text:
Store in a Quiet Place, or Tweeting in Tynt Meadow
September 1, 2018 November 5, 2018 / pursuitofabbeyness
I have just returned from my summer holiday, and for a week of that holiday I was living in one of a small cluster of cottages, set in a clifftop woodland, that had no telephone, no mobile phone reception and no wifi.
The cold blast of digital wind would hit me only when I emerged from that shelter, at the crest of a cliff or a hill, or on the right spot somewhere in an old fishing village. Then the data would rush in, blowing away cobwebs as it refreshed email inboxes and social media feeds.
But every evening, once the bicycles were propped against the fence, the hiking boots unlaced and the heavy doors locked, the howling binary storm was silenced – not even muffled, but silenced – and the familiar winking green and white and blue blips that demand our attention so urgently were blackened, like a coastline without a lighthouse.
As it happens, I had been away from Twitter for a couple of weeks already. Not in response to anything in particular – just because I don't like it much, and it feels like an assertion of my rational humanity to choose not to do something that I don't much like. I had started to recognise that I would get a pleasurable feeling each time Twitter would email me with its increasingly desperate attempts to draw me back into its cacophony. Each time I saw its icon on my screen and refused to prod it, I got a tingle of satisfaction.
My week in the isolated cottage was a deeper version of that abstinence. But it was deeper to such an extent that it changed the experience: with even the possibility of opening the windows to the gales of the digital world absent, the pleasure was no longer that of being aware of the cold and pulling one's collar up tight against it, but that of being warm enough to forget the cold and redirect one's attention to other things, wholly, purely and without distraction.
Many Twitter users appear to reach a point at which they want or need to take a break from it. Strangely, a good number of them feel the need to announce this on Twitter.
For many years I was the only living journalist without a social media account. I wasn't even on LinkedIn. In fact, I bought my first smartphone as recently as 2015, when I quit the magazine world. I joined LinkedIn a year or so later, in order to stay in touch with colleagues and other contacts from my time in journalism. I finally made it onto Twitter and Instagram in 2017, specifically to promote this "Pursuit of Abbeyness" blog.
If that seems back-to-front, I should point out that I did not primarily report news. I was a magazine section editor, and a writer of analytical feature articles.
For the reporter, Twitter is an essential tool: a lot of news breaks on Twitter long before it appears as a press release, and within seconds there is an archive of reaction to it. I always considered social media to be a dangerous place for the analyst, however. My preference was to talk to practitioners in my field unmediated by the noise. This was not about ignoring important trends. Rather, it was about not herding my contacts into addressing those trends. Instead, I wanted them to tell me what they really thought, and respond to questions that were genuinely my own.
Twitter has a peculiar effect on a majority of its users – I specify a majority because it is a community that self selects for the more-than-averagely opinionated and combative among us. It simultaneously makes its users more pliant (their views are shaped by what they read there, rather than by their own observation and reflection) and more strident (anonymity, distance and brevity are all more conducive to high-handed dismissals than to reasoning).
Does it also make them more informed? I suppose it depends upon what we understand by "informed". It certainly tells you what's going on, and it influences what its users do – indeed, that seems to be the entire basis of social media as a commercial enterprise. When we do things in a more "informed" way, are we still doing them authentically, in our own way? And if not, does that matter? By becoming more superficially "informed", do we risk losing the possibility of becoming somehow more profoundly informed?
"Cistercians esteem the value of simplicity," according to some words on the website of the Mount Saint Bernard Abbey in Leicestershire. "Simplicity doesn't stand for a thing done simply, or cheaply, but rather represents a distillation of complexity. It is about processing and ordering a rich, varied reality in such a way that the result seems self-evident: 'This is how it has to be!'"
Cistercians like this kind of paradox. I imagine that, for them, a paradox like this reflects something of the mystery of the Divine, and embracing it is part of what it means to be faithful.
"Despite living a life apart, monks are open to the world," it says a little further down the page. "We carry the world's anxieties and hopes in our prayers. We're always glad to welcome guests who turn up on our doorstep. The monks of old had a saying: Patet porta, cor magis. 'The door is open, the heart even more so.'"
Patet porta, cor magis.
This is a door that is open to allow the world in, rather than to allow the monks out. It is a passive sort of openness. It does not seek to impose. At the same time, however, it is assertive, in that it merely enables openness to co-exist with apartness.
The monks engage on their own terms: their hearts are open, but they also work with the assumption that those who come through their open door do so with their own hearts open to respect the unique identity of their vocation.
In July this year, Mount Saint Bernard became the latest Cistercian abbey, and the first in the UK, to brew a commercially-available beer certified as a Trappist product. It joined six in Belgium, two in the Netherlands, one in France, one in Austria, one in Italy, one in Spain and one in the US. The words above were written to describe this new Trappist Ale, which is called Tynt Meadow, after the local landscape in which their quiet work is done, and the only tweeting is done by the birds.
The value that Cistercians place upon work done in quietness is well-known. They regard it as a form of prayer or devotion. There is a certain hard-headedness about it, too. Work is not done for work's sake, or even just for God's sake. The monks receive no subsidy or stipend to support their day-to-day living and therefore commercial work and trade is necessary for sustenance. That abbey door is open for reasons other than good fellowship.
The monks of Mount Saint Bernard know this as well as any. Some work in carpentry, some in ceramics, others in bookbinding, candlemaking and miscellaneous lower-skilled crafts, still others in the guesthouse that welcomes pilgrims or retreatants. The most economically-significant work, however, used to be done in agriculture and dairy farming. Modern farming methods and low milk prices mean that this work is no longer profitable enough to support the abbey's "from each according to his ability, to each according to his need" economy. The decision to establish a brewery followed an urgent search for a more suitable source of income.
The door was kept open as the monks set about creating experimental recipes with a homebrewing kit.
"The monks of Norcia, Saint-Wandrille, and Zundert have taught us a lot, and were generous in sharing their own brewing experience," they write. "We've received invaluable advice from the eleven other Trappist breweries, and from the International Trappist Association. We've also benefited from the kindness and counsel of several local brewers."
They describe the beer they came up with as "a strong dark ale", at 7.4% ABV, brewed in a nod to the "great Trappist tradition". We would recognise it as a Belgian abbey Dubbel.
The monks need not have brewed a beer like this. The tradition the brewers refer to is not particularly old. The abbey ale styles of Belgium date from the early 20th century. Moreover, Mount Saint Bernard appears to have its own brewing tradition: while there are no documented recipes for the modern brewery to draw upon, the monks report that 19th-century visitors wrote about drinking the abbey's table beer.
And yet it would have been unsatisfying, somehow, had they chosen to make a modern version of that table beer, or some traditional English ale style – let alone something current like a hop-forward cloudy IPA. To be commercially viable the brewery needs to differentiate itself from the crowded modern beer market. The Trappist brand is its most valuable asset with which to do so, and the traditional Belgian abbey styles resonate most with that brand.
Most of the other Trappist breweries outside of the Low Countries at least allude to these styles in their range, despite their being no requirement to do so and no particularly long tradition of it. Even the most distinctive, the Spencer Brewery of Saint Joseph's Abbey in the US, backs up its range of "American Trappist Craft Beers" – including an Imperial Stout, an IPA and a Lager – with a Patersbier, a Christmas Spiced Ale and a Strong Dark Abbey Ale.
Still, the Monks of Saint Bernard are right to say that they have created an ale "with a clearly English character".
My first taste came in the estimable little bar run by the Six°North brewery in Aberdeen.
The latest certified Trappist ale, and the first made in the UK
Tynt Meadow has more dark malt character than a Belgian Dubbel, whose colour would come almost exclusively from candi sugar. It is sweeter and correspondingly less dry and "digestible", while staying just the right side of cloying or heavy. The brewers note flavours of dark chocolate, liquorice, "rich fruit", pepper and fig. I would agree, and add that the dark chocolate malt character is complemented by a certain smokiness in the finish, as well as other malt flavours on the palate such as biscuit and vanilla.
More notably, it was immediately obvious to me that this beer had been fermented with an English, rather than a Belgian strain of yeast. The shock of the high-gravity wort, and possibly a high fermentation temperature, has brought forth some very strong esters and fusels characteristic of English ales: particularly the rose aroma of phenethyl alcohol and a touch of the solvent-like bite of ethyl acetate.
I think this beer may need a few more batches before the fermentation temperature is exactly right to capture the signal Englishness of the yeast without unbalancing the overall flavour. At that point, the monks of Mount Saint Bernard will have created a Belgian abbey ale that speaks with a true English accent.
There is an alternative, however. The fresh beer could retain these rough edges and the brewers could instead call upon their consumers to learn the patience of the cenobite and allow those esters and fusels to soften and mellow with age and oxidation.
Indeed, perhaps the Mount Saint Bernard monks have already hinted at precisely this – and more besides. They tell us that their Tynt Meadow ale should be "stored in a cool, dark, quiet place". There is science behind those first two conditions, coolness and darkness. I do not know of the science behind storing an ale in a quiet place.
But it does seem right, somehow.
abbey, Aberdeen, beer, beer ageing, beer styles, Belgium, Cistercians, dubbel, esters, fusels, Mount Saint Bernard Abbey, Norcia, oxidation, Saint Joseph's Abbey, Saint-Wandrille, Six Degrees North, social media, Spencer Brewery, trappist, Tynt Meadow, UK, yeast, Zundert
← Brew Day!… "Espress Yourself"
A Tale of Two Breweries →
3 thoughts on "Store in a Quiet Place, or Tweeting in Tynt Meadow"
Gary Gillman
Hello, good post. I like you how examined goals and objects of Cistercian life in relation to the brewing project. After all there is a real and close connection and it shouldn't be overlooked. Plus, good taste notes, more qualified than others I've read.
Just two points.
In mid-2016, I documented the use of table beer at MSB in the 1800s. You may find it of interest as you referenced this subject.
Also, abbey beer as we know it today predates WW I and the Vandevelde law that restricted spirits (and did result in more strong beer being made).
The links below address these points, e.g., Chimay had a c.7% ABV beer in the 1870s and some strong beer was made by abbeys that "supported water" for everyday use. Sometimes or perhaps invariably this beer wasn't consumed by the monks but sold to the public.
Still there was a tradition to brew it, and one I argue came from England to start with. Hence the great suitability of MSB using an English brewing approach, which I argued for in mid-2016. Things come full circle, in other words.
http://www.beeretseq.com/how-strong-was-chimay-beer-139-years-ago/
http://www.beeretseq.com/brewing-at-mount-saint-bernard-abbey-in-england-19th-century/
Gary Gillman, Toronto
Twitter: @beeretseq
pursuitofabbeyness
Thanks for your comment, Gary!
I totally agree, and acknowledge the pre-20thC history of abbey brewing – after all, monks were probably the first industrial-scale brewers, who took the activity out of the hands of women in the home.


The company has been established in 2001.For this period we've become one of the leading companies trading with metal working machines.
17 years of experience in the trade of metalworking machines.
Our warehouse is situated on 4000 sq.m.
Our warehouse is situated on 4000 square meters, we have more than 300 machines in stock, like boring machines, gear machines, grinding machines, lathes, milling, drilling machines, CNC machines etc.


I recently received photos of the finished first and second grade classroom building of our Village Schoolhouse Project. Last July, I stood on this vacant land and saw only a pile of bricks. I am now amazed and overjoyed to see a completed structure in this space. Thanks to your generous giving, this September 5th, 200 young scholars from 21 villages will gather here to begin their education. The government has approved the building and is now adding the school to their system. They are ordering desks and supplies so classes can begin in September.
Our partners in Malawi asked me to provide a name for the school, which I did not consider lightly. After giving it much thought, I decided to name the school M'bwezela Junior Primary School. M'bwezela is the Chichewa (the native language of Malawi) word for Restoration. This school will restore dignity, health, opportunity and hope to the communities it serves.
This was all the confirmation I needed that the name is good.


When Elias Christeas piloted his 1989 Range Rover Classic to the front of the pack during the 2011 SCORE Baja 1000 race, it was the culmination of a lifetime spent driving off-road, a do-it-yourself attitude where anything seems possible, and more than a year's worth of teamwork. The vehicle didn't finish the race, but for Christeas and his team, called In Search of the Experience (ISOTE), their first taste of Baja was enough to never look back.
Though the Rover didn't make the finish, Christeas and the team only paid $400 for it from a junkyard lot. It had 80,000 miles on the odometer, and the drivetrain looked worthy of the effort to bring the car back to life. They rebuilt and re-engineered the Rover to get it race ready, using a friend's driveway as a shop. They were amateurs, reaching for a dream of competing in the hardest automobile race in North America.
Before the idea to run the Baja 1000 took shape in 2004, Christeas wanted to run a Rover in the Dakar Rally. As it turned out, far more accessible than the Dakar was the Baja 1000. For Christeas living in Southern California, the choice was clear: He was headed to Ensenada. That's when Christeas found Gerardo Barragan, multiple-time winner of the Baja 1000 Class 3 Trophy.
Barragan served as a mentor and inspiration for Christeas. In 2008, ISOTE was founded, the same year that Christeas released a documentary by the same name that covered Rover enthusiasts in the U.S., and brought national attention to him within the Rover and off-road communities.
In addition to sourcing Barragan's wealth of knowledge on the subject, Christeas found answers to his many questions on Internet forums, specifically those dedicated to Rovers. These forums, Christeas says, are invaluable tools. They're global resources of information that can address nearly any issue that might arise while re-engineering a vehicle for off-road racing.
After buying the $400 Rover, called Steve to this day, and assembling a crew of like-minded friends, Team ISOTE began the hunt for sponsors. Funding, obviously, is one of the biggest obstacles facing do-it-yourselfers. Sponsors, for Christeas and others in the same situation, are the key to footing the bill for such an undertaking.
Sponsors like ESSE Knives, Hella and Pelican supported the team and helped make the race possible. But Christeas makes it clear that if you decide to do something like re-engineer a dead Rover for the Baja 1000, it's going to cost some money.
To the team, Steve will always be the $400 truck. But after the initial purchase, the expenses are unknown. Everyone on the team contributed time and money. And after a year of hard laboring for the love of racing off-road, Steve was ready to run the Baja 1000.
Full of confidence, Team ISOTE charged hard from the beginning. They led the class for 411 miles, they passed trophy trucks worth $100,000, and they experienced the thrill of racing in Baja. Most of all, they were deeply impressed with the sense of community they found.
Coming out of the Ojo-Negros section of the race, which is the most heavily populated by spectators, Christeas and his co-driver Michael Rady noticed two kids "in an elevated position preparing something," Christeas says. As he and Rady raced by, the kids threw a package in the Rover. They had assumed the kids were throwing M-80s at them, but the package was actually a wad of micro-fiber towels.
And the second after the Rover passed the kids, it plunged into a mud obstacle. The kids were providing racers with towels to wipe mud from their visors.
The horror stories, though, are there. Racing Baja is no walk in the park, and the locals often provide unexpected adversity, sometimes in the form of booby traps. Like digging ditches or holes in the track and covering them with branches. Or running a strand of barbed wire across the track and lifting it as a dirt biker passes.
Team ISOTE withdrew from the 2011 race on the night of the second day. They were running out of time and the vehicle was worse for the wear. Their first clue that something was wrong with the suspension was the right strut poking through the hood. Although they didn't make the finish line, they ran their race.
But early on the team made a mistake. Which led to the suspension failure that prevented them from finishing the race. When they first purchased the vehicle they failed to notice that the frame had been twisted. The front left corner of the vehicle showed mild damage from an accident, but the team never gave the frame a close inspection—until the post-race teardown, when they noticed the left side of the frame was an inch and a half behind the right side.
The frame has since been straightened, and Team ISOTE is on track to race Baja again this year using the same chassis. When SCORE releases the course information in September, Christeas and the team will head south to recon the route. They'll gather intel on everything from ruts and ditches to refueling points. They'll also test Steve to ensure the vehicle is race ready.
The 2013 SCORE Baja 1000 is scheduled for November 14 – 17.
Great story! I chased for a Class 1 car for years and really miss it. Glory Days!


