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Some applications of the Jung-Kim-Srivastava integral operator.

1. Introduction

Let A(p, k) denote the class of functions of the form

f(z) = [z.sup.p] + [[infinity].summation over (m = k)] [a.sub.[p + m]] [z.sup.[p + m]] (p, k [member of] N = {1,2,3,...}) (1.1)

which are analytic in the unit disk E = {z:| z |< 1}. The Hadamard product or convolution ([f.sub.1] * [f.sub.2] (z) of two functions

[f.sub.j] (z) = [z.sup.p] + [[infinity].summation over (m = k)] [a.sub.[p + m,j]] [z.sup.[p + m]] [member of] A(p,k) (j = 1,2) (1.2)

is given by

([f.sub.1] * [f.sub.2]) (z) = [z.sup.p] + [[infinity].summation over (m = k)] [a.sub.[p + m,1]] [a.sub.[p + m,2]] [z.sup.[p + m]].

Let f (z) and g(z) be analytic in E. Then we say that the function g(z) is subordinate to f (z) if there exists an analytic function w(z) in E such that | w(z) |< 1 (z [member of] E) and g(z) = f (w(z)). For this relation the symbol g(z) [??] f (z) is used. In case f (z) is univalent in E we have that the subordination g(z) [??] f (z) is equivalent to g(0) = f (0) and g(E) [subset] f (E).

Recently, Jung, Kim, and Srivastava [2] introduced the following one-parameter family of integral operator

[I.sup.[sigma]] f(z) = [[2.sup.[sigma]]/[z [GAMMA]([sigma])]] [[[integral].sub.0.sup.z] (log z/t).sup.[[sigma] - 1]] f(t) dt (1.3)

for f (z) [member of] A(1,1) and [sigma] > 0. They showed that

[I.sup.[sigma]] f(z) = z + [[[[infinity].summation over (m = 1)](2/[m + 2])].sup.[sigma]] [a.sub.[m + 1]][z.sup.[m + 1]]. (1.4)

The operator [I.sup.[sigma]] is closely related to the multiplier transformations studied earlier by Flett (1). It follows from (1.4) that one can define the operator [I.sup.[sigma]] for any real number [sigma]. Certain properties of this operator have been investigated by Jung, Kim and Srivastava (2), Uralegaddi and Somanatha (6), Li (3) and the author (4).

Motivated essentially by some recent works (2), (3), (4), we now extend the operator [I.sup.[sigma]] to multivalent functions, which is given by the following

[I.sup.[sigma]] f(z) = [z.sup.p] + [[infinity].summation over (m = k)] [([p + 1]/[m + p + 1]).sup.[sigma]] [a.sub.[p + m]] [z.sup.[p + m]] (1.5)

for f (z) [member of] A(p, k) and any real number [sigma]. It is easily verified from the definition (1.5) that

z([I.sup.[sigma] + 1] f (z))' = (p +1) [I.sup.[sigma]] f (z) - [I.sup.[sigma] +1] f (z). (1.6)

In this note, we shall derive some interesting properties of the operator [I.sup.[sigma]].

2. Main Results

We begin by recalling the following result due to Miller and Mocanu [5], which we shall apply in proving our first theorem.

Lemma. Let h(z) be analytic and convex univalent in E, h(0) = 1, and let

g(z) =1 + [b.sub.k][z.sup.k] + [b.sub.k+1][z.sup.k+1] + be analytic in E. If g(z) + zg'(z) / c[??] h(z), (2.1)

then for c[not equal to] 0 and Rec[greater than or equal to] 0

g(z) [??] [c/k][z.sup.[ - c/k]][[integral].sub.0.sup.z] [t.sup.[c/k - 1]]h(t) dt. (2.2)

Theorem 1. Let -1 [greater than or equal to] B < A [greater than or equal to] 1 and 0 < [delta] < 1. Let f(z) = [z.sup.p] + [[infinity].summation over (m = k)] [a.sub.p+m][z.sup.p+m]

[member of] A(p, k). Suppose that

[[infinity].summation over (m = k)] [c.sub.m] | [a.sub.p+m] | [less than or equal to] 1, (2.3)

where

[c.sub.m] = [[1 - B]/[A - B]] * [(p + 1 + m(1 - d))/[(m + p + 1).sup.[[sigma] + 1]]]. (2.4)

(i) If -1 [less than or equal to] B [less than or equal to] 0, then

(1 - [delta]) [[[I.sup.[sigma]]f(z)]/[z.sup.p]] + [delta][[[I.sup.[[sigma] + 1]]f(z)]/[z.sup.p]] [??] [[1 + Az]/[1 + Bz]]. (2.5)

(ii) If -1 [less than or equal to] B [less than or equal to] 0, and [lambda] [greater than or equal to] 1 then for z [member of] E

Re {[([[I.sup.[[sigma] + 1]]f(z)]/[z.sup.p]).sup.[1/[lambda]]]} [omaga][{[[p + 1]/[k(1 - [delta])]] [[integral].sub.0.sup.1][u.sup.[(p + 1)/k(1 - [delta]) - 1]] ([1 - Au]/[1 - Bu])du}.sup.[1/[lambda]]]. (2.6)

The result is sharp.

