# Radon-type transforms for holomorphic and Hermitian monogenic
functions111Dedicated to Prof. Dr. Franciscus Sommen

Ren Hu<EMAIL_ADDRESS>Department of Basis Courses, Tianjin Sino-German
University of Applied Sciences  Pan Lian<EMAIL_ADDRESS>School of
Mathematical Sciences, Tianjin Normal University

###### Abstract

The standard Radon transform of holomorphic functions is not always well
defined, as the integration of such functions over planes may not converge. In
this paper, we introduce new Radon-type transforms of co-(real)dimension $2$
for harmonic and holomorphic functions on the unit ball. These transforms are
abstractly defined as orthogonal projections onto spaces of complex harmonic
and holomorphic plane waves, respectively. The inversion formulas are derived
based on the dual transform, while the latter is defined as an integration on
a complex Stiefel manifold. Our transforms are extended to the Fock space and
give rise to a new transform defined on the entire $L^{2}(\mathbb{R}^{n})$
through the Segal-Bargmann transform. Furthermore, we develop these transforms
for Hermitian monogenic functions on the unit ball, thereby refining the
Szegö-Radon transform for monogenic functions introduced by Colombo, Sabadini
and Sommen.

###### keywords:

Szegö-Radon transform , Dirac operator , Fock space , spherical harmonics ,
Clifford algebra , Hermitian monogenic

###### MSC:

[2020] 30G35 , 44A12

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Real spherical harmonics
    2. 2.2 Complex spherical harmonics
  3. 3 Radon-type transforms on the unit ball
    1. 3.1 The Szegö-Radon transform
    2. 3.2 The dual transform
    3. 3.3 Inversion formulas
  4. 4 Radon-type transform on Fock space
    1. 4.1 The Bargmann-Radon transform
    2. 4.2 Dual transform and inversion
    3. 4.3 Radon transform on $L^{2}(\mathbb{R}^{n})$ via Segal-Bargmann transform
  5. 5 Radon-type transform for Hermitian monogenic functions
    1. 5.1 Hermitian Clifford analysis
    2. 5.2 Hermitian Szegö-Radon transform
  6. 6 Conclusions

##  1 Introduction

The Radon transform, which was first introduced by J. Radon in 1917 in [27],
is one of the most important integral transforms. The original integral
transform established that a differential function defined on $\mathbb{R}^{3}$
can be precisely determined by integrals of the function itself over planes.
This mathematical foundation not only gave rise to the resolution of
reconstruction problems but also spurred the development of algorithms,
ultimately culminating in the invention of computed tomography (CT), see e.g.
[8].

Beyond its application in imaging, the Radon transform has proven to be an
important tool in diverse areas, such as the study of partial differential
equations, and distribution theory. Pioneering works by S. Helgason, D.
Ludwig, and many others (see e.g. [2, 16, 17, 18, 24, 34]) extended the Radon
transform to higher-dimensional Euclidean spaces and more broadly in the
context of integral geometry. These resulted in explicit inversion formulae
for Radon transforms in both even and odd dimensions, the characterization of
Radon transforms on the space of rapidly decreasing $\mathcal{C}^{\infty}$
functions and distributions, etc.

The study of the Radon transform was subsequently extended in [5] in
Quaternionic analysis and in [30, 31, 32] by F. Sommen into Clifford analysis
(see [11]), which is a higher-dimensional generalization of holomorphic
functions in the complex plane. Within the latter framework, the Clifford-
Radon transform was introduced and the boundary value representations for the
Radon transforms of functions in $\mathbb{R}^{m}$ were given. However, the
inversion formula has to take account into the parity of the dimension. In
fact, when the dimension is even, half-integer powers of the Laplace operator
are involved. Additionally, a Paley-Wiener-Schwartz type theorem was proven
for the radial part of the Clifford-Radon transform.

Note that when considering the standard Radon transform for holomorphic or
monogenic functions, a significant challenge arises: the integration of such
functions over hyper-planes may not converge. Addressing this problem, in the
seminal work [6] of F. Colombo, I. Sabadini and F. Sommen, the Radon-type
transform for monogenic functions on the unit ball is defined as the
orthogonal projection onto a closed submodule of monogenic functions that
depend only on two variables. Meanwhile, as the answer to the reconstruction
problem, the dual of the Radon-type transform, expressed as an integral over a
Stiefel manifold (see Figure 1), has been defined. Furthermore, an inversion
formula is provided, which does not depend on the parity of the dimension.
This approach by orthogonal projection in terms of Hilbert modules is indeed
the only alternative to the Radon transform for those functions [6].

This Radon-type transform was subsequently extended to other domains,
including the Lie ball and the entire $\mathbb{R}^{m}$, with different
kernels, see e.g. [7, 28]. Explicit calculations of these transforms were
later studied in e.g. [13, 14, 22]. All these Radon-type transforms which
abstractly defined as orthogonal projection, correspond to either the special
orthogonal group ${\rm SO}(n)$ or its double cover ${\rm Spin}(n)$. Thus they
are naturally related with Gegenbauer polynomials. Therefore these transforms
should be considered of co-dimension 1. Indeed, a partial dual transform can
be given by the integration over the great circle which perpendicular to the
given vector $t$, see Figure 1 and [6, Section 4]. Moreover, their approach is
closely related with the Fourier transform on the sphere introduced by Sherman
in [29].

$t$$s$ Figure 1: The tuple $(t,s)$ in the orthogonal setting
$(t\in\mathbb{S}^{n-1},s\in\mathbb{S}^{n-2},t\perp s)$

The main aim of this paper is to establish suitable Radon-type transforms for
holomorphic functions of several variables on some domain in
$\mathbb{C}^{n}\cong\mathbb{R}^{2n}$. It became evident that the role of the
orthogonal group ${\rm O}(2n)$ (resp. Gegenbauer polynomials) in prior
research should be replaced by the unitary group ${\rm U}(n)$ (resp. Jacobi
polynomials) in the current setting. This realization inspires us to initiate
the definition of the Radon-type transform for harmonic functions on the unit
ball as the orthogonal projection (see Definition 3) onto a closed submodule
spanned by the complex harmonic plane waves (see Definition 1).

Naturally, the dual transform is given as an integration over the complex
Stiefel manifold of order 2, which is isomorphic to the homogeneous space

$St^{(2)}(\mathbb{C}^{n}):={\rm U}(n)/{\rm U}(n-2).$

Consequently, the new transform is of co-(real)dimension 2. It is the
fundamental distinction from transforms in [6, 7], determined by the nature of
holomorphic and monogenic functions. The crucial ingredient in determining the
explicit dual transform is the plane wave representation for the reproducing
kernels of complex harmonics obtained in [9] by De Bie, Sommen and Wutzig.
Inversion formulas are provided based on the dual transform, which are much
subtler than those in orthogonal setting, see Section 3.3. This complexity is
indeed as anticipated, as can be deduced from the harmonic analysis of complex
harmonics. Additionally, since complex spherical harmonics (see subsequent
Section 2.2]) are regarded as a refinement of the ordinary spherical
harmonics, our transform emerges as a refined version of Sherman’s Fourier
transform on the sphere in [29].

Our approach is extended to the entire classical Segal-Bargmann-Fock space as
well, but with an exponential kernel (see subsequent Eq. (31)). Furthermore,
we explicitly determine the image projection under the Segal-Bargmann
transform. This leads to a new Radon-type transform defined on the whole space
$L^{2}(\mathbb{R}^{n})$. The explicit formula reveals a connection with the
standard Radon transform (see subsequent Eq. (39)), although they are distinct
transforms. This approach is further developed for the Hermitian monogenic
functions, providing a further refinement of the Szegö-Radon transform on the
unit ball for harmonic functions in Section 3 and monogenic functions in [6].
The computation involved is quite complicated but still yields compact
inversion formulas. An alternative method for defining the Radon-type
transform for Hermitian monogenic functions would be utilizing the matrix
operator, as done in [15]. It differs from the current method.

For readers’ convenience, we summarize our results in Table 1.

Table 1: Summary of the results | Szegö-Radon transform on unit ball | Bargmann-Radon transform
---|---|---
harmonic | holomorphic | Hermitian monogenic | Fock | $L^{2}(\mathbb{R}^{n})$
orthogonal projection | $\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})\to\mathcal{H}\mathcal{L}_{t,s}^{2}(\mathbb{B}_{n})$ | $\mathcal{H}^{2}(\mathbb{B}_{n})\to\mathcal{H}_{t,s}^{2}(\mathbb{B}_{n})$ | $\mathcal{M}_{h}\mathcal{L}(\mathbb{B}^{n})\to\mathcal{M}_{h}\mathcal{L}_{\underline{\tau}}(\mathbb{B}^{n})$ | $\mathcal{F}(\mathbb{C}^{n})\to\mathcal{F}_{t,s}(\mathbb{C}^{n})$ |  Eq.(37)
plane waves | $\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}$ | $\left\langle{z},{\overline{s+t}}\right\rangle^{p}$ |  $\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}$ $\times(\underline{t}-\underline{s})(\underline{t}^{\dagger}+\underline{s}^{\dagger})$ | $\left\langle{z},{\overline{s+t}}\right\rangle^{p}$ |  Th.14
Radon kernel | $\left(\frac{2}{2-\left\langle{z},{\overline{s+t}}\right\rangle\left\langle{\bar{u}},{s+t}\right\rangle-\left\langle{\bar{z}},{s-t}\right\rangle\left\langle{u},{\overline{s-t}}\right\rangle}\right)^{n}$ | $\left(\frac{2}{2-\left\langle{z},{\overline{s+t}}\right\rangle\left\langle{\bar{u}},{s+t}\right\rangle}\right)^{n}$ | $\frac{1}{4}\left(\frac{2}{2-\left\langle{z},{\overline{s+t}}\right\rangle\left\langle{\bar{u}},{s+t}\right\rangle-\left\langle{\bar{z}},{s-t}\right\rangle\left\langle{u},{\overline{s-t}}\right\rangle}\right)^{n}\underline{\tau}\underline{\tau}^{\dagger}$ | $\exp\left\\{\frac{\left\langle{z},{\overline{s+t}}\right\rangle\left\langle{\bar{w}},{s+t}\right\rangle}{2}\right\\}$ |  Eq.(38)
Inversion formula |  Th.9 |  Th.6 |  Th.21 |  Th.13 |  Th.16

Finally, it is interesting to remark the following. Whereas the Radon-type
transform in [6] corresponds to the orthogonal cases (co-dimension $1$), the
transforms in present work correspond to the unitary cases (co-dimension $2$).
It is reasonable to deduce that a Radon-type transform will exist for the
symplectic cases as well (co-dimension 3). Exploring these transforms (co-
dimension $k$) in a unified way would be an interesting avenue for further
consideration.

The remaining of this paper is organized as follows. In Section 2, we recall
basics of spherical harmonics and their reproducing kernels. Section 3 is
devoted to the Radon-type transform on the unit ball for holomorphic and
harmonic functions, which is the main results of this paper. In Section 4, we
investigate Radon-type transforms on the Fock space and the whole
$L^{2}(\mathbb{R}^{n})$. The subsequent Section 5 focuses on the Radon-type
transform for Hermitian monogenics. Conclusions and unsolved problems can be
found at the end of this paper.

##  2 Preliminaries

###  2.1 Real spherical harmonics

Let $\mathbb{R}^{n}$ be the $n$-dimensional Euclidean space with the standard
inner product $\langle x,y\rangle:=\sum_{j=1}^{n}x_{j}y_{j}$. The induced norm
of a vector $x$ is given by $|x|=\sqrt{\langle x,x\rangle}$.

Now, we consider the space $\mathcal{P}$ consisting of complex-valued
polynomials on $\mathbb{R}^{n}$. Using multi-index notation, any function in
$\mathcal{P}$ can be expressed as

$P(x)=\sum_{\left|{\alpha}\right|}c_{\alpha}x^{\alpha}=\sum_{\left|{\alpha}\right|}c_{\alpha}x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}\cdots
x_{n}^{\alpha_{n}},\qquad c_{\alpha}\in\mathbb{C},$

where $\alpha=(\alpha_{1},\alpha_{2},\ldots,\alpha_{n})\in\mathbb{N}^{n}$ is
an index vector with length
$\left|{\alpha}\right|=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}$.

Let $\mathbb{E}=\sum_{j=1}^{n}x_{j}\partial_{x_{j}}$ represent the Euler
operator acting on the space $\mathcal{P}$. A polynomial $P(x)$ is called
homogeneous of degree $k$ if it is an eigenfunction of the Euler operator with
the corresponding eigenvalue $k$. The set of homogeneous polynomials of degree
$k$ is denoted by

$\mathcal{P}_{k}:=\\{P(x):\mathbb{E}P(x)=kP(x)\\}.$

Clearly, it holds that
$\mathcal{P}=\mathop{\bigoplus}_{k=0}^{\infty}\mathcal{P}_{k}$.

The Fischer inner product, defined on $\mathcal{P}$, is given by

$\left\langle{P(x)},{Q(x)}\right\rangle_{\partial}:=\left[\overline{P(\partial)}Q(x)\right]_{x=0},$
(1)

where $P(\partial)$ is obtained by replacing each $x_{j}$ with
$\partial_{x_{j}}$, and $\overline{P(x)}$ denotes the usual complex
conjugation of $P(x)$. It is easy to verify that two homogeneous polynomials
of different degrees are orthogonal with respect the Fischer inner product
(1).

Furthermore, it is seen that $x_{j}$ and $\partial_{x_{j}}$ act as adjoint
operators on the space $\mathcal{P}$ with respect the Fischer inner product
(1). Consequently, the dual of $|x|^{2}$ is given by the Laplacian operator
$\Delta=\sum_{j=1}^{n}\partial^{2}_{x_{j}}$. More precisely, we have

$\left\langle{|x|^{2}P(x)},{Q(x)}\right\rangle_{\partial}=\left\langle{P(x)},{\Delta
Q(x)}\right\rangle_{\partial}.$

In this context, $|x|^{2}$ acts as a raising operator, while $\Delta$ acts as
a lowering operator. Together with the Euler operator $\mathbb{E}$, they
generate a Lie algebra, which is isomorphic with

$\mathfrak{sl}_{2}\,\cong\,\left\\{\frac{\Delta}{2},\,\frac{|x|^{2}}{2},\,\mathbb{E}+\frac{n}{2}\right\\}.$

The set of homogeneous polynomials of degree $k$ that are also harmonic,
namely they are null-solutions of the Laplace operator, is denote by

$\mathcal{H}_{k}=\\{P(x):P(x)\in\mathcal{P}_{k},\,\Delta P(x)=0\\}.$

It is well-known that the dimension of $\mathcal{H}_{k}$ is

$\dim\mathcal{H}_{k}=\frac{2k+n-2}{k+n-2}\binom{k+n-2}{n-2},$

when $n\geq 2$. It is noteworthy that each space $\mathcal{H}_{k}$ is
invariant under the usual action of the orthogonal group ${\rm O}(n)$. The
restriction of such a homogeneous harmonic polynomial to the unit sphere

$\mathbb{S}^{n-1}:=\left\\{(x_{1},\ldots,x_{n})\in\mathbb{R}^{n}:|x_{1}|^{2}+\cdots+|x_{n}|^{2}=1\right\\}$

is referred to as a spherical harmonic.

An important result in harmonic analysis is the Fischer decomposition, stated
as follows:

######  Lemma 1 (Fischer decomposition).

[1] For the space $\mathcal{P}_{k}$ of homogeneous polynomials, it holds that

$\mathcal{P}_{k}=\mathop{\bigoplus}_{j=0}^{\lfloor
k/2\rfloor}|x|^{2j}\mathcal{H}_{k-2j}.$

It is well understood nowadays that this decomposition is uniquely determined
by the action of the Howe dual pair ${\rm SO}(n)\times\mathfrak{sl}_{2}$.
Specifically, for each fixed $k$, the direct sum
$\bigoplus_{j}|x|^{2j}\mathcal{H}_{k}$ forms an irreducible
$\mathfrak{sl}_{2}$-module. The given sum represents a decomposition into an
infinite number of isomorphic copies of an irreducible ${\rm SO}(n)$-module
$\mathcal{H}_{k}$, see e.g. [3].

###  2.2 Complex spherical harmonics

When considering functions defined on an even-dimensional vector space, one
can introduce a complex structure to identify $\mathbb{R}^{2n}$ with the
complex vector space $\mathbb{C}_{n}$. In this case, the role of the
orthogonal group ${\rm O}(2n)$ in the previous section will be replaced by the
unitary group ${\rm U}(n)$.

Let $z=(z_{1},z_{2},\ldots,z_{n})\in\mathbb{C}^{n}$, where
$z_{j}=x_{j}+ix_{n+j}$ with $x_{j}\in\mathbb{R}$ for $j=1,\ldots,n$. The
complex conjugation is given by $\bar{z}_{j}=x_{j}-ix_{n+j}$. Let
$\bar{z}=(\bar{z}_{1},\bar{z}_{2},\ldots,\bar{z}_{n})$. For two complex
vectors $z,u\in\mathbb{C}^{n}$, the Hermitian inner product is defined as

$\left\langle{z},{\bar{u}}\right\rangle\,:=\,\sum_{j=1}^{n}z_{j}\,\bar{u}_{j}$
(2)

The complex derivatives are defined as usual by

$\partial_{z_{j}}:=\frac{1}{2}\left(\partial_{x_{j}}-i\partial_{x_{n+j}}\right)\quad\text{and}\quad\partial_{\bar{z}_{j}}:=\frac{1}{2}\left(\partial_{x_{j}}+i\partial_{x_{n+j}}\right).$
(3)

Now, we split the Euler operator $\mathbb{E}$ into two complex Euler
operators,

$\mathbb{E}_{z}:=\sum_{j=1}^{n}z_{j}\partial_{z_{j}}\qquad\text{and}\qquad\mathbb{E}_{\bar{z}}:=\sum_{j=1}^{n}\bar{z}_{j}\partial_{\bar{z}_{j}}.$

It is easy to see that $\mathbb{E}=\mathbb{E}_{z}+\mathbb{E}_{\bar{z}}$. This
allows us to refine the notion of a homogeneous polynomial of degree $k$ to a
bi-homogeneous polynomial of degree $(p,q)$ with $p+q=k$. More precisely, the
space of $(p,q)$-homogeneous polynomials is defined by

$\mathcal{P}_{p,q}:=\left\\{P(z,\bar{z}):\,\mathbb{E}_{z}P(z,\bar{z})=pP(z,\bar{z}),\,\mathbb{E}_{\bar{z}}P(z,\bar{z})=qP(z,\bar{z})\right\\}.$

The space of homogeneous polynomials $\mathcal{P}_{k}$ can be decomposed into
bi-homogeneous polynomials, i.e.

