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Caregiver: A family caregiver is someone who is responsible for attending to the daily needs of another person. Family caregivers are responsible for the physical, emotional and often financial support of another person who is unable to care for him/herself due to illness, injury or disability. The care recipient may be a family member, life partner or friend. (caregiver.org) A hired caregiver can be found through a qualified home health agency or independently, through online or social sources.
Coinsurance: The percentage of the cost a person pays after he or she has paid their deductible.
Copayment: The fixed price a person pays for a medical visit or service after their deductible has been paid.
Deductible: The amount the you have to pay out of pocket before your insurance plan kicks in and helps pay for medical costs.
Premium: The monthly cost of a person's health insurance.


Home News Lawyer Turned ''TAXI DRIVER'' Speaks On Why She Is A Taxi Drives.
Lawyer Turned ''TAXI DRIVER'' Speaks On Why She Is A Taxi Drives.
Voke Ejenavi is a Law graduate with a Masters Degree which she obtained to help enhance her chances of clinching her dream job in a reputable organization but, that was not to be as her efforts at securing a job proved very abortive and here she is today.
When like other graduates she didn't get the job she sought, she devised a means of helping herself out.
Voke's determination to make something out of her situation made her take one of the most difficult decisions of her life as a young person and a woman for that matter and she took up taxi driving, a rare profession for women in this part of the world. Operating as a taxi driver under an online transport company called Taxify in Lagos not only made her popular but took her to all the nooks and crannies of Lagos.


Why do you need to pack the bearings every 2 years or 2000 miles?
After a year of using (or sitting), condensation leads to rust. Rust leads to scratches. And scratches lead to friction and eventual bearing failure. To pack the bearings, you just jack up the axle and safely support it. Remove the wheel and tire. Remove the dust cap. Remove the cotter pin. Back the retaining nut off and wiggle the hub assembly to loosen the outer bearing on the axle. Remove the retaining nut, washer, and outer bearing. Slide the hub assembly off the axle stub. Clean the brake shoes, backing plate, and axle stub so they are ready for reassembly. Knock the inner bearing and grease seal out of the hub. Clean the brake dust from the brake drum, and the grease and dirt from the hub. Clean ALL, I repeat, ALL the old grease out of the bearings.
A single axle with 10″ brakes is $120. Double axle is, you guessed it, $240.


text:
Q: I'm trying to implement infinite scrolling into my UITableViewController and unsure how to use the database results returned Currently trying to implement infinite scrolling into my app using this plugin: https://github.com/pronebird/UIScrollView-InfiniteScroll
So far I've added this code to my tableview controller viewDidAppear and viewDidDisappear methods:
- (void)viewDidAppear:(BOOL)animated {
    [super viewDidAppear:animated];

    // setup infinite scroll
    // keep a weak reference to table view
    __weak UITableView *weakTableView = self.tableView;

    [self.tableView addInfiniteScrollWithHandler:^{
        // keep a strong reference to table view
        __strong UITableView *strongTableView = weakTableView;

        // seems like our table view didn't make it
        if(strongTableView == nil) return;

        //
        // fetch your data here, can be async operation,
        // just make sure to call finishInfiniteScroll in the end

        // finish infinite scroll animation
        [strongTableView finishInfiniteScroll];
    }];
}

- (void)viewDidDisappear:(BOOL)animated {
    [super viewDidDisappear:animated];

    // remove infinite scroll
    [self.tableView removeInfiniteScroll];
    [[self tableView] reloadData];
}

I drag the table and the spinner shows underneath the last row and disappears after a second or two. Now all I need to do is get the data from my array and add it to block in the viewDidAppear code.
This is how I currently get my parse.com data into an NSMuteableArray instance named "people":
- (void)populatePeopleArrayWithCloudData {
    // Grab data for datasource and store in people array
    NSLog(@"view did load");
    people = [[NSMutableArray alloc] init];

    PFQuery *query = [PFQuery queryWithClassName:@"People"];
    [query whereKey:@"active" equalTo:@1];
    [query orderByDescending:@"createdAt"];
    [query setLimit:10];
    [query findObjectsInBackgroundWithBlock:^(NSArray *objects, NSError *error) {

        if (!error) {
            for (PFObject *object in objects) {

                Person *person = [[Person alloc] init];
                [person setName:[object objectForKey:@"name"]];
                [person setNotes:[object objectForKey:@"notes"]];
                [person setAge:[[object objectForKey:@"age"] intValue]];
                [person setSince:[object objectForKey:@"since"]];
                [person setFrom:[object objectForKey:@"from"]];
                [person setReferenceNumber:[object objectForKey:@"referenceNumber"]];

                PFFile *userImageFile = object[@"image"];
                [userImageFile getDataInBackgroundWithBlock:^(NSData *imageData, NSError *error) {
                    if (!error) {
                        UIImage *image = [UIImage imageWithData:imageData];
                        [person setImage:image];
                    }
                }];

                [person setActive:[[object objectForKey:@"active"] intValue]];
                [person setObjectId:[object objectId]];
                [people addObject:person];
            }

        } else {
            // Log details of the failure
            NSLog(@"Error: %@ %@", error, [error userInfo]);
        }

        NSLog(@"Calling reloadData on %@ in viewDidLoad", self.tableView);
        [self.tableView reloadData];

    }];

}

I limit results to 10. Now what I wish to do is keep grabbing the next 10 results that haven't already been grabbed every time I scroll to the bottom of the table. This code that helps me do this needs to go in the block mentioned above.
The "people" instance is used by my tableviewdatasource methods:
- (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath
{

    static NSString *CellIdentifier = @"Cell";
    UITableViewCell *cell = [[self tableView] dequeueReusableCellWithIdentifier:CellIdentifier forIndexPath:indexPath];
    if (cell == nil) {
        cell = [[UITableViewCell alloc] initWithStyle:UITableViewCellStyleSubtitle reuseIdentifier:CellIdentifier];
    }

        // Configure the cell...

        Person *current;
        if (tableView == [[self searchDisplayController] searchResultsTableView]) {
            current = [searchResults objectAtIndex:indexPath.row];
        } else {
            current = [people objectAtIndex:[indexPath row]];
        }

        [[cell textLabel] setText: [current name]];
        [[cell imageView] setImage: [current image]];
        [[cell detailTextLabel] setText: [current notes]];

    return cell;
}

How do I use my database results with this plugin? As you can see I limit results to 10 and I need to grab the next 10 when I've scrolled to the bottom of the table and add them after the last row in the table.
Kind regards
UPDATE - my numbers of rows in section method as it stands:
- (NSInteger)tableView:(UITableView *)tableView numberOfRowsInSection:(NSInteger)section
{

    // Return the number of rows in the section.

    if (tableView == [[self searchDisplayController] searchResultsTableView]) {
        return [searchResults count];

    } else {
        return [people count];
    }

}

A: Happy to help, but you should probably give it a shot first for us to provide feedback. A few thoughts to get you going...
The general idea is to use the "skip" property on PFQuery to get the next 10. Each time you call it, you add 10.
So create your query as you do, keep it around in a property, but move your findObjectsInBackgroundWithBlock call to your infiniteScrollHandler,  adding 10 to skip each time after you call it. Then at the end of the handling (where you call table reload now), call [strongTableView finishInfiniteScroll]
In your numberOfRows, you'll have to provide the maximum number of people available at your source.

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text:
TechEngage » News » Security
Facebook HQ evacuated due to a bomb threat (updated)
Fazeel Ashraf December 12, 2018
Facebook has received bomb threats at one of their buildings. The company confirmed that the bomb threat came from their Menlo Park campus.
Update: People have been safely returned to the building. Facebook's spokesperson also clarified that not all of the building were evacuated.
Facebook has received bomb threats at one of their buildings. The company confirmed that the bomb threat came from their Menlo Park campus. Facebook sent an email confirming all their employees are safe.
Facebook Spokesperson, Genevieve Grdina, said,
"Late this afternoon, we received a bomb threat and took swift action to evacuate several buildings at our Menlo Park campus. We take the safety and security of our people at Facebook extremely seriously and are glad that everyone is safe. We are working closely with local authorities to investigate this threat and further monitor the situation. "
Some of the buildings at Facebook-owned Menlo Park were evacuated a few hours ago due to the bomb threat. Menlo Park authorities issued an alert at 5 PM local time, asking people to stay clear from block 200 of Jefferson Park.
Menlo Park Police Department confirmed they were conducting an investigation on the bomb threat. The Police Department confirmed this via their Twitter account. New York Police Department's Crimestoppers unit got the warning about the bomb threat first, which forwarded the information to local authorities.
The Police Department tweeted, "Update to the bomb threat at 200 Jefferson Drive: The building has been evacuated, and the bomb unit is on the scene."
Police respond to bomb threat; No suspicious package or device found https://t.co/uxIxHYbmyX
— Menlo Park PD (@MenloParkPD) December 12, 2018
San Mateo County allied agencies bomb units are still searching the building, and the search is still in progress. Till now, no device or suspicious package has been found. Law enforcement officers did not give out any other details regarding the threat.
Sadly this is the second time a bomb threat has been received in just a matter of days. A bomb threat forced CNN employees to evacuate their New York office. An "unsubstantiated telephone threat" came through to the Time Warner Center, according to NBC New York. The caller said there were five bombs in CNN's New York office.
After a thorough investigation by the law enforcement agencies, the building was deemed safe. The building was reopened soon afterward. It wasn't clear if the threat was made specifically to CNN, as there are other offices operating in the building as well.
It's still too early to call why or who made the Facebook bomb threat. There are a lot of parties that could benefit from terrorizing Facebook. There are Russians, Cambridge Analytica or it could even be a disgruntled user.
It could also be a fed up supporter of either of the two parties, who got upset because of some newly implemented policy or headline that they read on the site. It is still too early to call. As of yet, no bomb has been found.
The story is still developing, and we will keep you updated.
Related Tags: Bomb Threat Facebook Security
Britain's health chiefs directed to introduce safe limits for social media use
Is Facebook's "10 Year Challenge" more than a meme?
This European hotel has a "Social Media Sitter" to handle your Instagram
Incredible things your iPhone can do
Top 10 best calendar apps in 2023

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text:
YouTube VR Launches for Samsung Gear VR, Intros New Multi-User Experience
Tim July 25, 2018
YouTube announced today that it's bringing YouTube VR to Samsung's Gear VR headset by way of the Oculus store. By doing this, YouTube is opening up its extensive library of VR-supported videos to any Samsung device that is compatible with Gear VR, which is a...
YouTube App Being Rebuilt for Daydream, Google's VR Platform
Tim May 19, 2016 6
As announced during yesterday morning's keynote, Google is rebuilding its YouTube app, making it more enjoyable for those who plan on using the service with virtual reality. The announcement goes hand-in-hand with the unveiling of Daydream, the company's VR...


