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Dallas season finale recap: J.R.'s Masterpiece Unveiled
J.R.'s killer is revealed, as are two new alliances we didn't see coming
By Mandi Bierly
Updated April 16, 2013 at 06:31 AM EDT
Credit: Van Redin/TNT
Patrick Duffy told us to call him if the scene at J.R.'s grave, in which the truth about J.R.'s death was revealed, didn't make us cry. No need: I'm guessing we all got misty. As much as Bobby's breakdown got to me, it was John Ross' broken line delivery that hit me hardest: "I'm sorry, I'm confused. Who shot my father, Uncle Bobby?" he asked. "I shot your father, John Ross," Bum confirmed. J.R., it turns out, was dying of cancer, and his masterpiece was framing Cliff Barnes for his death. He'd asked Bum to steal Cliff's gun and shoot him in Mexico, where Cliff travels for an annual fishing tournament. "The only person that could take down J.R. was J.R.," John Ross said, shaking Bum's hand. Did you see that coming? Let's back up.
We opened with John Ross and Pamela on the Barnes Global jet marveling at her gumball machine wedding ring. Her father called, wanting her to come home so she could enjoy the takeover of Ewing Energies with him. She'd told him she was meeting with a casino in Vegas — he still didn't know about the nuptials. John Ross told Pamela whatever problems awaited them in Dallas they'd face together, including the one of taking down her daddy. She asked how much longer their flight was, and he looked at the plane's flight history and saw that Cliff had been in Nuevo Laredo, Mexico the night that J.R. had died. John Ross called Bobby, and he phoned Bum and told him it was time to put the plan in motion. Cue the music!
Back at Southfork, John Ross was ready to call the police, but Pamela and Bobby said they didn't have enough to pin this on Cliff — yet. In Zürich, Elena took a call from her mother, who was worried about Drew. Elena phoned him to check in. She wanted him to go straight to Mexico, but he wanted to play CSI on a pair of Harris' stolen shoes and find Roy so the Ewings can connect Harris to the rig explosion. Christopher came into the hotel room having talked to an investigator downstairs who gave him the address that his mother's bank account was registered to. The home is owned by an American who used to be a plastic surgeon, Dr. David Gordon. Elena was worried about Christopher because he hasn't seen his mother in over 25 years. But Christopher doesn't consider her his mother, he said. All he wants is her shares.
Bum found Ken Richards, and Sue Ellen surprised him in the Cayman Islands. She told him he had to give her hard evidence of the governor's involvement in the TESHA cover-up because now the governor is allowing Harris and Cliff to destroy the Ewings. Ken said he couldn't: The governor knows that Ken buried a report saying his son's startup tech company was polluting a river. Sue Ellen took a page out of J.R.'s book: If Ken doesn't help her and trust that she'll protect him, she'll out him and that buried report herself. Nice!
Christopher showed up at the house and Dr. David Gordon said Pam wasn't there. He clearly recognized Christopher's name. We heard a noise upstairs, and Christopher followed it. It wasn't Pam, it was another woman. Gordon said he was Pam's surgeon, they fell in love and married. She doesn't want to see Christopher. After the accident, Cliff had found her in Houston, he said, and asked her to come back with him, Bobby, and Christopher, and she'd said no. That part of her life was over. She didn't want any part of the Ewing/Barnes feud. She never wants to see him again. Christopher left in tears.
After we found out Emma had agreed to go to rehab to get out of jail, Christopher told Elena what Gordon had said. He fought back more tears, and Elena said they're not leaving until Christopher gets to confront his mom face-to-face. He was happy Elena was there. For now, Christopher!
NEXT: Shower time, again
Back at Southfork, Rhonda, the woman J.R. had met with the night he died, was brought in so Pamela could show her a picture of Cliff. She had seem him before: He was at the club that night, too, staring at J.R. and her. John Ross stormed off, convinced that was all the proof they needed. Bobby said he believed Cliff was the killer now, too, but he couldn't do what John Ross wanted: Sheriff Derrick had no jurisdiction since the crime took place in Mexico, and even if Carlos could help in Mexico, there wasn't enough evidence for a conviction if they could somehow get Cliff there. John Ross wanted the gun that J.R. had left him then. If Cliff wasn't going to jail, he'd put him in the ground, he said. Bobby reminded John Ross what J.R. had written in the letter: They take from Cliff what he wanted from them — his company — and then Bobby would know what to do. They had to get control of Barnes Global, which means Christopher needs Pam's shares. Then Bobby would need a day to get things ready, and they'd need Pamela's help, he said. I love Bobby and John Ross scenes.
Christopher and Elena take more showers than any other two people on television. While she was in the shower, he stumbled upon the cell she used to call Drew. He dialed, Drew answered, and Christopher knew Elena had been lying to him. He confronted her, and she tried to explain. But Christopher was having none of it. After a lot of shouting and a little lip quivering, he said she better be gone by the time he got back.
Things started moving fast: At Southfork, Emma was not at all happy about being weaned off her pills in preparation for rehab. In Zürich, Christopher wasn't taking Elena's calls. In the governor's office, Sue Ellen was showing McConaughey her copy of the initial TESHA report saying there were two explosions. Sue Ellen poured herself a drink, or so we thought. It was for the governor. She doesn't need it anymore. She was going to screw over Ken Richards and blackmail the governor sober. While they talked eminent domain, Drew went into a bar and told the barmaid to call him at his hotel room if she hears from her friend Roy. Drew can't be that stupid, right? He wanted her to call Roy and tell him where he was staying.
After sourpuss Emma showed up on her dad's doorstep and asked to come home, Bobby gave John Ross a replica of J.R.'s belt buckle. He had John Ross ask Pamela to help plant it on Cliff to tie him to the murder. The buckle had been stolen the night J.R. died. He told her they may not be able to prove Cliff's role in the rig explosion, but this can give them both justice. She agreed. (And that was J.R.'s real belt buckle, right? They were just lying because we weren't supposed to know Bum was there to take it.)
At Drew's motel, we got a masterful dual Honda and F-150 product placement shot as Roy pulled in next to Drew's bike. While Roy went upstairs to Drew's room, Drew planted the shoe, which dissolved into cocaine, in the back seat of his truck. We thought Roy was then chasing Drew, but when the cops pulled them over, the rider wasn't Drew — it was the dude we'd seen sitting at the bar when Drew talked to the barmaid. I guess he wanted to make a few quick bucks by allowing himself to be chased by a guy who wanted the man he was posing as dead? Not smart. But it worked out for him.
NEXT: The truth about Pam
After Sue Ellen reported her good news to Bobby while wearing a bold white suit — she's feeling anything but evil — Bobby got a call from Drew, who told him Roy had been arrested and that the shoes Harris was transporting were made from high-impact pressure-molded cocaine. Drew said the Ewings should be able to leverage the charges against Roy to get what they need on Harris, and, oh yeah, he's not turning himself in. Coward! But good riddance. That character was boring. Let's never hear about Papi's well again. (Wishful thinking!)
While Christopher slept off a hangover in Zürich, Elena was home in Dallas hearing her mother tell her that Drew had other options than to blow that rig. Elena needs to apologize to Christopher and give him time to forgive her, mom said. Um, mom, you're the one that made Elena help him in the first place!
At the bank, John Ross and Pamela prepared to slip the belt buckle into Cliff's safe deposit box. Won't there be a paper trail saying the two of them visited the box? That worries me. Meanwhile, Emma snuck into her dad's office presumably looking for pills — or was she helping the Ewings? Confused!
Christopher got a call that his mother was cleaning out her account. He ran after her and Gordon, in slow-motion, and it was the woman we had seen making the noise at the house earlier. He asked who she was, and my DVR cut off. Cliffhanger! Until I played the second hour. Pamela and John Ross found Pam's death certificate from Abu Dhabi in Cliff's safe deposit box — dated July 14, 1989. Sad trombone. No Victoria Principal for us!
Harris found Emma rifling through his things, and she told him she was looking for pills. Good thing he kept those in the safe… with whatever she was really interested in scoring.
Gordon told Christopher that Pam had come to him after the accident that burned over 60 percent of her body. She left Dallas because she didn't want to scare her little boy — she thought she was hideous. They got to the final round of surgeries, and she found out she had pancreatic cancer and was dying. She wanted to recover and see Christopher again, so he brought her to Abu Dhabi for an experimental treatment, but it didn't work. She died. When Cliff found out that Pam's shares would go to Christopher, he contacted the doctor and the woman who'd been Pam's nurse and asked them to keep her death a secret so that he could control her shares. In exchange, Cliff supported them. The nurse gave Pam's will to Christopher, along with a letter she'd written to him before her death. Cry, Christopher, cry! And cue the music.
NEXT: The truth about Cliff
Roy, we found out, wouldn't turn on Harris and Cliff. But he did recognize Pamela, who'd come with Bobby and John Ross to see if he'd struck a deal. Cliff called a meeting with Harris, who assured him that Roy would do his time like a man. The minute Harris got out of Cliff's car, Cliff called someone with a job (killing Roy in prison). Pamela visited Roy and played on his weakness as a father. She asked him if he knew she was having a boy and a girl. She told him to just tell her the truth, and he did: He told her to have other children and forget her father — there was nothing unintentional about the rig explosion. We'd seen how conflicted Roy was the day of the blast. The writers had earned that confession. And as expected, Roy, a grandfather, got knifed in the neck as he was calling his son. Cruel!
Harris told Emma he had nothing to do with Roy's death. She pretended she didn't care: She said she was happy Drew was gone and Roy was dead — now they can't hurt him. Still, the Ewings thought Harris orchestrated it, not Cliff. Regardless, Christopher came home and presented the copy of his mother's will. John Ross handed over the death certificate. Pamela announced they were two-thirds of the way in to Barnes Global, and John Ross told Christopher and Sue Ellen that he'd married Pamela. "Surprise," he said, awkwardly. Sue Ellen was elated, and Christopher was somewhat happy for them.
Now that they could run Barnes Global, Bobby said it was time to reveal what was in J.R.'s letter. He'd spent the last few months of his life working to protect the family, which meant taking down Cliff. Bobby took out the gun — and Pamela ID'd it as her father's. Bobby said it was used to kill J.R., and Bum explained that J.R. had caught wind that Cliff had followed him to Mexico, got worried, and called him. By the time Bum got to the hotel, J.R. was dying. J.R. told him Cliff was using the cartel to cover his tracks and to call Carlos. Carlos bought the gun back from the cartel. That would seem like a lot for a dying man to say. I should have suspected that was a load of crap.
John Ross was angry that Bum and Bobby hadn't told him the truth sooner, but they said J.R. wanted them to be patient to destroy Cliff. And also, John Ross would have killed Cliff. Bobby said Cliff would fight extradition to Mexico. Pamela said she'd get him there. After we saw Christopher NOT accept Elena's apology, we found out how Pamela planned to set her father up: She told her dad that Frank had been working on a casino deal in Mexico before his death, and she wanted to finish it. Cliff was happy to go: "Talk about dancing on J.R.'s grave," he said. Hubris!
It's nice that it all came down to the women, in the end: Sue Ellen worked over the governor. Police found the belt buckle and the gun Pamela had planted in that safe deposit box and Cliff's car, respectively. Emma drugged her dad while making him breakfast and found his ledger documenting all his dealings with the cartel and where he was hiding the money. She'd taken it to Ann, who turned it in to Derrick, who arrested Harris. Did Emma think that up herself? (No. Must have been John Ross.)
At J.R.'s grave ("The only deal he ever lost," his tombstone reads), his body was exhumed to recover the bullets. Ballistics confirmed they came from Cliff's gun.
NEXT: John Ross becomes J.R.
At their hotel in Mexico, Pamela hung up on a woman who called to warn Cliff about police searching his office. She disconnected the phone. There was a knock at the door, and in walked Bobby, Christopher, John Ross, and Sue Ellen. I cheered. Christopher told Cliff about his shares, and Pamela told him about her marriage to John Ross. We knew Cliff was lying when he told Pamela that he wasn't responsible for the rig, but we couldn't be sure that Cliff was fibbing when he said his gun was stolen before J.R.'s shooting and that someone had moved his plane (he says he was in Mexico, but it was Cabo). Cliff was arrested, repeating again and again that he didn't kill J.R. as he was escorted out. Cliff was convincing enough that at that point, I was willing to consider that J.R. had developed a terminal disease and decided to pay someone to end his life and frame Cliff for it. (And I was right!)
Bobby went to see Cliff, who still insisted that he was being framed for J.R.'s death, and Bobby made him an offer: He'd help him get back to an American prison if he admitted to sabotaging the rig, killing their grandchildren, and partnering with Harris. Cliff said he's never done what the Ewings wanted him to do and he's not starting now. As Bobby left, Cliff asked who killed J.R., and Bobby said Cliff would never know. WHAT? So Bobby knew Cliff didn't do it. Bad Bobby!
Back in the States, Bobby and Bum drove up to J.R.'s grave, and John Ross and Christopher seemed to surprise them there. They, too, now suspected that Cliff didn't kill J.R. and wanted the truth. They demanded to know what was in J.R.'s letter, and luckily, Bobby had it on him.
"Bobby, doctors say I've only got a few days left. Damn cancer." I WAS RIGHT!!! He didn't tell Bobby he was sick because he detested pity. He'd had Bum steal Cliff's gun because he knew Cliff came to Mexico every year for a fishing competition. He'd had Carlos smooth things out on that end. He wanted Bobby to end the Ewing/Barnes feud once and for all. The letter said Bum was the best friend J.R. didn't deserve to have — that's when I knew Bum pulled the trigger. It referenced a time when J.R. had framed Bobby as the one who'd borrowed their daddy's favorite shotgun by putting extra shells in Bobby's room. Bobby broke down and couldn't finish reading the letter. A single tear fell down his cheek as Christopher continued. J.R. hoped Bobby knew that his love for him outweighed his jealousy: "Goodbye, baby brother, I guess I'll be duck huntin' with daddy. I'll tell him I was the one who borrowed his gun."
If you weren't tearing up then, Josh Henderson's broken delivery of John Ross' response definitely got you. "I'm sorry, I'm confused. Who shot my father, Uncle Bobby?" he asked. "I shot your father, John Ross," Bum confirmed. He said that J.R.'s last act was one of love for his family and for his son. BUT DID HE HAVE TO DO IT WHEN HE WAS ON THE PHONE WITH JOHN ROSS???
"The only person that could take down J.R. was J.R.," John Ross said. And he held out his hand to Bum. Tears!
Bobby handed John Ross the letter and left him alone, kneeling at the grave. "Thank you, daddy, for watching over us. I love you," he said. We could have ended there, but we had to set up for season 3 with some serious craziness.
Elena found out that her father died for nothing. His land had no oil. As if on cue, papers were delivered to her from Cliff Barnes' lawyers. They were land records — one of them had J.R.'s name on it, and the other was her father's land. Cliff wanted her to come to the jail in Mexico to see him. WHAT?
Emma went to see her father in jail to gloat and assure him that she's exactly the daughter he raised. Harris told Ann she must be enjoying this. Ann didn't have much to do in these episode, but this could be the line of the night: "Actually, I'm not, Harris," she said, "because I'm not a sick, sadistic prick like you are." He was speechless.
At Southfork, Christopher drank a Miller Lite and read the letter from his mother — who had the handwriting of a fifth grader. (Oh, was that burn-related? Now I feel bad.) Pamela walked the grounds, unsure about whether she was moving back into Southfork but knowing she doesn't want to go home. She wants to start a new life with John Ross. Christopher said he was happy for her, and she told him he's always loved Elena. Even though he's mad at her, he should forgive her. All you get holding onto anger, she said, is wreckage like her father left. Who would've thought we'd end the season loving Pamela?
Elena went to see Cliff, who told her that J.R. had paid someone to switch the deeds so that Elena's father got J.R.'s worthless land, and J.R. got her father's goldmine. Elena told Cliff he must be making this up, but Cliff said to go check the records: J.R. had done the same thing to his father. He wants her to be his proxy for the third of Barnes Global he still owns and to make the Ewings pay for the sins against her family. J.R.'S SINS. And he's gone. Rise above it, Elena.
At the newly christened Ewing Global, a portrait was hung of J.R. as a reminder that no one goes against a Ewing and bourbon was poured. Ann came to the office, and she and Bobby kissed and made up. He told her that he'd only shut her out because he didn't want her to see him act like J.R. — but she said he's still the best man she knows. Ah.
After Sue Ellen told John Ross to treat his bride right (and he joked that he wasn't a scoundrel), he went off to buy champagne and flowers. We saw Christopher try to tell Elena he loved her, but he was talking to an empty room. And Elena, well she was at a compound owned by someone she and Drew grew up with in Mexico who is now, it appears, into some serious s—. That was a lot of armed guards. Please don't tell me Drew is there. What is she plotting? A coup?
John Ross arrived at a hotel room with his romantic gestures, found a pair of discarded panties, and went into the bedroom. I totally thought he and Pamela were celebrating their honeymoon. I was wrong. IT WAS EMMA. WHAT? She'd gotten the briefcase out of her dad's safe for him. He told her it looks like they hit the motherload. "That look's gonna cost you," she said. "Darlin' trust me," he replied. "I'm willing to pay the price. Just don't tell my wife." He picked her up, and they hit the sheets. Scoundrel! I really thought he cared about Pamela. I feel like a fool. When did he have time to orchestrate this with Emma? And how does he explain to his new wife that he's not spending the night with her? What happens if Pamela finds out Cliff didn't kill J.R.? Would she be fine (because he did kill her babies) or would she turn on John Ross again? And how soon will she find out he's cheating on her? Notice he didn't say he loved her when she said it. Are we going to see her fall back into Christopher's arms? How sick would that be? Theories? Go!
S3 E8 Recap
Dallas recap: Burning Down the House
Dallas recap: Striking Distance
Dallas recap: Like Father, Like Son
Dallas recap: The Family That Blackmails Together...
Dallas recap: Just Married (Again)
Dallas recap: Playing Chicken
Dallas recap: Trust Me
Dallas season premiere recap: Secrets and Lies
S2 E12 Recap
Dallas recap: Family Ties
Dallas recap: Conspiracy Theory
Dallas recap: Guilt and Innocence
Dallas recap: A Bang and a Blast
Dallas recap: The Long Goodbye
Dallas recap: The End of J.R. Ewing
Dallas recap: Hostage Crisis
Dallas recap: Trial and Error
Dallas recap: The Ultimate Betrayal
Dallas recap: Bullet to the Chest
Dallas season premiere recap: The Devil's Son
Dallas season finale recap: Revelations and Revenge
Dallas recap: The Return of J.R.
By Darren Franich
Dallas recap: Family Matters


