1. Introduction

Well known is the fundamental role of the classical Bessel potentials in the general theory of function spaces of fractional smoothness as well as applications to the theory of partial equations (see [1, 2]). The classical Bessel potentials are constructed on using the Fourier transform.

The goal of this article is to develop the theory of the space of generalized Bessel potentials \( \mathbf{B}^{\alpha}_{\gamma} \) constructed with the Hankel transform. This space was first introduced by Lyakhov in [3] who based on the Stein–Lizorkin approach. In [3], the \( B \)-hypersingular integrals and \( B \)-Riesz potentials, introduced earlier by Lyakhov in [4, 5], were applied the constructing the norm on \( \mathbf{B}^{\alpha}_{\gamma} \). In the present article, we use another approach to \( \mathbf{B}^{\alpha}_{\gamma} \) which is based on the works [6,7,8] by Aronszajn and Smith. The approach consists in introducing some norm on \( \mathbf{B}^{\alpha}_{\gamma} \) with the help of weighted Dirichlet integrals.

The spaces of generalized Bessel potentials of arbitrary order \( \alpha \) are necessary for defining the classes of solutions to the boundary value problem

$$ Au=f\text{ in }D,\quad B_{i}u=0\text{ on }\partial D, $$

where \( A \) is an elliptic operator containing the Bessel differential operators \( B_{\gamma_{i}}=\frac{\partial^{2}}{\partial x_{i}^{2}}+\frac{\gamma_{i}}{x_{i}}\frac{\partial}{\partial x_{i}} \); in particular, \( A \) can be the Laplace–Bessel operator \( \Delta_{\gamma}=\sum_{i=1}^{n}B_{\gamma_{i}} \).

2. Preliminaries

Let \( 𝕉^{n} \) be the \( n \)-dimensional Euclidean space, while

$$ \begin{gathered}\displaystyle 𝕉^{n}_{+}=\big{\{}x=(x_{1},\dots,x_{n})\in 𝕉^{n},\ x_{1}>0,\dots,x_{n}>0\big{\}},\\ \displaystyle\overline{𝕉}\,^{n}_{+}=\big{\{}x=(x_{1},\dots,x_{n})\in 𝕉^{n},\ x_{1}\geq 0,\,\dots,\,x_{n}\geq 0\big{\}}.\end{gathered} $$

Let \( \gamma=(\gamma_{1},\dots,\gamma_{n}) \) be the multi-index consisting of fixed positive reals \( \gamma_{i} \), with \( i=1,\dots,n \), and \( |\gamma|=\gamma_{1}+\dots+\gamma_{n} \).

Let \( \Omega \) be a finite or infinite open set in \( 𝕉^{n} \) symmetric with respect to each of the hyperplanes \( x_{i}=0 \), with \( i=1,\dots,n \), \( \Omega_{+}=\Omega\cap{𝕉}^{n}_{+} \), and \( \overline{\Omega}_{+}=\Omega\cap\overline{𝕉}^{n}_{+} \). We will work with the function class \( C^{m}(\Omega_{+}) \) consisting of \( m \) times differentiable functions on \( \Omega_{+} \). Denote by \( C^{m}(\overline{\Omega}_{+}) \) the subset of the functions in \( C^{m}(\Omega_{+}) \) whose all derivatives with respect to \( x_{i} \) extend continuously to \( x_{i}=0 \) for all \( i=1,\dots,n \). Let \( C^{m}_{ev}(\overline{\Omega}_{+}) \) consist of the functions in \( C^{m}(\overline{\Omega}_{+}) \) such that

$$ \frac{\partial^{2k+1}f}{\partial x_{i}^{2k+1}}\big{|}_{x_{i}=0}=0 $$

for all nonnegative integers \( k\leq\frac{m-1}{2} \) (see [9, p. 21]). In what follows, we use the abbreviation \( C^{m}_{ev} \) for \( C^{m}_{ev}(\overline{𝕉}^{n}_{+}) \) and put

$$ C^{\infty}_{ev}(\overline{\Omega}_{+})=\bigcap C^{m}_{ev}(\overline{\Omega}_{+}), $$

where the intersection is taken over all finite \( m \) and \( C^{\infty}_{ev}(\overline{𝕉}_{+})=C^{\infty}_{ev} \).

Let \( {\overset{\circ}{C}}\,^{\infty}_{ev}(\,\overline{\Omega}_{+}) \) be the space of compactly supported functions \( f\in C^{\infty}_{ev}(\,\overline{\Omega}_{+}) \). Put

$$ {\overset{\circ}{C}}\,^{\infty}_{ev}(\,\overline{\Omega}_{+})=\mathcal{D}_{+}(\,\overline{\Omega}_{+})\quad\text{and}\quad{\overset{\circ}{C}}\,^{\infty}_{ev}(\,\overline{𝕉}_{+})={\overset{\circ}{C}}\,^{\infty}_{ev}. $$

Let \( L_{p}^{\gamma}({𝕉}^{n}_{+})=L_{p}^{\gamma} \), with \( 1\leq p<\infty \), consists of the measurable functions on \( {𝕉}^{n}_{+} \) even in each of the variables \( x_{i} \), with \( i=1,\dots,n \), such that

$$ \int\limits_{{𝕉}^{n}_{+}}|f(x)|^{p}x^{\gamma}dx<\infty. $$

Here and in the sequel \( x^{\gamma}=\prod_{i=1}^{n}x_{i}^{\gamma_{i}} \). For every \( p\geq 1 \), the norm of \( f\in L_{p}^{\gamma} \) is defined as

$$ \|f\|_{L_{p}^{\gamma}({𝕉}^{n}_{+})}=\|f\|_{p,\gamma}=\Bigg{(}\,\int\limits_{{𝕉}^{n}_{+}}|f(x)|^{p}x^{\gamma}dx\Bigg{)}^{\frac{1}{p}}. $$

It is know that \( L_{p}^{\gamma} \) is a Banach space (see [9]).

The multidimensional Hankel transform of \( f\in L^{\gamma}_{1}(𝕉^{n}_{+}) \) is defined as

$$ \mathbf{F}_{\gamma}[f](\xi)=\mathbf{F}_{\gamma}[f(x)](\xi)=\widehat{f}(\xi)=\int\limits_{𝕉^{n}_{+}}f(x)\,\mathbf{j}_{\gamma}(x;\xi)x^{\gamma}dx, $$

where

$$ \mathbf{j}_{\gamma}(x;\xi)=\prod_{i=1}^{n}j_{\frac{\gamma_{i}-1}{2}}(x_{i}\xi_{i}). $$

Denote by \( j_{\nu} \) the normalized Bessel function of the first kind \( j_{\nu}(x)=\frac{2^{\nu}\Gamma(\nu+1)}{x^{\nu}}J_{\nu}(x) \), where \( J_{\nu} \) is a Bessel function of the first kind (see [10]).

Let \( f\in L_{1}^{\gamma}(𝕉_{+}) \) be a function of bounded variation in a neighborhood of a continuity point \( x \) of \( f \). Then for \( \gamma>0 \) the inversion formula for the transformation looks as

$$ \mathbf{F}^{-1}_{\gamma}\big{[}\widehat{f}(\xi)\big{]}(x)=f(x)=\frac{2^{n-|\gamma|}}{\prod\limits_{j=1}^{n}\Gamma^{2}\big{(}\frac{\gamma_{j}+1}{2}\big{)}}\,\int\limits_{𝕉^{n}_{+}}\mathbf{j}_{\gamma}(x,\xi)\widehat{f}(\xi)\,\xi^{\gamma}d\xi. $$

The Hankel transform reduces the Bessel operator to the multiplication of the square of the corresponding argument with the minus sign (see [9]):

$$ {F}_{\gamma_{i}}\big{[}(B_{\gamma_{i}})_{x_{i}}f\big{]}(\xi)=-|\xi_{i}|^{2}{F}_{\gamma_{i}}[f](\xi), $$
(1)

where \( (B_{\gamma_{i}})_{x_{i}}=\frac{\partial^{2}}{\partial x_{i}^{2}}+\frac{\gamma_{i}}{x_{i}}\frac{\partial}{\partial x_{i}} \) is the Bessel operator, \( i=1,\dots,n \).

