text:
\begin{document}
\maketitle

\begin{abstract}
Let $H$ be a real Hilbert space. In this short note, using some of the properties of bounded linear operators with closed range defined on $H$, certain bounds for a specific convex subset of the solution set of infinite linear complementarity problems, are established.
\end{abstract}

{\bf AMS Subject Classification(2010):} {47A99, 90C48.}

{\bf Keywords.} {Linear complementarity problem, Hilbert space, Closed range operator, Positive semidefinite operator.}

\newpage
\section{Introduction}\label{notation}
Let $H$ be a Hilbert space over the real field $\mathbb{R}$. Let $\mathcal{B}(H)$ denote the set of bounded linear operators on the Hilbert space $H$. A subset $K$ of a Hilbert space $H$ is said to be a cone, if $x, y \in K$ and $\lambda \geq 0$ imply that $x+ y \in K$, $\lambda x \in K$ and $K \cap (-K) = \{0\}.$ Let $H^{'}$ denote the space of all continuous linear functionals on $H$. The dual cone $K^*$ of a cone $K$ is defined as follows: $K^{*}=\{f \in H^{'} : f(x) \geq 0 , \forall x \in K \}$. Let $H_+$ be a cone in $H$. For a given vector $q \in H$ and an operator $T \in \cal{B}$ $(H)$ the linear complementarity problem, written as $LCP(T,q)$, is to find a vector $z \in H_+$ such that $Tz + q \in H_+^*$ and $\langle z, Tz+q \rangle = 0 $. For the case $H=\mathbb{R}^n$ and $H_+ =\mathbb{R}_+^n$ (entry-wise nonnegative vectors in $\mathbb{R}^n$), this problem is well studied \cite{cot-pan-sto}. The linear programming problem in $\mathbb{R}^n$ is defined as follows:  For a given vector $p \in \mathbb{R}^n$, find a vector  $x \in \mathbb{R}_+^n$ which minimizes $p^Tx$ subject to $Tx + q \in \mathbb{R}_+^n$, where $T$ is an $n \times n$ real matrix, $q$ is a vector in $\mathbb{R}^n$ and $p^T$ denotes the transpose of the vector $p$. In \cite{manga1}, the author proved that, under certain assumptions, each solution of linear programming problem is also a solution of the linear complementarity problem. In \cite{cry-demp}, the authors extended the results of \cite{manga1} to infinite dimensional Hilbert spaces. Also linear complementarity problems are closely related to the variational inequality problems. In \cite{cry-demp}, the authors established  that equivalence linear complementarity problem and variational inequality problem. In \cite{isac-numerical}, the author studied the boundedness of solution set of  linear complementarity problems over infinite dimensional Hilbert spaces. One of the main objectives of this article is to study some of the properties of closed range operators. Using these properties, we shall extend some of the bounds of the solution set of linear complementarity problems established in   \cite{isac-numerical}. Also, we provide an alternate simple proof for one of the main results (Theorem $3.1$) of  \cite{isac-numerical}.

This article is organized as follows: In section \ref{ndp}, we collect some of the known results. In section \ref{prop-clos}, we establish some of the properties of closed range operators. In Theorem \ref{posi-closed} and Theorem \ref{posi-closed-conv}, we derive an equivalent condition for an operator to be a closed range positive semidefinite operator.  In section \ref{bound_lcp}, we derive bounds for a convex subset of the solution set of linear complementarity problems.

\section{Notation, Definitions and Preliminary Results}\label{ndp}
For an operator $T \in \mathcal{B}(H)$,  let $T^{*}, R(T)$ and $N(T)$ denote the adjoint, range space and null space of $T$, respectively.  The \emph{Moore-Penrose inverse} of an operator $T  \in \mathcal{B}(H)$ is the unique operator, if it exists, $S \in \mathcal{B}(H)$ satisfying the following conditions: $(1)~ T = TST$,  $(2)~S = STS$, $(3)~(TS)^{*} = TS$ and $(4)~(ST)^{*} = ST$, and is denoted by $T^{\dagger}$.  For a subset $M$ of  $H$, $\overline{M}$ denotes the topological closure of $M$.  An operator $T \in  \mathcal{B}(H)$ is said be \emph{closed range operator}, if $R(T)$ is closed. It is well known that, an operator  $T \in \mathcal{B}(H)$ has Moore-Penrose inverse $S \in \mathcal{B}(H)$ if and only if $R(T)$ is closed \cite{benis, kato}. A self-adjoint operator $T  \in \mathcal{B}(H)$ is said to be \emph{positive semidefinite} if for all $x \in H$, $\langle Tx, x \rangle \geq 0$, and $T$ is said to be \emph{positive definite} if $\langle Tx , x \rangle > 0$ for all nonzero $x \in H$ \cite{isac-numerical}. The positive semidefinite operators are known as positive operators in the literature \cite{reed-simon}. If $T$ is a positive semidefinite operator, then there exist a unique  positive semidefinite operator $S$ such that $S^2 = T$ \cite[Lemma 3.2]{bott}. The operator $S$  is called the positive square root of the operator $T$ and is denoted by $T^\frac{1}{2}.$

If $T \in \mathcal{B}(H)$ and $R(T)$ is closed, then the following holds: $(a)~ R(T^*) = R(T^\dagger)$, $(b)~TT^\dagger y = y$ for all $y \in R(T)$ and $(c)~TT^\dagger = T^\dagger T$, whenever $R(T) = R(T^*)$. Further, if $T$ is positive semidefinite, then  $R(T) = R(T^\frac{1}{2}) , N(T) = N(T^\frac{1}{2})$.

Let $X$ be a real linear space. Then $X$ is called a {\it partially ordered vector space} if there is a partial order $"\leq"$ defined on $X$ such that the following compatibility conditions are satisfied: $(i)~ x \leq y \Longrightarrow x + z \leq y + z$ for all $z \in X$ and $(ii)~ x \leq y \Longrightarrow \alpha x \leq \alpha y$ for all $\alpha \geq 0$. A subset $X_+$ of a real linear space $X$ is said to be a {\it cone} if, $X_+ + X_+ \subseteq X_+,$  $\alpha X_+ \subseteq X_+ $ for all $ \alpha \geq 0,$  $X_+ \cap {-X_+} = \{0\}$ and $X_+ \neq \{0\}$. A vector $x \in X$ is said to be {\it nonnegative}, if $x \in X_+$. This is denoted by $x \geq 0$. We define  $x \leq y$ if and only if $y-x \in X_+$. Then $"\leq"$ is a partial order (induced by $X_+$) on $X$. Conversely, if $X$ is a partially ordered normed linear space with the partial order $"\leq"$, then the set $X_
+ =\{x \in X : x \geq 0\}$ is a cone, and it is called the {\it positive cone} of $X$. By a {\it partial ordered  real normed linear space} $X$ we mean a real normed linear space $X$ together with a {\it closed} positive cone $X_+$. Let $X'$ denote the space of all continuous linear functionals on $X$. The dual cone $X_+^*$ of $X_+$, is defined as follows: $X_+^{*}=\{f \in X' : f(x) \geq 0 , \forall x \in X_+ \}$.

A partially ordered real normed linear space which is also a Banach space is called a {\it partially ordered Banach space}. A partially ordered real normed linear space which is also a Hilbert space is called a {\it partially ordered Hilbert space}.
A cone $X_+$ on a real normed linear space $X$ is said to be solid if $\inte(X_+) \neq \emptyset $, where $\inte(X_+)$ denotes the set of all interior points of $X_+$. If $X$ is a Hilbert space, then a cone $X_+$ is said to be self-dual, if $X_+ = X_+^*$.

\section{Properties of closed range operators}\label{prop-clos}
In this section we study some of the properties of closed range operators.
\begin{defn}
For an operator $T \in \mathcal{B}(H)$ , define $M(T) = \sup \{\langle Tx, x \rangle : x \in H , ||x|| = 1\},$ $m(T) = \inf \{\langle Tx, x \rangle : x \in H , ||x|| = 1\}$ and $m_r(T) = \inf \{\langle Tx, x \rangle : x \in R(T^*) , ||x|| = 1\}.$
\end{defn}

\begin{pro}
Let $T  \in \mathcal{B}(H)$. Then the following statements hold:
\begin{enumerate}
\item[(i)] An operator $T$ is self adjoint if and only if $T^\dagger$ is self adjoint,
\item[(ii)]  For an operator $T$,  $m(T) \geq 0 $ if and only if $T$ is positive semidefinite,
\item[(iii)] A closed range operator $T$ is  self-adjoint positive semidefinite  if and only if $T^\dagger$ is self-adjoint  positive semidefinite.
\item[(iv)] For a closed range operator $T$, $m_r(T) > 0$ if and only if $m_r(T^{\dagger})>0$.
\end{enumerate}
\end{pro}
\begin{proof}
Proofs of part $(i)$ and part $(ii)$ are easy to verify.

