Characterization of the generalized Hausdorff and quasi Hausdorff operators on weighted integrable spaces

ABSTRACT

We examine the Hausdorff and Quasi Hausdorff matrices which act as an operator on Hardy spaces by its action on Taylor coefficients of an analytic function. Using the integral representation of the series on the Hardy spaces, we find the conditions on the weight functions on ${\mathbb{R}}^{+}$, to determine the boundedness of the operator on weighted integrable spaces and its adjoint operator on the respective domain.

KEYWORDS:

• Hausdorff matrix
• quasi Hausdorff matrix
• hardy spaces
• weighted integrable spaces

MATHEMATICS SUBJECT CLASSIFICATION:

• Primary 47B38
• Secondary 46E15

1. Introduction

Cesàro, Euler, Taylor and Hausdorff means have been studied in connection with summability of the series and assorted function spaces like integrable spaces, Hardy spaces, Bergman spaces see for example (Alexits, 1953; Hardy, 1943; Hung et al., 2006; Kalaivani & Youvaraj, 2013; Liflyand, 2013), and (Kalaivani & Madhu, 2024).

Hardy (Hardy, 1920) examined Cesàro matrix act as an operator on $\mathit{p}-$integrable spaces and determined the norm of the operator. Hardy’s result provides wide visibility to analyzing the behavior of Cesàro operator in various spaces like the Hardy spaces on the unit disk, Bergman spaces and so on, the reader is referred to see Hardy (1956), Hung et al. (2000) and Siskakis (1987).

Let $(h_{\mathit{n}\mathit{k}}{)}_{\mathit{n},\mathit{k}\in {\mathbb{Z}}^{+}}$ be an infinite matrix. For an analytic function f on unit disk $\mathbb{D}$ in the complex plane, consider the power series

$\begin{array}{l}\mathbb{H}(\mathit{f})(\mathit{z})=\sum _{\mathit{n}=0}^{\mathrm{\infty }}\left(\sum _{\mathit{k}=0}^{\mathit{n}}{h}_{\mathit{n}\mathit{k}}{c}_k\right)z^n\end{array}$

and for transpose matrix of $(h_{\mathit{n}\mathit{k}}{)}_{\mathit{n},\mathit{k}\in {\mathbb{Z}}^{+}},$ the power series

$\begin{array}{l}\mathbb{T}(\mathit{f})(\mathit{z})=\sum _{\mathit{k}=0}^{\mathrm{\infty }}\left(\sum _{\mathit{n}=\mathit{k}}^{\mathrm{\infty }}{h}_{\mathit{n}\mathit{k}}{c}_k\right)z^n\end{array}$

where $\mathit{f}(\mathit{z})=\sum _{\mathit{n}=0}^{\mathrm{\infty }}{c}_n{z}^n.$

It is natural to address the problem of finding the conditions on the coefficients or matrix, so that for an analytic function f, the series $\mathbb{H}(\mathit{f})(\mathit{z})$ and $\mathbb{T}(\mathit{f})(\mathit{z})$ defined and how the matrix induces as an operator on the corresponding domain.

For the Cesàro and Hankel matrix $(h_{\mathit{n}\mathit{k}}{)}_{\mathit{n},\mathit{k}\in {\mathbb{Z}}^{+}}$, numerous works are done on Hardy, Bergman spaces of the unit disk.

In this paper, we consider $(h_{nk}^{\alpha }{)}_{\mathit{n},\mathit{k}\in {\mathbb{Z}}^{+}}$ as generalized Hausdorff matrix, since it plays a significant part in summability theory for different cases of measure ϖ and $\mathit{\alpha }.$ In particular, for moment sequences, we obtain Cesàro, Euler, Riesz, Taylor, Holder matrices and so on as a special case of generalized Hausdorff matrix.

Generalized Hausdorff matrix: Let $\mathit{\alpha }\in \mathbb{R}$ and ϖ be of bounded variation on $(0,1].$ For $\mathit{n},\mathit{k}\in {\mathbb{Z}}^{+},$ we define

$\begin{array}{l}{h}_{nk}^{\alpha }:=\{\begin{array}{ll}(\genfrac{}{}{0ex}{}{\mathit{n}+\mathit{\alpha }}{\mathit{n}-\mathit{k}}){\displaystyle \underset{0}{\overset{1}{\int }}{t}^{\mathit{k}+\mathit{\alpha }}(1-\mathit{t}{)}^{\mathit{n}-\mathit{k}}\mathit{d\varpi }(\mathit{t})}& \text{for }0\le \mathit{k}\le \mathit{n},\\ \\ 0& \text{otherwise.}\end{array}\end{array}$

The operator obtained from $(h_{\mathit{n}\mathit{k}}{)}_{\mathit{n},\mathit{k}\in {\mathbb{Z}}^{+}}$ and its transpose of the matrix $(h_{\mathit{n}\mathit{k}}{)}_{\mathit{n},\mathit{k}\in {\mathbb{Z}}^{+}}$ are nothing but generalized Hausdorff operator ${\mathbb{H}}_{\mathit{\alpha }}$ and quasi Hausdorff operator ${\mathbb{T}}_{\mathit{\alpha }}$. In particular for $\mathit{\alpha }=0,$ we obtain Hausdorff matrix as a special case of generalized Hausdorff matrix.

P. Galanopoulos and A. G. Siskakis (Galanopoulos & Siskakis, 2001) studied Hausdorff operator ${\mathbb{H}}_0$ on the Hardy spaces of the unit disk and obtained the results in terms of composition operator. Later, P. Galanopoulos and M. Papadimitrakis (Galanopoulos & Papadimitrakis, 2006) considered Hausdorff and quasi Hausdorff matrices are act as an operator on the spaces of analytic functions on the unit disk $\mathbb{D}.$ Explicitly, they expressed the series in terms of integral as a composition function and addressed the problem on the Hardy spaces, Bergman spaces and Dirichlet space.

The action of the generalized Hardy-Cesàro operator on spaces of weighted integrable spaces have been studied by Pedersen (Pedersen, 2019). He briefly examined the boundedness and weakly compactness of an operator on the respective domain.

The aforementioned works motivate the study of generalised Hausdorff operator on Hardy spaces and using the integral representation of the operator in order to determine the condition on ϖ to corroborate the boundedness of the operator.

In section 2, we will examine the convergence of the power series ${\mathbb{H}}_{\mathit{\alpha }}(\mathit{f})(\mathit{z}),$ ${\mathbb{T}}_{\mathit{\alpha }}(\mathit{f})(\mathit{z})$ and identify the relation between the series and composition function for the generalized Hausdorff and quasi Hausdorff operator.

In section 3 and 4, we investigate the action of the generalized Hausdorff operator and quasi Hausdorff operator on weighted spaces of integrable functions and determine the boundedness of the operator under certain conditions on weighted functions w₁ and w₂ on ${\mathbb{R}}^{+}.$ Also, we address the problem for dual of the operators ${\mathbb{H}}_{\mathit{\alpha }}$ and ${\mathbb{T}}_{\mathit{\alpha }}$ on respective domain.

2. Integral representation of the generalized Hausdorff operator

Theorem 1.