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text:
Lipothyrea är ett släkte av skalbaggar. Lipothyrea ingår i familjen vivlar.

Kladogram enligt Catalogue of Life:

Källor

Vivlar
Lipothyrea

redpajama_set_name: RedPajamaWikipedia


text:
Викентий (Владислав) Данилович Герберский (Wincenty Herburt z Fulsztyna Herberski; 1785—1826) — российский , надворный советник, доктор медицины, ординарный профессор Виленского университета, член многих ученых обществ.

Биография

Викентий Герберский родился в 1785 (или 1784) году. Первоначальное образование получил в Слонимской школе, а среднее и высшее — в Виленских гимназии и университете.

5 апреля 1804 года он был послан Виленским университетом в английскую колонию в окрестностях Петербурга для изучения на практике сельского хозяйства и в Петербургский Педагогический институт на курс лекций по физике, химии и сельскому хозяйству.

В 1806 году он возвратился в Вильну и стал изучать медицину; в 1808 году был удостоен Виленским университетом степени кандидата философии, в 1811 году — магистра медицины и в 1812 году — доктора медицины за сочинение: «Dissertatio inauguralis medicopract. de hydrope specus vertebralis etc.» (Vilnae, 1812).

С 1813 по 1824 год Викентий Данилович Герберский был адъюнктом медицинской клиники и помощником профессора Йозефа Франка, в 1817 году стараниями последнего был отправлен за границу для изучения офтальмологии; пробыв за границей 6 лет и возвратившись в Вильну в 1823 году, он стал преподавать частную терапию и клинику.

В январе 1824 года В. Д. Герберский был назначен экстраординарным, а в ноябре ординарным профессором по кафедре частной терапии и патологии на место профессора Франка и директором клиники по внутренним болезням. Кроме того Герберский лечил и глазные болезни.

На публичных собраниях университета он читал доклады: «О госпиталях» и «Наука о глазных болезнях»; эти сочинения остались в рукописи.

Викентий Данилович Герберский умер 29 ноября 1826 года.

Среди прочих наград, Герберский был кавалером ордена Святого Владимира 4-й степени.

Примечания

Литература
 Змеев Л. Ф., "«Словарь русских врачей».
 «Kurier Lit.» 1826 г., № 151.
 «Wielka Encyklopedya powszechna illustrowana», t. XXVII—XXVIII, 795.

Выпускники Вильнюсского университета
Медики Вильнюса
Преподаватели Императорского Виленского университета
Офтальмологи Российской империи
Надворные советники


text:
Q: How to Convert to DateTime I have CSV files that has among other things, dates in them, that looks like:
2014-01-25 00:14:08.000

Now I parse the CSV file and get the dates. Then I run a regex to check if the date is valid(matches the format as above) then I convert the date from the CSV file to  DateTime.
This how ever gives me a ValueError: that states: need more than 3 values to unpack.
The CSV file can contain Thousands of records with dates like the above mentioned.
My regex:
date_re_time_norm = re.compile("""^(19|20)\d\d[- /.](0[1-9]|1[012])[- /.](0[1-9]|[12][0-9]|3[01])[ ](\d\d:\d\d:\d\d.\d\d\d)$""", re.I)

My convert method:
def convert_date(self, dstring):
    if self.date_re_time_norm.match(dstring):
        year, month, day, hour, minute, second = str(dstring).split('-')
    return datetime(int(year), int(month), int(day), int(hour), int(minute), int(second), 0, 0, 0, 0, 0, 0)

Method is shortened for readability.
Am I doing something wrong?

A: You are splitting on hyphens, and there are only two of those, so you only get three elements in the list:
['2014', '01', '25 00:14:08.000']

See @jonrsharpe's answer for a better way.

A: It would be much better to use datetime.strptime:
from datetime import datetime

d = datetime.strptime("2014-01-25 00:14:08.000", "%Y-%m-%d %H:%M:%S.%f")

This can replace both of your regular expressions, as it will parse the date if in an appropriate format or raise an error if not:
def convert_date(self, dstring, format="%Y-%m-%d %H:%M:%S.%f"):
    try:
        return datetime.strptime(dstring, format)
    except ValueError:
        # wrong format

(you could make the format string a class or instance attribute as necessary).

Edit: If you are determined to stick with the regexs, you will need to do more splitting:
date, time = dstring.split(" ")
year, month, day = date.split("-")
hour, minute, second = time.split(":")
second = second.split(".")[0]
return datetime(*map(int, (year, month, day, hour, minute, second)))

In Python, however, "There should be one-- and preferably only one --obvious way to do it" and that way is strptime.

redpajama_set_name: RedPajamaStackExchange


text:
Real EstateUK
Lendlease and CPP Investments strengthen UK BTR partnership with additional £85m investment
Henry ThomasJune 27, 2020
The partnership was first announced in 2018 when they agreed an initial target of investing £1.5bn in the sector
Lendlease and Canada Pension Plan Investment Board (CPP Investments) have strengthened their UK build-to-rent BTR partnership with an additional £85m investment.
The companies have each agreed to invest 50% of the capital to help fund 123 new BTR homes at Elephant Park in south London.
In February, the companies announced an £80m investment to begin construction of an 18-storey building with 118 BTR homes at Elephant Park.
The partnership was first announced in 2018 when they agreed an initial target of investing £1.5bn in the sector.
Today's announcement means that the partnership is now planning to provide over 900 BTR homes at Elephant Park, the companies said in a joint statement.
Construction is well underway on the first two buildings and later this year the partnership is expected to confirm timings for their completion and opening to residents. Lendlease will develop, construct and manage the build-to-rent homes on behalf of the partnership.
Victoria Quinlan, MD for investment management Europe at Lendlease, said: We're thrilled to have achieved another major milestone in our partnership with CPP Investments as we build momentum in our build-to-rent platform.
"Their investment demonstrates a vote of confidence in the build-to-rent sector in London and in Elephant Park."
Andrea Orlandi, MD, head of real estate Europe, CPP Investments, said: The build-to-rent sector in the UK continues to be supported by long term trends, particularly in London and other major UK cities.
"Private rentals are an increasingly attractive and flexible option for households. The Elephant Park project continues to move ahead towards completion for the first phase, and our, and our latest financial commitment will help to progress this landmark regeneration project."
Henry Thomas June 27, 2020
UK property funds stand by office allocations even as corporates encourage remote working
Prologis confirms sale of UK warehouse portfolio
The Hut Group submits redevelopment plans
Henry ThomasJanuary 20, 2021
LendInvest expects demand for complex lending to pick up
UK house prices rose 7.6 per cent in November
All about property investment crowdfunding
London house market set for £52bn post-lockdown splurge due to huge pent-up demand
Belfast is one of the cheapest places in UK to buy a home
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redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Getting an error attemping to use .NET Core dependency injection with NUnit test classes new at unit testing and .NET Core (1.0.1), looking for some help.
Im looking to unit test a few classes that are using the .NET Core dependency injection pattern. Application runs fine, but when trying to run NUnit tests with the classes that require DI, error outputs:

No handler for message 'TestExecution.TestResult' when at state
  'TestExecutionSentTestRunnerProcessStartInfo' An exception occurred
  while invoking executor 'executor://dotnetunittestexecutor/v1': Unable
  to write data to the transport connection: An existing connection was
  forcibly closed by the remote host. An existing connection was
  forcibly closed by the remote host

Haven't had a lot of luck with searching that error, or any documentation around using DI with NUnit. I have a feeling I'm not suppose to config like this. However, a few settings (connection strings etc) live in the appsettings.json file which get injected via Startup.cs. How would I access these in the NUnit classes?
My Startup.cs, gets the connection string and preps for DI
public void ConfigureServices(IServiceCollection services)
{
    //other Configurations happen here

    //get the connection string values from appsettings.json
    services.Configure<ConnectionStrings>(Configuration.GetSection("ConnectionStrings"));

    //Classes that get DI'd
    services.AddSingleton<IMyDAL, MyDal>();
    services.AddSingleton<MyBLL>();
}

MyDAL requires configurations to be injected, available via Startup.cs:
public class MyDAL : IMyDAL
{
    private readonly string _dbConnectionString;

    public MyDAL(IOptions<ConnectionStrings> connectionString)
    {
        _dbConnectionString = connectionString.Value.ConnectionString;
    }

    public bool DoDALStuff() { ... }
}

MyBLL class requires IMyDAL to be injected:
public class MyBLL
{
    public readonly IMyDAL _myDAL;

    public MyBLL(IMyDAL myDAL)
    {
        _myDAL = myDAL;
    }

    public bool DoBLLStuff() { ... }
}

My NUnit class requires the BLL injection to test:
[TestFixture]
public class BLLTest
{
    public MyBLL _myBLL;

    public BLLTest(MyBLL myBLL)
    {
        _myBLL = myBLL;
    }

    [Test]
    public void TestTheBLL()
    {
        Assert.IsTrue(_myBLL.DoBLLStuff() == true)
    }
}

This is how my regular program flow looks like, but I might be approaching it in the wrong way trying to implement this unit test in the same DI fashion. If I remove any DI/constructor code in the test class, the tests run fine, but I still need to test the methods that require DI inside the BLL. Anyone know how I can get the DI'd objects into my test class to use? Am I using an incorrect approach?