\textbf{$(iii)$} Let $T$  be a self adjoint positive semidefinite operator with  closed range. Then, $T^{\dagger}$ exists, and $T^\dagger$ is self adjoint. Let $x \in H$.  Then, $\langle T^\dagger x, x \rangle =  \langle T^\dagger T T^\dagger x , x \rangle=\langle T T^\dagger x , T^\dagger x \rangle \geq0$, since $T$ is positive semidefinite. Thus $T^\dagger$ is positive semidefinite. The converse can be proved in a similar way.

\textbf{$(iv)$} Let $x \in R(T^{*})$, then $x = T^{\dagger}y$ for some $y \in R(T)$. Now,  $\langle  Tx , x \rangle = \langle TT^\dagger y, T^\dagger y\rangle = \langle y, T^\dagger y\rangle$. Since,  $H$  is a Hilbert space over the field of real numbers, we have $ \langle y, T^\dagger y\rangle=  \langle T^\dagger y, y\rangle$. Thus, $\langle  Tx , x \rangle  =  \langle T^\dagger y, y\rangle$, for some $y \in R(T)$, and hence we have $m_r(T)>0$ if and only if $m_r(T^\dagger) >0$.

\end{proof}

The following lemma will be useful in the proof of the subsequent result.

\begin{lemma}\label{mrT-closure}
Let $T \in \mathcal{B}(H)$ be a self-adjoint operator. Then $m_r(T) = \inf \{\langle Tx, x \rangle : x \in \overline{R(T)} , ||x|| = 1\}.$
\end{lemma}
\begin{proof}
 Let $x \in \overline{R(T)}$ and $||x||=1$. Then, there exists a sequence of nonzero vectors $\{x_n\}$ in $R(T)$ such that $x_n$ converges to $x$. Now, since $||x|| =1 $, $x_n$ converges to $x$ implies that $\frac{x_n}{||x_n||}$ converges to $x$. So, without loss of generality, we can assume that $||x_n|| = 1$  for all $n$. Now, $|\langle Tx_n, x_n \rangle  - \langle Tx, x \rangle|=  |\langle  Tx_n, (x_n-x)\rangle + \langle T(x_n-x), x \rangle |\leq ||T|| ||x_n||||x_n-x|| + ||T||||x_n-x||||x||.$ Thus, $x_n$ converges to $x$ implies $\langle Tx_n, x_n \rangle$  converges to  $\langle Tx, x \rangle$, and hence $\langle Tx, x \rangle \geq m_r(T)$.
 \end{proof}

In the next theorem, we establish a sufficient condition for the positive semidefiniteness of the operator $T$ in terms of $m_r(T)$.

\begin{theorem}\label{posi-closed}
Let $T \in \mathcal{B}(H)$ be a self-adjoint operator. If  $m_r(T) > 0$, then $T$ is positive semidefinite and $R(T)$ is closed.
\end{theorem}
\begin{proof}
 By Lemma \ref{mrT-closure}, $\langle Tx, x \rangle \geq 0$ for all $x \in \overline{R(T)}$. Now, we shall prove that  $T$ is positive semidefinite. Let $x \in H$, then $x = x_1 + x_2$, where $x_1 \in N(T)$ and $x_2 \in N(T)^\perp =  \overline{R(T)}$. Now,  $\langle Tx, x \rangle =  \langle T(x_1+x_2), (x_1+x_2) \rangle  = \langle Tx_2, x_2 \rangle \geq 0$. Thus $T$ is positive semidefinite.
To complete the proof, let us show that $R(T)$ is closed. Assume that the sequence $\{Tx_n\}$  converges to $y$. We can assume that, the elements of the sequence $\{x_n\}$  are distinct and belong to $\overline{R(T)}$. Then, by Lemma \ref{mrT-closure}, we have $$m_r(T) \leq \frac{\langle Tx_n - T x_m, x_n-x_m \rangle}{||x_n - x_m||^2} \leq \frac{||Tx_n - T x_m||~||x_n-x_m||}{||x_n - x_m||^2} $$ so that  $$m_r(T) ||x_n - x_m|| \leq ||Tx_n - Tx_m||.$$ Since $\{Tx_n\}$ is a cauchy sequence, it follows that $\{x_n\}$ is also a cauchy  sequence and hence $\{x_n \}$ converges to some  $x \in H$. Thus $Tx_n$ converges to $Tx$, and hence $Tx=y$. Thus $R(T)$ is closed.
\end{proof}

Next, we show that the converse of Theorem \ref{posi-closed} is true.

\begin{theorem}\label{posi-closed-conv}
Let $T \in \mathcal{B}(H)$ be a self-adjoint operator. If $T$ is positive semidefinite and $R(T)$ is closed, then  $m_r(T) > 0$.
\end{theorem}
\begin{proof}
Suppose that $m_r(T) =0$. Then, there exists a sequence $\{x_n\}$ in $R(T)$ such that $||x_n|| =1$ and $\langle Tx_n, x_n \rangle  \rightarrow 0$. We have $\langle Tx_n , x_n \rangle =  \langle T^{\frac{1}{2}}x_n , T^{\frac{1}{2}}x_n \rangle = ||T^{\frac{1}{2}}x_n||$ and so $T^{\frac{1}{2}}x_n \rightarrow 0$ which, in turn implies that $Tx_n \rightarrow 0$. Since $T$ has closed range, $T^\dagger$ is bounded and hence $T^\dagger T x_n \rightarrow 0$. But $T^\dagger T x_n = x_n$ for all $n$. So $x_n$ converges to $0$, which is not possible. Thus $m_r(T) >0$.
\end{proof}

Next, let us establish a property of closed range operators with $m_r(T)>0$.

\begin{theorem}\label{norm_rel}
Let $T \in \mathcal{B}(H)$ be a self adjoint operator with $m_r(T) >0 $. If
\begin{itemize}
\item[(a)] $R_1 = \sup \{\langle Tx, x \rangle : x \in R(T)$ and $||x|| =1 \},$
\item[(b)] $R_2 = \sup \{\langle Tx, x \rangle : x \in R(T)$ and $||x|| \leq 1 \},$
\item[(c)] $R_3 = \sup \{\langle Tx, x \rangle : ||x|| =1 \},$ and
\item[(d)] $R_4 = \sup \{\langle Tx, x \rangle : ||x|| \leq 1 \}, $
\end{itemize}
then $R_1 =R_2 =R_3 =R_4$.
\end{theorem}
\begin{proof}
It is easy to verify that $R_1 = R_2$ and $R_3 =R_4$.To complete the proof, let us prove $R_2 = R_4$. From the definition, it is clear that $R_4 \geq R_2$. Let $x \in H$ such that $||x||\leq 1.$ Then, $x = x_1 + x_2$ with $x_1 \in N(T)^\perp = R(T)$, $x_2 \in N(T)$ such that $||x_1|| \leq 1$ and $||x_2|| \leq 1$. Thus, for any  $x \in H$, we have $\langle Tx , x \rangle = \langle Tx_1,  x_1\rangle$ for some $x_1 \in R(T)$ with $||x_1|| \leq 1$. Thus, we have $R_4 = R_2$.
\end{proof}

In the next theorem, we derive relationships between $M(T^\dagger)$ and $m_r(T)$, and $M(T)$ and $m_r(T^\dagger)$

\begin{theorem}\label{mMinv}
Let $T \in \mathcal{B}(H)$ be a self adjoint operator such that $m_r(T)>0$. Then the following holds:
 \begin{enumerate}
 \item $M(T^\dagger) = [m_r(T)]^{-1}$, and
 \item $m_r(T^\dagger) = [M(T)]^{-1}.$
\end{enumerate}
\end{theorem}
\begin{proof}
Let $y \in R(T)$ with $||y||=1$. Then $y = Tx$, for some $x \in R(T)$. Now, $ \langle T^\dagger y, y \rangle = \langle T^\dagger T x, T x \rangle = \langle Tx, x \rangle$. Since, $m_r(T) \leq \frac{1}{||x||^2} \langle Tx, x\rangle$, we have $ \frac{1}{\langle Tx, x\rangle} \leq \frac{1}{||x||^2 m_r(T)}$ and hence $\langle Tx, x \rangle \leq \frac{\langle Tx , x \rangle^2}{||x||^2 m_r(T)} \leq \frac{||Tx||^2}{m_r(T)}$. Thus $\langle T^\dagger y, y\rangle \leq \frac{1}{m_r(T)}$, and hence $\sup\{\langle T^\dagger y , y \rangle: y \in R(T), ||y|| = 1\} \leq \frac{1}{m_r(T)}$. Now, by Theorem \ref{norm_rel}, we get $M(T^\dagger) \leq \frac{1}{m_r(T)}.$