Let $(h_{nk}^{\alpha }{)}_{\mathit{n},\mathit{k}\in {\mathbb{Z}}^{+}}$ be an generalized Hausdorff matrix and $\mathit{f}\in H^p(\mathbb{D})$ for $1\le \mathit{p}<\mathrm{\infty }$. Then

1. ${\mathbb{H}}_{\mathit{\alpha }}(\mathit{f})(\mathit{z})$ is analytic on $\mathbb{D}.$

2. ${\mathbb{H}}_{\mathit{\alpha }}(\mathit{f})$ can be expressed as an integral form,

   (1) $\begin{array}{ll}{\mathbb{H}}_{\mathit{\alpha }}(\mathit{f})(\mathit{z})=& {\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \frac{t^{\mathit{\alpha }}}{{[(\mathit{t}-1)\mathit{z}+1]}^{\mathit{\alpha }+1}}}\mathit{f}\left({\displaystyle \frac{\mathit{tz}}{1-\mathit{z}(1-\mathit{t})}}\right)}\\ & \mathit{d\varpi }(\mathit{t}),\mathit{z}\in \mathbb{D}.\end{array}$(1)

Proof.

1. Let $\mathit{f}(\mathit{z})=\sum _{\mathit{k}=0}^{\mathrm{\infty }}{c}_k{z}^k,\mathit{z}\in \mathbb{D}$ and $A_n=\sum _{\mathit{k}=0}^{\mathit{n}}{h}_{nk}^{\alpha }{c}_k.$ Since Taylor coefficients of an analytic function $\mathit{f}\in H^p(\mathbb{D})$ are bounded by M > 0. Thus

   $\begin{array}{ll}|A_{\mathit{n}}|& =\left|\sum _{\mathit{k}=0}^{\mathit{n}}{h}_{nk}^{\alpha }{c}_k\right|\\ & \le \mathit{M}\left|\sum _{\mathit{k}=0}^{\mathit{n}}{h}_{nk}^{\alpha }\right|\le \mathit{M\varpi }(0,1]\\ & \text{as}\sum _{\mathit{k}=0}^{\mathit{n}}(\genfrac{}{}{0ex}{}{\mathit{n}+\mathit{\alpha }}{\mathit{n}-\mathit{k}})t^{\mathit{k}+\mathit{\alpha }}(1-\mathit{t}{)}^{\mathit{n}-\mathit{k}}\le 1\end{array}$

   which yields the radius of convergence for ${\mathbb{H}}_{\mathit{\alpha }}(\mathit{f})(\mathit{z})$ is atleast 1. Hence ${\mathbb{H}}_{\mathit{\alpha }}(\mathit{f})(\mathit{z})$ is an analytic function on the unit disk $\mathbb{D}$.

2. Consider

   $\begin{array}{ll}{\mathbb{H}}_{\mathit{\alpha }}(\mathit{f})(\mathit{z})& =\sum _{\mathit{n}=0}^{\mathrm{\infty }}\left(\sum _{\mathit{k}=0}^{\mathit{n}}{h}_{nk}^{\alpha }{c}_k\right)z^n\\ & =\sum _{\mathit{k}=0}^{\mathrm{\infty }}\sum _{\mathit{n}=\mathit{k}}^{\mathrm{\infty }}{\displaystyle \underset{0}{\overset{1}{\int }}(\genfrac{}{}{0ex}{}{\mathit{n}+\mathit{\alpha }}{\mathit{n}-\mathit{k}})}\\ & (1-\mathit{t}{)}^{\mathit{n}-\mathit{k}}{z}^{\mathit{n}-\mathit{k}}{c}_k{z}^k{t}^{\mathit{k}+\mathit{\alpha }}\mathit{d\varpi }(\mathit{t}).\end{array}$

   For each $\mathit{t}\in [0,1],$ the series $\sum _{\mathit{n}=\mathit{k}}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\mathit{n}+\mathit{\alpha }}{\mathit{n}-\mathit{k}})(1-\mathit{t}{)}^{\mathit{n}-\mathit{k}}{z}^{\mathit{n}-\mathit{k}}$ converges uniformly to ${\displaystyle \frac{1}{[1-\mathit{z}(1-\mathit{t}){]}^{\mathit{k}+\mathit{\alpha }+1}}}.$ Hence, by interchanging the sum and integral in the above equation,

   $\begin{array}{lll}{\mathbb{H}}_{\mathit{\alpha }}(\mathit{f})(\mathit{z})& =& \sum _{\mathit{k}=0}^{\mathrm{\infty }}{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \frac{1}{[1-\mathit{z}(1-\mathit{t}){]}^{\mathit{k}+\mathit{\alpha }+1}}}{c}_k{z}^k{t}^{\mathit{k}+\mathit{\alpha }}\mathit{d\varpi }(\mathit{t})}\\ & =& {\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \frac{t^{\mathit{\alpha }}}{{[(\mathit{t}-1)\mathit{z}+1]}^{\mathit{\alpha }+1}}}\mathit{f}\left({\displaystyle \frac{\mathit{tz}}{1-\mathit{z}(1-\mathit{t})}}\right)\mathit{d\varpi }(\mathit{t}).}\end{array}$

Theorem 2.

Let $(h_{nk}^{\alpha }{)}_{\mathit{n},\mathit{k}\in {\mathbb{Z}}^{+}}$ be an generalized Hausdorff matrix and $\mathit{f}\in H^p(\mathbb{D})$ for $1\le \mathit{p}<\mathrm{\infty }$. Then

1. ${\mathbb{T}}_{\mathit{\alpha }}(\mathit{f})(\mathit{z})$ is analytic on $\mathbb{D}.$

2. ${\mathbb{T}}_{\mathit{\alpha }}(\mathit{f})$ can be expressed as an integral form,

   (2) $\begin{array}{l}{\mathbb{T}}_{\mathit{\alpha }}(\mathit{f})(\mathit{z})=\underset{0}{\overset{1}{\int }}\frac{(\mathit{tz}+1-\mathit{t}{)}^{\mathit{\alpha }}}{z^{\mathit{\alpha }}}\mathit{f}(\mathit{tz}+1-\mathit{t})\mathit{d\varpi }(\mathit{t}).\end{array}$(2)

Proof.

1. Let $\mathit{f}(\mathit{z})=\sum _{\mathit{k}=0}^{\mathrm{\infty }}{c}_k{z}^k,\mathit{z}\in \mathbb{D}$ and $B_k=\sum _{\mathit{n}=\mathit{k}}^{\mathrm{\infty }}{h}_{nk}^{\alpha }{c}_n.$ Then

   $\begin{array}{ll}|B_{\mathit{k}}|& =\left|\sum _{\mathit{n}=\mathit{k}}^{\mathrm{\infty }}{h}_{nk}^{\alpha }{c}_n\right|\\ & \le \mathit{M}\left|\sum _{\mathit{n}=\mathit{k}}^{\mathrm{\infty }}{h}_{nk}^{\alpha }\right|\le \mathit{MC\varpi }(0,1]\\ & \text{as}\sum _{\mathit{n}=\mathit{k}}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\mathit{n}+\mathit{\alpha }}{\mathit{n}-\mathit{k}})t^{\mathit{k}+\mathit{\alpha }}(1-\mathit{t}{)}^{\mathit{n}-\mathit{k}}\le \mathit{C},|c_n|\le \mathit{M}\end{array}$

   which yields the radius of convergence for ${\mathbb{T}}_{\mathit{\alpha }}(\mathit{f})(\mathit{z})$ is atleast 1. Hence ${\mathbb{T}}_{\mathit{\alpha }}(\mathit{f})(\mathit{z})$ is an analytic on the unit disk $\mathbb{D}$.