A:
Am I using an incorrect approach?

Yes. You misunderstand the definition of unit testing. You want to test the BLL in isolation. not against actual implementations as that would be an integration test.

how can I get the DI'd objects into my test class to use?

You create fake implementations of your dependencies or use a mocking framework of your choice to mock the dependencies for the unit test.
[TestFixture]
public class BLLTest {
    private MyBLL _myBLL;

    public BLLTest() {
        //Arrange
        var dalMock = new Mock<IMyDAL>();
        dalMock.Setup(x => x.DoDALStuff()).Returns(true);//setup expected behavior
        //...setup other expected behavior of dependencies

        //create the target of unit test (class under test)
        _myBLL = new MyBLL(dalMock.Object); //manually injecting dependency
    }

    [Test]
    public void TestTheBLL() {
        //Act
        var result = _myBLL.DoBLLStuff();
        //Assert
        Assert.IsTrue(result);
    }
}

this is just a basic example to get you started. there are many other ways to do this. you should do some more research into unit testing

A: I'm using xUnit, I didn't have time yet to dive into Nunit.
Warning (source)

dotnet-test-nunit is still an alpha release

I believe you have an issue with your test runner. You should start by reading about Integration testing with ASP.NET Core. Examples are only with xUnit though.
Here is also a blog post how to perform this with Nunit.
What is important is that you run your test project, not directly your web project (even if this one is referenced and used by the test project).
Because of that, you need another JSON settings file inside your test project which will be used instead of the one in your web project. You certainly want other configuration in testing mode than in Dev or Prod.
In case if unit testing you would prefer calling your classes' methods directly but if you are in a context of integration testing you should call your controllers like any other client (JavaScript, mobile app...etc) would do.
To do so, you can create TestServer then a client from your and finally send normal HTTP request like this:
var webHostBuilder = new WebHostBuilder().UseEnvironment("Testing").UseStartup<Startup>();
var host = new TestServer(webHostBuilder);
var client = host.CreateClient();
var response = await client.PostAsync(endpoint, form);

In this example endpoint is simply this URL and form is a MultipartFormDataContent because I'm posting a file


Art Deco architecture packed with quirky bookstores and European-style cafes makes Tiong Bahru a great Saturday morning destination. Jacqui Gibson is your guide to one of Singapore's hippest neighbourhoods.
Tiong Bahru Market, Tiong Bahru, Singapore (c) Jacqui Gibson. Flight Network.
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Books Actually, Tiong Bahru, Singapore (c) Jacqui Gibson. Flight Network.
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Bikes of Tiong Bahru, Singapore (c) Jacqui Gibson. Flight Network.
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Street view of Tiong Bahru, Singapore (c) Jacqui Gibson. Flight Network.
Does Tanzania's Lake Natron Really Turn Live Animals to Stone?


Starting with the runway itself, which was set with vignettes of furniture from their brand-new home line, this was a season of firsts for Badgley Mischka. The ApparelMagic client is known around the world for their elegant gowns and cocktail dresses, but starting this fall, their loyal customers can wear Badgley Mischka all day long.
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Bridging the divide between comfortable and classy, the new line included everything from luxe velour sweatsuit separates to swing coats paired with easy slacks and leggings. Accessories were elevated with youthful details like embellishments on slip-on sneakers and fur pom-poms on beanie hats. Whether a fancy film premiere or a casual weekend movie at home, the new Badgley Mischka is a 24/7 lifestyle brand.


The Writing Center and the Liberman Graduate Center invite applications to participate in a dissertation writing retreat near the end of winter break. All Ph.D. candidates at any stage in the dissertation process are eligible to apply. Participants are required to commit to the entire retreat, pledge not to use the internet or email during retreat hours (unless necessary for their writing process), and have no other commitments during retreat hours. For more information, see below.
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text:
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Jerzy Stuhr, laureate of the Special Lifetime Achievement Award to the Polish Artist, is one of the most popular Polish actors. He also works as a screenplay writer, film director and drama professor. He serves as the Rector of Ludwik Solski Academy of the Dramatic Arts in Krakow. In the early 1970s Stuhr appeared in Polish theatre. Having met film director Krzysztof Kieslowski in the mid-1970s, he continued to work with him until Kieslowski's death in 1996. Talking about his film career Jerzy Stuhr said that "it all began in 1975 with Krzysztof Kieslowski". To an international audience Stuhr may be best known for his minor role as thick-witted hairdresser Jurek in Kieslowski's Three Colors: White.
Jerzy Stuhr in Kieślowski's Camera Buff
As well as appearing in a total of five of Kieślowski's films, Stuhr also was mentored by the older director when he moved from working in front of the camera to behind it. Moreover, he has drawn inspiration from Kieślowski's style and obsession with "moral concern." Stuhr's second film, Historie miłosnie (Love Stories, 1997), was dedicated to Kieślowski, and one of his latest features, Duże zwierzę (The Big Animal, 2000), even employs an early, previously unshot Kieślowski's screenplay.
Since his own directing debut in 1994 he has made 6 films in total, i.e.: Spis cudzołożnic (The List of Adultresses), Love Stories, Tydzień z życia mężczyzny (A Week in the Life of a Man), The Big Animal, Pogoda na jutro (Tomorrow's Weather) and Korowód (Procession).
Stuhr on the set of Procession
As an actor Stuhr also co-operated with Polish directing legends: Agnieszka Holland, Andrzej Wajda and Krzysztof Zanussi. In films as well as in theatre, he created many remarkable film characters. He is a recipient of many prestigious nominations and awards granted by Venice Film Festival, Polish Film Festival and Karlove Vary Film Festival among others, rewarding his achievements both as an actor and film director.
Selected filmography as an actor:
• Korowód (2007)
• Tomorrow's Weather (2003)
• Big Animal (2000)
• A Week in the Life of a Man (1999)
• Love Stories (1997)
• The Good Soldier Svejk (1999)
• The List of Adultresses (1993)
• Three Colors: White (1994)
• Deja vu (1989)
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• Kingsajz (1987)
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• O-Bi, O-Ba - The End of Civilization (1984)
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Timewarp presents a new release from InnerZone called The Inner Zone EP, we hope you enjoy the ride! Rei Albahari has been producing music since 2005 under the name "MadDoc". He has started his way around electronic music while co-producing & editing the Israeli radio show "BeatNik" (Brodcast on "Galey Tzahal" 2004-2006). After the show stopped Rei began to produce small parties & DJ sets around Israel, in 2008 forming the production "Ha'HaZit Ha'Amamit La'Trance", taking on promoting the GOA TRANCE & the Retro trend. "MadDoc" shows & DJ seta have been well known because of his unique sound and track selecting. Rei started producing music as InnerZone in 2011 and got his 1st release Analog Sunrise. Since then Rei released two E.P's and several tracks on compilations, remixed & collaborated with artist such as Astral Projection, Power Source, Radical Distortion, The Muses Rapt , Nervasystem and more. InnerZone would like to thank to his parents & family, Shanny Avitan, Steve Tzokas, Renato Brnic, Shahaf Shvarzman, Matan Ohana, Issac Sandman, Har-el Prusky, Avi & Lior, Juan Verdera, Drezz Nervasystem , Danny Marshal and all of you party people out there!


Unisex Red Tigers Eye bracelet can be worn cross season. Cord used is leather finish off with bronze on copper clasp. Length is 8inches/20cm with small extension chain. Other gemstone and size can be made to order. Sent securely wrapped within two days of payment received.
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Hope you're well today, glad to see Vaughan was able to assist here.


text:
\section*{Introduction}

In this paper, we extend the so called dual estimate
of~\cite{LewySt}
to the obstacle problem
for quasilinear elliptic equations
in the Heisenberg group.

The notation we use is the standard one:
for~$n\ge1$, we consider~$\R^{2n+1}$
endowed with the group law
\begin{eqnarray*} && \big( x^{(1)},y^{(1)},t^{(1)}
\big)\circ\big(x^{(2)},y^{(2)},t^{(2)}\big)
\\ && \qquad:=\Big(
x^{(1)}+x^{(2)},y^{(1)}+y^{(2)},
t^{(1)}+t^{(2)}+2(x^{(2)}\cdot y^{(1)}-
x^{(1)}\cdot y^{(2)} ) \Big),\end{eqnarray*}
for any~$(x^{(1)},y^{(1)},t^{(1)})$,
$(x^{(2)},y^{(2)},t^{(2)})\in\R^n\times\R^n\times\R$,
where the ``$\cdot$'' is the standard Euclidean scalar product.

Then, we
denote by~$\H^n$ the~$n$-dimensional Heisenberg
group, i.e.,~$\R^{2n+1}$ endowed with this group law.

The coordinates are usually written as~$(x,y,t)\in\R^n\times
\R^n\times\R$, and, as
customary, we introduce the {left invariant} vector fields~$(X,Y)$
induced by the group law
$$X_j:=\frac{\partial}{\partial x_j} +2y_j\frac{\partial}{\partial t}
{\mbox{
and
}}
Y_j:=\frac{\partial}{\partial y_j} -2x_j\frac{\partial}{\partial t},$$
for~$j=1,\dots,n$,
and the horizontal gradient~$\nabla_{\H^n}:=(X,Y)$.
The main issue of the
Heisenberg
group is that~$X$ and $Y$ do not commute, that is
\[[X,Y]=4\frac{\partial}{\partial t}\neq 0.\]
We are interested in studying the obstacle problem
in this framework.
For this, we consider a smooth
function~$\psi:\H^n\rightarrow\R$,
which will be our obstacle (more
precisely,~$\psi$ is supposed to
have continuous derivatives of second order in~$X$ and~$Y$).

Fixed a bounded open set~$\Omega$ with smooth boundary,
and~$p\in (1,+\infty)$,
we consider the space~$W^{1,p}_{\H^n}(\Omega)$ to
be the set of all functions~$u$ in~$L^p(\Omega)$
whose distributional horizontal derivatives~$X_ju$ and~$Y_ju$
belong to~$L^p(\Omega)$, for~$j=1,\dots,n$.

Such space is naturally endowed with the norm
$$ \|u\|_{ W^{1,p}_{\H^n}(\Omega) }:=
\|u\|_{L^p(\Omega)}+
\sum_{j=1}^n \Big( \|
X_j u\|_{L^p(\Omega)}+\|Y_ju\|_{L^p(\Omega)}\Big).$$
We call~$W^{1,p}_{\H^n,0}(\Omega)$ the closure of~$C^\infty_0(\Omega)$
with respect to this norm.

We fix
a smooth domain~$\Omega_\star\Supset {\Omega}$, $u_\star\in
W^{1,p}_{\H^n}(\Omega_\star)\cap L^\infty(\Omega_\star)$
and we introduce
the space
$$ {\mathcal{K}}:= \big\{ u\in W^{1,p}_{\H^n}(\Omega)
{\mbox{ s.t. }} u\le \psi, {\mbox{ and }} u-u_\star\in
W^{1,p}_{\H^n,0}(\Omega) \big\}.$$
Loosely speaking,~${\mathcal{K}}$ is the space of
all the functions having prescribed Dirichlet
boundary datum equal to~$u_\star$
along~$\partial\Omega$ and that stay below the obstacle~$\psi$.

Now we consider a parameter~$\varepsilon\ge0$
and we deal with
the variational problem
\begin{equation}\label{PB}
\inf_{u\in{\mathcal{K}}} {\mathcal{F}}_\varepsilon (u;\Omega),
{\mbox{ where }}
{\mathcal{F}}_\varepsilon (u;\Omega):=
\int_\Omega(\varepsilon+|\nabla_{\H^n}u|^2
)^{p/2}.\end{equation}
By direct methods, it is seen that such infimum is attained
(see, e.g., the compactness result
in~\cite{VSC, Dan}
or references therein)
and so we consider a minimizer~$\bar u_\varepsilon$.

Then,~$\bar u_\varepsilon$ is a solution of the variational
inequality\footnote{Formula~\eqref{ines} may be easily obtained
this way. Fixed~$v\in W^{1,p}_{\H^n}(\Omega)$ with~$v
\le \psi$, and~$v-\bar u_\varepsilon\in
W^{1,p}_{\H^n,0}(\Omega) $,
for any~$t\ge0$,
let~$u^{(t)}:=\bar u_\varepsilon+t(v-\bar u_\varepsilon)$.
Notice that
$$u^{(t)}:=(1-t)\bar u_\varepsilon+tv\le(1-t)\psi+t\psi\le\psi,$$
hence~$u^{(t)}\in {\mathcal{K}}$.
So, by the minimality of~$\bar u_\varepsilon$, we
have~${\mathcal{F}}_\varepsilon (u^{(0)};\Omega)
={\mathcal{F}}_\varepsilon (\bar u_\varepsilon;\Omega)
\le{\mathcal{F}}_\varepsilon (u^{(t)};\Omega)$ for any~$t\ge0$.
Consequently,
\begin{eqnarray*} 0&\le& \lim_{t\searrow0}\frac{{\mathcal{F}}_\varepsilon
(u^{(t)};\Omega)-{\mathcal{F}}_\varepsilon (u^{(0)};\Omega)}{t}
\\ &=&
\int_\Omega (\varepsilon+|\nabla_{\H^n}\bar u_\varepsilon|^2
)^{(p-2)/2} \nabla_{\H^n}
\bar u_\varepsilon\cdot \nabla_{\H^n}(v-\bar u_\varepsilon),
\end{eqnarray*}
that is~\eqref{ines}.}
\begin{equation}\label{ines}
\int_\Omega (\varepsilon+|\nabla_{\H^n}\bar u_\varepsilon|^2
)^{(p-2)/2} \nabla_{\H^n}
\bar u_\varepsilon\cdot \nabla_{\H^n}(v-\bar u_\varepsilon)\,
\ge\,0,\end{equation}
for any~$v\in W^{1,p}_{\H^n}(\Omega)$ with~$v
\le \psi$, and~$v-\bar u_\varepsilon\in
W^{1,p}_{\H^n,0}(\Omega) $.