If $x \in N(T)$ or $y \in N(T)$, then $\langle Tx, y \rangle = \langle x , Ty \rangle = 0.$ Let $x, y \in R(T)$, then, by Cauchy-Schwartz inequality for a semi inner product, we have $|\langle Tx, y \rangle|^2 \leq \langle Tx, x\rangle \langle Ty, y\rangle.$ If $y = T^\dagger x$, then $|\langle TT^\dagger x,  x \rangle|^2 \leq  \langle Tx, x\rangle \langle T^\dagger x, x\rangle$. Hence, we have $$\langle Tx, x \rangle \langle T^\dagger x, x \rangle \geq 1,$$ whenever $||x|| =1.$ Now, we have $\langle Tx, x \rangle  \geq \frac{1}{\langle T^\dagger x, x \rangle}$ and hence $\inf \{ \langle Tx, x \rangle: ||x||=1\}  \geq \inf \frac{1}{\{\langle T^\dagger x, x \rangle : ||x|| = 1\}} =  \frac{1}{\sup \{ \langle T^\dagger x, x \rangle: ||x|| = 1\}} $. Thus, we get $M(T^\dagger ) \geq \frac{1}{ m_r(T)}.$ The proof of the second assertion is similar.
\end{proof}

In the next corollary, we establish a relationship between $m_r(T)$ and norm of the operator $T^\dagger$.

\begin{cor}\label{comp-m-M}
If $T \in \mathcal{B}(H)$ is a self adjoint operator such that $m_r(T) >0 $, then $\displaystyle{||T^\dagger|| =\frac{1}{m_r(T)}}$.
\end{cor}
\begin{proof}
By the definition, $||T^\dagger|| =  \sup \{\displaystyle \langle T^\dagger x, x\rangle : x \in H, ||x|| =1 \}$.

 Thus, by Theorem  \ref{norm_rel}, we have $ ||T^\dagger||  = \displaystyle{\frac{1}{m_r(T)}}.$
\end{proof}

\section{Bounds for solution set of linear complementarity problems} \label{bound_lcp}
In this section, we establish bounds for a certain specific convex subset of the solution set of linear complementarity problems. For an operator $T \in \mathcal{B}(H)$ and  $b \in H$, the solution set of the associated linear complementarity problem, LCP$(T, b)$, is denoted by $SOL(T,b)$, is defined as the set of all solutions of the LCP$(T, b)$.

In general, the solution set of a linear complementarity problem need not be bounded. In the next theorem, we give a sufficient condition under which the solution is unbounded.

\begin{theorem}
Let $H$ be a real Hilbert space and let $K$ be a cone. Let $T \in \mathcal{B}(H)$  and $b \in H$. If  $ x \in N(T) \cap SOL(T, b)$ for some nonzero $x \in H$, then $SOL(T, b)$ is unbounded.
\end{theorem}
\begin{proof}
Let $ x \in N(T) \cap SOL(T, b)$ and $x  \neq 0$. Then $\alpha x \in N(T) \cap SOL(T, b)$, for all $ \alpha \geq 0.$ Hence $SOL(T, b)$ is unbounded.
\end{proof}

In the next theorem we establish a bound for those solutions which do not belong to the null space of the operator $T$.

\begin{theorem}\label{bounded1}
Let $H$ be a real Hilbert space and let $K$ be a cone. Let $T \in \mathcal{B}(H)$  and $b \in H$. If $m_r(T)>0$, then the solution set $SOL(T, b)\cap R(T^*)$ of $LCP(T, b)$ is a subset of $B(0, \frac{||b||}{m_r(T)}) \cap K$.
\end{theorem}
\begin{proof}For $x \in H$,
we have,
\begin{align*}
\langle Tx+b, x\rangle &=  \langle Tx, x\rangle + \langle b, x\rangle,      \\
 &\geq  m_r(T)||x||^2  - ||b|| || x||, \\
 & =  (m_r(T)||x|| - ||b||) || x||.
\end{align*}
Now, if $(m_r(T)||x|| - ||b||) >0 $, then $\langle Tx+b, x\rangle >0$. Thus $x \notin SOL(T, b)$.  Hence, if $ x \in SOL(T, b)\cap R(T^*) $, then $(m_r(T)||x|| - ||b||) \leq 0$. Thus $||x|| \leq  \frac{||b||}{m_r(T)}$,  which proves the claim.
\end{proof}

In the next theorem, we derive a bound for solution of the linear complementarity problem, whenever the solution vector belongs to the range space of the operator $T$.

\begin{theorem}
Let $H$ be a real Hilbert space and $K$ be a self-dual cone in $H$. If $T\in \mathcal{B}(H)$ is a self adjoint operator such that $m_r(T) >0 $, then for every solution $x \in R(T)$ of  $LCP(T, b)$ where $b \in R(T)$, one has $||x|| \leq \frac{M(T)}{m_r(T)} ||T^{\dagger}(b)||.$
\end{theorem}
\begin{proof}
Since $b\in R(T)$ and $T$ is self adjoint, we have $b=TT^\dagger b$ and $M(T)=\lvert \lvert T \rvert \rvert$. Thus $\lvert \vert b\rvert \rvert\leq \lvert \lvert T \rvert \rvert \lvert \lvert T^\dagger b \rvert \rvert$. Suppose that $x\in R(T)$ is a solution of $LCP (T,b)$. By Theorem \ref{bounded1},
\begin{align*}
\lvert \lvert x\rvert \rvert & \leq \frac{||b||}{m_r(T)},\\
&  \leq \frac{M(T)}{m_r(T)} ||T^{\dagger}(b)||.
\end{align*}
\end{proof}

The proof of the above theorem gives an alternate simple proof to \cite[Theorem 3.1]{isac-numerical}.

\begin{cor}
Let $H$ be a real Hilbert space and $K$ be a self-dual cone in $H$. If $T\in \mathcal{B}(H)$ is a self adjoint operator such that $m(T) >0 $, then for every solution $x$ of  $LCP(T, b)$, one has $||x|| \leq \frac{M(T)}{m(T)} ||T^{-1}(b)||.$
\end{cor}

\begin{theorem}\label{last}
Let $H$ be a real Hilbert space and let $T\in \mathcal{B}(H)$ be a self adjoint closed range operator. If $x$ is a  solution of $LCP(T, b)$, where $b \in R(T)$, then
\begin{center}
$\langle x-x_b, T(x-x_b)\rangle=\frac{1}{4}\langle b, T^\dagger b\rangle ,$
\end{center}
where $x_b=-\frac{1}{2} T^\dagger b$.
\end{theorem}
\begin{proof}
Let $b\in R(T)$ and $x_b=-\frac{1}{2}T^\dagger b$. Suppose that $x$ is a  solution of $LCP(T, b)$. Then
\begin{eqnarray}
\langle x-x_b, T(x-x_b)\rangle & =& \langle x,Tx\rangle + \frac{1}{2} \langle x,TT^\dagger b\rangle + \frac{1}{2} \langle T^\dagger b, T x\rangle + \frac{1}{4}\langle T^\dagger b, TT^\dagger b\rangle \nonumber \\
&=& \langle x,T x\rangle + \langle x, b\rangle +\frac{1}{4}\langle T^\dagger b, b\rangle \nonumber \\
&=&  \frac{1}{4}\langle b, T^\dagger b\rangle. \nonumber
\end{eqnarray}
\end{proof}

\begin{theorem}
Let $H$ be a real Hilbert space and  $T\in \mathcal{B}(H)$ be a self adjoint operator such that $m_r(T) >0 .$ Let $b \in R(T)$ and $x_b=-\frac{1}{2} T^\dagger b$.  Then for every solution $x\in R(T)$ of  $LCP(T, b)$, where $x\neq x_b$, one has
\begin{center}
$\frac{\| b \|}{2M(T)}\leq \| x-x_b \| \leq  \frac{\| b \|}{2m_r(T)}.$
\end{center}
\end{theorem}
\begin{proof}
By Theorem \ref{last}, we have $\langle x-x_b, T(x-x_b)\rangle=\frac{1}{4}\langle b, T^\dagger b\rangle.$
Also, $\frac{1}{4} m_r(T^\dagger)\|b\|^2\leq \|x-x_b\|^2M(T)$, by the definition of $M(T)$ and $m_r(T)$.
Now, by Theorem \ref{mMinv}, we have
$\frac{1}{4M(T)}\|b\|^2\leq \|x-x_b\|^2M(T).$
Thus, we can conclude that
\begin{equation}
\frac{1}{2M(T)}\|b\|\leq \|x-x_b\|. \label{1}
\end{equation}
Similarly, by Theorem \ref{mMinv} and \ref{last}, we have
$m_r(T)\|x-x_b\|^2\leq \frac{1}{4}M(T^{\dagger})\|b\|^2= \frac{1}{4m_r(T)}\|b\|^2$.
Thus \begin{equation}
\|x-x_b\|\leq \frac{1}{2m_r(T)}\|b\|. \label{2}
\end{equation}
From equation \eqref{1} and \eqref{2},
\begin{center}
$\frac{\| b \|}{2M(T)}\leq \| x-x_b \| \leq  \frac{\| b \|}{2m_r(T)}.$
\end{center}
\end{proof}

