text:
Film review: Isabelle Huppert stars in controversial rape-revenge comedy 'Elle'
Monica Reid
Tue 21st Mar 2017 09.00 GMT
It is understandable that Elle is a controversial film. Its central act, a woman being violently raped in her home, is approached in a distinctly and deliberately unusual way which is sure to offend at least a few members os every possible school of thought on the subject of rape, although some have called it, and the novel it was based on (Oh, by Phillippe Dijian) empowering to women.
The main character, Michelle (Isabelle Huppert), would normally be described as the victim of the rape; but Michelle is hard to categorise as a victim. Huppert describes her character as a woman who refuses to be a victim. Neither does Michelle take on the other typical role of a rape victim: that of a woman seeking justice or vengeance. "A revenge movie," director Paul Verhoeven joked, "would be the American way of doing it." The rape was horrifying, but Michelle refuses to let it impact her life – and not in the manner of a third type of standard movie rape victim, the woman who foolishly tries to repress her trauma, but successfully. Paul Verhoeven, calls Elle "my protest against genre."
The film begins with a blank screen and the sounds of what seems to be aggressive sex. After a second or two, it becomes clear that the female voice is crying out not in passion, but in pain or protest. The audio portion is brief, and the first visible scene is the aftermath of a rape, with a woman lying on the floor, her clothing is disarray, broken glass and shards of crockery all around her, while a man in a ski mask quickly leaves through an open window. Almost immediately, the film thwarts expectations: the woman sits up and, after a moment's reflection, begins calmly sweeping up the shattered dishes. She also throws her clothing in the trash and bathes, but these natural responses to a rape are done calmly. There is no call to the police to report the assault.
We follow Michelle to work the next day; she is founder and CEO of a video game company. She is not only successful, but confident and assertive, managing the sometimes resentful young game developers she oversees with ease. As Michelle continues to go about her daily business, we get a clearer idea of her character. She is firm and no-nonsense, even with her son, whom she loves but finds exasperating. She is by no means cold, as shown by her concern for friends, and for an injured bird she tries to revive; but she is cynical and cautious. The reasons for her attitude emerge gradually, leading us to slowly piece together information about a significant and traumatic event in her childhood.
We eventually see the actual assault in flashback. It is graphically violent and brutal, beginning with the masked man forcing his way into her home. This information makes it all the more shocking that Michelle is going about her business not only normally, but with a cynically humorous attitude to the events of her day. She finally mentions her attack to some friends at lunch; they are astonished that she is so calm, and that she hasn't gone to the police. Michelle's reluctance to involve police is an additional clue to her past; she refers more than once to having lost confidence in the police's ability to deal with such things appropriately. Michelle takes appropriate measures to make her home secure, has a STD scan, but these things are tossed in with her daily chores, performed with no strong emotion.
As Michelle deals with mundane issues, visiting her elderly but reprobate mother, helping her under-employed son find an apartment for himself and his pregnant girlfriend, her dry sense of humour continues to add comedy to scenes of her daily routine. So much casual humour in a film about rape seems inappropriate, even subversive. The film further pushes boundaries by having Michelle retain her normal interest in sex after the assault. The reality of the rape itself is not covered up, however: Michelle mentally revisits the scene, and we see it replayed, this time with Michelle, in her own fantasy, breaking free and beating her assailant unconscious with a blunt object. She never verbally identifies her feelings; the closest we come to such a statement is a scene in which Michelle walks into work, passing a group of test subjects playing a violent video game. The test supervisor, taking notes, asks them, "What do you feel? Fear or anger?" The groups replies in unison, "Anger."
The suspense steps up a bit when Michelle begins receiving threatening text messages, apparently from her attacker, and the story becomes something of a mystery, in which Michelle attempts to identify the rapist and track him down. At the same time, the mystery of Michelle's childhood trauma continues to develop, as we discover that her father has been in prison for some time, and more information emerges. From here, the double mysteries develop, incongruously side by side with Michelle's ordinary life with family and at work.
The final portion of the film may be the most difficult. Without revealing the details, or the highly problematic identity of the attacker, Michelle manages to track down and finally identify her rapist, and once again takes an approach which defies genre, one which is hard to pin down but could be interpreted as either revenge or self-victimisation. It seems to me that Michelle is risking a great deal to take control of her own assault; but her approach is unconventional in the extreme, revolting against any kind of common understanding of rape, violence, or control. She may be said to have turned the situation into a game, like one of her own violent video games, so that she can enter it and win on her own terms.
The film winds down as Michelle ties up loose ends regarding her mother, her son, and even the father who has been in prison all these years. She confronts and brings closure to the terrible events she went through as a girl. Finally, she also brings the matter of her assault to an end. In a deliberately ambiguous scene, Michelle deals with the threats from her rapist, but it is unclear whether the conclusion is accidental, or planned by her in advance.
Isabelle Huppert, commenting on Elle, has said that the simple rape revenge fantasy, common in film, is a caricature rather than a realistic portrayal. Elle's director has claimed he wanted to avoid seeing the event through "an artificial framework." Yet the realism of the rape victim's story in Elle is also in question. It may be best, as Huppert suggested, if taken as a fable rather than a realistic account. As for the comedic element, a risky choice in a film about rape, it is certainly no more callous or inappropriate than comedy in a story about cancer, or AIDS, neither of which are unusual or controversial – provided that the humour comes from the patient. In Elle, the humour is all Michelle's own. She doesn't lose her sarcasm or sense of irony when she is raped, any more than she loses her courage or sense of self, and that, at least, is a positive message.


Interested in learning more about Avaya Networking gear? Love baseball? If so, ScanSource Catalyst and Avaya have the event for you and your customers!
On September 12th, resellers and end users are invited to attend the Beer & Gear event in Miami, Florida. During the morning expo, attendees will learn about the exciting updates to Avaya's Networking portfolio centered around Shortest Path Bridging led by Avaya Director-Distinguished Engineer, Ed Koehler. ScanSource Catalyst will also provide information around the benefits offered through the Distribution channel, specifically around the tools and programs offered through ScanSource Catalyst. Special guest, Erick Gomez, Systems Support Manager at Miami-Dade County, will speak to the audience about plans for deploying Avaya SPB as the county's core network.
After the morning expo, attendees are invited to continue the event at the Marlins stadium for the Miami Marlins versus the Atlanta Braves baseball game. This will be a prime opportunity for resellers and end users to connect and network in a fun and casual environment.
Join ScanSource Catalyst and Avaya to explore the possibilities and economics that Shortest Path Bridging and IP Multicasting can bring to your business so you can reduce costs, optimize business capability, and improve customer service. Resellers and end users are both encouraged to register with Avaya as soon as possible to take advantage of this great opportunity for networking and fun!


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Tools for accessibility
HomeAccueil > Tools for accessibility
The Liberal Party of Canada and Disability Community Involvement
Accessibility is not only an issue of basic human rights, but also one of citizenship. In recognition of this, and to commemorate the International Day of Persons with Disabilities, the Liberal Party has set a new precedent for government in Canada.
By joining forces with eSSENTIAL Accessibility™, we are taking part in a global Disability Community Involvement Initiative launched in association with leading advocacy groups for the disabled.
If you have trouble typing, moving a mouse or reading, eSSENTIAL Accessibility™ can greatly enhance your experience at liberal.ca.
Click here to download keyboard and mouse replacement solutions free of charge.
Keyboard and mouse replacement solutions include a webcam-based hands-free movement tracking system, a page reader, predictive typing and other customizable options.
eSSENTIAL Accessibility™ tools can be used on a standard PC* by anyone with dexterity challenges or reading difficulties arising from a variety of conditions, including:
Stroke/paralysis
Limited proficiency in English
Literary deficiencies
Other learning issues
eSSENTIAL Accessibility™ is enhanced communication support for people who have trouble reading, typing or moving a mouse. It helps us meet customer service, employment, and information and communication standards mandated by the Accessibility for Ontarians with Disabilities Act (AODA).
The Liberal Party of Canada is proud to be a Founding Member of this Disability Community Involvement Initiative. It brings us one step closer to mirroring our society's diversity in the people we employ and serve.
* Using a Mac? Click here to learn about using eSSENTIAL Accessibility™ on an Apple computer.
Plan for Accessible Customer Services
The Liberal Party of Canada has developed a plan in accordance with the Accessibility for Ontarians with Disabilities Act to ensure that the services offered by our Party are done in an inclusive way in accordance with the Act. If you would like a copy of this plan, please feel free to contact us through any of the following means.
Phone: 1-888-LIBERAL (1-888-542-3725)
Email: assistance@liberal.ca
If there is a major service outage, a notice will be posted outlining alternative avenues of contact for the Party. This notice will be at: http://www.liberal.ca/essential-accessibility/
Let us know if you have any feedback on our accessible customer services. If follow-up is needed, we will get back to you as soon as possible.

redpajama_set_name: RedPajamaCommonCrawl


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text:
WENDY DIO WIFE OF LEGENDARY METAL VOICE RONNIE JAMES DIO CHATS ABOUT NEW AUTOBIOGRAPHY 'RAINBOW IN THE DARK' EXCLUSIVE ZOOM INTERVIEW WITH RAY SHASHO
EXCLUSIVE ZOOM INTERVIEW
W E N D Y D I O
WIFE & MANAGER
of LEGENDARY/METAL/ROCK/SINGER
R O N N I E J A M E S D I O
R a i n b o w i n t h e D a r k
The Autobiography
by Ronnie James Dio (Author), Mick Wall and Wendy Dio
WENDY DIO
EXCLUSIVE ZOOM TV INTERVIEW
WITH RAY SHASHO
CLICK BELOW TO WATCH IN ITS ENTIRETY
Ronnie James Dio was the most acclaimed heavy metal singer of his generation, and the only rock star to achieve multi-platinum success in not one, but three bands: Rainbow, Black Sabbath, and Dio. For Tenacious D's Jack Black, Dio was "the Pavarotti of heavy metal." For Metallica's Lars Ulrich, Dio was "one of the main reasons I made it onto the stage..." And for heavy metal fans all over the world, there were three words that summed him up: "Dio is God." He died in 2010.
Wendy Dio is the President and Owner of Niji Management. Over the past thirty years, she has been involved in many aspects of the music business, receiving awards from Performance and Pollstar for stage set design and concert video production, along with serving as executive producer on numerous gold and platinum albums. In 2010, she co-founded the Ronnie James Dio Stand Up and Shout Cancer Fund, which has raised over $2 million for research, education, and early detection screenings. She is known for her work on Dio: Live in London - Hammersmith Apollo 1993 (2014), Heaven & Hell: Neon Nights, Live in Europe (2010) and Heaven & Hell: Live from Radio City Music Hall (2007).
The long-awaited autobiography by one of heavy metal's most revered icons, treasured vocalists, and frontman for three legendary bands—Rainbow, Black Sabbath, and Dio.
Prior to his tragic death in 2010, Ronnie James Dio had been writing his autobiography, looking back on the remarkable life that led him from his hometown in upstate New York to the biggest stages in the world, including the arena that represented the pinnacle of success to him—Madison Square Garden, where this book begins and ends.
Prior to his tragic death in 2010, Ronnie James Dio had been writing his autobiography, looking back on the remarkable life that led him from his hometown in Upstate New York to the biggest stages in the world, including the arena that represented the pinnacle of success to him - Madison Square Garden, where this book begins and ends.
For more information about
www.ronniejamesdio.com
www.facebook.com/OfficialRonnieJamesDio
Very special thanks to Hadlee Wolfram Publicity Coordinator w/ Chipster PR & Consulting, Inc. for arranging this interview with Wendy Dio. If you have comments or suggestions for the show contact me at interviewingthelegends@gmail.com and please subscribe to my You Tube channel interviewing the legends with Ray Shasho for the very latest interviews … It's REAL NEWS!
1972 Elf
1974 Carolina County Ball
1975 Trying to Burn the Sun
1975 Ritchie Blackmore's Rainbow
1976 Rising UK: Gold (Original Edition)
UK: Silver (Deluxe Edition)
1978 Long Live Rock 'n' Roll UK: Silver
1977 On Stage UK: Silver
1986 Finyl Vinyl
1990 Live in Germany 1976
2006 Live In Munich 1977 (CD/DVD)
2006 Deutschland Tournee 1976 (Box-set)
2007 Live In Cologne 1976
Live In Düsseldorf 1976
Live In Nurnberg 1976
1980 Heaven and Hell
1981 Mob Rules
1992 Dehumanizer
2007 Black Sabbath: The Dio Years
2008 The Rules of Hell
1980 Black And Blue VHS
1982 Live Evil
2007 Live at Hammersmith Odeon
Holy Diver 1983
The Last in Line 1984
Sacred Heart 1985
Dream Evil 1987
Lock Up the Wolves 1990
Strange Highways 1993
Angry Machines 1996
Magica 2000
Killing the Dragon 2002
Master of the Moon 2004
with YOUR HOST
RAY SHASHO
on INTERVIEWING THE LEGENDS
THE CLASSIC ROCK MUSIC REPORTER (classicrockmusicwriter.com)
Music Tried And True
RAY'S BEST-SELLING BOOK!
ENTITLED
by Ray Shasho
CHRONICLES, TRUTHS, CONFESSIONS AND WISDOM FROM THE MUSIC LEGENDS THAT SET US FREE
…Order yours today on (Collector edition) Hardcover
or E-book
at bookbaby.com and amazon.com
Featuring over 45 intimate conversations with some of
the greatest rock legends the world will ever know.
CHRIS SQUIRE... DR. JOHN... GREG LAKE... HENRY MCCULLOUGH... JACK BRUCE … JOE LALA… JOHNNY WINTER... KEITH EMERSON... PAUL KANTNER... RAY THOMAS... RONNIE MONTROSE... TONY JOE WHITE... DAVID CLAYTON-THOMAS… MIKE LOVE... TOMMY ROE... BARRY HAY... CHRIS THOMPSON... JESSE COLIN YOUNG... JOHN KAY... JULIAN LENNON... MARK LINDSAY... MICKY DOLENZ… PETER RIVERA ...TOMMY JAMES… TODD RUNDGREN... DAVE MASON... EDGAR WINTER... FRANK MARINO... GREGG ROLIE... IAN ANDERSON... JIM "DANDY" MANGRUM... JON ANDERSON... LOU GRAMM... MICK BOX... RANDY BACHMAN… ROBIN TROWER... ROGER FISHER... STEVE HACKETT... ANNIE HASLAM… 'MELANIE' SAFKA... PETULA CLARK... SUZI QUATRO... COLIN BLUNSTONE… DAVE DAVIES... JIM McCARTY... PETE BEST
-By Literary Titan (5) STARS
The Rock Star Chronicles, by Ray Shasho, is a splendid book written by a music enthusiast who has poured their heart and soul into it. It's a story of a boy who loved rock music, and his obsessive passion of it earned himself the name Rock Raymond. He went to school but instead was schooled in all matters of music while his peers were buried chin-deep in coursework. He then became a radio DJ and has now compiled a book on all interviews he held with Rock gods who raided the airwaves back in the 70s and 80s. It's a compilation of interviews with outstanding vocalists, legendary guitarists and crazy drummers in the rock music scene. Each interview gives a reader an in-depth view into their personal lives and the philosophies that guide their lives which all serve to humanize these great icons. For readers who are old enough to call themselves baby boomers this book will bring old memories back to life. Millennials, on the other hand, may think of this book as a literal work of the Carpool Karaoke show.
The Rock Star Chronicles is a book I didn't know I was waiting for. To come across a book that will talk me into trying something new. One brave enough to incite me to venture into new frontiers. This book made me a believer- I am now a bona fide Rock and Roll music fan.
Ray Shasho masterfully gets the interviewees talking. He smartly coaxes answers from them with crafty questions designed to get a story rolling out of them. The artists talk about diverse issues ranging from music, politics, and their social engagements. Having been on the music seen all his life, Ray Shasho knows the buttons to press, how to get them comfortable about talking about their lives.
The book's cover is befitting of its subject matter with the leather look offering a royal background to the golden letter print. It speaks to how high a level rock music holds in the pecking order- arguably, modern music as we know it has originated from blues and rock music. The second noteworthy thing is the use of high-definition pictures to reference the musician being interviewed in every sub-chapter. This ensures that the book is for both original rock and roll lovers and aspiring new ones. Together is makes for a refreshing and consistently enjoyable read.
I recommend this book to rock music enthusiasts, aspiring musicians wondering what it takes and all readers curious to learn new things by going back in time.
https://literarytitan.com/2020/05/03/literary-titan-book-awards-may-2020/ Gold Award Winner
Posted by Classic Rock Here And Now at 11:51 PM
Labels: #DIO, #Heaven and Hell, #Interviewing the Legends, #Rainbow, #Ray Shasho, #Ronnie James Dio, #The Rock Star Chronicles, #Wendy Dio Interview
Bradenton,Florida Tampa, FL, USA