$\mathcal{P}_{k}=\mathop{\bigoplus}_{j=0}^{k}\mathcal{P}_{j,k-j}.$

For simplicity, we will use the notion $P(z)$ rather than $P(z,\bar{z})$ to
indicate a function that depends on both $z$ and $\bar{z}$ when there is no
confusion.

The Fischer inner product is defined as in (1), i.e.

$\left\langle{P(z)},{Q(z)}\right\rangle_{\partial}\,:=\,\left[\overline{P(\partial)}Q(z)\right]_{z=0}.$
(4)

Here $P(\partial)$ stands for the substitution of every $z_{j}$ by
$\partial_{\bar{z}_{j}}$ and every $\bar{z}_{j}$ by $\partial_{z_{j}}$ in
$P(z)$.

It is seen that with respect to the complex Fischer inner product, the
homogeneous polynomial (as a function in $z$)

$Z_{p,q}(z,u)=\frac{\left\langle{z},{\bar{u}}\right\rangle^{p}\left\langle{\bar{z}},{u}\right\rangle^{q}}{p!q!}$

acts as a reproducing kernel of $\mathcal{P}_{r,s}$. Namely, for any
homogeneous $P_{r,s}\in\mathcal{P}_{r,s}$, we have

$\left\langle
Z_{p,q}(z,u),P_{r,s}(z)\right\rangle_{\partial}=\delta_{pr}\delta_{qs}P_{r,s}(u),$

where $\delta_{pr}$ is the Kronecker delta symbol.

Let $\Delta_{z}:=\sum_{j=1}^{n}\partial_{\bar{z}_{j}}\partial_{z_{j}}$. It is
seen that the complex Laplacian is proportional to the one in real case, i.e.
$\Delta=4\Delta_{z}$. A spherical harmonic function of bi-degree $(p,q)$ is
defined as harmonic bi-homogeneous polynomials of the same order

$\mathcal{H}_{p,q}=\\{H(z):\,H(z)\in\mathcal{P}_{p,q},\,\Delta_{z}H(z)=0\\}$

with

$\dim\mathcal{H}_{p,q}=\frac{n+p+q-1}{n-1}\binom{q+n-2}{n-2}\binom{p+n-2}{n-2}.$
(5)

Together with the two complex Euler operators $\mathbb{E}_{z}$ and
$\mathbb{E}_{\bar{z}}$, as well as the squared norm $|z|^{2}$, they generate a
Lie algebra that is isomorphic to

$\mathfrak{gl}_{2}\,\cong\,\left\\{\Delta_{z},\,\|z\|^{2},\,\mathbb{E}_{z}+\frac{n}{2},\,\mathbb{E}_{\bar{z}}+\frac{n}{2}\right\\}.$

The spherical $L^{2}$ inner product on $\mathcal{H}_{p,q}$ is given by

$\langle
P(z),Q(z)\rangle_{\mathbb{S}}:=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\overline{P(z)}Q(z)\,{\rm
d}\sigma(z),$ (6)

where ${\rm d}\sigma(z)$ is the usual Lebesgue measure on the sphere
$\mathbb{S}^{2n-1}$ with surface $\omega_{2n-1}=2\pi^{n}/\Gamma(n)$.

The spherical and the complex Fischer inner products are related as follows.

######  Lemma 2.

[10] For a spherical harmonic $H_{p,q}\in\mathcal{H}_{p,q}$ and a homogeneous
polynomial $P_{p,q}\in\mathcal{P}_{p,q}$, it holds that

$(n)_{p+q}\left\langle{H_{p,q}(z)},{P_{p,q}(z)}\right\rangle_{\mathbb{S}}=\left\langle{H_{p,q}(z)},{P_{p,q}(z)}\right\rangle_{\partial},$
(7)

where $(n)_{p+q}$ is the Pochhammer symbol defined by
$(n)_{p+q}=\Gamma(n+p+q)/\Gamma(n)$.

The Howe dual pair ${\rm U}(n)\times\mathfrak{gl}_{2}$ leads to the following
decomposition in the complex setting (see e.g. [10, 23]).

######  Lemma 3.

For the space of homogeneous polynomial $\mathcal{P}_{p,q}$ of degree $(p,q)$,
we have

$\mathcal{P}_{p,q}=\mathop{\bigoplus}_{j=0}^{\min(p,q)}|z|^{2j}\mathcal{H}_{p-j,q-j},$

where
$|z|^{2}=\sum_{j=1}^{n}|z_{j}|^{2}=\left\langle{z},{\bar{z}}\right\rangle.$

It is worth noting that the unitary group ${\rm U}(n)$ acts irreducibly on
each space $\mathcal{H}_{p,q}$. Furthermore, representations of ${\rm U}(n)$
on different spaces $\mathcal{H}_{p,q}$ are inequivalent, and we have

$L^{2}(\mathbb{S}^{2n-1})=\bigoplus_{p,q=0}^{\infty}\mathcal{H}_{p,q}.$

##  3 Radon-type transforms on the unit ball

The main aim of this section is to establish Radon-type transforms for
harmonic and holomorphic functions on the unit ball. These transforms will be
abstractly defined as orthogonal projections from Hilbert spaces of harmonic
(resp. holomorphic) functions onto specific closed subspaces of plane waves
that depend only on two variables. Dual transforms and inversion formulas will
also be provided.

###  3.1 The Szegö-Radon transform

Let $\mathbb{B}_{n}$ be the open unit ball in $\mathbb{C}^{n}$,

$\mathbb{B}_{n}:=\left\\{(z_{1},\ldots,z_{n})\in\mathbb{C}^{n}:|z_{1}|^{2}+\cdots+|z_{n}|^{2}<1\right\\}.$

We denote $\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})$ for the Hilbert
space of complex harmonic functions $f(z):\mathbb{B}_{n}\to\mathbb{C}$, whose
extensions to the unit sphere are square integrable, i.e.

$\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\overline{f(z)}f(z)\,{\rm
d}\sigma(z)<\infty.$

We also need the following complex Stiefel manifold of order $2$, which is
isomorphic to the homogeneous space

$St^{(2)}(\mathbb{C}^{n}):={\rm U}(n)/{\rm U}(n-2).$

We consider the complex plane waves (see e.g. [10, 23]).

######  Definition 1.

Let $z,s,t\in\mathbb{C}^{n}$. We consider the following complex plane waves

$f_{t,s}^{(p,q)}(z):=\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}.$
(8)

Especially, when $(t,s)$ lies on the complex Stiefel manifold
$St^{(2)}(\mathbb{C}^{n})$, it is seen that $\Delta_{z}f_{t,s}^{(p,q)}(z)=0$.
In this case, it is called by the complex harmonic plane wave.

Here, the complex vector $t$ lies on the unit sphere $\mathbb{S}^{2n-1}$,
while $s$ is on the sphere $\mathbb{S}^{2n-3}$. The vectors $t$ and $s$ are
orthogonal with respect to the Hermitian inner product (2). While in the
orthogonal case, the tuple $(t,s)$ satisfies
$t\in\mathbb{S}^{n-1},s\in\mathbb{S}^{n-2}$ and $\langle t,s\rangle=0$, see
Figure 1.

It is obvious that the complex harmonic plane waves of different bi-degree are
orthogonal to each other under the spherical $L^{2}$ inner product (6). In the
following, we calculate their norms explicitly. It is more convenient to use
the Fischer inner product in (4) rather than computing the integral directly
over the sphere.

######  Theorem 1.

The complex harmonic plane waves $f_{t,s}^{(p,q)}(z)$ form an orthogonal
subset of $\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})$. In particular, we
have

$\begin{split}\left\langle
f_{t,s}^{(p,q)}(z),g_{t,s}^{(w,v)}(z)\right\rangle_{\mathbb{S}}=\left\\{\begin{array}[]{ll}\gamma_{p,q},&p=w,\,q=v,\\\
0,&\text{otherwise},\end{array}\right.\end{split}$ (9)

where $\gamma_{p,q}=2^{p+q}p!q!/(n)_{p+q}$.

For this purpose, we do some calculations first.

######  Lemma 4.

Let $\left\langle{z},{\bar{u}}\right\rangle=\sum_{j=1}^{n}z_{j}\bar{u}_{j}$,
then we have

$\begin{split}\left\langle{\partial_{z}},{a}\right\rangle^{p}\left\langle{\bar{z}},{b}\right\rangle^{v}=0,\qquad\qquad\left\langle{\partial_{\bar{z}}},{b}\right\rangle^{q}\left\langle{z},{a}\right\rangle^{w}=0,\end{split}$
(10)

and

$\begin{split}\left[\overline{\left\langle{\partial_{\bar{z}}},{a}\right\rangle^{p}}\left\langle{z},{a}\right\rangle^{w}\right]_{z=0}=\left\\{\begin{array}[]{rr}0,&p\neq
w,\\\ p!\left|{a}\right|^{2p},&p=w.\\\
\end{array}\right.\quad\left[\overline{\left\langle{\partial_{z}},{b}\right\rangle^{q}}\left\langle{\bar{z}},{b}\right\rangle^{v}\right]_{z=0}=\left\\{\begin{array}[]{rr}0,&q\neq
v,\\\ q!\left|{b}\right|^{2q},&q=v.\end{array}\right.\end{split}$ (11)

###### Proof.

Equations in (10) are easy to obtain. Indeed, since

$\displaystyle\partial_{z_{j}}\left(\sum_{i=1}^{n}\bar{z}_{i}b_{i}\right)^{v}$
$\displaystyle=\frac{1}{2}\left(\partial_{x_{j}}-i\partial_{x_{n+j}}\right)\left(\sum_{i=1}^{n}\left(x_{i}-ix_{n+i}\right)b_{i}\right)^{v}$
$\displaystyle=\frac{1}{2}\partial_{x_{j}}\left(\sum_{i=1}^{n}\left(x_{i}-ix_{n+i}\right)b_{i}\right)^{v}-\frac{1}{2}i\partial_{x_{n+j}}\left(\sum_{i=1}^{n}\left(x_{i}-ix_{n+i}\right)b_{i}\right)^{v}$
$\displaystyle=\frac{1}{2}v\left(\sum_{i=1}^{n}\left(x_{i}-ix_{n+i}\right)b_{i}\right)^{v-1}b_{j}-\frac{1}{2}v\left(\sum_{i=1}^{n}\left(x_{i}-ix_{n+i}\right)b_{i}\right)^{v-1}b_{j}$
$\displaystyle=0,$

it yields that

$\begin{split}\left\langle{\partial_{z}},{a}\right\rangle^{p}\left\langle{\bar{z}},{b}\right\rangle^{v}=\left(\sum_{j=1}^{n}a_{j}\partial_{z_{j}}\right)^{p}\left(\sum_{i=1}^{n}\bar{z}_{i}b_{i}\right)^{v}=0.\end{split}$

Similarly, it holds that
$\left\langle{\partial_{\bar{z}}},{b}\right\rangle^{q}\left\langle{z},{a}\right\rangle^{w}=0$.

Now we prove (11). Since

$\displaystyle\partial_{\bar{z}_{j}}\left\langle{\bar{z}},{b}\right\rangle^{v}$
$\displaystyle=\frac{1}{2}\left(\partial_{x_{j}}+i\partial_{x_{n+j}}\right)\left(\sum_{i=1}^{n}\left(x_{i}-ix_{n+i}\right)b_{i}\right)^{v}$
$\displaystyle=\frac{1}{2}\partial_{x_{j}}\left(\sum_{i=1}^{n}\left(x_{i}-ix_{n+i}\right)b_{i}\right)^{v}+\frac{1}{2}i\partial_{x_{n+j}}\left(\sum_{i=1}^{n}\left(x_{i}-ix_{n+i}\right)b_{i}\right)^{v}$
$\displaystyle=\frac{1}{2}vb_{j}\left(\sum_{i=1}^{n}\left(x_{i}-ix_{n+i}\right)b_{i}\right)^{v-1}+\frac{1}{2}vb_{j}\left(\sum_{i=1}^{n}\left(x_{i}-ix_{n+i}\right)b_{i}\right)^{v-1}$
$\displaystyle=vb_{j}\left(\sum_{i=1}^{n}\left(x_{i}-ix_{n+i}\right)b_{i}\right)^{v-1}$
$\displaystyle=vb_{j}\left\langle{\bar{z}},{b}\right\rangle^{v-1},$

it follows that

$\displaystyle\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle^{q}\left\langle{\bar{z}},{b}\right\rangle^{v}$
$\displaystyle=\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle^{q-1}\sum_{j=1}^{n}\bar{b}_{j}\partial_{\bar{z}_{j}}\left(\sum_{i=1}^{n}\bar{z}_{i}b_{i}\right)^{v}$
$\displaystyle=\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle^{q-1}\sum_{j=1}^{n}|b_{j}|^{2}v\left\langle{\bar{z}},{b}\right\rangle^{v-1}$
$\displaystyle=v\left|{b}\right|^{2}\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle^{q-1}\left\langle{\bar{z}},{b}\right\rangle^{v-1},$

in which $\left|{b}\right|^{2}=\left\langle{\bar{b}},{b}\right\rangle$.
Repeating this step by $q$ times, we obtain

$\begin{split}\left[\overline{\left\langle{\partial_{z}},{b}\right\rangle^{q}}\left\langle{\bar{z}},{b}\right\rangle^{v}\right]_{z=0}=\left\\{\begin{array}[]{rr}0,&q>v,\\\
q!\left|{b}\right|^{2q},&q=v,\\\ 0,&q<v.\end{array}\right.\end{split}$

Note that when $q>v$, the order of derivative is higher than the order of
variable $z$ which becomes to $0$ naturally; when $q<v$, the order of
derivative is lower than the variable $z$, the result is still $0$ when $z=0$.
The other equation can be obtained in a similar way. ∎

Now we are able to prove Theorem 1.

###### Proof of Theorem 1.

Let
$g^{(w,v)}_{t,s}(z)=\left\langle{z},{\overline{s+t}}\right\rangle^{w}\left\langle{\bar{z}},{s-t}\right\rangle^{v}$
with $\ell=w+v$, and $f^{(p,q)}_{t,s}(z)$ is defined similarly. By (7) in
Lemma 2, we have

$\displaystyle\left\langle
f^{(p,q)}_{t,s}(z),g^{(w,v)}_{t,s}(z)\right\rangle_{\mathbb{S}}$
$\displaystyle=\frac{1}{(n)_{p+q}}\left\langle{f^{(p,q)}_{t,s}(z)},{g^{(w,v)}_{t,s}(z)}\right\rangle_{\partial}$
$\displaystyle=\frac{1}{(n)_{p+q}}\left[\overline{f^{(p,q)}_{t,s}(\partial)}\,g^{(w,v)}_{t,s}(z)\right]_{z=0}.$

Now we consider

$\begin{split}\overline{f^{(p,q)}_{t,s}(\partial)}\,g^{(w,v)}_{t,s}(z)&=\left\langle{\partial_{z}},{s+t}\right\rangle^{p}\left\langle{\partial_{\bar{z}}},{\overline{s-t}}\right\rangle^{q}\left\langle{z},{\overline{s+t}}\right\rangle^{w}\left\langle{\bar{z}},{s-t}\right\rangle^{v},\end{split}$
(12)

when $z=0$. Denote $a=\overline{s+t}$ and $b=s-t$. Then (12) becomes

$\displaystyle\overline{f^{(p,q)}_{t,s}(\partial)}\,g^{(w,v)}_{t,s}(z)=\left\langle{\partial_{z}},{\bar{a}}\right\rangle^{p}\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle^{q}\left\langle{z},{a}\right\rangle^{w}\left\langle{\bar{z}},{b}\right\rangle^{v}.$
(13)

By Lemma 4, we have

$\displaystyle\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle^{q}\left\langle{z},{a}\right\rangle^{w}\left\langle{\bar{z}},{b}\right\rangle^{v}$
$\displaystyle=$
$\displaystyle\,\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle^{q-1}\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle\left(\left\langle{z},{a}\right\rangle^{w}\left\langle{\bar{z}},{b}\right\rangle^{v}\right)$
$\displaystyle=$
$\displaystyle\,\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle^{q-1}\left(\underbrace{\left(\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle\left\langle{z},{a}\right\rangle^{w}\right)}_{0}\left\langle{\bar{z}},{b}\right\rangle^{v}+\left\langle{z},{a}\right\rangle^{w}\left(\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle\left\langle{\bar{z}},{b}\right\rangle^{v}\right)\right)$
$\displaystyle=$
$\displaystyle\,\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle^{q-1}\left\langle{z},{a}\right\rangle^{w}\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle\left\langle{\bar{z}},{b}\right\rangle^{v}$
$\displaystyle=$
$\displaystyle\,\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle^{q-2}\left(\underbrace{\left(\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle\left\langle{z},{a}\right\rangle^{w}\right)}_{0}\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle\left\langle{\bar{z}},{b}\right\rangle^{v}+\left\langle{z},{a}\right\rangle^{w}\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle^{2}\left\langle{\bar{z}},{b}\right\rangle^{v}\right)$
$\displaystyle=$
$\displaystyle\,\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle^{q-2}\left\langle{z},{a}\right\rangle^{w}\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle^{2}\left\langle{\bar{z}},{b}\right\rangle^{v}$
$\displaystyle=$ $\displaystyle\,\ldots$ $\displaystyle=$
$\displaystyle\,\left\langle{z},{a}\right\rangle^{w}\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle^{q}\left\langle{\bar{z}},{b}\right\rangle^{v}$
$\displaystyle=$
$\displaystyle\,q!\left|{b}\right|^{q}\left\langle{z},{a}\right\rangle^{w}.$