text:
Official U.S. Navy website - Commander Naval Surface Force, U.S. Pacific Fleet
Commander, Naval Surface Force, U.S. Pacific Fleet
Pacific Fleet Surface Ships
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USS AMERICA
USS AMERICA (LHA 6)
Contact the Ship
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ALL HANDS UPDATE
USS America (LHA6) Facebook Page
SAN DIEGO (NNS) — The amphibious assault ship USS America (LHA 6) held a change of command ceremony while moored at Naval Base San Diego, June 8.
America Returns to Sea for Ammunition Offload
PACIFIC OCEAN - The amphibious assault ship USS America (LHA 6) completed a weeklong underway period March 2-9 to offload ammunition off the coast of southern California following a seven-month deployment to the 3rd, 5th, and 7th fleet areas of operation.
USS America 15th MEU Return from Deployment
The amphibious assault ship USS America (LHA 6), as part of the America Amphibious Ready Group (ARG), and embarked 15th Marine Expeditionary Unit (MEU) returned home to San Diego, Feb. 2.
America ARG returns to San Diego after deployment
Sailors and Marines of the USS America (LHA 6) Amphibious Ready Group (ARG) returned to their homeport at Naval Station San Diego, Feb. 2, after a successful seven-month deployment.
America Amphibious Ready Group
Sailors and Marines of the USS America (LHA 6) Amphibious Ready Group (ARG) are scheduled to return to their homeport at Naval Station San Diego
USS America and the 15th MEU Arrive in Guam
NAVAL BASE GUAM (Jan. 7, 2018) - The amphibious assault ship USS America (LHA 6) and embarked 15th Marine Expeditionary Unit (MEU) arrived in Guam for a scheduled port visit, Jan. 7.
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text:
Over 50 Organizations Demand a Fair Investigation in the Murder of Khalid Jabara
Posted on August 20, 2016 in Press Releases
Today, Muslim Advocates and the Arab American Institute (AAI) spearheaded a coalition letter of over 50 advocacy, civil rights, community and faith-based groups to demand that Tulsa Mayor Dewey Bartlett direct law enforcement to conduct a fair and thorough investigation of the heinous murder of Khalid Jabara, an Arab American who was killed by his neighbor, Stanley Majors. Majors had previously run over Mr. Jabara's mother with a car and had a history of calling him and his family "dirty Arabs," "filthy Lebanese," as well as "Mooslems" and "Aye-rabs." Signatories to the letter include the NAACP, National Council of Asian Pacific Americans, National LGBTQ Task Force, the ACLU of Oklahoma, and TULSA – The United League of Social Action.
"In the last year, hate crimes targeting Arabs, Muslims, and those perceived to be either have skyrocketed," said Madihha Ahussain, Muslim Advocates staff attorney and lead for the Program to Counter Anti-Muslim Hate. "Unfortunately, there has also been a pattern of law enforcement officials minimizing the possibility that these crimes may be motivated by bigotry, sending a dangerous message that hate violence is not taken seriously. We call on law enforcement officials to aggressively investigate crimes like these to ensure the safety of all communities. The Jabara family and other victims of hate violence deserve that much."
Even as the Jabara family had repeatedly asked law enforcement for protection from Majors, the Tulsa Police Department (TPD) did not closely monitor him. In fact, just minutes before the murder, after Mr. Jabara called to notify police that Mr. Majors had a gun, TPD officials said there was nothing they could do. Even after the horrific murder, TPD failed the family again, with TPD Sergeant Dave Walker saying in a national media interview, "this is an ongoing dispute with neighbors. Yeah, there is hate, but I don't think the only reason is because they are Middle Eastern."
"This attack like the other hate crimes that have taken place demonstrate both the threat our society is facing in terms of division and the need to come together as a nation to combat this bigotry," said Maya Berry, executive director of AAI. "Anti-Arab bias and discrimination is not new, but the current climate continues to make this type of hatred more prevalent. Khalid Jabara was a son, a brother, and an uncle who brought joy to his family and friends. We must do more to prevent tragedies like these from shattering the lives of other families."
Read the full text of the coalition letter here.
Muslim Advocates is a national legal advocacy and educational organization working on the frontlines of civil rights to guarantee freedom and justice for Americans of all faiths. Through high impact lawsuits, policy advocacy, and community education, Muslim Advocates serves as a resource to empower communities and ensures that the American Muslim community is heard by the courts and leaders at the highest level of government. Visit Muslim Advocates at http://www.muslimadvocates.org and follow @muslimadvocates.
Founded in 1985, the Arab American Institute (AAI) is a nonprofit organization committed to the civic and political empowerment of Americans of Arab descent. AAI has developed a host of services, from voter education to liaison with the national parties, to support the community's activities. AAI's programming provides communities with the platform and hard data necessary to strategically and successfully combat xenophobia, and anti-Arab and anti-Muslim bigotry. For more information please visit www.aaiusa.org and follow @AAIUSA.


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text:
Das Nauener Tor ist eines der drei erhaltenen Stadttore von Potsdam. Es wurde 1754/1755 erbaut und gilt als erstes Bauwerk neugotischen Stils in Kontinentaleuropa.

Im Jahr 1722 wurde etwa 400 Meter vom heutigen Standort entfernt während der sogenannten I. Barocken Stadterweiterung das erste Nauener Tor als eines der insgesamt fünf in der Stadtmauer befindlichen Stadttore gebaut. 1733 entstand während der II. Barocken Stadterweiterung ein neues Tor 20 Meter vor dem heutigen Platz. 1754/1755, also 20 Jahre vor dem eigentlichen Beginn der Neugotik auf dem Kontinent, entstand nach einer Skizze König Friedrichs II. durch den Architekten Johann Gottfried Büring mit dem Nauener Tor in Potsdam eine Architektur in gotischen Formen. 1867–1869 wurde dieses Tor durch König Wilhelm I. in seine heutige Gestalt umgebaut.

Auf der Suche nach einer konkreten, kopierten Vorlage ist man nicht fündig geworden. Das häufig angeführte Vorbild Inveraray Castle in Schottland wird nach neuesten Forschungen ausgeschlossen, da es in seiner ursprünglichen Form gar keine Kegeldächer besaß. Erst nach einem Brand im Jahr 1877 wurden dem Schloss Inveraray Castle (ursprüngliches Baujahr: 1457) bei den Reparaturarbeiten neue architektonische Elemente hinzugefügt: So erhielten die Ecktürme zum Beispiel jetzt erst die Kegelhelme.
Viel konkreter ist die charakteristische Gestalt des Nauener Tores hingegen an einem bisher unvermuteten Ort zu finden, für den zweifelsohne spricht, dass er Friedrich II. ganz sicher bekannt war: Für die erste Ausgabe seiner eigenen Schriftwerke ließ der König 1750 Illustrationen von dem bedeutenden Stecher Georg Friedrich Schmidt anfertigen. Die Vignette zur "Eloge de Jordan" des dritten Bandes zeigt im Hintergrund eine von Kolonnaden gesäumte Stadtmauer, die durch ein von Rundtürmen mit Spitzhelmen flankiertes Tor unterbrochen wird.

Vermutlich hat Friedrich II. zudem die Ähnlichkeit mit den Türmen von Schloss Rheinsberg inspiriert, in dem er seine Zeit als Kronprinz verbrachte, und die er zeit seines Lebens als die schönste bezeichnete.

Ursprünglich verband eine Stadtmauer das Nauener Tor mit den beiden anderen Toren, dem Jägertor und dem Brandenburger Tor am Luisenplatz. Anstelle der nicht erhaltenen Stadtmauer verbindet heute eine Promenade die drei noch verbliebenen Potsdamer Stadttore. Das Nauener Tor befindet sich in unmittelbarer Nähe zum Holländischen Viertel. Seine Nutzer waren das Militär und Händler, Handwerker und Verwaltungen bis zum Restaurant. Der Platz vor dem Nauener Tor ist seit der letzten Renovierung des Tores und Umgestaltung der Verkehrsführung 1996 mit der hohen Dichte von Cafés, Restaurants und Bars ein beliebter Treffpunkt der Potsdamer und ihrer Gäste.

Die Gleise der Straßenbahn  führen direkt durch das Nauener Tor. Im deutschsprachigen Raum gibt es nur noch zwei andere Städte mit derartigen Tordurchfahrten. Zum einen durchquert die Straßenbahn in Freiburg im Breisgau das Schwabentor, das Martinstor und das Günterstäler Tor, zum anderen fährt die Straßenbahn Bern durch den Käfigturm.

Literatur
 Paul Sigel, Silke Dähmlow, Frank Seehausen und Lucas Elmenhorst: Architekturführer Potsdam. Dietrich Reimer Verlag, Berlin 2006, ISBN 3-496-01325-7.
 Christiane Theiselmann: Potsdam und Umgebung. Von Preußens Arkadien zur Brandenburgischen Landeshauptstadt. Köln 1993.
 Friedrich Mielke: Potsdamer Baukunst – das klassische Potsdam, Propyläen Verlag; 2. Auflage 1991 Ullstein.

Weblinks

 Info der Stadt Potsdam
 Alfred Hagemann: Das Nauener Tor in Potsdam – Der "roi philosophe" und das Gothic Revival, Potsdam, 2012.

Einzelnachweis / Quellenangabe

Potsdam Nauener Tor
Bauwerk des Historismus in Potsdam
Stadttor in Brandenburg
Architektur (Preußen)
Erbaut in den 1750er Jahren
Baudenkmal in Potsdam

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text:
Senators race to seal infrastructure deal as pressure mounts
WASHINGTON — Senators are racing to seal a bipartisan infrastructure deal as soon as Monday, as pressure is mounting on all sides to show progress on President Joe Biden's top priority.Heading into a make-or-break week, key senators and staff spent the weekend trying to reach a final agreement. One major roadblock is how much money should go to public transit. But spending on highways, water projects, broadband and others areas remains unresolved, as is whether to take unspent COVID-19 relief funds to help pay for the infrastructure.The lead Republican negotiator, Ohio Sen. Rob Portman, said the two sides were "about 90% of the way there" on an agreement.A top Democrat, Sen. Mark Warner of Virginia, said he was hopeful a final bill would be ready Monday afternoon — though others were not so sure.The week ahead is crucial. Senate Majority Leader Chuck Schumer, D-N.Y., has said he wants to pass the nearly $1 trillion bipartisan package as well as the blueprint for a larger $3.5 trillion budget plan before the Senate leaves for its August recess. He held a procedural vote last week to begin debate on the bipartisan framework, but all 50 Senate Republicans voted against it, saying they needed to see the full details of the plan.The White House wants a bipartisan agreement for this first phase, but as talks drag on anxious Democrats, who have slim control of the House and Senate, could leave Republicans behind and try to go it alone. If it fails, it could be wrapped into the broader package of Biden's priorities that Democrats are hoping to pass later.The bipartisan package includes about $600 billion in new spending on public works projects. Democrats want to see more of the money go toward boosting public transportation, which includes subways, light-rail lines and buses, in line with Biden's original infrastructure proposal and the push to address climate change.The bipartisan group originally appeared to be moving toward agreement on more money for transit. But Pennsylvania Sen. Pat Toomey, the top Republican on the Senate Banking, Housing and Urban Affairs Committee, which oversees public transit, raised questions. He cited, in part, previous COVID-19 federal relief money that had already been allocated to public transit."Nobody's talking about cutting transit," Toomey said Sunday. "The question is, how many tens of billions of dollars on top of the huge increase that they have already gotten is sufficient? And that's where there is a little disagreement."Typically, spending from the federal Highway Trust Fund has followed the traditional formula of 80% for highways and 20% for transit. Democratic Sens. Sherrod Brown of Ohio and Tom Carper of Delaware say they will oppose the deal if transit funding falls below that.The White House has declined to say whether Biden would push for the additional funding for transit."Transit funding is obviously extremely important to the president — the 'Amtrak President,' as we may call him," White House press secretary Jen Psaki said Friday. "But we believe that members can get this work done and can work through these issues quite quickly."The final package would need the support of 60 senators in the evenly split 50-50 Senate to advance past a filibuster — meaning at least 10 Republicans along with every Democratic member. Last week's test vote failed along party lines.A Democratic aide granted anonymity to discuss the private talks said beyond transit, there are other remaining issues still unresolved around how to pay for it. For instance, details on broadband funding, as well as whether to tap into the leftover COVID relief funds previously passed by Congress, continue to be discussed, the aide said.Democrats are seeking a compromise to pay for the package after they rejected a hike in the gas tax drivers pay at the pump and Republicans dashed a plan to boost the IRS to go after tax scofflaws.Three rounds totaling nearly $70 billion in federal COVID-19 emergency assistance, including $30.5 billion that Biden signed into law in March, pulled transit agencies from the brink of financial collapse as riders steered clear of crowded spaces on subway cars and buses. That federal aid is expected to cover operating deficits from declining passenger revenue and costly COVID-19 cleaning and safety protocols through at least 2022.But Democrats and public transit advocates see expanded public transit systems as key to easing traffic congestion, combating climate change and curbing car pollution.Rep. Peter DeFazio, D-Ore., the chairman of the House Transportation and Infrastructure Committee, recently sent a letter with 30 Democrats on the panel warning that the Senate proposal was inadequate and that any deal should incorporate the House-passed $715 billion infrastructure bill, which includes more money for rail and transit."The historical share for public transit from the Highway Trust Fund is 20%," Paul Skoutelas, president of the American Public Transportation Association, said Sunday. "It is the absolute minimum acceptable level to help sustain our nation's public transportation systems. It is imperative that we make robust, forward looking investments to modernize and expand public transit that will assist in our economic recovery from the COVID pandemic and get Americans back to work."Portman appeared on ABC's "This Week," Toomey was on CNN's "State of the Union" and Warner spoke on "Fox News Sunday."
Divine transformation: Ex-Alaska strip club becomes church
Warming rivers in US West killing fish, imperiling industry


