TITLE: Generalization of $x^n-1$ following Fermat's Little theorem
QUESTION [0 upvotes]: Fermat's Little Theorem tells us that if $\gcd(x,p)=1$ and $p$ is prime, then $x^{p-1} = 1 \pmod p$. Equivalently, if $a(p)=x^{p-1}-1$, then $a(p-1)=0 \pmod p$ if $p$ is prime. When $p > 2$, $a(p-1)$ is reducible and its irreducible factors are cyclotomic polynomials, more specifically, its irreducible factors are polynomials defining a cyclotomic field. Let $q, q_2, q_3,... q_n$ denote all irreducible polynomial factors of $a(p-1)$. When $a(p-1)$ is factored into linear factors $\pmod p$, then all of its factors are distinct, meaning NO square or repeated factors, the factorization is squarefree!
Now the question is, does there exist a sequence $a(p)$ similar to $x^p-1$, such that all of its irreducible polynomial factors define a cyclotomic field, $a(p)$ does NOT contain any repeated factor (factors into distinct polynomials), and the factorization of $a(p-1)$ for any prime $p$ has $p-2$ distinct linear factors? Also, $a(p)$ cannot be of the form $a(p) = (x - a)^p - (y - b)^p$ for any integers $x, y, a, b$.
After repeatedly for days trying to solve this problem, my guess is, unfortunately no other sequence exists. I may be wrong though. Provide a counterexample sequence if so. Thanks for help!

REPLY [1 votes]: There is indeed a set of equations similar to this.  It is truly facinating.  I have been using these very equations for nigh on near next to forty years, if not more.  And yes, it generalises in a most spectular way.
An isoseries is a form t(n+1) = a.t(n) - t(n-1).  When t(0) = 0, t(1)=1, and t(2) is the short-chord (ie the base of a triangle of two edges) of a polygon, the isoseries makes the chords of a polygon.
The selection of members of an isoseries at a stepping of m, makes an isoseries itself, as t(n+m)=a(m)t(n)-t(m-n).  It follows directly that a(m) is itself an isoseries in a, in that a(m+1)=a.a(m)-a(m-1), and that a(0)=2, a(1)=a.
The equations break down like the base-form, into 'algebraic roots', the even ones are exactly the equations that the short-chord of a p/2 solve, the odd ones derive from the even ones in the manner of setting x to -x.
The general series corresponding to $b^n-1$ for these, is to start an isomorphic series at $0, \sqrt{n-2}$, with the shortchord $a = \sqrt{n+2}$.  Then the equations below will divide the result when the subscript divides the term number, so J4 divides every fourth term.
J4   x                a(2) = 0.0000000
J6   x-1              a(3) = 1.0000000
J3   x+1
J8   x2 - 2           a(4) = 1.41421356
J10  x2-x-1           a(5) = 1.618033988
J5   x2+x-1
J12  x2 - 3           a(6) = 1.73205080757
J14  x3-x2+2x+1       a(7) = 1.801937736
J7   x3+x2+2x-1
J18  x3-3x-1          a(9) = 1.879385241
J9   x3-3x-1
J16  x4-4x2+2         a(8) = 1.847759065
J20  x4-5x2+5         a(10) = 1.90211303259
J24  x4-4x2+1         a(12) = 1.93185165259

When one sets x=-3, these equations evaluate to the uniqie factors that occur in the fibonacci series.
The behaviour of this series, is that p can divide either a(p-1) or a(p+1), or some divisor of these (in the manner of fermat's little theorm), and there is an even or odd number of these according to gauss's law.
Cyclotomic Numbers
When a is taken to be the shortchord of a polygon, then the span of the isoseries is the span of chords, say Zn.  When this set is taken over the set of [1,w], where w^n = -1, as its simplest root, then this gives the cyclotomic numbers CZn.
The case for the heptagon
This web page http://www.os2fan2.com/p7flat.html  shows the heptagonal version of the fibonacci series.  It is two-dimensional.  Primes for which 7 divides p+1 or p-1, have periods dividing p-1.  The remainder have periods that divide p²+p+1.  For example, 2 has a 7-place period, and 5 has a 31-period, while 3 has a period of 13.  13, on the other hand, has a 12-place period.
The sevenly flat is generated by the iteration to the right, but in the convergant region, it's $\frac{a^xb^y}{-1+a+b}$.  The general form is to add two extra terms, where the numerator a,b are replaced by their isomorphs.


TITLE: Subgroups of a Finite Group containing two elements
QUESTION [0 upvotes]: Currently working on a Group Theory question:
Let $G$ be a cyclic group of order 100 and let $a \in G$ denote a generator.
Find two subgroups of $G$ which contain both $a^{20}$ and $a^{55}$.
Are there any other subgroups of $G$ containing these two elements?
So my attempt at this question is taking the given. The order of $G$ is 100 and I know that $a^m = n/gcd(m,n)$. But I'm unsure as to how this helps.  Any hints on how to proceed are appreciated.

REPLY [1 votes]: Hints: fill in details.
Observe that
$$ord(a^{20})=5\;,\;\;ord(a^{55})=20\;\implies\;\text{any subgroup containing this two elements will}$$
have at least order $\;20\;$ ....but also  order $\;20\;$ at most (why? Lagrange theorem), and since any finite cyclic group has one unique proper subgroup of any order dividing the group's order then there is only one such subgroup


TITLE: Definition of graph [Diestel's book]
QUESTION [2 upvotes]: A graph is a pair $G=(V,E)$ of sets such that $E\subseteq [V]^2$;
thus, the elements of $E$ are $2$-element subsets of $V$. To avoid
notational ambiguities, we shall always assume tacitly that $V\cap
 E=\varnothing$.

This is a definition of a graph from Diestel's book (fifth edition). By $[V]^2$ the author denotes the set of all $2$-element subsets of $V$. But we see that $V\cap E=\varnothing$ always because $V$ and $E$ are sets of distinct nature so they cannot have elements in common. Probably this a bad wording.
Can anyone explain it what the author meant? Or maybe I am wrong?

REPLY [5 votes]: I guess it could be possible to have vertex set $\{a,b,\{a,b\} \}$, but we are not allowing for such shenanigans to avoid confusion.


text:
\begin{document}
\title[The unboundedness of Hausdorff operators on Quasi-Banach spaces]
{The unboundedness of Hausdorff operators on Quasi-Banach spaces}

\author{WEICHAO GUO}
\address{School of Science, Jimei University, Xiamen, 361021, P.R.China}
\email{weichaoguomath@gmail.com}
\author{JIANMIAO RUAN}
\address{Department of Mathematics, Zhejiang International Studies University,
Hangzhou 310012, P. R. China}
\email{rjmath@163.com}

\author{GUOPING ZHAO}
\address{School of Applied Mathematics, Xiamen University of Technology, Xiamen, 361024, P.R.China}
\email{guopingzhaomath@gmail.com}
\thanks{}

\subjclass[2010]{Primary 42B35; Secondary 47G10.}

\keywords{Hausdorff operator, local Hardy space, modulation space, unboundedness.}

\begin{abstract}
  In this note, we show that the Hausdorff operator $H_{\Phi}$
  is unbounded on a large family of Quasi-Banach spaces, unless $H_{\Phi}$ is a zero operator.
\end{abstract} \maketitle

\section{Introduction and Motivation}
Hausdorff operator, in connection with some classical summation methods, has been studied for a long time in the field of real and complex analysis. For the historical background and recent developments of the boundedness on function spaces regarding Hausdorff operators,
we refer the reader to survey papers by Chen-Fan-Wang \cite{ChenFanWang2013AMJCUSB} and Liflyand \cite{Liflyand2011}.

For a suitable function $\Phi$, the corresponding Hausdorff operator $H_{\Phi}$ can be defined by
\begin{equation}
  H_{\Phi}f(x):=\int_{\mathbb{R}^n}\Phi(y)f\left(\frac{x}{|y|}\right)dy.
\end{equation}

Particularly, when $\Phi$ is taken suitably, Hausdorff operator contains some important operators in the field of harmonic analysis.
For instance, the Hardy operator, adjoint Hardy operator (see \cite{ChenFanLi2012CAMSB, ChenFanZhang2012AMSES, FanLin2014AB}),
and the Ces\`{a}ro operator \cite{Miyachi2004JFAA, Siskakis1990PAMS} in one dimension.
The Hardy-Littlewood-P\'{o}lya operator and the Riemann-Liouville fractional integral operator can also be derived from the Hausdorff operator.

In recent years, there is an increasing interest on
the boundedness of Hausdorff operators on function spaces,
one can see \cite{GaoWuGuo2015MIA,GaoZhong2014MIA} for the
the boundedness on Lebesgue spaces,
\cite{ZhaoFanGuo2018AFA} for the boundedness on modulation and Wiener amalgam spaces,
and \cite{FanLin2014AB, LiflyandMoricz2000PAMS, RuanFan2016JMAA} for the boundedness on $H^1$ and $h^1$.
However, since the argument of using Minkowski's inequality cannot be applied to Quasi-Banach spaces,
there are only a few boundedness results considering the boundedness of Hausdorff operators on Quasi-Banach spaces.

Among the previous results, the study of Hausdorff operators on $H^p(\rn)\ (0<p<1)$ has its special status, since
$H^p(\rn)$ is a Quasi-Banach space and also an important function space in the field of harmonic analysis.
The study of $H_{\Phi}f$ on $H^p(\rr)$ was first initiated by Kanjin \cite{Kanjin2001SM} and continued in
\cite{LiflyandMiyachi2009SM}.
Very recently, Liflyand-Miyachi \cite{LiflyandMiyachi2019TAMS} establishes the multidimensional boundedness results on $H^p(\rn)$.
For the $H^p(\rr)$ boundedness of another type of Hausdorff operator,
we refer the reader to \cite{ChenFanLinRuan2016AM, RuanFan2017MN}.

Based on the previous results,
an interesting problem is whether we can establish the boundedness result on local Hardy space $h^p(\rn)$,
or even on the general inhomogeneous frequency decomposition spaces such as  Triebel-Lizorkin spaces $F^{p,q}_s(\rn)$,
Besov spaces $B^{p,q}_s(\rn)$ and modulation spaces $M^{p,q}_s(\rn)$ with $p\in (0,1)$. In this paper, we will explore this problem and give an negative answer,
showing that the Hausdorff operator is unboundedness on a large family of
Quasi-Banach spaces.
In particular, we prove that all the nonzero Hausdorff operators are unbounded on
$F_s^{p,q}$, $B_s^{p,q}$ or $M^{p,q}_s$ with $p\in (0,1)$.

Our paper is organized as follows.
In Section 2, we collect some notations and basic properties of function spaces.
The unboundedness results and their proofs will be presented in Section 3.