TOWSON, MD. – Five Baltimore County Public Schools educators have been named finalists for the school system's annual Teacher of the Year award for 2017 – 2018.
From among the five finalists, one outstanding educator will be named Baltimore County's Teacher of the Year during a 6:30 p.m. ceremony on Wednesday, Apr. 26, 2017, at George Washington Carver Center for Arts and Technology, 938 York Road, Towson 21204.
The ceremony will feature student performances and remarks from 2016 – 2017 Baltimore County Teacher of the Year Corey Carter (Northwest Academy of Health Sciences), Elementary Principal of the Year Missy Fanshaw (Rodgers Forge Elementary School), and Secondary Principal of the Year Sandra Reid (Pikesville High School). During the event, all Teacher of the Year and Principal of the Year nominees and their schools will be recognized, and the new Elementary and Secondary Principals of the Year will be announced.
The Teacher of the Year finalists were selected by a panel of judges from across Baltimore County representing teachers, students, administrators and the Teachers Association of Baltimore County (TABCO). The Teacher of the Year will go on to participate in the state Teacher of the Year program, including recognition from the State Board of Education.


text:
Monkey King: Journey to the West
Imprint: Penguin Classics
Series: Penguin Clothbound Classics
Wu Ch'eng-en
Julia Lovell (Translator)
Pre-Order From
One of the world's greatest fantasy novels and a rollicking classic of Chinese literature, in a sparkling new translation and published in a Clothbound Classics edition.
A shape-shifting trickster on a kung-fu quest for eternal life, Monkey King is one of the most memorable superheroes in world literature. High-spirited and omni-talented, he can transform himself into whatever he chooses and turn each of his body's 84,000 hairs into an army of clones. But his penchant for mischief repeatedly gets him into trouble, and when he raids Heaven's Orchard of Immortal Peaches, the Buddha pins him beneath a mountain. Five hundred years later, Monkey King is finally given a chance to redeem himself: he must protect the pious monk Tripitaka on his journey in search of precious Buddhist sutras that will bring enlightenment to the Chinese empire.
Joined by two other fallen immortals - Pigsy, a rice-loving flying pig, and Sandy, a depressive river-sand monster - Monkey King does battle with Red Boy, Princess Jade-Face, the Monstress Dowager, and all manner of dragons, ogres, wizards and femmes fatales; navigates the perils of Fire-Cloud Cave, the River of Flowing Sand and the Water-Crystal Palace; and is serially captured, lacquered, sautéed, steamed and liquefied - but always hatches an ingenious plan to get himself and his fellow pilgrims out of their latest jam.
Comparable to The Canterbury Tales or Don Quixote, Monkey King is at once a gripping adventure, a comic satire and a spring of spiritual insight. With this new translation by the award-winning Julia Lovell, the irrepressible rogue hero of one of the Four Great Classical Novels of Chinese literature has the potential to vault, with his signature cloud-somersault, into the hearts of a whole new generation of readers.
A monument of world literature, Monkey King is also one of the funniest, most subversive satires ever written ... If you've not read Journey to the West, prepare yourself for the adventure of a lifetime and know that like Monkey himself, you are about to be transformed. Even if you have read it, Julia Lovell's magnificent new translation becomes its own cloud somersault, its own gold-hooped staff
A new translation of Monkey King is a cause for joy! Imaginative and mischievous, exhilarating and timeless, this sixteenth-century superhero saga is a delight to readers of all ages.
An exhilarating new translation of my favourite of all the classic Chinese novels - a great, wild epic that expands and fires one's imagination
Ha Jin
The Monkey King, one of Chinese literature's great characters, should add many new disciples to his existing fans through this compelling new version of his adventures. Julia Lovell here conveys a vibrant sense of the richness ? and also the sheer fun ? of this Ming dynasty text, a classic of world fiction
Craig Clunas, University of Oxford
An all-new translation of one of the greatest stories ever written . . . An out-and-out fantasy adventure that has captivated audiences and influenced creatives for centuries
Jam-packed with outrageous danger and outlandish transformations . . . Lovell does an admirable job condensing the original text . . . while capturing the essence of Chinese fantastical storytelling and parody. Readers who enjoy nutty adventures and nonsensical plots will get a kick out of this madcap fable
Uproarious and action-filled, this highly readable new translation captures the most beloved of Chinese characters in all his impossible charm. Irrepressible and irresistible, Monkey speaks to us across the centuries, and here makes us laugh anew
Gish Jen
Jane Austen: The Complete Works
Lady Chatterley's Lover
D H Lawrence
The Bhagavad Gita
William Makepeace Thackeray
The Mahabharata
John D. Smith
Major Works of Charles Dickens (Boxed Set)
Around the World in Eighty Days
Very little is known about Wu Ch-eng-en (c. 1505-80). He was a Chinese novelist and poet of the Ming Dynasty, and is considered by many to be the author of Journey to the West (Monkey King), one of the Four Great Classical Novels of Chinese literature.


The world has succumb to the zombie apocalypse and now most of your friends are dead. Somehow, you manage to live long, but you don't actually all that living since you are constantly hiding and running from the zombies. One day, while you are watching the television, other survivors tells that they can't hold the stronghold no more and urge everyone to get to them as they would leave for another stronghold. You know that this would be your last time to be rescued and if you can't get to them, then there is no telling how long can you live in such a forsaken world. Drive as fast as you can till you run out of fuel. At least, that is the idea, but there are times that your car can't move on not because it doesn't have any fuel left, but there are too much debris along the way. Anyway, you earn base on how far are you able to travel in this game. You can use the money to buy some upgrades for your car. That is the whole scenario and you would want to get closer inch by inch to your target. That is the real challenge that you need to overcome.


text:
https://www.myjoyonline.com/airtel-and-connect-to-learn-partners-connect-three-new-schools/-------https://www.myjoyonline.com/airtel-and-connect-to-learn-partners-connect-three-new-schools/
Airtel and Connect To Learn partners connect three new schools
9 November 2013 4:15pm
Airtel is helping Connect To Learn, a global education initiative of the Earth Institute, Ericsson and Millennium Promise, to provide mobile broadband connectivity to three new schools under the authority of the Accra Metropolitan Assembly (AMA) in Ghana.
This is a first step in an ambitious plan to provide mobile broadband connectivity to some 50 new schools being built by AMA.
This project is part of Airtel's ongoing support for Connect To Learn which uses Information and Communication Technology (ICT) to help improve the quality of learning in secondary education across schools in Africa.
Ericsson will provide access to its cloud-based solutions developed to support the global Connect To Learn initiative. Ericsson will also provide 3G routers as well as a minimum of 150 netbooks to be distributed among the three schools. Ericsson employees will also work hand-in-hand with teachers in the schools as part of a professional development program to improve their ability to use ICT in the classroom.
Deploying a cloud-based ICT solution in schools lowers both initial costs as well as the total cost of ownership for schools. It also significantly reduces the complexity of technology solutions for teachers and students alike. Cloud technology provides teachers with technology that is simpler to manage, so they can focus on improving the quality of education.
Andreas Karlsson, the Country Director of Ericsson Ghana says the project is part of Ericsson's long-standing support for the Millennium Development Goals. "Ericsson continues to use its technology for good to support people as well as the public and private sectors. We believe that such support can aid African governments in achieving their Goals. Technology can enhance education, improve health and support businesses all of which will serve to reduce poverty," says Karlsson.
For the project in Accra, Airtel will provide 3.75G internet connectivity at a subsidized rate of 50 percent. The AMA will provide each school with a secure ICT laboratory, as well as electricity.
Donald Gwira, Head of Corporate Communications and External Affairs at Airtel Ghana who spoke on behalf of the Managing Director Philip Sowah said education is a key pillar in the company's CSR initiatives so we thought it worthwhile to support Ericsson and the AMA to connect the schools to be fully ICT enabled.
"All these schools will have internet access to be powered by our award winning 3.75G network for the use of students and teachers," he added.
Ericsson and Airtel have worked closely together for a number of years to provide internet connectivity to schools across the region.
Airtel has supported connectivity for Connect To Learn in a number of secondary schools across Africa, including schools in the Millennium Village in Bonsasso, Ghana. This partnership supports both the Millennium VillagesProject and the Connect To Learn initiative, of which Ericsson is one founding partner. The Connect To Learn initiative has connected more than 33 schools and more than 35,000 students and their teachers with the contributions of Ericsson and other technology industry partners across sub-Saharan Africa. Connect To Learn has also provided 745 multi-year secondary school scholarships for girls to attend the schools it enables with broadband connectivity.
The Mayor was very grateful to Airtel and Ericsson for the kind gesture. It was his wish for better education in the country and that 97,000 pupils who studied under trees now have better classrooms to study. The AMA is building 85 Millennium City Schools for the metropolis and about five of them will be commissioned soon. Therefore this support came at an opportune time.


text:
\section{Introduction}

A perfect dyadic operator $T$ is defined by:
\begin{equation} \nonumber
\label{formula 1 T} Tf \; := \; \int K(x,y) \, f(y) \, dy \;\; \text{for} \;\; x \notin supp\, f,
\end{equation}
where kernel $K$ satisfies the following conditions:\\
standard size condition:
\begin{equation}
\label{formula 2 cond}
\mdl{K(x,y)} \; \leq \; \frac{1}{\mdl{x-y}}
\end{equation}
and perfect cancellation condition:
\begin{equation}
\label{formula 3 cond}
\mdl{K(x,y)-K(x,y^\prime)} +
\mdl{K(x,y)-K(x^\prime,y)} \; = \; 0
\end{equation}
whenever $x,x^\prime \in I \in D$, $y,y^\prime \in J \in D$, $I \cap
J = \varnothing$. Where $D$ is a set of dyadic intervals on the real line $D:= \{I = [2^j k, 2^j (k+1)): \; k, j \in \mathbb{Z}\}$.

Perfect dyadic operators appeared in the context of the local $T(b)$ theorems (see \cite{AHMTT}) and later were extended to spaces of homogeneous type (see \cite{AR}, \cite{LV}). Perfect dyadic operators are essentially Calder\'{o}n--Zygmund singular integrals with singularity adapted to the fixed dyadic grid. Their main property is that any function supported on a dyadic cube with zero mean is mapped into the function supported on the same cube. In \cite{AY} it was shown that in $L^2(dx)$ perfect dyadic operators are a good approximation of Calder\'{o}n--Zygmund singular integral operators.

