TITLE: What is a typical example of the tensor product of modules failing to be left exact?
QUESTION [5 upvotes]: I am looking for an example of an exact sequence of $R$-modules $$ 0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0 $$ and a $R$-module $N$, such that $$ 0 \rightarrow M' \otimes N \rightarrow M \otimes N \rightarrow M'' \otimes N \rightarrow 0 $$
fails to be exact.

REPLY [7 votes]: The typical example is made up with $M=M'=\mathbb Z$, and $N= \mathbb Z/ n \mathbb Z$, with the injection $\mathbb Z \to \mathbb Z, x \mapsto n x$.
When tensorized by $Id_{\mathbb Z/ n \mathbb Z}$, this map becomes zero.


TITLE: Why don't solid objects blend into each other?
QUESTION [0 upvotes]: For example, What prevents my feet from sinking and blending into the ground? Or my 2 hands from blending and morphing into one when I clap them? Is it merely the density of objects? Space between atoms?

REPLY [1 votes]: I'll give a go at an answer here.  Of course the bonds within a solid work to maintain the physical structure of the solid.  Also the electron clouds of the surface atoms extend away from the surface.  In general, when you bring two solid objects next to one another, these electron clouds begin to overlap and will produce a repulsive force.
I had a physics teacher one time strike a hammer against a table top.  He then turned to the class and said something along the lines of, "Electrons don't like to be forced next to one another, do they?"


TITLE: Any Borel Set can be approximated by closed and bounded set in probability measure.
QUESTION [1 upvotes]: $\forall \varepsilon>0,\forall A\in\mathscr{B}_\mathbf{R^n},\exists K\subseteq A$, which is closed and bounded, such that $P(A\setminus K)\le \varepsilon$, where $P$ is the probability measure.
There is also a similar proposition says something like any Borel set can be approximated by open sets.

REPLY [2 votes]: The very first theorem in Billngsley's book (Convergence of Probability measures, Theorem 1.1 of Chapter 1) says the following: For any Borel probability measure $P$ on  a metric space $S$, any Borel set $A$ and any $\epsilon >0$ there exists a closed set $C$ and an open set $U$ such that $C \subset A \subset U$ and $P(U\setminus C) <\epsilon$.
(The proof of the theorem is quite elementary).
If $S$ is a countable union of compact sets then choose compact sets $K_n$ increasing to $C$ and take $K=C\cap K_n$. For $n$ sufficiently large this compact set satisfies your requirement.
Of course $\mathbb R^{n}$ is  a countable union of compact sets, so we can apply this theorem.


TITLE: Find the absolute minimum and absolute maximum
QUESTION [1 upvotes]: Find the absolute minimum and absolute  maximum  in [-1,2]  of the function $ \ f(x)=|x| \ $. $$ $$ I have got absolute minimum at x=0 , ie., $ f(0)=0 $ is the absolute mimimum. The absolute maximum is $ f(2)=2 $ . Is this correct ?

REPLY [1 votes]: We can't differentiate normally when there is a modulus sign, so consider $g(x)=f(x)^2=|x|^2=x^2$
Then $g'(x) =2x$
Setting $g'(x)=0\implies f(x)^2=0\implies f(x)=0$ so clearly a minimum there.
Then we can maximise $g(x)$ by testing the upper and lower bounds of the given domain i.e. $g(-1)$ and $g(2)$.
$g(-1)=1<g(2) =4$. Thus $g(x)$ is maximised as $x=2$, which implies that $f(x)$ is maximised at $x=2$, so $f_{\text{max}}=f(2)=2$
Alternatively you could consider the separate cases when $f(x)$ is positive or negative and compute the derivatives accordingly, however the method I have used is often more helpful when dealing with more difficult questions.


TITLE: Integration by parts in 4 dimensions
QUESTION [1 upvotes]: I'm trying to understand why the following is true (assuming a vanishing boundary term). The integrals are over all space, $\phi=\phi(x)$.
$$\frac{1}{2}\int d^4x \left(\partial_\mu\phi \partial^\mu \phi-m^2\phi^2)\right) = -\frac{1}{2}\int d^4x \left( \phi \partial_\mu \partial^\mu \phi + m^2 \phi^2\right).$$
This is treated by book as trivial but I can't figure out the rule how to manipulate the partial derivatives in this case.

REPLY [1 votes]: First of all, the term $m^2\phi^2$ can be omitted on both sides.
The derivate of $\phi$ is integrated to $\phi$ and the rest of the expression is then differentiated.
This is exactly how integration by parts works.


TITLE: What is meant by "space is expanding"?
QUESTION [0 upvotes]: Is it just the distance between galaxies that increases?
Or is the distance increasing between the molecules in my body?
Does expansion affect universal constants?
Various sources I have heard on the topic of expansion seem to contradict each other. (at least in my mind) If a physicist says: "There's not enough gravity to prevent expansion" then it makes it seem like a simple matter of Newtonian physics; as if galaxies were flying off under their own momentum. But if a physicist says: "Nothing can travel faster than light EXCEPT for the expansion of space itself" then this makes it seem like the invisible fabric of space-time is stretching.

REPLY [1 votes]: The space that you inhabit is expanding, but it is expanding slowly enough (~70 km/s/Mpc) that it doesn't have any significant effect at that scale.  Even if you do think of this as "stretching" the bonds within molecules in your body, it's slow enough that everything can just go back to its equilibrium geometry.
To give you some idea of the rate of expansion, if molecules didn't go back to equilibrium geometry (they do), and everything within space also expanded (it doesn't), then over the course of a century, the distance between your head and feet would grow by about the length of a typical chemical bond (~$10^{-10} \text{m}$).  A bond between two atoms would grow by ~$10^{-20} \text{m}$, which is a distance ~10 orders of magnitude smaller than the bond itself.


TITLE: How to find $\sum_{k=1}^n k\binom{n}{k}?$
QUESTION [0 upvotes]: How to find the sum of series,
$$\sum_{k=1}^n k\binom{n}{k}?$$
I try solve using this fact
$$\sum_{k=1}^n \binom{n}{k} = 2^n$$
but it didn't help.
And now i haven't some ideas.
Maybe i can calculate $$\sum_{k=1}^n k\binom{n}{k}?$$
using $$\sum_{k=1}^{n-1} k\binom{n}{k}?$$
using this fact
$$\binom{n}{k}  = \binom{n - 1}{k - 1} + \binom{n - 1}{k}$$
I failed use it

REPLY [3 votes]: Combinatorial argument
To make a basket team, you need players, and among these players, you need one captain. So, $k\binom{n}{k}$ is the number of team with $k$ players you can make among $n$ people. Therefore, $\displaystyle\sum_{k=1}^nk\binom{n}{k}$ is the number of team you can make with $n$ people. But to make a team, you can also take one person among the $n$ people, and say that it's the captain. You have $n$ possibilities to choose it. After, you can decide if each of the $n-1$ other people are in the team or not. So at the end, you can make $n2^{n-1}$ teams. Therefore $$\sum_{k=1}^nk\binom{n}{k}=n2^{n-1}.$$

Using Binomial formula
$$\sum_{k=0}^n\binom{n}{k}x^k=(1+x)^n.$$
Therefore, \begin{align*}
\sum_{k=1}^nk\binom{n}{k}&=\left.\frac{\mathrm d }{\mathrm d x}\right|_{x=1}\sum_{k=0}^n\binom{n}{k}x^k\\
&=\left.\frac{\mathrm d }{\mathrm d x}\right|_{x=1}(1+x)^n\\
&=n2^{n-1}.
\end{align*}


TITLE: how to solve this graph theory question?
QUESTION [0 upvotes]: if G=(A,B,E) is a graph without directions.
and (i mean unite value) |A U B| =50
what is the max value of |E| ?
im very new at the topic and if there is anywebsite that also can explain more.. it will be very helpfull... thanks
WHAT I DID:
(probally its worng...)
if |A U B| =50 then max value of |A|=|B|=25 for each
what means the |E|=625
hopefully it correct?

REPLY [0 votes]: It seems that you mean that $G$ is an undirected bipartite graph with partite sets of vertices $A$ and $B$ and edge set $E$, with $|A \cup B| = 50$. So, you are asking for the maximum number of edges in a bipartite graph with 50 vertices.
The number of edges in a complete bipartite graph whose partite sets have size $m$ and $n$ respectively is $mn$. (Every edge contains exactly one vertex in the partite set of size $m$; for each of these $m$ vertices, there are edges to each of the $n$ vertices on the other side.)
And, given any bipartite graph $G$, obviously the complete bipartite graph on the same partite sets has at least as many edges as $G$. So your question boils down to finding $n$ between 0 and 50 so that $n(50-n)$ is maximized.
One way to do this is calculus: to find the maximum of $50n - n^2$,
solve for its derivative to be 0:
$$50 - 2n = 0$$
has the solution $n = 25$. This critical point is indeed a maximum (take the second derivative if you want to check).
Then the number of edges is $25 \cdot 25 = 625$.


TITLE: Band structure of a nanoparticle
QUESTION [1 upvotes]: Considering a crystal at such a small size (for example, a sphere with diameter about 20 atoms) that the periodic boundary condition assumed in Bloch Theorem is not really a good estimation, I wonder what are other possible ways to compute the band structure efficiently. Is there some quick and neat modification on Bloch Theorem that makes this happen?

REPLY [1 votes]: Without periodicity there's no band structure (in the sense of $E(\vec k)$), since quasiwavevector $\vec k$ is not conserved. Moreover, due to surface tension, such a nanocrystal will get distorted compared to a large version of this crystal, so even local periodicity will be broken.
Still, if you want to approximate the nanocrystal by simply a lump of elementary cells of a macrocrystal separated by the same lattice constant, and impose zero Dirichlet boundary conditions somewhere near its surface, you can actually use Bloch's theorem to calculate eigenstates of electron in this system, provided some conditions are fullfilled.
Consider the infinite crystal we would take as the macroscopic material of the same kind as the nanoparticle we are interested in. Bloch eigenfunctions of electron in this crystal are
$$\psi_{\vec k}(\vec r)=u_{\vec k}(\vec r)\exp(i\vec k\vec r),$$
where $u$ is periodic with the periodicity of the crystal's Bravais lattice. These functions are eigenfunctions of the translation operator.
But if the crystal potential has reflection symmetry$^\dagger$, so that e.g. $U(x,y,z)=U(-x,y,z)$, then the degenerate pairs of eigenfunctions corresponding to opposite $k_x$ can be linearly combined in such a way that they combinations will vanish at the reflection plane $x=0$. This would also make the wavefunctions purely real (up to constant phase factor), which then guarantees that they'll regularly attain zeros at some $x=x_0$ as we increase energy eigenvalue (keeping $|k_x|$, $k_y$ and $k_z$ constant). These zeros will define possible values of ($x$ contribution of) energy for which the wavefunction satisfies zero Dirichlet boundary conditions at both planes.
If we now, for e.g. a cubic crystal, define pairs of such planes in each of three symmetry directions and find the corresponding contributions to energy so that our wavefunctions vanish at all these planes, this will yield the eigenstates of our approximation of the (rectangular-shaped) nanocrystal.

$^\dagger$ I'm not sure if this is a necessary condition. It might be that the linear combination talked above can be found for any crystal. It's just easiest to prove for the reflectionally-symmetric potential (because the solutions can be classified by parity).


text:
\begin{definition}[Definition:Fractal]
A '''fractal''' is an [[Definition:Object|object]] that has the property of [[Definition:Self-Similarity|self-similarity]] on all [[Definition:Scale|scales]].
\end{definition}

config: wiki


text:
\begin{document}
\title{Positive entropy implies chaos along any infinite sequence}
\author{Wen Huang, Jian Li and Xiangdong Ye}

\dedicatory{Dedicated to the memory of Anatoly Mikhailovich Stepin (20.07.1940 -- 07.11.2020)}

\address[W.~Huang, X.~Ye]{CAS Wu Wen-Tsun Key Laboratory of Mathematics, School of Mathematical
Sciences, University of Science and Technology of China,
	Hefei, Anhui, 230026, P.R. China}
 \email{wenh@mail.ustc.edu.cn,
	yexd@ustc.edu.cn}
\address[J.~Li]{Department of Mathematics, Shantou University, Shantou, Guangdong 515063, P.R. China}
\email{lijian09@mail.ustc.edu.cn}
\subjclass[2010]{37B05,37B40,37A35}
\keywords{Li-Yorke chaos, topological entropy, measure-theoretic entropy,
	amenable group action}

\begin{abstract}
Let $G$ be an infinite countable discrete amenable group.
For any $G$-action on a compact metric space $(X,\rho)$, it turns out that
if the action has positive topological entropy, then for any
sequence $\{s_i\}_{i=1}^{+\infty}$ with pairwise distinct elements in $G$
there exists a Cantor subset $K$ of $X$ which is Li-Yorke chaotic along this sequence,
that is, for any two distinct points $x,y\in K$, one has
\[\limsup_{i\to+\infty}\rho(s_i x,s_iy)>0,\ \text{and}\ \liminf_{i\to+\infty}\rho(s_ix,s_iy)=0.\]
\end{abstract}

\maketitle

\section{Introduction}
Throughout this paper, let $G$ be a countable, discrete, infinite, amenable group
with the identity element $e_G$.
By a \emph{$G$-system} we mean a pair $(X,G)$,
where $X$ is a compact metrizable space with a metric $\rho$ and
a continuous action $\Gamma\colon G\times X\to X$
such that $\Gamma(e_G,x)=x$ and $\Gamma(g_1, \Gamma(g_2,x))=
\Gamma(g_1g_2,x)$ for all $g_1,g_2\in G$ and $x\in X$.
As usual, we let $gx=\Gamma(g,x)$ for simplicity.
In the case $G = \mathbb{Z}$,
we can describe the action as a homeomorphism $T\colon X\to X$
 which corresponds to the generator $1$ in $\mathbb{Z}$ and
generates an action $n\mapsto T^n$ through iterations.

Let $S$ be an infinite subset of $G$ and enumerate it as a sequence $\{s_i\}_{i=1}^\infty$,
and $\delta>0$ be a constant.
A pair $(x,y)\in X\times X$ is called \emph{$(S,\delta)$-scrambled}
if
\[\limsup_{i\to+\infty}\rho(s_ix,s_iy)>\delta\quad\text{ and }\quad
\liminf_{i\to+\infty}\rho(s_ix,s_iy)=0.\]
As $S$ is countable, it is easy to see that the above limits
do not depend on the order of elements of $S$.
A subset $K$ of $X$ with at least two points is called \emph{$(S,\delta)$-scrambled} if
every two distinct points $x,y\in K$ form an $(S,\delta)$-scrambled pair.
Following ideas in \cite{LY75}, we say that $(X,G)$ is \emph{$(S,\delta)$-Li-Yorke chaotic}
if there exists an uncountable $(S,\delta)$-scrambled subset of $X$.

Given a $G$-system $(X,G)$, one can define its topological entropy
$h_{top}(X,G)$ lying in $[0,+\infty]$ (see Section 2.2 for details).
One of fundamental questions in topological dynamics is:
if a dynamical system has positive topological entropy,
how complicated
its orbits can be.
There are many results in this line.
We refer the reader to a recent survey \cite{LY16}.

Here let us mention a few related results.
In \cite{BGKM02} using the measure-theoretical argument Blanchard \textit{et al.} showed that
if a $\mathbb{Z}$-action system $(X,T)$ has positive topological entropy
then there exists a $\delta>0$ and a Cantor $(\mathbb{Z}_+,\delta)$-scrambled subset of $X$.
For a while, people were seeking a proof which does not involve measures. It was done by
Kerr and Li who used a combinatorial method giving an alternative proof and  generalized it to
amenable group actions in \cite{KL07}, and  to sofic group actions in \cite{KL13}.
For a related work, see the recent paper by Li and Rong \cite{LR20}.

The main result of this paper is the following result:

\begin{thm} \label{thm:main-result-1}
If a $G$-system $(X,G)$ has positive topological entropy,
then for any infinite subset $S$ of $G$ there exists a $\delta>0$ and a Cantor $(S,\delta)$-Li-Yorke scrambled set.
\end{thm}

Note that this result is new even when $G=\mathbb{Z}$. After reviewing some auxiliary results of $G$-systems in Section 2,
we will prove the main result in Section 3.

\section{Preliminaries}
In this section, we review some notions and  properties  of $G$-systems which will be used later.
We refer the reader to \cite{EW} and \cite{Gl1}
for textbooks on ergodic theory.

