) defined by (1.2) is in the class L_{α}^{∗}(λ, β). Then

Re \{\frac{(f ∗ S_{α}(z))′}{λ(f ∗ S_{α}(z))′ + (1 − λ)}\} = Re \{\frac{1 − ∑_{n=2}^{∞} nC(α, n)a_{n}z^{n−1}}{1 − λ ∑_{n=2}^{∞} nC(α, n)a_{n}z^{n−1}} \} > β (2.2)

for z ∈ U. Choose values of z on the real axis so that \frac{(f ∗ S_{α}(z))′}{λ(f ∗ S_{α}(z))′ + (1 − λ)}

is real. Upon clearing the denominator in (2.2) and letting z → 1^{−} through real values, we obtain

1 − ∑_{n=2}^{∞} nC(α, n)a_{n} ⩾ β{1 − λ ∑_{n=2}^{∞} nC(α, n)a_{n}}

which gives (2.1). Finally, the result is sharp with the extremal function f(z) given by

f(z) = z − \frac{1 − β}{n(1 − λβ)C(α, n)}z^{n}  (n ⩾ 2).  (2.3)

Corollary 1. Let the function f(z) defined by (1.2) be in the class L_{α}^{∗}(λ, β). Then we have

a_{n} ⩽ \frac{1 − β}{n(1 − λβ)C(α, n)}  (n ⩾ 2).  (2.4)

The equality in (2.4) is attained for the function f(z) given by (2.3).

3. Some properties of the class L_{α}^{∗}(λ, β)

Theorem 2. Let 0 ⩽ α < 1, 0 ⩽ λ_{1} ⩽ λ_{2} < 1 and 0 ⩽ β < 1. Then L_{α}^{∗}(λ_{1}, β) ⊂ L_{α}^{∗}(λ_{2}, β).

Proof. It follows from Theorem 1 that

∑_{n=2}^{∞} n(1 − λ_{2}β)C(α, n)a_{n} ⩽ ∑_{n=2}^{∞} n(1 − λ_{1}β)C(α, n)a_{n} ⩽ 1 − β

for f(z) ∈ L_{α}^{∗}(λ_{1}, β). Hence f(z) is in L_{α}^{∗}(λ_{2}, β).

Theorem 3. Let 0 ⩽ α_{1} ⩽ α_{2} < 1, 0 ⩽ λ < 1 and 0 ⩽ β < 1. Then we have L_{α_{1}}^{∗}(λ, β) ⊂ L_{α_{2}}^{∗}(λ, β).

Proof. Since C(α, n) is a decreasing function in α, it follows from Theorem 1 that

∑_{n=2}^{∞} n(1 − λβ)C(α_{2}, n)a_{n} ⩽ ∑_{n=2}^{∞} n(1 − λβ)C(α_{1}, n)a_{n} ⩽ 1 − β

for f(z) ∈ L_{α_{1}}^{∗}(λ, β). Hence f(z) is in L_{α_{2}}^{∗}(λ, β).

86 M.K. Aouf, H.M. Hossen, A.Y. Lashin

4. Closure theorems We shall prove the following results for the closure of functions in the class

L_{α}^{∗}(λ, β).

Theorem 4. Let the functions f_{j}(z), j = 1, 2,…, m, defined by

f_{j}(z) = z − ∑_{n=2}^{∞} a_{n,j}z^{n} (a_{n,j} ⩾ 0) (4.1)

for z ∈ U, be in the class L_{α}^{∗}(λ, β). Then the function h(z) defined by

h(z) = z − ∑_{n=2}^{∞} b_{n}z^{n}

also belongs to the class L_{α}^{∗}(λ, β), where b_{n} = \frac{1}{m} ∑_{j=1}^{m} a_{n,j}.

Proof. Since f_{j}(z) ∈ L_{α}^{∗}(λ, β), it follows from Theorem 1 that

∑_{n=2}^{∞} n(1 − λβ)C(α, n)a_{n,j} ⩽ 1 − β (j = 1, 2,…, m).

Therefore

∑_{n=2}^{∞} n(1 − λβ)C(α, n)b_{n} = ∑_{n=2}^{∞} n(1 − λβ)C(α, n)(\frac{1}{m} ∑_{j=1}^{m} a_{n,j})

= \frac{1}{m} ∑_{j=1}^{m}\{ ∑_{n=2}^{∞} n(1 − λβ)C(α, n)a_{n,j}\} ⩽ 1 − β.

Hence by Theorem 1, h(z) ∈ L_{α}^{∗}(λ, β). Thus we have the theorem.