Parseval’s identity for the one-dimensional Hankel transform (see [9, p. 20]) and the fact that \( f\in L_{2}^{\gamma}(𝕉^{n}_{+}) \) yield \( \mathbf{F}_{\gamma}f\in L_{2}^{\gamma}(𝕉^{n}_{+}) \) and

$$ \int\limits_{𝕉^{n}_{+}}\big{|}\mathbf{F}_{\gamma}[f](\xi)\big{|}^{2}\xi^{\gamma}d\xi=2^{|\gamma|-n}\prod\limits_{j=1}^{n}\,\Gamma^{2}\Big{(}\frac{\gamma_{j}+1}{2}\Big{)}\int\limits_{𝕉^{n}_{+}}\big{|}f(x)\big{|}^{2}x^{\gamma}dx. $$
(2)

The multidimensional generalized shift is defined as

$$ (^{\gamma}\mathbf{T}^{y}_{x}f)(x)={\,{}^{\gamma}\mathbf{T}^{y}_{x}}f(x)=({{}^{\gamma_{1}}T_{x_{1}}^{y_{1}}}\dots{{}^{\gamma_{n}}T_{x_{n}}^{y_{n}}}f)(x), $$
(3)

where each one-dimensional generalized shift \( {\,{}^{\gamma_{i}}T_{x_{i}}^{y_{i}}} \), \( i=1,\dots,n \), acts by the formula

$$ (^{\gamma_{i}}T_{x_{i}}^{y_{i}}f)(x)=\frac{\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}{\sqrt{\pi}\,\Gamma\big{(}\frac{\gamma_{i}}{2}\big{)}}\int\limits_{0}^{\pi}f\big{(}x_{1},\dots,x_{i-1},\sqrt{x_{i}^{2}+\tau_{i}^{2}-2x_{i}y_{i}\cos\varphi_{i}},x_{i+1},\dots,x_{n}\big{)}\sin^{\gamma_{i}-1}\varphi_{i}\,d\varphi_{i}. $$

In what follows, put

$$ C(\gamma)=\pi^{-\frac{n}{2}}\prod^{n}_{i=1}\frac{\Gamma(\frac{\gamma_{i}+1}{2})}{\Gamma(\frac{\gamma_{i}}{2})}. $$

The generalized convolution generated by the multidimensional generalized shift \( {}^{\gamma}\mathbf{T}^{y}_{x} \) has the form

$$ (f*g)_{\gamma}(x)=(f*g)_{\gamma}=\int\limits_{𝕉^{n}_{+}}f(y)(^{\gamma}\mathbf{T}^{y}_{x}g)(x)\,y^{\gamma}dy. $$
(4)

The multidimensional Poisson operator \( \mathbf{P}^{\gamma}_{x} \) acts an integrable function \( f \) by the formula

$$ \mathbf{P}_{x}^{\gamma}f(x)=C(\gamma)\int\limits_{0}^{\pi}\dots\int\limits_{0}^{\pi}f(x_{1}\cos\alpha_{1},\dots,x_{n}\cos\alpha_{n})\prod\limits_{i=1}^{n}\sin^{\gamma_{i}-1}\alpha_{i}\,d\alpha_{i}. $$
(5)

3. The Weighted Dirichlet Integral

Let \( i=(i_{1},\dots,i_{m}) \) be a multi-index consisting of the integers from \( 1 \) to \( n \), with \( {|i|=i_{1}+\dots+i_{n}} \); while \( \xi^{i}=\prod_{k=1}^{m}\xi_{ik} \), \( \xi=(\xi_{1},\dots,\xi_{n}) \), and

$$ 𝔹_{i}=(B_{\gamma_{i_{m}}})_{x_{i_{m}}}\dots(B_{\gamma_{i_{1}}})_{x_{i_{1}}}, $$

where \( (B_{\gamma_{i_{k}}})_{x_{i_{k}}}=\frac{\partial^{2}}{\partial x_{i_{k}}^{2}}+\frac{\gamma_{i_{k}}}{x_{i_{k}}}\frac{\partial}{\partial x_{i_{k}}} \) is a Bessel operator for all \( k=1,\dots,m \). Given an integer \( \alpha\geq 0 \), define the weighted Dirichlet integral of order \( \alpha \) as

$$ d_{\alpha,\gamma}(u)=\sum\limits_{|i|=\alpha}\;\int\limits_{𝕉^{n}_{+}}|𝔹_{i}u|^{2}x^{\gamma}dx. $$

In the image of (1), we obtain

$$ d_{\alpha,\gamma}(u)=\int\limits_{𝕉^{n}_{+}}|\xi|^{4\alpha}\,\big{|}\mathbf{F}_{\gamma}[u](\xi)\big{|}^{2}\,\xi^{\gamma}d\xi. $$
(6)

Formula (6) can be used for defining the weighted Dirichlet integral \( d_{\alpha}^{\gamma} \) for an arbitrary \( \alpha\geq 0 \).

Lemma 1

Given \( 0<\alpha<1/2 \) and \( u\in{\overset{\circ}{C}}\,^{\infty}_{ev}(𝕉^{n}_{+}) \), we have

$$ \begin{gathered}\displaystyle d_{\alpha,\gamma}(u)=\int\limits_{𝕉^{n}_{+}}|\xi|^{4\alpha}\,\big{|}\mathbf{F}_{\gamma}[u](\xi)\big{|}^{2}\,\xi^{\gamma}d\xi\\ \displaystyle=\frac{1}{C(n,\gamma,\alpha)}\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\big{|}u(y)-u(x)\big{|}^{2}\Big{(}{\,{}^{\gamma}\mathbf{T}_{y}^{x}}\,\frac{1}{|y|^{n+|\gamma|+4\alpha}}\Big{)}x^{\gamma}y^{\gamma}dxdy,\end{gathered} $$
(7)

where

$$ C(n,\gamma,\alpha)=\frac{2^{1-|\gamma|-4\alpha}\pi}{\sin(2\alpha\pi)\,\Gamma(2\alpha+1)\,\Gamma\big{(}\frac{n+|\gamma|}{2}+2\alpha\big{)}\prod\limits_{i=1}^{n}\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}. $$

Proof

If

$$ \mathbf{I}=\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\,\frac{\big{|}{\,{}^{\gamma}\mathbf{T}_{x}^{y}}u(x)-u(x)\big{|}^{2}}{|y|^{n+|\gamma|+4\alpha}}\,x^{\gamma}y^{\gamma}dxdy, $$

then

$$ \begin{gathered}\displaystyle\mathbf{I}=C(\gamma)\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\int\limits_{0}^{\pi}\dots\int\limits_{0}^{\pi}\frac{\big{|}u\big{(}\sqrt{x_{1}^{2}-2x_{1}y_{1}\cos\beta_{1}+y_{1}^{2}},\dots,\sqrt{x_{n}^{2}-2x_{n}y_{n}\cos\beta_{1}+y_{n}^{2}}\,\big{)}-u(x)\big{|}^{2}}{|y|^{n+|\gamma|+4\alpha}}\\ \displaystyle\times\,\prod\limits_{i=1}^{n}\sin^{\gamma_{i}-1}\beta_{i}\,d\beta_{i}\,x^{\gamma}y^{\gamma}dxdy.\end{gathered} $$

Passing to the coordinates

$$ \begin{gathered}\displaystyle\widetilde{y}_{1}=y_{1}\cos\beta_{1},\quad\widetilde{y}_{2}=y_{1}\sin\beta_{1},\quad\widetilde{y}_{3}=y_{2}\cos\beta_{2},\\ \displaystyle\widetilde{y}_{4}=y_{2}\sin\beta_{2},\dots,\widetilde{y}_{2n-1}=y_{n}\cos\beta_{n},\quad\widetilde{y}_{2n}=y_{n}\sin\beta_{n},\end{gathered} $$

we infer that

$$ \begin{gathered}\displaystyle\mathbf{I}=C(\gamma)\int\limits_{𝕉^{n}_{+}}\int\limits_{\widetilde{𝕉}^{2n}_{+}}\frac{\big{|}u\big{(}\sqrt{(x_{1}-\widetilde{y}_{1})^{2}+\widetilde{y}_{2}^{\,2}},\dots,\sqrt{(x_{n}-\widetilde{y}_{2n-1})^{2}+\widetilde{y}_{2n}^{\,2}}\,\big{)}-u(x)\big{|}^{2}}{|\widetilde{y}\,|^{n+|\gamma|+4\alpha}}\,\prod\limits_{i=1}^{n}y_{2i}^{\gamma_{i}-1}x^{\gamma}d\widetilde{y}dx\\ \displaystyle=C(\gamma)\int\limits_{𝕉^{n}_{+}}\int\limits_{\widetilde{𝕉}^{2n}_{+}}\frac{\Big{|}u\Big{(}\sqrt{z_{1}^{2}+\widetilde{y}_{2}^{\,2}},\dots,\sqrt{z_{2n-1}^{2}+\widetilde{y}_{2n}^{\,2}}\,\Big{)}-u(x)\Big{|}^{2}}{\big{(}(x_{1}-z_{1})^{2}+\widetilde{y}_{2}^{\,2}+\dots+(x_{n}-z_{2n-1})^{2}+\widetilde{y}_{2n}^{\,2}\big{)}^{\frac{n+|\gamma|+4\alpha}{2}}}\\ \displaystyle\times\,\prod\limits_{i=1}^{n}y_{2i}^{\gamma_{i}-1}x^{\gamma}dz_{1}d\widetilde{y}_{2}\dots dz_{2n-1}\widetilde{y}_{2n}\,dx,\end{gathered} $$

where \( \{\widetilde{y}_{2i-1}-x_{i}=z_{2i-1},\ i=1,\dots,n\} \) and \( \widetilde{𝕉}^{2n}_{+}=\{\widetilde{y}\in 𝕉^{2n}:{\widetilde{y}}_{2i}>0,\ i=1,\dots,n\} \). Putting \( z_{2i-1}=y_{i}\cos\beta_{i} \) and \( \widetilde{y}_{2i}=y_{i}\sin\beta_{i} \), with \( i=1,\dots,n \), we obtain

$$ \begin{gathered}\displaystyle\mathbf{I}=C(\gamma)\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\int\limits_{0}^{\pi}\dots\int\limits_{0}^{\pi}\frac{\big{|}u(y)-u(x)\big{|}^{2}}{\big{(}x_{1}^{2}-2x_{1}y_{1}\cos\beta_{1}+y_{1}^{2}+\dots+x_{n}^{2}-2x_{n}y_{n}\cos\beta_{1}+y_{n}^{2}\big{)}^{\frac{n+|\gamma|+4\alpha}{2}}}\\ \displaystyle\times\,\prod\limits_{i=1}^{n}\sin^{\gamma_{i}-1}\beta_{i}\,d\beta_{i}\,x^{\gamma}y^{\gamma}dxdy=\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\big{|}u(y)-u(x)\big{|}^{2}\,\Big{(}{{}^{\gamma}\mathbf{T}_{y}^{x}}\frac{1}{|y|^{n+|\gamma|+4\alpha}}\Big{)}\,x^{\gamma}y^{\gamma}dxdy.\end{gathered} $$