Throughout this paper, we will adopt the following notations.
We use $X\lesssim Y$ to denote the statement that $X\leq CY$, with a positive constant $C$ that may depend on $n, \,p$,
but it might be different from line to line.
The notation $X\sim Y$ means the statement $X\lesssim Y\lesssim X$.
We use $X\lesssim_{\lambda}Y$ to denote $X\leq C_{\lambda}Y$,
meaning that the implied constant $C_{\lambda}$ depends on the parameter $\lambda$.
For $x=(x_{1},x_{2},...,x_{n})\in \rn$,
we denote $|x|_{\infty }:=\max\limits_{i=1,2,...,n}|x_{i}|$ and $\langle x\rangle:=(1+|x|^{2})^{{1}/{2}}.$

\section{Preliminary}
Let $\mathscr{S}:= \mathscr{S}(\mathbb{R}^{n})$ be the Schwartz space
and $\mathscr{S}':=\mathscr{S}'(\mathbb{R}^{n})$ be the space of tempered distributions.
We define the Fourier transform $\mathscr {F}f$ and the inverse Fourier transform $\mathscr {F}^{-1}f$ of $f\in \mathscr{S}(\mathbb{R}^{n})$ by
$$
\mathscr {F}f(\xi)=\hat{f}(\xi)=\int_{\mathbb{R}^{n}}f(x)e^{-2\pi ix\cdot \xi}dx
,
~~
\mathscr {F}^{-1}f(x)=\check{f}(x)=\int_{\mathbb{R}^{n}}f(\xi)e^{2\pi ix\cdot \xi}d\xi.
$$

The local Hardy space was introduced by Goldberg \cite{Goldberg1979DMJ}.
Let $0<p<\infty$ and let $\psi\in\mathscr{S}$ satisfy $\int_{\mathbb{R}^n}\psi(x)dx\neq0$.
Define $\psi_t(x): =t^{-n}\psi(x/t)$.
The \emph{local Hardy space} is defined by
\begin{equation*}
  h^p:=\{f\in\mathscr{S}': \|f\|_{h^p}=\|\sup_{0<t<1}|\psi_t\ast f|\|_{L^p}<\infty\}.
\end{equation*}
We note that the definition of the local Hardy spaces is independent of the choice of $\psi\in \mathscr{S}$.

To define Besov space and Triebel-Lizorkin space, we introduce  the dyadic decomposition of $\mathbb{R}^n$.
Let $\varphi$ be a smooth bump function supported in the ball $\{\xi: |\xi|<3/2\}$ and be equal to 1 on the ball $\{\xi: |\xi|\leq 4/3\}$.
Denote
\begin{equation}\label{decompositon function of Besov space}
\phi(\xi)=\varphi(\xi)-\varphi(2\xi),
\end{equation}
and a function sequence
\begin{equation}
\begin{cases}
\phi_j(\xi)=\phi(2^{-j}\xi),~j\in \mathbb{Z}^{+},
\\
\phi_0(\xi)=1-\sum\limits_{j\in \mathbb{Z}^+}\phi_j(\xi)=\varphi(\xi).
\end{cases}
\end{equation}
For integers $j\geq 0$, we define the Littlewood-Paley operators
\begin{equation}
\Delta_j=\mathscr{F}^{-1}\phi_j\mathscr{F}.
\end{equation}
Let $0<p,q\leq \infty$, $s\in \mathbb{R}$. For a tempered distribution  $\ f$,  we set the norm
\begin{equation}
\|f\|_{B^{p,q}_s}=\left(\sum_{j=0}^{\infty}2^{jsq}\|\Delta_jf\|_{L^p}^q \right)^{1/q},~

\end{equation}
with the usual modifications when $q=\infty$.
The (inhomogeneous) Besov  space ${B^{p,q}_s}$ is the space of all tempered distributions $f$ for which the quantity $\|f\|_{B^{p,q}_s}$ is finite.
Let $0<p<\infty$, $0<q\leq \infty$, $s\in \mathbb{R}$. For a tempered distribution  $\ f$,  we set the norm
\begin{equation}

\|f\|_{F^{p,q}_s}=\left\|\left(\sum_{j=0}^{\infty}2^{jsq}|\Delta_jf|^q \right)^{1/q}\right\|_{L^p}
\end{equation}
with the usual modifications when $q=\infty$.
The Triebel-Lizorkin space ${F^{p,q}_s}$ is the space of all tempered distributions $f$ for which the quantity $\|f\|_{F^{p,q}_s}$ is finite.
We recall that the local Hardy space $h^p$ is equivalent with the inhomogeneous Triebel-Lizorkin space $F^{p,2}_0$ for $p\in(0,\infty)$

Next, we introduce the modulation space.
Denote by $Q_{k}$ the unit cube with the center at $k$. Then the family $\{Q_{k}\}_{k\in\mathbb{Z}^{n}}$
constitutes a decomposition of $\mathbb{R}^{n}$.
Let
$\eta: \mathbb{R}^{n} \rightarrow [0,1]$ be a smooth function satisfying that $\eta(\xi)=1$ for
$|\xi|_{\infty}\leq {1}/{2}$ and $\eta(\xi)=0$ for $|\xi|_{\infty}\geq 3/4$. Let
\begin{equation*}
\eta_{k}(\xi)=\eta(\xi-k),  k\in \mathbb{Z}^{n}
\end{equation*}
be a translation of \ $\eta$.
Since $\eta_{k}(\xi)=1$ in $Q_{k}$, we have that $\sum_{k\in\mathbb{Z}^{n}}\eta_{k}(\xi)\geq 1$
for all $\xi\in\mathbb{R}^{n}$. Denote
\begin{equation*}
\sigma_{k}(\xi)=\eta_{k}(\xi)\left(\sum_{l\in\mathbb{Z}^{n}}\eta_{l}(\xi)\right)^{-1},  ~~~~ k\in\mathbb{Z}^{n}.
\end{equation*}
It is easy to know that $\{\sigma_{k}\}_{k\in\mathbb{Z}^{n}}$
constitutes a smooth partition of the unity, and $
\sigma_{k}(\xi)=\sigma(\xi-k)$. The frequency-uniform decomposition
operators can be defined by
\begin{equation*}
\Box_{k}:= \mathscr{F}^{-1}\sigma_{k}\mathscr{F}
\end{equation*}
for $k\in \mathbb{Z}^{n}$.
Now, we give the (discrete) definition of modulation space $M^{p,q}_s$.

\begin{definition}
Let $s\in \mathbb{R}, 0<p,q\leq \infty$. The modulation space $M^{p,q}_s$ consists of all $f\in \mathscr{S}'$ such that the (quasi-)norm
\begin{equation*}
\|f\|_{M^{p,q}_s}:=\left( \sum_{k\in \mathbb{Z}^{n}}\langle k\rangle^{sq}\|\Box_k f\|_{L^p}^{q}\right)^{1/q}
\end{equation*}
is finite.
Note that this definition is independent of the choice of $\{\sigma_k\}_{k\in\mathbb{Z}^n}$.
We also recall a basic fact that $C_c^{\fy}(\rn)\subset \mathscr{S}(\rn)\subset M^{p,q}_s$
for any $s\in \mathbb{R}, 0<p,q\leq \infty$.
\end{definition}

We say $X\hookrightarrow L^p_{0}(\rn)$, if for every $\va\in C_c^{\fy}(\rn)$ we have
\be
\|\va \ast f\|_{L^p(\rn)}\leq C\|f\|_X
\ee
for all $f\in X$, where the constant $C$ is only dependent on $\va$.

Following, we collect some basic embedding results of function spaces.
\begin{lemma}\label{lemma, ebd}
Let $0<p\leq 1, 0<q\leq \fy$, $s\in \mathbb{R}$.
\bn
\item For $X=h^p, M^{p,p}, B^{p,p}, F^{p,p}$,
we have $\|g\|_{L^p}\lesssim \|g\|_X$ for all measurable functions $g\in X$.
\item For $Y=M^{p,q}_s, \mathscr{F}M^{q,p}, B^{p,q}_s, F^{p,q}_s$, we have $Y\hookrightarrow L^p_{0}(\rn)$.
\en
\end{lemma}
\begin{proof}
  We first verify that $\|g\|_{L^p}\lesssim \|g\|_{h^p}$ for measurable functions $g\in h^p$.
  Take a $C_{c}^{\fy}$ function $\psi$ with $\int_{\mathbb{R}^n}\psi(x)dx=1$.
  We have
  \be
  \lim_{t\rightarrow 0}\psi_t\ast g=g\ \ \ \  a.e.\ \text{on}\  \rn.
  \ee
  Thus,
  \be
  \|g\|_{L^p}=\|\lim_{t\rightarrow 0}\psi_t\ast g\|_{L^p}
  \leq
\|\sup_{0<t<1}|\psi_t\ast g|\|_{L^p}=\|g\|_{h^p},
  \ee
 where we use the definition of $h^p$ in the last equality.

 For $X=M^{p,p}, B^{p,p}, F^{p,p}$, the inequality $\|g\|_{L^p}\lesssim \|g\|_X$ follows directly by
  the definition of function space and the triangle inequality, we only show the details for $B^{p,p}$:
  \be
  \begin{split}
    \|f\|_{L^p}
    =
    \bigg\|\sum_{j=0}^{\infty}\Delta_jf\bigg\|_{L^p}
    \leq
    \left(\sum_{j=0}^{\infty}\|\Delta_jf\|_{L^p}^p \right)^{1/p}
    =
    \|f\|_{B^{p,p}}=\|f\|_{F^{p,p}}.
  \end{split}
  \ee

 Next, we turn to the proof of statement (2).
 First, we deal with the case $Y=M^{p,q}_s$.
 Using the triangle inequality, we have
 \be
 \begin{split}
   \|\va\ast f\|_{L^p}
   =
   \bigg\|\sum_{k\in \mathbb{Z}^n}\Box_k (\va\ast f)\bigg\|_{L^p}
   \leq
   \left(\sum_{k\in \mathbb{Z}^n}\|\Box_k (\va\ast f)\|^p_{L^p}\right)^{1/p}.
 \end{split}
 \ee
Moreover, there exists a constant $c_n$ such that $\Box_k=\sum_{|l|\leq c_n}\Box_{k+l}\circ \Box_k$, then
 \be
 \begin{split}
 \|\Box_k (\va\ast f)\|_{L^p}
 =
 \bigg\|\sum_{|l|\leq c_n}\Box_{k+l}\va\ast \Box_k f\bigg\|_{L^p}
 \lesssim
 \sum_{|l|\leq c_n}\|\Box_{k+l}\va\|_{L^p}\|\Box_k f\|_{L^p}.
 \end{split}
 \ee
The above two estimates then yield that
\be
\begin{split}
  \|\va\ast f\|_{L^p}
  \lesssim &
  \left(\sum_{k\in \mathbb{Z}^n}\left(\sum_{|l|\leq c_n}\|\Box_{k+l}\va\|_{L^p}\|\Box_k f\|_{L^p}\right)^p\right)^{1/p}
  \\
  \leq &
  \sup_{k\in \mathbb{Z}^n}\langle k\rangle^s\|\Box_k f\|_{L^p}
  \left(\sum_{k\in \mathbb{Z}^n}\left(\langle k\rangle^{-s}\sum_{|l|\leq c_n}\|\Box_{k+l}\va\|_{L^p}\right)^p\right)^{1/p}
  \\
  \lesssim &
  \|f\|_{M^{p,\fy}_s}\|\va\|_{M^{p,p}_{-s}}\lesssim \|f\|_{M^{p,q}_s},
\end{split}
\ee
where we use the fact $\va\in M^{p,q}_{-s}$ and $M^{p,q}_s\subset M^{p,\fy}_s$.

