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##  eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2020; 60(1): 126-132

Published online March 31, 2020

### Hardy Spaces of Certain Convolution Operator

Rajbala and Jugal Kishore Prajapat∗

Department of Mathematics, Central University of Rajasthan, Ajmer-305817, Rajasthan, India
e-mail : rajbalachoudary9@gmail.com and jkprajapat@gmail.com

Received: January 14, 2018; Revised: March 20, 2019; Accepted: April 9, 2019

### Abstract

In this article, we determine sufficient conditions on the parameters of a generalized convolution operator to ensure that it belongs to the Hardy space and to the space of bounded analytic functions. We exhibit the utility of these results by deducing several interesting examples.

Keywords: analytic functions, bounded analytic functions, Hardy spaces.

### 1. Introduction

Let ℋ denote the class of analytic functions in the open unit disk , and be the subclass of ℋ consisting of normalized analytic functions of the form

f(z)=z+n=2anzn         (zD).

For the analytic functions f and g, we say that f is subordinate to g and write fg, if there exists an analytic function w in such that w(0) = 0 and f(z) = g(w(z)). In particular, if g is univalent in , then we have the following equivalence:

f(z)g(z)         (zD)[f(0)=g(0)         and         f(D)g(D)].

Further, for functions , given by fj(z)=z+n=2an,jzn (j = 1, 2), the Hadamard product (or Convolution) of f1 and f2 is defined by

(f1*f2)(z):=z+n=2an,1an,2zn         (zD).

Let ℛ(β) denote the class of functions such that , and let ℘(β) denote the class of functions f ∈ ℋ such that f(0) = 1 and . For β = 0, we denote ℛ(β) and ℘(β) simply by ℛ and ℘, respectively.

Let ℋ denote the space of all bounded analytic functions in . This is Banach algebra with respect to the norm . For f ∈ ℋ, we define

Mp(r,f)={(12π02πf(reiθ)pdθ)1/p(0<p<),maxzrf(z)(p=).

The function f ∈ ℋ belongs to the Hardy space ℋp (0 < p ≤ ∞), if Mp(r, f) is bounded for all r ∈ [0, 1). Clearly, we have

qp         for         0<p<q<

(see [3, p. 2]). For 1 ≤ p ≤ ∞, ℋp is a Banach space with the norm defined by

fp=limr1-Mp(r,f)         (1p)

(see [3, p. 23]). Following are two widely known results (see ) for the Hardy space ℋp:

(f(z))>0fp   for   all p<1fq/(1-q)   for   all   0<q<1.

In , Ponnusamy studied the Hardy space of hypergeometric functions. Further, Baricz  obtained the conditions for the generalized Bessel functions such that it belongs to Hardy space and Yagmur and Orhan  studied the same problem for the generalized Struve functions.

In  (see also ), Ibrahim studied the following generalized fractional integral operator in the complex plane ℂ:

Izα,μf(z)=(μ+1)1-αΓ(α)0z(zμ+1-ξμ+1)α-1ξμf(ξ)dξ         (α,μ,α>0,μ-1),

where the function f(z) is analytic in a simply connected region of ℂ containing the origin, and the multiplicity of (zμ+1ξμ+1)α−1 is removed by requiring log(zμ+1ξμ+1) to be real when (zμ+1ξμ+1) > 0. We observe that, if we take μ = 0 in (1.4), we arrive at the Srivastva-Owa fractional integral operator  (see also ) and when μ→−1+ in (1.4), we obtain the Hadamard fractional integral operator .

For α > 0 and μ > −1, we define a fractional integral operator Ωzα,μ:AA by

Ωzα,μf(z)=(μ+1)αΓ(1μ+1+α+1)Γ(1μ+1+1)z-α(μ+1)Izα,μf(z)=z+n=2Γ(1μ+1+α+1)Γ(nμ+1+1)Γ(nμ+1+α+1)Γ(1μ+1+1)anzn.

Note that Ωz0,0f(z),Ωz1,0f(z)=f(z) and Ωzα,0f(z)=L(2,α+2)f(z), where ℒ and L(a, c) denotes the Libera integral operator  and Carlson-Shaffer operator , respectively. Also, we observe that the operator Ωzα,μ satisfies the following recurrence relation

z(Ωzα,μf(z))=(1+α(μ+1))Ωzα-1,μf(z)-α(μ+1)Ωzα,μf(z)         (α>1,μ>-1).

Corresponding to fractional integral operator Ωzα,μf(z), a fractional differential operator Φα,μf(z) was studied in [6, 7] to obtain its boundedness in Bergman space and certain geometric properties were also discussed.

### 2. Main Results

In order to derive our main results, we recall here the following lemmas:

### Lemma 2.1.()

Let F and G be analytic functions in and F(0) = G(0). If H(z) = zG′(z) is starlike function in and zF′ (z) ≺ zG′ (z), then

F(z)G(z)=G(0)+0zH(t)tdt.

### Lemma 2.2.()

For α < 1, β < 1, we have ℘(α) * ℘(β) ⊂ ℘(δ), where δ = 1 − 2(1 − α)(1 − β). The value of δ is best possible.

Our first main result is given by Theorem 2.3 below.

### Theorem 2.3

Let α > 1 and μ > −1. If f(z) ∈ ℛ, thenΩzα,μf(z).

