# Local data inverse problem for the polyharmonic operator with anisotropic
perturbations

Sombuddha Bhattacharyya* and Pranav Kumar  *Department of Mathematics, Indian
Institute of Science Education and Research, Bhopal.
E-mail<EMAIL_ADDRESS>**Department of Mathematics, Indian Institute
of Science Education and Research, Bhopal.
E-mail<EMAIL_ADDRESS>
###### Abstract.

In this article, we study an inverse problem with local data for a linear
polyharmonic operator with several lower order tensorial perturbations. We
consider our domain to have an inaccessible portion of the boundary where
neither the input can be prescribed nor the output can be measured. We prove
the unique determination of all the tensorial coefficients of the operator
from the knowledge of the Dirichlet and Neumann map on the accessible part of
the boundary, under suitable geometric assumptions on the domain.

###### Key words and phrases:

Calderón problem, polyharmonic operator, Anisotropic perturbation, Tensor
tomography, Momentum ray transform.

###### 2020 Mathematics Subject Classification:

Primary 35R30, 31B20, 31B30, 35J40.

##  1\. Introduction

Let $\Omega$ be an open, bounded domain in $\mathbb{R}^{n}$ with smooth
boundary $\partial\Omega$, where $n\geq 3$. We consider the perturbed
polyharmonic operator $\mathcal{L}(x,D)$ of order $2m$ with perturbations up
to order $m$ of the form:

(1.1)
$\mathcal{L}(x,D):=(-\Delta)^{m}+\sum_{j=0}^{m-1}\sum_{i_{1},\cdots,i_{m-j}=1}^{n}A_{i_{1}\cdots
i_{m-j}}^{m-j}(x)D_{i_{1}\cdots i_{m-j}}^{m-j}+q(x);\quad m\geq 2,$

where $D_{i_{1}\dots
i_{j}}^{j}=\frac{1}{\mathrm{i}^{j}}\frac{\partial^{j}}{\partial
x_{i_{1}\dots}\partial x_{i_{j}}}$, $i_{1},\dots,i_{j}\in\\{1,2,\dots,n\\}$,
$1\leq j\leq m$, and the perturbations coefficients $A_{i_{1}\cdots
i_{j}}^{j}=A^{j}\in C_{c}^{\infty}(\Omega,\mathbb{C}^{n^{j}})$ symmetric
tensor fields of order $j$ for $1\leq j\leq m$ and $q(x)\in
L^{\infty}(\Omega,\mathbb{C})$. In this article, we solve a partial data
inverse problem related to the operator (1.1). We prove the unique recovery of
all the lower order perturbations $A^{j}\text{ for }1\leq j\leq m$, and $q$ in
$\Omega$ from the boundary data measured only on a part $\Gamma$ of the
boundary $\partial\Omega$ while the remaining part
$\partial\Omega\setminus\Gamma$ is part of a hyperplane.

Inverse problems for recovering unknown coefficients of PDEs were pioneered by
Alberto Calderón in his work [Cal80] on electrical impedance tomography. The
groundbreaking work by Sylvester and Uhlmann in [SU87] on the inverse problem
for the conductivity equation inspired subsequent developments. For a
comprehensive survey of Calderón-type inverse problems, see [Uhl09]. The study
of coefficient recovery problems for biharmonic and polyharmonic operators was
initiated by Krupchyk, Lassas, and Uhlmann in [KLU12b, KLU14]. It was soon
followed by [Yan14, GK16, BG19, BKS23, SS23], where coefficient recovery for
biharmonic and polyharmonic operators in a bounded domain with dimensions
$n\geq 3$ have been discussed. The higher order elliptic operators, as
represented by (1.1), have received significant attention in various areas,
such as physics and geometry, due to their connection with the theory of
elasticity equation on thin domains given by the Kirchhoff plate equation
(perturbed biharmonic operator) and the study of the Paneitz-Branson equation
in conformal geometry. For more applications of biharmonic and polyharmonic
operators, see [Sel00, GGS10].

To describe our setup, let us define the following notations. Define
$\mathcal{D}(\mathcal{L}(x,D)),$ where,

$\mathcal{D}(\mathcal{L}(x,D)):=\Big{\\{}u\in
H^{2m}(\Omega):u|_{\partial\Omega}=(-\Delta)u|_{\partial\Omega}=\cdots=(-\Delta)^{m-1}u|_{\partial\Omega}=0\Big{\\}.}$

The operator $\mathcal{L}(x,D)$ with the domain
$\mathcal{D}(\mathcal{L}(x,D))$ is an unbounded closed operator on
$L^{2}(\Omega)$ with discrete spectrum [Gru09].

######  Assumption 1.

We assume that $0$ is not an eigenvalue of the operator
$\mathcal{L}(x,D):\mathcal{D}(\mathcal{L}(x,D))\rightarrow L^{2}(\Omega)$.

Under Assumption 1, the following boundary value problem with Navier boundary
conditions:

(1.2) $\displaystyle\mathcal{L}(x,D)u$ $\displaystyle=0\quad\text{ in
}\Omega,$
$\displaystyle\left(u|_{\partial\Omega},(-\Delta)u|_{\partial\Omega},\cdots,(-\Delta)^{m-1}u|_{\partial\Omega}\right)$
$\displaystyle=\left(f_{0},f_{1},\cdots,f_{m-1}\right),$

has a unique solution $u\in H^{2m}(\Omega)$ for any
$f:=\left(f_{0},f_{1},\cdots,f_{m-1}\right)\in\prod_{i=0}^{m-1}H^{2m-2i-\frac{1}{2}}(\partial\Omega)$.
We state the existence and uniqueness of a solution to the problem (1.2).

######  Theorem 1 ([Tan17] Lemma 5.13).

Let the coefficients $A^{j}$ for $1\leq j\leq m$ belong to
$C_{c}^{\infty}(\Omega,\mathbb{C}^{n^{j}})$ and $q(x)\in
L^{\infty}(\Omega,\mathbb{C})$ in (1.1). Suppose Assumption 1 and
$f=\left(f_{0},f_{1},\ldots,f_{m-1}\right)\in\prod_{i=0}^{m-1}H^{2m-2i-\frac{1}{2}}(\partial\Omega)$.
Then there exists a unique solution $u\in H^{2m}(\Omega)$ to the problem (1.2)
satisfying

$\|u\|_{H^{2m}(\Omega)}\leq
C\|f\|_{\prod_{i=0}^{m-1}H^{2m-2i-\frac{1}{2}}(\partial\Omega)},$

where $C>0$ is a constant.

We define the Dirichlet-to-Neumann (D-N) map by

$\mathcal{N}:\prod_{i=0}^{m-1}H^{2m-2i-\frac{1}{2}}(\partial\Omega)\rightarrow\prod_{i=0}^{m-1}H^{2m-2i-\frac{3}{2}}(\partial\Omega),$
$\mathcal{N}(f)=\left(\partial_{\nu}u_{f}|_{\partial\Omega},\cdots,\partial_{\nu}(-\Delta)^{k}u_{f}|_{\partial\Omega},\cdots,\partial_{\nu}(-\Delta)^{m-1}u_{f}|_{\partial\Omega}\right),\quad
f=(f_{0},\cdots,f_{m-1});$

where $u_{f}\in H^{2m}(\Omega)$ is the unique solution to the problem (1.2)
corresponding to $f\in\prod_{i=0}^{m-1}H^{2m-2i-\frac{1}{2}}(\partial\Omega)$
and $\nu$ is the outward unit normal vector to the boundary $\partial\Omega$.
Let $\Gamma$ be a non-empty open subset of the boundary $\partial\Omega$, we
define the space

$\mathcal{H}_{\Gamma}:=\\{f\in\prod_{i=0}^{m-1}H^{2m-2i-\frac{1}{2}}(\partial\Omega):\operatorname{supp}{f}\subset\Gamma\\}.$

We define the partial D-N map as

(1.3)
$\displaystyle\mathcal{N}_{\Gamma}:\mathcal{H}_{\Gamma}\rightarrow\prod_{i=0}^{m-1}H^{2m-2i-\frac{3}{2}}(\Gamma),$
$\mbox{as
}\quad\mathcal{N}_{\Gamma}(f):=\left(\partial_{\nu}u_{f}|_{\Gamma},\partial_{\nu}(-\Delta)u_{f}|_{\Gamma},\cdots,\partial_{\nu}(-\Delta)^{m-1}u_{f}|_{\Gamma}\right).$

We prove unique recovery of the coefficients of (1.1) from the partial D-N map
$\mathcal{N}_{\Gamma}$, assuming that the inaccessible part of the boundary
$\partial\Omega\setminus\Gamma$ is a part of a hyperplane.

###  1.1. Statement of the main result

Throughout the article, we follow the standard tensor notations used in
[Sha94]. Given an integer $m\geq 0$, we write $T^{m}=T^{m}(\mathbb{R}^{n})$ to
denote the complex vector space of $\mathbb{R}$-multilinear functions from
$\underbrace{\mathbb{R}^{n}\times\cdots\times\mathbb{R}^{n}}_{\text{m-times}}$
to $\mathbb{C}$. Let $e_{1},\cdots,e_{n}$ be a basis for $\mathbb{R}^{n}$. We
write the components of the tensor $f\in T^{m}$ as $f_{i_{1}\cdots
i_{m}}:=f(e_{i_{1}},\cdots,e_{i_{m}})$. The notation
$S^{m}=S^{m}(\mathbb{R}^{n})$ denotes the subspace of $T^{m}$ that consists of
functions symmetric in all their arguments over the real vector space
$\mathbb{R}^{n}$.

For $f\in T^{m}$, $g\in T^{k}$, the tensor product $f\otimes g\in T^{m+k}$ is
defined by

$f\otimes
g(x_{1},\cdots,x_{m+k})=f(x_{1},\cdots,x_{m})g(x_{m+1},\cdots,x_{m+k}).$

Let $\sigma:T^{m}\to S^{m}$ be the symmetrization defined by

$\sigma
f(x_{1},\cdots,x_{m})=\frac{1}{m!}\sum_{\pi\in\Pi_{m}}f(x_{\pi(1)},\cdots,x_{\pi(m)}),$

where $\Pi_{m}$ is the set of all permutations of the set $\\{1,\cdots,m\\}$.

Given $f\in T^{m},g\in T^{k}$, the symmetrized tensor product $u\odot v$
defined by

$\displaystyle(f\odot g)(x_{1},\cdots x_{m+k}):=$
$\displaystyle\sigma(f\otimes g)(x_{1},\cdots x_{m+k})$ $\displaystyle=$
$\displaystyle\frac{1}{(m+k)!}\sum_{\pi\in\Pi_{m+k}}f(x_{\pi(1)},\cdots,x_{\pi(m)})g(x_{i_{\pi(m+1)}},\cdots,x_{i_{\pi(m+k)}}).$

We define the operator $i_{\delta}:S^{m}\to S^{m+2}$ as follows:

$(i_{\delta}f)_{i_{1}\cdots i_{m+2}}:=(\delta\odot f)_{i_{1}\cdots
i_{m+2}}=\sigma(f_{i_{1}\cdots i_{m}}\otimes\delta_{i_{m+1}i_{m+2}}).$

######  Definition 1.1.

We say a symmetric tensor $A^{j}\in S^{j}(\mathbb{R}^{n})$, $j\geq 2$, is
partial isotropic if there exists a symmetric $(j-2)$-tensor field
$\overline{A}^{j-2}\in S^{j-2}(\mathbb{R}^{n})$ in $\Omega$ such that
$A^{j}=i_{\delta}\overline{A}^{j-2}=\overline{A}^{j-2}\odot\delta$, where
$\delta$ is the Kronecker delta tensor and $\odot$ denotes symmetrized product
of tensors.

As an example, any $n$ dimensional partially isotropic $2$-tensors (matrix)
are of the form $a(x)I_{n}$ where $a(x)$ is a function ($0$-tensor) and
$I_{n}$ is the $n\times n$ identity matrix. We assume the leading order
perturbation $A^{m}$ of the operator (1.1) is to be partially isotropic, which
means $A^{m}=i_{\delta}\overline{A}^{m-2}$ for some $m-2$ order symmetric
tensor field $\overline{A}^{m-2}$. We now state the main result of this
article.

######  Theorem 2.

Let $\Omega\subset\left\\{\mathbb{R}^{n};x_{n}>0\right\\}$ be an open bounded
domain with smooth boundary, where $n\geq 3$. Let
$\Gamma_{0}=\partial\Omega\cap\left\\{x_{n}=0\right\\}$ be non-empty, open and
let $\Gamma=\partial\Omega\setminus\Gamma_{0}$. Let
$A^{j},\widetilde{A}^{j}\in C_{c}^{\infty}(\Omega)\text{ for }1\leq j\leq m$
and $q,\widetilde{q}\in L^{\infty}(\Omega)$. We consider a new operator
$\widetilde{\mathcal{L}}(x,D)$ by replacing $A^{j}$ by $\tilde{A}^{j}$ and $q$
by $\tilde{q}$ in (1.1). Let $\mathcal{N}_{\Gamma}$ and
$\widetilde{\mathcal{N}}_{\Gamma}$ be the partial D-N map corresponding to the
operators $\mathcal{L}(x,D)$ and $\widetilde{\mathcal{L}}(x,D)$ respectively.
Let the coefficients be such that Assumption 1 is satisfied for
$\mathcal{L}(x,D)$ and $\widetilde{\mathcal{L}}(x,D)$. We further assume that
for $j=m$, $A^{m},\widetilde{A}^{m}$ are partially isotropic. If

(1.4) $\mathcal{N}_{\Gamma}(f)=\widetilde{\mathcal{N}_{\Gamma}}(f),\quad\text{
for all }f\in\prod_{i=0}^{m-1}H^{2m-2i-\frac{1}{2}}(\partial\Omega)\text{ with
}\operatorname{supp}(f)\subset\Gamma,$

then for $1\leq j\leq m$, we have

$A^{j}=\widetilde{A}^{j}\quad\text{ and }\quad q=\widetilde{q}\quad\text{ on
}\overline{\Omega}.$

In the last few decades, partial data inverse problems, in which the boundary
data is available only on a part of the boundary, has attracted considerable
attention, see; [KS14]. For dimensions $n\geq 3$, Bukhgeim and Uhlmann in
[BU02] established the first uniqueness result for the Schrödinger operator
$(-\Delta+q)$ in a bounded domain for bounded potentials, assuming that the
Neumann data measured on slightly more than the half of the boundary. In
[KSU07], Kenig, Sjöstrand, and Uhlmann proved uniqueness for the Schrödinger
operator under the assumption that the Dirichlet-to-Neumann map measured on an
open subset of the boundary $\partial\Omega$ where the Dirichlet data is
supported on a complementary part of the boundary. In [FKSU07, Chu14, Bha18],
authors investigated partial data inverse problems for the magnetic
Schrödinger operator $(D+A)^{2}+q$ in a bounded simply-connected domain for
some vector field $A$ and potential $q$. In [KLU12b, KLU14] Krupchyk Lassas
and Uhlmann introduced Caldeón type inverse problems for biharmonic and
polyharmonic operators. They were soon followed by [Yan14, GK16, BG19, BG22,
BKS23, SS23], etc., where the authors have investigated uniqueness for inverse
boundary value problems for biharmonic and polyharmonic operators from full
and partial boundary data. So far, the question of recovering coefficients of
a higher order elliptic operator from partial boundary data where a part of
the boundary is inaccessible has been open in the literature. In [Yan14],
authors considered domains with inaccessible part of the boundary probed with
a biharmonic operator with just first and a zeroth order perturbation. On the
other hand, in [BKS23], the authors have considered polyharmonic operators
with several lower order tensorial perturbations while the entire boundary is
accessible. In this article, we generalise the existing results by considering
several tensorial perturbations as in (1.1) and an inaccessible part of the
boundary $\Gamma_{0}=\partial\Omega\setminus\Gamma$, assuming that
$\Gamma_{0}$ is part of a hyperplane.