Let $X$ be a compact metric space. A subset $K \subset X$ is called \emph{a Mycielski set}
if it is a union of countably many Cantor sets.
This definition was introduced in~\cite{BGKM02}.
For convenience we restate here a version of Mycielski's theorem (see~\cite[Theorem~1]{My})
which we shall use.
Recall that the \emph{diagonal} of $X\times X$ is
\[\Delta_X=\{(x,x)\in X\times X\colon x\in X\}.\]
\begin{thm}\label{thm:Myc}
	Let $X$ be a perfect compact metric space.
	Assume that $R$ is a dense $G_\delta$ subset of $X\times X$.
	Then  there exists a dense Mycielski subset $K$ of $X$  such that
\[K\times K\subset R\cup \Delta_X.\]
\end{thm}

Let $G$ be a countable, discrete, infinite group.
A sequence $\{F_n\}_{n=1}^{+\infty}$ of non-empty finite subsets  of $G$ is called a \emph{F{\o}lner sequence}
if for every $g\in G$,
\[\lim_{n\to +\infty} \frac{|gF_n\Delta F_n|}{|F_n|}=0,\]
where $|\cdot|$ denotes the cardinality of a set.
It is well known that
$G$ is amenable if and only if it admits a  F{\o}lner sequence.

\subsection{Invariant measures and the disintegration of measures}
Let $(X,G)$ be a $G$-system.
Denote by $\mathcal{B}$ the collection of all Borel subset of $ X$ and
$\mathcal{M}(X)$ the set of all Borel probability measures on $X$.
The \emph{support} of $\mu\in\mathcal{M}(X)$, denoted by $\supp(\mu)$,
is defined to be the set of all points $x\in X$ for which every open
neighborhood of $x$ has a positive measure.

A measure $\mu\in\mathcal{M}(X)$ is called \emph{$G$-invariant} if $\mu(g^{-1}A)=\mu(A)$
for all $g\in G$ and $A\in \mathcal{B}$,
and it is called \emph{ergodic} if it is $G$-invariant and
$\mu(\bigcup_{g\in G}gA)=1$ for any $A\in \mathcal{B}$ with $\mu(A)>0$.
Denote by $\mathcal{M}(X,G)$ (resp.\ $\mathcal{M}^{e}(X,G)$) the set of all $G$-invariant measures
(resp.\ ergodic measures) of $(X,G)$.

Let $\pi\colon (X,\mathcal{B},\mu)\to (Y,\mathcal{D},\nu)$ be a measure-preserving map between Borel probability spaces.
There exists a map
\[\mathbb{E}_{\mu}(\,\cdot\,|Y)\colon
L^1(X,\mathcal{B},\mu)\to L^1(Y,\mathcal{D},\nu),\]
called the \emph{conditional expectation}, such that
every $f\in L^1(X,\mathcal{B},\mu)$, $\mathbb{E}_\mu(f|Y)$
is the unique
element in $L^1(Y,\mathcal{D},\nu)$ satisfying for every $A\in\mathcal{D}$,
\begin{equation}
\int_A \mathbb{E}_\mu(f|Y)d\nu = \int_{\pi^{-1}(A)} f d\mu.
\end{equation}
There also exists a measurable map from $Y$ to $\mathcal{M}(X)$, denoted by $y\mapsto \mu_y$, which satisfies
for every $f\in L^1(X,\mathcal{B},\mu)$, one has $f \in L^1(X,\mathcal{B},\mu_y)$ for $\nu$-a.e. $y\in Y$ and
\begin{equation}
\mathbb{E}_{\mu}(f|Y)(y)=\int_X f\,d\mu_{y}\
\text{for $\nu$-a.e. } y\in Y. \label{eq:meas3}
\end{equation}
We shall write $\mu=\int \mu_yd\nu(y)$ and refer to this as the
\emph{disintegration of $\mu$ with respect to  $(Y,\mathcal{D},\nu)$}.
It is clear that for $\nu$-a.e. $y\in Y$, $\mu_y(\pi^{-1}(y))=1$.

Let $\mathcal{F}$ be a sub-$\sigma$-algebra of $\mathcal{B}$.
There exists a compact metric space $Y$ and a Borel probability measure
$\nu$ on $(Y,\mathcal{D})$, a measure-persevering map
$\pi\colon (X,\mathcal{B},\mu)\to (Y,\mathcal{D},\nu)$ such that
$\pi^{-1}(\mathcal{D})=\mathcal{F} \pmod \mu$.
For every $f\in L^1(X,\mathcal{B},\mu)$, the conditional expectation of $f$ with respect to $\mathcal{F}$,
denoted by $\mathbb{E}_\mu(f|\mathcal{F})$, is the unique element  in $L^1(X,\mathcal{F},\mu)$
such that $\mathbb{E}_\mu(f|\mathcal{F}) \circ \pi=\mathbb{E}_\mu(f|Y)$.

The following lemma is from \cite{R49} (see Lemma 3 in \S4 No. 2).
\begin{lem}\label{lem:Rohlin-measure}
Let $\mu\in \mathcal{M}(X)$ and
$\mu=\int  \mu_y d \nu(y)$ be the disintegration of $\mu$ with respect to  $(Y,\mathcal{D},\nu)$.
Suppose $\mu_y$ is non-atomic for $\nu$-a.e.\ $ y\in Y$.
If $0<r<1$ and $A$ is a measurable subset of $X$ with $\mu_y(A)\leq r$
for $\nu$-a.e.\ $y\in Y$,
then there exists a measurable subset $A'$ of $X$
such that $A \subset A'$ and $\mu_y(A')=r$ for $\nu$-a.e.\ $y\in X$.
\end{lem}

The \emph{relatively independent self-joining of $\mu$ with respect to  $(Y,\mathcal{D},\nu)$} is the Borel probability measure
$\mu\times_Y\mu=\int \mu_y\times \mu_y d\nu(y)$
on $X\times X$ in the sense that
\[\mu\times_Y\mu(A\times B)=\int_Y\mu_y (A)\mu_y(B)d\nu(y)\]
for all $A,B\in\mathcal{B}$.
Let $p\colon X\times X \to X$ be the canonical projection
to the first coordinate.
Then $\pi\circ p\colon (X\times X,\mathcal{B}\otimes\mathcal{B},\mu\times_Y\mu)\to (Y,\mathcal{D},\nu)$ is measure-preserving and
 the disintegration of $\mu\times_Y \mu$ with respect to $(Y,\mathcal{D},\nu)$ is
\[
(\mu\times_Y\mu)_y=\mu_y\times \mu_y.
\]
We have the following variation of Lemma~\ref{lem:Rohlin-measure}.
\begin{lem}\label{lem:Rohlin-measure-2}
Let $\mu\in \mathcal{M}(X)$ and
and
$\mu=\int  \mu_y d \nu(y)$ be the disintegration of $\mu$ with respect to  $(Y,\mathcal{D},\nu)$.
Suppose $\mu_y$ is non-atomic for $\nu$-a.e.\ $ y\in Y$.
If $0<r<1$ and
$B$ is a measurable subset of $X\times X$ with $\mu_y\times \mu_y(B)\leq r$
for $\nu$-a.e.\ $y\in Y$,
then there exists a measurable subset $B'$ of $X\times X$
such that $B\subset B'$ and $\mu_y\times \mu_y(B')=r$ for $\nu$-a.e.\ $y\in Y$.
\end{lem}

\subsection{Topological entropy and measure-theoretic entropy}
Let $(X,G)$ be a $G$-system.
A \emph{cover} of $X$ is a family of subsets of $X$ whose union is $X$.
Define the \emph{join} of two covers $\mathcal{U}$ and $\mathcal{V}$ of $X$ by
$\mathcal{U}\vee \mathcal{V}=\{U\cap V\colon U\in\mathcal{U}, V\in\mathcal{V}\}$.
Denote by $N(\mathcal{U})$ the number of sets in a subcover of $\mathcal{U}$ of minimal cardinality.

Let $\mathcal{U}$ be an open cover of $X$.
The \emph{entropy of $\mathcal{U}$ with respect to $G$} is defined by
\[h_{top}(G,\mathcal{U})=\lim_{n\to\infty}\frac{1}{|F_n|}
\log N\Bigl(\bigvee_{g\in F_n}g^{-1}\mathcal{U}\Bigr),\]
where $\{F_n\}$ is a F{\o}lner sequence for $G$.
It is well known that the limit exists and is independent of the choice of the F{\o}lner sequence.
The \emph{topological entropy of $(X,G)$} is then defined by
\[h_{top}(X,G)=\sup_{\mathcal{U}}h_{top}(G,\mathcal{U}),\]
where the supremum is taken over all finite open covers of $X$.

A \emph{partition} of $X$ is a family of measurable subsets of $X$
whose elements are pairwise disjoint and the union is $X$.
Given $\mu\in\mathcal{M}(X,G)$,
for a finite partition $\alpha=\{A_1,A_2,\dotsc,A_k\}$ of $X$,
define
\[H_\mu(\alpha)=-\sum_{i=1}^k \mu(A_i)\log\mu(A_i).\]
The \emph{measure-theoretic entropy of $\alpha$ with respect to $\mu$} is defined by
\[h_\mu(G,\alpha)=\lim_{n\to\infty}\frac{1}{|F_n|}
H_\mu\biggl(\bigvee_{g\in F_n}g^{-1}\alpha\biggr),\]
where $\{F_n\}$ is a F{\o}lner sequence for $G$.
It is well known that the limit exists and is independent of the choice of the F{\o}lner sequence.
The \emph{measure-theoretic entropy of $\mu$} is then defined by
\[h_{\mu}(X,G)=\sup_{\alpha}h_{\mu}(G,\alpha),\]
where the supremum is taken over all finite partitions of $X$.
By the well-known variational principle, we have
\[h_{top}(X,G)=\sup_{\mu\in \mathcal{M}(X,G)}h_\mu(X,G)=
\sup_{\mu\in \mathcal{M}^{e}(X,G)}h_\mu(X,G).\]

Let $\mathcal{F}$ be a sub-$\sigma$-algebra of $\mathcal{B}$.
For a finite partition $\alpha$ of $X$,
define
\begin{equation} \label{eq:conditional-entropy}
	H_\mu(\alpha|\mathcal{F})=-
	\int\sum_{A\in\alpha}\mathbb{E}_\mu(1_A|\mathcal{F})\log\mathbb{E}_\mu(1_A|\mathcal{F})d\mu.
\end{equation}
Similarly, we can define the \emph{conditional entropy of a partition $\alpha$ of $X$ given $\mathcal{F}$}
by
\[h_\mu(G,\alpha|\mathcal{F})=\lim_{n\to\infty}\frac{1}{|F_n|}
H_\mu\biggl(\bigvee_{g\in F_n}g^{-1}\alpha|\mathcal{F}\biggr),\]
where $\{F_n\}$ is a F{\o}lner sequence for $G$.

It is well known that if $\alpha$ is a partition of $X$ with $k$-atoms
then $H_\mu(\alpha)\leq \log k$.
We will use the following easy estimation.

\begin{lem}\label{lem:H-mu-alpha2}
Let $\alpha=\{A_1,A_2,\dotsc,A_k\}$ be a partition of $X$.
Then
\[H_\mu(\alpha)\leq \log 2+ (1-\mu(A_1))\log(k-1).\]
\end{lem}
\begin{proof}
Let $\beta=\{A_1,A_1^c\}$. Then
\begin{align*}
H_\mu(\alpha)=H_\mu(\alpha\vee\beta)
=H_\mu(\beta)+H_\mu(\alpha|\sigma(\beta))\leq
 \log 2+ (1-\mu(A_1))\log(k-1),
\end{align*}
where $\sigma(\beta)$ is the $\sigma$-algebra generated by $\beta$.
\end{proof}

\subsection{Pinsker \texorpdfstring{$\sigma$}{sigma}-algebra}
Given a $\mu\in \mathcal{M}^{e}(X,G)$,
\emph{the Pinsker $\sigma$-algebra of $(X,\mathcal{B},\mu,G)$}, denoted by $\mathcal{P}_\mu(G)$,
is the $G$-invariant sub-$\sigma$-algebra of $\mathcal{B}$
generated by all finite partitions $\alpha$ of $X$
with $h_\mu(G,\alpha)=0$.
The Pinsker $\sigma$-algebra $\mathcal{P}_\mu(G)$ corresponds
to a \emph{Pinsker factor} $(Y,\mathcal{P}_\mu(G),\nu,G)$ of $(X,\mathcal{B},\mu,G)$, which plays an important role in
the study of entropy in ergodic theory.

The following result is well known,
see e.g.\ \cite[Corollary 18.20]{Gl1} for $\mathbb{Z}$-actions,
\cite[Lemma 4.3]{HXY15} or \cite[Propsition 3.1]{WZ16} for amenable group actions.
\begin{lem} \label{lem:lambda}
Let $(X,G)$ be a $G$-system and $\mu\in \mathcal{M}^e(X,G)$ with $h_\mu(G)>0$.
Let
\[\mu=\int_Y \mu_y d \nu(y)\]
be the disintegration of $\mu$ with respect to the Pinsker factor $(Y,\mathcal{P}_\mu(G),\nu,G)$ and
$\lambda=\mu \times_{Y}\mu$.
Then $\mu_y$ is non-atomic for $\nu$-a.e.\ $ y \in Y$ and
 $\lambda\in \mathcal{M}^e(X\times X,G)$.
In particular, $\lambda(\Delta_X)=0$.
\end{lem}

The \emph{Pinsker $\sigma$-algebra of $(X,\mu,G)$ given $\mathcal{F}$},
denoted by $\mathcal{P}_\mu(G|\mathcal{F})$,
is the $G$-invariant sub-$\sigma$-algebra of $\mathcal{B}$
generated by all finite partitions $\alpha$ of $X$
with $h_\mu(G,\alpha|\mathcal{F})=0$.

The following result plays a key role
in our proof of the main result.
Note that it was proved in \cite[Theorem 2.13]{RW00} under the assumption that the action is free,
and in \cite[Theorem 0.1]{D01} in general.
Here we follow \cite[Theorem 6.10]{HYZ11} for this version.
\begin{thm} \label{thm:entorpy-UPE-extension}
Let $(X,G)$ be a $G$-system.
Assume that $\mu\in \mathcal{M}^e(X,G)$, $\alpha$ is a finite partition of $(X,\mu)$
and $\eps>0$.
Then there exists a finite subset $K$ of $G$ such that for every
finite subset $Q$ of $G$ with
 $(QQ^{-1}\setminus \{e_G\})\cap K=\emptyset$, one has
\begin{equation} \label{eq:partition-Pinsker}
\biggl|H_\mu\bigl(\alpha|\mathcal{P}_\mu(G)\bigr) - \frac{1}{|Q|}H_\mu\biggl(\bigvee_{g\in Q}g^{-1}\alpha\biggl|\mathcal{P}_\mu(G)\biggr)
\biggr|<\eps.
\end{equation}
\end{thm}
It is clear that
\[
\frac{1}{|Q|}H_\mu\biggl(\bigvee_{g\in Q}g^{-1}\alpha\biggl|\mathcal{P}_\mu(G)\biggr)
\leq \frac{1}{|Q|}\sum_{g\in Q} H_\mu\bigl(g^{-1}\alpha\bigl|\mathcal{P}_\mu(G)\bigr) = H_\mu\bigl(\alpha|\mathcal{P}_\mu(G)\bigr).
\]
So the term inside the absolute value in the formula \eqref{eq:partition-Pinsker} is non-negative.