Using the above and Lemma 4 again, we obtain

$\displaystyle\left\langle{\partial_{z}},{\bar{a}}\right\rangle^{p}\left\langle{\partial_{\bar{z}}},{\bar{b}}\right\rangle^{q}\left\langle{z},{a}\right\rangle^{w}\left\langle{\bar{z}},{b}\right\rangle^{v}$
$\displaystyle=\left\langle{\partial_{z}},{\bar{a}}\right\rangle^{p}q!\left|{b}\right|^{2q}\left\langle{z},{a}\right\rangle^{w}$
$\displaystyle=q!\left|{b}\right|^{2q}\left\langle{\partial_{z}},{\bar{a}}\right\rangle^{p}\left\langle{z},{a}\right\rangle^{w}$
$\displaystyle=p!q!\left|{a}\right|^{2p}\left|{b}\right|^{2q}.$

Note that now the complex vector $t$ is on the unit sphere $\mathbb{S}^{2n-1}$
of (real) dimension $2n-1$ and $s$ on the sphere $\mathbb{S}^{2n-3}$ that is
orthogonal to $t$ with respect to the Hermitian inner product (2). Therefore,
we have

$\displaystyle\left|{a}\right|^{2p}$
$\displaystyle=\left\langle{\overline{s+t}},{s+t}\right\rangle^{p}$
$\displaystyle=\left(\left\langle{\bar{s}},{s+t}\right\rangle+\left\langle{\bar{t}},{s+t}\right\rangle\right)^{p}$
$\displaystyle=\left(\left\langle{\bar{s}},{s}\right\rangle+\left\langle{\bar{s}},{t}\right\rangle+\left\langle{\bar{t}},{s}\right\rangle+\left\langle{\bar{t}},{t}\right\rangle\right)^{p}$
$\displaystyle=2^{p}.$

Similarly, we get
$\left|{b}\right|^{2q}=\left\langle{\overline{s-t}},{s-t}\right\rangle^{q}=2^{q}$.
Substituting these in (13), it yields

$\left[\overline{f^{(p,q)}_{t,s}(\partial)}g^{(p,q)}_{t,s}(z)\right]_{z=0}=2^{p+q}p!q!.$
(14)

By Lemma 2, we obtain the desired norm

$\displaystyle\left\langle
f^{(p,q)}_{t,s}(z),f^{(p,q)}_{t,s}(z)\right\rangle_{\mathbb{S}}=\frac{2^{p+q}p!q!}{(n)_{p+q}}.$

∎

Let us define the subspace spanned by the complex harmonic plane waves. It
will serve as the image space for our Radon transform later.

######  Definition 2.

For any given $t\in\mathbb{S}^{2n-1}$ and $s\in\mathbb{S}^{2n-3}$, which are
orthogonal with respect to the Hermitian inner product (2), the completion of
all finite linear combinations of complex plane waves of the form

$\sum_{p,q}\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}\alpha_{p,q},\qquad\alpha_{p,q}\in\mathbb{C},$

is a closed subspace of $\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})$ under
the spherical inner product in (6). We denote it by
$\mathcal{H}\mathcal{L}_{t,s}^{2}(\mathbb{B}_{n})$.

Now, we can define the Radon-type transform for harmonic functions on the unit
ball.

######  Definition 3.

Assume $n\geq 2$. The complex harmonic Szegö-Radon transform is defined as the
orthogonal projection

$\mathcal{R}_{t,s}:\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})\to\mathcal{H}\mathcal{L}_{t,s}^{2}(\mathbb{B}_{n}).$
(15)

An equivalent integral expression of this transform can be given. To this end,
we first consider the following series of complex harmonic plane waves.

######  Theorem 2.

The series

$K_{t,s}(z,u)=\sum_{p=0}^{\infty}\sum_{q=0}^{\infty}\frac{1}{\gamma_{p,q}}\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}\left\langle{\bar{u}},{s+t}\right\rangle^{p}\left\langle{u},{\overline{s-t}}\right\rangle^{q}$
(16)

converges when $\left|{z}\right|\leq 1$ and $\left|{u}\right|<1$, where
$\gamma_{p,q}$ is defined in (9). Moreover, it admits a closed form,

$K_{t,s}(z,u)=\left(\frac{2}{2-\left\langle{z},{\overline{s+t}}\right\rangle\left\langle{\bar{u}},{s+t}\right\rangle-\left\langle{\bar{z}},{s-t}\right\rangle\left\langle{u},{\overline{s-t}}\right\rangle}\right)^{n}.$

###### Proof.

Recall that the Appell’s hypergeometric function

$F_{2}\left(\alpha,\beta,\beta^{\prime},\gamma,\gamma^{\prime};x,y\right)=\sum_{m,n=0}^{\infty}\frac{(\alpha)_{m+n}(\beta)_{m}(\beta^{\prime})_{n}}{(\gamma)_{m}(\gamma^{\prime})_{n}}\frac{x^{m}y^{n}}{m!n!}$

converges when $\left|{x}\right|+\left|{y}\right|<1$, see [25]. Putting
$\displaystyle
x=\frac{\left\langle{z},{\overline{s+t}}\right\rangle\left\langle{\bar{u}},{s+t}\right\rangle}{2}$
and $\displaystyle
y=\frac{\left\langle{\bar{z}},{s-t}\right\rangle\left\langle{u},{\overline{s-t}}\right\rangle}{2}$
in $F_{2}$ and putting $m=p$ and $n=q$, it yields that

$\displaystyle
F_{2}\left(n,\beta,\beta^{\prime},\beta,\beta^{\prime};\frac{\left\langle{z},{\overline{s+t}}\right\rangle\left\langle{\bar{u}},{s+t}\right\rangle}{2},\frac{\left\langle{\bar{z}},{s-t}\right\rangle\left\langle{u},{\overline{s-t}}\right\rangle}{2}\right)$
$\displaystyle=$
$\displaystyle\,\sum_{p,q=0}^{\infty}\frac{(n)_{p+q}(\beta)_{p}(\beta^{\prime})_{q}}{(\beta)_{p}(\beta^{\prime})_{q}}\frac{\left(\left\langle{z},{\overline{s+t}}\right\rangle\left\langle{\bar{u}},{s+t}\right\rangle\right)^{p}\left(\left\langle{\bar{z}},{s-t}\right\rangle\left\langle{u},{\overline{s-t}}\right\rangle\right)^{q}}{2^{p}2^{q}p!q!}$
$\displaystyle=$
$\displaystyle\,\sum_{p,q=0}^{\infty}\frac{(n)_{p+q}}{2^{p}2^{q}p!q!}\left(\left\langle{z},{\overline{s+t}}\right\rangle\left\langle{\bar{u}},{s+t}\right\rangle\right)^{p}\left(\left\langle{\bar{z}},{s-t}\right\rangle\left\langle{u},{\overline{s-t}}\right\rangle\right)^{q}$
$\displaystyle=$ $\displaystyle\,K_{t,s}(z,u).$

Therefore, the series (16) converges when

$\left|{\frac{\left\langle{z},{\overline{s+t}}\right\rangle\left\langle{\bar{u}},{s+t}\right\rangle}{2}}\right|+\left|{\frac{\left\langle{\bar{z}},{s-t}\right\rangle\left\langle{u},{\overline{s-t}}\right\rangle}{2}}\right|<1.$

Note that $t\perp s$ and $|t|=|s|=1$, it follows that $(t+s)\perp(t-s)$ and
$\left|{t+s}\right|=\left|{t-s}\right|=\sqrt{2}$. Thus for any $|z|\leq 1$, we
have

$\left|{\left\langle{z},{\overline{s+t}}\right\rangle}\right|^{2}+\left|{\left\langle{z},{\overline{s-t}}\right\rangle}\right|^{2}\leq
2.$ (17)

Now, it follows from (17) that

$\displaystyle\left|{\frac{\left\langle{z},{\overline{s+t}}\right\rangle\left\langle{\bar{u}},{s+t}\right\rangle}{2}}\right|+\left|{\frac{\left\langle{\bar{z}},{s-t}\right\rangle\left\langle{u},{\overline{s-t}}\right\rangle}{2}}\right|$
$\displaystyle=$
$\displaystyle\,\frac{1}{2}\left(\left|{\left\langle{z},{\overline{s+t}}\right\rangle}\right|\left|{\left\langle{\bar{u}},{s+t}\right\rangle}\right|+\left|{\left\langle{\bar{z}},{s-t}\right\rangle}\right|\left|{\left\langle{u},{\overline{s-t}}\right\rangle}\right|\right)$
$\displaystyle=$
$\displaystyle\,\frac{1}{4}\left(\left|{\left\langle{z},{\overline{s+t}}\right\rangle}\right|^{2}+\left|{\left\langle{\bar{u}},{s+t}\right\rangle}\right|^{2}+\left|{\left\langle{\bar{z}},{s-t}\right\rangle}\right|^{2}+\left|{\left\langle{u},{\overline{s-t}}\right\rangle}\right|^{2}\right)$
$\displaystyle<$ $\displaystyle\,1,$

for any $|z|\leq 1$ and $|u|<1$. This proves the convergence claim.

The closed form follows from the hypergeometric expression, see [33, p.54
(9)]. This completes the proof. ∎

The series $K_{t,s}(z,u)$ in (16) will serve as the integral kernel of the
Szegö-Radon transform.

######  Definition 4.

Let the tuple $(t,s)$ lie on the Stiefel manifold $St^{(2)}(\mathbb{C}^{n})$.
The function $K_{t,s}(z,u)$ is called the complex harmonic Szegö-Radon kernel
associated to the tuple $(t,s)$.

We give some properties of this kernel function here.

######  Proposition 1.

The complex harmonic Szegö-Radon kernel $K_{t,s}(z,u)$ has the following
properties:

  1. 1.

$\overline{K_{t,s}(z,u)}=K_{t,s}(u,z)$;

  2. 2.

$K_{t,s}(z,u)$ is a reproducing kernel of
$\mathcal{H}\mathcal{L}_{t,s}^{2}(\mathbb{B}_{n})$;

  3. 3.

$\left\langle{K_{t,s}(u,z)},{K_{t,s}(v,z)}\right\rangle_{\mathbb{S}}=K_{t,s}(v,u)$.

###### Proof.

  1. 1.

Direct computations show that

$\displaystyle\overline{K_{t,s}(z,u)}$
$\displaystyle=\sum_{p=0}^{\infty}\sum_{q=0}^{\infty}\frac{1}{\gamma_{p,q}}\overline{f_{t,s}^{(p,q)}(z)\overline{f_{t,s}^{(p,q)}(u)}}$
$\displaystyle=\sum_{p=0}^{\infty}\sum_{q=0}^{\infty}\frac{1}{\gamma_{p,q}}\overline{\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}\left\langle{\bar{u}},{s+t}\right\rangle^{p}\left\langle{u},{\overline{s-t}}\right\rangle^{q}}$
$\displaystyle=\sum_{p=0}^{\infty}\sum_{q=0}^{\infty}\frac{1}{\gamma_{p,q}}\left\langle{\bar{z}},{s+t}\right\rangle^{p}\left\langle{z},{\overline{s-t}}\right\rangle^{q}\left\langle{u},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{u}},{s-t}\right\rangle^{q}$
$\displaystyle=K_{t,s}(u,z).$

  2. 2.

Let $f^{(w,v)}_{t,s}(z)$ be an element in
$\mathcal{H}\mathcal{L}_{t,s}^{2}(\mathbb{B}_{n})$, then we can write

$\displaystyle\left\langle{K_{t,s}(z,u)},{f^{(w,v)}_{t,s}(z)}\right\rangle_{\mathbb{S}}$
$\displaystyle=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\overline{K_{t,s}(z,u)}f^{(w,v)}_{t,s}(z)\,{\rm
d}\sigma(z)$
$\displaystyle=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}K_{t,s}(u,z)f^{(w,v)}_{t,s}(z)\,{\rm
d}\sigma(z),$

by the first property. Then by Definition 4, we have

$\displaystyle\left\langle{K_{t,s}(z,u)},{f^{(w,v)}_{t,s}(z)}\right\rangle_{\mathbb{S}}=$
$\displaystyle\,\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\sum_{p=0}^{\infty}\sum_{q=0}^{\infty}\frac{1}{\gamma_{p,q}}f^{(p,q)}_{t,s}(u)\overline{f^{(p,q)}_{t,s}(z)}f^{(w,v)}_{t,s}(z)\,{\rm
d}\sigma(z)$ $\displaystyle=$
$\displaystyle\,\sum_{p=0}^{\infty}\sum_{q=0}^{\infty}\frac{1}{\gamma_{p,q}}f^{(p,q)}_{t,s}(u)\underbrace{\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\overline{f^{(p,q)}_{t,s}(z)}f^{(w,v)}_{t,s}(z)\,{\rm
d}\sigma(z)}_{\left\langle
f^{(p,q)}_{t,s}(z),f^{(w,v)}_{t,s}(z)\right\rangle_{\mathbb{S}}}.$

Using Theorem 1, only the case $p=w$ and $q=v$ remains and

$\left\langle{f^{(p,q)}_{t,s}(z)},{f^{(w,v)}_{t,s}(z)}\right\rangle_{\mathbb{S}}=\gamma_{p,q}$

in this case, i.e.

$\begin{split}\left\langle{K_{t,s}(z,u)},{f^{(w,v)}_{t,s}(z)}\right\rangle_{\mathbb{S}}&=f^{(w,v)}_{t,s}(u).\end{split}$

  3. 3.

We obtain

$\displaystyle\left\langle{K_{t,s}(u,z)},{K_{t,s}(v,z)}\right\rangle_{\mathbb{S}}$
$\displaystyle=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\overline{K_{t,s}(u,z)}K_{t,s}(v,z)\,{\rm
d}\sigma(z)$
$\displaystyle=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}K_{t,s}(z,u)K_{t,s}(v,z)\,{\rm
d}\sigma(z)$
$\displaystyle=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\sum_{p=0}^{\infty}\sum_{q=0}^{\infty}\frac{1}{\gamma_{p,q}}f^{(p,q)}_{t,s}(z)\overline{f^{(p,q)}_{t,s}(u)}K_{t,s}(v,z)\,{\rm
d}\sigma(z)$
$\displaystyle=\sum_{p=0}^{\infty}\sum_{q=0}^{\infty}\frac{1}{\gamma_{p,q}}\overline{f^{(p,q)}_{t,s}(u)}\underbrace{\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\overline{K_{t,s}(z,v)}f^{(p,q)}_{t,s}(z)\,{\rm
d}\sigma(z)}_{\left\langle{K_{t,s}(z,v)},{f^{(p,q)}_{t,s}(z)}\right\rangle}$
$\displaystyle=\sum_{p=0}^{\infty}\sum_{q=0}^{\infty}\frac{1}{\gamma_{p,q}}f^{(p,q)}_{t,s}(v)\overline{f^{(p,q)}_{t,s}(u)}$
$\displaystyle=K_{t,s}(v,u).$

∎

We now show that the complex harmonic Szegö-Radon transform admits an
equivalent integral representation.

######  Theorem 3.

Let $n\geq 2$. The complex harmonic Szegö-Radon transform coincides with the
integral transform

$\mathcal{R}_{t,s}[f](z)=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}K_{t,s}(z,u)f(u)\,{\rm
d}\sigma(u),$ (18)

where $z\in\mathbb{B}_{n}$.

###### Proof.

Let $P$ denote the operator

$P:f(u)\to\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}K_{t,s}(z,u)f(u)\,{\rm
d}\sigma(u).$

We show that $P$ is an orthogonal projection operator from
$\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})$ to
$\mathcal{H}\mathcal{L}_{t,s}^{2}(\mathbb{B}_{n})$. For any
$f(u)\in\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})$, we write $P[f]$ as

$P[f](z)=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}K_{t,s}(z,u)f(u)\,{\rm
d}\sigma(u)=\left\langle{K_{t,s}(u,z)},{f(u)}\right\rangle_{\mathbb{S}}.$

By (16), we get

$P[f](z)=\sum_{p=0}^{\infty}\sum_{q=0}^{\infty}\frac{\left\langle{f^{(p,q)}_{t,s}(u)},{f(u)}\right\rangle_{\mathbb{S}}}{\gamma_{p,q}}f^{(p,q)}_{t,s}(z),$

which is clearly an element in
$\mathcal{H}\mathcal{L}_{t,s}^{2}(\mathbb{B}_{n})$. Meanwhile,

$\displaystyle\left\langle{P[f]},{g}\right\rangle_{\mathbb{S}}$
$\displaystyle=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\overline{P[f](z)}g(z)\,{\rm
d}\sigma(z)$
$\displaystyle=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\overline{\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}K_{t,s}(z,u)f(u)\,{\rm
d}\sigma(u)}g(z)\,{\rm d}\sigma(z)$
$\displaystyle=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}K_{t,s}(u,z)\overline{f(u)}\,{\rm
d}\sigma(u)g(z)\,{\rm d}\sigma(z)$
$\displaystyle=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}K_{t,s}(u,z)g(z)\,{\rm
d}\sigma(z)\overline{f(u)}\,{\rm d}\sigma(u)$
$\displaystyle=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\overline{f(u)}P[g](u)\,{\rm
d}\sigma(u)$ $\displaystyle=\left\langle{f},{P[g]}\right\rangle_{\mathbb{S}}.$

Thus, making use of Proposition 1, we have

$\displaystyle P^{2}[f](z)$
$\displaystyle=\left\langle{K_{t,s}(u,z)},{P[f](u)}\right\rangle_{\mathbb{S}}$
$\displaystyle=\left\langle{P[K_{t,s}](u,z)},{f(u)}\right\rangle_{\mathbb{S}}$
$\displaystyle=\left\langle{\left\langle{K_{t,s}(p,u)},{K_{t,s}(p,z)}\right\rangle_{\mathbb{S}}},{f(u)}\right\rangle_{\mathbb{S}}$
$\displaystyle=\left\langle{K_{t,s}(u,z)},{f(u)}\right\rangle_{\mathbb{S}}$
$\displaystyle=P[f](z),$

which proves that the operator $P$ is idempotent.