These kind of variational inequalities
has now receiving a considerable attention (see, e.g.,~\cite{Dan2}
and references therein), even when $p=2$ (notice
that in such a
case~$\varepsilon$
does not play any role).
We observe that, when~$p\ne2$, the operator
driving the variational inequality in~\eqref{ines}
is not linear anymore (in fact, it may be seen
as the Heisenberg group version of the $p$-Laplace
operator): for these kind of operators
even the regularity theory is more
problematic than expected at a first glance,
and many basic fundamental questions
are still open (see, e.g.,~\cite{Dom}, \cite{MinMan},
\cite{Zhong1} and~\cite{Zhong2}):
this is a crucial difference with respect to
the Euclidean case,
so we think it is worth dealing with the problem
in such a generality.

Now, we introduce the set of~$p$'s for
which our main result holds.
The definition we give is slightly technical,
but, roughly speaking, consists in taking
the set of all the $p$'s for
which a pointwise bound for the operator
of a sequence of minimal solutions
is stable under uniform limit.
The further difficulty of taking this
assumption is due to the lack
of a thoroughgoing regularity theory for
the quasilinear Heisenberg equation
(as remarked in
Lemma~\ref{p2P} at the end of this
paper, this technicality may be skipped when~$p=2$).

\begin{definition}\label{D} Let~$p\in(1,+\infty)$.
We say that~$p\in {\mathcal{P}}(\psi,\Omega)$
if the following
property holds true:

For any~$\varepsilon>0$, any~$v\in W^{1,p}_{\H^n}(\Omega)$,
any~$M>0$, any sequence~$F_k=F_k(r,\xi)\in
C([-M,M]\times\Omega)$,
with~$F_k(\cdot,\xi)\in
C^1([-M,M])$ and
\begin{equation}\label{1C4}\begin{split}
& 0\le \partial_r F_k\le
\left(
\div_{\H^n}\, \Big( (\varepsilon+|\nabla_{\H^n}\psi|^2)^{(p/2)-1}
\nabla_{\H^n}\psi\Big)
\right)^+ ,
\end{split}\end{equation}
if~$u_k:\Omega\rightarrow [-M,M]$ is a sequence
of minimizers of the functional
\begin{equation}\label{905} \int_\Omega
\frac{1}{p}(\varepsilon+|\nabla_{\H^n}
u(\xi)|^2)^{p/2}+F_k(u(\xi),\xi)\,d\xi\end{equation}
over the functions~$u\in W^{1,p}_{\H^n}(\Omega)$, $u-v\in
W^{1,p}_{\H^n,0}(\Omega)$,
with the property that~$u_k$ converges to some~$u_\infty$
uniformly in~$\Omega$, we have that
\begin{equation}\label{TX}\begin{split} & 0\le
\div_{\H^n}\, \Big(
(\varepsilon+|\nabla_{\H^n}u_\infty|^2)^{(p/2)-1}
\nabla_{\H^n}u_\infty\Big)
\\ &\qquad
\le
\left(
\div_{\H^n}\, \Big( (\varepsilon+|\nabla_{\H^n}\psi|^2)^{(p/2)-1}
\nabla_{\H^n}\psi\Big)
\right)^+.\end{split}\end{equation}
\end{definition}

As remarked\footnote{As
usual, the superscript~``$+$'' denotes the
positive part of a function, i.e.~$f^+(x):=\max\{f(x),0\}$.}
in Lemma~\ref{p2P} at the end of this paper,
we always have that $$ 2\in
{\mathcal{P}}(\psi,\Omega).$$
In particular, the main result of this paper (i.e., the
forthcoming Theorem~\ref{1})
always holds for~$p=2$ without any further restriction.
We think that it is an interesting open problem to
decide whether or not other
values of~$p$ belong to~${\mathcal{P}}(\psi,\Omega)$,
in general, or at least when the right hand side
of~\eqref{TX} is close to zero (e.g.,
when the obstacle is almost flat).
For instance, the property in Definition~\ref{D}
would be satisfied in presence of a
H\"older regularity theory for the horizontal
gradient for solutions of quasilinear
equations in the Heisenberg group -- namely,
if one knew that bounded solutions
of~$\div_{\H^n}\, ((\varepsilon+|\nabla_{\H^n}u|^2)^{(p/2)-1}
\nabla_{\H^n}u)=f$, with~$f$ bounded, have H\"older continuous
horizontal gradient, with interior estimates
(this would be the Heisenberg counterpart of classical
regularity results for
the Euclidean case, see, e.g.,
Theorem~1 in~\cite{Tolk}; notice also that
it would be a regularity theory for the equation,
not for the obstacle problem).
As far as we know, such a theory has not been
developed yet, not even for minimal solutions
(see, however,~\cite{Capogna, Zhong1, Zhong2}
for the case of homogeneous
equations).

The result we prove here is:

\begin{theorem}\label{1}
If~$p\in {\mathcal{P}}(\psi,\Omega)$ then
\begin{equation}\label{Tr}
\begin{split}
&0\le
\div_{\H^n}\, \Big( (\varepsilon+|\nabla_{\H^n}\bar u_\varepsilon|^2)^{(p/2)-1}
\nabla_{\H^n}\bar u_\varepsilon\Big)
\\ &\qquad\le \left(
\div_{\H^n}\, \Big( (\varepsilon+|\nabla_{\H^n}\psi|^2)^{(p/2)-1}
\nabla_{\H^n}\psi\Big)
\right)^+.\end{split}\end{equation}
\end{theorem}

The result in Theorem~\ref{1} is quite intuitive:
when~$\bar u_\varepsilon$ does not touch the obstacle, it is free
to make the operator vanish. When it touches and sticks to it,
the operator is driven by the one of the obstacle -- and
on these touching points the obstacle has to
bend in a somewhat convex fashion, which justifies the
first inequality in~\eqref{Tr} and
superscript~``$+$'' in the right hand side of~\eqref{Tr}.

Figure~1, in which
the thick curve represents the touching between~$\bar u_\varepsilon$
and the obstacle,
tries to describe this phenomena.
On the other hand, the set in which~$\bar u_\varepsilon$ touches
the obstacle may be very wild, so the actual proof
of Theorem~\ref{1} needs to me more technical than this.

In fact, the first inequality of~\eqref{Tr} is quite
obvious since it follows, for instance, by taking~$v:=
\bar u_\varepsilon-\varphi$
in~\eqref{ines}, with an arbitrary~$\varphi\in
C^\infty_0(\Omega,[0,+\infty))$), so the
core of~\eqref{Tr} lies on the second
inequality: nevertheless, we think it
is useful to write~\eqref{Tr} in this way to emphasize
a control from both the sides of the operator applied
to the solution.

We remark that the right hand side of~\eqref{Tr}
is always finite when~$\varepsilon>0$, and when~$\varepsilon=0$
and~$p\ge2$. In this case,~\eqref{Tr}
is an~$L^\infty$-bound and may be seen as
a regularity result for the solution
of the obstacle problem. It is worth
noticing that such regularity result holds
for~$\varepsilon=0$ as well, only
assuming that~$p\in
{\mathcal{P}}(\psi,\Omega)$,
which is a requirement on the problem when~$\varepsilon>0$.

On the other hand, if~$\varepsilon=0$ and~$p\in(1,2)$,
the right hand side of~\eqref{Tr}
may become infinite (in this
case~\eqref{Tr}
is true, but meaningless, stating that something is
less than or equal to an infinite quantity).

\begin{figure}[htbp]
\begin{center}
\resizebox{10cm}{!}{\input{obstacle.pstex_t}}
{\caption{\it Touching the obstacle}}
\label{FIG 1}
\end{center}
\end{figure}

In the Euclidean setting, the analogue
of~\eqref{Tr} was first obtained in~\cite{LewySt}
for the Laplacian case, and it is therefore often referred
to with the name of
Lewy-Stampacchia Estimate. It is also called Dual
Estimate, for it is, in a sense, obtained
by the duality expressed by the variational inequality~\eqref{ines}.
Other Authors refer to it
with the name of
Penalization Method, for the role played by~$\varepsilon$.

After~\cite{LewySt}, estimates of these type
became very popular and underwent
many important extensions
and strengthenings: see, among the others,~\cite{MoscoTro,
MoscoFre, DonatiF82, BocGallouet90, Mok}.
As far as we know, the estimate we prove is new in
the Heisenberg group setting, even for~$p=2$.

Hereafter, we deal with the proof of
Theorem~\ref{1}.
First, in~\S~\ref{S1}, we prove
Theorem~\ref{1} when~$\varepsilon>0$.

The proof when~$\varepsilon=0$ is contained
in~\S~\ref{S3} and it is based on a limit argument,
i.e., we consider the problem with~$\varepsilon>0$,
we use Theorem~\ref{1} in such a case, and then we
pass~$\varepsilon
\searrow0$. This procedure is quite delicate though,
because, as far as we know, it is not clear whether
or not the
Heisenberg group setting allows a complete H\"older regularity
theory for first derivatives (see~\cite{Dom}).
To get around this point, in~\S~\ref{S2},
we study the $L^p$-convergence of the solution~$\bar u_\varepsilon$
of the
$\varepsilon$-problem
to the solution~${\bar{u}_0}$ of the problem with~$\varepsilon=0$,
which, we believe, is of independent interest
(see, in particular Propositions~\ref{Lemma 2.3}
and~\ref{Lemma 1.23 p+}).

The paper ends with an Appendix that collects some
ancillary results needed in~\S~\ref{S2}.

\section{Proof of Theorem~\ref{1} when $\varepsilon>0$}\label{S1}

We prove~\eqref{Tr} in the simpler case~$\varepsilon>0$
(the case~$\varepsilon=0$ will be dealt with in~\S~\ref{S3}).
The technique used in this proof is a variation
of a classical penalized test function method
(see, e.g.,~\cite{MoscoTro, MoscoFre, DonatiF82, BocGallouet90,
Mok}
and references therein), and several steps
of this proof are inspired by
some estimates obtained by~\cite{Challal} in the Euclidean
case.

First of all, we set
$$ \mu := -1+\min \big\{ \inf_{\overline\Omega}\psi,
\,\inf_{\overline\Omega} u_\star\big\}\in\R$$
and we observe that
\begin{equation}\label{L1}
\bar u_\varepsilon \ge \mu
\end{equation}
a.e. in $\Omega$.
Indeed, let~$w:=\max\{ \bar u_\varepsilon, \mu\}$.
Since~$\psi$ and~$u_\star$ are below~$\mu$ in~$\Omega$,
we have that~$w\in {\mathcal{K}}$, thus
$$ 0\le {\mathcal{F}}_\varepsilon (w;\Omega)-
{\mathcal{F}}_\varepsilon (\bar u_\varepsilon;\Omega)
=-\int_{\Omega\cap \{ \bar u_\varepsilon<\mu\}
}(\varepsilon+|\nabla_{\H^n}\bar u_\varepsilon|^2)^{p/2}\le 0
,$$
and, from this,~\eqref{L1} plainly follows.

