text:
Networking News
F5 Networks Closes $670 Million NGINX Acquisition
May 09, 2019, 12:19 PM EDT
F5 Networks just closed its NGINX deal which will help the company on its evolution away from hardware in favor of software and services.
F5 Networks on Thursday closed its deal to buy one of its largest competitors, NGINX, for $670 million.
The deal comes as F5 continues to evolve from a hardware company to a more services-focused firm. The acquisition, according to F5, will help the company accelerate its "software and multi-cloud transformation."
Seattle, Wash.-based F5 first announced its intent to acquire the company on March 11.
[Related: F5 Networks Channel Chief: 'Channel Attitude' Will See Firm Through Digital Transformation ]
NGINX specializes in open source, software-based application delivery. The company's software powers more than 375 million websites, including for some well-known brands such as Netflix, Starbucks, and McDonald's, according to NGINX. The startup first announced a channel program earlier this year.
Together, F5 and NGINX will give customers multi-cloud application delivery services across all of their IT environments and flexibility for developers will also offering the scale, security, and reliability that network operations teams need, the two companies said in a statement.
F5 said in March that the two product lines will be integrated, with
F5 security adding protection to NGINX products, and NGINX load balancing software supporting F5's cloud-native solutions.
NGINX said it will continue to support the open source community.
Via the terms of the deal, F5 will keep the Nginx brand alive.
NGINX CEO Gus Robertson, along with the company's founders, Igor Sysoev and Maxim Konovalov and their team will join F5 to run the NGINX business from its headquarters in San Francisco. NGINX has about 250 employees today.
Learn More: Cloud Software | Mergers and Acquisitions
Gina Narcisi is a senior editor covering the networking and telecom markets for CRN.com. Prior to joining CRN, she covered the networking, unified communications and cloud space for TechTarget. She can be reached at gnarcisi@thechannelcompany.com.

redpajama_set_name: RedPajamaCommonCrawl


text:
Port Isabel-San Benito Navigation District
Navigation District
Port Isabel, TX 78578
http://www.portofportisabel.com
The Port Isabel-San Benito Navigation District is one of 12 deep water ports in Texas with a project depth of 36'. Formed in 1929 to lower the transportation costs of the agricultural interests of San… read more
The Port Isabel-San Benito Navigation District is one of 12 deep water ports in Texas with a project depth of 36'. Formed in 1929 to lower the transportation costs of the agricultural interests of San Benito and the marine interests of Port Isabel, the Port has broadened its scope to serve as the home base for more than 14 businesses that employ more than 500 persons. The principal mission of the Port is to lower transportation costs and to provide land for industrial development; focusing on the offshore oil and gas industry.
The Port is one of a select few classified Texas deepwater ports and is positioned at the southernmost tip of Texas to serve Mexico, Central and South America. Through the Texas Intracoastal Waterway, the Port can serve points as close as Corpus Christi and as far as St. Louis and Chicago. An estimated 45 acres of the Port's 726 acres of waterfront land and related facilities are available for lease on an annual basis in both large and small parcels at reasonable rates. An Industrial Development Corporation is also available at the Port to assist with development through the issuance of tax-free bonds.
The definition of a "Port" is the interface between the land and the water. This basic service results in reduced shipping costs and reliable transportation to the rest of America and to the world. Further than that, the Port is a hub of concentrated economic activity with benefits to be felt far beyond Port Isabel. It is difficult to quantify the effects of the Port on the Valley economy. But know this, the Port Board and the Port Staff have the econ
Steven Bearden
Port Director/General Manager


Clinical Resource Leader Certification Course© – Labor Management Institute, Inc.
This 4-day evidence based program is designed for staffing coordinators or managers of the resource center, directors of nursing administration, financial analysts or decision support personnel. It concentrates on leadership and strategic staffing, scheduling and labor management principles to improve workforce performance and enhance the workplace environment. Participants completing this program will receive a Clinical Resource Leader Certificate, 24.5 continuing education contact hours for RN's through the CA Board of Registered Nursing.


October was recognized as Mental Health Awareness month, but while that has passed us by, the discussion shouldn't be limited to a single time of the year. It is more important than ever to address this delicate subject and oft-misunderstood mental illness in the light of the Thousand Oaks Shooting, and other prevalent tragedies like Anthony Bourdian and Kate Spade, which are a sign of the ongoing mental illness crisis in this country. Only now the stigma surrounding mental health is clearing, and it's taken the deaths of iconic figures and repeated national tragedies to impact our communities' perceptions enough to begin healthier conversations about mental health.
But how long is too long to wait? Other cultures have centuries on us dealing with mental illness. Ancient Native Americans handled PTSD by aiding and purifying soldiers of war through rituals when they came home. And yet as a society today, we cannot sit down teenagers and help them, or even recognize their signs of depression or other mental health disorders.
That is why the work Smokey Point Behavioral Hospital is doing is so critical and important. Recently we have brought on CEO Sally Ann Schneider, who has over 30 years of extensive experience in the healthcare field, and an active history of involvement in community outreach and involvement. Schneider's involvement only adds weight to the already existing inpatient programs & outpatient care at Smokey Point Behavioral Hospital. They have two dedicated programs to helping teens with psychiatric treatment, one of which is a dual diagnosis inpatient program for mental health and substance abuse, with outreach programs in schools. Smokey Point Behavioral Hospital's outreach programs go above and beyond, extending aid to anyone in need, with additional specialized programs for veterans and their families through our Extra Mile Military Program, a specialized Women's Connection inpatient program, a senior program for aiding in mental health diseases and illnesses and outreach to nursing homes, and a free 24/7 crisis assessment for all.
The hospital works with both intensive inpatient and outpatient care programs for those who require immediate mental health assistance, with a 24/7 line anyone can call. Our programs range from depression to substance abuse, offering premier psychiatric services that are aimed to reinforce the idea that no one should bear this weight alone and that mental illness can be helped.


text:
OSA Publishing > JOSA A > Volume 24 > Issue 2 > Page 482
P. Scott Carney, Editor-in-Chief
Dynamics of the linearly polarized fundamental Gaussian light wave
S. R. Seshadri
S. R. Seshadri1
1109 North Whitney Way, Madison, Wisconsin 53705-2718 USA
S Seshadri
•https://doi.org/10.1364/JOSAA.24.000482
S. R. Seshadri, "Dynamics of the linearly polarized fundamental Gaussian light wave," J. Opt. Soc. Am. A 24, 482-492 (2007)
Physical Optics
Electromagnetic beams
Evanescent waves
First order optics
Gaussian beams
Phase conjugation
Paraxial wave optics (070.2580)
Propagation (350.5500)
Diffraction theory (260.1960)
Original Manuscript: February 27, 2006
Revised Manuscript: September 4, 2006
Manuscript Accepted: September 6, 2006
Equations (62)
A continuous planar array of dipoles that are oriented in a particular direction and have an amplitude distribution that is Gaussian in the paraxial limit is introduced as a source for the fundamental Gaussian light wave. The radiation intensity of the Gaussian light wave is determined and its characteristics are analyzed. The universal Gaussian beam factor is deduced and identified as the radiation intensity of the scalar Gaussian wave. The total radiated power, the mean center of the localized wave, and the beam widths of the intensity distribution are obtained. The ratio of the power in the Gaussian wave to that in the corresponding paraxial Gaussian beam is used as a measure of the quality of the paraxial beam approximation. A limiting factor for the power ratio is introduced as an indicator for the acceptability of the paraxial beam approximation. The cross section and the beam widths of the localized light wave are investigated in the large and small kw0 limits, where k is the wavenumber and w0 is the beam waist at the input plane. The beam width of the full Gaussian wave is found to be less than that of the corresponding paraxial Gaussian beam both for the scalar Gaussian wave and for the Gaussian light wave.
Linearly polarized anisotropic Gaussian light wave
J. Opt. Soc. Am. A 26(7) 1582-1587 (2009)
Quality of paraxial electromagnetic beams
Reactive power in the full Gaussian light wave
J. Opt. Soc. Am. A 26(11) 2427-2433 (2009)
Full-wave generalizations of the fundamental Gaussian beam
Partially coherent fundamental Gaussian wave generated by a fluctuating planar current source


text:
Morrissey claims Secret Service interviewed him over Donald Trump death threat
Morrissey is known for his choice actions and words, so it might not come as a surprise that he was interviewed by the Secret Service here stateside after saying he would press a button to have Trump dead "for the safety of the human race."
Read more: Audio from Morrissey interview is released, confirms he defended Spacey and Weinstein
In the original interview with German magazine Der Spiegel, he said: "Since he [Trump] was in power, he has exhausted the world. He grabs after everything like a little child. He is not a leader. He is a vermin. A huge vermin," as translated from German to English.
And, as NME reports, when asked: "If there was a button here and if you pressed on it, Trump would die dead – would you push it or not?," Morrissey replied: "I would, for the safety of the human race. It's not for my personal opinion of his face, or his life, or his family, but in the interest of the human race. I think he's a terrible, terrible scourge."
In response to those happenings, he released an eight-minute video, properly (or not so properly) titled "Morrissey's State Of The Union Address," that calls out the "haters" and explains how he had to chat with the secret service on his U.S. tour.
"As a further resort of Der Spiegl, I was cross-examined by the American Secret Service, who were very, very nice. And I do understand their position. So that went very, very well, and they assured me that they have no cause for concern. But it was a direct result of Der Spiegl that I was cross-examined, which is very, very sad," Morrissey says.
"So congratulations, Der Spiegl. You achieved everything that you set out to do. Whether again I'm allowed free access to America, I really don't know. I have to wait and see if I can enter the country again. If I can't, I have to say very sincere thanks to everyone who supported me in the U.S.
"In one week, Los Angeles announces Morrissey day, and then a week later I'm being cross-examined by the Secret Service."
Watch the full, er, State Of The Union address below:
Of course, Morrissey has had an interesting past few weeks, canceling his KROQ Almost Christmas Acoustic performance, bringing his total canceled or postponed number of shows to 127 since 2012. And the audio release from said Der Speigl that confirms he defended Spacey and Weinstein—even if he still denies it.
What are your thoughts on the Morrissey situation? Sound off in the comments below.
Written by Maggie Dickman


Aishwarya Rai Bachchan looks drop-dead gorgeous on the cover of Grazia India's latest issue, in which she leads the list of 100 most influential in fashion. The actress is said to be the hottest mommy in town.
She struck a pose on a white chair against a green backdrop. The timeless beauty is the face of the popular magazine's September cover and is a perfect combination of classy, sultry and 'don't mess with me' in black printed PJs by Sanchita.
Balancing the red, white and purple leaf and hearts prints on the outfit, she was co-styled by Pasham Alwani, who teamed the set with a pair of black, pointy heels from Jimmy Choo for that classic oomph, and accessorised the ensemble with trinkets from Isharya, and crystals from Swarovski. Bachchan's hair and makeup were done by the dynamic and sought after Daniel Bauer and Florian Hurel, respectively.
Grazia's issue will talk about Aishwarya Rai Bachchan as the "myth, legend, mother, actor, star, wife, girl-next-door and red carpet queen." Aishwarya looks phenomenal effortlessly and as we all know she's a hand-on mother to her five-year-old daughter Aaradhya and of course, her red carpet appearances are always much awaited – example Cannes Film Festival.
On the work front, Aishwarya will be next seen with Anil Kapoor in musical comedy Fanney Khan. Anil Kapoor and Aishwarya Rai Bachchan are finally sharing the screen space after 17 years with Fanney Khan. The film is being directed by debutant Atul Manjrekar and produced by Rakeysh Omprakash Mehra and KriArj Entertainment and T-Series.
In the film, both Anil Kapoor and Aishwarya Rai Bachchan will be seen playing the role of singers. The film is the official remake of the Oscar nominated Dutch 2000 film 'Everybody's Famous!' directed by Dominique Deruddere.The film also stars Rajkummar Rao opposite Aishwarya Rai Bachchan and is slated to release next year in April.
The post Aishwarya Rai Bachchan Owns The Cover Of Grazia appeared first on Koimoi.
Ranbir Kapoor & Aishwarya Rai Team Up For The Hottest Filmfare Cover Ever!


The Tampa Bay Buccaneers thrashed the San Francisco 49ers 34-17. The Buccaneers (3-3) emerged victorious over the 49ers (1-6). The 49ers have a six-game losing streak, and they could not snap out of it. On the other hand, before their match with the 49ers, the Buccaneers won against the Carolina Panthers.
Chip Kelly's team must come back after six consecutive losses. One of the biggest obstacles that the 49ers must address is their defense that cannot stop the run. On an average, the 49ers have an average of 174 yards per game. The problem with stopping the run leads to time-wasting drives, which likewise prevents the offense of the 49ers from having any kind of rhythm in the long run.


text:
Q: Visual Studio Code Inconsistency Running Python File I couldn't find an answer to this so it might just be how the program works, but I've been working in Visual Studio Code for a while now, in Python, and when I run the program, it doesn't always actually run. There's nothing wrong with the code, if I keep killing the terminal and running the file it'll eventually work, but I was wondering if there's a way to get it to do it the first time consistently.