text:
\section{\bf Introduction}

This paper, building upon the ideas in \cite{L}, continues the investigation of the boundedness properties of the bilinear Hilbert transform along curves. More precisely, if $\Gamma=(t, \, -\g(t))$ where here $\g$ is a suitable\footnote{In a sense that will be specified later.}
 smooth, non-flat curve near zero and infinity, we want to understand the behavior of the bilinear Hilbert transform along $\Gamma$ defined as
\beq\label{BHTc}
\eeq
$$H_{\G}: S(\R) \times S(\R)\longmapsto S'(\R)$$
$$H_{\G}(f,g)(x):= \textrm{p.v.}\int_{\R}f(x-t)g(x+\g(t))\frac{dt}{t}\:.$$
One can easily notice that taking $\g(t)=t$ we obtain the standard bilinear Hilbert transform. Thus, the problem considered in this paper is, in fact, the ``curved" analogue of the celebrated problem of providing $L^p$ bounds to the classical (``flat") bilinear Hilbert transform - the latter being solved in the seminal work of M. Lacey and C. Thiele (\cite{LT1}, \cite{LT2}).

It is worth noticing that similar studies regarding curved analogues of classical ``flat" objects arise naturally in harmonic analysis in various contexts. Indeed, recalling the discussion in the introduction of \cite{L}, a prominent such example  is given by the study of the boundedness of the linear Hilbert transform along curves given by
$$\H_{\G}:\: S(\R^n)\,\rightarrow\,S'(\R^n)\,,$$
$$\H_{\G}(f)(x):=p.v. \int_{\R} f(x-\G(t))\,\frac{dt}{t}\,,$$
where here $\Gamma:\:\R\,\rightarrow\,\R^n$, $n\geq 1$, is a suitable curve. This latter problem first appeared  in the work of Jones (\cite{Jo}) and  Fabes and Riviere (\cite{FR}) in connection with the analysis of the constant coefficient parabolic differential operators. The study of $H_{\G}$ was later extended to cover more general and diverse situations (\cite{NVWW} \cite{SW}, \cite{CNSW},\cite{C1}, \cite{C2}).

In the bilinear setting, the work on the ``curved" model referring to \eqref{BHTc} was initiated by X. Li in \cite{Li}. There, he showed that, for the particular case $\g(t)=t^d$ with $d\in\N$, $d\geq 2$, one has that
\beq\label{L2}
H_{\G}\:\:\textrm{maps}\:\:L^{2}(\R)\times L^{2}(\R)\,\rightarrow\,L^{1}(\R)\:.
\eeq
His proof relies on the concept of $\sigma-$uniformity, previously used in \cite{CLTT} and originating in the work of T. Gowers (\cite{G}).

In \cite{L}, the author proved that \eqref{L2} remains true for any curve $\g$ that belongs to $\n\f^{C}$ - a suitable  class\footnote{For its definition the reader is invited to consult Section 2. It is worth saying that $\n\f^{C}$ contains in particular the class of all polynomials without linear term - for more on this see Observation \ref{W}.} of smooth non-flat curves near zero and infinity. Our work greatly extended Li's result both qualitatively by significantly widening the class of curves, and quantitatively by revealing the scale type decay relative to the level sets of the multiplier's phase. Our proof is based on completely different methods, involving in a first instance a delicate analysis of the multiplier followed then by a special wave-packet discretization adapted to the two-directional oscillatory behavior of the phase. In the Appendix of the same paper, we also explained the main idea of how to upgrade the methods employed in \cite{Li} in order to be able to obtain the scale type decay.

Later, X. Li and L. Xiao, (\cite{LX}), relying heavily on \cite{L} in both the analysis treatment\footnote{See for example the decomposition of pp. 16 in \cite{LX} versus the one in Section 5.3. of \cite{L}, the treatment at pp. 29 in \cite{LX} versus the corresponding one at pp. 323 in \cite{L}, the perturbative strategy applied in analyzing the phase of the multiplier etc.} of the multiplier and the ``upgraded" $\sigma$-uniformity approach required for capturing the key necessary scale type decay revealed in \cite{L}, proved that if $P$ is any polynomial of degree $d\in\N$, $d\geq 2$ having no constant and no linear term \footnote{Any such polynomial $P$ is just a particular example of an element $\g\in\n\f$.} then taking $\g(t)=P(t)$ one has that $H_{\Gamma}$ maps boundedly $L^{p}(\R)\times L^{q}(\R)\:\:\mapsto\:\:L^{r}(\R)$ where $p,\,q,\,r$ obey $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}$, $1<p<\infty$, $1<q<\infty$ and $\frac{d}{d-1}<r<\infty$. (The bounds here depend only on the degree $d$ but not on the coefficients of $P$). Finally, more recently, in \cite{GX}, the authors take the proof in \cite{L} and translate it in the $\sigma-$uniformity language used initially in \cite{Li}.

The present paper should be regarded as a natural continuation of \cite{L}. Relying on the author's previous methods, we extend the earlier results by showing that for any $\g\in\n\f$ the bounds on $H_{\Gamma}$ can be extended to cover the natural Banach triangle case. Our result is optimal up to end points. For a precise statement of our result as well as for a reminder of the definition of the class $\n\f$ of ``smooth non-flat" curves near zero and infinity one is invited to consult the next section.
$\newline$
\textbf{Acknowledgement.} I would like to thank my wife Anca for drawing Figure 1 in this paper.
$\newline$
\section{\bf Main results}
$\newline$

We start by recalling from \cite{L} the definition of the set $\n\f_{0}$ of all curves $\g$ which are smooth non-flat functions near the origin:

\begin{itemize}

\item \textit{smoothness, no critical points, variation (near origin)}
\beq\label{nocrtic0}
\eeq
$\exists\: \d>0$ (possibly depending on $\g$) and $\V_{\d}:=(-\d,\d)\setminus\{0\}$ such that $\g\in C^N(\V_{\d})$ $(N\geq4)$ and $|\g'|>0$ on $\V_{\d}$;
moreover
\beq\label{variation0}
\sup_{\a\in\R_{+}}\#\{j\in\Z_{+}\,|\,|2^{-j}\,\g'(2^{-j})|\in[\a,2\a]\}<\infty\:,
\eeq
where here $\Z_{+}:=\{j\in\Z\,|\,j\geq 0\}$.

\item \textit{asymptotic behavior (near origin)}

There exists $\{a_j\}_{j\in\N}\subset\R_{+}$ with $\lim_{j\rightarrow\infty}a_j=0$ such that:
$\newline$

For any $t\in I:=\{s\,|\,\frac{1}{4}\leq|s|\leq 4\}$ and $j\in \Z_{+}$ we have
\beq\label{asymptotic0}
\frac{\g(2^{-j}\,t)}{2^{-j}\,\g'(2^{-j})}=Q(t)\,+\,Q_j(t)\,,
\eeq
with $Q,\,Q_j\in C^N(I)$ and $\|Q_j\|_{C^N(I)}\leq a_j$.

For $s\in J=Q'(I)$ we require
\beq\label{asymptotic0gaminv}
\frac{(\g')^{-1}(s\,\g'(2^{-j}))}{(\g')^{-1}(\g'(2^{-j}))}=r(s)\,+\,r_j(s)\,,
\eeq
where $r,\,r_j\in C^{N-1}(J)$ with $\|r_j\|_{C^{N-1}(J)}\leq a_j\:.$

(The existence of $(\g')^{-1}$, the inverse of $\g'$, will be a consequence of the next hypothesis.)

\item \textit{non-flatness (near origin)}

The main terms in the asymptotic expansion obey
\beq\label{fstterm0}
 \inf_{t\in I} |Q''(t)|,\:\inf_{t\in J} |r'(t)|>c_{\g}>0\:,
\eeq
and
\beq\label{convdualphase0}
 \inf_{{t_1,\,t_2\in J}\atop{t_1\not=t_2}}\frac{|t_1\,r'(t_1)-t_2\,r'(t_2)|}{|t_1-t_2|}>c_{\g}\:.
\eeq
\end{itemize}

In a similar fashion one can define $\n\f_{\infty}$ - the class of smooth, non-flat near infinity functions $\g$ having the following properties:

\begin{itemize}

\item \textit{smoothness, no critical points, variation (near infinity)}
\beq\label{nocrticinf}
\eeq
$\exists\: \d>0$ (possibly depending on $\g$) and $\tilde{\V}_{\d}:=(-\infty,-\d)\cup (\d,\infty)$ such that $\g\in C^N(\tilde{\V}_{\d})$ $(N\geq4)$ and $|\g'|>0$ on $\tilde{\V}_{\d}$;
moreover
\beq\label{variationinft}
\sup_{\a\in\R_{+}}\#\{j\in\Z_{-}\,|\,|2^{-j}\,\g'(2^{-j})|\in[\a,2\a]\}<\infty\:,
\eeq
where here $\Z_{-}:=\{j\in\Z\,|\,j\leq 0\}$.

\item \textit{asymptotic behavior (near infinity)}

 There exists $\{\tilde{a}_j\}_{j\in \Z_{-}}\subset\R_{+}$ with $\lim_{j\rightarrow-\infty}\tilde{a}_j=0$ such that:

For any $t\in I:=\{s\,|\,\frac{1}{4}\leq|s|\leq 4\}$ and $j\in \Z_{-}$ we have
\beq\label{asymptoticinfty}
\frac{\g(2^{-j}\,t)}{2^{-j}\,\g'(2^{-j})}=\tilde{Q}(t)\,+\,\tilde{Q}_j(t)\,,
\eeq
with $\tilde{Q},\,\tilde{Q}_j\in C^N(I)$ and $\|\tilde{Q}_j\|_{C^N(I)}\leq \tilde{a}_j$.

For $s\in \tilde{J}=\tilde{Q}'(I)$ we require
\beq\label{asymptoticinftygaminv}
\frac{(\g')^{-1}(s\,\g'(2^{-j}))}{(\g')^{-1}(\g'(2^{-j}))}=\tilde{r}(s)\,+\,\tilde{r}_j(s)\,,
\eeq
where $\tilde{r},\,\tilde{r}_j\in C^{N-1}(\tilde{J})$ with $\|\tilde{r}_j\|_{C^{N-1}(\tilde{J})}\leq \tilde{a}_j\:.$

\item \textit{non-flatness (near infinity)}

The main terms in the asymptotic expansion obey
\beq\label{fstterminfty}
 \inf_{t\in I} |\tilde{Q}''(t)|,\:\inf_{t\in \tilde{J}} |\tilde{r}'(t)|>c_{\g}>0\:,
\eeq
and
\beq\label{convdualphaseinfty}
 \inf_{{t_1,\,t_2\in \tilde{J}}\atop{t_1\not=t_2}}\frac{|t_1\,\tilde{r}'(t_1)-t_2\,\tilde{r}'(t_2)|}{|t_1-t_2|}>c_{\g}\:.
\eeq
\end{itemize}

With this done, we set $$\n\f:=C(\R\setminus\{0\})\cap\n\f_{0}\cap\n\f_{\infty}$$ and $\n\f^{C}:=\n\f\,+\,Constant$.

\begin{o0}\label{W} Following \cite{L}, we list here some interesting features of the class $\n\f$:
\begin{itemize}
\item Any real polynomial of degree $\geq 2$ with no constant and no linear term belongs to $\n\f$;

\item More generally, any finite linear combination over $\R$ of terms of the from $t^{\a}$ with $\a\in (0,\infty)\setminus\{1\}$ belongs to $\n\f$;

\item Even more, any finite linear combination over $\R$ of terms of the form
$|t|^{\a}\,|\log |t||^{\b}$ with $\a,\,\b\in\R$ and $\a\notin\{-1,\,0,\,1\}$ is in $\n\f$;

\item If $\g\in\n\f$ then there exist $K_2\geq K_1>0$ and $C_2\geq C_1>0$ (all constants are allowed to depend on $\g$) such that for any $t\in V(0)\setminus\{0\}$ or $t\in V(\infty)$, respectively
\beq\label{growth}
\eeq
\begin{itemize}
\item either $K_2^{-1}\,|t|^{C_2}<|\g'(t)|<K_1^{-1}\,|t|^{C_1}$;
\item or $\frac{K_1}{|t|^{C_1}}<|\g'(t)|<\frac{K_2}{|t|^{C_2}}$.
\end{itemize}
$\newline$

