text:
The PC Police Strike Again in the Form of Andy Duncan
by Amanda on November 27, 2018
I should be used to it by now. After all, it seems a week doesn't go by without someone claiming a "classic" book or movie is racist. We've watched publishers sanitize books like Huck Finn and Tom Sawyer, as well as others, by changing the "N"-word into something more socially acceptable. Forget about the fact the word was in common usage at the time when the book was written and set. Oh no, we must protect our poor little darlings from exposure to a word instead of using it as a teaching moment.
More recently, we have seen it with theaters refusing to show Gone With the Wind. Why? Because of language and the images it presents of slavery. Gone from much of the theater rounds is Showboat and for much the same reasons. If you do happen to see it, you will find it subtly changed from the original. Hell's bells, when they showed the live performance of Sound of Music on broadcast TV not long ago, even it had been changed.
And all to be more socially relevant and acceptable to those who don't want to face the harsh realities of our past and learn from them. I hate the term "teachable moment", but each of these titles present a number of those opportunities. Moments when we could discuss not only the "wrongness" of something but also how the characters did something right. You know, show our kids that we can learn from our mistakes and become better people.
But no. Better to shield them completely so they can make those very same mistakes.
The latest example of applying today's standards to something written decades ago comes to us via author Andy Duncan. If you looked at the name and asked "Who?", you aren't alone. I recognized the name but couldn't name a single title he'd written. Neither could most of those I spoke with yesterday after seeing the article that became the basis of today's post. A quick google search turned up he's a fantasy writer and he taught at Clarion for a couple of years. I guess that wasn't enough. Now he has to try to tear down one of the masters of fantasy, J. R. R. Tolkien and his classic The Lord of the Rings.
I'll admit, my first reaction when I saw the allegation that Tolkien was racist–and against Orcs no less–I didn't know whether to laugh or toss my iPad against the wall. Then, when I saw that Duncan had written "Senator Bilbo", the lightbulb went off. Here is yet another author criticizing a long dead master of the craft, and doing so in a way that the politically correct vocal minority in our field, would love him for. If it helps drum up sales for the collection that contains his short story, all the better.
And no, the Senator Bilbo he writes about is not our beloved Bilbo Baggins. Oh, the character he uses is named Bilbo Baggins. He is a senator, cut from the same cloth, allegedly, as real-life Senator Theodore Bilbo (who died more than a decade before LOTR was written). If I remember my history correctly, the real Senator Bilbo was a product of his time and place–the Deep South. Now those traits have been transferred to Bilbo Baggins of the Shire.
According to Duncan,
"'Senator Bilbo' is this parody in which you have this racist demagogue stomping around the world of the halflings, in a sort of desperate holding pattern to keep at bay all the change that is coming about as a result of what seems to have been the War of the Ring,"
But he doesn't leave it at that. He goes on to say,
"It's hard to miss the repeated notion in Tolkien that some races are just worse than others, or that some peoples are just worse than others," Duncan says. "And this seems to me—in the long term, if you embrace this too much—it has dire consequences for yourself and for society."
That sound you hear is my head banging against my desk–simply because I don't have Duncan's head to beat against it.
If that is how he feels, why doesn't he write his own ORIGINAL work, showing how such an attitude is wrong? Why doesn't he look at works by lesser known authors and use them for his parody? Silly me, the answer is simple. He couldn't make as much publicity hay by taking one Joe Blow author as he can by trying to drag down one of the most well-known fantasy authors of the modern age.
It seems the goal today of all too many authors is to destroy the foundation upon which our genres are built. These authors insist on applying today's rules and mores to works decades and more old. They refuse to look at the history of the author, unless they cherrypick actions or situations that fit the narrative. For example, in all his criticism of Tolkien and LOTR, does Duncan mention this little fact?
The criticism comes despite Tolkien being known as a fierce critic of racism, particularly the Nazi Germany regime in the 1930s and 40s.
And, yes, he brings in politics and Trump. You knew he would. Admit it. How many of these "woke" authors pass up a chance to take a swing, no matter how badly they'll whiff, at the current Administration.
It is easier to demonise one's opponents than to try to understand them and to understand the complex forces that are leading to, for example, refugees trying to cross the southern border [of the US] legally or illegally.
'It's easier to build walls and demonise them as "scum".
Duncan, and his supporters, make much of the fact that you have Light and Dark, good and bad, and certain "races" fall squarely into one or the other. Okay, life isn't like that. There are nuances to each of us, some good and some bad. However, this is fantasy. From the beginning of the first tale told around the fire, it has been good vs. evil, light vs. dark. Just because we are more "enlightened" and more "educated"–I won't say more mature because I don't believe it–that doesn't change. Tolkien wrote a nuanced series of books, and his characters were products of their environments.
Hmm, much like we are, no?
As someone who has read everything Tolkien published while he was alive and much of what Christopher Tolkien published of his father's work after his death, all I can say is that Duncan is full of it (and yes, there are two letters missing from the beginning of that word). Not that he or his supporters will ever admit it. Just as they have tried to remove so many from the history of our genre for bad think, now they are going after Tolkien. They can't allow anything to stand, even as a teaching moment, if it doesn't fall into lock step with with their agenda.
What they don't understand is they are opening the gates for folks to do the same thing to their work and it won't take decades or more for it to happen.
Tags: Andy Duncan, JRR Tolkien, Lord of the Rings, Orcs, racism
From: AMANDA, SCIENCE FICTION & FANTASY
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Greasing Mental Gears →
thewriterinblack	#
It is easier to demonise one's opponents
I guess self-awareness isn't a thing with him.
Is it ever with that mind set?
Christopher M Chupik	#
They do little else but demonize and dehumanize their opponents. Daily. And then turn around and lecture us about it . . .
herbn	#
Well, I do consider him an expert on taking the easy way, so…
It Isn't Demonization When It's True.
paladin3001	#
Read the story yesterday. Did a quick search on Amazon. No author page. One book (collection) released on the 16th of this month and another collection coming out on Monday. Looks like he's trying to drum up exposure.
I did that search this morning and found basically the same thing. Of course, the moment I saw he had a "Bilbo" story, I knew what he was doing. I don't begrudge anyone trying to drum up promo for their work. What I do begrudge–hell, I hate it–is doing it at the expense of someone else, especially someone who can write circles around that particular author.
Besides, I'm tired of the perpetually butthurt ones trying to apply today's standards on things written decades or more ago. It doesn't work. More importantly, we won't learn anything from the past's mistakes (and victories) if we erase or sanitize it.
Why is he teaching at Clarion with no books out?
My respect for Clarion just went down a notch.
Christopher M. Chupik	#
Authors writing books is so last century.
I suppose short stories are a possibility. Maybe.
So, the fact (in canon) that the orcs are the corrupt creations of a being who deliberately chose evil and corruption seems to have escaped Mr. Duncan's attention. That, I think, is more worthy of contemplation, especially since this year is also the anniversary of Mary Shelly's little book of science fiction.
But you can't really use that to flog modern politics, since it is the Chinese who are claiming to have created the first genetically modified (virus resistant) humans.
Makes no difference to him at all. We should take that into account and cut them slack. They have no control over their circumstances and there might be one who wants to be good. Or some such idiocy.
Like so many, he condemns what he doesn't understand and he tries to score points by showing how "woke" he is.
While Mr. Duncan is a moron and likely just trying to "get his 15 minutes of frame", Tolkien had problems with his orcs.
As a Christian, Tolkien believed in Redemption and Free Will so having a species that was completely evil and apparently unable to "chose the Light" bothered him.
While Morgoth might have created them as "evil", why didn't at least a few turn to the Light?
Oh, while it appears that Tolkien had the orcs as corrupted elves, in some of his notes, he had other ideas on their origins/nature.
Leigh Kimmel	#
The origins of the orcs became one of Tolkien's great mental struggles in his post-LOTR writings. In his youthful exuberance, he'd written the orcs as endless supplies of henchminions for his Big Bad, without really thinking about the ontological implications of an Always Evil race. As he tried to work backwards to first principles and understand the foundations of his subcreation, he shifted from one solution to another: orcs are corrupted elves. orcs are corrupted Men. orcs are not really intelligent, but a species that can sort of mimic speech and intelligence, when their master's will is upon them. orcs are in fact members of a order of the Maiar, similar to the one from which came Gandalf and the other Wizards, but were corrupted and chose evil, and when their bodies are destroyed, they are able to create new bodies for themselves, creating the appearance of endless hordes of orcs.
Also, there is the issue that, from an in-universe perspective, LOTR is written by hobbit authors, and reflects their perspective on the events, which is not necessarily the author's, who in this view is but a transcriber and translator.
That is truly fascinating.
Luke	#
If an orc ever turned to the light, he would quickly be detected and killed by his fellow orcs.
They were, after all, slaves and servants of an evil god that watched them closely. And the orcs actively sought his favor at each other's expense.
A story could be written about it, just like a story could be written about how The Prancing Pony was supplied with beer. But it would be completely incidental.
Of course, fellow Inkling C.S. Lewis addressed this very possibility in his works… I recall the same people complaining about Tolkien's orcs screaming bloody murder about Lewis's followers of Tash.
Yet, Tolkien never bought into your idea about the Orcs.
In the millenniums after Morgoth was finally defeated (and locked safely away from Middle Earth), the Orcs were still creatures of evil.
Sauron was much less powerful and couldn't prevent Saruman from gaining the apparent loyalty of some of the Orcs.
Yes, Orcs as a group might have worked to keep themselves "evil" but over the millenniums some groups of Orcs might (if they had Free Will) might have reformed.
Yeah, but look how many thousands of years tribal endemic warfare cultures can be stable over in the sense of fairly consistent behavior. That's not what Tolkein world-built, but it probably isn't as difficult to make it work as he thought it was.
The Orcs were "established" to be nasty beings even to members of the same "tribe".
Such a species IMO could not exist in the real world.
Yes, I could see tribes of Orcs seeing "outsiders" as enemies/untrustworthy but not internally in their own tribe.
How long would a tribe exist in the Real World if there was an extreme lack of trust among the members of the tribe?
This is my usual problem with a lot of fantasy. There HAS to be a basic level functionality in a society or children do not grow to adulthood. There also needs to be a very very large percentage of farmers and they have to manage to grow crops regularly or else absolutely everyone starves to death on relatively short order. "The Bad Evil wants to kill all life!" Oh, sure, that sounds like a fabulous way to "win".
OTOH, stories aren't always supposed to be realistic and an allegory is an allegory of what it is, not of whatever else someone wants to pretend that it is.
If orcs are corrupted elves, OTHO, that means that don't age and don't go to Heaven/Hell. Meaning that the orcs who are around in the time of LOTR are some of the same orcs that Morgoth first bred back at the beginning of the world. I could see Morgoth's programming being pretty hard to undo, even millenia after he was gone.
My opinion was that yes, there are good orcs, but no, the hobbits were not likely to meet up with them on their trip. Most would probably be killed as soon as they first asked, "Hey, is there a particular reason why we want to kill all the elves and men? Are we maybe the bad guys?" And while I could buy a band escaping and trying to be on the side of light, I also doubt they would advertise their existence.
It may not have been a good enough explanation for Tolkien, but it's good enough for me.
Mark	#
Drak, Zsuzsa hit the same thought I did. Evil Orcs would kill Good Orcs… period.
Eric Hinkle	#
Some of this talk about the evil orcs killing the good orcs reminds me of a scene from the Canadian movie 'Black Robes', which covers the early Jesuit missions to the Indians. One old Huron tells the Jesuit that he is a good man and means well, but if his people followed his faith and stopped torturing their enemies, killing them whenever they find them, and rather showed mercy to them, they would all die. 'Because no one will be afraid of us any more.'
And we get told at the end of the movie that a few years after the Hurons converted, the Iroquois did precisely that.
unpossible! all current history books say the Iroquois were a peaceful confederation!
What I think is telling is the author reads about orcs and the images that come to his mind are blacks.
Then he lectures us about how Tolkien was racist.
I think the person who instantly equates orcs with any race is the racist.
Mike Houst	#
Ahhh Melkor wasn't evil; he was just misunderstood and discriminated against.
It's not the first time that accusation has been raised.
Kind of like the accusations I see that Dr Seuss was racist, with half of the cartoons they show to prove it were from during the war, and the other half anti-racist cartoons.
Or using stills from "Der Furher's Face" to show that Disney lied the Nazis.
Yep. My responses to all of those is "who cares?". They were written at a time different from ours and we shouldn't condemn them for having values different from our own. Instead, we should learn from what they wrote, take the lessons–good and bad–to heart and move forward instead of trying to remove them from our memory.
Amazing ain't it though that folks have instant access to the knowledge of the ages at their fingertips yet insist on trying to relate everything to their perception of right now.
They don't read, they have no concept of real history, in other words ignorant and proud of it.
Yeah, but the LEFT is hell-bent on re-writing history to make the past look like a unicorn-fart paradise. The fact that they erase the very evidence of just how bad it was being the reason why they fight to eliminate it today totally escapes them.
I thought the modern Left spent most of their time arguing how awful it all was Back Then, so we have to create the Republic of Perfect Virtue (Perfect Virtue defined as 'what matters me me the most right this moment)' to get away from all that?
You expect consistency?
I agree – same goes for any Confederate Monument. I am definitely from the North, and am against slavery, racism, and bigotry. BUT – those monuments were erected to remind people of what other people had accomplished and should be remembered for.
Just because a few decades go by and values change does NOT invalidate the historical fact that so-and-so did such-and-such, and was highly regarded enough that people of his/her day and age felt it necessary to commemorate said deeds.
This is why I strongly object to colorizing movies, remakes, or this PC crap flying around these days. It is ALTERING THE HISTORICAL RECORD – and those who cannot remember the past…"
…or the nutters who write long theses on how racist the Charlie Chan movies were and how they denigrated Chinese…
Yeah. Wealthy, successful, competent, and internationally famous, that's RACIST!
Odd, they haven't targeted Peter Falk as Columbo yet. Obviously, the entire series was anti-Italisn racism…
I have seen a few Charlie Chan movies. In one of them, set early in WW2, Chan's Number One son is mentioned as being in the Army, and there's a scene where Chan mocks a racist who assumes he's there for the laundry.
RACIST!
And that was the movies. The original books are about as non-racist as you could get in the period.
But, Christopher, they have to be racist. After all, they didn't have a Chinese actor playing a Chinese role. (that sound you hear are my eyes rolling out of my head, across the floor and being batted around by the cats.)
at least you have happy, entertained kitties.
Ted Ung	#
Lost into the history memory hole: Although Texas City, Texas was segregated in the 1950s, my very Caucasian aunts and mother would have shunned me (at least) for a week, had I in a fit of bad behavior, used the word *nigger, even in private. Then again, my grandma, from Danzig/Gdansk environs, used *du Russe, you Russian, as the worst possible pejorative, with the exception of *faulpelz. Herself abhored laziness.
My mother would have pulled me up short and quite possibly knocked some sense in my head had I used the word. But, when it was used (in books, movies, etc), we discussed it. We discussed it in the historical context and in the context of the time when I heard it. I learned that there were things acceptable in the past that aren't acceptable now. I was raised to respect and remember "Never forget" because if we forget the past, we are doomed to repeat it.
My grandmother would go on about how she never discriminated against other races; but was horrified to learn I'd taken a couple of black gals on dates. An Asian girl would have been okay, as long as she had a PhD and came from a "good" family; but why couldn't I find a nice girl from around here (i.e. white.)
I forget whatever I did to keep from falling out of my chair laughing at her.
but was horrified to learn I'd taken a couple of black gals on dates.
I was out on a "home teaching" assignment (back when I was still active LDS) and the family we were visiting said they could prove I was "racist." Their proof was to ask me if I'd want my sister to marry "one".
They apparently didn't appreciate when I shrugged and said "who she marries is her choice and none of my business."
Huh. The old foggies would probably dismiss it as being concerned for the difficulties the couple and their children would encounter.
But never fear! Things have improved! /sarc
There was a whole documentary recently about how racist white people who married black people were and the conclusion, more or less, that a white parent wasn't able to be a good parent to a black child.
Including at least one part about how a mixed race kid loved her mom so much until she went to college and learned not to.
Frankly, I think this is one reason why the concept of "judgement" is so popular. You just want some people to have to answer for what they've done in a context where the consequences of their actions are fully revealed and they can't pretend anymore that they were the Good Guys.
I am waiting for that mindset of 'mixed ethnicity marriages= racist' to start biting them in the arse one day.
Then again, these are the dimwits who conflate culture and religion with race/ethnicity. They'd more likely be shocked the person they're making assumptions of victimhood on behalf of isn't worshipping them in gratitude.
Colored People is bad. People of Color is not. Go figure.
I remember in the 70s that "negro" was the appropriate term; not so much today.
Do memos come out updating this stuff?
If there were memos you could keep up to date and do no wrong. Since the point is to make what you do wrong, memos would defeat it.
Poodlehorde	#
My parents grew up in the Upper South (Arkansas and Oklahoma) between the Wars and my Dad managed several mixed race detachments during his navy career. I never remember him using the term " nigger", and he always emphasized that you judge everyone as indivduals and spacifically not by racial or class streotypes
My father used the word. But only about certain people (black crackers, as it were, and the kind of lazy parasites Steppin Fetchit made a mint portraying). And he never used it to a Negro's face.
adventuresfantastic	#
I had heard of Duncan a long time ago. I was buying the Golden Gryphon collections at the time, and he had one, Beluthahatchie. I've never read it. He won a Nebula for best novelette in 2013. I think I may have started that one and not finished.
I've heard all the arguments on Tolkien and race before. I don't care. Tolkien wrote fantasy in a different time, and I can handle that his ideas might be different than mine.
How open minded of you. You must now report to the re-education camp for reprogramming. The line forms behind the rest of us here.
OK, I admit I was wrong. It turns out I did read his first collection. I just don't remember anything about it. But the book has clearly been read, and I had shelved it with some other Golden Gryphon collections I know I read because I remember reading them. Must have suppressed the memory of it.
kamas716	#
I saw he said something, and my only real reaction was "Never heard of this blithering idiot. He must be looking for publicity." Because really that's all his screed is about, getting his name in front of people in the hopes he gets a following.
Yep. I wrote about him only because I am damned tired of the other side doing their best to tear down the great writers of yesterday who don't meet their artificial "woke" levels today.
Rusty	#
Then there was an incident from many years ago when the NAACP wanted to ban reruns of the "Amos 'n' Andy" show from being shown on TV. The NAACP wanted to ban the show because it made black folks look like fools. My reply to that was, "If you were worried that Amos 'n' Andy made blacks look like fools, then the Three Stooges must make whites look like fools! Folks, lighten up, it's just comedy!" And as comedy, Amos 'n' Andy was hilariously funny, especially with the "Kingfish" and Sapphire.
There's a British TV show called "Citizen Khan" about a Pakistani immigrant and his family. It's basically "All in the Family" except with a Pakistani instead of Archie Bunker; Mr. Khan is bigoted, racist, stubborn, sneaky, and cheap, which are the things that get him into trouble in every episode…
Of course "action groups" immediately materialized claiming the show was racist, there should be mass apologies, and of course it should be canceled. Which groups were countered by Pakistani immigrant groups who thought the show was hysterically funny and wanted more… the do-gooders were rather put out that the silly immigrants didn't appreciate the efforts on their behalf.
Matthew	#
The bastages may have succeeded with Apu from the Simpsons, thouhg.
midnightoildiary	#
Seems like lots of folks are equating "change" with "good" these days. Don't know why.
Socialism is a probably a religion. Modern socialists are practically all Marxists. As a religion, Marxism is sorta of an animism of curves fit to cherrypicked 'data'. A data science and as history, the curves that they primarily worship are complete garbage. They presume that all changes have desirable outcomes, hence more changes per unit time mean that their New Jerusalem will arrive sooner. This despite that most of them are not too mentally impaired for the following analysis.
"There are black motorists in America who have not been pulled over and killed by the police. Thus we can say that pulling over black motorists and killing them is not a rule always followed by every policeman. In other words, it is not the current state. Which means that circumstances where that rule is always followed by every policeman would be a change from the current state. Would that be a desirable change?"
thephantom182	#
Just change black motorist to white and you'll be right where the fringy ones are.
Re: Change.
I don't see much about "change" that isn't 100% about control. And no, objecting to control or to losing control is not objecting to "change". Because we all well know that my "change" is not the same at all as your "change".
There's another choice and that's choosing NOT to try to control others, one way or another.
You rarely if ever see Eddie Cantor movies anymore. You might encounter a book he wrote, or some of his singing, and perhaps his radio comedy. But not the movies, for there was always one Minstrel Show scene complete with 'blackface' which is Simply Not Done. But in Kid Millions there is a scene where he is preparing for that bit, putting on the black makeup… and commenting to the black servant, "You know, you're lucky" and gets SUCH A LOOK. It might have been meant as comedic, and it was, but it was perhaps just a bit more. And now people are unlikely to see that scene.
Sad. Unlike a lot of today's comics, Cantor was actually funny.
I vaguely recall reading the "Senator Bilbo" story in a Tolkien themed anthology once upon a very long time ago. I didn't much care for it then, and that was back when I was still a liberal.
"I'll admit, my first reaction when I saw the allegation that Tolkien was racist–and against Orcs no less–I didn't know whether to laugh or toss my iPad against the wall."
I saw this thing the other day, with the "racism against Orcs" and I didn't even care enough to find out what the shit head said about it so I could mock it. One more asshole with no talent, taking down the Master for the feeeelz and to get some eyeballs on his own scribblings.
I saw something else that was much more irritating, The Atlantic had an ode to the Peter S. Beagle book The Last Unicorn. It's been 50 years, apparently, since Peter S. Beagle got that thing published.
The Atlantic article, of course, sings the praises of the story. Blah blah blah. What it really is, is the same thing Andy Duncan is serving up. A cheesy, coarse mockery of Tolkien and high fantasy in general.
Good of them to self-identify like that.
60guilders	#
Are you sure about that? I seem to recall that The Last Unicorn was more of a love letter to high fantasy than a takedown.
He took all the hero and purtity tropes, turned them on their heads and called in some course language, is my recollection. I'm not going to say it was badly written at this distance, because I read it in the late 1970s and I don't remember.
But I do remember -detesting- the main characters and their actions, and wishing a nice big asteroid strike would take the lot of them. Spoiler, they all come to grief and live in misery at the end, but there's no satisfying asteriod.
One of the nice things I can do as an author these days is call down the thunderbolts of Jove on NPCs standing in for characters I have detested over the years. Beagle's MCs deserve that, IMHO.
Did you read a different book than I did? Because I recall the ending as being somewhat bittersweet, with the unicorn changed by her adventure, but the wizard and Molly Grue seemed to be doing well, and while Lir was heartbroken about the loss of his first love, there were strong hints that second love was on its way, running right behind him.
I was perturbed by the MC ditching the unicorn. It was so crass. If you're the hero you don't ditch the unicorn, particularly when its the LAST one. I'm not a fan of "feet of clay" heroes.
Everybody reads a different book. ~:D
While I don't need feet of clay heroes, having heroes with some identifiable human foibles I can relate to helps. If I want a hero with zero human faults, I'll read the Gospels.
Aragorn has human faults, and limitations. Which he overcomes, despite all.
What I'm talking about is Aragorn abandoning the quest and skulking off to live in penury with some frowzy bar maid he just met. And everybody else doing the same thing.
Even Sauron, who can't quite be bothered to exert himself for a proper Apocalypse. He only manages a kind of tepid unpleasantness that kind of merges with the pre-existing overall Suckage.
That was my experience of the Last Unicorn. YMMV.
John R. Ellis	#
….okay, I also need to point out that neither Schmendrick, Molly, OR Lir "ditch the Unicorn"…so I'm not sure which "Main Character" you mean. Are you sure you aren't referring to "Ariel" by Stephen R. Boyette? Because that's what it sounds like you're describing.
Coarse language? Where in The Last Unicorn did coarse language appear? Not even the harpy!
Okay, could be I'm wrong.
Willing to admit I could have the wrong book. I remember course language.
I plead the infirmity of age and my cranky nature.
Somebody surely wrote some kind of unicorn story that hacked me off though, unless that was a different life and this is The Matrix.
Backing down off my soapbox and slinking off to the bar.
sam57l0	#
I tried to read "The Last Unicorn"; couldn't do it.
You missed a whole lot of pain.
The Last Unicorn was, stylistically, an attempt to merge high fantasy, low fantasy, and picaresque adventure. It managed it about as well as one possibly could, and the ending was bittersweet. The only book that has done it better is Silverlock, but I think The Last Unicorn has more magic.
There is no shame in bouncing off a book. But I beseech you to reconsider your judgment.
"But I beseech you to reconsider your judgment."
Truthfully, I took the book as trolling Fantasy fans. Disrespectful.
If the hero you're looking for is Aragorn, the MC of The Last Unicorn is not that guy. Nobody in the book is that guy. They're not even Harold Shea. Although, memory fades since high school. Could be I forgot a bit player who showed some steel.
I would point out Beagle also wrote the forward to the first authorized PB version of the LotR in the US in which he lauded Tolkien and his creation.
Beagle might not have written fantasy to your taste (I enjoyed TLU), but to claim he was trying to take down Tolkien is 180 degrees from available evidence.
I have some points to make about Peter S. Beagle and 'The Last Unicorn'…while you are free to like or dislike it as you wish, I feel compelled to state a few things.
1) Back in the 1980s I received a boxed set of 'The Lord of the Rings' as a Christmas gift. (It was the editions with the Darrel K. Sweet covers.) The forward to the series was given by…Peter S. Beagle, who spoke with glowing praise and admiration for Tolkien and his work even as he had some very pointed, harsh words for the direction 20th century culture had taken (he believed in direct contradiction to Tolkien's message.)
2) Peter S. Beagle was a guest at Comic Con 2006 and I sat in the audience of his panel. Among other things, he went into great depth into the origins behind The Last Unicorn.
It was first written as a post-apocalyptic tale about a unicorn in a world completely inhabited by demons (mankind having long since vanished) but that idea was going nowhere. But the more he wrote, the more he realized he really wanted to write a fairy tale.
Not just a fairy tale, but one written in homage to the works of James Thurber's very dry, very witty "fractured" fairy tales, yet also having the emotional poignancy and bittersweet bite of…of…it's been many years and I've forgotten the name of the other author! But from what I recall, it was fairy tales written in the tone and style Hans Christian Anderson favored. Where yes, some of them ended unambiguously happy…but just as many end in regret, tragedy, or a very special, tender sort of melancholy.
This combined with characters from bedtime stories he was telling his kids (this is where Schemdrick the magician came from) and favorite poems, plays, and jokes from his life at the time. (This is where the Butterfly came from.)
I disagree strongly that the story ends in nothing but misery and loss. And Lir…he's not the Main Character, but he's definitely an important character.
Mocking Tolkien and High Fantasy didn't come into it…he was writing in a different subgenre. Not high fantasy, but fairy tale fantasy, which has very different rules and a very different tone.
(Full disclosure, I reread the book last year.)
…assuming you -are- actually referring to Stephen R. Boyette's "Ariel", that book definitely -does- mock Tolkien and has coarse language and Boyette himself has taken potshots at Tolkien. So I'm starting to think that's the actual book you're recalling. 🙂
I'm willing to be wrong about the title and author, but not willing enough to go re-read it to make sure.
Full disclosure, I read it as a teenager in the 1970s, and it pissed me off so bad I threw it away. One did not throw away books in my mother's house in the 1970s, but I did. I'm pretty sure it was The Last Unicorn, but then I was pretty sure I had my keys the other day when I didn't. 😡
I looked up Ariel, what I found indicates a 2008 publishing? Not that one.
So if I'm slandering Mr. Beagle unfairly, its because I'm old, bad tempered, crusty and can't remember shit. Wouldn't be the first time.
Ariel was originally published in the late 80s. It was post-apoc with a young man winding up with a unicorn companion (who I always thought was his school friend raped and murdered early on in the magic coming back apoc).
In the end he has sex with a female character and lost Ariel.
I enjoyed it enough, but wouldn't consider it a classic as I do the Lost Unicorn. Boyette went on to start a series that only got one book which was much more interesting. Then he disappeared only to show up in 2010 or so with a book in the Ariel universe which I never bothered reading. I have mostly read his rants about what happened to his raccoon universe series (the only book out was The Architect of Fate), but nothing he has said about Tolkien.
I have to second that I don't see misery at the end of TLU. Yes, the hero did lose the girl who turned out to be the titular character, but Schemdrick's point about how that was important. It is what allowed Lir to be a great hero, because all great heroes must have great sorrows and Lir's was to have loved a unicorn.
I think that concept fits well with Tolkien and parallels Frodo at the very least. It also finds a home in Lewis's "Screwtape Proposes a Toast" about the need for saints to have great sinners.
When racist is defined so broadly that anyone can be a racist, it is defined so loosely that there is little purpose in seeking to avoid meeting the definition.
Honestly, I think it's all about letting certain people completely off the hook. Racism isn't behavior, therefore no one need actually examine their own behavior. No self-reflection involved. Even "check your privilege" doesn't require any self-reflection other than to catalog how wonderful you are. Follow that with a bit of ritual public confession and *poof* you're done.
Geoff Withnell	#
Changing the language dramatically reduces the impact of Huck Finn's statement, when he decides to help his black companion evade recapture with "Well, then I'll just go to hell".
If that is how he feels, why doesn't he write his own ORIGINAL work, showing how such an attitude is wrong?
Because if you think the conventions of a genre could use shaking up, actually doing it is real work. Ask Michael Moorcock or Allen Moore.
Of course, if you do the work you will often become another icon of the field that people a generation will decide to tear down, but you won't get invited to the right parties tonight.
Not that you'd go if you did. You'd be too busy working.
Zachary Ricks	#
Note to self: maybe it's time to re-read LotR…
Yawn. So someone thinks the orcs are racist because they're Always Chaotic Evil. Gee, never heard that one before. It's not like fantasy fans and Tolkien himself have been debating the subject since the book came out.
I'm aware that there's a certain type of speculative fiction writer that can never be happy unless they're tearing down someone else while desperately trying to build themselves up, but Mr. Duncan should really stay out of the business until he can figure out something to say that wasn't already done to death half a century ago.
For Everybody's enjoyment, see: https://tvtropes.org/pmwiki/pmwiki.php/Main/OurOrcsAreDifferent
You can thank me about you manage to escape. [Very Big Evil Grin While Flying Away Very Fast]
allene	#
Oh, geez. I saw that headline and assumed it was a joke. I'm just surprised no-one's gone after Peter Pan yet. (That I've heard of, anyway.)
Um… the longer you can manage not to hear about that the better.
Yeah, don't investigate that.
You won't like what you find.
Evenstar	#
And yet fifty years from now, people will still be reading The Lord of the Rings and Mr. Duncan will likely be utterly forgotten.
Who? 😉
Duncan is getting more publicity over the past 48 hours than he's ever gotten in his whole career. It sucks that we have to give him attention at all, but sometimes you have to counter crap before it becomes dominant.
The good news is that after this week, his slide back into obscurity will hopefully begin.
Even worse, he might win a Hugo. Then he'll be actively avoided instead of just ignored.
Someone publicly sticking their foot in their mouth always gets good press for a few days; until the next fool comes around and does it.
David Foster	#
"when they showed the live performance of Sound of Music on broadcast TV not long ago, even it had been changed."
What did they change?
I saw one where the Mother Superior is black. Not impossible, just highly unlikely, and quite jarring. I had a dreadful feeling that she'd been cast because the producers insisted on having non-whites in the cast, not for her singing or acting skills. Also, the boyfriend in "I am Sixteen (Going on Seventeen)" was much, much too old for the role. It gave that scene an uncomfortable tone. I'm not certain if this was the same production mentioned above or not.
I have found that many American progressives have this delusion of European populations being as mixed as ours, (i.e. same proportion of minorities) and Europeans seem to have the same- our minorities are as rare as theirs are…
I have heard leftists claim, explicitly, that MEDIEVAL Europe was "diverse."
People complaining about racism in fantasy make me want to throw them into the wall, not the book.
What part of the word, "fantasy" don't they understand?
That just means it's an ALLEGORY of Real Life, not actually about another world. Everything's symbolic when it's not realistic.
snelson134	#
There are two things I'm not seeing anyone take into account here, one natural and one not.
Never looked at the actual dynamics of a wolf pack or a flock of chickens or tribal band of people for that matter? There's a rigid dominance hierarchy and as long as none of them tries to break it the organization works. Ask the lowest level if life sucks and they might agree…. But if that lowest level can be convincingly told they are still better / entitled to dominate the Other, that channels the rebellion.
And that leads us to the second factor, which I doubt even Tolkien fully explored: belief in a god or gods is a powerful reinforcement even when the gods aren't seen acting visibly and directly (this world). How much more powerful is it when they ARE?
David Weber shows that several times in his Bahzell books, where even the clergy / religious orders are disinclined to accept something…. until the deity or the deity's agent shows up and demonstrates through actual exercise of power that "Yes, this is how things are." And even if you don't have Morgoth or Sauron showing up in person, you have lesser agents such as the Nazgul or other lesser people who are actually able to call on them and something happens. Do you really break THAT conditioning?
The chickens and wolves do exist. But they do not exist among the orcish society. Remember Frodo and Sam escape once because of an intra-race murder.
Hey, if I was an Orc and had one of the Nazgul show up, I would do what he said – those guys are scary!!!
Earlier in this thread I had some harsh words about Peter S. Beagle and The Last Unicorn.
As many have surmised, I had the wrong book. It was bothering me, so I went and looked into it. Yep. Wrong.
Mea Culpa. I apologize to Mr. Beagle who I wrongly maligned.
wyrdbard	#
*offers tea and snickerdoodles*
Thank you, ma'am. I figure it's unlikely Mr. Beagle will hear of it, but there's no reason to raise the level of BS on the internet by letting my mistake stand.
As one of my characters is known to say, if all that gets hurt is your pride, its a good day. ~:D
Reziac on If You Give A Reader A Cookie…
The Phantom on An Alternate Australia?