A: You can try using the code runner extension and then using run python file option.
code runner
code runner button

redpajama_set_name: RedPajamaStackExchange


text:
\section{Introduction}

\numberwithin{equation}{section}

\subsection{ \! }

This note is motivated by problems in arithmetic combinatorics and related parts of Ramsey theory, and focuses on the
relation between two notions of pseudorandomness which appear in this context. The first notion is measured using the
\textit{Gowers uniformity norms} \cite{Go1,Go2}. These norms are very useful in order to  accurately count the number of copies
of certain ``patterns" in subsets of discrete structures; see, e.g., \cite[Lemma 11.4]{TV}. However, they are defined by estimating
the correlation of a function with shifts of itself, and so their dual norms are hopelessly difficult
to understand in full generality.

To compensate this problem, one adopts a functional analytic point of view. First one selects a class $\mathcal{D}$ of bounded functions
(the ``dual" functions), and then associates with $\mathcal{D}$ a norm defined by the rule
$\|f\|_{\mathcal{D}}\coloneqq \sup\big\{ |\langle f,g\rangle|: g\in\mathcal{D}\big\}$. If the set $\mathcal{D}$ is appropriately selected,
then the norm $\|\cdot\|_{\mathcal{D}}$ is comparable to the Gowers uniformity norm for \textit{bounded} functions. Unfortunately, in general,
the norm $\|\cdot\|_{\mathcal{D}}$ is significantly weaker, and this apparently excludes its applicability in the study of sparse sets like
the set of primes numbers.

Nevertheless, recently it was shown, first implicitly in \cite{CFZ} and then more explicitly in \cite{TZ1,TZ2}, that the Gowers uniformity norms
and their aforementioned weaker versions are essentially equivalent for a fairly large (and practically useful) family of unbounded functions.

We analyze further this phenomenon (both in the additive and in the hypergraph setting) and we show that it is more typical
than anticipated. Compared with the results in \cite{CFZ,TZ1,TZ2} which rely on the ``linear forms condition" (a pseudorandomness
hypothesis originating from \cite{GT1}), our approach is more intrinsic and is based exclusively on the properties of the Gowers
uniformity norms. In a nutshell, our main results (Propositions \ref{p2.1} and \ref{p3.1}) follow from the Gowers--Cauchy--Schwarz
inequality and a simple decomposition method introduced in \cite{DKK1}.

\subsection{ \! }

We present two applications of this equivalence. The first one is a variant of the Koopman--von Neumann decomposition
(Corollary \ref{c4.4} in the main text). It asserts that a real-valued function $f$ on a finite additive group can be approximated
in a Gowers uniformity norm by a bounded function, provided that $|f|$ is majorized by a function $\nu$ which satisfies a natural
norm-type pseudorandomness condition. It is important to note that, besides its separate theoretical interest, this approximation
is essential for further applications. Indeed, Corollary \ref{c4.4} together with an appropriate version of the generalized von
Neumann theorem---e.g., \cite[Proposition~5.3]{GT1}---provides yet another approach\footnote{See \cite{CFZ,Go3,Tao,Zh} for other
proofs of the relative Szemer\'{e}di theorem.} to the relative Szemer\'{e}di theorem \cite[Theorem 3.5]{GT1}, one of the main
two ingredients of the Green--Tao theorem \cite{GT1}.

The second application (which is presented in Section 5) is a proof of the relative inverse theorem for the Gowers $U^s[N]$-norm,
a result which is part of the \textit{nilpotent Hardy--Littlewood method} invented by Green and Tao \cite{GT2}. Our approach
is based on Corollary \ref{c4.4} and, as such, it shows that the relative inverse theorem can also be applied under a
norm-type pseudorandomness condition.

\subsection{Notation}

For every positive integer $n$ we set $[n]\coloneqq \{1,\dots,n\}$, and for every nonempty finite set $V$ by $|V|$ we denote its
cardinality. Moreover, for every function $f\colon V\to\rr$ by $\ave[f(v)\, |\, v\in V]$ we denote the average of~$f$, that~is,
\[ \ave[f(v)\, |\, v\in V] \coloneqq \frac{1}{|V|} \sum_{v\in V} f(v). \]
We also write $\ave_{v\in V} f(v)$ to denote the average of $f$, or simply $\ave[f]$ if the set $V$ is understood from the context.

We use the following $o(\cdot)$ and $O(\cdot)$ notation. If $a_1,\dots,a_k$ are parameters and $\eta$ is a positive real,
then we write $o_{\eta\to 0;a_1,\dots,a_k}(X)$ to denote a quantity bounded in magnitude by $X F_{a_1,\dots,a_k}(\eta)$
where $F_{a_1,\dots,a_k}$ is a function which depends on $a_1,\dots,a_k$ and goes to zero as $\eta\to 0$. Similarly, by
$O_{a_1,\dots,a_k}(X)$ we denote a quantity bounded in magnitude by $X C_{a_1,\dots,a_k}$ where $C_{a_1,\dots,a_k}$
is a positive constant depending on the parameters $a_1,\dots,a_k$; we also write $Y\ll_{a_1,\dots,a_k}\!X$ or
$X\gg_{a_1,\dots,a_k}\!Y$ for the estimate $|Y|=O_{a_1,\dots,a_k}(X)$.

\section{The Gowers uniformity norm versus its weak version}

\numberwithin{equation}{section}

\subsection{ \! }

Let $Z$ be a finite additive group and let $s\meg 2$ be an integer. Also let $f\colon Z\to\rr$ and recall
that the \textit{Gowers uniformity norm $\|f\|_{U^s(Z)}$ of $f$} is defined by the rule
\begin{equation} \label{e2.1}
\|f\|_{U^s(Z)} \coloneqq \ave\Big[ \prod_{\omega\in\{0,1\}^s}\!\! f(x+\omega\cdot \mathbf{h})\, \Big| \,
x\in Z, \mathbf{h}\in Z^s\Big]^{1/2^s}
\end{equation}
where $\omega\cdot \mathbf{h}\coloneqq \sum_{i=1}^s \omega_i\, h_i$ for every $\omega=(\omega_i)\in\{0,1\}^s$ and every $\mathbf{h}=(h_i)\in Z^s$.
One can also define these norms for complex-valued functions by appropriately inserting complex conjugation operations---see \cite{TV} for details.

As we have noted, there is a natural weak version of the $U^s(Z)$-norm. Specifically, let $f\colon Z\to\rr$ and
define\footnote{There is no standard terminology for these norms.} the \textit{weak uniformity norm $\|f\|_{w^s(Z)}$ of $f$} by setting
\begin{equation} \label{e2.2}
\|f\|_{w^s(Z)} \coloneqq \sup\Big\{ \ave\big[ f(x)\!\!\!\!\!\!\!\prod_{\omega\in\{0,1\}^s\setminus \{0^s\}}\!\!\!\!\!\!
h_\omega(x+\omega\cdot \mathbf{h})\, \big| \, x\in Z, \mathbf{h}\in Z^s\big]\Big\}
\end{equation}
where the above supremum is taken over all families $\langle h_\omega:\omega\in \{0,1\}^s\setminus \{0^s\}\rangle$ of
$[-1,1]\text{-valued}$ functions on $Z$, and $0^s=(0,\dots,0)\in \{0,1\}^s$ denotes the sequence of length $s$ taking the constant value $0$.
We observe that
\begin{equation} \label{e2.3}
\|f\|_{w^s(Z)}\mik \|f\|_{U^s(Z)}
\end{equation}
as can be seen by the Gowers--Cauchy--Schwarz inequality (see, e.g., \cite[(11.6)]{TV}).

\subsection{The main result}

By \eqref{e2.1} and \eqref{e2.2}, it follows readily that for every function $f\colon Z\to [-1,1]$ we have $\|f\|_{U^s(Z)}\mik \|f\|^{1/2^s}_{w^s(Z)}$.
The following proposition shows that the estimate \eqref{e2.3} can also be \textit{reversed} provided that $f$ is merely bounded in magnitude
by a function $\nu\colon Z\to\rr^+$ satisfying a norm-type pseudorandomness condition.
\begin{prop} \label{p2.1}
Let $Z$ be a finite additive group, let $s\meg 2$ be an integer and let $0<\eta\mik 1$. Also let $\nu\colon Z\to \rr^+$ such that
\begin{equation} \label{e2.4}
\|\nu-1\|_{U^{2s}(Z)}\mik \eta.
\end{equation}
Finally, let $f\colon Z\to\rr$ with $|f|\mik \nu$. If\, $\|f\|_{w^s(Z)}\mik \eta$, then
\begin{equation} \label{e2.5}
\|f\|_{U^s(Z)}= o_{\eta\to 0;s}(1).
\end{equation}
\end{prop}
Proposition \ref{p2.1} can be proved arguing as in \cite[Theorem 11]{TZ1} and using slightly stronger pseudorandomness
hypotheses (see also \cite[Proposition~3.7]{TZ2} for a variant of this argument). We will give a proof using as a main tool
the following simple consequence of the Gowers--Cauchy--Schwarz inequality for the $U^{2s}(Z)$-norm which was first
observed (in a slightly less general form) in the proof of Proposition 4 in~\cite{TZ1}.
\begin{fact} \label{f2.2}
Let $Z$ be a finite additive group and let $s\meg 2$ be an integer. Also let $g\colon Z\to \rr$, let
$\langle g^{(k)}_\omega \colon k\in\{1,2\}, \omega \in \{0,1\}^s\setminus\{0^s\}\rangle$ be a family of real-valued functions on $Z$, and set
\[ I\coloneqq \ave\Big[ g(x) \prod_{k\in\{1,2\}}\prod_{\omega\in \{0,1\}^s\setminus\{0^s\}} \!\!\!\!\!
g^{(k)}_\omega(x+\omega\cdot \mathbf{h}_k)\,\Big|\, x\in Z, \mathbf{h}_1, \mathbf{h}_2\in Z^s\Big].\]
Then we have
\[|I |\mik \|g\|_{U^{2s}(Z)}\, \cdot\! \prod_{k\in\{1,2\}}\prod_{\omega\in \{0,1\}^s\setminus\{ 0^s\}} \!\! \|g^{(k)}_\omega\|_{U^{2s}(Z)}.\]
\end{fact}
\begin{proof}
We identify $\{0,1\}^{2s}$ with $\{0,1\}^s\times \{0,1\}^s$ and we write every $\omega\in \{0,1\}^{2s}$ as $\omega=(\omega_1,\omega_2)$ where
$\omega_1,\omega_2\in\{0,1\}^s$. We define a family $\langle g_{\omega} \colon \omega\in \{0,1\}^{2s} \rangle$ of real-valued functions on $Z$
by setting: (i)  $g_{(0^s, 0^s)}=g$, (ii) $g_{(\omega, 0^s)}=g^{(1)}_\omega$ and  $g_{(0^s, \omega)}=g^{(2)}_\omega$
if $\omega\in \{0,1\}^s\setminus\{0^s\}$, and (iii) $g_{(\omega_1,\omega_2)}=1$ if $\omega_1, \omega_2\in \{0,1\}^s\setminus\{0^s\}$.
Noticing that
\[ I=\ave\Big[ \prod_{\omega\in \{0,1\}^{2s}} g_\omega(x+\omega\cdot \mathbf{h})\,\Big|\, x\in Z, \, \mathbf{h}\in Z^{2s}\Big], \]
the result follows from the Gowers--Cauchy--Schwarz inequality.
\end{proof}
We proceed to the proof of Proposition \ref{p2.1}.
\begin{proof}[Proof of Proposition \emph{\ref{p2.1}}]
We will show that for every nonempty subset $\Omega$ of $\{0,1\}^s$  and for every (possibly empty) family
$\langle h_\omega:\omega\in \{0,1\}^s\setminus \Omega\rangle$ of $[-1,1]$-valued functions on $Z$ we
have\footnote{In \eqref{e2.6} we follow the convention that the product of an empty family of functions is equal to the
constant function $1$.}
\begin{equation} \label{e2.6}
\ave\Big[ \prod_{\omega \in \Omega} f(x+\omega\cdot\mathbf{h})\!\!\! \prod_{\omega\in \{0,1\}^s\setminus \Omega}\!\!\!
h_\omega(x+\omega\cdot \mathbf{h}) \,\Big|\, x\in Z, \mathbf{h}\in Z^s  \Big]=o_{\eta\to 0;s}(1).
\end{equation}
Clearly, this is enough to complete the proof.