2. Consider

   $\begin{array}{ll}{\mathbb{T}}_{\mathit{\alpha }}(\mathit{f})(\mathit{z})& =\sum _{\mathit{k}=0}^{\mathrm{\infty }}\left(\sum _{\mathit{n}=\mathit{k}}^{\mathrm{\infty }}{h}_{nk}^{\alpha }{c}_n\right)z^k\\ & =\sum _{\mathit{k}=0}^{\mathrm{\infty }}(\sum _{\mathit{n}=\mathit{k}}^{\mathrm{\infty }}\underset{0}{\overset{1}{\int }}(\genfrac{}{}{0ex}{}{\mathit{n}+\mathit{\alpha }}{\mathit{n}-\mathit{k}})\\ & \phantom{\sum _{\mathit{n}=\mathit{k}}^{\mathrm{\infty }}\underset{0}{\overset{1}{\int }}(\genfrac{}{}{0ex}{}{\mathit{n}+\mathit{\alpha }}{\mathit{n}-\mathit{k}})}{t}^{\mathit{k}+\mathit{\alpha }}(1-\mathit{t}{)}^{\mathit{n}-\mathit{k}}\mathit{d\varpi }(\mathit{t})c_n)z^k\\ & =\sum _{\mathit{n}=0}^{\mathrm{\infty }}(\sum _{\mathit{k}=0}^{\mathit{n}}\underset{0}{\overset{1}{\int }}(\genfrac{}{}{0ex}{}{\mathit{n}+\mathit{\alpha }}{\mathit{n}-\mathit{k}})\\ & \phantom{\sum _{\mathit{k}=0}^{\mathit{n}}\underset{0}{\overset{1}{\int }}(\genfrac{}{}{0ex}{}{\mathit{n}+\mathit{\alpha }}{\mathit{n}-\mathit{k}})}{t}^{\mathit{k}+\mathit{\alpha }}(1-\mathit{t}{)}^{\mathit{n}-\mathit{k}}\mathit{d\varpi }(\mathit{t})z^k)c_n\\ & =\sum _{\mathit{n}=0}^{\mathrm{\infty }}(\sum _{\mathit{k}+\mathit{\alpha }=0}^{\mathit{n}+\mathit{\alpha }}\underset{0}{\overset{1}{\int }}(\genfrac{}{}{0ex}{}{\mathit{n}+\mathit{\alpha }}{\mathit{n}+\mathit{\alpha }-(\mathit{k}+\mathit{\alpha })})\\ & \phantom{\sum _{\mathit{k}+\mathit{\alpha }=0}^{\mathit{n}+\mathit{\alpha }}\underset{0}{\overset{1}{\int }}(\genfrac{}{}{0ex}{}{\mathit{n}+\mathit{\alpha }}{\mathit{n}+\mathit{\alpha }-(\mathit{k}+\mathit{\alpha })})}{t}^{\mathit{k}+\mathit{\alpha }}(1-\mathit{t}{)}^{\mathit{n}-\mathit{k}}\mathit{d\varpi }(\mathit{t})z^k)c_n\\ & =\sum _{\mathit{n}=0}^{\mathrm{\infty }}\underset{0}{\overset{1}{\int }}(\mathit{tz}+1-\mathit{t}{)}^{\mathit{n}+\mathit{\alpha }}{c}_n\mathit{d\varpi }(\mathit{t})z^{-\mathit{\alpha }}\end{array}$

   Thus

   $\begin{array}{lll}{\mathbb{T}}_{\mathit{\alpha }}(\mathit{f})(\mathit{z})& =& \underset{0}{\overset{1}{\int }}\frac{(\mathit{tz}+1-\mathit{t}{)}^{\mathit{\alpha }}}{z^{\mathit{\alpha }}}\mathit{f}(\mathit{tz}+1-\mathit{t})\mathit{d\varpi }(\mathit{t}).\end{array}$

3. Boundedness of the operator ${\mathbb{H}}_{\mathit{\alpha }}$ on weighted integrable spaces

In Section 2, we obtained the integral representation of ${\mathbb{H}}_{\mathit{\alpha }}(\mathit{f})(\mathit{z})$ and ${\mathbb{T}}_{\mathit{\alpha }}(\mathit{f})(\mathit{z})$ on the Hardy spaces of unit disk. Using the representation, we will study the action of generalized Hausdorff and quasi Hausdorff operator on weighted spaces of integrable functions on ${\mathbb{R}}^{+}$. Let w₁ be a non-negative continuous function on ${\mathbb{R}}^{+}$ and $L_1(w_1)$ be a space of measurable functions f on ${\mathbb{R}}^{+}$ with

$\begin{array}{l}\Vert \mathit{f}{\Vert }_{L_1(w_1)}={\displaystyle \underset{0}{\overset{1}{\int }}|\mathit{f}(\mathit{t})|w_1(\mathit{t})\mathit{dt}<\mathrm{\infty }.}\end{array}$

Denote $L_{\mathrm{\infty }}\left(\frac{1}{w_1}\right)$ be a space of measurable functions f on ${\mathbb{R}}^{+}$ with

$\begin{array}{l}\Vert \mathit{f}{\Vert }_{{\mathit{L}}_{\mathrm{\infty }}\left(\frac{1}{w_1}\right)}=\mathit{ess}\underset{\mathit{t}\in {\mathbb{R}}^{+}}{sup}{\displaystyle \frac{|\mathit{f}(\mathit{t})|}{w_1(\mathit{t})}}<\mathrm{\infty }.\end{array}$

Let $\mathfrak{M}(w_1)$ be a space of locally finite, complex Borel measures ϖ on ${\mathbb{R}}^{+}$ with

$\begin{array}{l}\Vert \mathit{\varpi }{\Vert }_{\mathfrak{M}(w)}=\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathit{w}(\mathit{t})\mathit{d}|\mathit{\varpi }|(\mathit{t}).\end{array}$

In the following theorem, we find the necessary and sufficient condition on weighted functions w₁ and w₂ for the boundedness of generalized Hausdorff operator ${\mathbb{H}}_{\mathit{\alpha }}$ on weighted integrable spaces on ${\mathbb{R}}^{+}.$

Theorem 3.

Let w₁ and w₂ be the non-negative weight functions on ${\mathbb{R}}^{+}.$ The operator ${\mathbb{H}}_{\mathit{\alpha }}$ is bounded linear operator from $L_1(w_1)$ to $L_1(w_2)$ iff

(3) $\begin{array}{l}{x}^{\mathit{\alpha }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{w_2(\mathit{y})}{y^{\mathit{\alpha }+1}}}\mathit{dy}\le \mathit{C}|1-\mathit{x}|w_1(\mathit{x})\text{ a.e. on }{\mathbb{R}}^{+}\text{ for some }\mathit{C}>0.\end{array}$(3)

Proof.