text:
For Christians, To be Born Again in Islam!
Common Between Christianity and Islam
Murdoch was sent 'hacking is rife' email
December 13, 2011 · Europe, France, Germany, Italy, UK, United States
Source / Courtesy: The Independent
Emails published today show that News International boss James Murdoch was sent details in 2008 of claims that phone-hacking was "rife" at the News of the World.
But Mr Murdoch told the House of Commons Culture Committee, which is investigating the hacking scandal, that he did not read the email exchange forwarded to him by the paper's then editor Colin Myler.
In an email dated Saturday June 7 2008, Mr Myler requested a meeting with Mr Murdoch to discuss the case being brought against the paper by Professional Footballers Association chief executive Gordon Taylor over claims reporters had eavesdropped on his messages.
The News of the World editor warned Mr Murdoch: "Unfortunately, it is as bad as we feared."
Attached to his message was a "chain" of emails detailing discussions between News International's legal adviser Julian Pike of Farrer & Co and Mark Lewis, who represented Mr Taylor.
Mr Murdoch and Mr Myler met three days later on June 10, along with Tom Crone, legal manager for the NotW's publishers News Group Newspapers. Mr Myler and Mr Crone say that they told Murdoch at that meeting about the discovery of the notorious "For Neville" email, which proved that phone-hacking was not limited to a single "rogue reporter" on the paper as the company had claimed.
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YEREVAN (Combined Sources) — The Office of the Prosecutor-General said Armenia's National Security Service (NSS) has ruled out the possibility that the explosion in a Yerevan commuter bus that left two people killed and seven injured was caused by a terrorist attack.
Investigators have found a SIM-card at the scene of the April 25 blast that belonged to a twice-convicted citizen of the Republic of Armenia. They also found explosives and detonators at his home in a search. They did not name the suspect and his whereabouts are unknown, but officials said it is possible the suspect was one of the two people killed in the blast.
The individual, investigators say, planned to hurt his relatives, but it is not clear whether he planned to do it on the bus or was just transporting the explosive materials when they apparently detonated. Probe is under way to check all versions and reveal the causes of the explosion.
Armenia's National police chief Vladimir Gasparian said earlier that the blast occurred on or under a second-row bus seat and that investigators were trying to identify a passenger thought to have been sitting there.
Nine of the ten people in the bus, including the driver, the seven injured passengers and one victim, have been identified. All of them are Armenian nationals.
The apartments of all of them have been searched as part of the investigation, no items of interest for the criminal case have been found, their involvement in the explosion has been ruled out.
The second victim of the blast is yet to be identified. The body is unrecognizable, and DNA test is needed. Probe is under way to find out whether the second victim was the owner of the SIM card found at the scene.