We proceed by induction on the cardinality of $\Omega$. Since the left-hand side of~\eqref{e2.6} is invariant under permutations of the cube,
the initial case $|\Omega|=1$ follows from our assumption that $\|f\|_{w^s(Z)}\mik \eta$. Next, let $m\in\{1,\dots, 2^s-1\}$ and assume that
\eqref{e2.6} has been proved for every $\Omega\subseteq \{0,1\}^s$ with $|\Omega|=m$. Fix $\Omega'\subseteq \{0,1\}^s$ with $|\Omega'|=m+1$.
By permuting the cube if necessary, we may assume that $0^s\in \Omega'$. Set $\Omega\coloneqq\Omega'\setminus \{0^s\}$ and notice that $|\Omega|=m$.
Also let $\langle h_\omega: \omega\in \{0,1\}^s\setminus \Omega'\rangle$ be an arbitrary family of $[-1,1]$-valued function on $Z$.
We have to show that
\[ \ave \Big[ f(x) \, \prod_{\omega \in \Omega} f(x+\omega\cdot\mathbf{h}) \!\!\!\! \prod_{\omega\in \{0,1\}^s\setminus \Omega'}\!\!\!\!
h_\omega(x+\omega\cdot\mathbf{h}) \,\Big|\, x\in Z, \mathbf{h}\in Z^s\Big]=o_{\eta\to 0;s}(1) \]
or, equivalently,
\begin{equation} \label{e2.7}
\ave[f(x) G(x) \,|\, x\in Z]=o_{\eta\to 0;s}(1)
\end{equation}
where $G\colon Z\to\rr$ is the marginal defined by the rule
\begin{equation} \label{e2.8}
G(x)=\ave \Big[ \prod_{\omega \in \Omega} f(x+\omega\cdot \mathbf{h}) \!\!\!\! \prod_{\omega\in \{0,1\}^s\setminus \Omega'}\!\!\!\!
h_\omega(x+\omega\cdot\mathbf{h})\,\Big|\, \mathbf{h}\in Z^s\Big].
\end{equation}
Since $|f|\mik \nu$ and $\ave [\nu]\mik \|\nu\|_{U^{2s}(Z)}\mik 1+\eta$, by the Cauchy--Schwarz inequality, it is enough to prove that
\begin{equation} \label{e2.9}
\ave[(\nu-1)G^2]=o_{\eta\to 0;s}(1) \ \text{ and } \ \ave[G^2]=o_{\eta\to 0;s}(1).
\end{equation}
The first estimate in \eqref{e2.9} follows from Fact \ref{f2.2}
and the fact that $\|\nu-1\|_{U^{2s}(Z)}\mik \eta$; indeed, observe that
\[ |\ave[(\nu-1)G^2]|\mik \|\nu-1\|_{U^{2s}(Z)} \cdot \|f\|^{2|\Omega|}_{U^{2s}(Z)}\, \cdot \!\!\!\!\!
\prod_{\omega\in \{0,1\}^s\setminus \Omega'}\!\!\! \|h_\omega\|^2_{U^{2s}(Z)} \ll_s \eta. \]
For the second estimate, as in \cite[Theorem 7.1]{DKK1}, we will use a simple decomposition. Specifically, let $\beta>0$ be a cut-off parameter
and write $G^2= \mathbf{1}_{[|G|\mik\beta]}G^2+\mathbf{1}_{[|G|>\beta]}G^2$. As we shall see, any value of $\beta$ greater than $1$ would suffice
for the proof; for concreteness we will use the value $\beta=2$. By linearity of expectation, it is enough to show that
\begin{equation} \label{e2.10}
\ave[\mathbf{1}_{[|G|\mik 2]}G^2]=o_{\eta\to 0;s}(1) \ \text{ and } \ \ave[\mathbf{1}_{[|G|>2]}G^2]=o_{\eta\to 0;s}(1).
\end{equation}
The first part of \eqref{e2.10} can be handled easily by our inductive assumptions. Indeed, set
$h_{0^s}=\boldsymbol{1}_{[|G|\mik 2]}(G/2)$ and notice that
\[ \ave[\mathbf{1}_{[G|\mik 2]}G^2]=2 \cdot \ave\Big[h_{0^s}\big(x) \prod_{\omega \in \Omega} f(x+\omega\cdot\mathbf{h})\!\!\!\!\!
\prod_{\omega\in \{0,1\}^s\setminus \Omega'} \!\!\! h_\omega(x+\omega\cdot\mathbf{h}) \,\Big|\, x\in Z, \mathbf{h}\in Z^s\Big] \]
which is $o_{\eta\to 0;s}(1)$ since $|h_{0^s}|\mik 1$ and $\Omega\subseteq \{0,1\}^s$ satisfies $|\Omega|=m$.
For the second part of \eqref{e2.10}, observe that
\begin{equation} \label{e2.11}
\ave[\mathbf{1}_{[|G|> 2]} G^2]\mik \ave[\mathbf{1}_{[\mathcal{N}> 2]}\mathcal{N}^2]
\end{equation}
where $\mathcal{N}\colon Z\to\rr$ is defined by
$\mathcal{N}(x)=\ave[\, \prod_{\omega \in \Omega} \nu(x+\omega\cdot\mathbf{h})\,|\,\mathbf{h}\in Z^s]$.
The function $\mathcal{N}$ satisfies the following moment estimate: for every $A\subseteq Z$ and every $k\in \{1,2\}$ we have
\begin{equation} \label{e2.12}
 \ave[\mathbf{1}_A\mathcal{N}^k]=\mathbf{P}(A)+o_{\eta\to 0;s}(1)
\end{equation}
where $\mathbf{P}(A)=\ave[\mathbf{1}_A]=|A|/|Z|$ is the probability of $A$ with respect to the uniform probability measure $\mathbf{P}$ on $Z$.
Indeed, since $|\ave[\mathbf{1}_A \mathcal{N}^k]-\mathbf{P}(A)|=|\ave[\mathbf{1}_A(\mathcal{N}^k-1)]|$, the estimate in \eqref{e2.12} follows
from Fact \ref{f2.2}, a telescopic argument and the fact that $\|\nu-1\|_{U^{2s}(Z)}\mik \eta$. Now, combining \eqref{e2.11} and \eqref{e2.12}
for $k=2$ and invoking Markov's inequality, we have
\begin{eqnarray*}
\ave[\mathbf{1}_{[|G|> 2]}G^2] & \mik & \mathbf{P}\big([\mathcal{N}>2]\big) + o_{\eta\to 0;s}(1)
\mik  \mathbf{P}\big(\big[|\mathcal{N}-1|>1\big]\big) + o_{\eta\to 0;s}(1) \\
& \mik & \ave\big[|\mathcal{N}-1|\big]+ o_{\eta\to 0;s}(1).
\end{eqnarray*}
On the other hand, by \eqref{e2.12} for $k=1$, we see that
\[ \ave\big[|\mathcal{N}-1|\big]=\ave[\mathbf{1}_{[\mathcal{N}\meg 1]}(\mathcal{N}-1)]+
\ave[\mathbf{1}_{[\mathcal{N}< 1]}(1-\mathcal{N})]=o_{\eta\to 0;s}(1).\]
Therefore, $\ave[\mathbf{1}_{[|G|>2]}G^2]=o_{\eta\to 0;s}(1)$ as desired.
\end{proof}
\begin{rem} \label{r1}
It is not hard to see that the proof of Proposition \ref{p2.1} in fact yields that for every $0<\ee\mik 1$, if $\nu\colon Z\to\rr^+$ satisfies
$\|\nu-1\|_{U^{2s}(Z)}\mik \eta$ for some $0<\eta\mik \ee$ and $f\colon Z\to\rr$ is such that $|f|\mik \nu$ and $\|f\|_{w^s(Z)}\mik \ee$,
then we have $\|f\|_{U^s(Z)}\ll_s \ee^{C}+o_{\eta\to 0}(1)$ where $C=(2^s\cdot 2^{2^s-1})^{-1}$.
\end{rem}
\begin{rem} \label{r2}
By appropriately modifying the proof of Proposition \ref{p2.1}, one can establish the equivalence between the $U^s(Z)$-norm and its weak
version using more general pseudorandomness hypotheses. In particular, we have the following proposition which is related to \cite[Theorem 7.1]{DKK1}.
\begin{prop} \label{p2.3}
Let $Z$ be a finite additive group, let $s\meg 2$ be an integer and let $0<\eta\mik 1$. Let $1<p\mik \infty$, let $q$ denote the
conjugate exponent of $p$ and set $\ell\coloneqq \min\{2n: n\in\nn \text{ and } 2n\meg 2q\}$. Also let $\nu\colon Z\to \rr^+$ such that
\begin{equation} \label{e2.13}
\|\nu-\psi\|_{U^{\ell s}(Z)}\mik \eta
\end{equation}
where $\psi\colon Z\to\rr$ satisfies\footnote{Here, the $L_p$-norm of $\psi$ is computed using the uniform probability measure on $Z$,
that is, $\|\psi\|_{L_p}\coloneqq \ave[\, |\psi(x)|^p\, |\, x\in Z]^{1/p}$.} $\|\psi\|_{L_p}\mik 1$ and $\|\psi\|_{U^{\ell s}(Z)}\mik 1$.
Finally, let $f\colon Z\to \rr$ with $|f|\mik \nu$. If\, $\|f\|_{w^s(Z)}\mik \eta$, then
\begin{equation} \label{e2.14}
\|f\|_{U^s(Z)}= o_{\eta\to 0;s,p}(1).
\end{equation}
\end{prop}
Observe that Proposition \ref{p2.1} corresponds to the case ``$p=\infty$" and ``$\psi=1$"\!. Also note that, by H\"{o}lder's inequality,
if $p$ is sufficiently large, then the estimate $\|\psi\|_{U^{\ell s}(Z)}\mik 1$ follows from the estimate $\|\psi\|_{L_p}\mik 1$.
\end{rem}

\section{The box norm versus the cut norm}

\numberwithin{equation}{section}

\subsection{ \! }

Let $V$ be a nonempty finite set and let $s\meg 2$ be an integer. Also let $F\colon V^s\to\rr$ and recall that the
\textit{box norm $\|F\|_{\square(V^s)}$ of~$F$} is defined by the rule
\begin{equation} \label{e3.1}
\|F\|_{\square(V^s)}\coloneqq \ave\Big[ \prod_{\omega\in [2]^s} F\big( \pi_\omega(x)\big)\,\Big|\, x\in V^{s\times 2} \Big]^{1/2^s}
\end{equation}
where for every $\omega=(\omega_i)\in [2]^s$ by $\pi_\omega\colon V^{s\times 2}\to V^s$ we denote the projection
$\pi_\omega\big((x_{ij})\big)=(x_{i\,\omega_i})_{i=1}^s$. These norms are the abstract versions of the Gowers uniformity norms;
indeed, notice that for every finite additive group $Z$ and every $f\colon Z\to\rr$ we have
\begin{equation} \label{e3.2}
\|f\|_{U^s(Z)}=\|f(x_1+\cdots+x_s)\|_{\square(Z^s)}.
\end{equation}
We will also work with the following slight variants of the box norms which first appeared in \cite{Ha}: for every even integer
$\ell\meg 2$ we define the \textit{$\ell$-box norm $\|F\|_{\square_\ell(V^s)}$ of\, $F$} by setting
\begin{equation} \label{e3.3}
\|F\|_{\square_\ell(V^s)}\coloneqq \ave\Big[ \prod_{\omega\in [\ell]^s} F\big( \pi_\omega(x)\big)\,\Big|\, x\in V^{s\times \ell} \Big]^{1/\ell^s}
\end{equation}
where, as above, for every $\omega=(\omega_i)\in [\ell]^s$ by $\pi_\omega\colon V^{s\times \ell}\to V^s$ we denote the projection
$\pi_\omega\big((x_{ij})\big)=(x_{i\,\omega_i})_{i=1}^s$. Clearly, the $\square_2(V^s)$-norm coincides with the $\square(V^s)$-norm.
As the parameter $\ell$ increases, the quantity $\|F\|_{\square_\ell(V^s)}$ also increases and measures the integrability of $F$.
In particular, for bounded functions all these norms are essentially equivalent. This fact, together with some basic properties
of the $\ell$-box norms, are discussed in the appendix.

The box norm also has a natural weak version which is known as the \textit{cut norm} and originates from \cite{FK}.
Specifically, let $V, s$ and $F$ be as above, and define\footnote{In several places in the literature, the cut norm is defined
by taking the supremum in \eqref{e3.4} over all families $\langle H_\omega:\omega\in [2]^s\setminus \{1^s\}\rangle$ of
$[0,1]\text{-valued}$ functions on $V^s$. However, it is clear that this more restrictive definition yields an equivalent norm.}
the \textit{cut norm $\|F\|_{\mathrm{cut}(V^s)}$ of\, $F$} by the rule
\begin{equation} \label{e3.4}
\|F\|_{\mathrm{cut}(V^s)}\coloneqq \sup\Big\{ \ave\big[ F\big(\pi_{1^s}(x)\big)\!\!\!\!\! \prod_{\omega\in [2]^s\setminus \{1^s\}}\!\!\!
H_\omega\big( \pi_\omega(x)\big)\,\big|\, x\in V^{s\times 2} \big] \Big\}
\end{equation}
where the above supremum is taken over all families $\langle H_\omega:\omega\in [2]^s\setminus \{1^s\}\rangle$ of
$[-1,1]\text{-valued}$ functions on $V^s$, and $1^s=(1,\dots,1)\in [2]^s$ denotes the sequence of length $s$ taking the constant value $1$.
By the Gowers--Cauchy--Schwarz inequality for the $\square(V^s)$-norm, we see that
\begin{equation} \label{e3.5}
\|F\|_{\mathrm{cut}(V^s)} \mik \|F\|_{\square(V^s)}.
\end{equation}
Also observe that if $F$ is $[-1,1]$-valued, then $\|F\|_{\square(V^s)}\mik \|F\|^{1/2^s}_{\mathrm{cut}(V^s)}$.