Furthermore, denote by $I$ the identity operator, we obtain that

$\displaystyle\left\langle{P[f]},{(I-P)[g]}\right\rangle_{\mathbb{S}}$
$\displaystyle=\left\langle{P[f]},{I[g]}\right\rangle_{\mathbb{S}}-\left\langle{P[f]},{P[g]}\right\rangle_{\mathbb{S}}$
$\displaystyle=\left\langle{P[f]},{g}\right\rangle_{\mathbb{S}}-\left\langle{P^{2}[f]},{g}\right\rangle_{\mathbb{S}}$
$\displaystyle=0,$

which shows the orthogonal direct sum

$\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})=P[\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})]+\left(I-P\right)[\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})].$

Therefore, $P$ is an orthogonal projection operator from
$\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})\to\mathcal{H}\mathcal{L}_{t,s}^{2}(\mathbb{B}_{n})$.
∎

###  3.2 The dual transform

In this subsection, we consider the dual transform of $\mathcal{R}_{t,s}$ in
(15), which involves integration over the complex Stiefel manifold
$St^{(2)}(\mathbb{C}^{n})={\rm U}(n)/{\rm U}(n-2)$. Its action will be
explicitly calculated.

######  Definition 5.

Let the tuple $(t,s)$ lie on the Stiefel manifold $St^{(2)}(\mathbb{C}^{n})$.
Then for any $F(z,t,s)\in\mathcal{H}\mathcal{L}_{t,s}^{2}(\mathbb{B}^{n})$,
the dual of the Szegö-Radon transform is defined by

$\widetilde{\mathcal{R}}[F](z):=\frac{1}{\omega_{2n-1}\omega_{2n-3}}\int_{St^{(2)}}F(z,t,s)\,{\rm
d}\sigma(s)\,{\rm d}\sigma(t).$

The dual transform can be calculated by employing the reproducing kernel of
$\mathcal{H}_{p,q}$, which admits an integral representation in terms of plane
waves (see e.g. [10, Theorem 13] ).

######  Theorem 4.

[10] The reproducing kernel of the space of homogeneous complex harmonics of
degree $(p,q)$ is given by

$K_{p,q}(z,u)=\frac{\lambda_{p,q}\dim\mathcal{H}_{p,q}}{\omega_{2n-1}\omega_{2n-3}}\int_{St(2)}\left\langle{z},{\overline{t+s}}\right\rangle^{p}\left\langle{\bar{z}},{t-s}\right\rangle^{q}\left\langle{u},{\overline{t-s}}\right\rangle^{q}\left\langle{\bar{u}},{t+s}\right\rangle^{p}\,{\rm
d}\sigma(s)\,{\rm d}\sigma(t),$

where $\displaystyle\lambda_{p,q}=\frac{(k+n-1)!}{2^{k}(n-1)!(k-\nu)!}$ with
$k=p+q$ and $\nu=\min(p,q)$, and $\dim\mathcal{H}_{p,q}$ is given in (5).

The above reproducing property implies the explicit action of the composition
of the Radon and its dual.

######  Theorem 5.

For any
$f(z)=\sum_{p,q}f_{p,q}(z)\in\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})$,
the dual transform of the complex Szegö-Radon transform of $f$ is given by

$\widetilde{\mathcal{R}}[\mathcal{R}_{t,s}[f]](z)=\sum_{p,q}\frac{1}{\gamma_{p,q}\lambda_{p,q}\dim\mathcal{H}_{p,q}}f_{p,q}(z),$
(19)

where $\gamma_{p,q}$ is defined in (9), $\dim\mathcal{H}_{p,q}$ and
$\lambda_{p,q}$ are the same constants given in Theorem 4.

###### Proof.

Since $\mathcal{H}\mathcal{L}_{s,t}^{2}(\mathbb{B}_{n})$ is the completion of
all finite linear combinations of complex plane waves, therefore, it is
sufficient to compute for each homogeneous component $f_{p,q}(z)$. According
to Definition 5, we have

$\widetilde{\mathcal{R}}[\mathcal{R}_{t,s}[f_{p,q}]](z)=\frac{1}{\omega_{2n-1}\omega_{2n-3}}\int_{St^{(2)}}\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}K_{t,s}(u,z)f_{p,q}(u)\,{\rm
d}\sigma(u)\,{\rm d}\sigma(s)\,{\rm d}\sigma(t).$

By Fubini’s theorem and the orthogonality of harmonics of distinct bi-degrees,
we have

$\displaystyle\widetilde{\mathcal{R}}[\mathcal{R}_{t,s}[f_{p,q}]](z)$
$\displaystyle=$
$\displaystyle\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\left(\frac{1}{\omega_{2n-1}\omega_{2n-3}}\int_{St^{(2)}}K_{t,s}(u,z)\,{\rm
d}\sigma(s)\,{\rm d}\sigma(t)\right)f_{p,q}(u)\,{\rm d}\sigma(u)$
$\displaystyle=$
$\displaystyle\int_{\mathbb{S}^{2n-1}}\underbrace{\left(\frac{1}{\omega_{2n-1}\omega_{2n-3}}\int_{St^{(2)}}\frac{1}{\gamma_{p,q}}\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}\left\langle{\bar{u}},{s+t}\right\rangle^{p}\left\langle{u},{\overline{s-t}}\right\rangle^{q}\,{\rm
d}\sigma(s)\,{\rm d}\sigma(t)\right)}_{J}$
$\displaystyle\times\frac{1}{\omega_{2n-1}}f_{p,q}(u)\,{\rm d}\sigma(u).$

We now consider the part $J$. Theorem 4 indeed implies

$J=\frac{1}{\gamma_{p,q}\lambda_{p,q}\dim\mathcal{H}_{p,q}}K_{p,q}(z,u).$

Since $K_{p,q}(z)$ is the reproducing kernel of $\mathcal{H}_{p,q}$, it
follows that

$\displaystyle\widetilde{\mathcal{R}}[\mathcal{R}_{t,s}[f_{p,q}]](z)$
$\displaystyle=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\frac{1}{\gamma_{p,q}\lambda_{p,q}\dim\mathcal{H}_{p,q}}K_{p,q}(z,u)f_{p,q}(u)\,{\rm
d}\sigma(u)$
$\displaystyle=\frac{1}{\gamma_{p,q}\lambda_{p,q}\dim\mathcal{H}_{p,q}}f_{p,q}(z).$

This completes the proof. ∎

###  3.3 Inversion formulas

The quantity $\nu=\min(p,q)$, which appears in (19), significantly complicates
the task of formulating an inversion formula. We begin by examining
holomorphic functions and subsequently address the general case.

####  3.3.1 Holomorphic case

We denote the subspace of $\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})$
consisting of holomorphic functions as $\mathcal{H}^{2}(\mathbb{B}_{n})$.
Thus, for each $f(z)\in\mathcal{H}^{2}(\mathbb{B}_{n})$, it has a Taylor
series expansion given by $f(z)=\sum_{p=0}^{\infty}f_{p,0}(z)$. In this case,
the quantity $\nu=\min(p,q)$ equals $0$ automatically, and the associated
constant in (19) becomes:

$\begin{split}\frac{1}{\gamma_{p,0}\lambda_{p,0}\dim\mathcal{H}_{p,0}}=\frac{\Gamma(n)\Gamma(p+1)}{\Gamma(n+p)}.\end{split}$

This observation suggests that we can derive the inversion of the Szegö-Radon
transform for holomorphic functions using the Euler operator, which is similar
with the inversion formula for the ordinary Radon transform using differential
operator.

######  Theorem 6.

For any $f(w)\in\mathcal{H}^{2}(\mathbb{B}_{n})$ with $n\geq 2$, it holds that

$f(z)=\left(T\circ\widetilde{\mathcal{R}}\circ\mathcal{R}_{t,s}\right)[f](z),$

where

$T[f](z):=\frac{1}{\Gamma(n)}(\mathbb{E}_{z}+1)(\mathbb{E}_{z}+2)\cdots(\mathbb{E}_{z}+n-1)f(z).$
(20)

###### Proof.

For each homogeneous component $f_{p,0}$ of $f$, using the fact

$\mathbb{E}_{z}f_{p,0}=pf_{p,0},$

we have

$\displaystyle(\mathbb{E}_{z}+1)(\mathbb{E}_{z}+2)\cdots(\mathbb{E}_{z}+n-1)f_{p,0}$
$\displaystyle=(\mathbb{E}_{z}+2)\cdots(\mathbb{E}_{z}+n-1)(p+1)f_{p,0}$
$\displaystyle=(\mathbb{E}_{z}+3)\cdots(\mathbb{E}_{z}+n-1)(p+1)(p+2)f_{p,0}$
$\displaystyle=\ldots$ $\displaystyle=(p+1)(p+2)\ldots(p+n-1)f_{p,0}$
$\displaystyle=\frac{\Gamma(p+n)}{\Gamma(p+1)}f_{p,0}.$

It follows that

$T[f_{p,0}](z)=\frac{\Gamma(p+n)}{\Gamma(n)\Gamma(p+1)}f_{p,0}(z).$

Combining with the dual transform given in (19), we obtain

$\displaystyle\left(T\circ\widetilde{\mathcal{R}}\circ\mathcal{R}_{t,s}\right)[f](z)$
$\displaystyle=\sum_{p=0}^{\infty}\frac{1}{\gamma_{p,0}\lambda_{p,0}\dim\mathcal{H}_{p,0}}T[f_{p,0}](z)$
$\displaystyle=\sum_{p=0}^{\infty}\frac{\Gamma(n)\Gamma(p+1)}{\Gamma(n+p)}\cdot\frac{\Gamma(p+n)}{\Gamma(n)\Gamma(p+1)}f_{p,0}(z)$
$\displaystyle=\sum_{p=0}^{\infty}f_{p,0}(z)=f(z),$

which completes the proof. ∎

Since we are now restricting ourselves to the space of holomorphic functions,
the Szegö-Radon kernel undergoes a further reduction as well.

######  Theorem 7.

For the space $\mathcal{H}^{2}(\mathbb{B}_{n})$ of holomorphic functions, the
Szegö-Radon kernel is given by

$\mathcal{K}_{t,s}(z,u)=\sum_{p=0}^{\infty}\frac{1}{\gamma_{p,0}}\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{u}},{s+t}\right\rangle^{p}.$
(21)

It admits a closed expression

$\mathcal{K}_{t,s}(z,u)=\left(\frac{2}{2-\left\langle{z},{\overline{s+t}}\right\rangle\left\langle{\bar{u}},{s+t}\right\rangle}\right)^{n}.$

###### Proof.

It is derived analogously to Theorem 2.

∎

####  3.3.2 General inversion formula

This subsection is devoted to the inversion formula for the general case, i.e.
when $q\neq 0$. The approach involves splitting the orthogonal projection into
two parts and then providing their inversion formulas separately, addressing
the challenge posed by $\nu=\min(p,q)$. The convergence issue, in this
context, becomes more complicated.

######  Definition 6.

The orthogonal projection

$\mathcal{R}_{t,s}:\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})\to\mathcal{H}\mathcal{L}_{t,s}^{2}(\mathbb{B}_{n})$

can be decomposed into two orthogonal projections

$\begin{split}&\mathcal{R}_{t,s}^{(1)}:\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})\to\mathcal{H}\mathcal{L}_{t,s}^{2}(\mathbb{B}_{n},p\geq
q),\\\
&\mathcal{R}_{t,s}^{(2)}:\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})\to\mathcal{H}\mathcal{L}_{t,s}^{2}(\mathbb{B}_{n},p<q),\end{split}$

where the space $\mathcal{H}\mathcal{L}_{t,s}^{2}(\mathbb{B}_{n},p\geq q)$
consists of functions of the form $f(z)=\sum_{p\geq q}f_{p,q}(z)$, and
$\mathcal{H}\mathcal{L}_{t,s}^{2}(\mathbb{B}_{n},p<q)$ is defined similarly.
It is easy to see that

$\mathcal{R}_{t,s}=\mathcal{R}_{t,s}^{(1)}+\mathcal{R}_{t,s}^{(2)},$

and
$\mathcal{H}\mathcal{L}_{t,s}^{2}(\mathbb{B}_{n})=\mathcal{H}\mathcal{L}_{t,s}^{2}(\mathbb{B}^{n},p\geq
q)\oplus\mathcal{H}\mathcal{L}_{t,s}^{2}(\mathbb{B}_{n},p<q).$

Similarly as done in Theorem 3, we can show that these two orthogonal
projections can be expressed equivalently as integral transforms. Hereto, we
calculate their integral kernels. Recall that the Horn function $H_{3}$ (see
[21]) is the hypergeometric function of order $2$ defined by

$H_{3}\left(\alpha,\beta,\gamma;z,w\right)=\sum_{m,n=0}^{\infty}\frac{(\alpha)_{2m+n}(\beta)_{n}}{(\gamma)_{m+n}}\frac{z^{m}w^{n}}{m!n!},$

which converges when $|z|<r,|w|<s,r+(s-1/2)^{2}=1/4,$ and $\gamma\neq
0,-1,-2,\ldots$. It admits an integral representation

$\displaystyle H_{3}\left(\alpha,\beta,\gamma;z,w\right)=$
$\displaystyle\frac{\Gamma(\gamma)}{\Gamma\left(\frac{\alpha}{2}\right)\Gamma(\beta)\Gamma\left(\gamma-\frac{\alpha}{2}-\beta\right)}\int_{0}^{1}\int_{0}^{1}u^{\beta-1}v^{\alpha/2-1}(1-u)^{\gamma-\beta-1}$
$\displaystyle(1-v)^{\gamma-\alpha/2-\beta-1}(1-uw)^{-\alpha}\left(1-\frac{4vz(1-u)}{(1-uv)^{2}}\right)^{-(\alpha+1)/2}\,{\rm
d}u\,{\rm d}v,$

when ${\rm Re}\,\alpha,{\rm Re}\,\beta,$ and ${\rm
Re}\,(\gamma-\alpha/2-\beta)>0$.

######  Theorem 8.

When $|z|<1/2$ and $|w|<1$, the integral kernel of $\mathcal{R}_{t,s}^{(1)}$
is given by

$\begin{split}K_{t,s}^{(1)}(z,w)&=\sum_{p=0}^{\infty}\sum_{q=0}^{p}\frac{1}{\gamma_{p,q}}f^{(p,q)}_{t,s}(z)\overline{f^{(p,q)}_{t,s}(w)}\\\
&=H_{3}\left(n,1,1;\frac{xy}{4},\frac{x}{2}\right),\end{split}$ (22)

and the integral kernel of $\mathcal{R}_{t,s}^{(2)}$ is given by

$\begin{split}K_{t,s}^{(2)}(z,w)&=\sum_{p=0}^{\infty}\sum_{q=p}^{\infty}\frac{1}{\gamma_{p,q}}f^{(p,q)}_{t,s}(z)\overline{f^{(p,q)}_{t,s}(w)}\\\
&=H_{3}\left(n,2,2;\frac{xy}{4},\frac{y}{2}\right)-{{}_{2}F_{1}}\left(\frac{n}{2},\frac{1+n}{2};1;xy\right),\end{split}$
(23)

where $\gamma_{p,q}$ is defined in (9), in which
$x=\left\langle{z},{\overline{s+t}}\right\rangle\left\langle{\bar{u}},{s+t}\right\rangle$
and
$y=\left\langle{\bar{z}},{s-t}\right\rangle\left\langle{u},{\overline{s-t}}\right\rangle$.

###### Proof.

(1) When $p\geq q$, the integral kernel is formally represented as

$\displaystyle K_{s,t}^{(1)}(z,w)$
$\displaystyle=\sum_{p=0}^{\infty}\sum_{q=0}^{p}\frac{1}{\gamma_{p,q}}f^{(p,q)}_{t,s}(z)\overline{f^{(p,q)}_{t,s}(w)}$
$\displaystyle=\sum_{p=0}^{\infty}\sum_{q=0}^{\infty}\frac{1}{\gamma_{p+q,q}}f^{(p+q,q)}_{t,s}(z)\overline{f^{(p+q,q)}_{t,s}(w)}.$

Explicitly, we have

$\displaystyle K_{s,t}^{(1)}(z,u)$
$\displaystyle=\sum_{q=0}^{\infty}\sum_{p=0}^{\infty}\frac{(n)_{p+2q}}{2^{p+q}2^{q}(p+q)!q!}x^{p+q}y^{q}$
$\displaystyle=\sum_{q=0}^{\infty}\sum_{p=0}^{\infty}\frac{(n)_{p+2q}}{2^{p}2^{2q}(p+q)!q!}x^{p}(xy)^{q}$
$\displaystyle=\sum_{q=0}^{\infty}\sum_{p=0}^{\infty}\frac{(n)_{p+2q}p!}{(p+q)!}\frac{(\frac{x}{2})^{p}}{p!}\frac{(\frac{xy}{4})^{q}}{q!}$
$\displaystyle=\sum_{q=0}^{\infty}\sum_{p=0}^{\infty}\frac{(n)_{p+2q}(1)_{p}}{(1)_{p+q}}\frac{(\frac{x}{2})^{p}}{p!}\frac{(\frac{xy}{4})^{q}}{q!}$
$\displaystyle=H_{3}\left(n,1,1;\frac{xy}{4},\frac{x}{2}\right).$

When $|z|<1/2$ and $|w|<1$, we have
$|x|=\left|\left\langle{z},{\overline{s+t}}\right\rangle\left\langle{\bar{w}},{s+t}\right\rangle\right|<\sqrt{2}/2\cdot\sqrt{2}=1$
and
$|y|=\left|\left\langle{\bar{z}},{s-t}\right\rangle\left\langle{u},{\overline{s-t}}\right\rangle\right|<1$.
It follows that $\left|xy/4\right|<1/4$ and $\left|x/2\right|<1/2$, which sits
in the convergence domain of $H_{3}$.

