# Generalized T-product Tensor Bernstein Bounds

Shih Yu Chang Yimin Wei Shih Yu Chang is with the Department of Applied Data
Science, San Jose State University, San Jose, CA, U. S. A. (e-mail:
shihyu.chang@sjsu.edu). 1This work was supported by the National Natural
Science Foundation of China (No. 11771099) and Innovation Program of Shanghai
Municipal Education Commission.Yimin Wei is with the School of Mathematical
Sciences, Shanghai Key Laboratory of Contemporary Applied Mathematics, Fudan
University, Shanghai, 200433, PR China(e-mail: ymwei@fudan.edu.cn).

###### Abstract

Since Kilmer et al. introduced the new multiplication method between two
third-order tensors around 2008 and third-order tensors with such
multiplication structure are also called as T-product tensors, T-product
tensors have been applied to many fields in science and engineering, such as
low-rank tensor approximation, signal processing, image feature extraction,
machine learning, computer vision, and the multi-view clustering problem, etc.
However, there are very few works dedicated to exploring the behavior of
random T-product tensors. This work considers the problem about the tail
behavior of the unitarily invariant norm for the summation of random symmetric
T-product tensors. Majorization and antisymmetric Kronecker product tools are
main techniques utilized to establish inequalities for unitarily norms of
multivariate T-product tensors. The Laplace transform method is integrated
with these inequalities for unitarily norms of multivariate T-product tensors
to provide us with Bernstein Bounds estimation of Ky Fan $k$-norm for
functions of the symmetric random T-product tensors summation.

Index terms— T-product tensors, T-eigenvalues, T-singular values, Bernstein
bound, Courant-Fischer theorem for T-product tensors.

##  1 Introduction

Since Kilmer et al. introduced the new multiplication method between two
third-order tensors (T-product tensors), many new algebraic properties about
such new multiplication rule between two third-order tensors are investigated
recently [1, 2]. For example, the singular value decomposition (SVD) for
third-order tensors via the tensor T-product is proposed in [3]. Some authors
suggest a new framework by treating third-order tensors as linear operators on
a space of matrices, see [4]. In [5], many useful tools of linear algebra are
extended to the third-order tensors, including the T-Jordan canonical form,
tensor decomposition theory, T-group inverse and T-Drazin inverse, and so on.
Moreover, the authors in [6] proposed a definition of tensor functions based
on the T-product of third-order F-square tensors, and Miao, Qi and Wei
generalized the tensor T-function from F-square third-order tensors to
rectangular tensors in [7]. These useful algebraic properties of T-product
tensors have been discovered as powerful tools in many science and engineering
fields: signal processing [8, 9], machine learning [10], computer vision [11,
12], image processing [13], low-rank tensor approximation [14, 15, 16], etc.

Although T-product tensors have attracted many practical applications, all of
these applications of T-product tensors assume that T-product tensors under
consideration are deterministic. This assumption is not practical in general
scientific and engineering applications based on T-product tensors. In [17,
18], the authors have tried to establish several new tail bounds for sums of
random T-product tensors. These probability bounds characterize large-
deviation behavior of the extreme T-eigenvalue of the sums of random T-product
tensors 111Definitions about T-eigenvalues and T-singular values associated to
T-product tensors are given in Section 2.1.. The authors first apply Lapalace
transform method and Lieb’s concavity theorem for T-product tensors obtained
from the work [17] to build several inequalities based on random T-product
tensors, then utilize these inequalities to generalize the classical bounds
associated with the names Chernoff, and Bernstein from the scalar to the
T-product tensor setting. Tail bounds for the norm of a sum of random
rectangular T-product tensors are also derived from corollaries of random
symmetric T-product tensors cases. The proof mechanism is also applied to
T-product tensor valued martingales and T-product tensor-based Azuma,
Hoeffding and McDiarmid inequalities are also derived [18].

In this work, we will apply majorization techniques to establish new Bernstein
bounds based on the summation of random symmetric T-product tensors. Compared
to the previous work studied in [17, 18], we make following generalizations:
(1) besides bounds related to extreme values of T-eigenvalues, we consider
more general unitarily invariant norm for T-product tensors; (2) the bounds
derived in [18] can only be applied to the identity map for the summation of
random symmetric T-product tensors, this work can derive new bounds for any
polynomial function raised by any power greater or equal than one for the
summation of random symmetric T-product tensors. In order to drive these new
bounds, we also establish Courant-Fischer min-max theorem for T-product
tensors in Theorem 3 and marjoization relation for T-singular values in Lemma
9. Our main theorem is provided below:

######  Theorem 1.1 (Generalized T-product Tensor Bernstein Bound)

Consider a sequence $\\{\mathcal{X}_{j}\in\mathbb{R}^{m\times m\times p}\\}$
of independent, random symmetric T-product tensors with random structure
defined by Definition 1. Let $g$ be a polynomial function with degree $n$ and
nonnegative coefficients $a_{0},a_{1},\cdots,a_{n}$ raised by power $s\geq 1$,
i.e., $g(x)=\left(a_{0}+a_{1}x+\cdots+a_{n}x^{n}\right)^{s}$ with $s\geq 1$.
Suppose following condition is satisfied:

$\displaystyle
g\left(\exp\left(t\sum\limits_{j=1}^{m}\mathcal{X}_{j}\right)\right)\succeq\exp\left(tg\left(\sum\limits_{j=1}^{m}\mathcal{X}_{j}\right)\right)\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{almost surely},$ (1)

where $t>0$, and we also have

$\displaystyle\mathcal{X}^{p}_{j}\preceq\frac{p!\mathcal{A}^{2}}{2}\mbox{\leavevmode\nobreak\
\leavevmode\nobreak\ almost surely for $p=2,3,4,\cdots$.}$ (2)

Then we have following inequality:

$\displaystyle\mathrm{Pr}\left(\left\|g\left(\sum\limits_{j=1}^{m}\mathcal{X}_{j}\right)\right\|_{(k)}\geq\theta\right)$
$\displaystyle\leq$ $\displaystyle(n+1)^{s-1}\inf\limits_{t>0}e^{-\theta
t}k\cdot$ (3)
$\displaystyle\left\\{a_{0}^{s}+\sum\limits_{l=1}^{n}a_{l}^{ls}\left[1+mlst\Phi(m,d_{1},d_{2})+\frac{(mlst)^{2}\sigma_{1}(\mathcal{A}^{2})}{2(1-mlst)}\right]\right\\}.$

The rest of this paper is organized as follows. In Section 2 , we review
T-product tensors basic concepts and introduce a powerful scheme about
antisymmetric Kronecker product for T-product tensors. In Section 3, we apply
a majorization technique to prove T-product tensor norm inequalities. We then
apply new derived T-product tensor norm inequalities to obtain random
T-product tensor Bernstein bounds for the extreme T-eigenvalues and Ky Fan
$k$-norm in Section 4. Finally, concluding remarks are given by Section 5.

##  2 T-product Tensors

In this section, we will introduce fundamental facts about T-product tensors
in Section 2.1. Several unitarily invariant norms about a T-product tensor are
defined in Section 2.2. A powerful scheme about antisymmetric Kronecker
product for T-product tensors will be provided by Section 2.3.

###  2.1 T-product Tensor Fundamental Facts

For a third order tensor $\mathcal{C}\in\mathbb{R}^{m\times n\times p}$, we
define bcirc operation to the tensor $\mathcal{C}$ as:

$\displaystyle\mbox{bcirc}(\mathcal{C})\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\left[\begin{array}[]{ccccc}\mathbf{C}^{(1)}&\mathbf{C}^{(p)}&\mathbf{C}^{(p-1)}&\cdots&\mathbf{C}^{(2)}\\\
\mathbf{C}^{(2)}&\mathbf{C}^{(1)}&\mathbf{C}^{(p)}&\cdots&\mathbf{C}^{(3)}\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\
\mathbf{C}^{(p)}&\mathbf{C}^{(p-1)}&\mathbf{C}^{(p-2)}&\cdots&\mathbf{C}^{(1)}\\\
\end{array}\right],$ (8)

where $\mathbf{C}^{(1)},\cdots,\mathbf{C}^{(p)}\in\mathbb{C}^{m\times n}$ are
frontal slices of tensor $\mathcal{C}$. The inverse operation of bcirc is
denoted as $\mbox{bcirc}^{-1}$ with relation
$\mbox{bcirc}^{-1}(\mbox{bcirc}(\mathcal{C}))\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\mathcal{C}$. Another operation to the tensor $\mathcal{C}$ is
unfolding, denoted as $\mbox{unfold}(\mathcal{C})$, which is defined as:

$\displaystyle\mbox{unfold}(\mathcal{C})\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\left[\begin{array}[]{c}\mathbf{C}^{(1)}\\\ \mathbf{C}^{(2)}\\\
\vdots\\\ \mathbf{C}^{(p)}\\\ \end{array}\right].$ (13)

The inverse operation of unfold is denoted as fold with relation
$\mbox{fold}(\mbox{unfold}(\mathcal{C}))\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\mathcal{C}$.

The multiplication between two third order tensors,
$\mathcal{C}\mathbb{R}^{m\times n\times p}$ and
$\mathcal{D}\mathbb{R}^{n\times l\times p}$, is via _T-product_ and this
multiplication is defined as:

$\displaystyle\mathcal{C}\star\mathcal{D}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\mbox{fold}\left(\mbox{bcirc}(\mathcal{C})\cdot\mbox{unfold}(\mathcal{C})\right),$
(14)

where $\cdot$ is the standard matrix multiplication. For given third order
tensors, if we apply _T-product_ to multiply them, we call them _T-product
tensors_. A T-product tensor $\mathcal{C}\in\mathbb{R}^{m\times n\times p}$
will be named as square T-product tensor if $m=n$.

For a symmetric T-product tensor $\mathcal{C}\in\mathbb{R}^{m\times m\times
p}$, we define Hermitian transpose of $\mathcal{C}$, denoted by
$\mathcal{C}^{\mathrm{H}}$ , as

$\displaystyle\mathcal{C}^{\mathrm{H}}=\mbox{bcirc}^{-1}((\mbox{bcirc}(\mathcal{C}))^{\mathrm{H}}).$
(15)

And a tensor $\mathcal{D}\in\mathbb{C}^{m\times m\times p}$ is called a
Hermitian T-product tensor if $\mathcal{D}^{\mathrm{H}}=\mathcal{D}$.
Similarly, for a symmetric T-product tensor $\mathcal{C}\in\mathbb{R}^{m\times
m\times p}$, we define _transpose_ of $\mathcal{C}$, denoted by
$\mathcal{C}^{T}$ , as

$\displaystyle\mathcal{C}^{T}=\mbox{bcirc}^{-1}((\mbox{bcirc}(\mathcal{C}))^{T}).$
(16)

And a tensor $\mathcal{D}\in\mathbb{R}^{m\times m\times p}$ is called a
symmetric T-product tensor if $\mathcal{D}^{T}=\mathcal{D}$.

The identity tensor $\mathcal{I}_{m,m,p}\in\mathbb{R}^{m\times m\times p}$ can
be defined as:

$\displaystyle\mathcal{I}_{m,m,p}=\mbox{bcirc}^{-1}(\mathbf{I}_{mp}),$ (17)

where $\mathbf{I}_{mp}$ is the identity matrix in $\mathbb{R}^{mp\times mp}$.
For a square T-product tensor, $\mathcal{C}\in\mathbb{R}^{m\times m\times p}$,
we say that $\mathcal{C}$ is nonsingular if it has an inverse tensor
$\mathcal{D}\in\mathbb{R}^{m\times m\times p}$ such that

$\displaystyle\mathcal{C}\star\mathcal{D}=\mathcal{D}\star\mathcal{C}=\mathcal{I}_{m,m,p}.$
(18)

A zero tensor, denoted as $\mathcal{O}_{mnp}\in\mathbb{C}^{m\times n\times
p}$, is a tensor that all elements inside the tensor as $0$.

For any circular matrix $\mathbf{C}\in\mathbb{R}^{m\times m}$, it can be
diagonalized with the normalized Discrete Fourier Transform (DFT) marix, i.e.,
$\mathbf{C}=\mathbf{F}^{\mathrm{H}}_{m}\mathbf{D}\mathbf{F}_{m}$, where
$\mathbf{F}_{m}$ is the Fourier matrix of size $m\times m$ defined as

$\displaystyle\mathbf{F}_{m}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\frac{1}{\sqrt{m}}\left[\begin{array}[]{ccccc}1&1&1&\cdots&1\\\
1&\omega&\omega^{2}&\cdots&\omega^{(m-1)}\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\
1&\omega^{(m-1)}&\omega^{2(m-1)}&\cdots&\omega^{(m-1)(m-1)}\\\
\end{array}\right],$ (23)

where $\omega=\exp(\frac{2\pi\iota}{m})$ with $\iota^{2}=-1$. This DFT matrix
can also be used to diagonalize a T-product tensor as [2]

$\displaystyle\mbox{bcirc}(\mathcal{C})=\left(\mathbf{F}^{\mathrm{H}}_{m}\otimes\mathbf{I}_{m}\right)\mbox{Diag}\left(\mathbf{C}_{i}:i\in\\{1,\cdots,m\\}\right)\left(\mathbf{F}_{m}\otimes\mathbf{I}_{m}\right),$
(24)

where $\otimes$ is the Kronecker Product and
$\mbox{Diag}\left(\mathbf{C}_{i}:i\in\\{1,\cdots,m\\}\right)\in\mathbb{C}^{mp\times
mp}$ is a diagonal block matrix with the $i$-th diagonal block as the matrix
$\mathbf{A}_{i}$.

The inner product between two T-product tensors
$\mathcal{C}\in\mathbb{C}^{m\times n\times p}$ and
$\mathcal{D}\in\mathbb{C}^{m\times n\times p}$ is defined as:

$\displaystyle\langle\mathcal{C},\mathcal{D}\rangle=\sum\limits_{i,j,k}c^{\ast}_{i,j,k}d_{i,j,k},$
(25)

where $\ast$ is the complex conjugate operation.

We say that a symmetric T-product tensor $\mathcal{C}\in\mathbb{R}^{m\times
m\times p}$ is a T-positive definite (TPD) tensor if we have

$\displaystyle\langle\mathcal{X},\mathcal{C}\star\mathcal{X}\rangle>0,$ (26)

holds for any non-zero T-product tensor $\mathcal{X}\in\mathbb{R}^{m\times
1\times p}$. Also, we said that a symmetric T-product tensor is a T-positive
semidefinite (TPSD) tensor if we have

$\displaystyle\langle\mathcal{X},\mathcal{C}\mathcal{X}\rangle\geq 0,$ (27)

holds for any non-zero T-product tensor $\mathcal{X}\in\mathbb{R}^{m\times
1\times p}$. Given two T-product tensors $\mathcal{C},\mathcal{D}$, we use
$\mathcal{C}\succ(\succeq)\mathcal{D}$ if
$\left(\mathcal{C}-\mathcal{D}\right)$ is a TPSD (TPD) T-product tensor.

We have the following theorem from Theorem 5 in [19].

###### Theorem 1

If a T-product tensor $\mathcal{C}\in\mathbb{R}^{m\times m\times p}$ can be
diagonalized as

$\displaystyle\mbox{bcirc}(\mathcal{C})=\left(\mathbf{F}^{\mathrm{H}}_{m}\otimes\mathbf{I}_{m}\right)\mbox{Diag}\left(\mathbf{C}_{i}:i\in\\{1,\cdots,m\\}\right)\left(\mathbf{F}_{m}\otimes\mathbf{I}_{m}\right),$
(28)

where $\mathbf{F}$ is the DFT matrix defined by Eq. (23); then $\mathcal{C}$
is symmetric, TPD (TPSD) if and only if all matrices $\mathbf{C}_{i}$ are
Hermitian, positive definite (positive semidefinite).

Let $\mathcal{C}\in\mathbb{R}^{m\times m\times p}$ can be block diagonalized
as Eq. (28). Then, a real number $\lambda$ is said to be a _T-eigenvalue_ of
$\mathcal{C}$, denoted as $\lambda(\mathcal{C})$, if it is an eigenvalue of
some $\mathbf{C}_{i}$ for $i\in\\{1,\cdots,m\\}$. The largest and smallest
T-eigenvalue of $\mathcal{C}$ are represented by $\lambda_{\max}(\mathcal{C})$
and $\lambda_{\min}(\mathcal{C})$, respectively. We use $\lambda_{i,j}$ for
the $j$-th largest T-eigenvalue of the matrix $\mathbf{C}_{i}$. We also use
$\sigma_{i,j}$, named as _T-singular values_ , for the $j$-th largest singular
values of the matrix $\mathbf{C}_{i}$.

We define the T-product tensor _trace_ for a tensor
$\mathcal{C}=(c_{ijk})\in\mathbb{C}^{m\times m\times p}$, denoted by
$\mathrm{Tr}(\mathcal{C})$, as following

$\displaystyle\mathrm{Tr}(\mathcal{C})\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\sum\limits_{i=1}^{m}\sum\limits_{k=1}^{p}c_{iik},$ (29)

which is the summation of all entries in f-diagonal components. Then, we have
the following lemma about trace properties.

###### Lemma 1

For any tensors $\mathcal{C},\mathcal{D}\in\mathbb{C}^{m\times m\times p}$, we
have

$\displaystyle\mathrm{Tr}(c\mathcal{C}+d\mathcal{D})=c\mathrm{Tr}(\mathcal{C})+d\mathrm{Tr}(\mathcal{D}),$
(30)

where $c,d$ are two contants. And, the transpose operation will keep the same
trace value, i.e.,

$\displaystyle\mathrm{Tr}(\mathcal{C})=\mathrm{Tr}(\mathcal{C}^{T}).$ (31)

Finally, we have

$\displaystyle\mathrm{Tr}(\mathcal{C}\star\mathcal{D})=\mathrm{Tr}(\mathcal{D}\star\mathcal{C}).$
(32)

Proof: Eqs. (30) and (31) are true from trace definition directly.

From T-product definition, the $i$-th frontal slice matrix of
$\mathcal{D}\star\mathcal{C}$ is

$\displaystyle\mathbf{D}^{(i)}\mathbf{C}^{(1)}+\mathbf{D}^{(i-1)}\mathbf{C}^{(2)}+\cdots+\mathbf{D}^{(1)}\mathbf{C}^{(i)}+\mathbf{D}^{(m)}\mathbf{C}^{(i+1)}+\cdots+\mathbf{D}^{(i+1)}\mathbf{C}^{(m)},$
(33)

similarly, the $i$-th frontal slice matrix of $\mathcal{C}\star\mathcal{D}$ is

$\displaystyle\mathbf{C}^{(i)}\mathbf{D}^{(1)}+\mathbf{C}^{(i-1)}\mathbf{D}^{(2)}+\cdots+\mathbf{C}^{(1)}\mathbf{D}^{(i)}+\mathbf{C}^{(m)}\mathbf{D}^{(i+1)}+\cdots+\mathbf{C}^{(i+1)}\mathbf{D}^{(m)}.$
(34)

Because the matrix trace of Eq. (33) and the matrix trace of Eq. (34) are same
for each slice $i$ due to linearity and invariant under cyclic permutations of
matrix trace, we have Eq. (32) by summing over all frontal matrix slices.
$\hfill\Box$

Below, we will define the _determinant_ of a T-product tensor
$\mathcal{C}\in\mathbb{R}^{m\times m\times p}$, represented by
$\det(\mathcal{C})$, as

$\displaystyle\det(\mathcal{C})$ $\displaystyle=$
$\displaystyle\prod\limits_{i=1,j=1}^{i=m,j=p}\lambda_{i,j}.$ (35)

We have the following theorem from Theorem 6 in [19] about symmetric T-product
tensor decomposition.