\subsection{The main result}

The following proposition is the analogue of Proposition~\ref{p2.1} and establishes the equivalence of the box norm with the cut norm.
\begin{prop} \label{p3.1}
Let $V$ be a nonempty finite set, let $s\meg 2$ be an integer and let $0<\eta\mik 1$. Also let $\nu\colon V^s\to \rr^+$ such that
\begin{equation} \label{e3.6}
\|\nu-1\|_{\square_4(V^s)}\mik \eta.
\end{equation}
Finally, let $F\colon V^s\to\rr$ with $|F|\mik \nu$. If\, $\|F\|_{\mathrm{cut}(V^s)}\mik \eta$, then
\begin{equation} \label{e3.7}
\|F\|_{\square(V^s)}= o_{\eta\to 0;s}(1).
\end{equation}
\end{prop}
It is possible to prove Proposition \ref{p3.1} arguing as in \cite[Theorem 2.17]{CFZ}. However, as the reader has probably already noticed,
Proposition \ref{p3.1} can be proved arguing precisely as in Proposition \ref{p2.1}, using instead of Fact \ref{f2.2} the
following elementary consequence of the  Gowers--Cauchy--Schwarz inequality for the $\square_4(V^s)$-norm.
\begin{fact} \label{f3.2}
Let $V$ be a nonempty finite set and let $s\meg 2$ be an integer. Also let $G\colon V^s\to \rr$, let
$\langle G^{(k)}_\omega \colon k\in\{1,2\}, \omega \in [2]^s\setminus\{1^s\}\rangle$ be a family of real-valued functions on $V^s$,
and set\footnote{Here, we identify $V^{s\times 2}$ with $V^s\times V^s$ via the bijection
$V^{s\times 2}\! \ni (x_{ij})\mapsto \big((x_{i1}),(x_{i2})\big)\!\in V^s\times V^s$. In particular, we write uniquely every $x\in V^{s\times 2}$
as $x=(y,z)\in V^s\times V^s$.}
\[ I\coloneqq \ave\Big[ G(y) \prod_{k\in\{1,2\}}\prod_{\omega\in [2]^s\setminus\{1^s\}}\!\!
G^{(k)}_\omega\big(\pi_\omega(y,z_k)\big)\,\Big|\, y,z_1, z_2\in V^{s}\Big].\]
Then we have
\[|I |\mik \|G\|_{\square_4(V^s)}\, \cdot \! \prod_{k\in\{1,2\}}\prod_{\omega\in [2]^s\setminus\{ 1^s\}} \!\! \|G^{(k)}_\omega\|_{\square_4(V^s)}.\]
\end{fact}
\begin{proof}
Define a map $\{1,3\}^s\setminus \{1^s\}\ni \omega=(\omega_i)\mapsto \omega'=(\omega'_i)\in [2]^s\setminus\{1^s\}$ by setting
$\omega'_i=1$ if $\omega_i=1$, and $\omega'_i=2$ if $\omega_i=3$. Then we may write
\[ I=\ave\Big[ G_{1^s}\big(\pi_{1^s}(x)\big) \!\! \prod_{\omega\in [2]^s\setminus\{1^s\}} \!\!\!\!\! G_\omega\big(\pi_{\omega} (x)\big) \!\!
\prod_{\omega\in \{1,3\}^s\setminus\{1^s\}} \!\!\!\!\! G_\omega\big(\pi_{\omega} (x)\big)\,\Big|\,  x\in V^{s\times 4}\Big] \]
where $G_{1^s}= G$, $G_\omega= G^{(1)}_\omega$ for every  $\omega\in [2]^s\setminus \{1^s\}$, and $G_\omega= G^{(2)}_{\omega'}$ for every
$\omega\in \{1,3\}^s\setminus\{1^s\}$. Thus, setting $G_{\omega}=1$ for all other $\omega\in [4]^s$, we see that
\[ I=\ave\Big[ \prod_{\omega\in [4]^s} G_\omega\big(\pi_\omega(x)\big)\,\Big|\, x\in V^{s\times 4}\Big] \]
and the result follows from the Gowers--Cauchy--Schwarz inequality.
\end{proof}
\begin{rem} \label{r3}
We point out that Proposition \ref{p2.3} can also be extended in the hypergraph setting. Specifically, we have the following
proposition; see \cite[Section 7]{DKK1} for further results in this direction.
\begin{prop} \label{p3.3}
Let $V$ be a nonempty finite set, let $s\meg 2$ be an integer and let $0<\eta\mik 1$. Let $1<p\mik \infty$, let $q$ denote the conjugate
exponent of $p$ and set $\ell\coloneqq \min\{2n: n\in\nn \text{ and } 2n\meg 2q+2\}$. Also let $\nu\colon V^s\to \rr^+$ such that
\begin{equation} \label{e3.8}
\|\nu-\psi\|_{\square_\ell(V^s)}\mik \eta
\end{equation}
where $\psi\colon V^s\to\rr$ satisfies\footnote{Here, as in Proposition \ref{p2.3}, the $L_p$-norm of $\psi$ is computed using the uniform probability
measure on $V^s$, that is, $\|\psi\|_{L_p}\coloneqq \ave[\, |\psi(x)|^p\, |\, x\in V^s]^{1/p}$.}
$\|\psi\|_{L_p}\mik 1$ and $\|\psi\|_{\square_\ell(V^s)}\mik 1$. Finally, let $F\colon V^s\to \rr$ with $|F|\mik \nu$.
If\, $\|F\|_{\mathrm{cut}(V^s)}\mik \eta$, then
\begin{equation} \label{e3.9}
\|F\|_{\square(V^s)}= o_{\eta\to 0;s,p}(1).
\end{equation}
\end{prop}
\end{rem}

\subsection{Transferring Proposition \ref{p3.1} to the additive setting}

There is an additive version of Proposition \ref{p3.1} which is somewhat distinct from Proposition \ref{p2.1} and is obtained
by transferring the $\ell$-box norms and the cut norm in the additive setting via formula \eqref{e3.2}. Specifically, let $Z$
be a finite additive group, let $s\meg 2$ be an integer and let $f\colon Z\to \rr$. For every even integer $\ell\meg 2$ we define
the \textit{$(s,\ell)$-uniformity norm $\|f\|_{U^s_\ell(Z)}$ of $f$} by
\begin{equation} \label{e3.10}
\|f\|_{U^s_\ell(Z)} \coloneqq \|f(x_1+\cdots+x_s)\|_{\square_\ell(Z^s)}.
\end{equation}
Respectively, we define the \textit{$s$-additive cut norm $\|f\|_{\mathrm{cut}^s(Z)}$ of $f$} by the rule
\begin{equation} \label{e3.11}
\|f\|_{\mathrm{cut}^s(Z)} \coloneqq \|f(x_1+\cdots+x_s)\|_{\mathrm{cut}(Z^s)}.
\end{equation}
(Notice that the additive cut norm is slightly stronger than the weak uniformity norm; in particular, we have
$\|f\|_{w^s(Z)}\mik \|f\|_{\mathrm{cut}^s(Z)}$.) Taking into account \eqref{e3.10} and \eqref{e3.11},
we see that Proposition \ref{p3.1} can be reformulated as follows.
\begin{cor} \label{c3.4}
Let $Z$ be a finite additive group, let $s\meg 2$ be an integer and let $0<\eta\mik 1$. Also let $\nu\colon Z\to \rr^+$ such that
\begin{equation} \label{e3.12}
\|\nu-1\|_{U^s_4(Z)}\mik \eta.
\end{equation}
Finally, let $f\colon Z\to\rr$ with $|f|\mik \nu$. If\, $\|f\|_{\mathrm{cut}^s(Z)}\mik \eta$, then
\begin{equation} \label{e3.13}
\|f\|_{U^s(Z)}= o_{\eta\to 0;s}(1).
\end{equation}
\end{cor}

\section{A variant of the Koopman--von Neumann decomposition}

\numberwithin{equation}{section}

\subsection{Overview}

The \emph{Koopman--von Neumann decomposition} is a circle of results asserting that, under certain circumstances, one can decompose
a function $f$ as $f=f_{\mathrm{bnd}}+f_{\mathrm{err}}$ where $f_{\mathrm{bnd}}$ is bounded in magnitude by $1$ and $f_{\mathrm{err}}$
has small uniformity norm\footnote{As we have already noted in the introduction, in applications it is not enough to control
the error-term $f_{\mathrm{err}}$ using a weaker norm.}. To see the relevance in this context of the equivalence between the
uniformity norms and their weaker versions, note that one can first approximate $f$ by a bounded function $f_{\mathrm{bnd}}$ such that
the difference $f-f_{\mathrm{bnd}}$ is small in a weaker norm, and then upgrade this information using the results in the previous sections.
This strategy (also used in \cite{TZ1,TZ2}) is quite effective partly because the aforementioned weaker approximation can be achieved
relatively easily using various methods. We will use one of these methods, the so-called \textit{dense model theorem}.

\subsection{Consequences of the dense model theorem}

We begin by recalling the dense model theorem; we will state the formulation which is closest to the purposes of this note (see
\cite[Theorem 1.1]{RTTV} or \cite[Theorem 3.5]{TZ2}).
\begin{prop} \label{p4.1}
Let $X$ be a finite set and let $\mathcal{F}$ be a family of $[-1,1]$-valued functions on $X$. Also let $0<\eta\mik 1$, and let
$\nu\colon X\to \rr^+$ such that $\ave[\nu]\mik 1+\eta$ and satisfying
\begin{equation} \label{e4.1}
\Big|\ave \Big[(\nu-1)\prod_{i=1}^k F_i \Big]\Big|\mik \eta
\end{equation}
for every $F_1,\dots,F_k\in\mathcal{F}$. Then for every $g\colon X\to \rr$ with $0\mik g\mik \nu$
there exists $w\colon X\to [0,1]$ such that
\begin{equation} \label{e4.2}
\sup\big\{ |\ave\big[(g-w)F]|: F\in\mathcal{F}\big\}=o_{\eta\to 0}(1).
\end{equation}
\end{prop}
We will need two consequences of Proposition \ref{p4.1}. The first one concerns functions defined on a finite additive group $Z$.
Recall that by $\|\cdot\|_{\mathrm{cut}^s(Z)}$ we denote the additive cut norm defined in \eqref{e3.11}.
\begin{cor} \label{c4.2}
Let $Z$ be a finite additive group and let $s\meg 2$ be an integer. Also let $0<\eta\mik 1$ and $\nu\colon Z\to \rr^+$ such that
$\|\nu-1\|_{\mathrm{cut}^s(Z)}\mik \eta$. Then for every $g\colon Z\to\rr$ with $0\mik g\mik \nu$ there exists $w\colon Z\to [0,1]$ such that
$\|g-w\|_{\mathrm{cut}^s(Z)}= o_{\eta\to 0}(1)$. Consequently, for every $f\colon Z\to\rr$ with $|f|\mik \nu$ there exists $h\colon Z\to [-1,1]$ such that
$\|f-h\|_{\mathrm{cut}^s(Z)}= o_{\eta\to 0}(1)$.
\end{cor}
The second consequence is the analogue of Corollary \ref{c4.2} for hypergraphs.
\begin{cor} \label{c4.3}
Let $V$ be a nonempty finite set and let $s\meg 2$ be an integer. Also let $0<\eta\mik 1$ and $\nu\colon V^s\to \rr^+$ such that
$\|\nu-1\|_{\mathrm{cut}(V^s)}\mik \eta$. Then for every $G\colon V^s\to\rr$ with $0\mik G\mik \nu$ there exists $W\colon V^s\to [0,1]$ such that
$\|G-W\|_{\mathrm{cut}(V^s)}= o_{\eta\to 0}(1)$. Consequently, for every $F\colon V^s\to\rr$ with $|F|\mik \nu$ there exists
$H\colon V^s\to [-1,1]$ such that $\|F-H\|_{\mathrm{cut}(V^s)}= o_{\eta\to 0}(1)$.
\end{cor}
Corollary \ref{c4.3} is a straightforward consequence of Proposition \ref{p4.1}. On the other hand, Corollary \ref{c4.2} follows
by applying Proposition \ref{p4.1} for the family $\mathcal{F}$ of all convex combinations\footnote{The need to convexify
the set of ``dual" functions is very natural from a functional analytic perspective; see, e.g., \cite{Go3}.} of functions
$D\colon Z\to  \rr$ of the form
\[ D(z)=\ave\Big[ \prod_{\omega\in [2]^s\setminus \{1^s\}}\!\!\! H_\omega\big( \pi_\omega(x)\big) \,\Big|\,
x=(x_{ij})\in Z^{s\times 2} \text{ with } \sum_{i=1}^s x_{i1}=z \Big] \]
where $H_\omega\colon Z^s\to [-1,1]$ for every $\omega\in  [2]^s\setminus \{1^s\}$. Indeed, it is not hard to see that
this family $\mathcal{F}$ is closed under multiplication (see, e.g., the proof of Lemma 3.3 in~\cite{Zh}).