In this paper we stick to the real line as a model case. In Section 2 we derive a very useful decomposition of the perfect dyadic operator into four well known operators: two selfadjoint operators, dyadic paraproduct and its adjoint. In particular, we show that a perfect dyadic operator $T$ and its adjoint formally can be written as

\begin{eqnarray}
\label{e:decompT}\nonumber Tf(x) &=& \frac{1}{4}\sum_{I\in
D}\left( K_I^+ + K_I^- \right) \mdl{I} \, m_If \, \chi_I -
 \frac{1}{4}\sum_{I\in
D}\left( K_I^+ - K_I^- \right)\sqrt{\mdl{I}}
\left\langle f;h_I \right\rangle \chi_I\\
&& +  \frac{1}{4}\sum_{I\in
D}\left( K_I^+ - K_I^- \right)
\mdl{I}^{3/2} \, m_If \, h_I -  \frac{1}{4}\sum_{I\in
D}\left( K_I^+
+ K_I^- \right)\mdl{I}\left\langle f;h_I \right\rangle h_I,
\end{eqnarray}
\begin{eqnarray}
\label{e:decompT*}\nonumber T^*f(x) &=& \frac{1}{4}\sum_{I\in D}\left( K_I^+ + K_I^- \right)
\mdl{I} \,m_If\, \chi_I(x) +  \frac{1}{4}\sum_{I\in
D}\left( K_I^+ -
K_I^- \right)\sqrt{\mdl{I}} \left\langle f;h_I \right\rangle
\chi_I(x)\\
&& -  \frac{1}{4}\sum_{I\in
D}\left( K_I^+ - K_I^- \right)
\mdl{I}^{3/2} m_If \, h_I(x) -  \frac{1}{4}\sum_{I\in
D}\left( K_I^+
+ K_I^- \right) \mdl{I} \left\langle f;h_I \right\rangle h_I(x).
\end{eqnarray}

For a given dyadic interval $I$, $I^+$ and $I^-$ are its left and right halves. Coefficients $K_I^+$ and $K_I^-$ are defined to be values of the kernel $K$ of the perfect dyadic operator $T$ on the dyadic cubes $I^+ \times I^-$ and $I^- \times I^+$ respectively. The notation $m_I f$ stands for the average of the function $f$ over the interval $I$, $m_I f := \frac{1}{|I|} \int_I f$.

In Section 3 using our decomposition we prove the $T(1)$ theorem and show several useful estimates on the kernel. We prove the following theorem.

\begin{theorem} \label{t:T(1)}
Let $T$ be a perfect dyadic operator that satisfies:

(i) BMO conditions: $\|T(1)\|_{BMO^d}$ and $\|T^*(1)\|_{BMO^d}$ are bounded by a finite numerical constant $Q$;

(ii) Testing conditions: $\langle T h_I; h_I \rangle$ are uniformly bounded by $Q$ for every dyadic interval $I\in D$.

Then $T$ is bounded on $L^2$.

Moreover, $T$ accepts decomposition (\ref{e:decompT}) with coefficients $K_I^\pm$ satisfying

\noindent size conditions:
\begin{equation}
\label{formula 19 size cond} \forall J\in D \;\;\;\;\; \mdl{K_J^+ +
K_J^-} \mdl{I} \; \leq \; 2 Q,
\end{equation}
$T(1)$ conditions:
\begin{equation}
\label{formula 20 T(1) cond} \forall J\in D \;\;\;\;\;
\frac{1}{\mdl{J}}\sum_{I\in D(J)}\left( K_I^+ - K_I^- \right)^2
\mdl{I}^3 \; \leq \; 16 Q
\end{equation}
and testing conditions:
\begin{equation}
\label{formula 21 test cond} \forall J\in D \;\;\;\;\;
\mdl{\frac{1}{\mdl{J}}\sum_{I\in D(J)}\left( K_I^+ + K_I^-
\right)\mdl{I}^2} \; \leq \; 8 Q.
\end{equation}
\end{theorem}

In Section 4 we `lift' the above $T(1)$ theorem to the one weight case, $L^2(w)$, and provide an elementary proof of the $A_2$ conjecture for such operators; in particular we trace how the weighted norm of perfect dyadic operator depends on the dyadic $BMO^d$ and testing constants of the perfect dyadic operator. We prove the following theorem.

\begin{theorem}
Let $T$ be a perfect dyadic operator on $\R$ such that
$$\|T(1)\|_{BMO^d} \leqslant Q \;\;\; \text{and} \;\;\; \|T^*(1)\|_{BMO^d} \leqslant Q $$
and
$$ \forall I\in D \;\;\; \langle T h_I , h_I \rangle \leqslant Q $$.
Then $T$ is bounded on $L^2(w)$ and
\begin{equation}\label{e:A2}
\|T f\|_{L^2(w)} \leqslant C Q [w]_{A_2^d}\|f\|_{L^2(w)},
\end{equation}
with some constant $C$ independent of the operator $T$.
\end{theorem}

In particular, by \cite{DGPP}, this implies that under assumptions of the theorem a perfect dyadic operator $T$ is also bounded in $L^p (w)$ for all $w\in A_p^d$, its norms are bounded by $C [w]_{A_p^d}^{\max\{1, \frac{1}{p}\}}$, and dependence $C(Q)$ can be traced as well.

To the best of our knowledge, such sharp weighted version of $T(1)$ theorem is new. In particular it is interesting that the constant in (\ref{e:A2}) depends only on the $BMO^d$ norms of $T(1)$ and $T^*(1)$ and on the constant in testing conditions. Moreover, the dependence is at most linear.

In Section 5 we go even further and give sufficient conditions for the two weight boundedness for a single perfect dyadic operator. Currently there are two schools of thought regarding the two weight problem. First, given one operator $T$ find necessary and sufficient conditions
on the weights to ensure boundedness of the operator on the appropriate spaces. Second, given a family of operators
find necessary and sufficient conditions on the weights to ensure boundedness of the family of operators. In both cases we are mostly interested in Calder\'on--Zygmund singular integral operators.
In the first case, the conditions are usually ``testing conditions" obtained from checking boundedness of the given operator on
a collection of test functions. In the second case, the conditions are more ``geometric", meaning they seem to only involve the weights and
not the operators, such as Carleson conditions or bilinear embedding conditions, Muckenhoupt $A_2$ type conditions or bumped conditions.
Operators of interest are the maximal function \cite{S1, Moe, PzR, V}, fractional and Poisson integrals \cite{S2,Cr}, the Hilbert transform \cite{CS1, CS2, KP, NazarovTreilVolberg:99, LSSU, L3}
and general Calder\'on--Zygmund singular integral operators and their commutators \cite{CrRV, CrMoe,CrMPz2, HoLWic,NRTV}, the square functions \cite{ LLi,LLi2}, paraproducts and their dyadic counterparts. Necessary and sufficient conditions are only known for the maximal function, fractional and Poisson integrals \cite{S1}, square functions \cite{LLi} and the Hilbert transform \cite{L3,LSSU}, and among the dyadic operators for the martingale transform, the dyadic square functions, positive and well localized dyadic operators  \cite{Wil87, NazarovTreilVolberg:99,NTV3, T, Ha, HaHLi, HL, LSU2, Ta, Vu}.

We prove the following sufficient conditions for the two weight boundedness of the perfect dyadic operators.

\begin{theorem} \label{t:2w-1}
Let $(v,u)$ be a pair of measurable functions, such that $u$ and $v^{-1}$, the reciprocal of $v$, are weights on $\R$ such that
\begin{equation}\label{e:uv-1}
(v,u)\in A_2^d \,,\text{i.e.} \;\; [v,u]_{A_2^d} := \sup_{I\in D} m_I (v^{-1}) m_I u <\infty,
\end{equation}
\begin{equation}\label{e:uv-2}
\forall J \in D \;\;\; \frac{1}{|J|} \sum_{I \in D(J)} (\Delta_I u)^2 m_I (v^{-1}) |I| \leqslant C m_I u,
\end{equation}
\begin{equation}\label{e:uv-3}
\forall J \in D \;\;\; \frac{1}{|J|} \sum_{I \in D(J)} (\Delta_I (v^{-1}))^2 m_I u |I| \leqslant C m_I (v^{-1}),
\end{equation}
and operator $T_0$ is bounded from $L^2(v)$ in $L^2(u)$, i.e.
\begin{equation}\label{e:uv-4}
\forall J \in D \;\;\; \frac{1}{|J|} \int_J \left(\sum_{I \in D(J)}  \left|\frac{\Delta_I u \Delta_I (v^{-1})}{m_Iu } \right| \chi_I(x) \right)^2 u(x) dx \leqslant C m_J (v^{-1})
\end{equation}
and
\begin{equation}\label{e:uv-5}
\forall J \in D \;\;\; \frac{1}{|J|} \int_J \left(\sum_{I \in D(J)}   \left|\frac{\Delta_I u \Delta_I (v^{-1})}{ m_I(v^{-1})}\right| \chi_I(x) \right)^2 v^{-1}(x) dx \leqslant C m_J u.
\end{equation}
Let $T$ be a perfect dyadic operator with $\|T(1)\|_{BMO^d}, \|T^*(1)\|_{BMO^d} \leqslant Q$ and for every $I \in D$ we have $\langle T h_I; h_I \rangle \leqslant Q$.

Then $T$ is bounded from $L^2 (v)$ to $L^2(u)$ whenever for any dyadic interval $J \in D$

\begin{equation}\label{e:Tuv-1}
\frac{1}{|J|} \sum_{I\in D(J)} (K_I^+ + K_I^-) |I|^2  m_I u \leqslant C m_J u,
\end{equation}
\begin{equation}\label{e:Tuv-2}
\frac{1}{|J|} \sum_{I\in D(J)} (K_I^+ + K_I^-) |I|^2 m_I(v^{-1}) \leqslant C m_J (v^{-1}),
\end{equation}
\begin{equation}\label{e:Tuv-3}
\frac{1}{|J|}\sum_{I\in {D}(J)} \frac{(K_I^+ - K_I^-)^2 |I|^{3}}{m_Iu} \leq C m_J (v^{-1}
\end{equation}
and
\begin{equation}\label{e:Tuv-4}
\frac{1}{|J|}\sum_{I\in {D}(J)} \frac{(K_I^+ - K_I^-)^2 |I|^{3}}{m_I (v^{-1})} \leq C m_J u.
\end{equation}
\end{theorem}

When both weights are in the Muckenhoupt class $A_\infty^d$ we simplify conditions of Theorem \ref{t:2w-1} to the following.

\begin{theorem}
Let $(v,u)$ be a pair of measurable functions, such that $u$ and $v^{-1}$, the reciprocal of $v$, are $A_\infty^d$ weights on $\R$ such that
\begin{equation}\label{e:uv-1}
(v,u)\in A_2^d \,,\text{i.e.} \;\; [v,u]_{A_2^d} := \sup_{I\in D} m_I (v^{-1}) m_I u <\infty,
\end{equation}
and operator $T_0$ is bounded from $L^2(v)$ in $L^2(u)$, i.e.
\begin{equation}\label{e:uv-4}
\forall J \in D \;\;\; \frac{1}{|J|} \int_J \left(\sum_{I \in D(J)}  \left|\frac{\Delta_I u \Delta_I (v^{-1})}{m_Iu } \right| \chi_I(x) \right)^2 u(x) dx \leqslant C m_J (v^{-1})
\end{equation}
and
\begin{equation}\label{e:uv-5}
\forall J \in D \;\;\; \frac{1}{|J|} \int_J \left(\sum_{I \in D(J)}   \left|\frac{\Delta_I u \Delta_I (v^{-1})}{ m_I(v^{-1})}\right| \chi_I(x) \right)^2 v^{-1}(x) dx \leqslant C m_J u.
\end{equation}
Let $T$ be a perfect dyadic operator with $\|T(1)\|_{BMO^d}, \|T^*(1)\|_{BMO^d} \leqslant Q$ and for every $I \in D$ we have $\langle T h_I; h_I \rangle \leqslant Q$.

Then $T$ is bounded from $L^2 (v)$ to $L^2(u)$ whenever for any dyadic interval $J \in D$

\begin{equation}\label{e:Tuv-3}
\frac{1}{|J|}\sum_{I\in {D}(J)} \frac{(K_I^+ - K_I^-)^2 |I|^{3}}{m_Iu} \leq C m_J (v^{-1}
\end{equation}
and
\begin{equation}\label{e:Tuv-4}
\frac{1}{|J|}\sum_{I\in {D}(J)} \frac{(K_I^+ - K_I^-)^2 |I|^{3}}{m_I (v^{-1})} \leq C m_J u.
\end{equation}
\end{theorem}

The main novelty of this paper is the decomposition, very strong quantitative estimates in the $A_2$ conjecture and sufficient conditions for the two weight boundedness of perfect dyadic operators.

\section{Perfect dyadic operator on $\R$: definition and formal decomposition.}

Let $T$ be a perfect dyadic singular integral operator, i.e an operator defined by:
\begin{equation}
\label{formula 1 T} Tf \; := \; \int K(x,y) \, f(y) \, dy \;\; \text{for} \;\; x \notin supp\, f,
\end{equation}
where kernel $K$ satisfies the following conditions:\\
standard size condition:
\begin{equation}
\label{formula 2 cond} \mdl{K(x,y)} \; \leq \; \frac{1}{\mdl{x-y}}
\end{equation}
and perfect cancelation condition:
\begin{equation}
\label{formula 3 cond}\mdl{K(x,y)-K(x,y^\prime)} +
\mdl{K(x,y)-K(x^\prime,y)} \; = \; 0
\end{equation}
whenever $x,x^\prime \in I \in D$, $y,y^\prime \in J \in D$, $I \cap
J = \varnothing$.

Note that, by perfect dyadic cancellation condition, $K(x,y)$ is
constant on $I^+ \times I^-$ and $I^- \times I^+$ for any dyadic interval $I \in D$.
We define
$$
K^+_I \; := \; K(x,y), \;\;\;\;\; x\in I^+, \;\; y\in I^-
$$
and
$$
K^-_I \; := \; K(x,y), \;\;\;\;\; x\in I^-, \;\; y\in I^+.
$$
Then
$$
K(x,y) = \sum_{I\in D} K^+_I \chi_{I^+\times I^-} + K^-_I
\chi_{I^-\times I^+}.
$$

Then perfect dyadic singular operator can be written as
\begin{equation}
\label{formula 4 Tf(x)}Tf(x) \; = \; \sum_{I\in D} K^+_I f_{I^-}
\chi_{I^+}(x) + K^-_I f_{I^+} \chi_{I^-}(x),
\end{equation}
where $f_I := \int_I f(x) dx$.