\subsection{Example}
Let $H = l^2(\mathbb{Z})$ and $H_+ = \{(\dots, x_{-2}, x_{-1},\fbox{$x_{0}$},x_{1},x_{2}, \dots): x_i \geq 0 ~\mbox{for all}~ i\}.$ Then $H_+$ is a self-dual cone.  Define $T : H \rightarrow H$ as $T((\dots, x_{-2}, x_{-1},\fbox{$x_{0}$},x_{1},x_{2}, \dots)) = (\dots, 0, 0, \fbox{$x_{0}$},x_{1},x_{2}, \dots)$. Then, $T$ is an orthogonal projection on to the space $l^2(\mathbb{N})$. So, $T = T^2 = T^*$ and $T = T^\dagger$. Consider  the vector $b = (\dots, 0, 0, \fbox{$0$}, \frac{-1}{3},1,\frac{1}{2}, \frac{1}{3}, \dots) \in R(T)$, it is easy to verify that $x = (\dots, 0, 0, \fbox{$0$}, \frac{1}{3},0,0, \dots)$ solves the associated LCP. It is clear that, $\Vert x \Vert \leq \Vert b \Vert$.  Also, $\Vert x - x_b \Vert = \frac{\sqrt{37}}{6} = \frac{\Vert b \Vert}{2}$. Indeed, in the vector $b$, if we may replace the entry $2$ by any real number.

It may be observed that, more generally, any orthogonal projection satisfies the assumptions of our theorem.\\

\textbf{Acknowledgement:}
Projesh Nath Choudhury  was partially supported by National Post-Doctoral Fellowship(PDF/2019/000275), the SERB, Department of Science and Technology, India,  and the NBHM Post-Doctoral Fellowship (0204/11/2018/R$\&$D-II/6437) from DAE (Govt. of India).
M. Rajesh Kannan would like to thank the SRIC, IIT Kharagpur, the SERB, Department of Science and Technology, India, for financial support through the projects ISIRD,  MATRICS (MTR/2018/000986) and Early Career Research Award (ECR/2017/000643).

\bibliographystyle{amsplain}
\bibliography{gen-inv-lcp}

\end{document}


TITLE: Numercial simulation of the double slit experiment
QUESTION [0 upvotes]: Originally I had a code that simulates single slit diffraction:
parameters for functions: f(start_of_a_slit, end_of_a_slit, start_of_a_screen, end_of_a_screen, number_of_points, distance_between_screen_and_slit, wavenumber, amplitude)

    double k = 2.0 * M_PI / lambda;
    double complexAmp[numPnt + 1];

    Real fReal(-a / 2.0, a / 2.0, c, d, numPnt, dDif, k, amp);
    Imaginary fIm(-a / 2.0, a / 2.0, c, d, numPnt, dDif, k, amp);
    // integration
    std::ofstream file("wave.bin");
    for (int i = 1; i <= numPnt; i++) {
        fReal.simpson(-a / 2.0, a / 2.0, fReal.y[i], dDif, k, amp, numInt);
        fIm.simpson(-a / 2.0, a / 2.0, fIm.y[i], dDif, k, amp, numInt);
        complexAmp[i] = fReal.integral * fReal.integral +
                fIm.integral * fIm.integral;
        file << fReal.y[i] << "\t" << complexAmp[i] << std::endl;
    }

Where method simpson calculates the integral of a function:
double Real::value (double x, double y, double d, double k, double amp) {
    double r = sqrt(d * d + (y - x) * (y - x));
    return amp * cos(k * r) / sqrt(r);
}

void Real::simpson(double a, double b, double y, double d, double k, double amp, double size) {
    double h = (b - a) / (2.0 * (double)(size));
    double simpson = .0;
    double x = a + h;
    int i = 1;
    while (i < size + 1) {
        simpson = simpson + (value(x + h, y, d, k, amp) +
                4.0 * value(x, y, d, k, amp) + value(x - h, y, d, k, amp))
                        *  h / 3.0;
        x = x + 2.0 * h;
        i = i + 1;
    }
    integral = simpson;
}

Now I want to change the program so that it simulates double-slit diffraction and I don't know how to do this. I need to create another set of functions, with a slit shifted by some distance. But how do I calculate the integral: do I need to add the values of the functions from two different slits first or do I integrate them first, and then add them together and raise them to the power of two to calculate the intensity?

REPLY [0 votes]: If you assume that the light has a sufficiently large coherence length, you have to calculate the amplitude of each slit on the detector, $A_1$ and $A_2$. Next, since most detectors measure the intensity, the quantity to calculate is proportional to $|A_1 + A_2|^2$.


text:
\subsection*{Organization of the paper}
The rest of the paper is organized as follows.
In~\S\ref{s:floer-setting} we briefly recall the necessary ingredients
from Lagrangian Floer and quantum homologies used in the sequel.
In~\S\ref{sb:more-discr} we also give more details on the
discriminant. \S\ref{s:lag-cubic-eq} is devoted to the Lagrangian
cubic equation. We prove in that section more general versions of
Theorems~\ref{t:cubic-eq} and~\ref{t:rel-to-discr}. Then
in~\S\ref{sb:prf-cubic-eq-S} we prove Theorem~\ref{t:cubic-eq-sphere}.
We also prove in~\S\ref{sb:further-res} additional corollaries derived
from these theorems. In~\S\ref{s:disc-lcob} we study the discriminant
in the realm of Lagrangian cobordism and prove Theorem~\ref{t:cob} and
Corollary~\ref{c:del_1=del_2}.  \S\ref{s:examples} is dedicated to
examples. We briefly explain how to construct Lagrangian spheres in
various homology classes on symplectic Del Pezzo surfaces and carry
out the calculation of the discriminants of those Lagrangians. We
discuss some higher dimensional examples too.  In~\S\ref{s:coeffs} we
explain an extension of the discriminant and the Lagrangian cubic
equation over a more general ring of coefficients that takes into
account the different homology classes of the holomorphic curves that
contribute to our invariants.  In~\S\ref{sb:examples-revisited} we
recalculate some of the examples from~\S\ref{s:examples} over this
ring. In~\S\ref{s:enum} we discuss the relation of the discriminant to
the enumerative geometry of holomorphic disks.  Finally,
in~\S\ref{s:non-monotone} we consider the non-monotone case and state
a version of Theorem~\ref{t:cubic-eq-sphere} for not necessarily
monotone Lagrangian 2-spheres.

config: arxiv


TITLE: If $[A,B], A \in \mathcal{L}$ then does this imply that $B \in \mathcal{L}$?
QUESTION [0 upvotes]: If $[A,B]$ and $A$ are in Lie algebra $\mathcal{L}$ then does this imply that $B \in \mathcal{L}$?

REPLY [0 votes]: An ideal $I$ of a Lie algebra $\mathfrak{g}$ is a subspace of $\mathfrak{g}$ such that
\begin{equation}
[\mathfrak{g},I] \subseteq I.
\end{equation}
The ideal $I$ forms a subalgebra of $\mathfrak{g}$.
Therefore if the Lie algebra $\mathcal{L}$ in the question is some ideal and one chooses any $A \in \mathcal{L}$ and some $B$ in the larger algebra then we would get
\begin{equation}
[A,B] \in \mathcal{L}.
\end{equation}
even for $B \notin \mathcal{L}$.
This provides a counter example and so the answer to the question is a (somewhat resounding) NO!


TITLE: Direct Proof (Basic Real Analysis)
QUESTION [0 upvotes]: $\forall y\in\Bbb Q\,, \exists n∈\Bbb Z~~\textrm{s.t.}~~n\leqslant y\leqslant n+1$.
I'm just unsure how to approach this. I tried applying the Euclidean Algorithm bu I haven't really gotten anywhere. I'm trying not to use the Completeness Axiom on this- just trying to do it as a basic direct proof.