###### Theorem 2

Every symmetric T-product tensor $\mathcal{C}\in\mathbb{R}^{m\times m\times
p}$ can be factored as

$\displaystyle\mathcal{C}=\mathcal{U}^{T}\star\mathcal{D}\star\mathcal{U},$
(36)

where $\mathcal{U}$ is an orthogonal tensor, i.e.,
$\mathcal{U}^{T}\star\mathcal{U}=\mathcal{I}_{m,m,p}$, and $\mathcal{D}$ is a
F-diagonal tensor, i.e., each frontal slice of $\mathcal{D}$ is a diagonal
matrix, such that diagonal entries of
$\left(\mathbf{F}_{m}\otimes\mathbf{I}_{m}\right)\mbox{bcirc}\left(\mathcal{D}\right)\left(\mathbf{F}^{\mathrm{H}}_{m}\otimes\mathbf{I}_{m}\right)$
are T-eigenvalues of $\mathcal{C}$. If $\mathcal{C}$ is a TPD (TPSD) tensor,
then all of its T-eigenvalues are positive (nonnegative).

From Theorem 2 and Lemma 1, we have the fact that

$\displaystyle\mathrm{Tr}(\mathcal{C})=\sum\limits_{i}\lambda_{i}(\mathcal{C}).$
(37)

If a symmetric T-product tensor $\mathcal{C}\in\mathbb{R}^{m\times m\times p}$
can be expressed as the format shown by Eq. (28), the T-eigenvalues of
$\mathcal{C}$ with respect to the matrix $\mathbf{C}_{i}$ are denoted as
$\lambda_{i,k_{i}}$, where $1\leq k_{i}\leq m$, and we assume that
$\lambda_{i,1}\geq\lambda_{i,2}\geq\cdots\geq\lambda_{i,m}$ (including
multiplicities). Then, $\lambda_{i,k_{i}}$ is the $k_{i}$-th largest
T-eigenvalue associated to the matrix $\mathbf{C}_{i}$. If we sort all
T-eigenvalues of $\mathcal{C}$ from the largest one to the smallest one, we
use $\tilde{k}$, a smallest integer between 1 to $m\times p$ (inclusive)
associated with $p$ given positive integers $k_{1},k_{2},\cdots,k_{p}$ that
satisfies

$\displaystyle\lambda_{\tilde{k}}=\min\limits_{1\leq i\leq
m}\lambda_{i,k_{i}},\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\
\lambda_{\tilde{k}}\geq\lambda_{i,k_{i}+1},$ (38)

and we set $\tilde{i}$ from $\lambda_{\tilde{k}}$ as

$\displaystyle\tilde{i}=\arg\min\limits_{i}\left\\{\lambda_{\tilde{k}}=\lambda_{i,k_{i}}\right\\}.$
(39)

Then, we will have the following Courant-Fischer theorem for T-product
tensors.

###### Theorem 3

Given a symmetric T-product tensor $\mathcal{C}\in\mathbb{R}^{m\times m\times
p}$ and $p$ positive integers $k_{1},k_{2},\cdots,k_{p}$ with $1\leq k_{i}\leq
m$, then we have

$\displaystyle\lambda_{\tilde{k}}$ $\displaystyle=$
$\displaystyle\max\limits_{\begin{subarray}{c}S\in\mathbb{R}^{m\times 1\times
p}\\\
\dim(\mathrm{S})=\\{k_{1},\cdots,k_{p}\\}\end{subarray}}\min\limits_{\mathcal{X}\in
S}\frac{\langle\mathcal{X},\mathcal{C}\star\mathcal{X}\rangle}{\langle\mathcal{X},\mathcal{X}\rangle}$
(40) $\displaystyle=$
$\displaystyle\min\limits_{\begin{subarray}{c}T\in\mathbb{R}^{m\times 1\times
p}\\\
\dim(T)=\\{n-k_{1},\cdots,n-k_{\tilde{i}-1},n-k_{\tilde{i}}+1,n-k_{\tilde{i}+1},\cdots,n-k_{p}\\}\end{subarray}}\max\limits_{\mathcal{X}\in
T}\frac{\langle\mathcal{X},\mathcal{C}\star\mathcal{X}\rangle}{\langle\mathcal{X},\mathcal{X}\rangle}$

where $\lambda_{\tilde{k}}$ and $\tilde{i}$ are defined by Eqs. (38) and (39).

Proof:

First, we have to express
$\langle\mathcal{X},\mathcal{C}\star\mathcal{X}\rangle$ by matrices of
$\mathbf{C}_{i}$ and $\mathbf{X}_{i}$ through the representation shown by Eq.
(28). It is

$\displaystyle\langle\mathcal{X},\mathcal{C}\star\mathcal{X}\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{p}\langle\mbox{bcirc}(\mathcal{X}),\mbox{bcirc}(\mathcal{C})\mbox{bcirc}(\mathcal{X})\rangle$
(41) $\displaystyle=$
$\displaystyle\frac{1}{p}\mathrm{Tr}\left(\mbox{bcirc}(\mathcal{X})^{\mathrm{H}}\mbox{bcirc}(\mathcal{C})\mbox{bcirc}(\mathcal{X})\right)$
$\displaystyle=$
$\displaystyle\frac{1}{p}\mathrm{Tr}\left(\mathbf{F}^{\mathrm{H}}_{p}\mbox{Diag}\left(\mathbf{x}^{\mathrm{H}}_{i}\mathbf{A}_{i}\mathbf{x}_{i}:i\in\\{1,\cdots,p\\}\right)\mathbf{F}_{p}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{p}\mathrm{Tr}\left(\mbox{Diag}\left(\mathbf{x}^{\mathrm{H}}_{i}\mathbf{A}_{i}\mathbf{x}_{i}:i\in\\{1,\cdots,p\\}\right)\right)=\frac{1}{p}\sum\limits_{i=1}^{p}\mathbf{x}^{\mathrm{H}}_{i}\mathbf{A}_{i}\mathbf{x}_{i}$

We will just verify the first characterization of $\lambda_{\tilde{k}}$. The
other is similar. Let $S_{i}$ be the projection of $S$ to the space with
dimension $k_{i}$ spanned by $\mathbf{v}_{i,1},\cdots,\mathbf{v}_{i,k_{i}}$,
for every $\mathbf{x}_{i}\in S_{i}$, we can write
$\mathbf{x}_{i}=\sum\limits^{k_{i}}_{j=1}c_{i,j}\mathbf{v}_{i,j}$. To show
that the value $\lambda_{\tilde{k}}$ is achievable, note that

$\displaystyle\frac{\langle\mathcal{X},\mathcal{C}\star\mathcal{X}\rangle}{\langle\mathcal{X},\mathcal{X}\rangle}$
$\displaystyle=$
$\displaystyle\frac{\frac{1}{p}\sum\limits_{i=1}^{p}\mathbf{x}^{\mathrm{H}}_{i}\mathbf{A}_{i}\mathbf{x}_{i}}{\frac{1}{p}\sum\limits_{i=1}^{p}\mathbf{x}^{\mathrm{H}}_{i}\mathbf{x}_{i}}=\frac{\sum\limits_{i=1}^{p}\sum\limits^{k_{i}}_{j=1}\lambda_{i,j}c_{i,j}^{\ast}c_{i,j}}{\sum\limits_{i=1}^{p}\sum\limits^{k_{i}}_{j=1}c_{i,j}^{\ast}c_{i,j}}$
(42) $\displaystyle\geq$
$\displaystyle\frac{\sum\limits_{i=1}^{p}\sum\limits^{k_{i}}_{j=1}\lambda_{\tilde{k}}c_{i,j}^{\ast}c_{i,j}}{\sum\limits_{i=1}^{p}\sum\limits^{k_{i}}_{j=1}c_{i,j}^{\ast}c_{i,j}}=\lambda_{\tilde{k}}$

To verify that this is the maximum, let $T_{\tilde{i}}$ be the projection of
$T$ to the space with dimension $k_{\tilde{i}}$ with dimension
$n-k_{\tilde{i}}+1$, then the intersection of $S$ and $T_{\tilde{i}}$ is not
empty. We have

$\displaystyle\min\limits_{\mathcal{X}\in
S}\frac{\langle\mathcal{X},\mathcal{C}\star\mathcal{X}\rangle}{\langle\mathcal{X},\mathcal{X}\rangle}$
$\displaystyle\leq$ $\displaystyle\min\limits_{\mathcal{X}\in S\cap
T}\frac{\langle\mathcal{X},\mathcal{C}\star\mathcal{X}\rangle}{\langle\mathcal{X},\mathcal{X}\rangle}.$
(43)

Any such $\mathbf{x}_{\tilde{i}}\in S\cap T_{\tilde{i}}$ can be expressed as
$\mathbf{x}_{\tilde{i}}=\sum\limits^{m}_{j=k_{\tilde{i}}}c_{\tilde{i},j}\mathbf{v}_{\tilde{i}j}$,
and any $i$ for $i\neq\tilde{i}$, we have $\mathbf{x}_{i}\in S\cap T_{i}$
expressed as
$\mathbf{x}_{i}=\sum\limits^{m}_{j=k_{i}+1}c_{i,j}\mathbf{v}_{i,j}$. Then, we
have

$\displaystyle\frac{\langle\mathcal{X},\mathcal{C}\star\mathcal{X}\rangle}{\langle\mathcal{X},\mathcal{X}\rangle}$
$\displaystyle=$
$\displaystyle\frac{\frac{1}{p}\sum\limits_{i=1}^{p}\mathbf{x}^{\mathrm{H}}_{i}\mathbf{A}_{i}\mathbf{x}_{i}}{\frac{1}{p}\sum\limits_{i=1}^{p}\mathbf{x}^{\mathrm{H}}_{i}\mathbf{x}_{i}}=\frac{\sum\limits_{i=1}^{p}\sum\limits^{m}_{\begin{subarray}{c}j=k_{i}+1;i\neq\tilde{i}\\\
j=k_{\tilde{i}};i=\tilde{i}\end{subarray}}\lambda_{i,j}c_{i,j}^{\ast}c_{i,j}}{\sum\limits_{i=1}^{p}\sum\limits^{m}_{\begin{subarray}{c}j=k_{i}+1;i\neq\tilde{i}\\\
j=k_{\tilde{i}};i=\tilde{i}\end{subarray}}c_{i,j}^{\ast}c_{i,j}}$ (44)
$\displaystyle\leq$
$\displaystyle\frac{\sum\limits_{i=1}^{p}\sum\limits^{m}_{\begin{subarray}{c}j=k_{i}+1;i\neq\tilde{i}\\\
j=k_{\tilde{i}};i=\tilde{i}\end{subarray}}\lambda_{\tilde{k}}c_{i,j}^{\ast}c_{i,j}}{\sum\limits_{i=1}^{p}\sum\limits^{m}_{\begin{subarray}{c}j=k_{i}+1;i\neq\tilde{i}\\\
j=k_{\tilde{i}};i=\tilde{i}\end{subarray}}c_{i,j}^{\ast}c_{i,j}}=\lambda_{\tilde{k}}.$

Therefore, for all subspaces $S$ of dimensions $\\{k_{1},\cdots,k_{p}\\}$, we
have $\min\limits_{\mathcal{X}\in
S}\frac{\langle\mathcal{X},\mathcal{C}\star\mathcal{X}\rangle}{\langle\mathcal{X},\mathcal{X}\rangle}\leq\lambda_{\tilde{k}}$
$\hfill\Box$

Given a symmetric T-product tensor $\mathcal{C}$ with associated matrices
$\mathbf{C}_{i}$ provided by Eq. (28), next theorem is the representation of
the summation of all the largest $k_{i}$ T-eigenvalues of $\mathbf{C}_{i}$ and
the summation of all the smallest $k_{i}$ T-eigenvalues of $\mathbf{C}_{i}$.

###### Theorem 4

Let $\mathcal{C}\in\mathbb{R}^{m\times m\times p}$ be a symmetric T-product
tensor with associated matrices $\mathbf{C}_{i}$ provided by Eq. (28), and we
sort T-eigenvalues of the matrix $\mathbf{C}_{i}$ as
$\lambda_{i,1}\geq\lambda_{i,2}\geq\cdots\geq\lambda_{i,k_{i}}$. Then, we have

$\displaystyle\sum\limits_{i=1}^{p}\max\limits_{\mathbf{U}_{i}\mathbf{U}^{\mathrm{H}}_{i}=\mathbf{I}_{k_{i}}}\mathrm{Tr}\left(\mathbf{U}_{i}\mathbf{C}_{i}\mathbf{U}^{\mathrm{H}}_{i}\right)$
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{p}\sum\limits_{j=1}^{k_{i}}\lambda_{i,j}(\mathbf{C}_{i});$
(45)

and

$\displaystyle\sum\limits_{i=1}^{p}\min\limits_{\mathbf{U}_{i}\mathbf{U}^{\mathrm{H}}_{i}=\mathbf{I}_{k_{i}}}\mathrm{Tr}\left(\mathbf{U}_{i}\mathbf{C}_{i}\mathbf{U}^{\mathrm{H}}_{i}\right)$
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{p}\sum\limits_{j=1}^{k_{i}}\lambda_{i,m-j+1}(\mathbf{C}_{i}),$
(46)

where $\mathbf{U}_{i}$ are $k_{i}\times m$ complex matrices.

Proof: From Theorem 1, we may assume that $\mathbf{C}_{i}$ are diagonal
matrices, denoted as $\mathbf{D}_{i}$, since $\mathbf{C}_{i}$ are symmetric
T-product matrices. Therefore, we have the expression
$\mathbf{C}_{i}=\mathbf{V}\mathbf{D}_{i}\mathbf{V}^{\mathrm{H}}$. Then, we
have

$\displaystyle\mathrm{Tr}\left(\mathbf{U}_{i}\mathbf{D}_{i}\mathbf{U}^{\mathrm{H}}_{i}\right)=\sum\limits_{j=1}^{k_{i}}\sum\limits_{l=1}^{m}u_{j,l}^{\ast}u_{j,l}\lambda_{i,l}(\mathbf{C}_{i})=\sum\limits_{j=1}^{k_{i}}\sum\limits_{l=1}^{m}p_{j,l}\lambda_{i,l}(\mathbf{C}_{i})=[\overbrace{1,1,\cdots,1}^{\mbox{$k_{i}$
terms}}]\mathbf{P}\left[\begin{array}[]{c}\lambda_{i,1}\\\ \lambda_{i,2}\\\
\vdots\\\ \lambda_{i,m}\\\ \end{array}\right],$ (51)

where $\mathbf{P}=(p_{j,l})$ is a $k_{i}\times m$ stochastic matrix. Then, we
can concatenate an $(m-k_{i})\times m$ matrix $\mathbf{Q}$ to the matrix
$\mathbf{P}$ to make the following matrix
$\left[\begin{array}[]{c}\mathbf{P}\\\ \mathbf{Q}\\\ \end{array}\right]$ as
doubly stochastic from 2.C.1(4) from [20]. Then, Eq. (51) can be expressed as

$\displaystyle\mathrm{Tr}\left(\mathbf{U}_{i}\mathbf{D}_{i}\mathbf{U}^{\mathrm{H}}_{i}\right)=[\overbrace{1,1,\cdots,1}^{\mbox{$k_{i}$
terms}},\overbrace{0,0,\cdots,0}^{\mbox{$m-k_{i}$
terms}}]\left[\begin{array}[]{c}\mathbf{P}\\\ \mathbf{Q}\\\
\end{array}\right]\left[\begin{array}[]{c}\lambda_{i,1}\\\ \lambda_{i,2}\\\
\vdots\\\ \lambda_{i,m}\\\ \end{array}\right]$ (58)

Given two lists of real numbers, $[a_{1},\cdots,a_{n}]$ and
$[b_{1},\cdots,b_{n}]$, we use $[a_{1},\cdots,a_{n}]\prec[b_{1},\cdots,b_{n}]$
to represent the following relationships:

$\displaystyle\sum\limits_{i=1}^{k}a_{i}\leq\sum\limits_{i=1}^{k}b_{i},$ (59)

holds for any $k$ between $1$ and $n$. From Eq. (58), we have
$[\lambda_{i,1},\cdots,\lambda_{i,m}][\mathbf{P}^{\mathrm{T}},\mathbf{Q}^{\mathrm{T}}]\prec[\lambda_{i,1},\cdots,\lambda_{i,m}]$
and 3.H.2.b from [20], we have

$\displaystyle\mathrm{Tr}\left(\mathbf{U}_{i}\mathbf{C}_{i}\mathbf{U}^{\mathrm{H}}_{i}\right)$
$\displaystyle\leq$
$\displaystyle\sum\limits_{j=1}^{k_{i}}\lambda_{i,j}(\mathbf{C}_{i});$ (60)

and

$\displaystyle\mathrm{Tr}\left(\mathbf{U}_{i}\mathbf{C}_{i}\mathbf{U}^{\mathrm{H}}_{i}\right)$
$\displaystyle\geq$
$\displaystyle\sum\limits_{j=1}^{k_{i}}\lambda_{i,m-j+1}(\mathbf{C}_{i}).$
(61)

Finally, this theorem is proved by applying $\sum\limits_{i=1}^{p}$ to both
sides of Eqs. (60) and (61) with respect to the index $i$, and note that
$\mathbf{U}_{i}\mathbf{V}_{i}=\left(\mathbf{I}_{k_{i}},\mathbf{O}\right)$ and
$\mathbf{U}_{i}\mathbf{V}_{i}=\left(\mathbf{O},\mathbf{I}_{k_{i}}\right)$,
respectively. $\hfill\Box$

###  2.2 Unitarily Invariant T-product Tensor Norms

Let us represent the T-eigenvalues of a symmetric T-product tensor
$\mathcal{H}\in\mathbb{R}^{m\times m\times p}$ in decreasing order by the
vector
$\vec{\lambda}(\mathcal{H})=(\lambda_{1}(\mathcal{H}),\cdots,\lambda_{m\times
p}(\mathcal{H}))$, where $m\times p$ is the total number of T-eigenvalues. We
use $\mathbb{R}_{\geq 0}(\mathbb{R}_{>0})$ to represent a set of nonnegative
(positive) real numbers. Let $\left\|\cdot\right\|_{\rho}$ be a unitarily
invariant tensor norm, i.e.,
$\left\|\mathcal{H}\star\mathcal{U}\right\|_{\rho}=\left\|\mathcal{U}\star\mathcal{H}\right\|_{\rho}=\left\|\mathcal{H}\right\|_{\rho}$,
where $\mathcal{U}$ is any unitary tensor. Let $\rho:\mathbb{R}_{\geq
0}^{m\times p}\rightarrow\mathbb{R}_{\geq 0}$ be the corresponding gauge
function that satisfies H$\ddot{o}$lder’s inequality so that

$\displaystyle\left\|\mathcal{H}\right\|_{\rho}=\left\||\mathcal{H}|\right\|_{\rho}=\rho(\vec{\lambda}(|\mathcal{H}|)),$
(62)

where $|\mathcal{H}|\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\sqrt{\mathcal{H}^{H}\star\mathcal{H}}$. The bijective
correspondence between symmetric gauge functions on $\mathbb{R}_{\geq
0}^{m\times p}$ and unitarily invariant norms is due to von Neumann [21].

Several popular norms can be treated as special cases of unitarily invariant
tensor norm. The first one is Ky Fan like $k$-norm [21] for tensors. For
$k\in\\{1,2,\cdots,m\times p\\}$, the Ky Fan $k$-norm [21] for tensors
$\mathcal{H}\mathbb{R}^{m\times m\times p}$, denoted as
$\left\|\mathcal{H}\right\|_{(k)}$, is defined as:

$\displaystyle\left\|\mathcal{H}\right\|_{(k)}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\sum\limits_{i=1}^{k}\lambda_{i}(|\mathcal{H}|).$ (63)

If $k=1$, the Ky Fan $k$-norm for tensors is the tensor operator norm, denoted
as $\left\|\mathcal{H}\right\|$. The second one is Schatten $p$-norm for
tensors, denoted as $\left\|\mathcal{H}\right\|_{p}$, is defined as:

$\displaystyle\left\|\mathcal{H}\right\|_{p}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}(\mathrm{Tr}|\mathcal{H}|^{p})^{\frac{1}{p}},$ (64)

where $p\geq 1$. If $p=1$, it is the trace norm.