Happy Friday! This week was a great week. It was so nice the sun was out.
Congratulations to our OBOB team who competed last Saturday and made the Top 8! They did an amazing job and it was so much fun watching them compete. A big shout out to Melody and Carol for leading this team of ROCKSTARS!
Our staffing is complete for next year and I am so happy to let you all know we will not be losing any of our teachers. Several schools lost teachers, and so we were very fortunate.
I look forward to seeing you all at the auction tomorrow night.
Dear Skyline School Community, It is with great pleasure that I announce BG Aguirre as the new principal for Skyline School. BG is currently the principal at St. Helens High School, a position she has held since 2013. She served as assistant principal for five years prior to taking over as principal.
BG earned her Bachelor's degree in sociology and Spanish from Oregon State University, her Masters of Arts in teaching from Concordia University and her administrator license from Portland State University. Following her graduation from Concordia, BG worked as an elementary school teacher in the Beaverton School District, first at Vose Elementary for five years, then Kinnaman Elementary for three years. She spent a year in the Lincoln County School District in Newport, before moving to the St. Helens School District to serve as assistant principal and principal for St. Helens High School in 2008.
BG has proven herself to be an effective school leader and is looking forward to leading Skyline for years to come. She has received International Baccalaureate MYP Level 1 training and IB Administration Level 2 training. BG is also well-versed in successfully running an AVID school. While at St. Helens, BG was an AVID district director, an AVID principal, attended multiple AVID strand trainings and successfully implemented AVID school-wide.
Throughout her time in both the St. Helens and Beaverton School Districts, BG has gone above and beyond her stated duties to ensure that her community thrives. She is a fluent Spanish speaker and is an advocate for all students. She has had great success writing grants and securing funding for intervention programs, professional development and students in need. She deeply values collaboration between teachers, families and administrative staff to ensure the success of the school.
BG is a dedicated wife and mom of three children ages 3, 5 and 7. In her free time, she loves watching her children participate in extra-curricular activities and spending time as a family. BG enjoys reading, playing golf and is an avid runner. BG tells me she is absolutely overjoyed to be back in elementary and to have the opportunity to work in her community. Please join me in congratulating BG in her new assignment as the principal for Skyline School!


text:
John F. Kennedy 55 Years On: Casting Light on the JFK Assassination, 9/11 and Other 21st Century Crimes
By Prof. Graeme McQueen
OffGuardian 22 November 2018
Theme: History, Law and Justice, Media Disinformation
Fifty-five years ago, on November 22, 1963, John F. Kennedy was assassinated. Although there has been a great deal written about this event over the years, I want to draw attention to one exceptionally important article, originally delivered as a talk on November 20, 1998. Vincent Salandria gave this talk in Dallas at the invitation of the Coalition on Political Assassinations. (See Sources.)
Salandria had been a high school teacher at the time of the assassination (he later became a lawyer) and was one of the first people in the US to write essays expressing dissent from the government narrative of lone gunman Lee Harvey Oswald, maverick leftist.
In his 1998 talk Salandria went through over a dozen of the famous obstacles to the government story—the grassy knoll witnesses, the "magic bullet," the testimony of the doctors at Parkland Hospital, and so on—but he did not let himself get sidetracked into detailed debates on any of these. By 1998 he had already seen, and participated in, 35 years of such debates. He had long ago concluded that, "the national security state at the very highest level of its power killed President John F. Kennedy for his efforts at seeking to develop a modus vivendi with the Soviets and with socialist Cuba."
In 1998 he felt it was time to warn researchers about the danger of wasting time in "false debates," where the essential facts had clearly been established and the wrangling served only the purposes of the assassins. Rather than repeat the debates, Salandria decided in 1998 to outline his basic approach. I will call this the Salandria Approach. I draw attention to it because I believe it helps us find our feet when we tackle not only the JFK killing but many of the killings in the 21st century's War on Terror.
Here are Salandria's words:
I began to sift through the myriad facts regarding the assassination which our government and the US media offered us. What I did was to examine the data in a different fashion from the approach adopted by our news media. I chose to assess how an innocent civilian-controlled US government would have reacted to those data. I also envisioned how a guilty US national security state which may have gained control of and may have become semi-autonomous to the civilian US governmental structure would have reacted to the data of the assassination."
He adds that,
only a guilty government seeking to serve the interests of the assassins would consistently resort to accepting one improbable conclusion after another while rejecting a long series of probable conclusions."
Let us take two cases from Salandria's list of over one dozen in order to see what he was getting at.
The Grassy Knoll
Dozens of witnesses thought there were shots from an extended grassy rise, containing several structures, situated west of the famous Texas School Book Depository Building. Salandria, refusing to get drawn into the familiar debate, says:
Let us assume arguendo [for the sake of argument] that all of the eyewitnesses who had concluded that shots were fired from the grassy knoll were dead wrong. But an innocent government could not and would not at that time have concluded that these good citizens were wrong and would not have immediately rushed to declare a far-fetched single assassin theory as fact."
Note that Salandria's emphasis is not on the details of the grassy knoll discussion but on the method the government followed in its investigation. And he is right, both about the immediate claim that Oswald acted alone— presented, as he explains, by a government representative on November 22 itself—and about the identical statement presented later by the Warren Commission.
In both cases the claim flew in the face of the eyewitness evidence. For example, despite the fact that there are references to dozens of witnesses to shots from the grassy knoll in the 26 volumes of evidence appended to the Warren Report, the Commission itself displayed little interest in them. And when the Commission dismissed every single one of the grassy knoll witnesses to protect its lone gunman theory it did so without bothering to make a sustained argument.
It chose instead to play a credibility game. It pronounced:
No credible evidence suggests that the shots were fired from the railroad bridge over the Triple Underpass, the nearby railroad yards or any place other than the Texas School Book Depository Building" Warren Report, p. 61
In other words, the Commission decided to gather together into one great agglomeration the credibility of its seven well dressed and high-ranking white men associated with government and use this to crush the credibility of the "good citizens" who were present in the Plaza and witnessed, with their senses, the unfolding of events.
It was a breathtaking move. But in what way could it be said to characterize an innocent government? How could any serious investigator pretend to solve an evidential problem by playing a credibility game? Standard practice in a homicide investigation would be to find all witnesses, to interview them, and to record their statements impartially, making sure to ask each one of them where they thought the shots came from and why they reached their conclusion. How would the opinions of congressmen, spies and the like possibly be relevant to the case when these gentlemen declined to offer adequate counter-evidence or to give a serious argument to support their peculiar conclusion?
Readers who have never had the opportunity to see and hear for themselves the good citizens in question may benefit from Mark Lane's documentary:
Well, where, in such a case, does the Salandria Approach lead us? We have no choice but to conclude that the Warren Commission's investigation was not what we would expect from "an innocent civilian-controlled US government."
It was more characteristic of "a guilty government seeking to serve the interests of the assassins." There was a predetermined perpetrator and an insistence on the guilt of this perpetrator, while evidence suggestive of a conspiracy was systematically ignored, distorted or suppressed.
Suppose we were to apply the Salandria Approach to events of the 21st century–to the eyewitnesses at the World Trade Center on September 11, 2001, for example? We have over 150 witnesses who reported that they saw, heard or felt explosions at the time of the beginning of destruction of the Twin Towers. (See Sources for assertions in this and the following paragraph.)
Their testimony constitutes very significant support for the theory that the Trade Center was blown up and did not undergo collapse from structural failure caused by airplane collision. We are not simply talking about loud sounds here. We are talking about sounds that experienced firefighters suspected were caused by bombs. We are talking about patterns of explosions seen pulverizing the buildings. We are talking, in some cases, about witnesses who say these explosions threw them through the air. Now, avoiding the debates about the details of this testimony, let us follow Salandria and ask: What did the government's 9/11 Commission do with these eyewitness accounts, all of which were in its possession?
The answer is that it called for no comprehensive search for eyewitnesses (neither did the FBI, as far as I can discover), nor did it have such witnesses asked the appropriate questions. It devoted to these witnesses a single line in the roughly 585 pages of its Report. And that single line is both dismissive and extremely misleading.
What about the National Institute of Standards and Technology, assigned by government the task of looking in detail at the destruction of the Trade Center and sorting out the reasons for its destruction? In the thousands of pages of its reports on the Twin Towers we find not a single mention of the explosion witnesses. Despite NIST's pride in its interviewing techniques, and despite its access to all the relevant information, it somehow missed over 150 witnesses. It made no attempt to find them, to sort out their testimony, or to discover how their words might illumine the mystery of the so-called "collapses."
We should recall that the efforts of the 9/11 Commission and NIST were mere follow-through. A strenuous attempt to promote the structural failure hypothesis was begun on the very day of September 11, 2001, in the absence of serious evidence in its favour and in bold contradiction to what large numbers of witnesses were saying. (Sources)
When we adopt the Salandria Approach we must, to paraphrase Salandria, conclude that, "an innocent government could not and would not at that time have concluded that these good citizens were wrong and would not have immediately rushed to declare a far-fetched [structural failure] theory as fact."
In his essay Salandria explains the absurdity of the single bullet ("magic bullet") theory, according to which one bullet passed entirely through the president's body and then caused all of Governor Connally's wounds, emerging after its adventure in near-pristine condition. This bullet evidently had no difficulty changing direction in mid-air, nor did it balk at losing mass in Connally's body and then regaining this mass at the end of its journey. Salandria concludes:
"our Cold War government in the context of the assassination had declared a moratorium on the science of physics."
Remember: the issue before us is not merely he single bullet theory itself but the behavior of government representatives in investigating this hypothesis. So it is in those moments when we read the Warren Commission transcripts and watch counsel Arlen Specter leading and pressuring witnesses into accepting the single bullet theory that we realize we are seeing the handiwork of a guilty state.
Now, what might we find if we were to apply the Salandria Approach to the destruction of the World Trade Center? To restrict ourselves, for the sake of this discussion, to World Trade 7, what would the approach of an innocent government to this building destruction look like? Would we not expect a thorough search for eyewitnesses?
Would not all of the recoverable steel be preserved carefully and made accessible to civilian experts? Would there not be a serious attempt to explain evidence of corrosion and vaporization of the steel? Would there not be the most rigorous examination of the Trade Center dust, searching for evidence that would allow ascertainment of temperatures reached during the building's destruction and searching as well for residue of explosives and incendiaries?
Would there not be frank astonishment at the fact that the descent of this 47-storey building, not hit by a plane, began rapidly, symmetrically, and at free fall acceleration? Would not physicists openly debate this astounding event, troubled by the fact that the vertical columns of this well constructed steel-framed high-rise offered no resistance whatsoever when, for mysterious reasons, the collapse began?
Surely an innocent government sincerely probing for the truth would not choose, instead of taking the path outlined above, to construct a computer simulation that, even with manipulation, could not replicate the historical event clearly preserved on video? Surely investigators would not bring the simulation to an abrupt end before it was able to represent total collapse, and surely they would not refuse to release the complete data set used in their simulation, claiming it might compromise national security? (Sources)
When we ask these questions and contemplate the answers we see at once what game NIST has been playing in its account of World Trade 7. In the 21st century there is, perhaps, no more obvious demonstration that the US government, for the sake of its War on Terror, has "declared a moratorium on the science of physics."
There is an entire organization, Architects & Engineers for 9/11 Truth, which has taken as its task for over a decade the pointing out of such violations of the laws of physics in the US government's account of the September 11, 2001 crime. The organization is to be praised for its creativity and persistence. Yet the false debate continues, and the intelligentsia continues to insist that the Emperor is well dressed, thank you very much.
Political Implications of Grassy Knolls and Magic Bullets
There is something I have always found arresting about the grassy knoll, and my concerns extend to the suppressed witnesses of September 11. In both cases we have ordinary folks—people like ourselves—who are, supposedly, citizens of a democracy. They are also, as far as we can tell, of sound mind and body, able to perceive with their senses and assess with their minds. Yet, all of a sudden, when their bodies and minds tell them something that conflicts with a government dictum, they are considered by government of no more political competence than cattle. I find it hard to think of a greater insult to these "good citizens" and to the notion of democracy, and I find it hard to think of a more brash assertion of the principle of authority.
This is why witnesses from the grassy knoll and the World Trade Center should be at the centre of the current debate about state deception and its relation to democracy.
As for magic bullets in Dealey Plaza and the mysterious collapse of World Trade 7, they are, I suggest, of comparable political importance to the abused witnesses. We face a collection of gentlemen in suits and ties (seven gentlemen in the Warren Commission and ten in the 9/11 Commission) telling us that their stories are more potent than the laws of the universe. How poor must be our self-confidence that we can put up with this guff? How defective must be our educational systems if they produce citizens who accept this?
Here we are, then, at the 55th anniversary of the murder of a president who was moving away from Cold War thinking and entering a different path. As we reflect on the direction in which his assassins have steered the United States of America, to the detriment of all of us, US citizens and otherwise, let us reflect on Salandria's words:
By coming to understand the true answer to the historical question of who killed President Kennedy and why, we will have developed a delicate and precisely accurate prism through which we can examine how power works in this militarized country. By understanding the nature of this monumental crime, we will become equipped to organize the struggle through which we can make this country a civilian republic in more than name only."
Graeme MacQueen is the former director of the Centre for Peace Studies at McMaster University. He is a member of the 9/11 Consensus Panel, former co-editor of the Journal of 9/11 Studies, and an organizer of the 2011 Toronto Hearings, the results of which have been published in book form as The 9/11 Toronto Report. He is a Research Associate of the Centre for Research on Globalization (CRG).
The Salandria essay that is the basis of my article, "The JFK Assassination: A False Mystery Concealing State Crimes," can be found here.
Both the Warren Report and the 26 volumes of evidence can be found at the Mary Ferrell Foundation website.
3. The list of 156 eyewitnesses to explosions in the Twin Towers can be found here. A discussion of the method used to arrive at the list as well as the treatment of these witnesses by the 9/11 Commission and NIST can be found in my article, "Eyewitness Evidence of Explosions in the Twin Towers" in The 9/11 Toronto Report, ed. James Gourley, International Center for 9/11 Studies, 2012.
For a discussion of the destruction of World Trade 7 see the website of Architects & Engineers for 9/11 Truth & especially Ted Walter's publication, Beyond Misinformation.
For the dismissal of evidence of controlled demolition from the earliest moment see Ted Walter's recent article, "Dick Cheney and Rudy Giuliani: The First Government Officials to Dismiss the Idea of Controlled Demolition on 9/11."
For a discussion of Kennedy's turn away from the Cold War see James Douglass's brilliant JFK and the Unspeakable: Why He Died and Why It Matters(New York: Simon and Schuster, 2008).
The original source of this article is OffGuardian
Copyright © Prof. Graeme McQueen, OffGuardian, 2018
Articles by: Prof. Graeme McQueen


