TITLE: Martingale property cannot hold for general random times
QUESTION [1 upvotes]: Let $\sigma \leq \tau$ be two random times that are no stopping times. I want to create a simple example that shows that for these random times $\mathbb{E}[M_\tau \mid \mathcal{F}_\sigma] = M_\sigma$ cannot hold.
I am aware of the fact that optional stopping does not work for all stopping times. However, all of this has to do with stopping times. I am a bit confused by the random times. Can anybody provide a clear example?

REPLY [0 votes]: Take a standars symmetric random walk. $P(X_i=1)=P(X_i=-1)=\frac{1}{2}$. Then $S_n = \sum_{i=1}^n X_k$ is a martingale.
Take $\sigma = \{n\geq 0, S_n = \alpha\}$ which is a stopping time. And take $\tau = \sigma +1$, which is not a stopping time. If
\begin{align}
\mathbb{E}[S_\tau \mid \mathcal{F}_\sigma] = S_\sigma.
\end{align}
This would imply that
\begin{align}
\mathbb{E}[S_\tau] = \mathbb{E}[S_\sigma] = \ldots = \mathbb{E}[S_0] = 0.
\end{align}
However,
\begin{align}
\mathbb{E}[S_\tau] = \frac{1}{2}(\alpha -1)+\frac{1}{2}(\alpha +1) = \alpha \neq 0 = \mathbb{E}[S_\sigma].
\end{align}


TITLE: On the continuity of functions
QUESTION [2 upvotes]: We have the following situation. Assume that $X$ a topological space, $u\in C(X)$ and $u\geq0$, then $U:=\{x\in X: u(x)>0\}$ is open in $X$. Moreover let $f\in C(U)$ and define for each $n\in\mathbb{N}$ the function $g_n$ on $X$ by $g_n(x)=\min\{f(x),n\cdot u(x)\}$ if $x\in U$ and $g_n(x)=0$ if $x\in X\setminus U$. The claim is that $g_n$ is continuous foe each $n$. Here C denote the space of continuous functions.
I think you can prove this by application of nets. Therefore assume $(x_a)_{a\in A}$ is a net in $X$ such that $x_a\rightarrow x$. We have to show that $g_n(x_a)\rightarrow g_n(x)$. This is clearly true if $x_a\in U$ for each $a\in A$ of if $x_a\in X\setminus U$ for each $a\in A$, but this has not to be the general case. How to prove this claim? Thank you very much.

REPLY [1 votes]: You need one more hypothesis: you must either assume that $f\ge 0$, or define
$$g_n(x)=\min\{|f(x)|,nu(x)\}$$
for $x\in U$. I’ll write this up using the latter assumption.
If $x\in U$, then there is an $a_0\in A$ such that $x_a\in U$ whenever $a\succeq a_0$, and if $x\in X\setminus\operatorname{cl}U$, there is an $a_0\in A$ such that $x_a\in X\setminus\operatorname{cl}U$ whenever $a\succeq a_0$; in both cases it’s clear that $g_n(x_a)\to g_n(x)$. Thus, we may assume that $x\in(\operatorname{cl}U)\setminus U$. Then $g_n(x)=0$, and we must show that $g_n(x_a)\to 0$.
Let $\epsilon>0$; $u(x)=0$, so $x$ has an open nbhd $V$ such that $0\le u(y)<\frac{\epsilon}n$ for all $y\in V$. Fix $a_0\in A$ such that $x_a\in V$ whenever $a\succeq a_0$; then for each $a\succeq a_0$ we have
$$g_n(x_a)=\begin{cases}
\min\{|f(x_a)|,nu(x_a)\}\le nu(x_a)<\epsilon,&\text{if }x_a\in U\\
0<\epsilon,&\text{if }x\in X\setminus U\;,
\end{cases}$$
so $g_n(x_a)\to 0$, as desired.


TITLE: Improper integral: $\int^{+\infty}_0e^{-(a^2t^2+b^2/t^2)}$
QUESTION [3 upvotes]: Consider the application $f: (\mathbb{R}^{+*})^2 \to \mathbb{R}$ given by
$f(a,b)=\int^{+\infty}_0e^{-(a^2t^2+b^2/t^2)}$
Calculate $f(a,b)$.

I thought to take the derivative inside the integral sign (this needs justification) I obtain:
$\frac{\partial^2 f}{\partial a\partial b}=4abf$
I don't know how to solve this differential equation, but two obvious solutions
are $f=Ke^{a^2+b^2}$ and $f=Ke^{-(a^2+b^2)}$, where $K$ is a constant. The first solution clearly doesn't work, because if we increase $a$ or $b$, the integral $f(a,b)$ should decrease.
So I have a good feeling that the answer is $f(a,b)=Ke^{-(a^2+b^2)}$, but to prove it rigorously I still need to

justify that we can take the derivative inside the integral sign
properly solve the differential equation
determine the constant $K$

Any hints?

REPLY [1 votes]: I think the variables $a$ and $b$ are positive to begin with.
The differential equation approach is valid, but not as you planned to do.
As you can see the answer by @pisco125, the form $Ke^{-a^2-b^2}$ is not the answer. This means that approaching from the following PDE
$$
\frac{\partial^2}{\partial a\partial b} f = 4ab f
$$
is impossible.
However, we can still approach from a differential equation. We can prove that
$$
\frac{\partial}{\partial b} f = -2a f.
$$
If this is true, then we integrate w.r.t. $b$ to obtain
$$
f(a,b)=\exp(-2ab+g(a)).
$$
To find $g(a)$, put $b=0$. This gives
$$
f(a,0)=\exp(g(a))=\int_0^{\infty} \exp(-a^2 t^2 ) dt = \frac{\sqrt \pi}{2a}
$$
yielding the correct answer.
For the justification of differentiation under integral sign, fix $a>0$, $b>0$ and $b_n$ be a sequence converging to $b$, not equal $b$.
Consider
$$
\frac{f(a,b_n)-f(a,b)}{b_n-b}=\int_0^{\infty} \exp(-a^2 t^2) \frac{ \exp\left( - \frac{b_n^2}{t^2}\right)-\exp\left(-\frac{b^2}{t^2}\right) }{b_n-b}dt.
$$
For sufficiently large $n$, $b/2\leq b_n\leq 2b$, and by Mean Value Theorem,
$$
\left| \frac{ \exp\left( - \frac{b_n^2}{t^2}\right)-\exp\left(-\frac{b^2}{t^2}\right) }{b_n-b}\right|\leq \frac1{t^2}\exp\left( -\frac{(b/2)^2}{t^2}\right).
$$
By Dominated Convergence Theorem, we have
$$
\lim_{n\rightarrow\infty} \frac{f(a,b_n)-f(a,b)}{b_n-b}=\int_0^{\infty} \frac{-2b}{t^2} \exp(-a^2 t^2-\frac{b^2}{t^2}) dt.
$$
The last integral can be shown to be equal $-2af$ by the substitution $atu=b$.


TITLE: What is the numerically greatest or magnitude wise greatest term in the expansion?
QUESTION [4 upvotes]: I am just a beginner to Binomial Theorem and I want to find out the numerically or magnitude wise greatest term in the expansion of $(3-5x)^{11} $ when $ x=\frac{1}{5} $
I know some methods, but they are quite confusing. And I want a simple method. So if anyone has any idea they are welcomed.

REPLY [2 votes]: When $x = \frac15$, we just have $(3-1)^{11}$. Then the question is basically "what is the largest $n$ such that $3\binom{11}{n-1}\leq\binom{11}{n}$", since the change from one term to the next is that the "denominator" of the binomial coefficient increases by $1$, and the exponent of $3$ decreases by $1$, and we want to know at what point the change in the binomial coefficient doesn't outweigh the change in the exponent of $3$ (ignoring sign).
(Alternatively, one may derive this inequality from "one term is larger than the previous if their ratio is larger than $1$" as the other answer does. Arguably, that's easier to follow than the above paragraph.)
We have that
$$
3\binom{11}{n-1}\leq\binom{11}{n}\\
3\leq\frac{\binom{11}{n}}{\binom{11}{n-1}}\\
3\leq\cfrac{\frac{11!}{(n)!(11-n)!}}{\frac{11!}{(n-1)!(11-n+1)!}}\\
3 \leq \frac{11-n+1}{n}\\
3n\leq12-n\\
4n\leq12\\
n\leq3
$$
So we actually get that $n = 3$ is the largest $n$ for which the previous term isn't larger (again, ignoring sign), and that term is
$$
3^{8}(-1)^3\binom{11}3 = -1\,082\,565
$$
Since the inequality gave $n\leq$ an integer, the previous term is actually equal in absolute value, and has positive sign, so numerically, that's the largest. In general, if one term is negative, but in absolute value the strictly largest (i.e. the inequality you get from the above argument is $n\leq$ some non-integer), then you will have to compare the term before and the term after to see which one is numerically larger, but it will necessarily be one of those two.


text:
\begin{document}

\title{The Role of Large-Scale Fading in Uplink Massive MIMO Systems}

\author{\IEEEauthorblockN {Ang Yang, Zunwen He, Chengwen Xing, Zesong Fei, and Jingming Kuang}
\thanks{The authors are with School of Information and Electronics,
Beijing Institute of technology, Beijing 100081, China
(email:taylorkingyang@163.com, hezunwen@bit.edu.cn,
chengwenxing@ieee.org, feizesong@bit.edu.cn, and
JMKuang@bit.edu.cn).}

}

\maketitle

\begin{abstract}
In this correspondence, we analyze the ergodic capacity of a large
uplink multi-user multiple-input multiple-output (MU-MIMO) system
over generalized-$K$ fading channels. In the considered scenario,
multiple users transmit their information to a base station equipped
with a very large number of antennas. Since the effect of fast
fading asymptotically disappears in massive MIMO systems,
large-scale fading becomes the most dominant factor for the ergodic
capacity of massive MIMO systems. Regarding this fact, in our work
we concentrate our attention on the effects of large-scale fading
for massive MIMO systems. Specifically, some interesting and novel
lower bounds of the ergodic capacity have been derived with both
perfect channel state information (CSI) and imperfect CSI.
Simulation results assess the accuracy of these analytical
expressions.
\end{abstract}

\begin{keywords}

Massive MIMO, large-scale fading, ergodic capacity.
\end{keywords}

\IEEEpeerreviewmaketitle

\section{Introduction}

In order to satisfy the ever-increasing demands from explosive
wireless data services, a breakthrough in spectral efficiency is
expected for the next generation wireless systems just as 5G.
Regarding the great success of MIMO technologies
\cite{yindijing_1,shijin_AF}, massive MIMO or large MIMO
technologies have attracted a lot of attention recently as a
promising enabling technology to greatly boost the spectrum
efficiency \cite{shijin_massive,Larsson_uplink}. Furthermore, for 5G
network densification is also of great importance to improve network
capacity by increasing frequency reuse factor. Therefore, wireless
designers are faced with a challenging task, which is how to
determine the coverage range of massive MIMO base stations (BSs). In
other words, the channel fading characteristics of massive MIMO
should be carefully investigated.