Following inequality is the extension of Hölder inequality to gauge function
$\rho$ which will be used later to prove majorization relations.

###### Lemma 2

For $n$ nonnegative real vectors with the dimension $r$, i.e.,
$\mathbf{b}_{i}=(b_{i_{1}},\cdots,b_{i_{r}})\in\mathbb{R}_{\geq 0}^{r}$, and
$\alpha>0$ with $\sum\limits_{i=1}^{n}\alpha_{i}=1$, we have

$\displaystyle\rho\left(\prod\limits_{i=1}^{n}b_{i_{1}}^{\alpha_{i}},\prod\limits_{i=1}^{n}b_{i_{2}}^{\alpha_{i}},\cdots,\prod\limits_{i=1}^{n}b_{i_{r}}^{\alpha_{i}}\right)\leq\prod\limits_{i=1}^{n}\rho(\mathbf{b}_{i})^{\alpha_{i}}$
(65)

Proof: This proof is based on mathematical induction. The base case for $n=2$
has been shown by Theorem IV.1.6 from [22].

We assume that Eq. (65) is true for $n=m$, where $m>2$. Let $\odot$ be the
component-wise product (Hadamard product) between two vectors. Then, we have

$\displaystyle\rho\left(\prod\limits_{i=1}^{m+1}b_{i_{1}}^{\alpha_{i}},\prod\limits_{i=1}^{m+1}b_{i_{2}}^{\alpha_{i}},\cdots,\prod\limits_{i=1}^{m+1}b_{i_{r}}^{\alpha_{i}}\right)=\rho\left(\odot_{i=1}^{m+1}\mathbf{b}_{i}^{\alpha_{i}}\right),$
(66)

where $\odot_{i=1}^{m+1}\mathbf{b}_{i}^{\alpha_{i}}$ is defined as
$\left(\prod\limits_{i=1}^{m+1}b_{i_{1}}^{\alpha_{i}},\prod\limits_{i=1}^{m+1}b_{i_{2}}^{\alpha_{i}},\cdots,\prod\limits_{i=1}^{m+1}b_{i_{r}}^{\alpha_{i}}\right)$
with $\mathbf{b}_{i}^{\alpha_{i}}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}(b_{i_{1}}^{\alpha_{i}},\cdots,b_{i_{r}}^{\alpha_{i}})$. Under such
notations, Eq. (66) can be bounded as

$\displaystyle\rho\left(\odot_{i=1}^{m+1}\mathbf{b}_{i}^{\alpha_{i}}\right)$
$\displaystyle=$
$\displaystyle\rho\left(\left(\odot_{i=1}^{m}\mathbf{b}_{i}^{\frac{\alpha_{i}}{\sum\limits_{j=1}^{m}\alpha_{j}}}\right)^{\sum\limits_{j=1}^{m}\alpha_{j}}\odot\mathbf{b}_{m+1}^{\alpha_{m+1}}\right)$
(67) $\displaystyle\leq$
$\displaystyle\left[\rho^{\sum\limits_{j=1}^{m}\alpha_{j}}\left(\odot_{i=1}^{m}\mathbf{b}_{i}^{\frac{\alpha_{i}}{\sum\limits_{j=1}^{m}\alpha_{j}}}\right)\right]\cdot\rho(\mathbf{b}_{m+1})^{\alpha_{m+1}}\leq\prod\limits_{i=1}^{m+1}\rho(\mathbf{b}_{i})^{\alpha_{i}}.$

By mathematical induction, this lemma is proved. $\hfill\Box$

###  2.3 Antisymmetric Kronecker Product for T-product Tensors

In this section, we will discuss a machinery of antisymmetric Kronecker
product for T-product tensors and this scheme will be used later for log-
majorization results. Let $\mathfrak{H}$ be an $m\times p$-dimensional Hilbert
space. For each $k\in\mathbb{N}$, let $\mathfrak{H}^{\otimes k}$ denote the
$k$-fold Kronecker product of $\mathfrak{H}$, which is the $(m\times
p)^{k}$-dimensional Hilbert space with respect to the inner product defined by

$\displaystyle\langle\mathbf{X}_{1}\otimes\cdots\otimes\mathbf{X}_{k},\mathbf{Y}_{1}\otimes\cdots\otimes\mathbf{Y}_{k}\rangle\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\prod\limits_{i=1}^{k}\langle\mathbf{X}_{i},\mathbf{Y}_{i}\rangle.$
(68)

For $\mathbf{X}_{1},\cdots,\mathbf{X}_{k}\in\mathfrak{H}$, we define
$\mathbf{X}_{1}\wedge\cdots\wedge\mathbf{X}_{k}\in\mathfrak{H}^{\otimes k}$ by

$\displaystyle\mathbf{X}_{1}\wedge\cdots\wedge\mathbf{X}_{k}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\frac{1}{\sqrt{k!}}\sum\limits_{\sigma}(\mbox{sgn}\sigma)\mathbf{X}_{\sigma(1)}\otimes\cdots\otimes\mathbf{X}_{\sigma(k)},$
(69)

where $\sigma$ runs over all permutations on $\\{1,2,\cdots,k\\}$ and
$\mbox{sgn}\sigma=\pm 1$ depending on $\sigma$ is even or odd. The subspace of
$\mathfrak{H}^{\otimes k}$ spanned by
$\\{\mathbf{X}_{1}\wedge\cdots\wedge\mathbf{X}_{k}\\}$, where
$\mathbf{X}_{i}\in\mathfrak{H}$, is named as $k$-fold antisymmetric Kronecker
product of $\mathfrak{H}$ and represented by $\mathfrak{H}^{\wedge k}$.

For each $\mathcal{C}\in\mathbb{R}^{m\times m\times p}$ and $k\in\mathbb{N}$,
the $k$-fold Kronecker product $\mathcal{C}^{\otimes
k}\in\mathbb{R}^{m^{k}\times m^{k}\times p^{k}}$ is given by

$\displaystyle\mathcal{C}^{\otimes
k}\star\left(\mathbf{X}_{1}\otimes\cdots\otimes\mathbf{X}_{k}\right)\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\left(\mathcal{C}\star\mathbf{X}_{1}\right)\otimes\cdots\otimes\left(\mathcal{C}\star\mathbf{X}_{k}\right).$
(70)

Because $\mathfrak{H}^{\wedge k}$ is invariant for $\mathcal{C}^{\otimes k}$,
the antisymmetric Kronecker product of $\mathcal{C}^{\wedge k}$ of
$\mathcal{C}$ can be defined as $\mathcal{C}^{\wedge
k}=\mathcal{C}^{\otimes}|_{\mathfrak{H}^{\wedge k}}$, then we have

$\displaystyle\mathcal{C}^{\wedge
k}\star\left(\mathbf{X}_{1}\wedge\cdots\wedge\mathbf{X}_{k}\right)=\left(\mathcal{C}\star\mathbf{X}_{1}\right)\wedge\cdots\wedge\left(\mathcal{C}\star\mathbf{X}_{k}\right).$
(71)

We will provide the following lemmas about antisymmetric Kronecker product.

###### Lemma 3

Let $\mathcal{A},\mathcal{B},\mathcal{C},\mathcal{E}\in\mathbb{R}^{m\times
m\times p}$ be T-product tensors , for any $k\in\\{1,2,\cdots,m\times p\\}$,
we have

  1. 1.

$(\mathcal{A}^{\wedge k})^{\mathrm{T}}=(\mathcal{A}^{\mathrm{T}})^{\wedge k}$.

  2. 2.

$(\mathcal{A}^{\wedge k})\star(\mathcal{B}^{\wedge
k})=(\mathcal{A}\star\mathcal{B})^{\wedge k}$.

  3. 3.

If
$\lim\limits_{i\rightarrow\infty}\left\|\mathcal{A}_{i}-\mathcal{A}\right\|\rightarrow
0$ , then $\lim\limits_{i\rightarrow\infty}\left\|\mathcal{A}^{\wedge
k}_{i}-\mathcal{A}^{\wedge k}\right\|\rightarrow 0$.

  4. 4.

If $\mathcal{C}\succeq\mathcal{O}$ (zero tensor), then $\mathcal{C}^{\wedge
k}\succeq\mathcal{O}$ and $(\mathcal{C}^{p})^{\wedge k}=(\mathcal{C}^{\wedge
k})^{p}$ for all $p\in\mathbb{R}_{>0}$.

  5. 5.

$|\mathcal{A}|^{\wedge k}=|\mathcal{A}^{\wedge k}|$.

  6. 6.

If $\mathcal{E}\succeq\mathcal{O}$ and $\mathcal{E}$ is invertible,
$(\mathcal{E}^{z})^{\wedge k}=(\mathcal{E}^{\wedge k})^{z}$ for all
$z\in\mathbb{E}$.

  7. 7.

$\left\|\mathcal{E}^{\wedge
k}\right\|=\prod\limits_{i=1}^{k}\lambda_{i}(|\mathcal{E}|)$.

Proof: Items 1 and 2 are the restrictions of the associated relations
$(\mathcal{A}^{H})^{\otimes k}=(\mathcal{A}^{\otimes k})^{H}$ and
$(\mathcal{A}\star\mathcal{B})^{\otimes k}=(\mathcal{A}^{\otimes
k})\star(\mathcal{B}^{\otimes k})$ to $\mathfrak{H}^{\wedge k}$. The item 3 is
true since, if
$\lim\limits_{i\rightarrow\infty}\left\|\mathcal{A}_{i}-\mathcal{A}\right\|\rightarrow
0$, we have $\lim\limits_{i\rightarrow\infty}\left\|\mathcal{A}^{\otimes
k}_{i}-\mathcal{A}^{\otimes k}\right\|\rightarrow 0$ and the asscoaited
restrictions of $\mathcal{A}_{i}^{\otimes k},\mathcal{A}^{\otimes k}$ to the
antisymmetric subspace $\mathfrak{H}^{k}$.

For the item 4, if $\mathcal{C}\succeq\mathcal{O}$, then we have
$\mathcal{C}^{\wedge k}=((\mathcal{C}^{1/2})^{\wedge
k})^{H}\star((\mathcal{C}^{1/2})^{\wedge k})\succeq\mathcal{O}$ from items 1
and 2. If $p$ is ratonal, we have $(\mathcal{C}^{p})^{\wedge
k}=(\mathcal{C}^{\wedge k})^{p}$ from the item 2, and the equality
$(\mathcal{C}^{p})^{\wedge k}=(\mathcal{C}^{\wedge k})^{p}$ is also true for
any $p>0$ if we apply the item 3 to approximate any irrelational numbers by
rational numbers.

Because we have

$\displaystyle|\mathcal{A}|^{\wedge
k}=\left(\sqrt{\mathcal{A}^{H}\mathcal{A}}\right)^{\wedge
k}=\sqrt{(\mathcal{A}^{\wedge k})^{H}\mathcal{A}^{\wedge
k}}=|\mathcal{A}^{\wedge k}|,$ (72)

from items 1, 2 and 4, so the item 5 is valid.

For item 6, if $z<0$, item 6 is true for all $z\in\mathbb{R}$ by applying the
item 4 to $\mathcal{E}^{-1}$. Since we can apply the definition
$\mathcal{E}^{z}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\exp(z\ln\mathcal{E})$ to have

$\displaystyle\mathcal{C}^{p}$ $\displaystyle=$
$\displaystyle\mathcal{E}^{z}\leavevmode\nobreak\ \leavevmode\nobreak\
\leftrightarrow\leavevmode\nobreak\ \leavevmode\nobreak\
\mathcal{C}=\exp\left(\frac{z}{p}\ln\mathcal{E}\right),$ (73)

where $\mathcal{C}\succeq\mathcal{O}$. The general case of any
$z\in\mathbb{C}$ is also true by applying the item 4 to
$\mathcal{C}=\exp(\frac{z}{p}\ln\mathcal{E})$.

For the item 7 proof, it is enough to prove the case that
$\mathcal{E}\succeq\mathcal{O}$ due to the item 5. Then, from Theorem 2, there
exists a set of orthogonal tensors
$\\{\mathcal{U}_{1},\cdots,\mathcal{U}_{r}\\}$ such that
$|\mathcal{E}|\star\mathcal{U}_{i}=\lambda_{i}\mathcal{U}_{i}$ for $1\leq
i\leq m\times p$. We then have

$\displaystyle|\mathcal{E}|^{\wedge
k}\left(\mathcal{U}_{i_{1}}\wedge\cdots\wedge\mathcal{U}_{i_{k}}\right)$
$\displaystyle=$
$\displaystyle|\mathcal{E}|\star\mathcal{U}_{i_{1}}\wedge\cdots\wedge|\mathcal{E}|\star\mathcal{U}_{i_{k}}$
(74) $\displaystyle=$
$\displaystyle\left(\prod\limits_{i=1}^{k}\lambda_{i}(|\mathcal{E}|)\right)\mathcal{U}_{i_{1}}\wedge\cdots\wedge\mathcal{U}_{i_{k}},$

where $1\leq i_{1}<i_{2}<\cdots<i_{k}\leq m\times p$. Hence,
$\left\||\mathcal{E}|^{\wedge
k}\right\|=\prod\limits_{i=1}^{k}\lambda_{i}(|\mathcal{E}|)$. $\hfill\Box$

##  3 Multivariate T-product Tensor Norm Inequalities

In this section, we will begin with the introduction of majorization
techniques in Section 3.1. Then, the majorization with integral average and
log-majorization with integral average will be introduced by Section 3.2 and
Section 3.3. These majorization results will be used to prove T-product tensor
norm inequalities in Section 3.4.

###  3.1 Majorization Basis

In this subsection, we will discuss majorization and several lemmas about
majorization which will be used at later proofs.

Let $\mathbf{x}=[x_{1},\cdots,x_{r}]\in\mathbb{R}^{m\times
p},\mathbf{y}=[y_{1},\cdots,y_{r}]\in\mathbb{R}^{m\times p}$ be two vectors
with following orders among entries $x_{1}\geq\cdots\geq x_{r}$ and
$y_{1}\geq\cdots\geq y_{r}$, _weak majorization_ between vectors
$\mathbf{x},\mathbf{y}$, represented by $\mathbf{x}\prec_{w}\mathbf{y}$,
requires following relation for vectors $\mathbf{x},\mathbf{y}$:

$\displaystyle\sum\limits_{i=1}^{k}x_{i}\leq\sum\limits_{i=1}^{k}y_{i},$ (75)

where $k\in\\{1,2,\cdots,r\\}$. _Majorization_ between vectors
$\mathbf{x},\mathbf{y}$, indicated by $\mathbf{x}\prec\mathbf{y}$, requires
following relation for vectors $\mathbf{x},\mathbf{y}$:

$\displaystyle\sum\limits_{i=1}^{k}x_{i}$ $\displaystyle\leq$
$\displaystyle\sum\limits_{i=1}^{k}y_{i},\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{for $1\leq k<r$;}$
$\displaystyle\sum\limits_{i=1}^{m\times p}x_{i}$ $\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{m\times p}y_{i},\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{for $k=r$.}$ (76)

For $\mathbf{x},\mathbf{y}\in\mathbb{R}^{m\times p}_{\geq 0}$ such that
$x_{1}\geq\cdots\geq x_{r}$ and $y_{1}\geq\cdots\geq y_{r}$, _weak log
majorization_ between vectors $\mathbf{x},\mathbf{y}$, represented by
$\mathbf{x}\prec_{w\log}\mathbf{y}$, requires following relation for vectors
$\mathbf{x},\mathbf{y}$:

$\displaystyle\prod\limits_{i=1}^{k}x_{i}\leq\prod\limits_{i=1}^{k}y_{i},$
(77)

where $k\in\\{1,2,\cdots,r\\}$, and _log majorization_ between vectors
$\mathbf{x},\mathbf{y}$, represented by $\mathbf{x}\prec_{\log}\mathbf{y}$,
requires equality for $k=r$ in Eq. (77). If $f$ is a single variable function,
$f(\mathbf{x})$ represents a vector of $[f(x_{1}),\cdots,f(x_{r})]$. From
Lemma 1 in [23], we have

###### Lemma 4

(1) For any convex function $f:[0,\infty)\rightarrow[0,\infty)$, if we have
$\mathbf{x}\prec\mathbf{y}$, then $f(\mathbf{x})\prec_{w}f(\mathbf{y})$.
(2) For any convex function and non-decreasing
$f:[0,\infty)\rightarrow[0,\infty)$, if we have
$\mathbf{x}\prec_{w}\mathbf{y}$, then $f(\mathbf{x})\prec_{w}f(\mathbf{y})$.

Another lemma is from Lemma 12 in [23], we have

###### Lemma 5

Let $\mathbf{x},\mathbf{y}\in\mathbb{R}^{m\times p}_{\geq 0}$ such that
$x_{1}\geq\cdots\geq x_{r}$ and $y_{1}\geq\cdots\geq y_{r}$ with
$\mathbf{x}\prec_{\log}\mathbf{y}$. Also let
$\mathbf{y}_{i}=[y_{i;1},\cdots,y_{i;r}]\in\mathbb{R}^{m\times p}_{\geq 0}$ be
a sequence of vectors such that $y_{i;1}\geq\cdots\geq y_{i;r}>0$ and
$\mathbf{y}_{i}\rightarrow\mathbf{y}$ as $i\rightarrow\infty$. Then, there
exists $i_{0}\in\mathbb{N}$ and
$\mathbf{x}_{i}=[x_{i;1},\cdots,x_{i;r}]\in\mathbb{R}^{m\times p}_{\geq 0}$
for $i\geq i_{0}$ such that $x_{i;1}\geq\cdots\geq x_{i;r}>0$,
$\mathbf{x}_{i}\rightarrow\mathbf{x}$ as $i\rightarrow\infty$, and

$\displaystyle\mathbf{x}_{i}\prec_{\log}\mathbf{y}_{i}\mbox{\leavevmode\nobreak\
\leavevmode\nobreak\ for $i\geq i_{0}$.}$ (78)

For any function $f$ on $\mathbb{R}_{\geq 0}$, the term $f(\mathbf{x}$ is
defined as $f(\mathbf{x})\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}(f(x_{1}),\cdots,f(x_{r}))$ with conventions $e^{-\infty}=0$ and
$\log 0=-\infty$.

###  3.2 Majorization with Integral Average

Let $\Omega$ be a $\sigma$-compact metric space and $\nu$ a probability
measure on the Borel $\sigma$-field of $\Omega$. Let
$\mathcal{C},\mathcal{D}_{\tau}\in\mathbb{R}^{m\times m\times p}$ be symmetric
T-product tensors. We further assume that tensors
$\mathcal{C},\mathcal{D}_{\tau}$ are uniformly bounded in their norm for
$\tau\in\Omega$. Let $\tau\in\Omega\rightarrow\mathcal{D}_{\tau}$ be a
continuous function such that
$\sup\\{\left\|D_{\tau}\right\|:\tau\in\Omega\\}<\infty$. For notational
convenience, we define the following relation:

$\displaystyle\left[\int_{\Omega}\lambda_{1}(\mathcal{D}_{\tau})d\nu(\tau),\cdots,\int_{\Omega}\lambda_{m\times
p}(\mathcal{D}_{\tau})d\nu(\tau)\right]\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\int_{\Omega^{m\times
p}}\vec{\lambda}(\mathcal{D}_{\tau})d\nu^{m\times p}(\tau).$ (79)

If $f$ is a single variable function, the notation $f(\mathcal{C})$ represents
a tensor function with respect to the tensor $\mathcal{C}$.

###### Theorem 5

Let $\Omega,\nu,\mathcal{C},\mathcal{D}_{\tau}$ be defined as the beginning
part of Section 3.2, and $f:\mathbb{R}\rightarrow[0,\infty)$ be a non-
decreasing convex function, we have following two equivalent statements:

$\displaystyle\vec{\lambda}(\mathcal{C})\prec_{w}\int_{\Omega^{m\times
p}}\vec{\lambda}(\mathcal{D}_{\tau})d\nu^{m\times
p}(\tau)\Longleftrightarrow\left\|f(\mathcal{C})\right\|_{\rho}\leq\int_{\Omega}\left\|f(\mathcal{D}_{\tau})\right\|_{\rho}d\nu(\tau),$
(80)

where $\left\|\cdot\right\|_{\rho}$ is the unitarily invariant norm defined in
Eq. (62).