Let
\[\mu=\int_Y \mu_y d \nu(y)\]
be the disintegration of $\mu$ with respect to the Pinsker factor $(Y,\mathcal{P}_\mu(G),\nu,G)$.
According to \eqref{eq:meas3} and \eqref{eq:conditional-entropy}, one has
\[
H_\mu\bigl(\beta|\mathcal{P}_\mu(G)\bigr) =
\int_Y H_{\mu_y}(\beta) d\nu(y)
\]
for a finite partition $\beta$ of $X$. Note that for any $y\in Y$,
\[
H_{\mu_y}
\biggl(\bigvee_{g\in Q} g^{-1} \alpha \biggr) \leq\sum_{g\in Q} H_{\mu_y}\bigl(g^{-1} \alpha\bigr),
\]
and for any $g\in G$,
\[
\int_Y H_{\mu_y}(g^{-1}\alpha)d\nu(y)  = H_\mu\bigl(g^{-1}\alpha|\mathcal{P}_\mu(G)\bigr)=H_\mu\bigl(\alpha|\mathcal{P}_\mu(G)\bigr)=
\int_Y H_{\mu_y}(\alpha)d\nu(y).
\]
Thus
\begin{align*}\label{eq:partition-Pinsker2}
 &\hskip0.5cm \int_Y\biggl|\frac{1}{|Q|} \sum_{g\in Q}H_{\mu_y}\bigl(g^{-1} \alpha\bigr)-\frac{1}{|Q|}H_{\mu_y}
 \biggl(\bigvee_{g\in Q} g^{-1}\alpha\biggr)
 \biggr| d\nu(y)\\
 &=\int_Y \big( \frac{1}{|Q|} \sum_{g\in Q}H_{\mu_y}\bigl(g^{-1} \alpha\bigr)-\frac{1}{|Q|}H_{\mu_y}
 \biggl(\bigvee_{g\in Q} g^{-1}\alpha\biggr)
 \big) d\nu(y)\\
 &=\frac{1}{|Q|} \sum_{g\in Q}\int_Y H_{\mu_y}\bigl(g^{-1} \alpha\bigr)d\nu(y)-\frac{1}{|Q|}\int_YH_{\mu_y}
 \biggl(\bigvee_{g\in Q} g^{-1}\alpha
 \biggr) d\nu(y)\\
 &=\int_Y H_{\mu_y}\bigl(\alpha\bigr)d\nu(y)-\frac{1}{|Q|}\int_YH_{\mu_y}
 \biggl(\bigvee_{g\in Q} g^{-1}\alpha
 \biggr) d\nu(y)\\
 &=H_\mu\bigl(\alpha|\mathcal{P}_\mu(G)\bigr) - \frac{1}{|Q|}H_\mu\biggl(\bigvee_{g\in Q}g^{-1}\alpha\biggl|\mathcal{P}_\mu(G)\biggr)\\
 &=\biggl|H_\mu\bigl(\alpha|\mathcal{P}_\mu(G)\bigr) - \frac{1}{|Q|}H_\mu\biggl(\bigvee_{g\in Q}g^{-1}\alpha\biggl|\mathcal{P}_\mu(G)\biggr)
\biggr|.
\end{align*}
Hence we can rewrite the formula \eqref{eq:partition-Pinsker}  as
\begin{equation}\label{eq:partition-Pinsker2}
 \int_Y\biggl|\frac{1}{|Q|} \sum_{g\in Q}H_{\mu_y}\bigl(g^{-1} \alpha\bigr)-\frac{1}{|Q|}H_{\mu_y}
 \biggl(\bigvee_{g\in Q} g^{-1}\alpha\biggr)
 \biggr| d\nu(y)<\eps.
\end{equation}
It should be noticed that the term inside the absolute value in the formula \eqref{eq:partition-Pinsker2} is non-negative,
but this form is convenient for later use.

We will use the following result on the Pinsker $\sigma$-algebra of
$(X\times X,\lambda,G)$.
It  was proved in \cite[Theorem 4]{GTW00} under the assumption that the action is free,
and  in \cite[Theorem 0.4]{D01} in general.
Here we follow \cite[Lemma 4.2]{HXY15} for this version.

\begin{lem}  \label{lem:relative-UPE}
Let $(X,G)$ be a $G$-system and $\mu\in \mathcal{M}^e(X,G)$.
Let
\[\mu=\int_Y \mu_y d \nu(y)\]
be the disintegration of $\mu$ with respect to the Pinsker factor $(Y,\mathcal{P}_\mu(G),\nu,G)$,
$\lambda=\mu \times_{Y}\mu$
and $p\colon X\times X \to X$ is the canonical projection to the first coordinate.
Then
\[\mathcal{P}_\lambda(G|p^{-1}(\mathcal{P}_\mu(G)))=p^{-1}(\mathcal{P}_\mu(G)) \pmod \lambda.\]
\end{lem}

By Lemma~\ref{lem:relative-UPE},
we know that $(X\times X,\lambda,G)$ is  $\mathcal{P}_\mu(G)$-relatively
complete positive entropy (see \cite{GTW00} or \cite{D01})
and the Pinsker $\sigma$-algebra $\mathcal{P}_\lambda(G)$ of $(X\times X,\lambda,G)$
is $p^{-1}(\mathcal{P}_\mu(G)) \pmod \lambda$.
Then
\[\lambda=\int_Y\mu_y\times \mu_y d\nu(y)\]
is the disintegration of $\lambda$ with respect to the Pinsker factor of $(X\times X,\lambda,G)$ and for any finite partition $\beta$ of $X\times X$, we have
\begin{align*}
 H_\lambda(\beta|\mathcal{P}_\lambda(G))
  &=\int_Y H_{\mu_y\times\mu_y}(\beta)d\nu(y).
\end{align*}

\section{Proof of the main result}
In fact, we will prove the following result, which is stronger
than Theorem~\ref{thm:main-result-1} stated in the introduction.
\begin{thm}\label{thm:Main-result-3}
Assume that $(X,G)$ is a $G$-system, $\mu \in \mathcal{M}^{e}(X,G)$,
$\pi$ is a factor map from $(X,\mathcal{B},\mu,G)$ to its Pinsker factor $(Y,\mathcal{P}_\mu(G),\nu,G)$ and $\mu=\int \mu_y d \nu(y)$
is the disintegration of $\mu$ with respect to $(Y,\mathcal{P}_\mu(G),\nu,G)$.
If $h_\mu(G)>0$, then
for any sequence $\{s_i\}_{i=1}^{+\infty}$ with pairwise distinct elements in $G$, one has the following: for $\nu$-a.e.\ $y\in Y$, there exists a constant
$\delta=\delta(y)>0$ and a dense Mycielski subset $K$ of  $\supp(\mu_y)$
such that $K$ is $(\{s_i\},\delta)$-scrambled, that is,
 for any two distinct points $x_1,x_2\in K$, we have
\begin{equation}
\limsup_{i\to+\infty}\rho(s_i x_1,s_ix_2)\geq\delta, \label{eq:R-delta}
\end{equation}
and
\begin{equation}
\liminf_{i\to+\infty}\rho(s_ix_1,s_ix_2)=0.  \label{eq:P}
\end{equation}
\end{thm}

The main idea of the proof of Theorem~\ref{thm:Main-result-3}
is that after constructing two proper partitions of  $X\times X$
we apply Theorem~\ref{thm:entorpy-UPE-extension}
to show that the collections of pairs satisfying
\eqref{eq:R-delta} and \eqref{eq:P} has a full $\mu_y\times \mu_y$-measure
for $y$ in a large measurable subset of $Y$,
see Lemmas \ref{lem:not-asy} and \ref{lem:proximal-r} below. Then we finish
the proof by applying Theorem~\ref{thm:Myc}.

\medskip

To do so, for every $r>0$, put
\[R_r(\{s_i\})=\Bigl\{(x_1,x_2)\in X\times X\colon
\limsup_{i\to+\infty}\rho(s_ix_1,s_ix_2)\geq r\Bigr\}.\]

\begin{lem}\label{lem:not-asy}
For every $\eps>0$ there exists a $r>0$
and a measurable subset $D$ of $Y$ with $\nu(D)>1-4\eps$
such that $\mu_y\times\mu_y(R_r(\{s_i\}))=1$ for all $y\in D$.
\end{lem}
\begin{proof}
For $t>0$, let
\[\Delta_t=\{(x_1,x_2)\in X\times X\colon \rho(x_1,x_2)<t\}.\]
Let $\lambda=\mu\times_{Y}\mu$.
By Lemma~\ref{lem:lambda}, $\lambda(\Delta_X)=0$.
Then $\lim_{t\to 0}\lambda(\Delta_t)=\lambda(\Delta_X)=0$.
Fix $\eps>0$.
There exists a $r>0$ such that
$\lambda(\Delta_r)<\frac{\eps}{2}$.
Note that
\[\lambda(\Delta_r)=\int_Y \mu_y\times \mu_y(\Delta_r)d\nu(y)
\geq \tfrac{1}{2}\mu(\{y\in Y\colon \mu_y\times \mu_y(\Delta_r)\geq \tfrac{1}{2}\}),\]
so $\nu(V_r)>1-\eps$, where
\[V_r=\{y\in Y\colon 0<\mu_y\times \mu_y(\Delta_r)<\tfrac{1}{2}\}.\]
By Lemma~\ref{lem:Rohlin-measure-2},
there exists a measurable set $\Delta_r\cap(\pi^{-1}(V_r)\times \pi^{-1}(V_r))\subset B\subset X\times X$
with $\mu_y\times\mu_y(B)=\frac{1}{2}$
for $\nu$-a.e.\ $y\in Y$.
Let $A_1=B\cap (\pi^{-1}(V_r)\times \pi^{-1}(V_r))$,
$A_2=(\pi^{-1}(V_r)\times \pi^{-1}(V_r))\setminus A_1$, $A_3=X\times X\setminus(\pi^{-1}(V_r)\times \pi^{-1}(V_r))$,
and  $$\alpha=\{A_1,A_2,A_3\}.$$

Let $Q_1=\{s_1\}$ and suppose that for some $m\in\mathbb{N}$,
we have defined the sets $Q_1,Q_2,\dotsc,Q_{m-1}$.
Now we apply Theorem~\ref{thm:entorpy-UPE-extension} to $(X\times X,\lambda,G)$,
the partition $\alpha$ and the constant $\bigl(\frac{\eps}{2^m}\bigr)^2$,
and we let  $K_m$ be the resulting set.
Choose $Q_m\subset\{s_{m},s_{m+1},\dotsc\}\setminus \bigcup_{i<m}Q_i$ such that
$(Q_mQ_m^{-1}\setminus\{e_G\})\cap K_m=\emptyset$ and $|Q_m|>|Q_{m-1}|$.
Then
\[ \int\biggl\vert
\frac{1}{|Q_m|}H_{\mu_y\times\mu_y}\biggl(\bigvee_{g\in Q_m} g^{-1}\alpha\biggr)-
\frac{1}{|Q_m|}\sum_{g\in Q_m} H_{\mu_y\times\mu_y}(g^{-1}\alpha)\biggr\vert d\nu(y)
<\Bigl(\frac{\eps}{2^m}\Bigr)^2.\]
Let
\[
D_m=\biggl\{y\in Y\colon\biggl\vert \frac{1}{|Q_m|}H_{\mu_y\times\mu_y}
\biggl(\bigvee_{g\in Q_m} g^{-1}\alpha\biggr)
-\frac{1}{|Q_m|}\sum_{g\in Q_m} H_{\mu_y\times\mu_y}(g^{-1}\alpha)\biggr\vert<
\frac{\eps}{2^m}\biggr\}.
\]
Then $\nu(D_m)>1- \frac{\eps}{2^m}$.
Let $D_0=\bigcap_{m=1}^{+\infty} D_m$. Then $\mu(D_0)>1-\eps$.

By \cite[Proposition 5.9]{F81} or \cite[Corollary 5.24]{EW}, for any $g\in G$,
$g\mu_y=\mu_{gy}$ holds for $\nu$-a.e.\ $y\in X$.
Then there exists a $G$-invariant measurable subset $Y_0$ of $Y$ with $\nu(Y_0)=1$
such that
\begin{enumerate}
\item for any $y\in Y_0\cap V_r$, $\mu_y\times\mu_y(A_1)=\frac{1}{2}$;
\item for any $y\in Y_0$ and $g\in G$,
\[H_{\mu_y\times\mu_y}(g^{-1}\alpha)= H_{\mu_{gy}\times\mu_{gy}}(\alpha);\]
\item for any $y\in Y_0$, $\mu_y(\pi^{-1}(y))=1$.
\end{enumerate}
For every $y\in Y_0$ and $g\in G$, we have the following two cases:
\begin{enumerate}
\item[Case 1]   if $gy\in V_r$, one has $\mu_{gy}\times\mu_{gy}(A_1)=\frac{1}{2}$,
$\mu_{gy}\times\mu_{gy}(A_2)=\frac{1}{2}$ and $\mu_{gy}\times\mu_{gy}(A_3)=0$,
then $H_{\mu_y\times\mu_y}(g^{-1}\alpha)=\log 2$;
\item[Case 2]  if $gy\not\in V_r$, one has $\mu_{gy}\times\mu_{gy}(A_1)=0$,
$\mu_{gy}\times\mu_{gy}(A_2)=0$ and $\mu_{gy}\times\mu_{gy}(A_3)=1$,
then $H_{\mu_y\times\mu_y}(g^{-1}\alpha)=0$.
\end{enumerate}
For any $y\in Y$, let $I_m(y)=\{g\in Q_m\colon gy\in V_r\}$ for $m\in \mathbb{N}$.
By conclusions of Cases 1 and 2, we have for any $y\in Y_0$ and $m\in \mathbb{N}$
\[\frac{1}{|Q_m|}\sum_{g\in Q_m} H_{\mu_y\times \mu_y}(g^{-1}\alpha)
=\frac{1}{|Q_m|} |I_m(y)| \log 2\]
and
\[
H_{\mu_y\times \mu_y}\biggl(\bigvee_{g\in Q_m}g^{-1}\alpha\biggr)
=H_{\mu_y\times \mu_y}\biggl(\bigvee_{g\in I_m(y)}g^{-1}\alpha \biggr).
\]
By the definition of $D_m$, we have
\begin{equation}
\biggl|\frac{1}{|Q_m|} H_{\mu_y\times\mu_y}
\biggl(\bigvee_{g\in I_m(y)} g^{-1}\alpha\biggr)-
\frac{1}{|Q_m|}|I_m(y)|\log2\biggr|<\frac{\eps}{2^m}, \label{eq:H-mux-1}
\end{equation}
for every $y\in D_0\cap Y_0$ and every $m\geq 1$.
For every $y\in Y$, define
\[f(y)
=\limsup_{m\to+\infty}\frac{1}{|Q_m|}|I_m(y)|
=\limsup_{m\to+\infty} \frac{1}{|Q_m|} \sum_{g\in Q_m} 1_{V_r}(gy).
\]
Then
\begin{align*}
\int_Y  f(y)d\nu(y)&=\int_Y  \limsup_{m\to+\infty}\frac{1}{|Q_m|} \sum_{g\in Q_m} 1_{V_r}(gy) d\nu(y) \\
&\geq \limsup_{m\to+\infty} \frac{1}{|Q_m|} \sum_{g\in Q_m}\int_Y 1_{V_r}(gy) d\nu(y), \quad \text{ by Fatou's Lemma} \\
&=\nu(V_r)>1-\eps,
\end{align*}
and $\nu(\{y\in Y\colon f(y)>\frac{1}{2}\})>1-2\eps$.
Let
\[D=V_r\cap D_0\cap Y_0\cap \{y\in Y\colon f(y)>\tfrac{1}{2}\}.\]
Then $\nu(D)>1-4\eps$.

Now we show that $D$ is as required. Fix $y\in D$ and $m\in\mathbb{N}$.
Let
\[a_m=\mu_y\times \mu_y\biggl(\bigcap_{g\in I_m(y)} g^{-1}A_1\biggr).\]
If $g \in I_m(y)$, i.e., $gy\in V_r$, then by the conclusion of Case 1 we have
\begin{equation}
  \mu_{gy}\times \mu_{gy}(A_3)=0 \label{eq:mu-A3-0}
\end{equation}
 and
\[ \bigvee_{g\in I_m(y)} g^{-1}\alpha =
\bigvee_{g\in I_m(y)} g^{-1}\{A_1,A_2\} \pmod{\mu_y\times\mu_y}.\]
By \eqref{eq:H-mux-1}, we have
\begin{align*}
\frac{1}{|Q_m|}|I_m(y)|\log2-\frac{\eps}{2^m}&\leq
\frac{1}{|Q_m|}H_{\mu_y\times \mu_y} \biggl(\bigvee_{g\in I_m(y)} g^{-1}\alpha\biggr)\\
&=\frac{1}{|Q_m|}H_{\mu_y\times \mu_y}\biggl(\bigvee_{g\in I_m(y)} g^{-1}\{A_1,A_2\}\biggr)\\
&\leq\frac{1}{|Q_m|}
\Bigl(\log 2+(1-a_m)\log\bigl(2^{|I_m(y)|}-1\bigr)\Bigr),
\quad \text{ by Lemma~\ref{lem:H-mu-alpha2}}\\
&\leq \frac{1}{|Q_m|}\Bigl( \log 2+(1-a_m)|I_m(y)|\log2\Bigr).
\end{align*}
Then
\[
a_m \cdot \frac{1}{|Q_m|}|I_m(x)| \leq \frac{1}{|Q_m|}+\frac{\eps}{2^m\log 2}.
\]
As $\limsup\limits_{m\to+\infty}\frac{1}{|Q_m|}|I_m(y)|=f(y)>\frac{1}{2}$,
we have $\liminf\limits_{m\to+\infty}a_m=0$.