Deploying very large numbers of antennas at the transmitter or
receiver, linear signal processing can substantially reduce the
jitters from various fast fading \cite{large_number}. However, as
the other part of the overall fading, the large-scale fading can not
be simply neglected in the massive MIMO \cite{Larsson_uplink}. It is
should be highlighted that to the best of the authors' knowledge, in
the existing works on massive MIMO systems only the effects of fast
fading have been investigated in detail \cite{Larsson_uplink}, and
the large-scale fading is simply assumed to be constant and known a
priori. In practical scenarios, the distributions of the large-scale
fading will largely vary in different scenarios, such as urban and
open areas. The work on the effect of the large-scale fading on
massive MIMO seems largely open up to date.

Motivated by this fact, in this correspondence we concentrate our
attention on  large-scale fading by considering the generalized-$K$
fading, which is a generic model that occurs when small-scale fading
is modeled via the Nakagami-$m$ distribution and large-scale fading
via the gamma distribution \cite{generalized_fading_1}. This model
has been demonstrated to effectively approximate most of the fading
and shadowing effects occurring in wireless channels, and also to be
analytically friendlier than the Nakagami-$m$/lognormal model
\cite{C_Zhong_Nakagami}. Moreover, we focus on the uplink of a
massive MU-MIMO system operating over generalized-$K$ fading, where
one BS equipped with $N$ antennas receives the information of $K$
single-antenna mobile users ($1 \ll K \ll N$). Different from
\cite{Larsson_uplink}, we focus on the following two fundamental
questions:

\noindent (1) What is the impact of large-scale
fading on the erogidc capacity of massive MIMO?

\noindent (2)
Can we provide analytical expressions for the ergodic capacity of
massive MIMO over generalized-$K$ fading?

To tackle these problems, new lower bounds for the ergodic capacity
of one user are derived for both perfect channel state information
(CSI) and imperfect CSI, as well as the average ergodic capacity of
all the users in the cell. Our analytical expressions are
substantiated via Monte Carlo simulations. It is shown by our result
that when the large-scale fading parameter $m<2$, the decrease of
$m$ will bring large reduction of the capacity of the system with
both perfect and imperfect CSI. This indicates that massive MIMO can
achieve high frequency reuse in the urban scenario.

\section{System model and Preliminaries}

In our work, we focus on  the uplink MU-MIMO system as shown in Fig.
\ref{system_model_central.eps}, which consists of one BS equipped
with $N$ antennas and $K$ single-antenna mobile users. The users
transmit their data to the BS in the same time-frequency resource
and the $N \times 1$ the received signal column vector at BS equals
to
\begin{align}
\label{received signals} {\bf{y}} = \sqrt {p_u} {\bf{G}}{\bf{x}} +
{\bf{w}},
\end{align}where the $K \times 1$ column vector $\sqrt {p_u}{\bf{x}}$ denotes the signal
transmitted by the $K$ users. Without loss of generality, the
average transmit power of each user is assumed to be $P$. In
addition, symbol ${\bf{G}}$ represents the $N \times K$ channel
matrix between the BS and the $K$ users and the $\{n,k\}^{\rm{th}}$
entry $g_{nk} \buildrel \Delta \over = [{\bf{G}}]_{nk}$ is the
channel coefficient between the $n$th antenna at the BS and the
$k$th user. The $K \times 1$ column vector ${\bf{w}}$ denotes the
additive zero-mean Gaussian white noise with unit variance.

In general, the channel matrix ${\bf{G}}$ is made of two kinds of
fading, i.e., fast fading and large-scale fading. Then each element
of ${\bf{G}}$ can be written as
\begin{align}\label{channel matrix}
{g_{nk}} = {h_{nk}}\sqrt {{\beta _{k}}} ,n = 1,2, \ldots ,N,k = 1,2,
\ldots ,K,
\end{align}where ${h_{nk}}$ is the fast fading coefficient between the $n$th antenna of the
BS and the $k$th user. Symbol $\sqrt {{\beta _{k}}}$ denotes the
large-scale fading from the $k$th user and the BS, which equals
\begin{align}\label{large-scale}
 {{\beta _{k}}}  =\mu_k /D_{k}^v, k = 1,2,
\ldots ,K,
\end{align}
where $D_{k}$ is the distance between the $k$th user to the BS, $v$
is the path-loss exponent with typical values ranging from 2 to 6. The
large-scale fading coefficient $\mu _{k}$ is modeled as independent
and identically distributed (i.i.d.) gamma random variable (RV), the
PDF of which can be expressed as
\begin{align}\label{large-scale 2}
p\left( {{\mu _k}} \right) = \frac{{{\mu _k}^{{m_k} - 1}}}{{\Gamma
\left( {{m_k}} \right){\Omega _k}^{{m_k}}}}\exp \left( { -
\frac{{{\mu _k}}}{{{\Omega _k}}}} \right),{\mu _k},{\Omega _k},{m_k}
> 0,
\end{align}where $\Gamma(\cdot)$ denotes the
gamma function. Moreover, denoting $E[\cdot]$ as the expectation of
a RV, $m_k$ and $\Omega _k= E[\mu_k]/{m_k}$ are the shape and scale
parameters of the gamma distribution, respectively. In practical
situations it is expected that  nonzero small and finite but large
values of $m$ for urban and open areas, respectively. The moderate
values of $m$ corresponds to suburban and rural areas.

\section{Ergotic capacity of the uplink MU-MIMO system}

In this section, we first derive the expressions of the ergotic capacity
of single user in the uplink MU-MIMO system to clarify the relationship between the
ergodic capacity and the large-scale fading clearly. Furthermore, the average ergotic capacity of all the
users in a cell is also analyzed to
describe the network performance. Since in
massive MIMO systems zero-forcing (ZF) detector tends to be optimal
\cite{Larsson_uplink}, it is adopted here as the beamformig strategy.

\subsection{Ergotic capacity of uplink MU-MIMO with perfect CSI}

\subsubsection{Ergotic capacity of the $k$th user}

With an ideal assumption that the BS has perfect CSI, the lower bound of the
capacity of the $k$th user in this case can be given by \cite[Eq.
(13)]{Larsson_uplink}
\begin{align}\label{capacity ref}
{C_{P,k}^L} = {\log _2}\left( {1 + {\beta _k}{p_u}(M-K)} \right).
\end{align} In the above expression (\ref{capacity ref}), the involved
large scale fading parameter ${\beta _k}$ is a variable, here we
take a further step to analyze the ergodic capacity over ${\beta
_k}$.

\begin{theorem}
Considering large-scale fading, the ergodic
capacity of the $k$th user with perfect CSI is
\begin{align}\label{capacity theorem}
E\left[ {{C_{P,k}^L}} \right] & = \frac{{\Omega _k^{}{{p_u}(M-K)}
{{m_k}  } }}{{\ln \left( 2 \right)D_k^v}} \ {}_3{F_1}\left(
{\begin{array}{*{20}{c}}
{{m_k} + 1,1,1}\\
2
\end{array}; - \frac{{\Omega _k^{}{p_u}(M-K)}}{{D_k^v}}} \right),
\end{align}where ${}_p{F_q}\left( {\begin{array}{*{20}{c}}
{{a_{1,}} \ldots ,{a_p}}\\
{{b_1}, \ldots ,{b_q}}
\end{array};z} \right)$ is the generalized hypergeometric
function.
\end{theorem}
\begin{proof}
See Appendix A.
\end{proof}

\subsubsection{Average ergotic capacity of all the users}
In the considered communication systems, the coverage area of a cell is model as a disc and the BS is located in
the center of the cell. The mobile users are located uniformly in the
cell with $R_0 < D_k <R$. The distribution of the users along the
radius of the cell is expressed as
\begin{align}\label{circle average 0}
f_d\left( x \right) = \frac{{2K}}{{{R^2-R_0^2}}}x.
\end{align}where $x$ is the distance between one user and the BS.

Exploiting Eqs. (\ref{capacity theorem}) and (\ref{circle average
0}), the lower bound of the average capacity of all the users in the
cell, noting as $\bar C_P^L$, can be averaged after deriving the
integral over the radius as
\begin{align}\label{circle average 1}
\bar C_P^L &= \frac{1}{K}\sum\limits_{k = 1}^K {E\left[ {C_{P,k}^L}
\right]} \notag \\
&\mathop  \to \limits^{a.s.} \frac{2}{{{R^2-R_0^2}}}\int_{{R_0}}^R
{x\frac{{\Omega _k^{}{p_u}(M - K) {{m_k} } }}{{\ln \left( 2
\right){x^v}}} \times {\;_3}{F_1}\left( {\begin{array}{*{20}{c}}
{{m_k} + 1,1,1}\\
2
\end{array}; - \frac{{\Omega _k^{}{p_u}(M - K)}}{{{x^v}}}} \right)}
dx.

\end{align}

Therefore the integral in the above expression (\ref{circle average
1}) becomes to the concern of the following work.

\begin{theorem}
Considering the distribution of large-scale fading, the average
ergodic capacity of all the users with perfect CSI is
\begin{align}\label{capacity average theorem}
\bar C_P^L &\mathop  \to \limits^{a.s.} \frac{{2\Omega _k^{}{p_u}(M
- K){m_k}}}{{\ln \left( 2 \right)\left( {{R^2} - R_0^2} \right)
\left( v - 2 \right)}}  \Bigg\{ \frac{1}{{{R_0}^{v -
2}}}{}_4{F_2}\left( {\begin{array}{*{20}{c}}
{\frac{{v - 2}}{v},{m_k} + 1,1,1}\\
{\frac{{v - 2}}{v} + 1,2}
\end{array};\frac{{ - \Omega _k^{}{p_u}(M - K)}}{{{R_0}^v}}} \right) \notag\\
& \quad - \frac{1}{{{R^{v - 2}}}}{}_4{F_2}\left(
{\begin{array}{*{20}{c}}
{\frac{{v - 2}}{v},{m_k} + 1,1,1}\\
{\frac{{v - 2}}{v} + 1,2}
\end{array};\frac{{ - \Omega _k^{}{p_u}(M - K)}}{{{R^v}}}}
\right)\Bigg\}.
\end{align}
\end{theorem}
\begin{proof}
See Appendix B.
\end{proof}

Perfect CSI is only an ideal assumption. In practice, CSI should be estimated via training sequences or pilots.  Due to limited length of training sequences and
time varying nature of wireless channels, channel estimation errors
are inevitable. In the following section, the performance with imperfect CSI is analyzed.