REPLY [4 votes]: Write $q=a/b$, with integer $a$ and $b$, $b>0$. The division algorithm gives you
$$
a=bn+r
$$
with integer $n$ and $r$, with $0\le r<b$. Then
$$
q=\frac{a}{b}=n+\frac{r}{b}
$$
Can you conclude?


text:
\begin{document}
\title{Robust Precoding for Network MIMO with Hierarchical CSIT}
\author{\IEEEauthorblockN{Paul de Kerret\IEEEauthorrefmark{1}, Richard Fritzsche\IEEEauthorrefmark{4}, David Gesbert\IEEEauthorrefmark{1}, and Umer Salim\IEEEauthorrefmark{3}}

\IEEEauthorblockA{\IEEEauthorrefmark{1}
Mobile Communication Department, Eurecom, {\tt{\{dekerret,gesbert\}@eurecom.fr}}}

\IEEEauthorblockA{\IEEEauthorrefmark{4}TU Dresden, Vodafone Chair Mobile Communications Systems, {\tt{richard.fritzsche@tu-dresden.de}}}

\IEEEauthorblockA{\IEEEauthorrefmark{3}Intel Mobile Communications, {\tt{umer.salim@intel.com}}}
}

\maketitle

\begin{abstract}
In this work\footnote{The research leading to these results has received funding from the European Union Seventh Framework Programme (FP7/2007-2013) under grant agreement n° 317941. The authors would like to acknowledge the contributions of their colleagues in iJOIN, although the views expressed are those of the authors and do not necessarily represent the project.} we consider a wireless network with~$K$ cooperating transmitters (TXs) serving jointly $K$~receivers (RXs). Due to the practical limitations of the backhaul network, it is relevant to consider a setting where each TX receives its own imperfect estimate of the multi-user channel state, denoted as the distributed channel state information (CSI) setting. We focus in this work on a particular distributed CSI configuration called \emph{hierarchical CSI} configuration in which the TXs can be ordered by increasing level of CSI. This scenario is particularly relevant for future networks with heterogeneous backhaul where the TXs connected with a weak backhaul link will receive only a coarse estimate while the TXs with a stronger backhaul will have a more accurate CSI. In that scenario, we formulate the optimal precoding as a \emph{team decision} problem. Solving optimally this problem is extremely challenging such that we propose a heuristic approach allowing to obtain a simple, yet efficient and practical, precoding algorithm. The proposed precoding algorithm exploits the hierarchical structure of the CSI to make the transmission more robust to the imperfect CSI knowledge at the TXs.
\end{abstract}

\IEEEpeerreviewmaketitle

\section{Introduction}

Network (or Multi-cell) MIMO methods, whereby multiple interfering TXs share user messages and allow for joint precoding, are currently considered for next generation wireless networks \cite{Gesbert2010}. With perfect message and CSI sharing, the different TXs can be seen as a unique virtual multiple-antenna array serving all RXs in a multiple-antenna broadcast channel (BC) fashion and well known precoding algorithms can be used \cite{Christensen2008}. However, this requires the feedback of a very accurate multi-user CSI to the TX side to achieve the desired high performances\cite{Caire2010}. As a consequence, there has been a large amount of works dealing with the feedback of the CSI (See \cite{Love2008} and references therein) and the design of robust precoders (See for example \cite{Vucic2009,Shenouda2007,Negro2012,Fritzsche2013a}).

However, the large literature on robust precoding typically assumes \emph{centralized} CSIT, i.e., that the precoding is based on the basis of a \emph{single} imperfect channel estimate. It implies that either the precoding is done in a central node or the CSI is \emph{perfectly shared} between the TXs. Although meaningful in the single TX case with multiple-antennas, this assumption is often unrealistic for distant cooperating TXs where the CSI obtained locally has to be exchanged between the distant TXs. This step introduces inevitably some delay and may also require further quantization. Hence, it is practically relevant for joint precoding across distant TXs to allow for the case where each TX receives its \emph{own} channel estimate, which we denote as the \emph{distributed CSI} configuration \cite{dekerret2013_thesis}.

With distributed CSI, the design of a joint precoder is particularly challenging and only few results are available. This problem belongs in fact to the category of \emph{Team Decision} problems \cite{Radner1962,Marschak1972,Ho1980} for which only particular cases have been solved. In \cite{dekerret2012_TIT}, the number of Degrees-of-freedom (DoF) obtained with conventional ZF precoding is derived and some precoding schemes improving the DoF are given. In \cite{Zakhour2010a}, a robust precoding algorithm is designed for the case of two TXs having distributed CSIT. However, the algorithm developed is computationally demanding and does not provide any insight. In \cite{Fritzsche2013b}, it is shown how to refine the precoder when more accurate CSI is available locally. However, this approach can only be used for a specific CSI configuration.

In this work, we consider a CSI configuration, denoted as \emph{hierarchical} (also called ``nested" \cite{Ho1972}) where the TXs can be ordered in such a way that a TX~$j$ has a more accurate channel estimate than a TX~$k$, if $k<j$. Specifically, our main contributions read as follows:
\begin{itemize}
\item We formulate the problem of precoding with hierarchical CSI. Although the hierarchical structure allows to simplify the optimization, it remains a difficult stochastic optimization problem \cite{Kall1994}.
\item Thus, we propose a simple robust hierarchical precoding algorithm exploiting the hierarchical structure of the CSI. The proposed algorithm outperforms precoding schemes from the literature. Furthermore, the approach developed can potentially be adapted to many other scenarios.
\end{itemize}

\section{System Model}

\subsection{Received Signal}

We study the transmission from $K$~TXs to $K$~RXs where the $i$-th TX is equipped with $M_i$~antennas and transmits $d_i$ streams to the $i$-th RX equipped with $N_i$ antennas. The total number of RX antennas, the total number of TX antennas and the total number of streams are respectively given by
\begin{equation}
N_{\Tot}\triangleq \sum_{i=1}^K N_i,\quad  M_{\Tot}\triangleq \sum_{i=1}^K M_i,\quad d_{\Tot}\triangleq \sum_{i=1}^K d_i.
\label{eq:SM_1}
\end{equation}
We further assume that the RXs have perfect CSI as we decide to focus primarily on the challenge of conveying CSI back to the TXs through some form of feedback. We consider that linear filtering are used and that the RXs treat interference as noise. The channel from the $K$~TXs to the $K$~RXs is represented by the channel matrix~$\mathbf{H}^{\He} \in \mathbb{C}^{N_{\Tot}\times M_{\Tot}}$ where $\bH_{ik}^{\He}\in \mathbb{C}^{N_i\times M_k}$ denotes the channel matrix from TX~$k$ to RX~$i$ and has all its elements i.i.d. as $\mathcal{CN}(0,\rho_{i,k}^2)$ and independent of the other channel matrices.

The transmission is then described as
\begin{equation}
\begin{bmatrix}
\by_1\\\vdots\\\by_K
\end{bmatrix}
=
\mathbf{H}^{\He}\xv
+
\bm{\eta}
=
\begin{bmatrix}
\bH^{\He}_1\xv\\\vdots\\
\bH^{\He}_K\xv
\end{bmatrix}
+
\begin{bmatrix}
\bm{\eta}_1\\\vdots\\
\bm{\eta}_K
\end{bmatrix}
\label{eq:SM_2}
\end{equation}
where $\by_i\in \mathbb{C}^{N_i\times 1}$ is the signal received at the $i$-th RX, $\bH^{\He}_i \in \mathbb{C}^{N_i\times M_{\Tot}}$ the channel from all TXs to the $i$-th RX, and $\bm{\eta}\triangleq[\bm{\eta}_1,\ldots,\bm{\eta}_K]^{\trans}\in \mathbb{C}^{N_{\Tot}\times 1}$ the normalized Gaussian noise with its elements i.i.d. as $\mathcal{CN}(0,1)$.