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A syndicated conservative columnist and cable-news commentator asserts her opinion on Barack Obama, his cabinet, and other members of his circle.
John Howard, asks: WHY? Why can't we display the Ten Commandments in public places? Why can't we say Merry Christmas, even though that's the holiday we are celebrating? Why do we all have to kowtow to the liberal Church of the Enlightened's dictates and lack of morals? Why can't children pray in school, if that is their choice? Why can't we find out the truth about our elected leaders who pay attorneys to cover up their past? And WHAT can the right-thinking Christian majority do about changing the messages that are sent out each day, seeking to dismiss them as crazy people that seek to destroy this great nation? The United States of America is the greatest country in the world and only when right-thinking foundational principles are brought to surface, can we truly have a nation of which to be proud.


text:
\section{Main definitions}

Let $T^2$ be a finite (but very deep) bi-tree. Bi-tree is the directed  graph  of all dyadic rectangles in the square $Q_0=[0,1]^2$. We assume that it terminates at small squares of size $2^{-N}\times 2^{-N}$, we call them generically by symbol $\omega$.  The boundary $\partial(T^2)$ is this collection of $\omega$'s. Often we identify $T^2$ with dyadic rectangles, whose family is called $\mathcal D$. When we write $E\subset (\partial T)^2$ we mean any subset of $\omega's$. It is convenient to think of $E$ as the union of ``$N$-coarse" dyadic rectangles.