Proof: We assume that the left statement of Eq. (80) is true and the function
$f$ is a non-decreasing convex function. From Lemma 4, we have

$\displaystyle\vec{\lambda}(f(\mathcal{C}))=f(\vec{\lambda}(\mathcal{C}))\prec_{w}f\left(\int_{\Omega^{m\times
p}}\vec{\lambda}(\mathcal{D}_{\tau})d\nu^{m\times p}(\tau)\right).$ (81)

From the convexity of $f$, we also have

$\displaystyle f\left(\int_{\Omega^{m\times
p}}\vec{\lambda}(\mathcal{D}_{\tau})d\nu^{m\times
p}(\tau)\right)\leq\int_{\Omega^{m\times
p}}f(\vec{\lambda}(\mathcal{D}_{\tau}))d\nu^{m\times
p}(\tau)=\int_{\Omega^{m\times
p}}\vec{\lambda}(f(\mathcal{D}_{\tau}))d\nu^{m\times p}(\tau).$ (82)

Then, we obtain $\vec{\lambda}(f(\mathcal{C}))\prec_{w}=\int_{\Omega^{m\times
p}}\vec{\lambda}(f(\mathcal{D}_{\tau}))d\nu^{m\times p}(\tau)$. By applying
Lemma 4.4.2 in [24] to both sides of
$\vec{\lambda}(f(\mathcal{C}))\prec_{w}=\int_{\Omega^{m\times
p}}\vec{\lambda}(f(\mathcal{D}_{\tau}))d\nu^{m\times p}(\tau)$ with gauge
function $\rho$, we obtain

$\displaystyle\left\|f(\mathcal{C})\right\|_{\rho}$ $\displaystyle\leq$
$\displaystyle\rho\left(\int_{\Omega^{m\times
p}}\vec{\lambda}(f(\mathcal{D}_{\tau}))d\nu^{m\times p}(\tau)\right)$ (83)
$\displaystyle\leq$
$\displaystyle\int_{\Omega}\rho(\vec{\lambda}(f(\mathcal{D}_{\tau})))d\nu(\tau)=\int_{\Omega}\left\|f(\mathcal{D}_{\tau})\right\|_{\rho}d\nu(\tau).$

Therefore, the right statement of Eq. (80) is true from the left statement.

On the other hand, if the right statement of Eq. (80) is true, we select a
function $f\stackrel{{\scriptstyle\mbox{\tiny def}}}{{=}}\max\\{x+c,0\\}$,
where $c$ is a positive real constant satisfying
$\mathcal{C}+c\mathcal{I}\geq\mathcal{O}$,
$\mathcal{D}_{\tau}+c\mathcal{I}\geq\mathcal{O}$ for all $\tau\in\Omega$, and
tensors $\mathcal{C}+c\mathcal{I},\mathcal{D}_{\tau}+c\mathcal{I}$. If the Ky
Fan $k$-norm at the right statement of Eq. (80) is applied, we have

$\displaystyle\sum\limits_{i=1}^{k}(\lambda_{i}(\mathcal{C})+c)\leq\sum\limits_{i=1}^{k}\int_{\Omega}(\lambda_{i}(\mathcal{D}_{\tau})+c)d\nu(\tau).$
(84)

Hence,
$\sum\limits_{i=1}^{k}\lambda_{i}(\mathcal{C})\leq\sum\limits_{i=1}^{k}\int_{\Omega}\lambda_{i}(\mathcal{D}_{\tau})d\nu(\tau)$,
this is the left statement of Eq. (80). $\hfill\Box$

Next theorem will provide a stronger version of Theorem 5 by removing weak
majorization conditions.

###### Theorem 6

Let $\Omega,\nu,\mathcal{C},\mathcal{D}_{\tau}$ be defined as the beginning
part of Section 3.2, and $f:\mathbb{R}\rightarrow[0,\infty)$ be a convex
function, we have following two equivalent statements:

$\displaystyle\vec{\lambda}(\mathcal{C})\prec\int_{\Omega^{m\times
p}}\vec{\lambda}(\mathcal{D}_{\tau})d\nu^{m\times
p}(\tau)\Longleftrightarrow\left\|f(\mathcal{C})\right\|_{\rho}\leq\int_{\Omega}\left\|f(\mathcal{D}_{\tau})\right\|_{\rho}d\nu(\tau),$
(85)

where $\left\|\cdot\right\|_{\rho}$ is the unitarily invariant norm defined in
Eq. (62).

Proof: We assume that the left statement of Eq. (85) is true and the function
$f$ is a convex function. Again, from Lemma 4, we have

$\displaystyle\vec{\lambda}(f(\mathcal{A}))=f(\vec{\lambda}(\mathcal{A}))\prec_{w}f\left(\left(\int_{\Omega^{m\times
p}}\vec{\lambda}(\mathcal{D}_{\tau})d\nu^{m\times
p}(\tau)\right)\right)\leq\int_{\Omega^{m\times
p}}f(\vec{\lambda}(\mathcal{D}_{\tau}))d\nu^{m\times p}(\tau),$ (86)

then,

$\displaystyle\left\|f(\mathcal{A})\right\|_{\rho}$ $\displaystyle\leq$
$\displaystyle\rho\left(\int_{\Omega^{m\times
p}}f(\vec{\lambda}(\mathcal{D}_{\tau}))d\nu^{m\times p}(\tau)\right)$ (87)
$\displaystyle\leq$
$\displaystyle\int_{\Omega}\rho\left(f(\vec{\lambda}(\mathcal{D}_{\tau}))\right)d\nu(\tau)=\int_{\Omega}\left\|f(\mathcal{D}_{\tau})\right\|_{\rho}d\nu(\tau).$

This proves the right statement of Eq. (85).

Now, we assume that the right statement of Eq. (85) is true. From Theorem 5,
we already have $\vec{\lambda}(\mathcal{C})\prec_{w}\int_{\Omega^{m\times
p}}\vec{\lambda}(\mathcal{D}_{\tau})d\nu^{m\times p}(\tau)$. It is enough to
prove $\sum\limits_{i=1}^{m\times
p}\lambda_{i}(\mathcal{C})\geq\int_{\Omega}\sum\limits_{i=1}^{m\times
p}\lambda_{i}(\mathcal{D}_{\tau})d\nu(\tau)$. We define a function
$f\stackrel{{\scriptstyle\mbox{\tiny def}}}{{=}}\max\\{c-x,0\\}$, where $c$ is
a positive real constant satisfying $\mathcal{C}\leq c\mathcal{I}$,
$\mathcal{D}_{\tau}\leq c\mathcal{I}$ for all $\tau\in\Omega$ and tensors
$c\mathcal{I}-\mathcal{C},c\mathcal{I}-\mathcal{D}_{\tau}$. If the trace norm
is applied, i.e., the sum of the absolute value of all eigenvalues of a
symmetric T-product tensor, then the right statement of Eq. (85) becomes

$\displaystyle\sum\limits_{i=1}^{m\times
p}\lambda_{i}\left(c\mathcal{I}-\mathcal{C}\right)\leq\int_{\Omega}\sum\limits_{i=1}^{m\times
p}\lambda_{i}\left(c\mathcal{I}-\mathcal{D}_{\tau}\right)d\nu(\tau).$ (88)

The desired inequality $\sum\limits_{i=1}^{m\times
p}\lambda_{i}(\mathcal{C})\geq\int_{\Omega}\sum\limits_{i=1}^{m\times
p}\lambda_{i}(\mathcal{D}_{\tau})d\nu(\tau)$ is established. $\hfill\Box$

###  3.3 Log-Majorization with Integral Average

The purpose of this section is to consider log-majorization issues for
unitarily invariant norms of TPSD T-product tensors. In this section, let
$\mathcal{C},\mathcal{D}_{\tau}\in\mathbb{R}^{m\times m\times p}$ be TPSD
T-product tensors with $m\times p$ nonnegative T-eigenvalues by keeping
notations with the same definitions as at the beginning of the Section 3.2.
For notational convenience, we define the following relation for logarithm
vector:

$\displaystyle\left[\int_{\Omega}\log\lambda_{1}(\mathcal{D}_{\tau})d\nu(\tau),\cdots,\int_{\Omega}\log\lambda_{m\times
p}(\mathcal{D}_{\tau})d\nu(\tau)\right]\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\int_{\Omega^{m\times
p}}\log\vec{\lambda}(\mathcal{D}_{\tau})d\nu^{m\times p}(\tau).$ (89)

###### Theorem 7

Let $\mathcal{C},\mathcal{D}_{\tau}$ be TPSD T-product tensors,
$f:(0,\infty)\rightarrow[0,\infty)$ be a continuous function such that the
mapping $x\rightarrow\log f(e^{x})$ is convex on $\mathbb{R}$, and
$g:(0,\infty)\rightarrow[0,\infty)$ be a continuous function such that the
mapping $x\rightarrow g(e^{x})$ is convex on $\mathbb{R}$ , then we have
following three equivalent statements:

$\displaystyle\vec{\lambda}(\mathcal{C})$ $\displaystyle\prec_{w\log}$
$\displaystyle\exp\int_{\Omega^{m\times
p}}\log\vec{\lambda}(\mathcal{D}_{\tau})d\nu^{m\times p}(\tau);$ (90)
$\displaystyle\left\|f(\mathcal{C})\right\|_{\rho}$ $\displaystyle\leq$
$\displaystyle\exp\int_{\Omega}\log\left\|f(\mathcal{D}_{\tau})\right\|_{\rho}d\nu(\tau);$
(91) $\displaystyle\left\|g(\mathcal{C})\right\|_{\rho}$ $\displaystyle\leq$
$\displaystyle\int_{\Omega}\left\|g(\mathcal{D}_{\tau})\right\|_{\rho}d\nu(\tau).$
(92)

Proof: The roadmap of this proof is to prove equivalent statements between Eq.
(90) and Eq. (91) first, followed by equivalent statements between Eq. (90)
and Eq. (92).

Eq. (90) $\Longrightarrow$ Eq. (91)

There are two cases to be discussed in this part of proof:
$\mathcal{C},\mathcal{D}_{\tau}$ are TPD tensors, and
$\mathcal{C},\mathcal{D}_{\tau}$ are TPSD T-product tensors. At the beginning,
we consider the case that $\mathcal{C},\mathcal{D}_{\tau}$ are TPD tensors.

Since $\mathcal{D}_{\tau}$ are positive, we can find $\varepsilon>0$ such that
$\mathcal{D}_{\tau}\geq\varepsilon\mathcal{I}$ for all $\tau\in\Omega$. From
Eq. (90), the convexity of $\log f(e^{x})$ and Lemma 4, we have

$\displaystyle\vec{\lambda}\left(f(\mathcal{C})\right)=f\left(\exp\left(\log\vec{\lambda}(\mathcal{C})\right)\right)$
$\displaystyle\prec_{w}$ $\displaystyle f\left(\exp\int_{\Omega^{m\times
p}}\vec{\lambda}(\mathcal{D}_{\tau})d\nu^{m\times p}(\tau)\right)$ (93)
$\displaystyle\leq$ $\displaystyle\exp\left(\int_{\Omega^{m\times p}}\log
f\left(\vec{\lambda}(\mathcal{D}_{\tau})\right)d\nu^{m\times p}(\tau)\right).$

Then, from Eq. (62), we obtain

$\displaystyle\left\|f(\mathcal{C})\right\|_{\rho}$ $\displaystyle\leq$
$\displaystyle\rho\left(\exp\left(\int_{\Omega^{m\times p}}\log
f\left(\vec{\lambda}(\mathcal{D}_{\tau})\right)d\nu^{m\times
p}(\tau)\right)\right).$ (94)

From the function $f$ properties, we can assume that $f(x)>0$ for any $x>0$.
Then, we have following bounded and continous maps on $\Omega$:
$\tau\rightarrow\log f(\lambda_{i}(\mathcal{D}_{\tau}))$ for
$i\in\\{1,2,\cdots,m\times p\\}$, and
$\tau\rightarrow\log\left\|f(\mathcal{D}_{\tau})\right\|_{\rho}$. Because we
have $\nu(\Omega)=1$ and $\sigma$-compactness of $\Omega$, we have
$\tau_{k}^{(n)}\in\Omega$ and $\alpha_{k}^{(n)}$ for $k\in\\{1,2,\cdots,n\\}$
and $n\in\mathbb{N}$ with $\sum\limits_{k=1}^{n}\alpha_{k}^{(n)}=1$ such that

$\displaystyle\int_{\Omega}\log
f(\lambda_{i}(\mathcal{D}_{\tau}))d\nu(\tau)=\lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^{n}\alpha_{k}^{(n)}\log
f(\lambda_{i}(\mathcal{D}_{\tau_{k}^{(n)}})),\mbox{for $i\in\\{1,2,\cdots
m\times p\\}$};$ (95)

and

$\displaystyle\int_{\Omega}\log\left\|f(\mathcal{D}_{\tau})\right\|_{\rho}d\nu(\tau)=\lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^{n}\alpha_{k}^{(n)}\log\left\|f(\mathcal{D}_{\tau_{k}^{(n)}})\right\|_{\rho}.$
(96)

By taking the exponential at both sides of Eq. (95) and apply the gauge
function $\rho$, we have

$\displaystyle\rho\left(\exp\int_{\Omega^{m\times p}}\log
f(\vec{\lambda}(\mathcal{D}_{\tau}))d\nu^{m\times
p}(\tau)\right)=\lim\limits_{n\rightarrow\infty}\rho\left(\prod\limits_{k=1}^{n}f\left(\vec{\lambda}\left(\mathcal{D}_{\tau_{k}^{(n)}}\right)\right)^{\alpha_{k}^{(n)}}\right).$
(97)

Similarly, by taking the exponential at both sides of Eq. (96), we have

$\displaystyle\exp\left(\int_{\Omega}\log\left\|f(\mathcal{D}_{\tau})\right\|_{\rho}d\nu(\tau)\right)=\lim\limits_{n\rightarrow\infty}\prod\limits_{k=1}^{n}\left\|f\left(\mathcal{D}_{\tau_{k}^{(n)}}\right)\right\|^{\alpha_{k}^{(n)}}_{\rho}.$
(98)

From Lemma 2, we have

$\displaystyle\rho\left(\prod\limits_{k=1}^{n}f\left(\vec{\lambda}\left(\mathcal{D}_{\tau_{k}^{(n)}}\right)\right)^{\alpha_{k}^{(n)}}\right)$
$\displaystyle\leq$
$\displaystyle\prod\limits_{k=1}^{n}\rho^{\alpha_{k}^{(n)}}\left(f\left(\vec{\lambda}\left(\mathcal{D}_{\tau_{k}^{(n)}}\right)\right)\right)$
(99) $\displaystyle=$
$\displaystyle\prod\limits_{k=1}^{n}\rho^{\alpha_{k}^{(n)}}\left(\vec{\lambda}\left(f\left(\mathcal{D}_{\tau_{k}^{(n)}}\right)\right)\right)$
$\displaystyle=$
$\displaystyle\prod\limits_{k=1}^{n}\left\|f\left(\mathcal{D}_{\tau_{k}^{(n)}}\right)\right\|_{\rho}^{\alpha_{k}^{(n)}}$

From Eqs. (97), (98) and (99), we have

$\displaystyle\rho\left(\exp\int_{\Omega^{m\times p}}\log
f(\vec{\lambda}(\mathcal{D}_{\tau}))d\nu^{m\times
p}(\tau)\right)\leq\exp\int_{\Omega}\log\left\|f(\mathcal{D}_{\tau})\right\|_{\rho}d\nu(\tau).$
(100)

Then, Eq. (91) is proved from Eqs. (94) and (100).

Next, we consider that $\mathcal{C},\mathcal{D}_{\tau}$ are TPSD T-product
tensors. For any $\delta>0$, we have following log-majorization relation:

$\displaystyle\prod\limits_{i=1}^{k}\left(\lambda_{i}(\mathcal{C})+\epsilon_{\delta}\right)$
$\displaystyle\leq$
$\displaystyle\prod\limits_{i=1}^{k}\exp\int_{\Omega}\log\left(\lambda_{i}(\mathcal{D}_{\tau})+\delta\right)d\nu(\tau),$
(101)

where $\epsilon_{\delta}>0$ and $k\in\\{1,2,\cdots r\\}$. Then, we can apply
the previous case result about TPD tensors to TPD tensors
$\mathcal{C}+\epsilon_{\delta}\mathcal{I}$ and
$\mathcal{D}_{\tau}+\delta\mathcal{I}$, and get

$\displaystyle\left\|f(\mathcal{C})+\epsilon_{\delta}\mathcal{I}\right\|_{\rho}$
$\displaystyle\leq$
$\displaystyle\exp\int_{\Omega}\log\left\|f(\mathcal{D}_{\tau})+\delta\mathcal{I}\right\|_{\rho}d\nu(\tau)$
(102)

As $\delta\rightarrow 0$, Eq. (102) will give us Eq. (91) for TPSD T-product
tensors.

Eq. (90) $\Longleftarrow$ Eq. (91)

We consider TPD tensors at first phase by assuming that $\mathcal{D}_{\tau}$
are TPD T-product tensors for all $\tau\in\Omega$. We may also assume that the
tensor $\mathcal{C}$ is a TPD T-product tensor. Since if this is a TPSD
T-product tensor, i.e., some $\lambda_{i}=0$, we always have following
inequality valid:

$\displaystyle\prod\limits_{i=1}^{k}\lambda_{i}(\mathcal{C})\leq\prod\limits_{i=1}^{k}\exp\int_{\Omega}\log\lambda_{i}(\mathcal{D}_{\tau})d\nu(\tau)$
(103)

If we apply $f(x)=x^{p}$ for $p>0$ and $\left\|\cdot\right\|_{\rho}$ as Ky Fan
$k$-norm in Eq. (91), we have

$\displaystyle\log\sum\limits_{i=1}^{k}\lambda^{p}_{i}\left(\mathcal{C}\right)\leq\int_{\Omega}\log\sum\limits_{i=1}^{k}\lambda_{i}^{p}\left(\mathcal{D}_{\tau}\right)d\nu(\tau).$
(104)

If we add $\log\frac{1}{k}$ and multiply $\frac{1}{p}$ at both sides of Eq.
(104), we have

$\displaystyle\frac{1}{p}\log\left(\frac{1}{k}\sum\limits_{i=1}^{k}\lambda^{p}_{i}\left(\mathcal{C}\right)\right)\leq\int_{\Omega}\frac{1}{p}\log\left(\frac{1}{k}\sum\limits_{i=1}^{k}\lambda_{i}^{p}\left(\mathcal{D}_{\tau}\right)\right)d\nu(\tau).$
(105)

From L’Hopital’s Rule, if $p\rightarrow 0$, we have

$\displaystyle\frac{1}{p}\log\left(\frac{1}{k}\sum\limits_{i=1}^{k}\lambda^{p}_{i}\left(\mathcal{C}\right)\right)\rightarrow\frac{1}{k}\sum\limits_{i=1}^{k}\log\lambda_{i}(\mathcal{C}),$
(106)

and

$\displaystyle\frac{1}{p}\log\left(\frac{1}{k}\sum\limits_{i=1}^{k}\lambda^{p}_{i}\left(\mathcal{D}_{\tau}\right)\right)\rightarrow\frac{1}{k}\sum\limits_{i=1}^{k}\log\lambda_{i}(\mathcal{D}_{\tau}),$
(107)

where $\tau\in\Omega$. Appling Eqs. (106) and (107) into Eq. (105) and taking
$p\rightarrow 0$, we have

$\displaystyle\sum\limits_{i=1}^{k}\lambda_{i}(\mathcal{C})\leq\int_{\Omega}\sum\limits_{i=1}^{k}\log\lambda_{i}(\mathcal{D}_{\tau})d\nu(\tau).$
(108)

Therefore, Eq. (90) is true for TPD tensors.