Therefore,

$$ \begin{gathered}\displaystyle\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\frac{\big{|}{{}^{\gamma}\mathbf{T}_{x}^{y}}u(x)-u(x)\big{|}^{2}}{|y|^{n+|\gamma|+4\alpha}}\,x^{\gamma}y^{\gamma}dxdy\\ \displaystyle=\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\big{|}u(y)-u(x)\big{|}^{2}\,\Big{(}{{}^{\gamma}\mathbf{T}_{y}^{x}}\frac{1}{|y|^{n+|\gamma|+4\alpha}}\Big{)}\,x^{\gamma}y^{\gamma}dxdy.\end{gathered} $$
(8)

If \( 0<\alpha<1/2 \); then, using Parseval’s identity for (2) and formula 3.170 in [11, p. 155], we get

$$ \begin{gathered}\displaystyle\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\frac{\big{|}{{}^{\gamma}\mathbf{T}_{x}^{y}}u(x)-u(x)\big{|}^{2}}{|y|^{n+|\gamma|+4\alpha}}\,x^{\gamma}y^{\gamma}dxdy\\ \displaystyle=\int\limits_{𝕉^{n}_{+}}\frac{y^{\gamma}dy}{{|y|^{n+|\gamma|+4\alpha}}}\int\limits_{𝕉^{n}_{+}}\big{|}{{}^{\gamma}\mathbf{T}_{x}^{y}}u(x)-u(x)\big{|}^{2}\,x^{\gamma}dx\\ \displaystyle=\frac{2^{n-|\gamma|}}{\prod\limits_{j=1}^{n}\Gamma^{2}\big{(}\frac{\gamma_{j}+1}{2}\big{)}}\,\int\limits_{𝕉^{n}_{+}}\frac{y^{\gamma}dy}{{|y|^{n+|\gamma|+4\alpha}}}\int\limits_{𝕉^{n}_{+}}\big{|}(\mathbf{F}_{\gamma})_{x}\big{[}{{}^{\gamma}\mathbf{T}_{x}^{y}}u(x)\big{]}(\xi)-\mathbf{F}_{\gamma}[u](\xi)\big{|}^{2}\,\xi^{\gamma}d\xi\\ \displaystyle=\frac{2^{n-|\gamma|}}{\prod\limits_{j=1}^{n}\Gamma^{2}\big{(}\frac{\gamma_{j}+1}{2}\big{)}}\,\int\limits_{𝕉^{n}_{+}}\frac{y^{\gamma}dy}{{|y|^{n+|\gamma|+4\alpha}}}\int\limits_{𝕉^{n}_{+}}\big{|}\mathbf{j}_{\gamma}(y;\xi)-1\big{|}^{2}\,\big{|}\mathbf{F}_{\gamma}[u](\xi)\big{|}^{2}\,\xi^{\gamma}d\xi\\ \displaystyle=\frac{2^{n-|\gamma|}}{\prod\limits_{j=1}^{n}\Gamma^{2}\big{(}\frac{\gamma_{j}+1}{2}\big{)}}\,\int\limits_{𝕉^{n}_{+}}\big{|}\mathbf{F}_{\gamma}[u](\xi)\big{|}^{2}\,\xi^{\gamma}d\xi\int\limits_{𝕉^{n}_{+}}\frac{|\mathbf{j}_{\gamma}(y;\xi)-1|^{2}}{{|y|^{n+|\gamma|+4\alpha}}}\,y^{\gamma}dy\\ \displaystyle=\int\limits_{𝕉^{n}_{+}}\mathcal{A}(\xi)\big{|}\mathbf{F}_{\gamma}[u](\xi)\big{|}^{2}\,\xi^{\gamma}d\xi,\end{gathered} $$

where

$$ \mathcal{A}(\xi)=\frac{2^{n-|\gamma|}}{\prod\limits_{j=1}^{n}\Gamma^{2}\big{(}\frac{\gamma_{j}+1}{2}\big{)}}\,\int\limits_{𝕉^{n}_{+}}\frac{|\mathbf{j}_{\gamma}(y;\xi)-1|^{2}}{{|y|^{n+|\gamma|+4\alpha}}}\,y^{\gamma}dy. $$

Since

$$ \begin{gathered}\displaystyle\mathcal{A}(\xi)=\frac{2^{n-|\gamma|}}{\prod\limits_{j=1}^{n}\Gamma^{2}\big{(}\frac{\gamma_{j}+1}{2}\big{)}}\,\int\limits_{𝕉^{n}_{+}}\frac{|\mathbf{j}_{\gamma}(y;\xi)-1|^{2}}{{|y|^{n+|\gamma|+4\alpha}}}\,y^{\gamma}dy=\bigg{\{}y=\frac{z}{|\xi|}\bigg{\}}\\ \displaystyle=|\xi|^{4\alpha}\,\frac{2^{n-|\gamma|}}{\prod\limits_{j=1}^{n}\Gamma^{2}\big{(}\frac{\gamma_{j}+1}{2}\big{)}}\,\int\limits_{𝕉^{n}_{+}}\frac{\big{|}\mathbf{j}_{\gamma}\big{(}z;\frac{\xi}{|\xi|}\big{)}-1\big{|}^{2}}{{|z|^{n+|\gamma|+4\alpha}}}\,z^{\gamma}dz=C(n,\gamma,\alpha)|\xi|^{4\alpha},\end{gathered} $$

where

$$ C(n,\gamma,\alpha)=\frac{2^{n-|\gamma|}}{\prod\limits_{j=1}^{n}\Gamma^{2}\big{(}\frac{\gamma_{j}{+}1}{2}\big{)}}\,\int\limits_{𝕉^{n}_{+}}\frac{\big{|}\mathbf{j}_{\gamma}\big{(}z;\frac{\xi}{|\xi|}\big{)}-1\big{|}^{2}}{{|z|^{n+|\gamma|+4\alpha}}}\,z^{\gamma}dz, $$
(9)

we see that \( \mathcal{A}(\xi) \) is a homogeneous function of order \( 4\alpha \) which is invariant under orthogonal transformations and

$$ \int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\frac{\big{|}{{}^{\gamma}\mathbf{T}_{x}^{y}}u(x)-u(x)\big{|}^{2}}{|y|^{n+|\gamma|+4\alpha}}\,x^{\gamma}y^{\gamma}dxdy=C(n,\gamma,\alpha)\int\limits_{𝕉^{n}_{+}}|\xi|^{4\alpha}\,\big{|}\mathbf{F}_{\gamma}[u](\xi)\big{|}^{2}\,\xi^{\gamma}d\xi. $$
(10)

Let us calculate \( C(n,\gamma,\alpha) \). Using the representation for \( \mathbf{j}_{\gamma} \) of the form \( \mathbf{j}_{\gamma}(x,\xi)=\mathbf{P}_{\xi}^{\gamma}\big{[}e^{-i\langle{}x,\xi\rangle{}}\big{]} \) (see [11, p. 137, formula 3.138]), where \( \mathbf{P}_{\xi}^{\gamma} \) is as in (5), we infer

$$ \int\limits_{𝕉^{n}_{+}}\frac{\big{|}\mathbf{j}_{\gamma}\big{(}z;\frac{\xi}{|\xi|}\big{)}-1\big{|}^{2}}{{|z|^{n+|\gamma|+4\alpha}}}\,z^{\gamma}dz=\int\limits_{𝕉^{n}_{+}}\frac{\Big{|}\mathbf{P}_{\xi}^{\gamma}\Big{[}e^{-i\big{\langle{}}z;\frac{\xi}{|\xi|}\big{\rangle{}}}\Big{]}-1\Big{|}^{2}}{{|z|^{n+|\gamma|+4\alpha}}}\,z^{\gamma}dz. $$

