text:
\section{Introduction and main results}\label{sb:intro}
Let $M$ be a symplectic manifold and consider  a collection $\mathcal{L}$ of Lagrangian submanifolds of $M$.
Given a Lagrangian submanifold $L \subset M$ we are interested in the ``splitting'' (or decomposition) of $L$ into
Lagrangian submanifolds picked from the collection $\mathcal{L}$. The type of splitting that we mainly
focus on is through Lagrangian cobordisms $V$ with a single positive end equal to $L$ and multiple negative ends, $V:L\cobto (L_{1},\ldots, L_{k})$ (the definition of cobordisms is recalled just below, in \S\ref{sb:intro-lcob}). This perspective on cobordism is natural not least because, as is known from previous work \cite{Bi-Co:cob1,Bi-Co:lcob-fuk} and under appropriate constraints on $\mathcal{L}$, such  cobordisms induce genuine (iterated cone) decompositions of $L$ with factors the negative ends $L_{i}$ in the derived Fukaya category of $M$.  Cobordisms $V$  are  Lagrangian
 submanifolds of $\R^{2}\times M$ and  they have an elementary measure called shadow \cite{Co-She:metric}, $\mathcal{S}(V)$, which is the area of their projection on $\R^{2}$ together with all the bounded regions bounded by this projection.

The central point of view for this paper is to view the shadow of a cobordism $V$ as an energy cost for the  splitting corresponding to $V$. We address three natural questions from this point of view:

\begin{itemize}
\item[I.] Assuming $L$ and $L_{1},\ldots, L_{k}$ fixed, find a lower bound for the minimal
energy cost required to split $L$ in  the factors $L_{i}$  (see Theorem \ref{t:main-A}) ?
\item[II.] Is there some form of Lagrangian intersections rigidity that is specific to low energy splittings (see Theorem \ref{t:main-B}) ?
\end{itemize}
A consequence of our answers to I and II is to define some new metrics and pseudo-metrics on certain classes of Lagrangian submanifolds. This leads to a third natural question.

\begin{itemize}
\item[III.] Are there some other types of energy costs that are relevant to this setting and, in particular, that also lead to associated pseudo-metrics but are independent from  cobordisms and their shadows (see \S\ref{sb:intro-wei-frag-metr})?
\end{itemize}

\subsection{Decomposition by Lagrangian cobordism}
\label{sb:intro-lcob}

Let $M$ be a symplectic manifold, compact or convex at infinity. A {\em Lagrangian cobordism}~\cite{Ar:cob-1}
(see~\cite{Bi-Co:cob1} for the formalism in use here) is a Lagrangian
submanifold $V \subset \mathbb{R}^2 \times M$ that, outside a compact
interval $\times \ \mathbb{R}$ looks like a finite disjoint union of
horizontal semi-infinite rays, each of which being multiplied by a
Lagrangian submanifold in $M$. More precisely, there exists a compact
interval $[a_-, a_+] \subset \mathbb{R}$ such that
$$V \setminus ([a_-,a_+] \times \mathbb{R} \times M) =
\Bigl(\coprod_{i=1}^k \ell_- \times \{i\} \times L_i \Bigr) \coprod \,
\Bigl(\coprod_{j=1}^{k'} \ell_+ \times \{j\} \times L'_j\Bigr),$$
where $\ell_- = (-\infty, a_-]$, $\ell_+ = [a_+, \infty)$ and the
$L_i$'s and $L'_j$'s are Lagrangian submanifolds of $M$.  The $L_i$'s
are called the negative ends of $V$ and the $L'_j$'s the positive ends.
We write $V: (L'_1, \ldots, L'_{k'}) \cobto (L_1, \ldots, L_k)$.  Note
that we allow any of $k'$ or $k$ to be $0$ in which case the
positive or negative end of the cobordism is void. In what follows we
will be mainly interested in cobordisms
$V: L \cobto (L_1, \ldots, L_k)$ between a single Lagrangian $L$ and a
tuple $(L_1, \ldots, L_k)$. We will view such a cobordism as a means to
decompose (or split) $L$ into the ``factors'' (or pieces) $L_1, \ldots, L_k$.

We need to restrict the class of
Lagrangian submanifolds as follows. Denote by $\mathcal{L}ag^*(M)$ either the
class of closed Lagrangian submanifolds in $M$ that are weakly
exact or the class of closed Lagrangians in $M$ that are
monotone (in the latter case there is an additional
constraint that will be explained later). Similarly, we will work with
cobordisms $V$ that are either weakly exact or monotone (where again,
in the monotone case there will be an additional constraint). We
denote the class of such cobordisms by
$\mathcal{L}ag^*(\mathbb{R}^2 \times M)$.

Our first theorem shows that the shadow of cobordisms
$V: L \cobto (L_1, \ldots, L_k)$ between fixed $L$ and
$(L_1, \ldots, L_k)$ cannot become arbitrarily small unless these
Lagrangians are placed in a very particular position.

\begin{mainthm} \label{t:main-A} Let $L, L_1, \ldots, L_k \subset M$
  be weakly exact Lagrangian submanifolds. Assume that
  $L \not \subset L_1 \cup \cdots \cup L_k$. Then there exists
  $\delta = \delta(L; S) > 0$ which depends only on $L$ and
  $S := L_1 \cup \cdots \cup L_k$, such that for every weakly exact
  Lagrangian cobordism $V : L \cobto (L_1, \ldots, L_k)$ we have
  \begin{equation} \label{eq:delta-L-S}
    \mathcal{S}(V) \geq \tfrac{1}{2}\delta.
  \end{equation}
\end{mainthm}
The proof is given~\S\ref{s:main-geom}. A non-technical outline of the
proof is presented in~\S\ref{sb:outline-prf-A}.

The next theorem establishes relations between $L$ and
$L_1, \ldots, L_k$ in case they are related by a Lagrangian cobordism
with small shadow.

\begin{mainthm} \label{t:main-B} Let $L, L_1, \ldots, L_k \subset M$
  be weakly exact Lagrangians and $S$ as in
  Theorem~\ref{t:main-A}. Let $N \subset M$ be another weakly exact
  Lagrangian and  assume that the family of Lagrangians $N,L, L_{1}\ldots, L_{k}$ are in general position.
  \begin{enumerate} [label=(\alph*),ref=\alph*]
  \item There
    exists $\delta' = \delta'(N, S)>0$ that depends on $N$ and
    $S$ (but not on $L$) such that for every weakly Lagrangian
    cobordism $V : L \cobto (L_1, \ldots, L_k)$ with
    $\mathcal{S}(V) < \tfrac{1}{2}\delta'$ we have:
    \begin{equation} \label{eq:N-cap-L-1} \# (N \cap L) \geq
      \sum_{i=1}^k \# (N \cap L_i).
    \end{equation}
  \item There exists $\delta'' = \delta''(N, S)>0$ that
    depends on $N$ and $S$ (but not on $L$) such that for every weakly
    exact Lagrangian cobordism $V: L \cobto (L_1, \ldots, L_k)$ with $\mathcal{S}(V) < \frac{1}{4}\ \delta''$ we
    have:
    \begin{equation} \label{eq:N-cap-L-2} \# (N \cap L) \geq
      \sum_{i=1}^k \dim_{\Lambda} HF(N, L_i).
    \end{equation}
    Here $HF(N,L_i)$ is the Floer homology of $(N,L_i)$ defined with
    coefficients in the Novikov field $\Lambda$.
  \end{enumerate}
\end{mainthm}
The numbers $\delta$, $\delta'$ and $\delta''$ are variants of the
Gromov width from~\cite{Bar-Cor:Serre}.  Namely, $\delta$ is the Gromov width of $L$ in the complement of $S$, $\delta'$ is a symplectic measure of the intersections $S\cap L$ and $\delta''$ is a symplectic measure of the double points of $S$ in the complement of $N$.
The precise definitions are given in~\S\ref{s:main-geom}, where more precise versions of
Theorems~\ref{t:main-A} and~\ref{t:main-B} are restated and proved as parts of a single statement,  Theorem \ref{thm:fission}.
A similar statement holds also in the monotone case,
see~\S\ref{s:main-geom}.

\subsection{From shadows to metrics on the space of
  Lagrangians} \label{sb:intro-shadow-metric}

Interestingly, the notion of shadow gives rise to a new class of
pseudo-metrics on the space of Lagrangians $\mathcal{L}ag^*(M)$. The
simplest of them, which we call the {\em shadow pseudo-metric}, is defined as
follows. Fix a family of Lagrangians
$\mathcal{F} \subset \mathcal{L}ag^*(M)$. The shadow pseudo-metric
$d^{\mathcal{F}}$ relative to $\mathcal{F}$ is defined
by
\begin{equation} \label{eq:dF-intro} d^{\mathcal{F}}(L,L'):= \inf_{V}
  \bigr\{ \mathcal{S}(V) \mid V : L \cobto (F_1, \ldots, F_{i-1}, L',
  F_i, \ldots, F_k), \, k \geq 0, F_i \in \mathcal{F} \bigr\},
\end{equation}
for every $L, L' \in \mathcal{L}ag^*(M)$. The infimum here is taken
only over cobordisms $V \in \mathcal{L}ag^*(\mathbb{R}^2 \times M)$,
i.e. the $V$'s are assumed to satisfy the same constraints as the
Lagrangians in $\mathcal{L}ag^*(M)$. These cobordisms are not assumed to be connected.
Here and in what follows we use
the convention that $\inf \emptyset = \infty$, so that
$d^{\mathcal{F}}(L,L') = \infty$ if there is no cobordism
$V \in \mathcal{L}ag^*(\mathbb{R}^2 \times M)$ as
in~\eqref{eq:dF-intro}.

It is easy to see that $d^{\mathcal{F}}$ is a pseudo-metric (possibly
with infinite values). However, unless $\mathcal{F} = \emptyset$,
$d^{\mathcal{F}}$ is generally degenerate (yet not identically zero).

Now fix a second family of Lagrangians
$\mathcal{F}' \subset \mathcal{L}ag^*(M)$. The preceding recipe
yields another pseudo-metric $d^{\mathcal{F}'}$. Consider now
\begin{equation} \label{eq:d-ff'} \widehat{d}^{\mathcal{F},
    \mathcal{F}'} = \max\{ d^{\mathcal{F}}, d^{\mathcal{F}'} \}.
\end{equation}

An easy consequence of Theorem~\ref{t:main-A} is that if
$(\overline{\cup_{K \in \mathcal{F}} K}) \cap (\overline{\cup_{K' \in
    \mathcal{F'}} K'})$ is totally disconnected (e.g. discrete), then
$\widehat{d}^{\mathcal{F}, \mathcal{F}'}$ is a non-degenerate metric
(possibly with infinite values) on $\mathcal{L}ag^*(M)$. We call it
the shadow metric associated to the pair of families $\mathcal{F}$,
$\mathcal{F}'$.

Shadow metrics can be viewed as generalizations of the ``simple
cobordism metric'' from~\cite{Co-She:metric} (which coincides with
$d^{\emptyset}$), which itself generalizes the Lagrangian Hofer
metric~\cite{Chek:finsler}. However, the shadow metrics are in general
different from these two metrics. In particular with appropriate
choices for the families $\mathcal{F}$, $\mathcal{F}'$, the shadow
metrics are finite for a larger collection of Lagrangians in
$\mathcal{L}ag^*(M)$. It is already known from \cite{Co-She:metric} that,
without appropriate rigidity constraints  (such as those imposed here) on the class of
Lagrangians under consideration, even the simple cobordism metric is degenerate.

\subsection{Weighted fragmentation pseudo-metrics on triangulated
  categories} \label{sb:intro-wei-frag-metr} The construction of the shadow pseudo-metrics
 can be generalized to a more abstract setting, as discussed in \S\ref{subsubsec:weights1}.
 In short, given a triangulated category $\mathcal{X}$ there is an associated category, $T^{S}\mathcal{X}$ \cite{Bi-Co:lcob-fuk}, whose  objects are ordered finite  families $(K_{1},\ldots, K_{r})$, $K_{i}\in \mathcal{O}b(\mathcal{X})$, and whose morphisms parametrize the  iterated cone-decompositions of the objects in $\mathcal{X}$.  A {\em weight} on $\mathcal{X}$ associates a value $\in [0,\infty]$ to the morphisms in $T^{S}\mathcal{X}$ in a way that is sub-additive with respect to composition. The weight of a morphism $\bar{\phi}\in \mor_{T^{S}\mathcal{X}}$
 can be viewed as a measure of the energy required for the (iterated cone)-decomposition represented by $\bar{\phi}$ to take place.
   Given such a weight $w$, and  for a fixed family  $\mathcal{F}\subset \mathcal{O}b(\mathcal{X})$, we can compare two objects
 $K,K'$ in $\mathcal{X}$ by  infimizing $w(\bar{\phi})$ among all morphisms $\bar{\phi}$ in $T^{S}\mathcal{X}$,  $\bar{\phi}: K\to (F_{1},\ldots, F_{i}, K', F_{i+1},\ldots, F_{r})$, $F_{i}\in \mathcal{F}$. The resulting measurement satisfies the triangle inequality and can be symmetrized formally, thus providing a pseudo-metric
 on the objects of $\mathcal{X}$. These pseudo-metrics are called {\em weighted fragmentation pseudo-metrics}. This construction recovers the shadow pseudo-metrics by using a functor $\widetilde{\Phi}:\mathcal{C}ob^{\ast}(M)\to T^{S}D\fuk^{\ast}(M)$ from
 \cite{Bi-Co:lcob-fuk} to define a shadow-weight, $w_{\mathcal{S}}$,  on  $D\fuk^{\ast}(M)$ - see (\ref{eq:shadow-weight}). The shadow pseudo-metrics are the weighted fragmentation pseudo-metrics associated to $w_{\mathcal{S}}$.

More interestingly, this abstract construction is applied in \S\ref{subsubsec:alg-we} to a different type of weight on $D\fuk^{\ast}(M)$ whose definition is based on algebraic measurements associated to filtered chain-complexes and modules and is independent of cobordism.  By applying the method above to this weight we deduce the  existence of additional pseudo-metrics on  $\mathcal{L}ag^{\ast}(M)$,  that are all more algebraic in nature compared to their shadow counterparts.
An additional key point, contained in Corollary \ref{cor:alg-weights}, is that, by carefully separating the geometric steps from the algebraic ones in the proof of Theorem \ref{t:main-A}, we show that an
 analogue of this theorem remains valid for the algebraic weights.  As a consequence, the algebraic weights can also be used to define actual metrics (possibly taking infinite values) on $\mathcal{L}ag^{\ast}(M)$.