For TPSD T-product tensors $\mathcal{D}_{\tau}$, since Eq. (91) is valid for
$\mathcal{D}_{\tau}+\delta\mathcal{I}$ for any $\delta>0$, we can apply the
previous case result about TPD tensors to
$\mathcal{D}_{\tau}+\delta\mathcal{I}$ and obtain

$\displaystyle\prod\limits_{i=1}^{k}\lambda_{i}(\mathcal{C})\leq\prod\limits_{i=1}^{k}\exp\int_{\Omega}\log\left(\lambda_{i}(\mathcal{D}_{\tau})+\delta\right)d\nu(\tau),$
(109)

where $k\in\\{1,2,\cdots,r\\}$. Eq. (90) is still true for TPSD T-product
tensors as $\delta\rightarrow 0$.

Eq. (90) $\Longrightarrow$ Eq. (92)

If $\mathcal{C},\mathcal{D}_{\tau}$ are TPD tensors, and
$\mathcal{D}_{\tau}\geq\delta\mathcal{I}$ for all $\tau\in\Omega$. From Eq.
(90), we have

$\displaystyle\vec{\lambda}(\log\mathcal{C})=\log\vec{\lambda}(\mathcal{C})\prec_{w}\int_{\Omega^{m\times
p}}\log\vec{\lambda}(\mathcal{D}_{\tau})d\nu^{m\times
p}(\tau)=\int_{\Omega^{m\times
p}}\vec{\lambda}(\log\mathcal{D}_{\tau})d\nu^{m\times p}(\tau).$ (110)

If we apply Theorem 5 to $\log\mathcal{C}$, $\log\mathcal{D}_{\tau}$ with
function $f(x)=g(e^{x})$, where $g$ is used in Eq. (92), Eq. (92) is implied.

If $\mathcal{C},\mathcal{D}_{\tau}$ are TPSD T-product tensors and any
$\delta>0$, we can find $\epsilon_{\delta}\in(0,\delta)$ to satisfy following:

$\displaystyle\prod\limits_{i=1}^{k}\left(\lambda_{i}(\mathcal{C})+\epsilon_{\delta}\right)\leq\prod\limits_{i=1}^{k}\exp\int_{\Omega}\log\left(\lambda_{i}(\mathcal{D}_{\tau})+\delta\right)d\nu(\tau).$
(111)

Then, from TPD T-product tensor case, we have

$\displaystyle\left\|g(\mathcal{C}+\epsilon_{\delta}\mathcal{I})\right\|_{\rho}\leq\int_{\Omega}\left\|g(\mathcal{D}_{\tau}+\delta\mathcal{I})\right\|_{\rho}d\nu(\tau).$
(112)

Eq. (92) is obtained by taking $\delta\rightarrow 0$ in Eq. (112).

Eq. (90) $\Longleftarrow$ Eq. (92)

For $k\in\\{1,2,\cdots,r\\}$, if we apply $g(x)=\log(\delta+x)$, where
$\delta>0$, and Ky Fan $k$-norm in Eq. (92), we have

$\displaystyle\sum\limits_{i=1}^{k}\log\left(\delta+\lambda_{i}\left(\mathcal{C}\right)\right)\leq\sum\limits_{i=1}^{k}\int_{\Omega}\log\left(\delta+\lambda_{i}(\mathcal{D}_{\tau})\right)d\nu(\tau).$
(113)

Then, we have following relation as $\delta\rightarrow 0$:

$\displaystyle\sum\limits_{i=1}^{k}\log\lambda_{i}\left(\mathcal{C}\right)\leq\sum\limits_{i=1}^{k}\int_{\Omega}\log\lambda_{i}(\mathcal{D}_{\tau})d\nu(\tau).$
(114)

Therefore, Eq. (90) ccan be derived from Eq. (92). $\hfill\Box$

Next theorem will extend Theorem 7 to non-weak version.

###### Theorem 8

Let $\mathcal{C},\mathcal{D}_{\tau}$ be TPSD T-product tensors with
$\int_{\Omega}\left\|\mathcal{D}_{\tau}^{-p}\right\|_{\rho}d\nu(\tau)<\infty$
for any $p>0$, $f:(0,\infty)\rightarrow[0,\infty)$ be a continuous function
such that the mapping $x\rightarrow\log f(e^{x})$ is convex on $\mathbb{R}$,
and $g:(0,\infty)\rightarrow[0,\infty)$ be a continuous function such that the
mapping $x\rightarrow g(e^{x})$ is convex on $\mathbb{R}$ , then we have
following three equivalent statements:

$\displaystyle\vec{\lambda}(\mathcal{C})$ $\displaystyle\prec_{\log}$
$\displaystyle\exp\int_{\Omega^{m\times
p}}\log\vec{\lambda}(\mathcal{D}_{\tau})d\nu^{m\times p}(\tau);$ (115)
$\displaystyle\left\|f(\mathcal{C})\right\|_{\rho}$ $\displaystyle\leq$
$\displaystyle\exp\int_{\Omega}\log\left\|f(\mathcal{D}_{\tau})\right\|_{\rho}d\nu(\tau);$
(116) $\displaystyle\left\|g(\mathcal{C})\right\|_{\rho}$ $\displaystyle\leq$
$\displaystyle\int_{\Omega}\left\|g(\mathcal{D}_{\tau})\right\|_{\rho}d\nu(\tau).$
(117)

Proof:

The proof plan is similar to the proof in Theorem 7. We prove the equivalence
between Eq. (115) and Eq. (116) first, then prove the equivalence between Eq.
(115) and Eq. (117).

Eq. (115) $\Longrightarrow$ Eq. (116)

First, we assume that $\mathcal{C},\mathcal{D}_{\tau}$ are TPD T-product
tensors with $\mathcal{D}_{\tau}\geq\delta\mathcal{I}$ for all
$\tau\in\Omega$. The corresponding part of the proof in Theorem 7 about TPD
tensors $\mathcal{C},\mathcal{D}_{\tau}$ can be applied here.

For case that $\mathcal{C},\mathcal{D}_{\tau}$ are TPSD T-product tensors, we
have

$\displaystyle\prod\limits_{i=1}^{k}\lambda_{i}(\mathcal{C})\leq\prod\limits_{i=1}^{k}\exp\int_{\Omega}\log\left(\lambda_{i}(\mathcal{D}_{\tau})+\delta_{n}\right)d\nu(\tau),$
(118)

where $\delta_{n}>0$ and $\delta_{n}\rightarrow 0$. Because
$\int_{\Omega^{m\times
p}}\log\left(\vec{\lambda}(\mathcal{D}_{\tau})+\delta_{n}\right)d\nu^{m\times
p}(\tau)\rightarrow\int_{\Omega^{m\times
p}}\log\vec{\lambda}(\mathcal{D}_{\tau})d\nu^{m\times p}(\tau)$ as
$n\rightarrow\infty$, from Lemma 5, we can find $\mathbf{a}^{(n)}$ with $n\geq
n_{0}$ such that $a^{(n)}_{1}\geq\cdots\geq a^{(n)}_{r}>0$,
$\mathbf{a}^{(n)}\rightarrow\vec{\lambda}(\mathcal{C})$ and
$\mathbf{a}^{(n)}\prec_{\log}\exp\int_{\Omega^{m\times
p}}\log\vec{\lambda}\left(\mathcal{D}_{\tau}+\delta_{n}\mathcal{I}\right)d\nu^{m\times
p}(\tau)$

Selecting $\mathcal{C}^{(n)}$ with
$\vec{\lambda}(\mathcal{C}^{(n)})=\mathbf{a}^{(n)}$ and applying TPD tensors
case to $\mathcal{C}^{(n)}$ and $\mathcal{D}_{\tau}+\delta_{n}\mathcal{I}$, we
obtain

$\displaystyle\left\|f(\mathcal{C}^{(n)})\right\|_{\rho}\leq\exp\int_{\Omega}\log\left\|f(\mathcal{D}_{\tau}+\delta_{n}\mathcal{I})\right\|_{\rho}d\nu(\tau)$
(119)

where $n\geq n_{0}$.

There are two situations for the function $f$ near $0$: $f(0^{+})<\infty$ and
$f(0^{+})=\infty$. For the case with $f(0^{+})<\infty$, we have

$\displaystyle\left\|f(\mathcal{C}^{(n)})\right\|_{\rho}=\rho(f(\mathbf{a}^{(n)}))\rightarrow\rho(f(\vec{\lambda}(\mathcal{C})))=\left\|f(\mathcal{C})\right\|_{\rho},$
(120)

and

$\displaystyle\left\|f(\mathcal{D}_{\tau}+\delta_{n}\mathcal{I})\right\|_{\rho}\rightarrow\left\|f(\mathcal{D}_{\tau})\right\|_{\rho},$
(121)

where $\tau\in\Omega$ and $n\rightarrow\infty$. From Fatou–Lebesgue theorem,
we then have

$\displaystyle\limsup\limits_{n\rightarrow\infty}\int_{\Omega}\log\left\|f(\mathcal{D}_{\tau}+\delta_{n}\mathcal{I})\right\|_{\rho}d\nu(\tau)\leq\int_{\Omega}\log\left\|f(\mathcal{D}_{\tau})\right\|_{\rho}.$
(122)

By taking $n\rightarrow\infty$ in Eq. (119) and using Eqs. (120), (121),
(122), we have Eq. (116) for case that $f(0^{+})<\infty$.

For the case with $f(0^{+})=\infty$, we assume that
$\int_{\Omega}\log\left\|f(\mathcal{D}_{\tau})\right\|_{\rho}d\nu(\tau)<\infty$
(since the inequality in Eq. (116) is always true for
$\int_{\Omega}\log\left\|f(\mathcal{D}_{\tau})\right\|_{\rho}d\nu(\tau)=\infty$).
Since $f$ is decreasing on $(0,\epsilon)$ for some $\epsilon>0$. We claim that
the following relation is valid: there are two constants $a,b>0$ such that

$\displaystyle
a\leq\left\|f(\mathcal{D}_{\tau}+\delta_{n}\mathcal{I})\right\|_{\rho}\leq\left\|f(\mathcal{D}_{\tau})\right\|_{\rho}+b,$
(123)

for all $\tau\in\Omega$ and $n\geq n_{0}$. If Eq. (123) is valid and
$\int_{\Omega}\log\left\|f(\mathcal{D}_{\tau})\right\|_{\rho}d\nu(\tau)<\infty$,
from Lebesgue’s dominated convergence theorem, we also have Eq. (116) for case
that $f(0^{+})=\infty$ by taking $n\rightarrow\infty$ in Eq. (119).

Below, we will prove the claim stated by Eq. (123). By the uniform boundedness
of tensors $\mathcal{D}_{\tau}$, there is a constant $\kappa>0$ such that

$\displaystyle
0<\mathcal{D}_{\tau}+\delta_{n}\mathcal{I}\leq\kappa\mathcal{I},$ (124)

where $\tau\in\Omega$ and $n\geq n_{0}$. We may assume that
$\mathcal{D}_{\tau}$ is TPD tensors because
$\left\|f(\mathcal{D}_{\tau})\right\|_{\rho}=\infty$, i.e., Eq. (123) being
true automatically, when $\mathcal{D}_{\tau}$ is TPSD T-product tensors. From
Theorem 2, we have

$\displaystyle f(\mathcal{D}_{\tau}+\delta_{n}\mathcal{I})$ $\displaystyle=$
$\displaystyle\sum\limits_{i^{\prime},\mbox{s.t.
$\lambda_{i^{\prime}}(\mathcal{D}_{\tau})+\delta_{n}<\epsilon$}}f(\lambda_{i^{\prime}}(\mathcal{D}_{\tau})+\delta_{n})\mathcal{U}_{i^{\prime}}\star\mathcal{U}^{H}_{i^{\prime}}+$
(125) $\displaystyle\sum\limits_{j^{\prime},\mbox{s.t.
$\lambda_{j^{\prime}}(\mathcal{D}_{\tau})+\delta_{n}\geq\epsilon$}}f(\lambda_{j^{\prime}}(\mathcal{D}_{\tau})+\delta_{n})\mathcal{U}_{j^{\prime}}\star\mathcal{U}^{H}_{j^{\prime}}$
$\displaystyle\leq$ $\displaystyle\sum\limits_{i^{\prime},\mbox{s.t.
$\lambda_{i^{\prime}}(\mathcal{D}_{\tau})+\delta_{n}<\epsilon$}}f(\lambda_{i^{\prime}}(\mathcal{D}_{\tau}))\mathcal{U}_{i^{\prime}}\star\mathcal{U}^{H}_{i^{\prime}}+$
$\displaystyle\sum\limits_{j^{\prime},\mbox{s.t.
$\lambda_{j^{\prime}}(\mathcal{D}_{\tau})+\delta_{n}\geq\epsilon$}}f(\lambda_{j^{\prime}}(\mathcal{D}_{\tau})+\delta_{n})\mathcal{U}_{j^{\prime}}\star\mathcal{U}^{H}_{j^{\prime}}$
$\displaystyle\leq$ $\displaystyle
f(\mathcal{D}_{\tau})+\sum\limits_{j^{\prime},\mbox{s.t.
$\lambda_{j^{\prime}}(\mathcal{D}_{\tau})+\delta_{n}\geq\epsilon$}}f(\lambda_{j^{\prime}}(\mathcal{D}_{\tau})+\delta_{n})\mathcal{U}_{j^{\prime}}\star\mathcal{U}^{H}_{j^{\prime}}.$

Therefore, the claim in Eq. (123) follows by the triangle inequality for
$\left\|\cdot\right\|_{\rho}$ and
$f(\lambda_{j^{\prime}}(\mathcal{D}_{\tau})+\delta_{n})<\infty$ for
$\lambda_{j^{\prime}}(\mathcal{D}_{\tau})+\delta_{n}\geq\epsilon$.

Eq. (115) $\Longleftarrow$ Eq. (116)

The weak majorization relation

$\displaystyle\prod\limits_{i=1}^{k}\lambda_{i}(\mathcal{C})\leq\prod\limits_{i=1}^{k}\exp\int_{\Omega}\log\lambda_{i}(\mathcal{D}_{\tau})d\nu(\tau),$
(126)

is valid for $k<m\times p$ from Eq. (90) $\Longrightarrow$ Eq. (91) in Theorem
7. We wish to prove that Eq. (126) becomes equal for $k=m\times p$. It is
equivalent to prove that

$\displaystyle\log\det(\mathcal{C})\geq\int_{\Omega}\log\det(\mathcal{D}_{\tau})d\nu(\tau),$
(127)

where $\det(\cdot)$ is defined by Eq. (35). We can assume that
$\int_{\Omega}\log\det(\mathcal{D}_{\tau})d\nu(\tau)\geq-\infty$ since Eq.
(127) is true for
$\int_{\Omega}\log\det(\mathcal{D}_{\tau})d\nu(\tau)=-\infty$. Then,
$\mathcal{D}_{\tau}$ are TPD tensors.

If we scale tensors $\mathcal{C},\mathcal{D}_{\tau}$ as
$a\mathcal{C},a\mathcal{D}_{\tau}$ by some $a>0$, we can assume
$\mathcal{D}_{\tau}\leq\mathcal{I}$ and $\lambda_{i}(\mathcal{D}_{\tau})\leq
1$ for all $\tau\in\Omega$ and $i\in\\{1,2,\cdots,m\times p\\}$. Then for any
$p>0$, we have

$\displaystyle\frac{1}{m\times
p}\left\|\mathcal{D}_{\tau}^{-\varrho}\right\|_{1}\leq\lambda^{-\varrho}_{r}(\mathcal{D}_{\tau})\leq(\det(\mathcal{D}_{\tau}))^{-\varrho},$
(128)

and

$\displaystyle\frac{1}{\varrho}\log\left(\frac{\left\|\mathcal{D}^{-\varrho}_{\tau}\right\|_{1}}{m\times
p}\right)\leq-\log\det(\mathcal{D}_{\tau}).$ (129)

If we use tensor trace norm, represented by $\left\|\cdot\right\|_{1}$, as
unitarily invariant tensor norm and $f(x)=x^{-\varrho}$ for any $\varrho>0$ in
Eq. (116), we obtain

$\displaystyle\log\left\|\mathcal{C}^{-\varrho}\right\|_{1}\leq\int_{\Omega}\log\left\|\mathcal{D}^{-\varrho}_{\tau}\right\|_{1}d\nu(\tau).$
(130)

By adding $\log\frac{1}{m\times p}$ and multiplying $\frac{1}{\varrho}$ for
both sides of Eq. (130), we have

$\displaystyle\frac{1}{\varrho}\log\left(\frac{\left\|\mathcal{C}^{-\varrho}\right\|_{1}}{m\times
p}\right)\leq\int_{\Omega}\frac{1}{\varrho}\log\left(\frac{\left\|\mathcal{D}_{\tau}^{-\varrho}\right\|_{1}}{m\times
p}\right)d\nu(\tau)$ (131)

Similar to Eqs. (106) and (107), we have following two relations as
$\varrho\rightarrow 0$:

$\displaystyle\frac{1}{\varrho}\log\left(\frac{\left\|\mathcal{C}^{-\varrho}\right\|_{1}}{m\times
p}\right)\rightarrow\frac{-1}{m\times p}\log\det(\mathcal{C}),$ (132)

and

$\displaystyle\frac{1}{\varrho}\log\left(\frac{\left\|\mathcal{D}_{\tau}^{-\varrho}\right\|_{1}}{m\times
p}\right)\rightarrow\frac{-1}{m\times p}\log\det(\mathcal{D}_{\tau}).$ (133)

From Eq. (129) and Lebesgue’s dominated convergence theorem, we have

$\displaystyle\lim\limits_{\varrho\rightarrow
0}\int_{\Omega}\frac{1}{\varrho}\log\left(\frac{\left\|\mathcal{D}_{\tau}^{-\varrho}\right\|_{1}}{m\times
p}\right)d\nu(\tau)=\frac{-1}{m\times
p}\int_{\Omega}\log\det(\mathcal{D}_{\tau})\nu(\tau)$ (134)

Finally, we have Eq. (127) from Eqs. (131) and (134).

Eq. (115) $\Longrightarrow$ Eq. (117)

First, we assume that $\mathcal{C},\mathcal{D}_{\tau}$ are TPD tensors and
$\mathcal{D}_{\tau}\geq\delta\mathcal{I}$ for $\tau\in\Omega$. From Eq. (115),
we can apply Theorem 6 to $\log\mathcal{C},\log\mathcal{D}_{\tau}$ and
$f(x)=g(e^{x})$ to obtain Eq. (117).

For $\mathcal{C},\mathcal{D}_{\tau}$ are TPSD T-product tensors, we can choose
$\mathbf{a}^{(n)}$ and corresponding $\mathcal{C}^{(n)}$ for $n\geq n_{0}$
given $\delta_{n}\rightarrow 0$ with $\delta_{n}>0$ as the proof in Eq. (115)
$\Longrightarrow$ Eq. (116). Since tensors
$\mathcal{C}^{(n)},\mathcal{D}_{\tau}+\delta_{n}\mathcal{I}$ are TPD T-product
tensors, we then have

$\displaystyle\left\|g(\mathcal{C}^{(n)})\right\|_{\rho}\leq\int_{\Omega}\left\|g(\mathcal{D}_{\tau}+\delta_{n}\mathcal{I})\right\|_{\rho}d\nu(\tau).$
(135)

If $g(0^{+})<\infty$, Eq. (117) is obtained from Eq. (135) by taking
$n\rightarrow\infty$. On the other hand, if $g(0^{+})=\infty$, we can apply
the argument similar to the portion about $f(0^{+})=\infty$ in the proof for
Eq. (115) $\Longrightarrow$ Eq. (116) to get $a,b>0$ such that

$\displaystyle
a\leq\left\|g(\mathcal{D}_{\tau}+\delta_{n}\mathcal{I})\right\|_{\rho}\leq\left\|g(\mathcal{D}_{\tau})\right\|_{\rho}+b,$
(136)

for all $\tau\in\Omega$ and $n\geq n_{0}$. Since the case that
$\int_{\Omega}\left\|g(\mathcal{D}_{\tau})\right\|_{\rho}d\nu(\tau)=\infty$
will have Eq. (117), we only consider the case that
$\int_{\Omega}\left\|g(\mathcal{D}_{\tau})\right\|_{\rho}d\nu(\tau)<\infty$.
Then, we have Eq. (117) from Eqs. (135), (136) and Lebesgue’s dominated
convergence theorem.