Box condition
\begin{equation}\label{e:Boxcond}
\sum_{Q\in T^2, \,Q\subset R}\mu^2(Q)\alpha_Q \leq C_{\mu}\mu(E),\quad \textup{for any}\; R\in T^2.
\end{equation}
Carleson condition
\begin{equation}\label{e:Carlcond}
\sum_{Q\in T^2,\,Q\subset E}\mu^2(Q)\alpha_Q \leq C_{\mu}\mu(E),\quad \textup{for any}\; E\subset (\partial T)^2.
\end{equation}
Restricted Energy Condition
\begin{equation}\label{e:REC}
\sum_{Q\in \mathcal{D}}\mu^2(Q\cap E)\alpha_Q \leq C\mu(E),\quad \textup{for any}\; E\subset (\partial T)^2
\end{equation}
Embedding
\begin{equation}\label{e:imbed}
\sum_{Q\in\mathcal{D}}\left(\int_{Q}\varphi\,d\mu\right)^2\alpha_Q \leq C\int_{Q_0}\varphi^2\,d\mu\quad \textup{for any}\; \varphi\in L^2(Q_0,d\mu).
\end{equation}
Each of the next statement implies the previous one. We are interested when the opposite direction implications hold, and whether they hold in general.

Embedding is the boundedness of embedding operator $L^2(Q_0, \mu)\to L^2(T^2, \alpha)$ acting as follows:
\[
f\in L^2(Q_0, \mu)\to \Big\{\int_R f\, d\mu\Big\}_{R\in T^2}\in L^2(T^2, \alpha).
\]
 Restricted energy condition (REC)  is the  boundedness of embedding operator $L^2(Q_0, \mu)\to L^2(T^2, \alpha)$ on characteristic functions ${\bf 1}_E$ for all $E\subset\partial(T^2)$ uniformly.

 \section{Examples having box condition but not Carleson condition}
 \label{boxNotC}

 In \cite{Car} Carleson constructed the families $\mathcal R$  of dyadic sub-rectangles of $Q=[0,1]^2$ having the following two  properties:
\begin{equation}
\label{boxCa}
\forall R_0 \in \mathcal D, \quad \sum_{R\subset R_0, R\in \mathcal R} m_2(R) \le C_0 m_2(R_0)\,,
\end{equation}
but for $U_\mathcal R:= \cup_{R\in \mathcal R} R$
\begin{equation}
\label{Ca}
 \sum_{ R\in \mathcal R} m_2(R) \ge C_1 m_2(U_\mathcal R)\,,
\end{equation}
where $C_1/C_0$ is as big as one wishes.

\medskip

 With {\it some} (rather wild) $\{\alpha_R\}_{R\in \mathcal D}$ Carleson's counterexample is readily provide the example of $(\mu, \alpha)$ such that box condition is satisfied but Carleson condition fails.
Indeed, let us put
$$
\alpha_R= \begin{cases} \frac{1}{m_2(R)}, \quad R\in \mathcal R,
\\
0, \quad \text{otherwise}
\end{cases}
$$
Measure $\mu$ is just planar Lebesgue measure $m_2$. Fix any dyadic rectangle $R_0$, then box condition is satisfied:
$$
\sum_{R\subset R_0}\mu(R)^2 \alpha_R = \sum_{R\subset R_0, R\in \mathcal R} m_2(R) \le m_2(R_0)=\mu(R_0)\,.
$$
But if $\Omega:= \cup_{R\in \mathcal R} R$, then
$$
\sum_{R\subset \Omega}\mu(R)^2 \alpha_R =\sum_{R\subset \Omega, R\in \mathcal R} m_2(R) =1\ge C\,m_2(\Omega)\,,
$$
where $C$ can be chosen as large as one wants. Hence, \eqref{Ca}  holds too with large constant.

\bigskip

The weight $\alpha=\{\alpha_R\}$. is rather wild here. But there is also an counterexample with $\alpha =1, 0$, see \cite{HPV}.

\section{Examples of having Carleson condition but not restricted energy condition REC.}
Our aim here is to show that if we do not restrict ourselves to the constant weights as in \cite{AMPS18}, \cite{AHMV18b}, \cite{AMVZ19}, then the Carleson condition \eqref{e:Carlcond} is no longer sufficient for the embedding \eqref{e:imbed}. In fact even the Restricted Energy Condition \eqref{e:REC} is not necessarily implied by \eqref{e:Carlcond}. Namely we prove the following statement.
\begin{proposition}\label{p:71}
For any $\delta>0$ there exists number $N$, a weight $\alpha: T^2_N\mapsto \mathbb{R}_+$ and a measure $\mu $ on $(\partial T)^2$ such that $\mu$ satisfies Carleson condition \eqref{e:Carlcond} with the constant $C_{\mu} = \delta$,
\begin{equation}\label{e:771}
\sum_{Q\subset E}\mu^2(Q)\alpha_Q \leq \delta\mu(E),\quad \textup{for any }E\subset (\partial T)^2,
\end{equation}
but one can also find a set $F$ such that
\begin{equation}\label{e:772}
\sum_{Q\in\mathcal{D}}\mu^2(Q\cap F)\alpha_Q \geq \mu(F),
\end{equation}
hence the constant in \eqref{e:REC} is at least $1$.
\end{proposition}
We intend to give two examples of this kind. Both of them rely on the fact that one can basically consider this problem on a cut bi-tree by letting $\alpha$ to be either $1$ or $0$.
This approach clearly does not work on the tree (see \cite{AHMV18a}[Theorem 1.1]), but the bi-tree has richer geometric structure.

While globally (i.e. for $\alpha\equiv 1$) it looks at least somewhat similar to the tree (this similarity is implicit in the proof of \cite{AHMV18b}[Theorem 1.5]), one can remove some vertices (which is what essentially happens when we put $\alpha_Q := 0$) in such a way that the remaining part looks nothing like the full bi-tree (or a tree for that matter). In particular this allows us to create a significant  between the amount of ``available" rectangles that lie inside $E$ or just intersect $E$ for a certain choice of the weight $\alpha$ and the set $E$ (this corresponds to the difference between Carleson and REC conditions).\par
The first example is quite simple and is inspired by the counterexample for $L^2$-boundedness of the biparameter maximal function. The weight $\alpha$ in this case cuts most of the bi-tree, and the resulting set differs greatly from the original graph.\\ The second example is somewhat more involved, on the other hand the weight there leaves a much bigger portion of the bi-tree, actually it has a certain monotonicity property: $\alpha_R \geq \alpha_Q$ for $R\supset Q$. The structure of the ``available" set is more rich in this case (it looks more like $\mathbb{Z}^2$ in a sense), nevertheless there is not enough rectangles to have the Carleson-REC equivalence.

\subsection{A simple example of having Carleson condition but not restricted energy condition}
Let $N\in\mathbb{N}$ be some large number (to be specified later), and let $T^2 = T_N^2$ be a bi-tree of depth $N$. We use the dyadic rectangle representation of $T^2$.\par
Let $\omega$ be our $[0,2^{-N}]^2$ left lower corner. Given $R = [a,b]\times[c,d]\in\mathcal{D}$ let $R^{++} := \left[\frac{a+b}{2},b\right]\times\left[\frac{c+d}{2},d\right]$ be the upper right quadrant of $R$. Consider
$Q_1= [0,1]\times [0, 2^{-N+1}]$ and its $Q_1^{++}$, and
$Q_2= [0,2^{-1}]\times [0, 2^{-N+2}]$ and its $Q_2^{++}$,
$Q_3= [0,2^{-2}]\times [0, 2^{-N+3}]$ and its $Q_3^{++}$,  et cetera\dots . In total, $N$ of them.

Put measure $\mu$ to have mass $\tau_0:=1/\sqrt{N}$ on $\omega$, and uniformly distribute mass $\tau_i$ on $Q_i^{++}$.

Now $\alpha_R$ is always zero except when $R= \omega, Q_1, Q_2, \dots$. For those $\alpha=1$, so we have $N+1$ alphas equal to $1$.

Now choose set $E=\omega$. When we calculate $\mathcal E[\mu|E]$ we sum up
$$
\tau_0^2 +\sum_{i=1}^N \mu(\omega\cap Q_i)^2 =( N+1)\tau_0^2 =\frac{N+1}{N}\ge \sqrt{N} \frac1{\sqrt{N}} =\sqrt{N}\mu(E)\,.
$$

 So REC condition has a big constant.

 \bigskip

Since
$$
\mathcal E_\Omega[\mu] =\sum_{R\subset \Omega,\, \alpha_R\neq 0} \mu(R)^2\,,
$$
then, denoting $Q_0:=\omega$, we have
$$
\mathcal E_\Omega[\mu] = \sum_{j: Q_j\subset \Omega} \mu(Q_j)^2 =:\sum_{j\in J( \Omega)} \mu(Q_j)^2\,.
$$

Let $\tau_0=1/\sqrt{N} \le \frac14=\tau_1=\tau_2=...=\tau_N$.
Then
$$
\mathcal E_\Omega[\mu] =\sum_{j\in J(\Omega)} (\tau_0+\tau_j)^2 \le 4\sum_{j\in J(\Omega)} \tau_j^2\,.
$$
And this is $\le \sum_{j\in J(\Omega)} \tau_j \le \mu(\Omega)$.
So Carleson condition holds with constant $1$.

\subsection{The lack of maximal principle matters}
\label{maxP}

All measures and dyadic rectangles  below will be $N$-coarse.

In this section we build another example when Carleson condition holds, but restricted energy condition fails.
But the example is more complicated (and more deep) than the previous one. In it the weight $\alpha$  again has values either $1$ or $0$, but the support $S$ of $\alpha$ is an {\it up-set}, that is, it contains every ancestor of every rectangle in $S$.

The example is based on the fact that potentials on bi-tree may not satisfy maximal principle. So we start with constructing $N$-coarse $\mu$ such that given a small $\delta>0$
\begin{equation}
\label{le1}
\mathbb V^\mu \lesssim \delta\quad \text{on}\,\, \operatorname{supp}\mu,
\end{equation}
but  with an absolute strictly positive $c$
\begin{equation}
\label{logN}
\max \mathbb V^\mu \ge \mathbb V^\mu(\omega_0) \ge c\, \delta\log N\,,
\end{equation}
where $\omega_0:=[0, 2^{-N}]\times [0, 2^{-N}]$.