Employing the techniques used earlier by Silverman [11], and with the aid of Theorem 1, we can prove the following

Theorem 5. The class L_{α}^{∗}(λ, β) is closed under convex linear combinations. As a consequence of Theorem 5, there exist extreme points of the class L_{α}^{∗}(λ, β). Theorem 6. Let f_{1}(z) = z and

f_{n}(z) = z − \frac{1 − β}{n(1 − λβ)C(α, n)} z^{n} (n ⩾ 2) (4.2)

for 0 ⩽ α < 1, 0 ⩽ λ < 1 and 0 ⩽ β < 1. Then f(z) is in the class L_{α}^{∗}(λ, β) if and only if it can be expressed in the form f(z) = ∑_{n=1}^{∞} μ_{n}f_{n}(z), where μ_{n} ⩾ 0 (n ⩾ 1) and ∑_{n=1}^{∞} μ_{n} = 1.

Corollary 2. The extreme points of the class L_{α}^{∗}(λ, β) are the functions f_{n}(z) (n ⩾ 1) given by Theorem 6.

5. Distortion theorems With the aid of Theorem 1, we may now find bounds of the modulus of f(z)

and f′(z) for f(z) ∈ L_{α}^{∗}(λ, β).

Theorem 7. If the function f(z) defined by (1.2) is in the class L_{α}^{∗}(λ, β), 0 ⩽ λ < 1, 0 ⩽ β < 1, and either 0 ⩽ α ⩽ 5/6 or |z| ⩽ 3/4, then

|f(z)| ⩾ max \{0, |z| − \frac{1 − β}{4(1 − λβ)(1 − α)} |z|^{2}\} ,

and |f(z)| ⩽ |z| + \frac{1 − β}{4(1 − λβ)(1 − α)} |z|^{2}. The bounds are sharp.

Proof. By virtue of Theorem 1, we note that

|f(z)| ⩾ max \{0, |z| − max_{n∈N\\{1\}} \frac{1 − β}{n(1 − λβ)C(α, n)} |z|^{n}\} ,
|f(z)| ⩽ |z| + max_{n∈N\\{1\}} \frac{1 − β}{n(1 − λβ)C(α, n)} |z|^{n}

for z ∈ U. Hence it suffices to deduce that

G(α, λ, β, |z|, n) = \frac{1 − β}{n(1 − λβ)C(α, n)} |z|^{n}

is a decreasing function of n (n ⩾ 2). Since C(α, n + 1) = \frac{n + 1 − 2α}{n}C(α, n), we can see that, for |z| ≠ 0, G(α, λ, β, |z|, n) ⩾ G(α, λ, β, |z|, n + 1) if and only if

H(α, |z|, n) = (n + 1)(n + 1 − 2α) − n^{2}|z| ⩾ 0.

It is easy to see that H(α, |z|, n) is a decreasing function of α for fixed |z|. Consequently it follows that

H(α, |z|, n) ⩾ H(5/6, |z|, n) = n^{2}(1 − |z|) + \frac{1}{3}(n − 2) ⩾ 0

for 0 ⩽ α ⩽ 5/6, z ∈ U and n ⩾ 2.

Further, since H(α, |z|, n) is decreasing in |z| and increasing in n, we obtain that H(α, |z|, n) > H(1, |z|, n) ⩾ H(1, 3/4, 2) = 0 for 0 ⩽ α ⩽ 1, |z| ⩽ 3/4 and n ⩾ 2. Thus max_{n∈N\{1}} G(α, λ, β, |z|, n) is attained at n = 2.

Finally, since the functions f_{n}(z) (n ⩾ 2) defined in Theorem 6 are extreme points of the class L_{α}^{∗}(λ, β), we can see that the bounds of Theorem 7 are attained by the function f_{2}(z), that is

f_{2}(z) = z − \frac{1 − β}{4(1 − λβ)(1 − α)} z^{2}.

Corollary 3. Let the function f(z) defined by (1.2) be in the class L_{α}^{∗}(λ, β), 0 ⩽ α ⩽ 5/6, 0 ⩽ λ < 1 and 0 ⩽ β < 1. Then f(z) is included in the disc with the

center at the origin and radius r given by r = 1 + \frac{1 − β}{4(1 − λβ)(1 − α)}.

Theorem 8. If the function f(z) defined by (1.2) is in the class L_{α}^{∗}(λ, β), 0 ⩽ λ < 1, 0 ⩽ β < 1, and either 0 ⩽ α ⩽ 1/2 or |z| ⩽ 1/2, then

1 − \frac{1 − β}{2(1 − λβ)(1 − α)} |z| ⩽ |f′(z)| ⩽ 1 + \frac{1 − β}{2(1 − λβ)(1 − α)} |z|.