Eq. (115) $\Longleftarrow$ Eq. (117)

The weak majorization relation

$\displaystyle\sum\limits_{i=1}^{k}\log\lambda_{i}(\mathcal{C})\leq\sum\limits_{i=1}^{k}\int_{\Omega}\log\lambda_{i}(\mathcal{D}_{\tau})d\nu(\tau)$
(137)

is true from the implication from Eq. (90) to Eq. (92) in Theorem 7. We have
to show that this relation becomes identity for $k=m\times p$. If we apply
$\left\|\cdot\right\|_{\rho}=\left\|\cdot\right\|_{1}$ and $g(x)=x^{-\varrho}$
for any $\varrho>0$ in Eq. (117), we have

$\displaystyle\frac{1}{\varrho}\log\left(\frac{\left\|\mathcal{C}^{-\varrho}\right\|_{1}}{m\times
p}\right)\leq\frac{1}{\varrho}\log\left(\int_{\Omega}\frac{\left\|\mathcal{D}_{\tau}^{-\varrho}\right\|_{1}}{m\times
p}d\nu(\tau)\right).$ (138)

Then, we will get

$\displaystyle\frac{-\log\det(\mathcal{C})}{m\times p}$ $\displaystyle=$
$\displaystyle\lim\limits_{\varrho\rightarrow
0}\frac{1}{\varrho}\log\left(\frac{\left\|\mathcal{C}^{-\varrho}\right\|_{1}}{m\times
p}\right)$ (139) $\displaystyle\leq$
$\displaystyle\lim\limits_{\varrho\rightarrow
0}\frac{1}{p}\log\left(\int_{\Omega}\frac{\left\|\mathcal{D}_{\tau}^{-\varrho}\right\|_{1}}{m\times
p}d\nu(\tau)\right)=_{1}\frac{-\int_{\Omega}\log\det(\mathcal{D}_{\tau})d\nu(\tau)}{m\times
p},$

which will prove the identity for Eq. (137) when $k=m\times p$. The equality
in $=_{1}$ will be proved by the following Lemma 6. $\hfill\Box$

###### Lemma 6

Let $\mathcal{D}_{\tau}$ be TPSD T-product tensors with
$\int_{\Omega}\left\|\mathcal{D}_{\tau}^{-p}\right\|_{\rho}d\nu(\tau)<\infty$
for any $p>0$, then we have

$\displaystyle\lim\limits_{p\rightarrow
0}\left(\frac{1}{p}\log\int_{\Omega}\frac{\left\|\mathcal{D}_{\tau}^{-p}\right\|_{1}}{m\times
p}d\nu(\tau)\right)$ $\displaystyle=$ $\displaystyle-\frac{1}{m\times
p}\int_{\Omega}\log\det(\mathcal{D}_{\tau})d\nu(\tau)$ (140)

Proof: Because
$\int_{\Omega}\left\|\mathcal{D}_{\tau}^{-p}\right\|_{\rho}d\nu(\tau)<\infty$,
we have that $\mathcal{D}_{\tau}$ are TPD tensors for $\tau$ almost everywhere
in $\Omega$. Then, we have

$\displaystyle\lim\limits_{p\rightarrow
0}\left(\frac{1}{p}\log\int_{\Omega}\frac{\left\|\mathcal{D}_{\tau}^{-p}\right\|_{1}}{m\times
p}d\nu(\tau)\right)$ $\displaystyle=_{1}$
$\displaystyle\lim\limits_{p\rightarrow
0}\frac{\int_{\Omega}\frac{-\sum\limits_{i=1}^{m\times
p}\log\lambda_{i}(\mathcal{D}_{\tau})}{m\times
p}d\nu(\tau)}{\int_{\Omega}\frac{\left\|\mathcal{D}_{\tau}^{-p}\right\|_{1}}{m\times
p}d\nu(\tau)}$ (141) $\displaystyle=$ $\displaystyle\frac{-1}{m\times
p}\int_{\Omega}\sum\limits_{i=1}^{m\times
p}\log\lambda_{i}(\mathcal{D}_{\tau})d\nu(\tau)$ $\displaystyle=_{2}$
$\displaystyle\frac{-1}{m\times
p}\int_{\Omega}\log\det(\mathcal{D}_{\tau})d\nu(\tau),$

where $=_{1}$ is from L’Hopital’s rule, and $=_{2}$ is obtained from $\det$
definition. $\hfill\Box$

###  3.4 T-product Tensor Norm Inequalities by Majorization

In this section, we will apply derived majorization inequalities for T-product
tensors to multivariate T-product tensor norm inequalities which will be used
to bound random T-product tensor concentration inequalities in later sections.
We will begin to present a Lie-Trotter product formula for tensors.

###### Lemma 7

Let $m\in\mathbb{N}$ and $(\mathcal{L}_{k})_{k=1}^{m}$ be a finite sequence of
bounded T-product tensors with dimensions
$\mathcal{L}_{k}\in\mathbb{R}^{m\times m\times p}$, then we have

$\displaystyle\lim_{n\rightarrow\infty}\left(\prod_{k=1}^{m}\exp(\frac{\mathcal{L}_{k}}{n})\right)^{n}$
$\displaystyle=$ $\displaystyle\exp\left(\sum_{k=1}^{m}\mathcal{L}_{k}\right)$
(142)

Proof:

We will prove the case for $m=2$, and the general value of $m$ can be obtained
by mathematical induction. Let $\mathcal{L}_{1},\mathcal{L}_{2}$ be bounded
tensors act on some Hilbert space. Define
$\mathcal{C}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\exp((\mathcal{L}_{1}+\mathcal{L}_{2})/n)$, and
$\mathcal{D}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\exp(\mathcal{L}_{1}/n)\star\exp(\mathcal{L}_{2}/n)$. Note we have
following estimates for the norm of tensors $\mathcal{C},\mathcal{D}$:

$\displaystyle\left\|\mathcal{C}\right\|,\left\|\mathcal{D}\right\|\leq\exp\left(\frac{\left\|\mathcal{L}_{1}\right\|+\left\|\mathcal{L}_{2}\right\|}{n}\right)=\left[\exp\left(\left\|\mathcal{L}_{1}\right\|+\left\|\mathcal{L}_{2}\right\|\right)\right]^{1/n}.$
(143)

From the Cauchy-Product formula, the tensor $\mathcal{D}$ can be expressed as:

$\displaystyle\mathcal{D}$ $\displaystyle=$
$\displaystyle\exp(\mathcal{L}_{1}/n)\star\exp(\mathcal{L}_{2}/n)=\sum_{i=0}^{\infty}\frac{(\mathcal{L}_{1}/n)^{i}}{i!}\star\sum_{j=0}^{\infty}\frac{(\mathcal{L}_{2}/n)^{j}}{j!}$
(144) $\displaystyle=$
$\displaystyle\sum_{l=0}^{\infty}n^{-l}\sum_{i=0}^{l}\frac{\mathcal{L}_{1}^{i}}{i!}\star\frac{\mathcal{L}_{2}^{l-i}}{(l-i)!},$

then we can bound the norm of $\mathcal{C}-\mathcal{D}$ as

$\displaystyle\left\|\mathcal{C}-\mathcal{D}\right\|$ $\displaystyle=$
$\displaystyle\left\|\sum_{i=0}^{\infty}\frac{\left([\mathcal{L}_{1}+\mathcal{L}_{2}]/n\right)^{i}}{i!}-\sum_{l=0}^{\infty}n^{-l}\sum_{i=0}^{l}\frac{\mathcal{L}_{1}^{i}}{i!}\star\frac{\mathcal{L}_{2}^{l-i}}{(l-i)!}\right\|$
(145) $\displaystyle=$
$\displaystyle\left\|\sum_{i=2}^{\infty}k^{-i}\frac{\left([\mathcal{L}_{1}+\mathcal{L}_{2}]\right)^{i}}{i!}-\sum_{m=l}^{\infty}n^{-l}\sum_{i=0}^{l}\frac{\mathcal{L}_{1}^{i}}{i!}\star\frac{\mathcal{L}_{2}^{l-i}}{(l-i)!}\right\|$
$\displaystyle\leq$
$\displaystyle\frac{1}{k^{2}}\left[\exp(\left\|\mathcal{L}_{1}\right\|+\left\|\mathcal{L}_{2}\right\|)+\sum_{l=2}^{\infty}n^{-l}\sum_{i=0}^{l}\frac{\left\|\mathcal{L}_{1}\right\|^{i}}{i!}\cdot\frac{\left\|\mathcal{L}_{2}\right\|^{l-i}}{(l-i)!}\right]$
$\displaystyle=$
$\displaystyle\frac{1}{n^{2}}\left[\exp\left(\left\|\mathcal{L}_{1}\right\|+\left\|\mathcal{L}_{2}\right\|\right)+\sum_{l=2}^{\infty}n^{-l}\frac{(\left\|\mathcal{L}_{1}\right\|+\left\|\mathcal{L}_{2}\right\|)^{l}}{l!}\right]$
$\displaystyle\leq$
$\displaystyle\frac{2\exp\left(\left\|\mathcal{L}_{1}\right\|+\left\|\mathcal{L}_{2}\right\|\right)}{n^{2}}.$

For the difference between the higher power of $\mathcal{C}$ and
$\mathcal{D}$, we can bound them as

$\displaystyle\left\|\mathcal{C}^{n}-\mathcal{D}^{n}\right\|$ $\displaystyle=$
$\displaystyle\left\|\sum_{l=0}^{n-1}\mathcal{C}^{m}(\mathcal{C}-\mathcal{D})\mathcal{C}^{n-l-1}\right\|$
(146) $\displaystyle\leq_{1}$
$\displaystyle\exp(\left\|\mathcal{L}_{1}\right\|+\left\|\mathcal{L}_{2}\right\|)\cdot
n\cdot\left\|\mathcal{L}_{1}-\mathcal{L}_{2}\right\|,$

where the inequality $\leq_{1}$ uses the following fact

$\displaystyle\left\|\mathcal{C}\right\|^{l}\left\|\mathcal{D}\right\|^{n-l-1}\leq\exp\left(\left\|\mathcal{L}_{1}\right\|+\left\|\mathcal{L}_{2}\right\|\right)^{\frac{n-1}{n}}\leq\exp\left(\left\|\mathcal{L}_{1}\right\|+\left\|\mathcal{L}_{2}\right\|\right),$
(147)

based on Eq. (143). By combining with Eq. (145), we have the following bound

$\displaystyle\left\|\mathcal{C}^{n}-\mathcal{D}^{n}\right\|$
$\displaystyle\leq$
$\displaystyle\frac{2\exp\left(2\left\|\mathcal{L}_{1}\right\|+2\left\|\mathcal{L}_{2}\right\|\right)}{n}.$
(148)

Then this lemma is proved when $n$ goes to infity. $\hfill\Box$

Below, new multivariate norm inequalities for T-product tensors are provided
according to previous majorization theorems.

###### Theorem 9

Let $\mathcal{C}_{i}\in\mathbb{R}^{m\times m\times p}$ be TPD tensors, where
$1\leq i\leq n$, $\left\|\cdot\right\|_{\rho}$ be a unitarily invaraint norm
with corresponding gauge function $\rho$. For any continous function
$f:(0,\infty)\rightarrow[0,\infty)$ such that $x\rightarrow\log f(e^{x})$ is
convex on $\mathbb{R}$, we have

$\displaystyle\left\|f\left(\exp\left(\sum\limits_{i=1}^{n}\log\mathcal{C}_{i}\right)\right)\right\|_{\rho}$
$\displaystyle\leq$
$\displaystyle\exp\int_{-\infty}^{\infty}\log\left\|f\left(\left|\prod\limits_{i=1}^{n}\mathcal{C}_{i}^{1+\iota
t}\right|\right)\right\|_{\rho}\beta_{0}(t)dt,$ (149)

where $\beta_{0}(t)=\frac{\pi}{2(\cosh(\pi t)+1)}$.

For any continous function $g(0,\infty)\rightarrow[0,\infty)$ such that
$x\rightarrow g(e^{x})$ is convex on $\mathbb{R}$, we have

$\displaystyle\left\|g\left(\exp\left(\sum\limits_{i=1}^{n}\log\mathcal{C}_{i}\right)\right)\right\|_{\rho}$
$\displaystyle\leq$
$\displaystyle\int_{-\infty}^{\infty}\left\|g\left(\left|\prod\limits_{i=1}^{n}\mathcal{C}_{i}^{1+\iota
t}\right|\right)\right\|_{\rho}\beta_{0}(t)dt.$ (150)

Proof: From Hirschman interpolation theorem [25] and $\theta\in[0,1]$, we have

$\displaystyle\log\left|h(\theta)\right|\leq\int_{-\infty}^{\infty}\log\left|h(\iota
t)\right|^{1-\theta}\beta_{1-\theta}(t)dt+\int_{-\infty}^{\infty}\log\left|h(1+\iota
t)\right|^{\theta}\beta_{\theta}(t)dt,$ (151)

where $h(z)$ be uniformly bounded on $S\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\\{z\in\mathbb{C}:0\leq\Re(z)\leq 1\\}$ and holomorphic on $S$. The
term $d\beta_{\theta}(t)$ is defined as :

$\displaystyle\beta_{\theta}(t)\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\frac{\sin(\pi\theta)}{2\theta(\cos(\pi t)+\cos(\pi\theta))}.$
(152)

Let $H(z)$ be a uniformly bounded holomorphic function with values in
$\mathbb{C}^{m\times m\times p}$. Fix some $\theta\in[0,1]$ and let
$\mathcal{U},\mathcal{V}\in\mathbb{C}^{m\times m\times p}$ be normalized
tensors such that
$\langle\mathcal{U},\mathcal{H}(\theta)\star\mathcal{V}\rangle=\left\|H(\theta)\right\|$.
If we define $h(z)$ as $h(z)\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\langle\mathcal{U},\mathcal{H}(z)\star\mathcal{V}\rangle$, we have
following bound: $\left|h(z)\right|\leq\left\|H(z)\right\|$ for all $z\in S$.
From Hirschman interpolation theorem, we then have following interpolation
theorem for tensor-valued function:

$\displaystyle\log\left\|H(\theta)\right\|\leq\int_{-\infty}^{\infty}\log\left\|H(\iota
t)\right\|^{1-\theta}\beta_{1-\theta}(t)dt+\int_{-\infty}^{\infty}\log\left\|H(1+\iota
t)\right\|^{\theta}\beta_{\theta}(t)dt.$ (153)

Let $H(z)=\prod\limits_{i=1}^{n}\mathcal{C}^{z}_{i}$. Then the first term in
the R.H.S. of Eq. (153) is zero since $H(\iota t)$ is a product of unitary
tensors. Then we have

$\displaystyle\log\left\|\left|\prod\limits_{i=1}^{n}\mathcal{C}_{i}^{\theta}\right|^{\frac{1}{\theta}}\right\|\leq\int_{-\infty}^{\infty}\log\left\|\prod\limits_{i=1}^{n}\mathcal{C}_{i}^{1+\iota
t}\right\|\beta_{\theta}(t)dt.$ (154)

From Lemma 3, we have following relations:

$\displaystyle\left|\prod\limits_{i=1}^{n}\left(\wedge^{k}\mathcal{C}_{i}\right)^{\theta}\right|^{\frac{1}{\theta}}=\wedge^{k}\left|\prod\limits_{i=1}^{n}\mathcal{C}^{\theta}_{i}\right|^{\frac{1}{\theta}},$
(155)

and

$\displaystyle\left|\prod\limits_{i=1}^{n}\left(\wedge^{k}\mathcal{C}_{i}\right)^{1+\iota
t}\right|=\wedge^{k}\left|\prod\limits_{i=1}^{n}\mathcal{C}^{1+\iota
t}_{i}\right|.$ (156)

If Eq. (154) is applied to $\wedge^{k}\mathcal{C}_{i}$ for $1\leq k\leq r$, we
have following log-majorization relation from Eqs. (155) and (156):

$\displaystyle\log\vec{\lambda}\left(\left|\prod\limits_{i=1}^{n}\mathcal{C}_{i}^{\theta}\right|^{\frac{1}{\theta}}\right)\prec\int_{-\infty}^{\infty}\log\vec{\lambda}\left|\prod\limits_{i=1}^{n}\mathcal{C}_{i}^{1+\iota
t}\right|^{\frac{1}{\theta}}\beta_{\theta}(t)dt.$ (157)

Moreover, we have the equality condition in Eq. (157) for $k=r$ due to
following identies:

$\displaystyle\det\left|\prod\limits_{i=1}^{n}\mathcal{C}_{i}^{\theta}\right|^{\frac{1}{\theta}}=\det\left|\prod\limits_{i=1}^{n}\mathcal{C}_{i}^{1+\iota
t}\right|=\prod\limits_{i=1}^{n}\det\mathcal{C}_{i}.$ (158)

At this stage, we are ready to apply Theorem 8 for the log-majorization
provided by Eq. (157) to get following facts:

$\displaystyle\left\|f\left(\left|\prod\limits_{i=1}^{n}\mathcal{C}_{i}^{\theta}\right|^{\frac{1}{\theta}}\right)\right\|_{\rho}$
$\displaystyle\leq$
$\displaystyle\exp\int_{-\infty}^{\infty}\log\left\|f\left(\left|\prod\limits_{i=1}^{n}\mathcal{C}_{i}^{1+\iota
t}\right|\right)\right\|_{\rho}\beta_{\theta}(t)dt,$ (159)

and

$\displaystyle\left\|g\left(\left|\prod\limits_{i=1}^{n}\mathcal{C}_{i}^{\theta}\right|^{\frac{1}{\theta}}\right)\right\|_{\rho}$
$\displaystyle\leq$
$\displaystyle\int_{-\infty}^{\infty}\left\|g\left(\left|\prod\limits_{i=1}^{n}\mathcal{C}_{i}^{1+\iota
t}\right|\right)\right\|_{\rho}\beta_{\theta}(t)dt.$ (160)

From Lie product formula for tensors given by Lemma 7, we have

$\displaystyle\left|\prod\limits_{i=1}^{n}\mathcal{C}_{i}^{\theta}\right|^{\frac{1}{\theta}}\rightarrow\exp\left(\sum\limits_{i=1}^{n}\log\mathcal{C}_{i}\right).$
(161)

By setting $\theta\rightarrow 0$ in Eqs. (159), (160) and using Lie product
formula given by Eq. (161), we will get Eqs. (149) and (150). $\hfill\Box$

##  4 Applications of T-product Tensor Norm Inequalities

The purpose of this section is to apply new derived T-product tensor norm
inequalities to obtain random symmetric T-product tensor Bernstein bounds. In
Section 4.1, Ky Fan $k$-norm inequalities for T-product tensors will be
provided and such Ky Fan $k$-norm inequalities will be utilized to establish
T-product tensor Bernstein bounds in Section 4.2 and Section 4.3.

###  4.1 Ky Fan $k$-norm Tail Bounds

We will present several lemmas required to prove Ky Fan $k$-norm tail bounds.

