TITLE: dual cone is closed
QUESTION [4 upvotes]: I have been trying to understand why dual of a cone is closed, no matter the cone is closed or not.
I know the proof is 'It is because dual cone is an intersection of closed halfspaces.'. I just do not understand how it is linked to the definition of the dual cone.

REPLY [6 votes]: Dua cone of $C$ is defined to be
$$C^*=\left\{ y \in X^*: \langle y,x\rangle \geq 0 , \forall x \in C  \right\}$$
For each $x$, $\left\{y :\langle y,x \rangle \geq 0\right\}$ is a closed half space. Hence $C^*$ is an intersection of closed halfspaces.


text:
/-
Copyright (c) 2019 Johannes Hölzl. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Johannes Hölzl, Mario Carneiro
-/

import data.rat.order
import data.int.sqrt
/-!
# Square root on rational numbers

This file defines the square root function on rational numbers `rat.sqrt`
and proves several theorems about it.

-/
namespace rat

/-- Square root function on rational numbers, defined by taking the (integer) square root of the
numerator and the square root (on natural numbers) of the denominator. -/
@[pp_nodot] def sqrt (q : ℚ) : ℚ := rat.mk (int.sqrt q.num) (nat.sqrt q.denom)

theorem sqrt_eq (q : ℚ) : rat.sqrt (q*q) = |q| :=
by rw [sqrt, mul_self_num, mul_self_denom, int.sqrt_eq, nat.sqrt_eq, abs_def]

theorem exists_mul_self (x : ℚ) : (∃ q, q * q = x) ↔ rat.sqrt x * rat.sqrt x = x :=
⟨λ ⟨n, hn⟩, by rw [← hn, sqrt_eq, abs_mul_abs_self],
λ h, ⟨rat.sqrt x, h⟩⟩

theorem sqrt_nonneg (q : ℚ) : 0 ≤ rat.sqrt q :=
nonneg_iff_zero_le.1 $ (mk_nonneg _ $ int.coe_nat_pos.2 $
nat.pos_of_ne_zero $ λ H, pos_iff_ne_zero.1 q.pos $ nat.sqrt_eq_zero.1 H).2
$ int.coe_nat_nonneg _

end rat

subset_name: curated


TITLE: Question about proving with Mathematical Induction (some confusions on the concept)
QUESTION [1 upvotes]: While proving a statement of $f(n)$ using mathematical induction we do the following-

we prove it for some natural number which satisfies the condition of $n$.
We assume it true for some $k$.
Then we try to prove it for $k+1$ by using the statement obtained by assuming it to be true for $k$.

Now,my question is that we see that the proof using $k+1$ is totally dependent on proof using $k$.But there is no guarantee that the statement is true for k.We have only assumed it.Our assumption maybe right or wrong.The how can be prove $k+1$ on the basis of k if it is not certain that k is true.
Please help.Thank You.

REPLY [1 votes]: Let' clarify the concept on a specific example.
We know that $$1+2+\cdots +n=\frac{n(n+1)}2.\tag 1$$
Prove this statement a fake naive way:

Observe that $(1)$ is true for $n=2$. Indeed, $$3=1+2=\frac {2\times 3}2.\tag 2$$
Observe that you can prove the same for $n=3$ based on $(2)$ now:  $$6=1+2+3=\frac{2\times3}2+3=\frac{2\times3+2\times3}2=\frac{3\times 4}{2}.\tag 3$$
You do it once again for $1+2+3+4$ based on $(3)$:
$$1+2+3+4=(1+2+3)+4=$$$$=\frac{3\times 4}2+4=\frac{3\times4+2\times 4}2=\frac{4\times5}2.$$
You notice the pattern now and try to do it in general:

$$1+2+\cdots n+n+1=\frac{n(n+1)}2+n+1=$$
$$=\frac{n(n+1)+2(n+1)}2=\frac{(n+1)(n+2)}2.$$
You generalize the idea now: If I could do it for $2$ and if I could do it for $n+1$ knowing that it is true for $n$ then it will be true for any natural number.
One more step ahead and you say that if this principle worked for the property $$1+2+\cdots +n=\frac{n(n+1)}{2}$$ then it will work for any property $P(n)$.
We've seen that every proof based on the principle of mathematical induction could be done step by step for any $n$. In that case you do not assume that the statement is true for an $n$ but you directly prove that. So, mathematical induction is a practical simplification of a special kind of inductive proof as long as we talk about the truth of a statement for finitely many $n$'s. The principle of mathematical induction becomes an unproven axiom if it comes to the statement that something is true for all $n$.


TITLE: Linear isomorphism determined by parallel transport
QUESTION [2 upvotes]: Suppose $ \nabla $ is an affine connection on a differentiable manifold $ M $, $
p, q\in M $, $ c $ is a smooth curve connecting $ p $ and $ q $, then the parallel transport along $ c $ gives us a linear isomorphism $$ P^{c}:T_{p}M\longrightarrow T_{q}M ,$$ and $ P^c $ is uniquely determined by $ \nabla $.

I read this paragraph from my textbook. What confused me is what exactly $ P^c $ looks like such that $ P^c $ is a linear isomorphism?
Edit: I want the explicit expression for $ P^c $.

REPLY [1 votes]: Let us take local coordinates at $p$ denoted by $x_1,\ldots,x_n$. Let $v\in T_pM$, and write $v=v^i\partial/\partial x^i$. Let $\Gamma_{ij}^k$ be the Christoffel symbols of $\nabla$ in these coordinates. Let $\gamma=(\gamma_1,\ldots,\gamma_n)$ be a smooth path starting at $p$. A vector field $X$ along $\gamma$ can be written as $$X(t)=X^i(t)\frac{\partial}{\partial x^i}.$$ Such a vector field is parallel with respect to $\nabla$ if it solves the equation
$$\begin{align}0&=\frac{DX}{dt}\\&=\dot{X}^k\frac{\partial}{\partial x^k}+\dot{\gamma}^iX^j\Gamma_{ij}^k\frac{\partial}{\partial x^k}\\&=\left(\dot{X}^k+\dot{\gamma}^iX^j\Gamma_{ij}^k\right)\frac{\partial}{\partial x^k}\;.\end{align}$$
Moreover, it is the parallel transport of $v$ if it satisfies, in addition, the initial value condition $X^i(0)=v^i$.
In general, it is practically impossible to write an explicit expression for parallel transport, as it is a solution to an ODE, and the coefficients may be selected arbitrarily. However, as demonstrated above, this is a linear ODE of the first order. Hence, the solution space is linear and defined for infinite time. In other words, parallel transport defines a linear isomorphism between the tangent spaces at the two endpoints of a path.


TITLE: Repeating something with (1/n)th chance of success n times
QUESTION [7 upvotes]: Is there anything that can be said about how many attempts it will take to correctly guess a random number out of 1000 numbers? If the number wouldn't change the probability would just increase every guess, giving in the end a probability of 1.
If the number changes every guess tough, you just have a 1/1000th chance everytime. Can it be calculated how many attempts it would probably take, on average?
(I'm sorry, i'm not very skilled in probability maths)

REPLY [8 votes]: This concept is the foundation of the Law of Large Numbers, which is behind most concepts of probability. By definition, if given 1000 unique random trials, a particular event (result) occurs once, then the observed probability of that event occurring is 1/1000. The inverse is basically the application of the calculated probability; if an event has a probability of 1/1000, then if you performed 1000 trials, you should expect the event to occur once. For an independent event (the results of past trials do not influence the probability of future ones), it's not guaranteed to happen exactly once per 1000 trials, but as the number of trials t grows large, the number of times an event with probability $1/n$ occurs will approach $t/n$.
Now, if you started with the first trial, and continued performing trials until the event occurred, and then counted how many trials it took, the probability of it occurring for the first time on exactly the Xth trial is equal to the probability that it doesn't occur on any of the X-1 trials, and that it does occur on the Xth. The probability that it doesn't happen is $\dfrac{n-1}{n}$ (or equivalently $1-\dfrac{1}{n})$ and it must happen X-1 times, followed by the event occurring the Xth time ($1/n$ probability), so the formula for the probability of it occurring on exactly the Xth trial is
$$P(X)=(1-\dfrac{1}{n})^{X-1}(\dfrac{1}{n})$$
Plugging in our known specific value for N, we get:
$$P(X)=(\dfrac{999}{1000})^{X-1}(\dfrac{1}{1000})$$
This is an asymptotically-reducing exponential function; it approaches zero as X grows infinitely large. As such, there isn't a point at which it starts becoming more likely to happen; it just becomes increasingly less likely that it doesn't happen. However, if we consider only how likely it is that it doesn't happen in X-1 trials ($(\dfrac{999}{1000})^{X-1}$), we can find an "over-under", where it is equally likely for it to take more than that many trials than fewer. This is our "expected value" for the number of trials it should take (similar to the "expected value" of one roll of the dice; the probability that it's greater than or equal to a certain face value is 1 at 1, but only 1/6 at 6, and at the expected value of 3.5, the odds are 50-50 either way). With a little graphing magic, we see that the number of trials after which the probability of your number having not occurred so far drops below .5 is 674.
Thus, before you begin choosing numbers, you can expect it to take, on average, about 674 trials for the number you want to pop up for the first time. However, there's still a 36% chance it won't happen in 1000 trials, and a 13% chance it won't even happen once in 2000 trials, and given that it hasn't happened in 674 trials (or 1000 or even 100,000), the probability the next number will be the one you want is still 1 in 1000.


text:
\begin{document}

\noindent
\begin{picture}(150,36)
\put(5,20){\tiny{Submitted to}}
\put(5,7){\textbf{Topology Proceedings}}
\put(0,0){\framebox(140,34){}}
\put(2,2){\framebox(136,30){}}
\end{picture}

\vspace{0.5in}

\renewcommand{\bf}{\bfseries}
\renewcommand{\sc}{\scshape}

\vspace{0.5in}

\date{June 23, 2018; revised August 22, 2018}

\title{Forcing among patterns with no block structure}

\author[A.~Blokh]{Alexander~Blokh}

\address[Alexander~Blokh]
{Department of Mathematics\\ University of Alabama at Birmingham\\
Birmingham, AL 35294\\ USA}
\email{ablokh@math.uab.edu}

\author[M.~Misiurewicz]{Micha{\l}~Misiurewicz}

\address[Micha{\l}~Misiurewicz]
{Department of Mathematical Sciences\\ Indiana University-Pur\-due
University Indianapolis\\ 402 N. Bla\-ck\-ford Street\\
Indianapolis, IN 46202\\ USA}
\email{mmisiure@math.iupui.edu}

\subjclass[2010]{Primary 37E15; Secondary 37E05, 37E45}

\keywords{Sharkovsky order; forcing relation; cyclic patterns}

\dedicatory{Dedicated to the memory of our colleague and dear friend
Sergiy Kolyada}

\thanks{Research of Micha{\l} Misiurewicz was partially
supported by grant number 426602 from the Simons Foundation.}

\thanks{Research by Alexander Blokh was partially
supported by NSF grant DMS--1201450}

\thanks{Both authors would like to thank the organizers of the 52nd
  Spring Topology and Dynamical Systems Conference at Auburn
  University, during which this paper was discussed.}

\begin{abstract}
Define the following order among all natural numbers except for 2 and
1:
\[
4\gg 6\gg 3\gg \dots \gg 4n\gg 4n+2\gg 2n+1\gg 4n+4\gg\dots
\]
Let $f$ be a continuous interval map. We show that if $m\gg s$ and $f$
has a cycle with no division (no block structure) of period $m$ then
$f$ has also a cycle with no division (no block structure) of period
$s$. We describe possible sets of periods of cycles of $f$ with no
division and no block structure.
\end{abstract}

\maketitle

\section{Introduction and statement of the results}\label{s:intro}

The simplest type of limit behavior of a trajectory is periodic;
studying \emph{periodic orbits} (\emph{cycles}) is one of the central
topics in one-dimension\-al dynamics. To some extent this can be
explained by a remarkable result, the Sharkovsky Theorem, proved by
A.~N.~Sharkovsky in the 1960s (see \cite{sha64} and~\cite{shatr} for
its English translation). To state it, let us first recall the
\emph{Sharkovsky order} of the set $\N$ of positive integers:
\begin{multline*}
3\sha 5\sha 7\sha\dots\sha 2\cdot3\sha 2\cdot5\sha 2\cdot7\sha\\ \dots
\sha 2^2\cdot3\sha 2^2\cdot5\sha 2^2\cdot7\sha\dots\sha 2^2\sha 2\sha 1.
\end{multline*}
Denote by $\Sh(k)$ the set of all integers $m$ such that $k\sha m$ or
$m=k$, and by $\Sh(2^\infty)$ the set $\{1,2,4,8,\dots\}$; denote by
$\Per(f)$ the set of periods of cycles of a map $f$ (by
the \emph{period} we mean the \emph{minimal} period). Below $I$ always
denotes a closed interval.

\begin{thmS}
If $g:I\to I$ is continuous, $m\sha n$ and $m\in\Per(g)$ then
$n\in\Per(g)$ and there exists $k\in\N\cup\{2^\infty\}$ with
$\Per(g)=\Sh(k)$. Conversely, if $k\in\N\cup\{2^\infty\}$ then there
exists a continuous map $f:I\to I$ with $\Per(f) =\Sh(k)$.
\end{thmS}

The Sharkovsky Theorem is important, in particular, because it
introduces a concept of \emph{forcing relation}: it states that if
$m\sha n$ then the fact that an interval map has a cycle of period $m$
\emph{forces} the presence of a cycle of period $n$.
Thus, it shows how various ``types'' of cycles (here by
``type'' one means ``period'') force each other. Another interpretation
of the Sharkovsky Theorem is that it fully describes all possible sets
of periods of cycles of interval maps. This leads to similar problems:
(a)~how the existence of cycles of certain types forces the existence
of cycles of certain other types, and (b) what possible sets of types
of cycles an interval map may have.