The bounds are sharp.

Proof. It is similar to the proof of Theorem 7.

6. Radii of close-to-convexity, starlikeness and convexity

Theorem 9. L_{α}^{∗}(λ, β) is a subclass of S if and only if 0 ⩽ α ⩽ 1/2.

Proof. Naote that the function f(z) defined by (1.2) is in the class S if ∑_{n=2}^{∞} n|a_{n}| ⩽ 1 (cf. [11]). Hence it suffices to prove that (1 − λβ)C(α, n) ⩾ 1 − β for 0 ⩽ α ⩽ 1/2, 0 ⩽ λ < 1, 0 ⩽ β < 1 and n ⩾ 2 by means of Theorem 1. Since C(α, n) ⩾ C(1/2, n) = 1 for 0 ⩽ α ⩽ 1/2, we can see that, for 0 ⩽ α ⩽ 1/2, 0 ⩽ λ < 1 and 0 ⩽ β < 1,

(1 − λβ)C(α, n) − (1 − β) ⩾ (1 − λβ) − (1 − β) ⩾ 0.

Conversely, if we assume α > 1/2, then lim_{n→∞} C(α, n) = 0. Taking the function f_{n}(z) given by (4.2), we have

f′_{n}(z) = 1 − \frac{1 − β}{(1 − λβ)C(α, n)} z^{n−1} = 0

for z^{n−1} = \frac{(1 − λβ)C(α, n)}{1 − β} which is less than one for n sufficiently large. Thus

f_{n}(z) is not univalent for α > 1/2 and n = n(α) sufficiently large.

By using Theorem 1, we can prove the following

Theorem 10. Let the function f(z) defined by (1.2) be in the class L_{α}^{∗}(λ, β), 0 ⩽ α ⩽ 1/2, 0 ⩽ λ < 1 and 0 ⩽ β < 1. Then f(z) is close-to-convex of order ρ (0 ⩽ ρ < 1) in |z| ⩽ R_{1}, where

R_{1} = inf_{n} \{\frac{(1 − ρ)(1 − λβ)C(α, n)}{1 − β}\}^{1/(n−1)} (n ⩾ 2).

The result is sharp, with extremal function f(z) given by (2.3).

Theorem 11. Let the function f(z) defined by (1.2) be in the class L_{α}^{∗}(λ, β), 0 ⩽ α ⩽ 1/2, 0 ⩽ λ < 1 and 0 ⩽ β < 1. Then f(z) is starlike of order ρ (0 ⩽ ρ < 1) in |z| ⩽ R_{2}, where

R_{2} = inf_{n} \{\frac{n(1 − ρ)(1 − λβ)C(α, n)}{(n − ρ)(1 − β)}\}^{1/(n−1)} (n ⩾ 2).

The result is sharp, with extremal function f(z) given by (2.3).

Corollary 4. Let the function f(z) defined by (1.2) be in the class L_{α}^{∗}(λ, β), 0 ⩽ α ⩽ 1/2, 0 ⩽ λ < 1 and 0 ⩽ β < 1. Then f(z) is convex of order ρ (0 ⩽ ρ < 1) in |z| ⩽ R_{3}, where

R_{3} = inf_{n} \{\frac{(1 − ρ)(1 − λβ)C(α, n)}{(n − ρ)(1 − β)}\}^{1/(n−1)} (n ⩾ 2).

The result is sharp, with extremal function f(z) given by (2.3).

7. Integral operators

Theorem 12. Let the function f(z) defined by (1.2) be in the class L_{α}^{∗}(λ, β), and let d be a real number such that d > −1. Then the function F (z) defined by

F (z) = \frac{d + 1}{z^{d}} ∫_{0}^{z} t^{d−1}f(t) dt  (7.1)

also belongs to the class L_{α}^{∗}(λ, β).

Proof. From the representation of F (z), it follows that F (z) = z − ∑_{n=2}^{∞} b_{n}z^{n}, where b_{n} = (\frac{d+1}{d+n})a_{n}. Therefore

∑_{n=2}^{∞} n(1 − λβ)C(α, n)b_{n} = ∑_{n=2}^{∞} n(1 − λβ)C(α, n) (\frac{d + 1}{d + n}) a_{n}

⩽ ∑_{n=2}^{∞} n(1 − λβ)C(α, n)a_{n} ⩽ 1 − β,

since f(z) ∈ L_{α}^{∗}(λ, β). Hence by Theorem 1, F (z) ∈ L_{α}^{∗}(λ, β).