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redpajama_set_name: RedPajamaGithub


text:
Q: Python/Excel - transpose row to column I am trying to format a print statement to align a specific column.
Here is my dataframe:
df = pd.DataFrame({"platform":['test3','test3','test3','test4','test4','test5','test5','test5','test6','test6'],
                   "ripenes type":['OverRipes', 'Ripes','UnRipes','Ripes','UnRipes','OverRipes', 'Ripes','UnRipes', 'Ripes', 'UnRipes'],
                   "total":[1,5,5,5,4,5,4,1,11,1]})

df

Currently my output is:

The output I am looking for is:

is there any available method to transpose my data or using python code to create this data frame?

A: You can use the df.pivot function.
Here is an example:
import pandas as pd

df = pd.DataFrame({"platform":['test3','test3','test3','test4','test4','test5','test5','test5','test6','test6'],
                   "ripenes type":['OverRipes', 'Ripes','UnRipes','Ripes','UnRipes','OverRipes', 'Ripes','UnRipes', 'Ripes', 'UnRipes'],
                   "total":[1,5,5,5,4,5,4,1,11,1]})

df2 = df.pivot(index="platform", columns="ripenes type", values="total").fillna(0)

# Provide the desired column order
column_order = ['Ripes', 'UnRipes', 'OverRipes']
df2 = df2.reindex(column_order, axis=1)
df2

OUTPUT:

Note: You can find some other useful pandas functions on their official cheatsheet


text:
100 MegaWatt wind farm project stuck in the Cabinet
September 9, 2018 Lankanewspapers 0 Comments
Danish wind turbine manufacturer has won a tender to build the 100 MegaWatt wind farm in Mannar but the project is held up by delays at Cabinet level.
The Ceylon Electricity Board Engineers' Union (CEBEU) has now urged Cabinet to speed up approval of the US$ 140 million (Rs 22,68 billion) project which Vestas Asia Pacific A/S, Denmark, secured, following evaluation of international competitive bids. The venture is expected to be completed by May 2020.
Approval of the Standing Cabinet Appointed Procurement Committee (SCAPC) and the Asian Development Bank, which is providing funds under a concessionary loan, has already been approved, but Cabinet approval is still pending. Electricity from this project can be generated at less than US 5 cents per kilowatt hour (kWh). This is just below Rs 8 per kWh at prevailing exchange rates, CEBEU Committee Member Athula Wanniarachchi said.
Construction of Wind Farm would be greatly influenced by climate conditions in Mannar, Mr Wanniarachchi warned. Therefore, timely awarding of the contract is crucial, to complete the project by the May 2020 deadline, as wind turbines need to be hoisted prior to the onset of the South-West monsoon. "If there is a considerable delay in commencing project activities, they may end up having to hoist up the turbines during strong monsoonal winds, which may not be feasible," he observed.
In such a scenario, the CEB estimates that, completion of the project may have to be pushed back to the first quarter of 2021. This would mean the plant would lose out on generating 280 million kWh of electricity, from a South-West monsoonal season. The CEBEU has urged the Cabinet to approve the project, so that, the contract can be awarded by next month.
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text:
\section*{Introduction}

Let $\Omega$ be a bounded strongly pseudoconvex open subset of $\mathbb{C}^n$.
Given $\phi \in {\mathcal C}^0(\partial \Omega)$ and $f \in L^p(\Omega)$,
we consider the Dirichlet problem
$$
MA(\Omega,\phi,f): \left\{ \begin{array}{l}
 u \in PSH(\Omega) \cap {\mathcal C}^0(\overline{\Omega}) \\
 u=\phi \text{ on } \partial \Omega \\
 (dd^c u)^n=f \beta_n \text{ in } \Omega.
 \end{array} \right.
$$
Here $\beta_n=dV$ denotes the euclidean volume form in $\mathbb{C}^n$, $d=\partial+\overline{\partial}$,
$d^c=i(\overline{\partial}-\partial)$, $PSH(\Omega)$ is the set of plurisubharmonic
(psh for short) functions
in $\Omega$ (the set of locally integrable functions $u$ such that $dd^c u \geq 0$
in the sense of currents), and $( dd^c \; \cdot )^n$ denotes the complex Monge-Amp\`ere operator:
this operator is well defined on the subset of bounded (in particular continuous) psh functions,
as follows from the work of E.Bedford and A.Taylor [BT 2]. We refer the reader to
[K 4] for a recent survey on its properties.

The equation $MA(\Omega,\phi,f)$ has been studied intensively during the last decades.
Let ${\mathcal B}(\Omega,\phi,f)$ denote the family of subsolutions to
$MA(\Omega,\phi,f)$, i.e. the set of bounded functions $v$ that are psh in $\Omega$ with
$v \leq \phi$ on $\partial \Omega$ (i.e. $\limsup_{\zeta \rightarrow z} v(\zeta) \leq \phi(z)$
when $z \in \partial \Omega$) and $(dd^c v)^n \geq f \beta_n$
in $\Omega$. It follows from the comparison principle [BT 2] that if a solution exists,
it must coincide with the Perron-Bremermann envelope,
$$
u(z)=u(\Omega,\phi,f)(z):=\sup \{ v(z) \, / \, v \in {\mathcal B}(\Omega,\phi,f) \}.
$$

It follows from the works of H.J.Bremermann [Br], J.Walsh [W]
and E.Bedford-A.Taylor [BT 1]
that $u(\Omega,\phi,f)$ is indeed
a solution to $MA(\Omega,\phi,f)$ when $f \in {\mathcal C}^0(\overline{\Omega})$.
S.Kolodziej has further shown in [K 2]
that the solution $u(\Omega,\phi,f)$
is still continuous when $f \in L^p(\Omega)$, for some
$p>1$ (it is easy to check that $u(\Omega,\phi,f)$ is not necessarily
locally bounded when the density is merely in $L^1(\Omega)$).