In dimension $n=3$, Isakov [Isa07] introduced the reflection approach for the
inverse boundary value problems for the Schrödinger equation and the
conductivity equation for dealing with inverse problems in domains with
inaccessible part of the boundary. Kenig and Salo [KS14] unified and improved
the approaches of [Isa07] and [KSU07] for the Schrödinger operator. In
[KLU12a], the authors considered the magnetic Schrödinger operator
$(-\Delta)+A\cdot D+q$ and proved that the magnetic field $dA$ and potential
$q$ can be uniquely recovered from the partial Dirichlet-to-Neumann map. In
[Yan14], the author proved recovery of first and zeroth order perturbations
for inverse boundary value problems for the biharmonic operator on a domain
from the partial Dirichlet-to-Neumann map. Some related works can also be
found in [COS09, Car11, LU10, LY21]. To the best of the author’s knowledge,
this is the first injectivity result for polyharmonic operators with higher
order tensorial perturbations, in a domain having an inaccessible part of the
boundary.

The novelty of this article lies in recovering tensorial perturbations having
a nontrivial inaccessible part of the boundary. We assume the inaccessible
part of the boundary is part of a hyperplane. Here we briefly sketch the idea
of the proof. In order to deal with the inaccessible part of the boundary, we
first extend the domain and the operator (1.1) by considering a suitable
reflection about the hyperplane where the inaccessible part is supported. For
$m=1,2$, similar types of reflections have been considered in [Isa07, Yan14,
PM17]. In the extended domain, we construct a class of Complex Geometric
Optics type solutions using appropriate Carleman estimates. The reflection and
the construction of solutions are done in such a way that the solution of the
extended operator satisfies the original boundary value problem in $\Omega$
upon folding the domain about the inaccessible part of the boundary. Thus we
obtain a class of solutions for the boundary value problem (1.2). Using these
solutions, a careful analysis entails having a class of integral equations of
the lower order tensorial perturbations. Upon further analysis of these
solutions, we reduce our problem to an uniqueness question for a series of
Momentum Ray Transforms (MRT) of the unknown tensorial perturbations. Recently
in [BKS23, SS23] MRT has been proved to be useful for recovering higher order
tensors from integral equations. We use uniqueness results of MRT from [BKS23,
SS23] to finally obtain uniqueness of the tensorial perturbations and thus
complete the proof of Theorem 2.

This paper is organized as follows. In Section 2, we provide the necessary
preliminary results for proving our main results. These include interior
Carleman estimates, constructions of Complex Geometric Optics (CGO) solutions,
and momentum ray transforms. In Section 3, we prove our main theorem by
constructing special CGO solutions and deriving appropriate integral
identities to recover the coefficients.

##  2\. Preliminary results

In this section, we start by recollecting the construction of the Complex
Geometric Optics (CGO) solutions for the operator (1.1) in a bounded domain.
We use Carleman estimates in the construction. We refer to [KLU12b, KLU14,
GK16, BKS23, SS23] for a detailed description of the construction.

###  2.1. Interior Carleman estimates

Following [KSU07], we start by recalling the definition of limiting Carleman
weights for the semiclassical Laplacian $(-h^{2}\Delta)$, where $h>0$ is a
small parameter. Let $\widetilde{\Omega}$ be an open set in $\mathbb{R}^{n}$
such that $\Omega\subset\subset\widetilde{\Omega}$ and $\varphi\in
C^{\infty}(\widetilde{\Omega},\mathbb{R})$ with $\nabla\varphi\neq 0$ in
$\overline{\Omega}$. We consider the conjugated semiclassical Laplacian
operator

(2.1)
$P_{0,\varphi}:=e^{\frac{\varphi}{h}}\left(-h^{2}\Delta\right)e^{-\frac{\varphi}{h}},\quad
0<h\ll 1,$

and its semiclassical principal symbol
${p}_{0,\varphi}(x,\xi)=|\xi|^{2}+2\mathrm{i}\nabla\varphi\cdot\xi-|\nabla\varphi|^{2}$.

######  Definition 2.1 ([KSU07]).

We say that $\varphi\in C^{\infty}(\widetilde{\Omega})$ is a limiting Carleman
weight for $P_{0,\varphi}$ in $\Omega$, if $\nabla\varphi\neq 0$ in
$\overline{\Omega}$ and
$\operatorname{Re}\left(p_{0,\varphi}\right),\operatorname{Im}\left(p_{0,\varphi}\right)$
satisfies

$\Big{\\{}\mathrm{Re}(p_{0,\varphi}),\mathrm{Im}(p_{0,\varphi})\Big{\\}}(x,\xi)=0\mbox{
whenever }p_{0,\varphi}(x,\xi)=0\mbox{ for
}(x,\xi)\in\widetilde{\Omega}\times(\mathbb{R}^{n}\setminus\\{0\\}),$

where $\\{\cdot,\cdot\\}$ denotes the Poisson bracket.

Examples of such $\varphi$ are linear weights $\varphi(x)=\mu_{1}\cdot x$,
where $0\neq\mu_{1}\in\mathbb{R}^{n}$ or logarithmic weights
$\varphi(x)=\log\left|x-x_{0}\right|$ with
$x_{0}\notin\overline{\widetilde{\Omega}}.$

We shall first introduce semiclassical Sobolev spaces. For any $h>0$, be a
small parameter and an open set $\Omega\subset\mathbb{R}^{n}$ and for any non-
negative integers $m$, the semiclassical Sobolev space
$H^{m}_{\mathrm{scl}}(\Omega)$ is the space $H^{m}(\Omega)$ endowed with the
following semiclassical norm

$\lVert u\rVert^{2}_{H^{m}_{\mathrm{scl}}(\Omega)}=\sum\limits_{|\alpha|\leq
m}\lVert(hD)^{\alpha}\,u\rVert^{2}_{L^{2}(\Omega)}.$

For $\Omega=\mathbb{R}^{n}$ we can equivalently define the semiclassical
Sobolev spaces $H_{scl}^{s}(\mathbb{R}^{n})$, $s\in\mathbb{R}$ as the space
$H^{s}(\mathbb{R}^{n})$ endowed with the semiclassical norm

$\|u\|_{H^{s}_{\mathrm{scl}}\left(\mathbb{R}^{n}\right)}^{2}=\left\|\langle
hD\rangle^{s}u\right\|_{L^{2}\left(\mathbb{R}^{n}\right)}^{2}=\int_{\mathbb{R}^{n}}(1+h^{2}|\xi|^{2})^{s}|\hat{u}(\xi)|^{2}\mathrm{d}\xi,$

here $\langle\xi\rangle=(1+|\xi|^{2})^{\frac{1}{2}}$. For a detailed
description of semiclassical analysis and Poisson bracket present here, see
[Mar02, Zwo12]. We recall the following Carleman estimate from [BKS23].

######  Proposition 2.1.

Let the coefficients $A^{j}$ belong to
$C_{c}^{\infty}(\Omega;\mathbb{C}^{n^{j}})$ for $1\leq j\leq m$ and $q\in
L^{\infty}(\Omega;\mathbb{C})$. Let $\varphi(x)$ be a limiting Carleman weight
for the conjugated semiclassical Laplacian. Then for $0<h\ll 1$ and $-2m\leq
s\leq 0$, we have

$h^{m}\|u\|_{H^{s+2m}_{\mathrm{scl}}(\Omega)}\leq
C\left\|h^{2m}e^{\frac{\varphi}{h}}\mathcal{L}(x,D)e^{-\frac{\varphi}{h}}u\right\|_{H^{s}_{\mathrm{scl}}(\Omega)},\text{
for all }u\in C^{\infty}_{c}(\Omega),$

where the constant $C=C_{s,\Omega,A^{j},q}$ is independent of $h$.

Let us denote

$\mathcal{L}_{\varphi}(x,D)=h^{2m}e^{\frac{\varphi}{h}}\mathcal{L}(x,D)e^{-\frac{\varphi}{h}}.$

Let $\mathcal{L}_{\varphi}^{*}(x,D)$ be the formal $L^{2}$ adjoint of
$\mathcal{L}_{\varphi}(x,D)$ given by
$(\mathcal{L}_{\varphi}u,v)_{L^{2}(\Omega)}=(u,\mathcal{L}_{\varphi}^{*}v)_{L^{2}(\Omega)}$
for all $u,v\in C^{\infty}_{c}(\Omega)$. It follows from a straightforward
computation using integration by parts that $\mathcal{L}^{*}(x,D)$ has a
similar form as $\mathcal{L}(x,D)$ with the same regularity of the
coefficients. Since $-\varphi$ is also a limiting Carleman weight if $\varphi$
is, the Carleman estimate derived in Proposition 2.1 holds true for
$\mathcal{L}_{\varphi}^{*}(x,D)$ as well. We convert the Carleman estimate for
$\mathcal{L}_{\varphi}^{*}(x,D)$ into a solvability result for
$\mathcal{L}_{\varphi}(x,D)$.

######  Proposition 2.2.

Let the coefficients $A^{j}$ for $1\leq j\leq m$ belong to
$C_{c}^{\infty}(\Omega,\mathbb{C}^{n^{j}})$ and $q\in
L^{\infty}(\Omega,\mathbb{C})$, and let $\varphi(x)$ be a limiting Carleman
weight the conjugated semiclassical Laplacian. Then for any $v\in
L^{2}(\Omega)$ and small enough $h>0$ there exists $u\in H^{2m}(\Omega)$ such
that

(2.2) $\mathcal{L}_{\varphi}(x,D)u=v\text{ in }\Omega,$

satisfying,

$h^{m}\|u\|_{H_{\mathrm{scl}}^{2m}}\leq C\|v\|_{L^{2}(\Omega)},$

where $C>0$ is independent of $h$ and depends only on $A^{j},\text{ for }1\leq
j\leq m,$ and $q$.

The proof of the above Proposition follows from Hahn–Banach extension theorem
and Riesz representation theorem (see [FKSU07, KLU12b, KLU14]).

###  2.2. Construction of Complex Geometric Optics solutions

We construct complex geometric optics solutions of the equation
$\mathcal{L}(x,D)u=0$ in $\Omega$, based on Proposition 2.2. Following [GK16,
BKS23], we propose the solution to be of the form:

(2.3)
$u(x)=u(x;h)=e^{\frac{\varphi+\mathrm{i}\psi}{h}}\left(a_{0}(x)+ha_{1}(x)+h^{2}a_{2}(x)\cdots+h^{m-1}a_{m-1}(x)+r(x;h)\right):=e^{\frac{\varphi+\mathrm{i}\psi}{h}}(\mathrm{a}(x;h)),$

where $h>0$ is a small parameter, $\varphi(x)$ is a limiting Carleman weight
for the semiclassical Laplacian, the real-valued phase function $\psi$ is
chosen such that $\psi$ is smooth near $\overline{\Omega}$ and solves the
Eikonal equation $p_{0,\varphi}(x,\nabla\psi)=0$ in $\widetilde{\Omega}$. The
functions $\\{a_{j}(x)\\}$, for $0\leq j\leq m-1$, are the complex amplitudes
solving certain transport equations and $r(x;h)$ is the correction term which
is determined later. We consider $\varphi$ and $\psi$ of the form

(2.4) $\varphi(x)=\mu^{(2)}\cdot x\quad\text{ and
}\quad\psi(x)=\left(\frac{h\xi}{2}+\sqrt{1-h^{2}\frac{|\xi|^{2}}{4}}\mu^{(1)}\right)\cdot
x,$

where $\xi,\mu^{(1)},\mu^{(2)}\in\mathbb{R}^{n}$ be such that
$|\mu^{(1)}|=|\mu^{(2)}|=1$ and
$\mu^{(1)}\cdot\mu^{(2)}=\mu^{(1)}\cdot\xi=\mu^{(2)}\cdot\xi=0$. One can
easily check that $\varphi$ and $\psi$ solves the Eikonal equation
$p_{0,\varphi}(x,\nabla\psi)$ in $\widetilde{\Omega}$, that is
$|\nabla\varphi|=|\nabla\psi|$ and $\nabla\varphi\cdot\nabla\psi=0$. We prove
that there exist $h_{0}>0$ such that $u(x)$ is a solution of (1.1) for each
fixed $h\in(0,h_{0})$. We now prove the following lemma.

######  Lemma 2.1.

Consider the equation

(2.5)
$\mathcal{L}(x,D)u=\Big{\\{}(-\Delta)^{m}+\sum_{j=0}^{m-1}\sum_{i_{1},\cdots,i_{m-j}=1}^{n}A_{i_{1}\cdots
i_{m-j}}^{m-j}(x)D_{i_{1}\cdots i_{m-j}}^{m-j}+q(x)\Big{\\}}u=0,$

where $A^{j}\in C_{c}^{\infty}(\Omega,\mathbb{C}^{n^{j}})\text{ for }1\leq
j\leq m$ and $q(x)\in L^{\infty}(\Omega,\mathbb{C})$. We assume that for
$j=m,$ $A^{j}$ is partially isotropic. Then for $h>0$ small enough, there
exists a solution $u\in H^{2m}(\Omega)$ of (2.5) of the form

(2.6) $\displaystyle u(x;h)=$ $\displaystyle
e^{\frac{\varphi+\mathrm{i}\psi}{h}}\left(a_{0}(x)+ha_{1}(x)+h^{2}a_{2}(x)\dots+h^{m-1}a_{m-1}(x)+r(x;h)\right):=e^{\frac{\varphi+\mathrm{i}\psi}{h}}(\mathrm{a}(x;h)),$

where $\varphi$ and $\psi$ are defined as in equation (2.4), and the functions
$a_{0},\cdots,a_{m-1}$ satisfy the transport equations (2.8) and (2.9), and
$r$ satisfies the estimate

$\|r(x;h)\|_{H_{\mathrm{scl}}^{2m}(\Omega)}\leq Ch^{m},$

where $C$ is independent of $h$.