Theorem 13. Let the function F (z) = z −∑_{n=2}^{∞} a_{n}z^{n} (a_{n} ⩾ 0) be in the class L_{α}^{∗}(λ, β), and let d be a real number such that d > −1. Then the function f(z) defined by (7.1) is univalent in |z| < R^{∗}, where

R^{∗} = inf_{n} \{\frac{(1 − λβ)C(α, n)(d + 1)}{(1 − β)(d + n)}\}^{1/(n−1)} (n ⩾ 2.)

The result is sharp.

Proof. From (7.1) we have

f(z) = \frac{z^{1−d}(z^{d}F (z))′}{d + 1} = z − ∑_{n=2}^{∞} (\frac{d + n}{d + 1}) a_{n}z^{n}.

In order to obtain the required result it suffices to show that |f′(z) − 1| < 1 in |z| < R^{∗}. Now

|f′(z) − 1| = |− ∑_{n=2}^{∞} n (\frac{d + n}{d + 1}) a_{n}z^{n−1}| ⩽ ∑_{n=2}^{∞} n (\frac{d + n}{d + 1}) a_{n}|z|^{n−1}.

Thus |f′(z) − 1| < 1 if

∑_{n=2}^{∞} n (\frac{d + n}{d + 1}) a_{n}|z|^{n−1} ⩽ 1.  (7.2)

But Theorem 1 confirms that ∑_{n=2}^{∞} \frac{n(1 − λβ)C(α, n)}{1 − β}a_{n} ⩽ 1. Hence (7.2) will be satisfied if

\frac{n(d + n)}{d + 1}|z|^{n−1} ⩽ \frac{n(1 − λβ)C(α, n)}{1 − β} (n ⩾ 2)

or if

|z| ⩽ \{\frac{(1 − λβ)C(α, n)(d + 1)}{(1 − β)(d + n)}\}^{1/(n−1)}  (n ⩾ 2).  (7.3)

The required result follows now from (7.3). The result is sharp for the function

f(z) = z − \frac{(1 − β)(d + n)}{n(1 − λβ)C(α, n)(d + 1)} z^{n} (n ⩾ 2).

8. Modified Hadamard products

Let the functions f_{j}(z) (j = 1, 2) be defined by (4.1). The modified Hadamard product of f_{1}(z) and f_{2}(z) is defined by

f_{1} ∗ f_{2}(z) = z − ∑_{n=2}^{∞} a_{n,1}a_{n,2}z^{n}.

Theorem 14. Let the functions f_{j}(z) (j = 1, 2) defined by (4.1) be in the class L_{α}^{∗}(λ, β) with 0 ⩽ α ⩽ 1/2, 0 ⩽ λ < 1 and 0 ⩽ β < 1. Then f_{1} ∗ f_{2}(z) ∈ L_{α}^{∗}(λ, γ(α, λ, β)) where

γ(α, λ, β) = 1 − \frac{(1 − λ)(1 − β)^{2}}{4(1 − λβ)^{2}(1 − α) − λ(1 − β)^{2}} .

The result is sharp.

Proof. Employing the technique used earlier by Schild and Silverman [10], we need to find the largest γ(α, λ, β) such that

∑_{n=2}^{∞} \frac{n(1 − λγ)C(α, n)}{1 − γ}a_{n,1}a_{n,2} ⩽ 1.

Since ∑_{n=2}^{∞} \frac{n(1 − λβ)C(α, n)}{1 − β}a_{n,1} ⩽ 1 and ∑_{n=2}^{∞} \frac{n(1 − λβ)C(α, n)}{1 − β}a_{n,2} ⩽ 1, by the

Cauchy- Schwarz inequality we have

∑_{n=2}^{∞} \frac{n(1 − λβ)C(α, n)}{1 − β}\sqrt{a_{n,1}a_{n,2}} ⩽ 1.

Thus it is sufficient to show that

\frac{n(1 − λγ)C(α, n)}{1 − γ}a_{n,1}a_{n,2} ⩽ \frac{n(1 − λβ)C(α, n)}{1 − β}\sqrt{a_{n,1}a_{n,2}} (n ⩾ 2),

that is that \sqrt{a_{n,1}a_{n,2}} ⩽ \frac{(1 − λβ)(1 − γ)}{(1 − λγ)(1 − β)}. Note that \sqrt{a_{n,1}a_{n,2}} ⩽ \frac{1 − β}{n(1 − λβ)C(α, n)}