Thus, if  $\g\in\n\f$, one has:
\beq\label{growth1}
\exists\:\:\lim_{{t\rightarrow 0}\atop{t\not=0}} |\g'(t)|\in\{0,\,\infty\}\:\:\:\textrm{and}\:\:\:\exists
\lim_{t\rightarrow \infty} |\g'(t)|\in\{0,\,\infty\}\:.
\eeq
\end{itemize}
\end{o0}

In \cite{L}, we proved the following result
$\newline$

\noindent\textbf{Theorem.} \textit{Let $\G=(t,-\gamma(t))$ be a curve such that $\g\in\n\f^{C}$. Recall the definition of the bilinear Hilbert transform $H_{\G}$ along the curve $\G$:
$$H_{\G}: S(\R) \times S(\R)\longmapsto S'(\R)$$
$$H_{\G}(f,g)(x):= \textrm{p.v.}\int_{\R}f(x-t)g(x+\g(t))\frac{dt}{t}\:.$$
Then $H_{\G}$ extends boundedly from $L^2(\R)\times L^2(\R)$ to $L^1(\R)$.}
$\newline$

In the present paper, we extend the boundedness range of the above theorem to the Banach triangle (see Figure 1):
$\newline$

\noindent\textbf{Main Theorem.} \textit{If $\g\in\n\f^{C}$ and $H_{\G}$ defined as above, we have that
\beq\label{bounds}
H_{\G}:\:\:\:\:\:L^{p}(\R)\times L^{q}(\R)\:\,\mapsto\:\, L^{r}(\R)\,,
\eeq
where the indices $p,\,q,\,r$ obey
\beq\label{holder}
\frac{1}{p}+\frac{1}{q}=\frac{1}{r}\,,
\eeq
with
\beq\label{ban}
1<p<\infty\,,\:\:\:1<q\leq\infty\,,\:\:\:\textrm{and}\:\:\:1\leq r<\infty\,.
\eeq}
$\newline$
\begin{o0}\label{optimal} The Banach range in our Main Theorem is optimal up to end-points. Indeed, let us define $\p_d$ be the class of all real polynomials of degree $d$ with no constant and no linear terms. Then, in \cite{Li} (pp. 9), it is shown that for any $d\in\N$, $d\geq 2$, there exists a polynomial $P_d\in \p_d$ such that for $\g=P_d$ one has that
\beq\label{cex}
\|H_{\G}\|_{L^{p}(\R)\times L^{q}(\R)\,\mapsto\,L^{r}(\R)}=\infty
\eeq
whenever $(p,\,q,\,r)$ obeys \eqref{holder} with $r<\frac{d-1}{d}$.

Since from Observation \ref{W} we deduce that for any $d\in\N$, $d\geq 2$, we have $\p_d\subset \n\f$ we conclude that in order for
\eqref{bounds} and \eqref{holder} to hold for \textit{any} $\g\in\n\f^{C}$ one must have $r\geq 1$.
\end{o0}

\begin{figure}[!h]
\begin{center}
\epsfig{file=Maulica311.eps,height=13cm}
\end{center}
\caption{\textbf{The boundedness range for the Bilinear Hilbert Transform $H_{\Gamma}$.}
In this figure we represent the bounds for our object viewed as a trilinear form defined by $\Lambda(f,g,h)=\int H_{\Gamma}(f,g)\,h$. Our Main Theorem states that $\Lambda$  maps boundedly $L^p\times L^q\times L^{r'}$ into $\C$ for all triples $(\frac{1}{p},\,\frac{1}{q},\,\frac{1}{r'})$ that belong to the region
$\Omega:=\textit{int}\,(\triangle ABC)\cup(AC)\cup(AB)\;.$
These bounds are optimal up to the bold boundaries defined by
$[BC]\cup\{A\}$.}
\end{figure}

\section{\bf Preparatives; isolating the main component of our operator}
$\newline$

In Section 3 of \cite{L}, after elaborated technicalities,  we proved that the study of our bilinear operator $H_{\G}(f,g)$, can be reduced to the corresponding study of the bilinear operator $T(f,g)$ defined as follows:
\beq\label{T}
T(f,g):=\sum_{j\in\Z}\sum_{m\in \N} T_{j,m}\,,
\eeq
with
\beq\label{pieces}
T_{j,m}(f,g)(x):= \int_{\R}\int_{\R}\hat{f}(\xi)\hat{g}(\eta)v_{j,m}(\xi,\eta) e^{i\xi x} e^{i\eta x} d\xi d\eta\:
 \eeq
and
\beq\label{vjm}
v_{j,m}(\xi,\eta):= 2^{-\frac{m}{2}}\,e^{i\,\varphi_{\xi,\eta}(t_c)}\,
\zeta\left(\frac{\eta\,\g'(2^{-j})}{2^{m+j}},\,\frac{\frac{\xi}{2^{m+j}}}{\frac{\eta\,\g'(2^{-j})}{2^{m+j}}}\right)\,
\phi\left(\frac{\xi}{2^{m+j}}\right)\,\phi\left(\frac{\eta\,\g'(2^{-j})}{2^{m+j}}\right)
 \eeq
where we have
\begin{itemize}
\item the function $\phi$ is smooth, compactly supported with
\beq\label{ssuport}
\textrm{supp}\,\phi\subset\{x\,|\,\frac{1}{10}<|x|<10\}\:.
\eeq

\item the phase of the multiplier is defined as
\beq\label{phase}
\varphi_{\xi,\eta}(t):=-\frac{\xi}{2^j}\,t+\eta\,\g(\frac{t}{2^j})\:.
 \eeq

\item for $\xi,\,\eta,\,j$ fixed (and based on the properties of $\g$) there exists exactly one critical point $t_c$
defined by
\beq\label{critpoint}
t_c=t_c(\xi\,,\eta,\,j)\in [2^{-k(\g)}, 2^{k(\g)}]\,\:\;\textrm{such that}\:\;\varphi'_{\xi,\eta}(t_c)=-\frac{\xi}{2^j}+\frac{\eta}{2^j}\,\g'(\frac{t_c}{2^j})=0\;,
\eeq
where here $k(\g)\in\N$ is an integer depending only on $\g$.

\item the function $\zeta$ obeys
\beq\label{zeta}
\zeta:\,[\frac{1}{10},10]\times[2^{-k(\g)},\,2^{k(\g)}]\,\rightarrow\,\R\:\:\:\textrm{with}\:\:\:\;\:\:\|\zeta\|_{C^{N-3}}\lesssim_{\g} 1\:.
\eeq
\end{itemize}

Now based on \eqref{growth1} and \eqref{growth}, wlog we can assume that

\beq\label{growth2}
\exists\:\:\lim_{{t\rightarrow 0}\atop{t\not=0}} \g'(t)=0\:\:\:\textrm{and}\:\:\:\exists
\lim_{t\rightarrow \infty} \g'(t)=\infty\:.
\eeq

The other cases can be treated similarly and we will not discuss them in detail here.

Setting
\beq\label{ph1}
\Psi_{\eta}(\xi)=-\varphi_{\xi,\eta}(t_c)\:,
\eeq
and following \cite{L}, we used the scaling symmetry in order to define the following operators:
\begin{itemize}

\item For $j>0$ (thus $2^{-j}\rightarrow 0$)

$$B_{j,m}(f(\cdot),g(\cdot))(x):=[\g'(2^{-j})]^{\frac{1}{2}}\,T_{j,m}\left( f(2^{m+j}\cdot), g(\frac{2^{m+j}}{[\g'(2^{-j})]}\cdot) \right)(\frac{x\,[\g'(2^{-j})]}{2^{m+j}})\:.$$
Remark that
\beq\label{poz0}
\eeq
$$B_{j,m}(f,g)(x)=2^{-\frac{m}{2}}\,[\g'(2^{-j})]^{\frac{1}{2}}\times$$
$$\int_{\R}\int_{\R}\hat{f}(\xi)\,\hat{g}(\eta)\,e^{i (\g'(2^{-j})\xi+\eta)\,x}\,e^{-i\,2^{m}\,2^j\,\Psi_{\frac{\eta}{\g'(2^{-j})}}(\xi)}\,
\zeta(\eta,\,\frac{\xi}{\eta})\,\phi(\xi)\,\phi(\eta)\,d\xi\,d\eta \:.$$

\item For $j<0$ (thus $2^{-j}\rightarrow \infty$)

$$B_{j,m}(f(\cdot),g(\cdot))(x):=
[\g'(2^{-j})]^{-\frac{1}{2}}\,T_{j,m}\left( f(2^{m+j}\cdot), g(\frac{2^{m+j}}{[\g'(2^{-j})]}\cdot) \right)(\frac{x}{2^{m+j}})\:.$$
As before, notice that
\beq\label{neg0}
\eeq
$$B_{j,m}(f,g)(x)=2^{-\frac{m}{2}}\,[\g'(2^{-j})]^{-\frac{1}{2}}\times$$
$$\int_{\R}\int_{\R}\hat{f}(\xi)\,\hat{g}(\eta)\,e^{i (\xi+\frac{\eta}{[\g'(2^{-j})]})\,x}\,e^{-i\,2^{m}\,2^j\,\Psi_{\frac{\eta}{\g'(2^{-j})}}(\xi)}\,
\zeta(\eta,\,\frac{\xi}{\eta})\,\phi(\xi)\,\phi(\eta)\,d\xi\,d\eta \:.$$
\end{itemize}
$\newline$

\begin{o0}\label{j}
In what follows we will focus on the case $j>0$ as the reasonings for the other case $j<0$ can be treated in a similar fashion.
\end{o0}
$\newline$
$\newline$
\textbf{Main reductions}
$\newline$

For $j\in\N$, using \eqref{growth2} and $\g\in\n\f$, one follows the reasonings from Section 5 in \cite{L} and successively simplifies the structure of $B_{j,m}$ as follows:
\begin{itemize}
\item for $s\in J$, recalling the definition of $r$ and $r_j$ in \eqref{asymptotic0gaminv}, we set
\beq\label{Psi}
R(s):=\int_{1}^{s} r(u)\, du\:\:\:\textrm{and}\:\:\:R_j(s):=\int_{1}^{s} r_j(u)\, du\,,
\eeq
and first notice that
\beq\label{Psi}
\begin{array}{rl}
&2^j\,\Psi_{\frac{\eta}{\g'(2^{-j})}}(\xi)=
 \eta\,R(\frac{\xi}{\eta})\,+\,\eta\,R_j(\frac{\xi}{\eta})\,.
\end{array}
\eeq
Based on the properties of $\g$ and on the assumptions on $\zeta$ it is enough to treat the term
\beq\label{Bjmintdec}
\eeq
$$B_{j,m}(f,g)(x)\approx
 2^{-\frac{m}{2}}[\g'(2^{-j})]^{\frac{1}{2}} $$
 $$\int_{\R}\int_{\R}\hat{f}(\xi)\,\hat{g}(\eta)\,e^{i (\g'(2^{-j})\xi+\eta)\,x}\,
e^{-i\,2^m\,\eta\,R(\frac{\xi}{\eta})}\,\phi(\xi)\,\phi(\eta)\,d\xi\,d\eta \:.$$
and thus
\beq\label{Tjm}
\eeq
$$T_{j,m}(f,g)(x)\approx$$
$$2^{-\frac{m}{2}}
 \int_{\R}\int_{\R}\hat{f}(\xi)\,\hat{g}(\eta)\,e^{i (\xi+\eta)\,x}\,
e^{-i\,\frac{\g'(2^{-j})}{2^{j}}\,\eta\,R(\frac{\xi}{\g'(2^{-j})\,\eta})}\,\phi(\frac{\xi}{2^{m+j}})\,
\phi(\frac{\g'(2^{-j})\,\eta}{2^{m+j}})\,d\xi\,d\eta \:.$$

\item Because the phase of the multiplier has roughly the size $2^m$ with the dominant factor in the variable $\eta$, we further divide the $\eta-$support in $\approx 2^{m}$ intervals of equal size. Thus, after running the same approximation algorithm as in \cite{L}, we deduce that the main component of $T_{j,m}$ is given by
\beq\label{mainTjm}
\eeq
$$\tilde{T}_{j,m}(f,g)(x):=$$
$$2^{-\frac{m}{2}}
\sum_{p_0=2^m}^{2^{m+1}} \int_{\R}\int_{\R}\hat{f}(\xi)\,\hat{g}(\eta)\,e^{i (\xi+\eta)\,x}\,
e^{-i\,p_0\,R(\frac{\xi}{2^j\,p_0})}\,\phi(\frac{\xi}{2^{m+j}})\,
\phi(\frac{\g'(2^{-j})\,\eta}{2^{j}}-p_0)\,d\xi\,d\eta \:.$$
\end{itemize}
$\newline$

For notational simplicity, we will allow a small abuse, and redenote $\tilde{T}_{j,m}$ as just simply $T_{j,m}$.
$\newline$

Let us introduce now several notations:
\beq\label{a}
\phi_{k}(\xi):=\phi(\frac{\xi}{2^{k}})\,,\:\:\:\:\:k\in\Z\,;
\eeq
\beq\label{b}
\phi_{j,p_0}(\eta):=\phi(\frac{\g'(2^{-j})\,\eta}{2^{j}}-p_0)\,,\:\:\:\:\:j\in\N,\:p_0\in [2^m,\,2^{m+1})\cap\N\,;
\eeq
\beq\label{c}
\psi_{m,p_0,j}(\xi):=2^{-\frac{m}{2}}\,e^{-i\,p_0\,R(\frac{\xi}{2^j\,p_0})}\,\phi(\frac{\xi}{2^{m+j}}),\:\:\:\:\:m\in\N,\:p_0\in [2^m,\,2^{m+1})\cap\N,\:j\in\N\,.
\eeq
It will be useful to record the following identity
\beq\label{ftr}
\check{\psi}_{m,p_0,j}(x)=2^{j}\,e^{i\,\vartheta_{p_0}(2^{j}\,x)}\,\phi^{*}(\frac{p_0}{2^m}\,r^{-1}(2^j\,x))\,+\,\textrm{Error Term}\:,
\eeq
where here
\beq\label{phase}
\vartheta_{p_0}(2^{j}\,x):=p_0\,\int_{0}^{r^{-1}(2^{j}\,x)} t\,r'(t)\,dt\,,
\eeq
$\phi^{*}\in C_0^{\infty}(\R)$ is a smooth function supported away from the origin with $\|\phi^{*}\|_{C^{100}}\lesssim 1$ and finally, the Error represents a smooth, fast decaying term relative to the $m$-parameter.