(2) Similarly, when $p<q$, we have

$\displaystyle K_{s,t}^{(2)}(z,w)$
$\displaystyle=\sum_{p=0}^{\infty}\sum_{q=p+1}^{\infty}\frac{1}{\gamma_{p,q}}f^{(p,q)}_{t,s}(z)\overline{f^{(p,q)}_{t,s}(w)}$
$\displaystyle=\sum_{p=0}^{\infty}\sum_{q=0}^{\infty}\frac{1}{\gamma_{p,p+q+1}}f^{(p,p+q+1)}_{t,s}(z)\overline{f^{(p,p+q+1)}_{t,s}(w)}$
$\displaystyle=H_{3}\left(n,2,2;\frac{xy}{4},\frac{y}{2}\right)-\sum_{p=0}^{\infty}\frac{(n)_{2p}}{(p!)^{2}}\left(\frac{xy}{4}\right)^{p}$
$\displaystyle=H_{3}\left(n,2,2;\frac{xy}{4},\frac{y}{2}\right)-{{}_{2}F_{1}}\left(\frac{n}{2},\frac{1+n}{2};1;xy\right).$

When $|z|<1/2$ and $|w|<1$, both the hypergeometric series above converge. ∎

Finally, we are able to give the inversion formula for the Szegö-Radon
transform.

######  Theorem 9.

For any $f(w,\bar{w})\in\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})$ with
$n\geq 2$, we have for any $|z|<1/2$,

$f(z,\bar{z})=\left(\mathcal{T}\circ\mathcal{T}_{1}\circ\widetilde{\mathcal{R}}\circ\mathcal{R}_{t,s}^{(1)}\right)[f](z,\bar{z})+\left(\mathcal{T}\circ\mathcal{T}_{2}\circ\widetilde{\mathcal{R}}\circ\mathcal{R}_{t,s}^{(2)}\right)[f](z,\bar{z}),$

where

$\begin{split}\mathcal{T}[f](z,\bar{z}):=&\,\frac{1}{\Gamma(n)\Gamma(n-1)}(\mathbb{E}+n-1)(\mathbb{E}_{z}+1)(\mathbb{E}_{z}+2)\cdots(\mathbb{E}_{z}+n-2)\\\
&\,\times(\mathbb{E}_{\bar{z}}+1)(\mathbb{E}_{\bar{z}}+2)\cdots(\mathbb{E}_{\bar{z}}+n-2)f(z,\bar{z}),\end{split}$

and

$\mathcal{T}_{1}[f](z,\bar{z})=\int_{0}^{\infty}f(z,\bar{z}t)e^{-t}\,{\rm
d}t,\qquad\mathcal{T}_{2}[f](z,\bar{z})=\int_{0}^{\infty}f(zt,\bar{z})e^{-t}\,{\rm
d}t.$

###### Proof.

We only prove the first case when $p<q$. The associated constant in (19) is

$\frac{1}{\gamma_{p,q}\lambda_{p,q}\dim\mathcal{H}_{p,q}}=\frac{(n-1)!(n-2)!}{(n+p+q-1)(q+1)_{n-2}(p+1)_{n-2}\Gamma(p+1)}.$
(24)

It is sufficient to verify for each homogeneous component $f_{p,q}$ when
$p<q$. Note that

$\int_{0}^{\infty}f_{p,q}(zt,\bar{z})e^{-t}\,{\rm
d}t=f_{p,q}(z,\bar{z})\int_{0}^{\infty}t^{p}e^{-t}\,{\rm
d}t=\Gamma(p+1)f_{p,q}(z,\bar{z}).$

It follows that

$\left(\mathcal{T}\circ\mathcal{T}_{2}\right)[f_{p,q}](z,\bar{z})=\frac{(n+p+q-1)(p+1)_{n-2}(q+1)_{n-2}\Gamma(p+1)}{\Gamma(n)\Gamma(n-1)}f_{p,q}(z,\bar{z}).$

Together with (24), we obtain the inversion formula for
$\mathcal{R}_{t,s}^{(2)}$. ∎

######  Remark 1.

When
$f\in\mathcal{L}^{2}_{\mathcal{H}}(\mathbb{B}_{n})\cap\mathcal{C}^{\infty}(\mathbb{B}_{n})$,
by the unique continuation property of harmonic function, we know that the
function $f(z)$ is uniquely determined by its value in $|z|<1/2$. Moreover,
the harmonic function can be given by

$f(z)=f\left(\frac{z_{0}|z|}{|z_{0}|}\right),\qquad 0<|z_{0}|<1/2.$

##  4 Radon-type transform on Fock space

This section is dedicated to the Radon-type transform on the classical Segal-
Bargmann-Fock space, which is defined in a similar way with the Szegö-Radon
transform studied in Section 3. One of the most intriguing aspects of this
transform is its ability to extend to the entire $L^{2}(\mathbb{R}^{n})$ via
the Segal-Bargmann transform, whereas the orthogonal Radon-type transform was
only defined on certain subspaces (e.g. the monogenic function space in [7])
of $L^{2}(\mathbb{R}^{n})$. This allows us to compare the Radon-type
transforms studied in this paper with the ordinary Radon transform.

###  4.1 The Bargmann-Radon transform

The holomorphic Segal-Bagmann-Fock space plays an important role in several
fields, including quantum mechanics, infinite dimensional analysis, and the
theory of stochastic distributions (see e.g. [12]).

######  Definition 7.

The complex Segal-Bargmann-Fock space $\mathcal{F}\left(\mathbb{C}^{n}\right)$
is the Hilbert space of entire functions in $\mathbb{C}^{n}$, which are square
integrable with respect to the $2n$-dimensional Gaussian density, i.e.

$\frac{1}{\pi^{n}}\int_{\mathbb{C}^{n}}\left|{f(z)}\right|^{2}\exp\left(-\left|{z}\right|^{2}\right)\,{\rm
d}z<\infty,$

and equipped with the inner product

$\left\langle
P(z),Q(z)\right\rangle_{\mathcal{F}}=\frac{1}{\pi^{n}}\int_{\mathbb{C}^{n}}\overline{P(z)}Q(z)\exp\left(-\left|{z}\right|^{2}\right)\,{\rm
d}z.$ (25)

We now introduce the entire plane waves.

######  Definition 8.

Let $z\in\mathbb{C}^{n}$, $(t,s)\in St^{(2)}(\mathbb{C}^{n})$, and

$f^{(p,q)}_{t,s}(z)=\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}.$
(26)

We see that $f^{(p,0)}_{t,s}(z)$ is holomorphic in $z$. Therefore they are
called by the entire plane waves.

Since we are now working on the Fock space, where only the holomorphic
functions are involved, we will use the shorthand notation $f^{(p)}_{t,s}(z)$
instead of $f^{(p,0)}_{t,s}(z)$ in the subsequent discussion. Note that the
Bargmann inner product (25) are related with the Fischer inner product, see
e.g. [26].

######  Lemma 5.

For a spherical harmonic $H_{p,q}\in\mathcal{H}_{p,q}$ and a homogeneous
polynomial $P_{p,q}\in\mathcal{P}_{p,q}$, it holds that

$\left\langle{H_{p,q}(z)},{P_{p,q}(z)}\right\rangle_{\mathcal{F}}=\left\langle{H_{p,q}(z)},{P_{p,q}(z)}\right\rangle_{\partial}.$
(27)

The above correspondence immediately leads to the following.

######  Theorem 10.

The entire complex plane waves $\\{f^{(p)}_{t,s}(z)\\}$ define an orthogonal
subset of $\mathcal{F}(\mathbb{C}^{n})$. In particular, we have

$\begin{split}\left\langle
f^{(p)}_{t,s}(z),g^{(u)}_{t,s}(z)\right\rangle_{\mathcal{F}}=\left\\{\begin{array}[]{ll}\mu_{p},&p=u.\\\
0,&\text{otherwise},\\\ \end{array}\right.\end{split}$ (28)

where $\mu_{p}=2^{p}p!$.

###### Proof.

The orthogonality is easy to see. We only need to compute their norms. By
(27), we have

$\begin{split}\left\langle
f^{(p)}_{t,s}(z),g^{(u)}_{t,s}(z)\right\rangle_{\mathcal{F}}&=\left\langle
f^{(p)}_{t,s}(z),g^{(u)}_{t,s}(z)\right\rangle_{\partial}\\\
&=\left[\overline{f^{(p)}_{t,s}(\partial)}g^{(u)}_{t,s}(z)\right]_{z=0}.\end{split}$

Similar to Theorem 1 and (14), we obtain

$\begin{split}\left\langle
f^{(p)}_{t,s}(z),f^{(p)}_{t,s}(z)\right\rangle_{\mathcal{F}}&=2^{p}p!.\end{split}$

∎

Similarly, the space of complex plane waves in $\mathcal{F}(\mathbb{C}^{n})$
is defined as follows.

######  Definition 9.

Let the tuple $(t,s)$ lie on the Stiefel manifold $St^{(2)}(\mathbb{C}^{n})$.
Then the completion under the inner product (25) of all finite linear
combinations of complex plane waves of the form

$\sum_{p=0}^{\ell}\left\langle{z},{\overline{s+t}}\right\rangle^{p}\alpha_{p},\quad\alpha_{p}\in\mathbb{C},$

is a closed subspace of $\mathcal{F}(\mathbb{C}^{n})$, which is denoted by
$\mathcal{F}_{t,s}(\mathbb{C}^{n})$.

The Randon-type transform on the Fock space $\mathcal{F}(\mathbb{C}^{n})$ is
defined as the orthogonal projection onto the subspace spanned by entire plane
waves. Precisely,

######  Definition 10.

Assume $n\geq 2$. The holomorphic Bargmann-Radon transform is defined as the
orthogonal projection

$R_{t,s}:\mathcal{F}(\mathbb{C}^{n})\to\mathcal{F}_{t,s}(\mathbb{C}^{n}).$
(29)

Furthermore, we have,

######  Theorem 11.

The holomorphic Bargmann-Radon transform in (29) can be equivalently written
as the following integral transform

$R_{t,s}[f](z)=\frac{1}{\pi^{n}}\int_{\mathbb{C}^{n}}\widetilde{K}_{t,s}(z,w)f(w)\exp\left(-\left|{w}\right|^{2}\right)\,{\rm
d}w,$ (30)

where $\widetilde{K}_{t,s}(z,w)$ is the Bargmann-Radon Kernel given by

$\begin{split}\widetilde{K}_{t,s}(z,w)&=\sum_{p=0}^{\infty}\frac{1}{\mu_{p}}f^{(p)}_{t,s}(z)\overline{f^{(p)}_{t,s}(w)}\\\
&=\exp\left\\{\frac{\left\langle{z},{\overline{s+t}}\right\rangle\left\langle{\bar{w}},{s+t}\right\rangle}{2}\right\\},\end{split}$
(31)

where the coefficient $\mu_{p}$ is defined in (28) and $n\geq 2$.

######  Remark 2.

The complex harmonic Bargmann-Radon kernel in (31) and the complex harmonic
Szegö-Radon kernel in (21) have a similar series expression but with different
coefficients (the norms are different).

###  4.2 Dual transform and inversion

In this subsection, we give the dual transform and inversion formula. The
proof follows a similar approach as done in the Szegö-Radon transform in
Section 3, and in [7] for the monogenic Bargmann-Radon transform. Therefore,
the proof is omitted.

######  Definition 11.

Let the tuple $(t,s)$ lie on the Stiefel manifold $St^{(2)}(\mathbb{C}^{n})$.
Suppose $F(z,t,s)\subset\mathcal{F}_{t,s}(\mathbb{C}^{n})$, the dual of the
Bargmann-Radon transform is given by

$\widetilde{R}_{t,s}[F](z):=\frac{1}{\omega_{2n-1}\omega_{2n-3}}\int_{St^{(2)}}F(z,t,s)\,{\rm
d}\sigma(s)\,{\rm d}\sigma(t).$

By Theorem 4, Lemma 2 and 5, the explicit action of the dual transform is
given as follows. We omit the computation details.

######  Theorem 12.

For any $f\in\mathcal{F}(\mathbb{C}^{n})$, the dual transform of the
holomorphic Bargmann-Radon transform is explicitly given by

$\widetilde{R}[R_{t,s}[f]](z)=\sum_{p=0}^{\infty}\frac{\Gamma(n)\Gamma(p+1)}{\Gamma(n+p)}f_{p,0}(z).$
(32)

Now the inversion formula follows from Theorem 12.

######  Theorem 13.

Let $n\geq 2$. For any $f\in\mathcal{F}(\mathbb{C}^{n})$, we have

$f(z)=\left(T\circ\widetilde{R}\circ R_{t,s}\right)[f](z),$

where $T$ is the operator defined in (20), i.e.

$T[f](z):=\frac{1}{\Gamma(n)}(\mathbb{E}_{z}+1)(\mathbb{E}_{z}+2)\cdots(\mathbb{E}_{z}+n-1)f(z).$

###  4.3 Radon transform on $L^{2}(\mathbb{R}^{n})$ via Segal-Bargmann
transform

It is well known that the Segal-Bargmann-Fock space
$\mathcal{F}(\mathbb{C}^{n})$ as well as the standard quantum mechanical
configuration space $L^{2}(\mathbb{R}^{n})$ carry irreducible representations
of the Weyl canonical commutation relations, and these representations are
unitarily equivalent by a theorem of Stone and von Neumann, see e.g. [12].
This unitary equivalence is realized explicitly by the Segal-Bargmann
transform. In this section, we use the Segal-Bargmann transform and the
holomorphic Bargmann-Radon transform in (29) to define a new Radon-type
transform on the space $L^{2}(\mathbb{R}^{n})$ of square integrable functions.

######  Definition 12.

The Segal-Bargmann transform
$\mathcal{B}:L^{2}(\mathbb{R}^{n})\to\mathcal{F}(\mathbb{C}^{n})$ of a
function $f(x)$ is defined by

$\mathcal{B}[f](z):=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}\exp\left(-\frac{z\cdot
z}{2}+x\cdot z-\frac{x\cdot x}{4}\right)f(x)\,{\rm d}x,$

where $x\cdot z=\sum_{j=1}^{n}x_{j}z_{j}$.

For that purpose, we recall some important properties of the one dimensional
Segal-Bargmann transform, with the extension to higher dimensions following a
similar approach.

######  Proposition 2.

An basis for $L^{2}(\mathbb{R})$ of square integrable functions is given by
$\\{\psi_{k}(x)\\}$, where

$\psi_{k}(x)=H_{k}(x)e^{-x^{2}/4},\qquad
H_{k}(x)=(-1)^{k}e^{x^{2}/2}\partial_{x}^{k}e^{-x^{2}/2}.$ (33)

Here $H_{k}(x)$ are the Hermite polynomials. This basis is orthogonal in the
sense that

$\langle\psi_{\ell},\psi_{k}\rangle:=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\overline{\psi_{\ell}(x)}\psi_{k}(x)\,{\rm
d}x=\delta_{\ell k}k!,$ (34)

where $\delta_{\ell k}$ is the Kronecker delta.

Furthermore, it is known that

$\mathcal{B}[\psi_{k}](z)=z^{k}.$

Let $F$ denote the ordinary Fourier transform and define

$G:\mathcal{F}(\mathbb{C})\to\mathcal{F}(\mathbb{C}):g(z)\mapsto g(-iz).$

$\begin{CD}L^{2}(\mathbb{R})@>{F}>{}>L^{2}(\mathbb{R})\\\
@V{\mathcal{B}}V{}V@A{}A{\mathcal{B}^{-1}}A\\\
\mathcal{F}(\mathbb{C})@>{G}>{}>\mathcal{F}(\mathbb{C})\end{CD}$ Figure 2:

$\begin{CD}L^{2}(\mathbb{R}^{n})@>{{\large?}}>{}>L^{2}_{t,s}(\mathbb{R}^{n})\\\
@V{\mathcal{B}}V{}V@A{}A{\mathcal{B}^{-1}}A\\\
\mathcal{F}(\mathbb{C}^{n})@>{R_{t,s}}>{}>\mathcal{F}_{t,s}(\mathbb{C}^{n})\end{CD}$
Figure 3:

The following diagram (Figure 3) commutes. We are interested in determining
explicitly the corresponding mapping of the Bargmann-Radon transform in (29)
under the Segal-Bargmann transform, as illustrated in Figure 3.

######  Definition 13.

Let the tuple $(t,s)$ lie on the Stiefel manifold $St^{(2)}(\mathbb{C}^{n})$.
We consider the set of orthogonal functions in $L^{2}(\mathbb{R}^{n})$ defined
by

$g^{(k)}_{t,s}(x):=\mathcal{B}_{n}^{-1}\left[\langle
z,\overline{s+t}\rangle^{k}\right](x),\qquad k\in\mathbb{N}\cup\\{0\\}.$ (35)

The completion of the form under the inner product (34)

$\sum_{k}\alpha_{k}\mathcal{B}_{n}^{-1}\left[\langle
z,\overline{s+t}\rangle^{k}\right](x),\quad\alpha_{k}\in\mathbb{C},$

is denoted by $L^{2}_{t,s}(\mathbb{R}^{n})$.