(n ⩾ 2). Consequently, we need only to prove that

\frac{1 − β}{n(1 − λβ)C(α, n)} ⩽ \frac{(1 − λβ)(1 − γ)}{(1 − λγ)(1 − β)} (n ⩾ 2),

or, equivalently, that γ ⩽ 1 − \frac{(1 − λ)(1 − β)^{2}}{n(1 − λβ)^{2}C(α, n) − λ(1 − β)^{2}} (n ⩾ 2). Since

A(n) = 1 − \frac{(1 − λ)(1 − β)^{2}}{n(1 − λβ)^{2}C(α, n) − λ(1 − β)^{2}}  (8.1)

is an increasing function of n (n ⩾ 2), for 0 ⩽ α ⩽ 1/2, 0 ⩽ λ < 1 and 0 ⩽ β < 1, letting n = 2 in (8.1), we obtain

γ ⩽ A(2) = 1 − \frac{(1 − λ)(1 − β)^{2}}{4(1 − λβ)^{2}C(α, n) − λ(1 − β)^{2}} ,

which completes the proof of Theorem 14.

Finally, by taking the functions f_{j}(z) given by

f_{j}(z) = z − \frac{1 − β}{4(1 − λβ)(1 − α)} z^{2}  (j = 1, 2),  (8.2)

we can see that the result is sharp.

Theorem 15. Let the function f_{1}(z) defined by (4.1) be in the class L_{α}^{∗}(λ, β) with 0 ⩽ α ⩽ 1/2, 0 ⩽ λ < 1 and 0 ⩽ β < 1, and the function f_{2}(z) defined by (4.1) be in the class L_{α}^{∗}(λ, τ) with 0 ⩽ α ⩽ 1/2, 0 ⩽ λ < 1 and 0 ⩽ τ < 1. Then f_{1} ∗ f_{2}(z) ∈ L_{α}^{∗}(λ, ζ(α, λ, β, τ)), where

ζ(α, λ, β, τ) = 1 − \frac{(1 − λ)(1 − β)(1 − τ)}{4(1 − λβ)(1 − λτ)(1 − α) − λ(1 − β)(1 − τ)}.

The result is sharp.

Proof. Proceeding as in the proof of Theorem 14, we get

ζ ⩽ B(n) = 1 − \frac{(1 − λ)(1 − β)(1 − τ)}{n(1 − λβ)(1 − λτ)C(α, n) − λ(1 − β)(1 − τ)} (n ⩾ 2). (8.3)

Since the function B(n) is an increasing function of n (n ⩾ 2), for 0 ⩽ α ⩽ 1/2, 0 ⩽ λ < 1 and 0 ⩽ τ < 1, letting n = 2 in (8.3), we obtain

ζ ⩽ B(2) = 1 − \frac{(1 − λ)(1 − β)(1 − τ)}{4(1 − λβ)(1 − λτ)(1 − α) − λ(1 − β)(1 − τ)},

which evidently proves Theorem 15.

Finally, the result is best possible for the functions

f_{1}(z) = z − \frac{1 − β}{4(1 − λβ)(1 − α)} z^{2} and f_{2}(z) = z − \frac{1 − τ}{4(1 − λτ)(1 − α)} z^{2}.

Corollary 4. Let the functions f_{j}(z) (j = 1, 2, 3) defined by (4.1) be in the class L_{α}^{∗}(λ, β) with 0 ⩽ α ⩽ 1/2, 0 ⩽ λ < 1 and 0 ⩽ β < 1. Then f_{1} ∗ f_{2} ∗ f_{3}(z) ∈ L_{α}^{∗}(λ, η(α, λ, β)), where

η(α, λ, β) = 1 − \frac{(1 − λ)(1 − β)^{3}}{16(1 − λβ)^{3}(1 − α)^{2} − λ(1 − β)^{3}} .

The result is best possible for the functions f_{j}(z) = z − \frac{1 − β}{4(1 − λβ)(1 − α)} z^{2} (j = 1, 2, 3).

Proof. From Theorem 14, we have f_{1} ∗ f_{2}(z) ∈ L_{α}^{∗}(λ, γ(α, λ, β)). We use now Theorem 15, and we get f_{1} ∗ f_{2} ∗ f_{3}(z) ∈ L_{α}^{∗}(λ, η(α, λ, β, γ)), where

η(α, λ, β, γ) = 1 − \frac{(1 − λ)(1 − β)(1 − γ)}{4(1 − λβ)(1 − λγ)(1 − α) − λ(1 − β)(1 − γ)}

= 1 − \frac{(1 − λ)(1 − β)^{3}}{16(1 − λβ)^{3}(1 − α)^{2} − λ(1 − β)^{3}} .