Also, throughout the paper, we have $p,\,q>1$, $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}$ with $1\leq r <\infty$ and $\frac{1}{r'}=1-\frac{1}{r}$.

Now for $f\in L^{p}(\R)$, $g\in L^{q}(\R)$ and $h\in L^{r'}(\R)$ (recall $j\in\N$) we set
\beq\label{ljm}
\ll_{j,m}(f,g,h):=\int_{\R} T_{j,m}(f,g)(x)\,h(x)\,dx\:.
\eeq
From the above considerations, using relation \eqref{growth2} and Observation \ref{j}, we deduce that\footnote{Throughout this paper we will ignore the contribution of the error term in \eqref{ftr} as its treatment was carefully presented in \cite{L}. Also, for notational convenience, we make a small abuse by letting the same function $\phi_{j,p_0}$ represent the Fourier localization of both $g$ and $h$; in reality the support of $\hat{h}$ should be slightly (e.g. twice) larger than the support of $\hat{g}$.}
\beq\label{ljmexp}
\ll_{j,m}(f,g,h)=\sum_{p_0=2^m}^{2^{m+1}}\int_{\R} (f*\check{\phi}_{m+j}*\check{\psi}_{m,p_0,j})(x)\,(g*\check{\phi}_{j,p_0})(x)\,(h*\check{\phi}_{j,p_0})(x)\,dx\:.
\eeq
We now define
\beq\label{Lambdamp}
\ll_{m}^{+}(f,g,h):=\sum_{j\in \N}\ll_{j,m}(f,g,h)\,.
\eeq
A similar decomposition can be done for treating the case $j<0$. In this situation the analogue of \eqref{Lambdamp} will read
\beq\label{Lambdamp1}
\ll_{m}^{-}(f,g,h):=\sum_{j\in \Z\setminus\N}\ll_{j,m}(f,g,h)\,,
\eeq
where for this case $\ll_{j,m}$ is an appropriate adaptation of \eqref{ljmexp} to the context of \eqref{neg0} instead of \eqref{poz0}. For the brevity of the exposition, we won't insist more on this.

With all these done, our main focus will be on understanding the main properties of the trilinear form
\beq\label{Lambda}
\ll(f,g,h):=\sum_{m\in\N}\ll_{m}(f,g,h)\,,
\eeq
with
\beq\label{Lambdam}
\ll_{m}(f,g,h):=\ll_{m}^{+}(f,g,h)\,+\,\ll_{m}^{-}(f,g,h)\:.
\eeq

Based on the reductions presented above, we deduce that for proving our Main Theorem it is enough to show
$\newline$

\begin{t1}\label{Ll1} The form $\ll$ initially defined on $S(\R) \times S(\R)\times S(\R)\longmapsto \C$ by \eqref{Lambda} obeys the bounds:
\beq\label{bound}
\ll:\:L^{p}(\R)\times L^{q}(\R)\times L^{r'}(\R)\:\:\rightarrow\:\:\C
\eeq
 where the indices $p,\,q,\,r'$ satisfy
$$\frac{1}{p}+\frac{1}{q}+\frac{1}{r'}=1\,,$$ with
$$1<p<\infty\,,\:\:\:1<q\leq\infty\,,\:\:\:\textrm{and}\:\:\:1\leq r<\infty\,.$$
\end{t1}
$\newline$

Give relation \eqref{Lambda}, we further notice that Theorem \ref{Ll1} follows from the theorem below\footnote{For a helpful geometric perspective, the reader is invited to consult Figure 1.} after applying real interpolation methods and a telescoping sum argument (see Section 5):
$\newline$

\begin{t1}\label{estimabac} \textit{Let $1<p<\infty$. Then the following estimates hold \footnote{Throughout the paper $p^*=\min\{p,\,p'\}\;.$}}:
\beq\label{ac}
\underline{\textrm{Edge}\:\:(AC)}\,:\:\:\:\:\:\:\:\:\:\:\:|\ll_{m}(f,g,h)|\lesssim_{p} (1+m^{1-\frac{2}{p}})\,2^{-\frac{m}{10}(1-\frac{1}{p^{*}})}\,\|f\|_{p}\,\|g\|_{\infty}\,\|h\|_{p'}\;,
\eeq
and
\beq\label{ab}
\underline{\textrm{Edge}\:\:(AB)}\,:\:\:\:\:\:\:\:\:\:\:\:|\ll_{m}(f,g,h)|\lesssim_{p} (1+m^{1-\frac{2}{p}})\,2^{-\frac{m}{10}(1-\frac{1}{p^{*}})}\,\|f\|_{p}\,\|g\|_{p'}\,\|h\|_{\infty}\;.
\eeq
\end{t1}
$\newline$

Finally, based on Observation \ref{j}, it will be enough to prove Theorem \ref{estimabac} above for the ``positive half-line" forms $\ll_{m}^{+}(f,g,h)$ with $m\in\N$.
$\newline$

\section{Controlled bounds for $\ll_{m}^{+}(f,g,h)$ on the edge $(AC)$.}
$\newline$

 In this section, we focus on providing ``slowly increasing" bounds\footnote{Relative to the $m-$parameter.} for our form $\ll_{m}^{+}(f,g,h)$ on the edge $(AC)$ (see Figure 1). More precisely, our goal is to get the following tamer result:
$\newline$

\begin{t1}\label{tame} Let $1<p<\infty$. Then the following estimate holds:
\beq\label{ac}
\underline{\textrm{Edge}\:\:(AC)}\,:\:\:\:\:\:\:\:\:\:\:\:|\ll_{m}^{+}(f,g,h)|\lesssim_p (1+m^{\frac{2}{p'}-1})\,\|f\|_{p}\,\|g\|_{\infty}\,\|h\|_{p'}\;.
\eeq
\end{t1}
$\newline$

\subsection{Estimates for the edge $(AB_1]$.} The main result of this subsection is

\begin{p1}\label{estimab1} If $1<p\leq 2$, the following holds:
\beq\label{pointAB1}
\underline{\textrm{Edge}\:\:(AB_1]}\,:\:\:\:\:\:\:\:\:\:\:\:|\ll_{m}^{+}(f,g,h)|\lesssim_{p}  \|f\|_{p}\, \|g\|_{\infty}\,\|h\|_{p'}\,.
\eeq
\end{p1}
$\newline$

\begin{proof}

As a consequence of \eqref{ljmexp}, we deduce that
\beq\label{shapelambda1}
\eeq
$$|\ll_{m}^{+}(f,g,h)|\lesssim\sum_{j\in\N}\int_{\R} |(f*\check{\phi}_{m+j}(y)|$$
$$\times\,\int_{\R}\left(\sum_{p_0=2^m}^{2^{m+1}}|(g*\check{\phi}_{j,p_0})(x)\,(h*\check{\phi}_{j,p_0})(x)|\right)
\,2^{j}\,|\phi^{*}(r^{-1}(2^j\,(x-y))|\,dx\,dy\:.$$

Now, letting $M$ stand for the standard Hardy-Littlewood maximal function, we deduce the key relation
\beq\label{key1}
|\ll_{m}^{+}(f,g,h)|\lesssim
\sum_{j\in\N}\int_{\R} |(f*\check{\phi}_{m+j}(y)|\:\:
M\left(\sum_{p_0=2^m}^{2^{m+1}}|(g*\check{\phi}_{j,p_0})\,(h*\check{\phi}_{j,p_0})|\right)(y)\,dy\:.
\eeq

Now, from \eqref{key1}, we deduce that
$$|\ll_{m}^{+}(f,g,h)|\lesssim$$
$$\int_{\R}\left( \sum_{j\in\N}|(f*\check{\phi}_{m+j}(y)|^2\right)^{\frac{1}{2}}\:
\left( \sum_{j\in\N} \left|M\left(\sum_{p_0=2^m}^{2^{m+1}}|(g*\check{\phi}_{j,p_0})\,(h*\check{\phi}_{j,p_0})|\right)(y)\right|^2\right)^{\frac{1}{2}}\,dy$$
$$\lesssim \left\| \left( \sum_{j\in\N}|(f*\check{\phi}_{m+j}(y)|^2\right)^{\frac{1}{2}}\right\|_{p}\,
\left\|\left( \sum_{j\in\N} \left|M\left(\sum_{p_0=2^m}^{2^{m+1}}|(g*\check{\phi}_{j,p_0})\,(h*\check{\phi}_{j,p_0})|\right)(y)\right|^2\right)^{\frac{1}{2}}\right\|_{p'}$$
$$\lesssim \|f\|_{p}\,\left\|\left( \sum_{j\in\N}\left(\sum_{p_0=2^m}^{2^{m+1}}|(g*\check{\phi}_{j,p_0})\,(h*\check{\phi}_{j,p_0})|\right)^2\right)^{\frac{1}{2}}\right\|_{p'}\:,$$
where for the last relation we used standard Littelwood-Paley theory (for providing bounds on the square function for $f$) and Fefferman-Stein's inequality (\cite{FS}) (for the term involving the functions $g$ and $h$).

Now we make use of the following observation:
\beq\label{L2infcont}
\sum_{p_0=2^m}^{2^{m+1}}|(g*\check{\phi}_{j,p_0})|^2\lesssim \|g\|_{\infty}^2\,.
\eeq

Indeed, this is a simple consequence of the fact that
\beq\label{conv}
(g*\check{\phi}_{j,p_0})(x)=\int_{\R} g(x-\frac{\g'(2^{-j})}{2^j}\,y)\,\check{\phi}(y)\,e^{i\,p_0\,y}\,dy\:.
\eeq

Thus applying Cauchy-Schwarz inequality we get
\beq\label{CS}
\left(\sum_{p_0=2^m}^{2^{m+1}}|(g*\check{\phi}_{j,p_0})\,(h*\check{\phi}_{j,p_0})|\right)^2\lesssim \|g\|_{\infty}^2\,\sum_{p_0=2^m}^{2^{m+1}}|(h*\check{\phi}_{j,p_0})|^2\,.
\eeq
Inserting \eqref{CS} in the last estimate on $|\ll_{m}^{+}(f,g,h)|$ we get:
\beq\label{L}
|\ll_{m}^{+}(f,g,h)|\lesssim  \|f\|_{p}\, \|g\|_{\infty}\,\left\|\left( \sum_{j\in\N}\sum_{p_0=2^m}^{2^{m+1}}|h*\check{\phi}_{j,p_0}|^2\right)^{\frac{1}{2}}\right\|_{p'}\:.
\eeq

Define now
$$A_{j,p_0}:=\left[\frac{p_0-10}{2^{-j}\,\g'(2^{-j})},\,\frac{p_0-\frac{1}{10}}{2^{-j}\,\g'(2^{-j})}\right]
\cup \left[\frac{p_0+\frac{1}{10}}{2^{-j}\,\g'(2^{-j})},\,\frac{p_0+10}{2^{-j}\,\g'(2^{-j})}\right]\,.$$
From \eqref{ssuport} and \eqref{b} we notice that
$$\textrm{supp} \:\phi_{j,p_0}\subseteq A_{j,p_0}\:.$$

Using now the key assumption \eqref{variation0}, we deduce that $\{A_{j,p_0}\}_{j,p_0}$ have the finite intersection property, that is
\beq\label{finint}
\sum_{j\in\N}\sum_{p_0=2^m}^{2^{m+1}}\chi_{A_{j,p_0}}\lesssim_{\g} 1\,.
\eeq
Now, due to \eqref{finint} and the fact that $p'\geq2$, we are precisely in the setting of Rubio de Francia's inequality (\cite{RF}):
\beq\label{RF}
\left\|\left( \sum_{j\in\N}\sum_{p_0=2^m}^{2^{m+1}}|h*\check{\phi}_{j,p_0}|^2\right)^{\frac{1}{2}}\right\|_{p'}\lesssim_{p'} \|h\|_{p'}\:.
\eeq
Thus, putting together \eqref{CS}, \eqref{L}, \eqref{RF} we conclude that \eqref{pointAB1} holds.
\end{proof}

$\newline$

\subsection{Estimates for the edge $[B_1 C)$.} In this subsection we will focus on proving

\begin{p1}\label{estimb1c} If $p\geq 2$, the following holds:
\beq\label{pointB1C}
\underline{\textrm{Edge}\:\:[B_1 C)}\,:\:\:\:\:\:\:\:\:\:\:\:|\ll_{m}^{+}(f,g,h)|\lesssim_{p} m^{\frac{2}{p'}-1}\,\|f\|_{p}\,\|g\|_{\infty}\,\|h\|_{p'}\;.
\eeq
\end{p1}
$\newline$

The proof of Proposition \ref{estimb1c} will be decomposed in two main steps encoded in the two propositions below:

\begin{p1}\label{Contf} Let $p\geq2$. Then, with the previous notations, we have
\beq\label{Cf}
\left\|\,\left(\sum_{j\in\N}\sum_{p_0=2^m}^{2^{m+1}}|f*\check{\phi}_{m+j}*\check{\psi}_{m,p_0,j}(x)|^2\right)^{\frac{1}{2}}\right\|_{p}\lesssim_{p} m^{\frac{2}{p'}-1}\, \|f\|_{p}\;.
\eeq
\end{p1}

\begin{p1}\label{Contgh} If $1<p<\infty$ and $\frac{1}{p}+\frac{1}{p'}=1$ then the following holds:
\beq\label{Cgh}
\left\|\left(\sum_{j\in\N}\sum_{p_0=2^m}^{2^{m+1}}|(g*\check{\phi}_{j,p_0})(x)\,(h*\check{\phi}_{j,p_0})(x)|^{2}\right)^{\frac{1}{2}}\right\|_{p'}
\lesssim_{p'}\|g\|_{\infty}\,\|h\|_{p'}\;.
\eeq
\end{p1}

Assuming for the moment that \eqref{Cf} and \eqref{Cgh} hold we pass to the
$\newline$

\noindent\textbf{Proof of Proposition \ref{estimb1c}.}
$\newline$

By Cauchy-Schwarz we have
$$|\ll_{m}^{+}(f,g,h)|\leq$$
$$\sum_{j\in\N}\int_{\R}\left(\sum_{p_0=2^m}^{2^{m+1}}|(g*\check{\phi}_{j,p_0})(x)\,(h*\check{\phi}_{j,p_0})(x)|^{2}\right)^{\frac{1}{2}}
\,\left(\sum_{p_0=2^m}^{2^{m+1}}|f*\check{\phi}_{m+j}*\check{\psi}_{m,p_0,j}(x)|^2\right)^{\frac{1}{2}}\,dx\:,$$
which after a second application of Cauchy-Schwarz and then H\"older inequality becomes:
\beq\label{key2}
\eeq
$$|\ll_{m}^{+}(f,g,h)|\leq$$
$$\left\|\left(\sum_{j\in\N}\sum_{p_0=2^m}^{2^{m+1}}|(g*\check{\phi}_{j,p_0})(x)\,(h*\check{\phi}_{j,p_0})(x)|^{2}\right)^{\frac{1}{2}}\right\|_{p'}
\left\|\,\left(\sum_{j\in\N}\sum_{p_0=2^m}^{2^{m+1}}|f*\check{\phi}_{m+j}*\check{\psi}_{m,p_0,j}(x)|^2\right)^{\frac{1}{2}}\right\|_{p}\:.$$

Now applying Propositions \ref{Contf} and \ref{Contgh} we conclude that \eqref{pointB1C} holds.