For $Y=\mathscr{F}M^{q,p}$. By the conclusion for $Y=M^{p,\fy}$,
we have $\|\va\ast f\|_{L^p}\lesssim \|f\|_{M^{p,\fy}}.$
Note that $\mathscr{F}M^{q,p}$ is equal to the Wiener amalgam space $W^{p,q}$ (see\cite[pp.10]{GuoZhao2020JFA}),
and recall the embedding relations (see \cite[Lemma 2.5]{GuoWuYangZhao2017JFA}):
\be
     \mathscr{F}M^{q,p}\subset \mathscr{F}M^{\fy,p}=W^{p,\fy}\subset M^{p,\fy}.
\ee
The conclusion follows by
\be
\|\va\ast f\|_{L^p}
  \lesssim \|f\|_{M^{p,\fy}}\lesssim \|f\|_{\mathscr{F}M^{q,p}}.
\ee

For $Y=B^{p,q}_s$, there exists a constant $\d(p,q)>0$ such that $B^{p,q}_s\subset M^{p,q}_{s-\d(p,q)}$ (see \cite[Theorem 1.2]{GuoFanZhao2018SCM}). This and the conclusion for $Y=M^{p,q}_{s-\d(p,q)}$ imply that
\be
\|\va\ast f\|_{L^p}\lesssim \|f\|_{M^{p,q}_{s-\d(p,q)}}\lesssim \|f\|_{B^{p,q}_s}.
\ee

For $Y=F^{p,q}_s$, take a constant $\d>0$, then $F^{p,q}_{s}\subset B^{p,q}_{s-\d}$ (see \cite[pp.47]{Triebel2010}).
This and the conclusion for $Y=B^{p,q}_{s-\d}$ imply that
\be
\|\va\ast f\|_{L^p}\lesssim \|f\|_{B^{p,q}_{s-\d}}\lesssim \|f\|_{F^{p,q}_s}.
\ee

\end{proof}

\section{Main Theorems}
In this section, we give our main theorems and their proofs.
Suppose $X$ is a (Quasi-)Banach space with
translation invariant: $\|T_yf\|_X=\|f\|_X$,
where
$T_yf(x):= f(x-y)$ is the translation of $f$.

\begin{theorem}\label{Thm, lp}
  Let $0<p<1$, $\Phi\in L_{loc}^{1}(\rn\bs\{0\})$. Suppose $C_c^{\infty}(\rn)\subset X$ and $\|g\|_{L^p}\lesssim \|g\|_X$ for all measurable function $g\in X$.
  We have
  \be
  H_{\Phi}\ \text{is bounded on}\ X\Longleftrightarrow H_{\Phi}=0.
  \ee
\end{theorem}
\begin{proof}
  The ``if'' part is trivial. We only present the proof for the ``only if'' part.

  We first point out that $H_{\Phi}f$ is pointwise well-defined for any smooth function $f$ supported away from the origin.
  In fact, in this case we denote $E_{x}:=\{y: f\big(\frac{x}{|y|}\big)\neq 0\}$.
  Observe that for any fixed $x\in \rn$, $E_x$ is a bounded measurable set away from the origin.
  Recalling that $\Phi\in L_{loc}^{1}(\rn\bs\{0\})$, then the following integral is convergent:
  \be
  H_{\Phi}f(x)=\int_{\mathbb{R}^n}\Phi(y)f\left(\frac{x}{|y|}\right)dy=\int_{E_x}\Phi(y)f\left(\frac{x}{|y|}\right)dy.
  \ee
  Moreover, $H_{\Phi}f$ is a measurable function on $\rn$.
  Using the polar coordinates, we write
  \be
  \begin{split}
    H_{\Phi}f(x)
    = &\int_{\mathbb{R}^n}\Phi(y)f\left(\frac{x}{|y|}\right)dy
    \\
    = &
    \int_{0}^{\fy}\int_{\mathbb{S}^{n-1}}\Phi(ry')f(x/r)d\sigma(y')r^{n-1}dr
    = : \int_{0}^{\fy}\phi(r)f(x/r)dr,
  \end{split}
  \ee
  where
  \be
  \phi(r):=\int_{\mathbb{S}^{n-1}}\Phi(ry')r^{n-1}d\sigma(y').
  \ee
It follows by $\Phi\in L_{loc}^{1}(\rn\bs\{0\})$ that $\phi\in L_{loc}^{1}(\mathbb{R}^+)$.
Since $\phi$ is a measurable function on $(0,\fy)$, almost every point in $(0,\fy)$ is
a Lebesgue point.
Hence, it suffices to verify that if there exists a Lebesgue point $r_0>0$ such that $\phi(r_0)\neq 0$,
then $H_{\Phi}$ is unbounded.

Without loss of  generality, we assume that $r_0=1$ is a Lebesgue point of $\phi$, satisfying
$\phi(1)=1$. The proof for other cases is similar.

Taking $\theta=\frac{1-p}{2}$, we set
\be
A_j=[1-2^{-\theta j}, 1+2^{-\theta j}].
\ee

Take $g$ to be a nonnegative smooth function supported on $B(0,2)$, satisfying $g=1$ on $B(0,1)$.
Denote by $g_j(x):=g(x-2^je_0)$ the translation of $g$, where $e_0=(1,0,0,\cdots,0)$ be the unit vector on $\rn$.
For sufficiently large $j$, we have
\be
(2^j+2)(1-2^{-\theta j})+1<(2^j-2)(1+2^{-\theta j})-1.
\ee
Denote
\be
E_{j}:= \bigcup_{a\in \mathbb{R}: (2^j+2)(1-2^{-\theta j})+1\leq a\leq (2^j-2)(1+2^{-\theta j})-1}B(ae_0,1/2).
\ee
We have $|E_j|\sim 2^{(1-\theta)j}$.
For $x\in E_{j}$, we have
\be
(2^j+2)(1-2^{-\theta j})\leq |x|\leq (2^j-2)(1+2^{-\theta j}),\ \ \ |x|\sim 2^{j}.
\ee
This implies that
\ben\label{pf, 2}
\frac{|x|}{2^j-2}\leq 1+2^{-\theta j},\ \ \  \frac{|x|}{2^j+2}\geq 1-2^{-\theta j}.
\een
Recall $\text{supp}g\subset B(0,2)$.  For every $x\in E_{j}$,
set
\be
E^0_{j,x}:=\{r: g_j\big(\frac{x}{r}\big)\neq 0\},\ \ \ \ \ E^1_{j,x}:=\{r: g_j\big(\frac{x}{r}\big)=1\}.
\ee
By a direct calculation and \eqref{pf, 2}, we deduce that for $x\in E_j$,
\ben
E_{j,x}^1\subset E_{j,x}^0\subset \left(\frac{|x|}{2^j+2}, \frac{|x|}{2^j-2}\right)\subset A_j.
\een
Since $|x|\sim 2^{j}$ for $x\in E_j$, we have the upper estimate of $|E_{j,x}^0|$:
\be
|E^0_{j,x}|\leq \left|\frac{|x|}{2^j-2}-\frac{|x|}{2^j+2}\right|\sim 2^{-j}.
\ee
Next, we turn to the lower estimate of $|E^1_{j,x}|$.
For $x\in E_j$, write $x=ae_0+y$ for $y\in B(0,1/2)$, we have
\be
\begin{split}
  r\in E^1_{j,x}
  \Longleftarrow
  \left|\frac{x}{r}-2^je_0\right|\leq 1
  & \Longleftrightarrow
  \left|\frac{ae_0+y}{r}-2^je_0\right|\leq 1
  \\
  & \Longleftarrow
  \left|\frac{ae_0}{r}-2^je_0\right|+\left|\frac{y}{r}\right|\leq 1
  \\
  & \Longleftarrow
  \left|\frac{a}{r}-2^j\right|\leq \frac{1}{4},\ \ \ (\text{if}\ r\geq \frac{2}{3})
  \\
  & \Longleftrightarrow
  r\in \left[\frac{a}{2^j+1/4}, \frac{a}{2^j-1/4}\right].
\end{split}
\ee
Observe that
\be
\lim_{j\rightarrow \fy}\frac{a}{2^j+1/4}=\lim_{j\rightarrow \fy}\frac{a}{2^j-1/4}=1
\ee
for $a\in [(2^j+2)(1-2^{-\theta j})+1, (2^j-2)(1+2^{-\theta j})-1]$.
For sufficiently large $j$, we deduce that
$\left[\frac{a}{2^j+1/4}, \frac{a}{2^j-1/4}\right]\subset [2/3,\fy)$.
Hence,
\be
\left[\frac{a}{2^j+1/4}, \frac{a}{2^j-1/4}\right]\subset E_{j,x}^1.
\ee
The lower estimate of $|E_{j,x}^1|$ follows by
\be
|E_{j,x}^1|\geq \left|\frac{a}{2^j-1/4}-\frac{a}{2^j+1/4}\right|\sim 2^{-j}.
\ee
The combination of lower estimate of $|E_{j,x}^1|$ and upper estimate of $|E_{j,x}^0|$ yields that
\be
2^{-j}\lesssim |E^1_{j,x}|\leq |E^0_{j,x}|\lesssim 2^{-j},\ \ \ \text{or equivalent:}\ \ |E^1_{j,x}|\sim |E^0_{j,x}|\sim 2^{-j}.
\ee
Next, we divide the Hausdorff operator into main term and error term by
  \be
  \begin{split}
    H_{\Phi}g_j(x)
    &=
    \int_{0}^{\fy}\phi(r)g_j(x/r)dr
    =
    \int_{0}^{\fy}\phi(1)g_j(x/r)dr+\int_{0}^{\fy}(\phi(r)-\phi(1))g_j(x/r)dr
    \\
    &= :
    H_{\Phi}^Mg_j(x)+H_{\Phi}^Eg_j(x).
  \end{split}
  \ee
Let us first turn to the estimate of main term.
For $x\in E_j$, we have
  \be
  \begin{split}
    H_{\Phi}^Mg_j(x)
    =
    \int_{0}^{\fy}g_j(x/r)dr
    \geq
    \int_{E_{j,x}^1}g_j(x/r)dr
    = |E_{j,x}^1|\sim 2^{-j}.
  \end{split}
\ee
Recalling $|E_j|\sim 2^{(1-\theta)j}$ and $\theta=\frac{1-p}{2}$, we have following estimate of the main term:
  \ben\label{estimate of main term}
  \begin{split}
    \|H_{\Phi}^Mg_j\|^p_{L^p(E_j)}
    \gtrsim &
    2^{-jp}|E_j|\sim 2^{-jp}2^{(1-\theta)j}=2^{\frac{(1-p)j}{2}}.
  \end{split}
\een

On the other hand,
\ben\label{pf, 3}
\begin{split}
  \|H_{\Phi}^Eg_j\|^p_{L^p(E_j)}
  \leq &
  \|H_{\Phi}^Eg_j\|^p_{L^1(E_j)}|E_j|^{1-p}
  \\
  \leq &
  \left(\int_{E_j}\int_{0}^{\fy}|\phi(r)-\phi(1)|g_j(x/r)drdx\right)^p|E_j|^{1-p}
  \\
  = &
  \left(\int_{E_j}\int_{E_{j,x}^0}|\phi(r)-\phi(1)|g_j(x/r)drdx\right)^p|E_j|^{1-p}.
\end{split}
\een