For example, given a cycle $P=\{x_1<x_2<\dots <x_n\}$ of an interval
map $f$, associate with it the (cyclic) permutation $\pi$ defined by
$f(x_i)=x_{\pi(i)}$, $i=1, 2, \dots, n$. Think of $\pi$ as the type of
$P$. The family of all cycles associated to $\pi$ is called an
\emph{oriented pattern} (see~\cite{alm}). If we identify oriented
patterns obtained from each other by a flip,
we get \emph{patterns} (we denote patterns with capital letters
$A,B,\dots$). Similar to the Sharkovsky Theorem, one can ask for an
interval map $f$ (a) how cycles of certain patterns force cycles of
other patterns, and (b) what possible sets of patterns of cycles $f$
may have.

A useful way of studying patterns is by decomposing them.

\begin{dfn}[Block structure]\label{d:no-bs}
Let $\pi$ be a permutation of the set $X=\{1, \dots, n\}$. Suppose that
for some $k>1$ and $m>1$ we have $n=km$ and the permutation $\pi$ maps
sets $Y_1=\{1, \dots, m\},$ $Y_2=\{m+1, \dots, 2m\},$ $\dots$,
$Y_k=\{n-m+1, \dots, n\}$ to one another. Then sets $Y_1, \dots, Y_k$
are called \emph{blocks} and the permutation $\pi$ is said to have
\emph{block structure}; if blocks are two-point sets, $\pi$ is said to
be a \emph{doubling}. As always, similar terminology is used for
patterns and cycles. Otherwise a permutation (a pattern, a cycle) is
said to have \emph{no block structure}.
\end{dfn}

The appropriate power of the map on a block can be viewed as a kind of
renormalization of a pattern; patterns with block structure admit a
renormalization like that. Consider an important particular case.

\begin{dfn}[No division]\label{d:no-div}
Let $\pi$ be a permutation of the set $\{1, \dots, 2m\}$ such that $\pi(i)\ge
m+1$ for each $i, 1\le i\le m$ (and, hence, $\pi(i)\le m$ for each
$i\ge m+1$). Then we say that $\pi$ (and the corresponding pattern and
cycles) \emph{has division}. Otherwise $\pi$ (and the corresponding
pattern and cycles) is said to have \emph{no division}.
\end{dfn}

Observe that a pattern of period 2 has no block structure but has a
division. Therefore we will treat period 2 separately.

Consider the family  $\nbs$ of patterns with no block structure and the
family $\nd$ of all patterns with no division. A pattern \emph{with}
block structure can be studied in two steps: study the factor-pattern
obtained if each block is collapsed to a point while the order among
blocks is kept, and then study the restriction of the pattern on
blocks. On the other hand, no division patterns constitute the ``core''
in the Sharkovsky Theorem. Thus, both patterns from $\nbs$ and $\nd$ are
important. To get uniformity, we consider only patterns of periods
larger than 2; patterns of periods 1 and 2 are discussed after the Main
Theorem.

Define the following order among all natural numbers larger than 2:
\begin{equation}\label{star}
4\gg 6\gg 3\gg \dots \gg 4n\gg 4n+2\gg 2n+1\gg 4n+4\gg\dots\tag{$*$}
\end{equation}
We get it by writing  even numbers in the natural order and
inserting odd numbers $n$ after $2n$. We regard $\gg$ as a strict
ordering (that is, it is not reflexive).

Let $\n_r$ be the set of all integers $s$ with $r\gg s$ and $r$ itself.
Given an interval map $f$, let $ND(f)$ be the set of periods (larger
than 2) of all $f$-cycles with no division, and let $NBS(f)$ be the set
of periods (larger than 2) of all $f$-cycles with no block structure.

\begin{thmM}
Let $f$ be a continuous interval map. If $m\gg s$ and
$f$ has a cycle with no division (no block structure) of period $m$
then $f$ has also a cycle with no division (no block structure) of
period $s$. The following are the only possible cases, and all of them
occur.
\begin{enumerate}
\item $ND(f)=NBS(f)=\emptyset$.
\item $ND(f)=NBS(f)=\n_r$, $r\ge 3$.
\item $ND(f)=\n_{4n+2}, NBS(f)=\n_{2n+1}$, $n\ge 1$.
\end{enumerate}
\end{thmM}

Complementing this theorem, we get additional information about the
structure of cycles if $NBS(f)=\n_{2n+1}$ (see
Proposition~\ref{stefan-only} and Remark~\ref{so}).

\begin{remark}\label{onetwo}
Consider patterns of period 1 and 2. A continuous interval map always
has a fixed point, so 1 should stand at the end of the
order~\eqref{star} both for both types of patterns. The situation with
2 is more complicated. Namely, there is only one pattern of period 2,
and it has a division, but not a block structure. Thus, for no division
patterns, 2 does not occur in the order. For no block structure
patterns, 2 should stand just before 1 because, by the Sharkovsky
Theorem, if $f$ has a cycle of period larger than 1, it has also a
cycle of period 2.
\end{remark}

\begin{remark}\label{refrem}
In view of the second part of Main Theorem, the first part can be
stated in a slightly stronger fashion. Namely, let $f$ be a continuous
interval map and $m,s\ge 3$. Suppose that $f$ has a cycle of period
$m$ with \emph{no division} and either (i)~$m\gg s$, or (ii)~$m=s$ and $m$ is
not of the form $4n+2$. Then $f$ has a cycle of period $s$ with
\emph{no block structure}.
\end{remark}

\begin{remark}\label{triod}
The order~\eqref{star} is
similar to the orders present for the continuous triod map
(see~\cite{almy}) and given by $5,8,4,11,$ $14,7,\dots$ and
$7,10,5,13,16,8,\dots.$ This makes interesting connections and allows
us to look at an interval as a ``diod.''
\end{remark}

\begin{remark}\label{minor}
In~\cite{mis94} it was proved that (a) patterns from $\nbs$ of period
$2n+1$ force patterns from $\nbs$ of period $4n+4$, and (b) patterns
from $\nbs$ of period $4n$ force patterns from $\nbs$ of period $4n+2$.
However our proofs here
are much simpler (because they use the \emph{rotation theory}). The
fact that patterns with no block structure of period $4n+2$ force
patterns with no block structure of period $2n+1$ is new; the order
~\eqref{star} was mentioned in~\cite{mis94} only for unimodal maps.
Finally, in our Main Theorem we take into account not only patterns
with no block structure but also patterns with no division.
\end{remark}

\noindent \textbf{Acknowledgments.} The authors are indebted to the referee for
useful remarks.

\section{Preliminaries}\label{s:prel}

We will be using standard tools of combinatorial dynamics. The reader
that is not acquainted with them can find details for instance
in~\cite{alm}, \cite{bc} or~\cite{mini}.

In particular, we will consider forcing among patterns. It is a
partial ordering on patterns (see~\cite{bal87}). Given a
pattern $A$ we will often consider a cycle $P$ and the $P$-linear
(``connect the dots'') map $f$. Patterns forced by $A$ are then
exactly the patterns of cycles of $f$. They can be found by looking at
the Markov graph of $(f,P)$, where vertices are the $P$-basic
intervals (intervals between consecutive points of $P$) and arrows
correspond to $f$-covering (there is an arrow from $J$ to $K$ if
$K\subset f(J)$). The loops in this graph correspond to cycles of $f$
(and, hence, they determine which patterns are forced by $A$).

We will also use extensively rotation theory for interval maps. Since
it is less known, we will present its basic notions and results
(see~\cite{blo95, bm97, bm97a, bm99, bs13}). We also prove some simple
lemmas that will be necessary later.

Let $f:I\to I$ be a continuous map with a cycle $P$ of period $q>1$. Let $m$
be the number of points $x\in P$ with $f(x)-x$ and $f^2(x)-f(x)$ of
different signs. Then the pair $(m/2,q)$ is called the
\emph{over-rotation pair} of $P$ and is denoted by $\orp(P)$; the
number $m/(2q)$ is called the \emph{over-rotation number} of the cycle
$P$ and is denoted by $\rho(P)$. The set of the over-rotation pairs of
all cycles of $f$ is denoted by $\ORP(f)$. Note that the number $m$
above is even, positive, and does not exceed $q/2$. Therefore in an
over-rotation pair $(p,q)$ both $p$ and $q$ are integers and $0<p/q\le
1/2$. We call over-rotation pairs $(p, q)$
\emph{coprime} if $p$ and $q$ are coprime. Clearly, we can speak
of over-rotation pairs and over-rotation numbers of patterns and
permutations.

\begin{dfn}\label{d:orp-order}
We write $(p,q)\gtrdot (r,s)$ if $p/q<r/s$, or $p/q=r/s=m/n$ with $m$
and $n$ coprime and $p/m\sha r/m$ (clearly, $p/m,r/m\in\mathbb N$).
\end{dfn}

The next lemma relates the fact that a pattern has a block structure to
the properties of the pattern's over-rotation pair.

\begin{lem}\label{bsdiv}
If a cycle $P$ with the over-rotation pair $(k,m)$ has block structure
with $q$ points in every block, then $q$ divides both $k$ and $m$. In
particular, if $k$ and $m$ are coprime then $P$ has no block structure.
\end{lem}

\begin{proof}
Clearly, $q$ divides $m$. To see that $q$ divides $k$, observe that if
we identify each block to a point to get a cycle $Q$ of period $m/q$,
the over-rotation number of $Q$ will be the same as for $P$, i.e.,
$k/m$. If $\orp(Q)=(k',m')$, then $k'/m'=k/m$ and $m'=m/q$, so $k=k'q$.
However, $k'$ is an integer, so $q$ divides $k$.
\end{proof}

\begin{dfn}\label{d:ovr}
Let $\mathbb M$ be the set consisting of 0, all irrational numbers
between 0 and $1/2$, and all pairs $(\alpha,n)$, where $\alpha$ is a
rational number from $(0,1/2]$ and $n\in\mathbb N\cup\{2^\infty\}$.
Then for $\eta\in\mathbb M$ the set $\Ovr(\eta)$ is equal to the
following.
\begin{enumerate}
\item If $\eta$ is an irrational number or 0, then $\Ovr(\eta)$ is the
  set of all integer pairs $(p,q)$ with $\eta<p/q\le 1/2$.
\item If $\eta=(r/s,n)$ with r,s coprime, then $\Ovr(\eta)$ is the
  union of the set of all integer pairs $(p,q)$ with $r/s<p/q\le 1/2$
  and the set of all integer pairs $(mr,ms)$ with $m\in \Sh(n)$.
\end{enumerate}
\end{dfn}

In case (2) of Definition~\ref{d:ovr} if $n\ne 2^\infty$ then
$\Ovr(\eta)$ is the set of all over-rotation pairs $(p,q)$ with
$(nr,ns)\gtrdot(p,q)$, plus $(nr,ns)$ itself.

\begin{thm}[Theorem 3.1 of \cite{bm97}]\label{t:new-order}
If $f:[0,1]\to [0,1]$ is continuous, $(p,q)\gtrdot(r,s),$ and
$(p,q)\in\ORP(f),$ then $(r,s)\in\ORP(f)$. Thus, $\ORP(f)=\Ovr(\eta)$
for some $\eta\in\mathbb M$. Conversely, if $\eta\in\mathbb M$ then
there exists a continuous map $f:[0,1]\to [0,1]$ such that
$\ORP(f)=\Ovr(\eta)$.
\end{thm}

For some patterns automatically we get cycles of all periods.

\begin{dfn}[Convergent/divergent patterns]\label{d:c-d}
A pattern (cycle) of period $n$ is \emph{convergent} if for the
corresponding permutation $\pi$ there is a number $m<n$ such that
$\pi(i)>i$ for $i\le m$ and $\pi(i)<i$ for $i>m$; otherwise a pattern
(cycle) is \emph{divergent}.
\end{dfn}

Observe that $P$ is a convergent cycle of a $P$-linear map $f$ if and
only if $f$ has only one fixed point.

\begin{lem}\label{divergent}
Any divergent pattern forces a pattern with no block structure of
period $n$ for every $n>1$. Moreover, if $f$ is an interval map with a
periodic orbit of divergent pattern then $\ORP(f)=\Ovr(0)$.
\end{lem}

\begin{proof}
Let $P$ be a cycle of divergent pattern $A$. By Lemma~3.2 of~\cite{bm97},
if $f$ is an interval map with a cycle of divergent pattern
then $\ORP(f)=\Ovr(0)$, which is exactly the second claim of the lemma.
It follows that $f$ has cycles of over-rotation pair $(1, n)$ for
every $n$. By Lemma~\ref{bsdiv} these cycles have no block structure.
Considering a $P$-linear map $f$
we see that $A$ forces patterns with no block structure of any period
$n$ as desired.
\end{proof}

{}From now on we consider only convergent patterns. Then we can use an
alternative way of computing over-rotation pairs. Let $P$ be a cycle of
a convergent pattern $A$ with $\orp(P)=(p, q)$. We will always denote
by $a_P=a$ the fixed point of the $P$-linear map $f$ (we may omit the
subscript $P$ if no ambiguity is possible). Then
$(x-f(x))(f(x)-f^2(x))<0$
if and only if $x$ is mapped to the other side of $a$ under $f$. Thus,
$p$ equals the number of times when a point in $P$ is mapped from the left
of $a$ to the right of $a$ (alternatively, from the right of $a$ to the
left of $a$); $p$ can also be computed if we count the number of times
in $P$ when a points maps from one side of $a$ to the other side of
$a$, and divide this number by $2$. We can think of $p$ as a
cumulative rotation of $P$ about $a$. This interpretation
helps, in particular, in the proof of the next lemma.

\begin{lem}\label{l:rot-no-div}
If $A$ is a convergent pattern, $\rho(A)=1/2$ if and only if
$P$ has division.
\end{lem}

\begin{proof}
If $A$ has division then $\rho(A)=1/2$. Now, if $\rho(A)=1/2$, then
$\orp(A)=(n, 2n)$ for some $n$. Let $P$ be a cycle of pattern $A$, and let
$f$ be a $P$-linear map. Then $P$ has $2n$ points and all of them are
mapped from one side of $a$ to the other side. Therefore, $P$ has a
division.
\end{proof}

Another concept related to Theorem~\ref{t:new-order} is that of a
\emph{twist pattern}.