Proof. (i) Let

J = (1 - [delta]) [[[I.sup.[sigma]] f(z)]/[z.sup.p]] + [delta][[[I.sup.[[sigma] + 1]]f(z)]/[z.sup.p]], (2.7)

then

J=1 + [[infinity].summation over (m = k)][[[[(p + 1)].sup.[sigma]](p + 1 + m(1 - [delta]))]/[[(m + p + 1)].sup.[[sigma] + 1]]][a.sub.[p + m]][z.sup.m]] (2.8)

For -1 [less than or equal to] B [less than or equal to] 0 and z [member of] E, it follows from (2.3) that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

which show that

(1-[delta])[[[I.sup.[sigma]]f(z)]/[z.sup.p]] + [delta][[[I.sup.[sigma]]f(z)]/[z.sup.p]][??][[1 + Az]/[1 + Bz]].

(ii) Put

g(z) = [I.sup.[sigma] + 1] f (z) / [z.sup.p]. (2.9)

Then the function g(z) = 1 + [b.sub.k][z.sup.k] + [b.sub.k + 1][z.sup.k + 1] + is analytic in E. Using (1.6) and

(2.9) we obtain

[I.sup.[sigma]] f (z)/[z.sup.p] = g (z) + 1/p + 1 zg' (z). (2.10)

Thus

(1-[delta]) [I.sup.[sigma]] f (z)/[z.sup.p] + [delta] [I.sup.[sigma] + 1] f (z)/[z.sup.p] = g(z)+ 1-[delta]/p + 1 zg' (z) [??] 1 + Az/1 + Bz.

Now an application of the lemma leads to

g(z)[??]p+1/k(1-[delta] [z.sup.-(p+1)/k(1-[delta])[[infinity].sub.0.sup.z][t.sup.(p+1)/k(1-[delta])-1](1+Az/1+Bz)dt

or

[I.sup.[[sigma]+1]f(z)/[z.sup.p] = p+1/k(1-delta]) [[infinity].sub.0.sup.1][u(p+1)/k(1-[delta])-1](1+Auw(z)/1+Buw(z))du, (2.11)

where w(z) is analytic in E with w(0) = 0 and | w(z) |< 1 (z [member of] E).

In view of -1 [less than or equal to] B < A [less than or equal to] 1, it follows from (2.11) that

Re [gamma] {[[[I.sup.[[sigma] + 1]]f(z)]/[z.sup.p]]} >[[p + 1]/[k(1 - [delta])]] [[integral].sub.0.sup.1] [u.sup.[(p + 1)/k(1 - [delta]) - 1]] ([1 - Au]/[1 - Bu]) du>0(z[member of]E).

Therefore, with the aid of the elementary inequality Re([w.sup.1/[[lambda]) [greater than or equal to] [(Rew).sup.1/ [[lambda] for

Rew < 0 and [lambda] [greater than or equal to] 1, the inequality (2.6) follows immediately.

To show the sharpness of (2.6), we take f (z) [member of] A(p, k) defined by

[[[I.sup.[[sigma] + 1]]f(z)]/[z.sup.p]] = [[p + 1]/[k(1 - [delta])]] [[integral].sub.0.sup.1] [u.sup.[(p + 1)/k (1 - [delta]) - 1]] ([1 + Au[z.sup.k]]/[1 + Bu[z.sup.k]]) du. (2.12)

For this function we find that

(1 - [delta])[[[I.sup.[sigma]]f(z)]/[z.sup.p]] + [delta] [[[I.sup.[sigma]]f(z)]/[z.sup.p]] = [[1 + A[z.sup.k]]/[1 + B [z.sup.k]]]

and

[[[I.sup.[[sigma] + 1]] f(z)]/[z.sup.P]][right arrow][[p + 1]/[k(1 - [delta])]][[integral].sub.0.sup.1] [u.sup.[(p + 1)/k(1 - [delta]) - 1]] ([1 - Au]/[1 - Bu]) du as z[right arrow][e.sup.[i[pi]/k]].

Hence the proof of the theorem is complete.