Now, let $\eta\in(0,1)$, to be taken arbitrarily small
in the sequel. Let also
\begin{equation}\label{hh}
h:=\left(
\div_{\H^n}\, \Big( (\varepsilon+|\nabla_{\H^n}\psi|^2)^{(p/2)-1}
\nabla_{\H^n}\psi\Big)\right)^+.
\end{equation}
Notice that
\begin{equation}\label{09890}
\| h\|_{L^\infty(\Omega)}<+\infty
,\end{equation}
because~$\varepsilon>0$.
For any $t\in \R$, we consider the truncation function
$$ H_\eta(t):=\begin{cases}
0 & {\mbox{ if $t\le 0$,}} \\
t/\eta & {\mbox{ if $0< t<\eta$,}} \\
1& {\mbox{ if $t\ge\eta$.}}
\end{cases}$$
Now, we take $u_\eta$ to be a weak solution
of
\begin{equation}\label{2..8} \begin{cases}
\div_{\H^n}\, \Big( (\varepsilon+|\nabla_{\H^n}
u_\eta|^2)^{(p/2)-1} \nabla_{\H^n}u_\eta\Big) =
h \cdot \big(1-H_\eta(\psi-u_\eta)\big) & {\mbox{ in $\Omega$,}}\\
u_\eta=\bar u_\varepsilon & {\mbox{ on $\partial \Omega$.}}
\end{cases}\end{equation}
where, as usual, the boundary datum is attained in the trace sense:
such a function $u_\eta$ may be obtained by the direct
method in the calculus of variations, by minimizing
the functional
$$ \int_\Omega\frac{1}{p}
(\varepsilon+|\nabla_{\H^n}
u(\xi)|^2)^{p/2}+F_\eta(u(\xi),\xi)\,d\xi$$
over~$u\in W^{1,p}_{\H^n}(\Omega)$, $u-\bar u_\varepsilon\in
W^{1,p}_{\H^n,0}(\Omega)$,
where
$$ F_\eta(r,\xi):=\int_0^r h(\xi) \cdot \big(1-H_\eta(\psi(\xi)-
\sigma)\big)
\,d\sigma.$$
Now, we claim that
\begin{equation}\label{A01}
u_\eta\le\psi {\mbox{ a.e. in $\Omega$}}.
\end{equation}
To establish this, we use the test function $(u_\eta-\psi)^+$
in~\eqref{2..8}.
Since, on $\partial \Omega$, we have $(u_\eta-\psi)^+=
(\bar u_\varepsilon-\psi)^+=0$, we obtain that
\begin{eqnarray*}
&& -\int_{\Omega}
\Big( (\varepsilon+|\nabla_{\H^n}
u_\eta|^2)^{(p/2)-1} \nabla_{\H^n}u_\eta\Big) \cdot \nabla_{\H^n}
(u_\eta-\psi)^+
\\ &=& \int_{\Omega}
h \cdot \big(1-H_\eta(\psi-u_\eta)\big)
(u_\eta-\psi)^+
= \int_{\Omega}
h \cdot (u_\eta-\psi)^+.
\end{eqnarray*}
Consequently, by~\eqref{hh},
\begin{eqnarray*}
&& \int_{\Omega}
\left[ \Big( (\varepsilon+|\nabla_{\H^n}
u_\eta|^2)^{(p/2)-1} \nabla_{\H^n}u_\eta\Big) \right.
\\ && \qquad \left.-
\Big( (\varepsilon+|\nabla_{\H^n}
\psi|^2)^{(p/2)-1} \nabla_{\H^n}\psi\Big)
\right]\cdot \nabla_{\H^n}
(u_\eta-\psi)^+
\\ &=&
\int_{\Omega} \left[
\div_{\H^n}\, \Big( (\varepsilon+|\nabla_{\H^n}
\psi|^2)^{(p/2)-1} \nabla_{\H^n}\psi\Big) -h
\right]\cdot (u_\eta-\psi)^+
\\ &\le& 0.
\end{eqnarray*}
By the strict monotonicity of the operator (i.e., by the strict convexity
of the function~$\R^{2n}\ni \zeta\mapsto
(\varepsilon+|\zeta|^2)^{p/2}$),
it follows that
$(u_\eta-\psi)^+$ vanishes almost everywhere in $\Omega$,
proving \eqref{A01}.

Now, we claim that
\begin{equation}\label{A02}
\bar u_\varepsilon\ge u_\eta
{\mbox{ a.e. in $\Omega$}}.
\end{equation}
To verify this, we consider the test function
$$ \tau:= \bar u_\varepsilon+(u_\eta-\bar u_\varepsilon)^+ .$$
We notice that
$$\tau=\begin{cases}
\bar u_\varepsilon & {\mbox{ in }} \{ u_\eta\le \bar u_\varepsilon\}, \\
u_\eta & {\mbox{ in }} \{ u_\eta> \bar u_\varepsilon\},
\end{cases}$$
hence $\tau
\le \psi$, due to \eqref{A01}.
Furthermore, on $\partial \Omega$, we have that
$\tau=
\bar u_\varepsilon$, due to the boundary datum
in~\eqref{2..8}. Therefore
the obstacle problem variational inequality~\eqref{ines}
gives that
\begin{equation}\label{2.10}
\begin{split}
& 0 \le
\int_{\Omega}
\Big( (\varepsilon+|\nabla_{\H^n}
\bar u_\varepsilon|^2)^{(p/2)-1} \nabla_{\H^n}
\bar u_\varepsilon\Big) \cdot \nabla_{\H^n}
(\tau-\bar u_\varepsilon)
\\ &\qquad=
\int_{\Omega}
\Big( (\varepsilon+|\nabla_{\H^n}
\bar u_\varepsilon
|^2)^{(p/2)-1} \nabla_{\H^n}\bar u_\varepsilon\Big) \cdot \nabla_{\H^n}
(u_\eta-\bar u_\varepsilon)^+.
\end{split}\end{equation}
On the other hand, from~\eqref{2..8},
\begin{equation}\label{2.11}
\begin{split}
& \int_{\Omega} \Big( (\varepsilon+|\nabla_{\H^n}
u_\eta|^2)^{(p/2)-1} \nabla_{\H^n}u_\eta\Big)\cdot
\nabla_{\H^n} (u_\eta-\bar u_\varepsilon)^+
\\ &\qquad= -\int_{\Omega}h\cdot
\big(1-H_\eta(\psi-u_\eta)\big)\cdot (u_\eta-\bar u_\varepsilon)^+
\le 0.
\end{split}
\end{equation}
By \eqref{2.10} and \eqref{2.11}, we obtain that
\begin{eqnarray*}
&& \int_{\Omega} \left[
\Big( (\varepsilon+|\nabla_{\H^n}
u_\eta|^2)^{(p/2)-1} \nabla_{\H^n}u_\eta\Big) \right. \\ &&\qquad
\left. - \Big( (\varepsilon+|\nabla_{\H^n}
\bar u_\varepsilon|^2)^{(p/2)-1} \nabla_{\H^n}\bar u_\varepsilon\Big)
\right]\cdot \nabla_{\H^n}
(u_\eta-\bar u_\varepsilon)^+
\le 0.
\end{eqnarray*}
This and the strict monotonicity of the operator implies that
$(u_\eta-\bar u_\varepsilon)^+$ vanishes almost everywhere in $\Omega$,
hence proving \eqref{A02}.

Now, we claim that
\begin{equation}
\label{A03}
\bar u_\varepsilon\le u_\eta+\eta {\mbox{ in $\Omega$}}.
\end{equation}
To do this, we set
$$ \theta:= \bar u_\varepsilon -(\bar u_\varepsilon-u_\eta-\eta)^+.$$
Notice that $\theta\le\bar u_\varepsilon\le \psi$, and also that,
on $\partial \Omega$, $\theta =\bar u_\varepsilon$. As a consequence,~\eqref{ines}
gives that
\begin{equation}\label{2.12}
\begin{split}
& 0 \le
\int_{\Omega}
\Big( (\varepsilon+|\nabla_{\H^n}
\bar u_\varepsilon|^2)^{(p/2)-1} \nabla_{\H^n}
\bar u_\varepsilon\Big) \cdot \nabla_{\H^n}
(\theta-\bar u_\varepsilon)
\\ &\qquad=
-\int_{\Omega}
\Big( (\varepsilon+|\nabla_{\H^n}
\bar u_\varepsilon|^2)^{(p/2)-1} \nabla_{\H^n}
\bar u_\varepsilon\Big) \cdot \nabla_{\H^n}
(\bar u_\varepsilon-u_\eta-\eta)^+.
\end{split}\end{equation}
On the other hand,~$(\bar u_\varepsilon-u_\eta-\eta)^+=0$
on~$\partial\Omega$, and
\begin{eqnarray*} \{ \bar u_\varepsilon-u_\eta -\eta >0\}
&\subseteq& \{ \psi-u_\eta >\eta\}\\
&\subseteq& \big\{
1-H_\eta(\psi-u_\eta)=0\big\},\end{eqnarray*}
and therefore, by~\eqref{2..8},
\begin{equation}\label{2.13}
\begin{split}
& \int_{\Omega}
\Big( (\varepsilon+|\nabla_{\H^n}
(u_\eta+\eta)|^2)^{(p/2)-1} \nabla_{\H^n}
(u_\eta+\eta) \Big) \cdot \nabla_{\H^n}
(\bar u_\varepsilon-u_\eta-\eta)^+
\\=& \int_{\Omega}
\Big( (\varepsilon+|\nabla_{\H^n}
u_\eta|^2)^{(p/2)-1} \nabla_{\H^n}
u_\eta \Big) \cdot \nabla_{\H^n}
(\bar u_\varepsilon-u_\eta -\eta)^+
\\ = &
-\int_{\Omega}
h \cdot \big(1-H_\eta(\psi-u_\eta)\big)
\cdot (\bar u_\varepsilon-u_\eta -\eta)^+ =0.
\end{split}\end{equation}
Then, \eqref{2.12} and \eqref{2.13} yield that
\begin{eqnarray*}
&& \int_{\Omega} \left[
\Big( (\varepsilon+|\nabla_{\H^n}
\bar u_\varepsilon|^2)^{(p/2)-1} \nabla_{\H^n}
\bar u_\varepsilon\Big) \right.
\\ &&\qquad-\left.
\Big( (\varepsilon+|\nabla_{\H^n}
(u_\eta+\eta)|^2)^{(p/2)-1} \nabla_{\H^n}
(u_\eta+\eta) \Big)
\right]\cdot \nabla_{\H^n}
(\bar u_\varepsilon-u_\eta -\eta)^+ \\ &&\qquad\qquad\le 0.
\end{eqnarray*}
Thus, in this case, the strict monotonicity of the operator
implies that $(\bar u_\varepsilon-u_\eta-\eta)^+$ vanishes
almost everywhere in $\Omega$, and so \eqref{A03}
is established.

In particular, by~\eqref{A01},
\eqref{A03} and~\eqref{L1},
\begin{equation}\label{L2} \|u_\eta\|_{L^\infty(\Omega)}
\le 2+\|\psi\|_{L^\infty(\Omega)}
+ \|u_\star\|_{L^\infty(\Omega)}.\end{equation}
Moreover,
by \eqref{A02} and \eqref{A03},
we have that
\begin{equation}\label{781}{\mbox{$u_\eta$
converges uniformly in $\Omega$
to $\bar u_\varepsilon$}}\end{equation} as $\eta\searrow0$.

Furthermore
\begin{equation*}0 \le
\partial_r F_\eta(r,\xi)\le
h(\xi)=\left(
\div_{\H^n}\, \Big( (\varepsilon+|\nabla_{\H^n}\psi|^2)^{(p/2)-1}
\nabla_{\H^n} \psi\Big)\right)^+
\end{equation*}
hence~\eqref{Tr} follows\footnote{It
is worth pointing out that this
is the only place in the paper where we use
the condition that~$p\in {\mathcal{P}}(\psi,\Omega)$.
In particular, all the estimates in~\S~\ref{S2}
are valid for all~$p\in(1,+\infty)$.}
from~\eqref{781}
and the fact that~$p\in{\mathcal{P}}(\psi,\Omega)$
(recall~\eqref{TX} in Definition~\ref{D}).~\qed

\section{Estimating the $L^p$-distance between
$\nabla_{\H^n}{\bar{u}_0}$ and
$\nabla_{\H^n}\bar u_\varepsilon$}\label{S2}

The purpose of this section
is to
consider
the solution~$\bar u_\varepsilon$
of the
$\varepsilon$-problem
and the solution~${\bar{u}_0}$ of the problem with~$\varepsilon=0$,
and to bound the~$L^p$-norm of $
|\nabla_{\H^n} {\bar{u}_0}-\nabla_{\H^n}\bar u_\varepsilon|$.
Such estimate is quite technical and it is different according
to the cases $p\in(1,2]$
and $p\in[2,+\infty)$: see the forthcoming Propositions~\ref{Lemma 2.3}
and~\ref{Lemma 1.23 p+}.

As a matter of fact, we think that the estimates proved
in Propositions~\ref{Lemma 2.3}
and~\ref{Lemma 1.23 p+} are of independent interest, since
they also allow to get around the more difficult (and in general
not available in the Heisenberg group)
H\"older-type estimates.

We recall the standard notation of balls in
the Heisenberg group (in fact, we deal with the so called
Folland-Kor\'any balls, but the
Carnot-Carath\'eodory balls would be good for our purposes too).
For all~$\xi:=(z,t)\in\R^{2n}\times\R$, we define
$$\|\xi\|_{\H^n}:=\sqrt[4]{|z|^4+t^2}.$$
Then, for any~$r>0$, we set
$$ \B_r:= \big\{ \xi\in \R^{2n+1} {\mbox{ s.t. }}
\|\xi\|_{\H^n}<r\big\}.$$
We denote by~${\mathcal{L}}$
the~$(2n+1)$-dimensional
Lebesgue measure, and we observe that~${\mathcal{L}}(\B_r)$
equals, up to a multiplicative constant, to~$r^Q$,
where~$Q:=2(n+1)$ is the homogeneous dimension of $\H^n$.
Also, for all~$g\in L^1(\B_r)$, we define the average of~$g$
in~$\B_r$ as
$$ (g)_r:= \frac{1}{ {\mathcal{L}} (\B_r)} \int_{\B_r} g.$$
In what follows, we focus on~$L^p$-estimates
around a fixed point, say~$\xi_\star$, of~$\Omega$.
Without loss of generality, we take~$\xi_\star$ to be the
origin, and we fix~$R\in(0,1)$ so small that~$\B_R
\Subset\Omega$.