We define a collection of rectangles
\begin{equation}\label{e:773}
Q_j :=  [0,2^{-2^j}] \times  [0,2^{-2^{-j}N}] ,\quad j=1\dots M\approx \log N,
\end{equation}
and we let
\begin{equation}\label{Qs}
\begin{split}
&Q_j^{++} :=  [2^{-2^j-1},2^{-2^j}]\times  [2^{-2^{-j}N-1},2^{-2^{-j}N}] \\
&Q_j^- := [0,2^{-2^j-1}] \times  [0,2^{-2^{-j}N}] ,\quad j=1\dots M\\
&Q_j^t := Q_j\setminus Q_j^-\\
&Q_j^r:= [2^{-2^j-1},2^{-2^j}]\times  [0,2^{-2^{-j}N}]\\
& Q_j^{--} := Q_j^-\setminus Q_j^r
\end{split}
\end{equation}
to be their upper right quadrants, lower halves, top halves, right halves, and lower quadrant respectively.
Now we put
\begin{equation}\label{e:774}
\begin{split}
&\mathcal{R} := \{R:\; Q_j\subset R\; \textup{for some }j=1\dots M\}\\
&\alpha_{Q} := \chi_{\mathcal{R}}(Q)\\
&\mu(\omega) := \frac{\delta}{N}\sum_{j=1}^M \frac{1}{|Q_j^{++}|}\chi_{Q_j^{++}}(\omega),\\
& P_j= (2^{-2^j}, 2^{-2^{-j}N})\,.
\end{split}
\end{equation}
here $|Q|$ denotes the total amount of points $\omega \in (\partial T)^2\cap Q$, i.e. the amount of the smallest possible rectangles (of the size $2^{-2N}$) in $Q$.\par
Observe that on $Q_j$ the measure is basically a uniform distribution of the mass $\frac{\delta}{N}$ over the upper right quarter $Q_j^{++}$ of the rectangle $Q_j$ (and these quadrants are disjoint).
\par

To prove \eqref{le1} we fix $\omega\in Q_j^{++}$ and split $\mathbb V^\mu(\omega)= \mathbb V^\mu_{Q_j^{++}}(\omega) +\mu(Q_j^t)+ \mu(Q_j^r)+ \mathbb V^\mu(Q_j^{++})$, where the first term sums up $\mu(Q)$ for $Q$ between $\omega$ and $Q_j^{++}$. This term obviously  satisfies $\mathbb V^\mu_{Q_j^{++}}(\omega) \lesssim \frac{\delta}{N}$. Trivially $\mu(Q_j^t)+ \mu(Q_j^r)\le  \frac{2\delta}{N} $. The non-trivial part is the estimate
\begin{equation}
\label{VQ}
\mathbb V^\mu(Q_j^{++}) \lesssim \delta\,.
\end{equation}
To prove \eqref{VQ}, consider the sub-interval of interval $[1, n]$ of integers. We assume that $j\in [m, m+k]$.  We call by $C^{[m, m+k]}_j$ the family of dyadic rectangles containing $Q_j^{++}$ along with all $Q_i^{++}$, $i\in [m, m+k]$ (and none of the others). Notice that $C^{[m, m+k]}_j$ are not disjoint families, but this will be no problem for us as we  wish to
estimate $\mathbb V^\mu(Q_j^{++})$ from above.

Notice that, for example, $C^{[m, m+1]}_j$ are exactly the dyadic rectangles containing point $P_j$. It is easy to calculate that the number of such rectangles
is
\[
(2^j+1)\cdot (2^{-j}N +1)\lesssim N\,.
\]

Analogously, dyadic rectangles in family $C^{[m, m+k]}_j$ have to contain  points $P_m, P_{m+k}$. Therefore,  each of such rectangles contains  point $(2^{-2^m}, 2^{-2^{-m-k}N})$.
The number of such rectangles is obviously at most $\lesssim 2^{-k}N$. The number of classes $C^{[m, m+k]}_j$ is at most $k+1$.

Therefore, $\mathbb V^\mu((Q_j^{++})$ involves at most $(k+1)2^{-k}N$ times the measure in the amount $k\cdot \frac{\delta}{N}$. Hence
\[
\mathbb V^\mu((Q_j^{++}) \le \sum_{k=1}^n k(k+1)2^{-k}N \cdot \frac{\delta}{N}\,,
\]
and \eqref{VQ} is proved. Inequality \eqref{le1} is also proved.

\medskip

We already denoted
\[
\omega_0:=[0, 2^{-N}]\times [0, 2^{-N}]\,,
\]
 calculate now $\mathbb V^\mu(\omega_0)$. In fact, we will estimate it from below. The fact that $C^{[m, m+k]}_j$ are not disjoint may represent the problem now because  we wish estimate $\mathbb V^\mu(\omega_0)$ from below.

 To be more careful for every $j$ we denote now by $c_j$ the family of dyadic rectangles containing the point $P_j$ but not containing any other point
$P_i, i\neq j$. Rectangles in $c_j$ contain $Q_j^{++}$ but do not contain any of $Q_i^{++}$, $i\neq j$. There are
$(2^{j-1} - 2^j-1)\cdot (2^{-j+1}N-2^{-j}N -1)$, $j=2, \dots, M-2$. This is at least $\frac18 N$.

But now families $c_j$ are disjoint, and rectangles of class $c_j$ contribute at least $\frac18 N\cdot \frac{\delta}{N}$ into the sum that defines $\mathbb V^\mu(\omega_0)$. W have $m_4$ such classes $c_j$, as $j=2, \dots, M-2$.
Hence,
\begin{equation}
\label{deltaM}
\mathbb V^\mu(\omega_0) \ge \frac18 N\cdot \frac{\delta}{N}\cdot (M-4)\ge\frac19 \delta M\,.
\end{equation}
Choose  $\delta$   to be a small absolute number $\delta_0$. Then we will have (see \eqref{le1})
\[
\mathbb V^\mu \le 1, \quad\text{on}\,\, \operatorname{supp}\mu\,.
\]
But \eqref{deltaM}  proves also \eqref{logN} as $M\asymp \log N$.

\medskip

\begin{remark}
\label{Mx}
Notice that in this example $\mathbb V^\mu\le 1$ on $\operatorname{supp}\mu$, and
\begin{equation}
\label{exp}
\text{cap} \{\omega: \mathbb V^\mu\ge \lambda\} \le c e^{-2\lambda}\,.
\end{equation}
Here capacity is the bi-tree capacity defined e. g. in \cite{AMPS18}. So there is no maximal principle for the bi-tree potential, but the set, where the maximal principle breaks down,  has small capacity.
\end{remark}

\bigskip

Now we construct the example of $\nu$ and $\alpha$ with $\alpha=1$ on an up-set (and zero otherwise), and such that Carleson condition is satisfied
but REC (restricted energy condition) is not satisfied. We use the same measure $\mu$ we have just constructed, and we put
\[
\nu := \mu +\nu|\omega_0,
\]
where $\nu|\omega_0$ is the uniformly distributed over $\omega_0$ measure of total mass $\frac1{MN}$. Weight $\alpha$ is chosen as in \eqref{e:774}.

\bigskip

\noindent{\bf Warning.} The meaning of $\mathbb V$ changes from now on. Before $\mathbb V^{\cdot}= \mathbb I \mathbb I^*(\cdot)$. Everywhere below,
\[
\mathbb V^\nu:= \mathbb I[\alpha \mathbb I^*(\nu)]\,.
\]

\bigskip

Let us first check that REC constant  is bad. We choose $F=\cup_j Q_j^-$. Then $\nu_F:=\nu|F = \nu|\omega_0$. On the hand, and this is the main feature,
\begin{equation}
\label{lies_in_M}
\omega_0 \,\,\text{lies in}\,\, M\,\, \text{rectangles} \,\, Q_j\,.
\end{equation}
Hence,
\begin{equation}
\label{number_of_rectangles}
\text{there are} \,\, \ge c NM\,\, \text{dyadic rectangles}\,\, R\,\,\text{ such that}\,\, \alpha_R=1\,\, \text{ and}\,\, \omega_0\subset R\,.
\end{equation}
 In fact, consider dyadic rectangles  in $\cup_j c_j$, where families $c_j$ were built above. For each
$R\in \cup c_j$ we have $\alpha_R=1$, see \eqref{e:774}. And there are at least $\frac18 NM$ of them.  We conclude
\begin{equation}
\label{bV_at_om0}
\mathbb V^{\nu_F}(\omega_0) \ge \frac18 MN\cdot \nu(\omega_0)\,.
\end{equation}
Therefore,
\[
\int \mathbb V^{\nu|F}\, d \nu|F \ge  \nu(\omega_0)^2\cdot \frac18 NM=\frac18 \frac1{MN} \ge c_0\,\nu(\omega_0)\,.
\]
This means that constant of REC is at least absolute constant $c_0$. Let us show that the Carleson constant is $\lesssim c\cdot \delta$. But $c_0$ has nothing to do with $\delta$ that can be chosen as small as we wish.
\begin{remark}
\label{inM}
We do not need the following claim  now, we will need it only later, but notice that in a fashion completely similar to the one that just proved \eqref{bV_at_om0}, one can also prove
\begin{equation}
\label{bVmu_at_om0}
\mathbb V^{\mu}(\omega_0) \ge \frac18 MN\cdot \frac{\delta}{N} \ge c\, \delta M\,.
\end{equation}
Moreover, we already proved it in \eqref{deltaM}. This holds because $\omega_0$ is contained in exactly $M$ rectangles $Q_j^{++}$.
\end{remark}

\bigskip

\begin{defin}
Dyadic rectangles whose left lower corner is $(0,0)$ will be called {\it hooked} rectangles.
\end{defin}

To check the Carleson condition with small constant we fix any finite family $\mathcal{A}$ of dyadic   rectangles, and let
\[
A=\cup_{R\in \mathcal{A}} R\,.
\]

We are interested in subfamily $\mathcal{A}'$ of $R$ such that $\alpha_R=1$. Other elements of $\mathcal{A}$ do not give any contribution to $\mathcal E_A[\nu]$ as $ \mu(Q)^2 \alpha_Q=\mu(Q)^2 \cdot  0=0$ for any $Q\subset Q', \, Q'\in \mathcal{A}\setminus \mathcal{A}',$  as the support of $\alpha$ is an up-set.

All rectangles from $\mathcal{A}'$ are hooked rectangles. As we noticed, we can think that $\mathcal{A}'=\mathcal{A}$.  In other words, without the loss of generality, we can think that $\mathcal{A}$ consists only of hooked rectangles. Any hooked rectangle generates  a closed interval $\mathcal J$ in the segment $[1, n]$ of integers: interval $\mathcal J$ consists of $j$, $1\le j\le n$, such that point $P_j$ lies in this hooked rectangle.  This is the same as to say that $Q_j$, $j\in \mathcal J$, is a subset of this hooked rectangle.

So family $\mathcal{A}$  generates the family of  closed intervals in the segment $[1, n]$ of integers. Let us call $\mathcal J_A$ this family of intervals in the segment $[1, n]$ of integers. Intervals of family $\mathcal J_A$ can be not disjoint. But we can do the following, if intervals intersect, or even if these closed intervals are adjacent, we unite them to a new interval. The new system (of disjoint and not even adjacent)  closed intervals corresponds to another initial system $\tilde \mathcal{A}$, and we can think that $\tilde \mathcal{A}$ consists of hooked rectangles.
We call a system of hooked rectangles  {\it a clean system}  if it gives rise to not adjacent disjoint family of closed intervals inside the set $[1,n]$ of integers
The relationship between rectangles in $\mathcal{A}$ and $\tilde\mathcal{A}$ is the following: each rectangle of $\tilde\mathcal{A}$ is a common ancestor of a group of rectangles in $\mathcal{A}$.