The multi-TX transmitted signal~$\xv\!\in \!\mathbb{C}^{M_{\Tot}\times 1}$ is obtained from the symbol vector $\bs\!\triangleq\![\bs_1^{\trans},\ldots,\bs_K^{\trans}]^{\trans}  \!\in\!  \mathbb{C}^{d_{\Tot}\times 1}$ with its elements i.i.d. $\mathcal{CN}(0,1)$ as
\begin{equation}
\xv=\bT \bs=
\begin{bmatrix}
\bT_1,
\hdots,
\bT_K
\end{bmatrix}
\begin{bmatrix}
\bs_1\\
\vdots\\
\bs_K
\end{bmatrix} =\sum_{k=1}^K\bT_k\bs_k
\label{eq:SM_3}
\end{equation}
with $\bT_j\in \mathbb{C}^{M_{\Tot}\times d_j}$ being the precoder serving user~$j$ and $\bT \in \mathbb{C}^{M_{\Tot}\times d_{\Tot}}$ being the multi-user precoder. We also introduce the matrix~$\bW_j^{\He}\in \mathbb{C}^{M_j\times d_{\Tot}}$ to denote the precoding coefficients at TX~$j$ such that the signal transmitted by TX~$j$, denoted by~$\xv_{j}\!\in \!\mathbb{C}^{M_{j}\times 1}$, is given by
\begin{equation}
\xv_j=\bW_j^{\He}\bs.
\label{eq:SM_4}
\end{equation}
The multi-user precoder~$\bT$ is then alternatively written as
\begin{equation}
\bT=
\begin{bmatrix}
\bW_1^{\He}\\
\vdots\\
\bW_K^{\He}
\end{bmatrix}.
\label{eq:SM_5}
\end{equation}

Finally, the received signals are further processed by the RX filter~$\bG^{\He} \in \mathbb{C}^{d_{\Tot}\times N_{\Tot}}$ equal to
\begin{equation}
\bG^{\He}\triangleq
\mathrm{blockdiag}(\bG^{\He}_1,\ldots,\bG_K^{\He})
\label{eq:SM_6}
\end{equation}
with $\bG_k^{\He}\in \mathbb{C}^{d_k\times N_k}$ being the RX filter at RX~$k$. It follows that we can define the mean square error (MSE) matrix at RX~$k$ for given precoders and RX filters, denoted by~$\bM_k\!\in\! \mathbb{C}^{d_k\times d_k}$, as
\begin{equation}
\begin{aligned}
\bM_k&\triangleq \E_{\bd_k}[(\bd_k-\bG_k^{\He}\by_k)(\bd_k-\bG_k^{\He}\by_k)^{\He}]\\
&= \I_{N_k}+\bG_k^{\He}\bG_k+\bG_k^{\He}\bH^{\He}_k\bT\bT^{\He}\bH_k\bG_k\\
&~~~~-\bG_k^{\He}\bH_k^{\He}\bT_{k}-\bT_k^{\He}\bH_k\bG_{k}.
\end{aligned}
\label{eq:SM_7}
\end{equation}
Following the assumption of Gaussian signaling, the rate of user~$i$ can be written as \cite{Cover2006}
\begin{equation}
R_k\triangleq \log_2\left| \bM_k^{-1}\right|,\qquad \qquad \forall k\in \{1,\ldots,K\}.
\label{eq:SM_8}
\end{equation}
Finally, we define the sum rate~$R$, which will be our main figure-of-merit, as
\begin{equation}
R\triangleq \sum_{k=1}^KR_k.
\label{eq:SM_8}
\end{equation}

\subsection{Distributed Precoding and Distributed CSIT}

In the distributed CSIT model considered here, each TX receives its own CSI based on which it designs its transmission parameters without any additional communication to the other TXs\cite{dekerret2012_TIT,dekerret2013_thesis}. The actual exchange mechanism based on which the TXs receive the multi-user channel estimate is out of the scope of its paper and a research topic in its own right.

TX~$j$ has then the knowledge of the global multi-user channel estimate~$(\hat{\bH}^{(j)})^{\He}\in \mathbb{C}^{N_{\Tot}\times M_{\Tot}}$. We define~$(\hat{\bH}^{(j)}_{i})^{\He}\in \mathbb{C}^{N_{i} \times M_{\Tot}}$ in a similar fashion as its counterpart~$\bH_{i}^{\He}\in \mathbb{C}^{N_{i} \times M_{\Tot}}$ with perfect CSIT.

Hence, TX~$j$ designs its transmit coefficient~$\xv_j\in \mathbb{C}^{M_j\times 1}$ as a function of~$\hat{\bH}^{(j)}$. The transmit signal~$\xv_j$ is then given by
\begin{equation}
\xv_j=\bW_j^{\He}\!(\hat{\bH}^{(j)})\bs
\label{eq:SM_9}
\end{equation}

Due to the assumption of distributed precoding, the actual precoder used for the transmission is equal to
\begin{equation}
\bT \triangleq
\begin{bmatrix}
\bW_1^{\He}\!(\hat{\bH}^{(1)})\\
\bW_2^{\He}\!(\hat{\bH}^{(2)})\\
\vdots\\
\bW_K^{\He}\!(\hat{\bH}^{(K)})
\end{bmatrix}.
\label{eq:SM_10}
\end{equation}

\subsection{Hierarchical Channel State Information}\label{se:SM:DCSI}

TX~$j$ receives an estimate~$\hat{\bH}^{(j)}$ of the multi-user channel~$\bH$. The estimate $\hat{\bH}^{(j)}$ can take a priori any form depending on the transmission scenarios considered, and we focus in this work on a particular CSI configuration called the \emph{hierarchical CSI configuration}.

In the hierarchical CSI configuration, the TXs can be ordered by increasing quality of CSI, i.e., such that the estimate of TX~$j$ is ``included" in the estimate of TX~$j\!+\!1$. This scenario is in particular obtained if a multi-level quantization scheme is used \cite{Ng2009,Zakhour2010a}. In this quantization scheme, the same codebook is used for all the TXs and each TX decodes the estimate up to a number of bits corresponding to the quality of its feedback channel. By decoding less bits, TX~$j$ is then able to reconstitute the channel estimate at TX~$k$, for $k<j$~\cite{Ho1972}.

In particular, we will model the imperfect estimate~$\hat{\bH}^{(j)}$ at TX~$j$ as
\begin{equation}
\{\hat{\bH}^{(j)}\}_{i,k}=\sqrt{1-(\sigma_{i,k}^{(j)})^2}\{\hat{\bH}\}_{i,k}+\sigma_{i,k}^{(j)}\{\bm{\Delta}\}^{(j)}_{i,k},\quad \forall i,k
\label{eq:SM_11}
\end{equation}
where $\{\bm{\Delta}\}^{(j)}_{i,k}\sim\mathcal{CN}(0,1)$ and $\sigma_{i,k}^{(j)}\in (0,1), \forall k,i$ represents the quality of the CSIT at TX~$j$. To model the hierarchical quantization, we hence assume that TX~$j$ has access to $\hat{\bH}^{(k)}$ for $k\leq j$.

\begin{remark}
Any other model for the imperfect knowledge of the channel state can be used in our approach. It is only critical to have the \emph{hierarchical} structure of the CSI. \qed
\end{remark}

\subsection{Precoding with Distributed CSIT: Team Decision Problem}\label{se:SM:TD}

With distributed CSIT, the precoding problem can then be formulated as the following Team Decision problem \cite{Radner1962,Marschak1972,Ho1980}:
\begin{equation}
\begin{aligned}
&(\bW_1^{\star},\ldots,\bW_K^{\star})\\
&=\argmax_{(\bW_1,\ldots,\bW_K)} \E[\Rate(\bW_1(\hat{\bH}^{(1)}),\ldots,\bW_K(\hat{\bH}^{(K)}))] \\
&~~~~~~~~~\text{, s.to $\|\bW_j(\hat{\bH}^{(j)})\|_{\Fro}^2\leq P_j,\forall j\in \{1,\ldots, K\}$.}
\end{aligned}
\label{eq:optimization_opt}
\end{equation}
A necessary condition for any optimal precoding strategy is that it should also be a \emph{best-response} strategy. This means that each TX applies the best strategy \emph{given the strategies of the others TXs}\cite{Nash1951}. Mathematically, a best-response power allocation policy~$(\bW_1^{\BR},\ldots,\bW_K^{\BR})$ satisfies, $\forall j$,
\begin{equation}
\begin{aligned}
&\bW_j^{\BR}(\hat{\bH}^{(j)})\\
&=\argmax_{\bW_j} \E_{|\hat{\bH}^{(j)}}[\Rate(\bW^{\BR}_1\!\!,\!\ldots,\bW^{\BR}_{j-1},\!\bW_j,\bW^{\BR}_{j+1},\!\ldots,\!\!\bW^{\BR}_K)] \\
&~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~  \text{, s.to $\|\bW_j(\hat{\bH}^{(j)})\|_{\Fro}^2\leq P_j$.}
\end{aligned}
\label{eq:optimization_best}
\end{equation}
Solving the best-response optimization is usually more intuitive and more tractable. It corresponds however only to a local optimum of the original optimization problem.

Coming back to hierarchical CSI, the fundamental property of the hierarchical CSI configuration lies in the fact that TX~$j$ is able to carry out the signal processing which was done at TX~$k$ for $k<j$ to obtain the precoding decision~$\bW_k(\hat{\bH}^{(k)})$. Hence, the precoding decisions~$\bW(\hat{\bH}^{(k)})$ for $k<j$ are already given when considering the optimization at TX~$j$. The best-response optimization problem is then simplified to
\begin{equation}
\begin{aligned}
&\bW_j^{\BR}(\hat{\bH}^{(j)})\\
&=\argmax_{\bW_j} \E_{|\hat{\bH}^{(j)}}[\Rate(\bW_j,\bW^{\BR}_{j+1}(\hat{\bH}^{(j+1)})\!,\!\ldots\!,\!\bW^{\BR}_K(\hat{\bH}^{(K)}))] \\
&~~~~~~~~~~~~~~~~~~~~ \text{, s.to $\|\bW_j(\hat{\bH}^{(j)})\|_{\Fro}^2\leq P_j$.}
\end{aligned}
\label{eq:optimization_hierarchical}
\end{equation}
Yet, this remains a difficult problem as it requires to \emph{estimate} the precoding decisions at the TX~$k$ for $k=j+1,\ldots,K$. Solving this problem optimally was not possible and we propose in the following a simple, yet efficient, heuristic precoding algorithm.