Let $\mathit{f}\in L_1(w_1)$ and assume the inequality holds a.e. on ${\mathbb{R}}^{+}$. Then

(4) $\begin{array}{ll}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{x^{\mathit{\alpha }}}{y^{\mathit{\alpha }+1}|1-\mathit{x}|}}|\mathit{f}(\mathit{x})|w_2(\mathit{y})\mathit{dydx}& =\underset{0}{\overset{\mathrm{\infty }}{\int }}|\mathit{f}(\mathit{x})|{\displaystyle \frac{x^{\mathit{\alpha }}}{|1-\mathit{x}|}}\\ & \underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}\frac{w_2(\mathit{y})}{y^{\mathit{\alpha }+1}}\mathit{dydx}\\ & \le \mathit{C}\underset{0}{\overset{\mathrm{\infty }}{\int }}|\mathit{f}(\mathit{x})|w_1(\mathit{x})\mathit{dx}\end{array}$(4)

by the assumptions on w₁ and w₂ in (3(3) $\begin{array}{l}{x}^{\mathit{\alpha }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{w_2(\mathit{y})}{y^{\mathit{\alpha }+1}}}\mathit{dy}\le \mathit{C}|1-\mathit{x}|w_1(\mathit{x})\text{ a.e. on }{\mathbb{R}}^{+}\text{ for some }\mathit{C}>0.\end{array}$(3) ). Note that

$\begin{array}{ll}\underset{0}{\overset{\mathrm{\infty }}{\int }}|{\mathbb{H}}_{\mathit{\alpha }}\mathit{f}(\mathit{y})|w_2(\mathit{y})\mathit{dy}& =\underset{0}{\overset{\mathrm{\infty }}{\int }}|\underset{0}{\overset{1}{\int }}{\displaystyle \frac{t^{\mathit{\alpha }}}{{[(\mathit{t}-1)\mathit{y}+1]}^{\mathit{\alpha }+1}}}\mathit{f}\\ & \phantom{\underset{0}{\overset{1}{\int }}{\displaystyle \frac{t^{\mathit{\alpha }}}{{[(\mathit{t}-1)\mathit{y}+1]}^{\mathit{\alpha }+1}}}}\left({\displaystyle \frac{\mathit{ty}}{(\mathit{t}-1)\mathit{y}+1}}\right)\mathit{dt}|w_2(\mathit{y})\mathit{dy}\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}|\underset{0}{\overset{\mathit{y}}{\int }}{\displaystyle \frac{x^{\mathit{\alpha }}}{y^{\mathit{\alpha }+1}(1-\mathit{x})}}\mathit{f}(\mathit{x})\mathit{dx}|\\ & w_2(\mathit{y})\mathit{dy}\\ & \le \underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathit{y}}{\int }}{\displaystyle \frac{x^{\mathit{\alpha }}}{y^{\mathit{\alpha }+1}}}{\displaystyle \frac{|\mathit{f}(\mathit{x})|}{|1-\mathit{x}|}}\mathit{dx}{w}_2(\mathit{y})\mathit{dy}\\ & \le \underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{x^{\mathit{\alpha }}}{y^{\mathit{\alpha }+1}|1-\mathit{x}|}}\\ & |\mathit{f}(\mathit{x})|w_2(\mathit{y})\mathit{dydx}\\ & \le \mathit{C}\Vert \mathit{f}{\Vert }_{L_1(w_1)},\end{array}$

obtained the inequality by using Fubini’s theorem and (4(4) $\begin{array}{ll}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{x^{\mathit{\alpha }}}{y^{\mathit{\alpha }+1}|1-\mathit{x}|}}|\mathit{f}(\mathit{x})|w_2(\mathit{y})\mathit{dydx}& =\underset{0}{\overset{\mathrm{\infty }}{\int }}|\mathit{f}(\mathit{x})|{\displaystyle \frac{x^{\mathit{\alpha }}}{|1-\mathit{x}|}}\\ & \underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}\frac{w_2(\mathit{y})}{y^{\mathit{\alpha }+1}}\mathit{dydx}\\ & \le \mathit{C}\underset{0}{\overset{\mathrm{\infty }}{\int }}|\mathit{f}(\mathit{x})|w_1(\mathit{x})\mathit{dx}\end{array}$(4) ). Thus ${\mathbb{H}}_{\mathit{\alpha }}(\mathit{f})$ is in $L_1(w_2)$ and $\Vert {\mathbb{H}}_{\mathit{\alpha }}{\Vert }_{L_1(w_1)\to L_1(w_2)}\le \mathit{C}$ for some C > 0. Conversely, assume ${\mathbb{H}}_{\mathit{\alpha }}$ is bounded linear operator from $L_1(w_1)$ to $L_1(w_2).$ Let $\mathit{g}\in C_0\left(\frac{1}{w_2}\right)$, then by usual inner product with ${\mathbb{H}}_{\mathit{\alpha }}(\mathit{f})$, we obtain

(5) $\begin{array}{ll}<\mathit{g},{\mathbb{H}}_{\mathit{\alpha }}(\mathit{f})>& =\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathit{g}(\mathit{y})\frac{1}{y^{\mathit{\alpha }+1}}\underset{0}{\overset{\mathit{y}}{\int }}{\displaystyle \frac{x^{\mathit{\alpha }}}{1-\mathit{x}}}\mathit{f}(\mathit{x})\mathit{dxdy}\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\mathit{g}(\mathit{y})}{y^{\mathit{\alpha }+1}}}{\displaystyle \frac{x^{\mathit{\alpha }}}{1-\mathit{x}}}\mathit{dyf}(\mathit{x})\mathit{dx}\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{x^{\mathit{\alpha }}}{w_1(\mathit{x})(1-\mathit{x})}}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\mathit{g}(\mathit{y})}{y^{\mathit{\alpha }+1}}}\mathit{dyf}(\mathit{x})w_1(\mathit{x})\mathit{dx}.\end{array}$(5)

Since $L_1(w_2)\subset \mathfrak{M}(w_2)$ which is closed and identify$\mathfrak{M}(w_2)$ is the dual of $C_0\left(\frac{1}{w_2}\right)$, then there exists a mapping $\mathit{\varpi }:{\mathbb{R}}^{+}\to \mathfrak{M}(w_2)$ such that $<\mathit{g},\mathit{\varpi }(\mathit{x})>$ is measurable and bounded on ${\mathbb{R}}^{+},$ for every $\mathit{g}\in C_0\left(\frac{1}{w_2}\right)$ with $\Vert {\mathbb{H}}_{\mathit{\alpha }}\Vert =\mathit{ess}\underset{\mathit{x}\in {\mathbb{R}}^{+}}{sup}\Vert \mathit{\varpi }(\mathit{x}){\Vert }_{\mathfrak{M}(w_2)}.$ Thus

(6) $\begin{array}{ll}<\mathit{g},{\mathbb{H}}_{\mathit{\alpha }}(\mathit{f})>& =\underset{0}{\overset{\mathrm{\infty }}{\int }}<\mathit{g},\mathit{\varpi }(\mathit{x})>\mathit{f}(\mathit{x})w_1(\mathit{x})\mathit{dx}\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathit{g}(\mathit{y})\mathit{d\varpi }(\mathit{x})(\mathit{y})\mathit{f}(\mathit{x})w_1(\mathit{x})\mathit{dx}.\end{array}$(6)