A very important geometric property of $\tilde \mathcal{A}$ is the following. Let $Q\in \tilde\mathcal{A}$ and let $R^1, \dots, R^s$ be all rectangles from $\mathcal{A}$ such that $R^i\subset Q$, $i=1,\dots, s$. Then
\begin{equation}
\label{tildeA}
\nu(Q\setminus \cup_{i=1}^m R^i) =0\,.
\end{equation}

In particular, \eqref{tildeA} implies
\begin{equation}
\label{tildeA1}
\nu(A)=\nu(\tilde A) \,.
\end{equation}

When checking the Carleson condition
\begin{equation}
\label{carlA}
\mathcal E_A[\nu] \lesssim \delta \nu(A),
\end{equation}
we can always think about $\mathcal{A}$ being replaced by $\tilde \mathcal{A}$ and $A$ being replaced by $\tilde A$  because in \eqref{carlA} the RHS stays the same, but the LHS can jump up
with passage from $A$ to $\tilde A$. Therefore, checking \eqref {carlA} for {\it clean systems} of rectangles is the same as to check it for all systems of rectangles. From now on $\mathcal{A}$ is supposed to be clean.

To prove \eqref{carlA} is the same as to prove (since $\nu_A(Q) = (\mu_A(Q) +\nu_{\omega_0}(Q)$)
\begin{equation}
\label{carlA1}
 \sum_{Q\subset A} \nu_A(Q)^2 \alpha_Q  + \sum_{Q\subset A} \nu_{\omega_0}(Q)^2 \alpha_Q  \lesssim \delta \nu(A),
\end{equation}

The first sum is bounded by $\int \mathbb V^{\mu|A}\, d\mu|A$. But by \eqref{le1} (which  follows from \eqref{VQ}) we have
\[
\sum_{Q\subset A} \nu_A(Q)^2 \alpha_Q = \int \mathbb V^{\mu|A}\, d\mu|A \lesssim \delta \|\mu_A\|= \delta \mu(A) \le \delta \nu(A)\,,
\]
and this means that the part of \eqref{carlA1} is proved.

To estimate $\sum_{Q\subset A} \nu_{\omega_0}(Q)^2 \alpha_Q= \frac1{(MN)^2} \sharp\{R: \alpha_R=1, R\subset A\} $  we take one interval $\mathcal J_k$ from the family generated by the clean $\mathcal{A}$ in $[1,n]$,  we denote
\begin{equation}
\label{mk}
m_k:=\sharp\mathcal J_k\,,
\end{equation}
and we estimate how many dyadic rectangles  $R$ contain one of $Q_j$, $j\in \mathcal J_k$. We even do not care now whether $R$ is a subset of $A$ or not.  The number of such rectangles in at most $m_k\cdot N$. On the other hand,
\[
\nu(A)\ge \sum_k m_k \cdot \frac{\delta}{N} + \frac1{MN} \sum_k m_k \ge \sum_k m_k \cdot \frac{\delta}{N} \,.
\]
So to prove the estimate for the second sum in \eqref{carlA1}, we need to see that
\[
 \frac1{(MN)^2}\sum_k m_k\cdot N \lesssim \frac1M\sum_k m_k \cdot \frac{\delta}{N},
 \]
 which is obviously true if we choose $\delta \ge \frac1{M}=\frac1{\log N}$. So \eqref{carlA1} is proved.

\section{Restricted energy condition holds but no embedding}
\label{RECnoEmb}

In this section we emulate the previous construction, we start with $\{Q_j\}$ and measure $\mu$ but instead of adding $\omega_0$ we will add a more sophisticated piece of measure.

Let us start with  recalling the system $\{Q_j\}, j=1, \dots, M$ and measure $\mu$ from the previous section. We continue with denoting
\[
Q_{0, j}:= Q_j,\quad \mu_0:=\mu\,\, \text{from the previous section}\,.
\]
Next we continue with defining a a sequence of collections $\mathcal{Q}_k,\; k=0\dots K,$ of dyadic rectangles as follows
\begin{equation}\label{e:874}
\begin{split}
\mathcal{Q}_k := \left\{Q_{k,j} = \bigcap_{i=j}^{j+2^k}Q_{0,i},\; j=1\dots M-2^k\right\},\; k=1\dots K.
\end{split}
\end{equation}
In other words, $\mathcal{Q}_0$ is the basic collection of rectangles, and
\begin{equation}
\label{lies_in_2k}
\mathcal{Q}_{k}\,\text{ consists of the intersections of} \,\,2^k\,\text{ consecutive elements of}\,\mathcal{Q}_0.
\end{equation}
 The total amount of rectangles in $\mathcal{Q}_k$ is denoted by $M_k = M-2^{k}+1$.

We also denoted by $\mathcal R$ the collection of rectangles lying above $\mathcal{Q}_0$
\begin{equation}\label{e:874.4}
\mathcal R := \{R:\; Q_{0,j}\subset R\;\textup{for some}\; 1\leq j\leq M\},
\end{equation}
and we let
\[
S_k := \bigcup_{Q\in \mathcal{Q}_k}Q.
\]
The weight $\alpha$ was defined as follows:
\begin{equation}\label{e:874.5}
\begin{split}
&\alpha_{Q} := 1,\quad \textup{if}\; Q\in \mathcal R \\
&\alpha_Q := 0 \quad \textup{otherwise}.
\end{split}
\end{equation}

Now we construct the measure $\mu$, whose main part will be already constructed $\mu_0$. Let
\[
Q_{k,j}^{++} := \left[2^{-2^{j+2^k}-1},2^{-2^{j+2^k}}\right]\times \left[2^{-2^{j}N-1},2^{-2^{j}N}\right]
\]
be the upper right quadrant of $Q_{k,j}$. For every $k = 0\dots K$ we distribute the mass $2^{-2k}\frac{M_k\delta}{N}$ over the rectangles $Q_{k,j}^{++}$. Namely, for every $j=1\dots M_k$ we attach a mass $2^{-2k}\frac{\delta}{N}$ to the rectangle $Q_{k,j}$ that is uniformly distributed over the quadrant $Q_{k,j}^{++}$. We note that all these quadrants $Q_{k,j}^{++}$ are disjoint.\par
Measure $\mu_0$ is the ``main" part of $\mu$, in the sense that $\mu_0$ is generated by the masses on $\mathcal{Q}_0$,
\[
\mu_0(\omega) := \frac{\delta}{N}\sum_{j=1}^M \frac{1}{|Q_{0,j}^{++}|}\chi_{Q_{0,j}^{++}}(\omega),\quad \omega \in (\partial T)^2,
\]
and let $\mu_k$ be the corresponding mass on $\mathcal{Q}_k$
\[
\mu_k(\omega) := \frac{2^{-2k}\delta}{N}\sum_{j=1}^{M_k} \frac{1}{|Q_{k,j}^{++}|}\chi_{Q_{0,j}^{++}}(\omega),\quad \omega \in (\partial T)^2,
\]
so that
\[
\mu = \mu_0 + \sum_{k=1}^{K}\mu_k.
\]
Finally we define the function $f$, and we do it in such a way that it is 'congruent' with the distribution of $\mu_0$ over $\Omega$, namely we let
\[
f(R) := \mu_0(R)\cdot \alpha_R.
\]

\subsection{Main idea}
\label{MI}

Notice that
\[
\mathbb V^{\mu_0} = \mathbb I f=\mathbb I[\alpha \mathbb I^*\mu_0], \, \,\,\int \mathbb V^{\mu_0}\, d\mu_0 =\sum\mathbb I^*\mu_0\cdot \mathbb I^*\mu_0 \cdot\alpha= \sum_{T^2} f^2\cdot \alpha\,.
\]
To prove that embedding has a bad constant, it is sufficient to show that the dual inequality has a bad constant:
\[
\int (\mathbb I f)^2 \, d\mu >> \sum f^2\cdot \alpha,
\]
which becomes
\begin{equation}
\label{dual}
\int (\mathbb V^{\mu_0})^2\, d\mu>> \int \mathbb V^{\mu_0} \, d\mu_0\,.
\end{equation}

Let us look at Remark \ref{inM}, at
\eqref{lies_in_M}, \eqref{number_of_rectangles},
\eqref{bVmu_at_om0} and compare \eqref{lies_in_M} with  \eqref{lies_in_2k}.
The conclusion is: since every $Q_{k,j}$ lies in $2^k$ of $Q_{0,j}$ (number $2^k$ replaces $M$ in \eqref{lies_in_M}, \eqref{number_of_rectangles}, \eqref{bVmu_at_om0}),  then
\begin{equation}
\label{bVmu0_on_muk}
\mathbb V^{\mu_0} \ge c\,2^k N \cdot \frac{\delta}{N}=c  \delta 2^k\,\,\text{on each}\,\, Q_{k, j}\,.
\end{equation}

We already saw that $\mathbb V^{\mu_0}\lesssim \delta$ on $\mu_0$, so
\begin{equation}
\label{RHS}
\int \mathbb V^{\mu_0} \, d\mu_0 \lesssim \delta^2 \frac{M}{N}\,.
\end{equation}

Now, using \eqref{bVmu0_on_muk}
we get
\begin{equation}
\label{LHS}
\int (\mathbb V^{\mu_0})^2\,d\mu =\sum_{k=1}^K \int (\mathbb V^{\mu_0})^2\,d\mu_k \ge c^2 \delta^2 \sum_{k=1}^K  2^{2k} \|\mu_k\| = c^2 \delta^3 \frac{M\log M}{N}
\end{equation}

 For example, let
 \begin{equation}\label{choice}
 \delta= \frac1{\log M}\,.
 \end{equation}
 Then the constant of  embedding is $\approx 1$.

 \subsection{REC condition holds with a small constant}
 \label{RECsmall}

 Let $\mathcal{A}$ be a collection of (hooked) rectangles, $A=\cap_{R\in \mathcal{A}} R$. Let $\nu_k:=\mu_k|A$, $k=1, \dots, K$, $\nu:= \mu |A=\sum \nu_k$.
 We need to  prove
 \begin{equation}
 \label{smREC}
 \mathcal E_A[\nu]\lesssim \delta \|\nu\|\,.
 \end{equation}
 Let $n>k$, we wish to estimate $\mathbb V^{\mu_n}(Q_{k, j})$. This is a certain sum over a system $\mathcal S$ of rectangles  of the form
 \[
 \sum_{R\in \mathcal S} \mu_n(R),
 \]
 where dyadic rectangles are a) contain $Q_{k, j}$, b) $\alpha_R=1$. Notice that this system depends on $Q_{k, j}$ but totally independent of $n$. So if we manage to estimate $\mu_n(R)$ via $\mu_0(R)$, then we compare $\mathbb V^{\mu_n}(Q_{k, j})$ to $\mathbb V^{\mu_0}(Q_{k, j})$.

 But let the number of $Q_{s, j}$ in $R$ be denoted by $m^s_R$. Then it is very easy to see that
 \[
 m^n_R \le m^0_R +2^n+1\,.
 \]
 Then
 \[
 2^{-2n}m^n_R \le 2^{-n}(m^0_R +1+2^{-n}) \le 3\cdot  2^{-n} m^0_R\,.
 \]
 Then
  \[
\mathbb V^{\mu_n}(Q_{k, j}) =\sum_{R\in \mathcal S} \mu_n(R) \le 3\cdot  2^{-n} \sum_{R\in \mathcal S} \mu_0(R) = 3\cdot 2^{-n} \mathbb V^{\mu_0}(Q_{k,j})\lesssim \delta 2^{k-n} \,.
 \]
 Therefore,
\[
\sum_{n\ge k} \mathbb V^{\nu_n} \, d\nu_k\lesssim \delta \sum_{n\ge k} 2^{k-n} \|\nu_k\|=2\delta \|\nu_k\|\,.
\]
And so
\[
\sum_k\sum_{n\ge k} \mathbb V^{\nu_n} \, d\nu_k\lesssim \delta \sum_k \|\nu_k\|=\delta\|\nu\|\,.
\]
Inequality \eqref{smREC} is proved.

\section{Strong dyadic maximal function and counterexamples}
\label{counter}

\noindent{\bf Definition.}
Let $\mathcal S$ be a family of dyadic sub-rectangles of $Q_0$ (may be $\mathcal S=\mathcal D$, the family of all dyadic sub-rectangles). We call the sequence of $\{\beta_Q\}_{Q\in \mathcal S}$ Carleson if
\begin{equation}
\label{bC}
\forall \mathcal S'\subset \mathcal S, \quad \sum_{Q\in \mathcal S'}\beta_Q \mu(Q) \le C\mu(\cup_{Q\in \mathcal S'} Q)\,.
\end{equation}
The best $C$ is called the Carleson norm of the sequence.