\subsection{Ergotic capacity of uplink MU-MIMO with imperfect CSI}

In this subsection we take a further step to derive several useful
expressions of the ergodic capacity of the massive MIMO systems with
imperfect CSI. In practice, the channel matrix $\bf{G}$ need to be
estimated  with the assistance of pilots \cite{gaofeifei_training}.
During the training phase in the coherence interval, mutually
orthogonal pilot signals of length $\tau$ are transmitted by
different users. The pilot sequences used by all the users are
denoted as a $\tau \times K$ matrix $\sqrt{\tau p_u} {\bf \Phi}$
($\tau \ge K$) with ${\bf \Phi}^{\rm H} {\bf \Phi}={\bf I}_K$.
Denoting ${\bf G}$ as the $M \times K$ channel matrix between the BS
and the $K$ users, the $M \times \tau$ received pilot matrix at the
BS is expressed as
\begin{align}\label{pilot 1}
{\bf Y}_p= \sqrt{\tau p_u} {\bf G} {\bf \Phi}^{\rm T} + {\bf N},
\end{align}where ${\bf N}$ is the $M \times \tau$ noise matrix with i.i.d. $CN (0, 1)$
elements. The MMSE estimate of $ {\bf G}$ is
\begin{align}\label{pilot 1}
{\bf{\hat G}} = \frac{1}{{\sqrt {\tau {p_u}} }}{{\bf{Y}}_p}{\bf{\Phi
\tilde D}} = \left( {{\bf{G}} + \frac{1}{{\sqrt {\tau {p_u}}
}}{\bf{W}}} \right){\bf{\tilde D}},
\end{align}where ${\bf{W}} \buildrel \Delta \over = {\bf N
\Phi^{*}}$ has i.i.d. $CN (0, 1)$ elements, as ${\bf \Phi}^{\rm H}
{\bf \Phi}={\bf I}_K$. Since ${\bf{D}} = diag\left\{ {{\beta _i}}
\right\},i \in \left[ {1,K} \right]$ denotes the favorable
propagation, we have matrix ${\bf{\tilde D}} \buildrel \Delta \over
= {\left( {\frac{1}{{\tau {p_u}}}{{\bf{D}}^{{\bf{ - 1}}}} +
{{\bf{I}}_K}} \right)^{ - 1}} = diag \left\{ {\frac{{\tau
{p_u}{\beta _i}}}{{\tau {p_u}{\beta _i} + 1}}} \right\}$, $i \in
\left[ {1,K} \right]$. Denoting the channel estimation error as
${{\bf{G}}_\Delta } = {\bf{\hat G}} - {\bf{G}}$, it can be seen that
the elements of the $i$th column of ${{\bf{G}}_\Delta } $ are RVs
with zero means and variances ${\frac{{\tau {p_u}{\beta _i}}}{{\tau
{p_u}{\beta _i} + 1}}}$, which will decrease the received signal to
noise ratio (SNR) at the BS.

\subsubsection{Ergotic capacity of the $k$th user}
With imperfect CSI, we begin with the derivation of the
ergodic capacity of the $k$th user by employing a lower bound given as
\cite[Eq. (42)]{Larsson_uplink}
\begin{align}\label{capacity ref im 1}
{C_{IP,k}} \ge {\log _2}\left( {1 + \frac{{\tau p_u^2\left( {M - K}
\right)\beta _k^2}}{{\left( {\tau p_u^{}\beta _k^{} + 1}
\right)\sum\limits_{i = 1}^K {\frac{{p_u^{}\beta _i^{}}}{{\tau
p_u^{}\beta _i^{} + 1}}}  + \tau p_u^{}\beta _k^{} + 1}}} \right).
\end{align}

However, the bound given in \cite[Eq. (42)]{Larsson_uplink} is
complex and not possible to calculate the integral over ${\beta
_k}$. Using the fact that $\frac{{p_u^{}\beta _i^{}}}{{\tau
p_u^{}\beta _i^{} + 1}} \le \frac{1}{\tau }$ and $1 + \frac{a}{b}
\ge \frac{{1 + a}}{b},a \ge 0,b \ge 1$, a novel lower bound is
proposed as
\begin{align}\label{capacity ref im}
{C_{IP,k}}

 \ge {\log _2}\left( {1 + \tau p_u^2\left( {M - K} \right)\beta _k^2} \right) - {\log _2}\left( {1 + \tau p_u^{}\beta _k^{}} \right) - {\log _2}\left( {1 + \frac{K}{\tau }} \right) \buildrel \Delta \over =
 C_{IP,k}^L.

\end{align}

Now we derive the integral over the large-scale fading ${\beta _k}$
in the above expression (\ref{capacity ref im}) and obtain the
following theorem.
\begin{theorem}
Taking the large-scale fading into account, the ergodic
capacity of the $k$th user with imperfect CSI is
\begin{align}\label{capacity theorem im}
&E\left[ {C_{IP,k}^L} \right]\notag\\& = \underbrace {\frac{{\tau
\Omega _k^2p_u^2\left( {M - K} \right){2^{{m_k} + 1}}\Gamma \left(
{\frac{{{m_k}}}{2} + 1} \right)\Gamma \left( {\frac{{{m_k}}}{2} +
\frac{3}{2}} \right)}}{{\sqrt \pi  \ln \left( 2 \right)\Gamma \left(
{{m_k}} \right)D_k^{2v}}}{\;_4}{F_1}\left( {\begin{array}{*{20}{c}}
{\frac{{{m_k}}}{2} + 1,\frac{{{m_k}}}{2} + \frac{3}{2},1,1}\\
2
\end{array}; - \frac{{4\tau \Omega _k^2p_u^2\left( {M - K} \right)}}{{D_k^{2v}}}} \right)}_{{{{\Xi _{\rm{1}}}\left( {{D_k},{m_k},{p_u},v} \right)}}}\notag\\
&\quad - \underbrace {\frac{{\tau \Omega _k^{}{p_u}{m_k}}}{{\ln
\left( 2 \right)D_k^v}}{\;_3}{F_1}\left( {\begin{array}{*{20}{c}}
{{m_k} + 1,1,1}\\
2
\end{array}; - \frac{{\tau \Omega _k^{}{p_u}}}{{D_k^v}}} \right)}_{{{{\Xi _{\rm{2}}}\left( {{D_k},{m_k},{p_u},v} \right)}}} - {\log _2}\left( {1 + \frac{K}{\tau }} \right).
\end{align}

\end{theorem}
\begin{proof}
See Appendix C.
\end{proof}

\subsubsection{Average ergotic capacity of all the users}
With the aid of Theorem 3, we are capable to derive the average
ergodic capacity of all the user with imperfect CSI, which can be
derived following the same logic of the perfect CSI case as
\begin{align}\label{average im}
\bar C_{IP}^L\mathop  \to \limits^{a.s.} \frac{2}{{{R^2} -
R_0^2}}\int_{R_0^2}^R x {\Xi _{\rm{1}}}\left( {x,{m_k},{p_u},v}
\right)dx - \frac{2}{{{R^2} - R_0^2}}\int_{R_0^2}^R x {\Xi _2}\left(
{x,{m_k},{p_u},v} \right)dx - {\log _2}\left( {1 + \frac{K}{\tau }}
\right).
\end{align}
Deriving the integral of Eq. (\ref{average im}) yields the following
theorem.
\begin{theorem}
Taking the effects of the large-scale fading into account, the
average ergodic capacity of all the users with imperfect CSI is
\begin{align}\label{capacity theorem im}
{{\bar C}_{IP}}&\mathop  \to \limits^{a.s.} \frac{{\tau \Omega _k^2p_u^2\left( {M - K} \right){2^{{m_k}}}\Gamma \left( {\frac{{{m_k}}}{2} + 1} \right)\Gamma \left( {\frac{{{m_k}}}{2} + \frac{3}{2}} \right)}}{{\left( {{R^2} - R_0^2} \right)\sqrt \pi  \ln \left( 2 \right)\Gamma \left( {{m_k}} \right)\left( v - 1\right)}}\notag \\
&\quad\times\Bigg\{ \frac{1}{{{R_0}^{2v - 2}}}{}_5{F_2}\left(
{\begin{array}{*{20}{c}}
{\frac{{v - 1}}{v},\frac{{{m_k}}}{2} + 1,\frac{{{m_k}}}{2} + \frac{3}{2},1,1}\\
{\frac{{v - 1}}{v} + 1,2}
\end{array};\frac{{ - 4\tau \Omega _k^2p_u^2\left( {M - K} \right)}}{{{R_0}^{2v}}}} \right)\notag\\
&\quad \quad \quad\quad- \frac{1}{{{R^{2v - 2}}}}{}_5{F_2}\left(
{\begin{array}{*{20}{c}}
{\frac{{v - 1}}{v},\frac{{{m_k}}}{2} + 1,\frac{{{m_k}}}{2} +  \frac{3}{2},1,1}\\
{\frac{{v - 1}}{v} + 1,2}
\end{array};\frac{{ - 4\tau \Omega _k^2p_u^2\left( {M - K} \right)}}{{{R^{2v}}}}} \right)\Bigg\}\notag \\
& \quad - \frac{{2\tau \Omega _k^{}{p_u}{m_k}}}{{\ln \left( 2
\right)\left( {{R^2} - R_0^2} \right)\left(v-2\right)}} \Bigg\{
\frac{1}{{{R_0}^{v - 2}}}{}_4{F_2}\left( {\begin{array}{*{20}{c}}
{\frac{{v - 2}}{v},{m_k} + 1,1,1}\\
{\frac{{v - 2}}{v} + 1,2}
\end{array};\frac{{ - \tau \Omega _k^{}{p_u}}}{{{R_0}^v}}} \right)\notag\\
& \quad \quad\quad\quad- \frac{1}{{{R^{v - 2}}}}{}_4{F_2}\left(
{\begin{array}{*{20}{c}}
{\frac{{v - 2}}{v},{m_k} + 1,1,1}\\
{\frac{{v - 2}}{v} + 1,2}
\end{array};\frac{{ - \tau \Omega _k^{}{p_u}}}{{{R^v}}}} \right)\Bigg\} - {\log _2}\left( {1 + \frac{K}{\tau }}
\right).
\end{align}

\end{theorem}
\begin{proof}
Due to space limitations, we will skip the proof, which can be
obtained following the similar logics of Appendices B and C.
\end{proof}

We note that all the expressions in the theorems of this section are
novel and can be used to predict the ergodic capacity of the large
uplink MU-MIMO systems over generalized-$K$ fading channels. It is
also worth noting that they are closed-form expressions, involving
only a finite number of basic operations such as summations of
logarithms, Gamma functions, and generalized hypergeometric
function, etc. This result can be used to determine the radius of the coverage of massive MIMO base stations.

\section{Numerical results}

In this section, simulation results are presented to
examine the impact of network parameters on the ergodic capacity of
the large uplink MU-MIMO systems. We assume that the users are
located uniformly in a cell of $1000$ meters while no user is closer
to the BS than $100$ meters, which means that $R=1000$ and
$R_0=100$. The distribution of the users is plotted in Fig.
\ref{User_location.eps}. The transmitted signals suffer from the
Nakagami-$m$ distributed fast fading (we set the fast fading
parameter $m=1$), and the gamma distributed large-scale fading (we
set $\Omega _k= 1/{m_k}$ and $m_1=m_2= \ldots =m_K =m$). We also
assume equal transmit power at each node and the value of $P$ is set
to be the received power at $500$ meters far from the transmitted
users. The length of the pilot is set as $\tau = K$. Furthermore,
the path loss exponent is set as $v = 3.6$ \cite{D_Tse}.