\subsection{The main results}

We are ready to state our first result in this section. It is a variant of \cite[Proposition~8.1]{GT1} (see also \cite[Proposition 10.3]{GT2}).
\begin{cor} \label{c4.4}
Let $Z$ be a finite additive group, let $s\meg 2$ be an integer and let $0<\eta\mik 1$. Also let $\nu\colon Z\to\rr^+$ such that
\begin{equation} \label{e4.3}
\|\nu-1\|_{U^{2s}(Z)}\mik \eta.
\end{equation}
Then for every $f\colon Z\to\rr$ with $|f|\mik \nu$ there exists $h\colon Z\to [-1,1]$ such that
\begin{equation} \label{e4.4}
\|f-h\|_{U^s(Z)}=o_{\eta\to 0;s}(1).
\end{equation}
Moreover, if $f$ is nonnegative, then $h$ is also nonnegative.
\end{cor}
Note that if $Z$ is a finite additive group and $f\colon Z\to \rr^+$ is a function which is approximated by a $[0,1]$-valued function
on $Z$ in the sense of \eqref{e4.4}---that is, there exists $h\colon Z\to [0,1]$ such that $\|f-h\|_{U^s(Z)}=o(1)$---then $f$ is
majorized by a function $\nu\colon Z\to \rr^+$ satisfying $\|\nu-1\|_{U^s(Z)}=o(1)$; indeed, simply take $\nu\coloneqq f+(1-h)$.
Thus we see that the pseudorandomness hypothesis \eqref{e4.3} is nearly optimal.
\begin{proof}[Proof of Corollary \emph{\ref{c4.4}}]
We first observe that, by \eqref{e4.3}, the monotonicity of the Gowers norms $\|\cdot\|_{U^s(Z)}\mik \|\cdot\|_{U^{2s}(Z)}$,
the identity \eqref{e3.2} and \eqref{e3.5}, we have that $\|\nu-1\|_{\mathrm{cut}^s(Z)}\mik \eta$. Therefore, by Corollary \ref{c4.2},
there exists $h\colon Z\to [-1,1]$ such that $\|f-h\|_{\mathrm{cut}^s(Z)}= o_{\eta\to 0}(1)$. Set $\nu'\coloneqq (\nu+1)/2$
and notice that $|f-h|/2\mik \nu'$ and $\|\nu'-1\|_{U^{2s}(Z)}\mik \eta$. By Proposition~\ref{p2.1}, the result follows.
\end{proof}
Our second result is a variant of \cite[Theorem 3.9]{Tao}.
\begin{cor} \label{c4.5}
Let $V$ be a nonempty finite set, let $s\meg 2$ be an integer and let $0<\eta\mik 1$. Also let $\nu\colon V^s\to\rr^+$ such that
\begin{equation} \label{e4.5}
\|\nu-1\|_{\square_4(V^s)}\mik \eta.
\end{equation}
Then for every $F\colon V^s\to\rr$ with $|F|\mik \nu$ there exists $H\colon Z\to [-1,1]$ such that
\begin{equation} \label{e4.6}
\|F-H\|_{\square(V^s)}=o_{\eta\to 0;s}(1).
\end{equation}
Moreover, if $F$ is nonnegative, then $H$ is also nonnegative.
\end{cor}
\begin{proof}
It is identical to the proof of Corollary \ref{c4.4}. Indeed, by \eqref{e4.5} and Corollary \ref{c4.3}, there exists
$H\colon V^s\to [-1,1]$ such that $\|F-H\|_{\mathrm{cut}(V^s)}= o_{\eta\to 0}(1)$. By Proposition \ref{p3.1}, the result follows.
\end{proof}

\section{On the relative inverse theorem for the Gowers $U^s[N]$-norm}

\numberwithin{equation}{section}

\subsection{Overview}

In order to put the main result of this section in a proper context, we begin with a brief discussion on the \textit{nilpotent Hardy--Littlewood method}
invented by Green and Tao \cite{GT2}. It is a powerful method for obtaining precise asymptotic estimates (as $N\to+\infty$)
for expressions of the form
\begin{equation} \label{e5.1}
\sum_{n\in K\cap \zz^{d}} \, \prod_{i=1}^t f_i\big(\psi_i(n)\big)
\end{equation}
where $f_1,\dots,f_t\colon \zz\to\rr$ are arithmetic functions supported on the set of positive integers,
$K\subseteq [-N,N]^d$ is a convex body and $\psi_1,\dots,\psi_t\colon \zz^d\to~\zz$ are affine linear forms no two of which are affinely dependent.
The first step of the method relies on the \textit{generalized von Neumann theorem}---see \cite[Proposition 7.1]{GT2}---which
reduces the estimation of the quantity in \eqref{e5.1} to a norm estimate
\begin{equation} \label{e5.2}
\|f_i-1\|_{U^s[N]}=o_s(1) \ \ \text{for every } s\meg 2 \text{ and every } i\in\{1,\dots,t\}
\end{equation}
where $\|\cdot\|_{U^s[N]}$ stands for the $s$-th Gowers uniformity norm on the interval $[N]$ which we will shortly recall.
This reduction can be performed provided that $|f_1|,\dots,|f_t|$ are simultaneously majorized by a function $\nu$ satisfying
the ``linear forms condition" (see \cite[Definition 6.2]{GT2}). The second (and more substantial) step of the method reduces
the estimate \eqref{e5.2} to a non-correlation estimate
\begin{equation} \label{e5.3}
\ave_{n\in [N]} \, (f_i(n)-1) F(g^n\cdot x) =o_{s,G/\Gamma,M}(1)
\end{equation}
where $G/\Gamma$ is an $(s-1)$-step nilmanifold equipped with a smooth Riemannian metric~$d_{G/\Gamma}$,
$F\colon G/\Gamma\to [-1,1]$ is a function with Lipschitz constant at most $M$, $g\in G$ and $x\in G/\Gamma$. (We recall the notion
of an $(s-1)$-step nilmanifold below.) For~\textit{bounded} functions, the equivalence between \eqref{e5.2} and \eqref{e5.3}
is a deep result which is known as the \textit{inverse theorem for the Gowers $U^s[N]$-norm} and is due to Green,
Tao and Ziegler \cite{GTZ}. One of the main steps in \cite{GT2} was to transfer the inverse theorem to the unbounded setting.
This was achieved with the \textit{relative inverse theorem for the Gowers $U^s[N]$-norm}---see \cite[Proposition 10.1]{GT2}---which
can be applied  provided that $|f_i|$ is majorized by a function $\nu$ satisfying the aforementioned linear forms condition
and an additional pseudorandomness condition known as the ``correlation condition" (see \cite[Definition 6.3]{GT2}).

Recently, a part of the proof of \cite[Proposition 10.1]{GT2} was revisited in \cite{TZ1}. One pleasant consequence of the approach
in \cite{TZ1} is that the relative inverse theorem (and, consequently, the whole nilpotent Hardy--Littlewood method) can be applied
assuming that the majorant $\nu$ satisfies only the linear forms condition\footnote{The possibility that one could dispense with the
need for the correlation condition entirely, was also noted in \cite[Appendix A]{FGKT}.}.

Our aim in this section is to give yet another proof of the relative inverse theorem using a norm-type pseudorandomness condition.
To this end, it is convenient at this point to properly introduce the concepts discussed so far.

\subsubsection{Uniformity norms on intervals}

Let $N\meg 1$ be an integer and let $f\colon [N]\to\rr$ be a function. We select an integer $N'>2N$ and we identify
(in the obvious way) the discrete interval $[N]$ with a subset of the cyclic group $\zz_{N'}\coloneqq \zz/N'\zz$.
The \textit{Gowers uniformity norm  $\|f\|_{U^s[N]}$ of\, $f$ on the interval $[N]$} is defined by setting
\begin{equation} \label{e5.4}
\|f\|_{U^s[N]}\coloneqq \|f\mathbf{1}_{[N]}\|_{U^s(\zz_{N'})}/ \|\mathbf{1}_{[N]}\|_{U^s(\zz_{N'})}
\end{equation}
where $\mathbf{1}_{[N]}\colon \zz_{N'}\to \{0,1\}$ stands for the indicator function of $[N]$. We note that the quantity $\|f\|_{U^s[N]}$
is, in fact, intrinsic and is independent of the choice of $N'$---see \cite[Appendix B]{GT2} for more details.

\subsubsection{Nilmanifolds}

Let $s\meg 2$ be an integer and recall that an \emph{$(s-1)$-step nilmanifold} is a homogeneous space $X\coloneqq G/\Gamma$
where $G$ is an $(s-1)$-step nilpotent, connected, simply connected Lie group, and $\Gamma$ is a discrete cocompact subgroup of~$G$.
The group $G$ acts on $G/\Gamma$ by left multiplication and this action will be denoted by $(g,x)\mapsto g\cdot x$. As in \cite{GT2},
we will assume that each nilmanifold $G/\Gamma$ is equipped with a smooth Riemannian metric $d_{G/\Gamma}$; in particular,
if $F\colon G/\Gamma\to\rr$ is a function, then its Lipschitz constant is computed using the metric $d_{G/\Gamma}$.

\subsection{The main result}

We are ready to state the main result in this section. As we have indicated, it is a refinement of \cite[Proposition 10.1]{GT2}.
\begin{thm} \label{t5.1}
For every integer $s\meg 2$, every $C\meg 20$ and every $0<\delta\mik 1$ there exist $\eta>0$, a constant $M>0$, a finite collection
$\calm$ of $(s-1)$-step nilmanifolds $($each equipped with a smooth Riemannian metric$)$, and a constant $c>0$ with the following property.
Let $N$ be a positive integer and let $N'\in [CN,2CN]$ be a prime. Also let $\nu\colon \zz_{N'}\to \rr^+$ satisfying
\begin{equation} \label{e5.5}
\|\nu-1\|_{U^{2s}(\zz_{N'})}\mik \eta.
\end{equation}
Finally, let $f\colon [N]\to\rr$ with $|f(n)|\mik \nu(n)$ for every $n\in [N]$. If\, $\|f\|_{U^s[N]}\meg \delta$, then there exist
$(G/\Gamma,d_{G/\Gamma})\in\calm$, a function $F\colon G/\Gamma\to [-1,1]$ with Lipschitz constant at most $M$,
$g\in G$ and $x\in G/\Gamma$ such that
\begin{equation} \label{e5.6}
|\ave_{n\in [N]}\, f(n) F(g^n\cdot x)| \meg c.
\end{equation}
\end{thm}
We notice that the estimate in \eqref{e5.5} follows if we assume that the function $\nu$ satisfies the $(4^s,4s,1)$-linear forms
condition in the sense of \cite[Definition 6.2]{GT2}, but~\eqref{e5.5} is certainly easier to grasp. It is likely that one can
follow a similar approach in other instances of the transfer method, and replace the linear forms condition with a norm estimate
of the form \eqref{e5.5} for a suitable uniformity norm\footnote{In this direction we recall (see also \cite{CFZ}) that it is not known
whether for every integer $k\meg 3$ there exists an integer $s\meg k-1$ such that the relative Szemer\'{e}di theorem for $k$-term
arithmetic progressions holds true under the condition $\|\nu-1\|_{U^s(\zz_N)}=o(1)$.}.
\begin{rem} \label{r4}
Using Corollary \ref{c3.4} instead of Proposition \ref{p2.1}, it is easy to verify that Theorem \ref{t5.1} also holds if the majorant
$\nu$ satisfies $\|\nu-1\|_{U^s_4(\zz_{N'})}\mik \eta$, a condition which is slightly different from \eqref{e5.5}. However, the use
of the $U^{2s}(\zz_{N'})\text{-norm}$ in Theorem \ref{t5.1} is conceptually more natural in the present arithmetic context.
\end{rem}

\subsection{Preliminary tools}

As in \cite{GT2}, the proof of Theorem \ref{t5.1} is based on three ingredients. The first one is the inverse theorem for the Gowers
$U^s[N]$-norm~\cite{GTZ}. It~gives a criterion for checking that a bounded arithmetic function has non-negligible uniformity norm.
\begin{thm} \label{t5.3}
For every integer $s\meg 2$ and every $0<\delta\mik 1$ there exist a constant $M>0$, a finite collection $\calm$ of $(s-1)$-step nilmanifolds
$($each equipped with a smooth Riemannian metric$)$, and a constant $c>0$ with the following property. Let $N$ be a positive integer, and let
$f\colon [N]\to [-1,1]$ be a function with $\|f\|_{U^s[N]}\meg \delta$. Then there exist $(G/\Gamma,d_{G/\Gamma})\in\calm$, a function
$F\colon G/\Gamma\to [-1,1]$ with Lipschitz constant at most $M$, $g\in G$ and $x\in G/\Gamma$ such that
\begin{equation} \label{e5.7}
|\ave_{n\in [N]}\, f(n) F(g^n\cdot x)| \meg c.
\end{equation}
\end{thm}
It is more natural to formulate Theorem \ref{t5.3} for complex-valued functions which are bounded in magnitude by $1$; however, we will not need the
complex version of Theorem \ref{t5.3} for the proof of Theorem \ref{t5.1}.