Note that
\be
\{(x,r): x\in E_j, r\in E_{j,x}^0\}\subset \{(x,r): r\in A_j, x\in \rn\}.
\ee
We deduce that
\be
\begin{split}
  &\int_{E_j}\int_{E_{j,x}^0}|\phi(r)-\phi(1)|g_j(x/r)drdx
  \\
  \leq &
  \int_{A_j}|\phi(r)-\phi(1)|\int_{\rn}g_j(x/r)dxdr
  \\
  = &
  \|g\|_{L^1}\int_{A_j}|\phi(r)-\phi(1)|r^ndr
  \lesssim  \int_{A_j}|\phi(r)-\phi(1)|dr=\ep_j|A_j|\lesssim \ep_j2^{-\theta j},
\end{split}
\ee
where $\ep_j\rightarrow 0^+$ as $j\rightarrow \fy$.
Combining this with \eqref{pf, 3}, we have
\ben\label{estimate of error term}
\begin{split}
  \|H_{\Phi}^Eg_j\|^p_{L^p(E_j)}
  \lesssim
  \ep_j^p2^{-\theta pj}|E_j|^{1-p}
  \lesssim
  \ep_j^p2^{-\theta pj}2^{(1-\theta)(1-p)j}=\ep_j^p2^{\frac{(1-p)j}{2}}.
\end{split}
\een
By \eqref{estimate of main term} and \eqref{estimate of error term}, there exist two constants $C_1$ and $C_2$ such that
\be
\|H_{\Phi}^Mg_j\|^p_{L^p(E_j)}\geq C_12^{\frac{(1-p)j}{2}},\ \ \ \|H_{\Phi}^Eg_j\|^p_{L^p(E_j)}\leq C_2\ep_j^p2^{\frac{(1-p)j}{2}}.
\ee
For sufficiently large $j$ such that $C_2\ep_j^p\leq C_1/2$, we have
\be
\begin{split}
  \|H_{\Phi}g_j\|^p_{L^p}
  \geq &
  \|H_{\Phi}g_j\|^p_{L^p(E_j)}
  \\
  \geq &
  \|H_{\Phi}^Mg_j\|^p_{L^p(E_j)}-\|H_{\Phi}^Eg_j\|^p_{L^p(E_j)}
  \\
  \geq &
  C_12^{\frac{(1-p)j}{2}}-C_2\ep_j^p2^{\frac{(1-p)j}{2}}\geq (C_1/2)2^{\frac{(1-p)j}{2}}.
\end{split}
\ee
However, the boundedness of $H_{\Phi}$ yields that
\be
\|H_{\Phi}g_j\|^p_{L^p}
\lesssim \|H_{\Phi}g_j\|^p_X
\lesssim \|g_j\|^p_X=\|g\|^p_X,
\ee
which leads to a contradiction.
\end{proof}
Recall that all the spaces $L^p,h^p, M^{p,p}, B^{p,p}, F^{p,p}$ are translation invariant.
The following corollary is a direct conclusion of Lemma \ref{lemma, ebd} and Theorem \ref{Thm, lp}.
\begin{corollary}
 Let $0<p<1$, $\Phi\in L_{loc}^{1}(\rn\bs\{0\})$. We have
  \be
  H_{\Phi}\ \text{is bounded on}\ X\Longleftrightarrow H_{\Phi}=0,
  \ee
  where $X=L^p,h^p, M^{p,p}, B^{p,p}, F^{p,p}$.
\end{corollary}

For more general frequency decomposition spaces such as $X=B^{p,q}_s$, $F^{p,q}_s$ or $M^{p,q}_s$,
the embedding condition $\|g\|_{L^p}\lesssim \|g\|_X$ is no longer valid.
We establish following theorem with the help of the modified embedding (see Lemma \ref{lemma, ebd}).

\begin{theorem}\label{Thm, local lp}
  Let $0<p<1$, $\Phi\in L_{loc}^{1}(\rn\bs\{0\})$. Suppose $C_c^{\infty}(\rn)\subset X$ and $X\hookrightarrow L^p_{0}(\rn)$. We have
  \be
  H_{\Phi}\ \text{is bounded on}\ X\Longleftrightarrow H_{\Phi}=0.
  \ee
\end{theorem}
\begin{proof}
The ``if'' part is trivial. We focus on the proof for the ``only if'' part.
  As the proof of Theorem \ref{Thm, lp}, we write
  $H_{\Phi}f(x)=\int_{0}^{\fy}\phi(r)f(x/r)dr$,
  and assume
  $r_0=1$ is the Lebesgue point of $\phi$ with
$\phi(1)=1$. Let $g_j, A_j, E_j, E_{j,x}^0, E_{j,x}^1, H_{\Phi}^Mg_j, H_{\Phi}^Eg_j$ be as in the proof of Theorem \ref{Thm, lp}.
 Take a $C_c^{\fy}(\rn)$ nonnegative function $\va$ satisfying $\va(x)= 1$ on $B(0,1)$ and $\text{supp}\va\subset B(0,2)$.
 By the assumption we have
\be
\|\va \ast H_{\Phi}g_j\|_{L^p(\rn)}\lesssim \|H_{\Phi}g_j\|_{X}\lesssim \|g_j\|_X=\|g\|_X.
\ee
Set
\be
\Xi_j=\{x: B(x, 2)\subset E_j\}.
\ee
We have
\be
|\Xi_j|\sim |E_j|\sim 2^{(1-\theta)j}\ \ \ \ (j\rightarrow \fy).
\ee
Recall that $H_{\Phi}^Mg_j(x)\geq 0$ for $x\in \rn$ and
$H_{\Phi}^Mg_j(x)\gtrsim 2^{-j}$ for $x\in E_j$.
For $x\in \Xi_j$ we have
\be
\begin{split}
  \va \ast H_{\Phi}^Mg_j(x)
  = &
  \int_{\rn}\va(x-z)H_{\Phi}^Mg_j(z)dz
  \\
  \geq &
  \int_{B(x,1)}H_{\Phi}^Mg_j(z)dz
  \gtrsim \int_{B(x,1)}2^{-j}dz\sim 2^{-j}.
\end{split}
\ee
From this and the fact $|\Xi_j|\sim 2^{(1-\theta)j}$, we deduce that
\be
\begin{split}
  \|\va \ast H_{\Phi}^Mg_j\|^p_{L^p(\Xi_j)}
  \gtrsim &
  2^{-jp}|\Xi_j|\gtrsim 2^{-jp}2^{(1-\theta)j}  =2^{\frac{(1-p)j}{2}}.
\end{split}
\ee
On the other hand, observing that $E_{j,z}^0\subset A_j$ for $z\in B(x,2)$ with $x\in \Xi_j$,
\be
\begin{split}
  \|\va \ast H_{\Phi}^Eg_j\|^p_{L^p(\Xi_j)}
  \leq &
  \|\va \ast H_{\Phi}^Eg_j\|^p_{L^1(\Xi_j)}|\Xi_j|^{1-p}
  \\
  \leq &
  \left(\int_{\Xi_j}\int_{B(x,2)}\va(x-z)\int_{0}^{\fy}|\phi(r)-\phi(1)|g_j(z/r)drdzdx\right)^p|\Xi_j|^{1-p}
  \\
  = &
  \left(\int_{\Xi_j}\int_{B(x,2)}\va(x-z)\int_{E_{j,z}^0}|\phi(r)-\phi(1)|g_j(z/r)drdzdx\right)^p|\Xi_j|^{1-p}
  \\
  \leq &
  \left(\int_{A_j}|\phi(r)-\phi(1)|\int_{\rn}g_j(z/r)\int_{\rn}\va(x-z)dxdz dr\right)^p|\Xi_j|^{1-p}
  \\
  \lesssim &
  \left(\int_{A_j}|\phi(r)-\phi(1)|dr\right)^p|\Xi_j|^{1-p}\sim \ep_j^p|A_j|^p|\Xi_j|^{1-p}\sim \ep_j^p2^{\frac{(1-p)j}{2}},
\end{split}
\ee
where $\ep_j\rightarrow 0^+$ as $j\rightarrow \fy$.
Now, we have finished the estimates of main term $\|\va \ast H_{\Phi}^Mg_j\|^p_{L^p(\Xi_j)}$
and error term $\|\va \ast H_{\Phi}^Eg_j\|^p_{L^p(\Xi_j)}$, the remainder of this proof is the same
as that of Theorem \ref{Thm, lp}.
\end{proof}

Using Theorem \ref{Thm, local lp} and Lemma \ref{lemma, ebd}, we have following corollary.
\begin{corollary}
 Let $0<q\leq \fy, s\in \mathbb{R}$, $\Phi\in L_{loc}^{1}(\rn\bs\{0\})$. If $0<p<1$, we have
  \be
  H_{\Phi}\ \text{is bounded on}\ X \Longleftrightarrow H_{\Phi}=0,
  \ee
  where $X=F_s^{p,q}$, $B_s^{p,q}$ or $M^{p,q}_s$.
\end{corollary}

In particular, due to the time-frequency symmetry of modulation space, we have following corollary.

\begin{corollary}
 Let $0<p,q\leq \fy$. Let $\Phi$ be a measurable function satisfying
   \be
  \int_{B(0,1)}|y|^n\Phi(y)dy<\infty,\\
  \ \ \ \ \
  \int_{B(0,1)^c}\Phi(y)dy<\infty.
\ee
  If $0<p<1$ or $0<q<1$, we have
  \be
  H_{\Phi}\ \text{is bounded on}\ M^{p,q}\Longleftrightarrow H_{\Phi}=0.
  \ee
\end{corollary}
\begin{proof}
The ``if'' part is trivial. We focus on the proof for the ``only if'' part.

If $0<p<1$, we have $M^{p,q}\hookrightarrow L^p_{0}(\rn)$, then the conclusion follows by Theorem \ref{Thm, local lp}.

If $p\geq 1$, $0<q<1$, we will use the Fourier transform to exchange the time and frequency space.
  It follows by Remark 1.4 in \cite{ZhaoFanGuo2018AFA} that
\begin{equation*}
  \widehat{H_{\Phi}f}=\widetilde{H_{\Phi}}\widehat{f},
\end{equation*}
where
\begin{equation*}
  \widetilde{H_{\Phi}}f(x)=\int_{\mathbb{R}^n}\Phi(y)|y|^nf(|y|x)dy.
\end{equation*}
Thus,
\begin{equation*}
    \|H_{\Phi}f\|_{M^{p,q}}=\|\widehat{H_{\Phi}f}\|_{\mathscr{F}M^{p,q}}
    =\|\widetilde{H_{\Phi}}\widehat{f}\|_{\mathscr{F}M^{p,q}}.
  \end{equation*}
If $H_{\Phi}$ is bounded on $M_{p,q}$, we have
  \begin{equation*}
    \|\widetilde{H_{\Phi}}f\|_{\mathscr{F}M^{p,q}}
    =
    \|H_{\Phi}\check{f}\|_{M^{p,q}}
    \lesssim
    \|\check{f}\|_{M^{p,q}}
    =
    \|f\|_{\mathscr{F}M^{p,q}}.
  \end{equation*}
Write
\be
\begin{split}
  \widetilde{H_{\Phi}}f(x)
  = &
  \int_{\mathbb{R}^n}\Phi(y)|y|^nf(|y|x)dy
  \\
  = &
    \int_{0}^{\fy}\int_{\mathbb{S}^{n-1}}\Phi(ry')r^{n}f(rx)d\sigma(y')r^{n-1}dr
  \\
  = &
  \int_{0}^{\fy}\int_{\mathbb{S}^{n-1}}\Phi(y'/r)r^{-1-2n}f(x/r)d\sigma(y')dr
  = : \int_{0}^{\fy}\widetilde{\phi}(r)f(x/r)dr,
\end{split}
\ee
where
\be
\widetilde{\phi}(r)=\int_{\mathbb{S}^{n-1}}\Phi(y'/r)r^{-1-2n}d\sigma(y').
\ee
Observe $\widetilde{\phi}\in L^1_{loc}(\mathbb{R}^+)$
and
recall $\mathscr{F}M^{p,q}\hookrightarrow L^q_{0}(\rn)$ with $q\in (0,1)$.
By the same argument as in the proofs of Theorem \ref{Thm, lp} and \ref{Thm, local lp}, we conclude that $\widetilde{\phi}=0$ and complete this proof.
\end{proof}

\subsection*{Acknowledgements}
This work was partially supported by the National Natural Foundation of China (Nos. 11701112, 11771388, 11771358, 11671414, 11601456),
Zhejiang Provincial Natural Science Foundation of China (No. LY18A010015)
and
Natural Science Foundation of Fujian Province (Nos. 2017J01723, 2018J01430).