Let
\[Asy_r(\{s_i\},m)=\{(x_1,x_2)\in X\times X\colon
\rho(s_i x_1,s_ix_2)< r,\ \forall i\geq m\}.\]
Note that $Asy_r(\{s_i\},m)$ is increasing as $m$ increases.
Let
\[Asy_r(\{s_i\})=\bigcup_{m=1}^{+\infty} Asy_r(\{s_i\},m).\]
Note that if $\rho(x_1,x_2)<r$, then $(x_1,x_2)\in A_1\cup A_3$.
As $Q_m\subset \{s_{m},s_{m+1},\dotsc\}$,
\begin{align*}
 Asy_r(\{s_i\},m) \subset \bigcap_{g\in Q_m}g^{-1}(A_1\cup A_3).
\end{align*}
So for any $y\in D$,
\begin{align*}
\mu_y\times\mu_y(Asy_r(\{s_i\},m))
&\leq \mu_y\times\mu_y
\biggl(\bigcap_{g\in Q_m}g^{-1}(A_1\cup A_3) \biggr)\\
&\leq \mu_y\times\mu_y
\biggl(\bigcap_{g\in I_m(y)}g^{-1}(A_1\cup A_3) \biggr)\\
&=\mu_y\times\mu_y \biggl(\bigcap_{g\in I_m(y)}g^{-1}A_1 \biggr),
\qquad \text{ by } \eqref{eq:mu-A3-0}\\
&=a_m.
\end{align*}
As $\liminf\limits_{m\to+\infty}a_m\to 0$,
we have $\mu_y\times\mu_y(Asy_r(\{s_i\}))=0$,
because $Asy_r(\{s_i\},m)$ is increasing as $m$ increases.
It is clear that
$X\times X\setminus Asy_r(\{s_i\})\subset R_r(\{s_i\})$.
Thus, $\mu_y\times\mu_y(R_r(\{s_i\}))=1$ for any $x\in D$. This ends the proof.
\end{proof}
For $r>0$, put
\[P_r(\{s_i\})=\Bigl\{(x_1,x_2)\in X\times X\colon \liminf_{i\to+\infty}\rho(s_ix_1,s_ix_2)\leq r\Bigr\}.\]

\begin{lem}\label{lem:proximal-r}
For every $r>0$ and $\eps>0$ there exists a measurable subset $E$ of $X$ with $\mu(E)>1-4\eps$ such that
for any $y\in E$, $\mu_y\times \mu_y(P_r(\{s_i\}))=1$.
\end{lem}
\begin{proof}
Fix $r>0$ and $\eps>0$.
As $\mu_y\times\mu_y(\Delta_r)>0$ for all $y\in Y$, there exists $L\in\mathbb{N}$ such that
$\nu(W_r)>1-\eps$, where $W_r=\{y\in Y\colon \mu_y\times \mu_y(\Delta_r)\geq \frac{1}{L}\}$.
By Lemma~\ref{lem:Rohlin-measure-2}, there exists
  $B_L\subset \Delta_r\cap (\pi^{-1}(W_r)\times \pi^{-1}(W_r))$
such that $\mu_y\times \mu_y(B_L)=\frac{1}{L}$ for $\nu$-a.e.\ $y\in W_r$.
By Lemma~\ref{lem:Rohlin-measure-2} again,
there exist $B_1,\dotsc,B_{L-1}\subset \pi^{-1}(W_r)\times \pi^{-1}(W_r)$ such that
$\{B_1,\dotsc,B_L\}$ is a partition of $\pi^{-1}(W_r)\times \pi^{-1}(W_r)$
and for  $\nu$-a.e.\
$y\in W_r$, $\mu_y\times \mu_y(B_i)=\frac{1}{L}$ for $i=1,2,\dotsc,L$.
Let $B_{L+1}=X\times X\setminus(\pi^{-1}(W_r)\times \pi^{-1}(W_r))$ and
$$\beta=\{B_1,B_2,\dotsc,B_L, B_{L+1}\}.$$

Let $Q_1=\{s_1\}$ and suppose that for some $m\in\mathbb{N}$,
we have define the sets $Q_1,Q_2,\dotsc,Q_{m-1}$.
Now we apply Theorem~\ref{thm:entorpy-UPE-extension} to
$(X\times X,\lambda,G)$, the partition $\beta$ and the constant $\bigl(\frac{\eps}{2^m}\bigr)^2$, and we let $K_m$ be the resulting set.
Choose $Q_m\subset\{s_{m},s_{m+1},\dotsc\}\setminus \bigcup_{i<m}Q_i$ such that
$(Q_mQ_m^{-1}\setminus\{e_G\})\cap K_m=\emptyset$ and $|Q_m|>|Q_{m-1}|$.
Then
\[ \int \biggl| \frac{1}{|Q_m|}H_{\mu_y\times\mu_y}\biggl(\bigvee_{g\in Q_m} g^{-1}\beta\biggr)-
\frac{1}{|Q_m|}\sum_{g\in Q_m} H_{\mu_y\times\mu_y}(g^{-1}\beta) \biggr| d\nu(y)<
\Bigl(\frac{\eps}{2^m}\Bigr)^2.\]
Let
\[
E_m=\biggl\{y\in Y\colon\biggl\vert \frac{1}{|Q_m|}H_{\mu_y\times\mu_y}
\biggl(\bigvee_{g\in Q_m} g^{-1}\beta\biggr)
-\frac{1}{|Q_m|}\sum_{g\in Q_m} H_{\mu_y\times\mu_y}(g^{-1}\beta)\biggr\vert<
\frac{\eps}{2^m}\biggr\}.
\]
Then $\mu(E_m)>1- \frac{\eps}{2^m}$.
Let $E_0=\bigcap_{m=1}^{+\infty} E_m$. Then $\mu(E_0)>1-\eps$.

There exists a $G$-invariant measurable subset $Y_0$ of $Y$ with $\nu(X_0)=1$
such that
\begin{enumerate}
	\item for any $y\in Y_0\cap W_r$,
	$\mu_y\times \mu_y(B_i)=\frac{1}{L}$ for $i=1,2,\dotsc,L$;
	\item for any $y\in Y_0$ and $g\in G$,
	\[H_{\mu_y\times\mu_y}(g^{-1}\beta)= H_{\mu_{gy}\times\mu_{gy}}(\beta);\]
    \item for any $y\in Y_0$, $\mu_y(\pi^{-1}(y))=1$.
\end{enumerate}
For every $y\in Y_0$ and $g\in G$, we have the following two cases:
\begin{enumerate}
	\item[Case 1]  if $gy\in W_r$, one has $\mu_{gy}\times\mu_{gy}(B_i)=\frac{1}{L}$
	for $i=1,2,\dotsc,L$, and $\mu_{gy}\times\mu_{gy}(B_{L+1})=0$,
	then $H_{\mu_y\times\mu_y}(g^{-1}\beta)=\log L$;
	\item[Case 2] if $gy\not\in W_r$, one has $\mu_{gy}\times\mu_{gy}(B_i)=0$
	for $i=1,2,\dotsc,L$, and
	$\mu_{gy}\times\mu_{gy}(B_{L+1})=1$
	then $H_{\mu_y\times\mu_y}(g^{-1}\beta)=0$.
\end{enumerate}
For $y\in Y$, let $I_m(y)=\{g\in Q_m\colon gy\in W_r\}$ for $m\in\mathbb{N}$.
By conclusions of Cases 1 and 2,
we have for any $y\in Y_0$ and $m\in \mathbb{N}$,
\[\sum_{g\in Q_m}H_{\mu_y\times\mu_y}(g^{-1} \beta)=|I_m(y)|  \log L,\]
and
\[H_{\mu_y\times\mu_y}\biggl(\bigvee_{g\in Q_m} g^{-1}\beta\biggr)=
H_{\mu_y\times \mu_y}\biggl(\bigvee_{g\in I_m(y)} g^{-1}\beta\biggr).\]
By the definition of $E_m$, we have
\begin{equation}
\biggl|\frac{1}{|Q_m|} H_{\mu_y\times\mu_y}
\biggl(\bigvee_{g\in I_m(y)} g^{-1}\beta\biggr)-
\frac{1}{|Q_m|}|I_m(y)|\log L\biggr|<\frac{\eps}{2^m}, \label{eq:H-mux-beta-1}
\end{equation}
for every $y\in E_0\cap Y_0$ and every $m\geq 1$.
Similarly as in the proof in Lemma~\ref{lem:not-asy},
the $\nu$-measure of the set
\[Y_1=\Bigl\{y\in Y\colon \limsup_{n\to+\infty} \frac{1}{|Q_m|}|I_m(y)|>\tfrac{1}{2}\Bigr\}\]
is at lease $1-2\eps$.
Let
\[E=W_r\cap E_0\cap Y_0 \cap Y_1.\]
Then $\nu(E)>1-4\eps$.

Now we show that $E$ is as required.
Fix $y\in E$ and $m\geq 1$.
Let
\[C_m=\bigcap_{g\in I_m(y)} g^{-1}(B_1\cup B_2\cup \dotsb\cup B_{L-1})\]
and $b_m=\mu_y\times\mu_y(C_m)$.
If $g\in I_m(y)$, i.e., $gy\in W_r$, then by the conclusion of Case 1
we have
\begin{equation}
\mu_y\times \mu_y(g^{-1}B_{L+1})=0 \label{eq:mu-x-B-L+1}
\end{equation}
 and
\begin{align*}
\bigvee_{g\in I_m(y)} g^{-1}\beta &=
\bigvee_{g\in I_m(y)} g^{-1}(\{B_1,B_2,\dotsc,B_L\}) \pmod{\mu_y\times\mu_y} \\
&=\bigvee_{g\in I_m(y)} g^{-1}(\{B_1,B_2,\dotsc,B_{L-1}\})   \\
&\qquad\cup \bigcup_{g_0\in I_m(y)} \biggl(\bigl\{g_0^{-1} B_L\bigr\}\bigvee
\bigvee_{g\in I_m(y)\setminus \{g_0\}} g^{-1}(\{B_1,B_2,\dotsc,B_{L-1},B_L\})\biggr)\\
&=\ (*)\ \cup\ (**)\
\end{align*}
Note that the part $(*)$ can be regarded as a partition of $C_m$  with at most $(L-1)^{|I_m(y)|}$ atoms
and the part $(**)$ as a partition of $X\times X \setminus C_m$ with less than $L^{|I_m(y)|}$ atoms.
By \eqref{eq:H-mux-beta-1}, we have
\begin{align*}
 \frac{1}{|Q_m|}|I_m(y)|\log L-\frac{\eps}{2^m} &\leq
  \frac{1}{|Q_m|} H_{\mu_y\times \mu_y}\biggl(\bigvee_{g\in I_m(y)} g^{-1}\beta\biggr)\\
  &=\frac{1}{|Q_m|}H_{\mu_y\times \mu_y}\bigl(\ (*)\ \cup\ (**)\ \bigr)  \\
&\leq \frac{1}{|Q_m|}\Bigl(\log 2+ b_m|I_m(y)| \log(L-1) \\
&\qquad\qquad\qquad+(1-b_m)|I_m(y)|\log L\Bigr), \quad
\text{by Lemma~\ref{lem:H-mu-alpha2}}.
\end{align*}
Then
\[ b_m\cdot\frac{1}{|Q_m|}|I_m(y)|\leq \frac{1}{\log L-\log(L-1)}\Bigl(\frac{\log2}{|Q_m|}+\frac{\eps}{2^m}\Bigr).\]
As $\limsup\limits_{m\to+\infty}\frac{1}{|Q_m|}|I_m(y)|>\tfrac{1}{2}$,
we have  $\liminf\limits_{m\to+\infty}b_m=0$.

For $m\geq 1$, let
\[J_r(\{s_i\},m)=\{(x_1,x_2)\in X\times X\colon \rho(s_ix_1,s_ix_2)\geq r,\ \forall i\geq m\},\]
and
\[J_r(\{s_i\})=\bigcup_{m=1}^{+\infty} J_r(\{s_i\},m)\]
Note that if $\rho(x_1,x_2)\geq r$ then $(x_1,x_2)\in B_1\cup B_2\cup \dotsb\cup B_{L-1}\cup B_{L+1}$.
As $Q_m\subset \{s_{m},s_{m+1},\dotsc\}$,
\begin{align*}
  J_r(\{s_i\},m)&\subset \bigcap_{g\in Q_m} g^{-1}(B_1\cup B_2\cup \dotsb\cup B_{L-1}\cup B_{L+1})
\end{align*}
and
\begin{align*}
\mu_y\times\mu_y(J_r(\{s_i\},m))&\leq
\mu_y\times\mu_y\biggl(\bigcap_{g\in Q_m} g^{-1}(B_1\cup B_2\cup \dotsb\cup B_{L-1}\cup B_{L+1})\biggr)\\
&=\mu_y\times\mu_y\biggl(\bigcap_{g\in I_m(x)} g^{-1}(B_1\cup B_2\cup \dotsb\cup B_{L-1})\biggr), \qquad\text{ by \eqref{eq:mu-x-B-L+1}}\\
&=b_m.
\end{align*}
Then $\mu_y\times\mu_y(J_r(\{s_i\}))=0$,
as $J_r(\{s_i\},m)$ is increasing as $m$ increases.
It is clear that
$X\times X\setminus J_r(\{s_i\})\subset P_r(\{s_i\})$.
Thus, $\mu_y\times\mu_y(P_r(\{s_i\}))=1$ for any $y\in E$.
This ends the proof of the Claim.
\end{proof}
Now we are ready to prove Theorem~\ref{thm:Main-result-3}
\begin{proof}[Proof of Theorem~\ref{thm:Main-result-3}]
By Lemma~\ref{lem:not-asy},
for every $k\in\mathbb{N}$, there exists a $r_k>0$ and a measurable set $D_k\subset Y$
with $\mu(D_k)>1-\frac{1}{k}$ such that for every $y\in D_k$,
$\mu_y\times \mu_y(R_{r_k}(s_i))=1$.
Let
\[P(\{s_i\})=\Bigl\{(x_1,x_2)\in X\times X\colon \liminf_{i\to+\infty}\rho(s_ix_1,s_ix_2)=0\Bigr\}.\]
It is clear that  $P(\{s_i\})=\bigcap_{k=1}^{+\infty} P_{\frac{1}{k}}(\{s_i\})$.
By Lemma~\ref{lem:proximal-r},
we know that for any $r>0$ and $\eps>0$,
there exists a measurable subset $E'$ of $Y$ with $\nu(E')>1-4\eps$ such that
for any $y\in E'$, $\mu_y\times \mu_y(P_r(\{s_i\}))=1$.
Then
\begin{align*}
	\lambda(P_r(\{s_i\}))&=\int\mu_y\times\mu_y (P_r(\{s_i\}))d\nu(y)\\
	&\geq \int_E\mu_y\times\mu_y (P_r(\{s_i\}))d\nu(y)\\
	&=\nu(E')> 1-4\eps.
\end{align*}
As $\eps>0$ is arbitrary, we have $\lambda(P_r(\{s_i\}))=1$ and thus $\lambda(P\{s_i\})=1$.
This implies that there exists a measurable set $E\subset Y$ with $\nu(E)=1$
such that for every $y\in E$, $\mu_y\times\mu_y(P(\{s_i\}))=1$.
Let
\[D=E\bigcap \biggl(\bigcup_{k=1}^{+\infty}D_k\biggr)\cap\{y\in Y\colon \mu_y\text{ is non-atomic}\}.\]
It is clear that $\nu(D)=1$.