###### Proof.

Consider the conjugated operator

$\displaystyle
e^{-\frac{\varphi+i\psi}{h}}\mathcal{L}(x,D)e^{\frac{\varphi+i\psi}{h}}\mathcal{A}(x;h)=$
$\displaystyle\left(-\frac{1}{h}T-\Delta\right)^{m}\mathcal{A}(x;h)-\sum_{i=1}^{n}\overline{A}_{i_{1}\ldots
i_{m-2}}^{m-2}\mathcal{I}_{i_{1}\ldots
i_{m-2}}^{m-2}\left(\frac{1}{h}T+\Delta\right)\mathcal{A}(x;h)$ (2.7)
$\displaystyle+\sum_{j=1}^{m-1}\sum_{i_{1},\cdots,i_{j}=1}^{n}A_{i_{1}\ldots
i_{j}}^{j}\mathcal{I}_{i_{1}\ldots
i_{j}}^{j}\mathcal{A}(x;h)+q(x)\mathcal{A}(x;h),$

where

$\mathcal{I}_{i_{1}\ldots
i_{j}}^{j}=e^{-\frac{\varphi+\mathrm{i}\psi}{h}}D_{i_{1}\ldots
i_{j}}^{j}e^{\frac{\varphi+\mathrm{i}\psi}{h}}=\prod_{k=1}^{j}\left(\frac{1}{h}D_{i_{k}}(\varphi+\mathrm{i}\psi)+D_{i_{k}}\right),$
$T=2\nabla(\varphi+\mathrm{i}\psi)\cdot\nabla+\Delta(\varphi+\mathrm{i}\psi)\quad\text{
and }\quad\mathcal{A}(x;h):=\sum_{j=0}^{m-1}h^{j}a_{j}(x).$

Now substituting (2.3) in (2.5), we get the following transport equations for
the amplitudes $a_{j}(x)$ with $0\leq j\leq m-1$ by equating the coefficient
of $h^{-m+j}$ to zero for $0\leq j\leq m-1$ as

(2.8) $\displaystyle T^{m}a_{0}(x)$ $\displaystyle=0,\quad\text{ in
}\quad\Omega.$ (2.9) $\displaystyle
T^{m}a_{j}(x)=-\sum_{k=1}^{j}M_{k}a_{j-k}(x),\quad\text{ in }\quad\Omega,\quad
1\leq j\leq m-1.$

From (2.2), we can calculate the differential operators $M_{j}$ of order $m+j$
where $1\leq j\leq m-1$. It is well-known that the solutions of (2.9) are
smooth; see [FKSU07]. We discuss an explicit form for the smooth solution
$a_{0}(x)$ solving (2.8) in the next section. Using
$a_{0}(x),\cdots,a_{m-1}(x)\in C^{\infty}(\Omega)$ satisfying (2.9), (2.8), we
see that

$e^{-\frac{\varphi+\mathrm{i}\psi}{h}}\mathcal{L}(x,D)e^{\frac{\varphi+\mathrm{i}\psi}{h}}\mathcal{A}(x;h)\simeq\mathcal{O}(1).$

Now if $u(x;h)$ as in (2.3) is a solution of $\mathcal{L}(x,D)u(x;h)=0$ in
$\Omega$, we see that

(2.10)
$0=e^{-\frac{\varphi+\mathrm{i}\psi}{h}}\mathcal{L}(x,D)u=e^{-\frac{\varphi+\mathrm{i}\psi}{h}}\mathcal{L}(x;D)e^{\frac{\varphi+\mathrm{i}\psi}{h}}(\mathcal{A}(x;h)+r(x;h)).$

This implies

$e^{-\frac{\varphi+\mathrm{i}\psi}{h}}\mathcal{L}(x,D)e^{\frac{\varphi+\mathrm{i}\psi}{h}}r(x;h)=F(x;h),\quad\text{
for some }F(x;h)\in L^{2}(\Omega),\quad\text{ for all }h>0\text{ small. }$

By our choices, $a_{j}(x)$ annihilates all the terms of order $h^{-m+j}$ in
$e^{-\frac{\varphi+\mathrm{i}\psi}{h}}\mathcal{L}(x;D)e^{\frac{\varphi+\mathrm{i}\psi}{h}}\mathcal{A}(x;h)$
in $\Omega$ for $j=0,\ldots,m-1$. Thus we get $\|F(x,h)\|_{L^{2}(\Omega)}\leq
C$, where $C>0$ is uniform in $h$ for $h\ll 1$. Using Proposition 2.2 we have
the existence of $r(x;h)\in H^{2m}(\Omega)$ solving

$e^{-\frac{\varphi+\mathrm{i}\psi}{h}}\mathcal{L}(x,D)e^{\frac{\varphi+\mathrm{i}\psi}{h}}r(x;h)=F(x;h),$

with the estimate

$\|r(x;h)\|_{H_{\mathrm{scl}}^{2m}(\Omega)}\leq Ch^{m},\quad\text{ for
}h>0\text{ small enough. }$

∎

Similarly, we can construct the CGO solution of the adjoint equation
$\widetilde{\mathcal{L}}^{*}(x,D)v(x;h)=0$ in $\Omega$.

###  2.3. Momentum ray transforms

Here we briefly introduce the ray transform and Momentum Ray Transforms (MRT)
of symmetric tensor fields and describe some injectivity results of these
transforms. MRT was introduced in [Sha94] and later studied by [KMSS19, AM19,
KMSS20, MS21, AKS22, MS23, BKS23, IKS23, JKKS24].

Recall that $S^{m}:=S^{m}(\mathbb{R}^{n})$ is the space of symmetric
$m$-tensor fields and $C^{\infty}(\mathbb{R}^{n};S^{m})$ denotes the space of
symmetric $m$-tensor fields in $\mathbb{R}^{n}$ whose components are smooth.
We write $C_{c}^{\infty}(\mathbb{R}^{n};S^{m})$ to denote tensors in
$C^{\infty}(\mathbb{R}^{n};S^{m})$ whose components are compactly supported.
Throughout the section, we assume the Einstein summation convention for
repeated indices.

######  Definition 2.2 ([Sha94], Chapter 2).

The ray transform is the bounded linear operator

(2.11) $I:C_{c}^{\infty}(\mathbb{R}^{n};S^{m})\to
C^{\infty}(\mathbb{R}^{n}\times\mathbb{R}^{n}\setminus\\{0\\})$

defined as

(2.12) $If(x,\xi)=\int_{-\infty}^{\infty}f_{i_{1}\cdots
i_{m}}(x+t\xi)\,\xi^{i_{1}}\,\cdots\xi^{i_{m}}\,\mathrm{d}t=\int_{-\infty}^{\infty}\langle{f(x+t\xi),\xi^{m}}\rangle\mathrm{d}t,$

for all $(x,\xi)\in\mathbb{R}^{n}\times\mathbb{R}^{n}\setminus\\{0\\}$ which
determines the lines $\\{x+t\xi\mid t\in\mathbb{R}\\}$. Hereafter
$\langle\cdot,\cdot\rangle$ is the standard dot-product on $\mathbb{R}^{n}$.
The Kernel description of the ray transform is given by the following Saint
Venant operator.

######  Definition 2.3 ([Sha94], Chapter 2).

The Saint Venant operator

$W:C^{\infty}(\mathbb{R}^{n};S^{m})\to C^{\infty}(\mathbb{R}^{n};S^{m}\otimes
S^{m})$

is defined by

$(Wu)_{i_{1}\cdots i_{m}j_{1}\dots j_{m}}=\sigma(i_{1}\cdots
i_{m})\sigma({j_{1}\dots
j_{m}})\sum_{p=0}^{m}(-1)^{p}\binom{m}{p}\frac{\partial^{m}u_{i_{1}\dots
i_{m-p}j_{1}\cdots j_{p}}}{\partial x_{j_{p+1}}\cdots\partial
x_{j_{m}}\partial x_{i_{m-p+1}}\cdots\partial x_{i_{m}}},$

where $\binom{m}{p}=\frac{m!}{p!(m-p)!}$ are binomial coefficients, $\otimes$
denotes the product of tensors, and $\sigma$ is the symmetrization operator.

Let us recall the injectivity result of the ray transform of symmetric tensor
fields.

######  Theorem 3 ([Sha94], Theorem 2.2.1).

Let $n\geq 2$ and $m$ be non-negative integers, $l=\text{ max }\\{m,2\\}$. For
a compactly supported field $f\in C^{l}(\mathbb{R}^{n};S^{m})$ the following
statements are equivalent:

  1. (1)

$If=0;$

  2. (2)

there exist a compactly-supported field $v\in C^{l+1}(\mathbb{R}^{n};S^{m-1})$
such that its support is contained in the convex hull of the support of $f$
and

$dv=f;$

  3. (3)

the equality $Wf=0$ is valid in $\mathbb{R}^{n}$.

From Theorem 3, we see that the operator $I$ has an infinite dimensional
kernel for $m>0$. A symmetric tensor field can be uniquely decomposed into a
solenoidal and a potential part in a bounded domain [Sha94], and the potential
part lies in the kernel of the ray transform. Therefore, unique recovery of
entire tensor fields $f$ is impossible from $If$ requires some additional
information, which leads to the study of the MRTs. For each integer $k\geq 0$,
Momentum Ray Transforms are bounded linear operators

$I^{k}:C_{c}^{\infty}(\mathbb{R}^{n};S^{m})\to
C^{\infty}(\mathbb{R}^{n}\times\mathbb{R}^{n}\setminus\\{0\\}),$

defined as

$I^{k}f(x,\xi):=\int_{\mathbb{R}}t^{k}\,f_{i_{1}\cdots
i_{m}}(x+t\xi)\,\xi^{i_{1}}\,\cdots\xi^{i_{m}}\,\mathrm{d}t\quad\text{ for all
}\quad(x,\xi)\in\mathbb{R}^{n}\times\mathbb{R}^{n}\setminus\\{0\\}.$

Notice that, for $k=0,I^{0}f=If$ is the classical ray transform (2.2). It has
been established that [Sha94, KMSS19] the knowledge of the MRTs $I^{k}$ for
$k=0,\cdots,m$ uniquely recovers a symmetric $m$-tensor, whereas having any
less information would lead to nontrivial kernels. For the purpose of our
result, we define the MRT on the unit sphere bundle. For $f\in
C_{c}^{\infty}(\mathbb{R}^{n};S^{m})$ and each integer $k\geq 0$, we define
the restricted MRT $J^{k}$ as

$J^{k}f(x,\xi):=\int_{\mathbb{R}}t^{k}\,f_{i_{1}\cdots
i_{m}}(x+t\xi)\,\xi^{i_{1}}\,\cdots\xi^{i_{m}}\,\mathrm{d}t=I^{k}f(x,\xi)|_{\mathbb{R}^{n}\times\mathbb{S}^{n-1}},\quad\mbox{for
all }(x,\xi)\in\mathbb{R}^{n}\times\mathbb{S}^{n-1}.$

In [Sha94, KMSS19, AM19], the authors showed the rank $m$ symmetric tensor
field $f$ can be uniquely determined by the function
$\mathcal{J}^{m}f:=(J^{0}f,J^{1}f,\dots,J^{m}f)$.

In our setup, we end up with a modified version of the above restricted MRTs,
where we deal with a sum consisting of MRTs of a function, a vector field, up
to a symmetric $m$-tensor field. Moreover, the setup is even more complicated
due to the presence of the complex vectors in place of the real vectors
$\xi\in\mathbb{R}^{n}$. In [BKS23], the authors define MRT of $F\in
C_{c}^{\infty}(\mathbb{R}^{n};\mathbf{S}^{m}\mathbb{R}^{n})$, where
$\mathbf{S}^{m}=S^{0}\oplus S^{1}\oplus\cdots\oplus S^{m}.$ Any such $F$ can
be described uniquely as

(2.13) $\displaystyle F$
$\displaystyle=\sum_{p=0}^{m}f^{(p)}=f^{(0)}_{i_{0}}+f^{(1)}_{i_{1}}\,\mathrm{d}x^{i_{1}}+\cdots+f^{(p)}_{i_{1}\cdots
i_{p}}\mathrm{d}x^{i_{1}}\cdots\mathrm{d}x^{i_{p}}+\cdots+f^{(m)}_{i_{1}\cdots
i_{m}}\mathrm{d}x^{i_{1}}\cdots\mathrm{d}x^{i_{m}}$ $\displaystyle\hskip
28.45274pt\quad=\left(f^{(0)}_{i_{0}},f^{(1)}_{i_{1}},\cdots,f^{(m)}_{i_{1}\cdots
i_{m}}\right),$

which can be viewed as the sum of a function, a vector field, and up to a
symmetric $m$-tensor field with $f^{(p)}\in S^{p}$ for each $0\leq p\leq m$.
Next, we recall the definition of MRT of $F\in C_{c}^{\infty}(\mathbb{R}^{n})$
from [BKS23]. For every integer $k\geq 0$ and for all
$(x,\xi)\in\mathbb{R}^{n}\times\mathbb{R}^{n}\setminus\\{0\\}$ we define

(2.14) $\displaystyle I^{m,k}F(x,\xi)$
$\displaystyle=\sum_{p=0}^{m}I^{k}f^{(p)}(x,\xi)$
$\displaystyle=\int_{\mathbb{R}}t^{k}\left(f^{(0)}_{i_{0}}(x+t\xi)+f^{(1)}_{i_{1}}(x+t\xi)\,\xi^{i_{1}}+\cdots+f^{(m)}_{i_{1}\cdots
i_{m}}(x+t\xi)\,\xi^{i_{1}}\,\cdots\xi^{i_{m}}\right)\mathrm{d}t.$

We conclude this section by stating the following injectivity result for the
MRT, which we will apply repeatedly in the proof of Theorem 2. For the proof
of the following result, [BKS23, Lemma 3.7].

######  Theorem 4.

Let $n\geq 3$ and $f$ be a smooth compactly supported symmetric $m$-tensor
field in $\mathbb{R}^{n}$. Suppose that for all unit vectors $\eta\perp
e_{1}$, we have that

(2.15)
$\sum_{i_{1},\ldots,i_{m}=1}^{n}\int_{\mathbb{R}}x_{2}^{k}f_{i_{1}\cdots
i_{m}}(0,x_{2},x^{\prime\prime})\left(\prod_{j=1}^{m}\left(e_{1}+\mathrm{i}\eta\right)_{i_{j}}\right)\mathrm{d}x_{2}=0,\quad\text{
for a.e. }x^{\prime\prime}\in\mathbb{R}^{n-2},$

for each  $0\leq k\leq m.$ Then,

$f(0,x^{\prime})=i_{\delta}v(0,x^{\prime}),\quad\text{ for }\quad m\geq 2,$

where $e_{1}$ is the standard basis vector in $\mathbb{R}^{n}$ and $v$ is a
symmetric $(m-2)$-tensor field compactly supported in $x^{\prime}$ variable
and

$f(0,x^{\prime})=0\quad\text{ for }\quad m=0,1.$

######  Remark 2.1.

One can rewrite the identity (2.15) as

$\int_{\mathbb{R}}x_{2}^{k}F(0,x_{2},x^{\prime\prime})\mathrm{d}x_{2}=0\quad\text{
for }\quad 0\leq k\leq m,$

where $F$ denotes the sum of symmetric tensors in $i_{j}=2,\cdots,n$ indices
for $1\leq j\leq m$ as

$F(0,x_{2},x^{\prime\prime})=\sum_{p=0}^{m}\mathrm{i}^{p}\tilde{f}^{p}_{i_{1}\cdots
i_{p}}(0,x_{2},x^{\prime\prime})(\eta)_{i_{1}}\cdots(\eta)_{i_{p}}\quad\text{
with }\quad\tilde{f}^{p}_{i_{1}\cdots i_{p}}=\binom{m}{p}f_{i_{1}\cdots
i_{p}1\cdots 1}.$

Also, note that the presence of non-linearity for the tensor $m\geq 2$
generates a nontrivial kernel $v$ in Theorem 4.

##  3\. Proof of the main result

In this section, we prove the Theorem 2. We divide this section into three
parts. In the first part we construct special solutions to the (1.1) with
Dirichlet data vanishing on $\Gamma_{0}$. Next, we use the boundary data to
obtain a series of momentum ray transforms of the unknown coefficients.
Finally, we determine the parameters using the injectivity in Theorem 4
repeatedly with particular choices of solutions.

###  3.1. Construction of special solutions

Our main goal of this section is to construct complex geometric optics
solutions $U\in H^{2m}(\Omega)$ for the polyharmonic equation
$\mathcal{L}(x,D)U=0\text{ in }\Omega$ with Dirichlet boundary conditions
$(-\Delta)^{k-1}U|_{\Gamma_{0}}=0\text{ for }1\leq k\leq m$. We use a
reflection argument as in [Isa07] to construct these solutions. We reflect
$\Omega$ with respect to the plane $x_{n}=0$ and denote this reflection by

$\Omega^{*}=\\{(x^{\prime},-x_{n}):x=(x^{\prime},x_{n})\in\Omega\\}\quad\text{
where }\quad x^{\prime}=\left(x_{1},\cdots,x_{n-1}\right)\in\mathbb{R}^{n-1}.$

We denote $x^{*}=\left(x^{\prime},-x_{n}\right)\in\Omega^{*}$ for any
$x=\left(x^{\prime},x_{n}\right)\in\Omega,\text{ and }f^{*}(x)=f(x^{*})$ for
any function $f$ on $\Omega$. We set

$O=\Omega\cup\text{int($\Gamma_{0}$) }\cup\Omega^{*},$

where int$(\Gamma_{0})\subset\\{x_{n}=0\\}$. We extend the coefficients $q$
and $A^{j}$ for $1\leq j\leq m$ from $\Omega$ to $O$ as

(3.1) $\displaystyle q(x)$ $\displaystyle:=\begin{cases}q(x)\quad&\text{if
}x\in\Omega,\\\ q(x^{*})&\text{if }x\in\Omega^{*},\end{cases}$
$\displaystyle\left(A^{j}\right)_{\underbrace{n\cdots
n}_{p-\text{times}}i_{1}\cdots i_{j-p}}(x)$
$\displaystyle:=\begin{cases}\left(A^{j}\right)_{nn\cdots ni_{1}\cdots
i_{j-p}}(x)\quad&\text{if }x\in\Omega,\\\ (-1)^{p}\left(A^{j}\right)_{nn\cdots
ni_{1}\cdots i_{j-p}}(x^{*})&\text{if }x\in\Omega^{*},\end{cases}$

where $1\leq i_{1},\dots,i_{j-p}\leq n-1,$ $p$ := number of indices are equal
to $n$. Note that, $A^{j},q\in C_{c}^{\infty}(O,\mathbb{C}^{n^{j}})$, for
$1\leq j\leq m$.