\bibliographystyle{abbrv}

config: arxiv


TITLE: Prove that $\kappa_F(A) \ge \sqrt{n}$
QUESTION [0 upvotes]: I reached at a point after solving it
$$\kappa_F(A) = \sqrt{\mbox{tr}(A^H A) \cdot \mbox{tr}(A^{-1}(A^{-1})^H}$$
Now I am stuck. How to proceed? Or, alternatively, a new approach is also appreciated.

REPLY [1 votes]: Hint. Every complex square matrix $X$ is unitarily triangulable and Frobenius norm is unitarily invariant. It follows that $\|X\|_F^2\ge\sum_i|\lambda_i(X)|^2$.


TITLE: Why $\Bbb E\{ \log_2(1+x)\} \approx \log_2(1+\Bbb E(x))$?
QUESTION [0 upvotes]: $\Bbb E\{ \log_2(1+x)\}  \approx \log_2(1+\Bbb E(x))$ ,where $\Bbb E$ is the notation of expectation. We think when $x$ is small or big enough, so the function $\log_2(1+x)$ seems to be linear, so the expectation and the function $log$ would change the order of calculation? I 'm confused about that. Any comments would appreciated!

REPLY [1 votes]: Do a Taylor series around $E(x)$, that is
$\mathbb{E}[\log(x)]\approx\log({E}[x])-\frac{{V}[x]}{2{E}[x]^2} \>$ according to this paper. Do the same thing for $\log(1+x)$ that is
$${E}[\log(1+x)]\approx\log(1+{E}[x])-\frac{\mathbb{V}[x]}{2(1+{E}[x])^2} \>.$$


TITLE: Stuck at Evaluating the Riemann-Stieltjes Integral
QUESTION [6 upvotes]: Let $$\alpha(x) = \left\lbrace \begin{array}{cc} 0 & x=0\\ \dfrac{1}{2^n} & \dfrac{1}{3^n} < x \leq \dfrac{1}{3^{n-1}}\quad n=1,2,...\end{array}\right.$$
Evaluate $$\int_{0}^{1}{x\mathrm{d}\alpha(x)}$$
Attempt: I dont know how to put this formally, but I know it is similar to problem of evaluating the R-S integral where $\alpha(x)$ is the jump function.
So If I evaluate the function $\alpha(x)$ for some values of $n$ then the function looks like this $$\alpha(x) = \left\lbrace \begin{array}{cc} 0 & x=0\\ \vdots & \vdots \\\dfrac{1}{8} & \dfrac{1}{27} < x \leq \dfrac{1}{9} \\ \dfrac{1}{4} & \dfrac{1}{9} < x \leq \dfrac{1}{3} \\ \dfrac{1}{2} & \dfrac{1}{3} < x \leq 1 \end{array}\right.$$ So $f(x)=x$ is continuous and $\alpha(x)$ is monotonically increasing. Hence the answer should be $$\int_{0}^{1}{x \mathrm{d}\alpha(x)} = \sum_{n=1}^{\infty}{\dfrac{1}{3^n}\dfrac{1}{2^{n+1}}}=\dfrac{1}{2}\sum_{n=1}^{\infty}{\left(\dfrac{1}{6}\right)^n}=\dfrac{1}{10}$$
So my questions are:
(a) Is the answer correct? If yes, then how can I write this formally?
(b) If the answer is wrong, what direction can you provide me to get the right answer?
Thanks!

REPLY [5 votes]: I think your answer is quite right.  I'm going to outline an alternative, wimpy approach using integration by parts:
$$\int_0^1 x \, d\alpha(x) = [x \, \alpha(x)]_0^1 - \int_0^1 dx \, \alpha(x)$$
The integrated term is $1/2$, and the resulting integral is straightforward:
$$\int_0^1 x \, d\alpha(x) = \frac12 - \sum_{n=1}^{\infty} \frac{1}{2^n} \left ( \frac{1}{3^{n-1}}-\frac{1}{3^n}\right) = \frac12 - 2 \sum_{n=1}^{\infty} \frac{1}{6^n} = \frac{1}{10}$$


TITLE: Ashcroft Mermin Eq. 17.47ff
QUESTION [2 upvotes]: In "Solid State Physics" by said authors, Eq. 17.46 is
$$ \rho^{ind}(\textbf{r}) = - e[n_0(\mu + e\phi(\textbf{r})) - n_0(\mu)]$$
and then the authors write

In the present case we assume that $\phi$ is small enough for Eq. 17.46
      to be expanded to give in leading order

$$\rho^{ind}(\textbf{r}) = -e^2 \frac{\partial n_0}{\partial \mu} \phi(\textbf{r}) \hspace{1cm} \mbox{[Eq. 17.47]}$$
I wonder how to formally exact write this expansion. To understand this, I would like to know

What is the independent variable in Eq. 17.46? Is it $\phi$, $\mu$ or $(e\phi + \mu)$?
Does it make sense to write it as
$$\rho^{ind}(\textbf{r}) = -e \left[ n_0(\mu) + \left .  e\phi(\textbf{r})\frac{\partial n_0(e\phi+\mu)}{\partial \mu}\right|_{e\phi=0} - n_0(\mu) \right]$$
But then, if I set $e\phi = 0$, would the middle term not collapse to zero?

REPLY [2 votes]: Your quotation of the equation is actually wrong. The correct version is given by
$\rho^{ind}(r)=-e[n_0(\mu+e\phi(r))-n_0(\mu)].$
Now you have to expand this expression around $\phi=0$
$\rho^{ind}(r)=\rho^{ind}(r)\mid_{\phi=0}+\frac{\partial}{\partial \phi}\rho^{ind}\mid_{\phi=0}\phi+\ldots$
Now we have to evaluate the zero and first order terms:
$\rho^{ind}(r)\mid_{\phi=0}=e[n_0(\mu)-n_0(\mu)]=0,$
$\frac{\partial}{\partial \phi}\rho^{ind}\mid_{\phi=0}\phi(r)=-e\frac{\partial n_0}{\partial(\mu+e\phi(r))}\frac{\partial}{\partial\phi}(\mu+e\phi(r))\mid_{\phi=0}\phi=-e^2\frac{\partial n_0}{\partial\mu}\phi(r).$
As you can see, the last line yields the desired expression by application of the chain rule.


TITLE: How to determine the spectrum of the following operator
QUESTION [0 upvotes]: Let $H$ a Hilbert space and $f \in H^{'}$. Fixed $z \in H$, $z \neq 0$, define $T: H \rightarrow H$ by $T(x) = f(x)z$. Then $T$ is a compact non-self-adjoint operator.
How to determine the spectrum of  $ T $?
I verified that $|| T || = || f |||| z || $ and I'm trying to verify that $ \sigma (T) = \{\lambda: | \lambda  | \leq || f |||| z || \} $, I have no idea how to make the inclusion $ \sigma (T) \supset \{ \lambda: | \lambda  | \leq || f |||| z || \} $. In fact, is it not necessary to place any restrictions on the $ f $ for the result to be true? Or does it work in general?

REPLY [1 votes]: Since  is compact its spectrum consists of eigenvalues, in addition to zero, which is always in the spectrum. To find the nonzero eigenvalues $\lambda$, write
$$
f(x)z=\lambda x,
$$
which implies that the eigenvector $x$ is a multiple of $z$, and hence can be taken to be $z$ itself, since eigenvalues are scalable.  The eigenvalue in turn is computed by solving the equation
$$
f(z)z=\lambda z,
$$
whose solution is clearly $\lambda=f(z)$.  In other words,
$$\sigma(T)=\{0,f(z)\}.$$


text:
\subsection{Problems involving edges}
Consider locally checkable problems where, for every edge, a certain condition comprising their endpoints and their consecutive edges must be satisfied.
We will show how to reduce these problems to {\rLCproblems{1}} in the jagged graph $J(G)$ of the input graph $G$.
We consider two kind of problems: when edges do not have requirements over other edges, and when they do.

For the first class of problems, the reduction is straightforward. We might need to duplicate the colors in order to distinguish the colors assigned to edges from the colors assigned to vertices.
As an example, we show how to model vertex cover:
\begin{itemize}
\item $L_{v} = \{0,1\}$ for all $v \in V(G)$ and\\
$L_{uv} = \{0\}$ for all $uv \in E(G)$;

\item $(\WeightSet, \wlesseq) = (\mathbb{R} \cup \{+\infty\}, \leq)$ and $\weightsSum = +$;

\item $\textsc{w}_{v', i} = i$ for all $v' \in V(J(G)), i \in L_{v'}$;

\item $check(v, c) = \True$ for all $v \in V(G)$, and\\
$check(uv, c) = (c(u) + c(v) \geq 1)$ for all $uv \in E(G)$.
\end{itemize}
And we give a constant partial neighborhood system:
\begin{itemize}
\item $\NeighborhoodSet{v'}{i} = \{0, 1\}$;

\item $n \neighborhoodSum{v'}{i} n' = \min(n + n', 1)$;

\item $\toNeighborhood{v'}{i}(u', j) = j$; and

\item $check_{v,i}(n) = \True$ for all $v \in V(G), i \in L_v$, and\\
$check_{uv,0}(n) = (n \geq 1)$ for all $uv \in E(G)$.
\end{itemize}

Edge cover is basically the same as vertex cover but interchanging vertices and edges.

As regards the second class of these problems, since two edges in $E(G)$ that share an endpoint are at distance two in $J(G)$, we can use the ideas from semitotal domination: the neighbors in $J(G)$ of the edges in $E(G)$ (which are vertices in $V(G)$) are the ones in charge of checking the requirements of the edges in $E(G)$.
We illustrate the maximum matching problem, for which we can set:
\begin{itemize}
\item $L_{v} = \{0\}$ for all $v \in V(G)$ and\\
$L_{uv} = \{0,1\}$ for all $uv \in E(G)$;

\item $(\WeightSet, \wlesseq) = (\mathbb{R} \cup \{-\infty\}, \geq)$ and $\weightsSum = +$;

\item $\textsc{w}_{v', i} = i$ for all $v' \in V(J(G)), i \in L_{v'}$;

\item $check(v, c) = \left(\sum_{u \in N_G(v)} c(uv) \leq 1\right)$ for all $v \in V(G)$, and\\
$check(uv, c) = \True$ for all $uv \in E(G)$.
\end{itemize}
And we give a constant partial neighborhood system:
\begin{itemize}
\item $\NeighborhoodSet{v'}{i} = \{0, 1\}$;

\item $n \neighborhoodSum{v'}{i} n' = \min(n + n', 1)$;

\item $\toNeighborhood{v'}{i}(u', j) = j$; and

\item $check_{v,0}(n) = (n \leq 1)$ for all $v \in V(G)$, and\\
$check_{uv,i}(n) = \True$ for all $uv \in E(G), i \in L_{uv}$.
\end{itemize}


text:
\section*{APPENDIX}

\appendix
\section{Proof of Homomorphic Property}
\label{sec:appendix-Homomorphic}

We formally prove the homomorphic property described in
\S\ref{subsec:homomorphism}.  We state the following theorem.

\begin{theorem}\label{th:homomorphic}
In the construction of the canonical stripe of STAIR codes, the encoding of
each chunk in the column direction via $\mc{C}_{col}$ is homomorphic, such
that each augmented row in the canonical stripe is a codeword of
$\mc{C}_{row}$.
\end{theorem}

\noindent
{\bf Proof:} We prove by matrix operations.  We define the matrices
$\mb{D} = [d_{i,j}]_{r\times(n-m)}$, $\mb{P} = [p_{i,k}]_{r\times m}$, and
$\mb{P}' = [p'_{i,l}]_{r\times m'}$.  Also, we define the generator matrices
$\mb{G}_{row}$ and $\mb{G}_{col}$ for the codes $\mc{C}_{row}$ and
$\mc{C}_{col}$, respectively, as:
\begin{eqnarray*}
\mb{G}_{row} & = & \left(\mb{I}_{(n-m)\times(n-m)} \mid
		\mb{A}_{(n-m)\times(m+m')} \right), \\
\mb{G}_{col} & = & \left(\mb{I}_{r\times r}
		\mid \mb{B}_{r\times e_{m'-1}}\right),
\end{eqnarray*}
where $\mb{I}$ is an identity matrix, and $\mb{A}$ and $\mb{B}$ are the
sub-matrices that form the parity symbols.  The upper $r$ rows of the stripe
can be expressed as follows:

\begin{eqnarray*}
\left(\mb{D} \mid \mb{P} \mid \mb{P}'\right) & = & \mb{D} \cdot \mb{G}_{row}.
\end{eqnarray*}

The lower $e_{m'-1}$ augmented rows are expressed as follows:

\begin{eqnarray*}
\left(\left(\mb{D} \mid \mb{P} \mid \mb{P}'\right)^T \cdot \mb{B}\right)^T
& = & \mb{B}^T \cdot \left( \mb{D} \cdot \mb{G}_{row} \right) \\
& = & \left(\mb{B}^T \cdot \mb{D} \right) \cdot \mb{G}_{row}
\end{eqnarray*}

We can see that each of the lower $e_{m'-1}$ rows can be calculated using
the generator matrix $\mb{G}_{row}$, and hence is a codeword of $\mc{C}_{row}$.