Introduce the new coordinates in the last integral on the right-hand side as follows:

$$ \begin{gathered}\displaystyle\widetilde{x}_{1}=z_{1}\cos\alpha_{1},\quad\widetilde{x}_{2}=z_{1}\sin\alpha_{1},\quad\widetilde{x}_{3}=z_{2}\cos\alpha_{2},\\ \displaystyle\widetilde{x}_{4}=z_{2}\sin\alpha_{2},\dots,\widetilde{x}_{2n-1}=z_{n}\cos\alpha_{n},\quad\widetilde{x}_{2n}=z_{n}\sin\alpha_{n}.\end{gathered} $$

In these coordinates,

$$ \int\limits_{𝕉^{n}_{+}}\frac{\big{|}\mathbf{j}_{\gamma}\big{(}z;\frac{\xi}{|\xi|}\big{)}-1\big{|}^{2}}{{|z|^{n+|\gamma|+4\alpha}}}\,z^{\gamma}dz=C(\gamma)\int\limits_{\widetilde{𝕉}^{2n}_{+}}\Big{|}e^{-i\big{\langle{}}\widetilde{x},\frac{\widetilde{\xi^{\prime}}}{|\xi|}\big{\rangle{}}}-1\Big{|}^{2}\,\prod\limits^{n}_{i=1}\widetilde{x}\,^{\gamma_{i}-1}_{2i}d\widetilde{x}, $$

where \( \widetilde{x}=(\widetilde{x}_{1},\dots,\widetilde{x}_{2n})\in 𝕉^{2n} \), \( \widetilde{x}_{2i}>0 \); \( i=\overline{1,n} \), \( {|\widetilde{x}|=|z|} \), \( \widetilde{\xi}^{\prime}=(\xi_{1},0,\xi_{2},0,\dots,\xi_{n},0)\in 𝕉^{2n} \), \( {|\widetilde{\xi}^{\prime}|=|\xi|} \), and \( \widetilde{𝕉}^{2n}_{+}=\{x\in 𝕉^{2n}:{x}_{2i}>0,\ i=1,\dots,n\} \). As the integrand in \( C(n,\gamma,\alpha) \), we have some function of the type of a planar wave. Choosing integration over \( \widetilde{x}_{1}=p \), noticing that \( \big{\langle{}}\widetilde{x},\frac{\widetilde{\xi}^{\prime}}{|\xi|}\big{\rangle{}}=p \), and putting \( \widetilde{x\,}^{\prime}=(\widetilde{x}_{2},\dots,\widetilde{x}_{2n}) \), we obtain

$$ \begin{gathered}\displaystyle\int\limits_{𝕉^{n}_{+}}\frac{\big{|}\mathbf{j}_{\gamma}\big{(}z;\frac{\xi}{|\xi|}\big{)}-1\big{|}^{2}}{{|z|^{n+|\gamma|+4\alpha}}}\,z^{\gamma}dz=C(\gamma)\int\limits_{\widetilde{𝕉}^{2n}_{+}}\frac{|e^{-ip}-1|^{2}}{\big{|}p^{2}+|\widetilde{x\,}^{\prime}|^{2}\big{|}^{\frac{n+|\gamma|+4\alpha}{2}}}\prod\limits^{n}_{i=1}\widetilde{x}\,^{\gamma_{i}-1}_{2i}dpd\widetilde{x\,}^{\prime}\\ \displaystyle=\big{\{}\widetilde{x\,}^{\prime}=pt\big{\}}=C(\gamma)\int\limits_{-\infty}^{\infty}\frac{|e^{-ip}-1|^{2}}{|p|^{4\alpha+1}}\,dp\int\limits_{{}_{\begin{subarray}{c}{𝕉}^{2n-1},\,t_{2i-1}>0\\ i=1,\dots,n\end{subarray}}}\frac{\prod\limits^{n}_{i=1}t\,^{\gamma_{i}-1}_{2i-1}\,dt}{(1+|t|^{2})^{\frac{n+|\gamma|+4\alpha}{2}}}.\end{gathered} $$

Since

$$ \int\limits_{-\infty}^{\infty}\frac{|e^{-ip}-1|^{2}}{|p|^{4\alpha+1}}\,dp=2^{3-4\alpha}\int\limits_{0}^{\infty}\frac{\sin^{2}p}{p^{4\alpha+1}}\,dp=-4\cos(2\pi\alpha)\Gamma(-4\alpha), $$

we have

$$ \begin{gathered}\displaystyle\int\limits_{{}_{\begin{subarray}{c}{𝕉}^{2n-1},\,t_{2i-1}>0\\ i=1,\dots,n\end{subarray}}}\frac{\prod\limits^{n}_{i=1}t\,^{\gamma_{i}-1}_{2i-1}\,dt}{(1+|t|^{2})^{\frac{n+|\gamma|+4\alpha}{2}}}=\int\limits_{0}^{\infty}\frac{\rho^{n+|\gamma|-2}}{(1+\rho^{2})^{\frac{n+|\gamma|+4\alpha}{2}}}\,d\rho\int\limits_{\widetilde{S}_{1}^{+}(2n-1)}\prod\limits^{n}_{i=1}\theta\,^{\gamma_{i}-1}_{2i-1}\,dS\\ \displaystyle\times\,\frac{\Gamma\big{(}2\alpha+\frac{1}{2}\big{)}\Gamma\big{(}\frac{n+|\gamma|-1}{2}\big{)}}{2\Gamma\big{(}\frac{n+|\gamma|}{2}+2\alpha\big{)}}\,\big{|}\widetilde{S}_{1}^{+}(2n-1)\big{|}_{|\gamma|-1}\end{gathered} $$

for \( t=\rho\theta \). Here \( \widetilde{S}_{1}^{+}(2n-1) \) is the part of the unit sphere centered at the origin in \( 𝕉^{2n-1} \) for \( \widetilde{x}_{2i}>0 \), with \( i=1,\dots,n \). Applying formula (107) in [11, p. 49], we set

$$ \int\limits_{\widetilde{S}_{1}^{+}(2n-1)}\prod\limits^{n}_{i=1}\theta\,^{\gamma_{i}-1}_{2i-1}\,dS=\frac{\Gamma^{n-1}\big{(}\frac{1}{2}\big{)}\prod\limits_{i=1}^{n}\Gamma\big{(}\frac{\gamma_{i}}{2}\big{)}}{2^{n-1}\Gamma\big{(}\frac{n+|\gamma|-1}{2}\big{)}}=\frac{\pi^{\frac{n-1}{2}}\prod\limits_{i=1}^{n}\Gamma\big{(}\frac{\gamma_{i}}{2}\big{)}}{2^{n-1}\Gamma\big{(}\frac{n+|\gamma|-1}{2}\big{)}}=\big{|}\widetilde{S}_{1}^{+}(2n-1)\big{|}_{|\gamma|-1}. $$

Consequently,

$$ \int\limits_{𝕉^{n}_{+}}\frac{\big{|}\mathbf{j}_{\gamma}\big{(}z;\frac{\xi}{|\xi|}\big{)}-1\big{|}^{2}}{{|z|^{n+|\gamma|+4\alpha}}}\,z^{\gamma}dz=C(\gamma)\int\limits_{-\infty}^{\infty}\frac{|e^{-ip}-1|^{2}}{|p|^{4\alpha+1}}\,dp\int\limits_{{}_{\begin{subarray}{c}{𝕉}^{2n-1},\,t_{2i-1}>0\\ i=1,\dots,n\end{subarray}}}\frac{\prod\limits^{n}_{i=1}t\,^{\gamma_{i}-1}_{2i-1}\,dt}{(1+|t|^{2})^{\frac{n+|\gamma|+4\alpha}{2}}}. $$

Finally, using Legendre’s duplication formula and Euler’s reflection formula, simplify \( C(n,\gamma,\alpha) \) to see that

$$ C(n,\gamma,\alpha)=\frac{2^{1-|\gamma|-4\alpha}\pi}{\sin(2\alpha\pi)\,\Gamma(2\alpha+1)\,\Gamma\big{(}\frac{n+|\gamma|}{2}+2\alpha\big{)}\prod\limits_{i=1}^{n}\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}. $$

Taking into account (8), (10), and the form of the constant \( C(n,\gamma,\alpha) \), we complete the proof of the lemma.  ☐

The constant \( C(n,\gamma,\alpha) \) possesses the properties

$$ \lim\limits_{\alpha\rightarrow 0+}\frac{1}{C(n,\gamma,\alpha)}=0,\quad\lim\limits_{\alpha\rightarrow\frac{1}{2}-0}\,\frac{1}{C(n,\gamma,\alpha)}=0. $$

Therefore,

$$ d_{\alpha,\gamma}(u)=\sum\limits_{|i|=\alpha}\ \int\limits_{𝕉^{n}_{+}}|𝔹_{i}u|^{2}\,x^{\gamma}dx\quad\text{if }\alpha\text{ is an integer}; $$

in the remaining cases,

$$ \begin{gathered}\displaystyle d_{\alpha,\gamma}(u)=\frac{1}{C(n,\gamma,\alpha-l)}\,\sum\limits_{|i|=l}\ \int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\big{|}𝔹_{i}u(y)-𝔹_{i}u(x)\big{|}^{2}\,\Big{(}{{}^{\gamma}\mathbf{T}_{y}^{x}}\,\frac{1}{|y|^{n+|\gamma|+4\alpha}}\Big{)}\,x^{\gamma}y^{\gamma},\end{gathered} $$

where \( l=[\alpha] \). The weighted integral \( d_{\alpha,\gamma} \) is continuous in \( \alpha \) and is independent of the orthogonal coordinates in \( 𝕉^{n}_{+} \).