Fig. \ref{Capacity_different_N.eps} plots both the simulated and
numerical average capacity of all the users with different number of
the BS antennas $N$. The number of the users is set as $K=9$ and the
large-scale fading parameter is set as $m=3.3$. It can be seen that
the numerical curves accurately predict the simulated ones in the
large number of $N$ with both perfect and imperfect CSI. By
observing these curves, it is evident that the capacity increases
with $N$, which indicates that increasing the number of the BS
antennas brings an improved performance of system throughput. For
example, increasing $N$ from $100$ to $300$ brings a capacity
advantage of almost $20\%$ at $P=10dB$ with perfect CSI. It is also evident
that the lack of CSI will decrease the capacity of the system. For
example, the imperfect CSI brings about a capacity loss of $12\%$ at
$P=10dB$ with $N=250$.

Figs. \ref{Capacity_different_naka_m_antennas.eps} and
\ref{Capacity_different_naka_m_power.eps} plot the average capacity
of all the users with different large-scale fading parameters $m$.
We set $K=9$. It can be observed that increasing $m$ will result in
a better average capacity of all the users in both figures. For
example, increasing $m$ from $0.1$ to $1$ brings about a capacity
advantage of $93\%$ at $N=128, P=20dB$ with perfect CSI. Increasing
$m$ from $2$ to $5$ brings about a capacity advantage of $2\%$ at
$N=128, P=20dB$ with perfect CSI. Similar observations can be
obtained by the curves with imperfect CSI. These observations
demonstrate that with both perfect CSI and imperfect CSI, the
average ergodic capacity of all the users increases obversely as $m$
goes large when $m<2$, which is the urban scenario with heavy
shadowing; larger $m$ will brings little improvement of the capacity
when $m>2$, which is the suburban, rural, or flatland scenario.

\section{Conclusions}
In this correspondence, we analyzed the ergodic capacity of a large
uplink MU-MIMO system where multiple single-antenna users transmit
their information to a base station equipped with a very large number
of antennas. Over generalized-$K$ fading channels, several novel
lower bounds of the ergodic capacity of a single user were derived
for both perfect CSI and imperfect CSI, as well as the average
ergodic capacity of all the users in the cell. Simulation results
were used to validate our analytical expressions. We figured out
that large-scale fading will have large effect on the capacity of
the system with both perfect and imperfect CSI in the urban
scenario. Our result is of importance to design the coverage of massive MIMO base stations.

\appendices

\section{Proof of Theorem 1} Substituting Eqs. (\ref{large-scale}) and
(\ref{large-scale 2}) into Eq. (\ref{capacity ref}), the ergodic
capacity of the $k$th user can be expressed as
\begin{align}\label{perfect proof 1}
E\left[ {{C_{P,k}^L}} \right] = &\int_0^\infty  {{{\log }_2}\left(
{1 + \frac{{{p_u}(M - K)}}{{D_k^v}}{\mu _k}} \right)} \frac{{{\mu
_k}^{{m_k} - 1}}}{{\Gamma \left( {{m_k}} \right){\Omega
_k}^{{m_k}}}}\exp \left( { - \frac{{{\mu _k}}}{{{\Omega _k}}}}
\right)d{\mu _k}.
\end{align}Employing ${\log _2}\left( x \right) = \ln \left( x \right)/\ln
\left( 2 \right)$ and $\ln \left( {1 + x} \right) = \sum\limits_{i =
0}^\infty  {{{\left( { - 1} \right)}^{i }}{x^{i+1}}/({i+1})}$
\cite[1.511]{Table_of_Integrals}, we can rewrite the above
expression (\ref{perfect proof 1}) as
\begin{align}\label{perfect proof 2}
E\left[ {C_{P,k}^L} \right] =  - \frac{1}{{\ln \left( 2
\right)\Gamma \left( {{m_k}} \right){\Omega
_k}^{{m_k}}}}\sum\limits_{i = 0}^\infty  {\frac{1}{{i + 1}}} {\left(
{ - \frac{{{{{p_u}(M - K)}}}}{{D_k^v}}} \right)^{i + 1}}
\int_0^\infty  {{\mu _k}^{{m_k} + i}\exp \left( { - \frac{{{\mu
_k}}}{{{\Omega _k}}}} \right)d} {\mu _k}.
\end{align}

Using the formula given by \cite[3.326.2]{Table_of_Integrals}, the
above expression (\ref{perfect proof 2}) can be expressed as
\begin{align}\label{perfect proof 3}
E\left[ {C_{P,k}^L} \right] &=  - \frac{1}{{\ln \left( 2
\right)\Gamma \left( {{m_k}} \right)}}\sum\limits_{i = 0}^\infty
{\frac{1}{{i + 1}}} \Gamma \left( {{m_k} + i + 1} \right)  {\left( {
- \frac{{\Omega _k^{}{p_u}(M - K)}}{{D_k^v}}} \right)^{i + 1}}.
\end{align}Multiplying the same items on both the numerators and denominators in the
fractions does not change the equality and thus Eq. (\ref{perfect proof 3}) can be rewritten as
\begin{align}\label{perfect proof 4}
E\left[ {C_{P,k}^L} \right] &= \frac{{\Omega _k^{}{p_u}(M - K)\Gamma \left( {{m_k} + 1} \right)}}{{\ln \left( 2 \right)\Gamma \left( {{m_k}} \right)D_k^v}}\sum\limits_{i = 0}^\infty  {\frac{{\Gamma \left( {{m_k} + 1 + i} \right)}}{{\Gamma \left( {{m_k} + 1} \right)}}} \frac{{\Gamma \left( {i + 1} \right)\Gamma \left( {i + 1} \right)\Gamma \left( 2 \right)}}{{\Gamma \left( 1 \right)\Gamma \left( 1 \right)\Gamma \left( {i + 2} \right)}}\frac{{{{\left( { - \frac{{\Omega _k^{}{p_u}(M - K)}}{{D_k^v}}} \right)}^i}}}{{\Gamma \left( {i + 1} \right)}}\notag \\
 &= \frac{{\Omega _k^{}{p_u}(M - K)\Gamma \left( {{m_k} + 1} \right)}}{{\ln \left( 2 \right)\Gamma \left( {{m_k}} \right)D_k^v}}\sum\limits_{i = 0}^\infty  {} \frac{{{{\left( {{m_k} + 1}
\right)}_i}{{\left( 1 \right)}_i}{{\left( 1 \right)}_i}}}{{{{\left(
2 \right)}_i}}}\frac{{{{\left( { - \frac{{\Omega _k^{}{p_u}(M -
K)}}{{D_k^v}}} \right)}^i}}}{{i!}},

\end{align}where ${\left( m
\right)_i} = \Gamma \left( {m + i} \right)/\Gamma \left( m \right)$
is named as the Pochmann symbol. Based on
\cite[16.2.1]{NIST_handbook_of_mathematical_functions}, Eq.
(\ref{perfect proof 4}) can be reformulated as Eq. (\ref{capacity
theorem}) and then the theorem can be achieved.

\section{Proof of Theorem 2}
Substituting Eq. (\ref{perfect proof 3}) into Eq. (\ref{circle
average 1}) and using the formula $\int {{x^a}} dx = {x^{a +
1}}/\left( {a + 1} \right)$, we can reformulate (\ref{circle average
1}) as
\begin{align}\label{average perfect proof 1}

\bar C_P^L &\mathop  \to \limits^{a.s.} \frac{{{\rm{2}}\Omega _k^{}{p_u}(M - K)\Gamma \left( {{m_k} + 1} \right)\Gamma \left( {\frac{{v - 2}}{v}} \right)}}{{v\ln \left( 2 \right)({R^2} - R_0^2)\Gamma \left( {{m_k}} \right)\Gamma \left( {\frac{{v - 2}}{v} + 1} \right)}}\sum\limits_{i = 0}^\infty  {\frac{{\Gamma \left( {i + 1} \right)\Gamma \left( {i + 1} \right)\Gamma \left( 2 \right)\Gamma \left( {i + \frac{{v - 2}}{v}} \right)}}{{\Gamma \left( 1 \right)\Gamma \left( 1 \right)\Gamma \left( {i + 2} \right)\Gamma \left( {\frac{{v - 2}}{v}} \right)}}}\notag \\
& \quad\times \frac{{\Gamma \left( {\frac{{v - 2}}{v} + 1}
\right)\Gamma \left( {{m_k} + i + 1} \right)}}{{\Gamma \left( {i +
\frac{{v - 2}}{v} + 1} \right)\Gamma \left( {{m_k} + 1}
\right)}}\left( {\frac{1}{{{R_0}^{v - 2}}}\frac{{{{\left( {\frac{{ -
\Omega _k^{}{p_u}(M - K)}}{{{R_0}^v}}} \right)}^i}}}{{i!}} -
\frac{{\rm{1}}}{{{R^{v - 2}}}}\frac{{{{\left( {\frac{{ - \Omega
_k^{}{p_u}(M - K)}}{{{R^v}}}} \right)}^i}}}{{i!}}} \right).
\end{align}Based on the definition of Pochmann symbol, the above expression
(\ref{average perfect proof 1}) can be expressed as
\begin{align}\label{average perfect proof 2}

\bar C_P^L &\mathop  \to \limits^{a.s.} \frac{{{\rm{2}}\Omega _k^{}{p_u}(M - K)\Gamma \left( {{m_k} + 1} \right)\Gamma \left( {\frac{{v - 2}}{v}} \right)}}{{v\ln \left( 2 \right)({R^2} - R_0^2)\Gamma \left( {{m_k}} \right)\Gamma \left( {\frac{{v - 2}}{v} + 1} \right)}}\notag\\
 &\quad\times \Bigg\{ \frac{1}{{{R_0}^{v - 2}}}\sum\limits_{i = 0}^\infty  {\frac{{{{\left( 1 \right)}_i}{{\left( 1 \right)}_i}{{\left( {\frac{{v - 2}}{v}} \right)}_i}{{\left( {{m_k} + 1} \right)}_i}}}{{{{\left( 2 \right)}_i}{{\left( {\frac{{v - 2}}{v} + 1} \right)}_i}}}} \frac{{{{\left( {\frac{{ - \Omega _k^{}{p_u}(M - K)}}{{{R_0}^v}}} \right)}^i}}}{{i!}} \notag\\
&\quad \quad\quad\quad- \frac{{\rm{1}}}{{{R^{v - 2}}}}\sum\limits_{i
= 0}^\infty {\frac{{{{\left( 1 \right)}_i}{{\left( 1
\right)}_i}{{\left( {\frac{{v - 2}}{v}} \right)}_i}{{\left( {{m_k} +
1} \right)}_i}}}{{{{\left( 2 \right)}_i}{{\left( {\frac{{v - 2}}{v}
+ 1} \right)}_i}}}} \frac{{{{\left( {\frac{{ - \Omega _k^{}{p_u}(M -
K)}}{{{R^v}}}} \right)}^i}}}{{i!}} \Bigg\},
\end{align}based on which and together with
\cite[16.2.1]{NIST_handbook_of_mathematical_functions}, Eq.
(\ref{capacity average theorem}) is obtained and then proof is
ended.