\subsection{Outline of the proof of
  Theorem~\ref{t:main-A}} \label{sb:outline-prf-A}
  We focus here on the proof of Theorem \ref{t:main-A}. The proof of Theorem \ref{t:main-B} makes use
of the similar ideas.  We consider a symplectic embedding
of a standard ball $e:B(r)\to M$ such that $e^{-1}(L)=B_{\R}(r)$, $e(B(r))\cap (L_{1}\cup L_{1}\cup\ldots \cup L_{k})=\emptyset$ and we put $P=e(0)$.
The purpose of the proof is to show that for any almost complex structure $J$ on $M$ there exists a $J$-holmorphic polygon $u$ in $M$ with a boundary edge on $L$ (and possibly on the other $L_{i}$'s) going through $P$ and of symplectic area smaller or equal than $\mathcal{S}(V)$.  By a suitable choice of $J,$ an application of the Lelong inequality, and the definition of $\delta=\delta(L;S)$ as in (\ref{eq:delta1}) this implies the claim.

To be able to control energy bounds in our considerations we first set up in \S\ref{s:wf-ai-theory} the
machinery of $A_{\infty}$-categories and modules in the (weakly) filtered setting.  Variants of this already appear in the literature,
for instance in \cite{FO3:book-vol1,FO3:book-vol2} (in somewhat different form), but we give enough details such as to be able to extend - in \S\ref{s:floer-theory} - the results from \cite{Bi-Co:lcob-fuk} to this setting. The wording {\em weakly} means that, to achieve regularity, we allow for small Hamiltonian perturbations in the definition of the various algebraic structures. As  a consequence, these structures are
filtered only up to a system of small, controllable errors. We also prove in \S\ref{sb:wf-ic} a structural result,  Theorem \ref{t:itcones}, concerning
iterated cones $\mathcal{K}$ of (weakly) filtered $A_{\infty}$ modules and, in particular, we show
that each such cone admits a quasi-isomorphic model $\mathcal{K}'$ which is an iterated cone with the same factors as $\mathcal{K}$ and such that $\mathcal{K}'$ has a filtration that is well controlled with respect  to that of $\mathcal{K}$
and, most importantly for us,  the $\mu_{1}$ operation of  $\mathcal{K}'$ can be written explicitly in terms of higher
$\mu_{k}$'s of the underlying $A_{\infty}$-category  - see (\ref{eq:aij}). This result is based on a (weakly) filtered version of the following property of the Yoneda embedding \cite{Se:book-fukaya-categ}: for an $A_{\infty}$-module $\mathcal{M}$ and an object $Y$ there is a natural quasi-isomorphism $\mathcal{M}(Y) \cong\mor (\mathcal{Y}, \mathcal{M})$ (where $\mathcal{Y}$ is  the Yoneda module of $Y$).
We prove  in \S\ref{sb:l-map} a weakly filtered version of this property which seems to be new
(and somewhat delicate to prove).

With this preparation, the proof of the theorem is given in \S\ref{s:main-geom}. By neglecting a number of technicalities, the
argument can be sketched as follows. We consider a new cobordism
$W:\emptyset \cobto (L,L_{1},\ldots, L_{k})$ obtained from $V$ by bending the end $L$ of $V$  clokwise half a turn, as in Figure \ref{f:cob-v-w}. The main result in
\cite{Bi-Co:lcob-fuk} implies that  the Yoneda modules $\mathcal{L}$, $\mathcal{L}_{i}$ associated to the negative
ends of the cobordism $W$ fit into an iterated cone of $A_{\infty}$-modules over the Fukaya category, $\fuk^{\ast}(M)$:
$$\mathcal{M}=\tcn(\mathcal{L}_{k} \stackrel{\varphi_{k}}{\longrightarrow} \tcn (\mathcal{L}_{k-1}\stackrel{\varphi_{k-1}}{\longrightarrow}\ldots \tcn(\mathcal{L}_{1}\stackrel{\varphi_{1}}{\longrightarrow} \mathcal{L})\ldots )~,~$$  and moreover the module $\mathcal{M}$ is acyclic. In view of our preparatory step
all the modules and structures involved here are filtered. The complex $\mathcal{M}(L)$ admits a geometric description
as follows. Let $\gamma$ be a curve in $\C$ that lies to left of the ``bulb'' of the cobordism $W$ and is crossing the negative ends of $W$ in a vertical way - as in
Figure \ref{f:gamma-gamma'}. Then,  by neglecting in this sketch the other choices (of perturbations etc) required, we have
$\mathcal{M}(L)=CF(\gamma\times L, W)$.  There is a Hamiltonian isotopy of energy very close
to $\mathcal{S}(W)$ that isotopes the curve $\gamma$ to a curve $\gamma'$ that passes to the right of the projection of $W$ onto $\C$ and thus $\gamma'\times L$ is disjoint from $W$ - see again Figure \ref{f:gamma-gamma'}. As a consequence,
by standard Floer theoretic considerations, we deduce that there exists a null-homotopy $\xi$ of the identity of $\mathcal{M}(L)$ that shifts filtrations by at most $\rho\leq \mathcal{S}(V)+\epsilon$ where we can take $\epsilon$ as small as desired.
A cycle $e_L$ in $CF(L,L)$ representing the fundamental class $[L] \in HF(L,L)$ still remains a cycle in $\mathcal{M}(L)$. We deduce that it has to be the boundary of some element in $\mathcal{M}(L)$ of filtration higher than that of $e_L$ by not more than $\rho$. In a more streamlined formalism, we say that the boundary depth (a notion introduced by Usher in \cite{Usher1} - see also \S\ref{s:filt-ch}) of
$e_L$ is at most $\rho$.  By suitable choices of perturbation data, we arrange it so that $e_L$ is the maximum point of a Morse function on $L,$ which is achieved at $P$. For the next step it is crucial that $\mathcal{M}$ is an iterated cone of (weakly) filtered {\em modules} (in contrast to  $\mathcal{M}(L)$ only being an iterated cone of filtered chain complexes). We now use the structural Theorem \ref{t:itcones} to associate to $\mathcal{M}$ the quasi-isomorphic module $\mathcal{M}'$. Because the filtrations
on $\mathcal{M}'$ and $\mathcal{M}$ are tightly related, we deduce that the boundary depth of $e_L$ in $\mathcal{M}'(L)$
is at most $\rho+\epsilon'$ where $\epsilon'>0$ can be taken arbitrarily small. From the special form  of the differential of $\mathcal{M}'(L)$ which involves the higher order $A_{\infty}$-operations $\mu_{d}$, we conclude that there
is a pseudoholomorphic polygon $u$ in $M$ with boundary on $L$ and on some of the $L_{i}$'s that  appears in the differential of $\mathcal{M}'(L)$ and that passes through $P$. Moreover, the area of this polygon is not more than $\rho+\epsilon'$. This concludes
 the proof by making $\epsilon,\epsilon'\to 0$ (in practice, the last step of finding the $J$-holomorphic curve $u$
 is a bit more complicated as, for the complex $CF(\gamma\times L, W)$ to be defined, some perturbations need to be introduced
 and these have to be sent to $0$ to deduce the existence of $u$).

\begin{rem}
From the fact that  $e_L$ becomes a boundary in $\mathcal{M}(L)$ one can directly deduce that there is a holomorphic curve
passing through $P$. However, this curve is in $\R^{2}\times M$ and its boundary
is along $\gamma\times L$ and $W$ - this is not the curve $u$ we are looking for. To obtain a curve $u$ in $M$ with the
desired properties, the algebraic Theorem \ref{t:itcones} is essential in our argument.
We do not know a geometric argument bypassing this theorem and
providing directly the specific form of differential of $\mathcal{M}'$ (except for the very special case of a
cobordism $W$ with at most three ends). Notice also that the curve $u$ that is identified in the proof can be shown to be
associated to an operation $\mu_{d}$ with $d\geq 2$. \end{rem}

\subsection{Structure of the
  paper} \label{sb:intro-further-rems}
  After two sections, \S \ref{s:wf-ai-theory} algebraic
  and \S\ref{s:floer-theory} that adjusts the Fukaya category constructions to
  the weakly filtered setting,  we assemble  in \S\ref{s:main-geom} all these tools to prove Theorem \ref{thm:fission} -  that combines the statements of both Theorems \ref{t:main-A} and \ref{t:main-B}.
The last section of the paper, \S\ref{sec:examples-metr}, contains a number of examples and calculations as well as the construction of the weighted fragmentation pseudo-metrics mentioned earlier in the introduction.
For the most part, this section can be read directly after the introduction. Except for Corollary \ref{cor:alg-weights}, it only uses the statements of Theorems \ref{t:main-A} and \ref{t:main-B} and not their proofs
 and just a few algebraic notions  from \S\ref{s:filt-ch}.

 \subsection*{Aknowledgements.} We thank Misha Khanevsky for mentioning to us the argument in Remark \ref{rem:Misha-etc} b.  The second author thanks Luis Diogo for useful comments. The last two authors thank
 the {\em Institute for Advanced Study} and Helmut Hofer for generously hosting them there for a part of this work. The second author also thanks the {\em Forschungsinstitut f\"ur Mathematik}  for support during repeated visits to Z\"urich.


TITLE: On the convergence of $\sum_{n=1}^{\infty} \frac{\lambda(n)}{n}$ and the Prime Number Theorem
QUESTION [3 upvotes]: Let $\lambda$ be the Lioville function of number theory.
I've heard several times that if $L=\sum_{n=1}^{\infty} \frac{\lambda(n)}{n} =O(1)$ then $L=0$ (the Prime Number Theorem). How can this be proved ? Or can someone kindly provide a reference ?

REPLY [10 votes]: $\sum_{n=1}^N\frac{\lambda(n)}{n}=O(1)$ can be proven in an elementary manner (with no analysis, real or complex). Indeed, observe we have
$$\sum_{n=1}^N\lambda(n)\left\lfloor\frac{N}{n}\right\rfloor=\sum_{n=1}^N\lambda(n)\sum_{k\leq N,n\mid k}1=\sum_{k=1}^N\sum_{n\mid k}\lambda(n)=\sum_{k=1}^NQ(n),$$
where $Q(n)=1$ if $n$ is a perfect square, $Q(n)=0$ otherwise. Hence the right-hand side is clearly $O(N)$, while the left-hand side is $N\sum_{n=1}^N\frac{\lambda(n)}{n}+O(N)$. It follows immediately $\sum_{n=1}^N\frac{\lambda(n)}{n}=O(1)$.
On the other hand, as you note, $\sum_{n=1}^\infty\frac{\lambda(n)}{n}=0$ is equivalent to the prime number theorem. Hence there likely isn't an easy deduction of convergence from $O(1)$ bound, since that would give us an equally easy proof of PNT.


TITLE: Multiplier norm vs cb norm
QUESTION [5 upvotes]: Let $f:G\to \mathbb{C}$ be a finitely supported functions and let $m_f$ denote the associated multiplier on $C^*_r(G)$, the reduced group $C^*$-algebra:
$$m_f(\alpha)(g)=f(g)\alpha(g)$$
for every $\alpha\in C^*_r(G)$.
We can look at two natural norms of $m_f$:

the completely bounded norm $\Vert m_f\Vert_{cb}$ of $m_f$ as a multiplier on $C^*_r(G)$
the multiplier norm $\Vert m_f\Vert_m$ of $m_f$ viewed as a bounded linear operator on $C^*_r(G)$.

In general,
$$\Vert m_f\Vert_m\le \Vert m_f\Vert_{cb}.$$
Is there anything else known about the relation of these two norms? For instance, are there examples of $f$ with $\Vert m_f\Vert_{cb}=1$ and $\Vert m_f \Vert_m\to 0$? Or should they be equal in some cases? I would be grateful for either an answer or direction where to look for this in the literature.

REPLY [2 votes]: The norms are equal if $G$ is amenable, and inequivalent if $G$ is e.g. the free group. Losert has stated that the two norms can be equivalent (in fact, equal IIRC) when $G=SL(2,R)$ but to my knowledge this has never been published
I can't give an explicit reference off the top of my head, but key words here are "Herz-Schur multiplier" and "multipliers of the Fourier algebra". A good place to find out some of the basic terminology, and to find links to the literature, is this set of minicourse notes by Todorov. See Theorem 3.6 in particular.


TITLE: Polynomials in Finite Field
QUESTION [0 upvotes]: Given $\Bbb F_{32}=\Bbb F_2[X]/(X^5+X^2+1)$ where the polynomial is irreducible over $\Bbb F_2$, how would I compute $(a^4+a^2)*(a^3+a+1)$ given $a=[X]$ is the congruence class of [X]?
Multiplying the polynomials acquires $a^7+2 a^5+a^4+a^3+a^2$, but how do I reduce that to fit the given conditions?