Higher regularity results have also been provided. E.Bedford and A.Taylor showed in
[BT 1] that if $\phi \in Lip_{2\alpha}(\Omega)$ and $f^{1/n} \in Lip_{\alpha}(\overline{\Omega})$,
then $u(\Omega,\phi,f) \in Lip_{\alpha}(\overline{\Omega})$.
The smoothness of the solution (assuming smoothness of $\phi$ and $f>0$) is established in [CKNS].
Our aim here is to establish H\"older-continuity of $u(\Omega,\phi,f)$ in the spirit
of [BT 1], only assuming the density $f$ belongs to $L^p(\Omega)$, $p >1$,
as in [K 2]. Our main result is the following:

\vskip.2cm \noindent
{\bf Theorem A.}
{\it Assume $f \in L^p(\Omega)$, for some $p>1$, and $\phi \in Lip_{2\alpha}(\partial \Omega)$,
with $\nabla u(\Omega,\phi,0) \in L^2(\Omega)$. Then
$$
u(\Omega,\phi,f) \in Lip_{\alpha'}(\overline{\Omega}),
\; \;
\text{ for all } \alpha'<\min(\alpha,2/[qn+2]),
$$
where $1/p+1/q=1$.}
\vskip.2cm

The condition $\nabla u(\Omega,\phi,0) \in L^2(\Omega)$ is automatically
satisfied if $\phi \in {\mathcal C}^{1,1}(\partial \Omega)$: in this
case $u(\Omega,\phi,0) \in Lip_1(\overline{\Omega})$, hence
$\nabla u(\Omega,\phi,0)$ is actually bounded in $\Omega$ (see [BT 1]).
What really matters here is that there should exist a subsolution
$v \in {\mathcal B}(\Omega,\phi,0)$ such that $\nabla v \in L^2(\Omega)$.
This implies (see Lemma 3.1), that
$u(\Omega,\phi,0)$ and $u(\Omega,\phi,f)$
both have gradient in $L^2(\Omega)$.

We could not avoid the use of this extra (technical ?) hypothesis on
the homogenous solution $u(\Omega,\phi,0)$.
Also the exponent $\alpha'$ is probably not optimal. We can get a better
exponent by assuming that $\Delta u(\Omega,\phi,0)$ has finite mass in $\Omega$
(this is automatically satisfied when $\phi \in {\mathcal C}^2 (\partial \Omega)$).

\vskip.2cm \noindent
{\bf Theorem B.}
{\it Assume $f \in L^p(\Omega)$, for some $p>1$, and
$\phi \in Lip_{2\alpha}(\partial \Omega)$ is such that
that $\Delta  u(\Omega,\phi,0)$ has finite mass in $\Omega$.
Then
$$
u(\Omega,\phi,f) \in Lip_{\alpha''}(\overline{\Omega}),
\; \;
\text{ for all } \alpha''<\min(\alpha,2/[qn+1]),
$$
where $1/p+1/q=1$.}
\vskip.2cm

The exponent $\alpha''$ is not far from being optimal as Example 4.2 shows.

\section{The stability estimate}

Our main tool is the following estimate which is proved in [EGZ]
in a compact setting (under growth -- but no boundary -- conditions, see
Proposition 3.3 in [EGZ]).
A similar -- but weaker -- estimate was established by S.Kolodziej in [K 3].

\begin{thm}
Fix $0 \leq f \in L^p(\Omega)$, $p>1$.
Let $\varphi,\psi$ be two bounded psh functions in $\Omega$ such that
$
(dd^c \varphi)^n=f \beta_n \text{ in } \Omega,
\text{ and } \varphi \geq \psi \text{ on } \partial \Omega.
$
Fix $r \geq 1$ and $0 \leq \gamma <r/[nq+r]$, $1/p+1/q=1$. Then
$$
\sup_{\Omega} (\psi-\varphi) \leq C ||(\max(\psi-\varphi,0)||_{L^r(\Omega)}^{\gamma},
$$
for some uniform constant $C=C(\gamma,||f||_{L^p(\Omega)},||\psi||_{L^{\infty}})>0$.
\end{thm}

The proof follows closely the one given in [EGZ], so we only
recall the main ingredients, for the reader's convenience.
The estimate is a simple consequence of the following result.

\begin{pro}
Fix $f \in L^p(\Omega)$, $p>1$, and let
$\varphi,\psi$ be bounded psh functions in $\Omega$ such that
$\varphi \geq \psi$ on $\partial \Omega$.
If $(dd^c \varphi)^n=f \beta_n$, then for all $\varepsilon>0,\tau>0$,
$$
\sup_{\Omega} (\psi-\varphi) \leq \varepsilon+C \left[ \text{Cap}(\varphi-\psi<-\varepsilon ) \right]^{\tau},
$$
for some uniform constant $C=C(\tau, ||f||_{L^p(\Omega)})$.
\end{pro}

Here $\text{Cap}(\cdot)$ denotes the Monge-Amp\`ere capacity introduced and studied
by E.Bedford and A.Taylor in [BT 2].
Recall that for $K \subset \Omega$,
$$
\text{Cap}(K):=\sup \left\{ \int_K (dd^c v)^n \, / \, v \in PSH(\Omega)
\text{ with } -1 \leq v \leq 0 \right\}.
$$

The proposition is a direct consequence of the following three lemmas:

\begin{lem}
Fix $\varphi,\psi \in PSH(\Omega) \cap L^{\infty}(\Omega)$ such that
$\overline{\lim}_{\zeta \rightarrow \partial \Omega} (\varphi-\psi) \leq 0$.
Then for all $t,s>0$,
$$
t^n \text{Cap}(\varphi-\psi<-s-t) \leq \int_{(\varphi-\psi<-s)} (dd^c \varphi)^n.
$$
\end{lem}

\begin{lem}
Assume $0 \leq f \in L^p(\Omega)$, $p>1$.
Then for all $\tau>1$, there exists $C_{\tau}>0$ such that
for all $K \subset \Omega$,
$$
0 \leq \int_K f dV \leq C_{\tau} \left[ \text{Cap}(K) \right]^{\tau}.
$$
\end{lem}

When $(dd^c \varphi)^n=f dV$, one can
combine Lemma 1.3 and Lemma 1.4 to control
$\text{Cap}(\varphi-\psi<-s)$ by $\text{Cap}(\varphi-\psi<-s-t)]^{\tau}$.
This has strong consequences since $\tau>1$, as the following
result shows, when applied to $g(t)=\text{Cap}(\varphi-\psi<-t-\varepsilon)^{1/n}$:

\begin{lem}
Let $g:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ be a decreasing right-continuous function.
Assume there exists $\tau,B>1$ such that
$g$ satisfies
$$
H(\alpha,B) \hskip1cm
tg(s+t) \leq B [ g(s) ]^{\tau}, \;
\forall s,t>0.
$$

Then there exists $S_{\infty}=S_{\infty}(\tau,B) \in \mathbb{R}^+$ such that
$g(s)=0$ for all $s \geq S_{\infty}$.
\end{lem}

One word about the proofs. Lemma 1.3 is a direct consequence
of the ``comparison principle'' of E.Bedford and A.Taylor [BT 2]
(see [K 4] p.32 for a detailed proof).
Using H\"older's inequality, one reduces the proof of Lemma 1.4 to
showing that the euclidean volume is bounded from above by the
Monge-Amp\`ere capacity. One can actually show that
$$
Vol(K) \lesssim \exp[-Cap(K)^{-1/n}] \; \;
(\text{see Theorem 7.1 in [Z]}),
$$
which is a much better control than what we actually need.
The last lemma is an elementary exercise, whose proof is given
in [EGZ], Lemma 2.3.

\section{H\"older continuous barriers}

For fixed $\delta >0$ we consider
$\Omega_{\delta}:=\{ z \in \Omega \, / \, \text{dist}(z,\partial \Omega) >\delta \}$ and set
$$
u_{\delta}(z):=\sup_{||\zeta|| \leq \delta } u(z+\zeta), \, z \in \Omega_{\delta}.
$$
This is a psh function in $\Omega_{\delta}$, when $u$ is psh in $\Omega$,
 which measures the modulus of continuity
of $u$. We would like to use Theorem 1.1 applied with $\psi=u_{\delta}$.
However $u_{\delta}$ is not globally defined in $\Omega$, so we need
to extend it with control on the boundary values.
This is the contents of our next result which makes heavy use of the
pseudoconvexity assumption.

\begin{pro} Let $u \in PSH (\Omega) \cap L^{\infty} (\Omega)$ be a psh
function such that $u_{|\partial \Omega}=\phi \in Lip_{2 \alpha}(\partial \Omega)$.
Then there exists a family $(\tilde{u}_{\delta})_{0 < \delta < \delta_0}$
of bounded psh functions on $\Omega$
such that $\tilde{u}_{\delta} \searrow u$ in $\Omega$ as $\delta \searrow 0$, with
$$
\tilde{u}_{\delta} = \left\{ \begin{array}{l}
\max (u_{\delta} - C \delta^{\alpha}, u)  \text{ in } \Omega_{\delta} \\
u  \text{ in } \Omega \setminus \Omega_{\delta}
\end{array} \right.
$$
\noindent In particular $\sup_{\Omega_{\delta}} \vert \tilde{u}_{\delta} - u_{\delta}\vert \leq C \delta^{\alpha}$
for $0 < \delta < \delta_0$.
\end{pro}

The proof relies on the construction of H\"older continuous plurisubharmonic
and plurisuperhamonic barriers for the Dirichlet problem $MA (\Omega,\phi,f)$:

\begin{lem} Fix $\phi \in Lip_{2 \alpha} (\partial \Omega)$, $f \in L^p(\Omega)$,
$p>1$, and set $u := u(\Omega,\phi,f)$.
Then there exists $v, w \in PSH (\Omega) \cap Lip_{\alpha} (\overline \Omega)$ such that
\begin{enumerate}
 \item $ v (\zeta) = \phi (\zeta) = - w (\zeta), \forall \zeta \in \partial \Omega$,\\

 \item $ v (z) \leq u (z) \leq - w (z), \forall z \in \Omega$.
\end{enumerate}
\end{lem}

\begin{proof}
Assume first that $\phi \equiv 0$. We are going to show that
there exists a weak barrier $b_f \in PSH (\Omega) \cap Lip_1 (\Omega)$ for the Dirichlet problem
$MA (0,f,\Omega)$, i.e. a psh function which satisfies
\begin{itemize}
\item $(i)$ $ b_f (\zeta) = 0, \  \forall \zeta \in \partial \Omega,$\\

\item $(ii)$ $ b_f \leq u (\Omega,0,f), $ in $\Omega$,\\

\item $(iii)$ $  \vert b_f (z) - b_f (\zeta) \vert \leq C_1 \vert z - \zeta \vert, \ \forall z \in \Omega, \ \forall \zeta \in \Omega$,
\end{itemize}
for some uniform constant  $C_1 > 0$.

In order to construct $b_f$, we set $u_0 := u (\Omega,0,f)$ and assume first that
the density $f$ is bounded near $\partial \Omega$: there
exists a compact subset $K \subset \Omega$ such that $0 \leq f \leq M$ on $\Omega \setminus K$.
Let $\rho$ be a ${\mathcal C}^2$ strictly plurisubharmonic defining function for $\Omega$.
Then for $A > 0$ large enough the function $b_f := A \rho$ satisfies the condition
$(dd^c b_f)^n \geq M \beta_n \geq f \beta_n$ on $\Omega \setminus K$.
Moreover taking $A$ large enough we also have $ A  \rho \leq  m \leq u_0$ on a neighborhood of $K$,
where $m := \min_{\Omega} u_0.$
Therefore the function $b_f$ is a ${\mathcal C}^2$ plurisubharmonic function on $\Omega$ satisfying the
conditions $(dd^c b_f)^n \geq (dd^c u_0)^n$ on $\Omega \setminus K$ and
$b_f \leq u_0$ on $\partial (\Omega \setminus K)$.
This implies, by the comparison principle [BT 2],
that $b_f \leq u_0$ in $\Omega \setminus K$, hence in $\Omega$.