Then, we denote by~${\bar{u}_0}:\Omega\rightarrow\R$ the minimizer
of problem~\eqref{PB} with~$\varepsilon=0$.
Then, for a fixed~$\varepsilon>0$, we take~$\bar
u_\varepsilon:\B_R\rightarrow\R$
to be the minimizer of~${\mathcal{F}}_\varepsilon (u;\B_R)$
among all the functions~$u
\in W^{1,p}_{\H^n}(\B_R)$, $u\le \psi$, and~$u-{\bar{u}_0}
\in W^{1,p}_{\H^n,0}(\B_R)$. We can then extend~$\bar u_\varepsilon$
also on~$\Omega\setminus\B_R$ by setting it equal to~${\bar{u}_0}$
in such a set.
By construction
\begin{equation}\label{ar}
\begin{split}
&\int_{\B_R} |\nabla_{\H^n} {\bar{u}_0}|^p
={\mathcal{F}}_0( {\bar{u}_0};\Omega)-
\int_{\Omega\setminus\B_R} |\nabla_{\H^n} {\bar{u}_0}|^p\\ &
\quad\le {\mathcal{F}}_0( \bar u_\varepsilon ;\Omega)-
\int_{\Omega\setminus\B_R} |\nabla_{\H^n} {\bar{u}_0}|^p
=\int_{\B_R} |\nabla_{\H^n} \bar u_\varepsilon|^p
\end{split}
\end{equation}
and
\begin{equation}\label{po}
\begin{split}&
\int_{\B_R} (\varepsilon+ |\nabla_{\H^n}\bar u_\varepsilon|^2)^{p/2}
= {\mathcal{F}}_\varepsilon (\bar u_\varepsilon;\B_R)
\\&\qquad
\le {\mathcal{F}}_\varepsilon ( {\bar{u}_0};\B_R)=
\int_{\B_R} (\varepsilon+ |\nabla_{\H^n} {\bar{u}_0}|^2)^{p/2}
.\end{split}\end{equation}

\begin{proposition}\label{Lemma 2.3}
Assume that
\begin{equation}\label{p}
p\in(1,2].
\end{equation}
Then,
there exists~$C>0$, only depending on~$n$ and~$p$, such that
\begin{equation}\label{15bis}
\int_{\B_R} |\nabla_{\H^n} {\bar{u}_0}-
\nabla_{\H^n}\bar u_\varepsilon|^p \le C\big( 1+ (|\nabla_{\H^n}
{\bar{u}_0}|^p)_R \big)^{1-(p/2)}\varepsilon^{(p/2)^2}
R^Q.\end{equation}
\end{proposition}

\begin{proof} The technique for this proof
is inspired by the one of Lemma~2.3
of~\cite{Mu}, where a similar result was obtained
in the quasilinear Euclidean case (however, our
proof is self-contained).
We have
\begin{equation}\label{po2}
\begin{split}
& |\nabla_{\H^n}\bar u_\varepsilon-\nabla_{\H^n} {\bar{u}_0}|^2\le
\big( |\nabla_{\H^n}\bar u_\varepsilon|+|\nabla_{\H^n} {\bar{u}_0}|\big)^2
\\ &\qquad\le C \big( |\nabla_{\H^n}\bar u_\varepsilon|^2 +|\nabla_{\H^n} {\bar{u}_0}|^2\big)
.\end{split}\end{equation} Here,~$C$ is a positive constant, which
is free to be different from line to line.
By~\eqref{p},
\eqref{po} and~\eqref{po2}, we obtain
\begin{equation}\label{po3}
\begin{split}
& \int_{\B_R} (\varepsilon+ |\nabla_{\H^n} {\bar{u}_0}|^2+
|\nabla_{\H^n}\bar u_\varepsilon|^2)^{(p/2)-1}|\nabla_{\H^n}\bar u_\varepsilon-\nabla_{\H^n} {\bar{u}_0}|^2
\\ \le \,&
C \int_{\B_R} \frac{
|\nabla_{\H^n}\bar u_\varepsilon|^2 +|\nabla_{\H^n} {\bar{u}_0}|^2
}{(\varepsilon+ |\nabla_{\H^n} {\bar{u}_0}|^2+
|\nabla_{\H^n}\bar u_\varepsilon|^2)^{1-(p/2)}}
\\ =\,&
C \left( \int_{\B_R} \frac{
|\nabla_{\H^n}\bar u_\varepsilon|^2
}{(\varepsilon+ |\nabla_{\H^n} {\bar{u}_0}|^2+
|\nabla_{\H^n}\bar u_\varepsilon|^2)^{1-(p/2)}}
\right.\\ &\qquad+\left.
\int_{\B_R} \frac{
|\nabla_{\H^n} {\bar{u}_0}|^2
}{(\varepsilon+ |\nabla_{\H^n} {\bar{u}_0}|^2+
|\nabla_{\H^n}\bar u_\varepsilon|^2)^{1-(p/2)}}\right)
\\ \le\, &
C \left( \int_{\B_R} \frac{
|\nabla_{\H^n}\bar u_\varepsilon|^2
}{(\varepsilon+
|\nabla_{\H^n}\bar u_\varepsilon|^2)^{1-(p/2)}}
+
\int_{\B_R} \frac{
|\nabla_{\H^n} {\bar{u}_0}|^2
}{(\varepsilon+ |\nabla_{\H^n} {\bar{u}_0}|^2)^{1-(p/2)}}\right)
\\ \le\,&
C\left(
\int_{\B_R} (\varepsilon+|\nabla_{\H^n}\bar u_\varepsilon|^2)^{p/2}
+
\int_{\B_R} (\varepsilon+|\nabla_{\H^n} {\bar{u}_0}|^2)^{p/2}\right)
\\ \le\,& C
\int_{\B_R} (\varepsilon+|\nabla_{\H^n} {\bar{u}_0}|^2)^{p/2}.
\end{split}
\end{equation}
Thus, \eqref{po3} and Lemma~\ref{from 10 to 11},
applied here with~$a:=\nabla_{\H^n}{\bar{u}_0}$ and~$b:=\nabla_{\H^n}\bar u_\varepsilon$,
yield that
\begin{equation}\label{11}
\begin{split}
& \int_{\B_R} (\varepsilon+ |\nabla_{\H^n} {\bar{u}_0}|^2+
|\nabla_{\H^n}\bar u_\varepsilon|^2)^{p/2}
\\ &\qquad\le
C\int_{\B_R}
(\varepsilon+ |\nabla_{\H^n} {\bar{u}_0}|^2+
|\nabla_{\H^n}\bar u_\varepsilon|^2)^{(p/2)-1}
|\nabla_{\H^n}\bar u_\varepsilon-\nabla_{\H^n}{\bar{u}_0}|^2+
\\ &\qquad\qquad+
C\int_{\B_R}(\varepsilon+ |\nabla_{\H^n} {\bar{u}_0}|^2)^{(p/2)}
\\ &\qquad\le
C \int_{\B_R} (\varepsilon+ |\nabla_{\H^n} {\bar{u}_0}|^2)^{(p/2)}.
\end{split}
\end{equation}
Now, from~\eqref{ar},
\begin{equation}\label{5}
\begin{split}
& \int_{\B_R} (\varepsilon+ |\nabla_{\H^n} {\bar{u}_0}|^2)^{(p/2)}
- \int_{\B_R} (\varepsilon+ |\nabla_{\H^n} \bar u_\varepsilon|^2)^{(p/2)}
\\ &\qquad\le
\int_{\B_R} (\varepsilon+ |\nabla_{\H^n} {\bar{u}_0}|^2)^{(p/2)}
- \int_{\B_R} |\nabla_{\H^n} \bar u_\varepsilon|^p
\\ &\qquad\le
\int_{\B_R} (\varepsilon+ |\nabla_{\H^n} {\bar{u}_0}|^2)^{(p/2)}
- \int_{\B_R} |\nabla_{\H^n} {\bar{u}_0}|^p.
\end{split}\end{equation}
Moreover, using~\eqref{p} and some elementary calculus,
we see that
$$ |(1+\tau)^{p/2}-\tau^{p/2}|\le C$$
for any~$\tau\ge 0$. Therefore, taking~$\tau:=\theta/\varepsilon$,
we obtain that
\begin{equation}\label{7bis0}
|(\varepsilon+\theta)^{p/2}-\theta^{p/2}|\le C\varepsilon^{p/2}
\end{equation}
for any~$\theta\ge 0$. Thus, using~\eqref{5} and~\eqref{7bis0}
with~$\theta:= |\nabla_{\H^n}{\bar{u}_0}|^2$,
we conclude that
\begin{equation}\label{6}
\int_{\B_R} (\varepsilon+ |\nabla_{\H^n} {\bar{u}_0}|^2)^{(p/2)}
- \int_{\B_R} (\varepsilon+ |\nabla_{\H^n} \bar u_\varepsilon|^2)^{(p/2)}
\le C \varepsilon^{p/2} R^Q.\end{equation}
Now, we estimate the left hand side of~\eqref{6} from below.
For this scope, we define
\begin{eqnarray*}
&& h := t \nabla_{\H^n}{\bar{u}_0}+(1-t)\nabla_{\H^n}\bar u_\varepsilon,\\
&& J:=
p\int_{\B_R} (\varepsilon+ |\nabla_{\H^n} \bar u_\varepsilon|^2)^{(p/2)-1}
\nabla_{\H^n} \bar u_\varepsilon\cdot (\nabla_{\H^n} {\bar{u}_0}-\nabla_{\H^n} \bar u_\varepsilon)
\\ {\mbox{and }} &&
\tilde J :=p \int_{\B_R}\Big[
\int_0^1(1-t)\frac{d}{dt}\Big( (\varepsilon+|h|^2)^{(p/2)-1}
h\cdot ( \nabla_{\H^n}{\bar{u}_0}-\nabla_{\H^n}\bar u_\varepsilon)
\Big)\,dt\Big].
\end{eqnarray*}
We observe that the variational inequality in~\eqref{ines}
for~$\bar u_\varepsilon$ gives that
\begin{equation}\label{J9}
J\ge 0.\end{equation}
Also, using
the Fundamental Theorem of Calculus, we
obtain
\begin{equation*}
\begin{split}
& \int_{\B_R} (\varepsilon+ |\nabla_{\H^n} {\bar{u}_0}|^2)^{(p/2)}
- \int_{\B_R} (\varepsilon+ |\nabla_{\H^n} \bar u_\varepsilon|^2)^{(p/2)}
\\ &\qquad= \int_{\B_R} \Big[ \int_0^1
\frac{d}{dt} (\varepsilon+ |t\nabla_{\H^n} {\bar{u}_0}+(1-t) \nabla_{\H^n}
\bar u_\varepsilon|^2)^{(p/2)}\,dt\Big]
\\ & \qquad= p \int_{\B_R} \Big[ \int_0^1
(\varepsilon+ |t\nabla_{\H^n} {\bar{u}_0}+(1-t) \nabla_{\H^n}
\bar u_\varepsilon|^2)^{(p/2)-1}
\\ &\qquad\qquad\times (t\nabla_{\H^n} {\bar{u}_0}+(1-t) \nabla_{\H^n}
\bar u_\varepsilon)\cdot(\nabla_{\H^n} {\bar{u}_0}-\nabla_{\H^n}
\bar u_\varepsilon)\,dt\Big]
\\ & \qquad=
p\int_{\B_R} \Big[ \int_0^1
(\varepsilon+ |h|^2)^{(p/2)-1}
h\cdot(\nabla_{\H^n} {\bar{u}_0}-\nabla_{\H^n}
\bar u_\varepsilon)\,dt\Big] .
\end{split}
\end{equation*}
Integrating by parts the
latter integral
in~$t$ (by writing~$dt=\frac{d}{dt} (t-1)\,dt$),
and exploiting~\eqref{J9},
we obtain
\begin{equation}\label{7}
\begin{split}
& \int_{\B_R} (\varepsilon+ |\nabla_{\H^n} {\bar{u}_0}|^2)^{(p/2)}
- \int_{\B_R} (\varepsilon+ |\nabla_{\H^n} \bar u_\varepsilon|^2)^{(p/2)}
\\ &\qquad= J+\tilde J\ge \tilde J.
\end{split}
\end{equation}
Making use of Lemma~\ref{sim-h} -- applied here
with~$a:=\nabla_{\H^n}{\bar{u}_0}$
and~$b:= \nabla_{\H^n}\bar u_\varepsilon$ --
we have that
\begin{equation*}
\tilde J\ge \frac1C\int_{\B_R}\Big[\int_0^1
(1-t) (\varepsilon+|t\nabla_{\H^n}{\bar{u}_0}+(1-t)\nabla_{\H^n} \bar u_\varepsilon|^2)^{(p/2)-1}
|\nabla_{\H^n}{\bar{u}_0}-\nabla_{\H^n}\bar u_\varepsilon|^2
\,dt\Big].
\end{equation*}
{F}rom this and Lemma~\ref{AT679} -- applied
here with~$\kappa:=1$
and~$\Psi(x):=x^{1-(p/2)}$, which is nondecreasing, thanks
to~\eqref{p} --
we deduce that
\begin{equation}\label{9}
\tilde J\ge \frac1C \int_{\B_R}
(\varepsilon+|\nabla_{\H^n}{\bar{u}_0}|^2+|\nabla_{\H^n} \bar u_\varepsilon|^2)^{(p/2)-1}
|\nabla_{\H^n}{\bar{u}_0}-\nabla_{\H^n}\bar u_\varepsilon|^2.
\end{equation}
In view of~\eqref{6}, \eqref{7}
and~\eqref{9}, we conclude that
\begin{equation}\label{10}
\int_{\B_R}
(\varepsilon+|\nabla_{\H^n}{\bar{u}_0}|^2+|\nabla_{\H^n} \bar u_\varepsilon|^2)^{(p/2)-1}
|\nabla_{\H^n}{\bar{u}_0}-\nabla_{\H^n}\bar u_\varepsilon|^2 \le C \varepsilon^{p/2} R^Q.
\end{equation}
Then,~\eqref{15bis} follows from~\eqref{11}, \eqref{10} and
Lemma~\ref{19pL}, applied here with~$f:=
\nabla_{\H^n}{\bar{u}_0}$ and~$g:=\nabla_{\H^n}\bar u_\varepsilon$.
\end{proof}