\section{Hierarchical Precoding Algorithm}

\subsection{Hierarchical Precoding Algorithm}

Even though TX~$j$ does not know the information obtained at TX~$k$ for $k>j$, it can use the statistical information (available at every TX) to obtain an estimate of the precoding strategy which will be used at TX~$k$ for $k>j$. Based on this statistical information, each TX should optimize the conditional expectation as in \eqref{eq:optimization_hierarchical}. This is a functional stochastic optimization problem \cite{Kall1994} and is out of the scope of this paper.

As a consequence, we use in the following the simplifying assumption that TX~$j$, when computing its precoding coefficient, implicitly assumes that TX~$k$ with $k>j$ shares the same channel estimate than he does (while TX~$k$ for $k>j$ has in fact a more accurate one, as described in Subsection~\ref{se:SM:DCSI}). Following this approximation, the optimization problem \eqref{eq:optimization_hierarchical} simplifies to
\begin{equation}
\begin{aligned}
&(\bW_j,\bV_{j+1}\ldots,\bV_K)\\
&=\argmax_{(\bW_j,\ldots,\bW_K)} \E[\Rate(\bW_j(\hat{\bH}^{(j)}),\ldots,\bW_K(\hat{\bH}^{(j)}))] \\
&~~~~~~~~~\text{, s.to $\|\bW_k(\bH^{(j)})\|_{\Fro}^2\leq P_k, \forall k\in \{j,\ldots, K\}$.}
\end{aligned}
\label{eq:optimization_approximation}
\end{equation}
\begin{remark}
Only~$\bW_j^{\He}(\bH^{(j)})$ is effectively used for the transmission since TX~$k$ with $k>j$ will use the more accurate information available locally to improve the precoding decision. This is why we have introduced the notation~$\bV_k$ to denote the precoding coefficients which will not be effectively used in the transmission. \qed
\end{remark}

\subsection{Precoding Algorithm for~$\bW_j^{\He}(\bH^{(j)})$}

We consider without loss of generality the optimization at TX~$j$. The precoding coefficients~$\bW_k^{\He}$ for $k<j$ are not part of the optimization problem since there have been already determined by the TXs having a less accurate channel estimate. Indeed, following the hierarchical CSI configuration, TX~$j$ can reconstitute the precoding decisions~$\bW_{k}^{\He}$ taken at TX~$k$ for $k<j$. Let us first introduce $M_{\In}$ and $M_{\Opt}$ as respectively the number of antennas with given precoding coefficients and the number of antennas with undetermined precoding coefficients:
\begin{equation}
\begin{aligned}
M_{\In}\triangleq \sum_{k=1}^{j-1}M_k,\qquad M_{\Opt}\triangleq \sum_{k=j}^{K} M_k.
\end{aligned}
\end{equation}
The multi-user precoder can then be written as
\begin{equation}
\bT=\begin{bmatrix}
\bW^{\He}_{\In}\\
\bW^{\He}_{\Opt}
\end{bmatrix}
\label{eq:Algo_2}
\end{equation}
where~$\bW^{\He}_{\In}\in \mathbb{C}^{M_{\In}\times d_{\Tot}}$ and~$\bW^{\He}_{\Opt}\in \mathbb{C}^{M_{\Opt}\times d_{\Tot}}$ are defined as
\begin{equation}
\bW^{\He}_{\In}\triangleq \begin{bmatrix}
\bW_1^{\He}\\
\vdots\\
\bW_{j-1}^{\He}
\end{bmatrix},\qquad \bW^{\He}_{\Opt}\triangleq \begin{bmatrix}
\bW_j^{\He}\\
\vdots\\
\bW_K^{\He}
\end{bmatrix} .
\label{eq:Algo_3}
\end{equation}
The precoding coefficients in~$\bW^{\He}_{\In}$ are fixed such that it remains only to maximize the sum rate according to~$\bW^{\He}_{\Opt}$. Following the same idea, we also split the multi-user channel~$\bH^{\He}$ into the two parts~$\bH_{\In}^{\He}\in \mathbb{C}^{N_{\Tot}\times M_{\In}}$ and $\bH_{\Opt}^{\He}\in \mathbb{C}^{N_{\Tot}\times M_{\Opt}}$ such that
\begin{equation}
\bH^{\He}=\begin{bmatrix}\bH_{\In}^{\He}&\bH_{\Opt}^{\He}\end{bmatrix}.
\label{eq:Algo_4}
\end{equation}
In the following, we consider first in the optimization a sum power constraint and we then show how it is possible to normalize the precoder so as to fulfill the per-TX power constraint in Sub-subsection~\ref{se:Algo:perTX}. Considering directly the per-TX power constraint requires finding the values of one Lagrangian variable per-TX, which is not practical.
\begin{remark}
In the following, we use a particular precoding algorithm to solve the optimization problem \eqref{eq:optimization_approximation}. However, our approach for the hierarchical CSI configuration can be adapted to other precoding algorithm or others figures-of-merit. The sole requirement being that having a part of the precoding coefficients fixed should not make the optimization intractable.\qed
\end{remark}

\subsubsection{Review of the Sum Rate Maximization Algorithm~\cite{Christensen2008,Negro2012,Fritzsche2013a}}

To optimize the conditional average sum rate given the precoders at TX~$k$ for $k<j$, we build upon the approach in \cite{Christensen2008} which is one of the most well known sum rate maximization algorithm. This algorithm exploits the relation between the MSE minimization and the sum rate maximization. Specifically, it is shown in \cite{Christensen2008} that a local optimum for the sum rate maximization is reached almost surely by solving the optimization problem:
\begin{equation}
\begin{aligned}
(\bG_k^{\star},\bT^{\star},\bm{\Omega}_k^{\star})&=\argmin_{\bG_k,\bT,\bm{\Omega}_k}  \sum_{k=1}^K\trace\LB \bm{\Omega}_k\bM_k\RB-\log_2|\bm{\Omega}_k| \\
&~~~~~~~~~~~~~~~~~~~~~~~~~~~\text{, s.to $\|\bT\|_{\Fro}^2\leq P$}
\end{aligned}
\label{eq:optimization_christensen}
\end{equation}
with $\bM_k$ being the MSE matrix defined in \eqref{eq:SM_7} and $\bm{\Omega}_k\in \mathbb{C}^{d_k\times d_k}$ being a weighting matrix left to be optimized. In fact, we will consider a robust precoding algorithm where the objective consists of the expected sum rate conditioned on the knowledge of the channel estimate at the TX. This comes down to replacing the MSE matrix~$\bM_k$ by the average MSE matrix~$\bar{\bM}_k$ defined as~\cite{Negro2012,Fritzsche2013a}
\begin{align}
\bar{\bM}_k&\triangleq \E_{\bm{\Delta}^{(j)}}[\bM_k]\\
&= \bM_k+\E[\bG_k^{\He}(\bm{\Delta}^{(j)})_k^{\He}\bT\bT^{\He}\bm{\Delta}^{(j)}_k\bG_k]\\
&=\bM_k+\sum_{\ell=1}^{N_k}\sum_{p=1}^{M_{\Tot}}\sum_{\ell'=1}^{N_k}\sum_{p'=1}^{M_{\Tot}}\E\bigg[ \bG_k^{\He} \be_{\ell} \{\bm{\Delta}^{(j)}_k\}^{*}_{p,\ell}\be_p^{\He}\bT\notag\\
&~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\bT^{\He}\be_{p'} \{\bm{\Delta}^{(j)}_k\}_{p',\ell'}\be_{\ell'}^{\He}\bG_k\bigg]\\

&=\bM_k+ \bG_k^{\He} \bm{\Phi}_k\bG_k
\label{eq:proof_prop_5}
\end{align}
with the matrix~$\bm{\Phi}_k\in \mathbb{C}^{N_k\times N_k}$ being a diagonal matrix defined such that~$\forall i \in \{1,\ldots, N_k\}$,
\begin{equation}
\begin{aligned}
\{\bm{\Phi}_k\}_{i,i}\triangleq \trace \LB \diag \LB (\sigma_{1,i}^{(j)})^2,\ldots,(\sigma_{M_{\Tot},i}^{(j)})^2\RB \diag\LB\bT\bT^{\He} \RB \RB.
\end{aligned}
\label{eq:proof_prop_6}
\end{equation}
The objective of the minimization can also be rewritten as
\begin{equation}
\begin{aligned}
&\sum_{k=1}^K\trace \LB \bm{\Omega}_k \bar{\bM}_k \RB\\