When $f$ is not bounded near $\partial \Omega$, we can proceed as follows.
Fix a large ball $\mathbb{B} \subset \mathbb{C}^n$ so that $\Omega \Subset \mathbb{B} \subset \mathbb{C}^n$.
Define $\tilde f := f$ in $\Omega$ and $\tilde f = 0$ in $\mathbb{B} \setminus \Omega$.
We can use our previous construction to find a barrier function
$b_{\tilde f} \in PSH (\mathbb{B}) \cap {\mathcal C}^2 (\mathbb{B})$ for the Dirichlet problem
$MA (\mathbb{B},0, \tilde f)$ for the ball $\mathbb{B}$.
Let  $h= u (\Omega,-b_{\tilde f},0)$ denote the Bremermann function
in $\Omega$ with boundary values $-b_{\tilde f}$, for the zero density.
Since $-b_{\tilde f} \in {\mathcal C}^2(\partial \Omega)$,
the psh function $h$ is Lipschitz in $\Omega$ (see [BT 1]),
therefore $b_f :=h + b_{\tilde f} \in PSH (\Omega) \cap Lip_1 (\Omega)$
is a barrier function for the Dirichlet problem $MA(\Omega,0,f)$.

It remains to construct the functions $v,w$ satisfying (1),(2) above.
It follows from [BT 1] that the psh
functions $u (\Omega,\pm \phi,0)$ are H\"older continuous of order $\alpha$.
We let the reader check that the functions $v := u (\Omega,\phi,0) + b_f$
and $w := u (\Omega,-\phi,0) + b_f$ do the job.
\end{proof}

We are now ready for the proof of the proposition.

\begin{proof}
It follows from Lemma 2.2 that
$$
|u(z) - u (\zeta)| \leq C |z - \zeta|^{\alpha},
\; \forall \zeta \in \partial \Omega, \forall z \in \Omega.
$$
The functions $u_{\delta}(z):=\sup_{||\zeta|| \leq \delta} u(z+\zeta)$
are psh in  $\Omega_{\delta}$.
Observe that if $z \in \partial \Omega_{\delta}$ and $\zeta  \in \mathbb{C}^n$ with
$||\zeta||\leq \delta$ then  $z + \zeta \in \partial \Omega$, hence
$u_{\delta} - C \delta^{\alpha} \leq u (z)$.
Thus the functions
$$
\tilde{u}_{\delta} (z) := \left\{ \begin{array}{l}
\sup \{u_{\delta} (z) - C \delta^{\alpha}, u (z)\} \text{ in } \Omega_{\delta} \\
u \text{ in } \Omega \setminus \Omega_{\delta} \end{array} \right.
$$
are psh and bounded in  $\Omega$
and decrease to $u$ as $\delta$ decreases to $0$.
\end{proof}

Our construction of barriers allows us to control the total
mass of the Laplacian of solutions to $MA(\Omega,\phi,f)$.
This will be important in section 4.

\begin{pro}
Fix $0 \leq f \in L^p(\Omega)$, $p>1$, and $\phi \in Lip_{2 \alpha}(\partial \Omega)$.

Then $\Delta u(\Omega,0,f)$ has finite
mass in $\Omega$.
In particular, if $\Delta u(\Omega,\phi,0)$ has finite mass in $\Omega$,
then $\Delta u(\Omega,\phi,f)$ has finite mass in $\Omega$.
\end{pro}

Note that $\Delta u(\Omega,\phi,0)$ has finite mass in $\Omega$ when
$\phi \in {\mathcal C}^2(\Omega)$,
as explained in the proof below.

\begin{proof}
Assume first that $\phi \in {\mathcal C}^2(\Omega)$.
Consider any smooth extension of $\phi$ and correct it by adding $A \rho$, $A>>1$,
in order to obtain a smooth plurisubharmonic extension $\hat{\phi}$
which is defined in
a neighborhood of $\overline{\Omega}$.
Since $\hat{\phi}$ is a subsolution to $MA(\Omega,\phi,0)$ whose
Laplacian has finite mass in $\Omega$, it follows from the comparison
principle that $\Delta u(\Omega,\phi,0)$ also has finite mass
in $\Omega$.

Let $\tilde{f}$ be the trivial extension of $f$ to a large ball $\mathbb{B}$
containing $\Omega$. Let $b_{\tilde{f}} \in {\mathcal C}^2(\mathbb{B})$
be a plurisubharmonic barrier for $MA(\mathbb{B},0,\tilde{f})$ (see the proof of Lemma 2.2).
Then $b_f:=u(\Omega,-b_{\tilde{f}},0)+b_{\tilde{f}}$ is
a plurisubharmonic barrier for $MA(\Omega,0,f)$. Its Laplacian has finite mass
in $\Omega$ since $b_{\tilde{f}}$ is smooth, so it follows from the comparison
principle that $\Delta u(\Omega,0,f)$ has finite mass in $\Omega$.

Set now $v:=u(\Omega,0,f)+u(\Omega,\phi,0)$.
This is a plurisubharmonic function in $\Omega$ such that
$v=\phi$ on $\partial \Omega$ and $(dd^c v)^n \geq f dV$ in $\Omega$.
If $\Delta u(\Omega,\phi,0)$ has finite mass in $\Omega$, then
$\Delta v$ has finite mass in $\Omega$, hence
$\Delta u(\Omega,\phi,f)$ also has finite mass in $\Omega$.
\end{proof}

\section{Gradient estimates}

This section is devoted to the proof of Theorem A.
Let $u=u(\Omega,\phi,f)$ be the unique solution to the
complex Monge-Amp\`ere equation
$$
(dd^c u)^n=f \beta_n \text{ in } \Omega,
$$
with boundary values $u=\phi \in Lip_{2\alpha}(\partial \Omega)$.
Since $f \in L^p(\Omega)$, $p>1$, it follows from [K 2] that
$u$ is a continuous plurisubharmonic function.
Our aim is to show that $u$ is H\"older continuous.

Let $\tilde{u}_{\delta}$ be the functions given by Proposition 3.1.
The stability estimate (Theorem 1.1) applied with $r=2$ yields
$$
\sup_{\Omega_{\delta}} (u_{\delta}-u) \leq
C_1 \delta^{\alpha}+\sup_{\Omega} (\tilde{u}_{\delta}-u)
\leq C_1 \delta^{\alpha}+ C_2 ||u_{\delta}-u||_{L^2(\Omega_{\delta})}^{\gamma},
$$
for $\gamma<2/(nq+2)$, $1/p+1/q=1$.
It remains to
show that $||u_{\delta}-u||_{L^2(\Omega_{\delta})} \leq C_4 \delta$ to conclude the proof.

It will be a consequence of Lemma 3.1 below that $\nabla u \in L^2(\Omega)$.
Assuming this for the moment, we derive the desired upper-bound
on $||u_{\delta}-u||_{L^2(\Omega_{\delta})}$.
Averaging the gradient of u on balls of radius $\delta$ yields, for all $||\zeta||<\delta$,
$$
\left( \int_{\Omega_{\delta}} |u(z+\zeta)-u(z)|^2 dV(z) \right)^{1/2}
\leq \delta \; || \nabla u||_{L^2(\Omega)}.
$$
By Choquet's lemma, there exists a sequence $\zeta_j$, $||\zeta_j|| <\delta$, such that
$u_{\delta}=( \sup_j u_j )^*$, in $\Omega_{\delta}$,
where $u_j(z):=u(z+\zeta_j)$.
Since $u_j-u \leq u_{\delta}-u$,
it follows from Lebesgue's dominated convergence theorem that
$$
\left( \int_{\Omega_{\delta}} |u_{\delta}(z)-u(z)|^2 dV(z) \right)^{1/2}
\leq \delta \; || \nabla u||_{L^2(\Omega)}.
$$
This ends the proof of Theorem A up to the fact, to be established now, that
$u$ has gradient in $L^2(\Omega)$.

\vskip.2cm

Since $u$ is plurisubharmonic and continuous, $\nabla u \in L_{loc}^2(\Omega)$.
It follows from Lemma 3.1 below that $\nabla u \in L^2(\Omega)$ as soon as
$u$ is bounded from below by a continuous psh function $v$ such that
$v \leq u \text{ in } \Omega$, $v=u=\phi$ on $\partial \Omega$, and $\nabla v \in L^2(\Omega)$.
Our extra assumption in Theorem A precisely yields such a function $v$.
Indeed set $v:=u(\Omega,\phi,0)+b_f$, where $b_f$ is the psh barrier
constructed in the proof of Lemma 2.2: this is a psh function such that
\begin{itemize}
\item $v=\phi+0=u$ on $\partial \Omega$;
\item $(dd^c v)^n \geq (dd^c b_f)^n \geq f \beta_n$ in $\Omega$, thus $v \leq u$ in $\Omega$;
\item $\nabla u(\Omega,\phi,0) \in L^2(\Omega)$
and $\nabla b_f \in L^{\infty}(\Omega)$, hence $\nabla v \in L^2(\Omega)$.
\end{itemize}

It is easy to check that $\nabla u(\Omega,\phi,0) \in L^{\infty}(\Omega) \subset L^2(\Omega)$
when $\phi \in {\mathcal C}^2(\partial \Omega)$.
We refer the reader to  [BT 1] for a proof of the more delicate result that this still holds
when $\phi \in {\mathcal C}^{1,1}(\partial \Omega)$.

\begin{lem} Let $u, v \in PSH (\Omega) \cap C^{0} (\overline{\Omega})$ such that
$v \leq u$ on $ \Omega$ and $v = u$ on $\partial \Omega$. Then
$ \int_{\Omega} d u \wedge d^c u \wedge \omega^{n - 1}
\leq  \int_{\Omega} d v \wedge d^c v \wedge \omega^{n - 1}.$
\end{lem}