REPLY [4 votes]: Use the Polynomial Division Algorithm (with Remainder) to write it as $\, q(a) (a^5\!+a^2\!+1)+ r(a),\,$ where $\,\deg r < 5.\,$ Or, equivalently, use the equation $\, \color{#c00}{a^5 = a^2+1}\,$ as a rewrite rule to reduce all exponents on $\,a\,$ till they are $\, < 5,\,$ e.g. $\,a^{k+5} = a^k \color{#c00}{a^5} = a^k(\color{#c00}{a^2+1}).\,$


text:
\begin{document}
\captionsetup[figure]{labelfont={default},labelformat={default},labelsep=period,name={Fig.}}

\title{Analog Versus Hybrid Precoding for Multiuser Massive MIMO with Quantized CSI Feedback}

\author{Yaqiong Zhao, \emph{Student Member, IEEE},
       Wei Xu, \emph{Senior Member, IEEE},
       Jindan Xu, \emph{Student Member, IEEE},
       Shi Jin, \emph{Senior Member, IEEE},
       Kezhi Wang, \emph{Member, IEEE},
      and Mohamed-Slim Alouini, \emph{Fellow, IEEE}
\thanks{Y. Zhao, J. Xu, and S. Jin are with the National Mobile Communications Research Laboratory, Southeast University, Nanjing 210096, China (email: $\lbrace$zhaoyaqiong, jdxu, jinshi$\rbrace$@seu.edu.cn). }
\thanks{W. Xu is with the National Mobile Communications Research Laboratory, Southeast University, Nanjing 210096, China, and is also with the Purple Mountain Laboratories, Nanjing 210000, China (wxu@seu.edu.cn).}
\thanks{K. Wang is with the Department of Computer and Information Sciences, Northumbria University, Newcastle, UK (email: kezhi.wang@northumbria.ac.uk).}
\thanks{M. -S. Alouini are with the Computer, Electrical and Mathematical  Science and Engineering Division, King Abdullah University of Science and  Technology, Thuwal, Saudi Arabia, 23955 (email: slim.alouini@kaust.edu.sa).}
}

\maketitle

\begin{abstract}
In this letter, we study the performance of a downlink multiuser massive multiple-input multiple-output (MIMO) system with sub-connected structure over limited feedback channels. Tight rate approximations are theoretically analyzed for the system with pure analog precoding and hybrid precoding. The effect of quantized analog and digital precoding is characterized in the derived expressions. Furthermore, it is revealed that the pure analog precoding outperforms the hybrid precoding using maximal-ratio transmission (MRT) or zero forcing (ZF) under certain conditions, and we theoretically characterize the conditions in closed form with respect to signal-to-noise ratio (SNR), the number of users and the number of feedback bits. Numerical results verify the derived conclusions on both Rayleigh
channels and mmWave channels.

\begin{IEEEkeywords}
Massive MIMO, analog precoding, hybrid precoding, limited feedback.
\end{IEEEkeywords}
\end{abstract}

\IEEEpeerreviewmaketitle
\section{Introduction}
\IEEEPARstart{M}{assive} multiple-input multiple-output (MIMO) has attracted increasing attention as a key technology in the fifth-generation (5G) network [1]. With its advantages for inter-user interference cancellation and noise supression, massive MIMO can achieve near-optimal performance with simple linear precoding schemes, such as  maximal-ratio transmission (MRT) and zero forcing (ZF) [2]. These fully-digital precoding schemes need each antenna to be driven by one dedicated radio-frequency (RF) chain, which imposes prohibitively high cost and power consumption [3]. To address this issue, two kinds of designs have been introduced, i.e., pure analog precoding and hybrid analog-and-digital precoding [4] [5].

Although the hybrid precoding has been widely considered in research, evidence has shown that it may not be the best choice for all cases compared with the pure analog precoding, especially when considering that analog precoding consumes less power than the hybrid precoding due to some implementation facts [4] [6]. In [7] [8], the analog precoding approximately transformed the effective channel into a diagonal matrix, which implies that digital processing is no longer needed for further multiuser interference cancellation. Specifically in [9], the authors found that pure analog processing could surpass the hybrid processing with maximal-ratio combination (MRC) or ZF in an uplink channel under the assumption of perfect channel state information (CSI). For the downlink channel, however, this compromise is still unclear. Moreover in practice, only quantized CSI, instead of perfect one, is available through limited feedback [10]-[13].

Against the above background, this letter investigates the performance of a downlink massive MIMO system with low-cost sub-connected architecture and quantized CSI feedback. Tight rate approximations are derived for the ZF/MRT-based hybrid precodings and the analog precoding in the large base station (BS) antenna regime. With the derived results, we explicitly characterize the performance comparison between the hybrid precoding and the analog precoding. In particular, we show that for all SNRs analog precoding outperforms the ZF-based hybrid precoding when $B_2 \leq B_2^0$ where $B_2$ denotes the number of feedback bits in digital precoding and $B_2^0$ is a constant in closed-form with respect to system parameters. Similar observation has also been obtained for the comparison of MRT-based hybrid precoding and pure analog precoding. For other cases, the superiority of different precoding schemes would be complicated and it depends on specific values of SNR, which has also been derived in closed forms.

The remainder of this paper is organized as follows. System model is introduced in Section \uppercase\expandafter{\romannumeral2}. In Section \uppercase\expandafter{\romannumeral3}, we derive the achievable rates of the system using  various precoding schemes. In Section \uppercase\expandafter{\romannumeral4}, we present the conditions under which the pure analog precoding can beat the hybrid precoding. Simulation results and conclusions are given in Section \uppercase\expandafter{\romannumeral5} and Section \uppercase\expandafter{\romannumeral6}, respectively.
\section{System Model}
We consider a multiuser massive MIMO downlink channel with sub-connected architecture. Multiple users are simultaneously served by a BS which is equipped with $M$ antennas and $K$ RF chains. Each RF chain is connected to a subset of $N$ antennas through dedicated phase shifters where $N=\frac{M}{K}$. Considering that the number of transmit streams should not exceed the number of RF chains, we assume that $K$ single-antenna users are scheduled from the user pool.
Assuming flat Rayleigh fading, the received signal at the $k$th user can be expressed by
\begin{equation}\label{eq:yk}
y_k =\beta_k{\bf h}_k^H{\bf A}{\bf W}{\bf s}+ n_k,\quad k=1,2,\cdots,K,
\end{equation}
where ${\bf h}_k^H \sim {\cal {CN}}({\bf 0}_M,{\bf I}_M)$ denotes the downlink channel from the BS to the $k$th user, and ${\bf s}\in {\mathbb C}^{K\times 1}$ is the data vector with ${\mathbb E}[{\bf s}{\bf s}^H]=\frac{P}{K}{\bf I}_K$ where $P$ is the total transmit power at the BS, $n_k\sim {\cal {CN}}(0,{\sigma ^{\rm{2}}})$ is the additive Gaussian noise, $\beta_k$ denotes the path loss of the $k$th user, and ${\bf A} = [{\bf a}_1,{\bf a}_2,\cdots,{\bf a}_K]\in {\mathbb C}^{M \times K}$ and ${\bf W} = [{\bf w}_1,{\bf w}_2,\cdots,{\bf w}_K]\in{\mathbb C}^{K\times K}$ respectively stand for the analog precoder and the digital precoder. Due to the hardware dissipation caused by the power dividers [14], the analog precoder is written as ${\bf{A}} = \frac{1}{{\sqrt N }}{\bf{F}}$ where ${\bf{F}}$ denotes the equivalent analog precoding implemented by the phase shifter network. To conduct equal power allocation for users, it holds that $\|{\bf F}{\bf w}_k\| = 1$ for $k=1,2,\cdots,K$.

In the current design of hybrid precoding for massive MIMO, the design of analog precoder can be accomplished in two ways, that is, by either utilizing a quantized CSI feedback from users, e.g., precoding matrix indicator (PMI) feedback, or by using uplink channel estimates, e.g., the sounding reference signal (SRS), as specified in the 5G New Radio (NR) specifications [15]. Based on this, the BS designs the analog precoder in terms of discretized phase shifters. Then, an analogly beamformed pilot will be sent by the BS for further equivalent channel estimation to compensate for the performance degradation caused by quantization error of the analog precoder. Each user estimates its own effective channel and feeds back a quantized version of the precoded CSI which is used by the BS to design the digital precoder. In this paper, phases of the $MK$ entries
of $\bf A$ are quantized up to $B_1$ bits, i.e., the codebook is designed as $\mathcal{A}=\left\{ {{e^{\frac{{j2\pi n}}{{{2^{{B_1}}}}}}},n = 0,1, \cdots ,{2^{{B_1}}-1}}\right\}$. Based on the minimum Euclidean distance criterion, a common way of analog precoder design in the
sub-connected structure follows [9] [10]
\begin{equation}\label{eq:analog}
{a_{k,i}} = \left\{ {\begin{array}{*{20}{c}}
{\frac{1}{N}\mathop {{\rm{argmax}}}\limits_{{e^{j{{\hat \varphi }_{k,i}}}} \in\mathcal{A}} {\mathcal{R}}[h_{k,i}^*{e^{j{{\hat \varphi }_{k,i}}}}],}&{N(k - 1) + 1 \le i \le Nk}\\
{0,}&{{\rm{otherwise}}}
\end{array}} \right.
\end{equation}
where $a_{k,i}$ and $h_{k,i}$ are respectively the $i$th element of ${\bf a}_k$ and ${\bf h}_k$, and $\mathcal{R}[x]$ returns the real part of $x$.
For the $k$th user, we define its effective channel as ${\bf g}_k^H \buildrel \Delta \over = {\bf h}_k^H\bf A.$ Then, the ergodic rate of the $k$th user can be written as
\begin{equation}\label{eq:rate}
R_k={\mathbb E}\left[ {{{\log }_2}\left( {1 + \frac{{\frac{{\gamma {\beta _k}}}{K}{{\left| {{\bf{g}}_k^H{{\bf{w}}_k}} \right|}^2}}}{{1 + \frac{\gamma }{K}\sum\limits_{j \ne k} {{\beta _j}{{\left| {{\bf{g}}_k^H{{\bf{w}}_j}} \right|}^2}} }}} \right)} \right],
\end{equation}
where $\gamma\triangleq \frac{P}{\sigma ^{\rm{2}}}$ is the SNR.

The digital part of the system is based on the effective channel. Since the effective channel is always correlated even for independent and identically distributed (i.i.d.) channels of $\bf H$, we adopt a channel statistics-based codebook $\mathcal{G}$ [10]-[12]. The $k$th user quantizes its effective channel ${\bf g}_k$ according to $\hat{{\bf g}}_k={\rm arg~max}_{\hat{\bf g}_{k}^i\in\mathcal{G}}\lvert{\bf g}_k^H\hat{\bf g}_{k}^i\rvert$, in which ${\hat{\bf g}}_{k}^i$ is defined as
\begin{equation}\label{eq:codebook}
{\hat{\bf g}}_{k}^i=\frac{{\bf R}_k^{\frac{1}{2}}{\bf v}_i}{\|{\bf R}_k^{\frac{1}{2}}{\bf v}_i\|},
\end{equation}
where ${\bf R}_k$ denotes the correlation matrix of user $k$'s effective channel and ${\bf v}_i\in {\mathbb C}^{K\times 1}$, $i\in\{1,2,\cdots,2^{B_2}\}$ is an i.i.d. complex Gaussian vector randomly chosen from random vector quantization (RVQ) codebook of size $2^{B_2}$. With the limited feedback of $B_2$ bits, the BS calculates the digital precoder by using the quantized CSI feedback $\hat{\bf g}_k$.
\section{Achievable Rate Analysis}
\subsection{MRT/ZF-based Hybrid Precoding with Limited Feedback}
For the MRT-based hybrid precoding, the digital precoder of user $k$ is determined as
\begin{equation}\label{eq:MRT}
{\bf w}_k={\hat{\bf g}_k}.
\end{equation}
With the design above, we derive the downlink rate in the following theorem.

\emph{Theorem~1:} Tight rate approximation of MRT-based hybrid precoding in the large BS antenna regime is obtained as
\begin{equation}\label{eq:rqmrt}
R_{{\rm H},k}^{{\rm{MRT}}}={\log _2}\left({1 +\frac{{\frac{{\gamma \beta_k }}{K}(\frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{4}+\frac{K}{N}- \frac{{{2^{ - \frac{{{B_2}}}{{K - 1}}}}}}{N})(\frac{{\pi N{\rm{sin}}{{\rm{c}}^2}(\delta )}}{4}+K)}}{{\frac{{\pi N{\rm{sin}}{{\rm{c}}^2}(\delta )}}{4}+K +\frac{{\gamma}}{K}\overline {\beta} _k(\frac{K}{N}+\frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{2})}}}\right),
\end{equation}
where $\delta\buildrel \Delta \over =\frac{\pi}{2^{B_1}}$, sinc$(x)\buildrel \Delta \over =\frac{{\rm sin}(x)}{x}$ and $\overline {\beta} _k  = \sum\limits_{j \ne k} {{\beta _j}}$.
\begin{proof} See Appendix A.\end{proof}
 From the above theorem, we can also get the achievable rate of the MRT-based hybrid system with perfect CSI as a special case by letting $B_1\to\infty$ and $B_2\to\infty$ in (\ref{eq:rqmrt}). It yields
\begin{equation}\label{eq:rmrt}
R_{{\rm H},k}^{\rm MRT_p}={\rm{lo}}{{\rm{g}}_2}\left( {1 + \frac{{\frac{{\gamma {\beta_k}N}}{K}{{(\frac{\pi }{4} + \frac{K}{N})}^2}}}{{\frac{{\pi N}}{4} + K + \frac{{\gamma {{\overline \beta  _k} }}}{K}(\frac{\pi }{2} + \frac{K}{N})}}} \right).
\end{equation}
Note that this result coincides with the result in [9, Eq. (15)] except for that the signal term and the interference term calculated in [9] have an additional factor of $\frac{{\rm{1}}}{N}$ if we let $\beta_i=1, i=1,2,\cdots,K$, which is because we consider the effect of the divider network.

For the ZF-based precoding, the digital precoder equals
\begin{equation}\label{eq:ZF}
{\bf W}^{\rm ZF}=\hat{\bf G}(\hat{\bf G}^H\hat{\bf G})^{-1},
\end{equation}
where $\hat{\bf G}=[\hat{\bf g}_1,\hat{\bf g}_2,\cdots,\hat{\bf g}_K]$ and ${\bf W}^{\rm ZF}=[{\bf w}_1^{\rm ZF},{\bf w}_2^{\rm ZF},\cdots,{\bf w}_K^{\rm ZF}]$. The digital precoder is then normalized as ${\bf w}_k=\frac{{\bf w}_k^{\rm ZF}}{\|{{\bf Fw}_k^{\rm ZF}}\|}$. A tractable and tight bound of its achievable rate is given in the following proposition.

\emph{Proposition~1:} The ergodic rate of the ZF-based hybrid precoding is asymptotically lower bounded by
\begin{equation}\label{eq:rqzf}
R_{{\rm H},k}^{\rm ZF}\geqslant{\rm{lo}}{{\rm{g}}_2}\left( {\frac{{4KN + \pi N\gamma {\beta _k}{\rm{sin}}{{\rm{c}}^2}(\delta )}}{{4KN + 4\gamma {{\overline \beta  }_k}{2^{ - \frac{{{B_2}}}{{K - 1}}}}}}} \right).
\end{equation}
\begin{proof} See Appendix B.\end{proof}
\subsection{Pure Analog Precoding}
When the pure analog precoding is applied, the digital precoding matrix reduces to an identity matrix, i.e., ${\bf W}={\bf I}_K$. Then we can easily get the following proposition.