Now we are going to rewrite $T$ in a more convenient form. It is
easy to see that
\begin{eqnarray}
\label{formula 5 K+K}\nonumber && K^+_I f_{I^-} \chi_{I^+}(x) +
K^-_I f_{I^+}
\chi_{I^-}(x)\\
\nonumber &=& \frac{K^+_I f_{I^-} + K^-_I f_{I^+}}{2}\left(
\chi_{I^+}(x) + \chi_{I^-}(x) \right) + \frac{K^+_I f_{I^-} - K^-_I
f_{I^+}}{2}\left( \chi_{I^+}(x) - \chi_{I^-}(x) \right)\\
&=& \frac{1}{2}\left( K_I^+ f_{I^-} + K_I^- f_{I^+} \right)
\chi_I(x) + \frac{1}{2}\left( K_I^+ f_{I^-} - K_I^- f_{I^+} \right)
\sqrt{\mdl{I}} \; h_I(x),
\end{eqnarray}
where $\{h_I\}_{I\in D}$ is the Haar system of functions, normalized
in $L^2$, $h_I := \mdl{I}^{-1/2}(\chi_{I^+} - \chi_{I^-})$. Thus,
$Tf(x)$ can be written as:
\begin{equation}
\label{formula 6 Tf(x)}Tf(x) = \sum_{I\in D} \frac{1}{2} \left( K_I^+
f_{I^-} + K_I^- f_{I^+} \right) \chi_I + \frac{1}{2} \left( K_I^+ f_{I^-} - K_I^- f_{I^+} \right) h_I \sqrt{\mdl{I}}.
\end{equation}
Now let us handle the coefficients $\left( K_I^+ f_{I^-} + K_I^-
f_{I^+} \right)$ and $\left( K_I^+ f_{I^-} - K_I^- f_{I^+} \right)$:
\begin{eqnarray}
\label{formula 7 K+K}\nonumber && K_I^+ f_{I^-} + K_I^- f_{I^+} \; =
\; K_I^+\int_{I^-}f + K_I^-\int_{I^+}f\\
\nonumber &=& \frac{1}{2}\left( K_I^+ + K_I^- \right)\left(
\int_{I^-}f + \int_{I^+}f \right) - \frac{1}{2}\left( K_I^+ - K_I^-
\right)\left( \int_{I^+}f - \int_{I^-}f \right)\\
& = & \frac{1}{2}\left( K_I^+ + K_I^- \right) \mdl{I} \, m_If -
\frac{1}{2}\left( K_I^+ - K_I^- \right)\sqrt{\mdl{I}}\left\langle
f;h_I \right\rangle
\end{eqnarray}
and, similarly, replacing $K_I^-$ by $-K_I^-$,
\begin{equation}
\label{formula 8 K-K}K_I^+ f_{I^-} - K_I^- f_{I^+} \; = \;
\frac{1}{2}\left( K_I^+ - K_I^- \right) \mdl{I} \, m_If -
\frac{1}{2}\left( K_I^+ + K_I^- \right)\sqrt{\mdl{I}}\left\langle
f;h_I \right\rangle.
\end{equation}
We plug (\ref{formula 7 K+K}) and (\ref{formula 8 K-K}) in
(\ref{formula 6 Tf(x)}) and obtain a formal representation of the perfect dyadic operator $T$:
\begin{eqnarray}
\label{formula 9 Tf(x)}\nonumber Tf(x) &=& \frac{1}{4}\sum_{I\in
D}\left( K_I^+ + K_I^- \right) \mdl{I} \, m_If \, \chi_I -
\left( K_I^+ - K_I^- \right)\sqrt{\mdl{I}}
\left\langle f;h_I \right\rangle \chi_I\\
&& + \left( K_I^+ - K_I^- \right)
\mdl{I}^{3/2} \, m_If \, h_I - \left( K_I^+
+ K_I^- \right)\mdl{I}\left\langle f;h_I \right\rangle h_I.
\end{eqnarray}

We can represent $T^*$ in a similar way:
\begin{eqnarray}
\label{formula 10 T*f(x)}\nonumber T^*f(x) &=& \sum_{I\in D}
K_I^-f_{I^-}\chi_{I^+}(x) + K_I^+f_{I^+}\chi_{I^-}(x)\\
\nonumber &=& \sum_{I\in D} \frac{1}{2}\left( K_I^-f_{I^-} +
K_I^+f_{I^+} \right) \chi_I(x) + \frac{1}{2}\left( K_I^-f_{I^-} -
K_I^+f_{I^+} \right)\sqrt{\mdl{I}} \, h_I(x)\\
\nonumber &=& \frac{1}{4}\sum_{I\in D}\left( K_I^+ + K_I^- \right)
\mdl{I} \,m_If\, \chi_I(x) + \left( K_I^+ -
K_I^- \right)\sqrt{\mdl{I}} \left\langle f;h_I \right\rangle
\chi_I(x)\\
&& - \left( K_I^+ - K_I^- \right)
\mdl{I}^{3/2} m_If \, h_I(x) - \left( K_I^+
+ K_I^- \right) \mdl{I} \left\langle f;h_I \right\rangle h_I(x).
\end{eqnarray}

\section{Decomposition of $T$, additional assumptions.}

In this section we prove Theorem \ref{t:T(1)} modulo the unweighted boundedness of the operator $T_1$, which is done in a more general weighted case in the next section.

We assume, in addition to $T$ being perfect dyadic, that $T(1)$ and $T^*(1)$ are both in the dyadic $BMO^d$, with the $BMO^d$ norm of at most $Q$. We will also assume that $T$ satisfies testing conditions $\langle T h_I, h_I \rangle \leqslant Q$ for all dyadic intervals $I$.

\subsection{The $BMO^d$ condition.} A function $b(x)$, or a sequence $\{b_I\}_{I\in D} = \{\langle b; h_I \rangle\}_{I\in D}$ is in the dyadic $BMO^d$ whenever the sequence $b_I^2$ is a Carleson sequence, i.e. there is a finite constant C such that for all dyadic intervals $J\in D$ we have
$$
\frac{1}{|J|} \sum_{I\in D(J)} b_{I}^2 \leqslant C.
$$
The best such constant C is called the dyadic $BMO^d$ norm of $b$.

Let $T(1)$, $T^*(1) \in BMO^d$. We want to show that
$$
\frac{1}{|J|} \sum_{I\in D(J)} (K_I^+ - K_I^-)^2 |I|^3 \leqslant 16Q.
$$

 First let us plug
$f(x)=1(x)$ in $T$ and $T^*$:
$$
T(1) \; = \; \frac{1}{4}\sum_{I\in D}\left( K_I^+ + K_I^- \right)
\mdl{I}\chi_I(x) + \left( K_I^+ - K_I^-
\right) \mdl{I}^{3/2}h_I(x)
$$
and
$$
T^*(1) \; = \; \frac{1}{4}\sum_{I\in D}\left( K_I^+ + K_I^- \right)
\mdl{I}\chi_I(x) - \left( K_I^+ - K_I^-
\right) \mdl{I}^{3/2}h_I(x).
$$

A function $f(x)$ belongs to the dyadic class $BMO^d$ whenever
$\sup_{J\in D}\frac{1}{\mdl{J}}\sum_{I\in D(J)}\langle f;h_I
\rangle^2 \leq Q$, the smallest such constant is the $BMO^d$-norm of
the function $f(x)$. So, let us find $\left\langle T(1);h_J
\right\rangle$ and $\left\langle T^*(1);h_J \right\rangle$:
\begin{eqnarray}
\label{formula 11 T(1);h}\nonumber \left\langle T(1);h_J
\right\rangle &=& \frac{1}{4}\sum_{I\in D}\left( K_I^+ + K_I^-
\right) \mdl{I}\left\langle \chi_I(x);h_J(x) \right\rangle +
\left( K_I^+ - K_I^- \right)
\mdl{I}^{3/2}\left\langle h_I(x);h_J(x) \right\rangle\\
\nonumber &=& \frac{1}{4}\sum_{I\in D(J^+)}\left( K_I^+ + K_I^-
\right) \mdl{I}\frac{\mdl{I}}{\sqrt{\mdl{J}}} -
\frac{1}{4}\sum_{I\in D(J^-)}\left( K_I^+ + K_I^- \right)\mdl{I}
\frac{\mdl{I}}{\sqrt{\mdl{J}}}\\
&& + \frac{1}{4}\left( K_J^+ - K_J^- \right) \mdl{J}^{3/2}.
\end{eqnarray}
Similarly,
\begin{eqnarray}
\label{formula 12 T*(1);h}\nonumber \left\langle T^*(1);h_J
\right\rangle &=& \frac{1}{4}\sum_{I\in D}\left( K_I^+ + K_I^-
\right) \mdl{I}\left\langle \chi_I(x);h_J(x) \right\rangle -
\left( K_I^+ - K_I^- \right)
\mdl{I}^{3/2}\left\langle h_I(x);h_J(x) \right\rangle\\
\nonumber &=& \frac{1}{4}\sum_{I\in D(J^+)}\left( K_I^+ + K_I^-
\right) \mdl{I}\frac{\mdl{I}}{\sqrt{\mdl{J}}} -
\frac{1}{4}\sum_{I\in D(J^-)}\left( K_I^+ + K_I^- \right)\mdl{I}
\frac{\mdl{I}}{\sqrt{\mdl{J}}}\\
&& - \frac{1}{4}\left( K_J^+ - K_J^- \right) \mdl{J}^{3/2}.
\end{eqnarray}
Let
$$
\alpha_J \; := \; \sum_{I\in D(J^+)}\left( K_I^+ + K_I^- \right)
\frac{\mdl{I}^2}{\sqrt{\mdl{J}}} - \sum_{I\in D(J^-)}\left( K_I^+ +
K_I^- \right) \frac{\mdl{I}^2}{\sqrt{\mdl{J}}}
$$
and
$$
\beta_J \; := \; \left( K_J^+ - K_J^- \right)\mdl{J}^{3/2},
$$
then
$$
\left\langle T(1);h_J \right\rangle \; = \; \frac{1}{4}\left( \alpha_J +
\beta_J \right), \;\;\;\;\; \left\langle T^*(1);h_J \right\rangle \; =
\; \frac{1}{4}\left( \alpha_J - \beta_J \right).
$$
If $T(1)\in BMO^d$ with $\norm{T(1)}{BMO^d} \leq Q$, then for every
dyadic interval $J\in D$
\begin{equation}
\label{formula 13 <Q} \frac{1}{\mdl{J}}\sum_{I\in D(J)}\left\langle
T(1);h_I \right\rangle^2 \; = \; \frac{1}{16\mdl{J}}\sum_{I\in D(J)}
\left( \alpha_I + \beta_I \right)^2 \; \leq \; Q.
\end{equation}
Similarly, $T^*(1)\in BMO^d$ with $\norm{T^*(1)}{BMO^d}\leq Q$
implies that
\begin{equation}
\label{formula 14 <Q} \forall J\in D \;\;\;\;\;
\frac{1}{\mdl{J}}\sum_{I\in D(J)} \left\langle T^*(1);h_I
\right\rangle^2 \; = \; \frac{1}{16\mdl{J}}\sum_{I\in D(J)} \left(
\alpha_I - \beta_I \right)^2 \; \leq \; Q.
\end{equation}
Let us see how conditions (\ref{formula 13 <Q}) and (\ref{formula 14
<Q}) imply that
$$
\forall J\in D \;\;\;\;\; \frac{1}{\mdl{J}}\sum_{I\in D(J)} \beta_I^2 \;
= \; \frac{1}{\mdl{J}}\sum_{I\in D}\left( K_I^+ - K_I^- \right)^2
\mdl{I}^3 \; \leq \; 16Q.
$$
The sum $\frac{1}{\mdl{J}}\sum_{I\in D(J)}\beta_I^2$ can be written as:
\begin{eqnarray*}
\frac{1}{\mdl{J}}\sum_{I\in D(J)}\beta_I^2 &=&
\frac{1}{\mdl{J}}\sum_{I\in D(J)}\left( \frac{\beta_I}{2} -
\frac{\alpha_I}{2} + \frac{\beta_I}{2} + \frac{\alpha_I}{2} \right)^2\\
&=& \frac{1}{\mdl{J}}\sum_{I\in D(J)}\left( \frac{\beta_I}{2} -
\frac{\alpha_I}{2} \right)^2 + \frac{2}{\mdl{J}}\sum_{I\in D(J)}\left(
\frac{\beta_I}{2} - \frac{\alpha_I}{2} \right)\left( \frac{\beta_I}{2} +
\frac{\alpha_I}{2} \right) + \frac{1}{\mdl{J}}\sum_{I\in D(J)}\left(
\frac{\beta_I}{2} + \frac{\alpha_I}{2} \right)^2\\
&\leq& 8Q + \frac{1}{2\mdl{J}}\sum_{I\in D(J)}\left( \beta_I - \alpha_I
\right) \left( \beta_I + \alpha_I \right).
\end{eqnarray*}
By Cauchy-Schwarz
$$
\frac{1}{\mdl{J}}\sum_{I\in D(J)}\left( \beta_I - \alpha_I \right) \left( \beta_I
+ \alpha_I \right) \; \leq \left( \frac{1}{\mdl{J}}\sum_{I\in D(J)}\left(
\beta_I - \alpha_I \right)^2 \right)^{\frac{1}{2}} \left(
\frac{1}{\mdl{J}}\sum_{I\in D(J)}\left( \beta_I + \alpha_I \right)^2
\right)^{\frac{1}{2}} \leq \; 16Q.
$$
Hence, for every dyadic interval $J\in D$
\begin{equation}
\label{formula 15 <9Q} \frac{1}{\mdl{J}}\sum_{I\in D(J)} \beta_I^2 \; =
\; \frac{1}{\mdl{J}}\sum_{I\in D}\left( K_I^+ - K_I^- \right)^2
\mdl{I}^3 \; \leq \; 16Q.
\end{equation}