\noindent{\bf Definition.} Abusing the language we say that the weight $\alpha:=\{\alpha_Q\}_{Q\in \mathcal S}$ satisfies Carleson condition if the sequence $\beta_Q:= \alpha_Q \cdot \mu(Q)$ is a Carleson sequence:
\begin{equation}
\label{alC}
\forall \mathcal S'\subset \mathcal S, \quad \sum_{Q\in \mathcal S'}\alpha_Q \mu(Q)^2 \le C\mu(\cup_{Q\in \mathcal S'} Q)\,.
\end{equation}

We already know, see \cite{AHMV18b} e.g., that bi-parameter Carleson embedding
\[
\sum_{R\in \mathcal D} \big(\int_R \psi\, d\mu\big)^2 \alpha_R =\sum_{R\in \mathcal D} [\mathbb I^*(\psi \mu)]^2 \alpha_R  \le C' \int_{Q_0} \psi^2\, d\mu
\]
is equivalent to the Carleson condition above if we have $\alpha_R=1,\hspace{0.1cm} \forall R \in \mathcal D$ .

Understanding the general two-weight bi-parameter situation (that is $\alpha \not\equiv 1$) seems to be super hard, as the examples above show. Notice that in two-weight  one-parameter situation the answer is known, see, e.g. \cite{NTV99}. And the answer is given in terms of Carleson condition.  However, in bi-parameter situation this is far from being so as the following theorem shows. First we give

\noindent{\bf Definition.} A finite positive measure $\mu$ on $Q_0$ is called a ``bad'' measure if there exists
weight $\alpha=\{\alpha_Q\}_{Q\in \mathcal D}$ that satisfies the Carleson condition but such that the embedding
\begin{equation}
\label{alEmb}
\sum_{R\in \mathcal D} \Big(\int_R \psi\, d\mu\Big)^2\alpha_R \le C' \int_{Q_0} \psi^2\, d\mu
\end{equation}
does not hold.

The strong maximal function with respect to $\mu$  is
\[
\mathcal M_\mu \psi (x) =\sup_{\substack{R\in \mathcal D \\ R\ni x}} \frac{1}{\mu(R)}\int_R |\psi|\, d\mu\,,
\]
where $0/0=0$. The supremum is taken over all dyadic sub-rectangles of $Q_0$.

\begin{theorem}
\label{Dor}
Let $\mu$ be atom free. Then the measure $\mu$  is bad if and only if $\mathcal M_\mu$ is not a bounded operator in $L^2(\mu)$.
\end{theorem}

We need some preparation, which is written down in \cite{Verb} and \cite{TH}, but we repeat it for the convenience of the reader.

If weight $\alpha$ satisfies Carleson condition with constant $1$, then $\beta_Q=\alpha_Q \mu(Q)$ is a Carleson sequence with Carleson constant $1$ and this means that for any family $\mathcal D'\subset \mathcal D$
\begin{equation}
\label{alCarl1}
\sum_{Q\in \mathcal D'} \alpha_Q \mu(Q)^2 \le \mu(\bigcup_{Q\in\mathcal D'} Q)\,.
\end{equation}

Using the idea of Igor Verbitsky \cite{Verb} this can be stated in terms of discrete Triebel-Lizorkin space $f^{1, \infty}(\mu)$:
\[
\{\alpha_Q\mu(Q)\}_{Q\in \mathcal D}\in f^{1, \infty}(\mu),
\]
and has norm at most $1$. But $( f^{1, \infty}(\mu))^* =  f^{ \infty, 1}(\mu)$, and this is the space of coefficients $\{\lambda_Q\}_{Q\in \mathcal D}$ given by the norm
\[
\|\{\lambda_Q\}_{Q\in \mathcal D}\|_{ f^{ \infty, 1}(\mu)} := \int\sup_{Q\in\mathcal D}\Big(\lambda_Q\chi_Q(x)\Big)\, d\mu\,.
\]
Thus, by duality \eqref{alCarl1} is equivalent to
\begin{equation}
\label{alCarlDual}
\forall \{\lambda_Q\}_{Q\in \mathcal D}, \quad\sum_{Q\in \mathcal D} \alpha_Q \mu(Q)^2\cdot \lambda_Q \le \int \sup_{Q\in\mathcal D}\Big(\lambda_Q\chi_Q(x)\Big)\, d\mu\,.
\end{equation}
 Without loss of generality we think that all $ \alpha_Q \mu(Q)>0$ (otherwise they are not in the LHS and we forget them). Then we can rewrite \eqref{alCarlDual} as

\begin{equation}
\label{alCarlDual2}
\forall \{b_Q\}_{Q\in \mathcal D}, \quad \sum_{Q\in \mathcal D} b_Q \le \int \sup_{Q\in\mathcal D}\Big(b_Q\frac{\chi_Q(x)}{\alpha_Q \mu(Q)^2}\Big)\, d\mu\,.
\end{equation}

In \cite{Verb} and in \cite{TH} now the theorem of L. Dor \cite{Dor} is used. Here it is
\begin{theorem}
\label{thDor}
Let $\mu$ be atom free. Let $g_i$ be a fixed sequence of measurable functions and suppose for all positive numbers $b_i$ that
\[
\sum b_i \le \int \sup_j[b_j g_j(x)] \, d\mu\,.
\]
Then there are disjoint sets $E_j$ such that
\[
1\le \int_{E_j} g_j\, d\mu\,.
\]
\end{theorem}

Using this theorem, we see that \eqref{alCarlDual2} implies the existence of measurable subsets $E_Q\subset Q$ with  $Q\in \mathcal D$, such that
\begin{equation}
\label{choice}
\alpha_Q\cdot \mu(Q)^2 \leq \mu(E_Q),\quad E_Q\cap E_{Q'} =\emptyset\,.
\end{equation}

We deduced \eqref{choice} from \eqref{alCarl1}. By now this has been done in various situations many times, see again \cite{Verb} (for cubes), \cite{TH} (for arbitrary Borel sets), and \cite{AB} for geometric proof of this geometric fact.

The deduction of \eqref{choice} from \eqref{alCarl1} is called ``sparsity  property from Carleson property deduction".

Now we are ready to give the proof of Theorem \ref{Dor}.

\begin{proof} $(\Rightarrow)$ Suppose measure $\mu$ has no atoms and it is a bad measure. Then we can find a sequence $\alpha=\{\alpha_Q\}_{Q\in\mathcal D}$ such that  condition \eqref{alC} is satisfied with constant $C=1$ (we can always normalize) but embedding \eqref{alEmb} is false. Then for every positive $K$ there exists a positive function $\psi$ such that
\[
\sum_{Q\in \mathcal D} \big(\int_Q \psi\, d\mu\big)^2 \alpha_Q\geq K\int \psi^2\, d\mu,
\]
or
\[
\sum_{Q\in \mathcal D} \langle \psi\rangle_{Q, \mu}^2 \alpha_Q\cdot \mu(Q)^2\ge K\int \psi^2\, d\mu
\]

Let $\mathcal F$ be the sub-collection of  $\mathcal D$ such that $\alpha_Q\neq 0$ for $Q\in \mathcal F$. We can use the sparsity condition \eqref{choice} to get

$$K\int \psi^2\, d\mu \leq \sum_{Q\in \mathcal D} \langle \psi\rangle_{Q, \mu}^2 \alpha_Q\cdot \mu(Q)^2 \leq \sum_{Q\in \mathcal F} \langle \psi\rangle_{Q, \mu}^2 \cdot \mu(E_Q)=$$

$$\sum_{Q\in \mathcal F} \Big(\frac{1}{\mu(Q)}\int_Q\psi\hspace{0.1cm}{d}\mu\Big)^2  \cdot\mu(E_Q)\leq \sum_{Q\in \mathcal F} \Big(\inf_{x\in Q}\mathcal M_{\mu}\psi(x)\Big)^2\cdot \mu(E_Q)\leq$$
$$\sum_{Q\in \mathcal F} \inf_{x\in Q}\Big(\mathcal M_{\mu}\psi(x)\Big)^2\cdot \mu(E_Q)\leq\sum_{Q\in \mathcal F} \inf_{x\in E_Q}\Big(\mathcal M_{\mu}\psi(x)\Big)^2\cdot \mu(E_Q)\leq$$

$$\sum_{Q\in \mathcal F} \int_{E_Q}\Big(\mathcal M_{\mu}\psi(x)\Big)^2\hspace{0.1cm}{d}\mu\leq\int_{[0,1]^2}\Big(\mathcal M_{\mu}\psi(x)\Big)^2{d}\mu$$

since the sets $E_Q$ are disjoint.

Hence
\begin{equation}
\label{MmuK}
\int \big(\mathcal M_\mu\psi \big)^2\, d\mu \geq K\int \psi^2\, d\mu\,.
\end{equation}
Since $K$ can be arbitrary large the operator $\mathcal M_\mu$ is not bounded in $L^2(\mu)$.

\bigskip

$(\Leftarrow)$
Now assume that operator $\mathcal M_\mu$ is not bounded in $L^2(\mu)$. Then, for an arbitrary $K>0$ there exists a positive function $\psi$ such that \eqref{MmuK} holds.

By approximating $\psi$ with an increasing sequence of positive, simple functions $\psi_n$ we can find $n_0\in \mathbb{N}$ such that \eqref{MmuK} holds with $\psi$ replaced by $\psi_{n_0}$. Also, we can find $N=N(n_0)\in\mathbb{N}$ so that the function $\psi_{n_0}$ is constant on each dyadic sub-square of $Q_0$ of size $2^{-N}\times 2^{-N}$. Given these, in the definition of $\mathcal M_\mu\psi_{n_0}$ we replace $\sup$ by $\max$. For simplicity we omit writing the sub-script $n_0$.

Now we group the elements of $Q_0=[0,1]^2$ into the sets $A_Q$, $Q\in\mathcal D$ as follows: $x\in A_Q$ if

$$\max_{\substack{R\in \mathcal D \\ R \ni x}}\frac{1}{\mu(R)}\int_R \psi {d}\mu = \frac{1}{\mu(Q)}\int_Q \psi {d}\mu$$

Of course there may be more than one set $A_Q$ for each $x$. We would like to make these sets disjoint. Let ${Q_1}$, ${Q_2}$,.. be an enumeration of the dyadic rectangles $Q$, $Q\in \mathcal D$. Then consider the sets $A_{Q_i}':=A_{Q_i}\setminus \bigcup\limits_{j=1}^{i}A_{Q_j}$. Obviously the sets $A_{Q_i}'$ are disjoint and their union is $Q_0=[0,1]^2$.

Hence, we have

$$\int_{Q_0}\Big(\mathcal M_{\mu}\psi(x)\Big)^2{d}\mu=\int_{Q_0}\Bigg(\max_{\substack{R\in \mathcal D \\ R \ni x}}\frac{1}{\mu(R)}\int_R \psi {d}\mu\Bigg)^2{d}\mu=\sum_{i}\int_{A_{Q_i}'}\Big(\frac{1}{\mu(Q_i)}\int_{Q_i} \psi {d}\mu\Big)^2 {d}\mu =$$

$$\sum_{i}\frac{\mu(A_{Q_i}')}{\mu(Q_i)^2}\Big(\int_{Q_i} \psi \hspace{0.1cm}{d}\mu\Big)^2:=\sum_{i}\alpha_{Q_i}\Big(\int_{Q_i} \psi \hspace{0.1cm}{d}\mu\Big)^2$$

In other words, we constructed a sequence $\alpha=\{\alpha_{Q_i}\}$, $Q_i\in \mathcal D$ for which the embedding \eqref{alEmb} does hold with a very large constant, but the Carleson condition holds with constant $1$. But, this sequence, as well as the function $\psi$ depend on $K$. To get a general sequence and function we do start with an $i\in\mathbb{N}$ and take $K=4^i$. Lets omit enumerating the dyadic rectangles in $[0,1]^2$ for now. Given the above,  $\exists \psi_i$ and $\{\alpha_{Q}^i\}_{Q\in\mathcal D}$ such that

$$4^i \int\limits_{[0,1]^2}\psi_i^2{d}\mu \hspace{0.1cm}\leq \hspace{0.1cm} \sum\limits_{Q\in \mathcal D}\alpha_{Q}^i\Big(\int_{Q} \psi_i \hspace{0.1cm}{d}\mu\Big)^2 $$ and re-normalizing we get

\begin{equation}\label{iemb}
2^i  \hspace{0.1cm}\leq \hspace{0.1cm} \sum\limits_{Q\in \mathcal D}\alpha_{Q}^i\Big(\int_{Q} \phi_i \hspace{0.1cm}{d}\mu\Big)^2
\end{equation}

for  $\phi_i=\dfrac{\psi_i}{2^{i/2}\|\psi_i\|_{L^2([0,1]^2)}}$.