We pass now to the
$\newline$

\noindent\textbf{Proof of Proposition \ref{Contf}.}
$\newline$

This proof will be decomposed in three main components.

The first one relies on the following key aspect: the
\textit{cancelation} offered by the phase of $\check{\psi}_{m,p_0,j}$; indeed, we have that
$\newline$

\begin{l1}\label{canc} The following holds:
\beq\label{claimc}
\sum_{p_0=2^m}^{2^{m+1}}|f*\check{\phi}_{m+j}*\check{\psi}_{m,p_0,j}(x)|^2\lesssim \int_{\R}|(f*\check{\phi}_{m+j})(y)|^2\, 2^j\,\nu(2^j\,(x-y))\,dy\:,
\eeq
where here $\nu\in C_0^{\infty}(\R)$ with $0\not\in\textrm{supp}\,\nu$ and $\|\nu\|_{C^{10}(\R)}\lesssim 1$.
\end{l1}

\begin{proof}

For fixed $m\in \N,\,j\in\N$ set now $w(y):=(f*\check{\phi}_{m+j})(\frac{y}{2^j})$. Then \eqref{claimc} turns into
\beq\label{claim1}
\sum_{p_0=2^m}^{2^{m+1}}|\int_{\R} w_x(y)\,e^{-i\,p_0\,\int_{0}^{r^{-1}(y)} t\,r'(t)\,dt}\,\phi^{*}(\frac{p_0}{2^m}\,r^{-1}(y))\,dy|^2\lesssim \int_{\R}|w_x(y)|^2\, \nu(y)\,dy\:.
\eeq
where here $x\in\R$ is considered fixed and we set $w_x(y):=w(2^j\,x-y)$.

Recall that from \eqref{phase}, we have $p_0\,\int_{0}^{r^{-1}(y)} t\,r'(t)\,dt=p_0\,\vartheta_{1}(y):=p_0\,\vartheta(y)$
Also recall that for expression \eqref{claim1} to make sense one should also have\footnote{The constant $C_\g$ are allowed to
change from line to line.}
\beq\label{cond}
\frac{1}{C_{\g}}<|y|<C_{\g}\:\:\:\textrm{for some fixed}\:C_{\g}>1\;.
\eeq
Set then $\mu\in C_{0}^{\infty}(\R)$ with $\textrm{supp}\,\mu\subseteq[\frac{1}{2 C_{\g}}, 2C_{\g}]$ and $\mu|_{[\frac{1}{ C_{\g}}, C_{\g}]}=1$.

Now in order to prove \eqref{claim1} it is enough to show the dual statement
\beq\label{claim2}
\int_{\R}|\sum_{p_0=2^m}^{2^{m+1}}\,a_{p_0}\,e^{i\,p_0\,\vartheta(y)}\,\phi^{*}(\frac{p_0}{2^m}\,r^{-1}(y))\,\mu (y)\,|^2\,dy\lesssim
\sum_{p_0=2^m}^{2^{m+1}} |a_{p_0}|^2 \:.
\eeq

For this we analyze the generic interaction
\beq\label{gen}
E_{p_0,q_0}:=\int_{\R}e^{i\,(p_0-q_0)\,\vartheta(y)}\,\phi^{*}(\frac{p_0}{2^m}\,r^{-1}(y))\,\overline{\phi^{*}}(\frac{q_0}{2^m}\,r^{-1}(y))\,\mu (y)\,\mu(y)\,dy\:.
\eeq

Now since $\vartheta'(y)=r^{-1}(y)$ has obviously no roots inside the support of $\mu$ we deduce that for $2^{m-10}<p_0, q_0< 2^{m+10}$ one has
\beq\label{gen1}
|E_{p_0,q_0}|\lesssim \frac{1}{1+|p_0-q_0|^2}\:.
\eeq
Thus
\beq\label{CSc}
\eeq
$$\int_{\R}|\sum_{p_0=2^m}^{2^{m+1}}\,a_{p_0}\,e^{i\,p_0\,\vartheta(y)}\,\phi^{*}(\frac{p_0}{2^m}\,r^{-1}(y))\,\mu (y)\,|^2\,dy$$
$$\lesssim\sum_{p_0,q_0=2^m}^{2^{m+1}} |a_{p_0}|\,|a_{q_0}|\,\frac{1}{1+|p_0-q_0|^2}\lesssim \sum_{p_0=2^m}^{2^{m+1}} |a_{p_0}|^2\:,$$
proving this way our claim \eqref{claim2} and hence \eqref{claimc}.
\end{proof}

We pass now to the second ingredient:
$\newline$

\begin{l1}\label{max}
Let $u\in L^1_{loc}(\R)$ and $\nu$ as before. Then, preserving the previous notations, we have:
\beq\label{claim4}
\begin{array}{cl}
I_{m,j}(u)(x):=\int|(u*\check{\phi}_{j+m})(y)|^2\,2^{j}\,\nu(2^j(x-y))\,dy\\\\
\lesssim \frac{1}{2^m}\,\sum_{l=2^{m-1}}^{2^{m+1}}\,|M(u)(x-\frac{l}{2^{m+j}})|^2\:.
\end{array}
\eeq
\end{l1}

\begin{proof}

 From the assumptions made in our hypothesis, using a smooth, compactly supported partition of unity, we have
\beq\label{dec}
\nu(y)=\sum_{l=2^{m-1}}^{2^{m+1}} \psi(2^m y-l)\:.
\eeq
From this, we deduce
\beq\label{desc}
\begin{array}{cl}
I_{m,j}(u)(x)=\int_{\R} \left(\int_{\R} u(s)\,2^{m+j}\,\check{\phi}(2^{m+j}(y-s))\,ds\right)\\\\
\times\left(\int_{\R}\overline{u(t)}\,2^{m+j}\,\overline{\check{\phi}(2^{m+j}(y-t))}\,dt\right)
\,\left(2^{j}\,\sum_{l=2^{m-1}}^{2^{m+1}} \psi(2^{m+j}\,(x-y)-l)\right)\,dy\\\\
=\frac{1}{2^m}\,\int_{\R\times\R} u(s)\,\overline{u(t)}\,\K_{m,j}(s,t)\,ds\,dt\,,
\end{array}
\eeq
where we set
\beq\label{defK}
\K_{m,j}(s,t):=2^{3(m+j)}\,\sum_{l=2^{m-1}}^{2^{m+1}}
\int_{\R}\check{\phi}(2^{m+j}(y-s))\,\overline{\check{\phi}(2^{m+j}(y-t))}\,\psi(2^{m+j}\,(x-y)-l)\,dy
\eeq
From the smooth behavior of $\check{\phi}$ we deduce
\beq\label{boundK}
|\K_{m,j}(s,t)|\lesssim 2^{2(m+j)}\,\sum_{l=2^{m-1}}^{2^{m+1}}
|\check{\phi}|(2^{m+j}(x-s)-l)\,|\check{\phi}|(2^{m+j}(x-t)-l)\,.
\eeq
Thus, from \eqref{desc} and \eqref{boundK} we have
\beq\label{cont}
|I_{m,j}(x)|\lesssim \frac{1}{2^m}\,\sum_{l=2^{m-1}}^{2^{m+1}}\,\left(\int_{\R} |u(s)|\,2^{m+j}\,
|\check{\phi}|(2^{m+j}(x-s)-l)\,ds\,\right)^2\,,
\eeq
from which we conclude
\beq\label{conc}
|I_{m,j}(x)|\lesssim \frac{1}{2^m}\,\sum_{l=2^{m-1}}^{2^{m+1}}\,[M (|u(\frac{\cdot}{2^{m+j}})|)(2^{m+j}x-l)]^2\,,
\eeq
which proves \eqref{claim4}.

\end{proof}

Combining now Lemmas \ref{canc} and \ref{max}, we deduce that
\beq\label{prop1}
\eeq
$$\left\|\,\left(\sum_{j\in\N}\sum_{p_0=2^m}^{2^{m+1}}\left|f*\check{\phi}_{m+j}*\check{\psi}_{m,p_0,j}(x)\right|^2\right)^{\frac{1}{2}}\right\|_{p}$$
$$\lesssim \left\|\,\left(\sum_{j\in\N}\frac{1}{2^m}\,\sum_{l\approx 2^m}\,\left|M(f*\check{\phi}_{m+j})(\cdot-\frac{l}{2^{m+j}})\right|^2\right)^{\frac{1}{2}}\right\|_{p}\:.$$
On the other hand, from Fefferman-Stein's inequality, (\cite{FS}), we have
\beq\label{FS}
\eeq
$$\left\|\,\left(\sum_{j\in\N}\frac{1}{2^m}\,\sum_{l\approx 2^m}\,\left|M(f*\check{\phi}_{m+j})(\cdot-\frac{l}{2^{m+j}})\right|^2\right)^{\frac{1}{2}}\right\|_{p}$$
$$\lesssim \left\|\,\left(\sum_{j\in\N}\frac{1}{2^m}\,\sum_{l\approx2^m}\,\left|(f*\check{\phi}_{m+j})(\cdot-\frac{l}{2^{m+j}})\right|^2\right)^{\frac{1}{2}}\right\|_{p}\;.$$

However, applying Fubini and Jensen's inequality (at this last point we use the key fact that $p\geq 2$)
$$\left\|\,\left(\sum_{j\in\N}\frac{1}{2^m}\,\sum_{l\approx2^m}\,\left|(f*\check{\phi}_{m+j})(\cdot-\frac{l}{2^{m+j}})\right|^2\right)^{\frac{1}{2}}\right\|_{p}^{p}$$
$$\lesssim\frac{1}{2^m}\,\sum_{l\approx2^m}\left\|\,\left(\sum_{j\in\Z}\,\left|(f*\check{\phi}_{m+j})(\cdot-\frac{l}{2^{m+j}})\right|^2\right)^{\frac{1}{2}}\right\|_{p}^{p}$$
Now, we appeal to the last ingredient in order to prove Proposition \ref{Contf}:
$\newline$

\begin{l1}\label{sqf}
Take $l\in\Z$ and define the $l-$shifted square function
\beq\label{Sqf}
S_{l}f(x):=\left(\sum_{j\in\Z}\left|(f*\check{\phi}_{j})(x-\frac{l}{2^{j}})\right|^2\right)^{\frac{1}{2}}\;.
\eeq
 Then, for any $1<q<\infty$, one has
\beq\label{Sqfp}
\|S_l f\|_q\lesssim_q (\log (|l|+10))^{\frac{2}{q^*}-1}\,\|f\|_q\;.
\eeq
\end{l1}
$\newline$

\begin{proof}

This statement relies on interpolation and Calderon-Zygmund theory adding to the classical theory of the $L^p$ boundedness of the (classical) square function a small extra twist in order to deal with the translation symmetry encoded in the $l$-parameter. In a different form, its statement/proof can be found in \cite{S} (see Chapter II, Section 5).  Shifted square functions (in various forms) appear naturally in analysis. One such instance refers to the case of the first Calderon commutator, whose approach involves a discrete version of \eqref{Sqf}. This theme is explained in detail in \cite{MS}, Ch.4, pp. 126, while the standard approach for the discretized analogue of \eqref{Sqf} is given at pp. 150.
$\newline$

\noindent\textbf{Step 1. $L^2$ boundedness}
$\newline$

This is a straightforward statement since $\{\phi_{m+j}\}_{j\in\N}$ can be split in at most $C$ disjoint families such that
the functions within each family have pairwise disjoint supports. Thus, from Parseval, we conclude

$$\|S_{l}f\|_2^2=\sum_{j\in\Z}\|(f*\check{\phi}_{j})(x-\frac{l}{2^{j}})\|_2^2$$
$$=\sum_{j\in\Z}\|\hat{f}(\xi)\,\hat{\phi}_{j}(\xi)\,e^{i\,\xi\,\frac{l}{2^{j}}}\|_2^2 \lesssim\|f\|_2^2\;.$$