\begin{dfn}[Twist patterns]\label{d:twist}
A pattern of over-rotation number $\rho$ is \emph{twist} if
it does not force any other pattern of over-rotation number $\rho$; we
use the same terminology for cycles and permutations.
\end{dfn}

By Lemma~\ref{divergent} a twist cycle must be convergent. In
particular, if $P$ is a twist cycle then the $P$-linear map has a
unique fixed point.

\begin{lem}[\cite{bm97a, bm99}]\label{l:twist}
Let $P$ be a twist cycle $P$ of the $P$-linear map $f$. Then, if
points $u, v\in P$ lie on the same side of $a$, map to the same side
of $a$, and $u$ is farther away from $a$ than $v$, then $f(u)$ is
farther away from $a$ than $f(v)$.
\end{lem}

\section{Proof of Main Theorem}\label{s:main}

We start by recalling the definition of a well known \v Stefan pattern.

\begin{dfn}[\v Stefan pattern]\label{d:stefan}
Consider the cyclic permutation $\sigma:\{1, 2, \dots, 2n+1\}\to \{1,
2, \dots, 2n+1\}$ ($n\ge 1$), defined as follows:
\begin{itemize}
\item $\sigma(1)=n+1$;
\item $\sigma(i)=2n+3-i$,\ \ if\ \ $2\le i\le n+1$;
\item $\sigma(i)=2n+2-i$,\ \ if\ \ $n+2\le i\le 2n+2$.
\end{itemize}
Then the pattern of this cyclic permutation is called the \emph{\v
  Stefan pattern}, and any cycle of this pattern is said to be a
\emph{\v Stefan cycle}.
\end{dfn}

The importance of those patterns is due to the following fact.

\begin{thm}[\cite{ste77}]\label{t:stefan}
Any pattern of period $2n+1$ forces the \v Stefan pattern of period
$2n+1$. Moreover, if a continuous interval map $f$ has a cycle of
period $2n+1$ and no cycles of period $2k+1$ with $1\le k<n$, then
every cycle of $f$ of period $2n+1$ is \v Stefan.
\end{thm}

Now we prove some preliminary results. If $P$ is a cycle of
period $n>1$ then for each point $x\in P$ we consider
\emph{germs} at $x$, i.e., small intervals with $x$ as one of the
endpoints. Each point of $P$ has two germs, except the leftmost and
rightmost points, which have one germ each. There is a natural map
induced on the set of germs by the $P$-linear map $f$, and if we start
at the germ of the leftmost point (or the rightmost point), we get
back exactly after $n$ applications of this map. Each germ is
contained in a $P$-basic interval, so this loop of germs gives us a
loop of $P$-basic intervals. These loops are called the
\emph{fundamental loop of germs} and the \emph{fundamental loop of
intervals}. Both loops correspond to the original periodic orbit $P$
Thus, the fundamental loop of intervals contains both
the leftmost and the rightmost $P$-basic intervals. Observe that, by
Lemma~\ref{l:twist}, if $P$ is a twist cycle then any germ at $x\in P$
that points toward $a$ maps to the germ at $f(x)\in P$ that points
toward $a$ too. Thus, if $P$ is a twist cycle, then the vertices of the
fundamental loop of germs form the set of germs pointing toward $a$.

In what follows we use the following notation. Denote by $I=[b_l, b_r]$
the $P$-basic interval containing the point $a$. Observe that the arrow
$I\to I$ is a part of the Markov graph $G$ of $P$. It follows that $I$
is repeated in the fundamental loop of intervals of $P$ twice while all
other $P$-basic intervals are repeated there once. Consider the set
$P'=P\cup\{a\}$. Though the germs at points of $P$ stay the same
whether we consider $P$ or $P'$, there is a change concerning
$P'$-basic intervals versus $P$-basic intervals: $I$ is now replaced by
two $P'$-basic intervals, $I_l=[b_l, a]$ and $I_r=[a, b_r]$. Notice
that the arrows $I_l\to I_r$ and $I_r\to I_l$ are in the Markov graph
of $P'$. Clearly, a germ at a point of $P'$ is contained in a
well-defined $P'$-basic interval. Hence the fundamental loop of germs
of $P$ gives rise to the \emph{fundamental loop of $P'$-basic
  intervals} and the \emph{fundamental loop of germs of $P'$}. Thus,
we get the following lemma.

\begin{lem}\label{fl}
If $P$ is a twist cycle of period larger than $1$ then the fundamental
loop of intervals of $P'$ passes exactly once through every $P'$-basic
interval.
\end{lem}

Now we investigate twist cycles close to the fixed point.

\begin{lem}\label{closetoa}
If $P$ is a twist cycle of period $n>2$ of a $P$-linear map $f$ then
at least one of the points $b_l,b_r$ is the image of a point of
$P$ that lies on the same side of $a$.
\end{lem}

\begin{proof}
Suppose that $b_l=f_P(c_l)$, $b_r=f(c_r)$, where
$c_l,c_r\in P$, and $c_r\le b_l<b_r\le c_l$. Since $n>2$,
either $c_r<b_l$ or $b_r<c_l$. We may assume that
$c_r<b_l$. However, then $c_r<b_l<a<b_r=f(c_r)<f(b_l),$
which contradicts Lemma~\ref{l:twist}.
\end{proof}

Twist patterns force other patterns with specific properties.

\begin{proposition}\label{mplus2}
If $P$ is a twist cycle of the $P$-linear map $f$ and $P$ has
over-rotation pair $(k,m)$ and over-rotation number $\rho(P)<\frac12$,
then $f$ has a cycle of over-rotation pair $(k+1,m+2)$, which
is not a doubling.
\end{proposition}

\begin{proof}
By Lemma~\ref{closetoa}, we may assume that $b_l=f(c_l)$ for some
$c_l\in P$ with $c_l<b_l$. Let $L$ be the fundamental loop of
intervals of $P'$. By Lemma~\ref{fl} it passes through $I_l$ exactly
once, so we can insert into $L$ the two arrows, $I_l\to I_r\to
I_l$, at that place. Denote by $M$ the loop of length $m+2$ obtained
in this way, and by $Q$ a corresponding periodic orbit of $f$. By the
construction, each $P'$-basic interval contains one element of $Q$,
except $I_l$ and $I_r$, which contain two elements each. This, in
particular, shows that the period of $Q$ is $m+2$.

Let $x\in Q$ be the point that belongs to the $P'$-basic interval
whose left endpoint is $c_l$. Then $f(x)\in I_l$, $f^2(x)\in
I_r$, and $f^3(x)\in I_l$. Since the fixed point $a$ is
repelling (because the interval $[b_l, b_r]$ is mapped linearly
onto a larger interval), we get $x<f^3(x)<f(x)<a<f^2(x)$. The
other point of $Q$ which is in $I_r$, is to the right of $f^2(x)$,
because otherwise its image would be the third point of $Q$ in $I_l$
(and by the construction there are two points of $Q$ in $I_l$ and
two points of $Q$ in $I_r$).

If $Q$ is a doubling, then $f(x)$ is paired with $f^3(x)$ or
$f^2(x)$. The first option is impossible because if it holds then
the pair of points mapped to the pair $\{f^3(x), f(x)\}$ must be
the pair $\{x, f^2(x)\}$ and the points $x$ and $f^2(x)$ are not
consecutive in space. The second option is impossible because then the
image pair $\{f^2(x), f^3(x)\}$ consists of points that are not
consecutive in space. Thus, $Q$ is not a doubling.

Finally, since we added two points that are mapped onto the opposite
side of $a$, and the rest of the points of $Q$ are mapped like the
analogous points of $P$, the over-rotation pair of $Q$ is $(k+1,m+2)$.
\end{proof}

{}From Lemma~\ref{bsdiv} and Proposition~\ref{mplus2} we get the
following lemma.

\begin{lem}\label{mp2}
Any pattern $A$ with $\orp(A)=(ks,ms)$ where $k$ and $m$ are coprime, and
$\rho(A)=k/m<1/2$ forces a pattern of over-rotation
pair $(k+1,m+2)$, which is not a doubling. In particular, a pattern
$A$ of over-rotation pair $(2n-1, 4n)$ forces a pattern of
over-rotation pair $(2n,4n+2)$ which has no block structure.
\end{lem}

\begin{proof}
Let $A$ be a pattern with $\orp(A)=(ks,ms)$ where $k$ and $m$ are coprime, and
$\rho(A)=k/m<1/2$. By Theorem~\ref{t:new-order} $A$
forces a twist pattern $A'$ of over-rotation pair $(k, m)$.
By Lemma~\ref{mplus2}, $A'$ forces a pattern of over-rotation pair
$(k+1, m+2)$ as desired. Now, let the over-rotation pair of $A$ be
$(2n-1, 4n)$.
By the above $A$ forces a pattern $B$ of over-rotation pair $(2n,
4n+2)$ that is not a doubling. Let us show that $B$ has no block
structure. Indeed, the only common divisor of $2n$ and $4n+2$ is $2$.
Hence by Lemma~\ref{bsdiv} the only way $B$ can have block structure is
when $B$ is a doubling, a contradiction.
\end{proof}

In what follows we will use the notation below: for every $m>2$ set
\[
\eta(m)=\begin{cases}
(s-1,2s)&\mbox{if\ \ } m=2s,\\
(n,2n+1)&\mbox{if\ \ } m=2n+1,
\end{cases}
\]
In particular, $\eta(4n)=(2n-1, 4n)$ and $\eta(4n+2)=(2n, 4n+2)$.

\begin{proposition}\label{stefan-only}
Let $n\ge 1$; then the following claims hold.

\begin{enumerate}

\item Let $g$ be a continuous interval map. Assume that $g$ has a
    cycle of period $2n+1$ with no block structure, and all cycles
    of $g$ of periods $2k+1$ with $1\le k<n$ have block structure.
    Then all cycles of $g$ of period $2n+1$ are \v Stefan and $g$ has no cycles of periods $2k+1$ with $1\le k<n$.

\item Let $f$ be a continuous interval map. Assume that $f$ has a
cycle of period $4n+2$ and no division, but all cycles of $f$ of period
$4n+2$ have block structure. Then all cycles of $f$ of period
$4n+2$ and no division are doublings of a \v Stefan cycle.

\end{enumerate}

\end{proposition}

\begin{proof}
(1) Suppose that $g$ has a cycle of period $2n+1$ which is not \v Stefan.
Then, by Theorem~\ref{t:stefan}, $g$ has a \v Stefan cycle of period
$2k+1$ for some $k$ with $1\le k<n$. By inspection, \v Stefan patterns
have no block structure, so we get a contradiction. Moreover, the same argument
shows that $g$ does not have cycles of periods $2k+1$ with $1\le k<n$ at all.

(2) Let $P$ be a cycle of $f$ of period $4n+2$ and no
division. Then, by Lemma~\ref{l:rot-no-div}, the over-rotation number of $P$ is
less than $\frac12$. It follows that the over-rotation pair of $P$ must be
$(2n, 4n+2)$, as otherwise
it is at most $\frac{2n-1}{4n+2}<\frac{2n-1}{4n}$, so, by
Theorem~\ref{t:new-order} and Lemma~\ref{mp2}, $f$ would have a cycle
of period $4n+2$ with no block structure. Thus, since $P$ has block
structure and the greatest common divisor of $2n$ and $4n+2$ is
$2$, the cycle $P$ is a doubling over a cycle, say, $Q$ of period $2n+1$.
If $Q$ is not \v Stefan, then by~(1) there must exist a cycle of $f$ of period $2n-1$.
Since $\frac{n-1}{2n-1}<\frac{2n-1}{4n}$, it again follows from
Theorem~\ref{t:new-order} and Lemma~\ref{mp2} that $f$ has a cycle
of period $4n+2$ with no block structure, a contradiction. Hence $Q$ is a \v Stefan cycle.
\end{proof}

We are ready to prove our Main Theorem. By Lemma~\ref{l:rot-no-div},
in the proof we can consider only convergent patterns.

\begin{proof}[Proof of Main Theorem]
Recall that because we are excluding the pattern of period 2, each
pattern with no block structure has no division. By
Lemma~\ref{l:rot-no-div}, patterns with no division have over-rotation
numbers less than $1/2$. Each integer larger than 2 is of one of the
three forms: $2n+1$, $4n$, $4n+2$, with $n\ge 1$. The largest possible
over-rotation numbers smaller than $1/2$ for patterns of those periods
are, respectively, $\frac{n}{2n+1}$, $\frac{2n-1}{4n}$,
$\frac{2n}{4n+2}$. Those numbers are ordered as follows:
\[
\dots<\frac{2n-1}{4n}<\frac{2n}{4n+2}=\frac{n}{2n+1}<\frac{2n+1}{4n+4}
<\dots.
\]
Thus, by the definition of the order $\gtrdot$, we get the following
order among over-rotation pairs associated with these over-rotation
numbers:
\[
\dots\gtrdot\eta(4n)\gtrdot\eta(4n+2)\gtrdot\eta(2n+1)\gtrdot
\eta(4n+4)\gtrdot\dots\tag{$**$}
\]

Observe that the over-rotation pairs $(2n-1, 4n)$ and $(n, 2n+1)$ are
coprime; on the other hand, the over-rotation pair $(2n, \break 4n+2)$ is not
coprime as the greatest common divisor of $2n$ and $4n+2$ is $2$.

Let $f$ be a continuous interval map. If all cycles of $f$ have
division then all cycles of $f$ have block structure as division is a
particular case of block structure. This means that case~(1) of Main
Theorem takes place. To proceed with less trivial cases of Main
Theorem, fix two integers, $m>2$ and $s$ such that $m\gg s$.

Consider first the case of cycles with no division. Assume that $f$
has a cycle $P$ of period $m>2$ with no division. This cycle has
over-rotation number less than $1/2$, so by Theorem~\ref{t:new-order}
the map $f$ has a cycle of over-rotation pair $\eta(m)$. If $m\gg s$
then $\eta(m)\gtrdot \eta(s)$, so again by Theorem~\ref{t:new-order},
$f$ has a cycle $Q$ of over-rotation pair $\eta(s)$. Since its
over-rotation number is smaller than $1/2$, $Q$ has no division. In
other words, if $f$ has a cycle of period $m$ with no division and
$m\gg s$ then $f$ must have a cycle of period $s$ with no division.
This proves, for cycles with no division, the first statement of Main
Theorem.