\emph{Proposition~2:} Tight rate approximation of the pure analog precoding in the large BS antenna regime is characterized as
\begin{equation}\label{eq:ra}
R_{{\rm A},k}={\rm{lo}}{{\rm{g}}_2}\left( {1 + \frac{{\gamma {\beta _k}(\frac{{\pi N{\rm{sin}}{{\rm{c}}^2}(\delta )}}{4} + 1)}}{{KN + \gamma {{\overline \beta  }_k}}}} \right).
\end{equation}
\begin{proof} With the pure analog precoding, ${\mathbb E}[{\left| {{{\bf{g}}_k^H}{{\bf{w}}_k}} \right|^2}]={\mathbb E}[{\left| {{g_{k,k}}} \right|^2}]=\frac{{\pi {\rm{sin}}{{\rm{c}}^2}{\rm{(}}\delta {\rm{)}}}}{4} + \frac{{{1} }}{N}-\frac{{\pi {\rm{sin}}{{\rm{c}}^2}{\rm{(}}\delta {\rm{)}}}}{4N}$ and ${\mathbb E}[{\left| {{{\bf{g}}_k^H}{{\bf{w}}_j}} \right|^2}]={\mathbb E}[{\left| {{g_{k,i}}} \right|^2}]=\frac{1}{N}$. By applying [16, Lemma 1], and substituting these two expressions into (\ref{eq:rate}) and using similar manipulations as that in the last step of (\ref{eq:signal}), the proof completes.\end{proof}
As a special case, we have the achievable rate of the system with perfect CSI by letting $B_1\to\infty$, which gives
\begin{equation}\label{eq:ra1}
R_{{\rm A},k}^{\rm p}={\rm{lo}}{{\rm{g}}_2}\left( {1 + \frac{{\frac{{\pi N\gamma {\beta _k}}}{4}}}{{KN + \gamma {{\overline \beta  }_k}}}} \right),
\end{equation}
which consists with the existing result in [9, Eq. (13)].
\section{Pure Analog vs Hybrid Precoding}
To compare the performance of the pure analog precoding and the hybrid precoding, we evaluate the rate gap as
\begin{equation}\label{eq:gap}
\Delta R=R_{{\rm H},k}^{X}-R_{{\rm A},k},
\end{equation}
where ${{X}} \in \{ {\rm{ZF}},{\rm{MRT}}\}$ represents the strategy of the digital precoding in the hybrid precoding.
\vspace{-10pt}
\subsection{MRT-based Hybrid Precoding}
By letting $\triangle R \geq 0$, and substituting (\ref{eq:rqmrt}) and (\ref{eq:ra}) into (\ref{eq:gap}), we get
\begin{equation}\label{eq:condition1}
\frac{{\gamma{\overline \beta  _k}}}{K}\left( {\frac{{K - 2 - {2^{ - \frac{{{B_2}}}{{K - 1}}}}}}{N} - \frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{4}} \right) > {2^{ - \frac{{{B_2}}}{{K - 1}}}} - K + 1.
\end{equation}
It implies that the MRT-based hybrid precoding outperforms the pure analog precoding if (\ref{eq:condition1}) holds. It is obvious that the right hand side of (\ref{eq:condition1}) is negative when $K>1$. Checking that $\frac{{K - 2 - {2^{ - \frac{{{B_2}}}{{K - 1}}}}}}{{{N}}}\geq\frac{{\pi {\rm{sin}}{{\rm{c}}^2}{\rm{(}}\delta {\rm{)}}}}{{4}}$, we can rewrite (\ref{eq:condition1}) as
\begin{equation}\label{eq:condition11}
K \geq \frac{{\pi N{\rm{sin}}{{\rm{c}}^2}(\delta )}}{4}{\rm{ + }}{2^{ - \frac{{{B_2}}}{{K - 1}}}} + 2,
\end{equation}
under which the MRT-bsed hybrid precoding outperforms the pure analog precoding for all SNR values. This is due to the fact that multiuser interference becomes a dominant factor when $K$ becomes large, and the hybrid precoding which utilizes digital processing to eliminate interference enjoys a better performance than the pure analog precoding.

Moreover from (\ref{eq:condition11}), it is revealed that $B_2$ has a marginal impact on the performance of the MRT-based hybrid precoding since ${{2^{ - \frac{{{B_2}}}{{K - 1}}}}} \ll1$. Therefore we can focus on the impact of $B_1$ on system performance.
By applying Taylor's expansion to ${\sin ^2}(\delta )$ with $\delta  = \frac{\pi }{{{2^{{B_1}}}}}$, we get ${\rm{sin}}{{\rm{c}}^2}(\frac{\pi }{{{2^{{B_1}}}}}) \approx 1 - \frac{{{{\left( {\frac{\pi }{{{2^{{B_1}}}}}} \right)}^2}}}{3}$, and (\ref{eq:condition11}) is further equivalent to
\begin{equation}\label{eq:condition12}
{B_1} \leq \frac{1}{2}{\log _2}\left( {\frac{{{\pi ^3}N}}{{3\pi N - 12\left( {K - 2 - {2^{ - \frac{{{B_2}}}{{K - 1}}}}} \right)}}} \right) \buildrel \Delta \over = B_1^0.
\end{equation}
Otherwise, when $B_1>B_1^0$, it is not difficult to get
\begin{equation}\label{eq:condition_mrt}
\Delta R\left\{ {\begin{array}{*{20}{c}}
{ > 0,}&{\gamma  < {\gamma _{0}}}\\
{ \le 0,}&{\gamma  \geq {\gamma _{0}}}
\end{array}} \right.
\end{equation}
where $\gamma_0\buildrel \Delta \over =\frac{{M\left( {{2^{ - \frac{{{B_2}}}{{K - 1}}}} - K + 1} \right)}}{{{\overline \beta  _k}\left( {K - 2 - {2^{ - \frac{{{B_2}}}{{K - 1}}}} - \frac{{\pi N{\rm{sin}}{{\rm{c}}^2}(\delta )}}{4}} \right)}}$.

\emph{Remark~1:} When $B_1\leq B_1^0$, the MRT-bsed hybrid precoding always outperforms the pure analog precoding in terms of the ergodic achievable rate for all SNRs. Otherwise the MRT-bsed hybrid precoding can beat the pure analog precoding only for low SNRs satisfying $\gamma<\gamma _0$.
\subsection{ZF-based Hybrid Precoding}
Similarly by letting $\triangle R \geq 0$ for $X=\rm ZF$ in (\ref{eq:gap}) and using (\ref{eq:rqzf}) and (\ref{eq:ra}), we obtain
\begin{equation}\label{eq:condition21}
{B_2} > (K - 1){\log _2}\left( {1 + \frac{{4({\beta _k} + {{\overline \beta  }_k})}}{{\pi N{\beta _k}{\rm{sin}}{{\rm{c}}^2}(\delta )}}} \right)\buildrel \Delta \over = B_{\rm{2}}^{\rm{0}},
\end{equation}which uses the fact that $\frac{{\pi N{\beta _k}{\rm{sin}}{{\rm{c}}^2}(\delta )\left( {{\rm{1 - }}{2^{ - \frac{{{B_2}}}{{K - 1}}}}} \right)}}{4} > {2^{ - \frac{{{B_2}}}{{K - 1}}}}({\beta _k} + {\overline \beta  _k})$. When $B_2>B_2^0$, it holds true that
\begin{equation}\label{eq:condition_zf}
\Delta R\left\{ {\begin{array}{*{20}{c}}
{ > 0,}&{\gamma> {\gamma _1}}\\
{ \le 0,}&{\gamma  \leq {\gamma _1}}
\end{array}} \right.
\end{equation}
where $\gamma_1\buildrel \Delta \over =\frac{{4M({{\overline \beta  }_k}{2^{ - \frac{{{B_2}}}{{K - 1}}}} + {\beta _k})}}{{{{\overline \beta  }_k}\left( {\pi N{\beta _k}{\rm{sin}}{{\rm{c}}^2}(\delta )\left( {{\rm{1 - }}{2^{ - \frac{{{B_2}}}{{K - 1}}}}} \right) - 4({\beta _k} + {{\overline \beta  }_k}){2^{ - \frac{{{B_2}}}{{K - 1}}}}} \right)}}$.

Otherwise, when $B_2\leq B_2^0$, the pure analog precoding always outperforms the ZF-based hybrid precoding for all SNRs. This is because the precoding vector of the $k$th user is supposed to lie in the null space of channels of other users and thus the ZF-based hybrid precoding is more dependent on the accuracy of channel estimation to cancel out interference.

\emph{Remark~2:} If $B_2\leq B_2^0$, the pure analog precoding beats the ZF-based hybrid precoding for all SNRs. Otherwise the ZF-based hybrid precoding takes over for high SNRs satisfying $\gamma\geq\gamma_1$.
\section{Simulation Results}
In this section, we compare the downlink performance of the analog precoding and hybrid precoding based on different linear processing schemes. The path loss factor $\beta_k,~k=1,2,\cdots,K$ is chosen uniformly in [0.5, 1.5].
\begin{figure}[!t]
\setlength{\abovecaptionskip}{0cm}
\setlength{\belowcaptionskip}{-0.5cm}
\centering
\subfigure[$M=120$, $K=6$, and $B_2=10.$]{
\begin{minipage}[t]{0.5\linewidth}
\centering
\includegraphics[width=3.9in]{./figs/MRT_BIT.eps}

\end{minipage}
}
\subfigure[$M=60$, $K=6$, and $B_1=2$.]{
\begin{minipage}[t]{0.5\linewidth}
\centering
\includegraphics[width=3.9in]{./figs/ZF.eps}

\end{minipage}
}
\centering
\caption{Achievable rates of analog precoding and MRT-based hybrid precoding in (a) and ZF-based hybrid precoding in (b)}
\end{figure}
\subsection{Rayleigh Fading Channels}

In Fig. 1(a) we illustrate the spectral efficiency achieved by pure analog precoding and the MRT-based hybrid precoding. It is observed that when $B_1\leq B_1^0=1.05$, the MRT-based hybrid precoding outperforms pure analog precoding for all SNRs, while when $B_1> B_1^0$ the pure analog precoding performs better at high SNRs, confirming the conclusions in Section \uppercase\expandafter{\romannumeral4}. This is because for small $B_1$, the system performance is severely affected by the quantization error, i.e., the interference becomes dominant, thus the hybrid precoding using digital processing to eliminate interference outperforms the pure analog precoding.  In addition, the corresponding threshold calculated from (14) is 17 for $B_1=5$, so for $K=6<17$, the pure analog precoding outperforms the MRT-based hybrid precoding at high SNRs.

Similar trends can be found when comparing pure analog precoding and the ZF-based hybrid precoding in Fig. 1(b). When the effective
channel is roughly quantized, i.e., $B_2\leq B_2^0=4.29$, the pure analog precoding performs better for all SNRs since the accuracy of channel estimation is insufficient to support effective beamforming for ZF, while the ZF-based hybrid precoding takes over at high SNRs when $B_2> B_2^0$, which verifies our observations in Section \uppercase\expandafter{\romannumeral4}.
\subsection{Large mmWave Multiuser Channels}
Apart from Rayleigh fading channels, hybrid/analog precoding can also be applied to mmWave communications. To capture the nature of high-frequency propagations, we adopt the widely-used geometric channel model [7] [9]
\begin{equation}\label{eq:mmwave}
{\bf{h}}_k^{\rm{H}} = \sqrt {\frac{{{M}}}{{{N_p}}}} \sum\limits_{l = 1}^{{N_p}} {\alpha _l^k{{\bf{a}}^H}} (\phi _l^k),
\end{equation}
where $N_p$ is the number of propagation paths from BS to user and ${\alpha _l^k}\sim {\cal {CN}}(0,1)$ denotes the complex gain of the $l$-th path. $\phi _l^k$ is the azimuth angle of departure drawn independently from the uniform distribution over [$0, 2\pi$]. ${\bf{a}}(\phi _l^k)$ is the array response vector of BS. Here we consider a uniform
linear array (ULA) whose array response vector is given by [9, eq. (29)].

Fig. 2 shows that conclusions on the superiority of the pure analog precoding over hybrid precodings also hold for the mm-Wave channels. Specifically, when $B_2$ is small, i.e., $B_2=3$, the analog precoding outperforms the ZF-based hybrid precoding for all SNRs, while for large $B_2$, i.e., $B_2=12$, the ZF-based hybrid precoding takes over at high SNRs, which verifies Remark 2.
\begin{figure}[!t]
\setlength{\abovecaptionskip}{0cm}
\setlength{\belowcaptionskip}{-0.5cm}
\centering
\includegraphics[width=6.0in]{./figs/mmwave.eps}
\caption{Achievable rates over mmWave channels with $M=40, K=5, B_1=2$, and $N_p=10$.}
\label{fig:0}
\end{figure}
\section{Conclusion}
In this paper, we derive tight rate approximations of a massive MIMO system using pure analog precoding and hybrid precoding with limited feedback in the large BS antenna regime. The effect of quantized analog and digital precodings are characterized in the obtained expressions. Furthermore, it is revealed that pure analog precoding outperforms hybrid precoding in terms of the ergodic achievable rate under certain conditions, which have been derived in closed forms with respect to the SNR, the number of users and the number of feedback bits. Numerical results verify the observed.
\begin{appendices}
\section{Proof Of Theorem 1}
To evaluate (\ref{eq:rate}), we calculate the terms ${\mathbb E}\left[{\left| {{{\bf{g}}_k^H}{{\bf{w}}_k}} \right|^2}\right]$ and ${\mathbb E}\left[{\left| {{{\bf{g}}_k^H}{{\bf{w}}_j}} \right|^2}\right]$. Let ${{\bf{g}}_k} = \left\| {{{\bf{g}}_k}} \right\|{\tilde {\bf{g}}_k}$ where ${\tilde {\bf{g}}_k}$ is the normalized ${\bf{g}}_k$, then we have ${\mathbb E}\left[{\left| {{{\bf{g}}_k^H}{{\bf{w}}_k}} \right|^2}\right]={\mathbb E}\left[\left\| {{{\bf{g}}_k}} \right\|_{\rm F}^2\right]{\mathbb E}\left[{\left| {{\tilde{\bf{g}}_k^H}{{\bf{w}}_k}} \right|^2}\right]$.