Equating (5(5) $\begin{array}{ll}<\mathit{g},{\mathbb{H}}_{\mathit{\alpha }}(\mathit{f})>& =\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathit{g}(\mathit{y})\frac{1}{y^{\mathit{\alpha }+1}}\underset{0}{\overset{\mathit{y}}{\int }}{\displaystyle \frac{x^{\mathit{\alpha }}}{1-\mathit{x}}}\mathit{f}(\mathit{x})\mathit{dxdy}\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\mathit{g}(\mathit{y})}{y^{\mathit{\alpha }+1}}}{\displaystyle \frac{x^{\mathit{\alpha }}}{1-\mathit{x}}}\mathit{dyf}(\mathit{x})\mathit{dx}\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{x^{\mathit{\alpha }}}{w_1(\mathit{x})(1-\mathit{x})}}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\mathit{g}(\mathit{y})}{y^{\mathit{\alpha }+1}}}\mathit{dyf}(\mathit{x})w_1(\mathit{x})\mathit{dx}.\end{array}$(5) ) and (6(6) $\begin{array}{ll}<\mathit{g},{\mathbb{H}}_{\mathit{\alpha }}(\mathit{f})>& =\underset{0}{\overset{\mathrm{\infty }}{\int }}<\mathit{g},\mathit{\varpi }(\mathit{x})>\mathit{f}(\mathit{x})w_1(\mathit{x})\mathit{dx}\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathit{g}(\mathit{y})\mathit{d\varpi }(\mathit{x})(\mathit{y})\mathit{f}(\mathit{x})w_1(\mathit{x})\mathit{dx}.\end{array}$(6) ), we obtain

$\begin{array}{ll}\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathit{g}(\mathit{y})\mathit{d\varpi }(\mathit{x})(\mathit{y})& ={\displaystyle \frac{x^{\mathit{\alpha }}}{w_1(\mathit{x})(1-\mathit{x})}}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\mathit{g}(\mathit{y})}{y^{\mathit{\alpha }+1}}}\mathit{dy}\text{ a.e. on }{\mathbb{R}}^{+}\end{array}$

for every $\mathit{g}\in C_0(\frac{1}{w_2}).$ As $\mathit{\varpi }(\mathit{x})$ is an element of $\mathfrak{M}(w_2),$ we get

$\begin{array}{lll}\mathit{d\varpi }(\mathit{x})(\mathit{y})& =& {\displaystyle \frac{x^{\mathit{\alpha }}}{y^{\mathit{\alpha }+1}{w}_1(\mathit{x})(1-\mathit{x})}}{\chi }_{\mathit{y}\ge \mathit{x}}\mathit{dz}\text{ a.e. on }{\mathbb{R}}^{+}\end{array}$

where χ is the characteristic function on the respective domain. Then

$\begin{array}{lll}\Rightarrow \Vert \mathit{\varpi }(\mathit{x}){\Vert }_{L_1(w_2)}& =& \underset{0}{\overset{\mathrm{\infty }}{\int }}{w}_2(\mathit{y})\mathit{d\varpi }(\mathit{x})(\mathit{y})\\ & =& \underset{0}{\overset{\mathrm{\infty }}{\int }}{w}_2(\mathit{y}){\displaystyle \frac{x^{\mathit{\alpha }}}{y^{\mathit{\alpha }+1}{w}_1(\mathit{x})|1-\mathit{x}|}}{\chi }_{\mathit{y}\ge \mathit{x}}\mathit{dy}\\ & =& {\displaystyle \frac{x^{\mathit{\alpha }}}{w_1(\mathit{x})|1-\mathit{x}|}}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{w_2(\mathit{y})}{y^{\mathit{\alpha }+1}}}\mathit{dy}.\end{array}$

By the last equality, we conclude the inequality (3(3) $\begin{array}{l}{x}^{\mathit{\alpha }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{w_2(\mathit{y})}{y^{\mathit{\alpha }+1}}}\mathit{dy}\le \mathit{C}|1-\mathit{x}|w_1(\mathit{x})\text{ a.e. on }{\mathbb{R}}^{+}\text{ for some }\mathit{C}>0.\end{array}$(3) ) holds a.e. on ${\mathbb{R}}^{+},$ if the operator ${\mathbb{H}}_{\mathit{\alpha }}$ is bounded on weighted spaces of the integrable functions from $L_1(w_1)$ to $L_1(w_2).$

Theorem 4.

Let ${\mathbb{H}}_{\mathit{\alpha }}$ be a bounded linear operator from $L_1(w_1)$ to $L_1(w_2)$ such that the condition (3(3) $\begin{array}{l}{x}^{\mathit{\alpha }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{w_2(\mathit{y})}{y^{\mathit{\alpha }+1}}}\mathit{dy}\le \mathit{C}|1-\mathit{x}|w_1(\mathit{x})\text{ a.e. on }{\mathbb{R}}^{+}\text{ for some }\mathit{C}>0.\end{array}$(3) ) holds a.e. on ${\mathbb{R}}^{+}.$ Then there exists an adjoint operator ${{\mathbb{H}}_{\mathit{\alpha }}}^{\ast }:L_{\mathrm{\infty }}(\frac{1}{w_2})\to L_{\mathrm{\infty }}(\frac{1}{w_1})$ such that ${{\mathbb{H}}_{\mathit{\alpha }}}^{\ast }\mathit{g}(\mathit{x})={\displaystyle \frac{x^{\mathit{\alpha }}}{1-\mathit{x}}}{\displaystyle \underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\mathit{g}(\mathit{y})}{y^{\mathit{\alpha }+1}}}\mathit{dy}}$ a.e. on ${\mathbb{R}}^{+}.$

Proof.

For $\mathit{f}\in L_1(w_1)$ and $\mathit{g}\in L_1(w_2),$ we have

$\begin{array}{ll}<{\mathbb{H}}_{\mathit{\alpha }}(\mathit{f}),\mathit{g}>& =\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{1}{\int }}{\displaystyle \frac{t^{\mathit{\alpha }}}{{[(\mathit{t}-1)\mathit{y}+1]}^{\mathit{\alpha }+1}}}\mathit{f}\\ & \left({\displaystyle \frac{\mathit{ty}}{(\mathit{t}-1)\mathit{y}+1}}\right)\mathit{dtg}(\mathit{y})\mathit{dy}\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathit{y}}{\int }}{\displaystyle \frac{x^{\mathit{\alpha }}}{y^{\mathit{\alpha }+1}(1-\mathit{x})}}\mathit{f}(\mathit{x})\mathit{dxg}(\mathit{y})\mathit{dy}\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{x^{\mathit{\alpha }}}{y^{\mathit{\alpha }+1}(1-\mathit{x})}}\mathit{f}(\mathit{x})\mathit{g}(\mathit{y})\mathit{dydx}\\ & =<\mathit{f},{{\mathbb{H}}_{\mathit{\alpha }}}^{\ast }(\mathit{g})>\end{array}$

where ${{\mathbb{H}}_{\mathit{\alpha }}}^{\ast }\mathit{g}(\mathit{x})={\displaystyle \frac{x^{\mathit{\alpha }}}{1-\mathit{x}}}{\displaystyle \underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\mathit{g}(\mathit{y})}{y^{\mathit{\alpha }+1}}}\mathit{dy}}$ a.e. on ${\mathbb{R}}^{+}.$ Thus ${{\mathbb{H}}_{\mathit{\alpha }}}^{\ast }$ is an adjoint operator of ${\mathbb{H}}_{\mathit{\alpha }}$ on $L_1(w_1).$ Also, for $\mathit{g}\in L_{\mathrm{\infty }}\left(\frac{1}{w_2}\right),$ we have