\subsection{Testing conditions.}
We assume that $T$ satisfies testing conditions
\begin{equation}
\label{formula 16 Th;h<Q} \forall J\in D \;\;\;\;\; \left\langle
Th_J;h_J \right\rangle \; \leq \; Q.
\end{equation}
Let us write $\left\langle Th_J;h_J \right\rangle$ as follows:
\begin{eqnarray*}
\mdl{\left\langle Th_J;h_J \right\rangle} &=& \frac{1}{4}\left|
\sum_{I\in D}\left( K_I^+ + K_I^- \right) \left\langle h_J;\chi_I
\right\rangle \left\langle \chi_I;h_J \right\rangle - \left( K_I^+ - K_I^- \right)\sqrt{\mdl{I}} \left\langle h_J;h_I
\right\rangle \left\langle \chi_I;h_J \right\rangle \right.\\
&& \left. + \left( K_I^+ - K_I^- \right)\sqrt{\mdl{I}}
\left\langle h_J;\chi_I \right\rangle \left\langle h_I;h_J
\right\rangle - \left( K_I^+ + K_I^- \right)\mdl{I}
\left\langle h_J;h_I \right\rangle \left\langle h_I;h_J
\right\rangle \right|\\
&=& \left| \frac{1}{4}\sum_{I\in D(J^+)}\left( K_I^+ + K_I^- \right)
\left( \frac{\mdl{I}}{\sqrt{\mdl{J}}} \right)^2 +
\frac{1}{4}\sum_{I\in D(J^-)}\left( K_I^+ + K_I^- \right) \left(
-\frac{\mdl{I}}{\sqrt{\mdl{J}}} \right)^2\right.\\
&& \left. - \frac{1}{4}\left( K_J^+ + K_J^- \right)\mdl{J} \right|\\
&=& \mdl{\frac{1}{4\mdl{J}}\sum_{I\in D(J)}\left( K_I^+ + K_I^-
\right)\mdl{I}^2 - \frac{1}{2}\left( K_J^+ + K_J^- \right) \mdl{J}}
\; \leq \; Q.
\end{eqnarray*}
Note that by standard size condition on the kernel $K$, we know that
$\mdl{K_J^+ + K_J^-}\mdl{J} \leq 2Q$, so
\begin{equation}
\label{formula 17 <8Q} \mdl{\frac{1}{\mdl{J}}\sum_{I\in D(J)}\left(
K_I^+ + K_I^- \right)\mdl{I}^2} \; \leq \; 8Q.
\end{equation}
So, a perfect dyadic Calder\'{o}n-Zygmund singular integral operator
$T$ satisfying the $T(1)$ conditions $\|T(1)\|_{BMO^2}, \|T^*(1)\|_{BMO^d} \leqslant Q$ and the dyadic testing conditions $\left( \left\langle
Th_J;h_J \right\rangle \leq Q \;\;\; \forall J\in D \right)$, can be
written as:
\begin{eqnarray}
\label{formula 18 Tf(x)} \nonumber Tf(x) &=& \sum_{I\in D}
K_I^+f_{I^-}\chi_{I^+}(x) + K_I^-f_{I^+}\chi_{I^-}(x)\\
\nonumber &=& \frac{1}{4}\sum_{I\in D}\left( K_I^+ + K_I^- \right)
\mdl{I} m_If\, \chi_I(x) - \left( K_I^+ -
K_I^- \right)\sqrt{\mdl{I}}\left\langle f;h_I \right\rangle
\chi_I(x)\\
&& +  \left( K_I^+ - K_I^- \right)
\mdl{I}^{3/2} m_If \, h_I(x) - \left( K_I^+
+ K_I^- \right) \mdl{I} \left\langle f;h_I \right\rangle h_I(x).
\end{eqnarray}
Moreover, coefficients $K_I^+$ and $K_I^-$ satisfy the following
conditions:\\
size conditions
\begin{equation}
\label{formula 19 size cond} \forall J\in D \;\;\;\;\; \mdl{K_J^+ +
K_J^-} \mdl{I} \; \leq \; 2 Q,
\end{equation}
$T(1)$ conditions
\begin{equation}
\label{formula 20 T(1) cond} \forall J\in D \;\;\;\;\;
\frac{1}{\mdl{J}}\sum_{I\in D(J)}\left( K_I^+ - K_I^- \right)^2
\mdl{I}^3 \; \leq \; 16 Q
\end{equation}
and testing conditions
\begin{equation}
\label{formula 21 test cond} \forall J\in D \;\;\;\;\;
\mdl{\frac{1}{\mdl{J}}\sum_{I\in D(J)}\left( K_I^+ + K_I^-
\right)\mdl{I}^2} \; \leq \; 8 Q.
\end{equation}
By (\ref{formula 18 Tf(x)}) operator $T$ can be written as

\begin{equation} \label{decomposition}
T = \frac{1}{4}
\left( T_1 - T_2 + T_3 - T_4 \right),
\end{equation}
where
\begin{equation}
\label{formula 22 T1} T_1f(x) \; = \; \sum_{I\in D}\left( K_I^+ +
K_I^- \right) \mdl{I} m_If\, \chi_I(x),
\end{equation}
\begin{equation}
\label{formula 23 T2} T_2f(x) \; = \; \sum_{I\in D}\left( K_I^+ -
K_I^- \right)\sqrt{\mdl{I}}\left\langle f;h_I \right\rangle
\chi_I(x),
\end{equation}
\begin{equation}
\label{formula 24 T3} T_3f(x) \; = \; \sum_{I\in D} \left( K_I^+ -
K_I^- \right) \mdl{I}^{3/2} m_If \, h_I(x),
\end{equation}
\begin{equation}
\label{formula 25 T4} T_4f(x) \; = \; \sum_{I\in D}\left( K_I^+ +
K_I^- \right) \mdl{I} \left\langle f;h_I \right\rangle h_I(x),
\end{equation}
since all four operators are known well-defined linear operators bounded on $L^2$. Operators $T_3$ and $T_2$ are dyadic paraproduct and its adjoint, that are both well defined and bounded in $L^2$ since the sequence $b_I = (K_I^+ - K_I^-) |I|^{3/2}$ satisfies $\sum_{I\in D(J)} b_I^2 \leqslant C |J|$ for any $J \in D$ by (\ref{formula 20 T(1) cond}). $T_4$ is a martingale transform with symbol bounded uniformly by (\ref{formula 19 size cond}). In the next section we show that $T_1$ is bounded on $L^2$ and on all weighted spaces for weights in $A_2^d$.

This completes the proof of Theorem \ref{t:T(1)}.

\section{Weighted $T(1)$ theorem for perfect dyadic operators}

Let $w$ be a weight, i.e. almost everywhere positive locally integrable function on the real line. Let a weight $w$ be such that $w^{-1}$ is a weight as well. Assume that $w$ is in the dyadic Muckenhoupt class $A_2^d$, i.e.
$$
[w]_{A_2^d} := \sup_{I\in D} m_I w \, m_I (w^{-1}) \; <\infty.
$$

In order to show that the weighted $L^2(w)$ norm of the operator $T$
depends on the $A_2$-constant of the weight $w$ at most linearly, it
is enough to show the linear bound for each of the operators $T_i$,
$i=1,2,3,4$.

First let us show that $T_1$ is bounded on $L^2(w)$ and its norm
depends on the $A_2$-constant of the weight $w$ at most linearly. We
will do it by duality, we will show that $\forall f\in L^2(w)$ and
$\forall g\in L^2(w^{-1})$
$$
\left\langle T_1 f;g \right\rangle \; = \; \sum_{I\in D} \left( K_I^+ +
K_I^- \right) \mdl{I}^2 m_If \, m_Ig \; \leq \; C
[{w}]_{A_2^d}\norm{f}{L^2(w)}\norm{g}{L^2(w^{-1})}
$$
or, alternatively, $\forall f,g\in L^2$
\begin{equation}
\label{formula 26 linear T1} \sum_{I\in D} \left( K_I^+ + K_I^-
\right) \mdl{I}^2 m_I\left( fw^{-1/2} \right)\, m_I\left( gw^{1/2}
\right) \; \leq \; C [{w}]_{A_2^d} \norm{f}{L^2} \norm{g}{L^2}.
\end{equation}
Without loss of generality we may assume that coefficients $(K_I^+ + K_I^-)$ are all non-negative. We are going to use the following version of the bilinear embedding
theorem from \cite{NazarovTreilVolberg:99}.
\begin{theorem}[Nazarov, Treil, Volberg]
\label{theorem 1 NTrVo2} Let $v$ and $w$ be weights. Let $\left\{
a_I \right\}$ be a sequence of nonnegative numbers, s.t. for
all dyadic intervals $J\in D$ the following three inequalities hold
with some constant $Q>0$:
\begin{equation}
\label{formula 27 theorem1} \frac{1}{\mdl{J}}\sum_{I\in
D(J)}a_I\, m_Iw \, m_Iv \; \leq \; Q,
\end{equation}
\begin{equation}
\label{formula 28 theorem1} \frac{1}{\mdl{J}}\sum_{I\in
D(J)}a_I\, m_Iw \; \leq \; Q \, m_Iw,
\end{equation}
\begin{equation}
\label{formula 29 theorem1} \frac{1}{\mdl{J}}\sum_{I\in
D(J)}a_I\, m_Iv \; \leq \; Q\, m_Iv,
\end{equation}
then for any two nonnegative functions $f,g\in L^2$
\begin{equation}
\label{formula 30 theorem1}\sum_{I\in D} a_I\, m_I\left(
fv^{1/2} \right) \, m_I\left( gw^{1/2} \right) \; \leq \; C\, Q\,
\norm{f}{L^2} \norm{g}{L^2}.
\end{equation}
\end{theorem}

Applying this theorem with $v=w^{-1}$ and $a_I =
\left( K_I^+ + K_I^- \right)\mdl{I}^2$, we can see that the desired
bound (\ref{formula 26 linear T1}) will hold if we can prove the
following three inequalities for every dyadic interval $J\in D$:
\begin{equation}
\label{formula 31 ineq1of3}\frac{1}{\mdl{J}}\sum_{I\in D(J)}\left(
K_I^+ + K_I^- \right)\mdl{I}^2 m_Iw \; m_I(w^{-1}) \; \leq \;
C[{w}]_{A_2^d},
\end{equation}
\begin{equation}
\label{formula 32 ineq2of3}\frac{1}{\mdl{J}}\sum_{I\in D(J)}\left(
K_I^+ + K_I^- \right)\mdl{I}^2 m_Iw \; \leq \; C[w]_{A_2^d} m_Jw,
\end{equation}
\begin{equation}
\label{formula 33 ineq3of3}\frac{1}{\mdl{J}}\sum_{I\in D(J)}\left(
K_I^+ + K_I^- \right)\mdl{I}^2 m_I(w^{-1}) \; \leq \; C[w]_{A_2^d}
m_J(w^{-1}).
\end{equation}
It turns out that all these inequalities follow from the testing
conditions (\ref{formula 21 test cond}). It is easy to see that
inequality (\ref{formula 31 ineq1of3}) follows from (\ref{formula 21
test cond}) and the definition of $A_2$-constant $\left( m_Iw \,
m_I(w^{-1}) \leq [w]_{A_2^d} \right)$.

In order to see that inequalities (\ref{formula 32 ineq2of3}) and
(\ref{formula 33 ineq3of3}) are true, we will need the following
lemma, which can be found, for example, in \cite{Beznosova:08}:
\begin{lemma}
\label{lemma1 Be} Let $v$ be a weight, such that $v^{-1}$ is a
weight as well, and let $\left\{ \lambda_I \right\}$ be a Carleson
sequence of nonnegative numbers, that is, there exists a constant
$Q>0$ s.t.
$$
\forall J\in D \;\;\;\;\; \frac{1}{\mdl{J}}\sum_{I\in D(J)}
\lambda_I \; \leq \; Q,
$$
then
$$
\forall J\in D \;\;\;\;\; \frac{1}{\mdl{J}}\sum_{I\in
D(J)}\frac{\lambda_I}{m_I(v^{-1})} \; \leq \; 4Q\, m_Jv
$$
and, therefore, if $v\in A_2$ then for any $J\in D$ we have
$$
\frac{1}{\mdl{J}}\sum_{I\in D(J)} m_Iv \, \lambda_I \; \leq \; 4Q\,
\norm{v}{A_2^d} \, m_Jv.
$$
\end{lemma}
Applying this lemma to $\lambda_I = \left( K_I^+ + K_I^- \right)
\mdl{I}^2$ and $v=w$, we obtain that (\ref{formula 32 ineq2of3})
follows from (\ref{formula 21 test cond}). If we take $v=w^{-1}$ and
observe that $\norm{w^{-1}}{A_2^d} = \norm{w}{A_2^d}$, we can see
that (\ref{formula 33 ineq3of3}) follows from (\ref{formula 21 test
cond}) as well.

So,
$$
\norm{T_1}{L^2(w)\rightarrow L^2(w)} \; \leq \; C \norm{w}{A_2^d}.
$$
Let us analyze operators $T_3$ and $T_2$ now.
$$
T_3f(x) \; = \; \sum_{I\in D} \left( K_I^+ - K_I^- \right)
\mdl{I}^{3/2} m_If \, h_I(x),
$$
$$
T_2f(x) \; = \; \sum_{I\in D}\left( K_I^+ - K_I^-
\right)\sqrt{\mdl{I}}\left\langle f;h_I \right\rangle \chi_I(x),
$$
We first note that $T_3$ is a paraproduct operator and $T_2$ is its
adjoint:
$$
T_3f(x) \; = \; \pi_b f(x) \; = \; \sum_{I\in D} m_If \; b_I \;
h_I(x),
$$
$$
T_2f(x) \; = \; \pi_b^* f(x) \; = \; \sum_{I\in D} b_I \left\langle
f;h_I \right\rangle \frac{1}{\mdl{I}} \chi_I(x)
$$
with the sequence $b_I = \left( K_I^+ - K_I^- \right)\mdl{I}^{3/2}$.

In order for dyadic paraproduct and its adjoint to be bounded in
$L^2$ we need $b$ to be in the $BMO^d$, i.e. we need $\left\{ b_I^2
\right\}_{I\in D}$ to be a Carleson sequence, which is
$$
\forall J\in D \;\;\;\;\; \frac{1}{\mdl{J}}\sum_{I\in D}\left( K_I^+
- K_I^- \right)^2\mdl{I}^{3} \; \leq \; Q
$$
and coincides with $T(1)$ conditions (\ref{formula 20 T(1) cond}).