From the definition of ${\bf{g}}_k$, its $k$th element equals
\begin{equation}\label{eq:gkk}
{g_{k,k}} = \frac{1}{{N}}\sum\limits_{l = N(k - 1) + 1}^{Nk} {{h_{k,l}}{e^{ - j{{\hat \varphi }_{k,l}}}}}\overset{(\rm a)}=\frac{1}{{N }}\sum\limits_{l = N(k - 1) + 1}^{Nk} {\lambda _l},
\end{equation}where (a) uses ${\lambda _l} \buildrel \Delta \over ={{h_{k,l}}{e^{ - j{{\hat \varphi }_{k,l}}}}}$. Defining ${\varepsilon _{k,l}} \buildrel \Delta \over = {\varphi _{k,l}} - {\hat \varphi _{k,l}}$, it yields ${\lambda _l} ={|{h_{k,l}}|{e^{j{{\varepsilon }_{k,l}}}}}$. Recall ${\bf h}_k^H \sim {\cal {CN}}({\bf 0}_M,{\bf I}_M)$, where ${\rm{\{ }}{{h}_{k,i}}\} $'s are i.i.d. complex Gaussian variables with zero mean and unit variance. It implies that $| {{h_{k,l}}}|$ follows the Rayleigh distribution with mean $\frac{{\sqrt \pi  }}{2}$ and variance $1 - \frac{\pi }{4}$. The quantization error ${\varepsilon _{k,l}}\sim U[-\delta,\delta)$ is a uniform distribution with $\delta\buildrel \Delta \over =\frac{\pi }{{{2^{{B_1}}}}}$. Then we have
\begin{equation}\label{eq:gkk_rm}
{\mathbb E}[\mathcal{R}[{\lambda _l}]]={\mathbb E}\left[\mathcal{R}[{|{h_{k,l}}|{e^{j{{\varepsilon }_{k,l}}}}}]\right]={\mathbb E}[| {{h_{k,l}}}|\cos ({\varepsilon _{k,l}})]
\overset{(\rm a)}={\mathbb E}[| {{h_{k,l}}}|]{\mathbb E}[\cos ({\varepsilon _{k,l}})]\overset{(\rm b)}=\frac{{\sqrt \pi  {\rm{sinc}}(\delta )}}{2},
\end{equation}
where (a) utilizes the independence between $| {{h_{k,l}}}|$ and ${\varepsilon _{k,l}}$, and (b) is due to ${\mathbb E}[| {{h_{k,l}}}|]=\frac{{\sqrt \pi  }}{2}$ and ${\mathbb E}[\cos ({\varepsilon _{k,l}})]=\frac{1}{{2\delta }}\int_{ - \delta }^\delta  {\cos {\varepsilon _{k,l}}} d{\varepsilon _{k,l}}={\rm{sinc}}(\delta )$ because of the given distributions of $| {{h_{k,l}}}|$ and ${\varepsilon _{k,l}}$. Analogously, we can also get ${\mathbb E}[(\mathcal{R}[{\lambda _l}])^2]={\mathbb E}[| {{h_{k,l}}}|^2]{\mathbb E}[\cos^2 ({\varepsilon _{k,l}})]=\frac{1+{\rm{sinc}}(\delta )\cos (\delta )}{2}$. Then,
\begin{equation}\label{eq:gkk_rv}
{\mathbb V}[\mathcal{R}[{\lambda _l}]]={\mathbb E}[(\mathcal{R}[{\lambda _l}])^2]-({\mathbb E}[\mathcal{R}[{\lambda _l}]])^2=w_1,
\end{equation}
where ${w_1}\buildrel \Delta \over =\frac{1+{\rm{sinc}}(\delta )\cos (\delta )}{2}-\frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{4}$. Applying the Central Limit Theorem to (\ref{eq:gkk}) and using (\ref{eq:gkk_rm}) and (\ref{eq:gkk_rv}), we get
\begin{equation}\label{eq:gkk_rd}
\mathcal{R}[g_{k,k}] \sim \mathcal{N}\left(\frac{{\sqrt \pi  {\rm{sinc(}}\delta {\rm{)}}}}{2},\frac{{{w_1}}}{N}\right).
\end{equation}
Following similar reasons, it can be readily proved that
\begin{equation}\label{eq:gkk_id}
\mathcal{I}[g_{k,k}] \sim \mathcal{N}(0,\frac{{{w_2}}}{N}),~\!
\mathcal{R}[g_{k,i}] \sim \mathcal{N}(0,\frac{{{1}}}{2N}),~\!\mathcal{I}[g_{k,i}] \sim \mathcal{N}(0,\frac{{{1}}}{2N}),
\end{equation}
where ${w_2}\buildrel \Delta \over = \frac{1-{\rm{sinc}}(\delta )\cos (\delta )}{2}$ and $\mathcal{I}[x]$ returns the imaginary part of $x$. Now we have
\begin{align}
\label{eq:gkk_val}{\mathbb E}[{\left| {{g_{k,k}}} \right|^2}]&={\mathbb E}[{\left| {{\cal R}[{g_{k,k}}]} \right|^2}]\!+\!{\mathbb E}[{\left| {{\cal I}[{g_{k,k}}]} \right|^2}]=\frac{{\pi {\rm{sin}}{{\rm{c}}^2}{\rm{(}}\delta {\rm{)}}}}{4}\!+\!\frac{{{w_1}\!+\!{w_2}}}{N}.\\
\label{eq:gki_val}{\mathbb E}[{\left| {{g_{k,i}}} \right|^2}]&={\mathbb E}[{\left| {{\cal R}[{g_{k,i}}]} \right|^2}]+{\mathbb E}[{\left| {{\cal I}[{g_{k,i}}]} \right|^2}]=\frac{1}{N}.
\end{align}

\indent According (\ref{eq:gkk_val}) and (\ref{eq:gki_val}), it gives
\begin{equation}\label{eq:gks}
{\mathbb E}\left[\left\| {{{\bf{g}}_k}} \right\|_{\rm F}^2\right]\!=\!{\mathbb E}[{\left| {{g_{k,k}}} \right|^2}]+\!\!\!\!\!\!{\sum\limits_{i = 1,i \ne k}^K\!\!\!\!{{\mathbb E}[\left| {{g_{k,i}}} \right|} ^2}]\!=\!\frac{{\pi {\rm{sin}}{{\rm{c}}^2}{\rm{(}}\delta {\rm{)}}}}{4}+\!\frac{K}{N}-\!\frac{{\pi {\rm{sin}}{{\rm{c}}^2}{\rm{(}}\delta {\rm{)}}}}{{4N}}.
\end{equation}
From [10, Lemma 2], ${\bf R}_k$ can be asymptotically written as ${\bf R}_k={\rm diag}{[r_{1,1},r_{2,2},\cdots,r_{k,k}]}$ where
\begin{equation}\label{eq:Rk}
\begin{split}
r_{i,i}=\left\{\begin{array}{lc}
{\frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{4} + \frac{1}{N} - \frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{{4N}}},\quad &i=k\\
\frac{1}{N},\quad &i\neq k\end{array}\right.
\end{split}
\end{equation}
and diag[$\cdot$] returns a diagonal matrix with the input as its elements. Then we follow the result in [11] and get
\begin{equation}\label{eq:gkss}
{\mathbb E}\left[{\left| {{\tilde{\bf{g}}_k^H}{\hat{\bf{g}}_k}} \right|^2}\right]  \approx  1-\frac{{\sigma _{k,2}^2}}{{\sigma _{k,1}^2}}{2^{-\frac{B_2}{K-1}}}=1 - \frac{{{2^{ - \frac{{{B_2}}}{{K - 1}}}}}}{{\frac{{\pi N{\rm{sin}}{{\rm{c}}^2}(\delta )}}{4} + 1 - \frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{4}}},
\end{equation}
where ${\sigma _{k,1}}$ and ${\sigma _{k,2}}$ respectively stand for the largest and the second largest singular value of ${\bf R}^{\frac{1}{2}}_k$.

Combining (\ref{eq:gks}) and (\ref{eq:gkss}), it yields
\begin{equation}\label{eq:signal}
{\mathbb E}[{\left| {{{\bf{g}}_k^H}{{\bf{w}}_k}} \right|^2}]\overset{(\rm a)}={\mathbb E}[\left\| {{{\bf{g}}_k}} \right\|_F^2]{\mathbb E}[{\left| {{\tilde{\bf{g}}_k^H}{\hat{\bf{g}}_k}} \right|^2}]\nonumber\\
\overset{(\rm b)}{\to}\frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{4} + \frac{K}{N} - \frac{{{2^{ - \frac{{{B_2}}}{{K - 1}}}}}}{N},
\end{equation}
where (a) utilizes ${{\bf{w}}_k} = {{\bf{g}}_k}$ and in (b) we consider the fact that $\frac{{\frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{4} + \frac{K}{N}}}{{\frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{4} + \frac{K}{N} - \frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{{4N}}}} \to 1$ and $\frac{{\frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{4} + \frac{K}{N} - \frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{{4N}}}}{{\frac{{\pi N{\rm{sin}}{{\rm{c}}^2}(\delta )}}{4} + 1 - \frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{4}}} \to \frac{1}{N}$ hold in massive MIMO with large $N$ and fixed $K$.

From (\ref{eq:codebook}), we get ${\hat {g}_{j,j}} = \frac{{{{({\bf{R}}_j^{\frac{1}{2}})}_{j,j}}{{{v}}_{i,j}}}}{\|{{\bf{R}}_j^{\frac{1}{2}}{{\bf{v}}_{i}}}\|}$, where ${\hat {g}_{j,j}}$ and ${v_{i,j}}$ are respectively the $j$th element of ${\hat {\bf{g}}_j}$ and ${{\bf{v}}_i}$, and $i$ denotes the index of the quantization vector of ${\bf{g}}_j$. Then we have
\begin{equation}\label{eq:g^jj_m}
{\mathbb E}[{{\hat g}}_{j,j}]= {\mathbb E}\left[\frac{{{{({\bf{R}}_j^{\frac{1}{2}})}_{j,j}}{{{v}}_{i,j}}}}{\|{{\bf{R}}_j^{\frac{1}{2}}{{\bf{v}}_{i}}}\|}\right]\overset{(\rm a)}={\mathbb E}[{{v_{i,j}}}]{\mathbb E}\left[\frac{{{{({\bf{R}}_j^{\frac{1}{2}})}_{j,j}}}}{{\| {{\bf{R}}_j^{\frac{1}{2}}{{\bf{v}}_i}}\|}}\right]\overset{(\rm b)}=0,
\end{equation}
where (a) uses the independence between ${v_{i,j}}$ and $\frac{{{{({\bf{R}}_j^{\frac{1}{2}})}_{j,j}}}}{{\| {{\bf{R}}_j^{\frac{1}{2}}{{\bf{v}}_i}}\|}}$, and (b) results from ${\bf v}_{i}\sim {\cal {CN}}({\bf 0}_M,{\bf I}_M)$. Analogously,
\begin{equation}\label{eq:g^jj_2}
{\mathbb E}[|{{\hat g}}_{j,j}|^2]= {\mathbb E}\left[\frac{{{{({\bf{R}}_j)}_{j,j}}{|{{v_{i,j}}}|^2}}}{\|{{\bf{R}}_j^{\frac{1}{2}}{{\bf{v}}_{i}}}\|^2}\right]={\mathbb E}[|{{v_{i,j}}}|^2]\frac{{{{({\bf{R}}_j)}_{j,j}}}}{{\mathbb E}\left[{\| {{\bf{R}}_j^{\frac{1}{2}}{{\bf{v}}_i}}\|^2}\right]}\overset{(\rm a)}=\frac{{\frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{4} + \frac{1}{N} - \frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{{4N}}}}{{\frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{4} + \frac{K}{N} - \frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{{4N}}}},
\end{equation}
where (a) uses ${\mathbb E}\left[{\| {{\bf{R}}_j^{\frac{1}{2}}{{\bf{v}}_i}} \|^2}\right]=\sum\limits_{l = 1}^K {{{({{\bf{R}}_j})}_{l,l}}{\mathbb E}[|{v_{i,l}}|^2]}=\sum\limits_{l = 1}^K {{{({{\bf{R}}_j})}_{l,l}}} ={\frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{4} + \frac{K}{N} - \frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{{4N}}}$ and ${\mathbb E}[|{{v_{i,l}}}|^2]=1$ according to the distribution of ${\bf v}_i$ and (\ref{eq:Rk}). Following trivially the above steps, we get ${\mathbb E}[{{\hat g}}_{j,i}]=0$, and ${\mathbb E}[|{{\hat g}}_{j,i}|^2]=\frac{1}{{\frac{{\pi N{\rm{sin}}{{\rm{c}}^2}(\delta )}}{4} + K - \frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{4}}}$.
Providing that ${\rm{\{ }}{{v}_{i,l}}\} $'s are i.i.d. complex Gaussian variables with zero mean and unit variance, then ${\hat {g}_{j,i}}$ and ${\hat {g}_{j,l}}$ are independent for any $i\neq l$, which implies ${\mathbb E}[{{\hat g}}_{j,i}{{\hat g}^*_{j,l}}]={\mathbb E}[{{\hat g}}_{j,i}]{\mathbb E}[{{\hat g}^*}_{j,l}]=0$ and ${\mathbb E}[{\left| {{{\hat g}}_{j,i}} \right|^2}{| {{{\hat g}^*_{j,l}}}|^2}]={\mathbb E}[{\left| {{{\hat g}_{j,i}}} \right|^2}]{\mathbb E}[{|{{{\hat g}^*_{j,l}}} |^2}]$. Then ${\mathbb E}\left[{\left| {{{\bf{g}}_k^H}{{\bf{w}}_j}} \right|^2}\right]$ can be calculated as
\begin{align}\label{eq:intf}
{\mathbb E}\left[{\left| {{{\bf{g}}_k^H}{{\bf{w}}_j}}\right|^2}\right]&={\mathbb E}\left[{\left| {{{\bf{g}}_k^H}{\hat{\bf{g}}_j}}\right|^2}\right]={\mathbb E}\left[{\left|\sum\limits_{i = 1}^K {{g^*_{k,i}}} {\hat g_{j,i}}\right|^2}\right]\overset{(\rm a)}=\sum\limits_{1 \le i \le l \le K}^K {{\mathbb E}[g_{k,i}^*{g_{k,l}}]{\mathbb E}[{{\hat g}_{j,i}}\hat g_{j,l}^*]}\nonumber\\
&\overset{(\rm b)}=\sum\limits_{i = 1}^K{\mathbb E}[{\left| {{{\hat g}_{j,i}}} \right|^2}]{\mathbb E}[{\left| {{g_{k,i}}} \right|^2}]\overset{(c)}{\to}\frac{{\frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{2} + \frac{K}{N}}}{{\frac{{\pi N{\rm{sin}}{{\rm{c}}^2}(\delta )}}{4} + K}},
\end{align}
where (a) is due to the independence between ${\bf{g}}_k$ and ${\hat{\bf{g}}_j}$, (b) holds because ${\mathbb E}[g_{k,i}^*{g_{k,l}}]{\mathbb E}[{{\hat g}}_{j,i}{{\hat g}^*_{j,l}}]=0$ for any $i\neq l$, and (c) is obtained by using similar manipulations as that in the last step of (\ref{eq:signal}). Finally, by applying [16, Lemma 1] and substituting (\ref{eq:signal}) and (\ref{eq:intf}) into (\ref{eq:rate}), the theorem is proved.
\section{Proof Of Proposition 1}
From the distributions of $g_{k,k}$ and $g_{k,i}$ in (\ref{eq:gkk_rd}) -- (\ref{eq:gkk_id}), we get $g_{k,k}\overset{a.s.}{\to}\frac{{\sqrt \pi  {\rm{sinc}}(\delta )}}{2}$ and $g_{k,i}\overset{a.s.}{\to}0$ since $\frac{{{w_1}}}{N} \to 0,\frac{{{w_2}}}{N} \to 0,$ and $\frac{1}{{2N}} \to 0$ for the massive MIMO with large $N$, thus ${\bf G}\overset{a.s.}{\to}\frac{{\pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{4}{{\bf{I}}_K}$. Now the asymptotic rate of the ZF-based hybrid system with perfect CSI can be represented as
\begin{equation}\label{eq:ZF}
R_{{\rm H},k}^{\rm ZF_p}\overset{a.s.}{\to}{\rm{lo}}{{\rm{g}}_2}\left( {1 + \frac{{\gamma\beta_k \pi {\rm{sin}}{{\rm{c}}^2}(\delta )}}{{4K}}} \right).
\end{equation}
On the other hand, it is known from [12] that the rate loss caused by the imperfect CSI can be upper bounded by ${\rm{lo}}{{\rm{g}}_2}\left( {1 + \frac{{\gamma{\overline \beta _k}{\mathbb E}\left[{{\left| {{\bf{g}}_k^H{{\bf{w}}_j}} \right|}^2}\right]}}{K}} \right)$. Thanks to the orthogonality between $\hat{\bf g}_k$ and ${\bf w}_j$ in the ZF precoding, we further have
\begin{equation}\label{eq:orth}
{\mathbb E}\left[{\left| {{\bf{g}}_k^{\rm{H}}{{\bf{w}}_j}} \right|^2}\right] \le {\mathbb E}\left[\left\| {{{\bf{g}}_k}} \right\|_{\rm F}^2\right]{\mathbb E}\left[1 - {\left| {\tilde {\bf{g}}_k^H{{\hat {\bf{g}}}_k}} \right|^2}\right] = \frac{{{2^{ - \frac{{{B_2}}}{{K - 1}}}}}}{N}.
\end{equation}
Therefore, we have
\begin{equation}\label{eq:rloss}
R_{\rm loss}\leq{\log _2}(1 + \frac{{\gamma{\overline \beta  _k} {2^{ - \frac{{{B_2}}}{{K - 1}}}}}}{{KN}}).
\end{equation}
Combining (\ref{eq:ZF}) and (\ref{eq:rloss}), the proposition is directly proved.
\end{appendices}