Now, assume that $f$ has a cycle $P$ of period $m>2$ with no block
structure. Then, in particular, $P$ has no division. As before,
this implies that $f$ has a cycle of over-rotation pair $\eta(m)$
and, again, $f$ has some cycles of over-rotation pair $\eta(s)$.
To prove the first statement of Main Theorem for cycles with no block
structure we need to show that a cycle of over-rotation pair
$\eta(s)$, forced by $P$, can be chosen to be with no block structure. By
Lemma~\ref{bsdiv} and by the analysis of over-rotation pairs
$\eta(4n),$ $\eta(4n+2),$ and $\eta(2n+1),$ any cycle $Q$ of
over-rotation pair $\eta(s)$ with $s=4n$ or $s=2n+1$
automatically has no block structure.
If $s=4n+2$, then either $m=4n$ or $m\gg 4n$, so
$f$ must have a cycle of over-rotation pair $\eta(4n)=(2n-1,4n)$. Then by
Lemma~\ref{mp2} the map $f$ must have a cycle $Q$ of
over-rotation pair $(2n, 4n+2)$ and no block structure. This
completes the proof of the first statement of Main Theorem for cycles
with no block structure.

This also proves that $ND(f)$ and $NBS(f)$ are either empty or of the
form $\n_r$, i.e., $ND(f)=\n_{r_{nd}}$ and $NBS(f)=\n_{r_{nbs}}$ for
some numbers $r_{nd}$ and $r_{nbs}$. Consider all the
cases in more detail. Any cycle with no block structure has no
division. Hence in general $ND(f)\supset NBS(f)$, so
$r_{nd}\gg r_{nbs}$ or $r_{nd}=r_{nbs}$. If $r_{nd}=2n+1$ then, as before, $f$ must have a
cycle of over-rotation pair $\eta(2n+1)$ which is coprime. It follows
that this cycle has no block structure and, hence, in this case
$ND(f)=NBS(f)=\n_{2n+1}$. This covers case~(2) of Main Theorem for $r=2n+1$. If
$r_{nd}=4n$, then, again, $f$ must have a cycle of over-rotation pair
$\eta(4n)$ which is coprime, this cycle has no block structure, and
$ND(f)=NBS(f)=\n_{4n}$. This covers case~(2) of Main Theorem for $r=4n$.

Suppose now that $r_{nd}=4n+2$. Then $f$ must have a cycle of
over-rotation pair $\eta(4n+2)$. If $f$ has a cycle of over-rotation
pair $\eta(4n+2)$ with no block structure, then $ND(f)=NBS(f)=
\n_{4n+2}$, which corresponds to case~(2) of Main Theorem for $r=4n+2$. Suppose now
that all cycles of over-rotation pair $\eta(4n+2)$ have block
structure. Then, while $ND(f)=\n_{4n+2}$, the set $NBS(f)$ is strictly
smaller than $ND(f)$. The first statement of Main Theorem implies that
$f$ has a point of period $2n+1$; we may assume that its over-rotation
pair is $\eta(2n+1)$, which is coprime, so the corresponding periodic
orbit has no block structure. We conclude that in this case
$ND(f)=\n_{4n+2}$ while $NBS(f)=\n_{2n+1}$. This covers case~(3) of
Main Theorem.

To prove that all cases (1)-(3) can occur, we first note that a
constant map is an example for case~(1).

To give an example of case~(2) for a given $r\ge 3$, we observe that
there exists a pattern of period $r$ with no block structure, and
there are only finitely many such patterns. Since the forcing relation
is a partial order, there is a pattern $A$ of period $r$ with no block
structure which is minimal, in the sense that it forces no other such
pattern. Let $P$ be a cycle of pattern $A$ of the $P$ linear map $f$.
By the part of the theorem already proven, $\n_r\subset NBS(f)$.

If $f$ has a cycle $Q$ of period $m\ge 3$, $m\notin\n_r$, with no
block structure, then $m\gg r$, so the pattern $B$ of $Q$ forces some
pattern $A'$ of period $r$ with no block structure, and thus, $f$ has
a cycle of pattern $A'$. Since $f$ is $P$-linear, it follows that $A$
forces $A'$, and by minimality of $A$ we get $A'=A$. Hence $A$ forces
$B$ and $B$ forces $A$, a contradiction. This proves that
$NBS(f)=\n_r$.

By the part of the theorem already proven, either~(2) holds or
$r=2n+1$ and $ND(f)=\n_{2r}$. In the latter case, by
Proposition~\ref{stefan-only}, $A$ is a \v Stefan pattern that forces
its own doubling which is impossible. This proves that~(2) holds.

Finally, to give an example of case~(3) for a given $n\ge 1$, we take
a $P$-linear map $f$, where the pattern $A$ of $P$ is a doubling of
the \v Stefan pattern of period $2n+1$. By the theorems on forcing
extensions of patterns (see~\cite{alm}), if $B$ is a pattern and $A$
is a doubling of $B$, then $A$ forces $B$ and the only pattern forced
by $A$ but not by $B$ is $A$ itself. Since the \v Stefan pattern does
not force any other pattern of the same period,~(3) holds.
\end{proof}

\begin{remark}\label{so}
It follows from Main Theorem and Proposition~\ref{stefan-only} that
the case~(3) of Main Theorem occurs if and only if a cycle of period
$4n+2$ with no division exists and all such cycles are doublings of \v
Stefan cycles.
\end{remark}

config: arxiv


TITLE: Gradient of $X^{T}AX$ wrt matrix X
QUESTION [0 upvotes]: I need to calculate the gradient of $X^{T}AX$ with respect to $X$, where $X$ and $A$ are $nxn$ matrixes. Using the matrix cookboox here (pag 9), I see that the gradient is obtained for each element $ij$ of $X$ denoted by $X_{ij}$ as
$$X^{T}AJ_{ij}+J_{ij}AX$$
where $J_{ij}$ denotes the single-entry matrix with a $1$ at position $ij$ and $0$ elsewhere. However,I find that for each element $X_{ij}$ the resulting partial derivative is a matrix, so that the gradient of the whole $X$ would be a set of matrixes. Which in my head seems strange. Could anyone help on this?

REPLY [2 votes]: Index notation might help you see what's going on.
$$\eqalign{
F &= X^TAX \\
F_{il} &= \sum_{j}\sum_{k} X_{ij}^TA_{jk}X_{kl}
 &= X_{ji}A_{jk}X_{kl} \\
}$$
where the last expression uses the summation convention, i.e. the sigma's are implicit rather than explicit for repeated indices. It also drops the superscript indicating the transpose, instead the order of the indices is swapped.
The next thing to note is that the gradient of a matrix with respect to itself can be written in terms of Kronecker deltas.
$$\eqalign{
\frac{\partial X_{kl}}{\partial X_{mn}} &= \delta_{km}\delta_{ln} \\
}$$
Putting these two things together, you can calculate the gradient
$$\eqalign{
G_{ilmn} &= \frac{\partial F_{il}}{\partial X_{mn}} \\
&= (\delta_{jm}\delta_{in})A_{jk}X_{kl} + X_{ji}A_{jk}(\delta_{km}\delta_{ln}) \\
&= \delta_{in}A_{mk}X_{kl} + X_{ji}A_{jm}\delta_{ln} \\
}$$
and see that it requires 4 indices, instead of the two indices that are required for a matrix.
Thus the name fourth-order tensor.


TITLE: Prove Graph is connected
QUESTION [0 upvotes]: Prove that every graph $G= (V, E)$ where $\forall$ v $\in$ V, $d(v) \ge |V|- 3$
and $|V| > 4$ is connected.
I was able to prove that this graph would be cyclic as well as
$|E| \ge \frac{n^2-3}{2}$
But how do i prove that the graph must be connected

REPLY [1 votes]: Write $n=|V|$. Every vertex has at least $n-3$ neighbors, i.e. every connected component is at least of size $n-2$. If you have two such components of size $a$ and $b$, we have
$$n=a+b\ge n-2+n-2=2n-4\quad\implies\quad n\le 4.$$
Similar for more components.


TITLE: Are these two function convex?
QUESTION [2 upvotes]: Screenshot here
$$f(x)=\left\{\begin{array}{ll}x \log x, & x>0 \\ 1, & x=0\end{array}\right.
\\
f(x)=\left\{\begin{array}{c}x^{2}, \qquad x \geq 0 \\ 1+x^{2}, \quad x<0\end{array}\right.$$
first func is for $x\geq 0$
second func is for any $x$
How do I prove that these functions are convex or not?
Thanks

REPLY [3 votes]: The second function is not convex because of the jump at 0. You can look for a secant line that crosses the graph near 0. For instance you can try $x_0=-1/2, x_1=1/2$ and any point $tx_0+(1-t)x_1$ with $t<1/2$:

For the first, its easy to check that its convex as a function on $(0,\infty)$ by differentiating twice. (a $C^2$ function is convex iff $f'' \ge 0.$) What this means is that if you form the secant line segments joining any two points on the graph with $x=x_0,x_1>0$ then the line lies above the graph. It remains for you to check the case where one of the points is at $x=0$, i.e. we need to know if
$$ f((1-t)0 + tx) \le (1-t)f(0) + tf(x)
\\
\iff tx\log tx \le (1-t) + tx\log x$$
but $ tx\log tx  = tx \log x + tx \log t$ so we just need to know if for any $t\in (0,1)$, and any $x\ge 0$,
$$ tx\log t \le 1-t$$
and this is true because for $t\in(0,1)$, $\log t<0$, so in fact
$$ tx\log t \le 0 \le 1-t.$$


TITLE: Vitali set of outer-measure exactly $1$.
QUESTION [25 upvotes]: I know that for any $\varepsilon\in (0,1]$ we can find a non-measurable subset (w.r.t Lebesgue measure) of $[0,1]$ so that its outer-measure equals exactly $\varepsilon$. It is done basicly with the traditional Vitali construction inside the interval $[0,\varepsilon]$ and noticing that such a set carries zero inner-mass, and thus its complement in $[0,\varepsilon]$ (being non-measurable as well) must carry the full outer-mass of $[0,\varepsilon]$.
However, this resulting non-measurable set is a complement of the traditional Vitali constructed set. My question asks if the Vitali construction itself can yield a non-measurable set with outer-measure of exactly $1$ (or any before-hand decided number from $(0,1]$). Some modifications can be done inside the construction of course, but in particular I would like to stay away from taking complements. Maybe someone knows how this could be done?
Any references and input is appreciated. Thanks in advance.

REPLY [11 votes]: $\newcommand{\c}{\mathfrak{c}}$
Let $\c$ denote the cardinal of the continuum and wellorder the Borel subsets of $[0,1]$ as $(B_\alpha)_{\alpha < \c}$.  We build by transfinite recursion a sequence $(x_\alpha)_{\alpha < \c}$ of elements of $[0,1]$ such that:
(a) $x_\alpha$ is Vitali inequivalent to $x_\beta$ for all $\beta < \alpha$, and
(b) $x_\alpha \in [0,1] \setminus B_\alpha$ if $[0,1] \setminus B_\alpha$ is uncountable.
Note that this process can't get stuck, since if the complement of $B_\alpha$ is uncountable then it has cardinality $\c$, and thus it meets an unused Vitali equivalence class (since at most $|\alpha| < \c$ have been used so far).  Then by setting $X = \{x_\alpha : \alpha < c\}$ we obtain a set such that whenever $B$ is a Borel set with $X \subseteq B$, then $B$ has countable complement (and in particular has measure $1$).  So $X$ has outer measure $1$ as desired.


TITLE: Steklov averages and negative parabolic sobolev spaces
QUESTION [1 upvotes]: Suppose one is given a function
$$
w \in L^p(0,T;W^{1,p}(\Omega)) \qquad \text{and} \qquad \frac{dw}{dt} \in L^{p'}(0,T; W^{-1,p'}(\Omega))
$$
I am interested if the following holds:
Denote the Steklov average to be $[f]_h(x,t) := \frac{1}{2h}\int_{t-h}^{t+h} f(x,s)\ ds$, then does the following hold:
$$
\int_a^b \left< \frac{d[w]_h}{dt}, \varphi \right>_{(W^{-1,p'}(\Omega),W_0^{1,p}(\Omega)}(t) \ dt = \int_a^b \left[ \left<\frac{dw}{dt}, \varphi \right>_{(W^{-1,p'}(\Omega),W_0^{1,p}(\Omega)} \right]_h(t) \ dt
$$
Edit: My question was not framed correctly. I ask what class of functions allow for such an equality? Smooth functions are admissible for $w$, how much can I relax the condition further? Here $\varphi \in C_c^{\infty}(\Omega)$ only as a function of $x$.
Edit: The interval $(a,b) \subset (0+h,T-h)$ for any fixed small $h$.

REPLY [1 votes]: The question seems to be if the Steklov average commutes with $\langle \cdot, \phi \rangle$ and the derivative by t. Then you could pull the Steklov average inside and the two sides are equal. I dont see any reason why this shoulden hold for all functions where everything is defined, i.e. for all functions in your domain.


TITLE: Is there anything I could read that talks about dimensionality of prime/composite numbers?
QUESTION [2 upvotes]: Is there anything out there that talks about how primes are one dimensional numbers and composites can only be in dimensions greater than 1?
What I mean is, 4 would be a two dimensional number (2x2) same with 6 (2x3).
27 would be three dimensional (3x3x3) and so on. While primes can only be made up of themselves therefore they are one dimension.