This completes the proof of Corollary 4.

Theorem 16. Let the functions f_{j}(z) (j = 1, 2) defined by (4.1) be in the class L_{α}^{∗}(λ, β) with 0 ⩽ α ⩽ 1/2, 0 ⩽ λ < 1 and 0 ⩽ β < 1. Then the function

h(z) = z − ∑_{n=2}^{∞}(a_{n,1}^{2} + a_{n,2}^{2})z^{n}

belongs to the class L_{α}^{∗}(λ, ϕ(α, λ, β)), where

ϕ(α, λ, β) = 1 − \frac{(1 − λ)(1 − β)^{2}}{2(1 − λβ)^{2}(1 − α) − λ(1 − β)^{2}} .

The result is sharp for the functions f_{j}(z) (j = 1, 2) defined by (8.2).

Proof. By virtue of Theorem 1, we obtain

∑_{n=2}^{∞} [\frac{n(1 − λβ)C(α, n)}{1 − β}]^{2} a_{n,1}^{2} ⩽ [∑_{n=2}^{∞} \frac{n(1 − λβ)C(α, n)}{1 − β}a_{n,1}]^{2} ⩽ 1  (8.4)

and

∑_{n=2}^{∞} [\frac{n(1 − λβ)C(α, n)}{1 − β}]^{2} a_{n,2}^{2} ⩽ [∑_{n=2}^{∞} \frac{n(1 − λβ)C(α, n)}{1 − β}a_{n,2}]^{2} ⩽ 1.  (8.5)

It follows from (8.4) and (8.5) that

∑_{n=2}^{∞} \frac{1}{2} [\frac{n(1 − λβ)C(α, n)}{1 − β}]^{2} [a_{n,1}^{2} + a_{n,2}^{2}] ⩽ 1.

Therefore, we need to find the largest ϕ = ϕ(α, λ, β) such that

\frac{n(1 − λϕ)C(α, n)}{1 − ϕ} ⩽ \frac{1}{2} [\frac{n(1 − λβ)C(α, n)}{1 − β}]^{2} (n ⩾ 2),

that is ϕ ⩽ 1 − \frac{2(1 − λ)(1 − β)^{2}}{n(1 − λβ)^{2}C(α, n) − 2λ(1 − β)^{2}} (n ⩾ 2). Since

D(n) = 1 − \frac{2(1 − λ)(1 − β)^{2}}{n(1 − λβ)^{2}C(α, n) − 2λ(1 − β)^{2}}

is an increasing function of n (n ⩾ 2), for 0 ⩽ α ⩽ 1/2, 0 ⩽ λ < 1 and 0 ⩽ β < 1, we readily have

ϕ ⩽ D(2) = 1 − \frac{(1 − λ)(1 − β)^{2}}{2(1 − λβ)^{2}(1 − α) − λ(1 − β)^{2}} ,

and Theorem 15 follows at once.

9. Fractional calculus operators

The object of this section is to obtain several growth and distortion properties of functions in the class L_{α}^{∗}(λ, β) involving a family of operators of fractional calculus (that is, fractional integral and fractional derivative).

First of all, in terms of Gauss hypergeometric function

2F_{1}(δ, τ; γ; z) = ∑_{k=0}^{∞} \frac{(δ)_{k}(τ)_{k}}{(γ)_{k}} \frac{z^{k}}{k!} (z ∈ U; δ, τ, γ ∈ C; γ ≠ 0, −1, −2,…),

where (m)_{k} = \frac{Γ(m + k)}{Γ(m)} denotes the Pochhammer symbol, we recall the definitions

of fractional integral operator I_{0,z}^{μ,ν,η} and the fractional derivative operator J_{0,z}^{μ,ν,η} as follows (cf., e.g., [4] and [14], see also [13]).

Definition 1. The fractional integral of order μ is defined, for a function f(z), by

I_{0,z}^{μ,ν,η}f(z) = \frac{z^{−μ−ν}}{Γ(μ)} ∫_{0}^{z}(z − ζ)^{μ−1}_{2}F_{1} (μ + ν, −η; μ; 1 − \frac{ζ}{z} ) f(ζ) dζ (μ > 0),

where f(z) is an analytic function in a simply-connected region of the z-plane containing the origin, and the multiplicity of (z − ζ)^{μ−1} is removed by requiring log(z − ζ) to be real when z − ζ > 0, provided further that

f(z) = O(|z|^{ε})  (z → 0; ε > max{0, ν − η} − 1).  (9.1)

Definition 2. The fractional derivative of order μ is defined, for a function f(z), by