Fix $y\in D$. There exists a $k\in\mathbb{N}$ with $y\in D_k$.
Then $\mu_y\times\mu_y(R_{r_k}(\{s_i\})\cap P(\{s_i\})=1$.
It is easy to see that both $R_{r_k}(\{s_i\})$ and $P(\{s_i\})$ are $G_\delta$ subsets of $X\times X$.
So
\[R_{r_k}(\{s_i\})\cap P(\{s_i\}\cap \bigl( \supp(\mu_y)\times\supp(\mu_y)\bigr)\]
 is a dense $G_\delta$
subset of $\supp(\mu_y)\times\supp(\mu_y)$.
As $\mu_y$ is non-atomic, $\supp(\mu_y)$ is perfect.
Applying Theorem~\ref{thm:Myc},
we get a dense Mycielski
subset $K$ of $\supp(\mu_y)$ such that
$K\times K \subset (R_{r_k}(\{s_i\})\cap P(\{s_i\}) \cup \Delta_X$.
Then $K$ is $(\{s_i\},r_k)$-scrambled, which ends the proof by setting $\delta=r_k$.
\end{proof}

\section{Final remarks}
Finally we make several remarks.
\begin{rem}
For a $\mathbb{Z}$-action system $(X,T)$
there is another approach to the proof of Theorem~\ref{thm:Main-result-3}
instead of using Lemma~\ref{lem:proximal-r}.
Let $\{s_i\}$ be a  sequence of pairwise distinct integers.
Without loss of generality, we can assume that $\{s_i\}$
is an increasing sequence of positive integers.
By \cite[Lemma 3.1]{HLY14}, for $\nu$-a.e.\ $y\in Y$,
\[ \overline{W_{\mathbb{Z}_+}^s(x,T)\cap\supp(\mu_y) }=\supp(\mu_y),\]
where $W_{\mathbb{Z}_+}^s(x,T)=\{y\in X\colon \lim_{i\to\infty}\rho(T^ix,T^y)=0\}$.
In particular,
$$Asy_{\mathbb{Z}_+}(X,T)\cap \bigl(\supp(\mu_y)\times\supp(\mu_y)\bigr)$$
is dense in $\supp(\mu_y)\times\supp(\mu_y)$.
It is clear that $Asy_{\mathbb{Z}_+}(X,T)\subset P(\{s_i\})$, then
\[ P(\{s_i\}) \cap \bigl( \supp(\mu_y)\times\supp(\mu_y)\bigr)\]
is a dense $G_\delta$
subset of $\supp(\mu_y)\times\supp(\mu_y)$.
\end{rem}

\begin{rem} \label{rem:n-chaos}
Let $n\geq 2$.
An $n$-tuple $(x_1,x_2,\dotsc,x_n)\in X^n $ is called
\emph{$(\{s_i\},\delta)$-$n$-scrambled}
if
\[\limsup_{i\to+\infty} \min_{1\leq j<k\leq n} \rho(s_i x_j,s_ix_k)\geq\delta
\quad\text{ and }\quad
\liminf_{i\to+\infty}
\max_{1\leq j<k\leq n} \rho(s_ix_j,s_ix_k)=0.\]
Following ideas in \cite{X05},
we say that a subset $K$ of $X$ is \emph{$(\{s_i\},\delta)$-$n$-scrambled}
if for every  $n$ pairwise distinct points $x_1,x_2,\dotsc,x_n\in K$,
$(x_1,x_2,\dotsc,x_n)$ is $(\{s_i\},\delta)$-$n$-scrambled.

In fact, we can require $K$ be to $(\{s_i\},\delta)$-$n$-scrambled
in Theorem~\ref{thm:Main-result-3}.
As the proof is almost the same as in Section 3,
we only outline the ideas.
Assume $\mu\in \mathcal{M}^e(X,G)$ with $h_\mu(X,G)>0$ and
$\mu=\int  \mu_y d \nu(y)$ be the disintegration of $\mu$ with respect
to the the Pinsker factor  $(Y,\mathcal{P}_\mu(G),\nu,G)$.
For $n\geq 2$, let
\[ \lambda_n=\int  \underbrace{\mu_y\times\mu_y\times\dotsb\times\mu_y}
_{n \text{ times}}d\nu(y)=
\int\mu_y^{(n)}d\nu(y).\]
Similarly as in Lemmas~\ref{lem:lambda} and \ref{lem:relative-UPE},
we have $\lambda_n\in \mathcal{M}^e(X^n,G)$,
$P_{\lambda_{n}}(G)=p^{-1}(\mathcal{P}_\mu(G))$ and
$ \lambda_n=
\int\mu_y^{(n)}d\nu(y)$ can be regard as
the disintegration of $\lambda_n$ with respect to the Pinsker factor of $(X^n,\lambda_n,G)$.

Let
\[
\Delta^{(n)}=\{(x_1,x_2,\dotsc,x_n)\in X^n\colon
\exists 1\leq j<k\leq n,\ \textrm{s.t.}\ x_j=x_k\}
\]
As  $\mu_y$ is non-atomic for $\nu$-a.e.\ $y\in Y$,
 $\lambda_n(\Delta^{(n)})=0$.
For $r>0$, let
\[\Delta^{(n)}_r=\{(x_1,x_2,\dotsc,x_n)\in X^n\colon
\exists 1\leq j<k\leq n,\ \textrm{s.t.}\
\rho(x_j,x_k)<r\}
\]
It is clear that $\Delta^{(n)}=\bigcap_{k=1}^{+\infty} \Delta^{(n)}_{\frac{1}{k}}$.
In the proof of Lemma~\ref{lem:not-asy},
using $\lambda_n$ and  $\Delta^{(n)}_r$ instead of $\lambda$ and $\Delta_r$,
we can show that for every $\eps>0$ there exists a $r>0$
and a measurable subset $D$ of $Y$ with $\nu(D)>1-4\eps$
such that
\[\mu_y^{(n)}\bigl(R_r^{(n)}(\{s_i\})\bigr)=1\]
for all $y\in D$, where
\[R_r^{(n)}(\{s_i\})=\Bigl\{(x_1,x_2,\dotsc,x_n)\in X^n\colon
\limsup_{i\to+\infty} \min_{1\leq j<k\leq n}
\rho(s_i x_j,s_ix_k)\geq r\Bigr\}.\]
Let
\[
\Delta^{n}=\{(x,x,\dotsc,x)\in X^n\colon x\in X
\}
\]
For $r>0$, let
\[\Delta^{n}_r=\{(x_1,x_2,\dotsc,x_n)\in X^n\colon
\rho(x_j,x_k)<r, 1\leq j<k\leq n\}
\]
In the proof of Lemma~\ref{lem:proximal-r},
using $\lambda_n$ and  $\Delta^{n}_r$ instead of $\lambda$ and $\Delta_r$,
we can show that
\[\lambda_n\bigl(P^{(n)}(\{s_i\})\bigr)=1,\]
where
\[P^{(n)}(\{s_i\})=\Bigl\{(x_1,x_2,\dotsc,x_n)\in X^n\colon
\liminf_{i\to+\infty}
\max_{1\leq j<k\leq n} \rho(s_ix_j,s_ix_k)=0
\Bigr\}.\]
\end{rem}

\begin{rem}\label{rem:sofic}
Sofic groups were introduced by Gromov in \cite{G99}
as a common generalization of amenable and residually finite groups.
Initiated in a breakthrough of Bowen in \cite{B10},
a substantial amount of progress has been made in expanding
the entropy theory for actions of discrete amenable groups to sofic groups.
We refer the reader to a recent book \cite{KL16} for this topic.
It is natural to ask that whether a similar result of Theorem~\ref{thm:main-result-1} holds for sofic group actions.
Since our proof depends on Theorem~\ref{thm:entorpy-UPE-extension},
it is not clear how to extend it to sofic group actions.
\end{rem}

\begin{rem}\label{rem:proof}
As we can see, Theorem~\ref{thm:main-result-1} is a purely topological result.
But our proof heavily relies on ergodic theory.
It would be interesting to know whether
there is a topological or combinatorial proof.
In \cite{KL07},  using a local analysis of combinatorial independence
of topological entropy,
Kerr and Li showed that
if an amenable group action has positive topological entropy
then it is Li-Yorke chaotic, see \cite{KL13} for sofic gorup actions.
But it is not clear how to adapt their method to our setting.
\end{rem}

\subsection*{Acknowledgements}
Part of this work was done during a visit of W. Huang and J. Li to
the Chinese University of Hong Kong.
They would like to thank Prof.\ De-Jun Feng for his warm hospitality.
We are grateful to Prof.\ Hanfeng Li  for helpful suggestions,
which lead to the Remarks \ref{rem:n-chaos}, \ref{rem:sofic} and~\ref{rem:proof}.
The authors would also like to thank the referee who made significant comments and fixed many English
errors. This research was supported in part by NNSF of China (11731003, 11771264, 12090012, 12031019) and NSF of Guangdong Province (2018B030306024).

config: arxiv


text:
\begin{document}

\setcounter{page}{1}

\title{Weak Unbounded Norm Topology and  Dounford-Pettis Operators}
\author[M. Matin]{Mina Matin}
\author[K. Haghnejad Azar]{Kazem Haghnejad Azar}
\author[R. Alavizadeh]{Razi Alavizadeh}
\address{Department  of  Mathematics  and  Applications, Faculty of  Sciences, University of Mohaghegh Ardabili, Ardabil, Iran.}
\email{\textcolor[rgb]{0.00,0.00,0.84}{minamatin1368@yahoo.com}}
\email{\textcolor[rgb]{0.00,0.00,0.84}{haghnejad@uma.ac.ir}}
\email{\textcolor[rgb]{0.00,0.00,0.84}{ralavizadeh@uma.ac.ir}}
\subjclass[2010]{Primary 46B42; Secondary 46A40, 46C05.}
\keywords{ $un$-dual, $wun$-topology, $wun$-Dounford-Pettis.
\newline \indent $^{*}$Corresponding author}
\begin{abstract}
	In this paper, we study $un$-dual (in symbol, $\ud{E}$) of Banach lattice $E$ and compare it with topological dual $E^*$. If $E^*$ has order continuous norm, then $E^* = \ud{E}$. We introduce and study weakly unbounded norm topology ($wun$-topology) on Banach lattices and compare it with weak topology and $uaw$-topology. In the final, we introduce and study $wun$-Dunford-Pettis opertors from a Banach lattice $E$ into Banach space $X$ and we investigate  some of its properties and its relationships with others known operators.
\end{abstract}
\maketitle

\section{Introduction}
In \cite{5}, authors shows that $uo$-convergence need not be given by a topology, but $un$-convergence is topological. We will refer to this topology as $un$-topology. The smallest topology $\tau$ that each $un$-continuous functional $f:E\rightarrow \mathbb{R}$ is continuous with respect to that topology is called weakly unbounded norm topology (for short, $wun$-topology), and we denote it by $\tau_{wun}$.
First  we will ours motivate  to write this article.
\begin{enumerate}
	\item We have defined $un$-continuous operators between two Banach lattices $E$ and $F$ in \cite{12o}, and so we introduced the $un$-dual space for a Banach lattice $E$ and we study some of its properties.
	\item Such as the definition of weak topology for a normed space,  we  define  $wun$-topology  for  a Banach lattice and compare it  with  $uaw$-topology  which is introduced in \cite{15}. We show that in general $un$-topology and $wun$-topology are different and by some conditions both topologies coincide.
	\item By studying of $uaw$-Dounford-Pettis operators in \cite{zabeti1},  it is  interested to define a new generation of operators as $wun$-Dounford-Pettis operators. We study some of its properties and compare  with other known classifications of operators.
\end{enumerate}
Here we bring some definitions of need.\\
Let $E$ be a vector lattice and $x\in E$.
A net $(x_{\alpha})_{\alpha \in A}\subseteq E$ is said to be order convergent to $x$ if there is a net
	$(z_{\beta})_{\beta \in B} $ in $ E $ such that
	$ z_{\beta} \downarrow 0 $ and for every $ \beta \in B$,
	there exists $\alpha_{0} \in A$ such that
	$ | x_{\alpha} - x |\leq z_{\beta}$ whenever $ \alpha \geq \alpha_{0}$.
	We denote this convergence by $ x_{\alpha} \xrightarrow{o} x $ and write
	that $ (x_{\alpha})_{\alpha} $ is $o$-convergent to $x$. In vector lattice $E$ we write
$x_\alpha\uoto x$ and say that ($x_\alpha$) is $uo$-convergent to $x$ if $|x_\alpha - x|\linf u\oto 0$
for every $u \in E^+$.
In Banach lattice $E$ we write
$x_\alpha\unto x$ and say that ($x_\alpha$) is $un$-convergent  to $x$ if $|x_\alpha - x|\linf u\nto 0$
for every $u \in E^+$.

It  was  observed  in  \cite{5} that  $un$-convergence  is
topological. Let $x_0\in E$ be arbitrary. For every $\epsilon > 0$ and non-zero $u\in X^+$, put
$$V_{\epsilon,u} =\left\{ x \in X : \norm{\abs{x-x_0}\linf u} < \epsilon \right\}.$$
The collection of all sets of this form is a base of $x_0$ neighbourhoods for a topology, and the convergence in
this topology agrees with $un$-convergence. We will refer to this topology as $un$-topology.\\
 Recall of \cite{15}, a net $(x_\alpha)$ in Banach lattice $E$ is unbounded absolutely weakly convergente ($uaw$-convergent) to $x$ if $(|x_\alpha - x| \wedge u)$ converges to zero weakly for every $u\in E^+$; we write $x_\alpha \xrightarrow{uaw}x$. It  was  observed  in  \cite{15} that  $uaw$-convergence  is  topological. By Theorem 4 of \cite{15}, if Banach lattice $E$ has order continuous norm, then  $(x_\alpha)\subseteq E$ is $un$-null iff it is $uaw$-null in $E$.\\
Let $E$ be a vector lattice and $e\in E^+$. $e$ is weak unit, if band $B_e$ generated by $e$ is equal  with $E$; equivalently, $x \wedge ne \uparrow x$ for every $x\in E^+$, and $e$ is strong unite when ideal $I_a$ generated by $e$ is equal $E$; equivalently, for every $x\geq 0$ there exists $n\in \mathbb{N}$ such that $x\leq ne$. A positive non-zero vector $a$ in a vector lattice $E$ is an atom if the ideal $I_a$ generated by $a$ coincides with span $a$.
We say that $E$ is non-atomic if it has no atoms. We say that $E$ is atomic if $E$ is the band generated by all the atoms.\\
Let $E$ be a vector lattice, $E^\sim$  be the space of all order continuous functionals on $E$ and $(E^{\sim})^\sim$ be the order bidual of $E$.
Recall that a subset $A$ of $E$ is $b$-order bounded in $E$
if $A$ is order bounded in $(E^{\sim})^\sim$. If each $b$-order bounded subset $A$ of vector lattice $E$ is order bounded, we say that $E$ has property $(b)$.\\
Let  $X$, $Y$ be two Banach spaces, then  the continuous operator $T:X\rightarrow Y$ is said to be:
\begin{itemize}
	\item
	 \textbf{Dounford-Pettis}
	whenever $(x_n)\subseteq X$ and $x_n \xrightarrow{w}0$, then $T(x_n)\xrightarrow{\|.\|}0$.

	\item
	  \textbf{weakly compact}
	whenever $T$ carries the closed unit ball of $X$ to a relatively weakly compact subset of $Y$.\\
\end{itemize}

Let  $E$ be a Banach lattice and  $X$  Banach space, then  the continuous operator $T:E\rightarrow X$ is said to be:
\begin{itemize}
	\item
	\textbf{$M$-weakly compact}
 if $ \lim \|Tx_{n}\| = 0$ holds for
every norm bounded disjoint sequence $(x_{n}) $ of $ E $.

\item
 \textbf{order weakly compact}
whenever $T[0,x]$ is relatively weakly compact subset of $X$ for each $x\in E^+$.

\item
\textbf{b-weakly compact}
whenever for each $b$-order bounded subset $A$ of $E$, $T(A)$ is a relatively weakly compact. The class of $b$-weakly compact operators was firstly  introduced by  Alpay, Altin and Tonyali \cite{Alpay1}, the class  of  all  $b$-weakly compact operators between $E$ and $X$ will be denoted by
 $W_b(E,X)$.   One of the interesting properties of  the class of $b$-weakly compact operators is that it satisfies the domination property. Some more investigations on $W_b(E,X)$ were done by   \cite{Alpay2, 2c, Hag}.

 \item
\textbf{$uaw$-Dunford-Pettis} if for every norm bounded  sequence $(x_{n})$ in $E$,  $ x_{n}\xrightarrow{uaw} 0$ in $ E $ implies
$\| Tx_{n}\|\rightarrow 0$ in $X$. These operators are introduced and examined in \cite{zabeti1}, the class  of  all  $uaw$-Dounford-Pettis operators on $E$  will be denoted by
$B_{UDP}(E)$. This is continued in \cite{Hui}.\\

Moreover if $F$ is a Banach lattice, a continuous operator $T:E\rightarrow F$ is said to be
\item
\textbf{$un$-continuous} if for every norm bounded and $un$-null net $(x_\alpha)\subseteq E$, $T(x_\alpha)\xrightarrow{un}0$ in $F$. These operators are introduced and examined in \cite{12o}.
\end{itemize}
Recall that a Banach lattice $E$ is said to have dual positive Schur property if every $w^*$-null positive sequence in $E^*$ is norm null.\\

Throughout this article $E$ and $X$ will be assumed to be Banach lattice and Banach space, respectively, and  $(e_n)$ is the sequence of real numbers whose $n^{th}$ term is one and the rest are zero, i.e. $e_n := (0,0,...,0,1,0,0,...)$ unless specified otherwise. For a normed space $X$,  $A\subseteq X$ and $B\subseteq X^*$,  $\sigma (A, B)$ is the smallest topology for $A$ such that each $f\in B$ is continuous on $A$ with respect to this topology.