######  Proposition 3.1.

Let $A^{j}\in C_{c}^{\infty}({\Omega},\mathbb{C}^{n^{j}})$ be symmetric for
$1\leq j\leq m$ with $A^{m}$ to be partially isotropic, $q\in
L^{\infty}\left(\Omega,\mathbb{C}\right)$ and $h>0$ be small enough. Then
there exist a solution $U\in H^{2m}\left(\Omega\right)$ to the equation
$\mathcal{L}(x,D)U=0$ in $\Omega$ with $(-\Delta)^{k-1}U|_{\Gamma_{0}}=0\text{
for }1\leq k\leq m$ of the form

(3.2) $U(x;h)=\mathcal{U}(x;h)-\mathcal{U}(x^{*};h),\quad x\in\Omega,$

where $\mathcal{U}\in H^{2m}(O)$ solving $\mathcal{L}(x,D)\mathcal{U}(x;h)=0$
in $O$ and has the form

(3.3) $\displaystyle\mathcal{U}(x;h)=$ $\displaystyle
e^{\frac{\varphi+\mathrm{i}\psi}{h}}\left(a_{0}(x)+ha_{1}(x)+h^{2}a_{2}(x)\dots+h^{m-1}a_{m-1}(x)+r(x;h)\right),\quad
x\in O.$

Here $\varphi$ and $\psi$ are defined as in (2.4), and the functions
$a_{0},\cdots,a_{m-1}$ satisfy the transport equations (2.8) and (2.9) in $O$,
and $r$ satisfies the estimate

(3.4) $\|r(x;h)\|_{H_{\mathrm{scl}}^{2m}(O)}\leq Ch^{m},$

where $C$ is independent of $h$.

###### Proof.

We extend the operator $\mathcal{L}(x,D)$ over $O$ by extending $q$ and
$A^{j}$ for $1\leq j\leq m$ as (3.1). Applying Lemma 2.1 to $A^{j}\in
C_{c}^{\infty}(O)$ and $q\in L^{\infty}(O)$ we get a CGO solution
$\mathcal{U}(x;h)\in H^{2m}(O)$ of $\mathcal{L}(x,D)\mathcal{U}(x;h)=0$ in $O$
as

(3.5)
$\mathcal{U}(x;h)=e^{\frac{\varphi+\mathrm{i}\psi}{h}}\left(a_{0}(x)+ha_{1}(x)+h^{2}a_{2}(x)\dots+h^{m-1}a_{m-1}(x)+r(x;h)\right),\text{
in }O,$

with the decay estimate (3.4). A direct calculation shows

$\mathcal{L}(x,D)U(x;h)=\mathcal{L}(x,D)\left(\mathcal{U}(x;h)-\mathcal{U}(x^{*};h)\right)=0,\quad
x\in\Omega\subset\left\\{\mathbb{R}^{n};x_{n}>0\right\\},$

and $(-\Delta)^{k-1}U|_{\Gamma_{0}}=0$ for all $k=1,2,\cdots m$. ∎

We recall our choices of phase functions

$\varphi(x)=\mu^{(2)}\cdot x\quad\text{ and
}\quad\psi(x)=\left(\frac{h\xi}{2}+\sqrt{1-h^{2}\frac{|\xi|^{2}}{4}}\mu^{(1)}\right)\cdot
x,$

where $\xi,\mu^{(1)},\mu^{(2)}\in\mathbb{R}^{n}$ be such that
$|\mu^{(1)}|=|\mu^{(2)}|=1$ and
$\mu^{(1)}\cdot\mu^{(2)}=\mu^{(1)}\cdot\xi=\mu^{(2)}\cdot\xi=0$ (here we use
that $n\geq 3$). We define $\zeta_{1}$, $\zeta_{2}\in\mathbb{C}^{n}$ by

(3.6)
$\displaystyle\zeta_{1}=\frac{ih\xi}{2}+i\sqrt{1-h^{2}\frac{|\xi|^{2}}{4}}\mu^{(1)}+\mu^{(2)},$
$\displaystyle\zeta_{2}=-\frac{ih\xi}{2}+i\sqrt{1-h^{2}\frac{|\xi|^{2}}{4}}\mu^{(1)}-\mu^{(2)}.$

Using Lemma 2.1 we write the CGO solutions to the equations
$\mathcal{L}(x,D)\mathcal{U}=0$ and
$\widetilde{\mathcal{L}}^{*}(x,D)\mathcal{V}=0$ in $O$ respectively

(3.7) $\displaystyle\mathcal{U}(x;h)=$ $\displaystyle e^{\frac{\zeta_{1}\cdot
x}{h}}\left(a_{0}(x)+ha_{1}(x)+h^{2}a_{2}(x)\dots+h^{m-1}a_{m-1}(x)+r(x;h)\right)=e^{\frac{\zeta_{1}\cdot
x}{h}}(\mathrm{a}(x;h)),$ $\displaystyle\mathcal{V}(x;h)=$ $\displaystyle
e^{\frac{\zeta_{2}\cdot
x}{h}}\left(b_{0}(x)+hb_{1}(x)+h^{2}b_{2}(x)\dots+h^{m-1}b_{m-1}(x)+\widetilde{r}(x;h)\right)=e^{\frac{\zeta_{2}\cdot
x}{h}}(\mathrm{b}(x;h)).$

We have the following decay estimates for the remainder terms

(3.8) $\|r\|_{H^{2m}_{\mathrm{scl}(O)}}=\mathcal{O}\left(h^{m}\right)\text{
and
}\|\widetilde{r}\|_{H^{2m}_{\mathrm{scl}(O)}}=\mathcal{O}\left(h^{m}\right).$

Following Proposition 3.1, we write down the special solutions:

(3.9) $\displaystyle U(x;h)=e^{\frac{\zeta_{1}\cdot
x}{h}}\left(\mathrm{a}(x;h)\right)-e^{\frac{\zeta_{1}\cdot
x^{*}}{h}}\left(\mathrm{a}(x^{*};h)\right),$ (3.10) $\displaystyle
V(x;h)=e^{\frac{\zeta_{2}\cdot
x}{h}}\left(\mathrm{b}(x;h)\right)-e^{\frac{\zeta_{2}\cdot
x^{*}}{h}}\left(\mathrm{b}(x^{*};h)\right),$

where $U(x;h)\text{ and }V(x;h)$ solves the polyharmonic equations
$\mathcal{L}(x;D)U=0$, $\widetilde{\mathcal{L}}^{*}(x,D)V=0$ in $\Omega$, with
the required boundary conditions
$(-\Delta)^{k-1}U|_{\Gamma_{0}}=0=(-\Delta)^{k-1}V|_{\Gamma_{0}}\text{ for
}1\leq k\leq m$.

###  3.2. The integral identity

We recall our given operator (1.1)

$\mathcal{L}(x,D)=(-\Delta)^{m}+\sum_{j=0}^{m-1}\sum_{i_{1},\cdots,i_{m-j}=1}^{n}A_{i_{1}\cdots
i_{m-j}}^{m-j}(x)D_{i_{1}\cdots i_{m-j}}^{m-j}+q(x),$

where $A^{j}\in C_{c}^{\infty}(\Omega)\text{ for }1\leq j\leq m$ and $q\in
L^{\infty}({\Omega})$. By using integration by parts, we get

(3.11) $\displaystyle\int_{\Omega}$
$\displaystyle\left(\mathcal{L}(x,D)u\right)\overline{v}\,\mathrm{d}x-\int_{\Omega}u\,\overline{\mathcal{L}^{*}(x,D)v}\,\mathrm{d}x$
$\displaystyle=$
$\displaystyle\sum_{l=1}^{m}\int_{\partial\Omega}\left((-\Delta)^{(m-l)}u\overline{\left(\partial_{\nu}(-\Delta)^{l-1}v\right)}-(\partial_{\nu}(-\Delta)^{(m-l)}u)\overline{\left((-\Delta)^{l-1}v\right)}\right)\,\mathrm{d}S$
$\displaystyle+\mathrm{i}\sum_{j=1}^{m}\sum_{l=1}^{j}\sum_{i_{1},\cdots,i_{j}=1}^{n}\int_{\partial\Omega}(-1)^{l}\nu_{i_{j-l+1}}\left(D^{j-l}_{i_{1}\dots
i_{j-l}}u\right)D^{l-1}_{i_{j-l+2}\dots i_{j}}\left(A^{j}_{i_{1}\dots
i_{j}}\overline{v}\right)\,\mathrm{d}S,$

where $\mathrm{d}S$ is the surface measure on $\partial\Omega$ and for all
$u,v\in H^{2m}(\Omega)$.

Let us take $u=U-\widetilde{U}$ where $U\in H^{2m}(\Omega)$ solves

$\displaystyle\mathcal{L}(x,D)U=0,\quad\text{ in }\Omega,\quad\mbox{with
}(-\Delta)^{k}U|_{\Gamma_{0}}=0,\quad\mbox{for }k=0,1,\ldots,m-1;$

and $\widetilde{U}$ solves

(3.12) $\widetilde{\mathcal{L}}(x,D)\widetilde{U}=0\quad\mbox{in
}\Omega,\quad\mbox{with
}(-\Delta)^{k}\widetilde{U}|_{\partial\Omega}=(-\Delta)^{k}U|_{\partial\Omega},\quad\mbox{for
}k=0,1,\ldots,m-1.$

From the assumptions of Theorem 2 on the boundary measurements, we must have

(3.13)
$\partial_{\nu}(-\Delta)^{k}U|_{\Gamma}=\partial_{\nu}(-\Delta)^{k}\widetilde{U}|_{\Gamma},\quad\mbox{for
}k=0,1,2,\ldots,m-1.$

We choose $V\in H^{2m}(\Omega)$ to be any solution satisfying

(3.14) $\widetilde{\mathcal{L}}^{*}(x,D)V=0\text{ in }\Omega,\quad\mbox{with
}(-\Delta)^{k}V|_{\Gamma_{0}}=0,\quad\mbox{for }k=0,1\cdots m-1.$

From (3.12), (3.13) and (3.14), using a similar argument as in (3.11) we
obtain

(3.15) $\displaystyle
0=\int_{\Omega}\left(\widetilde{\mathcal{L}}(x,D)(U-\widetilde{U})\right)\overline{V}\mathrm{d}x-\int_{\Omega}(U-\widetilde{U})\overline{\widetilde{\mathcal{L}}^{*}(x,D)V}\mathrm{d}x=\int_{\Omega}\left(\widetilde{\mathcal{L}}(x,D)(U-\widetilde{U})\right)\overline{V}\mathrm{d}x.$

A straightforward calculation entails

$\displaystyle\widetilde{\mathcal{L}}(x,D)(U-\widetilde{U})=$
$\displaystyle(\widetilde{\mathcal{L}}(x,D)-\mathcal{L}(x,D))U$
$\displaystyle=$
$\displaystyle\sum_{j=0}^{m-1}\sum_{i_{1},\cdots,i_{m-j}=1}^{n}\left(\widetilde{A}_{i_{1}\cdots
i_{m-j}}^{m-j}-A_{i_{1}\cdots i_{m-j}}^{m-j}\right)D_{i_{1}\cdots
i_{m-j}}^{m-j}U+(\widetilde{q}-q)U.$

Therefore, the identity (3.15) implies,

(3.16) $\displaystyle
0=\sum_{j=0}^{m-1}\sum_{i_{1},\cdots,i_{m-j}=1}^{n}\int_{\Omega}\left(\widetilde{A}_{i_{1}\cdots
i_{m-j}}^{m-j}-A_{i_{1}\cdots i_{m-j}}^{m-j}\right)D_{i_{1}\cdots
i_{m-j}}^{m-j}U\overline{V}\mathrm{d}x+\int_{\Omega}(\widetilde{q}-q)U\overline{V}\mathrm{d}x.$

For simplicity, we denote $W^{j}=\widetilde{A}^{j}-A^{j}$ for
$j=1,2,\ldots,m\text{ and }W^{0}=\widetilde{q}-q$. Note that, using
Proposition 3.1, we can choose $U$ and $V$ to be of the form (3.9), (3.10). We
compute

(3.17) $\displaystyle
e^{x\cdot\zeta_{1}/h}e^{x\cdot\overline{\zeta}_{2}/h}\,=$ $\displaystyle
e^{ix\cdot\xi}$
$\displaystyle\vspace{1ex}e^{x\cdot\zeta_{1}/h}e^{(x^{*})\cdot\overline{\zeta}_{2}/h}\,=$
$\displaystyle e^{ix\cdot\xi_{+}+2\mu^{(2)}_{n}x_{n}/h}$
$\displaystyle\vspace{1ex}e^{(x^{*})\cdot\zeta_{1}/h}e^{x\cdot\overline{\zeta}_{2}/h}\,=$
$\displaystyle e^{ix\cdot\xi_{-}-2\mu^{(2)}_{n}x_{n}/h}$
$\displaystyle\vspace{1ex}e^{(x^{*})\cdot\zeta_{1}/h}e^{(x^{*})\cdot\overline{\zeta}_{2}/h}\,=$
$\displaystyle e^{i(x^{*})\cdot\xi}$

where $\xi_{+}$ , $\xi_{-}\in\mathbb{R}^{n}$ are

$\xi_{+}=\left(\xi^{\prime},+\frac{2}{h}\sqrt{1-h^{2}\frac{|\xi|^{2}}{4}}\mu^{(1)}_{n}\right),\qquad\xi_{-}=\left(\xi^{\prime},-\frac{2}{h}\sqrt{1-h^{2}\frac{|\xi|^{2}}{4}}\mu^{(1)}_{n}\right),\quad\mbox{where
}\xi^{\prime}=(\xi_{1},\cdots,\xi_{n-1}).$

To eliminate the unwanted terms, we further assume that $\mu^{(2)}_{n}=0$ and
$\mu^{(1)}_{n}\neq 0$. So, by the Riemann–Lebesgue lemma and the fact that
$|\xi_{+}|,|\xi_{-}|\rightarrow\infty$ as $h\rightarrow 0$, we have the
following four limiting identities,

(3.18) $\lim_{h\to
0}\int_{\Omega}(W^{0})e^{(x^{*})\cdot\zeta_{1}/h}e^{x\cdot\overline{\zeta}_{2}/h}\mathrm{a}(x^{*};h)\overline{\mathrm{b}(x;h)}\mathrm{d}x=0.$
(3.19) $\displaystyle\lim_{h\to
0}\sum_{j=0}^{m-1}\sum_{i_{1},\cdots,i_{m-j}=1}^{n}h^{m-j}\int_{\Omega}$
$\displaystyle\left(W_{i_{1}\cdots
i_{m}-j}^{m-j}\right)e^{(x^{*})\cdot\zeta_{1}/h}e^{x\cdot\overline{\zeta}_{2}/h}$
$\displaystyle\times\left(\prod_{k=1}^{m-j}\left(\frac{-\mathrm{i}}{h}(\mu^{(2)}+\mathrm{i}\mu^{(1)})_{i_{k}}+D_{i_{k}}\right)\mathrm{a}(x^{*};h)\right)\overline{\mathrm{b}(x;h)}\mathrm{d}x=0.$
(3.20) $\lim_{h\to
0}\int_{\Omega}(W^{0})e^{x\cdot\zeta_{1}/h}e^{(x^{*})\cdot\overline{\zeta}_{2}/h}\mathrm{a}(x;h)\overline{\mathrm{b}(x^{*};h)}\mathrm{d}x=0.$
(3.21) $\displaystyle\lim_{h\to
0}\sum_{j=0}^{m-1}\sum_{i_{1},\cdots,i_{m-j}=1}^{n}h^{m-j}\int_{\Omega}$
$\displaystyle\left(W_{i_{1}\cdots
i_{m}-j}^{m-j}\right)e^{x\cdot\zeta_{1}/h}e^{(x^{*})\cdot\overline{\zeta}_{2}/h}$
$\displaystyle\times\left(\prod_{k=1}^{m-j}\left(\frac{-\mathrm{i}}{h}(\mu^{(2)}+\mathrm{i}\mu^{(1)})_{i_{k}}+D_{i_{k}}\right)\mathrm{a}(x;h)\right)\overline{\mathrm{b}(x^{*};h)}\mathrm{d}x=0.$