$\begin{array}{lll}\left|{\displaystyle \frac{x^{\mathit{\alpha }}}{1-\mathit{x}}}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\mathit{g}(\mathit{y})}{y^{\mathit{\alpha }+1}}}\mathit{dy}\right|& \le & \Vert \mathit{g}{\Vert }_{{\mathit{L}}_{\mathrm{\infty }}\left(\frac{1}{w_2}\right)}{\displaystyle \frac{x^{\mathit{\alpha }}}{|1-\mathit{x}|}}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{w_2(\mathit{y})}{y^{\mathit{\alpha }+1}}}\mathit{dy}\\ & \le & \Vert \mathit{g}{\Vert }_{{\mathit{L}}_{\mathrm{\infty }}\left(\frac{1}{w_2}\right)}{w}_1(\mathit{x})\text{ a.e. on }{\mathbb{R}}^{+}\end{array}$

the last inequality obtained by the boundedness of ${\mathbb{H}}_{\mathit{\alpha }},$ which concludes the proof the theorem.

4. Boundedness of the operator ${\mathbb{T}}_{\mathit{\alpha }}$ on weighted integrable spaces

Theorem 5.

Let w₁ and w₂ be the non-negative weight functions on ${\mathbb{R}}^{+}.$ The operator ${\mathbb{T}}_{\mathit{\alpha }}$ is bounded linear operator from $L_1(w_1)$ to $L_1(w_2)$ iff

(7) $\begin{array}{l}{x}^{\mathit{\alpha }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{w_2(\mathit{y})}{y^{\mathit{\alpha }}|1-\mathit{y}|}}\mathit{dy}\le \mathit{C}{w}_1(\mathit{x})\text{ a.e. on }{\mathbb{R}}^{+}\text{ for some }\mathit{C}>0.\end{array}$(7)

Proof.

$\begin{array}{ll}\underset{0}{\overset{\mathrm{\infty }}{\int }}|{\mathbb{T}}_{\mathit{\alpha }}\mathit{f}(\mathit{y})|w_2(\mathit{y})\mathit{dy}& =\underset{0}{\overset{\mathrm{\infty }}{\int }}|\underset{0}{\overset{1}{\int }}\frac{(\mathit{ty}+1-\mathit{t}{)}^{\mathit{\alpha }}}{y^{\mathit{\alpha }}}\\ & \phantom{\underset{0}{\overset{1}{\int }}\frac{(\mathit{ty}+1-\mathit{t}{)}^{\mathit{\alpha }}}{y^{\mathit{\alpha }}}}\mathit{f}(\mathit{ty}+1-\mathit{t})\mathit{dt}|w_2(\mathit{y})\mathit{dy}\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}|\underset{1}{\overset{\mathit{y}}{\int }}\frac{x^{\mathit{\alpha }}}{y^{\mathit{\alpha }}}\mathit{f}(\mathit{x}){\displaystyle \frac{\mathit{dx}}{1-\mathit{y}}}|w_2(\mathit{y})\mathit{dy}\\ & \le \underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{1}{\overset{\mathit{y}}{\int }}\frac{x^{\mathit{\alpha }}}{y^{\mathit{\alpha }}}\mathit{f}(\mathit{x}){\displaystyle \frac{\mathit{dx}}{|1-\mathit{y}|}}{w}_2(\mathit{y})\mathit{dy}\\ & \le \underset{0}{\overset{1}{\int }}{x}^{\mathit{\alpha }}\left(\underset{0}{\overset{\mathit{x}}{\int }}\frac{1}{y^{\mathit{\alpha }}}{\displaystyle \frac{w_2(\mathit{y})}{|1-\mathit{y}|}}\mathit{dy}\right)\\ & |\mathit{f}(\mathit{x})|\mathit{dx}\\ & +\underset{1}{\overset{\mathrm{\infty }}{\int }}{x}^{\mathit{\alpha }}\left(\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}\frac{1}{y^{\mathit{\alpha }}}{\displaystyle \frac{w_2(\mathit{y})}{|1-\mathit{y}|}}\mathit{dy}\right)\\ & |\mathit{f}(\mathit{x})|\mathit{dx}\\ & \le \mathit{C}\underset{0}{\overset{1}{\int }}|\mathit{f}(\mathit{x})|w_1(\mathit{x})\mathit{dx}\\ & +\mathit{C}\underset{1}{\overset{\mathrm{\infty }}{\int }}|\mathit{f}(\mathit{x})|w_1(\mathit{x})\mathit{dx}\end{array}$

by the assumption (7(7) $\begin{array}{l}{x}^{\mathit{\alpha }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{w_2(\mathit{y})}{y^{\mathit{\alpha }}|1-\mathit{y}|}}\mathit{dy}\le \mathit{C}{w}_1(\mathit{x})\text{ a.e. on }{\mathbb{R}}^{+}\text{ for some }\mathit{C}>0.\end{array}$(7) ), we obtained ${\mathbb{T}}_{\mathit{\alpha }}$ is bounded operator from $L_1(w_1)$ to $L_1(w_2).$ Conversely, assume ${\mathbb{T}}_{\mathit{\alpha }}$ is bounded linear operator on the respective domain. Let $\mathit{g}\in C_0(\frac{1}{w_2}),$ then

(8) $\begin{array}{ll}<\mathit{g},{\mathbb{T}}_{\mathit{\alpha }}(\mathit{f})>& =\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathit{g}(\mathit{y}){\mathbb{T}}_{\mathit{\alpha }}\mathit{f}(\mathit{y})\mathit{dy}\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathit{g}(\mathit{y}){\displaystyle \frac{1}{{\mathit{y}}^{\mathit{\alpha }}(1-\mathit{y})}}\underset{1}{\overset{\mathit{y}}{\int }}{x}^{\mathit{\alpha }}\mathit{f}(\mathit{x})\mathit{dxdy}\\ & =(\underset{0}{\overset{1}{\int }}\underset{0}{\overset{\mathit{x}}{\int }}+\underset{1}{\overset{\mathrm{\infty }}{\int }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }})\mathit{g}(\mathit{y}){\displaystyle \frac{1}{{\mathit{y}}^{\mathit{\alpha }}(1-\mathit{y})}}\\ & x^{\mathit{\alpha }}\mathit{f}(\mathit{x})\mathit{dydx}\\ & =\underset{0}{\overset{1}{\int }}{x}^{\mathit{\alpha }}\underset{0}{\overset{\mathit{x}}{\int }}{\displaystyle \frac{\mathit{g}(\mathit{y})}{{\mathit{y}}^{\mathit{\alpha }}(1-\mathit{y})}}\mathit{dyf}(\mathit{x})\mathit{dx}\\ & +\underset{1}{\overset{\mathrm{\infty }}{\int }}{x}^{\mathit{\alpha }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\mathit{g}(\mathit{y})}{{\mathit{y}}^{\mathit{\alpha }}(1-\mathit{y})}}\mathit{dyf}(\mathit{x})\mathit{dx}\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathit{K}(\mathit{x},\mathit{y})\mathit{g}(\mathit{y})\mathit{dyf}(\mathit{x})w_1(\mathit{x})\mathit{dx}\end{array}$(8)