config: arxiv


TITLE: Series basic question
QUESTION [2 upvotes]: $$\sum^N_{n=1}\liminf_{k \to \infty} f_k(n) = \lim_{k \to \infty} \sum_{n=1}^N \inf_{j \ge k} f_j(n)$$
I am not sure that equation true. Is that equation true? Then why is it?

REPLY [3 votes]: It is known that for any sequence $\{a_k:k\in\mathbb{N}\}\subset\mathbb{R}$ we have
$$
\liminf\limits_{k\to\infty}a_k=\lim\limits_{k\to\infty}\inf\limits_{j\geq k}a_j
$$
so
$$
\sum\limits_{n=1}^N\liminf\limits_{k\to\infty}f_k(n)=
\sum\limits_{n=1}^N\lim\limits_{k\to\infty}\inf\limits_{j\geq k}f_j(n)=
\lim\limits_{k\to\infty}\sum\limits_{n=1}^N\inf\limits_{j\geq k}f_j(n)
$$


TITLE: Let $G$ be a group and $g\in G$. Prove or Disprove: If $o(g^3)=2$ then $o(g)=6$.
QUESTION [3 upvotes]: Let $G$ be a group and $g\in G$.
Prove or Disprove: If $o(g^3)=2$ then $o(g)=6$.
I tried to disprove with $g=(14)(32)$ then $g^3=(14)(32)$ so that $o(g^3)=2$ but $o(g)=2\neq6$.
Is it right?

REPLY [1 votes]: (I). Example. Let $G=\{1,g\}$ where $g^2=1\ne g.$ Then $o(g)=2.$ And $g^3=g(g^2)=g\cdot 1=g\ne 1$ but $(g^3)^2=(g^2)^3=1^3=1,$ so $o(g^3)=2.$
(II). If $G$ is any group with identity $1$ and if $o(g^3)=2$ then $[o(g)=6\iff g^3\ne g].$


TITLE: Prove that $(P \leftrightarrow Q) \wedge (R \leftrightarrow S) \implies (P \vee R) \leftrightarrow (Q \vee S).$
QUESTION [2 upvotes]: Let $P, Q, R$ and $S$ be statements. Prove that
$$(P \leftrightarrow Q) \wedge (R \leftrightarrow S) \implies (P \vee R) \leftrightarrow (Q \vee S).$$
My attempt:
I already check that this logical implication is true, by showing that the conditional statement $(P \leftrightarrow Q) \wedge (R \leftrightarrow S) \to (P \vee R) \leftrightarrow (Q \vee S)$ is a tautology.
Although, I am trying to come up with a derivation of this implication. Here is what I have
$$(P \leftrightarrow Q) \wedge (R \leftrightarrow S) \implies (P \to Q) \wedge (Q \to P) \wedge (R \to S) \wedge (S \to R).$$
Using the Material Implication, I deduce that
$$(P \to Q) \wedge (Q \to P) \wedge (R \to S) \wedge (S \to R)  \implies (\neg P \vee Q) \wedge (\neg Q \vee P) \wedge (\neg R \vee S) \wedge (\neg S \vee R).$$
I thought that maybe I could use the Distributive Law to continue to simplifying this expression. By doing this, I had to rotate my sheet, because the expression was really big.
I can continue this until I got something, but the expression is becoming extremely length.
Is there any hint of what I should do in order to prove this logical implication?
Or any simpler idea of what to do?
Thank you for your attention!

REPLY [1 votes]: Let $P, Q, R$ and $S$ be statements, and consider the following statement
$$(P \leftrightarrow Q) \wedge (R \leftrightarrow S) \tag{1}$$
Using the rule of inference that states that for any statements $P$ and $Q$ $P \leftrightarrow Q \iff (P \to Q) \wedge (Q \to P),$ we get the following
$$(P \to Q) \wedge (Q \to P) \wedge (R \to S) \wedge (S \to R) \tag{2}$$
Applying the Material Implication to each conditional statement in $(2)$ yields the following expression
$$(\neg P \vee Q) \wedge (\neg Q \vee P) \wedge (\neg R \vee S) \wedge (\neg S \vee R) \tag{3}$$
Using the Rule of inference of Simplification, we get the following four disjunctions
$$\neg P \vee Q \tag{4}$$
$$\neg Q \vee P \tag{5}$$
$$\neg R \vee S \tag{6}$$
$$\neg S \vee R \tag{7}$$
Next, we apply the rule of Addition to each of the previous expressions. Namely, we will add $S$ to $(4),$ $R$ to $(5),$ $Q$ to $(6)$ and $P$ to $(7).$
$$\neg P \vee Q \vee S \tag{8}$$
$$\neg Q \vee P \vee R \tag{9}$$
$$\neg R \vee S \vee Q \tag{10}$$
$$\neg S \vee R \vee P \tag{11}$$
By Adjunction, we get the following expression
$$(\neg P \vee Q \vee S) \wedge (\neg Q \vee P \vee R) \wedge (\neg R \vee S \vee Q) \wedge (\neg S \vee R \vee P) \tag{12}$$
Using the Distributive Law yields that
$$\big( (\neg P \wedge \neg R) \vee (Q \vee S) \big) \wedge \big( (\neg Q \wedge \neg S) \vee (P \vee R) \big) \tag{13}$$
Which, by the De Morgan’s Law, is equivalent to
$$\big( \neg(P \vee R) \vee (Q \vee S) \big) \wedge \big( \neg(Q \vee S) \vee (P \vee R) \big) \tag{14}$$
Using, again, the Material Implication we write our previous expression as
$$\big( (P \vee R) \to (Q \vee S) \big) \wedge \big( (Q \vee S) \to (P \vee R)\big) \tag{15}$$
Finally, using the first rule of inference used here to deduce $(2)$ from $(1),$ we get
$$(P \vee R) \leftrightarrow (Q \vee S) \tag{16}$$
Hence, we have proved that $(1) \implies (16),$ i.e.
$$(P \leftrightarrow Q) \wedge (R \leftrightarrow S) \implies (P \vee R) \leftrightarrow (Q \vee S) \ \text{.  } \text{ } \text{ }\square$$


TITLE: Determine the busiest day from a list of dates.
QUESTION [1 upvotes]: There are N date ranges which include check-in date and check-out date. How can I find out the busiest day?
Example:
2016-10-12 to 2016-10-18
2016-10-11 to 2016-10-15
2016-10-13 to 2016-10-14
2016-10-12 to 2016-10-13

From these dates, it is obvious that 2016-10-12 was booked on all 4 times. How can I mathematically solve this?
Using two dates we can find out overlaps like this.

REPLY [0 votes]: Hint -
I think you can create 4 sets for given 4 date range's. Then you can find intersection of these 4 sets.


TITLE: Showing that $f(x)$ is a probability density function
QUESTION [3 upvotes]: I'm having trouble showing the following $f(x)$ is a probability density function. The following information is given:
$$f(x) = \frac{1}{\sum_{k=x}^\infty {k-1\choose x-1}(1-p)^{k-x-1}}; x = 0, 1, 2, \ldots$$
I'm having trouble in particular getting the denominator to something that I can take the integral of. I've attempted to get it to be equal to a binomial series but I eventually end up with an integral of $p^x$ which is zero.
Thanks for any help or tips.

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{x = 0}^{\infty}\bracks{1 \over
\bbox[1px,#ffd]{\sum_{k = x}^{\infty}{k - 1 \choose x - 1}
\pars{1 - p}^{k - x - 1}}} = 1:\ {\large ?}}$.

\begin{align}
&\mbox{Note that}\quad\bbox[10px,#ffd]{\ds{%
\sum_{k = x}^{\infty}{k - 1 \choose x - 1}\pars{1 - p}^{k - x - 1}}} =
\sum_{k = 0}^{\infty}{k + x - 1 \choose x - 1}\pars{1 - p}^{k - 1}
\\[5mm] = &\
{1 \over 1 - p}\sum_{k = 0}^{\infty}{k + x - 1 \choose k}\pars{1 - p}^{k} =
{1 \over 1 - p}\sum_{k = 0}^{\infty}{-x \choose k}\pars{-1}^{k}\pars{1 - p}^{k} \\[5mm] = &\
{1 \over 1 - p}\sum_{k = 0}^{\infty}{-x \choose k}\pars{-1 + p}^{k} =
{1 \over 1 - p}\bracks{1 + \pars{-1 + p}}^{\,-x} = \bbox[10px,#ffd]{\ds{{1 \over \pars{1 - p}p^{x}}}}
\end{align}

Then,
\begin{align}
\sum_{x = 0}^{\infty}\bracks{1 \over
\bbox[1px,#ffd]{\sum_{k = x}^{\infty}{k - 1 \choose x - 1}
\pars{1 - p}^{k - x - 1}}} & =
\sum_{x = 0}^{\infty}{1 \over
\bbox[#ffd]{\ds{1/\bracks{\pars{1 - p}p^{x}}}}}
\\[5mm] & =
\pars{1 - p}\sum_{x = 0}^{\infty}p^{x} = \pars{1 - p}\,{1 \over 1 - p} = \bbx{1}
\end{align}


TITLE: Lemma 1.4.3 A First Course in Harmonic Analysis
QUESTION [2 upvotes]: In Lemma 1.4.3 of Anton Deitmar's A First Course in Harmonic Analysis, he proves that the Fourier series of a periodic function $f:\mathbb R\to\mathbb R$ such that $\left.f\right|_{\left[0,1\right]}$ is a Riemann step function converges to $f$ in $L^2\left(\left[0,1\right]\right)$.
He makes extensive use of the equality
$$\left\|f\right\|_2^2=\sum_{n=-\infty}^\infty\left|c_n\right|^2,\tag{$\star$}$$
where the $c_n$s are the Fourier coefficients of $f$.
He first proves it for the case where $f\left(x\right)=\mathbf1_{\left[0,a\right]}\left(x\right)$, where $a\in\left[0,1\right]$, and then extends the proof to the case where $f\left(x\right)=\mathbf1_I\left(x\right)$, where $I$ is an open, closed, or half-open subinterval of $\left[0,1\right]$. I understand these proofs.
However, at the end he says that the proof extends, "by linearity", to the case where
$$f\left(x\right)=\sum_{n=1}^m\alpha_n\mathbf1_{I_n}\left(x\right),$$
where the $\alpha_n$s are constants. How is this so? It does not look like I can use "linearity" with $\left(\star\right)$.

REPLY [2 votes]: You need to show that if $(*)$ holds for $f$ and $g$, then it holds for
$\alpha f$ (obvious) and $f+g$ (not obvious). Let's stick to real-valued
functions $f$ and $g$. Let $f$ have Fourier coefficients $c_n$
and $g$ have Fourier coefficients $d_n$.
We need to show that $\|f+g\|^2=\sum_n (c_n+d_n)^2$. From Bessel's inequality, $\|f+g\|^2\ge\sum_n (c_n+d_n)^2$
and $\|f-g\|^2\ge\sum_n (c_n-d_n)^2$.
But
$$\|f+g\|^2+\|f-g\|^2=2\|f\|^2+2\|g\|^2=
2\sum_n(c_n^2+d_n^2)=\sum_n(c_n+d_n)^2
+\sum_n(c_n-d_n)^2.$$
Using Bessel's inequality for $f-g$ gives
$$\|f+g\|^2\le\sum_n(c_n+d_n)^2.$$
Another application of Bessel gives
$$\|f+g\|^2=\sum_n(c_n+d_n)^2$$
so $(*)$ holds for $f+g$.
For complex valued functions use $\|f+g\|^2+\|f+ig\|^2+\|f-g\|^2+\|f-ig\|^2
=4\|f\|^2+4\|g\|^2$.