\section{Proof of Theorem 3}

Following a similar logic as the derivation of Eq. (\ref{perfect
proof 3}), the expectation of first item of Eq.
(\ref{capacity ref im}) over $\beta_k$ can be derived to be
\begin{align}\label{imperfect proof 1}
&{{\Xi _{\rm{1}}}\left(
{{D_k},{m_k},{p_u},v} \right)}\notag\\ &\buildrel \Delta \over = E\left[ {{{\log }_2}\left( {1 + \tau p_u^2\left( {M - K} \right)\beta _k^2} \right)} \right]\notag \\
& =  - \frac{1}{{\ln \left( 2 \right)\Gamma \left( {{m_k}} \right){\Omega _k}^{{m_k}}}}\sum\limits_{i = 0}^\infty  {\frac{1}{{i + 1}}} {\left( { - \frac{{\tau p_u^2(M - K)}}{{D_k^{2v}}}} \right)^{i + 1}} \int_0^\infty  {{\mu _k}^{{m_k} + 2i+1}\exp \left( { - \frac{{{\mu _k}}}{{{\Omega _k}}}} \right)d} {\mu _k}\notag \\
& = \frac{{\tau \Omega _k^2p_u^2(M - K)}}{{\ln \left( 2
\right)\Gamma \left( {{m_k}} \right)D_k^{2v}}}\sum\limits_{i =
0}^\infty  {\frac{1}{{i + 1}}}\Gamma \left( {{m_k} + 2i + 2}
\right){\left( { - \frac{{\tau \Omega _k^2p_u^2(M -
K)}}{{D_k^{2v}}}} \right)^i},

\end{align}where the last step is obtained by using
\cite[3.326.2]{Table_of_Integrals}. We apply the expression given by \cite[8.335.1]{Table_of_Integrals}
to derive the integral in (\ref{imperfect proof 1}), which can be
written as
\begin{align}\label{imperfect proof 2}
\Gamma \left( {2x} \right) = \frac{{{2^{2x - 1}}}}{{\sqrt \pi
}}\Gamma \left( x \right)\Gamma \left( {x + \frac{1}{2}} \right).
\end{align}Substituting (\ref{imperfect proof 2}) into (\ref{imperfect proof
1}), $I_1$ can be reformulated as
\begin{align}\label{imperfect proof 3}
{{\Xi _{\rm{1}}}\left(
{{D_k},{m_k},{p_u},v} \right)}& = \frac{{\tau \Omega _k^{2}p_u^2\left( {M - K} \right){2^{{m_k}+1}}}}{{\sqrt \pi  \ln \left( 2 \right)\Gamma \left( {{m_k}} \right)D_k^{2v}}}\sum\limits_{i = 0}^\infty  {\frac{1}{{i + 1}}} \Gamma \left( {i + \frac{{{m_k}}}{2} + 1} \right)\notag \\
& \quad\times \Gamma \left( {i + \frac{{{m_k}}}{2} +  \frac{3}{2}}
\right){\left( { - \frac{{4\tau \Omega _k^{2}p_u^2\left( {M - K}
\right)}}{{D_k^{2v}}}} \right)^i}.
\end{align} Multiplying the same items on both the numerators and denominators in the
fractions, the equality does not change and Eq. (\ref{imperfect proof 3}) can be further rewritten as
\begin{align}\label{imperfect proof 4}
{{\Xi _{\rm{1}}}\left(
{{D_k},{m_k},{p_u},v} \right)}& = \frac{{\tau \Omega _k^2p_u^2\left( {M - K} \right){2^{{m_k} + 1}}\Gamma \left( {\frac{{{m_k}}}{2} + 1} \right)\Gamma \left( {\frac{{{m_k}}}{2} + \frac{3}{2}} \right)}}{{\sqrt \pi  \ln \left( 2 \right)\Gamma \left( {{m_k}} \right)D_k^{2v}}}\notag \\
& \quad\times \sum\limits_{i = 0}^\infty  {} \frac{{\Gamma \left( {i + \frac{{{m_k}}}{2} + 1} \right)\Gamma \left( {i + \frac{{{m_k}}}{2} +  \frac{3}{2}} \right)\Gamma \left( {i + 1} \right)\Gamma \left( {i + 1} \right)}}{{\Gamma \left( {\frac{{{m_k}}}{2} + 1} \right)\Gamma \left( {\frac{{{m_k}}}{2} +  \frac{3}{2}} \right)\Gamma \left( 1 \right)\Gamma \left( 1 \right)}}\notag \\
& \quad\times \frac{{\Gamma \left( 2 \right)}}{{\Gamma \left( {i + 2} \right)}}\frac{{{{\left( { - \frac{{4\tau \Omega _k^{2}p_u^2\left( {M - K} \right)}}{{D_k^{2v}}}} \right)}^i}}}{{\Gamma \left( {i + 1} \right)}}\notag \\
& = \frac{{\tau \Omega _k^2p_u^2\left( {M - K} \right){2^{{m_k} + 1}}\Gamma \left( {\frac{{{m_k}}}{2} + 1} \right)\Gamma \left( {\frac{{{m_k}}}{2} + \frac{3}{2}} \right)}}{{\sqrt \pi  \ln \left( 2 \right)\Gamma \left( {{m_k}} \right)D_k^{2v}}}\notag \\
& \quad\times \sum\limits_{i = 0}^\infty  {} \frac{{{{\left(
{\frac{{{m_k}}}{2} + 1} \right)}_i}{{\left( {\frac{{{m_k}}}{2} +
\frac{3}{2}} \right)}_i}{{\left( 1 \right)}_i}{{\left( 1
\right)}_i}}}{{{{\left( 2 \right)}_i}}}\frac{{{{\left( { -
\frac{{4\tau \Omega _k^{2}p_u^2\left( {M - K} \right)}}{{D_k^{2v}}}}
\right)}^i}}}{{i!}}.
\end{align}Applying \cite[16.2.1]{NIST_handbook_of_mathematical_functions} to
Eq. (\ref{imperfect proof 4}), the expectation of first term over
large-scale fading can be expressed as ${{\Xi _{\rm{1}}}\left(
{{D_k},{m_k},{p_u},v} \right)}$ of Eq. (\ref{capacity theorem im}).
Since ${\log _2}\left( {1 + \tau p_u^{}\beta _k^{}} \right)$ is
similar to Eq. (\ref{capacity ref}), the expectation of second term
of Eq. (\ref{capacity ref im}) over $\beta_k$ can be also expressed
as ${{\Xi _{\rm{2}}}\left( {{D_k},{m_k},{p_u},v} \right)}$ of Eq.
(\ref{capacity theorem im}). Thus the proof is ended.

config: arxiv


TITLE: How to show that $1-\zeta$ is prime in the order $\{ 1, \zeta, \ldots, \zeta^{l-2} \}$?
QUESTION [8 upvotes]: I am trying to prove the following:
Let $l$ be a prime and let $\zeta$ be a $l$th root of unity.  Show that,  in the order $\{ 1, \zeta, \ldots, \zeta^{l-2}  \}$ of the field $\mathbb{Q}(\zeta)$, if a product $\alpha \beta$ is divisible by $1-\zeta$, then $\alpha$ or $\beta$ must be divisible by $1-\zeta$.
I know that $1-\zeta$ is irreducible in the maximal order $\mathbb{Z}[ \zeta ]$, and I am trying to mimic the proof of that statement, but I'm stuck.
Can someone please help?
I see that all the products I am dealing with are of the form $\zeta^k$.
Also does the order contain $\zeta^{l-1}$?
I have a feeling that "the order ${ 1, \zeta, \ldots, \zeta^{l-2}  }$" is actually just $\mathbb{Z}[ \zeta ]$.  Is this true?

REPLY [0 votes]: Hint: Look at the constant of the minimal polynomial of $1-\zeta$. This has something to do with the norm.


TITLE: Quantum Computing Power Advantages
QUESTION [9 upvotes]: Currently, the world's fastest supercomputer runs at 17.59 Petaflops, which consumes 9 megawatts of electricity. A qubit-based quantum computer has the potential to operate much more quickly for some operations than a classical computer would be able to (as noted in this question). In fact, classical computers face a major hurdle in computation speed due to the inherent capacitance of components. My question is: would a quantum computer able to compute things faster than a classical computer do so with less, about the same, or greater power draw for equivalent operations? Would that depend on its implementation?

REPLY [4 votes]: If a quantum computer can be implemented at all, it will inherently be many orders of magnitude more power efficient for arriving at a given solution.
One way to explain why is that a quantum computer doesn't rely on entropy in the conventional sense at all. Instead, it sort of "settles down" to a solution that undermines standard (entropic) definitions of past and future. That is the "entanglement" part of quantum computing, the part that just cannot be modeled by any classical methods. As a very rough analogy, that part is a bit like moving towards a solution every so slightly, then going back in time a wee bit and starting the computation over again with that slightly improved result. Do that a gazillion times and even an intractable factoring problem becomes solvable.
Here's a physical comparison: Imagine trying to erase a small granite block by using only pencil erasers. Doesn't sound very easy, does it? Sure, you can rub a few molecules off of the block with a pencil eraser before the eraser gives out, but to do the entire job would take untold numbers of centuries, at an unbelievable cost in terms of erasers (the "entropic" cost).
But now imagine that every time you finish rubbing a few molecules off of the granite block, you get to go back in time and repeat the erasing process, this time using that slightly smaller granite block. You will still need to repeat the process an unimaginable number of times to finish erasing the granite block, but two things have changed dramatically: You no longer need centuries to do it (it's incredibly faster), and you no longer need a nearly infinite number of new pencil erasers to do it (it's incredibly less entropic, that is, the resources needed have become almost negligible).
If all of that sounds way too good to be true, you now are getting a small inkling of why in the decades since @PeterShor first proposed a quantum computing algorithm, the very possibility has riveted the minds and resource of many physics, security, and computational researchers around the world. Factoring a truly huge number actually by known methods turns out to be harder than erasing granite blocks using pencil erasers, not easier. In fact, it's pretty easy to come up with factoring problems that to complete using conventional computing would require time spans that are many times larger than the entire history of the universe.
I should mention that the second "spice" in quantum computing is its massive parallelism, which is a somewhat different aspect of quantum mechanics that is often presented as a full explanation. That part relies on the idea that instead of running just a single value through a computer (e.g., sending a "1" bit into an input port), quantum mechanics allows you to run multiple inputs through the same port at the same time (e.g, both "0" and "1" bits, a combination that is referred to as a qubit). Do that with enough inputs and you end up getting a huge and combinatorial increase in computational power, an increase that is nothing to be laughed at.
This is often thought to be the truly "quantum" part of quantum computing because the idea that one location can have multiple states at the same time is a very quantum idea. It is called quantum superpositioning of states, and it's sort of the defining feature of quantum physics, since you don't have any superpositioning of states, your system can be described by ordinary classical physics.
But you know what? If a qubit is used only that way, it's not nearly enough.
One way to understand why is to think of such multiple states running through a computer as providing parallelism in space. That is, instead of having to use two computers in parallel to process "0" and "1" inputs at the same time, you use just one computer with a the "0" and "1" bits superimposed on the same spot in space. Voila! You've saved all the space and power requirement of potentially many, many computers running in parallel, performing the same tasks on only one.
But think back to the example I gave earlier. The power for erasing the granite block came not from being parallel in space, but from being parallel in time. That is, each iteration of the eraser was "piled up" so that it took place during the same fragment of time, not just of space.
Time parallelism emerges only when you start considering the curious effect of entanglement, which plays havoc with ordinary concepts of time by providing effects that are "instantaneous" across distances in space. Relativity says very odd things about anything that works like that, since it makes space and time interchangeable.
So, the bottom line is this:

Any kind of quantum computer will be more efficient because it will use forms of parallelism that don't invoke the entropic costs of very large numbers of computers.
The most powerful designs for quantum computer are parallel in both space and time, a combination that allows pretty much unimaginable levels of computation to take place within quite tiny volumes of space-and-time.