\section{weakly unbounded norm topology}
 Let $E$ be a Banach lattice. A functional $f:E\rightarrow \mathbb{R}$ is $un$-continuous, if $x_\alpha \xrightarrow{un}0$ implies $f(x_\alpha)\rightarrow 0$ for each norm bounded net $(x_\alpha)\subseteq E$.

We denote the vector space of all $un$-continuous functionals on $E$ by $\ud{E}$ and we call it $un$-dual of Banach lattice $E$.\\
It is clear that $\ud{E}$ is a subspace of $E^*$. The functional $f:\ell^1\rightarrow \mathbb{R}$ defined by $f(x_1,x_2,x_3,...)= \sum_{i=1}^{n}x_i$ is continuous but  is not $un$-continuous. Therefore $\ud{E} \neq E^*$.\\

Let $f\in \ud{E}$, we define $\|f\|_{\ud{E}} = \sup\{|f(x)|: x\in E,\  \|x\|_E\leq 1\}$. It is clear that $\ud{E}$ is a normed space.

\begin{theorem}\label{po}
	Let $E$ be a Banach lattice. Then we have the following assertions.
	\begin{enumerate}
	\item If  $E^*$ has order continuous norm, then $E^* = \ud{E}$.
		\item $\ud{E}$ is an ideal in $E^*$.

\item If $E$ is $AM$-space, then $\ud{E}$ is $AL$-space.
\end{enumerate}
\end{theorem}
\begin{proof}
		\begin{enumerate}
		\item Proof follows by   Theorem 6.4 of \cite{5}.
			\item Proof has similar argument of Proposition 5.3 of \cite{12}
			\item Let $E$ be an $AM$-space. Since $E^*$ is an $AL$-space, so $E^*$ has order continuous norm and therefore by part (1), we have  $\ud{E}= E^*$ and it is an $AL$-space.

\end{enumerate}
\end{proof}

\begin{example}\label{Elin}
Let $c$ be the sublattice of $\ell^\infty$ consisting of all convergent sequences.
Since $c^*=\ell^1$ has order continuous norm, therefore by preceding theorem, we have  $\ud{c}= c^* =\ell^1$.
\end{example}

Theorem \ref{po} shows that $\ud{E}$ is a normed sublattice of a Banach lattice of $E$. Thus we  define the second $un$-dual of Banach lattice $E$ which  show by
 $\udd{{E}}$.
 $\udd{{E}}$ in general is not equal with topological  second dual of $E$, $E^{**}$. Set $E=c$. It is obvious that
 $\ud{c^{*}}=\udd{{c}}\neq \ell^\infty=c^{**}$
On the other hand $\ud{E}$ is neither norm closed nor order closed, since $E=\ell_2$, then $\ud{\ell_2}=c_{00}$.

\begin{proposition}\label{abc}
	Let $E$ be a Banach lattice and $G$ be a sublattice of $E$ such that one of the following conditions hold.
	\begin{enumerate}
		\item $ G $ is majorizing in $ E $;
		\item $ G $ is norm dense in $ E $;
		\item $ G $ is a projection band in $ E $.
	\end{enumerate}
	If $\ud{E} = E^*$, then $G^* = \ud{G}$.
	\end{proposition}
	\begin{proof}
		We know that $\ud{G} \subseteq G^*$. Now assume that $ f \in G^*$ and $(x_{\alpha})\subseteq G$ is norm bounded and $un$-null in $G$. By Theorem 3.6 of \cite{14a}, there exists $g \in E^*$ such that $f= g $ on $G$. Note that by assumption $g\in \ud{E}$. By Theorem 4.3  \cite{12}, $x_{\alpha}\xrightarrow{un}0$ in $E$.  We obvious  that $(x_\alpha)$ is norm bounded in $E$. Therefore  $f(x_{\alpha}) = g(x_{\alpha})\xrightarrow{\|.\|}0$ in $\mathbb{R}$. It follows that  $f\in \ud{G}$.
	\end{proof}

Naturally, we can define weakly unbounded norm topology ($wun$-topology) as follows.
\begin{definition}
The smallest topology $\tau$ that each $f\in\ud{E}$  is continuous with respect to that topology is called weakly unbounded norm topology (for short, $wun$-topology), and we denote  by $\tau_{wun}$. In the other words,
$$\tau_{wun}:=\bigcap \left\{ \tau : \text{ each } f\in\ud{E} \text{ is } \tau\text{-countinuous}\right\}.$$
\end{definition}

For every $\epsilon>0$ and each $f\in\ud{E}$, put
$$V_{\epsilon,f} =\left\{ x \in X : \abs{f(x)} < \epsilon \right\}.$$
It easily follows from Definition 2.4.2 of \cite{13} that
the collection of all $V_{\epsilon,f}$ is a subbasis for $wun$-topology at zero.\\
Let $x , y \in V_{\epsilon,f}$ and $ 0 \leq \lambda \leq 1$. We have $\abs{f(\lambda x + (1-\lambda)y)} = \lambda |f(x)| + (1-\lambda)|f(y)| \leq \lambda \epsilon + (1-\lambda) \epsilon =  \epsilon$. Therefore $ \lambda x + ( 1 - \lambda) y \in V_{\epsilon,f}$. Hence $\tau_{wun}$ is a locally convex in Banach lattice $E$.\\
Let $E$ be a Banach lattice. It is obvious that  for each norm bounded net $(x_\alpha)\subseteq E$,   $x_\alpha \xrightarrow{wun}0$ if and only if for each $f \in \ud{E}$, $f(x_\alpha)\xrightarrow{}0$ in $\mathbb{R}$.\\

It is obvious that every $un$-null net is $wun$-null in a Banach lattice, but in general the converse not holds. The following example shows that in general  both topologies $un$ and $wun$-topology are not the same.
On the other hand by Proposition 3.5 of \cite{12}, every norm bounded and disjoint net in order continuous Banach lattice $E$ is $wun$-null. Thus if we set $E=\ell^1$, then  $(e_n)$ is $wun$-null in $\ell^1$.

\begin{example}\label{piu}
 Consider the sequence $(e_n)$ in the sublattice $c$ of $\ell^{\infty}$. By Example \ref{Elin}, $\ud{c}=\ell^1$. For each $f=(x_1,x_2,...,x_n,...)\in\ud{c}$, $f(e_n) = x_n \xrightarrow{\|.\|}0$, therefore $e_n \xrightarrow{wun}0$ in $c$, but $(e_n)$ is not $un$-null in $c$. Consider $u = (1,1,1,...)\in c^+$. We have $\|e_n \wedge u\| = \|e_n\| = 1 \nrightarrow 0$.
\end{example}

The following facts are in $wun$-topology that will be used throughout the paper.

\begin{lemma}\label{tr}
	Let $E$ be a Banach lattice and $(x_\alpha)\subseteq E$, then
	\begin{enumerate}
		\item $x_\alpha \xrightarrow{wun}x$ iff $(x_\alpha - x)\xrightarrow{wun}0$;
		\item $wun$-limits are unique;

		\item If $x_\alpha \xrightarrow{wun}x$ and $y_\beta \xrightarrow{wun}y$, then $ax_\alpha + b y_\beta \xrightarrow{wun} ax + by$, for any scalars $a,b$;

		\item If $x_\alpha \xrightarrow{wun}x$, then $y_\beta \xrightarrow{wun}x$, for every subnet $(y_\beta)$ of $(x_\alpha)$.
	\end{enumerate}
\end{lemma}
\begin{proof}
	\begin{enumerate}
	\item The proof is clear.
	\item Let $(x_\alpha)\subseteq E$ with $x_\alpha \xrightarrow{wun}x$ and $x_\alpha\xrightarrow{wun}y$. For each $
	f \in \ud{E} \subseteq E^*$ we have $\|f (x-y)\| = \| f (x-x_\alpha + x_\alpha -y)\| \leq \| f (x-x_\alpha)\| + \| f (x_\alpha - y)\| $. Therefore $f(x-y)= 0$. Since $E^*$ separates the point of $E$, so $x=y$.
	\item Let $f \in \ud{E}$. We have $\| f (ax_\alpha + b y_\beta - ax - by)\| \leq |a|\|f (x_\alpha -x )\| + |b| \| f (y_\beta - y)\|  $, and proof follows.
	\item Assume that $ (x_\alpha)\subseteq E$ and $x_\alpha \xrightarrow{wun}x$ in $E$.
Because for each $f \in \ud{E}$, we have $\| f (x_\alpha)\| \rightarrow 0$. It is clear that each subnet $(f( y_\beta))$ of $(f (x_\alpha))$ is norm-null.
	\end{enumerate}
\end{proof}
\begin{remark}
Since $wun$-limits are unique, therefore for each $x\in E$, $\{x\}$ is $wun$-closed. By condition 2  of Lemma \ref{tr}, $\tau_{wun}$ is a vector topology on $E$, and $(E,\tau_{wun} )$  is a topological vector space. By Theorem 1.12 of \cite{14a}, $\tau_{wun}$ is a Hausdorff topology.
\end{remark}

Note that $wun$-topology is different with  weak toplogy (in short $w$-topology).  Consider the sequence $(e_n)$ in $\ell^1$.  Since $(e_n) $ is norm bounded and disjoint in $\ell^1$, then by Example \ref{piu}(1), $e_n \xrightarrow{wun}0$ in $\ell^1$. On the other hand,  $(e_n) $ is not  $w$-null in $\ell^1$. Therefore $\tau_w \neq \tau_{wun}$.
Since   ${\ud{E}}$ is a subspace of $E^*$,  $w$-topology is weaker then $wun$-topology. Thus every $w$-null net is $wun$-null  for each Banach lattice $E$. \\

\begin{proposition}\label{ert}
 Let $E$ be a Banach lattice. If $E$ has strong unit, then $w$-topology and $wun$-topology coincide.
\end{proposition}
\begin{proof}
 Since $\ud{E}\subseteq E^*$, follows that $wun$-topology is weaker than  $w$-topology in Banach lattice $E$. When $E$ has strong unit then by Theorem 2.3 of \cite{12}, $\ud{E}=E^*$ and therefore,  $\tau_w=\tau_{wun}$ in $E$.
 \end{proof}

O. Zabti in \cite{15} has been introduced unbounded absolutely weakly topology (in shorn $uaw$-topology) and investigated some of its properties.
Note that $wun$-topology is different with $uaw$-topology.  By Lemma 2 of \cite{15}, the sequence $(e_n)\subseteq \ell^\infty$ is $uaw$-null in $\ell^{\infty}$, but $(e_n)$ is not weak convergent to zero in $\ell^\infty$, and so by Proposition \ref{ert}, it is not $wun$-null in $\ell^\infty$.

\begin{remark}
Note that if Banach lattice $E$ has order continuous norm and $(x_\alpha)\subseteq E$ is norm bounded and $uaw$-null, then by Proposition 5 of \cite{15}, $x_\alpha \xrightarrow{wun}0$ in $E$. If $E$ is an $AM$-space with strong unit and $(x_n)\subseteq E$ with $x_n \xrightarrow{wun}0$, then $x_n\xrightarrow{w}0$ and by Exercise 5 of page 355 of \cite{1}, $|x_n|\xrightarrow{w}0$. It follows that  $x_n \xrightarrow{uaw}0$ in $E$.
\end{remark}
\begin{proposition}\label{samedual}
If a Banach lattice $E$ is atomic with order continuous norm, then $(E,\tau_{wun})^*=\ud{E}$.
\end{proposition}
\begin{proof}
It follows from Theorem 5.2 of \cite{12} that $un$-topology is locally convex. Therefore, $\ud{E}$ separates the points of $E$. Thus, by Theorem 3.10 of \cite{14a} we have $(E,\tau_{wun})^*=\ud{E}$.
\end{proof}

Let $G$ be a sublattice of $E$ and $(x_\alpha)\subseteq G$. By Theorem 3.6 of \cite{14a}, $x_\alpha \xrightarrow{w}0$ in $G$ iff  $x_\alpha\xrightarrow{w}0$ in $E$. The situation is different for $wun$-convergence.
\begin{example}
	\begin{enumerate}
	\item Consider the sequence $(e_n)$ of $\ell^1$. It is $wun$-null in $\ell^1$ while  is not $wun$-nul in $\ell^{\infty}$.
\item Note that $\ell^1$ is an order continuous  Banach lattice with a weak unit $e$. It is known that $\ell^1$ can be represented as an order and norm dense ideal in $L_1(\mu)$ for some finite measures $\mu$. Consider the sequence $( x_n) = (\dfrac{1}{2}(e_1 - e_n))\subseteq \ell^1$.  Since $L_1(\mu)$ has order continuous norm and  is non-atomic, therefore by Corollary 5.4 of \cite{12}, $x_n \xrightarrow{wun}0$ in $L_1(\mu)$. On the other hand $(x_n)$ is not $wun$-null in $\ell^1$. Since $(x_n)$ is $un$-convergent to $\dfrac{1}{2}e_1$ in $\ell^1$ and therefore is $wun$-convergent to $\dfrac{1}{2}e_1$ in $\ell^1$.
\end{enumerate}
\end{example}
Let $G$ be a sublattice of a Banach lattice $E$.
In the following,  we bring the conditions that if a net $(x_\alpha)\subseteq G$ is $wun$-null in  $G$,  then is $wun$-null in $E$ and vic versa.
\begin{theorem}\label{norm}
	Let  $G$ be a sublattice of Banach lattice $E$ and  $(x_{\alpha})\subseteq G$.
	 \begin{enumerate}
\item	If $E^* = \ud{E}$ and $x_{\alpha}\xrightarrow{wun}0 $ in $E$, then  $x_{\alpha}\xrightarrow{wun}0$ in $G$.
	\item  If $G^* = \ud{G}$ and $x_{\alpha}\xrightarrow{wun} 0$ in $G$, then $x_{\alpha}\xrightarrow{wun}0$ in $E$.

	\end{enumerate}
\end{theorem}
\begin{proof}
	\begin{enumerate}
\item	Let $(x_{\alpha})\subseteq G$ and $ x_{\alpha}\xrightarrow{wun}0$ in $E$. By $E^* = \ud{E}$, $x_{\alpha} \xrightarrow{w}0$ in $E$. By Theorem 3.6 of \cite{14a}, $x_{\alpha} \xrightarrow{w}0$ in $G$ and therefore $x_{\alpha} \xrightarrow{wun}0$ in $G$.
	\item If $x_{\alpha}\xrightarrow{wun}0$ in $G$, then $x_\alpha \xrightarrow{w}0$ in $G$ and therefore $x_\alpha \xrightarrow{w}0$ in $E$. So $x_\alpha \xrightarrow{wun}0$ in $E$.

	\end{enumerate}
\end{proof}
\begin{corollary}
	Let $G$ be a sublattice of Banach lattice $E$ and $(x_\alpha)\subseteq G$. If $E^* = \ud{E}$ and $G$ is a  majorizing or norm dense or
		  band projection in $E$, then $x_\alpha \xrightarrow{wun}0$ in $E$ iff $x_\alpha\xrightarrow{wun}0$ in $G$.
\end{corollary}

Let $\func{T}{E}{F}$ be a continuous operator between two Banach lattices.
Then $T$ has a $un$-adjoint if there exists the unique operator $\ud{T}:\ud{F}\rightarrow\ud{E}$ satisfying
$$
<\ud{T}\ud{y},x> = <\ud{y},Tx> = \ud{y}(Tx),\qquad \forall \ud{y}\in \ud{F}, \forall x\in E.
$$
It easily follows from $\ud{F}\subseteq F^*$ that $\ud{T}=T^*|_{\ud{F}}$.

\begin{theorem}\label{mf}
	Let $\func{T}{E}{F}$ be an operator between two Banach lattices. If $T$ is $un$-continuous, then
	$T$ has $un$-adjoint.