To state the second ingredient, we first recall some definitions. Let $s\meg 2$ be an integer. Also let $N$ be a positive integer,
let $F\colon [N]\to\rr$ be a function, and define the \textit{dual uniformity norm $\|F\|_{U^s[N]^*}$ of\, $F$} by the rule
\begin{equation} \label{e5.8}
\|F\|_{U^s[N]^*}\coloneqq \sup\big\{ |\ave_{n\in [N]}\, f(n)F(n)|: \|f\|_{U^s[N]}\mik 1\big\}.
\end{equation}
We will need the following result which follows from \cite[Proposition 11.2]{GT2}.
\begin{prop} \label{p5.4}
Let $s\meg 2$ be an integer, let $(G/\Gamma,d_{G/\Gamma})$ be an $(s-1)$-step nilmanifold, and let $M>0$.
Also let $F\colon G/\Gamma\to [-1,1]$ be a function with Lipschitz constant at most $M$, $g\in G$ and $x\in G/\Gamma$.
Finally, let $N$ be a positive integer and let $0<\ee\mik 1$. Then there exists a decomposition
\begin{equation} \label{e5.9}
F(g^n\cdot x)= F_1(n) + F_2(n) \ \ \text{for every } n\in [N]
\end{equation}
where the functions $F_1,F_2\colon [N]\to \rr$ obey the estimates
\begin{equation} \label{e5.10}
\|F_1\|_{\ell_\infty}=O(\ee) \ \text{ and } \ \|F_2\|_{U^s[N]^*}=O_{s,M,\ee,G/\Gamma}(1).
\end{equation}
\end{prop}
We point out that, by \cite[Proposition 11.2]{GT2}, one can additionally ensure that the function $F_2$ in the above decomposition
is an ``averaged nilsequence" in the sense of \cite[Definition~11.1]{GT2}. We also note that the proof of \cite[Proposition 11.2]{GT2}
is non-effective and yields no estimate for the dual uniformity norm of $F_2$. However, explicit estimates can be obtained
by combining \cite[Lemmas A.2 and A.3]{M}---see \cite[Appendix~A]{M} for more details on this approach.

The last ingredient needed for the proof of Theorem \ref{t5.1} is the following version of Corollary \ref{c4.4} which concerns functions
defined on intervals of $\zz$.
\begin{cor} \label{c5.5}
For every integer $s\meg 2$, every $C\meg 20$ and every $0<\ee\mik 1$ there exist a positive integer $N_0$ and $\eta>0$ with the
following property. Let $N\meg N_0$ be an integer and let $N'\in [CN,2CN]$ be a prime. Also let $\nu\colon \zz_{N'}\to\rr^+$ satisfying
\begin{equation} \label{e5.11}
\|\nu-1\|_{U^{2s}(\zz_{N'})}\mik \eta.
\end{equation}
Finally, let $f\colon [N]\to\rr$ with $|f(n)|\mik \nu(n)$ for every $n\in [N]$. Then there exists a function $h\colon [N]\to [-1,1]$ such that
\begin{equation} \label{e5.12}
\|f-h\|_{U^s[N]} \mik \ee.
\end{equation}
Moreover, if $f$ is nonnegative, then $h$ is also nonnegative.
\end{cor}
\begin{proof}
It is a consequence of Corollary \ref{c4.4} and a standard truncation argument. Specifically, fix the parameters $s,C$ and $\ee$, and set
\begin{equation} \label{e5.13}
\alpha=\Big(\frac{\ee}{32 C}\Big)^{2^s} \ \text{ and } \ N_0=\lceil 2/\alpha\rceil.
\end{equation}
Moreover, by Corollary \ref{c4.4}, we select $0<\eta\mik 1$ such that for every finite additive group $Z$, every $\nu'\colon Z\to \rr^+$
satisfying $\|\nu'-1\|_{U^{2s}(Z)}\mik \eta$ and every $g\colon Z\to \rr$ with $|g|\mik \nu'$ there exists $w\colon Z\to [-1,1]$ such that
$\|g-w\|_{U^s(Z)}\mik \ee\alpha/(32C)$. We will show that $N_0$ and $\eta$ are as desired.

So, let $N, N', \nu$ and $f$ be as in the statement of the corollary, and let $\tilde{f}\colon \zz_{N'}\to\rr$ be the extension of $f$
obtained by setting $\tilde{f}(n)=0$ if $n\notin [N]$. By the choice of $\eta$, there exists $H\colon \zz_{N'}\to [-1,1]$ satisfying
\begin{equation} \label{e5.14}
\|\tilde{f}-H\|_{U^s(\zz_{N'})}\mik \frac{\ee\alpha}{32C}.
\end{equation}
We claim that $\|f-h\|_{U^s[N]}\mik \ee$ where $h\colon [N]\to [-1,1]$ is the restriction of $H$ on~$[N]$. Indeed, set
$l=\lfloor \alpha N\rfloor$ and let $2L$ be the least even integer greater than or equal to $N$; notice that $N\meg L\meg l\meg 2$
and $\alpha/2\mik l/N\mik \alpha$. Next, write $N'=2k+1$ and identify $\zz_{N'}$ with the interval $\{-k,\dots,k\}$. Let
$\varphi\colon \zz_{N'}\to [0,1]$ be the cut-off function which is nonzero on the set $\{-l+2,\dots,2L+l-1\}$, increases linearly
from $0$ to~$1$ between $-l+1$ and $1$, is equal to $1$ on $[2L]$, and decreases linearly from $1$ to $0$ between $2L$ and $2L+l$.
Observe that $\tilde{f}\varphi=\tilde{f}$ and so, setting $\tilde{h}\coloneqq H\mathbf{1}_{[N]}$, we have
\begin{equation} \label{e5.15}
\tilde{f}-\tilde{h}= (\tilde{f}-H)\varphi + H(\varphi-\mathbf{1}_{[N]}).
\end{equation}
Also note that the Fourier transform $\widehat{\varphi}$ of $\varphi$ satisfies the estimate $\|\widehat{\varphi}\|_{\ell_1(\zz_{N'})}\mik 4L/l$
(see, e.g., the proof of Lemma A.1 in \cite{FH} where this is explained in some detail). Hence, by the triangle inequality and \cite[(11.11)]{TV},
we have\footnote{Note that here we work with the complex version of the Gowers uniformity norm.}
\begin{equation} \label{e5.16}
\|(\tilde{f}-H)\varphi\|_{U^s(\zz_{N'})} \mik \|\widehat{\varphi}\|_{\ell_1(\zz_{N'})} \cdot \|\tilde{f}-H\|_{U^s(\zz_{N'})} \mik
\frac{4N}{l}\, \|\tilde{f}-H\|_{U^s(Z_{N'})}.
\end{equation}
On the other hand, since $H (\varphi-\mathbf{1}_{[N]})$ is bounded in magnitude by $1$ and is supported on a subset of $\zz_{N'}$
of cardinality at most $2l+1$, we obtain that
\begin{equation} \label{e5.17}
\|H(\varphi-\mathbf{1}_{[N]})\|_{U^s(Z_{N'})}\mik \Big(\frac {2l+1}{N'}\Big)^{1/2^s}\mik  \Big(\frac {3l}{C N}\Big)^{1/2^s}.
\end{equation}
Finally, note that $\|\mathbf{1}_{[N]}\|_{U^s(Z_{N'})}\!\meg \ave[\mathbf{1}_{[N]}] =N/N'\meg 1/2C$.  Thus,
by \eqref{e5.15}--\eqref{e5.17}, the triangle inequality and the definition of the $U^s[N]$-norm, we see that
\[ \|f-h\|_{U^s[N]}\mik 2C\, \Big( \frac{4N}{l}\, \|\tilde{f}-H\|_{U^s(Z_{N'})}+ \Big(\frac {3l}{C N}\Big)^{1/2^s}\Big). \]
By the previous inequality and taking into account the choice of $\alpha,l$ and the estimate \eqref{e5.14}, we conclude
that $\|f-h\|_{U^s[N]}\mik \ee$.
\end{proof}

\subsection{Proof of Theorem \ref{t5.1}}

We follow the proof from \cite[Proposition 10.1]{GT2} quite closely\footnote{Actually, there is a minor oversight
in the proof of \cite[Proposition 10.1]{GT2} which is fixed in the present paper. Specifically, the appeal to
Proposition 8.2 at the top of \cite[page 1796]{GT2} is invalid without appeal to the material from \cite[Section 11]{GT2}.}.
We first observe that, by compactness, for every positive integer $d$ there exists a constant $D\meg 1$ such that for every
$N\in [d]$ and every $f\colon [N]\to\rr$ we have $\|f\|_{U^s[N]}\mik D\|\widehat{f}\|_{\ell_{\infty}}$. (Here, we identify
$[N]$ with $\zz_N$.) Therefore, if $N\in [d]$, then Theorem \ref{t5.1} follows using as nilmanifold the torus $\rr/\zz$.
Thus, at the cost of worsening the constants, it is enough to prove Theorem \ref{t5.1} for every sufficiently large
positive integer $N$.

So, fix the parameters $s,C$ and $\delta$, and let $M,\mathcal{M}$ and $c$ be as in Theorem \ref{t5.3} when applied for $\delta/2$.
Next, by Proposition \ref{p5.4}, we select $K\meg 1$ such that for every $(G/\Gamma,d_{G/\Gamma})\in \mathcal{M}$, every function
$F\colon G/\Gamma\to [-1,1]$ with Lipschitz constant at most $M$, every $g\in G$, every $x\in G/\Gamma$ and every positive integer $N$
we have the decomposition \eqref{e5.9} with $\|F_1\|_{\ell_{\infty}}\mik c/12$ and $\|F_2\|_{U^s[N]^*}\mik K$. Finally, let $N_0$ and~$\eta$
be as in Corollary \ref{c5.5} when applied for $\ee\coloneqq\min\{\delta/2,c/(4K)\}$. We claim that Theorem \ref{t5.1} holds true for
$\eta,M,\mathcal{M}$ and $c/2$ provided that $N\meg N_0$. Indeed, let $N$ be an arbitrary positive integer with $N\meg N_0$, and let
$N',\nu$ and $f$ be as in the statement of the theorem. By Corollary \ref{c5.5}, there exists $h\colon [N]\to [-1,1]$ such that
$\|f-h\|_{U^s[N]}\mik \ee$; in particular, we have $\|h\|_{U^s[N]}\meg \delta /2$ and so, by Theorem \ref{t5.3}, there exist
$(G/\Gamma,d_{G/\Gamma})\in \mathcal{M}$, a function $F\colon G/\Gamma\to [-1,1]$ with Lipschitz constant at most $M$, $g\in G$
and $x\in G/\Gamma$ such that
\begin{equation} \label{e5.18}
|\ave_{n\in [N]}\, h(n)F(g^n\cdot x)|\meg c.
\end{equation}
Write $F(g^n\cdot x)=F_1(n)+F_2(n)$ with $\|F_1\|_{\ell_{\infty}}\mik c/12$ and $\|F_2\|_{U^s[N]^*}\mik K$, and notice that,
by \eqref{e5.18} and the triangle inequality, it suffices to show that
\begin{equation} \label{e5.19}
|\ave_{n\in [N]}\, \big(f(n)-h(n)\big)F_1(n)|\mik \frac{c}{4} \ \text{ and } \
|\ave_{n\in [N]}\, \big(f(n)-h(n)\big)F_2(n)|\mik \frac{c}{4}.
\end{equation}
The first part of \eqref{e5.19} follows from the fact that $\ave\big[|f-h|\big]\mik \ave[\nu+1]\mik 3$ and  $\|F_1\|_{\ell_{\infty}}\mik c/12$.
On the other hand, by the choice of $\ee$ and $h$, we have
\[ |\ave_{n\in [N]}\, \big(f(n)-h(n)\big)F_2(n)|\mik \|f-h\|_{U^s[N]}\cdot \|F_2\|_{U^s[N]^*}\mik \ee K \mik \frac{c}{4} \]
and the proof is completed.