PLEASE NOTE: Peter Shor most definitely participates in this group, and I rubbed his magic lamp earlier when I put an ampersand in front of his name.
So, if Professor Shor should show up (Poof!), I hereby abdicate any and all of what I just said to any and all of what Professor Shor may have to say. Scorn and ridicule are in fact fully welcome if appropriate, and if dished out will simply cause me to say "cool dude, thanks!"


TITLE: Why is the $W$ boson (and none other elementary boson) the only with EM charge?
QUESTION [4 upvotes]: I have read this question:
Working out the charge of a W Boson
The role of W bosons in the weak nuclear force and beta decay
The W, Z, together with the photon comprise the four gauge bosons of the electroweak interaction.
In the SM, the W, Z, photon are produced through spontaneous symmetry breaking, of the electroweak symmetry SU(2) × U(1)Y to U(1)em affected by the Higgs mechanism, that rearranges degrees of freedom.

The electric charge arises as a (nontrivial) linear combination of Y (weak hypercharge) and the T3 component of weak isospin ${\displaystyle Q=T_{3}+{\tfrac {1}{2}}Y_{\mathrm {W} }}{\displaystyle Q=T_{3}+{\tfrac {1}{2}}Y_{\mathrm {W} }}$ that does not couple to the Higgs boson – that is to say, the Higgs and the electromagnetic field have no effect on each other at the level of the fundamental forces ("tree level"), while any other linear combination of the hypercharge and the weak isospin will interact with the Higgs. This causes an apparent separation between the weak force, which interacts with the Higgs, and electromagnetism, which does not. Mathematically, the electric charge is a specific combination of the hypercharge and T3 outlined in the figure.

https://en.wikipedia.org/wiki/Electroweak_interaction

The gauge group of the electroweak part of the standard model is SU(2) × U(1). The group SU(2) is the group of all 2-by-2 unitary matrices with unit determinant; all the orthonormal changes of coordinates in a complex two dimensional vector space.
This combination of generators (a z rotation in the SU(2) and a simultaneous U(1) rotation by half the angle) preserves the vacuum, and defines the unbroken gauge group in the standard model, namely the electric charge group. The part of the gauge field in this direction stays massless, and amounts to the physical photon.

So basically we get why the photon remains EM neutral too.
Now the Z is EM neutral too. This is where it got its name from, zero EM charge.

The
Z
boson mediates the transfer of momentum, spin and energy when neutrinos scatter elastically from matter (a process which conserves charge). Such behavior is almost as common as inelastic neutrino interactions and may be observed in bubble chambers upon irradiation with neutrino beams. Whenever an electron is observed as a new free particle suddenly moving with kinetic energy, it is inferred to be a result of a neutrino interacting directly with the electron, since this behavior happens more often when the neutrino beam is present. In this process, the neutrino simply strikes the electron and then scatters away from it, transferring some of the neutrino's momentum to the electron.[a]

https://en.wikipedia.org/wiki/W_and_Z_bosons
Now the W boson got its name after the weak interaction.

The two
W
bosons are verified mediators of neutrino absorption and emission. During these processes, the
W±
boson charge induces electron or positron emission or absorption, thus causing nuclear transmutation. The
Z
boson is not involved in the absorption or emission of electrons and positrons.

The emission of a
W+
or
W−
boson either raises or lowers the electric charge of the emitting particle by one unit, and also alters the spin by one unit. At the same time, the emission or absorption of a
W±
boson can change the type of the particle – for example changing a strange quark into an up quark. The neutral Z boson cannot change the electric charge of any particle, nor can it change any other of the so-called "charges" (such as strangeness, baryon number, charm, etc.).

Now what I do not understand is, where does the W get its EM charge from. Is it because the interaction itself, when the particle emits or absorbs a neutrino, this is mediated by a W boson, and this induces electron or positron emission or absorption, and so the emitting (that emits the W) gains or loses EM charge? Is this the functionality of the W?
None other elementary boson has EM charge, not the Z, photon, gluon (color, but not EM), graviton, Higgs.
I do understand that the math only describes the reality of the particles, but is there a way in the math that shows better why the W is special? Or is the functionality of the weak interactions that makes the W special in terms of EM charge?
Question:

Why is the W the only elementary boson with EM charge? Is there an explanation in the functionality of the W during the weak interaction?

REPLY [5 votes]: It is a coincidence of presently recognized facts, not a cosmic logical necessity. You may cook up grand unified theories with more charged elementary bosons.
Actually, SUSY models also have charged bosons, such as selectrons, etc.
So, why? Fact of nature… ours. So far.


TITLE: Connected set in $C[0,1]$
QUESTION [2 upvotes]: Let $X=C([0,1])$ be the space of all continuous real valued function on $[0,1]$. Equip $X$ with the $\sup$ metric, $d(f,g)=\sup_{x\in[0,1]}|f(t)-g(t)|$
Let,
$S=\{f\in X| f(0)\neq f(1)\}$
$T=\{f\in X|f(0)\leq f(1)\}$
Are $S$ and $T$ connected?
This is a completely odd problem for me. So far I only came across problems where $X$ was some subset of $\mathbb{R}^n$. How to deal with function spaces?
Thanks for reading.

REPLY [2 votes]: In this kind of situation, remember the definition of a connected set :
A topological space $X$ is said to be disconnected if it is the disjoint union of two open sets (its equivalent to be the disjoint union of two closed sets, and it is also equivalent to have a continuous surjective application from X to $\{0,1\}$) ; it is said to be connected otherwise.
It is not obvious to understand why such a definition encapsulates the intuitive idea of connexity. The basic idea is the following lemma, linked to the definition : if $X$ is connex included in $Y$, if $X$ meets a set $C$ and its complementary, it meets its boundary. So you cannot use a set $C$ to "separate" $X$ in two parts. It would be possible to show this : $X$ is connex if and only if for all embedding of $X$ in a topological space $Y$, for all set $C$ included in $Y$, the preceding property is valid. Intuitively this is clearer, but the first definition is simpler in the sense you do not have to check all the embeddings $X$ in $Y$ and all the subsets $C$ of $Y$...
With this definition, it is quite easy, as we will see, to show that a set is disconnected.
To show the connectedness, a nice way is to show the path-connectedness. It is a more intuitive notion and easier to prove.
A topological set $X$ is path-connected if for all $x,y \in X$ there exists $\gamma$ from $[0,1]$ to $X$ continuous such that $\gamma(0) = x, \gamma(1) = y$ (you trace a path in $X$ linking $x$ and $y$).
you can show that path-connectedness imply connectedness.
You can also define the notion of connex component of a $x\in X$ : it is the biggest connex set containing $x$ (this is well define because a union of connex sets which intersects non trivially is connex ; hence, the connex component of $x$ is simply the union of all the connex sets containing $x$). Quite useful if you want to be more precise.
Remember also some properties (you can try to show these by using definitions) :
- The direct image of a connected set by a continuous application if a connected set.
- More generally, the direct image of a topological X set by a continuous application has less connex components than X.
- The reciprocal image of a space $Y$ by a continuous application has more connex components than $Y$.
It is quite powerful to prove results about connexity : you just need to exhibit some continous application to have information.
For the first example, there is several ways to reason.
- You can remark that S_+ = { f(1) > f(0) } and $S_- = \{ f(1) < f(0) \}$ are open and disjoint ; so you're done.
- You can use the continuous application $f \rightarrow f(1) - f(0)$ as in the José Carlos Santos' post.
You can even show that $S_+$ and $S_-$ are (path-)connex.
In the second example, you also prove that it is (path-)connex as in Carlos Santos' post.


text:
\section{Characterization of Optimization Landscape}
\label{sec:landscape}

In this section, we analyze the optimization landscape for the objective \eqref{eqn:obj} for Tucker decomposition. In particular, we establish the following result.
\begin{theorem} \label{thm:exact}
For any fixed $\lambda > 0$, all local minima of the objective function $f= L + \lambda R$ as in Equation \eqref{eqn:obj} have loss 0.

\end{theorem}

Note that the theorem would not hold for $\lambda = 0$ (when there is no regularizer). A counter-example is when $\ten T = \bm a^*\otimes \bm b^*\otimes \bm c^*$ for some unit vectors $\bm a^*, \bm b^*, \bm c^*$, and $\ten S = \bm 0, \bm A = \bm a^\top, \bm B = \bm b^\top, \bm C = \bm c^\top$ where $\bm a,\bm b,\bm c$ are unit vectors that are orthogonal to $\bm a^*, \bm b^*, \bm c^*$ respectively. A local change will have no effect if the new $\bm S$ is still $\bm 0$, and will make the objective function larger if the new $\bm S$ is nonzero.

In order to prove this theorem, we demonstrate a direction of improvement for all points $(\tup)$ that don't achieve the global optimum.
A direction of improvement is a tuple $(\Delta\ten S, \Delta\bm A, \Delta\bm B, \Delta\bm C)$ such that
\[
f(\ten S+\epsilon \Delta\ten S,\bm  A + \epsilon \Delta\bm  A,\bm  B + \epsilon \Delta\bm  B, \bm C + \epsilon \Delta\bm  C) < f(\tup)
\]
for all sufficiently small $\epsilon > 0$. Clearly, if a point $(\tup)$ has a direction of improvement, then it cannot be a local minimum.

Throughout the section,
let $\bm P_1$ ($\bm P_2, \bm P_3$) be the projection onto the column span of $\ten T_{(1)}$ ($\ten T_{(2)}, \ten T_{(3)}$).
Let $\bm A_1 = \bm A\bm P_1$ and $\bm A_2 = \bm A(\bm I-\bm P_1)$ (similarly for $\bm B,\bm C$). The proof works in the following 4 steps:

\vspace*{0.1in}
{\noindent \bf Bounding the regularizer} First we show that when $\nabla f = \bm 0$, the regularizer $R$ must be equal to 0 (Lemm~\ref{lem:nonzeroreg} in Section~\ref{sec:nonzeroreg}). At a high level, this is because the gradient of regularizer $R$ is always orthogonal to the gradient of main term $L$. Therefore if the gradient of the entire objective is $\bm 0$, the gradient of $R$ must also be $\bm 0$. We complete the proof by showing that $\nabla R = \bm 0$ implies $R = 0$.

\vspace*{0.1in}
{\noindent \bf Removing extraneous directions}
Next, we show that when $\nabla f = \bm 0$, the projection in the wrong subspaces $\bm A_2,\bm B_2, \bm C_2$ are all equal to $\bm 0$.
This is because the direction of directly removing the projection in the wrong subspace $\bm A_2$ is a direction of improvement (see Lemma~\ref{lem:a21}).