Now, inserting the CGO solutions (3.9) and (3.10) into the identity (3.16) and
using (3.17) we get

$\displaystyle 0=$
$\displaystyle\sum_{j=0}^{m-1}\sum_{i_{1},\cdots,i_{m-j}=1}^{n}\int_{\Omega}\left(W_{i_{1}\cdots
i_{m-j}}^{m-j}\right)\prod_{k=1}^{m-j}\left(\frac{-\mathrm{i}}{h}(\mu^{(2)}+\mathrm{i}\mu^{(1)})_{i_{k}}+D_{i_{k}}\right)\mathrm{a}(x;h)\overline{\mathrm{b}(x;h)}e^{ix\cdot\xi}\mathrm{d}x$
$\displaystyle+\int_{\Omega}(W^{0})e^{ix\cdot\xi}\mathrm{a}(x;h)\overline{\mathrm{b}(x;h)}\mathrm{d}x$
$\displaystyle-\sum_{j=0}^{m-1}\sum_{i_{1},\cdots,i_{m-j}=1}^{n}\int_{\Omega}\left(W_{i_{1}\cdots
i_{m}-j}^{m-j}\right)e^{x\cdot\zeta_{1}/h}e^{(x^{*})\cdot\overline{\zeta}_{2}/h}\prod_{k=1}^{m-j}\left(\frac{-\mathrm{i}}{h}(\mu^{(2)}+i\mu^{(1)})_{i_{k}}+D_{i_{k}}\right)\mathrm{a}(x;h)$
$\displaystyle\hskip
85.35826pt\times\overline{\mathrm{b}(x^{*};h)}\mathrm{d}x-\int_{\Omega}(W^{0})e^{x\cdot\zeta_{1}/h}e^{(x^{*})\cdot\overline{\zeta}_{2}/h}\mathrm{a}(x^{*};h)\overline{\mathrm{b}(x;h)}\mathrm{d}x$
$\displaystyle-\sum_{j=0}^{m-1}\sum_{i_{1},\cdots,i_{m-j}=1}^{n}\int_{\Omega}\left(W_{i_{1}\cdots
i_{m}-j}^{m-j}\right)e^{(x^{*})\cdot\zeta_{1}/h}e^{x\cdot\overline{\zeta}_{2}/h}$
$\displaystyle\hskip
85.35826pt\times\left(\prod_{k=1}^{m-j}\left(\frac{-\mathrm{i}}{h}(\mu^{(2)}+i\mu^{(1)})_{i_{k}}+D_{i_{k}}\right)\mathrm{a}(x^{*};h)\right)\overline{\mathrm{b}(x;h)}\mathrm{d}x$
$\displaystyle-\int_{\Omega}(W^{0})e^{(x^{*})\cdot\zeta_{1}/h}e^{x\cdot\overline{\zeta}_{2}/h}\mathrm{a}(x;h)\overline{\mathrm{b}(x^{*};h)}\mathrm{d}x$
$\displaystyle+\sum_{j=0}^{m-1}\sum_{i_{1},\cdots,i_{m-j}=1}^{n}\int_{\Omega}\left(W_{i_{1}\cdots
i_{m-j}}^{m-j}\right)\prod_{k=1}^{m-j}\left(\frac{-\mathrm{i}}{h}(\mu^{(2)}+\mathrm{i}\mu^{(1)})_{i_{k}}+D_{i_{k}}\right)\mathrm{a}(x^{*};h)$
$\displaystyle\hskip
85.35826pt\times\overline{\mathrm{b}(x^{*};h)}e^{i(x^{*})\cdot\xi}\mathrm{d}x+\int_{\Omega}(W^{0})e^{i(x^{*})\cdot\xi}\mathrm{a}(x^{*};h)\overline{\mathrm{b}(x^{*};h)}\mathrm{d}x.$

Making the change of variables $x\mapsto x^{*}$ we obtain the integral
identity in $O$ as

(3.22) $\displaystyle 0=$
$\displaystyle\sum_{j=0}^{m-1}\sum_{i_{1},\cdots,i_{m-j}=1}^{n}\int_{O}\left(W_{i_{1}\cdots
i_{m-j}}^{m-j}\right)\left(\prod_{k=1}^{m-j}\left(\frac{-\mathrm{i}}{h}(\mu^{(2)}+\mathrm{i}\mu^{(1)})_{i_{k}}+D_{i_{k}}\right)\mathrm{a}(x;h)\right)$
$\displaystyle\times\overline{\mathrm{b}(x;h)}e^{ix\cdot\xi}\mathrm{d}x-\sum_{j=0}^{m-1}\sum_{i_{1},\cdots,i_{m-j}=1}^{n}\int_{O}\left(W_{i_{1}\cdots
i_{m}-j}^{m-j}\right)e^{x\cdot\zeta_{1}/h}e^{(x^{*})\cdot\overline{\zeta}_{2}/h}$
$\displaystyle\times\left(\prod_{k=1}^{m-j}\left(\frac{-\mathrm{i}}{h}(\mu^{(2)}+i\mu^{(1)})_{i_{k}}+D_{i_{k}}\right)\mathrm{a}(x;h)\right)\overline{\mathrm{b}(x^{*};h)}\mathrm{d}x$
$\displaystyle-\int_{O}(W^{0})e^{(x^{*})\cdot\zeta_{1}/h}e^{x\cdot\overline{\zeta}_{2}/h}\mathrm{a}(x^{*};h)\overline{\mathrm{b}(x;h)}\mathrm{d}x+\int_{O}(W^{0})e^{ix\cdot\xi}\mathrm{a}(x;h)\overline{\mathrm{b}(x;h)}\mathrm{d}x,$

for all $\xi,\mu^{(1)},\mu^{(2)}\in\mathbb{R}^{n}$ with

(3.23)
$\mu^{(1)}\cdot\mu^{(2)}=\xi\cdot\mu^{(1)}=\xi\cdot\mu^{(2)}=0,\quad|\mu^{(1)}|=|\mu^{(2)}|=1,\quad\mu^{(2)}_{n}=0,\quad\mu^{(1)}_{n}\neq
0.$

Recall the form of $a(x;h)$ and $b(x;h)$ from (3.9), (3.10). Here
$a_{0}(x),b_{0}(x)\in C^{\infty}(\overline{O},\mathbb{C})$ satisfy the
transport equations

(3.24)
$\left(2(\mu^{(2)}+i\mu^{(1)})\cdot\nabla\right)^{m}a_{0}(x)=\left(2(-\mu^{(2)}+i\mu^{(1)})\cdot\nabla\right)^{m}b_{0}(x)=0\text{
in }O\text{ for }m\geq 2.$

Next, we solve the transport equations (3.24) following [BKS23]. Observe that,
up to a rotation, we can take $\mu^{(1)}=e_{n}$ and $\mu^{(2)}=e_{j}$, $j\neq
n$ where $\\{e_{1},\cdots,e_{n}\\}$ is the standard orthonormal basis for
$\mathbb{R}^{n}$. Then the transport equations (3.24) become

(3.25)
$2^{m}\left(\partial_{x_{j}}+\mathrm{i}\partial_{x_{n}}\right)^{m}a_{0}=0\text{
and
}2^{m}\left(\partial_{x_{j}}+\mathrm{i}\partial_{x_{n}}\right)^{m}\overline{b_{0}}=0\quad\text{
in }O\text{ for }m\geq 2.$

For $1\leq j\leq n-1$, denoting $z=x_{j}+\mathrm{i}x_{n}$ the above transport
equations reduce to

(3.26) $\partial_{\bar{z}}^{m}a_{0}=0\quad\text{ and
}\quad\partial_{\bar{z}}^{m}\overline{b_{0}}=0.$

The general solution of equation (3.26) can be given by

(3.27) $a_{0}(x)=g(x^{\prime\prime})\sum_{k=0}^{m-1}(z-\bar{z})^{k}f_{k}(z),$

where $f_{k}$ is a holomorphic function for all $0\leq k\leq m-1$, $g$ is any
smooth function and $x^{\prime\prime}=x-x_{j}e_{j}-x_{n}e_{n}$.

###  3.3. Recovering the coefficients

In this section, we closely follow [BKS23] to recover the coefficients from
the integral equation (3.22). We multiply (3.22) by suitable powers of $h$ and
take $h\to 0$ to obtain momentum ray transforms of the coefficients. We recall
the definition of partially isotropic tensors. We say a tensor $W^{j}$, $j\geq
2$ is partially isotropic if there exists a symmetric $(j-2)$-tensor field
$\overline{W}^{j-2}$ in $O$ such that
$W^{j}=i_{\delta}\overline{W}^{j-2}=\overline{W}^{j-2}\odot\delta$, where
$\delta$ is the Kronecker delta tensor and $\odot$ denotes the symmetric
product of tensors. We assume that $A^{m}=i_{\delta}\overline{A}^{m-2}$
(respectively for $\widetilde{A}^{m}$ ) and
$\operatorname{denote}\overline{W}^{m-2}=\overline{\widetilde{A}}^{m-2}-\overline{A}^{m-2}$.

###### Proof of Theorem 2.

Step 1. We start with a rotation of coordinates in (3.22) by taking $e_{n}$ to
$e_{1}$ and denote $\eta$ as any unit vector perpendicular to $e_{1}$. So,
denoting the new coordinates by $y$ (w.r.t. some orthonormal basis
$\\{e_{1},\eta,\cdots\\}$), we get
$z=y_{1}+\mathrm{i}(y\cdot\eta)=y_{1}+\mathrm{i}y_{2}$ and
$y=(y_{1},y^{\prime})=(y_{1},y_{2},y^{\prime\prime})$. We multiply (3.22) by
$h^{m-1}$ and let $h\to 0$ along with (3.18), (3.19), (3.20), (3.21), and
(3.8) we have

(3.28)
$\displaystyle\sum_{i_{1},\dots,i_{m-2}=1}^{n}\int_{O}\overline{W}_{i_{1}\cdots,i_{m-2}}^{m-2}\left(\prod_{k=1}^{m-2}(e_{1}+\mathrm{i}\eta)_{i_{k}}\right)\left[-Ta_{0}(y)\right]\overline{b_{0}(y)}\mathrm{~{}d}y$
$\displaystyle+\sum_{i_{1},\dots,i_{m-1}=1}^{n}\int_{O}W_{i_{1}\cdots
i_{m-1}}^{m-1}\left(\prod_{k=1}^{m-1}(e_{1}+\mathrm{i}\eta)_{i_{k}}\right)a_{0}(y)\overline{b_{0}(y)}\mathrm{~{}d}y=0.$

We choose the amplitudes $a_{0}(y)\text{ and }b_{0}(y)$ as

$a_{0}(y)=1\mbox{ and
}\overline{b_{0}(y)}=(z-\bar{z})^{k}f(z)g(y^{\prime\prime}),\quad\text{ for
all }k\text{ with }0\leq k\leq m-1,$

for any holomorphic function $f(z)$ and any smooth function
$g(y^{\prime\prime})$. For $\lambda\in\mathbb{R}$ we choose

$f(z)=e^{-\mathrm{i}\lambda z}.$

Note that, using compact support of $W^{j}$’s we can extend
$W^{m-1}(\cdot,\cdot,y^{\prime\prime})=0$ over $\mathbb{R}^{2}\backslash O$,
therefore, for each $0\leq k\leq m-1$ we obtain,

$\int_{\mathbb{R}^{n-2}}\left(\int_{\mathbb{R}^{2}}\sum_{i_{1},\cdots,i_{m-1}=1}^{n}W_{i_{1}\cdots
i_{m-1}}^{m-1}\left(\prod_{k=1}^{m-1}(e_{1}+\mathrm{i}\eta)_{i_{k}}\right)y_{2}^{k}e^{-\mathrm{i}\lambda
z}\mathrm{~{}d}y_{1}\mathrm{~{}d}y_{2}\right)g(y^{\prime\prime})\mathrm{~{}d}y^{\prime\prime}=0.$

Varying $g$ to be any smooth function in the $y^{\prime\prime}$ variable, we
obtain

$\int_{\mathbb{R}^{2}}\sum_{i_{1},\cdots,i_{m-1}=1}^{n}W_{i_{1}\cdots
i_{m-1}}^{m-1}\left(\prod_{k=1}^{m-1}(e_{1}+\mathrm{i}\eta)_{i_{k}}\right)y_{2}^{k}e^{-\mathrm{i}\lambda
z}\mathrm{~{}d}y_{1}\mathrm{~{}d}y_{2}=0.$

After performing the integration in the $y_{1}$ variable we obtain

(3.29)
$\int_{\mathbb{R}}\sum_{i_{1},\cdots,i_{m-1}=1}^{n}\widehat{W}_{i_{1}\cdots
i_{m-1}}^{m-1}\left(\prod_{k=1}^{m-1}(e_{1}+\mathrm{i}\eta)_{i_{k}}\right)y_{2}^{k}e^{\lambda
y_{2}}\mathrm{~{}d}y_{2}=0,\quad\text{ for a.e.
}y^{\prime\prime}\in\mathbb{R}^{n-2}\text{ and for all }\lambda\in\mathbb{R},$

where we denote $\widehat{.}$ to be the partial Fourier transform in the
$y_{1}$ variable. We set $\lambda=0$ in above to obtain

$\int_{\mathbb{R}}\sum_{i_{1},\cdots,i_{m-1}=1}^{n}\widehat{W}_{i_{1}\cdots
i_{m-1}}^{m-1}\left(\prod_{k=1}^{m-1}(e_{1}+\mathrm{i}\eta)_{i_{k}}\right)y_{2}^{k}\mathrm{~{}d}y_{2}=0.$

Since the above identity can be interpreted as the vanishing of a certain MRT,
by using Lemma 4, we see that there exists a symmetric $(m-3)$-tensor field
$\widehat{V}^{m-1,1,0}(0,y^{\prime})$, compactly supported in $y^{\prime}$
such that

$\widehat{W}^{m-1}(0,y^{\prime})=\begin{cases}i_{\delta}\widehat{V}^{m-1,1,0}(0,y^{\prime}),\quad&\mbox{when
}m\geq 3\\\ 0&\mbox{when }m=2\end{cases},\quad\mbox{for a.e.
}y^{\prime}\in\mathbb{R}^{n+1}.$

Next, for $m\geq 3$, we differentiate (3.29) with respect to $\lambda$ and set
$\lambda=0$ to get

(3.30) $\displaystyle 0=$
$\displaystyle\int_{\mathbb{R}}\sum_{i_{1},\cdots,i_{m-1}=1}^{n}\left(\widehat{W}_{i_{1}\cdots
i_{m-1}}^{m-1}(0,y^{\prime})\prod_{k=1}^{m-1}(e_{1}+\mathrm{i}\eta)_{i_{k}}\right)y_{2}^{k+1}\mathrm{~{}d}y_{2}$
$\displaystyle+\sum_{i_{1},\cdots,i_{m-1}=1}^{n}\int_{\mathbb{R}}y_{2}^{k}\frac{\mathrm{d}}{\mathrm{d}\lambda}\left(\widehat{W}_{i_{1}\cdots
i_{m-1}}^{m-1}(0,y^{\prime})\prod_{k=1}^{m-1}(e_{1}+\mathrm{i}\eta)_{i_{k}}\right)\mathrm{~{}d}y_{2}=0.$