###### Lemma 8

Given a symmetric T-product tensor $\mathcal{C}\in\mathbb{R}^{m\times m\times
p}$ which can be expressed as the format shown by Eq. (28), the T-eigenvalues
of $\mathcal{C}$ with respect to the matrix $\mathbf{C}_{i}$ are denoted as
$\lambda_{i,k_{i}}$, where $1\leq k_{i}\leq m$, and we assume that
$\lambda_{i,1}\geq\lambda_{i,2}\geq\cdots\geq\lambda_{i,m}$ (including
multiplicities). We have following relation about T-eigenvalues summation
representation:

$\displaystyle\sum\limits_{i=1}^{p}\max\limits_{\mathbf{U}_{i}\mathbf{U}^{\mathrm{H}}_{i}=\mathbf{I}_{k_{i}}}\mathrm{Tr}\mathbf{U}_{i}\mathbf{C}_{i}\mathbf{U}^{\mathrm{H}}_{i}=\sum\limits_{i=1}^{p}\sum\limits_{j=1}^{k_{i}}\lambda_{i,j}(\mathbf{C}_{i}),$
(162)

and

$\displaystyle\sum\limits_{i=1}^{p}\min\limits_{\mathbf{U}_{i}\mathbf{U}^{\mathrm{H}}_{i}=\mathbf{I}_{k_{i}}}\mathrm{Tr}\mathbf{U}_{i}\mathbf{C}_{i}\mathbf{U}^{\mathrm{H}}_{i}=\sum\limits_{i=1}^{p}\sum\limits_{j=1}^{k_{i}}\lambda_{i,m-j+1}(\mathbf{C}_{i}).$
(163)

Proof: From Theorem 1, each matrix $\mathbf{C}_{i}$ associated to
$\mathcal{C}$ based on the format shown by Eq. (28) is Hermitian, then the
matrix $\mathbf{C}_{i}$ can be diagonalized as $\mathbf{D}_{i}$ by the unitary
matrix $\mathbf{V}_{i}$. Without loss of generality, we may assume that
$\mathbf{C}_{i}$ are diagonal matrices. Then, we have

$\displaystyle\mathrm{Tr}\mathbf{U}_{i}\mathbf{D}_{i}\mathbf{U}^{\mathrm{H}}_{i}=\sum\limits_{j=1}^{k_{i}}\sum\limits_{l=1}^{m}u_{i_{k,l}}u^{\ast}_{i_{k,l}}\lambda_{i,l}=\sum\limits_{j=1}^{k_{i}}\sum\limits_{l=1}^{m}p_{i_{k,l}}\lambda_{i,l}=[\overbrace{1,\cdots,1}^{k_{i}\mbox{terms}}]\mathbf{P}_{i}\left[\begin{array}[]{c}\lambda_{i,1}\\\
\lambda_{i,2}\\\ \vdots\\\ \lambda_{i,m}\\\ \end{array}\right],$ (168)

where the superscript $\ast$ is the operation of a complex conjugate, and
$\mathbf{P}_{i}=(p_{i_{k,l}})$ is a $k_{i}\times m$ stochastic matrix. From
the fact provided by 2.C.1 in [20], there exists an $(m-k_{i})\times m$ matrix
$\mathbf{Q}_{i}$ such that $\left[\begin{array}[]{c}\mathbf{P}_{i}\\\
\mathbf{Q}_{i}\\\ \end{array}\right]$ is a doubly stochastic matrix. Then, Eq.
(168) can be expressed as

$\displaystyle\mathrm{Tr}\mathbf{U}_{i}\mathbf{D}_{i}\mathbf{U}^{\mathrm{H}}_{i}$
$\displaystyle=$
$\displaystyle[\overbrace{1,\cdots,1}^{k_{i}\mbox{terms}},\overbrace{0,\cdots,0}^{m-k_{i}\mbox{terms}}]\left[\begin{array}[]{c}\mathbf{P}_{i}\\\
\mathbf{Q}_{i}\\\ \end{array}\right]\left[\begin{array}[]{c}\lambda_{i,1}\\\
\lambda_{i,2}\\\ \vdots\\\ \lambda_{i,m}\\\ \end{array}\right].$ (175)

Because we have

$\displaystyle[\lambda_{i,1},\cdots,\lambda_{i,m}][\mathbf{P}^{\mathrm{T}}_{i}\mathbf{Q}^{\mathrm{T}}_{i}]\preceq[\lambda_{i,1},\cdots,\lambda_{i,m}],$
(176)

then, we can apply the fact 3.H.2 about majorization in [20] to get

$\displaystyle\mathrm{Tr}\mathbf{U}_{i}\mathbf{D}_{i}\mathbf{U}^{\mathrm{H}}_{i}\leq\sum\limits_{j=1}^{k_{i}}\lambda_{i,j}(\mathbf{C}_{i}),$
(177)

and

$\displaystyle\mathrm{Tr}\mathbf{U}_{i}\mathbf{D}_{i}\mathbf{U}^{\mathrm{H}}_{i}\geq\sum\limits_{j=1}^{k_{i}}\lambda_{i,m-j+1}(\mathbf{C}_{i}),$
(178)

where $1\leq k_{i}\leq m$. We have to note that Eqs (177) and (178) are
achieved with equalities for
$\mathbf{U}_{i}\mathbf{V}_{i}=(\mathbf{I}_{k_{i}},\mathbf{O})$ and
$\mathbf{U}_{i}\mathbf{V}_{i}=(\mathbf{O},\mathbf{I}_{k_{i}})$, respectively.
This lemma is proved by taking summation with respect to the index $i$ at both
sides of Eqs (177) and (178). $\hfill\Box$

Following lemma will apply Lemma 8 to prove majorization relation between
T-product tensors summation.

###### Lemma 9

Given two symmetric T-product tensors
$\mathcal{C},\mathcal{D}\in\mathbb{R}^{m\times m\times p}$. We have following
majorization relation about T-singular values:

$\displaystyle\sigma(\mathcal{C}+\mathcal{D})\prec_{w}\sigma(\mathcal{C})+\sigma(\mathcal{D}).$
(179)

Proof: Since we have

$\displaystyle\sum\limits_{i=1}^{p}\sum\limits_{j=1}^{k_{i}}\sigma_{i,j}(\mathcal{C}+\mathcal{D})$
$\displaystyle=_{1}$
$\displaystyle\max\limits_{\mathbf{U}_{i}\mathbf{U}^{\mathrm{H}}_{i}=_{1}\mathbf{I}_{k_{i}}}\Re\left(\sum\limits_{i=1}^{p}\mathrm{Tr}\mathbf{U}_{i}(\mathbf{C}_{i}+\mathbf{D}_{i})\mathbf{U}^{\mathrm{H}}_{i}\right)$
(180) $\displaystyle\leq$
$\displaystyle\max\limits_{\mathbf{U}_{i}\mathbf{U}^{\mathrm{H}}_{i}=_{1}\mathbf{I}_{k_{i}}}\Re\left(\sum\limits_{i=1}^{p}\mathrm{Tr}\mathbf{U}_{i}\mathbf{C}_{i}\mathbf{U}^{\mathrm{H}}_{i}\right)+\max\limits_{\mathbf{U}_{i}\mathbf{U}^{\mathrm{H}}_{i}=_{1}\mathbf{I}_{k_{i}}}\Re\left(\sum\limits_{i=1}^{p}\mathrm{Tr}\mathbf{U}_{i}\mathbf{D}_{i}\mathbf{U}^{\mathrm{H}}_{i}\right)$
$\displaystyle=_{2}$
$\displaystyle\sum\limits_{i=1}^{p}\sum\limits_{j=1}^{k_{i}}\sigma_{i,j}(\mathcal{C})+\sum\limits_{i=1}^{p}\sum\limits_{j=1}^{k_{i}}\sigma_{i,j}(\mathcal{D})$

where $\Re$ is the operation to take the real part, and the equalities $=_{1}$
and $=_{2}$ come from Lemma 8. $\hfill\Box$

We are ready to introduce the following two lemmas about Ky Fan $k$-norm
inequalities for the product of tensors (Lemma 10) and the summation of
tensors (Lemma 11).

###### Lemma 10

Let $\mathcal{C}_{i}\in\mathbb{R}^{m\times m\times p}$ be symmetric T-product
tensorsand let $p_{i}$ be positive real numbers satisfying
$\sum\limits_{i=1}^{m}\frac{1}{p_{i}}=1$. Then, we have

$\displaystyle\left\|\left|\prod\limits_{i=1}^{m}\mathcal{C}_{i}\right|^{s}\right\|_{(k)}\leq\prod\limits_{i=1}^{m}\left(\left\|\left|\mathcal{C}_{i}\right|^{sp_{i}}\right\|_{(k)}\right)^{\frac{1}{p_{i}}}\leq\sum\limits_{i=1}^{m}\frac{\left\|\left|\mathcal{C}_{i}\right|^{sp_{i}}\right\|_{(k)}}{p_{i}}$
(181)

where $s\geq 1$ and $k\in\\{1,2,\cdots,m\times p\\}$.

Proof: Since we have

$\displaystyle\left\|\left|\prod\limits_{i=1}^{m}\mathcal{C}_{i}\right|^{s}\right\|_{(k)}=\sum\limits_{j=1}^{k}\lambda_{j}\left(\left|\prod\limits_{i=1}^{m}\mathcal{C}_{i}\right|^{s}\right)=\sum\limits_{j=1}^{k}\lambda^{s}_{j}\left(\left|\prod\limits_{i=1}^{m}\mathcal{C}_{i}\right|\right)=\sum\limits_{j=1}^{k}\sigma^{s}_{j}\left(\prod\limits_{i=1}^{m}\mathcal{C}_{i}\right),$
(182)

where we have orders for eigenvalues as
$\lambda_{1}\geq\lambda_{2}\geq\cdots$, and singular values as
$\sigma_{1}\geq\sigma_{2}\geq\cdots$.

From Lemma 3, we have

$\displaystyle\left\|\left(\prod\limits_{i=1}^{m}\mathcal{C}_{i}\right)^{\wedge
k}\right\|=\prod\limits_{j=1}^{k}\sigma_{j}\left(\prod\limits_{i=1}^{m}\mathcal{C}_{i}\right).$
(183)

Apply Theorem H.1. in [20] to each matrix at block diagonal of
$\mbox{bcirc}(\mathcal{C}_{i})$ by Eq. (28), we will have

$\displaystyle\sum\limits_{j=1}^{k}\sigma^{s}_{j}\left(\prod\limits_{i=1}^{m}\mathcal{C}_{i}\right)\leq\sum\limits_{j=1}^{k}\left(\prod\limits_{i=1}^{m}\sigma^{s}_{j}(\mathcal{C}_{i})\right).$
(184)

Then, we can apply Hölder’s inequality to Eq. (184) and obtain

$\displaystyle\sum\limits_{j=1}^{k}\left(\prod\limits_{i=1}^{m}\sigma^{s}_{j}(\mathcal{C}_{i})\right)$
$\displaystyle\leq$
$\displaystyle\prod_{i=1}^{m}\left(\sum\limits_{j=1}^{k}\sigma^{sp_{i}}_{j}(\mathcal{C}_{i})\right)^{\frac{1}{p_{i}}}=\prod_{i=1}^{m}\left(\sum\limits_{j=1}^{k}\lambda^{sp_{i}}_{j}(\left|\mathcal{C}_{i}\right|)\right)^{\frac{1}{p_{i}}}$
(185) $\displaystyle=$
$\displaystyle\prod_{i=1}^{m}\left(\sum\limits_{j=1}^{k}\lambda_{j}(\left|\mathcal{C}_{i}\right|^{sp_{i}})\right)^{\frac{1}{p_{i}}}=\prod\limits_{i=1}^{m}\left(\left\|\left|\mathcal{C}_{i}\right|^{sp_{i}}\right\|_{(k)}\right)^{\frac{1}{p_{i}}}$

The second inequality in Eq. (181) is obtained by applying Young’s inequality
to numbers $\left\|\left|\mathcal{C}_{i}\right|^{sp_{i}}\right\|_{(k)}$ for
$1\leq i\leq m$. This completes the proof. $\hfill\Box$

###### Lemma 11

Let $\mathcal{C}_{i}\in\mathbb{R}^{m\times m\times p}$ be symmetric T-product
tensors, then we have

$\displaystyle\left\|\left|\sum\limits_{i=1}^{m}\mathcal{C}_{i}\right|^{s}\right\|_{(k)}\leq
m^{s-1}\sum\limits_{i=1}^{m}\left\|\left|\mathcal{C}_{i}\right|^{s}\right\|_{(k)}$
(186)

where $s\geq 1$ and $k\in\\{1,2,\cdots,m\times p\\}$.

Proof: Since we have

$\displaystyle\left\|\left|\sum\limits_{i=1}^{m}\mathcal{C}_{i}\right|^{s}\right\|_{(k)}=\sum\limits_{j=1}^{k}\lambda_{j}\left(\left|\sum\limits_{i=1}^{m}\mathcal{C}_{i}\right|^{s}\right)=\sum\limits_{j=1}^{k}\lambda^{s}_{j}\left(\left|\sum\limits_{i=1}^{m}\mathcal{C}_{i}\right|\right)=\sum\limits_{j=1}^{k}\sigma^{s}_{j}\left(\sum\limits_{i=1}^{m}\mathcal{C}_{i}\right).$
(187)

where we have orders for eigenvalues as
$\lambda_{1}\geq\lambda_{2}\geq\cdots$, and singular values as
$\sigma_{1}\geq\sigma_{2}\geq\cdots$.

From Lemma 9, we have

$\displaystyle\sum\limits_{j=1}^{k}\sigma_{j}(\sum\limits_{i=1}^{m}\mathcal{C}_{i})\leq\sum\limits_{j=1}^{k}\left(\sum\limits_{i=1}^{m}\sigma_{j}(\mathcal{C}_{i})\right),$
(188)

where $k\in\\{1,2,\cdots,m\times p\\}$. Then, we have

$\displaystyle\sum\limits_{j=1}^{k}\sigma^{s}_{j}(\sum\limits_{i=1}^{m}\mathcal{C}_{i})$
$\displaystyle\leq$
$\displaystyle\sum\limits_{j=1}^{k}\left(\sum\limits_{i=1}^{m}\sigma_{j}(\mathcal{C}_{i})\right)^{s}\leq
m^{s-1}\sum\limits_{j=1}^{k}\left(\sum\limits_{i=1}^{m}\sigma^{s}_{j}(\mathcal{C}_{i})\right)$
(189) $\displaystyle=$ $\displaystyle
m^{s-1}\sum\limits_{j=1}^{k}\left(\sum\limits_{i=1}^{m}\sigma^{s}_{j}(\left|\mathcal{C}_{i}\right|)\right)=m^{s-1}\sum\limits_{j=1}^{k}\left(\sum\limits_{i=1}^{m}\sigma_{j}(\left|\mathcal{C}_{i}\right|^{s})\right)$
$\displaystyle=$ $\displaystyle
m^{s-1}\sum\limits_{i=1}^{m}\left\|\left|\mathcal{C}_{i}\right|^{s}\right\|_{(k)}$

$\hfill\Box$

Now, we are ready to present our main theorem about Ky Fan $k$-norm
probability bound for a function of tensors summation.

###### Theorem 10

Consider a sequence $\\{\mathcal{X}_{j}\in\mathbb{R}^{m\times m\times p}\\}$
of independent, random, symmetric T-product tensors. Let $g$ be a polynomial
function with degree $n$ and nonnegative coefficients
$a_{0},a_{1},\cdots,a_{n}$ raised by power $s\geq 1$, i.e.,
$g(x)=\left(a_{0}+a_{1}x+\cdots+a_{n}x^{n}\right)^{s}$. Suppose following
condition is satisfied:

$\displaystyle
g\left(\exp\left(t\sum\limits_{j=1}^{m}\mathcal{X}_{j}\right)\right)\succeq\exp\left(tg\left(\sum\limits_{j=1}^{m}\mathcal{X}_{j}\right)\right)\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{almost surely},$ (190)

where $t>0$. Then, we have

$\displaystyle\mathrm{Pr}\left(\left\|g\left(\sum\limits_{j=1}^{m}\mathcal{X}_{j}\right)\right\|_{(k)}\geq\theta\right)$
$\displaystyle\leq$
$\displaystyle(n+1)^{s-1}\inf\limits_{t,p_{j}}\exp\left(-\theta t\right)$
(191)
$\displaystyle\cdot\left(ka_{0}^{s}+\sum\limits_{l=1}^{n}\sum\limits_{j=1}^{m}\frac{a^{ls}_{l}\mathbb{E}\left\|\exp\left(p_{j}lst\mathcal{X}_{j}\right)\right\|_{(k)}}{p_{j}}\right),$

where $\sum\limits_{j=1}^{m}\frac{1}{p_{j}}=1$ and $p_{j}>0$.

Proof: Let $t>0$ be a parameter to be chosen later. Then

$\displaystyle\mathrm{Pr}\left(\left\|g\left(\sum\limits_{j=1}^{m}\mathcal{X}_{j}\right)\right\|_{(k)}\geq\theta\right)=\mathrm{Pr}\left(\left\|\exp\left(tg\left(\sum\limits_{j=1}^{m}\mathcal{X}_{j}\right)\right)\right\|_{(k)}\geq\exp\left(\theta
t\right)\right)$ $\displaystyle\leq_{1}\exp\left(-\theta
t\right)\mathbb{E}\left(\left\|\exp\left(tg\left(\sum\limits_{j=1}^{m}\mathcal{X}_{j}\right)\right)\right\|_{(k)}\right)$
$\displaystyle\leq_{2}\exp\left(-\theta
t\right)\mathbb{E}\left(\left\|g\left(\exp\left(t\sum\limits_{j=1}^{m}\mathcal{X}_{j}\right)\right)\right\|_{(k)}\right)$
(192)

where $\leq_{1}$ uses Markov’s inequality, $\leq_{2}$ requires conditions
provided by Eq. (190).

We can further bound the expectation term in Eq. (191) as

$\displaystyle\mathbb{E}\left(\left\|g\left(\exp\left(t\sum\limits_{j=1}^{m}\mathcal{X}_{j}\right)\right)\right\|_{(k)}\right)$
$\displaystyle\leq_{3}$
$\displaystyle\mathbb{E}\int_{-\infty}^{\infty}\left\|g\left(\left|\prod\limits_{j=1}^{m}e^{(1+\iota\tau)t\mathcal{X}_{j}}\right|\right)\right\|_{(k)}\beta_{0}(\tau)d\tau$
(193) $\displaystyle\leq_{4}$
$\displaystyle(n+1)^{s-1}\left(ka_{0}^{s}\right.$
$\displaystyle\left.+\sum\limits_{l=1}^{n}a^{ls}_{l}\mathbb{E}\int_{-\infty}^{\infty}\left\|\left|\prod\limits_{j=1}^{m}e^{(1+\iota\tau)t\mathcal{X}_{j}}\right|^{ls}\right\|_{(k)}\beta_{0}(\tau)d\tau\right),\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ $

where $\leq_{3}$ from Eq. (150) in Theorem 9, $\leq_{4}$ is obtained from
function $g$ definition and Lemma 11. Again, the expectation term in Eq. (193)
can be further bounded by Lemma 10 as

$\displaystyle\mathbb{E}\int_{-\infty}^{\infty}\left\|\left|\prod\limits_{j=1}^{m}e^{(1+\iota\tau)t\mathcal{X}_{j}}\right|^{ls}\right\|_{(k)}\beta_{0}(\tau)d\tau$
$\displaystyle\leq$
$\displaystyle\mathbb{E}\int_{-\infty}^{\infty}\sum\limits_{j=1}^{m}\frac{\left\|\left|e^{t\mathcal{X}_{j}}\right|^{p_{j}ls}\right\|_{(k)}}{p_{j}}\beta_{0}(\tau)d\tau$
(194) $\displaystyle=$
$\displaystyle\sum\limits_{j=1}^{m}\frac{\mathbb{E}\left\|e^{p_{j}lst\mathcal{X}_{j}}\right\|_{(k)}}{p_{j}}.$

Note that the final equality is obtained due to that the integrand is
independent of the variable $\tau$ and
$\int_{-\infty}^{\infty}\beta_{0}(\tau)d\tau=1$.

Finally, this theorem is established from Eqs. (4.1), (193), and (194).
$\hfill\Box$

Remarks: The condition provided by Eq. (190) can be achieved by normalizing
tensors $\mathcal{X}_{j}$ through scaling.

###  4.2 T-product Tensor Bernstein Bound

In this section, we will present a tensor Bernstein bound for the maximum and
the minimum T-eigenvalue for summation of random symmetric T-product tensors.
We will provide the following definition to define a random structure for the
T-product tensor $\mathcal{X}\in\mathbb{R}^{m\times m\times p}$.