TITLE: Minimize area of polygon
QUESTION [1 upvotes]: Let us consider given $n$ segments with length and tangent($0 < \theta < 90)$. Starts with the origin(i.e. $(0, 0)$), we connect the segments with end to end. And calculate the area between the segments and $x$-axis. The problem is to minimize this area.
I already know the answer; connect the segment with smallest tangent value, but how can I prove it(In algebraic way)?
I apologize for my bad English.
--------Addition---------

I apologize to make an ambiguity; all segments have different and fixed length and angle. We also choose the arbitrary order to connect. Also, the segments cannot be rotated(i.e. the endpoints are increase both x-coordinate and y-coordinate), and our goal is minimize the area covered by the segments, the x-axis, and the perpendicular line from end of last segment.

REPLY [1 votes]: Your problem is very clear now. Your suggested solution is correct: We should arrange the given vectors ${\bf a}_k=\rho_k(1,\tan\theta_k)$ in order of  increasing $\theta_k$.
Proof. Assume that the $\theta_k$ are not increasing with $k$. Then there is a $k$ such that $\theta_k>\theta_{k+1}$. Draw a figure showing this situation. You can then easily see that interchanging ${\bf a}_k$ and ${\bf a}_{k+1}$ leaves most of the polygon unaltered. Only the two vertical strips coming from ${\bf a}_k$ and ${\bf a}_{k+1}$ are affected: We see that a parallelogram spanned by these two nonparallel vectors, hence having positive area, is subtracted.
Since there are only $n!$ ways of ordering the vectors there has to be an optimal ordering.  The above argument shows that an optimal ordering has to have increasing $\theta_k$.


TITLE: Electromagnetic current for interaction with Dirac spinors
QUESTION [3 upvotes]: The covariant form of the Dirac equation is given by
$$(i\gamma^{\mu}\partial_{\mu} - M) \Psi(x) = 0 $$
Einstein's summation is implied here, $x=(x^0,x^1,x^2,x^3)^T$.
I am simply looking for the Dirac theory equivalent to the classic electromagnetic interaction energy
$$ W = \int_V \rho \phi - \frac{1}{c} \mathbf{j\cdot A} \ dV $$
I have shown the Dirac equation's covariance and current conservation, but I can't seem to make the transition to the interaction energy.
Also, I have made myself familiar with the Klein-Gordon equation
$$(\square + M^2) \Psi(x) = 0$$
In that case, the interaction energy was "brought into" the equation by making the transition $ p^{\mu} \rightarrow p^{\mu} - \frac{e}{c} A^{\mu} $, where $p$ denotes the momentum operator, so the equation transforms to the form
$$(i\hbar\partial_{\mu} - \frac{e}{c}A_{\mu})(i\hbar\partial^{\mu} - \frac{e}{c}A^{\mu})\Psi(x) = M^2 \Psi(x)$$
Could it be that simple for the Dirac case, too?

REPLY [1 votes]: The equation of motion that you have is for a free Dirac field. That is, there is no photon (electromagnetic) field for the Dirac field to interact with. You would need to add an interaction term to your Lagrangian.
Have you done interacting QFT?
I believe adding a Maxwell Lagrangian
$$ \mathcal{L}_M = −\frac{1}{4} \mathcal{F}_{\mu \nu} \mathcal{F}^{\mu \nu} $$
and promoting your partial derivative to a gauge covariant derivative
$$ \partial_{\mu}  \rightarrow \mathcal{D}_{\mu} = \partial_{\mu} − i e A_{\mu}(x) $$
should give you what you want.

REPLY [1 votes]: The probability current for the Dirac equation is $$j_p^\mu = \bar\psi \gamma^\mu \psi$$
and therefore the electromagnetic current is $$j^\mu =  -e\bar\psi\gamma^\mu\psi$$ since the charge of the Dirac field is $-e$. So to the Lagrangian we just add the usual term coupling current to vector potential $$\mathcal L = \bar\psi(i\gamma^\mu\partial_\mu - m)\psi - e\bar\psi\gamma^\mu\psi A_\mu.$$
Now the Euler-Lagrange equation for $\bar\psi$ can be read off as $$(i\gamma^\mu \partial_\mu - m -e \gamma^\mu A_\mu)\psi = 0.$$
(Different conventions will change some signs here but that's not really important.)


TITLE: A quick question on in logs
QUESTION [2 upvotes]: I was solving indefinite integrals
$$\int_0^22^x x \,dx$$
I use ILATE as a clue to consider the first function and second function.
$2^x$ is a algebraic function or logarithmic function?

REPLY [1 votes]: For:
 $$\int\limits_{0}^{2} x\cdot 2^x\:dx$$
Take $2^x =t$, then we get $2^x \cdot \ln(2) \cdot dx=dt$, and, $x=\log_2(t)$. Thus:
\begin{align}
\int\limits_{0}^{2} x \cdot 2^x\:dx&=\frac{1}{\ln(2)}\int\limits_{1}^{4}\log_2(t)\: dt \\
&=\frac{1}{(\ln(2))^2}\int\limits_{1}^{4}\ln(t) \: dt
\end{align}
Now, I leave it for you. Use by parts (Hint: $\ln(2)= 1\cdot\ln(2)$).


TITLE: derivative of cost function for Neural Network classifier
QUESTION [0 upvotes]: I am following Andrew NG's Machine Learning course on Coursera.
The cost function without regularization used in the Neural network course is:
$J(\theta) = \frac{1}{m} \sum ^{m}_{i=1}\sum ^{K}_{k=1} [-y_{k}^{(i)}log((h_{\theta}(x^{(i)}))_{k}) -(1-y_{k}^{(i)})log(1-(h_{\theta}(x^{(i)}))_{k})]$
, where $m$ is the number of examples, $K$ is the number of classes, $J(\theta)$ is the cost function, $x^{(i)}$ is the i-th training example, $\theta$ are the weight matrices and $h_{\theta}(x^{(i)})$ is the prediction of the neural network for the i'th training example.
I understand intuitively that the backpropagation error associated with the last layer(h) is h-y. Nevertheless, I want to be able to prove this formally.
For simplicity, I considered m = K = 1:
$J(\theta) = -y \log(h_{\theta}) - (1-y) \log(1-h_{\theta})$
and tried to prove this to myself on paper but wasn't able to.
Neural Network Definition:
This neural network has 3 layers. (1 input, 1 hidden, 1 output).
It uses the sigmoid activation function,
$\sigma(z) = \frac{1}{1+e^{-z}}$.
The input is $x$.
Input layer: $a^{(1)} = x$. (add bias $a_{0}^{(1)}$).
Hidden Layer: $z^{(2)} = \Theta^{(1)}a^{(1)}$ , $a^{2} = \sigma(z^{(2)})$, (add bias $a_{0}^{(2)}$).
Output layer: $z^{(3)} = \Theta^{(2)}a^{(2)}$ , $a^{3} = \sigma(z^{(3)}) = h_{\theta}(x)$.
During backpropagation, $\delta^{(3)}$ is the error associated with the output layer.
Question:

Why is it that:

$\delta^{(3)} = h_{\theta} - y$ ?

Shouldn't:

$\delta^{(3)} = \frac{\partial {J}} {\partial {h_{\theta}}}$ ?

REPLY [0 votes]: First, since your cost function is using the binary cross-entropy error $\mathcal{H}$ with a sigmoid activation $\sigma$, you can see that:
\begin{align}
\frac{\partial J}{\partial h_\theta}
&= \frac{1}{m}\sum_i\sum_k\frac{\partial }{\partial h_\theta}\mathcal{H}\left(y_k^{(i)},h_\theta(x^{(i)})_k\right) \\
&= \frac{1}{m}\sum_i\sum_k \left[ \frac{-y_k^{(i)}}{h_\theta(x^{(i)})_k} + \frac{1-y_k^{(i)}}{1-h_\theta(x^{(i)})_k} \right] \\
&= \frac{1}{m}\sum_i\sum_k \frac{h_\theta(x^{(i)})_k - y_k^{(i)}}{ h_\theta(x^{(i)})_k(1-h_\theta(x^{(i)})_k) }
\end{align}
Hence, for $m=K=1$, as a commenter notes $$ \frac{\partial J}{\partial h_\theta}
= \frac{h_\theta - y}{ h_\theta(1-h_\theta) } $$
But this is not so useful, as it computes how the error changes as the final output changes. What you really want is how the cost changes as the weights $\theta^{(\ell)}_{ij}$ are varied, so you can do gradient descent on them.
An intermediate calculation is to compute the variation with respect to the activation $ h_\theta=\sigma(z)$.
Let the last layer be $s$. Then the output layer error is:
\begin{align}
\delta^{(s)}_j
&= \frac{\partial J}{\partial z_j^{(s)}}\\
&= \frac{1}{m}\sum_i\sum_k \frac{\partial }{\partial z_j^{(s)}} \mathcal{H}\left(y_k^{(i)},h_\theta(x^{(i)})_k\right) \\
&= \frac{-1}{m}\sum_i\sum_k y_k^{(i)} \frac{1}{h_\theta(x^{(i)})_k}\frac{\partial h_\theta(x^{(i)})_k}{\partial z_j^{(s)}}  + (1-y_k^{(i)})\frac{1}{1-h_\theta(x^{(i)})_k}\frac{\partial h_\theta(x^{(i)})_k}{\partial z_j^{(s)}} \\
&= \frac{-1}{m}\sum_i\sum_k [1-h_\theta(x^{(i)})_k]y_k^{(i)} - h_\theta(x^{(i)})_k[1-y_k^{(i)}]\\
&= \frac{1}{m}\sum_i\sum_k h_\theta(x^{(i)})_k -y_k^{(i)}
\end{align}
using the fact that
$$
\frac{\partial h_\theta(x^{(i)})_k}{\partial z_j^{(s)}}
= \sigma'(z_j^{(s)})
= \sigma(z_j^{(s)})[1-\sigma(z_j^{(s)})]
= h_\theta(x^{(i)})_k[1-h_\theta(x^{(i)})_k]
$$
So in the case that $m=K=1$ and $s=3$, we have:
$$
\delta^{(3)} = h_\theta - y
$$


TITLE: Proof of continuity of Thomae Function at irrationals.
QUESTION [34 upvotes]: In Thomae's  Function:
$$
\begin{align}
t(x) = \begin{cases}
0 & \text{if $x$ is irrational}\\
\frac{1}{n} & \text{if $x = \frac{m}{n}$ where $\gcd(m,n) = 1$}
\end{cases}
\end{align}
$$
I can prove the discontinuity at rational $b$ by taking a sequence of irrationals $x_n$ which converge to $b$.
But while going through an argument for continuity at irrationals.
I found this in a book.

On the other hand if $b$ is an irrational number and $\epsilon > 0$
  then there is a     natural number $n_0$ such that $1/n_0 < \epsilon$.
  There are only finite number of rationals with denominator less than
  $n_0$ in the interval $(b-1,b+1)$. Hence we can find a $\delta > 0$
  such that $\delta$ neighbourhood of $b$ contains no rational with
  denominator less than $n_0$.

I understand the rest of the proof. But I am unable to prove the emphasized text. Although I find it intuitive.

REPLY [0 votes]: The natural extension of Thomae's function to the hyperreals is defined by the same formula:
$$
\begin{align}
t(x) = \begin{cases}
\frac{1}{n} & \text{if $x = \frac{m}{n}$ where $m,n\in{}^\ast\mathbb N$ and $\gcd(m,n) = 1$}
\\
0 & \text{otherwise.}\\
\end{cases}
\end{align}
$$
To show that $t(x)$ is continuous at $c\in\mathbb R\setminus \mathbb Q$, suppose $q$ is a hyperrational infinitely close to $c$.  Clearly $q\not\in\mathbb R$. Since $q$ is appreciable (i.e., finite but not infinitesimal), its denominator $n$ is necessarily an infinite hyperinteger.  Hence $t(q)=\frac{1}{n}\approx 0$ where $\approx$  is the relation of infinite proximity.  Thus $t(x)$ is infinitesimal at all points infinitely close to $c$, proving the continuity of $t(x)$ at $c$.


TITLE: The set of lines in $\mathbb{R}^2$ is a Möbius band?
QUESTION [11 upvotes]: I encountered a hard-to-believe and hard-to-understand statement in this problem:
Let $\mathcal{L} = \{\textrm{lines in}\ \mathbb{R}^2\}$. Consider the 2-to-1 map $f: S^1 \times \mathbb{R} \rightarrow \mathcal{L}$ given by $(\theta,x) \mapsto L_{\theta,x} = \{t(\mathrm{cos}\,\theta,\mathrm{sin}\,\theta)+x(-\mathrm{sin}\,\theta,\mathrm{cos}\,\theta): t\in\mathbb{R}\}$.
Show that for almost every $(\theta,x) \in S^1\times\mathbb{R}$, we have that $f^{-1}(L_{\theta,x})$ is a smooth submanifold of $X$.
Remark: The set of lines in $\mathbb{R}^2$ is a Möbius band.
So my question is, why the set is a Möbius band? And I really hope to get some help to understand the function $f$ - how it work: why $t(\mathrm{cos}\,\theta,\mathrm{sin}\,\theta)+x(-\mathrm{sin}\,\theta,\mathrm{cos}\,\theta)$
gives lines in $\mathbb{R}^2$? And why it is 2-to-1? Thank you very much for your guidance.

REPLY [1 votes]: I feel like the existing answers, while mathematically correct, may be a bit much for someone new to topology. Let me try a less formal explanation why "The Möbius strip is equivalent (homeomorphic) to the set of non-oriented lines in the plane".
Consider the standard construction of a Möbius strip by gluing the vertical ends of a flat strip of paper of width $[0,2\pi)$ and height $(-1,1)$ "with a 180° twist".
Clearly you can specify any point on the strip uniquely by the "angle" $\theta\in [0,2\pi)$ and the "height" $x\in (-1,1)$; this is simply the parametrization of the paper strip before you glue it.
Similarly you can specify any line in the plane uniquely by an angle $\theta \in [0,\pi)$ (where does the line point?) and a height $(-\infty,\infty)$ (how far do we translate the line from the origin towards where it points?). This is what your $f$ does, except it does it "twice" by allowing $\theta \in [0,2\pi)$.
Note that $[0,\pi)$ and $(-\infty, \infty)$ are homeomorphic (can be stretched or shrunk in)to $[0,2\pi)$ and $(-1,1)$. Since they have the same parametrizations (up to homeomorphism) it seems the Möbius strip and the set of lines in the plane are really the same. However, to really be able to conclude that, we still need to talk about the "edges" of our parametrizations. To define the Möbius strip by gluing the edges of a strip means that we map the limiting angle $\theta\to2\pi$ to the same points as $\theta=0$ but with the heights $x\in(-1,1)$ inverted.
Is that property true for our parametrization of the lines too? Yes! The horizontal lines that you get for $\theta=0$ are exactly those that you get if you let $\theta\to\pi$, but the former diverge upwards as you increase the "height" parameter, while the latter diverge downwards.