J_{0,z}^{μ,ν,η}f(z) =

= \{ \begin{matrix}\frac{1}{Γ(1 − μ)} \frac{d}{dz} \{z^{μ−ν} ∫_{0}^{z}(z − ζ)^{−μ}_{2}F_{1}(ν − μ, 1 − η; 1 − μ; 1 − (ζ/z))f(ζ) dζ\} \\ (0 ⩽ μ < 1) \\ \frac{d^{n}}{dz^{n}} J_{0,z}^{μ−n,ν,η}f(z) (n ⩽ μ < n + 1; n ∈ N), \end{matrix}

where f(z) is constrained, and the multiplicity of (z − ζ)^{−μ} is removed, as in Defintion 1, and ε is given by the order estimate (9.1).

It follows from Definitions 1 and 2 that

I_{0,z}^{μ,−μ,η}f(z) = D_{z}^{−μ}f(z)  (μ > 0)  (9.2)

and

J_{0,z}^{μ,μ,η}f(z) = D_{z}^{μ}f(z)  (0 ⩽ μ < 1),  (9.3)

where D_{z}^{μ}f(z) (μ ∈ R) is the fractional calculus operator considered by Owa [3] and subsequently by Owa and Srivastava [5] and in many other works (cf., e.g., [12] and [13]). Furthermore, in terms of Gamma functions Definitions 1 and 2 readily yield

Lemma 1. (cf. Srivastava et al. [14]) The (generalized) fractional integral and the (generalized) fractional derivative of a power function are given by

I_{0,z}^{μ,ν,η}z^{ρ} = \frac{Γ(ρ + 1)Γ(ρ − ν + η + 1)}{Γ(ρ − ν + 1)Γ(ρ + μ + η + 1)} z^{ρ−ν} (μ > 0; ρ > max\{0, ν − η\} − 1)

(9.4) and

J_{0,z}^{μ,ν,η}z^{ρ} = \frac{Γ(ρ + 1)Γ(ρ − ν + η + 1)}{Γ(ρ − ν + 1)Γ(ρ − μ + η + 1)} z^{ρ−ν} (0 ⩽ μ < 1; ρ > max\{0, ν −η\}−1).

(9.5)

Theorem 17. Let the function f(z) defined by (1.2) be in the class L_{α}^{∗}(λ, β), with 0 ⩽ α ⩽ 1/2, 0 ⩽ λ < 1 and 0 ⩽ β < 1. Then

\frac{Γ(2 − ν + η)}{Γ(2 − ν)Γ(2 + μ + η)}|z|^{1−ν} \{1 − \frac{(1 − β)(2 − ν + η)}{2(2 − ν)(2 + μ + η)(1 − λβ)(1 − α)}|z|\}

⩽ |I_{0,z}^{μ,ν,η}f(z)| ⩽

⩽ \frac{Γ(2 − ν + η)}{Γ(2 − ν)Γ(2 + μ + η)}|z|^{1−ν} \{1 + \frac{(1 − β)(2 − ν + η)}{2(2 − ν)(2 + μ + η)(1 − λβ)(1 − α)}|z|\} (9.6)

(z ∈ U_{0}; μ > 0, max{ν, ν − η, −μ − η} < 2; ν(μ + η) ⩽ 3μ), and

\frac{Γ(2 − ν + η)}{Γ(2 − ν)Γ(2 − μ + η)}|z|^{1−ν} \{1 − \frac{(1 − β)(2 − ν + η)}{2(2 − ν)(2 − μ + η)(1 − λβ)(1 − α)}|z|\} ⩽

⩽ |J_{0,z}^{μ,ν,η}f(z)|

⩽ \frac{Γ(2 − ν + η)}{Γ(2 − ν)Γ(2 − μ + η)}|z|^{1−ν} \{1 + \frac{(1 − β)(2 − ν + η)}{2(2 − ν)(2 − μ + η)(1 − λβ)(1 − α)}|z|\} (9.7)

(z ∈ U_{0}; 0 ⩽ μ < 1, max{ν, ν − η, μ − η} < 2; ν(μ − η) ⩾ 3μ), where

U_{0} = \{ \begin{matrix}U, & (ν ⩽ 1), \\ U \ \{0\}, & (ν > 1). \end{matrix} Each of these results is sharp for the function f(z)

given by

f(z) = z − \frac{1 − β}{4(1 − λβ)(1 − α)} z^{2}.  (9.8)

Proof. First of all, since the function f(z) defined by (1.2) is in the class L_{α}^{∗}(λ, β), 0 ⩽ α ⩽ 1/2, 0 ⩽ λ < 1 and 0 ⩽ β < 1, we can apply Theorem 1 to deduce that