We now introduce two new class of functions related to the tuple $(t,s)$,

$H^{(k)}_{t,s}(x):=(-1)^{k}e^{|x|^{2}/2}\langle\nabla_{x},\overline{s+t}\rangle^{k}e^{-|x|^{2}/2},\qquad\psi^{(k)}_{t,s}(x):=H^{(k)}_{t,s}(x)e^{-|x|^{2}/4}.$
(36)

######  Example 1.

Suppose $t=(1,0,0,\ldots,0)$ and $s=(0,1,0,\ldots,0)$, then we have

$H^{(k)}_{t,s}(x)=\sum_{j=0}^{k}\binom{k}{j}H_{j}(x_{1})H_{k-j}(x_{2}),$

where $H_{k}$ is the Hermite polynomial given in (33).

Generally, we have,

######  Lemma 6.

Let $t=(t_{1},t_{2},\ldots,t_{n})$ and $s=(s_{1},s_{2},\ldots,s_{n})$, then we
have

$H^{(k)}_{t,s}(x)=\sum_{k_{1}+k_{2}+\cdots+k_{n}=k}\binom{k}{k_{1},k_{2},\ldots,k_{n}}C_{(k_{1},\ldots,k_{n})}H_{k_{1}}(x_{1})H_{k_{2}}(x_{2})\ldots
H_{k_{n}}(x_{n}),$

where

$C_{(k_{1},\ldots,k_{n})}=\prod_{j=1}^{n}\overline{(s+t)_{j}}^{k_{j}}.$

######  Theorem 14.

For any tuple $(t,s)$ on the Stiefel manifold $St^{(2)}(\mathbb{C}^{n})$, we
have

$g^{(k)}_{t,s}(x)=\mathcal{B}_{n}^{-1}\left[\langle
z,\overline{s+t}\rangle^{k}\right](x)=\psi^{(k)}_{t,s}(x),$

where $\psi^{(k)}_{t,x}(x)$ is defined in (36).

###### Proof.

We show the opposite direction. By the definition (36), we have

$\displaystyle\mathcal{B}\left[\psi^{(k)}_{t,s}(x)\right](z)$
$\displaystyle=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}\exp\left(-\frac{z\cdot
z}{2}+x\cdot z-\frac{x\cdot x}{4}\right)\psi^{(k)}_{t,s}(x)\,{\rm d}x$
$\displaystyle=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}\langle\nabla_{x},\overline{s+t}\rangle^{k}\left(\exp\left(-\frac{z\cdot
z}{2}+x\cdot z\right)\right)\exp\left(-\frac{x\cdot x}{2}\right)\,{\rm d}x$
$\displaystyle=\frac{1}{(2\pi)^{n/2}}\langle
z,\overline{s+t}\rangle^{k}\int_{\mathbb{R}^{n}}\exp\left(-\frac{z\cdot
z}{2}+x\cdot z\right)\exp\left(-\frac{x\cdot x}{2}\right)\,{\rm d}x$
$\displaystyle=\frac{1}{(2\pi)^{n/2}}\langle
z,\overline{s+t}\rangle^{k}\int_{\mathbb{R}^{n}}\exp\left(-\sum_{j=1}^{n}\frac{(z_{j}-x_{j})^{2}}{2}\right)\,{\rm
d}x$ $\displaystyle=\langle z,\overline{s+t}\rangle^{k}.$

This completes the proof. ∎

Now, we can consider the Radon-type transform on the usual space of square
integrable functions, which is the image of the Bargmann-Radon transform (29)
under the Segal-Bargmann transform.

######  Definition 14.

Assume $n\geq 2$. The Radon-type transform on $L^{2}(\mathbb{R}^{n})$ is
defined as the orthogonal projection

$\widehat{\mathcal{R}}_{t,s}:L^{2}(\mathbb{R}^{n})\to
L^{2}_{t,s}(\mathbb{R}^{n}).$ (37)

######  Remark 3.

The most appealing feature of this transformation is that it is defined on the
entire $L^{2}(\mathbb{R}^{n})$ space, rather than being limited to its
harmonic or monogenic subspace, which can be overly restrictive in practical
applications.

######  Theorem 15.

Let the tuple $(t,s)$ lie on the Stiefel manifold $St^{(2)}(\mathbb{C}^{n})$.
We dentoe $t=(t_{1},t_{2},\ldots,t_{n})$ and $s=(s_{1},s_{2},\ldots,s_{n})$.
Then the Radon transform in (37) can be equivalently written as an integral
transform. More precisely, for any $f\in L^{2}(\mathbb{R}^{n})$, we have

$\widehat{\mathcal{R}}_{t,s}[f](x)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}\widehat{K}_{t,s}(x,y)f(y)\,{\rm
d}y,$

where

$\begin{split}\widehat{K}_{t,s}(x,y)&=\sum_{k=0}^{\infty}\frac{1}{2^{k}k!}\psi^{(k)}_{t,s}(x)\overline{\psi^{(k)}_{t,s}(y)}\\\
&=\prod_{j=1}^{n}\frac{1}{\sqrt{1-\rho_{j}^{2}}}\exp\left(-\frac{(1+\rho_{j}^{2})(x_{j}^{2}+y_{j}^{2})-4\rho
x_{j}y_{j}}{2(1-\rho_{j}^{2})}\right),\end{split}$ (38)

in which $\rho_{j}=|(t+s)_{j}|^{2}$.

###### Proof.

It follows from the multinomial theorem, Lemma 6 and the Mehler’s formula

$\frac{1}{\sqrt{1-\rho^{2}}}\exp\left(-\frac{\rho^{2}(x^{2}+y^{2})-2\rho
xy}{1-\rho^{2}}\right)=\sum_{k=0}^{\infty}\frac{\rho^{k}}{2^{k}k!}H_{k}(x)H_{k}(y),$

where $H_{k}(x)$ is the Hermite polynomial. ∎

######  Remark 4.

When $\rho_{j}=1$, the kernel should be considered as the Dirac delta
function.

######  Example 2.

Suppose $t=(1,0,0,\ldots,0)$ and $s=(0,1,0,\ldots,0)$, we have

$\widehat{\mathcal{R}}_{t,s}[f](x)=\frac{1}{(2\pi)^{n/2-1}}\int_{\mathbb{R}^{n-2}}f(x_{1},x_{2},y_{3},\ldots,y_{n})\,{\rm
d}y_{3}\ldots\,{\rm d}y_{n},$ (39)

which has the similar form with the ordinary Radon transform.

The dual transform is again defined as the integral on the Stiefel manifold
$St^{(2)}(\mathbb{C}^{n})$, which formally shares the same form as in
Definition 11. We will not repeat here. At the end of this subsection, we give
the inversion formula for this new defined Radon-type transform.

######  Theorem 16.

For any $f\in L^{2}(\mathbb{R}^{n})$ with $n\geq 2$, we have

$f(x)=\left(\mathcal{T}\circ\widetilde{\mathcal{R}}\circ\widehat{\mathcal{R}}_{t,s}\right)[f](x),$

where

$\mathcal{T}[f](x):=\frac{1}{\Gamma(n)}(E+1)(E+2)\cdots(E+n-1)f,$

in which $E=(-\Delta_{x}+|x|^{2}-n)/2$.

###### Proof.

It is obtained by the correspondence of the raising and lowering operators in
the Fock space and $L^{2}(\mathbb{R}^{n})$. ∎

##  5 Radon-type transform for Hermitian monogenic functions

In this section, we define the Radon-type transforms specifically for
Hermitian monogenic functions. This formulation is intended as a refinement of
both the Radon-type transform (15) for complex harmonic functions and the
Szegö-Radon transform for monogenic functions studied in [6]. The refinement
aims to enhance and extend existing Radon-type transform methodologies,
aligning them with the distinctive characteristics inherent in Hermitian
monogenic functions.

###  5.1 Hermitian Clifford analysis

In this subsection, we recall some fundamental concepts in Hermitian Clifford
analysis. More details can be found in e.g. [3, 4]. Hermitian Clifford
analysis stands as a refinement of Euclidean Clifford analysis in [11], and
has attracted considerable attention in the past two decades.

Let $\\{{e}_{1},{e}_{2},\ldots,{e}_{m}\\}$ be an orthonormal basis for the
real vector space $\mathbb{R}^{m}$. We denote by $\mathbb{R}_{0,m}$ the real
Clifford algebra generated by the orthonormal basis under the multiplication
relations ${e}_{i}{e}_{j}+{e}_{j}{e}_{i}=-2\delta_{ij}$. The Euclidean vector
space $\mathbb{R}^{m}$ is naturally embedded into $\mathbb{R}_{m}$ by means of
the identification
$(x_{1},x_{2},\ldots,x_{m})\mapsto\underline{x}:=\sum_{i=1}^{m}{e}_{i}x_{i}.$
Meanwhile, $\mathbb{R}_{0,m}$ has the construction
$\mathbb{R}_{0,m}=\bigoplus_{k=0}^{m}\mathbb{R}_{0,m}^{k}$, where
$\mathbb{R}_{0,m}^{k}$ is the subspace of $k$-vectors spanned by the products
of $k$ different basis vectors. The so-called Dirac operator in Euclidean
Clifford analysis is thus defined as
$\partial_{\underline{x}}=\sum_{j=1}^{m}e_{j}\partial_{x_{j}}$, which is the
Fischer dual of the Clifford vector variable
$\underline{x}=\sum_{j=1}^{m}e_{j}x_{j}$.

The complex Clifford algebra $\mathbb{C}_{m}$ is the complexification of the
real Clifford algebra, i.e. $\mathbb{C}_{m}=\mathbb{R}_{0,m}\oplus
i\mathbb{R}_{0,m}$. Let $m=2n$ be an even dimension, then any Clifford number
$\lambda\in\mathbb{C}_{2n}$ can be written as $\lambda=a+ib$,
$a,b\in\mathbb{R}_{0,2n}$. The complex Clifford algebra $\mathbb{C}_{2n}$ is
equipped with the Hermitian conjugation
$\lambda^{\dagger}=\left(a+ib\right)^{\dagger}=\bar{a}-i\bar{b}$, where
$\overline{e_{j}}=-e_{j}$ is the unique anti-involution in
$\mathbb{R}_{0,2n}$. The Hermitian inner product and the induced norm can thus
be given using the conjugation,

$\left(\lambda,\mu\right)=\left[\lambda^{\dagger}\,\mu\right]_{0},\qquad\qquad\left|{\lambda}\right|=\sqrt{\left[\lambda^{\dagger}\,\lambda\right]_{0}},$

where $\left[\,\cdot\,\right]_{0}$ stands for the scalar part.

For the complex Clifford algebra $\mathbb{C}_{2n}$, the so-called Witt basis
is given by

$f_{j}=\frac{1}{2}\left(e_{j}-ie_{n+j}\right),\qquad\qquad
f_{j}^{\dagger}=-\frac{1}{2}\left(e_{j}+ie_{n+j}\right),$ (40)

where $j=1,2,\ldots,n$. These new Witt basis are isotropic, i.e.

$f_{j}^{2}=\left(f_{j}^{\dagger}\right)^{2}=0,$

and satisfy the Grassmann identities

$f_{j}f_{k}+f_{k}f_{j}=f_{j}^{\dagger}f_{k}^{\dagger}+f_{k}^{\dagger}f_{j}^{\dagger}=0,\qquad
j,k=1,\cdots,n,$

as well as the duality identities

$f_{j}f_{k}^{\dagger}+f_{k}^{\dagger}f_{j}=\delta_{jk},\qquad j,k=1,\cdots,n.$
(41)

By employing the Witt basis, we identify the Clifford vector $\underline{X}$
and $\left(x_{1},\ldots,x_{n},y_{1},\ldots,y_{n}\right)$ in $\mathbb{R}^{2n}$,
where

$\underline{X}:=\sum_{j=1}^{n}(x_{j}e_{j}+y_{j}e_{n+j})=\sum_{j=1}^{n}f_{j}z_{j}-\sum_{j=1}^{n}f_{j}^{\dagger}\bar{z}_{j},$

in which $z_{j}=x_{j}+iy_{j}$ and $\bar{z}_{j}=x_{j}-iy_{j}$. This allows us
to introduce the Hermitian vectors as

$\underline{z}:=\sum_{j=1}^{n}f_{j}z_{j},\qquad\qquad\underline{z}^{\dagger}:=\sum_{j=1}^{n}f_{j}^{\dagger}\overline{z}_{j}.$

Meanwhile, the Hermitian Dirac operators are defined by

$\partial_{\underline{z}}:=\sum_{j=1}^{n}f_{j}^{\dagger}\partial_{z_{j}},\qquad\qquad\partial_{\underline{z}^{\dagger}}:=\left(\partial_{z}\right)^{\dagger}=\sum_{j=1}^{n}f_{j}\partial_{\overline{z}_{j}},$

where the complex derivatives $\partial_{z_{j}}$ and
$\partial_{\overline{z}_{j}}$ are defined as in (3).

Applying the Hermitian Dirac operator $\partial_{\underline{z}}$ from the left
to the vector $\underline{z}$ results in a constant Clifford number, denoted
as $\beta$, given by

$\beta:=\partial_{\underline{z}}\underline{z}=\sum_{j=1}^{n}\sum_{k=1}^{n}f_{j}^{\dagger}\partial_{z_{j}}f_{k}z_{k}=\sum_{j=1}^{n}f_{j}^{\dagger}f_{j}.$
(42)

Now, we introduce the key concept in this framework.

######  Definition 15 (Hermitian monogenic functions).

A function $f\in\mathbb{C}_{m}$ is called Hermitian monogenic, or h-monogenic,
if it belongs to the kernel of the Hermitian Dirac operators, i.e.

$\partial_{\underline{z}}f=0\qquad{\rm
and}\qquad\partial_{\underline{z}^{\dagger}}f=0.$

The spinor space $\mathcal{S}_{n}$ is the $2^{n}$-dimensional minimal left
ideal, defined as

$\mathcal{S}_{n}=\mathbb{C}_{2n}I,$

where $I$ is the primitive idempotent

$I=f_{1}f_{1}^{\dagger}f_{2}f_{2}^{\dagger}\ldots f_{n}f_{n}^{\dagger}.$

It can be decomposed in terms of the homogeneous spinors as

$\mathcal{S}_{n}=\bigoplus_{j=0}^{n}\mathbb{S}^{(j)}.$

The Clifford number $\beta$ can be understood as a multiplicative operator
that measures the degree of a spinor, i.e.

$\beta f^{(j)}=jf^{(j)},\qquad\forall\,f^{(j)}\in\mathbb{S}^{(j)},$

and is therefore called the spin-Euler operator in [3].

Thus the spinor-valued polynomial space can be decomposed as

$\mathcal{P}\otimes\mathcal{S}_{n}=\bigoplus_{p,q=0}^{\infty}\bigoplus_{j=1}^{n}\mathcal{P}^{(j)}_{p,q},$

where $\mathcal{P}^{(j)}_{p,q}:=\mathcal{P}_{p,q}\otimes\mathbb{S}^{(j)}$ is
the space of $\mathbb{S}^{(j)}$-valued homogeneous polynomials of degree
$(p,q)$.

######  Definition 16.

The space

$\mathcal{M}_{p,q}^{(j)}=\left\\{M(z)|M(z)\in\mathcal{P}_{p,q}^{(j)},\,\partial_{\underline{z}}M(z)=\partial_{\underline{z}}^{\dagger}M(z)=0\right\\}$

of h-monogenic $\mathbb{S}^{(j)}$-valued polynomials of homogeneity $(p,q)$ is
called the space of $\mathbb{S}^{(j)}$-valued spherical h-monogenics.

Naturally, we can adapt the Fischer inner product in the Hermitian setting.
For any $P(z),Q(z)\in\mathcal{P}^{(j)}_{p,q}$, it is defined by

$\left\langle{P(z)},{Q(z)}\right\rangle_{\partial}:=\left[P(\partial)^{\dagger}Q(z)\right]_{z=0},$
(43)

where the symbol $P(\partial)$ is the replacement of the complex variable
$z_{j}=x_{j}+ix_{n+j}$ by
$\partial_{\bar{z}_{j}}=\left(\partial_{x_{j}}+i\partial_{x_{n+j}}\right)/2$,
see e.g. [35].

Let $\mathcal{H}_{p,q}^{(j)}=\mathcal{H}_{p,q}\otimes\,\mathbb{S}^{(j)}$ be
the space of $\mathbb{S}^{(j)}$-valued spherical harmonics. The Fischer
decomposition leads to a further refinement of complex spherical harmonics

$\begin{split}\mathcal{H}_{p,q}^{(j)}=\mathcal{M}_{p,q}^{(j)}\,\oplus&\,\underline{z}\,\mathcal{M}_{p-1,q}^{(j-1)}\,\oplus\,\underline{z}^{\dagger}\mathcal{M}_{p,q-1}^{j+1}\\\
\,\oplus&\,\left(\frac{\underline{z}\underline{z}^{\dagger}}{q-1+j}+\frac{\underline{z}^{\dagger}\underline{z}}{p-1+n-j}\right)\mathcal{M}_{p-1,q-1}^{(j)},\end{split}$

where we put $\mathcal{M}_{p,q}^{(j)}=\\{0\\}$ whenever $p<0,q<0,j<0$ or $j>n$
and moreover, the last summand reduces to
$\underline{z}\,\underline{z}^{\dagger}\mathcal{M}_{p-1,q-1}^{(j)}$ when
$q-1+j=0$, while, it reduces to
$\underline{z}^{\dagger}\underline{z}\mathcal{M}_{p-1,q-1}^{(j)}$ when
$p-1+n-j=0$.

###  5.2 Hermitian Szegö-Radon transform

Suppose the tuple $(t,s)$ lies on the Stiefel manifold
$St^{(2)}(\mathbb{C}^{n})$, let

$\underline{t}=\sum_{j=1}^{n}f_{j}t_{j},\qquad\qquad\underline{s}=\sum_{j=1}^{n}f_{j}s_{j}.$

Straightforward computation shows that
$\left|{\underline{t}}\right|=\left|{\underline{s}}\right|=\sqrt{\left[\underline{s}^{\dagger}\underline{s}\right]_{0}}=\sqrt{2}/2.$
It should be keep in mind that the notations $t$ and $\underline{t}$ now stand
for different vectors.