redpajama_set_name: RedPajamaArXiv


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text:
Innovate UK awards £50,000 each to 15 female entrepreneurs
Written by Jo Cook on 18 November 2016 in News
Female entrepreneurs are brought to the forefront with a tailored business support package
Innovate UK is continuing its mission to challenge disproportionately low numbers of women entrepreneurs in the UK, with the announcement of the 15 winners of its infocus Women in Innovation competition.
Each has been awarded £50,000, a tailored business support package and the chance to be paired with a business mentor.
More than half of businesses admit 'widespread problem' with UK's gender pay gap
Government's focus on numbers puts skills reform at risk
Government needs a new digital approach, says think tank
Government must improve careers provision to combat skills gap
Innovate UK is a government-backed agency that aims to drive the science and technology innovations that will grow the UK economy. The competition was launched in May 2016 and set out to address the problem that very few funding applications to Innovate UK come from women. Boosting female entrepreneurship could deliver approximately £180bn to the UK economy.
As well as providing much-needed funding, Innovate UK's award will help break down a major barrier to female entrepreneurship – having limited access to relevant role models and identification with entrepreneurs as 'women like me'.
Following an exceptional number of applicants, the 15 winners (who were selected from 34 finalists), were announced at London's Royal Society of Arts on 15 November 2016.
Dr Ruth McKernan CBE and chief executive of Innovate UK said: "It is very clear that harnessing the talent of women entrepreneurs could significantly enhance UK economic growth. I am delighted that we are taking action; supporting and funding female entrepreneurs to help them succeed and inspire other women to come forward, apply for funding and turn their ideas into successful business."
The winners' creations certainly ticked the 'innovative' box. "From air quality monitors, to fully autonomous drone software, affordable solar power technology and a digital education portal which engages school children with the molecular world, the calibre of winners is an eye-opening view into the talent and vision of the UK's female entrepreneurs," said Dr McKernan.
Asked how we can inspire more female entrepreneurs to come forward and apply for funding, winner Natwilai Utoomprukporn of Gettrik Ltd said, "My advice to any women with an idea is to dream big! You have the opportunity to turn this innovation into a business, you just need the support, but without actually starting you don't know what your full potential is."
Science, technology and research
Society and welfare
Six steps to support neurodiverse talent at work
In another article on supporting wellbeing at work Cass Coulston and Ricardo Twumasi examine neurodiversity
As lack of motivation and engagement grips organisations, Giora Morein suggests that agile methodology could be the way to re-energise flagging employees
Internal Recruitment: How can you overcome the potential pitfalls?
Vincent Belliveau, Senior Vice President & General Manager EMEA at Cornerstone OnDemand, explores the benefits of internal recruitment
Building fulfilling careers for young people
We need to do a better job of preparing young people for the world of work, so they can make informed choices and build fulfilling careers.
Free resource: Apprenticeship update 2018
Download TJ's 2016 apprenticeships guide here.
White paper: Save your company from boring elearning
Download this white paper to discover how to save your company from boring elearning.


Job security is on the minds of many with the current state of our economy and political climate, but add the threat of artificial intelligence (AI) and there is good reason why you may be concerned for your future livelihood.
I will admit that I started out writing this piece with slight bias as I am what you might consider 'technologically-delayed'. I don't use an app to order coffee, I avoid self-checkout situations at all costs, I am uneasy about a future with driverless vehicles and I detest automated call support. I still use a pen sometimes and I, for the most part, prefer human contact. I typically resist technology until I'm forced to use it. But over time I have come to rely on my Roomba to vacuum, I often attend online courses in virtual classrooms, I depend on my Google Home to turn on music or adjust the thermostat, and I Google anything and everything. I still have a job because my job is people centric, but it looks very different from what it did 20 years ago.
The truth is, AI has been around for 70 years and we are now experiencing what they call the 'Fourth Industrial Revolution'. AI is changing the way humans exchange and distribute value and it is predicted to spark societal transformation on a global scale.
There is nothing new about continual change. The Canadian farming industry is a perfect example. In the year of confederation Canada already had hundreds of agricultural patents for improvements of farm equipment and machinery to improve productivity. Back then around 90% of Canadians lived in rural communities (now 18.9%). A decade ago agriculture was the single most common occupation, employing 1 million Canadians and accounting for 33% of jobs in the country (now 1.8%). Today, less than 2% of Canada's population operate farms. There are way fewer farms, but the farms that exist have greatly expanded in size. The equipment has become exceedingly efficient, albeit incredibly expensive. So, case in point: fewer farms (and operators) but farms larger in area; reduced exhausting manual labour; greater crop production, unprecedented efficiencies; but significantly fewer jobs. Yet, Canada remains one of the top agricultural exporters in the world. Is this good or bad or both? Depends on who you ask.
What's new is the PACE of change. Think of the music industry and how quickly it has changed. How many ways in your lifetime have you gone about listening to and buying music? How different is the way that musicians and producers earn a living? Do album cover designers have the same skills now that they did 50 years ago? How many record stores exist? How about the telecommunications industry…will your grandchildren know what a land-line is? Up until a few years ago we all had one. What about a pager, a flip-phone, a car phone, a Blackberry? There will always be sales people (however, there is virtual sales) but they have to adapt with the times and learn new technologies in order to sell the products. The examples are endless, but when thinking of any industry, company, product or service consider what jobs have stayed the same (probably very few), consistently changed or been completely eradicated.
It is the pace of change that concerns the adversaries of AI the most. This pace is unprecedented and some predict AI will be the detriment of society as we know it…disrupting labour markets, displacing workers and intensifying the disparity of wealth.
Just a few months before Stephen Hawking died, he spoke at a conference and he believed that the emergence of AI could be the "worst event in the history of our civilization" unless society finds ways to control its development. Although he realized the potential of AI to help reverse damage done to the environment and its potential to eradicate poverty and disease, he felt that unless we are responsible with AI management and identify the dangers, he could foresee "powerful autonomous weapons, or new ways for the few to oppress the many. It could bring great disruption to our economy." Similarly, according to Elon Musk, AI is a "fundamental risk to the existence of human civilization", this coming from the CEO who clearly speaks from both sides of the fence. However, he touts a consistent message…that AI will cause widespread job losses and potentially start World War III.
Some proponents see limitless opportunity in AI and believe that even though jobs will be lost, new ones will emerge from new technologies and this will allow new opportunities for people to develop new skills and create safe and rewarding jobs. Some see AI as a great way to let technology perform the jobs that people don't really want to and allow us to do more with less.
Bill Gates sees AI as "just the latest in technologies that allow us to produce a lot more goods and services with less labor. And overwhelmingly, over the last several hundred years, that has been great for society". However, he acknowledges that it will be problematic for labour markets to keep up with the pace as AI develops and the key challenge will be how to figure out how to retrain workers and distribute benefits in a new economy.
Some suggest that a universal income will be the only way to sustain society when a huge portion of workers will be displaced and for those who become 'unemployable', mass unemployment insurance payouts won't make sense in the new economy. This idea is already being tossed around in Canada, but here they refer to it as 'National Basic Income' which would guarantee a minimum income for all. A pilot project is currently underway in Ontario with the hopes that it will prove to increase food and housing security, improve stress and anxiety, mental health and healthcare and assist with the employment labour market.
My research shows that in the future it is certain that most unskilled jobs will be replaced by AI. Those are the jobs that are manual or process-driven, for example manufacturing or transportation. But AI can't replace creativity, human interactions, team work, negotiations or people care. Occupations that require these skills will survive. The good news is that many new jobs will evolve or emerge because of AI in the future, jobs we can't even fathom right now. It takes a whole lot of people to develop this' machine intelligence', in fact Canada has become a global leader in AI and has seen a 500% growth in AI job opportunities since 2015. Some say that children entering primary school today will end up working in jobs that don't even exist right now. Just think, jobs like Social Media Manager, Uber Driver, App Developer, Cloud Computing Specialist, Driverless Car Engineer and Drone Operator didn't even exist until recently!
I believe that most companies will have plenty of jobs to offer but they will be redesigned jobs. Savvy companies will retain their best employees and invest in reskilling or upskilling them to prepare them for the new wave of AI. There is no doubt that educational institutions, industries and governments will be forced to transform in order to prepare us for sustainability, it's already happening. But we also need to prepare ourselves and our children for the rapid changes on the horizon.
How can you determine if you will be "technologically unemployed"?
Yes, 'technologically unemployed' is a new and real buzz phrase. First off, I want to point out that jobs have vanished throughout history…think: Milkman, Lamplighter, Switchboard Operator, Movie Rental Store Worker, Typesetter, Elevator Operator, etc. And we are now seeing less: Receptionists, Assembly Line Workers, Farm Workers, Store Clerks, Tram Drivers, Fast Food Attendants, etc. We book our own travel, enter our own expenses and manage our own health benefits in the name of efficiency. Jobs becoming obsolete is nothing new. Products becoming obsolete is nothing new. However, a 2017 report by KPMG forecasted that technology will perform the job equivalent of 120 million employees by 2025 (over 3 times the population of Canada). Other studies, including research from McKinsey & Company, estimate that 45 to 47% of today's jobs are at risk of being automated.
There is an interesting website called www.willrobotstakemyjob.com where you can enter your occupation and it will provide you with your risk level, projected growth and median annual wage. It also provides a rankings list of jobs with the highest/lowest risk of being replaced by a robot. The two founders of the site have been featured in Fortune and Business Insider. I can't verify the exact science or accuracy behind the site, but you could ask a robot if it's legit 😉.
What you need to do right now is assess your job and your skills. Where do they fall? In the process-driven, manual, manufacturing type category? If so then start to develop new skills in the areas that will be less affected. Be open to development no matter what or you will fall behind.
What if your employer has no training and development plan? What if you are unemployed and trying to navigate this revolution? What if you've dedicated yourself to the same career, trade or job for decades? Or, what if you realize that you are in the unfortunate position of losing your job to a robot?
Experts suggest that lifelong learning has now become a necessity, rather than an option. Prepare yourself by developing new skills, become an expert in a certain area, keep up with news and trends, plan on investing in learning throughout your lifetime. It is said that those who are able to adopt new skills and continually learn and change with the times will be less likely to be replaced by AI. Remember the saying "survival of the fittest"?
It's hard to predict the future, but I think it's safe to say that yes, robots and AI will continue displace our jobs. New technologies will continue to alter our planet and our way of life. I will leave you with a quote by the Founder and Executive Chairman of the World Economic Forum, "I am convinced of one thing—that in the future, talent, more than capital, will represent the critical factor of production." – Klaus Schwab.


text:
HOME --> CLASS MAMMALIA --> ORDER RODENTIA --> SUBORDER MYOMORPHA --> SUPERFAMILY Muroidea --> FAMILY Cricetidae --> SUBFAMILY Arvicolinae --> GENUS Arvicola
SPECIES Arvicola sapidus
Author: Miller, 1908.
Citation: Ann. Mag. Nat. Hist., ser. 8, 1: 195.
Common Name: Southwestern Water Vole
Type Locality: Spain, Burgos Prov., Santo Domingo de Silos.
Distribution: Portugal, Spain, and France.
Status: IUCN – Lower Risk (nt).
Although arranged as a subspecies of terrestris by Ellerman and Morrison-Scott (1951), A. sapidus is strongly differentiated in morphological, morphometric, and chromosomal traits (Panteleyev, 1996; Reichstein, 1982a; Ventura and Sans-Fuentes, 1997). Phylogenetic analyses of cytochrome>b segregated A. sapidus from European samples of A. amphibius at a high level of percent sequence divergence (Taberlet et al., 1998, as terrestris).
Reviewed by Hinton (1926a), Corbet et al. (1970), Corbet (1978c), Reichstein (1982a), and Mitchell-Jones et al. (1999). New information covering morphology, discrimination from A. amphibius, and ecology supplied by Garde et al. (1993), Garde and Escala (1994, 1996, 1999), Ventura (2000), Ventura and Gosalbez (1990, 1992a), and Ventura et al. (1989, 1994). Population in NE Spain documented by owl pellet remains (Torre et al., 1996); distribution in N Spain and considerable range overlap with A. amphibius covered by Castien and Gosalbez (1992).
musiniani (Lataste, 1884)
tenebricus Miller, 1908