\end{theorem}
\begin{proof}
	Assume that $T$ is $un$-continuous. It is enough to prove that
	$\ud{T}(\ud{F})\subseteq \ud{E}$. Let $\ud{y}\in\ud{F}$ and $(x_\alpha)$ be norm bounded and $un$-null net in $E$. Since $T$ is $un$-continuous, we have $(Tx_\alpha)$ is norm bounded and $un$-null in $F$. As $\ud{y}$ is $un$-continuous, $\ud{T}\ud{y}(x_\alpha)=\ud{y}(Tx_\alpha)\unto 0$. Thus, $\ud{T}\ud{y}\in\ud{E}$.
\end{proof}

\begin{example}\label{uy}
	The operator $T:C[0,1] \rightarrow c_0$, given by
	$$ T(f) = (\int_0 ^1 f(x)\sin x dx, \int_0 ^1 f(x)\sin2x dx,...)$$ is a $un$-continuous. By Theorem \ref{mf}, $T$ has $un$-adjoint. We have $\ud{T} : \ud{c_0} \rightarrow \ud{(C[0,1])}$, given by\\
	$$  \langle\ud{T}(x_1,x_2,...),f\rangle = \langle(x_1,x_2,...),Tf\rangle = \sum_{n=1}^\infty x_n \int_0^1 f(x)\ sinnx\ dx,$$
	for all 	$ (x_1,x_2,...)\in \ud{c_0}$  and  $f\in C[0,1].$\\
\end{example}

Now, assume that $Q_E$ be a natural mapping from $E$ into $E^{**}$ where $\langle x^\prime, Q_E(x)\rangle=\langle x, x^\prime\rangle=x^\prime (x)$ for all $x\in E$ and $x^\prime\in E^*$.
Since $\udd{{E}}$ is a subspace of $E^{**}$,  we have the following lemma.

\begin{lemma}\label{lm1}
Let $E$ be a Banach lattice. Then $Q_E(E)\subseteq \udd{{E}}$.
\end{lemma}
\begin{proof}
Let $(x^\prime_\alpha)\subseteq \ud{E}$ be a norm bounded and $un$-null net in $\ud{E}$. By Theorem \ref{po} we know that    $\ud{E}$ is an ideal in $E^*$,  and so  $x^\prime_\alpha \wedge y^\prime \in \ud{E}$ for all $y^\prime\in E^*$.
It follows that $x^\prime_\alpha \wedge (x^\prime_\alpha \wedge y^\prime )\xrightarrow{\Vert .\Vert}0$ in $\ud{E}$. Thus for each $y^\prime \in E^*$, we have
$x^\prime_\alpha \wedge y^\prime\xrightarrow{\Vert .\Vert}0$ in $\ud{E}$.\\
Let $x\in E$. If $x=0$, $Q_E(0)\in  \udd{{E}}$, and proof holds. Now assume that  $x\neq 0$. Then there exists $y^\prime\in E^*$ such that $y^\prime (x)=1$.
Then we have $(y^\prime)^+(x)\geq1$.
Since
$x^\prime_\alpha (x) \wedge (y^\prime)^+(x)\xrightarrow{\Vert .\Vert}0$ in $\ud{E}$,  follows that $x^\prime_\alpha(x)\xrightarrow{\Vert .\Vert}0$ in $\ud{E}$, and so $Q_E(x)(x^\prime_\alpha )\rightarrow 0$. It follows that $Q_E(x)\in \udd{{E}}$ and proof follows.
\end{proof}

 Now the Lemma \ref{lm1} make motivation to us for definition  a new  topology  for $\ud{E}$,  that is,    the smallest topology on $\ud{E}$ such that each $Q_E(x)$ is continuous with respect to it where $x\in  E$.  This topology is called weak$^*$ unbounded topology (for short $w^*un$-topology). In this way,  $ x_{\alpha}\xrightarrow{w^*un}0$  if and only if $x^\prime (x)\rightarrow 0$ for all $x\in E$. It is clear that the $w^*un$-topology in $\ud{E}$ is a subtopology of $w^*$-topology in $E^*$,  and $w^*un$-topology  is a subset of $wun$-topology in $\ud{E}$, that is,
 $\sigma (\ud{E},Q_E(E))\subseteq\sigma (\ud{E}, \udd{{E}})\subseteq  \sigma (\ud{E},(\ud{E})^*)\subseteq \sigma (\ud{E}, E^{**})$.

\begin{theorem}
Let $E$ be a Banach lattice. Then $B_{\ud{E}}=\{x^\prime \in \ud{E}:~\Vert x\Vert\leq 1\}$ is $w^*un$-compact.
\end{theorem}
\begin{proof}
It is obviously that $A\subseteq \ud{E}$ is $w^*un$-closed in $\ud{E}$ if and only if there exists $B\subseteq E^*$ which is $w^*$-closed in $E^*$  and $A=B\cap\ud{E}$.  Since $B_{\ud{E}}=B_{E^*}\cap\ud{E}$ and $B_{E^*}$ is $w^*$-compact, proof follows.
\end{proof}

In the following we have some facts for  $wun$-topology and $w^*un$-topology in $\ud{E}$ which  theirs proofs  has similar arguments such as classical studying for $w^*$ and $w$-topologies in $E^*$.

\begin{corollary} Suppose that
$E$
and
$F$
are Banach lattices. Then we have the following assertions.
\begin{enumerate}
\item  If
$T\in B(E, F)$,
then
$\ud{T}$
is $w^*un$-$w^*un$ continuous. Conversely, if
$S$
is a $w^*un$-$w^*un$ continuous linear operator from
$\ud{F}$
into
$\ud{E}$,
then there is a
$T$
in
$B(E, F)$
such that
$\ud{T}=S$.
\item If
$T\in B(E, F)$, then
$\udd{{T}}Q_E(E)\subseteq Q_F(F)$
and
$Q^{-1}_F\udd{{T}}Q_E =T$.
\item A bounded linear operator from a Banach lattice into a Banach lattice is $wun$-compact if and only if its adjoint is $wun$-compact.
\item Suppose that
$T\in B(E, F)$
and
$Q_F$
is the natural map from
$F$
into
$\udd{{F}}$.
Then
$T$
is $wun$-compact if and only if
$\udd{{T}}(\udd{{E}})\subseteq Q_F(F)$.

\end{enumerate}

\end{corollary}

	\begin{definition}
		Let $E$ and $F$ be two Banach lattices.
		A continuous operator $T: E \rightarrow F $ is said to be, weak unbounded norm continuous (or, $wun$-continuous for short),
		if $ x_{\alpha} \xrightarrow{wun} 0$ in $ E $ implies
		$ Tx_{\alpha} \xrightarrow{wun} 0 $ in $F$ for each norm bounded net $(x_\alpha)\subseteq E$. The collection of all $wun$-continuous operators from $E$ to $F$,  will be denoted by
		$ L_{wun}(E,F) $.
	\end{definition}

	\begin{example}
		\begin{enumerate}
			\item	Let $G$ be a majorizing or norm dense or band projection of $\ell^{\infty}$. Then each continuous operator from $G$ to $\ell^{\infty}$ is $wun$-continuous.
			\item Consider the functional $f :\ell^{\infty} \rightarrow \mathbb{R}$ defined with
			$$f(x_1,x_2,...) = \lim_{n\rightarrow\infty} x_n.$$
		 Since $f$ is positive, 	$f$ is continuous. Now if $(x_n)\subseteq \ell^\infty$ is norm bounded and $x_n \xrightarrow{wun}0$ then $x_n\xrightarrow{w}0$ and therefore $f(x_n)\xrightarrow{w}0$. Hence $f(x_n)\xrightarrow{wun}0$ in $\mathbb{R}$.

		\end{enumerate}
	\end{example}
\begin{remark}
		Note that the operator  $T:\ell^{1} \rightarrow \ell^{\infty} $ defined by
		\[
		T(x_{1},x_{2},\ldots) = (\sum_{i=1}^\infty x_i , \sum_{i=1}^\infty x_{i},\ldots)
		\]
		is continuous, while  is not $wun$-continuous.
\end{remark}

	\begin{theorem}\label{ytr}
A		functional $f:E\rightarrow \mathbb{R}$ is $un$-continuous if and only if is $wun$-continuous.
	\end{theorem}
	\begin{proof}
		Let $f$ is $un$-continuous and $(x_{\alpha})\subseteq E$ is norm bounded with $x_{\alpha}\xrightarrow{wun}0 $ in $E$. Therefore for each $\ud{x} \in \ud{E}$, we have $\ud{x}(x_{\alpha})\xrightarrow{\|.\|}0 $ in $\mathbb{R}$. Since $f$ is $un$-continuous, therefore by Theorem \ref{mf}, $f$ has $un$-adjoint. Hence $\ud{f}(\ud{\mathbb{R}})\subseteq \ud{E}$. Therefore for all $\ud{y} \in \ud{\mathbb{R}}$, we have $\ud{y}(fx_{\alpha}) = \ud{f} \ud{y} (x_{\alpha}) \xrightarrow{\|.\|}0 $ in $\mathbb{R}$. Hence $f(x_{\alpha}) \xrightarrow{wun}0$ in $\mathbb{R}$.\\
		Conversally, let $(x_\alpha)\subseteq E$ is norm bounded and $x_\alpha \xrightarrow{un}0$ in $E$. It is clear that $x_\alpha \xrightarrow{wun}0$ in $E$ and therefore $f(x_\alpha)\xrightarrow{wun}0 $ in $\mathbb{R}$. So $f(x_\alpha)\xrightarrow{\|.\|}0$ in $\mathbb{R}$.
	\end{proof}
\begin{corollary}
	Let $E$ and $F$ be two Banach lattices. Similar to Therem \ref{ytr}, if operator $T:E\rightarrow F$ is $un$-continuous, then  is $wun$-continuous.
\end{corollary}

\begin{remark} Note that, if Banach lattice $E$ is an atomic $KB$-space, then by Theorem 7.5 of \cite{12}, $B_E$ is $un$-compact. Since $\tau_{wun} \subseteq \tau_{un}$, therefore $B_E$ is $wun$-compact.
	\end{remark}
\begin{theorem}
	Let $E$ be an atomic Banach lattice with order continuous norm. If $A\subseteq E$ is a convex set, then $wun$-closure of $A$ is the same as its $un$-closure, that is; $\overline{A}^{wun}=\overline{A}^{un}$.
\end{theorem}
\begin{proof}
	Since $\tau_{wun}\subseteq\tau_{un}$, $\overline{A}^{un}\subseteq\overline{A}^{wun}$.
	On the other hand, by Theorem 5.2 of \cite{12}, $un$-topology is locally convex, hence if $x\notin\overline{A}^{un}$ then by Theorem 3.13 \cite{2}
	there exists some $f\in\ud{E}$, $\alpha\in\RR$ and an $\epsilon>0$
	such that
	$$f(a)\leq \alpha < \alpha + \epsilon < f(x),$$
	for all $a\in \overline{A}^{un}$.
	Therefore, $\overline{A}^{un}\subseteq B=\{y: f(y)\leq \alpha\}$. By Proposition \ref{samedual} $f$ is $wun$-continuous, thus $B$ is $wun$-closed. Hence $\overline{A}^{wun}\subseteq B$. Therefore, $x\notin \overline{A}^{wun}$. Consequently, $\overline{A}^{un}=\overline{A}^{wun}$.

\end{proof}
\section{$wun$-Dunford-Pettis operators}
	A continuous operator $T$ from Banach lattice $E$ into Banach space $X$ is a $wun$-Dunford-Pettis whenever $x_n \xrightarrow{wun}0$ in $E$ implies $Tx_n \xrightarrow{\|.\|}0$ in $X$ for each norm bounded sequence $(x_n)\subseteq E$.
\begin{example}\label{we}
Operator $T:C[0,1] \rightarrow \ell^1$, given by
$$ T(f) = (\dfrac{\int_0 ^1 f(x)\sin x dx}{1^2}, \dfrac{\int_0 ^1 f(x)\sin2x dx}{2^2},...)$$
is a $wun$-Dunford-Pettise operator. Let $(f_n)\subseteq C[0,1]$ is norm bounded and $f_n \xrightarrow{wun}0$. Since $(C[0.1])^* = \ud{(C[0,1])}$, so $f_n \xrightarrow{w}0$ in $C[0,1]$. By continuity of $T$, we have $T(f_n)\xrightarrow{w}0$ in $\ell^1$ and by Schur property of $\ell^1$, $T(f_n)\xrightarrow{\|.\|}0$ in $\ell^1$.
\end{example}
\begin{remark} Let $E$ and $F$ be two Banach lattices and $X$ be a Banach space. If $T:E\rightarrow F$ and $S:F\rightarrow X$ are $wun$-Dounford-Pettis, then $SoT$ is $wun$-Dounford-Pettis.
\end{remark}
It is clear that if $T$ is $wun$-Dunford-Pettis, then it is Dunford-Pettis and $\sigma$-$un$-continuous. The identity operator $I:\ell^1\rightarrow \ell^1$ is a $\sigma$-$un$-continuous, but it is not $wun$-Dounford-Pettis operator.\\
 Here we give an example to illustrate the difference between Dunford-Pettis and $wun$-Dunford-Pettis operators.
\begin{example}\label{poi}
	The operator $T:\ell^{1} \rightarrow \ell^{\infty} $ defined by
	\[
	T(x_{1},x_{2},\ldots) = (\sum_{i=1}^\infty x_i , \sum_{i=1}^\infty x_{i},\ldots)
	\]
	is a Dunford-Pettis operator
	($\ell^{1}$ has Schur property and $ T $ is a continuous operator).
	Consider $(e_n)\subseteq \ell^1$. $e_n \xrightarrow{wun}0$ in $\ell^1$. We have $T(e_{n})=(1,1,1,\ldots)$,
	therefore $ (Te_n) $ is not convergent to zero in norm topology. Thus $ T $
	is not $wun$-Dunford-Pettis operator.
\end{example}
\begin{remark}
It is clear that if $E^* = \ud{E}$, then operator $T:E\rightarrow X$ is Dunford-Pettis iff it is $wun$-Dunford-Pettis.
\end{remark}

\begin{proposition}\label{p:4.55}
A linear operator from a Banach lattice into a Banach lattice is $wun$-Dounford-Pettis if and only if it is $wun$-norm\index{$wun$-norm} sequentially continuous.
\end{proposition}
\begin{proof}
The forward implication is an easy consequence of the $wun$-$wun$ continuity of $wun$-Dounford-Pettis operators along with the fact that a subset of a Banach lattice consisting of the terms and limit of a $wun$-convergent sequence is $wun$-compact. The converse follows directly from the fact that $wun$-compact subsets of a normed space are $wun$-sequentially compact.
\end{proof}