TITLE: $L^1$-Cauchy sequence of step functions that converges a.e. also converges in $L^1$
QUESTION [0 upvotes]: Let $X$ be a measure space and $E$ be a Banach space.  A step function from $X$ to $E$ is a measurable function ($E$ takes the Borel measure) with finite image and whose support has finite measure.
If $(f_n)$ is a Cauchy sequence of step functions that converge a.e. to a function $f$, then do they also converge to $f$ in the $L^1$ seminorm?
The context for this question is that Lang seems to make this leap in his proof of Theorem 3.4 on page 133 of Real and Functional Analysis, and I want to know if the proof is still correct.

REPLY [0 votes]: As Kavi Rama Murthy points out, if the sequence is Cauchy in $L^1$, then it converges to $f$ in $L^1$.  He has provided a counterexample in the case that the sequence $\{ f_n \}$ is merely pointwise Cauchy.  In his example, the pointwise limit $f$ is not in $L^1$.  However, it is possible to construct a counterexample in which the limit function is in $L^1$.  One such example is given by setting $f_n = \frac{1}{n}I_{[0,n]}$, where $I_{[0,n]}$ is the indicator function for the interval $[0,n]$.  The sequence $f_n$ is (uniformly) Cauchy because if $n,m \geq N$, then $|f_n(x) - f_m(x)| \leq \frac{2}{N}$ for all $x$.  The sequence tends pointwise to the zero function but does not converge to $0$ in $L^1$ because $\| f_n \|_1 = 1$ for al $n$.


TITLE: Find $\iint_Ay\,dA$, where $A$ is defined by $z=x+y^2$,$0\le x\le 1$ and $0\le y\le2$
QUESTION [3 upvotes]: Can anyone help me to find the $\iint_Ay\,dA $ where $A$ is defined by $z=x+y^2$ and $0\le x\le1$ and $0\le y\le2$
What I was thinking to do is to put $y=\sqrt{z-x}$ but I am not sure that I am doing right because I cannot find the limits of the integral.

REPLY [0 votes]: Running the following code in Mathematica:
z[x_, y_] := x + y^2
dA = Sqrt[1 + D[z[x, y], x]^2 + D[z[x, y], y]^2];
Integrate[y*dA, {x, 0, 1}, {y, 0, 2}]

we can get the following results:
(13 Sqrt[2])/3


TITLE: How prove this inequality $ \sum_{i=1}^{m} \sum_{j=1}^{m}|A_i|\cdot |A_i \cap A_j|\geq \frac{1}{mn}\left(\sum_{i=1}^{m}|A_i|\right)^3$
QUESTION [5 upvotes]: Let $A_1,A_2,\cdots,A_m$  be $m$ subsets of the set of size $n$  . Prove that $$\sum_{i=1}^{m} \sum_{j=1}^{m}|A_i|\cdot |A_i \cap A_j|\geq \frac{1}{mn}\left(\sum_{i=1}^{m}|A_i|\right)^3.$$

Someone know what this inequality background? The article studies such problems? Thank you.
Idea:
it seem this inequality with The sum can be written in terms of indicator functions as solve it? How to it?

REPLY [3 votes]: Hint. Let $\{1,2,\dots,n\}$ be the given set of size $n$ and let $X_k:=\{i:k\in A_i\}$ for $k=1,\dots,n$. Then show that
$$\sum_{i=1}^{m} \sum_{j=1}^{m}|A_i|\cdot |A_i\cap A_j|=\sum_{i=1}^{m}\sum_{k\in A_i}|A_i||X_k|=\sum_{k=1}^{n}\sum_{i\in X_k}|A_i||X_k|.$$
Finally, the conclusion follows by Hölder's inequality with three functions,
$$ \underbrace{\left(\sum_{i=1}^{m} \sum_{k\in A_i}\frac{1}{|A_i|}\right)}_{m}\cdot\underbrace{\left(\sum_{k=1}^{n}\sum_{i\in X_k}\frac{1}{|X_k|}\right)}_n\cdot\underbrace{\left(\sum_{i=1}^{m} \sum_{k\in A_i}|A_i||X_k|\right)}_{\sum_{i=1}^{m} \sum_{j=1}^{m}|A_i|\cdot |A_i\cap A_j|}\geq \underbrace{\left(\sum_{i=1}^{m} \sum_{k\in A_i}1\right)^3}_{\left(\sum_{i=1}^{m}|A_i|\right)^3}.$$


text:
\section{Belief propagation and Bethe free energy}
\label{sec:bp}

For a model with states $\vec{\sigma}$ having weight $p(\vec{\sigma})$
(as in (\ref{prob})), the convex functional
\begin{eqnarray}
{\cal F}\{b(\vec{\sigma})\}\equiv
\sum_{\vec{\sigma}}b(\vec{\sigma})\ln\frac{b(\vec{\sigma})}{p(\vec{\sigma})}
\label{Gibbs}
\end{eqnarray}
has a single minimum, at $b(\vec{\sigma})=p(\vec{\sigma})/Z$ (under the
normalization condition $\sum_{\vec{\sigma}} b(\vec{\sigma})=1$), and the
corresponding value of the functional ${\cal F}$ is the free energy, $-\ln Z$.  As
shown in \cite{YFW:constructing}, the Bethe free energy approximation and BP
equations stem from (\ref{Gibbs}) by considering an ansatz of the form
\begin{eqnarray}
 b(\vec{\sigma})\approx
 \frac{\prod_i b_i(\vec{\sigma}_i)\prod_j b^j(\vec{\sigma}^j)}{
 \prod_{(i,j)} b_i^j(\sigma_i^j)}\,.
\label{BP_Belief}
\end{eqnarray}
Vectors $\vec{\sigma}_i\equiv\{\sigma_i^j|j=1,\cdots,N\}$ and
$\vec{\sigma}^j\equiv\{\sigma_i^j|i=1,\cdots,N\}$ are allowed to take any of the $N$
possible values $(0,\cdots,0,1,0,\cdots,0)$ with exactly one nonzero entry. Beliefs
$b_i^j(\sigma_i^j),b_i(\vec{\sigma}_i),b^j(\vec{\sigma}^j)$ satisfy for any $i$ and
$j$ the consistency relation for marginal probabilities: $b_i^j(\sigma_i^j)=
\sum_{\vec{\sigma}_i\setminus\sigma_i^j}b_i(\vec{\sigma}_i)=
\sum_{\vec{\sigma}^j\setminus\sigma_i^j}b^j(\vec{\sigma}^j)$, where
$\sum_{\vec{\sigma}_i\setminus\sigma_i^j}$ denotes the sum over all possible values
of the vector $\vec{\sigma}_i$ keeping fixed the value of the component
$\sigma_i^j$.  Eq.~(\ref{BP_Belief}) is exact for a tree and serves as an
approximation for graphs with loops, e.g. for the fully connected bi-partite graph
of our matching problem. Beliefs, as approximations for probabilities, should also
satisfy the normalization conditions: $ \forall (i,j), \quad b_i^j(1)+b_i^j(0)=1$.
Using the normalization and consistency conditions, we can express all beliefs via
$\beta_i^j\equiv b_i^j(1)$ and obtain for the Bethe free energy and the
normalization conditions
\begin{eqnarray}
  && {\cal F}_{BP}\{\beta\}=\sum_{(i,j)}\left(\beta_i^j\ln
\frac{\beta_i^j}{p_i^j}-
      (1-\beta_i^j)\ln(1-\beta_i^j)\right),\label{FE}\\
  && \forall i:\quad \sum_j \beta_i^j=1;\qquad \forall j:\quad \sum_i
  \beta_i^j=1. \label{short_cond}
\end{eqnarray}

A simple argument shows that constrained minima of (\ref{FE}) are
either a perfect matching (beliefs are all zeros and $N$ of them are
unity) or they are attained in the interior of the domain. The latter is the
case generally encountered in the situations of interest to us, i.e.
where no statistically dominant matching is present. Minima in the
interior are stationary points of ${\cal F}_{BP}$,
corresponding to the following set of equations:
\begin{equation}
 \forall (i,j):\quad
 \beta_i^j(1-\beta_i^j)=p_i^j\exp\left(\mu_i+\mu^j\right),\label{BP1}
\end{equation}
where the $2N$ Lagrangian multipliers $\mu$ (chemical potentials) are determined by
Eqs.~(\ref{short_cond}). Note that the Bethe free energy is not convex, and thus
multiple minima in the interior of the domain might be possible. Empirically, we
never found more than one though. In the limit where
only the Maximum Likelihood configuration is of interest, entropy terms are
discarded and (\ref{FE}) reduces to Linear Programming, yielding optimal integer
solution in accordance with \cite{08BSS,08Che}. Another relevant remark is that
convexity is restored for a modified expression of the free energy, where the minus
sign of the second term in Eq.~(\ref{FE}) is reversed. The latter expression follows
from an integral representation for $Z$, approximated in a saddle-point way. This
approximation overestimates the diffusion coefficient and we use its unique solution
(easy to find numerically) as initial condition to the following iterative version
of Eqs.~(\ref{short_cond},\ref{BP1}):
\begin{eqnarray}
 && \hspace{-1cm}\forall (i,j):\ \ \beta_i^j(n\!+\!1)\!=\!\lambda\beta_i^j(n)\!+\!
 \frac{(1-\lambda)p_i^j}{p_i^j+(\sum_k\beta_k^j(n)/2+\sum_k
\beta_i^k(n)/2-\beta_i^j(n))^2/(u_i(n)v^j(n))},
 \label{beta_n}\\
 && \forall i:\ \
 u_i(n+1)=\frac{1-\sum_j(\beta_i^j(n))^2}{\sum_k p_i^k v^k(n)},\quad
\forall j:\ \ v^j(n+1)=\frac{1-\sum_i(\beta_i^j(n))^2}{\sum_k p_k^j u_i(n)},
 \label{Vn}
\end{eqnarray}
where the arguments of the $\beta$'s indicate the order of the iterations,
$u_i=\exp(\mu_i)$ and $v^j=\exp(\mu^j)$.  The damping parameter $\lambda$ (typically
chosen $0.4\div0.5$) helps with convergence. To ensure appropriate accuracy for
solutions with $\beta$'s close to zero or unity we also insert a normalization step
after Eqs.~(\ref{beta_n}) but prior to Eqs.~(\ref{Vn}), making the following two
transformations consequently, (a) $\forall (i,j)$: $\beta_i^j\to
\beta_i^j/\sum_k\beta_i^k$, and (b) $\forall (i,j)$: $\beta_i^j\to
\beta_i^j/\sum_k\beta_k^j$.  Numerical experiments show that this procedure
converges to a stationary point of the Bethe free energy (\ref{FE}).


TITLE: Does an extreme electric/magnetic field dilate time rates for a charged/magnetized clock?
QUESTION [1 upvotes]: I want to know if general relativity (or any defined metric in general relativity) predicts any change for the clock rates in the electromagnetic fields (potentials) for some charged or magnetic clocks?

REPLY [1 votes]: The Reissner–Nordström (RN) metric described the spacetime outside a charged and non-rotating spherical source. It has a static EM field, and an EM field has a stress energy tensor, so the metric is not a vacuum metric and hence it differs from the Schwarzschild spacetime without violating Birkhoff’s theorem.
In the RN spacetime the metric is different from the standard Schwarzschild spacetime. In particular, to your question it has different gravitational time dilation. $$\gamma=\frac{dt}{d\tau}=\left( 1- \frac{R_S}{r}+\frac{R_Q^2}{r^2}\right)^{-1/2}$$ where $R_S$ is a parameter that parameterizes the curvature from the mass and $R_Q$ parameterizes the curvature from the charge.
As you see, this differs from the standard Schwarzschild gravitational time dilation. The resulting time dilation applies to all clocks of any construction.
A clock that is also charged would have a force acting on it. Depending on the design of the clock, that force may distort the structure of the clock or otherwise result in a reading that deviates from the above equation. If so, that would not be described as additional time dilation, but as a deviation of such clocks from ideal behavior. In other words, such a clock would no longer be recording proper time.


TITLE: Upper Half "Plane" not diffeomorphic to the whole plane
QUESTION [1 upvotes]: maybe the title isnt the best but I have the following excercise, lets define
$$\cal{H}^n = [0,\infty)^n$$
I have to proove that an open set in $\mathcal{H}^n$  in general is not diffeomorphic to an open set in $\mathbb{R}^n$, I understand the idea of the proof but I don't know how to write it.
My attempt of a solution:
I have the following theorem: Given an open set $U$ in $\mathbb{R}^n$ and an arbitrary set $V$ in $\mathbb{R}^n$, if there is a diffeomorphism $f:U\rightarrow V$ then $V$ is open.
if $V$ is a subset of $(0,\infty)^n$ there is no problem because $V$ will be open in $\mathbb{R}^n$. However, if $V$ intersects one of the axis then $V$ wont be open in $\mathbb{R}^n$ and we get a contradiction.
Is this right? is there a better way to write it?

REPLY [1 votes]: So you have to pick a point on $V$ that has one of its coordinates zero.  Now if there is a diffeomorphism from that point to an open set in $\mathbb R^n$ - well, you know that map has a derivative, and that is an $n\times n$ matrix that is invertible.  So locally it is almost linear, and so is its inverse.  But a map that is almost linear cannot map an open set in $\mathbb R^n$ to the point we started with.
Let's make it more rigorous.  So let $f:U \to V$ be the purported map, and suppose $f(x_0) = y_0$ where $y_0$ has one of its coordinates equal to zero.  Let $L$ be the derivative of $f$ at $x_0$.  By applying $L$ to $U$, without loss of generality we can assume $L$ is the identity map.  Now there exists $\epsilon>0$ such that $f(x) = y_0 + (x-x_0) + g(x-x_0)$ where $|g(x-x_0)| \le |x-x_0|/10$ if $|x-x_0|<\epsilon$.  (This follows from the definition of being differentiable.)
So now consider $x_0$ with one of the coordinates having plus or minus $\epsilon/2$ added to it.  One of those $2n$ points must map to a point outside of $\mathcal H^n$.