Maria V. Snyder has done it again with another brilliant series. For those who may not know, Maria wrote the Study Series which followed a young woman, Yelena, who discovered she has magical talent. Well, since I am talking about the new series, I don't want to get too into that one. If you stop by her website, you can find information and book descriptions for her first three books from her first series.
Her new series still takes place in the same world as Yelena. A few years later, Yelena even stops by in the last few chapters. Along with a few other characters from her first trilogy.
Those are really GREAT reads. Maria's books brings you into this whole other magical world. All books draw you into their world and most of the time, the ones that I read, are already in our current world and time, things are just different with vampires lurking about. There are no vampires in this series, this is a strictly magical series.
Going back to the Glass Series, Opal is a glass-maker and a magician in training, but she feels like maybe she is only a one trick magician, meaning she can only do one thing really well. Opal has the power to do amazing things with glass blowing, she is even able to trap bad guys' souls into them or create glass object so other magicians can communicated to one another from long distances. No need for cell phones!
But something is going wrong with the Stormdancer clan's glass orbs, powerful magicians are dying. The clan needs Opal's help, especially with her particular talents. But the things go wary when she reaches the clan and naturally things are a lot worse than she was originally led to believe.
Opal soon discovers that she has a different kind of power within her and that it can be dangerous, to the people around her and to herself if the wrong people were to find out about it. So now she must keep a secret from her most trusted adviser and few friends.
To make things even MORE interesting and intriguing there is a conflict of love interests going on the book. Opal is torn between two men, both who have the power to steal her heart. She cannot determine who would not break it if she were to hand it over and the results are shocking! You will never see it coming, unless you look (read) very closely. I caught on pretty quickly and was mostly right about the turnout. I won't delve further into my guesses because then that would ruin the book for you.
Maria is a wonderful author and I love reading her books. I even had the pleasure of meeting with her last year at the Romantic Times Book Convention 2008. I was able to talk with her about her writing career and how she got started since I want to be a writer myself someday soon. She had to go through several rejections before finally getting published by Mira books. Determination is the key, you have to keep going for your dream no matter how many rejections you get. I have taken this advice to heart and think about it constantly as I am trying to write my first novel. I again say, that she is a wonderful and thoughtful person who I still talk to through email every now and then.
So go now, read her books!!
HBO's True Blood series is actually based on Charlaine Harris' Southern Vampire Mystery Series?
It is probably common knowledge by now, what with the DVD release today and Season 2 kicking off next month. I don't have HBO but bought the DVD today and plan on watching it all tomorrow since I don't have to work.
I already know, from hearing others talk about what it's like that it is really like the books. Some of the characters are vastly different in the show than they are in the book. I heard someone mention Tara being really "crazy" on the show and in the book, she is nothing like that. She was once into the whole vampire thing, but after a serious incident she steered clear of them and went back to being plain old human Tara, instead of the Tara who hangs out with vampires.
Why is it that Hollywood, to put it in basic terms, always messes with the book when making it into a TV show or movie? I understand editing scenes out for time purposes, although, I can say for myself if the movie would be exactly like the book, keeping in all the important and not-as-important scenes, I would stay in the theater no matter how long the movie is. Everyone knows by now that the Harry Potter movies are nothing like the books, I mean come on where the hell has Dobby been since the second movie? He has a critical role in the last book, a whole chapter is dedicated to him. What is it going to be Neville instead?
But that's Hollywood. I am fully prepared for there to be extreme differences from the actual series, which are wonderful to read by the way. If you loved the show, check the books out, they are fabulous. They even have a pack of the first 7 books in paperback. Charlaine is up to book 9 that released just a few weeks ago. There is to be another set of paperbacks releasing books 1-8 later this year. Or if you really love the show, they are revamping the covers to have pictures from the show on it instead of the cartoon drawings that are currently on the covers.
Speaking of vampires and televison shows along the line of books, since this is a reading blog, there is very vague news that L.J. Smith, another fantastic vampire author, this one for YA, who wrote the Vampire Diaries which had recently been republished with a brand new book starting a new Vampire Diaries series, so to speak, called The Vampire Diaries: The Return: Nightfall. This series is supposed to focus more on the character Damon, apparently. At least that was what I heard. I read the series years ago, back when they were 10 years old and now they're over 20, wow! It's a good series, high school and vampires pre-Twilight. I read this long before I read Twilight, and I liked Twilight.
Anyway back to the news, there is supposedly a possible TV show in the making of this book series. The CW is up for making the pilot to pitch it to the people on top, whatever that means, and if it goes okay, we might have ourselves another vampire television show based on a vampire book series! Yea! I am super excited to hear about this and hope whoever it is that makes the BIG decision gives it the okay. There's a chance it might since Twilight has become so big. Ian Somerhandler is supposed to be playing Damon, the bad guy. I think he's perfect for that role, because he does have a sort of rugged bad boy look. Don't ya think?
Again if you haven't read the first or second one yet, do NOT read this post any further for there are more spoilers!!!
The third and current installment in the Night Huntress Series. The fourth novel is due out this July. Yea!
In this one Cat and Bones celebrate the holidays, with a battle that takes place in the dead of night. Zombies attack! Seriously!
A woman from Bones' past has come to reek havoc on his life and those around him. Which puts Cat into danger and not even her training she learned as a special agent can save her. She is going to have to embrace the vampire half of her life.
On subplots this book is pretty good. Cat and Bones are married in the eyes of vampires, and Bones is now set to make the union legal in the eyes of humans. Yes, Cat and Bones are planning their wedding, although the plans are put on hold when chaos erupts.
Tate, once Cat's best guy pal, is still vying for her affections despite Bones' many warnings and despite the fact that she loves Bones and is married to him, in one way. Tate takes on certain obstacles and changes his life around in order to make himself more appealing to Cat. He is determined to win her affections once and for all.
While reading the book, you can tell what Cat is going to decide. So be prepared. Cat's father makes another appearance in the book and the results from that are shocking. Not to mention the fact that Cat must tell her mother, who hates vampires with a fiery passion, that she is going to marry Bones.
Her mother has a better role in this novel than she did in book 1. We see more of her as Cat feels she would rather be the one to protect her at her side while this thing with the woman from Bones' past takes place.
Out on shelves on July 28!
Okay if you haven't read the first book, don't continue reading this, for there will be some spoilers.
This second installment of the Night Huntress series takes place a few years after the first one. Cat has joined up with the special unit in the FBI and hunts down more of the bad vampires and other supernatural bad guys. Although she still cannot forget the vampire that she loved and left behind in order to protect; Bones.
It doesn't help that Tate, one of her men on her team is starting to show more affectionate feelings towards her. She doesn't want to be romantically involved with him, but he is determined to win her.
When Cat attends a wedding of one of her friends, who does she standing across from her on the groom's side. Why none other than Bones. Cat begins to panic, wondering how he was able to find her after so long.
But that is nowhere near the beginning of Cat's problems. Someone wants her dead. Permanently. And they're willing to pay big for it. Bones is there to protect the woman he loves who left him without a word.
Cat and Bones' relationship grows stronger than ever when an old flame comes to aid Cat at Bones' request. Cat learns some things about Bones' past. She is also introduced to a large group of vampires and takes part in an important ceremony among the vampires. She also comes face to face with her father.
While a whole lot more happens, I do not want to tell the whole story here. Even the last bit might have been too much of a spoiler, but never fear, I end my rambling here.
Okay so I am super behind on blogging on my favorite books and such. I've been busy.
So I want to talk about some books I have read and this is one of them. Halfway to the Grave is the first in the Night Huntress series and is a fabulous read!
In it is Cat who we learn right away is half vampire and half human. Her mother was raped by a newbie vamp and it left her mother hating vampires. Vampires are still a secret to the majority of society and when Cat learns of her father's true identity she begins hunting vampires. She was only a teenager when she killed her first one and is determined to hunt down her father and kill him for what he did to her mother.
Along the way she kills any vampire she can get to. Her mother is pleased by this. But then there is one vampire, Bones, who outsmarts her and is about to kill her, but then he discovered her secret, that she is half vamp. The two then form a union of sorts to hunt down a different vampire that is bad, obviously, but Bones promises that he will use his sources to find her father.
Since this is found in the romance section of the bookstore, you can probably guess what happens between the two. The ending is the ultimate shocker, one that I will not reveal. You'll just have to buy the book for yourself!


This is the second time I visit Paris and I feel it's like the first time. I remember when I went back in 2009 it was around the same time of the year, cold and rainy... but it didn't matter to me because I would go there regardless of what weather!
Everything looked very new to me; I'm not talking about the monuments and museums, but the shops have a totally different feel. It's a city that renews itself day after day. I imagine that because of the huge amount of tourists it gets yearly.
This is some of the pictures I took. All of them make me so happy. I hope they make you happy too!
Christmas decorations are a must in Paris and they are to die for!
Can you believe that you can order this as a floral arrangement?
This cute drawer cabinet was cover with a floral tapestry fabric. What a great idea to use our stash!
They even make these grapes look more delicious. Artistic decorations everywhere!
I'm glad you had a wonderful trip!! I love all the photos -- I'm feeling inspired!!
Thank you my lovely friend, for being by my side!!
Beautiful pictures! Such wonderful inspiration too! Congratulations on your new blog! I love your fabric!
Oh what an inspiration in these photos - thank you so much! LOVE that you have a personal blog too now and will enjoy following you!
Thank you so much, hope you can follow me along in this fun journey!
I live near Limoges, at Christmas the flowers are lovely, they dip the tips in gold or spray them slightly with gold... so pretty.
How fun, I saw this in Paris too!!
Beautiful, inspirational pics, can't wait to see more!!!So glad you had a wonderful time!!!
Beautiful is the only thing I can think of.
How nice to read you like it Margaret! This will be my little nook where I share my inspiration to design and make things for you all. Thanks for your love!
aahh Paris! My husband whisked me away for Christmas there when we'd only been dating for 3 months. Wonderfully special place for me - we bought our wedding rings while we were there and married the next February. Truly the city of love!
Oh Lucy, what a romantic story!! You will treasure these memories all your life, and... who knows... maybe you can go back again to celebrate one of your anniversaries:) Thank you for commenting!!
Congrats on your new blog. These pics reminded me of my trip to Paris, which was over 20 years ago. I hope to go back one day myself. Such a beautiful place.
That's true Janet, Paris is a city made for tourists to come back over and over... There are so many things to see!
Oh my gosh, how beautiful. I can only dream about ever going on such a wonderful trip. I'm so glad to see you with you own sweet little blog home.
Congratulations on your "own" blog! I think we're very lucky to be able to get a little bit closer to you in this way!!
These are great pictures and looking at them made me very happy!
I love Paris and I've been there many, many times ( not to much of a distance from where I live). I feel very much at home there, but that feeling is very different from the at-home-feeling I have here in the Netherlands. I totally agree with you saying it inspires you the moment you're there. The energy, the dynamic of the city is contagious, at least it works on me that way. The importance of beauty and culture is evident in the french society and it can be found in the most simple details like the grapes on a death trunk of a tree at market. The French value their history and their culture (attitude heavily criticized in this country ) and I think this inspires them to live up to it, to maintain that reputation of doing and creating memorable things, which does not necessarily mean continuing in the same way. I think it's this energy, this creative urge that artists like you, plug into when visiting this beautiful city.
These images literally took my breath away. I tend to get that way when I see beauty. You are so very, very fortunate to be able to do something that you obviously love so much, and are clearly good at. Thanks for sharing so that I can live vicariously through your travels and lovely photos!
Absolutely lovely eye-candy Thankyou! ... don't the French have great linens....too nice to wipe the dishes.
Cant wait to get your blog updates...I'm sure they will all be so pretty.
Hi Salley, welcome to my blog! The dish towels are to die for... They sell so many because they don't use that much paper towels as we americans do, lol! Is very funny how each country have different habits, right?
Fabulous pictures Pat!! I love all the details you captured. We just went to Paris in October - my first time in 19 years! I was even more fabulous than I remembered. We only had a day and a half and now I am dying to go back again and spend more time!
I can see where you get your inspiration from. Beautiful photos.
WOW...beautiful pictures! I love how many different textiles there are. Oh, and the roses are like no other I've ever seen before. Great pictures of inspiration indeed!
Beautiful, I am so glad you have made your own blog, you create such wonderful colors in your collections. I am visiting from Maureen Cracknell, I am so glad she sent me.
Beautiful pics...brings back fun memories of visiting there with my DH...thanks for sharing!
totally amazing images!!!!!! thank you for making my day!


ARTIVIST. CHARANGOPUNK. MUSICIAN. SONGWRITER. LGBTQ+.
MONIQUE "HONEYBIRD" MIZRAHI – Multi-instrumentalist (charango, bass, guitar, vocals), singer-songwriter, and LGBTQ+ artivist. Born and raised in Los Angeles she found her musical calling in Italy, and is currently based in Bologna. After graduating Berklee College of Music she worked in Hollywood recording studios with Primus, Rage Against the Machine, Elliott Smith and countless others. She has studied charango with Argentinean maestro Patricio Sullivan. As Honeybird she has brought her music around Europe, Latin America, Canada and the US. With trio honeybird & the birdies, she performed at Primavera Sound (Spain), Eurosonic (Holland), Cherchez La Femme (France) as well as tours in Italy including a performance at Primo Maggio in Rome.
In 2015 she recorded a solo album Out Comes Woman chronicling her coming out experience as bisexual, produced by Marc Urselli (John Zorn, Laurie Anderson, Mike Patton). In the US she received the steve ben israel commission to create a yiddish pacifist ritual for BRIC; an unique invite to present Bi at the Brooklyn Museum for Women's History Month; and a prestigious invite to perform at the US White House under the Obama Administration for the Bisexual Community Briefing. Two wild tours and an EP release (WM) with charango-punk trio Honeybird & The Monas. She spent much of 2016 touring with popular Italian rock band Tre Allegri Ragazzi Morti playing charango, guitar and singing backing vocals in an extensive summer and winter tour. Performances took place the following year in Vancouver, Canada at Too Queer: A Bi-Visibility Cabaret and around California. In 2017 and 2018 she was visiting artist at Berklee College Music, Valencia, Spain, and presented Artivism and LGBTQ+ themed topics to faculty, staff and students. Her mission is to listen to intersectional LGBTQ+ community voices and transform the daily struggle into song.