In the degenerate case $p\in[2,+\infty)$
the estimate obtained in Proposition~\ref{Lemma 2.3}
for the singular case $p\in(1,2]$
needs to be modified according to the
following result:

\begin{proposition}\label{Lemma 1.23 p+}
Suppose that \begin{equation}\label{p+}
p\in[2,+\infty).\end{equation}
Then, there exists $C>0$,
only depending on $n$ and $p$, such that
$$ \int_{\B_R} |{\nabla_{\H^n}} {\bar{u}_0}-{\nabla_{\H^n}} \bar u_\varepsilon|^p\le C
\big( 1+
(|\nabla_{\H^n}{\bar{u}_0}|^p)_R \big)^{1-(1/p)}\varepsilon
R^Q.$$
\end{proposition}

\begin{proof} The variational inequalities~\eqref{ines}
for ${\bar{u}_0}$ and $\bar u_\varepsilon$ imply that
\begin{eqnarray*}
&& \int_{\B_R} |{\nabla_{\H^n}} {\bar{u}_0}|^{p-2}{\nabla_{\H^n}} {\bar{u}_0}
\cdot ({\nabla_{\H^n}}\bar u_\varepsilon-{\nabla_{\H^n}} {\bar{u}_0})\ge 0
\\ {\mbox{and }} &&
\int_{\B_R} (\varepsilon+
|{\nabla_{\H^n}} \bar u_\varepsilon|^2)^{(p/2)-1}{\nabla_{\H^n}} \bar u_\varepsilon
\cdot ({\nabla_{\H^n}} {\bar{u}_0}-{\nabla_{\H^n}} \bar u_\varepsilon)\ge 0.
\end{eqnarray*}
Consequently,
$$ \int_{\B_R} \Big( |{\nabla_{\H^n}} {\bar{u}_0}|^{p-2}{\nabla_{\H^n}} {\bar{u}_0}
-(\varepsilon+
|{\nabla_{\H^n}} \bar u_\varepsilon|^2)^{(p/2)-1}{\nabla_{\H^n}} \bar u_\varepsilon
\Big)\cdot ({\nabla_{\H^n}} {\bar{u}_0}-{\nabla_{\H^n}} \bar u_\varepsilon)\le 0.$$
Using this and \eqref{Fu1} of Lemma~\ref{FFU}, applied here
with $A:={\nabla_{\H^n}} {\bar{u}_0}$
and $B:={\nabla_{\H^n}} \bar u_\varepsilon$, we obtain
\begin{gather*}
\int_{\B_R} |{\nabla_{\H^n}} {\bar{u}_0}-{\nabla_{\H^n}}
\bar u_\varepsilon|^p\\ \le C
\int_{\B_R} \Big( |{\nabla_{\H^n}}
{\bar{u}_0}|^{p-2}{\nabla_{\H^n}} {\bar{u}_0}
-|{\nabla_{\H^n}}\bar u_\varepsilon|^{p-2}{\nabla_{\H^n}} \bar u_\varepsilon
\Big)\cdot ({\nabla_{\H^n}} {\bar{u}_0}-{\nabla_{\H^n}} \bar u_\varepsilon)
\\ \le C
\int_{\B_R} \Big( (\varepsilon+
|{\nabla_{\H^n}} \bar u_\varepsilon|^2)^{(p/2)-1}
{\nabla_{\H^n}} \bar u_\varepsilon
-  |{\nabla_{\H^n}} \bar u_\varepsilon|^{p-2}
{\nabla_{\H^n}} \bar u_\varepsilon
\Big)\cdot ({\nabla_{\H^n}} {\bar{u}_0}-{\nabla_{\H^n}} \bar u_\varepsilon)
.\end{gather*}
This and Corollary \ref{FuC}, applied here with
$a:={\nabla_{\H^n}}\bar u_\varepsilon$, give
\begin{gather*}
\int_{\B_R} |{\nabla_{\H^n}} {\bar{u}_0}-{\nabla_{\H^n}}
\bar u_\varepsilon|^p\\ \le C
\int_{\B_R} \Big( (\varepsilon+
|{\nabla_{\H^n}} \bar u_\varepsilon|^2)^{(p/2)-1}
-  |{\nabla_{\H^n}} {\bar u_\varepsilon}|^{p-2}
\Big)\,|{\nabla_{\H^n}} \bar u_\varepsilon| \, |{\nabla_{\H^n}} {\bar{u}_0}-
{\nabla_{\H^n}} \bar u_\varepsilon|
\\ \le C \varepsilon \int_{\B_R}
(\varepsilon+|{\nabla_{\H^n}}\bar u_\varepsilon|^2)^{(p-2)/2}
\big( |{\nabla_{\H^n}} {\bar{u}_0}|+|{\nabla_{\H^n}} \bar u_\varepsilon|\big).
\end{gather*}
Therefore, recalling \eqref{p+}, noticing that
$$ \frac{p-2}p+\frac1p+\frac1p=1$$
and using the Generalized H\"older Inequality with
the three exponents $p/(p-2)$, $p$ and $p$, we obtain
\begin{gather*}
\int_{\B_R} |{\nabla_{\H^n}} {\bar{u}_0}-{\nabla_{\H^n}} \bar u_\varepsilon|^p\\ \le
C \varepsilon \left( \int_{\B_R}
(\varepsilon+|{\nabla_{\H^n}}\bar u_\varepsilon|^2)^{p/2}\right)^{(p-2)/p}
\left( \int_{\B_R}
\big( |{\nabla_{\H^n}} {\bar{u}_0}|^p+|{\nabla_{\H^n}} \bar u_\varepsilon|^p\big)\right)^{1/p}
R^{Q/p}
.\end{gather*}
Then, by the minimal property of ${\bar{u}_0}$
in~\eqref{ar},
\begin{gather*}
\int_{\B_R} |{\nabla_{\H^n}} {\bar{u}_0}-{\nabla_{\H^n}} \bar u_\varepsilon|^p\\ \le
C \varepsilon \left( \int_{\B_R}
(\varepsilon+|{\nabla_{\H^n}}\bar u_\varepsilon|^2)^{p/2}\right)^{(p-2)/p}
\left( \int_{\B_R}
|{\nabla_{\H^n}} \bar u_\varepsilon|^p \right)^{1/p}
R^{Q/p}
\\ \le
C \varepsilon \left( \int_{\B_R}
(\varepsilon+|{\nabla_{\H^n}}\bar u_\varepsilon|^2)^{p/2}\right)^{(p-1)/p}
R^{Q/p}
\\ \le
C \varepsilon \left( R^Q +\int_{\B_R} |{\nabla_{\H^n}}\bar u_\varepsilon|^p\right)^{(p-1)/p}
R^{Q/p}
\\ \le
C \varepsilon \left( R^Q +\int_{\B_R} |{\nabla_{\H^n}}{\bar{u}_0}|^p
\right)^{(p-1)/p}
R^{Q/p},\end{gather*}
from which the desired result follows.
\end{proof}

\begin{corollary}
For all~$p\in(1,+\infty)$, we have that
\begin{equation}\label{C1}
\lim_{\varepsilon\searrow0}
\| {\nabla_{\H^n}}\bar u_\varepsilon -
{\nabla_{\H^n}}{\bar{u}_0}\|_{L^p(\B_R)}=0.
\end{equation}
Also, there exist a subsequence of~$\varepsilon$'s
and a function~$G\in L^p(\B_R)$ such
that
\begin{equation}\label{C2}
|{\nabla_{\H^n}}\bar u_\varepsilon (x)|\le G(x)
\end{equation}
for almost every~$x\in\B_R$.

Furthermore, if we set
\begin{equation}\label{C4}
\Gamma_\varepsilon:=
(\varepsilon+|\nabla_{\H^n}\bar u_\varepsilon|^2)^{(p/2)-1}
\nabla_{\H^n}\bar u_\varepsilon,\end{equation}
then there exist a subsequence of~$\varepsilon$'s
and a function~$G_\star \in L^1(\B_R)$
such that
\begin{equation}\label{C3}
|\Gamma_\varepsilon(x)|\le G_\star(x)
\end{equation}
for almost every~$x\in\B_R$.
\end{corollary}

\begin{proof} We obtain~\eqref{C1} from
Propositions~\ref{Lemma 2.3}
and~\ref{Lemma 1.23 p+}, according to whether~$p\in(1,2)$ or~$p\in
[2,+\infty)$.

{F}rom~\eqref{C1}, one deduces~\eqref{C2}
(see, e.g., Theorem~4.9(b) in~\cite{Brezis}).

Now, we define~$G_\star:= 2^{(p/2)}(G+G^{p-1})$.
We observe that~$G_\star\in L^1(\B_R)$, since~$G\in L^p(\B_R)
\subseteq L^1(\B_R)$ and~$G^{p-1} \in L^{p/(p-1)}(\B_R)
\subseteq L^1(\B_R)$ . So, in order to obtain the
desired result, we have only to show that the inequality
in~\eqref{C3} holds true.

For this, we notice that, if~$p\in(1,2)$,
\begin{eqnarray*}
&& |\Gamma_\varepsilon|=
\frac{|\nabla_{\H^n}\bar u_\varepsilon|}{
(\varepsilon+|\nabla_{\H^n}\bar u_\varepsilon|^2)^{1-(p/2)} }
= \frac{|\nabla_{\H^n}\bar u_\varepsilon|^{p-1}
|\nabla_{\H^n}\bar u_\varepsilon|^{2-p}}{
(\varepsilon+|\nabla_{\H^n}\bar u_\varepsilon|^2)^{1-(p/2)} }
\\ &&\quad\le
\frac{|\nabla_{\H^n}\bar u_\varepsilon|^{p-1}
(\varepsilon+|\nabla_{\H^n}\bar u_\varepsilon|^2)^{(2-p)/2}}{
(\varepsilon+|\nabla_{\H^n}\bar u_\varepsilon|^2)^{1-(p/2)} }
=|\nabla_{\H^n}\bar u_\varepsilon|^{p-1}
\le G^{p-1},
\end{eqnarray*}
which implies~\eqref{C3} in this case.

On the other hand, if~$p\in[2,+\infty)$,
\begin{eqnarray*}
&& |\Gamma_\varepsilon|\le 2^{(p/2)-1}
\big(\varepsilon^{(p/2)-1}
+|\nabla_{\H^n}\bar u_\varepsilon|^{p-2}
\big)
\,|\nabla_{\H^n}\bar u_\varepsilon|
\\ &&\quad\le 2^{(p/2)-1}
(1+G^{p-2}) G,
\end{eqnarray*}
which implies~\eqref{C3} in this case too.
\end{proof}

\section{Proof of Theorem~\ref{1} when~$\varepsilon=0$}\label{S3}

By Theorem~\ref{1} (for~$\varepsilon>0$,
which has been proved in~\S~\ref{S1}), we know
that, for a sequence~$\varepsilon\searrow0$,
\begin{equation}\label{C5}
0\le \int_{\B_R} \Gamma_\varepsilon\cdot\nabla\varphi
\le \int_{\B_R}\left(
\div_{\H^n}\,
\Big( (\varepsilon+|\nabla_{\H^n}\psi|^2)^{(p/2)-1}
\nabla_{\H^n}\psi\Big)
\right)^+ \varphi,
\end{equation}
for any~$\varphi\in C^\infty_0(\B_R,[0,+\infty))$, as long as~$\B_R
\subset\Omega$, where~$\Gamma_\varepsilon$ is as in~\eqref{C4}.

By possibly taking subsequences,
in the light of~\eqref{C1} and~\eqref{C3},
we have that
$$ \lim_{\varepsilon\searrow0}\Gamma_\varepsilon=
|\nabla_{\H^n}{\bar{u}_0}|^{p-2}
\nabla_{\H^n}{\bar{u}_0}$$
almost everywhere in~$\B_R$, and that~$\Gamma_\varepsilon$
is equidominated in~$L^1(\B_R)$. Consequently,
we can pass to the limit in~\eqref{C5} via the
Dominated Convergence Theorem and obtain~\eqref{Tr}
for~${\bar{u}_0}$.
This completes the proof of Theorem~\ref{1} also
when~$\varepsilon=0$.~\qed

\section*{Appendix}

In this appendix, we collect some technical estimates of
general interest that will be used in the proofs of
the main results of this paper.

We start with some classical estimates (see, e.g. Lemma~3
in~\cite{Fuchs} and references therein), which
turns out to be quite useful to deal with nonlinear
operators of degenerate type:

\begin{lemma}\label{FFU}
Let $M\in\N$, $M\ge 1$, and $p\in[2,+\infty)$. Then, there
exists $C>1$, only depending on $M$ and $p$, such that,
for any $A$, $B\in \R^M$,
\begin{equation}\label{Fu1}
|A-B|^p \le C
\Big( |A|^{p-2} A-|B|^{p-2}B\Big)
\cdot(A-B)
\end{equation}
and
\begin{equation}\label{Fu2}
\Big|  |A|^{p-2} A-|B|^{p-2}B\Big|\le C
|A-B|\,\Big( |A|^{p-2} +|B|^{p-2} \Big)
.\end{equation}
\end{lemma}

\begin{corollary}\label{FuC}
Let $N\in\N$ and
and $p\in[2,+\infty)$. Then, there
exists $C>1$, only depending on $N$ and $p$, such that
for any $\varepsilon>0$ and any $a\in\R^N$
$$\big( (\varepsilon+|a|^2)^{(p/2)-1}-|a|^{p-2}\big) \,|a|
\le C\varepsilon  (\varepsilon+|a|^2)^{(p-2)/2}.$$
\end{corollary}

\begin{proof} We let $A:=(a,\varepsilon)$ and $B:=(a,0)\in\R^{N+1}$
and we exploit \eqref{Fu2}. We obtain
\begin{equation*}\begin{split}
& 2C\varepsilon  (\varepsilon+|a|^2)^{(p-2)/2}
\\ &\qquad\ge
C \varepsilon \Big( (\varepsilon+|a|^2)^{(p-2)/2}+|a|^{p-2}\Big)
\\ &\qquad=
C |A-B|\,\Big( |A|^{p-2} +|B|^{p-2} \Big)
\\ &\qquad \ge
\Big|  |A|^{p-2} A-|B|^{p-2}B\Big|
\\ &\qquad=
\Big| (\varepsilon+|a|^2)^{(p-2)/2}(a,\varepsilon)-|a|^{p-2}(a,0)\Big|
\\ &\qquad=
\Big|  \Big(  \big( (\varepsilon+|a|^2)^{(p-2)/2}-|a|^{p-2}\big) a,\,
(\varepsilon+|a|^2)^{(p-2)/2} \varepsilon
\Big) \Big|\\ &\qquad
\ge
\big( (\varepsilon+|a|^2)^{(p-2)/2}-|a|^{p-2}\big) \,|a|,
\end{split}
\end{equation*}
as desired.
\end{proof}