Proof

From the definition of operator Ωzα,μf(z), we have

{(Ωzα,μf(z))}=(μ+1)Γ(1μ+1+α+1)Γ(α)Γ(1μ+1+1)01uμ(1-uμ+1)α-1{f(zu)}du.

By hypothesis f(z) ∈ ℛ, hence it follows that Ωzα,μf(z). By first implication of (1.3), we have (Ωzα,μf(z))q for all q < 1. Further, by second implication of (1.3), we have Ωzα,μfq/(1-q) for all 0 < q < 1, or equivalently, Ωzα,μfp for all 0 < p < ∞.

Since f ∈ ℛ, then using the well known bound for Caratheodory functions in , we have |an| ≤ 2/n, n ≥ 2. Hence

Ωzα,μf(z)z+Γ(1μ+1+α+1)Γ(1μ+1+1)n=2Γ(nμ+1+1)Γ(nμ+1+α+1)anzn<1+2n=2θ(n),

where θ(n) is given by

θ(n)=Γ(1μ+1+α+1)Γ(nμ+1+1)nΓ(nμ+1+α+1)Γ(1μ+1+1).

Evidently, θ(1) = 1 and θ(n) > 0 for all n ∈ ℕ. It is well known that the digamma function Ψ(x) = Γ′(x)/Γ(x) is increasing for all x > 0 (see [12, II, sec. 3/eqn. 4 on p. 723]). Therefore Ψ(x + ɛ) > Ψ(x) for ɛ > 0. The auxiliary function Γ̃(x) := Γ(x + ɛ)/Γ(x) has a positive derivative as

Γ˜(x)=Γ(x+ɛ)Γ(x)-Γ(x+ɛ)Γ(x)[Γ(x)]2=Γ(x+ɛ)Γ(x)(Γ(x+ɛ)Γ(x+ɛ)-Γ(x)Γ(x))>0         (x>0,ɛ>0).

Thus Γ̃(x) is an increasing function, so that

Γ(x+ɛ)Γ(x)Γ(y+ɛ)Γ(y)         whenever         xy>0.

Hence

θ(n)θ(n+1)=(n+1)Γ(1+nμ+1)Γ(1+α+n+1μ+1)nΓ(1+n+1μ+1)Γ(1+α+nμ+1)1,

which shows that 0 < θ(n + 1) < θ(n) ≤ θ(2) for each n ≥ 2. Hence θ(n) is a non-increasing function of n ≥ 2.

Now, we shall show that limn |θ(n)|1/n = 1. Using the asymptotic formula for the Gamma function [4, sec. 1.18, (4)], we have

limnθ(n)1/n=limn[Γ(1+α+1μ+1)Γ(1+1μ+1)]1/n·limn[Γ(1+nμ+1)nΓ(1+α+nμ+1)]1/n~limn[(nμ+1)-α]1/n=limn(n1/n)-α[1(μ+1)α]1/n=1.

This implies that the series in (2.2) converges for |z| < 1. Further, applying Raabe’s test for convergence, we deduce that the power series for Ωzα,μf converges absolutely for |z| = 1 for all α > 0 and μ > −1.

Also, using the known result [3, Theorem 3.11], (Ωzα,μf)q implies continuity of Ωzα,μf on the compact set . Since the continuous function Ωzα,μf on the compact set is bounded, thus Ωzα,μf is a bounded analytic function in . Therefore Ωzα,μf and this completes the proof.

### Remark 2.4

If we consider Ψ(z)=-z-2log(1-z)=z+2n=2znn, then from the geometrical descriptions of image domains of functions in the class ℛ, we can easily verify that Ψ ∈ ℛ and not belongs to ℋ, but as per the Theorem 2.3, function Ωzα,μΨ(z), where

Ωzα,μΨ(z)=z+2n=2Γ(1μ+1+α+1)Γ(nμ+1+1)nΓ(nμ+1+α+1)Γ(1μ+1+1)zn         (α>1,μ>-1,zD).

### Theorem 2.5

Let , α > 1 and μ > −1. If

Ωzα-1,μf(z)-Ωzα,μf(z)z<11+α(μ+1)         (zD),

then Ωzα,μf(z)z.

Proof

The inequality (2.6) is equivalent to

Ωzα-1,μf(z)-Ωzα,μf(z)zz1+α(μ+1)         (zD),

which in view of (1.6), can be written as z(Ωzα,μf(z)z)z(1+z). Now applying Lemma 2.1, for F(z)=Ωzα,μf(z)z and G(z) = 1+z, the desired result follows.

### Theorem 2.6

Let , α > 1 and μ > −1. If g ∈ ℛ(1/2) andΩzα,μfsatisfy the inequality (2.6), thenΩzα,μf(z)*g(z).

Proof

Let u(z)=Ωzα,μf(z)*g(z), then u(z)=Ωzα,μf(z)z*g(z). In view of the hypotheses and Theorem 2.5, we have Ωzα,μf(z)z and g′(z) ∈ ℘(1/2). Now using Lemma 2.2, we obtain that u′(z) ∈ ℘, which is equivalent to u(z) ∈ ℛ, hence the desired result follows.

### Acknowledgements

The authors wish to thank the referee for her/his valuable suggestions, which have improved the paper. The work was supported by the University Grant Commission, India (Project Number: MRP-MAJOR-MATH-2013-19114).

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