where

$\begin{array}{l}\mathit{K}(\mathit{x},\mathit{y})=\{\begin{array}{ll}{\displaystyle \frac{x^{\mathit{\alpha }}}{w_1(\mathit{x})}}{\displaystyle \frac{1}{{\mathit{y}}^{\mathit{\alpha }}(1-\mathit{y})}}{\chi }_{0<\mathit{y}<\mathit{x}}& 0<\mathit{x}<1,\\ {\displaystyle \frac{x^{\mathit{\alpha }}}{w_1(\mathit{x})}}{\displaystyle \frac{1}{{\mathit{y}}^{\mathit{\alpha }}(1-\mathit{y})}}{\chi }_{\mathit{x}<\mathit{y}<\mathrm{\infty }}& 1<\mathit{x}<\mathrm{\infty }\end{array}\end{array}$

Similar to the steps as in Theorem 3, we obtain

(9) $\begin{array}{l}<\mathit{g},{\mathbb{T}}_{\mathit{\alpha }}(\mathit{f})>=\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathit{g}(\mathit{y})\mathit{d\varpi }(\mathit{x})(\mathit{y})\mathit{f}(\mathit{x})w_1(\mathit{x})\mathit{dx}\end{array}$(9)

for every $\mathit{g}\in C_0(\frac{1}{w_2}).$ Equating (8(8) $\begin{array}{ll}<\mathit{g},{\mathbb{T}}_{\mathit{\alpha }}(\mathit{f})>& =\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathit{g}(\mathit{y}){\mathbb{T}}_{\mathit{\alpha }}\mathit{f}(\mathit{y})\mathit{dy}\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathit{g}(\mathit{y}){\displaystyle \frac{1}{{\mathit{y}}^{\mathit{\alpha }}(1-\mathit{y})}}\underset{1}{\overset{\mathit{y}}{\int }}{x}^{\mathit{\alpha }}\mathit{f}(\mathit{x})\mathit{dxdy}\\ & =(\underset{0}{\overset{1}{\int }}\underset{0}{\overset{\mathit{x}}{\int }}+\underset{1}{\overset{\mathrm{\infty }}{\int }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }})\mathit{g}(\mathit{y}){\displaystyle \frac{1}{{\mathit{y}}^{\mathit{\alpha }}(1-\mathit{y})}}\\ & x^{\mathit{\alpha }}\mathit{f}(\mathit{x})\mathit{dydx}\\ & =\underset{0}{\overset{1}{\int }}{x}^{\mathit{\alpha }}\underset{0}{\overset{\mathit{x}}{\int }}{\displaystyle \frac{\mathit{g}(\mathit{y})}{{\mathit{y}}^{\mathit{\alpha }}(1-\mathit{y})}}\mathit{dyf}(\mathit{x})\mathit{dx}\\ & +\underset{1}{\overset{\mathrm{\infty }}{\int }}{x}^{\mathit{\alpha }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\mathit{g}(\mathit{y})}{{\mathit{y}}^{\mathit{\alpha }}(1-\mathit{y})}}\mathit{dyf}(\mathit{x})\mathit{dx}\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathit{K}(\mathit{x},\mathit{y})\mathit{g}(\mathit{y})\mathit{dyf}(\mathit{x})w_1(\mathit{x})\mathit{dx}\end{array}$(8) ) and (9(9) $\begin{array}{l}<\mathit{g},{\mathbb{T}}_{\mathit{\alpha }}(\mathit{f})>=\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathit{g}(\mathit{y})\mathit{d\varpi }(\mathit{x})(\mathit{y})\mathit{f}(\mathit{x})w_1(\mathit{x})\mathit{dx}\end{array}$(9) ), we obtain

(10) $\begin{array}{ll}\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathit{g}(\mathit{y})\mathit{d\varpi }(\mathit{x})(\mathit{y})& =\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathit{g}(\mathit{y})\mathit{K}(\mathit{x},\mathit{y})\mathit{dy}\\ \Rightarrow \mathit{d\varpi }(\mathit{x})(\mathit{y})& =\mathit{K}(\mathit{x},\mathit{y})\text{ a.e. on }{\mathbb{R}}^{+}.\end{array}$(10)

As $\mathit{\varpi }\in L_1(w_2),$ we have

$\begin{array}{lll}\Rightarrow {\displaystyle \frac{x^{\mathit{\alpha }}}{w_1(\mathit{x})}}\underset{0}{\overset{\mathit{x}}{\int }}{\displaystyle \frac{w_2(\mathit{y})}{y^{\mathit{\alpha }}(1-\mathit{y})}}\mathit{dy}& \le & \mathit{C}\text{ for }0\le \mathit{x}<1\\ \text{and}{\displaystyle \frac{x^{\mathit{\alpha }}}{w_1(\mathit{x})}}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{w_2(\mathit{y})}{y^{\mathit{\alpha }}|1-\mathit{y}|}}\mathit{dy}& \le & \mathit{C}\text{ for }1\le \mathit{x}<\mathrm{\infty }\end{array}$

from the notation of $\mathit{K}(\mathit{x},\mathit{y}),$ which concludes the proof of the theorem.

Theorem 6.

Let ${\mathbb{T}}_{\mathit{\alpha }}$ be a bounded linear operator from $L_1(w_1)$ to $L_1(w_2)$ such that the condition $x^{\mathit{\alpha }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{w_2(\mathit{y})}{y^{\mathit{\alpha }}|1-\mathit{y}|}}\mathit{dy}\le w_1(\mathit{x})$ holds a.e. on ${\mathbb{R}}^{+}.$ Then there exists an adjoint operator ${{\mathbb{T}}_{\mathit{\alpha }}}^{\ast }:L_{\mathrm{\infty }}(\frac{1}{w_2})\to L_{\mathrm{\infty }}(\frac{1}{w_1})$ such that

$\begin{array}{l}{{\mathbb{T}}_{\mathit{\alpha }}}^{\ast }\mathit{g}(\mathit{x})=\{\begin{array}{ll}{x}^{\mathit{\alpha }}{\displaystyle \underset{0}{\overset{\mathit{x}}{\int }}{\displaystyle \frac{1}{{\mathit{y}}^{\mathit{\alpha }}(\mathit{y}-1)}}\mathit{g}(\mathit{y})\mathit{dy}}& \text{for }0\le \mathit{x}<1,\\ x^{\mathit{\alpha }}{\displaystyle \underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{1}{{\mathit{y}}^{\mathit{\alpha }}(\mathit{y}-1)}}\mathit{g}(\mathit{y})\mathit{dy}}& \text{for }1\le \mathit{x}<\mathrm{\infty }.\end{array}\end{array}$

Proof.