\begin{proof}
First assume  that $u = v$ near the boundary $\partial \Omega$.
Then we can approximate by smooth plurisubharmonic functions $u_{\delta}\downarrow u$ and
$v_{\delta}  \downarrow v$ with $u_{\delta} \leq v_{\delta}$ on $\Omega_{\delta}$ and
$u_{\delta} = v_{\delta}$
in a neighborhood of $\partial  \Omega_{\varepsilon},$ for $0 < \delta < \varepsilon$ and $\varepsilon$ small enough.
Integrating by parts, we get
$$
\int_{\Omega_{\varepsilon}} d v_{\delta} \wedge d^c v_{\delta} \wedge \omega^{n - 1} =
\int_{\partial \Omega_{\varepsilon}} v_{\delta} d^c v_{\delta} \wedge \omega^{n - 1} -
\int_{\Omega_{\varepsilon}} v_{\delta} dd^c v_{\delta} \wedge \omega^{n - 1} .
$$
Since $v_{\delta} = u_{\delta}$ on a neighbourhood of $\partial \Omega_{\varepsilon}$ we conclude that
$v_{\delta} d^c v_{\delta} = u_{\delta} d^c u_{\delta}$ on this neighbourhood and
then using again integration by parts we get
$$
\int_{\Omega_{\varepsilon}} d v_{\delta} \wedge d^c v_{\delta} \wedge \omega^{n - 1} =
\int_{\Omega_{\varepsilon}}  d u_{\delta} \wedge d^c u_{\delta} \wedge \omega^{n - 1}
+ \int_{\Omega_{\varepsilon}} (u_{\delta} dd^c u_{\delta} - v_{\delta} dd^c v_{\delta}) \wedge \omega^{n - 1}.
$$
On the other hand, since $v_{\delta} = u_{\delta}$ on a neighbourhood of
$\partial \Omega_{\varepsilon}$ we conclude that
$v_{\delta} \wedge d^c u_{\delta} = u_{\delta} \wedge d^c v_{\delta}$ on this neighbourhood and then
\begin{eqnarray*}
\int_{\Omega_{\varepsilon}} u_{\delta} dd^c v_{\delta} - v_{\delta} dd^c u_{\delta}) \wedge \omega^{n - 1}
& = &\int_{\partial \Omega_{\varepsilon}} (u_{\delta} d^c v_{\delta} - v_{\delta} d^c u_{\delta}) \wedge \omega^{n - 1} \\
& - & \int_{\Omega_{\varepsilon}} (d u_{\delta} \wedge d^c v_{\delta} - d v_{\delta}
\wedge d^c u_{\delta}) \wedge \omega^{n - 1} = 0.
\end{eqnarray*}
Therefore
$$
\int_{\Omega_{\varepsilon}} d v_{\delta} \wedge d^c v_{\delta} \wedge \omega^{n - 1}
=  \int_{\partial \Omega_{\varepsilon}} d u_{\delta} \wedge d^c u_{\delta} \wedge \omega^{n - 1}
 + \int_{\Omega_{\varepsilon}} (u_{\delta} - v_{\delta}) (dd^c u_{\delta} + dd^c v_{\delta}) \wedge \omega^{n - 1}.
$$
 Since $v_{\delta} \leq u_{\delta}$ on $\Omega_{\delta},$ we obtain
$$
\int_{\Omega_{\varepsilon}} d v_{\delta} \wedge d^c v_{\delta} \wedge \omega^{n - 1}
\geq \int_{\partial \Omega_{\varepsilon}} d u_{\delta} \wedge d^c u_{\delta} \wedge \omega^{n - 1}.
$$
We know by Bedford and Taylor's convergence theorem that
$d v_{\delta} \wedge d^c v_{\delta} \wedge \omega^{n - 1}
\rightarrow d v \wedge d^c v \wedge \omega^{n - 1}.$
Taking the limit when $\delta \searrow 0$ we get
$$
\int_{\overline{\Omega}_{\varepsilon}} d v \wedge d^c v \wedge \omega^{n - 1}
\geq \int_{\Omega_{\varepsilon}} d u \wedge d^c u \wedge \omega^{n - 1}.
$$
Taking the limit when $\varepsilon \downarrow 0$ we get the required inequality.

 Now if we only know that $u = v,$ we can define for $\varepsilon > 0$
$u_{\varepsilon} := \sup \{u - \varepsilon , v\}.$ Then $v \leq u_{\varepsilon}$ on $\Omega$ and $u_{\varepsilon} = v$
near the boundary of $\Omega$. Therefore we have for $\delta > 0$ small enough
$$
\int_{\Omega} d v \wedge d^c v \wedge \omega^{n - 1} \geq \int_{\Omega} d u_{\varepsilon}
\wedge d^c v_{\varepsilon} \wedge \omega^{n - 1}.
$$
 Now by Bedford and Taylor's convergence theorem, we know that
$d u_{\varepsilon} \wedge d^c u_{\varepsilon} \wedge \omega^{n - 1} \to d u \wedge d^c u \wedge \omega^{n - 1}$
as $\varepsilon \downarrow 0$. Therefore we have
$$
\int_{\Omega} d v \wedge d^c v \wedge \omega^{n - 1} \geq \int_{\Omega}
d u \wedge d^c u \wedge \omega^{n - 1},
$$
which proves the required inequality.
\end{proof}

\section{Laplacian estimates}

This section is devoted to the proof of Theorem B.
We use the same method as above. The finiteness of the
total mass of $\Delta u(\Omega,\phi,0)$
allows a good control (see Lemma 4.2)
on the terms $\hat u_{\delta} - u$,
where
$$
 \hat u_{\delta} (z) := \frac{1}{\tau_{n} \delta^{2 n}} \int_{\vert
 \zeta - z\vert \leq \delta} u (\zeta) d V (\zeta), z \in \Omega_{\delta},
$$
where $\tau_n$ denotes the volume of the unit ball in $\mathbb{C}^n$.
We shall compare $\hat u_{\delta}$ to $u_{\delta}$ in Lemma 4.1 below.

It follows from the construction of plurisubharmonic H\"older continuous barriers
that the solution $u = u (\Omega,\phi,f)$ is H\"older continuous near the boundary,
i.e. for $\delta > 0$ small enough, we have
\begin{equation}
u (z) - u (\zeta) \leq c_0 \delta^{\alpha},
\end{equation}
for $z, \zeta \in \overline{\Omega}$ with
$\hbox{dist} (z, \partial \Omega) \leq \delta, \hbox{dist} (\zeta, \partial \Omega) \leq \delta $
and $ \vert z - \zeta \vert \leq \delta.$
\vskip.1cm

The link between $u_{\delta}$ and $\hat{u}_{\delta}$ is made by the
following lemma.

\begin{lem} Given $\alpha \in ]0, 1[$, the following two conditions are equivalent.

$(i)$ There exists $\delta_0, A > 0$ such that for any
$0 < \delta \leq \delta_0$,
$$
u_{\delta} - u \leq A \delta^{\alpha} \, \hbox{ on } \, \Omega_{\delta}.
$$

$(ii)$ There exists $\delta_1, B > 0$ such that for any $0 <\delta< \delta_1$,
$$
\hat u_{\delta} - u \leq  B \delta^{\alpha} \, \hbox{ on }  \Omega_{\delta}.
$$
\end{lem}

\begin{proof}
Observe  that $\hat u_{\delta} \leq u_{\delta}$ in $\Omega_{\delta}$, hence
$(i) \Longrightarrow (ii)$ follows immediately.

We now prove that $(ii) \Longrightarrow (i)$.
We need to show that there exists $A,\delta_0>0$ such that
for $0 < \delta \leq \delta_0$,
$$
\omega (\delta) := \sup_{z \in \Omega_{\delta}} [u_{\delta} (z) - u (z)] \leq A \delta^{\alpha}.
$$
Fix $\delta_{\Omega}>0$ small enough so that $\Omega_{\delta} \neq \emptyset$ for
$\delta \leq 3 \delta_{\Omega}$.
Since $u$ is uniformly continuous, for any fixed
$0< \delta <\delta_{\Omega}$,
$$
\nu (\delta) := \sup_{\delta < t \leq \delta_{\Omega}}
\omega (t)  t^{- \alpha} < + \infty.
$$

We claim that there exists $\delta_0  > 0$  small enough so that for any
$0 < \delta  \leq \delta_0$,
$$
\omega (\delta) \leq  A \delta^{\alpha},
\text{ with }
A= (1+4^{\alpha}) c_0+2^{\alpha} 4^n B+\nu(\delta_{\Omega}),
$$
where $c_0$ is the constant arising in inequality (1), while
$B$ is the constant from condition (ii).
Assume this is not the case.
Then there exists $0 < \delta  < \delta_{\Omega}$ such that
\begin{equation}
\omega (\delta)> A \delta^{\alpha}.
\end{equation}
Set $\delta:=\sup \{ t <\delta_{\Omega} \, / \, \omega(t)> A t^{\alpha} \}$.
Then
\begin{equation}
\frac{\omega(\delta)}{\delta^{\alpha}} \geq A \geq \frac{\omega(t)}{t^{\alpha}}
\text{ for all } t \in [\delta,\delta_{\Omega}].
\end{equation}
Since $u$ is continuous, we can find
$z_0 \in \overline{\Omega_{\delta}}$, $  \zeta_0 \in \overline{\Omega}$ with
$ \vert z_0 - \zeta_0 \vert  \leq  \delta$ s.t.
$$
\omega(\delta) = \sup_{z \in \Omega_{\delta}} \left[ \sup_{w \in B(z,\delta)} u(w)  -u(z)
\right]=u (\zeta_0) - u (z_0).
$$
\vskip.2cm

We first derive a contradiction if $z_0$
is close enough to the boundary of $\Omega$.
Assume that $\hbox{dist} (z_0,\partial \Omega) \leq 3  \delta$.
Take $z_1 \in \partial \Omega$ such that
$\hbox{dist} (z_0,\partial \Omega)  = \hbox{dist} (z_0, z_1) \leq 4   \delta$.
It follows from $(1)$ that
$$
\omega(\delta) =  u (\zeta_0) - u (z_0) =
[u (\zeta_0) - u (z_1)] + [u (z_1) - u (z_0)] \leq [1+4^{\alpha}] c_0 \delta^{\alpha}.
$$
This contradicts $(3)$.

Thus we can assume that $\hbox{dist} (z_0,\partial \Omega) >  3  \delta$.
Fix $b>1$ so that $\hbox{dist} (z_0,\partial \Omega) >  (2b+1)  \delta$.
Thus any $z \in \mathbb{B}(\zeta_0,b\delta)$ satisfies
$z \in \mathbb{B}(z_0,[b+1]\delta)$, hence $z \in \Omega_{b \delta}$.
By using inequality (3) with $t=b\delta$, we get
$u(\zeta_0)-u(z) \leq b^{\alpha} \omega(\delta)$, hence
\begin{equation}
u(z) \geq u(\zeta_0)-b^{\alpha} \omega(\delta),
\, \text{ for all } z \in \mathbb{B}(\zeta_0,b\delta).
\end{equation}

Observe now that $\mathbb{B}(\zeta_0,\delta) \subset \mathbb{B}(z_0,[b+1]\delta)$, hence
\begin{eqnarray*}
\hat u_{(b+1) \delta} (z_0)
& = & \left(\frac{b}{b+1} \right)^{2n} \hat u_{b \delta} (\zeta_0)  +
\frac{1}{\tau_n (b+1)^{2n} \delta^{2 n} }
\int_{ \mathbb{B} (z_0, (b+1) \delta) \setminus  \mathbb{B} (\zeta_0, b \delta)} u d V \\
& \geq &
\left(\frac{b}{b+1} \right)^{2n}  u(\zeta_0)  +
\left[(1 - \frac{b^{2n}}{(b+1)^{2 n}}\right]
[u (\zeta_0) -  b ^{\alpha} \omega (\delta)] \\
&=& u(\zeta_0) -b^{\alpha} \left[1 - \frac{b^{2n}}{(b+1)^{2 n}}\right] \omega(\delta),
\end{eqnarray*}
where we have used the subharmonicity of $u$ together with inequality (4).
Since $u(\zeta_0)=u(z_0)+\omega(\delta)$, we infer, letting $b \rightarrow 1$,
$$
\hat u_{2 \delta} (z_0)  \geq u(z_0)
+4^{-n}\omega(\delta).
$$

We now use assumption (ii), only considering small enough values of $\delta>0$:
since $\hat{u}_{2 \delta}(z_0) \leq u(z_0)+B 2^{\alpha} \delta^{\alpha}$, we get
$$
\omega(\delta) \leq 4^n 2^{\alpha} B \delta^{\alpha} < A \delta^{\alpha}.
$$
This contradicts the definition of $\delta$, hence we have proved that
$(ii) \Rightarrow (i)$.
\end{proof}

It is straightforward to check that if (i) is satisfied, then
$u$ belongs to $Lip_{\alpha}(\overline{\Omega})$.
Thus Theorem B will be proved if we can establish (ii).
It follows from Theorem 1.1 that is suffices to get control
on the $L^1$-average of $\hat{u}_{\delta}-u$.
This is the contents of our next result.