It has been shown in \cite{Beznosova:08} that the norm of the dyadic
paraproduct (and, by symmetry of $A_2$-constant, of its adjoint as
well) depends on the $A_2$-constant of the weight $w$ at most
linearly, i.e.
$$
\norm{T_3}{L^2(w)\rightarrow L^2(w)} + \norm{T_2}{L^2(w)\rightarrow
L^2(w)} \; \leq \; C\, Q^{1/2} [w]_{A_2^d}.
$$

And, finally, the linear bound on the operator $T_4$,
$$
T_4f(x) \; = \; \sum_{I\in D}\left( K_I^+ + K_I^- \right) \mdl{I}
\left\langle f;h_I \right\rangle h_I(x),
$$
is implied by the uniform boundedness of the symbol (i.e. size conditions (\ref{formula 19 size cond})) and
Wittwer's weighted linear bound on the martingale transform (see \cite{Wittwer:00}).

Since all parts of $T$ obey linear bounds on their $L^2(w)$ norms
with respect to the $A_2$-constant of the weight $w$, bound on the
norm of $T$ is linear as well,
$$
\norm{T}{L^2(w)\rightarrow L^2(w)} \; \leq \; C \norm{w}{A_2},
$$
where constant $C$ depends on the constant in the size condition
(\ref{formula 2 cond}), dyadic $BMO^d$-norms of $T(1)$ and $T^*(1)$
(at most linearly) and the constant in testing conditions
(\ref{formula 16 Th;h<Q}) (at most linearly as well). Therefore, we have the following theorem.

\begin{theorem} \label{t:A2conjecture}
Let $T$ be a perfect dyadic operator on $\R$ such that
$$\|T(1)\|_{BMO^d} \leqslant Q \;\;\; \text{and} \;\;\; \|T^*(1)\|_{BMO^d} \leqslant Q $$
and
$$ \forall I\in D \;\;\; \langle T h_I , h_I \rangle \leqslant Q $$.
Then $T$ is bounded on $L^2(w)$ and
$$
\|T f\|_{L^2(w)} \leqslant C Q \|f\|_{L^2(w)},
$$
with some constant $C$ independent of the operator $T$.
\end{theorem}

Theorem \ref{t:A2conjecture} in the case $w(x)=1$ is a $T(1)$ theorem, so we can view it as a weighted version of the $T(1)$ theorem. It is also interesting that the $L^2(w)$ norm of the operator $T$ after proper normalization (we assume that the decay constant of the kernel is $1$) only depends the $BMO^d$ norms of $T(1)$ and $T^*(1)$ and the constant in testing conditions. It also depends on these constants at most linearly.

\section{Two weight boundedness of perfect dyadic operators.}

We start with the decomposition (\ref{decomposition}),
$$
T f(x) = \frac{1}{4} (T_1 f(x) + T_2 f(x) + T_3 f(x) + T_4 f(x)).
$$
First we consider
$$
T_1 f = \sum_{I \in D} (K_I^+ + K_I^-) |I| m_If \chi_I(x).
$$
Similarly to the one weight case, by duality using Theorem \ref{theorem 1 NTrVo2}, we know it is bounded from $L^2(v)$ in $L^2(u)$ (with the norm bounded by $C Q_1$) whenever for every dyadic interval $J \in D$ the following three conditions hold simultaneously
\begin{equation}\label{e:T1-1}
\frac{1}{|J|} \sum_{I\in D(J)} (K_I^+ + K_I^-) |I|^2 m_I(v^{-1}) m_I u \leqslant Q_1,
\end{equation}
\begin{equation}\label{e:T1-2}
\frac{1}{|J|} \sum_{I\in D(J)} (K_I^+ + K_I^-) |I|^2  m_I u \leqslant Q_1 m_J u,
\end{equation}
\begin{equation}\label{e:T1-3}
\frac{1}{|J|} \sum_{I\in D(J)} (K_I^+ + K_I^-) |I|^2 m_I(v^{-1}) \leqslant Q_1 m_J (v^{-1}).
\end{equation}

 Second, consider the paraproduct $T_3$ and its adjoint $T_2$. In \cite{BCMP} we have sufficient conditions for the two weight boundedness for the dyadic paraproduct:

\begin{theorem}\label{T:para1} Let $\pi_b$ be the dyadic paraproduct operator associated to the sequence $b = \{b_I\}_{I\in D}$:
$$
\pi_bf := \sum_{I\in {D}} m_If \,b_I\, h_I.
$$
Let $(v,u)$ be a pair of measurable functions on $\R$ such that $u$ and
$v^{-1}$, the reciprocal of $v$, are
weights  on $\R$ and such that

(i) $(v,u)\in {A}_2^d$, that is
$ \; [v,u]_{{A}_2^d} := \sup_{I\in {D}}m_I (v^{-1})\, m_I u <\infty.$

(ii)
Assume that there is a constant $\mathcal{C}_{v,u}>0$ such that
\[ \sum_{I\in {D}(J)} |\Delta_I u|^2|I|\, m_I(v^{-1}) \leq \mathcal{C}_{v,u} u(J) \quad\quad \mbox{for all} \; J\in {D},\]
where $\Delta_I u :=  m_{I_+}u-m_{I_-}u$, and $I_{\pm}$ are the right and left children of $I$.

Assume that $b\in Carl_{v,u}$, i.e.
there is a constant $\mathcal{B}_{v,u}>0$ such that
\[ \sum_{I\in {D}(J)} \frac{|b_I|^2}{m_Iu} \leq \mathcal{B}_{v,u} v^{-1}(J)\quad\quad \mbox{for all} \; J\in {D}.\]

Then $\pi_b$, the dyadic paraproduct associated to $b$, is bounded from
$L^2(v)$ in $L^2(u)$.
Moreover,  there exist $C>0$ such that
$$\|\pi_bf\|_{L^2(u)}\leq C\sqrt{[v,u]_{{A}_2^d}\mathcal{B}_{v,u}}
\Big(\sqrt{[v,u]_{{A}_2^d}}+\sqrt{\mathcal{C}_{v,u}}\,\Big)\|f\|_{L^2(v)}\,.$$
\end{theorem}

Therefore, in order to be able to bound the paraproduct $T_3$ that has symbol $b_I = (K_I^+ - K_I^-) |I|^{3/2}$, we need the following three conditions to hold:

\begin{equation}\label{e:T3-1}
 [v,u]_{{A}_2^d} := \sup_{I\in {D}}m_I (v^{-1})\, m_I u <\infty,
\end{equation}
\begin{equation}\label{e:T3-2}
\sum_{I\in {D}(J)} |\Delta_I u|^2|I|\, m_I(v^{-1}) \leq \mathcal{C}_{v,u} u(J) \quad\quad \mbox{for all} \; J\in {D},
\end{equation}
\begin{equation}\label{e:T3-3}
\sum_{I\in {D}(J)} \frac{(K_I^+ - K_I^-)^2 |I|^{3}}{m_Iu} \leq \mathcal{B}_{v,u} v^{-1}(J)\quad\quad \mbox{for all} \; J\in {D}.
\end{equation}

The operator $T_2$ is adjoint to $T_3$, $T_2 = T_3^*$, so $T_2$ is bounded from $L^2(v)$ in $L^2 (u)$ whenever the paraproduct $T_3$ is bounded from $L^2(u^{-1})$ in $L^2(v^{-1})$. Therefore, in order for $T_2$ to be bounded from $L^2(v)$ in $L^2 (u)$ we need the following three conditions to hold:

\begin{equation}\label{e:T2-1}
 [u^{-1},v^{-1}]_{{A}_2^d} := \sup_{I\in {D}}m_I (u)\, m_I (v^{-1}) <\infty,
\end{equation}
\begin{equation}\label{e:T2-2}
\sum_{I\in {D}(J)} |\Delta_I (v^{-1})|^2|I|\, m_I u \leq \mathcal{C}_{u^{-1},v^{-1}} v^{-1}(J) \quad\quad \mbox{for all} \; J\in {D},
\end{equation}
\begin{equation}\label{e:T2-3}
\sum_{I\in {D}(J)} \frac{(K_I^+ - K_I^-)^2 |I|^{3}}{m_I (v^{-1})} \leq \mathcal{B}_{u^{-1},v^{-1}} u(J)\quad\quad \mbox{for all} \; J\in {D}.
\end{equation}

Finally, consider $T_4$, the martingale transform with symbol $\sigma = (K_I^+ +
K_I^-) |I|$, which is uniformly bounded by (\ref{formula 19 size cond}),
$$
T_4f(x) \; = \; \sum_{I\in D}\left( K_I^+ +
K_I^- \right) \mdl{I} \left\langle f;h_I \right\rangle h_I(x).
$$
Necessary and sufficient conditions were obtained in \cite{NazarovTreilVolberg:99}:
\begin{theorem}\label{t:NTV1} Let $\sigma = \{\sigma_I\}_{I\in D}$ be a sequence of signs $\pm$.
The family of operators $T_\sigma$ of martingale transforms with symbol $\sigma$ is  uniformly bounded from $L^2(v)$ to $L^2(u)$ if and only if the following assertions hold simultaneously:
\begin{equation}\label{e:T4-1}
\forall J \in D \;\;\; m_J u \; m_J v^{-1} \leqslant C <\infty,
\end{equation}
\begin{equation}\label{e:T4-2}
\forall J \in D \;\;\; \frac{1}{|J|} \sum_{I \in D(J)} (\Delta_I u)^2 m_I (v^{-1}) |I| \leqslant C m_I u,
\end{equation}
\begin{equation}\label{e:T4-3}
\forall J \in D \;\;\; \frac{1}{|J|} \sum_{I \in D(J)} (\Delta_I (v^{-1}))^2 m_I u |I| \leqslant C m_I (v^{-1}),
\end{equation}

and an operator $T_0$, defined by $T_0 f := \sum_{I\in D} \left |\frac{\Delta_I (v^{-1}) \Delta_I u}{m_I(v^{-1}) m_I u}\right | m_I f \chi_I (x)$, is bounded from $L^2(v)$ to $L^2(u)$, or, equivalently

\begin{equation}\label{e:T4-4}
\forall J \in D \;\;\; \frac{1}{|J|} \int_J \left(\sum_{I \in D(J)}  \left|\frac{\Delta_I u \Delta_I (v^{-1})}{m_Iu } \right| \chi_I(x) \right)^2 u dx \leqslant C m_J (v^{-1})
\end{equation}
and
\begin{equation}\label{e:T4-5}
\forall J \in D \;\;\; \frac{1}{|J|} \int_J \left(\sum_{I \in D(J)}   \left|\frac{\Delta_I u \Delta_I (v^{-1})}{ m_I(v^{-1})}\right| \chi_I(x) \right)^2 v^{-1} dx \leqslant C m_J u.
\end{equation}
\end{theorem}

Now we need to put all conditions we obtained for the operators $T_{1,2,3,4}$, conditions (\ref{e:T1-1}), (\ref{e:T1-2}), (\ref{e:T1-3}), (\ref{e:T3-1}), (\ref{e:T3-2}), (\ref{e:T3-3}), (\ref{e:T2-1}), (\ref{e:T2-2}), (\ref{e:T2-3}), (\ref{e:T4-1})(\ref{e:T4-2})(\ref{e:T4-3})(\ref{e:T4-4})(\ref{e:T4-5}), together.
Note that (\ref{e:T3-1}), (\ref{e:T2-1}) and (\ref{e:T4-1}) are all the same joint dyadic $A_2^d$ condition, while condition (\ref{e:T1-1}) follows from the joint $A_2^d$ and (\ref{formula 21 test cond}) under no additional assumptions.
Note  also that conditions (\ref{e:T3-2}) and (\ref{e:T4-2}) are the same, conditions (\ref{e:T2-2}) and (\ref{e:T4-3}) also coincide. Therefore, we obtain the following theorem.

\begin{theorem}\label{t:Tuv}
Let $(v,u)$ be a pair of measurable functions, such that $u$ and $v^{-1}$, the reciprocal of $v$, are weights on $\R$ such that
\begin{equation}\label{e:uv-1}
(v,u)\in A_2^d \,,\text{i.e.} \;\; [v,u]_{A_2^d} := \sup_{I\in D} m_I (v^{-1}) m_I u <\infty,
\end{equation}
\begin{equation}\label{e:uv-2}
\forall J \in D \;\;\; \frac{1}{|J|} \sum_{I \in D(J)} (\Delta_I u)^2 m_I (v^{-1}) |I| \leqslant C m_I u,
\end{equation}
\begin{equation}\label{e:uv-3}
\forall J \in D \;\;\; \frac{1}{|J|} \sum_{I \in D(J)} (\Delta_I (v^{-1}))^2 m_I u |I| \leqslant C m_I (v^{-1}),
\end{equation}
and operator $T_0$ is bounded from $L^2(v)$ in $L^2(u)$, i.e.
\begin{equation}\label{e:uv-4}
\forall J \in D \;\;\; \frac{1}{|J|} \int_J \left(\sum_{I \in D(J)}  \left|\frac{\Delta_I u \Delta_I (v^{-1})}{m_Iu } \right| \chi_I(x) \right)^2 u(x) dx \leqslant C m_J (v^{-1})
\end{equation}
and
\begin{equation}\label{e:uv-5}
\forall J \in D \;\;\; \frac{1}{|J|} \int_J \left(\sum_{I \in D(J)}   \left|\frac{\Delta_I u \Delta_I (v^{-1})}{ m_I(v^{-1})}\right| \chi_I(x) \right)^2 v^{-1}(x) dx \leqslant C m_J u.
\end{equation}
Let $T$ be a perfect dyadic operator with $\|T(1)\|_{BMO^d}, \|T^*(1)\|_{BMO^d} \leqslant Q$ and for every $I \in D$ we have $\langle T h_I; h_I \rangle \leqslant Q$.