In the subsequent Lemmata~\ref{sim-h} and~\ref{from 10 to 11},
we collect some simple, but interesting,
estimates that are
used in Proposition~\ref{Lemma 2.3}:

\begin{lemma}\label{sim-h}
Let~$N\in\N$, $N\ge 1$, $t\in[0,1]$, $\varepsilon>0$, and~$a$, $b\in\R^N$.
Let~$h(t) := t a+(1-t) b$.
Then, there exists~$C>1$, only depending on~$N$
and~$p$, such that
$$ \frac{d}{dt} \Big(  (\varepsilon+|h|^2)^{(p/2)-1} h\cdot
(a-b)\Big)\ge \frac{1}{C}
(\varepsilon+|ta+(1-t)b|^2)^{(p/2)-1}
|a-b|^2.$$
\end{lemma}

\begin{proof} We have
\begin{equation*}\begin{split}
& \frac{d}{dt} \Big(  (\varepsilon+|h|^2)^{(p/2)-1} h\cdot
(a-b)\Big)
=
\frac{d}{dt} \Big(  (\varepsilon+|h|^2)^{(p/2)-1} h \Big)\cdot
(a-b)
\\ &\qquad
= (\varepsilon+|h|^2)^{(p/2)-2} \big( \varepsilon+(p-1)|h|^2\big)
\frac{dh}{dt}\cdot (a-b)
\\ &\qquad
\ge \frac{1}{C} (\varepsilon+|h|^2)^{(p/2)-1}
|a-b|^2
\\ &\qquad=\frac{1}{C}
(\varepsilon+|ta+(1-t)b|^2)^{(p/2)-1}
|a-b|^2,\end{split}\end{equation*}
as desired.\end{proof}

\begin{lemma}\label{from 10 to 11}
Let
\begin{equation}\label{22p}
p\in(1,2].\end{equation}
Let~$N\in\N$, $N\ge 1$, $\varepsilon>0$, and~$a$, $b\in\R^N$.
Then, there exists~$C>1$, only depending on~$N$ and~$p$, such that
$$
(\varepsilon+ |a|^2+
|b|^2)^{p/2} \le C\Big[
(\varepsilon+ |a|^2+
|b|^2)^{(p/2)-1}
|b-a|^2
+(\varepsilon+ |a|^2)^{(p/2)} \Big].$$
\end{lemma}

\begin{proof}
We have
$$ |b|^2=
|b -a+a|^2
\le \big(
|b-a| +|a|
\big)^2
\le C \big( |b -a|^2
+|a|^2\big) $$
and so
\begin{eqnarray*}
&& (\varepsilon+ |a|^2+
|b|^2)^{p/2} \\ &=& (\varepsilon+ |a|^2+
|b|^2)^{(p/2)-1}
(\varepsilon+ |a|^2+
|b|^2)
\\ &\le& C (\varepsilon+ |a|^2+
|b|^2)^{(p/2)-1}
(\varepsilon+ |a|^2+
|b-a|^2)
\\ &=&
 C (\varepsilon+ |a|^2+
|b|^2)^{(p/2)-1}|b-a|^2
+ C (\varepsilon+ |a|^2+
|b|^2)^{(p/2)-1}
(\varepsilon+ |a|^2).
\end{eqnarray*}
Therefore, by~\eqref{22p},
\begin{equation*}\begin{split} &
(\varepsilon+ |a|^2+
|b|^2)^{p/2} \\ &\qquad\le C
(\varepsilon+ |a|^2+
|b|^2)^{(p/2)-1}
|b-a|^2+C(\varepsilon+ |a|^2)^{(p/2)},
\end{split}\end{equation*}
that is the desired claim.\end{proof}

The following result
deals with some technical
estimates on monotone integrands.

\begin{lemma}\label{AT679}
Let~$N\in\N$, $N\ge1$. Let~$\kappa\in \{0,1\}$.
Let~$\varepsilon$, $\varepsilon'>0$. Let~$a$, $b\in
\R^{N}$.
Let~$\Psi:[\varepsilon,+\infty)\rightarrow
[\varepsilon',+\infty)$ be a
measurable and nondecreasing function. Then
\begin{equation}\label{DES}
\int_0^1 \frac{(1-t)^\kappa}{\Psi
(\varepsilon+|ta+(1-t)b|^2)}\,dt
\ge \frac{1}{ 2 \Psi (\varepsilon+|a|^2+|b|^2)}.
\end{equation}
\end{lemma}

\begin{proof} If~$|a|\le |b|$, for any~$t\in[0,1]$,
\begin{eqnarray*} && |ta+(1-t)b|^2\le
t^2 |a|^2 +(1-t)^2 |b|^2+2t(1-t)|a||b|
\\ &&\qquad \le t^2 |b|^2 +(1+t^2-2t) |b|^2+2t(1-t)|b|^2= |b|^2.
\end{eqnarray*}
On the other hand, if~$|a|\ge |b|$, for any~$t\in[0,1]$,
\begin{eqnarray*} && |ta+(1-t)b|^2\le
t^2 |a|^2 +(1-t)^2 |b|^2+2t(1-t)|a||b|
\\ &&\qquad \le t^2 |a|^2 +(1+t^2-2t) |a|^2+2t(1-t)|a|^2= |a|^2.
\end{eqnarray*}
In any case,
$$ \varepsilon+|ta+(1-t)b|^2\le \varepsilon+|a|^2+|b|^2$$
and the claim follows from the monotonicity of~$\Psi$.
\end{proof}

The next is a useful H\"older/$L^p$ type estimate,
that is exploited
in Proposition~\ref{Lemma 2.3}.

\begin{lemma}\label{19pL}
Let~$N\in\N$, $N\ge 1$. Let~$f$, $g\in L^p (\B_R,\R^N)$.
Suppose that
\begin{equation}\label{p56}
p\in(1,2].\end{equation}
Then
\begin{equation*}\begin{split}
& \int_{\B_R}|f-g|^p
\\
&\qquad\le \left(\int_{\B_R}
(\varepsilon+|f|^2+|g|^2)^{(p/2)-1}
|f-g|^2\right)^{p/2}
\\ &\qquad\qquad\times \left(\int_{\B_R}
(\varepsilon+|f|^2+|g|^2)^{p/2}
\right)^{(2-p)/2}.
\end{split}\end{equation*}
\end{lemma}

\begin{proof}
We observe that
\begin{eqnarray*}
&& |f-g|^p\\ &
=& \Big[
(\varepsilon+|f|^2+|g|^2)^{(p/2)-1}
|f-g|^2\Big]^{p/2}
\Big[
(\varepsilon+|f|^2+|g|^2)^{p/2}
\Big]^{(2-p)/2},\end{eqnarray*}
and so the desired result follows
from the H\"older Inequality
with exponents~$2/p$ and~$2/(2-p)$, which can be used here
due to~\eqref{p56}.
\end{proof}

To end this paper, we remark that
Definition~\ref{D} is always nonvoid
(independently of~$\psi$ and~$\Omega$), in the
sense that

\begin{lemma}\label{p2P}
$ 2\in {\mathcal{P}}(\psi,\Omega)$.
\end{lemma}

\begin{proof} The functional in~\eqref{905}
when~$p=2$ boils down to
\begin{equation}\label{905a} \int_\Omega
\frac{1}{2} |\nabla_{\H^n}
u(\xi)|^2+F_k(u(\xi),\xi)\,d\xi,\end{equation}
up to an additive constant that does not play any
role in the minimization.
Hence,
if~$u_k$ minimizes this functional, we have that
$$-
\int_\Omega \nabla_{\H^n} u_k(\xi)\cdot\nabla_{\H^n}\varphi(\xi)
\,d\xi
=\int_\Omega \partial_r F_k(u_k(\xi),\xi)\,\varphi(\xi)\,d\xi$$
for any~$\varphi\in C^\infty_0(\Omega)$.

Accordingly, if also~$u_k$ approaches some~$u_\infty$
uniformly in~$\Omega$, it follows that
\begin{equation}\label{905bb}
\begin{split}
& \int_\Omega u_\infty \Delta_{\H^n}\varphi=
\lim_{k\rightarrow+\infty} \int_\Omega u_k \Delta_{\H^n}\varphi
\\ &\qquad= \lim_{k\rightarrow+\infty} -\int_\Omega
\nabla_{\H^n}u_k \cdot \nabla_{\H^n}\varphi
= \lim_{k\rightarrow+\infty} \int_\Omega
\partial_r F_k(u_k,\xi)\,\varphi
\end{split}
\end{equation}
for any~$\varphi\in C^\infty_0(\Omega)$.

Also, from~\eqref{1C4},
$$ 0\le \partial_r F_k \le
\left(
\Delta_{\H^n}
\psi\right)^+$$
and so~\eqref{905bb} gives that
\begin{equation}\label{905b}
0\le \int_\Omega u_\infty \Delta_{\H^n}\varphi
\le \int_\Omega\left(
\Delta_{\H^n}
\psi\right)^+\varphi
\end{equation}
for any~$\varphi\in C^\infty_0(\Omega,[0,+\infty))$.

On the other hand, since~$u_k$ is a minimizer
for~\eqref{905a}, we have that
$$ \sup_{k\in\N} \| \nabla_{\H^n}u_k\|_{L^2(\Omega)}<+\infty$$
and so, up to a subsequence, we may suppose
that~$\nabla_{\H^n}u_k$ converges to some~$\nu\in {L^2(\Omega)}$
weakly in~${L^2(\Omega)}$. It follows from the
uniform convergence of~$u_k$ that
\begin{eqnarray*}
&&
-\int_\Omega\nu\cdot\nabla_{\H^n}\varphi=
-\lim_{k\rightarrow+\infty}
\int_\Omega \nabla_{\H^n}u_k\cdot \nabla_{\H^n}\varphi
\\ &&\qquad=\lim_{k\rightarrow+\infty}
\int_\Omega u_k \,\Delta_{\H^n}\varphi
=\int_\Omega u_\infty\,\Delta_{\H^n}\varphi
\end{eqnarray*}
for any~$\varphi\in C^\infty_0(\Omega)$. That is,
$\nabla_{\H^n}u_\infty=\nu$ in the sense of distributions,
and so as a function. In particular,~$
\nabla_{\H^n}u_\infty\in {L^2(\Omega)}$, and therefore~\eqref{905b}
yields that
\begin{eqnarray*}
0\le
\int_\Omega \nabla_{\H^n} u_\infty \cdot
\nabla_{\H^n}\varphi
\le \int_\Omega \left(\Delta_{\H^n}
\psi\right)^+\varphi,
\end{eqnarray*}
for any~$\varphi\in C^\infty_0(\Omega,[0,+\infty))$. This shows
that~$u_\infty$ satisfies~\eqref{TX}, in the distributional sense,
hence as a function.
\end{proof}

\section*{Acknowledgmens}

We thank Luca Capogna, Juan J. Manfredi,
Xiao Zhong and
William Ziemer for some interesting observations.
EV is supported by FIRB project ``Analysis
and Beyond''.

redpajama_set_name: RedPajamaArXiv


text:
Nov 24, 2020 - Health
OxyContin maker Purdue Pharma pleads guilty to three criminal charges
Marisa Fernandez
Purdue Pharma headquarters in Stamford, Connecticut. Photo: Drew Angerer/Getty Images
Purdue Pharma, the maker of the highly addictive painkiller OxyContin, pleaded guilty on Tuesday to three criminal charges, AP reports.
Why it matters: The plea signals the company's admission it played a role in the opioid epidemic that has contributed to nearly half a million deaths over the past two decades.
In a virtual hearing on Tuesday, the Stamford, Connecticut company admitted:
It impeded the U.S. Drug Enforcement Administration's efforts to combat the addiction crisis.
It had not maintained an effective program in preventing any black market sales. Purdue also gave misleading information to the DEA in order to boost manufacturing quotas.
It also admitted paying doctors through a speakers program as an incentive to prescribe painkillers.
The big picture: The plea is one part of a slew of legal and financial troubles for the opioid manufacturer.
Purdue earlier agreed to pay $225 million out of the $8.3 billion in penalties and forfeitures owed to the federal government.
Members of the wealthy Sackler family who own the company have also agreed to pay $225 million to the federal government to settle civil claims.
Yes, but: Advocates and several attorneys general say the plea does little for the lives lost and the ongoing opioid crisis, which has worsened throughout the coronavirus pandemic, early data shows.


The alleged second-in-command of the Medellin-based crime syndicate Oficina de Envigado has been captured in northern Colombia, police said Sunday.
Alias "El Indio," real name Dairo Alberto Muñoz, was captured in Envigado, a town south of Medellin. The 43-year-old Medellin native is suspected of extortion and theft, said police.
Radio station RCN reported the suspect had been captured on Friday, but that the arrest was not confirmed until Sunday.
Muños was allegedly the replacement of another Oficina de Envigado commander, alias "Beto," who was captured in May 2012.
The Oficina de Envigado was originally a group of hit men hired by drug lord Pablo Escobar and his Medellin Cartel. When Escobar was killed in 1993, "the Office" converted itself into a powerful crime syndicate engaged in drug trafficking, extortion and assassinations.
The group, currently led by "Sebastian," controls most of Medellin's neighborhoods and underworld.