\vspace*{0.1in}
{\noindent \bf Adding missing directions} After the previous steps, we know that the rows of $\bm A$ are in the column span of $\ten T_{(1)}$. However, the row span of $\bm A$ might be smaller. In this case, there exist directions $\bm a, \bm b, \bm c$ such that $\ten T(\bm a,\bm b,\bm c) > 0$, and $\bm A\bm a = \bm 0$. We will show that in this case we can always add the missing directions into $\bm A$ and $\ten S$.

This is the most technical part of our proof, and high order stationary points may appear when $\bm B\bm b$ or $\bm C\bm c$ are also $\bm 0$. See Sections~\ref{sec:missingdirections}.

\vspace*{0.1in}
{\noindent \bf Fixing $\ten S$} Finally, we know that the components $\bm A,\bm B, \bm C$ must span the correct subspaces. Our final step shows that in this case, if $L > 0$ then it is easy to find a direction of improvement, see Section~\ref{sec:handlings}.

\subsection{Direction of Improvement for Points with Nonzero Regularizer}\label{sec:nonzeroreg}
We show any point with nonzero regularizer must also have a nonzero gradient, therefore the (negative) gradient itself is a direction of improvement.

\begin{lemma} \label{lem:nonzeroreg}
For any $\tup$, if $R(\tup) > 0$ then $\|\nabla f\| > 0$.
\end{lemma}

To prove this, we first show that if the regularizer is nonzero, then its gradient is nonzero.
\begin{lemma}
\label{lem:reg_gradient}
The function $\reg$ satisfies
\[
4\reg(\tup) = \langle \nabla_{\bm A} \reg, \bm A\rangle + \langle \nabla_{\bm B} \reg, \bm B\rangle + \langle \nabla_{\bm C} \reg, \bm C\rangle + \langle \nabla_{\ten S} \reg, \ten S\rangle
\]
\end{lemma}
\begin{proof}
Note the following calculations:
\begin{align*}
\langle \nabla_{\bm A} \reg, \bm A\rangle &=  \langle   4(\bm A \bm A^\top- \ten S_{(1)}\ten S_{(1)}^\top)\bm A,\bm A\rangle\\
&= 4\langle \bm A \bm A^\top - \ten S_{(1)}\ten S_{(1)}^\top,\bm A\bm A^\top\rangle\\
\langle 4\ten S(\ten S_{(1)}\ten S_{(1)}^\top - \bm A\bm A^\top,\bm I,\bm I), \ten S\rangle &= -4\langle \bm A\bm A^\top - \ten S_{(1)}\ten S_{(1)}^\top,\ten S_{(1)}\ten S_{(1)}^\top\rangle
\end{align*}
The left-hand side above is one of the terms in $\nabla_{\ten S}\reg$. Doing the same calculation for the other modes and then adding everything together yields the result.
\end{proof}

We next show that the gradient of the regularizer is always orthogonal to the gradient of the main term (i.e. the tensor loss $L$).
\begin{lemma}
\label{lem:reg_orthogonality}
For any $\tup$, $\langle \nabla L(\tup), \nabla R(\tup)\rangle = 0$.
\end{lemma}
\begin{proof}
We start by calculating the partial gradients for $L$ and $r$. We have
\begin{align*}
&\begin{aligned}[c]
\nabla_{\bm A} L &= 2\ten S_{(1)}(\bm B\otimes \bm C)(\diff)_{(1)}^\top\\
\nabla_{\bm B} L &= 2\ten S_{(2)}(\bm A\otimes \bm C)(\diff)_{(2)}^\top\\
\nabla_{\bm C} L &= 2\ten S_{(3)}(\bm A\otimes \bm B)(\diff)_{(3)}^\top\\
\end{aligned}\quad
\begin{aligned}[c]
\nabla_{\bm A} \reg &= 4(\bm A\bm  A^\top- \ten S_{(1)}\ten S_{(1)}^\top)\bm A\\
\nabla_{\bm B} \reg &= 4(\bm B \bm B^\top - \ten S_{(2)}\ten S_{(2)}^\top)\bm B\\
\nabla_{\bm C} \reg &= 4(\bm C\bm C^\top - \ten S_{(3)}\ten S_{(3)}^\top)\bm C\\
\end{aligned}\\
&\nabla_{\ten S} L = 2(\tuk{S}{A}{B}{C}-\ten T)(\bm A^\top,\bm B^\top,\bm C^\top)\\
&\nabla_{\ten S} \reg = 4\ten S(\ten S_{(1)}\ten S_{(1)}^\top - \bm A\bm A^\top,\bm I,\bm I) + 4\ten S(\bm I,\ten S_{(2)}\ten S_{(2)}^\top-\bm B\bm B^\top,\bm I)\\
&\qquad\quad+4\ten S(\bm I,\bm I,\ten S_{(3)}\ten S_{(3)}^\top-\bm C\bm C^\top)
\end{align*}
We now compute the following:
\begin{align*}
\langle \nabla_{\bm A} L, \nabla_{\bm A} \reg \rangle &=
 8\langle  \ten S_{(1)}(\bm B\otimes \bm C)(\diff)_{(1)}^\top, (\bm A\bm  A^\top- \ten S_{(1)}\ten S_{(1)}^\top)\bm A 	 \rangle\\
&= 8\langle (\diff)(\bm A^\top,\bm B^\top,\bm C^\top), \ten S(\bm A\bm A^\top - \ten S_{(1)}\ten S_{(1)}^\top,\bm I,\bm I)\rangle

\end{align*}
From here it is easy to see that
$\langle \nabla_{\ten S} L, \nabla_{\ten S} \reg\rangle = - \langle \nabla_{\bm A} L, \nabla_{\bm A} \reg\rangle - \langle \nabla_{\bm B} L, \nabla_{\bm B} \reg\rangle - \langle \nabla_{\bm C} L, \nabla_{\bm C} \reg\rangle$, therefore $\langle \nabla L, \nabla \reg\rangle = 0$.
Since $\nabla R = 2\reg\nabla \reg$, the result follows.
\end{proof}

Now we are ready to prove Lemma~\ref{lem:nonzeroreg}:

\begin{proof}
By Lemma \ref{lem:reg_orthogonality}, we know
$\|\nabla f\|_F^2 = \|\nabla L\|_F^2 + \|\nabla R\|_F^2$.
On the other hand, by Lemma \ref{lem:reg_gradient} and an application of the Cauchy-Schwarz inequality, we see that
\[
\|\nabla \reg\|_F\|(\tup)\|_F \geq 4\reg(\tup),
\]
which means that $\|\nabla \reg\|_F > 0$ whenever $R = \reg^2 > 0$.
But $\nabla R = 2\reg\nabla \reg$, so we have that $\|\nabla R\|_F > 0$, whence $\nabla f \neq \bm 0$.
\end{proof}

To facilitate later proofs, we will also show a fact that if one perturbs a solution with 0 regularizer, then the regularizer remains very small.

\begin{lemma}\label{lem:reg_perturb}
If $R = 0$, and $\|\Delta \bm A\|_F+\|\Delta \bm B\|_F + \|\Delta \bm C\|_F+\|\Delta \ten S\|_F \le O(1)$, then $R(\ten S+\epsilon\Delta \ten S, \bm A+\epsilon \Delta \bm A, \bm B+\epsilon \Delta \bm B, \bm C+\epsilon \Delta \bm C) = O(\epsilon^4)$ for sufficiently small $\epsilon$.
\end{lemma}

\begin{proof}
It suffices to check that the term $\|(\bm A+\epsilon \Delta \bm A)(\bm A+\epsilon \Delta \bm A)^\top - (\ten S+\epsilon\Delta \ten S)_{(1)}(\ten S+\epsilon\Delta \ten S)_{(1)}^\top\|_F = O(\epsilon)$, as other terms are symmetric, and the final $R$ is degree 4 over these terms. This is clear as
we know $\|\bm A\bm A^\top - \ten S_{(1)}\ten S_{(1)}^\top\|_F = 0$ because $R=0$, and all the remaining terms are bounded by $O(\epsilon)$.
\end{proof}

\subsection{Removing Extraneous Directions}\label{sec:extradirections}
In this section, we show that if  $\bm A$ (respectively $\bm B$, $\bm C$) has a direction in its row-space that is perpendicular to the column-space of $\ten T_{(1)}$ (respectively $\ten T_{(2)}$, $\ten T_{(3)}$), then we have a direction of improvement.
In particular, our goal is to show $\bm A_2 = \bm 0$ for all local minima (symmetric arguments will then show $\bm B_2 = \bm C_2 = \bm 0$). We first show that $\ten S(\bm A_2,\bm B,\bm C) = \bm 0$.

\begin{lemma} \label{lem:a21}
Assume that $R(\tup) = 0$. If $\ten S(\bm A_2,\bm B,\bm C) \neq \bm 0$, then $\Delta \bm A = -\bm A_2$ is a direction of improvement.
\end{lemma}
\begin{proof}
Set $\Delta \bm A =  - \bm A_2$.
Then for $\epsilon > 0$
\begin{align*}
L(\ten S,\bm A+\epsilon \Delta \bm A,\bm B,\bm C) &= \|\diff + \epsilon \ten S(\Delta \bm A,\bm B,\bm C)\|_F^2\\
&= L(\tup) -2\epsilon\|\ten S(\bm A_2,\bm B,\bm C)\|_F^2 + O(\epsilon^2),
\end{align*}
since $\langle \diff, \ten S(\bm A_2,\bm B,\bm C)\rangle = \langle \ten S(\bm A_2,\bm B,\bm C), \ten S(\bm A_2,\bm B,\bm C)\rangle$.
Hence, for all sufficiently small $\epsilon$, $L(\ten S,\bm A+\epsilon \Delta \bm A,\bm B,\bm C) < L(\tup)$.
By Lemma~\ref{lem:reg_perturb} we know $R(\ten S,\bm A+\epsilon \Delta \bm A, \bm B, \bm C) = O(\epsilon^4)$.
Hence, for sufficiently small $\epsilon$, the decrease in $L$ will exceed any increase in $R$.
This shows that $\Delta \bm A$ is a direction of improvement.
\end{proof}

We next establish that $R(\tup) =0$ and $\ten S(\bm A_2,\bm B,\bm C) = \bm 0$ together imply that $\bm A_2 = \bm 0$.