Then $T$ is bounded from $L^2 (v)$ to $L^2(u)$ whenever for any dyadic interval $J \in D$

\begin{equation}\label{e:Tuv-1}
\frac{1}{|J|} \sum_{I\in D(J)} (K_I^+ + K_I^-) |I|^2  m_I u \leqslant C m_J u,
\end{equation}
\begin{equation}\label{e:Tuv-2}
\frac{1}{|J|} \sum_{I\in D(J)} (K_I^+ + K_I^-) |I|^2 m_I(v^{-1}) \leqslant C m_J (v^{-1}),
\end{equation}
\begin{equation}\label{e:Tuv-3}
\frac{1}{|J|}\sum_{I\in {D}(J)} \frac{(K_I^+ - K_I^-)^2 |I|^{3}}{m_Iu} \leq C m_J (v^{-1}
\end{equation}
and
\begin{equation}\label{e:Tuv-4}
\frac{1}{|J|}\sum_{I\in {D}(J)} \frac{(K_I^+ - K_I^-)^2 |I|^{3}}{m_I (v^{-1})} \leq C m_J u.
\end{equation}
\end{theorem}

\section{Sufficient conditions for the two weight boundedness of perfect dyadic operators under additional assumptions that weighted are in $A_\infty^d$}

In this section we will assume that $v^{-1}$ and $u$ are $A^d_\infty$ weights and show that conditions of Theorem \ref{t:Tuv} can be reduced in this case. A weight $w$ belongs to the class $A_\infty^d$ whenever
$$
[w]_{A_\infty^d} := \sup_{I\in D} m_I w e^{-m_I(\ln w)} \,<\, \infty.
$$
From \cite{mydisser} we have the following lemma

\begin{lemma} \label{t:littleoo}
Let $w$ be a weight, such that $\log w \in L^1_{loc}$, let $\{\lambda_I\}_{I\in D}$ be a Carleson sequence:
$$
\exists Q >0 \;\;s.t.\;\; \forall J\in D \;\;\; \frac{1}{|J|} \sum_{I \in D(J)} \lambda_I \leqslant Q.
$$
Then for every dyadic interval $J \in D$
$$
\frac{1}{|J|} \sum_{I\in D(J)} e^{m_I (\log w)} \lambda_I \leqslant 4 Q m_J w
$$
and therefore, if $w \in A_\infty^d$ then for every dyadic interval $J\in D$ we have:
$$
\frac{1}{|J|} \sum_{I\in D(J)} m_Iw \;\lambda_I \leqslant 4 Q [w]_{A_\infty^d} m_I w.
$$
\end{lemma}

In particular, $u,v^{-1} \in A_\infty^d$ implies that $u$ and $v^{-1}$ are in $RH_1^d$ (see \cite{BeRez}), where $RH_1^d$ is defined as follows
$$
w \in RH_1^d \;\;\;\Longleftrightarrow \;\;\; [w]_{RH_1^d} := \sup_{I\in D} m_I \left(\frac{w}{m_I w}\log \frac{w}{m_I w}\right) \;<\,\infty.
$$
We also know from \cite{BeRez} that $[w]_{RH_1^d} \leqslant \log 16 [w]_{A_\infty^d}$ and we have the following theorem that first appeared in \cite{Bu} without the sharp constant and in \cite{BeRez} it was stated in stronger form and the constant was traced. Here we state the strong form from \cite{BeRez}.
\begin{theorem} \label{t:Buckley}
Let $w$ be almost everywhere positive locally integrable function on the real line and $J$ be any interval. Then up to a numerical constant
$$
m_J \left(w \log w\right) - m_J w \log m_J w \approx \frac{1}{|J|} \sum_{I\in D(J)} \left(\frac{\Delta_I w}{m_I w}\right)^2 m_I w |I|.
$$
In particular case when $w$ is a weight in $RH_1^d$, we have that for every dyadic interval $J \in D$
\begin{equation}
\frac{1}{|J|} \sum_{I\in D(J)} \left(\frac{\Delta_I w}{m_I w}\right)^2 m_I w |I| \leqslant C [w]_{RH_1^d} m_J w.
\end{equation}
\end{theorem}

Firstly, note that conditions (\ref{e:uv-2}) and (\ref{e:uv-3}) follow from the fact that $u$ and $v^{-1}$ are in $A_\infty^d$, Theorem \ref{t:Buckley} and the joint $A_2^d$ condition (\ref{e:uv-1}).

Secondly, conditions (\ref{e:Tuv-1}) and (\ref{e:Tuv-2}) by Lemma \ref{t:littleoo} follow from the fact that $u, v^{-1} \in A_\infty^d$ and \ref{formula 21 test cond}. Therefore, under additional assumptions that $u,v^{-1} \in A_\infty^d$, we can simplify Theorem \ref{t:Tuv} as follows.

\begin{theorem}\label{t:Tuv-oo}
Let $(v,u)$ be a pair of measurable functions, such that $u$ and $v^{-1}$, the reciprocal of $v$, are $A_\infty^d$ weights on $\R$ such that
\begin{equation}\label{e:uv-1}
(v,u)\in A_2^d \,,\text{i.e.} \;\; [v,u]_{A_2^d} := \sup_{I\in D} m_I (v^{-1}) m_I u <\infty,
\end{equation}
and operator $T_0$ is bounded from $L^2(v)$ in $L^2(u)$, i.e.
\begin{equation}\label{e:uv-4}
\forall J \in D \;\;\; \frac{1}{|J|} \int_J \left(\sum_{I \in D(J)}  \left|\frac{\Delta_I u \Delta_I (v^{-1})}{m_Iu } \right| \chi_I(x) \right)^2 u(x) dx \leqslant C m_J (v^{-1})
\end{equation}
and
\begin{equation}\label{e:uv-5}
\forall J \in D \;\;\; \frac{1}{|J|} \int_J \left(\sum_{I \in D(J)}   \left|\frac{\Delta_I u \Delta_I (v^{-1})}{ m_I(v^{-1})}\right| \chi_I(x) \right)^2 v^{-1}(x) dx \leqslant C m_J u.
\end{equation}
Let $T$ be a perfect dyadic operator with $\|T(1)\|_{BMO^d}, \|T^*(1)\|_{BMO^d} \leqslant Q$ and for every $I \in D$ we have $\langle T h_I; h_I \rangle \leqslant Q$.

Then $T$ is bounded from $L^2 (v)$ to $L^2(u)$ whenever for any dyadic interval $J \in D$

\begin{equation}\label{e:Tuv-3}
\frac{1}{|J|}\sum_{I\in {D}(J)} \frac{(K_I^+ - K_I^-)^2 |I|^{3}}{m_Iu} \leq C m_J (v^{-1}
\end{equation}
and
\begin{equation}\label{e:Tuv-4}
\frac{1}{|J|}\sum_{I\in {D}(J)} \frac{(K_I^+ - K_I^-)^2 |I|^{3}}{m_I (v^{-1})} \leq C m_J u.
\end{equation}
\end{theorem}

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


text:
Rockaway Beach Boulevard, opened in 1886, was the first major east-west thoroughfare on the Rockaway Peninsula in the Borough of Queens in New York City. Much of its route parallels the Rockaway Freeway and the IND Rockaway Line above the Freeway. The boulevard first forks off at its eastern end from Beach Channel Drive at Beach 35th Street in Edgemere and merges once again with Beach Channel Drive by Jacob Riis Park shortly before the Marine Parkway–Gil Hodges Memorial Bridge.

Rather curiously, the Boulevard runs north of the Freeway from its eastern end, while Edgemere Avenue runs exactly south of the Freeway, until Beach 56th Street in Arverne, where the northern flank becomes Arverne Boulevard and Edgemere Avenue abruptly turns into Rockaway Beach Boulevard for the remainder of its run.

While the Boulevard served as the heart of a bustling business and entertainment district in the heart of the Rockaway Beach neighborhood, which included numerous hotels and amusement attractions, including the famous Rockaways' Playland, it was found that its original width could barely cope with the expanding traffic in its heyday. Much of the area in Arverne served by Rockaway Beach Boulevard was leveled through abortive urban renewal efforts in the 1960s.

Transportation
The IND Rockaway Line parallels Rockaway Beach Boulevard east Beach 116th Street. The Q22 runs on Rockaway Beach Boulevard east of Beach 73rd Street. The Q52+ Select Bus Service and Q53+ Select Bus Service run on Rockaway Beach Boulevard between Cross Bay Parkway and Beach 73rd Street and Beach 116th Street, respectively. The QM16 and QM17 express bus between Jacob Riis Park and Midtown runs on Rockaway Beach Boulevard west of Cross Bay Parkway and Beach 73rd Street, respectively.

References

Sources
Vincent Seyfried and William Asadorian, Old Rockaway, New York in Early Photographs, Dover Publications, Mineola, NY, 2000.