&=\trace \LB \bm{\Omega}\bM \RB +\trace\LB \bT^{\He}\bm{\Psi} \bT\RB\\
\end{aligned}
\label{eq:proof_prop_5}
\end{equation}
with $\bm{\Omega}\triangleq\mathrm{blockdiag}(\bm{\Omega}_1,\ldots,\bm{\Omega}_K)$ and the matrix~$\bm{\Psi}\in \mathbb{C}^{M_{\Tot}\times M_{\Tot}}$ being a diagonal matrix such that $\forall i \in \{1,\ldots, M_{\Tot}\}$,
\begin{equation}
\begin{aligned}
\{\bm{\Psi}\}_{i,i}\triangleq \trace \LB \bG  \bG^{\He} \diag \LB (\sigma^{(j)}_{i,1})^{2},\ldots,(\sigma^{(j)}_{i,N_{\Tot}})^{2} \RB\RB.
\end{aligned}
\label{eq:proof_prop_6}
\end{equation}
This minimization is convex in each of the optimization variables. It follows that updating each of the optimization variables alternatively, the algorithm converges to a local maximum. For brevity and to focus on the specificity of the hierarchical algorithm, the details of the update of the optimization variables is relegated to Appendix~\ref{app:centralized}.

The RX filters are updated as the minimum MSE
\begin{equation}
\bG_k^{\He}=\bT_k^{\He} \bH_k \LB{\bH}_k\bT\bT^{\He}{\bH}_k^{\He}+ \bI_{N_k}+\bm{\Phi}_k\RB^{-1}.
\label{eq:algo_5}
\end{equation}
and the weighting matrix~$\bm{\Omega}_k$ as
\begin{equation}
\bm{\Omega}_k=\bar{\bM}_k^{-1},\qquad \forall k.
\label{eq:algo_6}
\end{equation}
Furthermore, it is shown the optimal update of~$\bT$ is given by
\begin{equation}
\bT\!=\!\LB \bH\bG\bm{\Omega}\bG^{\He}\bH^{\He}\!+\!\frac{\trace(\bm{\Omega}\bG^{\He}\bG)}{P_{\Tot}}\I_{M_{\Tot}}\!+\!\bm{\Psi}\RB^{-1}\!\!\!\!\bH \bG\bm{\Omega}.
\label{eq:algo_7}
\end{equation}

\subsubsection{Hierarchical Sum Rate Maximization}

In the case of hierarchical precoding, it can easily be seen that the update of the RX filters and the weighting matrices is the same as in the centralized algorithm since the only difference between the two optimization problems comes from the additional constraints over the precoder~$\bT$. Indeed, the update of the precoder is different as the first $M_{\In}$ rows of~$\bT$ are fixed. For the precoder update, we take the derivative of the Lagrangian according to~$\bW_{\Opt}$ with a sum power constraint. The precoding matrix canceling this derivative is then given by \cite{Fritzsche2013b}
\begin{align}
&\bW_{\Opt}^{\He}=\LB\bH_{\Opt}\bG\bm{\Omega}\bG^{\He}\bH_{\Opt}^{\He}+\lambda \I_{M_{\Opt}}\RB^{-1}\notag\\
&~~~~~~~~~~~~~~~~ ~~~ \cdot\bH_{\Opt}\bG\bm{\Omega}\LB \I_{d_{\Tot}}- \bG ^{\He}\bH_{\In}^{\He}\bW_{\In}^{\He} \RB.
\label{eq:Algo_8}
\end{align}
The detailed derivation of this formula is available in Appendix~\ref{app:hierarchical}. The value of~$\lambda_{\Opt}$ is then obtained by bisection to fulfill the sum power constraint.

However, combined with the iterative update of the RX filters and the weighting matrices, this makes the algorithm relatively demanding. As an alternative, we propose to set~$\lambda_{\Opt}=0$ and simply proceed by clipping, i.e., normalizing the power used if it is larger than the power constraint. Mathematically, this is written as
\begin{equation}
\begin{aligned}
\bW_{\Opt}^{\He}&=\frac{\bW_{\Opt}^{\He}}{\|\bW_{\Opt}\|_{\Fro}} \LB \min\LB \|\bW_{\Opt}\|_{\Fro}, \sqrt{P_{\Tot}-\|\bW_{\In}\|_{\Fro}^2}\RB\RB.
\end{aligned}
\label{eq:Algo_10}
\end{equation}

\subsubsection{Per-TX Power Constraint}\label{se:Algo:perTX}

The TXs being in fact not collocated, we consider a per-TX power constraint. It is thus necessary to scale down the precoder obtained according to the algorithm described above, to ensure that all the per-TX power constraints are fulfilled. This is obtained by setting:
\begin{equation}
\bW_{\Opt}^{\He}= \frac{\bW_{\Opt}^{\He}}{\max_{j=1}^K \|\bW_{j}\|_{\Fro}^2/P_j}.
\label{eq:Algo_9}
\end{equation}
\begin{remark}
Applying this normalization at each update of the TX has for consequence that the convergence of the algorithm is no longer guaranteed. Finding the optimal solution with the per-TX antenna power constraints taken into account is an ongoing research topic.\qed
\end{remark}

\section{Simulation Results}

To evaluate the performance of our algorithm, we average the performance over~$1000$ channel realizations via Monte-Carlo simulations. We start with a simple configuration with $K=4$~TX/RX pairs equipped each with a single antenna and with all the wireless links being unit-variance. We furthermore assume that each TX has the same power constraint. We compare our hierarchical precoding algorithm to the maximum sum rate algorithm from \cite{Christensen2008} using perfect CSIT. To fulfill the per-TX power constraint also in this perfect CSIT configuration, we normalize the precoder according to \eqref{eq:Algo_9} at each TX update. Finally, we compare our hierarchical precoding algorithm to the ``naive" \emph{distributed} use of the robust maximum sum rate algorithm from \cite{Negro2012,Fritzsche2013a} at each TX using the imperfect CSIT locally available.

We show in Fig.~\ref{Rate_SPC}, the average sum rate as a function of the per-TX SNR in the following simple CSI configuration
\begin{equation}
\begin{aligned}
(\sigma_{ik}^{(j)})^2&=0.25,&\forall i,k,\qquad j=1,2\\
(\sigma_{ik}^{(j)})^2&=0,&\forall i,k,\qquad j=3,4.
\label{eq:simu_1}
\end{aligned}
\end{equation}
It can be seen that both hierarchical precoding schemes outperform very significantly the naive distributed precoding scheme. In particular, a positive DoF (slope in the SNR) is achieved. This is a consequence of TX~$3$ and TX~$4$ having perfect CSI.

\section{Conclusion}

We have developed in this work a robust precoding algorithm for the hierarchical CSI configuration which outperforms conventional precoding schemes by taking explicitly the hierarchical CSI structure into account. Finding the optimal precoder being too difficult, we have proposed an approximate solution consisting in considering during the optimization at TX~$j$ that all the TXs with a more accurate CSI have in fact received the same CSI as TX~$j$. This approach is very general and can be applied in many other wireless settings with a hierarchical information structure.

The hierarchical CSI structure is particularly relevant for the next generation wireless networks with a partially centralized architecture and an heterogeneous backhaul. How to approach the optimal precoder in this hierarchical CSI configuration is a challenging topic which will be tackled in the future. Another interesting direction of research consists in evaluating analytically the performance obtained with a hierarchical architecture.

\begin{figure}[htp!]
\centering
\includegraphics[width=1\columnwidth]{main_rate_robust}
\caption{Average sum rate as a function of the SNR~$P$.}
\label{Rate_SPC}
\end{figure}


TITLE: Naming subfunctions of a piecewise function
QUESTION [0 upvotes]: How do I name the subfunctions of a piecewise function? Is it like this?
\begin{align*}f(x) &\;=\; \begin{cases}f_{1}(x) = x - 1 & \text{if } x \leqslant 1 \\ f_{2}(x) = x^{2} & \text{if } x > 1\end{cases}\end{align*}

REPLY [2 votes]: This is a pretty subjective question. The answer is more about style than anything. I'd do something like:

Let $f$ be defined by
\begin{align*}f(x) &\;=\; \begin{cases}f_{1}(x) & \text{if } x \leqslant 1 \\ f_{2}(x) & \text{if } x > 1\end{cases}\end{align*}
where $f_1(x) = x-1$ and $f_2(x) = x^2$.

But it's clear what you mean from how you've expressed it too.