$\newline$

\noindent\textbf{Step 2. $L^{1}$ to $L^{1,\infty}$ boundedness}
$\newline$

Our intention is to show that there exists $C>0$ absolute constant such that
\beq\label{Sqf1}
\forall\:\:\l>0\:\:\:\:\;\;\:\:\:\:|\{x\,|\,|S_l f(x)|>\l\}|\leq \frac{C}{\l}\,\log (|l|+10)\,\|f\|_1\;.
\eeq
As expected, we will make use of the Calderon-Zygmund decomposition of $f$ at level $\l$:

Thus, for simplicity, we consider $M_d f$ - the \textit{dyadic} Hardy-Littlewood maximal function associated to $f$. Then,
for $\l$ as above, we set $$E_{\l}:=\{x\,|\,M_d f(x)> \l\}\:.$$
  From the definition of $E_{\l}$ we know that there exists a unique collection $\I$ of maximal (disjoint) dyadic intervals such that
\beq\label{El}
E_{\l}=\bigcup_{J\in \I} J\:.
\eeq
  Of course, we retain for later that
\beq\label{meas}
|E_{\l}|=\sum_{J\in \I} |J|\leq \frac{1}{\l}\,\sum_{J\in \I} \int_{J} |f|\leq \frac{1}{\l}\,\|f\|_1\:,
\eeq
and that
\beq\label{Linfb}
|f(x)|\leq \l\:\:\:\:\:\:\:\forall\:\:x\in\R\setminus E_{\l}\;.
\eeq
Next, we decompose the function $f$ as follows:
\beq\label{CZ}
f=g+b\,,
\eeq
with
\beq\label{g}
g:=f\,\chi_{E_{\l}}\,+\sum_{J\in\I}\left(\frac{1}{|J|}\,\int_{J}f\right)\,\chi_{J},
\eeq
and
\beq\label{bbad}
b:=\sum_{J\in\I} b_{J}\:\:\:\:\textrm{with}\:\:\:b_{J}:= \left(f-\frac{1}{|J|}\,\int_{J}f\right)\,\chi_{J}\,.
\eeq
Notice that from the above decomposition we have the following important properties
\beq\label{prop}
\eeq
\begin{itemize}
\item $\|g\|_{\infty}\lesssim \l\,;$
\item $\|g\|_{1}\leq \|f\|_1\,;$
\item $\textrm{supp}\,b_J\subseteq J$ for every $J\in\I$;
\item $\int_{\R} b_{J}=0$ for every $J\in\I$;
\item $\|b_J\|_1\lesssim \l\,|J|$ for every $J\in\I$.
\end{itemize}
Now as in the classical theory approach we continue with the remark
\beq\label{levset}
|\{S_{l}f>\l\}|\leq |\{S_{l}g>\frac{\l}{2}\}|\,+\,|\{S_{l}b>\frac{\l}{2}\}|\;.
\eeq
For the first term we just use the $L^2$ boundedness of $S$ and then the $L^{\infty}-$control on $g$:
\beq\label{gg}
\eeq
$$|\{S_{l}g>\frac{\l}{2}\}|\lesssim \frac{1}{\l^2}\,\|S_{l}g\|_{L^2}^2\lesssim\frac{\|g\|_2^2}{\l^2}\lesssim \frac{1}{\l}\|f\|_1\;.$$
In order to treat the second term, we adopt the following convention: given an interval $J$ with center $c(J)$ and $a>0$, we refer to $a\,J$ as the interval with the same center $c(J)$ and of length $a\,|J|$.

With these done, we have
\beq\label{b1}
\eeq
$$|\{x\,|\,S_{l}b(x)>\frac{\l}{2}\}|$$
$$=|\{x\in\bigcup_{J\in\I}100 J\,|\,S_{l}b(x)>\frac{\l}{2}\}|\,+\,
|\{x\in\R\setminus\bigcup_{J\in\I}100 J\,|\,S_{l}b(x)>\frac{\l}{2}\}|=A\,+\,B\,.$$
Now, on the one hand
\beq\label{A}
A\lesssim \sum_{J\in\I} |J|\leq \frac{\|f\|_1}{\l}\;,
\eeq
while on the other hand
\beq\label{B}
B\lesssim \frac{1}{\l}\,\int_{\R\setminus\bigcup_{J\in\I}100 J}\,S_{l}b\lesssim\frac{1}{\l}\sum_{J\in\I}\int_{\R\setminus100J}
S_{l} b_J\;.
\eeq
It will be thus enough to show that
\beq\label{S}
\int_{\R\setminus100J}S_{l} b_J\lesssim \log (|l|+10)\,\l\,|J|\;.
\eeq
In fact we will show the stronger statement
\beq\label{Ss}
I_{b_J}:=\int_{\R\setminus100J}\sum_{j\in\Z}
|(b_{J}*\check{\phi}_{j})(x-\frac{l}{2^{j}})|\,dx\lesssim \log(|l|+10)\,\l\,|J|\;.
\eeq
For this, we first use the mean zero condition for $b_J$ and write
\beq\label{meanz}
\eeq
$$I_{b_J}^{l,j}(x):=|(b_{J}*\check{\phi}_{j})(x-\frac{l}{2^{j}})|\leq$$
$$\left|\int_{\R}\left(\check{\phi}_{j}(x-c(J)-s-\frac{l}{2^{j}})-\check{\phi}_{j}(x-c(J)-\frac{l}{2^{j}})\right)\,b_{\tilde{J}}(s)\,ds\right|\,,$$
where here we denoted with $c(J)$ the center of $J$, $\tilde{J}=J-c(J)$ and $b_{\tilde{J}}(s):=b_{J}(s+c(J))$.

As a consequence, we have
\beq\label{Ibj}
\begin{array}{cl}
I_{b_J}^{l,j}(x)\lesssim\\\\
\int_{\R}\min\left\{\max_{|u|\leq 2^j\,|J|}\,\frac{2^{2j} |J|}{|2^{j}(x-c(J))-l-u|^2+1},\,\frac{2^{j} }{|2^{j}(x-c(J)-s-\frac{l}{2^{j}})|^2+1}\right\}\,|b_{\tilde{J}}(s)|\,ds\,.
\end{array}
\eeq
Then, implementing \eqref{Ibj} into the LHS of \eqref{Ss}, we have
\beq\label{ctrl}
\eeq
$$|I_{b_J}|\lesssim \l\,|J|\,\sum_{{j\in\Z}\atop{2^{j} |J|\leq 1}} 2^{j} |J|\,\int_{\R\setminus100J}
\frac{2^{j} }{|2^{j}(x-c(J))-l|^2+1}\,dx$$
$$+\,\sum_{{j\in\Z}\atop{2^{j} |J|>1}} \int_{\R\setminus100J}\int_{\tilde{J}}
\frac{2^{j} }{|2^{j}(x-c(J)-s-\frac{l}{2^{j}\,|J|}|J|)|^2+1}\,|b_{\tilde{J}}(s)|\,dx\,ds$$
$$\lesssim  \l\,|J|\,\log (|l|+10 )\,.$$
This finishes Step 2.
$\newline$

\noindent\textbf{Step 3. $L^{q}$ boundedness, $1<q<\infty$}
$\newline$

The fact that \eqref{Sqfp} holds for $1<q\leq 2$ is just a consequence of real interpolation between the $q=2$ case (Step 1)
and $q=1$ case (Step 2).

The other part of the range, \textit{i.e.} $2<q<\infty$, follows from duality. This is again a standard argument, an application of Hincin inequality (probabilistic method). For completeness, we give here the details:

 For $\{\o_{j,l}\}_{{j\in\Z}}$ a sequence of i.i.d. random variables on $[0,1]$ with each $\o_{j}$ taking the values $\{\pm 1\}$ with equal probability (the probability measure here can be taken to be the Lebesgue measure on $[0,1]$) we define for any function $h\in L^1_{loc}(\R)$ the linear operator
\beq\label{Ll}
\L_{l,\o(t)}h(x):=\sum_{j\in\Z}\o_{j}(t)\,(h*\check{\phi}_{j})(x-\frac{l}{2^{j}})\:.
\eeq
  Then, for any $f\in L^q(\R)$, we have
$$\|S_{l}f\|_q^q\approx\int_{[0,1]}\int_{\R}|\L_{l,\o(t)}f(x)|^q\,dt\,dx$$
Taking now any function $g\in L^{q'}(\R)$ with $\|g\|_{q'}=1$ we define
$$V_{l,t}(f,g):=\int_{\R}g(-x)\,\L_{l,\o(t)}f(x)\,dx\,,$$
and notice that
$$V_{l,t}(f,g)=\int_{\R}f(x)\,\L_{l,\o(t)}g(-x)\,dx\,,$$
from which we deduce
\beq\label{defV}
|V_{l,t}(f,g)|\lesssim \|f\|_q\,\|\L_{l,\o(t)}g\|_{q'}\;.
\eeq
Finally, we notice that following exactly the same procedure as that presented at the Step 1 and 2, we have that
\beq\label{Some}
\|\L_{l,\o(t)}h\|_{2}\lesssim \|h\|_{2}\:\:\:\textrm{and}\:\:\:\|\L_{l,\o(t)}h\|_{1,\infty}\lesssim \log(|l|+10)\,\|h\|_{1}\;.
\eeq
Thus from interpolation we get that for any $1< q'\leq2$ we have
\beq\label{Ssq}
\|\L_{l,\o(t)}g\|_{q'}\lesssim_{q'} \log(|l|+10)^{\frac{2}{q'}-1} \|g\|_{q'}\;,
\eeq
with the constant in the above inequality \textit{independent} of $t$.

Now it only remains to observe that we can choose a function $g_t(\cdot)\in L^{q'}(\R)$ such
that $(\int_{[0,1]\times\R}|g_t(x)|^{q'}\,dx\,dt)^{\frac{1}{q'}}=1$ and
\beq\label{Ssqfin}
\|S_{l}f\|_q\approx\int_{[0,1]}\int_{\R}g_t(x)\L_{l,\o(t)}f(x)\,dt\,dx\lesssim_{q'} \log(|l|+10)^{\frac{2}{q'}-1} \|f\|_{q} \|g_t\|_{L_{t,x}^{q'}}\;.
\eeq
\end{proof}

Proposition \ref{Contf} follows now trivially from combining Lemmas \ref{canc}, \ref{max} and \ref{sqf}.
$\newline$
$\newline$
\noindent\textbf{Proof of Proposition \ref{Contgh}.}
$\newline$

The proof of this result follows from Fefferman-Stein's inequality (\cite{FS}) and the result below
$\newline$

\begin{l1}\label{dual} The following holds:
\beq\label{claim}
\sum_{p_0=2^m}^{2^{m+1}}|(g*\check{\phi}_{j,p_0})(x)\,(h*\check{\phi}_{j,p_0})(x)|^{2}\lesssim \|g\|_{\infty}^2\, M(h*\check{\phi}_{\g,m,j})^2(x)\:,
\eeq
where here we set
\beq\label{gmj}
\phi_{\g,m,j}(\eta):=\phi(\frac{\g'(2^{-j})}{2^{j+m}}\,\eta)\;.
\eeq
\end{l1}

\begin{proof}
Notice first that, recalling \eqref{L2infcont}, we have
\beq\label{l2}
\sum_{p_0=2^m}^{2^{m+1}}|(g*\check{\phi}_{j,p_0})(x)|^{2}\lesssim \|g\|_{\infty}^2:.
\eeq
We also know that
\beq\label{MAx}
\begin{array}{cl}
\sup_{p_0\approx2^m}|h*\check{\phi}_{j,p_0}|(x)\\\\
=\sup_{p_0\approx2^m}\left|\int_{\R} (h*\check{\phi}_{\g,m,j})(x-y)\,\frac{2^j}{\g'(2^{-j})}\,\check{\phi}(\frac{2^j}{\g'(2^{-j})}\,y)\,
e^{i\frac{2^j}{\g'(2^j)}\,p_0\,y}\,dy\right|\\\\
\lesssim M(h*\check{\phi}_{\g,m,j})(x)\;.
\end{array}
\eeq
Thus combining \eqref{MAx} and \eqref{l2} we deduce that \eqref{claim} holds.
\end{proof}
$\newline$

\section{Controlled bounds for $\ll_{m}^{+}(f,g,h)$ on the edge $(AB)$.}
$\newline$

 As in the previous section, we only intend here to get $\log$-type upper bounds for our form $\ll_{m}^{+}(f,g,h)$ when restricted to edge $(AB)$ (see Figure 1). It is now simple to notice that taking advantage on the symmetry of our form, by just switching the role of $g$ and $h$ in Theorem \ref{tame}, one gets the desired result:
$\newline$

\begin{t1}\label{tameAB} Let $1<p<\infty$. Then the following holds:
\beq\label{AB}
\underline{\textrm{Edge}\:\:(AB)}\,:\:\:\:\:\:\:\:\:\:\:\:|\ll_{m}^{+}(f,g,h)|\lesssim_{p}  (1+m^{\frac{2}{p'}-1})\,\|f\|_{p}\,\|g\|_{p'}\,\|h\|_{\infty}\,\;.
\eeq
\end{t1}
$\newline$

\section{The proof of our Main Theorem}
$\newline$

It now remains to prove our Theorem \ref{estimabac}. We will do this, by applying real interpolation methods between the tame bounds obtained in Theorems \ref{tame} (and respectively \ref{tameAB}) and the corresponding bounds obtained in \cite{L} for the $L^2$ case.

\subsection{Estimates for $\ll_{m}^{+}(f,g,h)$ at points $B_1$ and $C_1$}

In this subsection we only quote the previous results on the author addressing the (local) $L^2$ case.

As in the previous section, it is useful to notice that because of the symmetry between the role of $g$ and $h$ we only need to refer to one of the mid points of the segments $(AB)$ and $(AC)$ respectively.