REPLY [2 votes]: Assuming you meant that 27 would be "three dimensional" but 9 would be "two dimensional", then the "dimension" of a number $n$ is the total number of prime factors, denoted $\Omega(n)$ (often read "big omega"). It is not usually called the dimension of a number, but if you really like to think geometrically, you can think about it as the greatest dimension in which the number is the "hypervolume" of a hyperrectangle with side lengths greater than 1.
See the following links for some information.

https://oeis.org/A001222
https://en.wikipedia.org/wiki/Prime_factor#Omega_functions
https://en.wikipedia.org/wiki/Arithmetic_function#.CE.A9.28n.29.2C_.CF.89.28n.29.2C_.CE.BDp.28n.29_.E2.80.93_prime_power_decomposition


TITLE: How to sum this series and what is the sum?
QUESTION [2 upvotes]: We have the series $$ \sum_{n=2}^{\infty} \frac{ \log [ (1 + 1/n)^n (n+1)] }{ (\log n^n) [ \log (n+1)^{n+1} ] }, $$ which we can express as $$ \sum_{n=2}^{\infty} \frac{ \log (n+1)^{n+1} - \log n^n }{ (\log n^n) [ \log (n+1)^{n+1} ] } = \sum_{n=2}^{\infty} [ \frac{ 1}{\log n^n} - \frac{1}{\log (n+1)^{n+1}} ], $$ which is a telescoping series whose sum is $1/\log4$. Is it correct?

REPLY [3 votes]: This is correct and very nice, with one caveat.  To justify a telescoping series converging, you need to show the final term (i.e. the term in the partial sum) goes to zero.  So in this case you need to show that
$$
\frac{1}{\log (k^k)}
$$
goes to zero.  This should be easy, but in some cases you have to be careful.
EDIT: In general the telescoping series converges as long as this final term goes to anything, i.e. as long as the final term converges.  But if it converges to something other than zero, then you have to factor that into your calculation of the sum.  For instance, if we found that $\frac{1}{\log(k^k)}$ converged to $.03$, then we would calculate our final answer to be $\frac14 - .03$.


TITLE: What is the relationship between Faraday cage mesh size and attenuation of cell phone reception signals?
QUESTION [24 upvotes]: This is related to the question how can electromagnetic waves reach a cell phone in faraday cage?, where in the answer it was stated that the holes (=size of the mesh) would need to be smaller than that of the wavelengths of the EM radiation used for cell phone signals for attenuation to 'start'.
Also, in the question Should a Faraday cage block a radio's signal? where the focus of the question is on AM radio frequencies (which I know are not the same used by cell phones), but the idea was using baking trays, effectively making a Faraday 'box'.  However, it lead to the question Building a Faraday cage for mobile networks that is transparent for optical wavelengths, which looked at blocking cell phone signals using a cage, but still allowing optical wavelengths in.
However, I am seeking to know the answer to is there is a relationship between a grounded Faraday cage mesh hole size (relative to the cellphone signal used) and cell phone reception attenuation?

REPLY [11 votes]: The most important concept relating Faraday cage hole size to cell phone signal attenuation is the idea of a cutoff frequency. For round holes, you would model them as cylindrical waveguides. For simplicity, we'll consider rectangular waveguides instead.
Matching the boundary conditions at the metal wall, you get so called transverse electric (TE) and transverse magnetic (TM) modes. These look like partially standing waves, with a traveling wave component for the third. For TE modes, they're of the form (for polarization in the y direction):
$$E = E_0 \sin(k_x x) \cos(k_y y) e^{i(k_z z-\omega t)}$$
There's a multitude of standing wave modes. These are described by different values of $k_x$ and $k_y$, which are solved by setting the above expression to zero at the walls of the waveguide (for the sine portion), or derivative zero (for the cosine portion). The solutions:
$$k_x = \frac{m\pi}{w}$$
$$k_y = \frac{n\pi}{h}$$
Where $w$ and $h$ are the width and height of the waveguide, and $m$ and $n$ are integers. Substituting the above expression into the wave equation, we get the relation between the different $k$ components and frequency.
$$\left( \frac{\omega}{c} \right)^2 = {k_x}^2+{k_y}^2+{k_z}^2 $$
The lowest possible such frequency is when $k_z = 0$ and $m=1,n=0$
$$f_c= \frac{c}{2w}$$
This is the cutoff frequency. Below this frequency, the signal exponentially decays as it propagates through the structure. To show this, solve for $k_z$, and write it in terms of cutoff frequency.
$$k_z = \frac{2 \pi}{c} \sqrt{f^2-{f_c}^2}$$
Evidently, at frequencies below cutoff, $k_z$ becomes imaginary. Substituting this into our travelling wave expression, it becomes exponential decay.
$$\alpha :=  \frac{2 \pi}{c} \sqrt{{f_c}^2-f^2}, f < f_c$$
$$E = E_0 \sin(k_x x) \cos(k_y y) e^{-\alpha z-i\omega t}$$
Notice that for our rectangular waveguide, the cutoff frequency depended only on the width. In general,
$$\lambda_c \approx \textrm{largest feature size}*2 $$
(This is exactly true for a rectangular waveguide, and should hold approximately for other shapes.)
For Faraday cages with openings in the size of centimeters, the cutoff frequency is around 20GHz, which is quite large compared cell phone signals in the range of 2GHz. We can approximate the decay constant $\alpha$
$$\alpha \simeq \frac{2 \pi f_c}{c} = \frac{2\pi}{\lambda _c} $$
To figure out the amount of decay, we need to assume some length $l$ to the opening (equivalently, thickness of the cage material), and then substitute $l$ for $z$ in the wave expression. Converting this to a decibel scale, we get the following power loss:
$$\frac{l}{\textrm{largest feature size}} (48.6 \textrm{dB})$$
Another important point is that the signal will mostly negatively interfere with itself in the cage, except at a few spots within the cage where it is effectively amplified (probably the center). If the cage features are fairly large, you might be able to notice this signal hotspot.
Edit: There are also complex effects where the fields in one hole can induce fields in another hole. The above analysis is a simplified description of a complex field problem, but I expect the general principles to hold.


TITLE: Suppose $\ \pi+e\ $ is transcendental. What about $\ \pi-e\ $?
QUESTION [5 upvotes]: Suppose $\ \pi+e\ $ is a transcendental number.

Can we conclude whether $\ \pi-e\ $ is rational, algebraic irrational or transcendental ?

If I understood the consequences of Schanuel's conjecture correctly, it implies that $\ \pi-e\ $ is trancendental. But can we also say something without any further unproven conjecture ?

REPLY [2 votes]: I claim that that statement "if $\pi+e$ is transcendental, then $\pi-e$ is transcendental" implies that $\pi-e$ is transcendental. This shows that the knowledge that $\pi+e$ is transcendental is of no help in concluding that $\pi-e$ is transcendental.
Assume that if $\pi+e$ is transcendental, then $\pi-e$ is transcendental. Separately, we can see directly that if $\pi+e$ is algebraic, then $\pi-e = (\pi+e)-2e$ is transcendental (since we know $e$ is transcendental). Since $\pi+e$ is either transcendental or algebraic, we conclude that $\pi-e$ is transcendental.


text:
\subsection{Weakly filtered iterated cones coming from cobordisms}
\label{sb:wf-icones-cobs}

Let $V \in \widetilde{\mathcal{C}}$ be a Lagrangian cobordism and
denote by $L_0, \ldots, L_r \in \mathcal{C}$ its negative ends. (In
contrast to~\S\ref{sb:ext-lcob} as well as~\cite{Bi-Co:lcob-fuk}, in
this section we index the negative ends from $0$ to $r$ rather than
from $1$ to $r$.)

Let $\gamma \subset \mathbb{R}^2$ be the curve depicted in
Figure~\ref{f:gamma-cob-c}. Let $p \in E'_{\text{reg}}$ and
$h \in \mathcal{H}'_{\text{prof}}(\gamma)$ be such that
$l := \# (\phi^h_1)^{-1}(\gamma) \cap \gamma = 2r+5$.
\begin{figure}[htbp]
   \begin{center}
     \includegraphics[scale=0.70]{gamma-cob-c}
   \end{center}
   \caption{The curves $\gamma$, $(\phi_1^h)^{-1}(\gamma)$ and the
     cobordism $V$.}
   \label{f:gamma-cob-c}
\end{figure}
Denote by $\mathcal{V}$ the Yoneda module of $V$, which we view here
as an $A_{\infty}$-module over the category
$\fukcob(\widetilde{\mathcal{C}}_{1/2};\iota_{\gamma}(p,h))$. Consider
now the pullback module
\begin{equation} \label{eq:pull-back-V} \mathcal{M}_{V; \gamma, p,h}
  := \mathcal{I}_{\gamma; p,h}^* \mathcal{V},
\end{equation}
which is a $\fuk(\mathcal{C}; p)$-module. Since
$\mathcal{I}_{\gamma; p, h}$ is a weakly filtered functor the module
$\mathcal{M}_{V; \gamma, p, h}$ is weakly filtered.
\begin{prop} \label{p:wf-M_V} The weakly filtered module
  $\mathcal{M}_{V; \gamma, p, h}$ has the following properties.
  \begin{enumerate}
  \item For every $N \in \mathcal{C}$ and $\alpha \in \mathbb{R}$ we
    have
    \begin{equation*}
      \begin{aligned}
        & \mathcal{M}_{V; \gamma, p, h}^{\leq \alpha}(N) = \\
        & CF^{\leq \alpha-h(O_3)}(N,L_0; p) \oplus CF^{\leq \alpha -
          h(O_5)}(N,L_1;p) \oplus \cdots \oplus CF^{\leq \alpha -
          h(O_{2r+3})}(N,L_r;p),
      \end{aligned}
    \end{equation*}
    where the last equality is of $\Lambda_0$-modules (but not
    necessarily of chain complexes). Here $CF(N, L_i; p)$ stands for
    $CF(N,L_i; \mathscr{D}_{N, L_i}) $, where $\mathscr{D}_{N, L_i}$
    is the Floer datum prescribed by $p \in E'_{\textnormal{reg}}$.
  \item $\mathcal{M}_{V; \gamma, p, h}$ has discrepancy
    $\leq \bme^{\mathcal{M}_{V; \gamma, p, h}}$, where
    \begin{equation} \label{eq:discrep-M_V}
      \epsilon_d^{\mathcal{M}_{V; \gamma, p, h}} \leq (d-1) \max\{
      h(O_k) \mid 1 \leq k = \text{odd} \leq 2r+5\} +
      \epsilon_d^{\fuk_{\text{cob}}(\widetilde{\mathcal{C}}_{1/2};
        \iota_{\gamma}(p,h))}.
    \end{equation}
  \end{enumerate}
\end{prop}
\begin{proof}
  The second statement follows immediately from
  Proposition~\ref{p:wf-inc-functors} and Lemma~\ref{l:pull-back-M}
  together with the fact that the higher terms of
  $\mathcal{I}_{\gamma; p,h}$ vanish. The first statement can be
  verified by a straightforward calculation.
\end{proof}

\begin{rem} \label{r:discrep-M_V-2} An inspection of the arguments
  from~\cite[Section~4.4]{Bi-Co:lcob-fuk} shows that the estimate for
  the discrepancy $\epsilon_d^{\mathcal{M}_{V; \gamma, p, h}}$
  in~\eqref{eq:discrep-M_V} can be slightly improved by replacing the
  ``$\max$'' term from~\eqref{eq:discrep-M_V} with
  $\max\{ h(O_k) \mid 3 \leq k = \text{odd} \leq 2r+3\}$. We will not
  go into details on that since this improvement will not play any
  role in our applications.
\end{rem}

Recall from~\cite[Section~4.4]{Bi-Co:lcob-fuk} that the module
$\mathcal{M}_{V; \gamma, p, h}$ is naturally isomorphic to an iterated
cone with attaching objects corresponding to the ends
$L_0, \ldots, L_r$ of $V$. More precisely, denote by $\mathcal{L}_j$
the Yoneda module corresponding to $L_j$. Then
\begin{equation*} \mathcal{M}_{V; \gamma, p, h} \cong \tcn
  (\mathcal{L}_r \xrightarrow{\; \phi_r \;} \tcn(\mathcal{L}_{r-1}
  \xrightarrow{\; \phi_{r-1} \;} \tcn( \cdots \tcn(\mathcal{L}_2
  \xrightarrow{\; \phi_2 \;} \tcn(\mathcal{L}_1 \xrightarrow{\; \phi_1
    \;} \mathcal{L}_0 )) {\cdot}{\cdot}{\cdot}))),
\end{equation*}
where $\phi_j$ is a module homomorphism between $\mathcal{L}_j$ and
the intermediate iterated cone involving the attachment of only the
first $j+1$ objects $\mathcal{L}_0, \ldots, \mathcal{L}_j$.

As we will see shortly, the module homomorphisms $\phi_j$ are weakly
filtered (and obviously the $\mathcal{L}_i$'s too) and consequently
the iterated cone $\mathcal{M}_{V; \gamma, p, h}$ can be endowed with
a weakly filtered structure by the algebraic recipe
of~\S\ref{sb:wf-mc} and~\S\ref{sb:wf-ic}. At the same time, we have
just seen that $\mathcal{M}_{V; \gamma, p, h}$ has another weakly
filtered structure being the pull back module by an inclusion functor,
as described in Proposition~\ref{p:wf-M_V}. Our goal now is to compare
these two weakly filtered structures and show that they are
essentially the same.

Consider the following collection of curves
$\gamma_1, \ldots, \gamma_r \subset \mathbb{R}^2$ with horizontal
ends, as depicted in Figure~\ref{f:gamma-j-c}. We assume that
$\gamma_r = \gamma$, the curve involved in the definition of
$\mathcal{M}_{V; \gamma, p, h}$.