Now let $\alpha_Q:=\sum\limits_{i=1}^{\infty}2^{-i}\alpha_Q^i$ and $\phi=\sum\limits_{i=1}^{\infty}\phi_i$. Notice that $\phi\in L^2([0,1]^2)$ and the sequence $\alpha=\{\alpha_Q\}_{Q\in \mathcal D}$ satisfies the Carleson condition: Let any sub-collection $\mathcal S$ of $\mathcal D$. There is an indexing set $J_{\mathcal S}=\{j_1,j_2,...\}$ which enumerates the dyadic rectangles in $\mathcal S$. By the disjointness of $A_{Q_j}'^{i}$'s (for each fixed $i$) we have

$$\sum_{Q\in \mathcal S}\alpha_{Q} \mu(Q)^2=\sum_{j\in J_{\mathcal S}}\alpha_{Q_j} \mu(Q_j)^2=\sum_{j\in J_{\mathcal S}} \sum\limits_{i=1}^{\infty}2^{-i}\alpha_{Q_j}^i \mu(Q_j)^2 = \sum\limits_{i=1}^{\infty}2^{-i} \sum_{j\in J_{\mathcal S}} \mu(A_{Q_j}'^{i})= $$

$$\sum\limits_{i=1}^{\infty}2^{-i}\mu(\bigcup\limits_{j\in J_{\mathcal S}} A_{Q_j}'^{i}) \leq \sum\limits_{i=1}^{\infty}2^{-i} \mu(\bigcup\limits_{Q \in {\mathcal S}} {Q})=\mu(\bigcup\limits_{Q \in {\mathcal S}} {Q})  $$

Finally, using \eqref{iemb} we get

$$\sum\limits_{Q\in\mathcal D}\alpha_Q \Big(\int_{Q} \phi \hspace{0.1cm}{d}\mu\Big)^2=\sum\limits_{Q\in\mathcal D}\sum\limits_{i=1}^{\infty}2^{-i}\alpha_Q^i\Big(\int_{Q} \phi \hspace{0.1cm}{d}\mu\Big)^2 \ge \sum\limits_{i=1}^{\infty}2^{-i}\sum\limits_{Q\in\mathcal D}\alpha_Q^i\Big(\int_{Q} \phi_i  \hspace{0.1cm}{d}\mu\Big)^2 = \infty$$

\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
Q: Xamarin-CrossDownloadManager - waiting for download file I use Xamarin-CrossDownloadManager (https://github.com/SimonSimCity/Xamarin-CrossDownloadManager) and I need waiting for download a file. I have this code:
private static async Task<bool> FileDownloadedTest(string LinkToFile, string PathFile)
    {
        var downloadManager = CrossDownloadManager.Current;
        CrossDownloadManager.Current.PathNameForDownloadedFile = new System.Func<IDownloadFile, string>(file => {
            return PathFile;
        });
        {
            await DeleteFile(PathFile);
            var file = downloadManager.CreateDownloadFile(LinkToFile);
            await Task.Run(() => downloadManager.Start(file, true)); //why this not wait???
        }
        bool FileExist = await IsFileExist(PathFile);
        return FileExist;
    }

Why it not wait for finish download action? How to do it?
On library site they wrote, that I can watch the IDownloadManager.Queue to get information when the file is downloaded. But, I don't know how to use this in my method... Can you help me?
PS: Sorry for my english, I'm still learning it ;)

A: With that library, there is no callback or event published for when a file is finished downloading, but you can do a simple check and wait some more loop.
await Task.Run(async () =>
{
    var downloadManager = CrossDownloadManager.Current;
    var file = downloadManager.CreateDownloadFile(someFileBasedUrl);
    downloadManager.Start(file);
    bool isDownloading = true;
    while (isDownloading)
    {
        await Task.Delay(10 * 1000);
        isDownloading = IsDownloading(file);
    }
});

The IsDownloading method:
bool IsDownloading(IDownloadFile file)
{
    if (file == null) return false;

    switch (file.Status)
    {
        case DownloadFileStatus.INITIALIZED:
        case DownloadFileStatus.PAUSED:
        case DownloadFileStatus.PENDING:
        case DownloadFileStatus.RUNNING:
            return true;

        case DownloadFileStatus.COMPLETED:
        case DownloadFileStatus.CANCELED:
        case DownloadFileStatus.FAILED:
            return false;
        default:
            throw new ArgumentOutOfRangeException();
    }
}

Re: https://github.com/SimonSimCity/Xamarin-CrossDownloadManager/blob/develop/Sample/DownloadExample/Downloader.cs#L46

A: I don't know why IDownloadFile.Status = COMPLETED not working, but i found solve problem:
await Task.Run(() =>
        {
            var downloadManager = CrossDownloadManager.Current;
            var file = downloadManager.CreateDownloadFile(LinkToFile);
            downloadManager.Start(file);
            while (file.Status == DownloadFileStatus.INITIALIZED)
            {
                while (file.TotalBytesExpected > file.TotalBytesWritten)
                {
                    Debug.WriteLine(file.Status);
                }
            }
        });

Somebody know why DownloadFileStatus.INITIALIZED working, but DownloadFileStatus.COMPLETED not?

A: SushiHangover's answer above worked great, and I combined it with ACR User Dialog's package (https://github.com/aritchie/userdialogs) to show a nice loading progress screen to the user while waiting for the download to complete.  This works nicely on Android (I couldn't test iOS yet).
   ...
            var downloadManager = CrossDownloadManager.Current;
            var fileToDownload = downloadManager.CreateDownloadFile(args.Url);
            downloadManager.Start(fileToDownload, true);
            args.Cancel = true;
            UserDialogs.Instance.Progress("Downloading").Show();
            bool isDownloading = true;
            while (isDownloading)
            {
                await Task.Delay(100);
                if (fileToDownload.TotalBytesExpected > 0)
                {
                    UserDialogs.Instance.Progress().PercentComplete = (int)(fileToDownload.TotalBytesWritten / fileToDownload.TotalBytesExpected * 100);
                    Console.WriteLine(("DOWNLOAD PROGRESS: " + fileToDownload.TotalBytesWritten / fileToDownload.TotalBytesExpected * 100).ToString() + "%");
                }

                isDownloading = IsDownloading(fileToDownload);
            }
             UserDialogs.Instance.Progress().Hide();

            //Below code opens the download location after download has completed.
            Intent intent = new Intent(DownloadManager.ActionViewDownloads);
            intent.AddFlags(ActivityFlags.NewTask);
            Android.App.Application.Context.StartActivity(intent);
            return;
        }
    }

}

bool IsDownloading(IDownloadFile file)
{
    if (file == null) return false;

    switch (file.Status)
    {
        case DownloadFileStatus.INITIALIZED:
        case DownloadFileStatus.PAUSED:
        case DownloadFileStatus.PENDING:
        case DownloadFileStatus.RUNNING:
            return true;

        case DownloadFileStatus.COMPLETED:
        case DownloadFileStatus.CANCELED:
        case DownloadFileStatus.FAILED:
            return false;
        default:
            throw new ArgumentOutOfRangeException();
    }
}

redpajama_set_name: RedPajamaStackExchange


If you're planning a trip to central Florida check out the best family attractions at Universal Orlando Resort. If you're visiting while expecting, I also included what to do at Universal Orlando Resort when you're pregnant.
Disclosure: My family and I were given complimentary passes to Universal Orlando Resort. All opinions are my own.
I haven't even been back for a week yet, but I have been so excited to tell all of you about my Orlando trip that I'm going to jump right into it.
We LOVED Universal Orlando Resort. My family and I had such an incredible time, we can't wait to go back and visit after the baby is born, though we're actually planning on waiting until she's two. My husband enjoyed the fact that there was something to do for everyone. He had fun on the family rides, like Despicable Me Minion Mayhem, but he really had a blast on the big boy rides like The Incredible Hulk Coaster and Hollywood Rip Ride RockIt. Me? I was pregnant so I didn't go on those…well, that's the story I'm sticking to anyway.
A great addition to Universal Orlando Resort for families is that there are a lot of opportunities for the littles to burn through energy! There are multiple areas for them to play like Fievel's Playground, Curious George Goes to Town, and over at the Dr. Suess area in Islands of Adventure.
Naturally, this list will change for your family depending on the age range of your children. At the time of our visit, our son was five.
I'm sure Jurassic Park River Adventure would have made the list (it was our favorite at Hollywood Studios) but it was closed for refurbishment and updates during our visit.
Randa's Tip: To get the most of both parks, splurge for the Express Pass. Yes, it costs extra but you get to skip the line and almost right straight on to the ride. It's a great time saver and a feature Universal Orlando Resort offers that we won't pass up on future visits.
Since I was 23 weeks pregnant during my visit, I learned on the fly what rides are doable for pregnant women. Many of the attractions at Universal Orlando warn pregnant women that they should think about skipping the ride, it's always better to play it safe. However, I did find that some were not as bad as the signs made them out to be. I personally found that anything with a lap bar was easier for me to brace and control against the shaking of the ride, but seat belts were a no. Obviously, if I was carrying differently, a lap bar may not have been okay. Use your best judgment and when in doubt, skip it. It's always best to play it safe.
5. Despicable Me Minion Mayhem – warning if you get motion sickness, I'd skip it.
There were a lot of rides that I ended up skipping, most of the rides in fact, but that didn't change the amount of fun I had. While my family rode, I browsed the shops and took in the scenery.
I encouraged my family to take their time while visiting The Wizarding World of Harry Potter because I could have stayed there all day. I had a great time just taking it in. I especially loved watching the dragon breathe fire over Gringotts in Diagon Alley or simply staring in awe at Hogwarts over at Hogsmeade (all while sipping on a hot butterbeer of course).
If you can, purchase an interactive wand over at Ollivanders. While my husband was off riding all the big rides, my son and I explored the Wizarding World of Harry Potter and tried out the spells. He was so excited to make magic happen with his oak wand!
Randa's Tip: Keep in mind that both Diagon Alley and Hogsmeade are in separate parks: Universal Orlando Resort and Universal Islands of Adventure. To access both in one day you will need a Park-to-Park ticket. Having this ticket also allows you to ride the Hogwarts Express, which is more than just a train ride, it's a magical experience that my family and I adored.
If you could, what one would you choose? These milkshakes are bloody insane! Not my thing but super fun to look at and watch being made.
One last thing: You have to check out Toothsome Chocolate Emporium. We stopped in for lunch (it was so good) and I also grabbed a few treats for later. I highly suggest the Key Lime Macaron!
If you're planning a trip to Universal Orlando Resort, consider saving time and staying on site. We stayed nearby at Cabana Bay so be sure to read my Cabana Bay Beach Resort Review.
Have you visited Universal Orlando Resort? Tell me what some of your favorites were in the comments.
Skull Island and ALL of the Wizarding World of Harry Potter are musts in my book. They're so exciting.
We still have not been here. It's on our list for sure. Every time I see an ad, it makes me want to hurry up and get there. Looks like so much fun!
It sounds like being pregnant did not keep you from enjoying the trip. We have not been to Universal for a few years but had a great tme when we went and our kids were at the perfect ages.
Not at all! There was still so much to do!
This really makes me want to plan another trip to Universal Orlando. Quite a few of those newer rides sound fun and the Harry Potter scenery is so cool!
I haven't been to Universal for years and I'm dying to go back. Those milkshakes are amazing!
You definitely pegged this list right on. I couldn't agree more with your choices. Now I wish I were going!!
These are also my favorites!! I can't wait to go back!
Number one Harry Potter and The Escape from Gringotts is my personal favorite. I so want to go back and try some of the newer rides. Those milkshakes are awesome! Sharing Thank you for sharing on Merry Monday! Hope to see ya next week!