Since we have
$\widehat{W}^{m-1}(0,y^{\prime})=i_{\delta}\widehat{V}^{m-1,1,0}(0,y^{\prime})$,
so the first term in (3.30) vanishes using the fact that
$(e_{1}+\mathrm{i}\eta)\cdot(e_{1}+\mathrm{i}\eta)=0.$ Now, repeating the same
steps as before, we see that there exists a symmetric $(m-3)$-tensor field
$\widehat{V}^{m-1,1,1}(0,y^{\prime})$ such that

$\left.\frac{\mathrm{d}}{\mathrm{d}\lambda}\right|_{\lambda=0}\widehat{W}^{m-1}(0,y^{\prime})=i_{\delta}\widehat{V}^{m-1,1,1}(0,y^{\prime}),\quad\text{
for a.e. }y^{\prime}\text{. }$

Repeating the same technique we obtain

(3.31)
$\left.\frac{\mathrm{d}^{r}}{\mathrm{d}\lambda^{r}}\right|_{\lambda=0}\widehat{W}^{m-1}(0,y^{\prime})=i_{\delta}\widehat{V}^{m-1,1,r}(0,y^{\prime})\quad\text{
for a.e. }y^{\prime}.$

Since $W^{m-1}$ is compactly supported in $\Omega$, we have that
$\widehat{W}^{m-1}$ is analytic in $\lambda$ variable. Using Payley–Weiner’s
theorem, we have

$\widehat{W}^{m-1}(\lambda,y^{\prime})=i_{\delta}\widehat{V}^{m-1,1}(\lambda,y^{\prime})\text{
a.e. in }\Omega,\quad m\geq 3,$

where $\widehat{V}^{m-1,1}$ is symmetric $(m-3)$-tensor field and

$\widehat{V}^{m-1,1}(\lambda,y^{\prime})=\sum_{r=0}^{\infty}\frac{{\lambda}^{r}}{r!}\widehat{V}^{m-1,1,r}(0,y^{\prime}).$

Hence, by using the inverse Fourier transform, we get

(3.32)
$W^{m-1}(y_{1},y^{\prime})=\begin{cases}i_{\delta}V^{m-1,1}(y_{1},y^{\prime})\quad&\mbox{for
}m\geq 3,\\\ 0\quad&\mbox{for }m=2.\end{cases}$

By using (3.32) in (3.28), we obtain

(3.33)
$\displaystyle\sum_{i_{1},\dots,i_{m-2}=1}^{n}\int_{\mathbb{R}}\overline{W}_{i_{1}\cdots
i_{m-2}}^{m-2}\left(\prod_{k=1}^{m-2}(e_{1}+\mathrm{i}\eta)_{i_{k}}\right)\left[-Ta_{0}(y)\right]\overline{b_{0}(y)}\mathrm{~{}d}y=0.$

Note that, using compact support of $W^{j}$’s we can smoothly extend
$W^{m-2}(\cdot,\cdot,y^{\prime\prime})=0$ over $\mathbb{R}^{2}$. Let us now
choose

$a_{0}(y)=(z-\bar{z})\quad\text{ then }\quad Ta_{0}(y)=-1.$

By choosing $\overline{b_{0}(y)}=(z-\bar{z})^{k}e^{-\mathrm{i}\lambda
z}g(y^{\prime\prime})$ for all $0\leq k\leq m-2$ and $g$ is any smooth
function in $y^{\prime\prime}\in\mathbb{R}^{n-2}$, we are in the same set-up
as before. Repeating the same steps as above, we see that

(3.34)
$\overline{W}^{m-2}\left(y_{1},y^{\prime}\right)=\begin{cases}i_{\delta}V^{m,2}\left(y_{1},y^{\prime}\right)\quad&\mbox{for
}m\geq 4\\\ 0&\mbox{for }m=2,3\end{cases},$

where $V^{m,2}$ is a symmetric, smooth, compactly supported, $(m-4)$-tensor
field. Since we have $W^{m}=i_{\delta}\overline{W}^{m-2}$, therefore, by using
(3.34) we obtain

(3.35) $W^{m}=\begin{cases}i_{\delta}^{2}V^{m,2}\quad&\mbox{if }m\geq 4,\\\
0&\mbox{if }m=2,3,\end{cases}\quad\mbox{and}\quad
W^{m-1}=\begin{cases}i_{\delta}V^{m-1,1},\quad&\mbox{for }m\geq 3,\\\
0&\mbox{for }m=2.\end{cases}$

The rest of the proof follows from the mathematical induction given below.

Step 2. We assume, for $j\geq 1$,

(3.36) $W^{m-r}=\begin{cases}i_{\delta}^{j+1-r}V^{m-r,j+1-r},\quad&\mbox{for
}2j+2-m\leq r\leq j,\\\ 0&\mbox{for }0\leq r<2j+2-m,\end{cases}$

for some compactly supported, smooth, symmetric tensor fields $V^{m-r,j+1-r}$
of order $m+r-2j-2$. For $j=1$, the initial induction step (3.36) follows from
(3.35). We want to show that (3.36) holds for $j+1$, assuming it holds for
$j$. That is, assuming (3.36), we prove

(3.37) $W^{m-r}=\begin{cases}i_{\delta}^{j+2-r}V^{m-r,j+2-r},\quad&\mbox{for
}2j+4-m\leq r\leq j+1,\\\ 0&\mbox{for }0\leq r<2j+4-m.\end{cases}$

We multiply (3.22) by $h^{m-j-1}$ and assuming induction hypothesis (3.36)
along with (3.19), (3.20) we obtain

(3.38) $\displaystyle
0=\sum_{r=0}^{j+1}\sum_{i_{1},\cdots,i_{m-2j-2+r}=1}^{n}\int_{O}V^{m-r,j+1-r}_{i_{1}\dots
i_{m-2j-2+r}}\left(\prod_{k=1}^{m-2j-2+r}(e_{1}+\mathrm{i}\eta)_{i_{k}}\right)\left[(-T)^{j+1-r}a_{0}(y)\right]\,\overline{b_{0}(y)}\mathrm{~{}d}y,$

as $h\to 0$. Now we choose

$a_{0}(y)=1\mbox{ and
}\overline{b_{0}(y)}=(z-\bar{z})^{k}e^{-\mathrm{i}\lambda
z}g(y^{\prime\prime})\quad\text{ for all }0\leq k\leq m,$

where $g$ is an arbitrary smooth function in
$y^{\prime\prime}\in\mathbb{R}^{n-2}$ and $\lambda\in\mathbb{R}$. We notice
that

$T^{j+1-r}a_{0}(y)=0\quad\text{ for all }\quad 0\leq r\leq j.$

Therefore, we get,

$\displaystyle
0=\sum_{i_{1},\cdots,i_{m-j-1}=1}^{n}\int_{O}V^{m-j-1,0}_{i_{1}\dots
i_{m-j-1}}\left(\prod_{k=1}^{m-j-1}(e_{1}+\mathrm{i}\eta)_{i_{k}}\right)y_{2}^{k}e^{-\mathrm{i}\lambda
z}g(y^{\prime\prime})\mathrm{~{}d}y_{2},$

for all $0\leq k\leq m$ and $\lambda\in\mathbb{R}.$ Now, proceeding similarly
as we have done in step 1, we have

(3.39) $W^{m-j-1}=i_{\delta}V^{m-j-1,1}.$

Note that, using (3.39) in (3.38) we obtain

(3.40) $\displaystyle
0=\sum_{r=0}^{j}\sum_{i_{1},\cdots,i_{m-2j-2+r}=1}^{n}\int_{O}V^{m-r,j+1-r}_{i_{1}\dots
i_{m-2j-2+r}}\left(\prod_{k=1}^{m-2j-2+r}(e_{1}+\mathrm{i}\eta)_{i_{k}}\right)\left[(-T)^{j+1-r}a_{0}(y)\right]\,\overline{b_{0}(y)}\,\mathrm{~{}d}y.$

Now, we choose $a_{0}(y)=(z-\bar{z})\mbox{ and
}\overline{b_{0}(y)}=(z-\bar{z})^{k}e^{-\mathrm{i}\lambda
z}g(y^{\prime\prime}),\text{ for all }0\leq k\leq m$ where $g$ is an arbitrary
smooth function in $y^{\prime\prime}$ variables and $\lambda\in\mathbb{R}$.
Simplifying (3.40) we get,

$\displaystyle
0=\sum_{i_{1},\cdots,i_{m-j-2}=1}^{n}\int_{O}V^{m-j,1}_{i_{1}\dots
i_{m-j-2}}\left(\prod_{k=1}^{m-j-2}(e_{1}+\mathrm{i}\eta)_{i_{k}}\right)y_{2}^{k+1}e^{-\mathrm{i}\lambda
z}g(y^{\prime\prime})\,\mathrm{~{}d}y_{2}.$

Again, we use the same technique as in Step 1; to obtain

(3.41) $V^{m-j,1}=\begin{cases}i_{\delta}V^{m-j,2}\quad&\mbox{for }m-j\geq
4.\\\ 0&\mbox{for }m-j<4.\end{cases}$

Since, for $m-j\geq 4$ we have $W^{m-j}=i_{\delta}V^{m-j,1}$ in the induction
assumption, therefore, using (3.41) we get

$W^{m-j}=i_{\delta}^{2}V^{m-j,2},\quad\mbox{for }m-j\geq 4.$

Repeating this argument for $l$-times, for $0\leq l\leq j+1$ and using
amplitudes

$a_{0}(y)=(z-\overline{z})^{l}\quad\mbox{and}\quad\overline{b_{0}(y)}=(z-\bar{z})^{k}e^{-\mathrm{i}\lambda
z}g(y^{\prime\prime}).$

We obtain

$W^{m-j+l-1}=\begin{cases}i_{\delta}^{l+1}V^{m-j+l-1,l+1}&\mbox{ for}\quad
m-j-l\geq 3\\\ 0&\mbox{ for}\quad m-j-l<3\end{cases},\quad 0\leq l\leq
j+1,\quad j\geq 1.$

The proof is now complete by observing that, setting $j$ large and $0\leq
l\leq j+1$ suitably, we can make $m-j-l<3$ for any fixed $m\geq 2$. Hence, by
mathematical induction, we get all the coefficients $W^{m-k}$, $0\leq k\leq m$
are zero, that is,

$A^{m-k}=\widetilde{A}^{m-k},\quad\mbox{for }0\leq k\leq m.$

∎

## Acknowledgements

The authors want to express their gratitude to the Indian Institute of Science
Education and Research, Bhopal, and the Department of Mathematics at IISER-B
for their invaluable support. S.B. was partially supported by the grant:
SRG/2022/00129 from the Science and Engineering Research Board, India.

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# On the Rényi Cross-Entropy∗ ††thanks: *This work was supported in part by
NSERC of Canada.

Ferenc Cole Thierrin, Fady Alajaji, and Tamás Linder  Department of
Mathematics and Statistics
Queen’s University
Kingston, ON K7L 3N6, Canada
Emails: {14fngt, fa<EMAIL_ADDRESS>
###### Abstract

The Rényi cross-entropy measure between two distributions, a generalization of
the Shannon cross-entropy, was recently used as a loss function for the
improved design of deep learning generative adversarial networks. In this
work, we examine the properties of this measure and derive closed-form
expressions for it when one of the distributions is fixed and when both
distributions belong to the exponential family. We also analytically determine
a formula for the cross-entropy rate for stationary Gaussian processes and for
finite-alphabet Markov sources.

###### Index Terms:

Rényi information measures, cross-entropy, exponential family distributions,
Gaussian processes, Markov sources.

##  I Introduction

The Rényi entropy [1] of order $\alpha$ of a discrete distribution
(probability mass function) $p$ with finite support $\mathbb{S}$, defined as

$H_{\alpha}(p)=\frac{1}{1-\alpha}\ln\sum_{x\in\mathbb{S}}p(x)^{\alpha}$

for $\alpha>0,\alpha\neq 1$, is a generalization of the Shannon entropy,111For
ease of reference, a table summarising the Shannon entropy and cross-entropy
measures as well as the Kullback-Liebler (KL) divergence is provided in
Appendix A. $H(p)$, in that $\lim_{\alpha\to 1}H_{\alpha}(p)=H(p)$. Similarly,
the Rényi divergence (of order $\alpha$) between two discrete distributions
$p$ and $q$ with common finite support $\mathbb{S}$, given by

$D_{\alpha}(p||q)=\frac{1}{\alpha-1}\ln\sum_{x\in\mathbb{S}}p(x)^{\alpha}q(x)^{1-\alpha},$

reduces to the KL divergence, $D(p\|q)$, as $\alpha\to 1$.

Since the introduction of these measures, several other Rényi-type information
measures have been put forward, each obeying the condition that their limit as
$\alpha$ goes to one reduces to a Shannon-type information measure (e.g., see
[2] and the references therein for three different order $\alpha$ extensions
of Shannon’s mutual information due to Sibson, Arimoto and Csiszár.)

Many of these definitions admit natural counterparts in the (absolutely)
continuous case (i.e., when the involved distributions have a probability
density function (pdf)), giving rise to information measures such as the Rényi
differential entropy for pdf $p$ with support $\mathbb{S}$,

$h_{\alpha}(p)=\frac{1}{1-\alpha}\ln\int_{\mathbb{S}}p(x)^{\alpha}\,dx,$

and the Rényi (differential) divergence between pdfs $p$ and $q$ with common
support $\mathbb{S}$,

$D_{\alpha}(p||q)=\frac{1}{\alpha-1}\ln\int_{\mathbb{S}}p(x)^{\alpha}q(x)^{1-\alpha}\,dx.$

The Rényi cross-entropy between distributions $p$ and $q$ is an analogous
generalization of the Shannon cross-entropy $H(p;q)$. Two definitions for this
measure have been suggested. In [3], mirroring the fact that Shannon’s cross-
entropy satisfies $H(p;q)=D(p\|q)+H(p)$, the authors define Rényi cross-
entropy as

$\displaystyle\tilde{H}_{\alpha}(p;q):=D_{\alpha}(p||q)+H_{\alpha}(p).$ (1)

In contrast, prior to [3], the authors of [4] introduced the Rényi cross-
entropy in their study of the so-called shifted Rényi measures (expressed as
the logarithm of weighted generalized power means). Specifically, upon
simplifying Definition 6 in [4], their expression for the Rényi cross-entropy
between distributions $p$ and $q$ is given by

$H_{\alpha}\left(p;q\right):=\frac{1}{1-\alpha}\ln\sum_{x\in\mathbb{S}}p\left(x\right)q\left(x\right)^{\alpha-1}.$
(2)

For the continuous case, the definition in (2) can be readily converted to
yield the Rényi differential cross-entropy between pdfs $p$ and $q$:

$h_{\alpha}(p;q):=\frac{1}{1-\alpha}\ln\int_{\mathbb{S}}p(x)q(x)^{\alpha-1}\,dx.$
(3)

As the Rényi differential divergence and entropy were already calculated for
numerous distributions in [5] and [6], respectively, determining the Rényi
differential cross-entropy using the definition in (1) is straightforward. As
such, this paper’s focus is to establish closed-form expressions of the Rényi
differential cross-entropy as defined in (3) for various distributions, as
well as to derive the Rényi cross-entropy rate for two important classes of
sources with memory, Gaussian and Markov sources.

Motivation for determining formulae for the Rényi cross-entropy extends beyond
idle curiosity. The Shannon differential cross-entropy was used as a loss
function for the design of deep learning generative adversarial networks
(GANs) [7]. Recently, the Rényi differential cross-entropy measures in (3) and
(1), were used in [8, 9] and [3], respectively, to generalize the original GAN
loss function. It is shown that in [8] and [9] that the resulting Rényi-
centric generalized loss function preserves the equilibrium point satisfied by
the original GAN based on the Jensen-Rényi divergence [10], a natural
extension of the Jensen-Shannon divergence [11]. In [3], a different Rényi-
type generalized loss function is obtained and is shown to benefit from
stability properties. Improved stability and system performance are shown in
[8, 9] and [3] by virtue of the $\alpha$ parameter that can be judiciously
used to fine-tune the adopted generalized loss functions which recover the
original GAN loss function as $\alpha\to 1$.