$\hfill\Box$

\section{Explicit Expressions of $P_{str}$ for Various Erasure Codes}
\label{sec:appendix-expressions}

\subsection{Reed-Solomon Codes}

The explicit expression of $P_{str}$ for Reed-Solomon codes is as follows:
\begin{equation}
    P_{str} = 1 - P_{chk(0)}^{n-m}.
\end{equation}

\subsection{STAIR Codes}

Explicit expressions of $P_{str}$ for some STAIR codes with special $\mb{e}$'s are as follows:

\begin{enumerate}
  \item For a STAIR code with $\mb{e}=(s)$ for $s \geq 1$,
\begin{equation}
    P_{str} = 1 - P_{chk(0)}^{n-m} -  {n-m \choose 1} \cdot \sum\limits_{i=1}^{s} P_{chk(i)} \cdot P_{chk(0)}^{n-m-1}.
\end{equation}
  \item For a STAIR code with $\mb{e}=(1,s-1)$ for $s \geq 2$,
\begin{equation}
  \begin{split}
    P_{str} = & 1 - P_{chk(0)}^{n-m} -  {n-m \choose 1} \cdot \sum\limits_{i=1}^{s-1} P_{chk(i)} \cdot P_{chk(0)}^{n-m-1} -  \\
                & {n-m \choose 2} \cdot P_{chk(1)}^2 \cdot P_{chk(0)}^{n-m-2} - {n-m \choose 1} \cdot {n-m-1 \choose 1} \cdot \sum\limits_{i=2}^{s-1} P_{chk(i)} \cdot P_{chk(1)}  \cdot P_{chk(0)}^{n-m-2}.
  \end{split}
\end{equation}
  \item For a STAIR code with $\mb{e}=(2,s-2)$ for $s \geq 4$,
\begin{equation}
  \begin{split}
    P_{str} = & 1 - P_{chk(0)}^{n-m} -  {n-m \choose 1} \cdot \sum\limits_{i=1}^{s-2} P_{chk(i)} \cdot P_{chk(0)}^{n-m-1} - \\
                & {n-m \choose 2} \cdot P_{chk(1)}^2 \cdot P_{chk(0)}^{n-m-2} - {n-m \choose 1} \cdot {n-m-1 \choose 1} \cdot \sum\limits_{i=2}^{s-2} P_{chk(i)} \cdot P_{chk(1)} \cdot P_{chk(0)}^{n-m-2} - \\
                & {n-m \choose 2} \cdot P_{chk(2)}^2 \cdot P_{chk(0)}^{n-m-2} - {n-m \choose 1} \cdot {n-m-1 \choose 1} \cdot \sum\limits_{i=3}^{s-2} P_{chk(i)} \cdot P_{chk(2)} \cdot P_{chk(0)}^{n-m-2}.
  \end{split}
\end{equation}
  \item For a STAIR code with $\mb{e}=(1,1,s-2)$ for $s \geq 3$,
\begin{equation}
  \begin{split}
    P_{str} = & 1 - P_{chk(0)}^{n-m} -  {n-m \choose 1} \cdot \sum\limits_{i=1}^{s-2} P_{chk(i)} \cdot P_{chk(0)}^{n-m-1} - \\
                & {n-m \choose 2} \cdot P_{chk(1)}^2 \cdot P_{chk(0)}^{n-m-2} -  {n-m \choose 1} \cdot {n-m-1 \choose 1} \cdot \sum\limits_{i=2}^{s-2} P_{chk(i)} \cdot P_{chk(1)} \cdot P_{chk(0)}^{n-m-2} - \\
                & {n-m \choose 3} \cdot P_{chk(1)}^3 \cdot P_{chk(0)}^{n-m-3} -  {n-m \choose 2} \cdot {n-m-2 \choose 1} \cdot \sum\limits_{i=2}^{s-2} P_{chk(i)} \cdot P_{chk(1)}^2 \cdot P_{chk(0)}^{n-m-3}.  \\
  \end{split}
\end{equation}
  \item For a STAIR code with $\mb{e}=(\overset{s}{\overbrace{1,1,\cdots,1}})$ for $s \geq 1$,
\begin{equation}
    P_{str} = 1 - \sum\limits_{i=0}^{s} \left({n-m \choose i} \cdot
			P_{chk(1)}^{i} \cdot P_{chk(0)}^{n-m-i} \right).
\end{equation}
\end{enumerate}

\subsection{SD Codes}

Explicit expressions of $P_{str}$ for SD codes with $s \leq 3$ \cite{Plank13b,Plank13c} are as follows:

\begin{enumerate}
  \item For an SD code with $s=1$,
\begin{equation}
    P_{str} = 1 - P_{chk(0)}^{n-m} -  {n-m \choose 1} \cdot P_{chk(1)} \cdot P_{chk(0)}^{n-m-1}.
\end{equation}
  \item For an SD code with $s=2$,
\begin{equation}
    P_{str} = 1 - P_{chk(0)}^{n-m} -  {n-m \choose 1} \cdot \sum\limits_{i=1}^{2} P_{chk(i)} \cdot P_{chk(0)}^{n-m-1} - {n-m \choose 2} \cdot P_{chk(1)}^2 \cdot P_{chk(0)}^{n-m-2}.
\end{equation}
  \item For an SD code with $s=3$,
\begin{equation}
  \begin{split}
    P_{str} = & 1 - P_{chk(0)}^{n-m} -  {n-m \choose 1} \cdot \sum\limits_{i=1}^{3} P_{chk(i)} \cdot P_{chk(0)}^{n-m-1} - \\
                & {n-m \choose 2} \cdot P_{chk(1)}^2 \cdot P_{chk(0)}^{n-m-2} - {n-m \choose 1} \cdot {n-m-1 \choose 1} \cdot P_{chk(2)} \cdot P_{chk(1)} \cdot P_{chk(0)}^{n-m-2} - \\
                & {n-m \choose 3} \cdot P_{chk(1)}^{3} \cdot P_{chk(0)}^{n-m-3}.
  \end{split}
\end{equation}
\end{enumerate}


TITLE: A nonexample of a prime submodule?
QUESTION [1 upvotes]: Let $M$ be an $A$-module. A submodule $P\subset M$ is called a prime submodule if it is proper and $am\in P$ implies $aM\subset P$ or $m\in P$. It is easy to see that if $P\subset M$ is a prime submodule, then the ideal
$$(P:M):=\{a\in A \mid aM\subset P\}$$
is prime. I'm trying to find a counterexample of the converse; that is, a submodule $N\subset M$ which is not prime yet $(N:M)\subset A$ is a prime ideal.

REPLY [1 votes]: Let $A=K[x,y]$ be the polynomial ring in two variables over a field $K$ and $M=A \oplus A$. We let $P = (x,y) \oplus (x) \subseteq M$, where $(x)$ and $(x,y)$ are the ideals generated by $x$ and $x,y$ respectively. Clearly $(P:M)=(x)$ is prime, but $P$ is not a prime submodule: Let $a=y$ and $m=(1,x) \in M$. We have $a \not\in (P:M)$ and $m \not\in P$, but $am \in P$.


TITLE: Why these series have the same radius of convergence?
QUESTION [1 upvotes]: I'm trying to explain why these two series
$$\sum_{n=1}^{\infty}nc_{n}(z-a)^{n}$$ and$$\sum_{n=1}^{\infty}nc_{n}(z-a)^{n-1}$$
have the same radius of convergence.
I know that $\sum_{n=1}^{\infty}nc_{n}(z-a)^{n}=(z-a)\sum_{n=1}^{\infty}nc_{n}(z-a)^{n-1}$ and knowing that I should know how to conclude that the set of points that both converge are the same, but I don't.
On the other hand, in most of the tests that I use I know that if some limit exist then the radius of convergence is this limit.
In those tests mentioned it doesn't require more than $c_n$ (which both series have in common) but if there is a radius of convergence I am not sure that it implies that the test should give it as a consequence, and because of that I don't know how to proceed. Any ideas? Thanks in advance!!

REPLY [3 votes]: If $\lambda\in\mathbb C\setminus\{0\}$, then a series $\sum_{n=1}^\infty a_n$ converges if and only if the seres $\sum_{n=1}^\infty\lambda a_n$ converges. Therefore,$$z\neq a\implies\sum_{n=1}^\infty nc_n(z-a)^n\text{ converges if and only if }\sum_{n=1}^\infty nc_n(z-a)^{n-1}\text{ converges.}$$Besides, both series converge when $z=a$. So$$\left\{z\in\mathbb C\,\middle|\,\sum_{n=1}^\infty nc_n(z-a)^n\text{ converges}\right\}=\left\{z\in\mathbb C\,\middle|\,\sum_{n=1}^\infty nc_n(z-a)^{n-1}\text{ converges}\right\}$$and therefore both series have the same radius of convergence.


TITLE: Why is there no such forest
QUESTION [2 upvotes]: Let $ G = (V, E) $ be a graph such that $ |V| = |E| = 14 $. I need to prove $ G $ is not a forest. I tried showing that if $ |E| = 14 $, then the sum of all degress is $ 28 $ and therefore the degree of each $ vertex $ is at least $ 2 $, which means there is a cycle.