###### Definition 1

Random structure for random symmetric T-product tensor
$\mathcal{X}\in\mathbb{R}^{m\times m\times p}$

  1. 1.

There are $p$ Hermitian matrices with size $m\times m$, denoted as
$\mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{p}$, obtained from Eq. (28).
The entries for the matrix $\mathbf{X}_{i}$ are denoted by $(x_{i_{j,k}})$,
where $x_{i_{j,k}}$ is a complex number.

  2. 2.

For each $\mathbf{X}_{i}$, the random variables $x_{i_{j,j}}$, $\Re
x_{i_{j,k}}$ for $j<k$, and $\Im x_{i_{j,k}}$ for $j<k$, are independent.

  3. 3.

For each $\mathbf{X}_{i}$, the random variables $x_{i_{j,j}}$ follow Gaussian
distribution with zero mean and variance as $\frac{1}{m}$.

  4. 4.

For each $\mathbf{X}_{i}$, the random variables $\Re x_{i_{j,k}}$ for $j<k$,
and $\Im x_{i_{j,k}}$ for $j<k$, follow Gaussian distribution with zero mean
and variance as $\frac{1}{2m}$.

Following lemma is about the expectation of the largest T-eigenvalue of
symmetric T-product tensor $\exp(\gamma\mathcal{X})$, where $\gamma$ is a real
number.

###### Lemma 12

Given a random symmetric T-product tensor $\mathcal{X}\in\mathbb{R}^{m\times
m\times p}$ satisfying Definition 1 and any real number $\gamma$, we have

$\displaystyle\mathbb{E}\lambda_{1}\left(\exp(\gamma\mathcal{X})\right)\leq\frac{3mc_{1}c_{2}}{2}\int^{\infty}_{-\infty}(y-2)^{1/2}\exp\left[\gamma
y+c_{2}m(y-2)^{3/2}\right]dy\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\Psi(m,\gamma,c_{1},c_{2})$ (195)

where $\lambda_{1}$ is the largest T-eigenvalue, and $c_{1},c_{2}$ are
constants related to the bound of cumulative distribution function of the
largest eigenvalue of the random Hermitian matrix $\mathbf{X}$.

Proof: From random structure of discussed random symmetric T-product tensor
$\mathcal{X}\in\mathbb{R}^{m\times m\times p}$, all random Hermitian matrix
$\mathbf{X}_{i}$ have same probability density distributions. The maximum
T-eigenvalue of $\mathcal{X}$ will be equal to the maximum eigenvalue of
$\mathbf{X}_{i}$, and we use $\mathbf{X}$ to represent any random Hermitian
matrix $\mathbf{X}_{i}$ since they share same distribution.

From Eq. (2) in [26], given a $m\times m$ random Hermitian matrix
$\mathbf{X}$, we have

$\displaystyle\mathrm{Pr}\left(\lambda_{1}(\mathbf{X})\leq y\right)\leq
1-c_{1}\exp(-c_{2}m(y-2)^{3/2})$ (196)

where $c_{1},c_{2}$ are constants related to the bound of cumulative
distribution function of the largest eigenvalue of any random Hermitian matrix
$\mathbf{X}$. Then, we have

$\displaystyle\mathbb{E}\lambda_{1}(\exp(\gamma\mathcal{X}))$
$\displaystyle=_{1}$
$\displaystyle\mathbb{E}\exp(\gamma\lambda_{1}(\mathcal{X}))=\int^{\infty}_{-\infty}\exp(\gamma
y)d(\mathrm{Pr}\left(\lambda_{1}(\mathbf{X})\leq y\right))$ (197)
$\displaystyle\leq$ $\displaystyle\int^{\infty}_{-\infty}\exp(\gamma
y)d\left\\{1-c_{1}\exp\left[c_{2}m(y-2)^{3/2}\right]\right\\}$
$\displaystyle=$
$\displaystyle\frac{3mc_{1}c_{2}}{2}\int^{\infty}_{-\infty}(y-2)^{1/2}\exp\left[\gamma
y+c_{2}m(y-2)^{3/2}\right]dy,$

where $=_{1}$ comes from the spectral mapping theorem. $\hfill\Box$

We are ready to present our theorem about the maximum and the minimum of
T-eigenvalue for the summation of random symmetric T-product tensors.

######  Theorem 11 (T-product Tensor Bernstein Bound for T-eigenvalue)

Consider a sequence $\\{\mathcal{X}_{j}\in\mathbb{R}^{m\times m\times p}\\}$
of independent, random, symmetric T-product tensors with random structure
defined by Definition 1. Then we have following inequalities: given
$\theta_{1}>0$, we have

$\displaystyle\mathrm{Pr}\left(\lambda_{\max}\left(\sum\limits_{j=1}^{m}\mathcal{X}_{j}\right)\geq\theta_{1}\right)$
$\displaystyle\leq$
$\displaystyle\inf\limits_{t>0}\left[\frac{\exp(-\theta_{1}t)}{m}\sum\limits_{j=1}^{m}\Psi(m,mt,c_{1},c_{2})\right],$
(198)

and, given $\theta_{2}<0$, we have

$\displaystyle\mathrm{Pr}\left(\lambda_{\min}\left(\sum\limits_{j=1}^{m}\mathcal{X}_{j}\right)\leq\theta_{2}\right)$
$\displaystyle\leq$
$\displaystyle\inf\limits_{t>0}\left[\frac{\exp(\theta_{2}t)}{m}\sum\limits_{j=1}^{m}\Psi(m,-mt,c_{1},c_{2})\right].$
(199)

The $\Psi$ function is defined by Eq. (195).

Proof: Since we have

$\displaystyle\mathrm{Pr}\left(\lambda_{\max}\left(\sum\limits_{j=1}^{m}\mathcal{X}_{j}\right)\geq\theta_{1}\right)$
$\displaystyle=_{1}$
$\displaystyle\mathrm{Pr}\left(\sigma_{\max}\left(\sum\limits_{j=1}^{m}\mathcal{X}_{j}\right)\geq\theta_{1}\right)$
(200) $\displaystyle\leq_{2}$
$\displaystyle\inf\limits_{t,p_{j}}\exp(-\theta_{1}t)\left(\sum\limits_{j=1}^{m}\frac{\mathbb{E}\sigma_{\max}\left(\exp(p_{j}t\mathcal{X}_{j})\right)}{p_{j}}\right)$
$\displaystyle\leq_{3}$
$\displaystyle\inf\limits_{t,p_{j}}\exp(-\theta_{1}t)\left(\sum\limits_{j=1}^{m}\frac{\Psi(m,p_{j}t,c_{1},c_{2})}{p_{j}}\right)$
$\displaystyle\leq_{4}$
$\displaystyle\inf\limits_{t>0}\exp(-\theta_{1}t)\left(\sum\limits_{j=1}^{m}\frac{\Psi(m,mt,c_{1},c_{2})}{m}\right)$
$\displaystyle=$
$\displaystyle\inf\limits_{t>0}\left[\frac{\exp(-\theta_{1}t)}{m}\sum\limits_{j=1}^{m}\Psi(m,mt,c_{1},c_{2})\right],$

where $=_{1}$ comes from that maximum singular value equals to the maximum
absolute value of an T-eigenvalue and the maximum and the minimum of
T-eigenvalue has same distribution due to the symmetry of random structure
given by Definition 1; the inequality $\leq_{2}$ comes from Theorem 10 when
$g$ is the identity function; the equality $\leq_{3}$ comes from Lemma 12 and
$\sigma_{\max}\left(\exp(p_{j}t\mathcal{X}_{j})\right)=\lambda_{\max}\left(\exp(p_{j}t\mathcal{X}_{j})\right)$
due to TPD of $\exp(p_{j}t\mathcal{X}_{j})$; the inequality $\leq_{4}$ is
obtained by selecting $p_{j}=m$. Therefore, we have Eq. (198).

For the minimum T-eigenvalue, we also have

$\displaystyle\mathrm{Pr}\left(\lambda_{\min}\left(\sum\limits_{j=1}^{m}\mathcal{X}_{j}\right)\leq\theta_{2}\right)$
$\displaystyle=_{1}$
$\displaystyle\mathrm{Pr}\left(\lambda_{\max}\left(\sum\limits_{j=1}^{m}-\mathcal{X}_{j}\right)\geq-\theta_{2}\right)$
(201) $\displaystyle=_{2}$
$\displaystyle\mathrm{Pr}\left(\sigma_{\max}\left(\sum\limits_{j=1}^{m}-\mathcal{X}_{j}\right)\geq-\theta_{2}\right)$
$\displaystyle\leq_{3}$
$\displaystyle\inf\limits_{t,p_{j}}\exp(\theta_{2}t)\left(\sum\limits_{j=1}^{m}\frac{\mathbb{E}\sigma_{\max}\left(\exp(-p_{j}t\mathcal{X}_{j})\right)}{p_{j}}\right)$
$\displaystyle\leq_{4}$
$\displaystyle\inf\limits_{t,p_{j}}\exp(\theta_{2}t)\left(\sum\limits_{j=1}^{m}\frac{\Psi(m,-p_{j}t,c_{1},c_{2})}{p_{j}}\right)$
$\displaystyle\leq_{5}$
$\displaystyle\inf\limits_{t>0}\left[\frac{\exp(\theta_{2}t)}{m}\sum\limits_{j=1}^{m}\Psi(m,-mt,c_{1},c_{2})\right],$

where $=_{1}$ comes from Theorem 3; $=_{2}$ is true since the maximum singular
value equals to the maximum absolute value of an T-eigenvalue and the maximum
and the minimum of T-eigenvalue has same distribution due to the symmetry of
random structure given by Definition 1; the inequality $\leq_{3}$ comes from
Theorem 10 again when $g$ is an identity map; the equality $\leq_{4}$ comes
from Lemma 12 and
$\sigma_{\max}\left(\exp(p_{j}t\mathcal{X}_{j})\right)=\lambda_{\max}\left(\exp(p_{j}t\mathcal{X}_{j})\right)$
due to TPD of $\exp(p_{j}t\mathcal{X}_{j})$; the inequality $\leq_{5}$ is
obtained by selecting $p_{j}=m$. Hence, we have Eq. (199). $\hfill\Box$

###  4.3 Generalized T-product Tensor Bernstein Bound

In this section, we will present a generalized tensor Bernstein bound for Ky
Fan $k$-norm, and we will begin with a lemma to bound exponential of a random
T-product tensor.

###### Lemma 13

Suppose that $\mathcal{X}\in\mathbb{R}^{m\times m\times p}$ is a random
symmetric T-product tensor that satisfies

$\displaystyle\mathcal{X}^{p}\preceq\frac{p!\mathcal{A}^{2}}{2}\mbox{\leavevmode\nobreak\
\leavevmode\nobreak\ almost surely for $p=2,3,4,\cdots$,}$ (202)

where $\mathcal{A}$ is a fixed TPD tensor. Then, we have

$\displaystyle
e^{t\mathcal{X}}\preceq\mathcal{I}+t\mathcal{X}+\frac{t^{2}\mathcal{A}^{2}}{2(1-t)}\mbox{\leavevmode\nobreak\
\leavevmode\nobreak\ almost surely,}$ (203)

where $0<t<1$.

Proof: From Tayler series of the tensor exponential expansion, we have

$\displaystyle e^{t\mathcal{X}}$ $\displaystyle=$
$\displaystyle\mathcal{I}+t\mathcal{X}+\sum\limits_{p=2}^{\infty}\frac{t^{p}(\mathcal{X}^{p})}{p!}\preceq\mathcal{I}+t\mathcal{X}+\sum\limits_{p=2}^{\infty}\frac{t^{p}\mathcal{A}^{2}}{2}=\mathcal{I}+t\mathcal{X}+\frac{t^{2}\mathcal{A}^{2}}{2(1-t)}.$
(204)

Therefore, this Lemma is proved. $\hfill\Box$

###### Lemma 14

Given a random symmetric T-product tensor $\mathcal{X}\in\mathbb{R}^{m\times
m\times p}$ satisfying Definition 1, we have

$\displaystyle\mathbb{E}\sigma_{1}\left(\mathcal{X}\right)\leq\int^{\infty}_{-2}d_{1}\exp(-d_{2}mz^{3/2})dz\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\Phi(m,d_{1},d_{2}).$ (205)

where $\sigma_{1}$ is the largest T-singular value, and $d_{1},d_{2}$ are
constants related to the upper bound of the largest eigenvalue of the random
Hermitian matrix $\mathbf{X}$.

Proof: From random structure of discussed random symmetric T-product tensor
$\mathcal{X}\in\mathbb{R}^{m\times m\times p}$, all random Hermitian matrix
$\mathbf{X}_{i}$ have same probability density distributions. The maximum
T-singular value of $\mathcal{X}$ will be equal to the maximum singular value
of $\mathbf{X}_{i}$, and we use $\mathbf{X}$ to represent any random Hermitian
matrix $\mathbf{X}_{i}$ since they share the same distribution.

From Eq. (8) in [26], given a $m\times m$ random Hermitian matrix
$\mathbf{X}$, we have

$\displaystyle\mathrm{Pr}\left(\sigma_{1}(\mathbf{X})>y\right)\leq
d_{1}\exp(-d_{2}m(y-2)^{3/2})$ (206)

where $d_{1},d_{2}$ are constants related to the upper bound of the largest
(or smallest) eigenvalue of any random Hermitian matrix $\mathbf{X}$. Then, we
have

$\displaystyle\mathbb{E}\sigma_{1}(\mathcal{X})$ $\displaystyle=$
$\displaystyle\int^{\infty}_{0}\mathrm{Pr}\left(\sigma_{1}(\mathcal{X})>y\right)dy$
(207) $\displaystyle\leq$
$\displaystyle\int^{\infty}_{0}d_{1}\exp\left[-d_{2}m(y-2)^{3/2}\right]dy$
$\displaystyle=$ $\displaystyle\int^{\infty}_{-2}d_{1}\exp(-d_{2}mz^{3/2})dz$

$\hfill\Box$

Following lemma is about Ky Fan $k$-norm bound for the exponential of a random
T-product tensor with subexponential constraints.

###### Lemma 15

Given a symmetric random T-product tensor $\mathcal{X}\in\mathbb{R}^{m\times
m\times p}$ with random structure defined by Definition 1 and

$\displaystyle\mathcal{X}^{p}\preceq\frac{p!\mathcal{A}^{2}}{2}\mbox{\leavevmode\nobreak\
\leavevmode\nobreak\ almost surely for $p=2,3,4,\cdots$,}$ (208)

where $\mathcal{A}$ is a TPD T-product tensor. Then, we have following bound
about the expectation value of Ky Fan $k$-norm for the random T-product tensor
$\exp(\theta\mathcal{X})$

$\displaystyle\mathbb{E}\left\|\exp(\theta\mathcal{X})\right\|_{(k)}$
$\displaystyle\leq$ $\displaystyle
k\left[1+\theta\Phi(m,d_{1},d_{2})+\frac{\theta^{2}}{2(1-\theta)}\sigma_{1}\left(\mathcal{A}^{2}\right)\right].$
(209)

Proof: From Lemma 13, we have

$\displaystyle\mathbb{E}\left\|\exp(\theta\mathcal{X})\right\|_{(k)}$
$\displaystyle=$
$\displaystyle\sum\limits_{l=1}^{k}\mathbb{E}\sigma_{l}\left(\exp(\theta\mathcal{X})\right)$
(210) $\displaystyle\leq$
$\displaystyle\sum\limits_{l=1}^{k}\mathbb{E}\sigma_{l}\left(\mathcal{I}+\theta\mathcal{X}+\frac{\theta^{2}\mathcal{A}^{2}}{2(1-\theta)}\right)\leq
k\mathbb{E}\sigma_{1}\left(\mathcal{I}+\theta\mathcal{X}+\frac{\theta^{2}\mathcal{A}^{2}}{2(1-\theta)}\right)$

where $\sigma_{l}(\cdot)$ is the $l$-th largest T-singular value.

From Lemma 9, we have
$\sigma_{1}(\mathcal{A}+\mathcal{B})\leq\sigma_{1}(\mathcal{A})+\sigma_{1}(\mathcal{B})$
for two symmetric T-product tensors $\mathcal{A}$ and $\mathcal{B}$. Then, we
can bound
$\mathbb{E}\sigma_{1}\left(\mathcal{I}+\theta\mathcal{X}+\frac{\theta^{2}\mathcal{A}^{2}}{2(1-\theta)}\right)$
as

$\displaystyle\mathbb{E}\sigma_{1}\left(\mathcal{I}+\theta\mathcal{X}+\frac{\theta^{2}\mathcal{A}^{2}}{2(1-\theta)}\right)$
$\displaystyle\leq$ $\displaystyle
1+\theta\mathbb{E}\sigma_{1}(\mathcal{X})+\frac{\theta^{2}}{2(1-\theta)}\sigma_{1}\left(\mathcal{A}^{2}\right)$
(211) $\displaystyle\leq$ $\displaystyle
1+\theta\mathbb{E}\Phi(m,d_{1},d_{2})+\frac{\theta^{2}}{2(1-\theta)}\sigma_{1}\left(\mathcal{A}^{2}\right)$

where we use $\Phi(m,d_{1},d_{2})$ from Lemma 14 to bound
$\mathbb{E}\sigma_{1}(\mathcal{X})$ in the last inequality. This Lemma is
proved by multiplying $k$ at Eq. (211) $\hfill\Box$

We are ready to present our main theorem about the generalized T-product
tensor Bernstein bound.

See 1.1 Proof: Since we have

$\displaystyle\mathrm{Pr}\left(\left\|g\left(\sum\limits_{j=1}^{m}\mathcal{X}_{j}\right)\right\|_{(k)}\geq\theta\right)\leq_{1}(n+1)^{s-1}\inf\limits_{t,p_{j}}e^{-\theta
t}\left(ka_{0}^{s}+\sum\limits_{l=1}^{n}\sum\limits_{j=1}^{m}\frac{a^{ls}_{l}\mathbb{E}\left\|\exp\left(p_{j}lst\mathcal{X}_{j}\right)\right\|_{(k)}}{p_{j}}\right)$
$\displaystyle\leq_{2}(n+1)^{s-1}\inf\limits_{t,p_{j}}e^{-\theta
t}\left\\{ka_{0}^{s}+\sum\limits_{l=1}^{n}\sum\limits_{j=1}^{m}\frac{a^{ls}_{l}k\left[1+p_{j}lst\Phi(m,d_{1},d_{2})+\frac{(p_{j}lst)^{2}\sigma_{1}(\mathcal{A}^{2})}{2(1-p_{j}lst)}\right]}{p_{j}}\right\\}$
$\displaystyle\leq_{3}(n+1)^{s-1}\inf\limits_{t>0}e^{-\theta
t}k\left\\{a_{0}^{s}+\sum\limits_{l=1}^{n}a_{l}^{ls}\left[1+mlst\Phi(m,d_{1},d_{2})+\frac{(mlst)^{2}\sigma_{1}(\mathcal{A}^{2})}{2(1-mlst)},\leavevmode\nobreak\
\leavevmode\nobreak\ \right]\right\\}.$ (212)

where the inequality $\leq_{1}$ comes from Theorem 10; the inequality
$\leq_{2}$ comes from Lemma 15; the inequality $\leq_{3}$ is obtained by
setting $p_{j}=m$. $\hfill\Box$

##  5 Conclusions

This work extend previous work in [18] by making following generalizations via
majorization techniques: (1) besides bounds related to extreme values of
T-eigenvalues, this works considers more general unitarily invariant norm for
T-product tensors; (2) this work derives new bounds for any polynomial
function raised by any power greater or equal than one for the summation of
random symmetric T-product tensors. We also establish the Courant-Fischer min-
max theorem for T-product tensors and marjoization relation for T-singular
values which are by-products of our procedure to prove the generalized random
T-product Bernstein bounds. Possible future work about this research is to
consider tail bounds behaviors for the summation of random symmetric T-product
tensors equipped with other random structures different from random structure
provided by Definition 1.

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