\begin{lemma} \label{lem:zeroa2}
If $R(\tup) = 0$ and $\ten S(\bm A_2,\bm B,\bm C) = \bm 0$, then $\bm A_2 = \bm 0$.
\end{lemma}
\begin{proof}
Since $R(\tup) = 0$, we have $\bm B\bm B^\top = \ten S_{(2)} \ten S_{(2)}^\top$ and $\bm C\bm C^\top = \ten S_{(3)} \ten S_{(3)}^\top$. This means the column span of $\ten S_{(2)}$ ($\ten S_{(3)}$) is the same as column span of $\bm B$ ($\bm C$).
Let $\bm B^+$ and $\bm C^+$ denote the pseudoinverses of $\bm B$ and $\bm C$.
Note that the orthogonal projections onto the column-space of $\bm B$ and $\bm C$ are given by
$\bm P_{\bm B} := \bm B\bm B^+$ and $\bm P_{\bm C} := \bm C\bm C^+$, respectively.
Using these facts along with $\ten S(\bm A_2,\bm B,\bm C) =\bm 0$, we have
\begin{align*}
\bm 0 &= \ten S(\bm A_2,\bm B,\bm C)(\bm I,\bm B^+,\bm C^+) = \ten S(\bm A_2,\bm P_{\bm B},\bm P_{\bm C}) = \ten S(\bm A_2,\bm I,\bm I).
\end{align*}
Using the fact that $\ten S_{(1)}\ten S_{(1)}^\top = \bm A \bm A^\top = \bm A_1 \bm A_1^\top  + \bm A_2 \bm A_2^\top$, we have
\[
\|\bm A_2\bm A_2^\top\|_F^2 \le \langle\bm A_2\bm A_2^\top, \ten S_{(1)}\ten S_{(1)}^\top\rangle = \|\ten S(\bm A_2, \bm I,\bm I)\|_F^2 = 0,
\]
which, in particular, means that $\bm A_2 = \bm 0$.
\end{proof}

\subsection{Adding Missing Directions}
\label{sec:missingdirections}

We now consider the case where the row-spans of $\bm A$, $\bm B$, and $\bm C$ are not equal to the column-spans of $\ten T_{(1)}$,$ \ten T_{(2)}$, and $\ten T_{(3)}$, respectively. Again by symmetry, we focus on the case when row-span of $\bm A$ is not equal to column-span of $\ten T_{(1)}$.

\begin{lemma}\label{lem:missingdirections} If the row-span of $\bm A$ is a strict subset of the column-span of $\ten T_{(1)}$ and $R = 0$, then there is a direction of improvement.
\end{lemma}

\begin{proof}
If the row-span of $\bm A$ is a strict subset of column-span of $\ten T_{(1)}$, we must have a vector $\bm a$ that is in the column-span of $\ten T_{(1)}$, but $\bm A \bm a = \bm 0$. For this vector we know $\ten T(\bm a, \bm I, \bm I) \ne \bm 0$, therefore there must exist vectors $\bm b,\bm c$ such that $\ten T(\bm a,\bm b, \bm c) > 0$.
This is true even if we restrict $\bm b$ to be either in the row span of $\bm B$ or to satisfy $\bm B b = 0$ (and similarly for $\bm c$), as we can partition the matrix into 4 subspaces based on the projections of its columns to row span of $\bm B$ (and its rows to row span of $\bm C$).
In particular, if we let $\bm b_1$ be the projection of $\bm b$ onto the row-span of $\bm B$ and $\bm c_1$ be the projection of $\bm c$ onto the row-span of $\bm C$, and set $\bm b_2 = \bm b - \bm b_1$ and $\bm c_2 = \bm c - \bm c_1$, then we have $\ten T(\bm a, \bm b, \bm c) = \sum_{i,j\in \{1,2\}} \ten T(\bm a, \bm b_i, \bm c_j).$
Hence, $\ten T(\bm a, \bm b_i, \bm c_j) > 0$ for some choice of $i, j \in \{1,2\}$.

\vspace*{0.1in}
{\noindent \bf One missing direction}

In this case  $\bm b$ and $\bm  c$ are in row span of $\bm B,\bm C$ respectively.
Choose unit vectors $\bm u,\bm v,\bm w\in \mathbb{R}^r$ such that $\bm A^\top \bm u = \bm 0$, $\bm B^\top \bm v = \alpha_1 \bm b$, and $\bm C ^\top \bm w = \alpha_2\bm  c$, where $\alpha_1$ and $\alpha_2$ are positive real numbers.
Consider the directions $\Delta \bm A =\bm u\bm a^\top$, $\Delta \ten S =  \bm u\otimes \bm v\otimes \bm w$.
Observe that $\Delta \point = \bm A^\top\bm u\otimes \bm B^\top\bm v\otimes \bm C^\top\bm w = \bm 0$ and $\ten S(\Delta \bm A,\bm  B, \bm C) = \bm 0$ since the column-space of $\bm S_{(1)}$ is equal to the column-space of $\bm A$.
Moreover, $\Delta \ten S(\Delta \bm A, \bm B, \bm C) = \bm a\otimes \bm B^\top \bm v\otimes \bm C^\top \bm w = \alpha_1\alpha_2\bm a\otimes \bm b \otimes \bm c$.
Hence,  for $\epsilon > 0$, we have
\begin{align*}
L(\ten S + \epsilon \Delta \ten S, \bm A + \epsilon \Delta \bm A, \bm B, \bm C) &= \|\diff + \epsilon^2\Delta\ten S(\Delta \bm A, \bm B, \bm C)\|_F^2\\

&= L(\tup)-2\epsilon^2\alpha_1\alpha_2\ten T(\bm a,\bm b,\bm c) + O(\epsilon^4).
\end{align*}
On the other hand, by Lemma~\ref{lem:reg_perturb} $R(\ten S + \epsilon\Delta\ten S, \bm A + \epsilon \Delta \bm A, \bm B, \bm C) = O(\epsilon^4)$ since $R(\tup) = 0$.
Hence, for small enough $\epsilon$, the improvement in the tensor loss dominates all other perturbations, so we have a direction of improvement.

\vspace*{0.1in}
{\noindent \bf Two missing directions}

Now assume that $\bm A\bm a = \bm B\bm b = \bm 0$, and $\bm c$ is in the row span of $\bm C$.
Choose unit vectors $\bm u,\bm v,\bm w \in \mathbb{R}^r$ such that $\bm A^\top \bm u = \bm B^\top \bm v =\bm 0$ and $\bm C^\top \bm w = \alpha \bm c$ where $\alpha > 0$.
Consider the directions $\Delta \bm A =   \bm u\bm a^\top$, $\Delta \bm B = \bm v\bm b^\top$, $\Delta \ten S = \bm u\otimes \bm v\otimes \bm w$.
Through a very similar calculation as in the previous case,
\[
L(\ten S+\epsilon \Delta \ten S, \bm A+\epsilon \Delta \bm A, \bm B+\epsilon \Delta \bm B,\bm C) = L(\tup) - 2\epsilon^3\alpha \ten T(\bm a,\bm b,\bm c) + \epsilon^6\alpha^2.
\]
As before, by Lemma~\ref{lem:reg_perturb} $R(\ten S + \epsilon\Delta\ten S, \bm A + \epsilon \Delta \bm A, \bm B +\epsilon \Delta \bm B, \bm C) = O(\epsilon^4)$.
Hence, the decrease in the tensor loss dominates all other perturbations for sufficiently small $\epsilon$, and so this is a direction of improvement. Note that in this case the amount of improvement is $\Theta(\epsilon^3)$, so the point is a 3rd order saddle point.

The case where $\bm C \bm c = \bm 0$ and $\bm b$ is in the row-span of $\bm B$ is similar, and likewise yields a direction of improvement.

\vspace*{0.1in}
\noindent
{\bf Three missing directions}
Now assume that $\bm A\bm a = \bm B\bm b = \bm C\bm c = \bm 0$, and choose unit vectors $\bm u,\bm v,\bm w \in \mathbb{R}^r$ such that $\bm A^\top \bm a  = \bm B^\top \bm v = \bm C^\top \bm w = \bm 0$.
Consider the  directions $\Delta \bm A = \bm u\bm a^\top$, $\Delta \bm B = \bm v\bm b^\top$, $\Delta \bm C = \bm w\bm c^\top$, and $\Delta \ten S = \bm u\otimes\bm  v\otimes \bm w$.
Once again, most perturbations in the tensor loss vanish, and we have
\[
L(\ten S+\epsilon \Delta \ten S, \bm A+\epsilon\Delta \bm A, \bm B+\epsilon \Delta \bm B,\bm C+\epsilon \Delta \bm C) = L(\tup) - 2\epsilon^4 \ten T(\bm a,\bm b,\bm c) + \epsilon^8.
\]
In this case, the regularizer doesn't change at all,
since $\Delta \ten S_{(i)}\ten S_{(i)}^\top = \bm 0$ for $i = 1,2,3$, $\Delta \bm A \bm A^\top = \Delta \bm B \bm B^\top = \Delta \bm C \bm C^\top =\bm  0$,
and $\Delta \bm A\Delta \bm A^\top - \Delta \ten S_{(1)}\Delta \ten S_{(1)}^\top  = \bm u\bm u^\top - \bm u \bm u^\top =\bm 0$ (and the two other analogous terms likewise vanish).
Hence, for sufficiently small $\epsilon$, the objective function decreases, so this is a direction of improvement. This point is a 4th order saddle point.

\end{proof}

\subsection{Improving the core tensor}
\label{sec:handlings}
We finally consider the case where the matrices $\bm A, \bm B, \bm C$ have the correct row-spaces but $\point \neq\ten T$. In this situation, we can make
progress by changing only $\ten S$.

\begin{lemma}\label{lem:handlings}
If $R = 0$, row spans of $\bm A,\bm B,\bm C$ are equal to column span of $\ten T_{(1)},\ten T_{(2)}, \ten T_{(3)}$ respectively, but $L > 0$, then there exists a direction of improvement.
\end{lemma}

\begin{proof}
Since the spans of $\bm A,\bm B, \bm C$ are already correct, let $\bm A^+$ be the pseudoinverse of $\bm A$, then if we let $\ten S' = \ten T(\bm A^+, \bm B^+, \bm C^+)$, we have $\ten S'(\bm A,\bm B,\bm C) = \ten T$.

Consider the direction $\Delta \ten S = \ten S' - \ten S$.

\begin{align*}
L(\ten S+\epsilon \Delta \ten S, \bm A, \bm B, \bm C) &= \|(1-\epsilon)\point - (1-\epsilon)\ten T\|_F^2\\
&= (1-\epsilon)^2L(\tup).
\end{align*}

For the regularizer $R$, again by Lemma~\ref{lem:reg_perturb} we have $R(\ten S+\epsilon \Delta \ten S, \bm A, \bm B, \bm C) = O(\epsilon^4)$. Hence, this is a direction of improvement.
\end{proof}

\subsection{Proof of Main Theorem}

Now with all the lemmas we are ready to prove the main theorem:

\begin{proof}[Proof of Theorem~\ref{thm:exact}] The Theorem follows immediately from the sequence of lemmas.

First, by Lemma~\ref{lem:nonzeroreg}, we know any local minima must satisfy $R = 0$. Next, by Lemma~\ref{lem:zeroa2} and Lemma~\ref{lem:a21}, we know the row spans of $\bm A$, $\bm B$, $\bm C$ must be subsets of column spans of $\ten T_{(1)}, \ten T_{(2)}, \ten T_{(3)}$ respectively. In the third step, by Lemma~\ref{lem:missingdirections}, we further show that  the row spans of $\bm A$, $\bm B$, $\bm C$ must be exactly equal to column spans of $\ten T_{(1)}, \ten T_{(2)}, \ten T_{(3)}$ respectively. Finally, by Lemma~\ref{lem:handlings} we know the loss function must be equal to 0.
\end{proof}