4. Finding a Suitable Norm for \( {\overset{\circ}{C}}\,^{\infty}_{ev}(𝕉^{n}_{+}) \)

From the practical and theoretical standpoints, it is important to find out in what space and with what norm \( {\overset{\circ}{C}}\,^{\infty}_{ev}(𝕉^{n}_{+}) \) is dense.

In [3], there was introduced some function space that is connected with multiplication by \( |x|^{-\alpha} \) in the images of the Hankel transform. This was called the space of Riesz \( B \)-potentials. Recall that in the theory of \( B \)-potentials a Riesz \( B \)-potential has the form (see [4])

$$ (U^{\alpha}_{\gamma}f)(x)=u(x)=C_{n,\gamma}\int\limits_{𝕉^{n}_{+}}f(y)\big{(}{{}^{\gamma}\mathbf{T}_{x}^{y}}|x|^{\alpha-n-|\gamma|}\big{)}\,y^{\gamma}dy,\quad\alpha>0. $$

An analog of Sobolev’s Theorem holds for \( U^{\alpha}_{\gamma} \) (see [4, Theorem 1]). Namely, for \( 0<\alpha<\frac{n+|\gamma|}{p} \), \( p>1 \), the operator \( U^{\alpha}_{\gamma} \) with density \( f\in L^{\gamma}_{p} \) is bounded from \( L_{p}^{\gamma} \) into \( L_{q}^{\gamma} \), where \( \frac{1}{q}=\frac{1}{p}-\frac{\alpha}{n+|\gamma|} \). For \( \alpha\geq\frac{n+|\gamma|}{p} \), the potential \( U^{\alpha}_{\gamma} \) can be defined in the sense of weighted distributions. As a consequence of this fact, in [3, Theorem 5], \( {\overset{\circ}{C}}\,^{\infty}_{ev}(𝕉^{n}_{+}) \) was shown to be dense in the space of Riesz \( B \)-potentials only for \( 0<\alpha<\frac{n+|\gamma|}{p} \). Therefore, it is inconvenient to use the norm that bases on a Riesz \( B \)-potential in differential problems since these problems need potentials of arbitrarily large order.

We will prove that since norm convergence does not imply the pointwise convergence of a sequence in the spaces with the norm \( \sqrt{d_{\alpha,\gamma}} \); therefore, \( {\overset{\circ}{C}}\,^{\infty}_{ev}(𝕉^{n}_{+}) \) is not a function space for \( \alpha\geq\frac{n+|\gamma|}{4} \).

Theorem 1

If \( \alpha\geq\frac{n+|\gamma|}{4} \) then the space \( {\overset{\circ}{C}}\,^{\infty}_{ev}(𝕉^{n}_{+}) \) normalized by \( \sqrt{d_{\alpha,\gamma}} \) is not a function space with respect to any exceptional class.

Proof

Let \( u\in{\overset{\circ}{C}}\,^{\infty}_{ev}(𝕉^{n}_{+}) \) and let \( u \) be identically 1 in a neighborhood of zero in \( 𝕉^{n}_{+} \). Put \( u_{\rho}=u(x/\rho) \). We have

$$ \begin{gathered}\displaystyle d_{\alpha,\gamma}(u_{\rho})=\int\limits_{𝕉^{n}_{+}}|\xi|^{4\alpha}\,\big{|}\mathbf{F}_{\gamma}[u]\big{(}\xi/\rho\big{)}\big{|}^{2}\,\xi^{\gamma}d\xi=\big{\{}\xi/\rho=y\big{\}}\\ \displaystyle=\rho^{n+|\gamma|-4\alpha}\int\limits_{𝕉^{n}_{+}}|y|^{4\alpha}\,\big{|}\mathbf{F}_{\gamma}[u](y)\big{|}^{2}\,y^{\gamma}dy=\rho^{n+|\gamma|-4\alpha}\,d_{\alpha,\gamma}(u).\end{gathered} $$

Consequently, for \( \alpha>\frac{n+|\gamma|}{4} \), we see that

$$ \lim\limits_{\rho\rightarrow\infty}d_{\alpha,\gamma}(u_{\rho})=0,\quad\text{but }\lim\limits_{\rho\rightarrow\infty}u_{\rho}(x)=1. $$

This shows that the space under consideration cannot be a function space (since otherwise the whole of \( 𝕉^{n}_{+} \) must be an exceptional set).

Now, consider the case of \( \alpha=\frac{n+|\gamma|}{4} \). Choosing \( \varepsilon\in(0,\alpha) \) and \( v\in{\overset{\circ}{C}}\,^{\infty}_{ev}(𝕉^{n}_{+}) \), for the bilinear norm corresponding to the quadratic form \( d_{\alpha,\gamma}(u) \), we see that

$$ \begin{gathered}\displaystyle d_{\alpha,\gamma}(u_{\rho},v)\\ \displaystyle=\int\limits_{𝕉^{n}_{+}}|\xi|^{4\alpha}\,\mathbf{F}_{\gamma}[u_{\rho}](\xi)\,\overline{\mathbf{F}_{\gamma}[v](\xi)}\,\xi^{\gamma}d\xi=\int\limits_{𝕉^{n}_{+}}|\xi|^{2\alpha+2\varepsilon}\,\mathbf{F}_{\gamma}[u_{\rho}](\xi)\,|\xi|^{2\alpha-2\varepsilon}\,\overline{\mathbf{F}_{\gamma}[v](\xi)}\,\xi^{\gamma}d\xi\\ \displaystyle\leq\Bigg{(}\,\int\limits_{𝕉^{n}_{+}}|\xi|^{4\alpha+4\varepsilon}\,\big{|}\mathbf{F}_{\gamma}[u_{\rho}](\xi)\big{|}^{2}\,\xi^{\gamma}d\xi\Bigg{)}^{\frac{1}{2}}\Bigg{(}\,\int\limits_{𝕉^{n}_{+}}|\xi|^{4\alpha-4\varepsilon}\,\big{|}{\mathbf{F}_{\gamma}[v](\xi)}\big{|}^{2}\,\xi^{\gamma}d\xi\Bigg{)}^{\frac{1}{2}}\\ \displaystyle=\sqrt{d_{\alpha+\varepsilon,\gamma}(u_{\rho})}\,\sqrt{d_{\alpha-\varepsilon,\gamma}(v)}.\end{gathered} $$

Here we used the Cauchy–Bunyakovsky inequality. Since

$$ \lim_{\rho\rightarrow\infty}d_{\alpha+\varepsilon,\gamma}(u_{\rho})=0, $$

the equality \( \lim_{\rho\rightarrow\infty}d_{\alpha,\gamma}(u_{\rho},v)=0 \) is fulfilled for every \( v\in{\overset{\circ}{C}}\,^{\infty}_{ev}(𝕉^{n}_{+}) \). We consider the Hilbert space that is a complete metric space with respect to the distance function induced by the inner product \( d_{\alpha,\gamma}(u,v) \). This space is the abstract completion of \( {\overset{\circ}{C}}\,^{\infty}_{ev}(𝕉^{n}_{+}) \) under the norm \( \sqrt{d_{\alpha,\gamma}} \). Thus, \( u_{\rho} \) converges weakly to \( 0 \) as \( \rho\rightarrow\infty \) in this Hilbert space because \( d_{\alpha,\gamma}(u_{\rho}) \) is bounded. Then there exists a sequence \( \rho_{k}\rightarrow\infty \) such that the arithmetic means of \( \{u_{\rho_{k}}\} \) converges strongly to \( 0 \) (see [12]). But the equality \( \lim_{\rho\rightarrow\infty}u_{\rho}(x)=1 \) implies that the sequence of the arithmetic means pointwise converges everywhere to \( 1 \). Thus, \( {\overset{\circ}{C}}\,^{\infty}_{ev}(𝕉^{n}_{+}) \) normalized by \( \sqrt{d_{\alpha,\gamma}} \) cannot be a function space.  ☐

One of the simplest norms on \( {\overset{\circ}{C}}\,^{\infty}_{ev}(𝕉^{n}_{+}) \) equivalent to \( \sqrt{d_{\alpha,\gamma}} \) looks as

$$ \big{\|}u\big{\|}^{2}_{\alpha,\gamma}=\int\limits_{𝕉^{n}_{+}}\big{(}1+|\xi|^{2}\big{)}^{\alpha}\,\big{|}\mathbf{F}_{\gamma}[u](\xi)\big{|}^{2}\,\xi^{\gamma}d\xi. $$
(11)

Below we prove that (11) is representable by the convolution kernels generating a generalized Bessel potential.