Theorem 2. Let f(z) [z.sup.p] + [[infinity].summation over (m = k)] [member of] A(p, k), [s.sub.1] (z) = [z.sup.p] and

[s.sub.n](z) = [z.sup.p] + [k + n-2.summation over (m = k)] [a.sub.p + m][z.sup.p + m] (n [greater than or equal to] 2). If the sequence {[c.sub.m]} (m [greater than or equal to] k) is

nondecreasing with [c.sub.k] > 1, where [c.sub.m] is given by (2.4) and satisfies the condition (2.3), then

Re[gamma] {[[f(z)]/[[s.sub.n](z)]]} > [[[c.sub.[k + n - 1]]-1]/[c.sub.[k + n - 1]]] (2.13)

and

Re[gamma] {[[[s.sub.n](z)]/[f(z)]]} > 1-[1/[1 + [c.sub.[k + n - 1]]]]. (2.14)

Each of the bounds in (2.13) and (2.14) is best possible for n [member of] N.

Proof. Under the hypothesis of the theorem, we have

[[k + n - 2].summation over (m = k)]|[a.sub.[p + m]]| + [c.sub.[k + n - 1]][[infinity].summation over (m = k + n - 1)]|[a.sub.[p + m]]|[less than or equal to][[infinity].summation over (m = k)][c.sub.m]|[a.sub.[p + m]]|[less than or equal to]1. (2.15)

Let

[g.sub.1](z) = [c.sub.[k + n - 1]]{[[f(z)]/[[s.sub.n](z)]] - [[[c.sub.[k + n - 1]] - 1]/[c.sub.[k + n - 1]]]},

then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

and it follows from (2.15) that

|[[[g.sub.1](z) - 1]/[[g.sub.1](z) + 1]]|[less than or equal to] [[[c.sub.[k + n - 1]][[infinity].summation over (m = k + n - 1)]|[a.sub.[p + m]]|]/[2 - 2[[k + n - 2].summation over (m = k)]|[a.sub.[p + m]]| - [c.sub.[k + n - 1]][[infinity].summation over (m = k + n - 1)]|[a.sub.[p + m]]|]]

< or equal to] 1 (z [member of] E),

which readily yields the inequality (2.13).

If we take

f (z) = [z.sup.p] - [[z.sup.[p + k + n - 1]]/[c.sub.[k + n - 1]]], (2.16)

then

[[f(z)]/[[s.sub.n](z)]] = 1 - [[z.sup.[k + n - 1]]/[c.sub.[k + n - 1]]] [right arrow] 1 - [1/[c.sub.[k + n - 1]]] as z [right arrow] [1.sup. - ].

This show that the bound in (2.13) is best possible for each n.

Similarly, if we put

[g.sub.2] (z) = (1 + [c.sub.[k + n - 1]]) {[[[s.sub.n](z)]/[f(z)]] - (1 - 1/[c.sub.[k + n - 1]])},

then we deduce that

|[[[g.sub.2](z) - 1]/[[g.sub.2](z) + 1]] | [less than or equal to] [[(1 + [c.sub.[k + n - 1]])[[infinity].summation over (m = k + n - 1)] | [a.sub.[p + m]]|]/[2 - 2[[k + n - 2].summation over (m = k)] | [a.sub.[p + m]] | + (1 + [c.sub.[k + n - 1]]) [[infinity].summation over (m = k + n - 1)] | [a.sub.[p + m]]|]]

[less than or equal to] 1 (z [member of] E),

which yields (2.14). The estimate (2.14) is sharp for each n with the extremal function f (z) given by (2.16). The proof is now complete.

References

(1) T. M. Flett, The dual of an inequality of Hardy and Littlewood and some related inequalities, J. Math. Anal. Appl. 38(1972), 746-765.

(2) I. B. Jung, Y. C. Kim, and H. M. Srivastava, The Hardy space of analytic functions associated with certain one-parameter families of integral operators, J. Math. Anal. Appl. 176(1993), 138-147.

(3) J. L. Li, Some properties of two integral operators, Soochow J. Math. 25(1999), 91-96.

(4) J. -L. Liu, A linear operator and strongly starlike functions, J. Math. Soc. Japan, 54(2002),975-981.

(5) S. S. Miller and P. T. Mocanu, Differential subordinations and univalent functions, Michigan Math. J. 28(1981), 157-171.

(6) B. A. Uralegaddi and Somanatha, Certain integral operators for starlike functions, J. Math. Res. Expo. 15(1995), 14-16.

Some Applications of the Jung-Kim-Srivastava Integral Operator *

Jin-Lin Liu [dagger]

Department of Mathematic,Yangzhou University, Yangzhou 225002, Jiangsu, People's Republic of China

Received October 24, 2008, Accepted February 17, 2009.

* 2000 Mathematics Subject Classification. Primary 30C45.

[dagger] E-mail: jlliucn@yahoo.com.cn
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Author:Liu, Jin-Lin
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Date:Dec 15, 2010
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