∑_{n=2}^{∞} a_{n} ⩽ \frac{1 − β}{4(1 − λβ)(1 − α)}.  (9.9)

Next, making use of the assertion 9.4 of Lemma 1, we find from (1.2) that

F (z) = \frac{Γ(2 − ν)Γ(2 + μ + η)}{Γ(2 − ν + η)} z^{ν}I_{0,z}^{μ,ν,η}f(z) = z − ∑_{n=2}^{∞} Φ(n)a_{n}z^{n},  (9.10)

where, for convenience,

Φ(n) = \frac{(1)_{n}(2 − ν + η)_{n−1}}{(2 − ν)_{n−1}(2 + μ + η)_{n−1}} (n ∈ N \ \{1\}).  (9.11)

The function Φ(n) defined by (9.11) can easily be seen to be nonincreasing under the parametric constraints stated already after (9.6), and thus we have

0 < Φ(n) ⩽ Φ(2) = \frac{2(2 − ν + η)}{(2 − ν)(2 + μ + η)} (n ∈ N \ \{1\}).  (9.12)

Now the assertion (9.6) of the theorem follows readily from (9.9), (9.10) and (9.12).

The assertion (9.7) of the theorem can be proven similarly by noting from (9.5) that

G(z) = \frac{Γ(2 − ν)Γ(2 − μ + η)}{Γ(2 − ν + η)} z^{ν}J_{0,z}^{μ,ν,η}f(z) = z − ∑_{n=2}^{∞} Ψ(n)a_{n}z^{n},

where

0 < Ψ(n) = \frac{(1)_{n}(2 − ν + η)_{n−1}}{(2 − ν)_{n−1}(2 − μ + η)_{n−1}} ⩽ Ψ(2) = \frac{2(2 − ν + η)}{(2 − ν)(2 − μ + η)},

(n ∈ N \ {1}) under the parametric constraints stated already after (9.7).

Finally, by observing that the equalities in each of the assertions (9.6) and (9.7) are attained by the function f(z) given by (9.8), we complete the proof of the theorem.

In view of the relationships (9.2) and (9.3), by setting ν = −μ and ν = μ in our assertions (9.6) and (9.7), respectively, we obtain

Corollary 5. Let the function f(z) defined by (1.2) be in the class L_{α}^{∗}(λ, β), 0 ⩽ α ⩽ 1/2, 0 ⩽ λ < 1 and 0 ⩽ β < 1. Then

\frac{|z|^{1+μ}}{Γ(2 + μ)} \{1 − \frac{1 − β}{2(2 + μ)(1 − λβ)(1 − α)} |z|\} ⩽ |D_{z}^{−μ}f(z)| ⩽

⩽ \frac{|z|^{1+μ}}{Γ(2 + μ)} \{1 + \frac{1 − β}{2(2 + μ)(1 − λβ)(1 − α)} |z|\} (z ∈ U; μ > 0) (9.13)

and

\frac{|z|^{1−μ}}{Γ(2 − μ)} \{1 − \frac{1 − β}{2(2 − μ)(1 − λβ)(1 − α)} |z|\} ⩽ |D_{z}^{μ}f(z)| ⩽

⩽ \frac{|z|^{1−μ}}{Γ(2 − μ)} \{1 + \frac{1 − β}{2(2 − μ)(1 − λβ)(1 − α)} |z|\} (z ∈ U; 0 ⩽ μ < 1). (9.14)

Each of these results is sharp for the function f(z) given by (9.8).

The assertions (9.13) and (9.14) of Corollary 5 can indeed be applied further in order to deduce the following interesting results for functions in the class L_{α}^{∗}(λ, β).

Corollary 6. Under the hypotheses of Corollary 5, D_{z}^{−μ}f(z) (μ > 0) is included in the disc with its center at the origin and radius r_{1} given by

r_{1} = \frac{1}{Γ(2 + μ)} \{1 + \frac{1 − β}{2(2 + μ)(1 − λβ)(1 − α)}\} .

Corollary 7. Under the hypotheses of Corollary 5, D_{z}^{μ}f(z) (0 ⩽ μ < 1) is included in the disc with its center at the origin and radius r_{2} given by

r_{2} = \frac{1}{Γ(2 − μ)} \{1 + \frac{1 − β}{2(2 − μ)(1 − λβ)(1 − α)}\} .

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(received 03.08.2002)

Department of Mathematics, Faculty of Science, Mansoura University, Mansoura, Egypt E-mail: sinfac@mum.mans.eun.eg