5. The Class of Generalized Bessel Potentials

In this section, we obtain a special representation of (11) which is most convenient for classes of generalized Bessel potentials. The generalization of the spaces of Bessel potentials has a rich history. In [13], for functions in the space of generalized Bessel potentials which is constructed from rearrangement invariant spaces, some equivalent description was obtained for the cone of decreasing rearrangements. In [14], some equivalent characterizations were established for the cones of decreasing rearrangements for the spaces of generalized Riesz and Bessel potentials.

A generalized Bessel potential is defined by the relation (see [15, 16])

$$ u=(\mathbf{G}^{\alpha}_{\gamma}\varphi)(x)=\int\limits_{𝕉^{n}_{+}}G_{\alpha}^{\gamma}(y)\big{(}{{}^{\gamma}\mathbf{T}_{x}^{y}}\varphi(x)\big{)}\,y^{\gamma}dy, $$
(12)

where

$$ G_{\alpha}^{\gamma}(x)=\mathbf{F}_{\gamma}^{-1}\Big{[}\big{(}1+|\xi|^{2}\big{)}^{-\frac{\alpha}{2}}\Big{]}(x) $$
(13)

is a generalization of the Bessel kernel. The two forms of the inverse operator of (12) were constructed in [16].

In [3], the space \( \mathbf{B}^{\alpha}_{\gamma}(L_{p}^{\gamma})=\{u:u=\mathbf{G}^{\alpha}_{\gamma}\varphi,\varphi\in L_{p}^{\gamma}\} \) with the norm \( \|u\|_{\mathbf{B}^{\alpha}_{\gamma}(L_{p}^{\gamma})}=\|\varphi\|_{L_{p}^{\gamma}} \) was introduced by using \( B \)-hypersingular integrals.

It was shown in [15] that

$$ G_{\alpha}^{\gamma}(x)=\frac{2^{\frac{n-|\gamma|-\alpha}{2}+1}}{|x|^{\frac{n+|\gamma|-\alpha}{2}}\,\Gamma\left(\frac{\alpha}{2}\right)\prod\limits^{n}_{i=1}\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}\,K_{\frac{n+|\gamma|-\alpha}{2}}(|x|), $$
(14)

where \( K_{\frac{n+|\gamma|-\alpha}{2}} \) is a modified Bessel function of the second kind (see [10]).

Since \( G_{\alpha}^{\gamma} \) is integrable with the weight \( x^{\gamma} \) (see [15, 16]), its Hankel transform exists for every \( \xi \). The kernel \( G_{\alpha}^{\gamma} \) is an analytic function of \( \alpha \) for \( \alpha>0 \). Thus, from (13), by analytic continuation, we conclude that the Hankel transform of the generalized Bessel kernel for \( \alpha>0 \) is equal to

$$ \mathbf{F}_{\gamma}[G_{\alpha}^{\gamma}](\xi)=\big{(}1+|\xi|^{2}\big{)}^{-\frac{\alpha}{2}}. $$
(15)

Moreover, the kernel \( G_{\alpha}^{\gamma} \) satisfies the properties

$$ \int\limits_{𝕉^{n}_{+}}G_{\alpha}^{\gamma}(x)x^{\gamma}dx=1\quad\text{and}\quad(G_{\alpha}^{\gamma}*G_{\beta}^{\gamma})_{\gamma}=G_{\alpha+\beta}^{\gamma},\ \alpha>0,\ \beta>0 $$

(see [15, 16]), where \( (G_{\alpha}^{\gamma}*G_{\beta}^{\gamma})_{\gamma} \) is the generalized convolution; see (4).

To obtain (11) for \( 0<\alpha<1/2 \), we first introduce the function

$$ \begin{aligned} \omega_{\alpha,\gamma}(|x|)&=\frac{2^{\frac{n-|\gamma|-\alpha}{2}+1}}{\Gamma\left(\frac{\alpha}{2}\right)\prod\limits^{n}_{i=1}\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}\,|x|^{\frac{n+|\gamma|-\alpha}{2}}\,K_{\frac{n+|\gamma|-\alpha}{2}}(|x|)\\ &=\frac{2^{n-\alpha+2}}{\Gamma\left(\frac{\alpha}{2}\right)\prod\limits^{n}_{i=1}\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}\int\limits_{0}^{\infty}t^{\frac{n+|\gamma|-\alpha}{2}-1}e^{-t-\frac{|x|^{2}}{4t}}\,dt.\end{aligned} $$
(16)

Then the generalized Bessel potential in (12) is representable as the generalized convolution operator of (4):

$$ \big{(}\mathbf{G}^{\alpha}_{\gamma}\varphi\big{)}(x)=\left(\frac{\omega_{\alpha,\gamma}(|x|)}{|x|^{n+|\gamma|-\alpha}}*\varphi\right)_{\gamma},\quad\alpha>0. $$

Below we will need \( \omega_{-4\alpha,\gamma}(|x|) \) for \( 0<\alpha<1/2 \). The asymptotic properties of the modified Bessel function \( K_{\nu} \) guarantee that the kernel function \( \omega_{-4\alpha,\gamma}(|x|) \) decreases exponentially at infinity and turns into a constant at the origin:

$$ \begin{gathered}\displaystyle\lim\limits_{|x|\rightarrow\infty}\omega_{-4\alpha,\gamma}(|x|)=\frac{2^{\frac{n-|\gamma|+4\alpha}{2}+1}}{\Gamma\left(-2\alpha\right)\prod\limits^{n}_{i=1}\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}\,\lim\limits_{|x|\rightarrow\infty}\,|x|^{\frac{n+|\gamma|+4\alpha}{2}}\,K_{\frac{n+|\gamma|+4\alpha}{2}}(|x|)\\ \displaystyle=\frac{2^{\frac{n-|\gamma|+4\alpha}{2}+1}}{\Gamma\left(-2\alpha\right)\prod\limits^{n}_{i=1}\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}\,\sqrt{\frac{\pi}{2}}\,\lim\limits_{|x|\rightarrow\infty}\,|x|^{\frac{n+|\gamma|+4\alpha}{2}-\frac{1}{2}}\,e^{-|x|}=0,\end{gathered} $$
$$ \begin{gathered}\displaystyle\lim\limits_{|x|\rightarrow 0}\omega_{-4\alpha,\gamma}(|x|)=\frac{2^{\frac{n-|\gamma|+4\alpha}{2}+1}}{\Gamma\left(-2\alpha\right)\prod\limits^{n}_{i=1}\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}\,\lim\limits_{|x|\rightarrow 0}\,|x|^{\frac{n+|\gamma|+4\alpha}{2}}\,K_{\frac{n+|\gamma|+4\alpha}{2}}(|x|)\\ \displaystyle=\frac{2^{\frac{n-|\gamma|+4\alpha}{2}+1}}{\Gamma\left(-2\alpha\right)\prod\limits^{n}_{i=1}\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}\,\lim\limits_{|x|\rightarrow 0}\,|x|^{\frac{n+|\gamma|+4\alpha}{2}}\,\frac{\Gamma\big{(}\frac{n+|\gamma|+4\alpha}{2}\big{)}}{2^{1-\frac{n+|\gamma|+4\alpha}{2}}}\,|x|^{-\frac{n+|\gamma|+4\alpha}{2}}\\ \displaystyle=\frac{2^{n+4\alpha}\,\Gamma\big{(}\frac{n+|\gamma|+4\alpha}{2}\big{)}}{\Gamma\big{(}-2\alpha\big{)}\prod\limits^{n}_{i=1}\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}.\end{gathered} $$

Consequently,

$$ \lim\limits_{|x|\rightarrow\infty}\omega_{-4\alpha,\gamma}(|x|)=0,\quad\lim\limits_{|x|\rightarrow 0}\omega_{-4\alpha,\gamma}(|x|)=\frac{2^{n+4\alpha}\,\Gamma\big{(}\frac{n+|\gamma|+4\alpha}{2}\big{)}}{\Gamma\big{(}-2\alpha\big{)}\prod\limits^{n}_{i=1}\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}. $$
(17)

Theorem 2

The norm \( \|u\|_{\alpha,\gamma} \) admits the representation

$$ \begin{gathered}\displaystyle\big{\|}u\big{\|}_{\alpha,\gamma}^{2}=2^{|\gamma|-n+1}\prod\limits_{i=1}^{n}\Gamma^{2}\Big{(}\frac{\gamma_{i}+1}{2}\Big{)}\\ \displaystyle\times\,\Bigg{(}\,\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\frac{\big{|}{{}^{\gamma}\mathbf{T}_{x}^{y}}u(x)+u(y)\big{|}^{2}}{{|x|^{n+|\gamma|+4\alpha}}}\,\big{(}\omega_{-4\alpha,\gamma}(|x|)-\omega_{-4\alpha,\gamma}(0)\big{)}\,x^{\gamma}dx\,y^{\gamma}dy\\ \displaystyle-\,\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\frac{\big{|}{{}^{\gamma}\mathbf{T}_{x}^{y}u(x)-u(y)\big{|}^{2}}}{{|x|^{n+|\gamma|+4\alpha}}}\,\big{(}\omega_{-4\alpha,\gamma}(|x|)+\omega_{-4\alpha,\gamma}(0)\big{)}\,x^{\gamma}dx\,y^{\gamma}dy\Bigg{)}.\end{gathered} $$
(18)