\begin{proposition}\label{ky}
	If each  Dunford-Pettis operator $T:E \rightarrow F$ between two Banach lattices is $wun$-Dunford-Pettis, then the norm of $E^*$ is order continuous or $F=\{0\}$.
\end{proposition}
\begin{proof}
The proof  has the similar argument   of Theorem 3.1 of \cite{Hui}.
\end{proof}
\begin{theorem}
	Let  $F\neq \{0\}$ be a reflexive Banach lattice. The zero operator is the only  $wun$-Dounford-Pettis positive operator $T:\ell^1\rightarrow F$.
\end{theorem}
\begin{proof}
	Let $T:\ell^1\rightarrow F$ be a positive operator. Since $F$ is reflexive, then by Theorem 5.29 of \cite{1}, $T$ is a weakly compact operator. By Theorem 5.85 of \cite{1}, $\ell^1$ has Dounford-Pettis property. Therefore by Theorem 5.82 of \cite{1}, $T$ is Dounford-Pettis. Since the norm of $(\ell^1)^*$ is not order continuous and $F\neq \{0\}$, so by Proposition \ref{ky}, $T$ is not $wun$-Dounford-Pettis.
\end{proof}
\begin{remark}
	It is known that every compact operator between Banach lattices is Dunford-Pettis. In the case of a $wun$-Dunford-Pettis operator, the situation is
	different. The Example \ref{poi} is compact while it is not $wun$-Dounford-Pettis.
\end{remark}
Here we give an example that it illustrate $uaw$-Dunford-Pettis operators differ from  $wun$-Dunford-Pettis operators.
\begin{example}\label{elin}
	For each continuous operator $T:C[0,1] \rightarrow c_0$,  the adjoint operator $T^* : \ell^1 \rightarrow (C[0,1])^*$ is a $uaw$-Dunford-Pettis. Indeed let $(x_n)\subseteq \ell^1$ is norm bounded and $x_n\xrightarrow{uaw}0$. By Proposition 5 of \cite{15}, $x_n \xrightarrow{w^*}0$ in $\ell^1$ and therefore $T^*(x_n)\xrightarrow{w^*}0$ in $(C[0,1])^*$. Since $C[0,1]$ has dual positive Schur property, so we have $T^*(x_n)\xrightarrow{\|.\|}0$ in $(C[0,1])^*$. Note that for each continuous operator $T:C[0,1]\rightarrow c_0$, the adjoint operator $T^*: \ell^1 \rightarrow (C[0,1])^*$ is Dunford-Pettis (we know that $T^*$ is continuous and $\ell^1$ has Schur property). Since $(\ell^l)^* = \ell^\infty $ does not has order continuous norm and $(C[0,1])^* \neq 0$, therefore by Proposition \ref{ky}, there exists some $T: C[0,1]\rightarrow c_0$ such that $T^*$ is not $wun$-Dunford-Pettis.
\end{example}
\begin{remark}
	If $T:E\rightarrow X$ is a $wun$-Dounford-Pettis where $E$ has order continuous norm, then $T$ is a $M$-weakly compact and therefore by Theorem 1 of \cite{zabeti1}, it is a $uaw$-Dounford-Pettis.
\end{remark}
\begin{theorem}\label{elia}
	Let $T:E\rightarrow X$ be an operator from $AM$-space $E$  to Banach space $X$. Then the following assertions are equivalent:
	\begin{enumerate}
		\item $T$ is $M$-weakly compact.
		\item $T$ is weakly compact.
\item $T$ is Dunford-Pettis.
	\item $T$ is $wun$-Dunford-Pettis.
\item $T$ is $uaw$-Dunford-Pettis.
\item $T$ is $b$-weakly compact.
\item Moreover if $E$ has property $(b)$, $T$ is order weakly compact.
\end{enumerate}
\end{theorem}
\begin{proof}
	\begin{enumerate}
\item[]	$(1)\Leftrightarrow (2)$ By Theorem 5.62 of \cite{1}, the proof is complete.
\item[]	$(2)\Rightarrow (3)$ By Theorems 5.85 and 5.82 of
\cite{1}, $ T $ is a Dunford-Pettis operator.
\item[] $(3)\Rightarrow(1)$ Since $E$ is an $AM$-space, then by Theorem 4.23 of \cite{1}, $E^*$ is an $AL$-space and therefore $E^*$ has order continous norm. By Theorem 3.7.10 of \cite{14}, $T$ is $M$-weakly compact.
\item[] $(3)\Leftrightarrow (4)$ Since $E^* = \ud{E}$, therefore $x_n\xrightarrow{w}0$ in $E$ iff $x_n \xrightarrow{wun}0$ in $E$. Hence the proof is clear.
\item[]	$(3)\Leftrightarrow (5)$ Follows from Corollary 3.7 of \cite{Hui}.
\item[] $(2)\Rightarrow (6)$ Since each $b$-order bounded set in Banach lattice is norm bounded, hence the proof is clear.
\item[] $(6)\Rightarrow (2)$ Let $B_E$ be a closed unit ball of $E$. Since $E$ has strong unit, so $B_E$ is an order interval. Therefore $(TB_E)$ is a relatively weakly compact subset of $X$.
\item[] $(6)\Leftrightarrow (7)$ By property $(b)$ of $E$, $A\subseteq E$ is order bounded if and only if it is $b$-order bounded, hence the proof is clear.

	\end{enumerate}
\end{proof}
	Let $S,T:E\rightarrow F$ be two positive operators satisfying $0\leq S\leq T$ with $T$ is $wun$-Dounford-Pettis, it is clear that $S$ is $wun$-Dounford-Pettis.

We  give an example that an operator $T$ is $wun$-Dounfore-Pettis while its adjoint is not $wun$-Dounford-Pettis and vic versa. In the following  with under certain conditions,  an operator  $T$ is $wun$-Dounford-Pettis iff $T^*$ is $wun$-Dounford-Pettis.

\begin{example}\label{as}
	\begin{enumerate}
		\item  Consider the operator $T:C[0,1] \rightarrow c_0$, given by
$$ T(f) = (\int_0 ^1 f(x)\sin x dx, \int_0 ^1 f(x)\sin2x dx,...).$$
If $ (f_n)\subseteq C[0,1]$ is norm bounded and $f_n \xrightarrow{wun}0$ then $f_n\xrightarrow{w}0$.  We have \  \  $\|Tf_n\| = \sup_{m\geq 1}|\int_0^1 f_n(t) \ sin mt\  dt | \leq \int_0^1 |f_n(t)|dt\rightarrow 0$. Hence $T$ is a $wun$-Dunforde-Pettise. Similar to Example 3 of \cite{zabeti1}, adjoint of $T$, $T^*$  is not $wun$-Dounford-Pettis.
\item The functional $f:\ell^1 \rightarrow \mathbb{R}$ defined by
$$f(x_1,x_2,...)= \sum_1^n x_i$$
is not $wun$-Dounford-Pettis, but $f^*$ is $wun$-Dounford-Pettis.
\end{enumerate}
\end{example}
\begin{theorem}
	Let $E$ and $F$ be two Banach lattices such that $E$ and $F^*$ have strong unit. Then $T:E \rightarrow F$ is $wun$-Dounford-Pettis iff $T^*$ is $wun$-Dounford-Pettis.
\end{theorem}
\begin{proof}
	\begin{enumerate}
\item	Let $T: E \rightarrow F$ be a $wun$-Dounford-Pettis. Since $E$ has strong unit, then by Theorem \ref{elia}, $T$ is $b$-weakly compact operator. because $E$ has strong unit therefore it ia an $AM$-space. By Theorem 4.23   of \cite{1}, $E^*$ is an $AL$-space. Each $AL$-space is a $KB$-space. Therefore $E^*$ is a $KB$-space. Hence by Theorem 3.1 of \cite{2c}, $T^*$ is $b$-weakly compact. Since $F^*$ has strong unit, hence by Theorem \ref{elia}, $T^* $ is $wun$-Donuford-Pettis.
\item The Proof has similar argument of $(1)$ and by  Theorem 3.5 of \cite{2c}, proof follows.
\end{enumerate}
\end{proof}
\begin{theorem}
	Let $F$ be a Banach lattic. If for each arbitrary Banach lattice $E$, operator $T:E\rightarrow F$ is $wun$-Dounford-Pettis, then
	\begin{enumerate}
		\item $F$ is $KB$-space.
		\item $T$ is $b$-weakly compact.
		\end{enumerate}
\end{theorem}
\begin{proof}
	\begin{enumerate}
		\item 	Let $c_0$ be embeddable in $F$ and $T:c_0 \rightarrow F$ be this emmbeding. Then there exist two positive constants $K$ and $M$ satisfying
	$$K\|(x_n)\| \leq \|T(x_n)\| \leq M\|(x_n)\| \ \ \ for \ all \ (x_n)\subseteq c_0.$$
	Consider the sequence $(e_n)\subseteq c_0$. $e_n \xrightarrow{wun}0$ and norm bounded in $c_0$ but $\| T(e_n)\| \geq K\|e_n\| = K> 0$ which contradicts with assumption. Therefore $c_0$ is not embeddable in $F$. Hence by Theorem 4.61 of \cite{1}, $F$ is a $KB$-space.
	\item By past part we have $F$ is $KB$-space. Since $c_0$ is not embeddable in $F$, then by Theorem 4.63 of \cite{1}, there exist a $KB$-space $H$, a lattice homomorphism $Q:E\rightarrow H$ and a continuous operator $S:H\rightarrow F$ such that $T = SoQ$. Let $(x_n)$ be a $b$-order bounded disjoint sequence in $E$. It is clear that $(Q(x_n))$ is also $b$-order bounded and disjoint sequence in $H$. By Lemma 2.1 of \cite{2c}, $Q(x_n)\xrightarrow{\|.\|}0$ in $H$. Thererore $T(x_n) = SoQ(x_n)\xrightarrow{\|.\|}0$. So $T$ is $b$-weakly compact.
	\end{enumerate}
\end{proof}
\begin{remark}
	Note that if $F$ is $KB$-space, then every operator $T$ from a Banach lattice $E$ into $F$, in general,  is not $wun$-Dounford-Pettis. By Example \ref{as} there exists adjoint operator $T^*$ from $\ell^1$ into $KB$-space $(C[0,1])^*$ such that it is not $wun$-Dounford-Pettis.
\end{remark}
\begin{remark}
	Let $E$ and $F$ be two Banach lattices. If  an operator $T:E\rightarrow F$ is $b$-weakly compact, in general, $T$ is not $wun$-Dounford-Pettis necessarily. We know that $\ell^1$ is  $KB$-space. Therefore by Theorem 4.61 of \cite{1}, $c_0$ is not embeddable in $\ell^1$. By Proposition 2.2 of \cite{2c}, for any Banach lattice $E$, each operator from $E$ into $\ell^1$ is $b$-weakly compact. On the other hand,  the identity operator $I:\ell^1\rightarrow \ell^1$ is not $wun$-Dounford-Pettis.
\end{remark}


TITLE: An abstract a priori estimate in finite element method
QUESTION [2 upvotes]: Let $V$ and $K$ be Banach spaces (with norms $\|\cdot\|_V$ and $\|\cdot\|_K$ resp.) and suppose that there is a compact linear embedding $K\hookrightarrow V$. Furthermore, let $P_n$ be a family of projections on $V$ whose rank is finite and that for all $u\in V$, $\lim_{n\rightarrow\infty} \|P_n u-u\|_V=0$.
Is there a bounded sequence $s_n$ that tends to zero and
\begin{equation}
\|P_n u-u\|_V \leq s_n\|u\|_K
\end{equation}
for all $u\in K$?
The motivation to this question is to generalize
the usual finite element interpolation estimate "$\|I_h u -u\|_1 \leq C h^{k-1}|u|_k$", where $k$ is the order polynomial order and $h$ is the mesh size of the FE space.

REPLY [0 votes]: I suppose you want to prove the following statement: For all $\epsilon>0$ there exists $N$ such that for all $n>N$
$$
\|P_nu-u\|_V\le \epsilon \|u\|_K \quad\forall u\in K.
$$
Let me argue by contradiction.
Suppose the claim does not hold: Then there exists $c>0$, such that for all $n$ there is $u_n\in K$ with
$$
\|P_nu_n-u_n\|_V >c \|u_n\|_K .
$$
Wlog assume $\|u_n\|_K=1$. Then the set of $u_n$ is bounded in $K$ and pre-compact in $V$ by the compact embedding. And there exists $u\in V$ such that after extracting a subsequence $u_n\to u$ in $V$.
Now consider the difference
$$
P_n u_n - u = P_n(u_n-u) + (P_n-Id)u.
$$
The term $P_nu-u$ vanishes by assumption.
By the principle of uniform boundedness the operators $P_n$ are uniformly bounded (stability), hence $P_n(u_n-u)$ vanishes as well.
This contradicts the inequality $\|P_nu_n-u_n\|_V>1$. And the claim is proven.


TITLE: Cohomology of a quotient manifold
QUESTION [2 upvotes]: Assume one has a finitedimensional vector space $V$ over the complex numbers and a discrete subgroup $G$ of translations on $V$, so that the quotient is a complex manifold $X=V/G$.
My question: can one express the cohomology group $H^1(X,\mathcal{O_X})$ in terms of the data $V$ and $G$?

REPLY [2 votes]: This answer has been corrected per Torsten's remark.
If $V/G = Y$ is a complex torus, i.e. $G$ has maximal rank, then Hodge symmetry says $H^1(Y,\mathcal O) = \overline{H^0(Y,\Omega)}$. But the cotangent bundle of $Y$ is trivial by translation on the group, so $H^0(Y,\Omega) = H^0(Y,\mathcal O) \otimes (T_0 Y)^\ast$ (the cotangent space at the identity). Now $T_0Y \cong V$, so $H^1(Y,\mathcal O) \cong \overline V^\ast$. If $Y=V/G$ is an algebraic torus, i.e. a product $(\mathbf C^\ast)^p$, then $H^1(Y,\mathcal O)$ is trivial. One way to see this is through the exponential sequence and the triviality of the Picard group of $Y$.
This also takes care of any product of complex tori, algebraic tori and affine spaces. But in general I see no reason that a quotient $V/G$ couldn't be a nontrivial extension of a complex torus by an algebraic torus. (In the algebraic setting this would be a semiabelian variety.) I don't know what happens to the cohomology then.


TITLE: Question envolving bilinear forms (symmetric and skew symmetric).
QUESTION [1 upvotes]: Let $f: V \times V \to F $ be a bilinear form and $f$ has the follow property: Whenever $f(u,v)=0$ we have $f(v,u)=0$. The goal is to prove that $f$ is symmetric or $f$ is skew - symmetric, for that prove the follow statements.
$a)$ Let's $u,v,w \in V$, prove that $f(u, f(u,w)v - f(u,v)w) =0$. Then show that $f(u,w)f(v,u)= f(w,u)f(u,v)$.
$b)$ Now  prove that $w=u$ implies $f(u,u)=0$ if $f(u,v) \neq f(v,u)$.
$c)$ Show that if $f$ is now symmetric then $f(w,w)=0, \forall w \in V$. For that, choose  $u,v$ s.t $f(u,v) \neq f(v,u)$ then observe the cases $f(u,w) \neq f(w,u)$ and $f(u,w) \neq f(w,u)$
$d)$Show that if $f(w,w)=0 \forall w \in V$ then $f$ is skew symmetric.
$\textbf{My attempt:}$
a) I don't know how to prove that $f(u, f(u,w)v - f(u,v)w) =0$, but if that was true then it is easy to show that $f(u,w)f(v,u)= f(w,u)f(u,v)$.
b) if $f(u,w)f(v,u)= f(w,u)f(u,v)$ and $w=u$ it is easy to show that $f(u,u)=0$ by $f(u,v) \neq f(v,u)$.
The other itens I don't know how to continue, can you help me?

REPLY [0 votes]: Hint
for the very first one, you want to prove
$$
f(u, f(u,w)v - f(u,v)w) =0 \tag{1}
$$
All you know about $f$ is that it's bilinear and has the special property. The special property doesn't look all that relevant here, so what happens if you expand the left-hand side of equation 1 using bilinearity?
Also: in working on problems like this, it's good to have an example in mind. In this case, I recommend thinking about $V = \Bbb R^2$, and $f(u, v) = u \cdot v$. Then it's easy to see that the thing your attempted solution was based on can't possibly be true in general, so that's a blind lead.


TITLE: Systems of linear equations: solving by addition
QUESTION [1 upvotes]: I have not understood this question. it confused me a lot. I have tried to solve it but the solution was senseless.
$$\begin{cases}
3x + 4y - 5 = 2x + 3y - 1\\
6x - 2y + 2 = 4x - 3y - 5
\end{cases}$$
Can anyone help me with this?

REPLY [0 votes]: HINT:
First turn the top and bottom equations into standard form, so you end up with a system that looks like this.
$$\begin{cases}
Ax + By  = C\\
Dx + Ey = F
\end{cases}$$
This can be done by combining the $x$ and $y$-terms on the left and moving the constant terms to the right. Then solve this "normal-looking" system using addition.


TITLE: Is quasi-isomorphism an equivalence relation?
QUESTION [8 upvotes]: Let $E^\bullet$ and $F^\bullet$ be complexes on an abelian category; what does it mean to say that $E^\bullet$ and $F^\bullet$ are quasi-isomorphic?
Does it only mean that there is a map of complexes $f:E^\bullet \to F^\bullet$ that induces isomoprhisms between the cohomology objects?
Or does it also guarantee the existence of a map of complexes $g:F^\bullet \to E^\bullet$ inducing the inverses of $H^pf:H^p(E^\bullet)\to H^p(F^\bullet)$?
Put in another way: is quasi-isomorphism an equivalence relation?

REPLY [11 votes]: $\def\ZZ{\mathbb Z}$The relation $E \sim F$ defined by «there exists a morphism $E\to F$ inducing an isomorphism in homology» is not an equivalence relation because it is not symmetric (it is relfexive and transitive)
For example, there is a morphism from
$$\cdots 0\to \ZZ\xrightarrow2\ZZ\to0\to\cdots$$
to the complex
$$\cdots 0\to 0\to\ZZ/2\ZZ\to0\to\cdots$$
inducing an isomorphism in homology, but there is no non-zero morphism in the other direction.
The useful relation is the symmetric closure of this relation.