The rest of this paper is organised as follows. In Section II, basic
properties of the Rényi cross-entropy are examined. In Section III, the Rényi
differential cross-entropy for members of the exponential family is
calculated. In Section IV, the Rényi differential cross-entropy between two
different distributions is obtained. In Section V, the Rényi differential
cross-entropy rate is derived for stationary Gaussian sources. Finally in
Section VI, the Rényi cross-entropy rate is established for finite-alphabet
time-invariant Markov sources.

##  II Basic Properties of the Rényi cross-entropy and differential cross-
entropy

For the Rényi cross-entropy $H_{\alpha}(p;q)$ to deserve its name it would be
preferable that it satisfies at least two key properties: it reduces to the
Rényi entropy when $p=q$ and its limit as $\alpha$ goes to one is the Shannon
cross-entropy. Similarly, it is desirable that the Rényi differential cross-
entropy $h_{\alpha}(p;q)$ reduces to the Rényi differential entropy when $p=q$
and its limit as $\alpha$ tends to one yields the Shannon differential cross-
entropy. In both cases, the former property is trivial, and the latter
property was proven in [9] for the continuous case under some finiteness
conditions (in the discrete case, the result holds directly via L’Hôpital’s
rule).

It is also proven in [9] that the Rényi differential cross-entropy
$h_{\alpha}(p;q)$ is non-increasing in $\alpha$ by showing that its derivative
with respect to $\alpha$ is non-positive. The same monotonicity property holds
in the disrcrete case.

Like its Shannon counterpart, the Rényi cross-entropy is non-negative
($H_{\alpha}(p;q)\geq 0$); while the Rényi differential cross-entropy can be
negative. This is easily verified when, for example, $\alpha=2$ and $p$ and
$q$ are both Gaussian (normal) distributions with zero mean and variance
$1/(8\sqrt{\pi})$, and parallels the same lack of non-negativity of the
Shannon differential cross-entropy.

We close this section by deriving the cross-entropy limit,
$\lim_{\alpha\to\infty}H_{\alpha}(p;q)$. To begin with, for any non-zero
constant $\tilde{c}$, we have

$\displaystyle\lim_{\alpha\to\infty}\frac{1}{1-\alpha}\ln\sum_{x\in\mathbb{S}}\tilde{c}q\left(x\right)^{\alpha-1}$
$\displaystyle=\lim_{\alpha\to\infty}\frac{1}{1-\alpha}\ln\tilde{c}+\lim_{\alpha\to\infty}\frac{1}{1-\alpha}\ln\sum_{x\in\mathbb{S}}q\left(x\right)^{\alpha-1}$
$\displaystyle=\lim_{\beta\to\infty}\frac{1-\beta}{-\beta}\frac{1}{1-\beta}\ln\sum_{\mathbb{S}}q\left(x\right)^{\beta}\qquad(\beta=\alpha-1)$
$\displaystyle=\lim_{\beta\to\infty}H_{\beta}(q)=-\ln q_{M},$ (4)

where $q_{M}:=\max_{x\in\mathbb{S}}q(x)$ and where we have used the fact that
for the Rényi entropy, $\lim_{\alpha\to\infty}H_{\alpha}(q)=-\ln q_{M}$. Now,
denoting the minimum and maximum values of $p(x)$ over $\mathbb{S}$ by $p_{m}$
and $p_{M}$, respectively, we have that for $\alpha>1$,

$\displaystyle\frac{1}{1-\alpha}\ln\sum_{x\in\mathbb{S}}p_{m}q\left(x\right)^{\alpha-1}$
$\displaystyle\leq\frac{1}{1-\alpha}\ln\sum_{x\in\mathbb{S}}p(x)q\left(x\right)^{\alpha-1}$
and
$\displaystyle\frac{1}{1-\alpha}\ln\sum_{x\in\mathbb{S}}p(x)q\left(x\right)^{\alpha-1}$
$\displaystyle\leq\frac{1}{1-\alpha}\ln\sum_{x\in\mathbb{S}}p_{M}q\left(x\right)^{\alpha-1},$

and hence by (4) we obtain

$\lim_{\alpha\to\infty}H_{\alpha}\left(p;q\right)=-\ln q_{M}.$ (5)

##  III Rényi Differential Cross-Entropy for Exponential Family Distributions

A probability distribution on $\mathbb{R}$ or $\mathbb{R}^{n}$ with parameter
$\theta$ is said to belong to the exponential family (e.g., see [12]) if on
its support $\mathbb{S}$ it admits a pdf of the form

$f(x)=c(\theta)b(x)\exp\left(\eta(\theta)\cdot T(x)\right),\qquad
x\in\mathbb{S},$ (6)

for some real-valued (measurable) functions $c$, $b$, $\eta$ and $T$.222Note
that $\theta$ and consequently $T(x)$ can be vectors in cases where the
distribution admits multiple parameters. Here $\eta$ is known as the natural
parameter of the distribution, $T(x)$ is the sufficient statistic and
$c(\theta)$ is the normalization constant in the sense that for all $\theta$
within the parameter space

$\int_{\mathbb{S}}b(x)\exp\left(\eta(\theta)\cdot
T(x)\right)dx=c(\theta)^{-1}.$

The pdf in (6) can also be written as

$\displaystyle{f(x)=b(x)\exp\left(\eta\cdot T(x)+A(\eta)\right)},$ (7)

where $A\left(\eta(\theta)\right)=\ln c(\theta)$. Examples of distributions in
the exponential family include the Gaussian, Beta, and exponential
distributions.

######  Lemma 1.

Let $f_{1}(x)$ and $f_{2}(x)$ be pdfs of the same type in the exponential
family with natural parameters $\eta_{1}$ and $\eta_{2}$, respectively. Define
$f_{h}(x)$ as being of the same type as $f_{1}$ and $f_{2}$ but with natural
parameter $\eta_{h}=\eta_{1}+(\alpha-1)\eta_{2}$. Then

$h_{\alpha}\left(f_{1};f_{2}\right)=\frac{A\left(\eta_{1}\right)-A\left(\eta_{h}\right)+\ln
E_{h}}{1-\alpha}-A\left(\eta_{2}\right),$ (8)

where $E_{h}=\mathbb{E}_{f_{h}}\left[b(X)^{\alpha-1}\right]=\int
b(x)^{\alpha-1}f_{h}(x)\,dx$

###### Proof.

Using (7), we have

$\displaystyle f_{1}$
$\displaystyle\left(x\right)f_{2}\left(x\right)^{\alpha-1}$
$\displaystyle=b\left(x\right)\exp\big{(}\eta_{1}\cdot
T\left(x\right)+A\left(\eta_{1}\right)\big{)}$ $\displaystyle\hskip
10.84006pt\cdot\bigg{(}b\left(x\right)\exp\big{(}\eta_{2}\cdot
T\left(x\right)+A\left(\eta_{2}\right)\big{)}\bigg{)}^{\alpha-1}$
$\displaystyle=b\left(x\right)^{\alpha}\exp\left((\eta_{1}+(\alpha-1)\eta_{2})\cdot
T\left(x\right)\right)$ $\displaystyle\hskip
10.84006pt\cdot\exp\left(A\left(\eta_{1}\right)+\left(\alpha-1\right)A\left(\eta_{2}\right)\right)$
$\displaystyle=b\left(x\right)^{\alpha}\exp\left(\eta_{h}\cdot
T\left(x\right)+A\left(\eta_{h}\right)\right)$ $\displaystyle\hskip
10.84006pt\cdot\exp\left(A\left(\eta_{1}\right)+\left(\alpha-1\right)A\left(\eta_{2}\right)-A\left(\eta_{h}\right)\right)$
$\displaystyle=b\left(x\right)^{\alpha-1}f_{h}\left(x\right)\exp\left(A\left(\eta_{1}\right)+\left(\alpha-1\right)A\left(\eta_{2}\right)-A\left(\eta_{h}\right)\right).$

Thus,

$\displaystyle\int_{\mathbb{S}}$ $\displaystyle
f_{1}\left(x\right)f_{2}\left(x\right)^{\alpha-1}dx$
$\displaystyle=\int_{\mathbb{S}}b\left(x\right)^{\alpha-1}f_{h}\left(x\right)dx$
$\displaystyle\hskip
10.84006pt\cdot\exp\left(A\left(\eta_{1}\right)+\left(\alpha-1\right)A\left(\eta_{2}\right)-A\left(\eta_{h}\right)\right)$
$\displaystyle=\exp\left(A\left(\eta_{1}\right)+\left(\alpha-1\right)A\left(\eta_{2}\right)-A\left(\eta_{h}\right)\right)E_{h},$

and therefore,

$h_{\alpha}\left(f_{1};f_{2}\right)=\frac{A\left(\eta_{1}\right)-A\left(\eta_{h}\right)+\ln
E_{h}}{1-\alpha}-A\left(\eta_{2}\right).$

∎

######  Remark.

If $b(x)=b$ is a constant for all $x\in\mathbb{S}$, then

$\frac{\ln E_{h}}{1-\alpha}=-\ln b.$

In many cases, we have that $b(x)=1$ on $\mathbb{S}$, and thus the $\frac{\ln
E_{h}}{1-\alpha}$ term disappears in (8).

Table I lists Rényi differential cross-entropy expressions we derived using
Lemma 1 for some common distributions in the exponential family (which we
describe in Appendix B for convenience). In the table, the subscript of $i$ is
used to denote that a parameter belongs to pdf $f_{i}$, $i=1,2$.

TABLE I: Rényi Differential Cross-Entropies for Common Continuous Distributions Name | $h_{\alpha}(f_{1};f_{2})$
---|---
Beta | $\displaystyle{\ln{B(a_{2},b_{2})}+\frac{1}{\alpha-1}\ln\frac{B(a_{h},b_{h})}{B(a_{1},b_{1})}}$
|  $a_{h}:=a_{1}+(\alpha-1)(a_{2}-1)$,
| $b_{h}:=b_{1}+(\alpha-1)(b_{2}-1)$
$\boldsymbol{\chi^{2}}$ | $\displaystyle{\frac{1}{1-\alpha}\left(\frac{\nu_{1}}{2}\ln\left(\alpha\right)-\ln\Gamma\left(\frac{\nu_{1}}{2}\right)+\ln\Gamma\left(\frac{\nu_{h}}{2}\right)\right)}$
| $+\displaystyle{\frac{2-\nu_{2}}{2}\ln\left(\alpha\right)+\ln
2\Gamma\left(\frac{\nu_{2}}{2}\right)}$
| $\nu_{h}:=\nu_{1}+(\alpha-1)(k-2)$
Exponential | $\displaystyle{\frac{1}{1-\alpha}\ln\frac{\lambda_{i}}{\lambda_{h}}-\ln\lambda_{2}}$
| $\lambda_{h}:=\lambda_{1}+(\alpha-1)\lambda_{2}$
Gamma | $\displaystyle{\ln\Gamma(k_{2})+k_{2}\ln\theta_{2}}$
|
$+\displaystyle{\frac{1}{1-\alpha}\left(\ln\frac{\Gamma(k_{h})}{\Gamma(k_{1})}-k_{h}\ln\theta_{h}-k_{1}\ln\theta_{1}\right)}$
|
$\theta_{\alpha}^{*}:=\frac{\theta_{1}+(a-1)\theta_{2}}{(\alpha-1)\theta_{1}\theta_{1}}$,
$k_{h}:=k_{i}+(\alpha-1)k_{2}$
Gaussian | $\displaystyle{\frac{1}{2}\left(\ln(2\pi\sigma_{2}^{2})+\frac{1}{1-\alpha}\ln\left(\frac{\sigma_{2}^{2}}{(\sigma^{2})_{h}^{*}}\right)+\frac{(\mu_{1}-\mu_{2}^{2}}{(\sigma^{2})_{h}^{*}}\right)}$
| $(\sigma^{2})_{h}^{*}:=\sigma_{2}^{2}+(\alpha-1)\sigma_{1}^{2}$
$\begin{array}[]{c}\textbf{Laplace}\\\ \end{array}$ | $\displaystyle{\ln(2b_{2})+\frac{1}{1-\alpha}\ln\left(\frac{b_{2}}{2b_{h}}\right)}$
($\mu_{1}=\mu_{2}$) | $\displaystyle{b_{h}:=b_{2}+(1-\alpha)b_{1}}$

##  IV Rényi differential Cross-Entropy between different distributions

Let $p$ and $q$ be pdfs with common support $\mathbb{S}\subseteq\mathbb{R}$.
Below are some general formulae for the differential Rényi cross-entropy
between one specific (common) distribution and any general distribution. If
$\mathbb{S}$ is an interval below, then $|\mathbb{S}|$ denotes its length.

###  IV-A Distribution $q$ is uniform

Let $q$ be uniformly distributed on $\mathbb{S}$. Then

$\displaystyle h_{\alpha}(p;q)$
$\displaystyle=\frac{1}{1-\alpha}\ln\int_{\mathbb{S}}p(x)q(x)^{\alpha-1}dx=\ln|\mathbb{S}|.$

###  IV-B Distribution $p$ is uniform

Now suppose $p$ is uniformly distributed on $\mathbb{S}$. Then

$\displaystyle h_{\alpha}(p;q)$
$\displaystyle=\frac{1}{1-\alpha}\ln\int_{\mathbb{S}}p(x)q(x)^{\alpha-1}dx$
$\displaystyle=\frac{1}{1-\alpha}\ln\frac{1}{|\mathbb{S}|}-h_{\alpha-1}(q).$

###  IV-C Distribution $q$ is exponentially distributed

Suppose the $\mathbb{S}=\mathbb{R}^{+}$ and $q$ is exponential with parameter
$\lambda$. Suppose also that the moment generating function (MGF) of $p$,
$M_{p}(t)$ exists. We have

$\displaystyle h_{\alpha}(p;q)$
$\displaystyle=\frac{1}{1-\alpha}\ln\int_{\mathbb{S}}p(x)q(x)^{\alpha-1}dx$
$\displaystyle=\frac{1}{1-\alpha}\ln\mathbb{E}_{p}\left[q(x)^{\alpha-1}\right]$
$\displaystyle=\frac{1}{1-\alpha}\ln\mathbb{E}_{p}\left[\left(\lambda\exp\left(-\lambda
x\right)\right)^{\alpha-1}\right]$
$\displaystyle=-\ln\lambda+\frac{1}{1-\alpha}\ln
M_{p}\left(\lambda(1-\alpha)\right).$

###  IV-D Distribution $q$ is Gaussian

Now assume that $q$ is a (normal) Gaussian $\mathcal{N}(\mu,\sigma^{2})$
distribution and that the MGF of $Y:=(X-\mu)^{2}$, $M_{Y}$, exists, where $X$
is a random variable with distribution $p$. Then

$\displaystyle h_{\alpha}(p;q)$
$\displaystyle=\frac{1}{1-\alpha}\ln\mathbb{E}_{p}\left[q(X)^{\alpha-1}\right]$
$\displaystyle=\frac{1}{1-\alpha}\ln\sigma(\sqrt{2\pi})^{1-\alpha}\mathbb{E}\left(\exp\left((1-\alpha)\frac{Y}{2\sigma^{2}}\right)\right)$
$\displaystyle=\ln\sigma\sqrt{2\pi}+\frac{1}{1-\alpha}\ln
M_{Y}\left(\frac{1-\alpha}{2\sigma^{2}}\right).$

The case where $q$ is a half-normal distribution can be directly derived from
the above. Given $q$ is a half-normal distribution, on its support its pdf is
the same as that of a normal $\mathcal{N}(0,\sigma^{2})$ distribution times 2.
Hence if $p$’s support is $\mathbb{R}^{+}$, then
$h_{\alpha}(p;q)=\ln\sigma\sqrt{\frac{\pi}{2}}+\frac{1}{1-\alpha}\ln
M_{Y}\left(\frac{1-\alpha}{2\sigma^{2}}\right)$.