We also choose profile functions
$h_1, \ldots, h_r: \mathbb{R}^2 \longrightarrow \mathbb{R}$ with
$h_j \in \mathcal{H}'_{\text{prof}}(\gamma_j)$ and such that the
following holds (see Figure~\ref{f:gamma-j-c}):
\begin{enumerate}
\item $h_r = h$. \label{pp:h_j-functions}
\item
  $(\phi_1^{h})^{-1}(\gamma) \cap \gamma = \{O_1, \ldots, O_{2r+5}\}$
\item
  $(\phi_1^{h_j})^{-1}(\gamma_j) \cap \gamma_j = \{O^j_1, \ldots,
  O^j_{2j+5}\}$, where $O^j_k = O_k$ for all $1 \leq k \leq
  2j+3$. Thus only the last two intersection points
  $O^j_{2j+4}, O^j_{2j+5}$ do not belong to the $\gamma_l$'s for
  $l>j$.
\item $h_j$ coincides with $h$ over the half-plane
  $\{y \leq y_{2j+3} + \tfrac{1}{100} \}$, where $y_{2j+3}$ is the
  $y$-coordinate of $O_{2j+3}$.
\end{enumerate}

\begin{figure}[htbp]
   \begin{center}
     \includegraphics[scale=0.80]{gamma-j-c}

   \end{center}
   \caption{A closer look at the curves $\gamma$, $\gamma_{j-1}$, and
     $\gamma_j$ near the $(j-1)$'th and $j$'th ends of $V$.}
   \label{f:gamma-j-c}
\end{figure}

We denote the space of all tuples of profile functions
$(h_1, \ldots, h_r)$ satisfying these conditions by
$\mathcal{H}'_{\text{prof}}(\gamma_1, \ldots, \gamma_r)$ and denote
elements of this space by $\upsilon = (h_1, \ldots, h_r)$. With this
notation, it is possible to choose maps
$\iota_{\gamma_j}: E'_{\textnormal{reg}} \times
\mathcal{H}'_{\text{prof}}(\gamma_j) \longrightarrow
\widetilde{E}'_{\textnormal{reg},1/2}$ for $j=1, \ldots, r$, as
in~\eqref{eq:iota-gamma} satisfying the following. For every
$(p, \upsilon) \in E'_{\text{reg}} \times
\mathcal{H}'_{\text{prof}}(\gamma_1, \ldots, \gamma_r)$ the choice of
data $\iota_{\gamma_j}(p,h_j) \in \widetilde{E}'_{\text{reg},1/2}$ has
the properties listed for $\iota_{\gamma}(p,h)$ on
page~\pageref{pp:iota-gamma-h} but with $\gamma$ replaced by
$\gamma_j$ and $h$ by $h_j$. (Consequently, for every
$\widetilde{p}_0 \in \mathcal{N}$ we have
$\lim \iota_{\gamma_j}(p, h_j) \in \widetilde{\mathcal{N}}_{1/2}$ as
$p \longrightarrow p_0$ and
$\upsilon = (h_1, \ldots, h_r) \longrightarrow (0, \ldots, 0)$.)
Moreover, we require that for every $j$ and
$p \in E'_{\textnormal{reg}}$,
$\upsilon = (h_1, \ldots, h_r) \in
\mathcal{H}'_{\text{prof}}(\gamma_1, \ldots, \gamma_r)$ the data
prescribed by $\iota_{\gamma_j}(p, h_j)$ is compatible with that
prescribed by $\iota_{\gamma_{j-1}}(p,h_{j-1})$ (in the obvious sense,
similar to $h_j$ being compatible with
$h_{j-1}$). By~\S\ref{sb:inc-functors}, the curves $\gamma_j$ and the
maps $\iota_{\gamma_j}$ induce a family of inclusion functors
$$\mathcal{I}_{\gamma_j; p, h_j}: \fuk(\mathcal{C}; p)
\longrightarrow \fuk(\widetilde{\mathcal{C}}_{1/2};
\iota_{\gamma_j}(p, h_j)),$$ parametrized by
$p \in E'_{\textnormal{reg}}$,
$\upsilon = (h_1, \ldots, h_r) \in
\mathcal{H}'_{\text{prof}}(\gamma_1, \ldots, \gamma_r)$.  We will use
below the notation
$\mathcal{I}_{\gamma_j; p, \upsilon} := \mathcal{I}_{\gamma_j; p,
  h_j}$, where $h_j$ is the $j$'th entry in the tuple $\upsilon$ since
it reflects better the parameters $(p, \upsilon)$ parametrizing this
family of functors. We will also write sometimes
$\iota_{\gamma_j}(p, \upsilon)$ for $\iota_{\gamma_j}(p, h_j)$.

Consider now the pullback $\fuk(\mathcal{C};p)$-modules
$$\mathcal{M}_{V; \gamma_j, p, \upsilon} =
\mathcal{I}^*_{\gamma_j; p, \upsilon} \mathcal{V}, \quad j=1, \ldots,
r.$$ We endow each of these modules with its weakly filtered structure
as defined at the beginning of~\S\ref{sb:wf-icones-cobs} and further
described by Proposition~\ref{p:wf-M_V} (where $l = 2j+5$, and
$\gamma$ should be replaced by $\gamma_j$ and $h$ by $h_j$). Next, for
every $0 \leq j \leq r$ denote by $\mathcal{L}_j$ the Yoneda module
associated to $L_j$, endowed with its weakly filtered structure
induced from $\fuk(\mathcal{C}; p)$. Finally, recall that for a weakly
filtered module $\mathcal{M}$ and $\nu \in \mathbb{R}$,
$S^{\nu}\mathcal{M}$ stands for the weakly filtered module obtained
from $\mathcal{M}$ by an action-shift of $\nu$
(see~\S\ref{sbsb:action-shifts}).

\begin{prop} \label{p:icones-M_j} For every
  $(p, \upsilon) \in E'_{\textnormal{reg}} \times
  \mathcal{H}'_{\text{prof}}(\gamma_1, \ldots, \gamma_r)$ there exist
  weakly filtered module homomorphisms
  $\phi_1: \mathcal{L}_1 \longrightarrow \mathcal{L}_0$ and
  $\phi_j: \mathcal{L}_j \longrightarrow S^{h(O_3)}\mathcal{M}_{V;
    \gamma_{j-1}, p, \upsilon}$ for $j=2, \ldots, r$ such that the
  following holds for every $1 \leq j \leq r$:
  \begin{enumerate}
  \item $\phi_j$ shifts action by $\leq 0$.
  \item The discrepancy of $\phi_j$ is $\leq \bm{\delta}^{\phi_j}$,
    where
    \begin{equation} \label{eq:delta-phi-j} \delta_d^{\phi_j} :=
      (d-1)\max_{\substack{1 \leq k \leq 2j+3 \\ k \text{ odd}}}
      h(O_k) + \epsilon_d^{\fukcob(\widetilde{\mathcal{C}}_{1/2};
        \iota_{\gamma_j}(p,\upsilon))} + h(O_{2j+3})-h(O_3).
    \end{equation}
  \item For every $1\leq j \leq r$,
    $S^{h(O_3)}\mathcal{M}_{V; \gamma_j, p, \upsilon} = \tcn(\phi_j;
    0, \bm{\delta}^{\phi_j})$ as a weakly filtered
    modules. (See~\S\ref{sb:wf-mc} for our conventions for weakly
    filtered cones.) In other words, the weakly filtered module
    $S^{h(O_3)}\mathcal{M}_{V; \gamma_j, p, \upsilon}$ coincides with
    the weakly filtered mapping cone over $\phi_j$.
  \end{enumerate}
\end{prop}

Recalling that
$\mathcal{M}_{V; \gamma, p,h} = \mathcal{M}_{V; \gamma_r, p,
  \upsilon}$, the above Proposition implies that
\begin{equation}\label{eq:filtered-cob-it-cone}
  \begin{aligned}
    S^{h(O_3)} & \mathcal{M}_{V; \gamma, p, h} \\ = \; & \tcn
    (\mathcal{L}_r \xrightarrow{\; \overline{\phi}_r \;}
    \tcn(\mathcal{L}_{r-1} \xrightarrow{\; \overline{\phi}_{r-1} \;}
    \tcn( \cdots \tcn(\mathcal{L}_2 \xrightarrow{\; \overline{\phi}_2
      \;} \tcn(\mathcal{L}_1 \xrightarrow{\; \overline{\phi}_1 \;}
    \mathcal{L}_0 )) {\cdot}{\cdot}{\cdot}))),
  \end{aligned}
\end{equation}
where $\overline{\phi}_j := (\phi_j; 0, \bm{\delta}^{\phi_j})$ and the
cones in~\eqref{eq:filtered-cob-it-cone} are endowed with the
filtrations as defined in~\S\ref{sb:wf-mc}. In other words, up to a
small action-shift, $\mathcal{M}_{V; \gamma, p, h}$ can be viewed as a
weakly filtered iterated cone by the very same recipe described at the
beginning of~\S\ref{sb:wf-ic} (with $\rho_j=0$ and
$\mathcal{K}_j = S^{h(O_3)} \mathcal{M}_{V; \gamma_j, p,
  \upsilon}$). Consequently, we can apply Theorem~\ref{t:itcones} with
$\mathcal{K}_r = S^{h(O_3))}\mathcal{M}_{V; \gamma, p, h}$.

\begin{proof}[Proof of Proposition~\ref{p:icones-M_j}]
  The proof is based on two main ingredients. The first one is the
  theory developed in~\cite[Sections~4.2,~4.4]{Bi-Co:lcob-fuk} from
  which it follows that, ignoring action-filtrations, we have
  $\mathcal{M}_{V; \gamma_j, p, \upsilon} = \tcn(\mathcal{L}_j
  \longrightarrow \mathcal{M}_{V; \gamma_{j-1}, p, \upsilon})$. The
  second one comprises direct action-filtration calculations for the
  modules $\mathcal{M}_{V; \gamma_j, p, \upsilon}$ and the
  homomorphisms $\phi_j$.

  Before we go on, we should remark a notational difference
  between~\cite{Bi-Co:lcob-fuk} and the present
  paper. In~\cite{Bi-Co:lcob-fuk} the negative ends of the cobordism
  $V$ are indexed from $1$ to $r$, whereas in the present text the
  indexing runs between $0$ and $r$. This results in several other
  indexing differences between the two texts. For example, the curves
  $\gamma_j$ in the present text are the same as $\gamma_{j+1}$
  in~\cite{Bi-Co:lcob-fuk}. In the present text, the number of
  intersection points between $\phi_1^{h_j}(\gamma_j)$ and $\gamma_j$
  is $2j+5$, whereas in~\cite{Bi-Co:lcob-fuk} this number is $2j+3$,
  etc.

  We start by adding to the collection of curves
  $\gamma_1, \ldots, \gamma_r$ another curve $\gamma_0$, defined in
  the same way as the $\gamma_j$'s only that it is adapted to the
  $L_0$-end of $V$ in the sense that the negative end of $\gamma_0$
  goes above the $L_0$-end and below the $L_1$ end. We also choose
  $h_0 \in \mathcal{H}'_{\text{prof}}(\gamma_0)$ satisfying the same
  conditions as the $h_j$'s (see page~\pageref{pp:h_j-functions}) only
  for $j=0$. We write
  $(\phi_1^*{h_0})^{-1}(\gamma_0) \cap \gamma_0 = \{O_1, O_2, O_3,
  O_4^0, O_5^0\}$. To simplify the notation we also extend the tuple
  $\upsilon = (h_1, \ldots, h_r)$ to contain also $h_0$ and write
  $\upsilon = (h_0, \ldots, h_r)$. As before we have an inclusion
  functor associated to $\gamma_0, p, h_0$ and we consider the
  pullback module
  $\mathcal{M}_{V; \gamma_0, p, h_0} := \mathcal{I}_{\gamma_0; p,
    h_0}^* \mathcal{V}$. To be consistent with the previous notation,
  we will denote this module also by
  $\mathcal{M}_{V; \gamma_0, p, \upsilon}$.

  We first claim that there exist module homomorphisms
  $\phi_j: \mathcal{L}_j \longrightarrow \mathcal{M}_{V; \gamma_{j-1},
    p, h_{j-1}}$ for all $1 \leq j \leq r$, such that
  \begin{equation} \label{eq:M_V-cone} \mathcal{M}_{V; \gamma_j, p,
      h_j} = \tcn(\mathcal{L}_j \xrightarrow{\; \phi_j \;}
    \mathcal{M}_{V; \gamma_{j-1}, p, h_{j-1}}),
  \end{equation}
  where at the moment we ignore the action filtrations. This statement
  is not explicitly stated in~\cite[Section~4.4.2]{Bi-Co:lcob-fuk},
  but it follows easily from the arguments in that paper. More
  specifically, what is stated explicitly
  in~\cite[Section~4.4.2]{Bi-Co:lcob-fuk} is that there exists an
  exact triangle (in the derived category $D\fuk(\mathcal{C}; p)$) of
  the form
  $\mathcal{L}_j \longrightarrow \mathcal{M}_{V; \gamma_{j-1}, p,
    h_{j-1}} \longrightarrow \mathcal{M}_{V; \gamma_j, p, h_j}$. Here
  however, we claim a stronger statement, namely
  that~\eqref{eq:M_V-cone} holds at the chain level. We will now
  explain how to deduce~\eqref{eq:M_V-cone} from the theory developed
  in~\cite{Bi-Co:lcob-fuk}. In doing that we will mostly follow the
  notation from that paper.