REPLY [1 votes]: Your attempt doesn't work, unfortunately. We can say that the average degree of the vertices is exactly $2$, but this doesn't say much about the individual degrees. For example, if you take a cycle $C_{13}$, then add an extra vertex connected by a single edge to one of the vertices of $C_{13}$, you get a graph of $14$ vertices and edges, but where one of the vertices has degree $1$.
You can prove this by induction: show that if $G = (V, E)$ satisfies $|V| \le |E|$, then $G$ is not a forest. We proceed by induction on $|V|$, and prove it for $|V| \ge 3$ (otherwise we'd have to deal with repeated edges and/or loops).
For $|V| = 3$, then we have $3$ vertices and at least $3$ edges. Every graph on $3$ vertices has at most $\binom{3}{2} = 3$ edges, and the only graph with $3$ edges would be the complete graph $K_3$. This graph must be $G$. Note it is also a cycle, so it is not a forest.
Suppose $n \ge 3$ is such that no graph with $n$ vertices and at least $n$ edges is a forest. Suppose $G = (V, E)$ where $|V| = n + 1 \le |E|$. For the sake of contradiction, suppose $G$ is a forest.
If every vertex in $G$ had degree at least $2$, then $G$ contains a cycle, and hence is not a forest. Thus, $G$ contains a vertex of degree $1$ or $0$. Let $G' = (V', E')$ be the result of removing this vertex, and its incident edge (if it has one). Note that a single vertex was removed, so $|V'| = |V| - 1 = n$, and at most one edge was removed, so $|E'| \ge |E| - 1 \ge n$.
By our induction hypothesis, $G'$ must not be a forest, and hence contains a cycle. But, since $G$ is simply $G'$ with an extra vertex and possibly an extra edge, the cycle in $G'$ must also be contained in $G$, contradicting our assumption that $G$ is a forest.


TITLE: ${\rm Aut}\,\mathbb Z_p\simeq \mathbb Z_{p-1}$
QUESTION [8 upvotes]: I'm trying to prove that ${\rm Aut}\,\mathbb Z_p\simeq \mathbb Z_{p-1}$ (p prime).
I know that ${\rm Aut}\,\mathbb Z_p$ has $p-1$ elements because $\mathbb Z_p$ has $p-1$ possiblities of generators, so intuitively I see ${\rm Aut}\,\mathbb Z_p\simeq \mathbb Z_{p-1}$, but I couldn't prove it formally.
I'm trying to build an isomorphic function, but I don't how to do it.
Am I in the right way? Maybe I'm forgetting some trick or something.
Thanks

REPLY [0 votes]: Use one lemma: Give $F$ a field, finite subgroup (in multiplicative operation in $F$) of $F^{\times}$ is a multiplicative cyclic group.
Since $Z_{p}$ is a field, we know that there exists m in $Z_{p}$ such that $$Z_{p}^{\times} = {1,m,m^{2},...,m^{p-2}}$$
now, define $\sigma \in Aut(Z_{p})$, $\sigma(1)=m$, then for every $\tau \in Aut(Z_{p}), \tau(1)=n=m^k$
Hence all $\tau = \sigma^k,$ which gives $Aut(Z_{p})$ isomorphic to $<\tau>$


TITLE: given a positive integer $N$, prove there always exist integers $x, y, z,$ such that $\{\sqrt{x}\}+\{\sqrt{y}\} = 1+\{\sqrt{z}\}$ with $x, y, z > N$
QUESTION [0 upvotes]: Given a real number $a$, $\{a\}$ denotes the fractional part of $a$, i.e. the only real number with $0\leq \{a\} < 1$ such that $a = \{a\}+k$ for some integer $k$. I have tried this problem but can't seem to get anywhere. How should one approach such a problem?

REPLY [0 votes]: So in particular, $\sqrt x+\sqrt y-\sqrt z$ is an integer, which is unusual as "essentially different" square roots are linearly independent over the rationals.
This means that except for trivial cases (i.e., except when at least one of $x,y$ is a perfect square) the desired integer can only be $0$.
This means that we do not need to look for too elaborate and exotic cases.
In fact, the simplest examples to achieve $\sqrt x+\sqrt y=\sqrt z$ are $x=y$ and $z=4x$. The remaining obstacle is that we need $\{\sqrt x\}+\{\sqrt y\}>1$, i.e., $\{\sqrt x\}>\frac12$. Now note that $\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}<\frac1{2\sqrt n}<\frac12$ for $n\ge1$ so that the desired condition holds in particular when $x$ is one less than a perfect square $>1$.
Conclusio. Given $N$ (where wlog $N>1$), let $x=y=N^2-1$ and $z=4N^2-4$.

If you are unhappy with $x=y$, experiment with $\sqrt x+2\sqrt x=3\sqrt x$, i.e., $y=4x$, $z=9x$. We just have to be slightly more careful to really have $\{\sqrt x\}+\{\sqrt y\}=1+\{\sqrt z\}$ instead of $\{\sqrt x\}+\{\sqrt y\}=\{\sqrt z\}$: If we (again) take $x=N^2-1$, so $y=4N^2-4$ and $z=9N^2-9$, we have
$$N-\sqrt x=\frac1{N+\sqrt x}\quad\text{and}\quad 2N-\sqrt y=\frac2{N+\sqrt x}$$
and each of these is so kind to be  $<\frac12$ for $N\ge3$.


TITLE: What is the classical counterpart of an eigenstate?
QUESTION [2 upvotes]: Does this question make sense for every system or just some?
If it makes sense, it is a periodic orbit?

REPLY [1 votes]: As noted in some of the comments already, the answer to your question depends on what properties of quantum eigenstates you want your classical analogues to have.
If you're thinking of eigenstates of arbitrary observables, quantum eigenstates have the property that the value of that observable is precisely defined. In classical mechanics, individual points in phase space have this property, since the positions and momenta of all particles are precisely defined. One of the interesting things about quantum mechanics is that there are observables (like position and momentum) for which no quantum state can simultaneously be an eigenstate of both.
When a quantum state is not the eigenstate of a particular observable, the value of that observable is not precisely defined. Instead, the quantum state defines a probability distribution over possible values of that observable. You can have the same thing in classical mechanics, if you allow the state of your system to be probability distributions $\rho(\vec{q},\vec{p})$ over points in phase space instead of only single points. This brings us to another possible property of eigenstates: time stationarity.
A quantum mechanical eigenstate of the Hamiltonian will not evolve in time. You might ask yourself what kinds of classical states don't evolve with time. Well, you can see that certain points in phase space will be like this if you set $\dot{q}$ and $\dot{p}$ equal to zero in Hamilton's equations of motion
\begin{align}
\dot{q}&=\frac{\partial H}{\partial p}=\{q,H\}, \\
\dot{p}&=-\frac{\partial H}{\partial q}=\{p,H\}.
\end{align}
(Here the curly braces are Poisson brackets.) These states might not be very interesting (for a 1d simple harmonic oscillator, this state will be where the mass is sitting in the rest position with no velocity). However, if we let our classical states again be distributions over phase space instead of only points, we can get perhaps more interesting states! The evolution of distributions over phase space is governed by the Liouville equation
\begin{align}
\frac{\partial\rho}{\partial t}=-\{\rho,H\}.
\end{align}
Therefore, any distribution $\rho$ over phase space for which $\{\rho,H\}=0$ is an analogue of a stationary state in quantum mechanics. For a 1d simple harmonic oscillator these states include uniform distributions over circles centered at the origin (that is, periodic orbits) in phase space.
The condition $\{\rho,H\}=0$ is very similar to the condition $[\rho,H]=0$ in quantum mechanics, where $[A,B]:=AB-BA$ is the commutator. If $\rho=|\psi\rangle\langle\psi|$ (that is, the quantum state is a pure state and not a probability distribution over pure states), $[\rho,H]=0$ means that $|\psi\rangle$ is an eigenstate of $H$, so we've come full-circle.
As one last thing to leave you with, the equation for describing the time evolution of distributions over quantum states (the von Neumann equation) gives another look at the close relationship between the Poisson bracket in classical mechanics and the commutator in quantum mechanics:
\begin{align}
i\hbar\frac{\partial\rho}{\partial t}=-[\rho,H]
\end{align}


TITLE: How to find the second general solution of this equation?
QUESTION [1 upvotes]: $\frac{dx}{y(x+y)+az} = \frac{dy}{x(x+y)-az} = \frac{dz}{z(x+y)}$
By comparing the first two ratios (adding them), I found one solution. How do find the second solution?

REPLY [2 votes]: $$\frac{dx}{y(x+y)+az} = \frac{dy}{x(x+y)-az} = \frac{dz}{z(x+y)}$$
$\frac{dx}{y(x+y)+az} = \frac{dy}{x(x+y)-az} =\frac{dx+dy}{(x+y)^2}$
$$\frac{dx+dy}{(x+y)^2}= \frac{dz}{z(x+y)}$$
$$\frac{dx+dy}{x+y}= \frac{dz}{z}$$
$$\boxed{\frac{z}{x+y}=c_1}$$
$\frac{dx}{y(x+y)+az} = \frac{dy}{x(x+y)-az}=\frac{dx}{(y+ac_1)(x+y)} = \frac{dy}{(x-ac_1)(x+y)}$
$$\frac{dx}{y+ac_1} = \frac{dy}{x-ac_1}$$
$$(x-ac_1)dx-(y+ac_1)dy=0$$
$$\frac12(x^2-y^2)-ac_1(x+y)=c_2$$
$$\boxed{\frac12(x^2-y^2)-az=c_2}$$
The PDE is certainly
$$(y(x+y)+az)\frac{\partial z}{\partial x}+(x(x+y)-az)\frac{\partial z}{\partial y}=(x+y)z$$
So, the general solution expressed on implicit form $c_1=F(c_2)$ is :
$$\boxed{z=(x+y)\:F\left(\frac12(x^2-y^2)-az \right)}$$
$F$ is an arbitrary function, to be determined according to some boundary condition (presently not specified in the wording of the question).
AN ALTERNATIVE WAY TO SOLVE THE PDE :
Change of function
$$z=(x+y)u\quad\implies\quad \begin{cases} z_x=u+(x+y)u_x \\z_y=u+(x+y)u_y \end{cases}$$
$(y(x+y)+a(x+y)u)(u+(x+y)u_x) +(x(x+y)-a(x+y)u)(u+(x+y)u_y)=(x+y)^2u$
After simplification :
$$(y+au)\frac{\partial u}{\partial x}+(x-au)\frac{\partial u}{\partial y}=0$$
Solving with method of characteristics leads to :
$$u=F\left(\frac12(x^2-y^2)-a(x+y)u \right)$$


TITLE: Does tensoring with a module preserve this injection even if the module is not flat?
QUESTION [1 upvotes]: Suppose $R$ is a domain and let $Q$ be its field of fractions. Then $ 0 \to R \to Q $ is exact. Now suppose that $M$ is a torsion free $R$ module.

Is it necessary that $ 0 \to M \to Q \otimes M $ is exact ?

This is true when $R$ is a PID. But I am not sure whether assuming that it is a domain suffices. If it is not please give a counter example. Thanks

REPLY [2 votes]: Maybe this works:
It's easy to see that both $Q(R) \otimes_{R}(-)$ and $(R-0)^{-1}(-)$ are two functors that works as left-adjoint to the forgetful functor from $Q(R)$ modules to $R$ modules.
This immediately tells you that the two functors are isomorphic. So this tells you that $Q(R) \otimes_{R}(M)$ and $(R-0)^{-1}(M)$ are naturally isomorphic.
Now you have by construction that $\frac{m}{1}=0$ in $(R-0)^{-1}M$ iff there is $a \neq 0$ in R such that $am=0$. That is, iff m is a torsion element.
So for torsion free elements, your map is injective.
Hope it helps and hope it works


TITLE: Proving $\mathbb{P}(\sum_{n=1}^{\infty} S_n = \infty) = 1$ where $S_i \sim \exp(\lambda_i)$ and $\sup_n \lambda_n < \infty$.
QUESTION [1 upvotes]: I was working through J.R. Norris Markov Chains and got stuck at exercise 2.3.1. The question is the following:

Let $S_1, S_2, \ldots$ be independent exponential random variables of parameters $\lambda_1, \lambda_2, \ldots$ respectively. Show that $\lambda_1 S_1$ is exponential with parameter $1$.
Use the strong law of large numbers to show, first in the special case $\lambda_n = 1$ for all $n$, and then subject only to the condition $\sup_n \lambda_n < \infty$, that
  $$
 \mathbb{P}\left(\sum_{n=1}^{\infty} S_n = \infty\right) = 1
$$
  Is the condition $\sup_n \lambda_n < \infty$.