Streets in Queens, New York
Transportation in Rockaway, Queens

redpajama_set_name: RedPajamaWikipedia


text:
Home Movie Top 10 guy who plays robbie on victorious They Hide From You
Top 10 guy who plays robbie on victorious They Hide From You
Extra Information About guy who plays robbie on victorious That You May Find Interested
Matt Bennett | Victorious Wiki – Fandom
Matt Bennett | Victoria Justice Wiki – Fandom
Where Is Robbie From Victorious Now? – Looper
Who Is Matt Bennett? The Former 'Victorious' Star Is Still …
Matt Bennett biography: Age, movies and TV shows, and is …
Actor Matt Bennett – American Profile
Frequently Asked Questions About guy who plays robbie on victorious
Who played Robbie in Victorious now?
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Video About guy who plays robbie on victorious
Below is information and knowledge on the topic guy who plays robbie on victorious gather and compiled by the show.vn team. Along with other related topics like: Matt Bennett, Matt Bennett Ariana Grande, Matt Bennett Dynasty, Robbie Shapiro, Matt Bennett now, Matt Bennett wife, Matt Bennett DJ.
Matt Bennett
Robbie Shapiro
Actor, Writer, Musician, Singer-Songwriter
Years Active:
Matthew Hank "Matt" Bennett is an American actor and writer. He is best known for playing the role of Robbie Shapiro in the Nickelodeon sitcom, Victorious. His alter ego is Mott Bonnott.
Matt began his acting career with an appearance in one episode of the Comedy Central, Michael & Michael Have Issues as Greg the Intern.
Matt with Ariana Grande.
He also had a film debut by starring in a 2010 "mockumentary" called The Virginity Hit (produced by Adam Mckay and Will Ferrell) where he played a nerd attempting to lose his virginity while his friends film the events.
Matt co-starred as Robbie Shapiro in the Nickelodeon sitcom, Victorious. He co-wrote a short film, Text Me, with Bayou Bennett and Daniel Lir in 2010.[1] He also starred in the feature film The Virginity Hit, which was released on September 24, 2010. He had a very brief cameo in the film Bridesmaids.
On July 19, 2011, he released a single "I Love My Guitar" under the name Mott Bonnott (which is also his YouTube username).
On May 20, 2012, he released his debut EP, Warm Fuzzies Vol. 1 on his bandcamp under Mott Bonnott. He sold it for $5 and half the proceeds went to St. Jude Children's Hospital.
His star sign is Scorpio.
It is unknown whether Matt is a real ventriloquist or not. Nick.com said that Matt had special training in ventriloquism before he played Robbie, but it has been stated by cast and crew members that Rex's voice is dubbed by an uncredited Jake Farrow. At the very least, Bennett did the puppetry for Rex.
Unlike his co-stars, who began acting at young ages, Matt did not officially start acting until he was 18 years old.
Him and Avan
He lived with Avan Jogia for a year.
He is known as the class clown of the Victorious cast, which is more or less true to his character in the show.
He loves to do stand-up comedy.
He loves coconuts.
He loves laser tag.
His glasses in real life are larger than Robbie's.
Sundays are his favorite day of the week.
He's known for looking a lot like Andy Samberg.
His height is 5'10" (1.78 m).
He was the BrainSurge champion in the episode of BrainSuge that he appeared in.
He likes a variety of music, including Bob Dylan and screamo bands (according to costar Victoria Justice).
His alter ego is Mott Bonnott.
He has said in an interview that he is an 'adventurous eater' and that he has consumed stingray, snails, shark, alligator, sea urchin with raw quail eggs on top, and salmon roe (salmon eggs).
According to his Tumblr, his favorite veteran actors are Gene Wilder, Groucho Marx, David Cross, and Robin Williams.
He has a younger sister named Ally.
Has been said to have risque humor by the extra who confirmed Tori & Jade's Play Date. She said that Matt was making jokes about Avan Jogia's "package" due to the tightness of his pants in her blog.
He often tweets to much about his "Gizmo", he brings him everywhere and takes pictures of him to tweet them.
Ironically, even though Robbie appears to be one of the younger members of the group, Matt is actually the oldest member of the main cast (not including Daniella Monet), being the only one born in 1991, while Avan Jogia was born in '92, and Ariana Grande, Victoria Justice, Leon Thomas, and Elizabeth Gillies were born in '93.
He has been slimed 21 times so far.
He was a celebrity guest panelist in 'Figure It Out'.
The Hebrew name Matthew is from the name Matthias and means "Gift of god".
His favorite Robin Williams movies are Mrs. Doubtfire, Jumanji, Aladdin, Hook, and World's Greatest Dad. (This is confirmed on his tumblr.)
A lot of people believe that he is Jewish in real life, but he confirmed on his Tumblr that he's not.[citation needed]
His celebrity crush is Kyary Pamyu Pamyu.
The song "I Think You're Swell" was originally made before Victorious.
During a Twitch livestream in February 2021, where he reacted to Robarazzi and Survival of the Hottest, he said that Season 1's Cat was his favorite season of Cat. He further said that Cat's character Season 2 onward completely changed and was not the same.
Read More: Top 10 yucko the clown movies and tv shows They Hide From You
2010 The Virginity Hit Matt Film debut, main role
2011 Bridesmaids Helen's Stepson Minor role
2011 The Death and Return of Superman Metropolis Kid Short film
2009 Totally for Teens Jamie Pilot
Michael & Michael Have Issues Greg the Intern "Greg the Intern" (Season 1, Episode 1)
2010–2013 Victorious Robbie Shapiro Main role
Television guest appearances
Episode(s)
2011 BrainSurge Himself Contestant
iCarly Robbie Shapiro Guest Star in "iParty with Victorious"
(Season 4, Episodes 11-13)
2012 Figure It Out Himself Judge Panelist
Click here to see the photo gallery for Matt Bennett.
Matt Bennett – IMDb
Matt Bennett – Twitter
Matt Bennett Fanmail Address
Matt Bennett Tumblr
Matt Bennett Youtube
Matt Bennett ask.fm
Wikipedia has an article related to:
v • e • dVictorious Cast
Victoria Justice • Leon Thomas lll • Ariana Grande • Avan Jogia • Elizabeth Gillies • Matt Bennett • Daniella Monet
Recurring Cast
Eric Lange • Lane Napper • Michael Eric Reid • Jim Pirri • Jennifer Carta
Perez Hilton • Ke$ha • Jerry Trainor • Kenan Thompson • Jake Farrow • Yvette Nicole Brown • Jennette McCurdy • Jillian Clare • Shirley Jones • Tristin Mays • Ryan Rottman • Alexandra Shipp • Halston Sage • Mara Hall • Jamie Snow • Amber Urban
Author: victorious.fandom.com
Sumary: Matthew Hank "Matt" Bennett is an American actor and writer. He is best known for playing the role of Robbie Shapiro in the Nickelodeon sitcom, Victorious. His alter ego is Mott Bonnott. Matt began his acting career with an appearance in one episode of the Comedy Central, Michael & Michael Have Issues as Greg the Intern. He also had a film debut by starring in a 2010 "mockumentary" called The Virginity Hit (produced by Adam Mckay and Will Ferrell) where he played a nerd attempting to lose his vi
Matching Result: Matthew Hank "Matt" Bennett is an American actor and writer. He is best known for playing the role of Robbie Shapiro in the Nickelodeon sitcom, Victorious.
Intro: Matt Bennett Matt Bennett Gender: Male Age: 31 Born: November 13th, 1991 Character: Robbie Shapiro Occupation: Actor, Writer, Musician, Singer-Songwriter Years Active: 2009 – present Matthew Hank "Matt" Bennett is an American actor and writer. He is best known for playing the role of Robbie Shapiro in the Nickelodeon sitcom, Victorious. His alter ego is Mott Bonnott. Career Matt began his acting career with an appearance in one episode of the Comedy Central, Michael & Michael Have Issues as Greg the Intern. Matt with Ariana Grande. He also had a film debut by starring in a 2010 "mockumentary" called…
Source: https://victorious.fandom.com/wiki/Matt_Bennett
Author: victoria-justice.fandom.com
Sumary: Matthew H. "Matt" Bennett (born November 13, 1991) is an American actor, singer and screenwriter. He is best known for playing the role of Robbie Shapiro in the Nickelodeon sitcom Victorious and for starring in the Will Ferrell-produced film The Virginity Hit. Victoria Justice first met Bennett in late 2009, when filming of Victorious started. Bennett plays Robbie Shapiro, a weird, nerdy teenage boy who is one of Tori Vega's (Justice) friends.
Matching Result: Matthew H. "Matt" Bennett (born November 13, 1991) is an American actor, singer and screenwriter. He is best known for playing the role of Robbie Shapiro in …
Intro: Matt Bennett Matt Bennett Gender: Male Age: 20 Born: November 13, 1991 Character: Robbie Shapiro Occupation: Actor, singer Years Active: 2009-present Matthew H. "Matt" Bennett (born November 13, 1991) is an American actor, singer and screenwriter. He is best known for playing the role of Robbie Shapiro in the Nickelodeon sitcom Victorious and for starring in the Will Ferrell-produced film The Virginity Hit. Working With Victoria Victoria Justice first met Bennett in late 2009, when filming of Victorious started. Bennett plays Robbie Shapiro, a weird, nerdy teenage boy who is one of Tori Vega's (Justice) friends. This article is a…
Source: https://victoria-justice.fandom.com/wiki/Matt_Bennett
Read More: Top 10 how much is method man worth They Hide From You
Author: looper.com
Sumary: Here's what Matt Bennett, who played Robby on Victorious, has been doing since the show's last episode.
Matching Result: Robbie was played in the 2010-2013 Nickelodeon series by actor Matt Bennett. In the nine years since the series' last episode aired (it …
Intro: Where Is Robbie From Victorious Now? Nickelodeon By Helen A. Lee/Updated: Aug. 23, 2022 5:20 pm EST Of the group of Hollywood Arts High School students who made up the main cast in "Victorious," Robbie Shapiro is by far the most awkward. He's a ventriloquist who often carries around a scary-looking puppet, Rex, who is somehow suaver than Robbie could hope for. He has trouble interacting with girls, occasionally shows off some tech skills, has major ham boning skills, and can get carried away sometimes with his selfish actions. Fans may remember that time he drank more than his share of water when…
Source: https://www.looper.com/734969/where-is-robbie-from-victorious-now/
Author: suggest.com
Sumary: Get the latest on Victorious co-star Matt Bennett—and his friendship with Ariana Grande.
Matching Result: His most notable role to date was on Victorious. The actor played Robbie Shapiro from the show's 2010 inception to its sudden 2013 finale.
Intro: Who Is Matt Bennett? The Former 'Victorious' Star Is Still Friends With Ariana Grande Anyone familiar with Victorious knows that it helped launch Ariana Grande into superstardom. But what about fellow cast members? The musical sitcom, which ran on Nickelodeon from 2010 to 2013, starred a number of performers who have also gone on to have successful careers. The list includes actor and musician Matt Bennett. Find out who he was on the show and what he's up to today. (You may be surprised to learn that a Matt Bennett and Ariana Grande connection still remains.) Matt Bennett was born…
Source: https://www.suggest.com/who-is-matt-bennett/307/
Author: legit.ng
Sumary: MATT BENNETT is an American actor and singer known for playing Robbie Shapiro in Victorious. Find out more about his age, career, net worth, and sexuality.
Matching Result: MATT BENNETT is an American actor and singer known for playing Robbie Shapiro in Victorious. Find out more about his age, career, net worth, …
Intro: Matt Bennett biography: Age, movies and TV shows, is he gay? Matthew Hank Bennett, commonly known by his stage name Matt Bennett, is a sensational American-born actor, singer, and writer. He became a household name after his role as Robbie Shapiro in the Nickelodeon sitcom known as Victorious. Matt Bennett attends the LA Premiere of Gravitas Ventures. Photo: Rodin EckenrothSource: Getty ImagesMatt made his television debut in 2009 with the role of Jamie in the TV short Totally for Teens. In 2011, he was nominated for a Nickelodeon Kids' Choice Award (UK) in the category of Nick UK's Funniest Person….
Source: https://www.legit.ng/1383561-matt-bennett-biography-age-movies-tv-shows-gay.html
Author: americanprofile.com
Sumary: Does the 'Victorious' star do the voice of Rex?
Matching Result: Does Matt Bennett, who plays Robbie Shapiro on the Nickelodeon sitcom Victorious, really do the voice for his puppet Rex?
Intro: Actor Matt Bennett Celebrity Q&A, People https://americanprofile.com/articles/matt-bennett-rex/ By American Profile on September 26, 2011 https://americanprofile.com/wp-content/uploads/2011/09/matt_bennett_with_puppet_rex-150×150.jpg Does Matt Bennett, who plays Robbie Shapiro on the Nickelodeon sitcom Victorious, really do the voice for his puppet Rex? —Brian McCreary, Upper Sandusky, Ohio Wisecracking dummy Rex is voiced by writer-actor Jake Farrow, 32, but Farrow's role is uncredited to help enhance the idea that Rex thinks and talks for itself. Bennett, 19, has taken ventriloquism lessons, and his coach, Jay Johnson, 62, was influential in the creation of Rex's character. Johnson played Chuck Campbell in the 1977-1981 sitcom Soap, where he had a…
Source: https://americanprofile.com/articles/matt-bennett-rex/
If you have questions that need to be answered about the topic guy who plays robbie on victorious, then this section may help you solve it.
Matt Bennett (X) Matt Bennett was born on in Massapequa, New York, USA. He is an actor an
As of 2022, he currently tours as a DJ, playing songs from Disney and Nickelodeon series at several venues, including House of Blues chains. The shows are billed as "iParty with DJ Matt Bennett."
Matt Bennett/Appears in
In the episode, Robbie does not understand why he was referred to as a "ventriloquist", which either jokingly indicates that Rex has a mind of his own and is not controlled by Robbie at all or (in a more serious matter) that Robbie suffers from schizophrenia, therefore considering Rex a real person.
Ariana Grande, aka Cat Valentine, is now one of the biggest pop stars in the world, while Elizabeth Gillies (Jade West) and Victoria Justice (Tori Vega) have starred in a handful of notable movies and shows.
The Nickelodeon show "Victorious" was first broadcast on 27 March 2010, when Ariana Grande, in the role of Cat Valentine, was 16 years old.
Ariana said that if it were not an actor making a solo tour and having a cast tour instead, the authorities would have sanctioned the future Victorious seasons. By saying so, Ariana Grande shifted the burden of getting the show cancelled to her co-star, Victoria Justice.
However, in a more recent video, she explains that her brother thought she was an intruder when coming home late one night and smashed her head with a vase, resulting in her head bleeding. The blood seeped into her hair and stained it red.
Ariana is notably close friends with the rest of the Victorious cast. Liz Gillies, Daniella Monet and Matt Bennett all still hang out with her often, both in and out of public, and they even starred in her 'thank u, next' video.
why is victorious rated 12? Hi there, Season 2 of Victorious is rated 12 for uses of discriminatory language in the episode Helen Back Again. When we classify seasons, we rate the whole series on the basis of the highest classification.
Justin was on the set of Victorious in February 2011 for Nickelodeon's "Crush Week".
In "Tori Goes Platinum," Jade's facial expressions when she's watching Tori jump into Beck's arms in the hall and when Tori refuses to kiss Beck because she and Jade are friends. She's so devastated and it's heart-wrenching.
This show is amazing, really funny and appropriate for kids. Yes there are some kissing scenes, but kids see their parents kissing, so why not see it in a show. The show is about teens but very kid friendly and funny and sweet. It is hilarious and always gives a good laugh.
How much did Ariana Grande make on Victorious? Grande reportedly made $9,000 per episode for Season 1 of Victorious, likely getting pay increases for subsequent seasons. If we assume her pay stayed at $9,000 per episode, we can calculate that she made, at minimum, $513,000.


text:
Home Sports GAA Collins insists there's more in Clare
Collins insists there's more in Clare
Clare manager Colm Collins. Picture credit: Barry Cregg / SPORTSFILE
A place in the quarter-finals of the All-Ireland Senior Football Championship is on the line as Clare prepare to take on Roscommon.
Manager of the Clare Senior Footballers, Colm Collins has overseen some of the greatest days for the footballers in the County's recent history. He still feels there is more ability within the panel that has yet to be seen.
Collins is adamant several of the team would make their place on any football team in the country and is hopeful their journey won't come to an end in -.
"I think these players still have more in them, I think that a lot of our players would get on any team so I think there's a lot more in them, hopefully we're not finished yet".
Scan over recent championship campaigns for the County's footballers and a place in the last twelve would satisfy most followers of the game. However Colm is more ambitious and wants to make the quarter-finals.
"It's a game it's there now, we've to go out all hell-bent to give a really good performance and hopefully get over that, that's it there's only twelve left now so we're one of the twelve so we're not happy with that now at this stage".
All-Ireland Senior Football Championship
Colm Collins
Inagh/Kilnamona V Sixmilebridge SHC semi-final LIVE
Crusheen GAA Club launches first annual magazine
Newmarket-on-Fergus GAA release annual magazine
GAA star Collins announced as Elevate ambassador
Clare GAA fans asked to drive safely
Senior team presents gear to development squad
Fundraising event for Clare hurling teams
Devine goal pivotal as Clare in relegation dogfight
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Golden Triangle Tour – If you are traveling to India, and do not have much time to spare, you can have the gist of this heavenly country in just five days. Yes, opt for a golden triangle tour. Visit the capital of the country-Delhi, city of the monument of Love- Agra and a mute evidence of Rajput valor and chivalry- Jaipur. And all this can be enjoyed in just 5 Nights and 6 Days.
After your arrival to Delhi, you will be assisted by our representative and will be transferred to the hotel.
The morning will be spent in a majestic way by visiting some of the royal emblems of the country namely the Parliament House, President's House, Parliament House, Government Secretariat Buildings including some of the cremation sites of some of our freedom fighters. The grandeur of the Mughal emperors can be relived by viewing the sky high monuments of the era like Qutab Minar and the Humayun's Tomb. Your shopping spree will be best taken care of by the Chandni Chowk and connaught Place shopping areas some where in the afternoon. The eveninig will be spent in a religious way by visiting the most famous Lakshmi Narayan Temple; popularly known as Birla Mandir. Return to the hotel for an over night stay.
A four hours drive form here will take you to Agra where you can see the eighth wonder of the World- The Taj mahal and rekindle the flame of your love for your beloved. Built by Shah Jehan in 1630, this eternal symbol of love enshrines the mortal remains of both the forer and his beloved empress Mumtaz Mahal. From here you will move to Agra Fort to view Pearl Mosque, the halls of Public and Private Audience. You will also visit the tomb of Itmad-ud-Daullah built by Empress Noorjehan in memory of her father will also be on your list. Return to the hotel in the evening for an overnight stay in this city of Love.
Breakfast at the hotel and drive to Jaipur visiting Fatehpur Sikri also including the Jama Masjid, Tomb of Salim Chisti, Panch Mahal and other palaces en route.
Check in at the hotel and get prepared for an afternoon tour of the pink city.
On your visit to this 260 years old city, you will be first taken to the City Palace of Maharaja Jai Singh, the founder of the city. From here you will move on to the Observatory & Ram Niwas Gardens. From here, you will drive past Jal Mahal and Hawa Mahal through the pink rose residential and commercial areas. In the evening, you can pick your choose from the Johri Bazar and Bapu Market.
As you are in the royal city, enliven the Princely grandeur by visiting Amber Fort on the back of an Elephant that will take you to the main entrance of the fort that served as the ancient capital of the State. From here you will be moved to Sheesh Mahal or the Hall of Victory scintillating with mirrors. The lunch can again be an extravagant affair at the ethnically built Chokhi Dhani resort.
The evening can be leisurely enjoyed by watching the folk dances performed by the native artists.
Visit the ancient temple of Lord Ganesha – the elephant headed god of Hindus and seek his blessings for your journey backwards.
In the afternoon drive to Delhi and get transferred to the international airport to board flight for your onward journey.


Watch the Philippine men's team vs. the Brazil team today June 9, 2018. Game will be held at the Philippine Arena in Bocaue Bulacan. Game will start at 3:10PM.
This is the first game of both team in the tournament.