  By~\cite[Proposition~4.4.1]{Bi-Co:lcob-fuk} for every
  $0 \leq j \leq r$ we have the following:
  \begin{enumerate}
  \item $A_{\infty}$-categories $\mathcal{B}_j$ and $\mathcal{B}'_j$
    (depending on $\gamma_j, p$ and $h_j$).
  \item Quasi-isomorphisms of $A_{\infty}$-categories:
    $e_j: \fuk(\mathcal{C}; p) \longrightarrow \mathcal{B}_j$,
    $p_j: \mathcal{B}_j \longrightarrow \mathcal{B}'_j$,
    $\sigma_j: \mathcal{B}'_j \longrightarrow \fuk(\mathcal{C}; p)$
    and $q_j: \mathcal{B}'_j \longrightarrow \mathcal{B}'_{j-1}$, for
    $j \geq 1$, all with vanishing higher order terms. Moreover, they
    satisfy:
    \begin{equation} \label{eq:sigma_j-p_j-e_j} \sigma_j \circ p_j
      \circ e_j = \id, \; \; \forall \; j\geq 0, \quad \text{and}
      \quad q_j \circ p_j \circ e_j = p_{j-1} \circ e_{j-1} \; \;
      \forall \; j\geq 1.
    \end{equation}
  \item A $\mathcal{B}_j$-module $\overline{\mathcal{M}}_j$ and a
    $\mathcal{B}'_j$-module $\mathcal{M}'_j$ such that:
    \begin{equation} \label{eq:M_j-M'_j}
      \begin{aligned}
        & \mathcal{M}_{V; \gamma_j, p, h_j} =
        e_j^*\overline{\mathcal{M}}_j, \quad p_j^* \mathcal{M}'_j =
        \overline{\mathcal{M}}_j, \;\; \forall \; j\geq 0 \\
        & \mathcal{M}'_j = \tcn (\sigma_j^* \mathcal{L}_j
        \xrightarrow{\; \varphi_j \;} q_j^*\mathcal{M}'_{j-1}), \; \;
        \forall \; j\geq 1,
      \end{aligned}
    \end{equation}
    for some module homomorphism $\varphi_j$. (This homomorphism was
    denoted by $\phi_j$
    in~\cite[Proposition~4.4.1]{Bi-Co:lcob-fuk}. We have denoted it
    here by $\varphi_j$ since $\phi_j$ is already used for a slightly
    different homomorphism.)
  \item For $j=0$ we have: $\mathcal{M}'_0 =
    \sigma_0^*\mathcal{L}_0$. \label{i:M'_0}
  \end{enumerate}
  We now pull back the second line of~\eqref{eq:M_j-M'_j} by the
  functor $p_j \circ e_j$. The desired equality~\eqref{eq:M_V-cone}
  now follows by using~\eqref{eq:sigma_j-p_j-e_j} together with the
  fact that $A_{\infty}$-functors pull back mapping cones to mapping
  cones (at the chain level). Note that for $j=0$, pulling back the
  equality from point~\eqref{i:M'_0} above yields:
  $\mathcal{M}_{V;\gamma_0, p, h_0} = \mathcal{L}_0$.

  We now turn to the weakly filtered setting. Throughout the rest of
  the proof it is useful to keep in mind that $h_j(O_k) = h(O_k)$ for
  every $0 \leq j \leq r$ and $1 \leq k \leq 2j+3$.

  We claim that in the weakly filtered setting the correct version
  of~\eqref{eq:M_V-cone} has the form:
  \begin{equation} \label{eq:SMV-1}
    \begin{aligned}
      & S^{h(O_3)}\mathcal{M}_{V;\gamma_j, p, \upsilon} =
      \tcn(\mathcal{L}_j \xrightarrow{\; (\phi_j; 0,
        \bm{\delta}^{\phi_j}) \;} S^{h(O_3)}
      \mathcal{M}_{V;\gamma_{j-1}, p, \upsilon}), \; \; \forall \; 1\leq j\leq r, \\
      & S^{h(O_3)}\mathcal{M}_{V; \gamma_0, p, \upsilon} = \mathcal{L}_0.
    \end{aligned}
  \end{equation}

  Of course, by Lemma~\ref{l:action-shift-cone}, the first line
  of~\eqref{eq:SMV-1} is equivalent to:
  \begin{equation} \label{eq:SMV-2} \mathcal{M}_{V;\gamma_j, p,
      \upsilon} = \tcn(\mathcal{L}_j \xrightarrow{\; (\phi_j; 0,
      \bm{\delta}^{\phi_j}+h(O_3)) \;} \mathcal{M}_{V;\gamma_{j-1}, p,
      \upsilon}), \; \; \forall \; 1\leq j\leq r.
  \end{equation}

  To prove~\eqref{eq:SMV-2} one needs to go over the arguments in the
  proof of~\cite[Proposition~4.4.1]{Bi-Co:lcob-fuk} and take
  action-filtrations into consideration. An inspection of these
  arguments shows that the categories $\mathcal{B}_j$,
  $\mathcal{B}'_j$ and functors $e_j$, $p_j$, $\sigma_j$, $q_j$ are
  all weakly filtered, and so are the modules $\mathcal{M}'_j$ and
  $\overline{\mathcal{M}}_j$. Moreover, we have:
  \begin{enumerate}
  \item The discrepancies of both $\mathcal{B}_j$ and $\mathcal{B}'_j$
    are
    $\leq \bme^{\fukcob(\widetilde{\mathcal{C}}_{1/2};
      \iota_{\gamma_j}(p, h_j))}$.
  \item Both functors $p_j$ and $q_j$ are filtered, i.e. have
    discrepancies $\leq \bm{0}$.
  \item $e_j$ has discrepancy $\leq \bme^{e_j}$, where
    $\epsilon_1^{e_j} = \max \{ h_j(O^j_k) \mid 1 \leq k = \text{odd}
    \leq 2j+5\}$ and $\epsilon_d^{e_j} = 0$ for all $d \geq 2$.
  \item $p_j \circ e_j$ has discrepancy $\leq \bme^{p_j \circ e_j}$,
    where
    $\epsilon_1^{p_j \circ e_j} = \max \{ h(O_k) \mid 1 \leq k =
    \text{odd} \leq 2j+3\}$ and $\epsilon_d^{p_j \circ e_j} = 0$ for
    all $d \geq 2$.
  \item $\sigma_j$ has discrepancy $\leq \bme^{\sigma_j}$, where
    $\epsilon_1^{\sigma_j} = -h(O_{2j+3})$ and
    $\epsilon_d^{\sigma_j} = 0$ for all $d \geq 2$.
  \item The module homomorphism
    $\varphi_j: \sigma_j^*\mathcal{L}_j \longrightarrow q_j^*
    \mathcal{M}'_{j-1}$ shifts action by $\leq 0$ and has discrepancy
    $\leq \bme^{\varphi_j}$, where
    $$\epsilon_d^{\varphi_j} =
    \epsilon_d^{\fukcob(\widetilde{\mathcal{C}}_{1/2};
      \iota_{\gamma_j}(p, \upsilon))} + h(O_{2j+3}).$$
  \item The modules $\mathcal{M}'_j$ and $\overline{\mathcal{M}}'_j$
    have discrepancies
    $\leq \bme^{\fukcob(\widetilde{\mathcal{C}}_{1/2};
      \iota_{\gamma_j}(p, h_j))}$.
  \item The equalities (or identifications) from~\eqref{eq:M_j-M'_j}
    hold also in the weakly filtered sense, where the cone over
    $\varphi_j$ on the 2'nd line of~\eqref{eq:M_j-M'_j} is now taken
    over $(\varphi_j; 0, \bme^{\varphi_j})$.
  \item $\mathcal{M}'_0 = S^{-h(O_3)}\sigma_0^*\mathcal{L}_0$ as
    weakly filtered modules. \label{i:M'_0-wf}
  \end{enumerate}
  To conclude the proof of~\eqref{eq:SMV-2} we pull back the weakly
  filtered version of the 2'nd line of~\eqref{eq:M_j-M'_j} by
  $p_j \circ e_j$ and use
  Lemmas~\ref{l:pull-back-cone},~\ref{l:pull-back-M}
  and~\ref{l:pull-back-f} (recall that $p_j$, $e_j$ do not have higher
  order terms). The assertion that
  $S^{h(O_3)}\mathcal{M}_{V; \gamma_0, p, \upsilon} = \mathcal{L}_0$
  follows in a similar way.
\end{proof}


TITLE: Correct proof of multiplicative limit law?
QUESTION [0 upvotes]: I want to prove that if  $\lim_{x \to a} f\left(x \right) = l$ and
$\lim_{x \to a} g(x) = m$ we have $\lim_{x \to a} f(x)g(x) = l\cdot m$.
Proof:
Let $\epsilon   > 0$ arbitrary. We have
\begin{align*}
  &|f(x)g(x)-l\cdot m| = |f(x)g(x) - l\cdot m + f(x)\cdot m - f(x)\cdot m|
  \\[15pt]
  &=  |f(x)(g(x)-m) + m(f(x)-l)| \le  |f(x)(g(x)-m)| + |m(f(x)-l)|
  \\[15pt]
  &= |f(x)|\cdot  |g(x)-m| + |m|\cdot  |f(x)-l|
 .\end{align*}
By assumption there exists some $\delta _1> 0$ s.t. $|f(x)-m| < \epsilon _1$
for $\epsilon _1 = \min\left\{ \dfrac{\epsilon}{|m|},\epsilon \right\}  $
and $x \in (a-\delta _1, a+\delta _1)$.
We can estimate the factor $|f(x)|$ as follows:
\begin{align}
  |f(x)| = |f(x) - l + l| \le |f(x) - l| + |l|
 .\end{align}
and  hence obtain
\begin{align*}
  |f(x)|\cdot  |g(x)-m| + |m|\cdot  |f(x)-l|
  \le  &|f(x) - l|\cdot |g(x)-m|
  \\[15pt]
  &+ |l|\cdot |g(x)-m|+
  |m|\cdot  |f(x)-l|
 .\end{align*}
As before, there exists some $\delta _2 > 0$ s.t.
$|g(x)-m| < \epsilon _2$ with $\epsilon _2=
 \min\left\{\dfrac{\epsilon}{|l|},\epsilon \right\} $
so choosing $\delta_3 = \min\left\{\delta_1, \delta_2 \right\} $
we get for $x \in (a-\delta_3, a+\delta_3)$
\begin{align*}
  \underbrace{|f(x)-l|}_{\displaystyle < \epsilon}
  \underbrace{|g(x)-m|}_{\displaystyle < \epsilon}
  + \underbrace{|l|\cdot |g(x)-m|}_{\displaystyle < \epsilon}
  + \underbrace{|m|\cdot |f(x)-l|}_{\displaystyle < \epsilon}
  < \epsilon^2 +  2\epsilon
 .\end{align*}
Since $\epsilon$ was arbitrary the corollary follows.
Does my proof seem correct?

REPLY [1 votes]: Your proof is mostly fine, but here are some issues:

In your definition of limits, is it a punctured neighborhood ($0<|x-a|<\delta$) or is it a full neighborhood ($|x-a|<\delta$, i.e equivalently $x\in (a-\delta,a+\delta)$)? I suggest you check again with your textbook, because different books have different definitions (though punctured neighborhoods make more sense to me).

You introduce $\frac{\epsilon}{|m|}$ and $\frac{\epsilon}{|l|}$, but the issue is the denominators could be $0$, so you should introduce something like $\frac{\epsilon}{1+|m|}$ and $\frac{\epsilon}{1+|l|}$. This is fine because $|m|\frac{\epsilon}{1+|m|}<\epsilon$, so all your estimates still work.

If you have proven that $x\mapsto x^2$ is a continuous function (at the origin), then your proof is fine, because you finished off with $\epsilon^2+2\epsilon$, which approaches $0$ as $\epsilon\to 0^+$. On the other hand, if you havent proven that $x\mapsto x^2$ is continuous at the origin, then I would suggest you be more explicit and say in the beginning that we may WLOG assume $0<\epsilon<1$. Then, in the end you can finish off with $\cdots <\epsilon^2+\epsilon < \epsilon+2\epsilon=3\epsilon$

Also, it looks like you're not entirely concerned with having "a perfect $<\epsilon$" at the end of your calculation. I assume you've already proven some theorems which assert that this is fine. If yes, then I would suggest the following shortening of the proof:

Let $\epsilon>0$. By our assumption there exist $\delta_1>0$ such that for all $0<|x-a|<\delta_1$, we have $|f(x)-l|<\min(1,\epsilon)$. Similarly, there is a $\delta_2>0$ such that $0<|x-a|<\delta_2$ implies $|g(x)-m|<\min(1,\epsilon)$. Now, let $0<|x-a|<\delta:=\min(\delta_1,\delta_2)$. Then,
\begin{align}
|f(x)g(x)-lm|&\leq |f(x)|\cdot|g(x)-m|+|m|\cdot|f(x)-l|\\
&\leq (1+|l|)\cdot\epsilon+|m|\cdot \epsilon\\
&=(1+|l|+|m|)\epsilon.
\end{align}
Since $\epsilon>0$ was arbitrary, this completes the proof.


TITLE: Is $\mathbb R$ with usual euclidean topology, homeomorphic with some topological field of positive characteristic?
QUESTION [4 upvotes]: Does there exists a topological field of positive characteristic which is homeomorphic with $\mathbb R$ with the usual topology ?
By homeomorphism here, I mean just topological homeomorphism, not necessarily preserving any algebraic structure.

REPLY [5 votes]: Consider the map $\psi:x\mapsto x+1$ on a topological field $K$.
It's a homeomorphism from $K$ to itself, and it has no fixed points.
If $K$ has characteristic $p$ then $\psi$'s $p$-th power is the identity.
But on $\Bbb R$ every homeomorphism $\psi:\Bbb R\to \Bbb R$ without fixed points
is strictly increasing and either $\psi(x)-x$ is always positive
or $\psi(x)-x$ is always negative. So either $0<\psi(0)<\psi^2(0)<\psi^3(0)<
\cdots$ or $0>\psi(0)>\psi^2(0)>\psi^3(0)>
\cdots$. Either way, $\psi^p$ cannot be the identity.


TITLE: Determining the modulus of a complex number from equation
QUESTION [0 upvotes]: $|2z+3|=|z+6|$ where $z$ is a complex number represents a circle with center $(0,0)$ and radius $3$. It is fairly easy to prove it by letting $z=x+iy$ and we get the standard equation of circle in cartesian form. But is there any way we can deduce $|z|=3$ without exploiting $z=x+iy$ e.g in cartesian form? The reason being the cartesian calculation is a tad bit tedious. So it will be helpful if someone suggests a clever way of manipulating this.