However, the estimate for  $\ll_{m}(f,g,h)$ for the point $C_1$ is precisely the content of the author's paper (\cite{L}). Thus, based on Propositions 1 and 2 in \cite{L} together with the symmetric role played by the points $C_1$ and $B_1$ we conclude the following

\begin{t1}\label{estimLblc} \textit{The following estimates hold}:
\beq\label{pointLc}
\underline{\textrm{Point}\:\:C_1(\frac{1}{2},\,\frac{1}{2},\,0)}\,:\:\:\:\:\:\:\:\:\:\:\:|\ll_{m}(f,g,h)|\lesssim 2^{-\frac{m}{16}}\,\|f\|_{2}\,\|g\|_{2}\,\|h\|_{\infty}\;,
\eeq
and
\beq\label{pointLb}
\underline{\textrm{Point}\:\:B_1(\frac{1}{2},\,0,\,\frac{1}{2})}\,:\:\:\:\:\:\:\:\:\:\:\:|\ll_{m}(f,g,h)|\lesssim 2^{-\frac{m}{16}}\,\|f\|_{2}\,\|g\|_{\infty}\,\|h\|_{2}\;.
\eeq
\end{t1}

\subsection{Interpolation: Proof of the main result}

The remaining task for proving our Main Theorem relies on the standard multi-linear interpolation theory.

Indeed, we first interpolate
\begin{itemize}
\item along the edge $(AB)$ between the mid point $C_1$ and a generic point $P\in(AB)$, that is between the bounds obtained
in Theorem \ref{estimLblc} and Theorem \ref{tameAB}.
\item along the edge $(AC)$ between the mid point $B_1$ and a generic point $P\in(AC)$, that is between the bounds obtained
in Theorem \ref{estimLblc} and Theorem \ref{tame}.
\end{itemize}
Through this process we get precisely the content of Theorem \ref{estimabac}. Applying one more time real interpolation between the newly better bounds just obtained on $(AB)$ and $(AC)$, we get the final global result

\begin{t1}\label{final} \textit{Let $P(\frac{1}{p},\,\frac{1}{q},\,\frac{1}{r'})$ be any point in the plane determined by the points $A,\,B,\,C$ such that $P\in \Omega=\overline{\textrm{int}\,\triangle\,ABC}\setminus \{[BC]\cup \{A\}\}$. Then there exists $\a(P)>0$ such that
for any $m\in\N$ the following holds}:
\beq\label{fm}
\underline{\textrm{Point}\:\:P(\frac{1}{p},\,\frac{1}{q},\,\frac{1}{r'})}\,:\:\:\:\:\:\:\:\:\:\:\:|\ll_{m}^{+}(f,g,h)|\lesssim_{P} 2^{-\a(P)\,m}\,\|f\|_{p}\,\|g\|_{q}\,\|h\|_{r'}\;.
\eeq
\end{t1}

Now as mentioned at the end of Section 2, the same reasonings applied to the form
\beq\label{frm}
\ll_m^{-}(f,g,h)=\sum_{j\in\Z\setminus\N}\ll_{j,m}(f,g,h)\,.
\eeq
will give the corresponding analogue of the bound \eqref{fm}.

Finally, recalling that
\beq\label{frm}
\ll(f,g,h)=\sum_{m\in\N}\ll_m(f,g,h)\,
\eeq
and using the geometric summation in the parameter $m$ as resulting from bound \eqref{fm} we deduce that our Main Theorem is true.

redpajama_set_name: RedPajamaArXiv


text:
Q: Malformed Json Error on http request Hey all I am getting an error and cannot for the life of me figure out why.  Thanks in advance for help!
Error is:

Failed to invoke future method 'public static void pushToLead2(String)' on class 'SomethingAPI' for job id 'something'

caused by: System.JSONException: Malformed JSON: Expected '{' at the beginning of object

Class.System.JSON.deserialize: line 15, column 1
Class.SomethingAPI.pushToLead2NotFuture: line 153, column 1
Class.SomethingAPI.pushToLead2: line 102, column 1

Code:
@future(callout=true)
public static void pushToLead(String accountId) {
    auth(false);
    Account account = [SELECT Id, PersonEmail, Name, Phone, Utm_Source__pc, Utm_Campaign__pc,
                    Utm_Term__pc, Utm_Medium__pc, Utm_Content__pc, Something_Lead_Id__c FROM Account WHERE Id =: accountId];
    HttpRequest req2 = new HttpRequest();
    req2.setEndpoint('https://wwwSomething.com/api/v1/lead' + (account.Something_Lead_Id__c != null ? '/' + account.Something_Lead_Id__c : ''));
    req2.setMethod(account.Something_Lead_Id__c == null ? 'POST' : 'PUT');
    req2.setHeader('x-api-token', auth);
    req2.setHeader('Content-Type', 'application/json');
    Map<String, Object> bodyMap = new Map<String, Object>();

    SomethingLead ll = new SomethingLead();
    if (lead.Something_Lead_Id__c != null) {
        //   ll.id = Integer.valueOf(account.Something_Lead_Id__c);
        // ll.skip_salesforce_update = true;
    }
    ll.email = account.PersonEmail;
    ll.source = 'Salesforce';
    ll.name = account.Name;
    ll.phone = account.Phone;
    ll.utm_source = account.Utm_Source__pc;
    ll.utm_campaign = account.Utm_Campaign__pc;
    ll.utm_term = account.Utm_Term__pc;
    ll.utm_medium = account.Utm_Medium__pc;
    ll.utm_content = account.Utm_Content__pc;

    bodyMap.put('lead', ll);

    req2.setBody(JSON.serialize(bodyMap));

    Http http2 = new Http();
    HttpResponse res2;
    if (!Test.isRunningTest()) {
        res2 = http2.send(req2);
    } else {
        res2 = new HttpResponse();
        Integer randomNumber = Integer.valueof((Math.random() * 10000000));
        res2.setBody('{"status":"ok","lead":{"id":' + randomNumber + '}}');
    }

    System.debug('ALL: ' + res2.getBody());

    // Attempt relogin on 500
    if (500 == res2.getStatusCode()) {
        auth(true);
        req2.setHeader('x-api-token', auth);
        res2 = http2.send(req2);
    }

    LeadResponse resBody = (LeadResponse) JSON.deserialize(res2.getBody(), LeadResponse.class);
    Account updAcc = new Account(Id = account.Id);

    if (resBody.status == 'ok' && account.Something_Lead_Id__c == null) {
        ll = resBody.lead;
        updAcc.Something_Lead_Id__c = ll.id;
        runOnce = false;

    }

    updAcc.Push_To_Something__c = false;
    update updAcc;
}

redpajama_set_name: RedPajamaStackExchange


Hello all! TGIF (Thank God It's Friday!)! I came across a video on my Facebook news feed a few days ago and upon watching it I realised it was a poem talking about immigration. It was beautifully written and read by Deasy Bamford whose video became viral almost overnight after posting it.
As I can imagine most of you have probably heard about the news about the UK leaving the EU (European Union) because they do not want to further accept any immigrants or foreigners into their country. I believe that it is ridiculous because even if we believe it or not, most of us are products of (legal and illegal) immigration one way or another. People who spew hate and say they do not want immigrants or foreigners in their country are pretty ignorant.
Prev Equal rights also includes equal slaps.


text:
December 24, 2018 12:10PM ET
The Farm to Closet Movement: Connecting Cotton Sourcing to Consumer Storytelling
By Marci Zaroff
It's no secret that fashion has piggybacked on the storytelling strategy started by the farm-to-table movement.
EPA Approves Key Pesticide for Use on Cotton Plants After Extensive Testing
Young DTC Shoppers Are Fundamentally Different From Their Elders—Here's How
Consumers have moved beyond asking what's in their food and now want to understand what's in their closet. Motivated by a need to get to the source of who we are, and where our products come from, humanity's internal awakening parallels the industry's need for transparency. Unsatisfied by surface interaction, people want more meaning in every aspect of their lives.
This thirst for transparency has evolved the way we think about our products and their lifecycles. Through the lens of design, we can connect people to the story of our products in a manner that brings everyone closer together—not just physically, but spiritually and emotionally, too. Our designs are an extension of ourselves. When we maintain a holistic view of the economic, environmental and social aspects of business, our products have a more profound purpose. Embedding that story into product design adds both value and values, validates transparency and empowers consumers to vote with their dollars.
The current state of making textiles is both challenging and disconcerting. Most brands and companies do not own their supply chains, so what happens in the farming and milling of their product is conveniently out of their jurisdiction. Unfortunately, this is often where some of the worst human rights offenses take place. Then there are issues like subcontracting, where a supplier cannot meet the demands of its overseas client, so it farms out some of the work to local factories that may not have the same code of ethics.
Undocumented overtime, worker abuse, and fire and safety code non-compliance are just a few more of the issues plaguing fashion's supply chains, exacerbated by time and money constraints. Fashion's supply chains are much more complicated and spread out than the food industry, which is perhaps why transparency is more difficult. But our clothes tell a deeper story, one of dedication, science, craftsmanship and culture.
When it comes to fiber, mainstream consumers often view conventional cotton as "natural," not realizing its agricultural and social impacts. The dark side to conventional cotton involves excessive water use, GMO and chemical dependence, pollution, and heartbreaking stories of farmer suicide. Because many farmers rely on bank loans against their land to afford expensive genetically modified seed and pesticides, the pressure to keep up with rising costs can feel like running on a treadmill that keeps getting faster and faster. Organic cotton, on the other hand, is not only better for soil and human health, and less dependent on water due to greater resilience and a productive relationship with rainwater, but it is also truly better for farmer livelihoods.
Symbiotic solutions
After being asked repeatedly by brands and designers how I have managed to achieve visibility throughout supply chains, I decided to share the love and make sustainability "easy" by creating MetaWear in 2013. MetaWear is a turnkey, vertical solution provider, connecting the dots between farmer, mill, dyehouse, manufacturer and distributor. My long history of building organic cotton supply chains from the ground up drives both authenticity and transparency—from farm to finished fashion, while creating win-win-win opportunities to lift vulnerable smallholder farming communities in India.
MetaWear recently launched a new project called RESET (Regenerate Environment, Society, Economy thru Textiles). RESET aims to educate and empower cotton famers in India's tribal cotton belt to convert to regenerative farming practices. This type of low-cost, low risk farming can sequester atmospheric carbon (a key solution to climate change mitigation), while restoring soil health and ecosystems. RESET can save conventional cotton farmers from perpetuating damaging and unsustainable conditions, by showing them that there is a viable alternative—demonstrating the potential for true industry transformation. But that's not all: this initiative is also an example of a powerful story, that once translated to the shelf, can empower consumers to take action, with little to no added effort. As they have done in driving the success of companies like Tom's shoes and FEED bags.
Positive storytelling
It's not fair to bombard people with doom and gloom if we can't arm them with the tools to affect positive change, by sharing the narratives of those doing well by doing good. The more we communicate our journey to the source of our supply chains, especially stories of the long-term benefits of innovation and redesign, the more customers will expect being a part of solutions—even demanding that their products have a meaningful purpose
Just like knowing that the origin of our food can literally make it taste better, bringing the stories to the forefront of fashion makes our style feel better—touching us on levels beyond the surface. The farm to closet movement is defined by this shift, as fashion is no longer one-dimensional. Sustainable style makes us look good, feel good and do good as global citizens.
We are entering a new era of fashion—from source to story, where we can wear the change we wish to see in the world.
Marci Zaroff coined the term "ECOfashion" and is an internationally recognized ECOlifestyle expert, educator, innovator and serial ecopreneur. Founder/CEO of leading sustainable fashion manufacturer MetaWear.
Too Many Brands Are Missing Opportunities with 'Silver Consumers'
In 2019, Supply Chains Are Still Too Far Behind the Curve

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Can I test Ajax functions locally? I have a simple call to replace the contents of a <div> with those of another HTML file using the .load() function:

$('#content').show().load('01.html');

01.html is in the same folder as this page. But the function isn't being executed. Is there a problem with my code? Or is it that the .load() function won't work until it's being served?
In the latter case, what's a good way to go about testing it?

A: You'll need to do this instead:
$('#content').load('01.html', function() {
  $(this).show();
});

It will show the content after a successful load (under the presumption that content was hidden to start with).

A: If you're already using jQuery why don't you just go for mocking or simulating ajax requests using jquery-mockjax ?


Krav Maga Illinois is a registered USA Boxing Gym specializing in boxing classes near Deerfield for beginners as well as offering boxing lessons for individuals wanting to advance in the sport. Boxing has become the fitness sport of choice for many over the past few years due to the conditioning and extensive boxing training required. KMI boxing instructors will teach you everything you need to know to be a confident competitor in the ring or provide the boxing lessons that will get you into shape. If you are serious about getting in shape and not afraid of hard work, boxing classes from KMI boxing gym may be for you.
Approximately 25 miles north of Chicago is the village of Deerfield, located mainly in Lake County and has a small portion extended into Cook County, Illinois. The village is a thriving community of 18,225 residents that enjoy access to parks, shopping and convenient access to Chicago's downtown via the two Metra Milwaukee road train stations located in Deerfield. Providing one-hour long boxing classes and boxing training is the conveniently located boxing gym of KMI. Our boxing lessons focus on teaching proper techniques of the sweet science. For boxing classes near Deerfield, KMI is the one to call!
If you like a challenge, then boxing lessons may be for you. KMI offers boxing classes that will help you perfect the art of boxing. The KMI boxing gym provides boxing lessons on different nights of the week to accommodate most schedules. Our boxing training staff take the boxing lessons offered very seriously. You will receive first-rate boxing lessons that align with your goals and your fitness journey. At Krav Maga Illinois, we offer beginner boxing classes as well as advanced boxing training for those who want to progress beyond basic boxing lessons. Find out if our boxing gym is right for you, sign up for a free trial today!


text:
Q: Migrating the model from Client that refers a Company to Client that can refer multiple Companies? I'm thinking of allowing more flexibility in my Sales app. Up until this moment the Clients table had a company_id column essentially allowing a client to only belong to one company at a time. I want to allow the same client data to be visible in multiple companies partener list.
How would a migration to this scenario look like? I essentially have to construct another table to store this more complex relationship as rows of client_id and company_id so they can both appear multiple times.
One would need:

*

*Create a ClientCompanyJoinTable.

*Move current Clients table company_id column data for each Client to the join table.

*Remove the Clients table company_id column.