For $\mathit{f}\in L_1(w_1)$ and $\mathit{g}\in L_1(w_2),$ we have

$\begin{array}{ll}<{\mathbb{T}}_{\mathit{\alpha }}(\mathit{f}),\mathit{g}>& =\underset{0}{\overset{\mathrm{\infty }}{\int }}{\mathbb{T}}_{\mathit{\alpha }}(\mathit{f})(\mathit{y})\mathit{g}(\mathit{y})\mathit{dy}\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{1}{\int }}{\displaystyle \frac{(\mathit{ty}+1-\mathit{t}{)}^{\mathit{\alpha }}}{y^{\mathit{\alpha }}}}\mathit{f}((\mathit{t}-1)\mathit{y}+1)\mathit{dt}\\ & \mathit{g}(\mathit{y})\mathit{dy}\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{1}{\overset{\mathit{y}}{\int }}{\displaystyle \frac{x^{\mathit{\alpha }}}{y^{\mathit{\alpha }}(\mathit{y}-1)}}\mathit{f}(\mathit{x})\mathit{dx}\mathit{g}(\mathit{y})\mathit{dy}\\ & =\underset{0}{\overset{1}{\int }}\underset{0}{\overset{\mathit{x}}{\int }}{\displaystyle \frac{x^{\mathit{\alpha }}}{y^{\mathit{\alpha }}(\mathit{y}-1)}}\mathit{g}(\mathit{y})\mathit{dyf}(\mathit{x})\mathit{dx}\\ & +\underset{1}{\overset{\mathrm{\infty }}{\int }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{x^{\mathit{\alpha }}}{y^{\mathit{\alpha }}(\mathit{y}-1)}}\mathit{g}(\mathit{y})\mathit{dyf}(\mathit{x})\mathit{dx}\\ & =<\mathit{f},{{\mathbb{T}}_{\mathit{\alpha }}}^{\ast }\mathit{g}>\end{array}$

where

$\begin{array}{l}{{\mathbb{T}}_{\mathit{\alpha }}}^{\ast }\mathit{g}(\mathit{x})=\{\begin{array}{ll}{x}^{\mathit{\alpha }}{\displaystyle \underset{0}{\overset{\mathit{x}}{\int }}{\displaystyle \frac{1}{{\mathit{y}}^{\mathit{\alpha }}(\mathit{y}-1)}}\mathit{g}(\mathit{y})\mathit{dy}}& 0\le \mathit{x}<1\\ x^{\mathit{\alpha }}{\displaystyle \underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{1}{{\mathit{y}}^{\mathit{\alpha }}(\mathit{y}-1)}}\mathit{g}(\mathit{y})\mathit{dy}}& 1\le \mathit{x}<\mathrm{\infty }.\end{array}\end{array}$

Thus ${{\mathbb{T}}_{\mathit{\alpha }}}^{\ast }$ is an adjoint operator of ${\mathbb{T}}_{\mathit{\alpha }}$ on $L_1(w_1).$ Also, for $\mathit{g}\in L_{\mathrm{\infty }}\left(\frac{1}{w_2}\right)$ and $0\le \mathit{x}<1,$ we have

(11) $\begin{array}{ll}\left|x^{\mathit{\alpha }}\underset{0}{\overset{\mathit{x}}{\int }}{\displaystyle \frac{\mathit{g}(\mathit{y})}{{\mathit{y}}^{\mathit{\alpha }}(\mathit{y}-1)}}\mathit{dy}\right|& \le \Vert \mathit{g}{\Vert }_{{\mathit{L}}_{\mathrm{\infty }}\left(\frac{1}{w_2}\right)}{x}^{\mathit{\alpha }}\underset{0}{\overset{1}{\int }}{\displaystyle \frac{w_2(\mathit{y})}{y^{\mathit{\alpha }}|1-\mathit{y}|}}\mathit{dy}\\ & \le \Vert \mathit{g}{\Vert }_{{\mathit{L}}_{\mathrm{\infty }}\left(\frac{1}{w_2}\right)}{w}_1(\mathit{x})\text{ a.e. on }{\mathbb{R}}^{+}\end{array}$(11)

For $1\le \mathit{x}<\mathrm{\infty },$ we have

(12) $\begin{array}{ll}\left|x^{\mathit{\alpha }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\mathit{g}(\mathit{y})}{{\mathit{y}}^{\mathit{\alpha }}(\mathit{y}-1)}}\mathit{dy}\right|& \le \Vert \mathit{g}{\Vert }_{{\mathit{L}}_{\mathrm{\infty }}\left(\frac{1}{w_2}\right)}{x}^{\mathit{\alpha }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{w_2(\mathit{y})}{y^{\mathit{\alpha }}|1-\mathit{y}|}}\mathit{dy}\\ & \le \Vert \mathit{g}{\Vert }_{{\mathit{L}}_{\mathrm{\infty }}\left(\frac{1}{w_2}\right)}{w}_1(\mathit{x})\text{ a.e. on }{\mathbb{R}}^{+}\end{array}$(12)

the last two inequalities (11(11) $\begin{array}{ll}\left|x^{\mathit{\alpha }}\underset{0}{\overset{\mathit{x}}{\int }}{\displaystyle \frac{\mathit{g}(\mathit{y})}{{\mathit{y}}^{\mathit{\alpha }}(\mathit{y}-1)}}\mathit{dy}\right|& \le \Vert \mathit{g}{\Vert }_{{\mathit{L}}_{\mathrm{\infty }}\left(\frac{1}{w_2}\right)}{x}^{\mathit{\alpha }}\underset{0}{\overset{1}{\int }}{\displaystyle \frac{w_2(\mathit{y})}{y^{\mathit{\alpha }}|1-\mathit{y}|}}\mathit{dy}\\ & \le \Vert \mathit{g}{\Vert }_{{\mathit{L}}_{\mathrm{\infty }}\left(\frac{1}{w_2}\right)}{w}_1(\mathit{x})\text{ a.e. on }{\mathbb{R}}^{+}\end{array}$(11) ) and (12(12) $\begin{array}{ll}\left|x^{\mathit{\alpha }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\mathit{g}(\mathit{y})}{{\mathit{y}}^{\mathit{\alpha }}(\mathit{y}-1)}}\mathit{dy}\right|& \le \Vert \mathit{g}{\Vert }_{{\mathit{L}}_{\mathrm{\infty }}\left(\frac{1}{w_2}\right)}{x}^{\mathit{\alpha }}\underset{\mathit{x}}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{w_2(\mathit{y})}{y^{\mathit{\alpha }}|1-\mathit{y}|}}\mathit{dy}\\ & \le \Vert \mathit{g}{\Vert }_{{\mathit{L}}_{\mathrm{\infty }}\left(\frac{1}{w_2}\right)}{w}_1(\mathit{x})\text{ a.e. on }{\mathbb{R}}^{+}\end{array}$(12) ) obtained by the boundedness of ${\mathbb{T}}_{\mathit{\alpha }}.$

Disclosure statement

No potential conflict of interest was reported by the author(s).

Additional information

Funding

The work was supported by the VIT University.