Proof

Let \( \gamma_{n+1}\geq 0 \) be arbitrary and

$$ \mathbf{J}=\int\limits_{0}^{\infty}\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\frac{\big{|}j_{\frac{\gamma_{n+1}-1}{2}}(z_{0})u(x)-u(y)\big{|}^{2}}{\big{[}{{}^{\gamma}\mathbf{T}_{x}^{y}}\,|x|^{2}+z_{0}^{2}\big{]}^{\frac{n+|\gamma^{\prime}|+1+4\alpha}{2}}}\,x^{\gamma}dx\,y^{\gamma}dy\,z_{0}^{\gamma_{n+1}}dz_{0}, $$

where

$$ j_{\nu}(x)=\frac{2^{\nu}\Gamma(\nu+1)}{x^{\nu}}\,J_{\nu}(x), $$

and \( J_{\nu} \) is the Bessel function of the first kind. Acting as in proving (8) and applying (2) to the integral over \( y \), we infer

$$ \begin{gathered}\displaystyle\mathbf{J}=\int\limits_{0}^{\infty}\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\frac{\big{|}j_{\frac{\gamma_{n+1}-1}{2}}(z_{0})\,^{\gamma}\mathbf{T}_{x}^{y}u(x)-u(y)\big{|}^{2}}{\big{[}|x|^{2}+z_{0}^{2}\big{]}^{\frac{n+|\gamma^{\prime}|+1+4\alpha}{2}}}\,x^{\gamma}dx\,y^{\gamma}dy\,z_{0}^{\gamma_{n+1}}dz_{0}\\ \displaystyle=\int\limits_{0}^{\infty}z_{0}^{\gamma_{n+1}}dz_{0}\int\limits_{𝕉^{n}_{+}}\frac{x^{\gamma}dx}{\big{[}|x|^{2}+z_{0}^{2}\big{]}^{\frac{n+|\gamma|+1+2\alpha}{2}}}\,\int\limits_{𝕉^{n}_{+}}\Big{|}j_{\frac{\gamma_{n+1}-1}{2}}(z_{0})\,^{\gamma}\mathbf{T}_{x}^{y}u(x)-u(y)\Big{|}^{2}\,y^{\gamma}dy\\ \displaystyle=\frac{2^{n-|\gamma|}}{\prod\limits_{j=1}^{n}\Gamma^{2}\big{(}\frac{\gamma_{j}{+}1}{2}\big{)}}\,\int\limits_{0}^{\infty}dz_{0}\int\limits_{𝕉^{n}_{+}}\frac{x^{\gamma}dx}{\big{[}|x|^{2}+z_{0}^{2}\big{]}^{\frac{n+|\gamma|+1+2\alpha}{2}}}\,\int\limits_{𝕉^{n}_{+}}\Big{|}j_{\frac{\gamma_{n+1}-1}{2}}(z_{0})\,\mathbf{j}_{\gamma}(x;\xi)-1\Big{|}^{2}\,\big{|}\mathbf{F}_{\gamma}[u](\xi)\big{|}^{2}\,\xi^{\gamma}d\xi.\end{gathered} $$

Putting \( (x,z_{0})=\widetilde{z} \), \( \widetilde{\xi}=(\xi,1) \), and \( \gamma^{\prime}=(\gamma,\gamma_{n+1}) \), we have

$$ \begin{gathered}\displaystyle\mathbf{J}=\frac{2^{n-|\gamma|}}{\prod\limits_{j=1}^{n}\Gamma^{2}\big{(}\frac{\gamma_{j}{+}1}{2}\big{)}}\int\limits_{0}^{\infty}z_{0}^{\gamma_{n+1}}dz_{0}\\ \displaystyle\times\int\limits_{𝕉^{n}_{+}}\frac{x^{\gamma}dx}{\big{[}|x|^{2}+z_{0}^{2}\big{]}^{\frac{n+|\gamma^{\prime}|+1+4\alpha}{2}}}\int\limits_{𝕉^{n}_{+}}\Big{|}j_{\frac{\gamma_{n+1}-1}{2}}(z_{0})\mathbf{j}_{\gamma}(x;\xi)-1\Big{|}^{2}\big{|}\mathbf{F}_{\gamma}[u](\xi)\big{|}^{2}\xi^{\gamma}d\xi\\ \displaystyle=\frac{2^{n-|\gamma|}}{\prod\limits_{j=1}^{n}\Gamma^{2}\big{(}\frac{\gamma_{j}{+}1}{2}\big{)}}\,\int\limits_{𝕉^{n}_{+}}\big{|}\mathbf{F}_{\gamma}[u](\xi)\big{|}^{2}\,\xi^{\gamma}d\xi\int\limits_{𝕉^{n+1}_{+}}\frac{\big{|}\mathbf{j}_{\gamma^{\prime}}\big{(}\widetilde{z};\widetilde{\xi}\,\big{)}-1\big{|}^{2}}{|\widetilde{z}\,|^{{n+|\gamma^{\prime}|+1+4\alpha}}}\,\widetilde{z}\,^{\gamma^{\prime}}d\widetilde{z}\\ \displaystyle=\int\limits_{𝕉^{n}_{+}}\mathcal{B}(\xi)\,\big{|}\mathbf{F}_{\gamma}[u](\xi)\big{|}^{2}\,\xi^{\gamma}d\xi.\end{gathered} $$

Performing the change of variables \( \widetilde{z}=z/|\widetilde{\xi}| \), we obtain

$$ \begin{gathered}\displaystyle\mathcal{B}(\xi)=\frac{2^{n-|\gamma|}}{\prod\limits_{j=1}^{n}\Gamma^{2}\big{(}\frac{\gamma_{j}{+}1}{2}\big{)}}\,\int\limits_{𝕉^{n+1}_{+}}\frac{|\mathbf{j}_{\gamma^{\prime}}(\widetilde{z};\widetilde{\xi})-1|^{2}}{|\widetilde{z}\,|^{{n+|\gamma^{\prime}|+1+4\alpha}}}\,\widetilde{z}\,^{\gamma^{\prime}}d\widetilde{z}\\ \displaystyle=\frac{2^{n-|\gamma|}}{\prod\limits_{j=1}^{n}\Gamma^{2}\big{(}\frac{\gamma_{j}{+}1}{2}\big{)}}\,\big{(}1+|\xi|\big{)}^{4\alpha}\int\limits_{𝕉^{n+1}_{+}}\frac{\Big{|}\mathbf{j}_{\gamma^{\prime}}\Big{(}z;\frac{\widetilde{\xi}}{|\widetilde{\xi}\,|}\Big{)}-1\Big{|}^{2}}{|z|^{{n+|\gamma^{\prime}|+1+4\alpha}}}\,z^{\gamma^{\prime}}dz\end{gathered} $$

and

$$ \mathbf{J}=D(n,\gamma^{\prime},\alpha)\int\limits_{𝕉^{n}_{+}}\big{(}1+|\xi|\big{)}^{4\alpha}\,\big{|}\mathbf{F}_{\gamma}[u](\xi)\big{|}^{2}\,\xi^{\gamma}d\xi. $$

It follows from (9) that

$$ \int\limits_{𝕉^{n+1}_{+}}\frac{\left|\mathbf{j}_{\gamma^{\prime}}\left(z;\frac{\widetilde{\xi}}{|\widetilde{\xi}|}\right)-1\right|^{2}}{|z|^{{n+|\gamma^{\prime}|+1+4\alpha}}}\,z^{\gamma^{\prime}}dz=\frac{\prod\limits_{j=1}^{n+1}\Gamma^{2}\big{(}\frac{\gamma_{j}{+}1}{2}\big{)}}{2^{n+1-|\gamma^{\prime}|}}\,C(n+1,\gamma^{\prime},\alpha) $$

and

$$ D(n,\gamma^{\prime},\alpha)=\frac{\Gamma^{2}\big{(}\frac{\gamma_{n+1}{+}1}{2}\big{)}}{2^{1-\gamma_{n+1}}}\,C(n+1,\gamma^{\prime},\alpha), $$

where

$$ C(n+1,\gamma^{\prime},\alpha)=\frac{2^{1-|\gamma^{\prime}|-4\alpha}\pi}{\sin(2\alpha\pi)\,\Gamma(2\alpha+1)\,\Gamma\big{(}\frac{n+|\gamma^{\prime}|+1}{2}+2\alpha\big{)}\prod\limits_{i=1}^{n+1}\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}. $$

Hence, for every \( \gamma_{n+1}\geq 0 \), we infer

$$ \big{\|}u\big{\|}_{\alpha,\gamma}^{2}=\frac{1}{D(n,\gamma^{\prime},\alpha)}\int\limits_{0}^{\infty}\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\frac{\big{|}j_{\frac{\gamma_{n+1}-1}{2}}(z_{0})u(x)-u(y)\big{|}^{2}}{\big{[}{{}^{\gamma}\mathbf{T}_{x}^{y}}\,|x|^{2}+z_{0}^{2}\big{]}^{\frac{n+|\gamma|+1+2\alpha}{2}}}\,x^{\gamma}dx\,y^{\gamma}dy\,z_{0}^{\gamma_{n+1}}dz_{0}. $$