A: You should have 2 models then, Client model and Company model.
You can use associations.
In the Client model:
    belongs_to :company

In the Company model:
    has_many :clients


text:
Home / Tennessee Titans 1979
Tennessee Titans 1979
Nola McConnan
Artist Nola McConnan
The outer mat is a rich textured black acid-free mat with a decorative inset white v-groove, while the inner mat is a complimentary colored acid-free mat reflecting one of the team's primary colors. The website image of this framed piece shows the mat color that we suggest (Medium Blue), but since each piece is custom framed, we are happy to use whatever color mat you wish (depending on availability) – our standard mat colors are:
This original, one-of-a-kind watercolor painting of the 1979 Houston Oilers (now Tennessee Titans) uniform is the original artwork that was used in the creation of this Tennessee Titans uniform evolution print and tens of thousands of other Tennessee Titans products that have been sold across North America. This original piece of art was painted by artist Nola McConnan for Maple Leaf Productions Ltd.
Among the many coaches who have led successful Oilers teams were Lou Rymkus, who guided them to their first championship in 1960 and Wally Lemm, who coached the 1961 AFL title team. "Bum" Phillips, who joined the team in 1975, led the Oilers to three straight playoff appearances between 1979 and 1981.The Oilers finish in second place this season with a 11-5 record, and one again see themselves make it to the playoffs. The first team to feel the wrath of the Oilers was Denver. In the AFC Wild Card game, Houston defeated the Broncos 13-7 in front of 48,776 fans at the Astrodome. they're second victims would be the San Diego Chargers, whom they played a close game with, winning 17-14 in front of 51,192 fans at San Diego Stadium. The last game for the Oilers in this season would be the AFC Championship Game played at Three Rivers Stadium, in front of 50,475 fans. The Oilers would lose the game 27 - 13.This light blue jersey adds the player numbers to the sleeves above the red, blue and white stripes. The helmet is once again white, and the logo of the oil rig is outlined. If you look closely, you will see the change of the mask.
Houston Astros 1962


Here is an old post I never finished. With the benefits of the Nexus 2000 and the FEX architecture (a earlier post), scalability, simplified management, flexibility, Cisco extended its use further into the servers all the way up to the virtual hosts.This allows much greater control and flexibility. After all network guys should look after all aspects of networking, server guys should look after the servers and today virtual hosts.
Enter your email address to subscribe to Routing-Bits and receive notifications of new posts by email.


text:
New York Giants Punter Steve Weatherford Apologizes For Posting Video
By Ed Valentine@Valentine_Ed Aug 19, 2012, 7:07pm EDT
Share All sharing options for: New York Giants Punter Steve Weatherford Apologizes For Posting Video
New York Giants punter Steve Weatherford has issued an apology via Twitter for posting a video online of Jason Pierre-Paul rather violently dunking teammate Prince Amukamara into a cold tub.
"I want to apologize to the fans... The video I posted was distasteful. Our team is a family, and we love each other. I am sorry to the fan," wrote Weatherford on Twitter Sunday evening.
Earlier in the day, head coach Tom Coughlin said he was just learning about the issue, the he would "speak to the parties involved" and that "anything that occurs within this family or within our group should not be a part of any social media aspect."
Boys will be boys, of course. Put two-and-two together here, however, and you have to guess that Coughlin had some not-so-nice words for the parties involved in 'Cold Tub-Gate.'
Weatherford is not the first athlete to post something on social media that he ended up regretting. He certainly won't be the last.


text:
Seattle Seahawks finalize deal with free agent DE Cliff Avril
RENTON, Wash. – The Seattle Seahawks have finalized their contract with free agent defensive end Cliff Avril, adding another needed pass rusher along their defensive line.
The contract, reportedly for two years, was finalized on Thursday.
Avril immediately addresses Seattle's need to add someone who can pressure the quarterback. Avril had 9½ sacks last season, but the best of his five years with Detroit came in 2011. Avril had 11 sacks — good for seventh in the NFC — along with six forced fumbles and an interception. His long, lean frame fits the mold of the type of defensive ends that Seattle coach Pete Carroll wants.
On a conference call Thursday, Carroll said, "I don't feel like you can have enough speed pass rushers on your club."


Mexico City is one of my favorite and most surprising cities that I lived in on the trip. I have taken many trips to the beach towns like Cabo, Rosarito, and Ensenada but have never really spent time in the city in Mexico. I was blown away at how much I loved it and here are a few reasons why.
The nightlife in Mexico City is pretty intense. The bars and clubs are open almost all night and there is a constant flow of people on the streets especially Saturdays and Sundays.
Lucha Libre… Mexican Wrestling is fantastic. The outfits they wear are unbelievable and the energy from the crowd is insane. We all went out before and bought Lucha Libre wrestling masks and wore them to match. I figured we would be some of the only people wearing the mask but I was wrong. There was a little boy sitting in front of us during the match with his mom so half way through I decided to give him my mask. He was thrilled.
The music in Mexico City is great. Every city we visited before was playing the same American music that we listened to in the United States. I loved Mexico City because they did not play any of our music at all and had their own culture and style. Which is refreshing because I didn't have to listen to Justin Bieber – Sorry for the nine millionth time.


I confess – I went to the Met last night with a why-the-fuss attitude. Widely mediated Russian soprano Anna Netrebko starring Bellini's Puritani covered with benign reviews of usually fearsome critics. This is always something to be suspicious about. What is beyond debate is she more than meets the look for the role. And moves gracefully and, even with apparently no stage direction (at least that is the way everybody else seemed to behave), has imagination for some beautiful stage gestures. That is already something. She also had a concept to the role – and now that is pretty commendable and even rare. Although the opera is named I Puritani, her Elvira had nothing cold about her. Her unusually sensuous passionate attitude bridled by noblesse-oblige modesty made her sudden insanity more believable than usual: sexual frustration allied to romantic disappointment could do that. What about the voice then? Netrebko's slightly dark-tinted creamy-all-the-way homogeneous soprano is per se something to marvel. Hers is indeed an admirable voice – and used with unfailing technique. The problem remains if Giulia Grisi's Elvira is a role fit to that voice. If you have Sutherland, Gruberová or Mariella Devia in mind, the answer is obviously "no". Netrebko is no soprano coloratura – but there is nothing to be ashamed about that. Neither was Caballé – and she recorded the part (among many other bel canto roles) for Muti, with Alfredo Kraus. And, as much as Caballé, she adopts this cleverest motto (and that's valid for everything in life): when you have to do something you are not comfortable with, do it your way. So she does – using all her LYRIC soprano powers, she pulled out an expressive touching Elvira, a true Romantic heroine. It cannot be denied that her passagework is not athletic, but slower tempi and solid legato made everything sound dependable and musicianly. Some may complain her in alts were short-lived, but all her excursions above top c were solidly produced. As for the much denounced pitch situation, yes, a couple of notes were not true to the expected frequency, but ultimately… who cares?! Netrebko proved to master the most difficult element in bel canto: she colours her voice with unending variety and is an expert in playing with tempo for expressive purposes. Let's take for example the cabaletta in her mad scene. Instead of trying to dazzle the audience with pyrotechnics (unavailable to her, truth be said), she delicately handled Bellini's strings of notes to depict the wanderings of Elvira's mind with the expertise of an actress. I had read that she would do something like that, but I could only believe it when I saw that. Amazing.
American tenor Eric Cutler took the part of Arturo and, as much as his leading lady, he knows the art of expressive phrasing and, what is more, his Italian sounds really legit. His voice is not exquisite in itself, but it is capably used – and he has physique de rôle. Franco Vassallo tackled Riccardo's act I aria impressively – his is a solid dark baritone – and produced bright forceful high notes with commendable ease. Later he would prove somewhat awkward when Bellinian lines revealed themselves a bit more tangled. As for John Relyea, he is a reliable singer, but his bass sounds curdled too often in this kind of role. That said, how many truly commendable basses have appeared in a Bellini opera? Pity. Patrick Summers accommodates the needs of his singers as this repertoire requires, but does not command his orchestra to produce the graceful and light sounds this score asks for. Sandro Sequi's uninspired staging is supposed to take second place in the proceedings, but the truth is a beautiful prima donna such as Netrebko deserved costumes that took more advantage of her figure and that made her look more diaphanous (as one could see otherwise in the similarly uninspired production of Lucia from the same theatre) and scenery that could anticipate the beauties reserved by the composer.


text:
const fs = require('fs');
const strftime = require('strftime');
const AbstractModule = require('../core/abstract-module');
const Log = require('../entity/log');

/** logger module */
class Logger extends AbstractModule {

    constructor(directory, consoleLevels) {

        super();

        this._directory = directory;
        this._consoleLevels = consoleLevels;

        if(!fs.existsSync(this._directory)) fs.mkdirSync(this._directory);

        if(!fs.existsSync(this._directory + 'debug.log')) fs.writeFileSync(this._directory + 'debug.log', '');
        if(!fs.existsSync(this._directory + 'info.log')) fs.writeFileSync(this._directory + 'info.log', '');
        if(!fs.existsSync(this._directory + 'error.log')) fs.writeFileSync(this._directory + 'error.log', '');

    }

    debug(data, meta) {
        this._log(data, meta, 'debug');
    }

    info(data, meta) {
        this._log(data, meta, 'info');
    }

    error(data, meta) {
        this._log(data, meta, 'error');
    }

    _log(data, meta, type) {

        switch(typeof data) {
            case 'string':
                data = data.split("\n");
                break;
            case 'object':
                data = [JSON.stringify(data)];
                break;
            default:
                data = [String(data)];
                break;
        }

        switch(typeof meta) {
            case 'object':
                meta = JSON.stringify(meta);
                break;
            case 'undefined':
                meta = null;
                break;
            default:
                meta = String(meta);
                break;
        }

        for (let line of data) {
            if(line) {

                let log = new Log(line, type, new Error().stack, meta);

                this.modules.entityManager.persist(log);
                this.modules.entityManager.flush();

                switch(type) {
                    case 'debug':

                        if(-1 !== this._consoleLevels.indexOf(type)) console.log(log.longMessage);

                        fs.appendFileSync(
                            this._directory + 'debug.log',
                            log.longMessage + '\n'
                        );
                        break;
                    case 'info':

                        if(-1 !== this._consoleLevels.indexOf(type)) console.info(log.longMessage);

                        fs.appendFileSync(
                            this._directory + 'debug.log',
                            log.longMessage + '\n'
                        );

                        fs.appendFileSync(
                            this._directory + 'info.log',
                            log.longMessage + '\n'
                        );

                        break;
                    case 'error':

                        if(-1 !== this._consoleLevels.indexOf(type)) console.error(log.longMessage);

                        fs.appendFileSync(
                            this._directory + 'debug.log',
                            log.longMessage + '\n'
                        );

                        fs.appendFileSync(
                            this._directory + 'info.log',
                            log.longMessage + '\n'
                        );

                        fs.appendFileSync(
                            this._directory + 'error.log',
                            log.longMessage + '\n'
                        );
                        break;
                }
            }
        }
    }
}

module.exports = Logger;

redpajama_set_name: RedPajamaGithub


1.5 in. amorphous silicon TFT color LCD With variable-angle function.
OS: Microsoft Windows 98 (including Second Edition)/Me/2000/XP.
CPU: Pentium 150MHz or higher processor (Windows 98/ Me/ 2000), Pentium 300MHz or higher processor (Windows XP).
Interface: USB (Windows 98/Me/2000/XP pre-install model only).
*2 Canon's standard conditions of measuring playback time are as follows: Normal temperature (23 degrees Celsius). Repeat playback automatically at a speed of 1 image per 5 seconds.


text:
Chip Somodevilla/Getty Images News/Getty Images
Tuition-Free College Is Apparently Being Proposed In Congress & Here's What It Would Do
By Hannah Golden
Democrats in the House have introduced a higher education bill that may change the game for those who have yet to embark on their college careers. Tuition-free college is apparently on the table in the new bill. The Aim Higher Act was rolled on Tuesday, July 24 on Capitol Hill by a group of House Democrats.
Collectively, Americans hold over $1.5 trillion in student loan debt, according to May figures from the Federal Reserve. The Democrats' plan lays out how Congress would reauthorize the Higher Education Act of 1965, which is due for reauthorization every five years.
The Aim Higher Act, per the official House page, seeks to make college more affordable for students and their families in a few ways. It would push states invest more in higher education via student aid, expand the loan forgiveness program, and crack down on for-profit institutions. Additionally, the act aims to help students graduate in a timely manner with a "quality" degree. Per the bill summary, it would also invest in programs like career counseling and on-campus child care, streamline the financial aid application, and target access for "traditionally underrepresented students" including homeless and foster students and "Dreamers," or undocumented students protected by the Deferred Action for Childhood Arrivals (DACA) program.
Notably, the bill would create a partnership between states and the federal government to "make an associate's degree at the state's public two-year colleges free for every student."
While the act appears to be a step in the right direction, it may be too soon to pull out the party hats. Substantively, it appears that the bill is more a symbolic gesture than a practical one, as the bill is almost certain to fail in a GOP-majority House. But according to higher education professor Robert Kelchen, quoted in The Washington Post, it's seen as "a litmus test" ahead of the upcoming elections.
The issue of college tuition may be more on voters' radar this year as the November midterm elections approach and House members are up for reelection. Some candidates like Alexandria Ocasio-Cortez have succeeded — even against members of their own party — on a platform that includes a $15 minimum wage, universal health care, and free college, among other policy priorities.
The idea of reducing student loan debt and enacting free education is a hot topic on the hill right now, as the Congressional Progressive Caucus also addressed the issue, rolling out its "People's Budget" this week.
Meanwhile, a bill proposed by House Republicans, per the Post, sought to scale back some of those aid programs, resulting in what the Congressional Budget Office estimated would be a loss of $15 billion in federal aid given to students over the next decade. The GOP's "PROSPER Act" was advanced in December 2017.
It remains to be seen how the Aim Higher Act and any reauthorization of the Higher Education Act will play out in this Congress, given what appear to be conflicting priorities for student loan debt, and what that might mean for student's balances.