REPLY [0 votes]: Instead of working with the modulus directly, you can start with the two expressions $2z+3$ and $z+6$ and multiply each by their complex conjugate. Then you get:
$$4|z^2|  + 12|z| +9 = |z^2| +12 |z| + 36$$
Simplifying leads to: $$|z^2| = 9$$


text:
\section{Mathematical Preliminaries}
\label{sec:Preliminaries}
\subsection{Discrete prolate spheroidal sequences and Slepian basis vectors}
\label{sec:DPSS}
We begin by defining the discrete prolate spheroidal sequences (DPSSs) and Slepian basis vectors and reviewing some of their key properties. For a discrete signal $x \in \ell_2(\Z)$, we define its discrete time Fourier transform (DTFT) $\widehat{x} \in L_2([-\tfrac{1}{2},\tfrac{1}{2}])$ by $$\widehat{x}(f) = \sum_{n = -\infty}^{\infty}x[n]e^{-\j 2\pi fn} \quad \text{for} \quad f \in [-\tfrac{1}{2},\tfrac{1}{2}],$$ where we use the notation $\j = \sqrt{-1}$. The inverse DTFT is given by $$x[n] = \int_{-1/2}^{1/2}\widehat{x}(f)e^{\j 2\pi fn}\,df \quad \text{for} \quad n \in \Z.$$ With these definitions, any $x,x' \in \ell_2(\Z)$ satisfy the Parseval-Plancherel identity $\inner{x,x'}_{\ell_2(\Z)} = \inner{\widehat{x},\widehat{x}'}_{L_2([-1/2,1/2])}$. For any $N \in \N$, we say that $x \in \ell_2(\Z)$ is timelimited to $n \in \{0,\ldots,N-1\}$ if $x[n] = 0$ for $n \in \Z \setminus \{0,\ldots,N-1\}$. Also, for any $W \in (0,\tfrac{1}{2})$, we say that $x \in \ell_2(\Z)$ is bandlimited to $|f| \le W$ if $\widehat{x}(f) = 0$ for $|f| > W$.

We can now ask the question ``what signal bandlimited to $|f| \le W$ has a maximum concentration of energy over the time indices $n \in \{0,\ldots,N-1\}$?'', i.e., $$\maximize_{x \in \ell_2(\Z)} \sum_{n = 0}^{N-1}|x[n]|^2 \quad \text{subject to} \quad \|x\|_{\ell_2(\Z)}^2 = 1 \quad \text{and} \quad \widehat{x}(f) = 0 \ \text{for} \ |f| > W.$$

To help answer this question, we define two self-adjoint operators. For a given $N \in \N$ we define a timelimiting operator $\calT_N : \ell_2(\Z) \to \ell_2(\Z)$ by $$(\calT_N x)[n] = \begin{cases}x[n] & \text{if} \ n \in \{0,\ldots,N-1\} \\ 0 & \text{if} \ n \in \Z\setminus\{0,\ldots,N-1\} \end{cases}.$$ For a given bandwidth parameter $W \in (0,\tfrac{1}{2})$, we define a bandlimiting operator $\calB_W : \ell_2(\Z) \to \ell_2(\Z)$ by $$(\calB_W x)[n] = \sum_{m = -\infty}^{\infty}\dfrac{\sin[2\pi W(m-n)]}{\pi(m-n)}x[m] \quad \text{for} \quad n \in \Z.$$ Note that the DTFT of $\calB_W x$ satisfies $\widehat{\calB_W x}(f) = \widehat{x}(f)$ for $|f| \le W$ and $\widehat{\calB_W x}(f) = 0$ for $|f| > W$.

For bandlimited signals $x \in \ell_2(\Z)$, we can write $$\sum_{n = 0}^{N-1}|x[n]|^2 = \inner{x,\calT_Nx}_{\ell_2(\Z)} = \inner{\calB_Wx,\calT_N\calB_Wx}_{\ell_2(\Z)} = \inner{x,\calB_W\calT_N\calB_Wx}_{\ell_2(\Z)}.$$ Subject to the constraint $\|x\|_{\ell_2(\Z)}^2 = 1$, this is maximized by the eigensequence of $\calB_W\calT_N\calB_W$ corresponding to the largest eigenvalue. Slepian defined the discrete prolate spheroidal sequences (DPSSs) $s_0,\ldots,s_{N-1} \in \ell_2(\Z)$ as the $N$ orthonormal eigensequences of $\calB_W\calT_N\calB_W$ corresponding to non-zero eigenvalues. The corresponding eigenvalues $1 > \lambda_0 > \lambda_1 > \cdots > \lambda_{N-1} > 0$ are referred to as the DPSS eigenvalues and are sorted in descending order. Note that the notation $s_k$ and $\lambda_k$ hides the dependence on $N$ and $W$. When it is necessary to make this dependence explicit, we will use the expanded notation $s_k(N,W)$ and $\lambda_k(N,W)$ respectively.

Slepian \cite{SlepianV} showed that these eigenvalues are all distinct and strictly between $0$ and $1$. In addition to $s_0$ being the bandlimited sequence in $\ell_2(\Z)$ with a maximal concentration of energy in $\{0,\ldots,N-1\}$, it is also true that for each $k = 1,\ldots,N-1$, $s_k$ is the bandlimited sequence in $\ell_2(\Z)$ with a maximal concentration of energy in $\{0,\ldots,N-1\}$ subject to the additional constraint of being orthogonal to $s_0,\ldots,s_{k-1}$. Furthermore $\lambda_k$ is equal to the amount of energy $s_k$ has in the time interval $\{0,\ldots,N-1\}$.

We can also ask the question ``what signal timelimited to $n \in \{0,\ldots,N-1\}$ has a maximum concentration of energy in the frequency band $|f| \le W$?'', i.e. $$\maximize_{x \in \ell_2(\Z)}\int_{-W}^{W}\left|\widehat{x}(f)\right|^2\,df \quad \text{subject to} \quad \|x\|_{\ell_2(\Z)}^2 = 1 \quad \text{and} \quad x[n] = 0 \ \text{for} \ n \in \Z\setminus\{0,\ldots,N-1\}.$$

For timelimited signals $x \in \ell_2(\Z)$, we can write $$\int_{-W}^{W}\left|\widehat{x}(f)\right|^2\,df = \inner{\widehat{x},\widehat{\calB_Wx}}_{L_2([-1/2,1/2])} = \inner{x,\calB_Wx}_{\ell_2(\Z)} = \inner{\calT_Nx,\calB_W\calT_Nx}_{\ell_2(\Z)} = \inner{x,\calT_N\calB_W\calT_Nx}_{\ell_2(\Z)}.$$ Since $\calT_N\calB_W\calT_N$ is self-adjoint, the sequence $x \in \ell_2(\Z)$ which solves the above maximization problem is the eigensequence of the operator $\calT_N\calB_W\calT_N$ corresponding to the largest eigenvalue.

Clearly, the range of $\calT_N\calB_W\calT_N$ and the orthogonal complement of the kernel of $\calT_N\calB_W\calT_N$ is the $N$-dimesional space of timelimited signals. Hence, we can reduce this eigenproblem on $\ell_2(\Z)$ to an eigenproblem on $\R^N$. With respect to the Euclidean basis for the space of timelimited signals, the matrix representation of $\calT_N\calB_W\calT_N$ is given by
\begin{equation}
\mB[m,n] = \dfrac{\sin[2\pi W(m-n)]}{\pi(m-n)} \quad \text{for} \quad m,n = 0,\ldots,N-1.
\label{eq:Prolate}
\end{equation}
This matrix $\mB \in \R^{N \times N}$ is known in the literature as the prolate matrix \cite{Varah93,Bojanczyk95}. The Slepian basis vectors $\vs_0,\ldots,\vs_{N-1} \in \R^N$ are the orthonormal eigenvectors of $\mB$, where again the eigenvalues $1 > \lambda_0 > \lambda_1 > \cdots > \lambda_{N-1} > 0$ are sorted in descending order. Note that the eigenvalues of $\mB$ are the same as the eigenvalues of $\calT_N\calB_W\calT_N$, which are the same as the eigenvalues of $\calB_W\calT_N\calB_W$. Hence, we can reuse the notation $\lambda_k$ for $k = 0,\ldots,N-1$ to denote the eigenvalues of $\mB$. The eigensequences $s'_0,\ldots,s'_{N-1} \in \ell_2(\Z)$ of $\calT_N\calB_W\calT_N$ are then given by $s'_k[n] = \vs_k[n]$ for $n \in \{0,\ldots,N-1\}$ and $s'_k[n] = 0$ for $n \in \Z\setminus\{0,\ldots,N-1\}$. Note that in addition to $s'_0$ being the timelimited sequence in $\ell_2(\Z)$ whose DTFT has a maximal concentration of energy in $[-W,W]$, it is also true that for each $k = 1,\ldots,N-1$, $s'_k$ is a timelimited sequence in $\ell_2(\Z)$  whose DTFT has a maximal concentration of energy in $[-W,W]$ subject to the additional constraint of being orthogonal to $s'_0,\ldots,s'_{k-1}$. Furthermore, the eigenvalue $\lambda_k$ is equal to the amount of energy that the DTFT of $s'_k$ has in the frequency band $[-W,W]$.

\subsection{Prolate spheroidal wave functions}
\label{sec:PSWF}
The concentration problems in the previous subsection can also be posed in the continuous case. For continuous signals in $L_2(\R)$, we define the continuous time Fourier transform and inverse Fourier transform by $$\widehat{y}(\omega) = \int_{-\infty}^{\infty}y(t)e^{-\j \omega t}\,dt \quad \text{for} \quad \omega \in \R \quad \text{and} \quad y(t) = \dfrac{1}{2\pi}\int_{-\infty}^{\infty}\widehat{y}(\omega)e^{\j \omega t}\,d\omega \quad \text{for} \quad t \in \R.$$ With these definitions, any $y,y' \in L_2(\R)$ satisfy the Parseval-Plancherel identity $\inner{y,y'}_{L_2(\R)} = \tfrac{1}{2\pi}\inner{\widehat{y},\widehat{y}'}_{L_2(\R)}$.

Again, for a given bandlimit $\Omega > 0$ and duration $T > 0$, we can ask ``what signal bandlimited to $|\omega| \le \Omega$ has a maximum concentration of energy in the time interval $t \in [-\tfrac{T}{2},\tfrac{T}{2}]$?'', i.e. $$\maximize_{y \in L_2(\R)}\int_{-T/2}^{T/2}\left|y(t)\right|^2\,dt \quad \text{subject to} \quad \|y\|_{L_2(\R)}^2 = 1 \quad \text{and} \quad \widehat{y}(\omega) = 0 \ \text{for} \ |\omega| > \Omega.$$

Just as was done with the discrete case, we can define a self-adjoint timelimiting operator $\calTc_T : L_2(\R) \to L_2(\R)$ and a self-adjoint bandlimiting operator $\calBc_{\Omega} : L_2(\R) \to L_2(\R)$ by $$(\calTc_Ty)(t) = \begin{cases} y(t) & \text{if} \ |t| \le \tfrac{T}{2} \\ 0 & \text{if} \ |t| > \tfrac{T}{2}\end{cases} \quad \text{and} \quad (\calBc_{\Omega}y)(t) = \int_{-\infty}^{\infty}\dfrac{\sin[\Omega(t-t')]}{\pi(t-t')}y(t')\,dt' \quad \text{for} \quad t \in \R,$$ and write $$\int_{-T/2}^{T/2}\left|y(t)\right|^2\,dt = \inner{y,\calTc_{T}y}_{L_2(\R)} = \inner{\calBc_{\Omega}y,\calTc_{T}\calBc_{\Omega}y}_{L_2(\R)} = \inner{y,\calBc_{\Omega}\calTc_T\calBc_{\Omega}y}_{L_2(\R)}$$ for any bandlimited signal $y \in L_2(\R)$.

The result is that the solution to this concentration problem is top eigenfunction of the operator $\calBc_{\Omega}\calTc_T\calBc_{\Omega}$. The orthonormal eigenfunctions $\psi_0, \psi_1, \ldots \in L_2(\R)$ of the self-adjoint operator $\calBc_{\Omega}\calTc_T\calBc_{\Omega}$ are known as the prolate spheroidal wave functions (PSWFs), and the corresponding eigenvalues $1 > \widetilde{\lambda}_0 > \widetilde{\lambda}_1 >  \cdots > 0$ are known as the PSWF eigenvalues, and are sorted in decreasing order.

It is easy to check that the timelimited PSWFs $\calTc_T\psi_k$ for $k \in \N$ are the eigenfunctions of the operator $\calTc_T\calBc_{\Omega}\calTc_T$ with the same corresponding eigenvalues $\widetilde{\lambda}_k$. The action of the operator $\calTc_T\calBc_{\Omega}\calTc_T$ on a signal $y \in L_2(\R)$ is given by $$(\calTc_T\calBc_{\Omega}\calTc_Ty)(t) = \int_{-T/2}^{T/2}\dfrac{\sin[\Omega(t-t')]}{\pi(t-t')}y(t')\,dt' \quad \text{for} \quad t \in [-\tfrac{T}{2},\tfrac{T}{2}].$$ With a simple change of variable, it can be shown that the eigenvalues of the above kernel integral operator only depend on the product $\Omega T$. When it is necessary to denote this dependence, we use the notation $\widetilde{\lambda}_k(c)$ to denote $\widetilde{\lambda}_k$ for values of $\Omega, T > 0$ which satisfy $\Omega T/2 = c$.

As in the discrete case, $\psi_0$ is the signal bandlimited to $|\omega| \le \Omega$ whose energy is maximally concentrated in the time interval $t \in [-\tfrac{T}{2},\tfrac{T}{2}]$, and for each integer $k \ge 1$, $\psi_k$ is the signal bandlimited to $|\omega| \le \Omega$ whose energy is maximally concentrated in the time interval $t \in [-\tfrac{T}{2},\tfrac{T}{2}]$ subject to the additional constraint of being orthogonal to $\psi_0,\ldots,\psi_{k-1}$. Furthermore, the eigenvalue $\widetilde{\lambda}_k$ is the equal to the energy of $\psi_k$ in the time interval $t \in [-\tfrac{T}{2},\tfrac{T}{2}]$. Also, $\calTc_T\psi_0$ is the signal timelimited to $t \in [-\tfrac{T}{2},\tfrac{T}{2}]$ whose Fourier transform is maximally concentrated in the frequency band $|\omega| \le \Omega$, and for each integer $k \ge 1$, $\calTc_T\psi_k$ is the signal timelimited to $t \in [-\tfrac{T}{2},\tfrac{T}{2}]$ whose Fourier transform is maximally concentrated in the frequency band $|\omega| \le \Omega$ subject to the additional constraint of being orthogonal to $\calTc_T\psi_0,\ldots,\calTc_T\psi_{k-1}$. Furthermore, the eigenvalue $\widetilde{\lambda}_k$ is the equal to the fraction of energy that $\calTc_T\psi_k$ has in the frequency band $|\omega| \le \Omega$.