##  V Rényi Differential Cross-Entropy Rate for Stationary Gaussian Processes

######  Lemma 2.

The Rényi differential cross-entropy between two zero-mean multivariate
dimension-$n$ Gaussian distributions with invertible covariance matrices
$\Sigma_{1}$ and $\Sigma_{2}$, respectively, is given by

$h_{\alpha}(p;q)=\frac{\ln|\Sigma_{1}||S|}{2\alpha-2}+\frac{1}{2}\ln|\Sigma_{2}|+\frac{n}{2}\ln
2\pi,$ (9)

where $S:=\Sigma_{1}^{-1}+(\alpha-1)\Sigma_{2}^{-1}$.

###### Proof.

Recall that the pdf of a multivariate Gaussian with mean
$\textbf{0}=(0,0...,0)^{T}$ and invertible covariance matrix $\Sigma$ is given
by:

$f(\textbf{x})=\frac{\exp(\frac{-1}{2}\textbf{x}^{T}\Sigma^{-1}\textbf{x})}{(2\pi)^{k/2}|\Sigma|^{1/2}}$

for $\textbf{x}\in\mathbb{R}^{n}$. Note that this distribution is a member of
the exponential family, where $T(\textbf{x})=\textbf{x}$,
$\eta=\frac{1}{2}\Sigma^{-1}$, $A(\eta)=\frac{1}{2}\ln|-2\eta|$ and
$b(\textbf{x})=(2\pi)^{\frac{-n}{2}}$. Hence the Rényi differential cross-
entropy between two zero-mean multivariate Gaussian distributions with
covariance matrices $\Sigma_{1}$ and $\Sigma_{2}$, respectively, is

$\displaystyle h_{\alpha}(p;q)$
$\displaystyle=\frac{1}{1-\alpha}\left(\frac{1}{2}\ln\left|2\frac{\Sigma_{1}^{-1}}{2}\right|\right.$
$\displaystyle\qquad\left.-\frac{1}{2}\ln\left|2\frac{\Sigma_{1}^{-1}+(\alpha-1)\Sigma_{2}^{-1}}{2}\right|\right)$
$\displaystyle\qquad-\frac{1}{2}\ln\left|2\frac{\Sigma_{2}^{-1}}{2}\right|-\ln(2\pi)^{\frac{-n}{2}}$
$\displaystyle=\frac{\ln|\Sigma_{1}||S|}{2\alpha-2}+\frac{1}{2}\ln|\Sigma_{2}|+\frac{n}{2}\ln
2\pi.$

∎

Let $\\{X_{j}\\}_{j=1}^{\infty}$ and $\\{Y_{j}\\}_{j=1}^{\infty}$ be
stationary zero-mean Gaussian processes. For a given $n$,
$X^{n}:=(X_{1},X_{2},...,X_{n})$ and $Y^{n}:=(Y_{1},Y_{2},...,Y_{n})$ are
multivariate Gaussian random variables with mean 0 and covariance matrices
$\Sigma_{X^{n}}$ and $\Sigma_{Y^{n}}$, respectively. Since $\\{X_{j}\\}$ and
$\\{Y_{j}\\}$ are stationary, their covariance matrices are Toeplitz.
Furthermore, $B^{n}:=\Sigma_{Y^{n}}+(\alpha-1)\Sigma_{X^{n}}$ is Toeplitz.

######  Lemma 3.

Let $\tilde{f}(\lambda)$, $\tilde{g}(\lambda)$ and $\tilde{h}(\lambda)$ be the
power spectral densities of $\\{X_{j}\\}$, $\\{Y_{j}\\}$ and the zero-mean
Gaussian process with covariance matrix $B^{n}$, respectively.

Then the Rényi differential cross-entropy rate between $\\{X_{j}\\}$ and
$\\{Y_{j}\\}$, $\lim_{n\to\infty}\frac{1}{n}h_{\alpha}(X^{n};Y^{n})$, is given
by

$\frac{\ln
2\pi}{2}+\frac{1}{4\pi(1-\alpha)}\int_{0}^{2\pi}\left[(2-\alpha)\ln\tilde{g}(\lambda)-\ln\tilde{h}(\lambda)\right]\,d\lambda.$

###### Proof.

From Lemma 2, we first note that
$S=\Sigma_{X^{n}}^{-1}B_{n}\Sigma_{Y^{n}}^{-1}$. With this in mind the Rényi
differential cross-entropy can be rewritten using (9) as

$\displaystyle\frac{1}{n}\bigg{(}\frac{\ln|\Sigma_{X^{n}}||\Sigma_{X^{n}}^{-1}B^{n}\Sigma_{Y^{n}}^{-1}|}{2(\alpha-1)}+\frac{1}{2}\ln|\Sigma_{Y^{n}}|+\frac{n}{2}\ln
2\pi\bigg{)}$ $\displaystyle=\frac{\ln
2\pi}{2}+\frac{1}{2n}\bigg{(}\frac{\ln|\Sigma_{X^{n}}||\Sigma_{X^{n}}^{-1}||B^{n}||\Sigma_{Y^{n}}^{-1}|}{(\alpha-1)}+\ln|\Sigma_{Y^{n}}|\bigg{)}$
$\displaystyle=\frac{\ln
2\pi}{2}+\frac{1}{2n}\bigg{(}\frac{\ln|B^{n}|-\ln|\Sigma_{Y^{n}}|}{(\alpha-1)}+\ln|\Sigma_{Y_{n}}|\bigg{)}$
$\displaystyle=\frac{\ln
2\pi}{2}+\frac{1}{2n(1-\alpha)}\left(\left(2-\alpha\right)\ln|\Sigma_{Y^{n}}|-\ln|B^{n}|\right).$

It was proven in [13] that for a sequence of Toeplitz matrices $T_{n}$ with
spectral density $t(\lambda)$ such that $\ln t(\lambda)$ is Reimann
integrable, one has

$\lim_{n\to\infty}\ln|T^{n}|=\frac{1}{2\pi}\int_{0}^{2\pi}\ln
t(\lambda)\,d\lambda.$

We therefore obtain that the Rényi differential cross-entropy rate is given by

$\displaystyle\frac{\ln
2\pi}{2}+\frac{1}{4\pi(1-\alpha)}\int_{0}^{2\pi}\left[(2-\alpha)\ln\tilde{g}(\lambda)-\ln\tilde{h}(\lambda)\right]\,d\lambda.$

Note that
$\tilde{h}(\lambda)=\tilde{g}(\lambda)+(\alpha-1)\tilde{f}(\lambda)$. ∎

##  VI Rényi cross-entropy rate for Markov sources

Consider two time-invariant Markov sources $\\{X_{j}\\}_{j=1}^{\infty}$ and
$\\{Y_{j}\\}_{j=1}^{\infty}$ with common finite alphabet $\mathbb{S}$ and with
transition distribution $P(\cdot|\cdot)$ and $Q(\cdot|\cdot)$, respectively.
Then for any $i^{n}=(i_{1},\ldots,i_{n})\in\mathbb{S}^{n}$, their
$n$-dimensional joint distributions are given by

$p^{(n)}(i^{n})=P(i_{n}|i_{n-1})P(i_{n-1}|i_{n-2})...P(i_{2}|i_{1})q(i_{1})$

and

$q^{(n)}(i^{n})=Q(i_{n}|i_{n-1})Q(i_{n-1}|i_{n-2})...Q(i_{2}|i_{1})p(i_{1}),$

respectively, with arbitrary initial distributions, $p(i_{1})$ and $q(i_{1})$,
$i_{1}\in\mathbb{S}$. For simplicity, we assume that $p(i)$, $q(i),Q(j|i)>0$
for all $i,j\in\mathbb{S}$. Define the Rényi cross-entropy rate between
$\\{X_{j}\\}$ and $\\{Y_{j}\\}$ as

$\displaystyle\lim_{n\to\infty}\frac{1}{n}H_{\alpha}(X^{n};Y^{n})$
$\displaystyle=$
$\displaystyle\lim_{n\to\infty}\frac{1}{n}\frac{1}{1-\alpha}\ln\left(\sum_{i^{n}\in\mathbb{S}^{n}}p^{(n)}(i^{n})q^{(n)}(i^{n})^{\alpha-1}\right).$

Note that by defining the matrix $R$ using the formula

$R_{ij}=P(j|i)Q(j|i)^{\alpha-1}$

and the row vector s as having components $s_{i}=p(i)q(i)^{\alpha-1}$, the
Rényi cross-entropy rate can be written as

$\lim_{n\to\infty}\frac{1}{n}\frac{1}{1-\alpha}\ln\textbf{s}R^{n-1}\textbf{1},$
(10)

where 1 is a column vector whose dimension is the cardinatliy of the alphabet
$\mathbb{S}$ and with all its entries equal to 1.

A result derived by [14] for the Rényi divergence between Markov sources can
thus be used to find the Rényi cross-entropy rate for Markov sources.

######  Lemma 4.

Let $P$, $Q$, s and R be defined as above. If $R$ is irreducible, then

$\lim_{n\to\infty}\frac{1}{n}H_{\alpha}(X^{n};Y^{n})=\frac{\ln\lambda}{1-\alpha},$
(11)

where $\lambda$ is the largest positive eigenvalue of $R$.

###### Proof.

Since the non-negative matrix $R$ is irreducible, by the Frobenius theorem
(e.g., cf. [15, 16]), it has a largest positive eigenvalue $\lambda$ with
associated positive eigenvector b. Let $b_{m}$ and $b_{M}$ be the minimum and
maximum elements, respectively, of b. Then due to the non-negativity of s,

$\lambda^{n-1}\textbf{s}\cdot\textbf{b}=\textbf{s}R^{n-1}\textbf{b}\leq\textbf{s}R^{n-1}\textbf{1}b_{M},$

where $\cdot$ denotes the Euclidean inner product. Similarly,
$\lambda^{n-1}\textbf{s}\cdot\textbf{b}\geq\textbf{s}R^{n-1}\textbf{1}b_{m}.$
As a result,

$\frac{1}{n}\ln\frac{\lambda^{n-1}\textbf{s}\cdot\textbf{b}}{b_{M}}\leq\frac{1}{n}\ln\textbf{s}R^{n-1}\textbf{1}\leq\frac{1}{n}\ln\frac{\lambda^{n-1}\textbf{s}\cdot\textbf{b}}{b_{m}}.$

Note that for all $n$, $\frac{\textbf{s}\cdot\textbf{b}}{b_{M}}$ is a
constant. Thus

$\displaystyle\lim_{n\to\infty}\frac{1}{n}\ln\frac{\lambda^{n-1}\textbf{s}\cdot\textbf{b}}{b_{M}}$
$\displaystyle=\lim_{n\to\infty}\frac{n-1}{n}\ln\lambda+\lim_{n\to\infty}\frac{1}{n}\ln\frac{\textbf{s}\cdot\textbf{b}}{b_{M}}$
$\displaystyle=\ln\lambda.$

Similarly, we have
$\lim_{n\to\infty}\frac{1}{n}\ln\frac{\lambda^{n-1}\textbf{s}\cdot\textbf{b}}{b_{m}}=\ln\lambda.$
Hence,

$\lim_{n\to\infty}\frac{1}{n}H_{\alpha}(X^{n};Y^{n})=\lim_{n\to\infty}\frac{1}{n}\ln\frac{\lambda^{n-1}\textbf{s}\cdot\textbf{b}}{(1-\alpha)b_{m}}=\frac{\ln\lambda}{1-\alpha}.$

∎

Another technique can be borrowed from [14] to generalize Lemma 4 to the case
where $R$ is reducible. First $R$ is rewritten in the canonical form detailed
in Proposition 1 of [14]. Let $\lambda_{k}$ be the largest positive eigenvalue
of each self-communicating sub-matrix of $R$, and let $\tilde{\lambda}$ be the
maximum of these $\lambda_{k}$’s. For each inessential class $C_{i}$, let
$\lambda_{j}$ be the largest positive eigenvalue of the sub-matrix of each
class $C_{j}$ that is reachable from $C_{i}$, and let $\lambda^{\dagger}$ be
the maximum of these $\lambda_{j}$’s. Define
$\lambda=\max\\{\tilde{\lambda},\lambda^{\dagger}\\}$. Then (11) holds.

## Appendix A: Shannon-type information measures

Name | Definition
---|---
$\begin{array}[]{c}\textbf{Shannon}\\\ \textbf{Entropy}\\\ \end{array}$ | $\displaystyle{H(p)=-\sum_{x\in\mathbb{S}}p(x)\ln p(x)}$
$\begin{array}[]{c}\textbf{Shannon}\\\ \textbf{Differential}\\\ \textbf{Entropy}\\\ \end{array}$ | $\displaystyle{h(p)=-\int_{\mathbb{S}}p(x)\ln p(x)\,dx}$
$\begin{array}[]{c}\textbf{Shannon}\\\ \textbf{Cross-Entropy}\\\ \end{array}$ | $\displaystyle{H(p;q)=-\sum_{x\in\mathbb{S}}p(x)\ln q(x)}$
$\begin{array}[]{c}\textbf{Shannon}\\\ \textbf{Differential}\\\ \textbf{Cross-Entropy}\\\ \end{array}$ | $\displaystyle{h(p;q)=-\int_{\mathbb{S}}p(x)\ln q(x)dx}$
$\begin{array}[]{c}\textbf{KL Divergence,}\\\ \textbf{(Discrete)}\\\ \end{array}$ | $\displaystyle{D(p\|q)=-\sum_{x\in\mathbb{S}}p(x)\ln\frac{p(x)}{q(x)}}$
$\begin{array}[]{c}\textbf{KL Divergence,}\\\ \textbf{(Continuous)}\\\ \end{array}$ | $\displaystyle{D(p\|q)=-\int_{\mathbb{S}}p(x)\ln\frac{p(x)}{q(x)}\,dx}$

## Appendix B: Distributions listed in Table I

Name |  PDF $f(x)$
---|---
(Parameters) | (Support)
Beta | $\displaystyle{{B(a,b)}x^{a-1}(1-x)^{b-1}}$
($a>0$, $b>0$) | $\mathbb{S}=(0,1)$
$\boldsymbol{\chi^{2}}$ | $\displaystyle{\dfrac{1}{2^{\frac{\nu}{2}}\Gamma\left(\frac{\nu}{2}\right)}x^{\frac{\nu}{2}-1}e^{-\frac{x}{2}}}$
($\nu\in\mathbb{Z}^{+}$) | $\mathbb{S}=\mathbb{R^{+}}$
Exponential | $\displaystyle{\lambda e^{-\lambda x}}$
($\lambda>0$) | $\mathbb{S}=\mathbb{R^{+}}$
Gamma | $\displaystyle{\dfrac{1}{\theta^{k}\Gamma\left(k\right)}x^{k-1}e^{-\frac{k}{\theta}}}$
($k>0$, $\theta>0$) | $\mathbb{S}=\mathbb{R^{+}}$
Gaussian | $\displaystyle{\frac{1}{\sqrt{2\pi\sigma^{2}}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}}}$
($\mu$, $\sigma^{2}>0$) | $\mathbb{S}=\mathbb{R}$
Laplace | $\displaystyle{\frac{1}{2b}e^{-\frac{|x-\mu|}{b}}}$
($\mu$, $b^{2}>0$) | $\mathbb{S}=\mathbb{R}$

Notes

  * •

${\displaystyle\mathrm{B}(a,b)=\int_{0}^{1}t^{a-1}(1-t)^{b-1}\,dt}$ is the
Beta function.

  * •

${\displaystyle\Gamma(z)=\int_{0}^{\infty}x^{z-1}e^{-x}\,dx}$ is the Gamma
function.

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