To define Radon transform for the Hermitian monogenic functions, we first
introduce the null-vector associated to the tuple $(t,s)\in
St^{(2)}(\mathbb{C}^{n})$ in the Hermitian Clifford setting,

$\underline{\tau}:=(\underline{t}-\underline{s})(\underline{t}^{\dagger}+\underline{s}^{\dagger}).$

######  Proposition 3.

The following properties of the null-vector $\underline{\tau}$ hold

  1. 1.

$\underline{\tau}^{2}=0$,

  2. 2.

$\underline{\tau}\,\underline{\tau}^{\dagger}\underline{\tau}=4\underline{\tau}.$

###### Proof.

  1. 1.

Since $t$ and $s$ are orthogonal, we have
$\\{\underline{t}^{\dagger},\underline{s}\\}=0$. Together with the facts
$\\{\underline{z},\underline{z}^{\dagger}\\}=|z|^{2}$,
$\\{\underline{t},\underline{s}\\}=0$ and
$\underline{t}^{2}=\underline{s}^{2}=0$, we have

$\displaystyle\underline{\tau}^{2}$
$\displaystyle=(\underline{t}-\underline{s})(\underline{t}^{\dagger}+\underline{s}^{\dagger})(\underline{t}-\underline{s})(\underline{t}^{\dagger}+\underline{s}^{\dagger})$
$\displaystyle=(\underline{t}\,\underline{t}^{\dagger}-\underline{s}\,\underline{t}^{\dagger}+\underline{t}\,\underline{s}^{\dagger}-\underline{s}\,\underline{s}^{\dagger})(\underline{t}-\underline{s})(\underline{t}^{\dagger}+\underline{s}^{\dagger})$
$\displaystyle=(\underline{t}\,|t|^{2}-\underline{s}\,\underline{t}^{\dagger}\underline{t}-\underline{s}\,\underline{s}^{\dagger}\,\underline{t}-\underline{t}\,\underline{t}^{\dagger}\,\underline{s}-\underline{t}\,\underline{s}^{\dagger}\underline{s}-|s|^{2}\underline{s})(\underline{t}^{\dagger}+\underline{s}^{\dagger})$
$\displaystyle=(\underline{t}\,|t|^{2}-\underline{s}\,|t|^{2}+\underline{s}\,\underline{t}\,\underline{t}^{\dagger}-\underline{s}\,\underline{s}^{\dagger}\underline{t}-\underline{t}\,\underline{t}^{\dagger}\underline{s}-\underline{t}\,|s|^{2}+\underline{t}\,\underline{s}\,\underline{s}^{\dagger}-|s|^{2}\underline{s})(\underline{t}^{\dagger}+\underline{s}^{\dagger})$
$\displaystyle=(\underline{t}\,|t|^{2}-\underline{s}\,|t|^{2}-\underline{t}\,|s|^{2}-|s|^{2}\underline{s})(\underline{t}^{\dagger}+\underline{s}^{\dagger}).$

Now the property $\underline{\tau}^{2}=0$ follows from $|t|=|s|=1$.

  2. 2.

Note that
$\underline{\tau}^{\dagger}=(\underline{t}+\underline{s})(\underline{t}^{\dagger}-\underline{s}^{\dagger})=-(\underline{t}^{\dagger}-\underline{s}^{\dagger})(\underline{t}+\underline{s})$.
We have

$\displaystyle\underline{\tau}\,\underline{\tau}^{\dagger}\underline{\tau}=$
$\displaystyle(\underline{t}-\underline{s})(\underline{t}^{\dagger}+\underline{s}^{\dagger})(\underline{t}+\underline{s})(\underline{t}^{\dagger}-\underline{s}^{\dagger})(\underline{t}-\underline{s})(\underline{t}^{\dagger}+\underline{s}^{\dagger})$
$\displaystyle=$
$\displaystyle(\underline{t}\,\underline{t}^{\dagger}-\underline{s}\,\underline{t}^{\dagger}+\underline{t}\,\underline{s}^{\dagger}-\underline{s}\,\underline{s}^{\dagger})(\underline{t}\,\underline{t}^{\dagger}+\underline{s}\,\underline{t}^{\dagger}-\underline{t}\,\underline{s}^{\dagger}-\underline{s}\,\underline{s}^{\dagger})(\underline{t}\,\underline{t}^{\dagger}-\underline{s}\,\underline{t}^{\dagger}+\underline{t}\,\underline{s}^{\dagger}-\underline{s}\,\underline{s}^{\dagger})$
$\displaystyle=$
$\displaystyle(\underline{t}\,\underline{t}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}-\underline{s}\,\underline{t}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}+\underbrace{\underline{t}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}}_{0}-\underline{s}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}+\underbrace{\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{t}^{\dagger}}_{0}-\underbrace{\underline{s}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{t}^{\dagger}}_{0}+\underline{t}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{t}^{\dagger}-\underline{s}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{t}^{\dagger}$
$\displaystyle-$
$\displaystyle\underline{t}\,\underline{t}^{\dagger}\,\underline{t}\,\underline{s}^{\dagger}+\underline{s}\,\underline{t}^{\dagger}\,\underline{t}\,\underline{s}^{\dagger}-\underbrace{\underline{t}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{s}^{\dagger}}_{0}+\underbrace{\underline{s}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{s}^{\dagger}}_{0}-\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}+\underbrace{\underline{s}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}}_{0}-\underline{t}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}+\underline{s}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger})$
$\displaystyle\times$
$\displaystyle(\underline{t}\,\underline{t}^{\dagger}-\underline{s}\,\underline{t}^{\dagger}+\underline{t}\,\underline{s}^{\dagger}-\underline{s}\,\underline{s}^{\dagger})$
$\displaystyle=$
$\displaystyle(\underline{t}\,\underline{t}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}-\underline{s}\,\underline{t}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}-\underline{s}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}+\underline{t}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{t}^{\dagger}-\underline{s}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{t}^{\dagger}-\underline{t}\,\underline{t}^{\dagger}\,\underline{t}\,\underline{s}^{\dagger}+\underline{s}\,\underline{t}^{\dagger}\,\underline{t}\,\underline{s}^{\dagger}$
$\displaystyle-$
$\displaystyle\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}-\underline{t}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}+\underline{s}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger})(\underline{t}\,\underline{t}^{\dagger}-\underline{s}\,\underline{t}^{\dagger}+\underline{t}\,\underline{s}^{\dagger}-\underline{s}\,\underline{s}^{\dagger}).$

Using the fact
$\underline{t}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}=\underline{t}^{\dagger}(|t|-\underline{t}^{\dagger}\underline{t})=|t|$,
it becomes

$\displaystyle\underline{\tau}\,\underline{\tau}^{\dagger}\underline{\tau}$
$\displaystyle=(\underline{t}\,\underline{t}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}-\underline{t}\,\underline{t}^{\dagger}\,\underline{t}\,\underline{s}^{\dagger}-\underline{s}\,\underline{t}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}-\underline{s}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}+\underline{t}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{t}^{\dagger}+\underline{s}\,\underline{t}^{\dagger}\,\underline{t}\,\underline{s}^{\dagger}-\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}$
$\displaystyle\quad-\underline{s}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{t}^{\dagger}-\underline{t}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}+\underline{s}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger})(\underline{t}\,\underline{t}^{\dagger}-\underline{s}\,\underline{t}^{\dagger}+\underline{t}\,\underline{s}^{\dagger}-\underline{s}\,\underline{s}^{\dagger})$
$\displaystyle=(\underline{t}\,\underline{t}^{\dagger}|t|-|t|\underline{t}\,\underline{s}^{\dagger}+|t|\underline{t}^{\dagger}\,\underline{s}-\underline{s}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}+\underline{t}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{t}^{\dagger}+\underline{s}\,\underline{t}^{\dagger}\,\underline{t}\,\underline{s}^{\dagger}-\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}$
$\displaystyle\quad+|s|\underline{t}^{\dagger}\,\underline{s}-|s|\underline{t}\,\underline{s}^{\dagger}+|s|\underline{s}\,\underline{s}^{\dagger})(\underline{t}\,\underline{t}^{\dagger}-\underline{s}\,\underline{t}^{\dagger}+\underline{t}\,\underline{s}^{\dagger}-\underline{s}\,\underline{s}^{\dagger})$
$\displaystyle=(\underline{t}\,\underline{t}^{\dagger}+\underline{s}\,\underline{s}^{\dagger}-2|t|\underline{t}\,\underline{s}^{\dagger}+2|t|\underline{t}^{\dagger}\,\underline{s}-\underline{s}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}+\underline{t}(|s|-\underline{s}\,\underline{s}^{\dagger})\underline{t}^{\dagger}$
$\displaystyle\quad+\underline{s}(|t|-\underline{t}\underline{t}^{\dagger})\underline{s}^{\dagger}-\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger})(\underline{t}\,\underline{t}^{\dagger}-\underline{s}\,\underline{t}^{\dagger}+\underline{t}\,\underline{s}^{\dagger}-\underline{s}\,\underline{s}^{\dagger})$
$\displaystyle=(2\underline{t}\,\underline{t}^{\dagger}+2\underline{s}\,\underline{s}^{\dagger}-2\underline{t}\,\underline{s}^{\dagger}+2\underline{t}^{\dagger}\,\underline{s}-4\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger})(\underline{t}\,\underline{t}^{\dagger}-\underline{s}\,\underline{t}^{\dagger}+\underline{t}\,\underline{s}^{\dagger}-\underline{s}\,\underline{s}^{\dagger}).$

Distributing the last terms, we get

$\displaystyle\underline{\tau}\,\underline{\tau}^{\dagger}\underline{\tau}$
$\displaystyle=2\underline{t}\,\underline{t}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}+2\underline{s}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}-2\underline{t}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}+2\underline{t}^{\dagger}\,\underline{s}\,\underline{t}\,\underline{t}^{\dagger}-4\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}$
$\displaystyle\quad-2\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{t}^{\dagger}-2\underline{s}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{t}^{\dagger}+2\underline{t}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{t}^{\dagger}+2\underline{t}^{\dagger}\,\underline{s}\,\underline{s}\,\underline{t}^{\dagger}+4\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{t}^{\dagger}$
$\displaystyle\quad+2\underline{t}\,\underline{t}^{\dagger}\,\underline{t}\,\underline{s}^{\dagger}+2\underline{s}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{s}^{\dagger}-2\underline{t}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{s}^{\dagger}+2\underline{t}^{\dagger}\,\underline{s}\,\underline{t}\,\underline{s}^{\dagger}-4\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{s}^{\dagger}$
$\displaystyle\quad-2\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}-2\underline{s}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}+2\underline{t}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}+2\underline{t}^{\dagger}\,\underline{s}\,\underline{s}\,\underline{s}^{\dagger}+4\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}$
$\displaystyle=2|t|\underline{t}\,\underline{t}^{\dagger}+2\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}-\underbrace{2\underline{t}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}}_{0}-2|t|\underline{s}\,\underline{t}^{\dagger}-4|t|\underline{s}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{t}^{\dagger}$
$\displaystyle\quad-\underbrace{2\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{t}^{\dagger}}_{0}+2|s|\underline{t}^{\dagger}\,\underline{s}+2\underline{t}\,\underline{t}^{\dagger}\,\underline{s}^{\dagger}\,\underline{s}+\underbrace{2\underline{t}^{\dagger}\,\underline{s}\,\underline{s}\,\underline{t}^{\dagger}}_{0}+\underbrace{4|s|\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{t}^{\dagger}}_{0}$
$\displaystyle\quad+2|t|\underline{t}\,\underline{s}^{\dagger}+\underbrace{2\underline{s}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{s}^{\dagger}}_{0}-\underbrace{2\underline{t}\,\underline{s}^{\dagger}\,\underline{t}\,\underline{s}^{\dagger}}_{0}-2\underline{t}^{\dagger}\,\underline{t}\,\underline{s}\,\underline{s}^{\dagger}-\underbrace{4\underline{t}\,\underline{t}^{\dagger}\,\underline{t}\,\underline{s}\,\underline{s}^{\dagger}\,\underline{s}^{\dagger}}_{0}$
$\displaystyle\quad-2\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}-2|s|\underline{s}\,\underline{s}^{\dagger}+2|s|\underline{t}\,\underline{s}^{\dagger}+\underbrace{2\underline{t}^{\dagger}\,\underline{s}\,\underline{s}\,\underline{s}^{\dagger}}_{0}+4|s|\underline{t}\,\underline{t}^{\dagger}\,\underline{s}\,\underline{s}^{\dagger}$
$\displaystyle=2\underline{t}\,\underline{t}^{\dagger}+4\underline{t}^{\dagger}\,\underline{s}+2\underline{t}\,\underline{t}^{\dagger}\,\underline{s}^{\dagger}\,\underline{s}-2\underline{u}\,t\underline{t}\,\underline{s}\,\underline{s}^{\dagger}-2\underline{s}\,\underline{s}^{\dagger}+4\underline{t}\,\underline{s}^{\dagger}.$

Finally, we obtain

$\displaystyle\underline{\tau}\,\underline{\tau}^{\dagger}\underline{\tau}$
$\displaystyle=2\underline{t}\,\underline{t}^{\dagger}+4\underline{t}^{\dagger}\,\underline{s}+2\underline{t}\,\underline{t}^{\dagger}\,\underline{s}^{\dagger}\,\underline{s}-2(|t|-\underline{t}\,\underline{t}^{\dagger})(|s|-\underline{s}^{\dagger}\,\underline{s})-2\underline{s}\,\underline{s}^{\dagger}+4\underline{t}\,\underline{s}^{\dagger}$
$\displaystyle=2\underline{t}\,\underline{t}^{\dagger}+4\underline{t}^{\dagger}\,\underline{s}+2\underline{t}\,\underline{t}^{\dagger}\,\underline{s}^{\dagger}\,\underline{s}-2+2\underline{t}\,\underline{t}^{\dagger}+2\underline{s}^{\dagger}\,\underline{s}-2\underline{t}\,\underline{t}^{\dagger}\,\underline{s}^{\dagger}\,\underline{s}-2\underline{s}\,\underline{s}^{\dagger}+4\underline{t}\,\underline{s}^{\dagger}$
$\displaystyle=2\underline{t}\,\underline{t}^{\dagger}+4\underline{t}^{\dagger}\,\underline{s}-2+2\underline{t}\,\underline{t}^{\dagger}+2(|s|-\underline{s}\,\underline{s}^{\dagger})-2\underline{s}\,\underline{s}^{\dagger}+4\underline{t}\,\underline{s}^{\dagger}$
$\displaystyle=4\underline{t}\,\underline{t}^{\dagger}+4\underline{t}^{\dagger}\,\underline{s}-4\underline{s}\,\underline{s}^{\dagger}+4\underline{t}\,\underline{s}^{\dagger}$
$\displaystyle=4((\underline{t}-\underline{s})\underline{t}^{\dagger}+(\underline{t}-\underline{s})\underline{s}^{\dagger})$
$\displaystyle=4(\underline{t}-\underline{s})(\underline{t}^{\dagger}+\underline{s}^{\dagger})$
$\displaystyle=4\underline{\tau}.$

∎

By employing the null-vection $\underline{\tau}$ defined in (5.2), we proceed
to introduce a significant category of Hermitian monogenic plane waves.

######  Lemma 7.

Suppose the tuple $(t,s)$ lies on the Stiefel manifold
$St^{(2)}(\mathbb{C}^{n})$, then the homogeneous polynomial
$f^{[p,q]}_{\underline{\tau}}(z)$ of degree $p,q$ given by

$f^{[p,q]}_{\underline{\tau}}(z):=\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}\underline{\tau}$
(44)

is Hemitian monogenic.

###### Proof.

We verify that these functions reside within the kernel of the two Hermitian
Dirac operators. First, we have

$\begin{split}\partial_{\underline{z}^{\dagger}}f^{[p,q]}_{\underline{\tau}}(z)&=\sum_{j=1}^{n}f_{j}\partial_{\overline{z}_{j}}\left(\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}(\underline{t}-\underline{s})(\underline{t}^{\dagger}+\underline{s}^{\dagger})\right)\\\
&=q\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q-1}(\underline{s}-\underline{t})(\underline{t}-\underline{s})(\underline{t}^{\dagger}+\underline{s}^{\dagger})\\\
&=0.\end{split}$

The second equation can be verified similarly,

$\displaystyle\partial_{\underline{z}}f^{[p,q]}_{\underline{\tau}}(z)$
$\displaystyle=\sum_{j=1}^{n}f_{j}^{\dagger}\partial_{z_{j}}\left(\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}(\underline{t}-\underline{s})(\underline{t}^{\dagger}+\underline{s}^{\dagger})\right)$
$\displaystyle=p\left\langle{z},{\overline{s+t}}\right\rangle^{p-1}\left\langle{\bar{z}},{s-t}\right\rangle^{q}(\underline{s}^{\dagger}+\underline{t}^{\dagger})(\underline{t}-\underline{s})(\underline{t}^{\dagger}+\underline{s}^{\dagger})$
$\displaystyle=p\left\langle{z},{\overline{s+t}}\right\rangle^{p-1}\left\langle{\bar{z}},{s-t}\right\rangle^{q}[(\underline{s}^{\dagger}\underline{t}+\underline{t}^{\dagger}\underline{t}-\underline{s}^{\dagger}\underline{s}-\underline{t}^{\dagger}\underline{s})(\underline{t}^{\dagger}+\underline{s}^{\dagger})]$
$\displaystyle=p\left\langle{z},{\overline{s+t}}\right\rangle^{p-1}\left\langle{\bar{z}},{s-t}\right\rangle^{q}[\underline{s}^{\dagger}\underline{t}\,\underline{t}^{\dagger}+|t|^{2}\underline{t}^{\dagger}-\underline{s}^{\dagger}\underline{s}\,\underline{t}^{\dagger}$
$\displaystyle\quad+\underline{t}^{\dagger}\underline{t}\,\underline{s}^{\dagger}-|s|^{2}\underline{s}^{\dagger}-\underline{t}^{\dagger}\underline{s}\,\underline{s}^{\dagger}]$
$\displaystyle=p\left\langle{z},{\overline{s+t}}\right\rangle^{p-1}\left\langle{\bar{z}},{s-t}\right\rangle^{q}[\underline{s}^{\dagger}|t|^{2}+|t|^{2}\underline{t}^{\dagger}-\underline{s}^{\dagger}\underline{s}\,\underline{t}^{\dagger}$
$\displaystyle\quad-|s|^{2}\underline{s}^{\dagger}+\underline{t}^{\dagger}\underline{s}^{\dagger}\underline{s}-\underline{t}^{\dagger}|s|^{2}]$
$\displaystyle=0.$

∎

Let $\mathcal{M}_{h}\mathcal{L}(\mathbb{B}^{n})$ be the space consisting of
$\mathcal{S}_{n}$-valued $h$-monogenic functions on the unit ball, equipping
with the $L^{2}$ inner product given by

$\langle
P(z),Q(z)\rangle_{\mathcal{M}_{h}}:=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}P(z)^{\dagger}Q(z)\,{\rm
d}\sigma(z).$ (45)

Similarly, the space of $h$-monogenic $\mathbb{S}^{(j)}$-valued functions on
the unit ball with the inner product (45) is denoted by
$\mathcal{M}_{h}^{(j)}\mathcal{L}(\mathbb{B}^{n})$.