I was able to prove that $\lambda_i S_i \sim \exp(1)$, but I was not able to prove that $\mathbb{P}\left(\sum_{n=1}^{\infty} S_n = \infty\right) = 1
$, even for the special case $\lambda_n = 1 \, \forall n$.
Intuitively I have concluded the following: Since $\mathbb{E}(S_n) = \frac{1}{\lambda_n}$ and $\sup_n \lambda_n < \infty$, we can conclude that
$$
\mathbb{E}(\sum_{n=1}^{\infty} S_n) = \sum_{n=1}^{\infty} \mathbb{E}(S_n) = \sum_{n=1}^{\infty} \frac{1}{\lambda_n} = \infty,
$$
since $\sup_n \lambda_n < \infty$.
Is there any way to build on this intuition? How do I exactly use the strong law of large numbers to solve the exercise?
Thanks in advance!

REPLY [1 votes]: First let $\lambda_n = 1$. Then, by the strong law of large numbers, $$\frac{1}{N} \sum_{n=1}^{N} S_n \to 1$$ almost surely. In particular, the sequence of partial sums $(\sum_{n=1}^{N} S_n)_{N\geq 1}$ is almost surely unbounded (otherwise the limit would be $0$ on a set of nonzero probability).
Now assume $\Lambda:=\sup_n \lambda_n < \infty$. By what we've just shown and because $\lambda_i S_i \sim \exp(1)$, we know that $$\mathbb{P}\left(\sum_{n=1}^{\infty}\lambda_n S_n = \infty\right) = 1.$$
But if $\sum_{n=1}^{\infty}\lambda_n S_n (\omega) = \infty$, then also $\frac{1}{\Lambda}\sum_{n=1}^{\infty}\lambda_n S_n (\omega) = \infty$, and
$$\infty = \sum_{n=1}^{\infty}\frac{\lambda_n}{\Lambda} S_n \leq \sum_{n=1}^{\infty} S_n .$$


TITLE: Proposition 1.16 in Lee's Introduction to Smooth Manifolds
QUESTION [0 upvotes]: I'm having some trouble with one of the statements in the proof for Prop. 1.16 in Lee's Introduction to Smooth Manifolds. A countable collection $\mathcal{B}$ of coordinate balls covering a manifold $M$ is given, and Lee asserts:
"For any pair of coordinate balls $B,B' \in \mathcal{B}$, the intersection $B \cap B'$ has at most countably many components, each of which is path-connected."
I'm trying to figure out why this is true. My thoughts for the "countably many components" part is as follows: If there were uncountably many components, then we could find a basis element in each of them, and since the components are disjoint, these basis elements must be distinct. This would then violate second countability. The hole here is that we can only find these basis elements if the components are open in $M$, and I don't know if this is true. Similarly I don't see why the components are path connected.
The fact that $M$, being a manifold, is locally path connected means that the components of $M$ are open and path connected, but I don't know if this is true of the components of $B \cap  B'$. If open subspaces inherit local path connectedness then I think all this would be resolved, but is that the case?

REPLY [2 votes]: It all follows from standard general topological facts. I need that the coordinate balls are open, separable and locally (path-)connected:
The coordinate balls are locally connected, and this means (by a standard theorem) that all components of open subsets are open, so this applies to $B \cap B'$ (which is open in $B$). And open subsets of locally path-connected subsets are still locally path-connected (and so this holds of for all components of $B \cap B'$ in particular).
Also, another standard theorem is that a connected, locally path-connected space is path-connected, so all these (open) components of $B \cap B'$ are path-connected as well.
That there are only at most countably many of such components follows from the fact that in a separable space there can be at most countably many disjoint open sets (each has to contain a distinct member from a counatble dense set) (in general, a separable space is ccc) and $B$ (and $B \cap B'$) is certainly separable.


TITLE: Simplify multiple integration of an exponential function
QUESTION [1 upvotes]: I need to simplify the following integration: $I=\int_a^b\int_a^b...\int_a^b\exp\{-\frac{1}{2}x^TAx\}dx_1dx_2...dx_n$, where $x=(x_1,x_2,...,x_n)^T$ and $A$ is a positive definite matrix of size $n$, $a,b$ - fixed real numbers. Is there any way to make it simple? Thanks !

REPLY [1 votes]: We can try to use the rule for multidimensional substitution to make this integral look simpler (Spoiler alert: This will not result in a solvable integral, however I'm going through the steps for the substitution anyway):
$$\int_{\varphi(U)} f(\mathbf{x})\, d \mathbf{x} = \int_U f(\varphi(\mathbf{y})) \left|\det(\operatorname{D}\varphi)(\mathbf{y})\right| \,d \mathbf{y}$$
Here, we're looking for a function $\varphi$ which simplifies $f(\mathbf{x}) = \exp(-\mathbf{x}^TA\mathbf{x}/2)$ to $f(\varphi(\mathbf y))= \exp(-\mathbf{y}^T \mathbf{y}/2)$.
Staring at this formula suggests $\mathbf y = A^{1/2} \mathbf x$, and in fact $A^{1/2}$ is well-defined since $A$ is symmetric (and hence diagonaliable) and all of its eigenvalues are positive. Positivity ensures that this matrix is invertible and we get $\varphi(\mathbf y) = A^{-1/2} \mathbf y$ and hence $$f(\varphi(\mathbf y))=\exp\left(-\frac{\mathbf{y^T} A^{-1/2}AA^{-1/2}\mathbf y }{2}\right)= \exp\left(-\frac{\mathbf{y^T}\mathbf y }{2}\right).$$
Since $\varphi$ is linear, its derivative is constant and precisely its representing matrix, i.e. $$\left|\det(\operatorname{D}\varphi)(\mathbf{y})\right|=|\det A^{-1/2}|=\left( \prod_{i=1}^n  \lambda_i \right)^{-1/2}$$ where $\lambda_i$ are the eigenvalues of $A$ (and the determinant is the product of the inverse square roots of them).
Finally for the domain of integration, we are given $\varphi(U) = [a,b]^n$, i.e.
$U = A^{1/2}([a,b]^n)$.
To put it together, our integral now looks like
$$\left(\prod_{i=1}^n  \lambda_i \right)^{-1/2} \int_{A^{1/2}([a,b]^n)}\exp\left(-\frac{\mathbf y ^T \mathbf y}{2}\right) d \mathbf y.$$
This is slightly nicer than the original integral, but does remind of the distribution function of a standard normal distribution $\exp(-x^2/2)$. In fact, we are trying to compute a probability for a multivariate normal distribution. Unfortunately, no nice substitution can help us to solve this integral analytically.


TITLE: Necessity of LUB property to prove the nested interval theorem
QUESTION [2 upvotes]: Is least upper bound property of the set of real numbers necessary to prove the nested interval theorem ? I know that it is sufficient, but I doubt whether it is necessary.

REPLY [2 votes]: It is not necessary, in the following sense. There exists an ordered field $S$ where the nested interval theorem holds, but the least upper bound property does not. To prove the nested interval theorem for $S$, one need not assume LUB; indeed one cannot assume LUB, since it isn't true!
As usual for these questions, my source is James Propp's "Real Analysis in Reverse".


text:
theory Rose_Tree
imports Main "HOL-Library.Sublist"
begin

text \<open>For theory \<open>Incredible_Trees\<close> we need rose trees; this theory contains
the generally useful part of that development.\<close>

subsubsection \<open>The rose tree data type\<close>

datatype 'a rose_tree = RNode (root: 'a) (children:  "'a rose_tree list")

subsubsection \<open>The set of paths in a rose tree\<close>

text \<open>Too bad that @{command inductive_set} does not allow for varying parameters...\<close>

inductive it_pathsP :: "'a rose_tree \<Rightarrow> nat list \<Rightarrow> bool"  where
   it_paths_Nil: "it_pathsP t []"
 | it_paths_Cons: "i < length (children t) \<Longrightarrow> children t ! i = t' \<Longrightarrow> it_pathsP t' is \<Longrightarrow> it_pathsP t (i#is)"

inductive_cases it_pathP_ConsE: "it_pathsP t (i#is)"

inductive_cases it_pathP_RNodeE: "it_pathsP (RNode r ants) is"

definition it_paths:: "'a rose_tree \<Rightarrow> nat list set"  where
  "it_paths t = Collect (it_pathsP t)"

lemma it_paths_eq [pred_set_conv]: "it_pathsP t = (\<lambda>x. x \<in> it_paths t)"
 by(simp add: it_paths_def)

lemmas it_paths_intros [intro?] = it_pathsP.intros[to_set]
lemmas it_paths_induct [consumes 1, induct set: it_paths] = it_pathsP.induct[to_set]
lemmas it_paths_cases [consumes 1, cases set: it_paths] = it_pathsP.cases[to_set]
lemmas it_paths_ConsE = it_pathP_ConsE[to_set]
lemmas it_paths_RNodeE = it_pathP_RNodeE[to_set]
lemmas it_paths_simps = it_pathsP.simps[to_set]

lemmas it_paths_intros(1)[simp]

lemma it_paths_RNode_Nil[simp]: "it_paths (RNode r []) = {[]}"
  by (auto elim: it_paths_cases)

lemma it_paths_Union: "it_paths t \<subseteq> insert [] (Union (((\<lambda> (i,t). ((#) i) ` it_paths t) ` set (List.enumerate (0::nat) (children t)))))"
  apply (rule)
  apply (erule it_paths_cases)
  apply (auto intro!: bexI simp add: in_set_enumerate_eq)
  done

lemma finite_it_paths[simp]: "finite (it_paths t)"
  by (induction t) (auto intro!:  finite_subset[OF it_paths_Union]  simp add: in_set_enumerate_eq)

subsubsection \<open>Indexing into a rose tree\<close>

fun tree_at :: "'a rose_tree \<Rightarrow> nat list \<Rightarrow> 'a rose_tree" where
  "tree_at t [] = t"
| "tree_at t (i#is) = tree_at (children t ! i) is"

lemma it_paths_SnocE[elim_format]:
  assumes "is @ [i] \<in> it_paths t"
  shows "is \<in> it_paths t \<and> i < length (children (tree_at t is))"
using assms
by (induction "is" arbitrary: t)(auto intro!: it_paths_intros elim!: it_paths_ConsE)

lemma it_paths_strict_prefix:
  assumes "is \<in> it_paths t"
  assumes "strict_prefix is' is"
  shows "is' \<in> it_paths t"
proof-
  from assms(2)
  obtain is'' where "is = is' @ is''" using strict_prefixE' by blast
  from assms(1)[unfolded this]
  show ?thesis
    by(induction is' arbitrary: t) (auto elim!: it_paths_ConsE intro!: it_paths_intros)
qed

lemma it_paths_prefix:
  assumes "is \<in> it_paths t"
  assumes "prefix is' is"
  shows "is' \<in> it_paths t"
using assms it_paths_strict_prefix strict_prefixI by fastforce

lemma it_paths_butlast:
  assumes "is \<in> it_paths t"
  shows "butlast is \<in> it_paths t"
using assms prefixeq_butlast by (rule it_paths_prefix)

lemma it_path_SnocI:
  assumes "is \<in> it_paths t"
  assumes "i < length (children (tree_at t is))"
  shows "is @ [i] \<in> it_paths t"
  using assms
  by (induction t arbitrary: "is" i)
     (auto 4 4  elim!: it_paths_RNodeE intro: it_paths_intros)

end

subset_name: curated