\begin{lem}
For $\delta > 0$ small enough, we have
$$
\int_{\Omega_{\delta}} [\hat u_{\delta} (z) - u (z)] d V_{2 n} (z) \leq c_n
\Vert \Delta u\Vert \delta^2,
$$
where $c_n > 0$ is a uniform constant.
\end{lem}

\begin{proof}
It follows from Jensen's formula that for $z \in \Omega_{\delta}$ and $0 < r < \delta$,
$$
\frac{1}{\sigma_{2 n - 1}} \int_{\vert \xi\vert = 1} u (z + r \xi) d S_{2 n - 1} = u (z) +
\int_0^{r} t^{1 - 2 n} (\int_{\vert \zeta\vert \leq t} dd^c u \wedge \beta_{n - 1}) d t.
$$
Using polar coordinates we get, for $z \in \Omega_{\delta}$,
$$
 \hat u_{\delta} (z) - u (z)  = \frac{1}{\sigma_{2 n - 1}
\delta^{2 n}} \int_{0}^{\delta} r^{2 n - 1} d r \int_0^{r} t^{1 - 2 n}
(\int_{\vert \zeta - z\vert \leq t} dd^c u \wedge \beta_{n - 1}) d t.
$$
Finally Fubini's theorem yields
\begin{eqnarray*}
\int_{\Omega_{\delta}} (\hat u_{\delta} - u) d V_{2 n} &\leq&
 a_n \delta^{- 2 n}\int_{0}^{\delta} r^{2 n - 1} d r
\int_0^{r} t^{1 - 2 n} (\int_{\vert \zeta\vert \leq t} (\int_{\Omega} \Delta u) d t \\
&\leq& c_n  \delta^2 \Vert \Delta u\Vert.
\end{eqnarray*}
\end{proof}

This ends the proof of Theorem B since by proposition 2.3,
$\Delta u=\Delta u(\Omega,\phi,f)$ has finite mass in $\Omega$.
\vskip.2cm

We now give a simple example which shows that one can not expect a better
exponent than $\alpha=2/nq$, for $1/p+1/q=1$.

\begin{exa}
Consider $u(z_1,\ldots,z_n):=|z_1|^{\alpha} \cdot ||(z_2,\ldots,z_n)||^2$.
This is a plurisubharmonic function in $\mathbb{C}^n$ which is H\"older-continuous
of exponent $\alpha \in ]0,1[$. We let the reader check that
$$
(dd^c u)^n=f dV, \;
\text{ with } f(z)=\frac{1}{|z_1|^{2-n\alpha}} g(z_2,\ldots z_n),
$$
where $g>0$ is a smooth density.

Given $p>1$, $f$ belongs to  $L_{loc}^p(\mathbb{C}^n)$ whenever
$\alpha=\varepsilon+2/nq$, for some $\varepsilon>0$.
This shows that we cannot get a better exponent than
$2/nq$ in Theorems A,B.
\end{exa}

redpajama_set_name: RedPajamaArXiv


Wealthy Investors Answer The Question "Are You Wealthy?"
"Be proud of who you are".
That expression of positive mental attitude is aimed at improving the psychological health of anyone suffering from self-doubt. But such phrases can also be aimed at those people who have it made and must deal with the pressures of being financially secure.
Many people would love to have the problems that come with great wealth, but that does not mean the problems aren't real and don't require attention.
Spectrem's new study on the population of investors who have a net worth of $25 million or more examines the attitudes those investors have about their wealth level. The Wealthiest Americans asks those investors to discuss the pitfalls of their wealth and the perception of wealth they carry with them.
Such pitfalls and perceptions can impact how a wealthy investor feels about investing, from attitudes towards risk, wealth transfer and philanthropy.
The perception of wealth is a multi-faceted influence on how investors proceed with investing and personal finances. These investors with a net worth of $25 million or more were asked to consider the question "are you wealthy?" and place their response on a 0-to-100 scale, with a yes response represented by the 100 end of the scale. On average, those investors rated their average answer at 80.49, which is definitely an affirmative response but still indicates some level of uncertainty about the response, which is hard for anyone not worth $25 million to understand.
The uncertainty level of investors with this level of net worth can impact investment decisions and should be understood by advisors for handling investment recommendations accordingly.
The first step in understanding the mindset of wealthy investors is understanding their overall opinion about the value of great wealth on their lives. It is revealing that 79 percent of $25 million plus investors admit that "I attribute my happiness in large part to the wealth I have accumulated." That is the sort of statement some people might shy away from, and perhaps some of the 21 percent who did not respond in the affirmative do so just because it does not sound appropriate, but apparently great wealth does provide happiness for most people.
This fact is important to advisors to understand to what lengths investors will go to protect their wealth. Investment decisions may be made to protect wealth rather than extend it because of the value the wealth level has for the investor.
While investors are happy because of their wealth, it also causes them unhappiness, or could cause them unhappiness, based on what happens to that money in the future. Seventy-four percent of $25 million plus investors worry about how the next generation in their family will handle the money they will receive an an inheritance or in a wealth transfer over time.
There is no question this concern can color how an investor approaches wealth transfer and estate decisions. The level of concern about the impact of wealth on the next generation will determine the path the investor takes in passing their wealth along (and might prompt greater charitable donations as a preventative measure).
On a slightly amusing data point, wealthy investors did point out one of the drawbacks of significant wealth. Two-thirds of all investors report being hit up for money by friends and family, and that includes almost 90 percent of investors currently under the age of 38.
Yes, wealthy people have "first-world" problems, but problems they still are. More importantly from an advisor's standpoint, an investor's perception and personal attitude toward their wealth plays a role in investment decisions. Sometimes the effect of the problems is subtle, but sometimes it is right out in the open and needs to be considered when making recommendations.


text:
In der Mathematik werden mit dem Querschnitt bestimmte Mengen bezeichnet. Wenn  ein Mengensystem über der Grundmenge  ist, dann heißt  ein Querschnitt von , falls  alle Mengen in  schneidet.

Die kleinstmögliche Mächtigkeit eines Querschnitts von  heißt Querschnittszahl  des Mengensystems .

Beispiele
1.  sei die Menge der Mitglieder einer Partei mit zwei Vorsitzenden.  sei die Menge der Interessengruppen in dieser Partei, beispielsweise die vier Gruppen Gewerkschafter, Konservative, Frauen und Mediziner. Jede dieser vier Gruppen stellt eine Teilmenge der Menge aller Mitglieder in der Partei dar. Wenn jeder dieser Gruppen einer der beiden Vorsitzenden angehört, dann bilden diese einen Querschnitt von . Dazu müsste zum Beispiel der Vorsitz aus einem konservativen Mediziner und einer Gewerkschafterin bestehen.
Findet sich unter allen Mitgliedern der Partei kein solches Paar, das zusammen alle Interessengruppen bedient, so ist die Querschnittszahl größer als 2. Gibt es ein Mitglied, das zugleich konservativ, weiblich und Mediziner ist und zusätzlich der Gewerkschaft angehört, dann ist die Querschnittszahl sogar nur 1.

2. Sei  die euklidische Ebene und  die Menge aller abgeschlossenen Einheitsquadrate, die die -Achse schneiden. Ein Querschnitt dieses Mengensystems wäre z. B. , denn jedes abgeschlossene Einheitsquadrat, das die -Achse schneidet, enthält auch einen solchen Punkt. Dieser Querschnitt ist abzählbar unendlich, die Querschnittszahl  somit höchstens abzählbar unendlich. Man überlegt sich leicht, dass es in diesem Mengensystem keinen endlichen Querschnitt geben kann, also ist  abzählbar unendlich.

Mengenlehre


Dhangkhar lhakhang is 7km south of the Zhemgang Dzong and can be reached by a farm road. The temple is located on the right side of Dhangkhar village. The farm road leads to Dhangkhar temple first and ends at the village.
Dhangkhar lhakhang is a two-storey traditional Bhutanese house built right beside the farm road. The lhakhang is enclosed within an old wall. There is an old cypress tree in front of the temple, and a stone next to the temple that contains the footprint of Guru Rinpoche. And on the left side of the temple at the entrance is a big prayer wheel.
The date of the construction and the founder of the temple are unknown, but the local sources say that it was the seat of Pema Lingpa (1450-1521) who visited and blessed many places in Zhemgang. The community still practices the Peling tradition of Buddhism (Peling Choelug).
According to oral sources, the village was named Dhangkhar (dhang means clear, khar means clear/immaculate in the Kheng language) because the early morning sun rays from the Rewotala peak fall on the village, hence the name.
Dhangkhar lhakhang is one of the oldest temples under Trong gewog. It is believed that in the mid 8th century, Guru Rinpoche is said to have visited Dhangkhar on this return from Ugyen Drak in Nabji after settling the dispute between the kings, Nawoche and Sindhu Raja. He left his footprint on the stone which is on the right side of the temple.
The oral source said that earlier the temple was below the farm road and it was a one-storey temple. In the 1980s the temple was shifted to the present location upon the advice of Kunzang Thinley, the former Lam Neten of Zhemgang because there was not enough space for religious and other social functions. The whole community was involved in the dismantling and in the reconstruction of the new temple. Ap Sangay says that the temple was rebuilt around 1982.
Dhangkhar lhakhang is a two-storey temple built in the traditional Bhutanese style and it is surrounded by a wall. The main altar is on the first floor and the ground floor is used as a storeroom during the funeral rites. There is a separate room attached to the right side of the temple and this room is used as a kitchen and also as a torkhang (room for preparation of ritual cakes) during religious festivals.
The sacred objects in the temple are a statue of Guru Nangsi Zilngon in the centre, a statue of the Zhabdrung Ngawang Namgyal (1592-1651) on the right side, and the historical Buddha (Shakyamuni) on the left side. The Guru Nangsi Zilngon's statue was already there in the old temple, but the other two statues were added after the temple was shifted to its present location. Inside the temple there are no wall paintings, and the temple is more like a traditional Bhutanese house.
Ap Sangay is the current caretaker of the temple and he has been for over 50 years. He takes care of the daily offerings and the community takes care of other big rituals and offerings. The rituals are conducted by the gomchens (lay-practitioners) of Dangkhar village. The community practices Peling tradition and performs the following activities.
From the 13th – 15th day of the 1st month of the Bhutanese calendar, Nyungney (fasting retreat and prayers) is sponsored by the community.
From the 8th – 10th day of the 4th month of the Bhutanese calendar, Kanjur (Buddhist Canon) recitation is conducted.
From the 8th – 10th day of the 7th month of the Bhutanese calendar, Nyungney (fasting retreat and prayers) is sponsored by the community.
From the 13th – 15th day of the 7th month of the Bhutanese calendar, Baza Guru Dungdrup/ Mani Dungdrup recitation is organized.
Pema Youden, Asst. Lecturer, College of Language and Culture Studies, Taktse, Trongsa, Royal University of Bhutan, 2017.


text:
Sound system may refer to:
Technology[edit]
Sound reinforcement system, a system for amplifying audio for an audience
High fidelity, a sound system intended for accurate reproduction of music in the home
Public address, an institutional speech-reinforcement or public safety announcement system
Shelf stereo, a compact sound system for personal use
Arts and entertainment[edit]
Sound system (DJ), a group of disc jockeys performing together
Sound system (Jamaican), a group of disc jockeys, engineers and MCs playing ska, rocksteady or reggae music
Sound System (album), a compilation album by The Clash
Sound-System (album), a 1984 album by Herbie Hancock
Soundsystem (311 album), 1999
Sound System Records, an Australian record label
"Sound System", a song by Operation Ivy
Phonology, the study of sound systems of languages
This disambiguation page lists articles associated with the title Sound system.
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