Passing to \( \gamma_{n+1}=0 \) and putting

$$ E(n,\gamma,\alpha)=D(n,\gamma^{\prime},\alpha)\big{|}_{\gamma_{n+1}=0}=\frac{\pi\sqrt{\pi}\,2^{-|\gamma|-4\alpha}}{\sin(2\alpha\pi)\,\Gamma(2\alpha+1)\,\Gamma\big{(}\frac{n+|\gamma|+4\alpha+1}{2}\big{)}\prod\limits_{i=1}^{n}\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}\,, $$

we write

$$ \begin{aligned} \big{\|}u\big{\|}_{\alpha,\gamma}^{2}&=\frac{1}{E(n,\gamma,\alpha)}\int\limits_{-\infty}^{\infty}\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\frac{\big{|}e^{iz_{0}}u(x)-u(y)\big{|}^{2}}{\big{[}{{}^{\gamma}\mathbf{T}_{x}^{y}}\,|x|^{2}+z_{0}^{2}\big{]}^{\frac{n+|\gamma|+1+2\alpha}{2}}}\,x^{\gamma}dx\,y^{\gamma}dydz_{0}\\ &=\frac{1}{E(n,\gamma,\alpha)}\int\limits_{-\infty}^{\infty}\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\frac{\big{|}e^{\frac{iz_{0}}{2}}u(x)-e^{-\frac{iz_{0}}{2}}u(y)\big{|}^{2}}{\big{[}{{}^{\gamma}\mathbf{T}_{x}^{y}}\,|x|^{2}+z_{0}^{2}\big{]}^{\frac{n+|\gamma|+1+2\alpha}{2}}}\,x^{\gamma}dx\,y^{\gamma}dydz_{0}\\ &=\frac{1}{E(n,\gamma,\alpha)}\int\limits_{-\infty}^{\infty}\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\frac{\big{|}{{}^{\gamma}\mathbf{T}_{x}^{y}u(x)-u(y)|^{2}\cos^{2}\frac{z_{0}}{2}}}{\big{[}|x|^{2}+z_{0}^{2}\big{]}^{\frac{n+|\gamma|+1+2\alpha}{2}}}\,x^{\gamma}dx\,y^{\gamma}dydz_{0}\\ &+\frac{1}{E(n,\gamma,\alpha)}\int\limits_{-\infty}^{\infty}\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\frac{\big{|}{{}^{\gamma}\mathbf{T}_{x}^{y}}u(x)+u(y)\big{|}^{2}\sin^{2}\frac{z_{0}}{2}}{\big{[}|x|^{2}+z_{0}^{2}\big{]}^{\frac{n+|\gamma|+1+2\alpha}{2}}}\,x^{\gamma}dx\,y^{\gamma}dydz_{0}.\end{aligned} $$

Using the Wolfram Mathematica and (17), we get

$$ \begin{gathered}\displaystyle\int\limits_{-\infty}^{\infty}\frac{\cos^{2}\frac{z_{0}}{2}\,dz_{0}}{\big{[}|x|^{2}+z_{0}^{2}\big{]}^{\frac{n+|\gamma|+1+2\alpha}{2}}}\\ \displaystyle=\frac{\sqrt{\pi}\,2^{1-n-4\alpha}\,\Gamma(-2\alpha)\prod\limits^{n}_{i=1}\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}{|x|^{n+|\gamma|+4\alpha}\,\Gamma\big{(}\frac{n+|\gamma|+4\alpha+1}{2}\big{)}}\,\big{(}\omega_{-4\alpha,\gamma}(0)+\omega_{-4\alpha,\gamma}(|x|)\big{)},\end{gathered} $$
$$ \begin{gathered}\displaystyle\int\limits_{-\infty}^{\infty}\frac{\sin^{2}\frac{z_{0}}{2}dz_{0}}{\big{[}|x|^{2}+z_{0}^{2}\big{]}^{\frac{n+|\gamma|+1+2\alpha}{2}}}\\ \displaystyle=\frac{\sqrt{\pi}\,2^{1-n-4\alpha}\,\Gamma(-2\alpha)\prod\limits^{n}_{i=1}\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}{|x|^{n+|\gamma|+4\alpha}\,\Gamma\big{(}\frac{n+|\gamma|+4\alpha+1}{2}\big{)}}\,\big{(}\omega_{-4\alpha,\gamma}(0)-\omega_{-4\alpha,\gamma}(|x|)\big{)}.\end{gathered} $$

Since

$$ \begin{gathered}\displaystyle\frac{1}{E(n,\gamma,\alpha)}\;\frac{\sqrt{\pi}\,2^{1-n-4\alpha}\,\Gamma(-2\alpha)\prod\limits^{n}_{i=1}\,\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}{|x|^{n+|\gamma|+4\alpha}\,\Gamma\big{(}\frac{n+|\gamma|+4\alpha+1}{2}\big{)}}\\ \displaystyle=\frac{\sin(2\alpha\pi)\,\Gamma(2\alpha+1)\,\Gamma\big{(}\frac{n+|\gamma|+4\alpha+1}{2}\big{)}\prod\limits_{i=1}^{n}\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}{\pi\sqrt{\pi}\,2^{-|\gamma|-4\alpha}}\;\frac{\sqrt{\pi}\,2^{1-n-4\alpha}\,\Gamma(-2\alpha)\prod\limits^{n}_{i=1}\Gamma\big{(}\frac{\gamma_{i}+1}{2}\big{)}}{|x|^{n+|\gamma|+4\alpha}\,\Gamma\big{(}\frac{n+|\gamma|+4\alpha+1}{2}\big{)}}\\ \displaystyle=-\frac{2^{|\gamma|-n+1}}{|x|^{n+|\gamma|+4\alpha}}\,\prod\limits_{i=1}^{n}\Gamma^{2}\Big{(}\frac{\gamma_{i}+1}{2}\Big{)},\end{gathered} $$

we have

$$ \begin{gathered}\displaystyle\big{\|}u\big{\|}_{\alpha,\gamma}^{2}=2^{|\gamma|-n+1}\prod\limits_{i=1}^{n}\Gamma^{2}\Big{(}\frac{\gamma_{i}+1}{2}\Big{)}\\ \displaystyle\times\,\Bigg{(}\,\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\frac{\big{|}{{}^{\gamma}\mathbf{T}_{x}^{y}}u(x)+u(y)\big{|}^{2}}{{|x|^{n+|\gamma|+4\alpha}}}\,\big{(}\omega_{-4\alpha,\gamma}(|x|)-\omega_{-4\alpha,\gamma}(0)\big{)}\,x^{\gamma}dx\,y^{\gamma}dy\\ \displaystyle-\,\int\limits_{𝕉^{n}_{+}}\int\limits_{𝕉^{n}_{+}}\frac{\big{|}{{}^{\gamma}\mathbf{T}_{x}^{y}}u(x)-u(y)|^{2}}{{|x|^{n+|\gamma|+4\alpha}}}\,\big{(}\omega_{-4\alpha,\gamma}(|x|)+\omega_{-4\alpha,\gamma}(0)\big{)}\,x^{\gamma}dx\,y^{\gamma}dy\Bigg{)}.\end{gathered} $$

Thus, we have validated (18).  ☐

6. Conclusion

In conclusion, we observe that the two approaches are possible to defining the class of generalized Bessel potentials \( \mathbf{B}^{\alpha}_{\gamma}(L_{p}^{\gamma}) \) of order \( \alpha \) in \( 𝕉^{n}_{+} \). The first is that \( u\in\mathbf{B}^{\alpha}_{\gamma}(L_{p}^{\gamma}) \) if \( u \) is a generalized convolution \( (G_{\alpha}^{\gamma}*\varphi)_{\gamma} \) for some \( \varphi\in L^{\gamma}_{p}(𝕉^{n}_{+}) \). This approach was presented in [3], which used B-hypersingular integrals. The second approach consists in endowing \( \mathbf{B}^{\alpha}_{\gamma}(L_{p}^{\gamma}) \) with the norm

$$ \big{\|}u\big{\|}_{\alpha,\gamma}^{2}=\int\limits_{𝕉^{n}_{+}}\big{(}1+|\xi|^{2}\big{)}^{\alpha}\,\big{|}\mathbf{F}_{\gamma}[u](\xi)\big{|}^{2}\,\xi^{\gamma}d\xi, $$

which can be written down on using the convolution kernel that generates the generalized Bessel potential. This expression shows that the quadratic interpolation between \( \|u\|_{\alpha,\gamma} \) and \( \|u\|_{\beta,\gamma} \) gives \( \|u\|_{\delta,\gamma} \), where \( \delta \) is the interpolation order \( \alpha(l-t)+\beta t \). The norm \( \|u\|_{\alpha,\gamma} \) is most convenient for studying the class of generalized Bessel potentials in \( 𝕉^{n}_{+} \).