######  Theorem 17.

It holds that

$\begin{split}\left\langle
f_{\underline{\tau}}^{[p,q]}(z),g_{\underline{\tau}}^{[u,v]}(z)\right\rangle_{\mathcal{M}_{h}}=\left\\{\begin{array}[]{ll}\gamma_{p,q}\underline{\tau}^{\dagger}\underline{\tau},&p=u,\,q=v,\\\
0,&\text{otherwise},\\\ \end{array}\right.\end{split}$ (46)

where $\gamma_{p,q}=2^{p+q}p!q!/(n)_{p+q}$.

###### Proof.

By the definition (45) of the Hermite inner product, we have

$\displaystyle\quad\left\langle{f_{\underline{\tau}}^{[p,q]}(z)},{g_{\underline{\tau}}^{[u,v]}(z)}\right\rangle_{\mathcal{M}_{h}}$
$\displaystyle=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\left(f_{\underline{\tau}}^{[p,q]}(z)\right)^{\dagger}g_{\underline{\tau}}^{[u,v]}(z)\,{\rm
d}\sigma(z)$
$\displaystyle=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\left(\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}\underline{\tau}\right)^{\dagger}\left(\left\langle{z},{\overline{s+t}}\right\rangle^{u}\left\langle{\bar{z}},{s-t}\right\rangle^{v}\underline{\tau}\right)\,{\rm
d}\sigma(z)$
$\displaystyle=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\overline{\left(\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}\right)}\left(\left\langle{z},{\overline{s+t}}\right\rangle^{u}\left\langle{\bar{z}},{s-t}\right\rangle^{v}\right)\,{\rm
d}\sigma(z)\underline{\tau}^{\dagger}\underline{\tau}$
$\displaystyle=\left\langle
f_{t,s}^{(p,q)}(z),g_{t,s}^{(w,v)}(z)\right\rangle_{\mathbb{S}}\underline{\tau}^{\dagger}\,\underline{\tau},$

where $\left\langle
f_{t,s}^{(p,q)}(z),g_{t,s}^{(w,v)}(z)\right\rangle_{\mathbb{S}}$ is calculated
in (9).

∎

######  Definition 17.

The completion of all finite linear combinations of Hermitian monogenic plane
waves of the form

$\sum_{p,q}\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}\underline{\tau}\,\alpha_{p,q},\qquad\alpha_{p,q}\in\mathcal{S}_{n},$

is a closed submodule of $\mathcal{M}_{h}\mathcal{L}(\mathbb{B}^{n})$, denoted
by $\mathcal{M}_{h}\mathcal{L}_{\underline{\tau}}(\mathbb{B}^{n})$.

Now, the Hermitian Radon transform is defined as the orthogonal projection on
the $\underline{\tau}$-submodule. More precisely, we have

######  Definition 18.

Assume $n\geq 2$. The orthogonal projection

$\mathcal{R}_{\underline{\tau}}:\mathcal{M}_{h}\mathcal{L}(\mathbb{B}^{n})\to\mathcal{M}_{h}\mathcal{L}_{\underline{\tau}}(\mathbb{B}^{n})$
(47)

is called the Szegö-Radon transform transform for Hermitian monogenic
functions or the Hermitian Szegö-Radon transform.

The Hermitian Szegö-Radon transform in (47) can be expressed equivalently
through an integral expression. The proof closely follows that of Theorems 2
and 3; therefore, it is omitted.

######  Theorem 18.

Let $n\geq 2$. The Hermitian Szegö-Radon transform coincides with the integral
operator

$\mathcal{R}_{\underline{\tau}}[f](z)=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}K_{\underline{\tau}}(z,u)f(u)\,{\rm
d}\sigma(u),$ (48)

where the Hermitian Szegö-Radon kernel is given by

$\begin{split}K_{\underline{\tau}}(z,u)=&\sum_{p=0}^{\infty}\sum_{q=0}^{\infty}\frac{1}{4\gamma_{p,q}}\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}\left\langle{\bar{u}},{s+t}\right\rangle^{p}\left\langle{u},{\overline{s-t}}\right\rangle^{q}\underline{\tau}\underline{\tau}^{\dagger}.\end{split}$

Moreover, this series admits a closed form,

$\begin{split}K_{\underline{\tau}}(z,u)=&\frac{1}{4}\left(\frac{2}{2-\left\langle{z},{\overline{s+t}}\right\rangle\left\langle{\bar{u}},{s+t}\right\rangle-\left\langle{\bar{z}},{s-t}\right\rangle\left\langle{u},{\overline{s-t}}\right\rangle}\right)^{n}\underline{\tau}\underline{\tau}^{\dagger}.\\\
\end{split}$ (49)

The Hermitian Szegö-Radon kernel $K_{\underline{\tau}}(z,u)$ has the following
properties:

######  Proposition 4.

It holds that

  1. 1.

$K_{\underline{\tau}}(z,u)^{\dagger}=K_{\underline{\tau}}(u,z)$;

  2. 2.

$K_{\underline{\tau}}(z,u)$ is a reproducing kernel of
$\mathcal{M}_{h}\mathcal{L}_{\underline{\tau}}(\mathbb{B}^{n})$;

  3. 3.

$\left\langle{K_{\underline{\tau}}(u,z)},{K_{\underline{\tau}}(v,z)}\right\rangle_{\mathcal{M}_{h}}=K_{\underline{\tau}}(v,u)$.

###### Proof.

We only prove (2) here. Indeed, we have

$\displaystyle\quad\left\langle
K_{\underline{\tau}}(z,u),g_{\underline{\tau}}^{[u,v]}(z)\right\rangle_{\mathcal{M}_{h}}$
$\displaystyle=\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\left(K_{\underline{\tau}}(z,u)\right)^{\dagger}g_{\underline{\tau}}^{[u,v]}(z)\,{\rm
d}\sigma(z)$
$\displaystyle=\int_{\mathbb{S}^{2n-1}}\left(\sum_{p,q}\frac{1}{4\gamma_{p,q}}\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}\left\langle{\overline{u}},{s+t}\right\rangle^{p}\left\langle{u},{\overline{s-t}}\right\rangle^{q}\underline{\tau}\underline{\tau}^{\dagger}\right)^{\dagger}$
$\displaystyle\quad\times\frac{1}{\omega_{2n-1}}g_{\underline{\tau}}^{[u,v]}(z)\,{\rm
d}\sigma(z)$
$\displaystyle=\sum_{p,q}\sum_{u,v}\frac{1}{\gamma_{p,q}}\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\overline{\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}}\,\overline{\left\langle{\bar{u}},{s+t}\right\rangle^{p}\left\langle{u},{\overline{s-t}}\right\rangle^{q}}$
$\displaystyle\quad\times\left\langle{z},{\overline{s+t}}\right\rangle^{u}\left\langle{\bar{z}},{s-t}\right\rangle^{v}\,{\rm
d}\sigma(z)\underline{\tau}$
$\displaystyle=\sum_{p,q}\sum_{u,v}\underbrace{\frac{1}{\omega_{2n-1}}\int_{\mathbb{S}^{2n-1}}\overline{\left\langle{z},{\overline{s+t}}\right\rangle^{p}\left\langle{\bar{z}},{s-t}\right\rangle^{q}}\,\left\langle{z},{\overline{s+t}}\right\rangle^{u}\left\langle{\bar{z}},{s-t}\right\rangle^{v}\,{\rm
d}\sigma(z)}_{\left\langle
f_{t,s}^{(p,q)}(z),g_{t,s}^{(w,v)}(z)\right\rangle_{\mathbb{S}}}\underline{\tau}$
$\displaystyle\quad\times\frac{1}{\gamma_{p,q}}\left\langle{u},{\overline{s+t}}\right\rangle^{p}\left\langle{\overline{u}},{s-t}\right\rangle^{q}$
$\displaystyle=\left\langle{u},{\overline{s+t}}\right\rangle^{p}\left\langle{\overline{u}},{s-t}\right\rangle^{q}\underline{\tau}$
$\displaystyle=g_{\underline{\tau}}^{[u,v]}(u),$

which shows the reproducing property of $K_{\underline{\tau}}(z,u)$. ∎

Similarly as the scalar case, the dual transform is defined as an integral on
the Stiefel manifold $St^{(2)}(\mathbb{C}^{n})$.

######  Definition 19.

Suppose the tuple $(t,s)$ lies on the Stiefel manifold
$St^{(2)}(\mathbb{C}^{n})$,
$F\in\mathcal{M}_{h}\mathcal{L}_{\underline{\tau}}(\mathbb{B}^{n})$, the dual
of the Szegö-Radon transform $\mathcal{R}_{\underline{\tau}}$ in (47) is
defined by

$\widetilde{\mathcal{R}}_{\underline{\tau}}[F](z)=\frac{1}{\omega_{2n-1}\omega_{2n-3}}\int_{St^{(2)}}F(z,t,s)\,{\rm
d}\sigma(s)\,{\rm d}\sigma(t).$

To calculate the dual transform and construct the inversion formula, we recall
the plane wave representation for the Hermitian monongenics obtained in [35].
Recall that $\underline{s}=\sum_{j=1}^{n}f_{j}s_{j}$ and
$\underline{t}=\sum_{j=1}^{n}f_{j}t_{j}$ denote the Hermitian Clifford vectors
corresponding to the complex vectors $s,t\in\mathbb{C}^{n}$.

######  Theorem 19.

[35] The reproducing kernel of the space $\mathcal{M}_{p,q}^{(j)}$, which
consists of $\mathbb{S}^{(j)}$-valued $h$-monogenics of degree $(p,q)$, can be
expressed as

$\begin{split}\mathcal{K}_{p,q}(z,u)=&\,d_{p,q}(\beta)\int_{St(2)}\langle
z,\overline{t+s}\rangle^{p}\langle\overline{z},t-s\rangle^{q}\langle
u,\overline{t-s}\rangle^{q}\langle\overline{u},t+s\rangle^{p}\\\
&\times\frac{\widetilde{\lambda}_{p,q}\,{\rm
dim}\mathcal{H}_{p+1,q+1}}{\omega_{2n-1}\omega_{2n-3}}\underline{\tau}\underline{\tau}^{\dagger}\,{\rm
d}\sigma(s)\,{\rm d}\sigma(t),\end{split}$

where
$\widetilde{\lambda}_{p,q}=\frac{(p+1)^{2}(q+1)^{2}(k+n+1)!}{2^{k+2}(n-1)!(k-\nu)!}$,
$k=p+q$, $\nu=\min(p,q)$, $\dim\mathcal{H}_{p,q}$ as in (5) and
$d_{p,q}(\beta)$ satisfying

$(n+p+q+1)^{2}(n-\beta+q)(\beta+p)d_{p,q}(\beta)=1$

with $\beta$ the Clifford-vector given in (42).

Theorem 19 yields the following explicit results for the dual transform. The
proof is again omitted.

######  Theorem 20.

For any $f\in\mathcal{M}_{h}^{(j)}\mathcal{L}(\mathbb{B}^{n})$, its dual
transform of the Szegö-Radon transform is given by

$\widetilde{\mathcal{R}}_{\underline{\tau}}[\mathcal{R}_{\underline{\tau}}[f]](z)=\sum_{p,q}\frac{(n+p+q+1)^{2}(n-\beta+q)(\beta+p)}{4\gamma_{p,q}\widetilde{\lambda}_{p,q}\dim\mathcal{H}_{p+1,q+1}}f_{p,q}^{(j)}(z),$
(50)

where $\beta$, $\widetilde{\lambda}_{p,q}$ and $\dim\mathcal{H}_{p,q}$ as in
Theorem 19, and $\gamma_{p,q}=2^{p+q}p!q!/(n)_{p+q}$.

By the property $\beta f^{(j)}=jf^{(j)}$ for any $f^{(j)}\in\mathbb{S}^{(j)}$,
we can give the inversion formula as in the scalar cases. The projections
$\mathcal{R}_{\underline{\tau}}^{(1)}$ and
$\mathcal{R}_{\underline{\tau}}^{(2)}$ are defined similarly for $p\geq q$ and
$p<q$ respectively as in Definition 6.

######  Theorem 21.

For any $f\in\mathcal{M}_{h}^{(j)}\mathcal{L}(\mathbb{B}^{n})$ with $1\leq
j<n$, it holds that for any $|z|<1/2$,

$f(z,\bar{z})=\mathcal{T}^{(1)}\circ\widetilde{\mathcal{R}}_{\underline{\tau}}\left[\mathcal{R}_{\underline{\tau}}^{(1)}[f(w,\bar{w})]\right]+\mathcal{T}^{(2)}\circ\widetilde{\mathcal{R}}_{\underline{\tau}}\left[\mathcal{R}_{\underline{\tau}}^{(2)}[f(w,\bar{w})]\right],$

where

$\begin{split}\mathcal{T}^{(1)}[f](z,\bar{z}):=&\,\frac{1}{\Gamma(n)\Gamma(n-1)}(\mathbb{E}+n)(\mathbb{E}_{\bar{z}}+1)\underbrace{(\mathbb{E}_{\bar{z}}+n-j+1)\cdots(\mathbb{E}_{\bar{z}}+n+1)}_{\mathcal{J}_{1}}\\\
&\times(\mathbb{E}_{z}+1)\underbrace{(\mathbb{E}_{z}+1)\cdots(\mathbb{E}_{z}+j-1)(\mathbb{E}_{z}+j+1)\cdots(\mathbb{E}_{z}+n-1)}_{\mathcal{J}_{2}}\\\
&\times\int_{0}^{\infty}f(z,\bar{z}t)t^{n-j-1}e^{-t}\,{\rm d}t,\end{split}$

and

$\begin{split}\mathcal{T}^{(2)}[f](z,\bar{z}):=&\,\frac{1}{\Gamma(n)\Gamma(n-1)}(\mathbb{E}+n)(\mathbb{E}_{z}+1)\underbrace{(\mathbb{E}_{z}+j+1)\cdots(\mathbb{E}_{z}+n-1)}_{\mathcal{J}_{3}}\\\
\times&(\mathbb{E}_{\bar{z}}+1)\underbrace{(\mathbb{E}_{\bar{z}}+1)\cdots(\mathbb{E}_{\bar{z}}+n-j-1)(\mathbb{E}_{\bar{z}}+n-j+1)\cdots(\mathbb{E}_{\bar{z}}+n-1)}_{\mathcal{J}_{4}}\\\
\times&\int_{0}^{\infty}f(zt,\bar{z})t^{j-1}e^{-t}\,{\rm d}t.\end{split}$

######  Remark 5.

For any $P_{p,q}\in\mathcal{P}_{p,q}$, it holds that

$\begin{split}&\mathcal{J}_{1}P_{p,q}(z)=(q+n-j+1)_{j+1}P_{p,q}(z),\qquad\mathcal{J}_{2}P_{p,q}(z)=\frac{(p+1)_{n-1}}{p+j}P_{p,q}(z),\\\
&\mathcal{J}_{3}P_{p,q}(z)=(p+j+1)_{n-j-1}P_{p,q}(z),\qquad\mathcal{J}_{4}P_{p,q}(z)=\frac{(q+1)_{n-1}}{n-j}P_{p,q}(z).\end{split}$

######  Remark 6.

When $j=0$ or $n$, the inversion formula can be obtained similarly.

##  6 Conclusions

In this paper, we have introduced Radon-types transforms for the holomorphic,
harmonic and Hermitian monogenic functions on the unit ball. Additionally, the
dual transform and explicit inversion formulas are provided. The transform has
also been extended to the Fock space, resulting in a new transform on
$L^{2}(\mathbb{R}^{n})$. The latter reveals the connections and differences
with the standard Radon transform.

It is of great interests for us to establish an alternative dual transform
based only on integration over $\mathbb{S}^{2n-3}$, in contrast to the present
approach relying on integration on the complex Stiefel manifold
$St^{(2)}(\mathbb{C}^{n})$. This question is, in certain sense, equivalent to
deriving a new plane wave type integral expression for the reproducing kernel
of the complex harmonics, which is given in terms of Jacobi polynomials. This
shall be obtained through the well-known Pizzetti’s formula. However, the
computations involved are unexpectedly much more complicated than the current
approach. We will address it in a separate work.

## Acknowledgment

We thank Paula Cerejeiras, Denis Constales and Uwe Kähler for interesting
discussions on this topic. The second author was supported by NSFC Grant
No.12101451.

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