We are sitting in port at the end of a long day. This city has proven a surprise to all, not the least of which is the fact that it is under local control. Not a hint of the world government anywhere.
I wonder how long that will last.
We have filled the holds and will top off tomorrow by securing a few containers on deck. The Captain is pleased.
Elijah and I bid goodbye and God speed to the Raj family, who will depart from here upon their journey and mission.
They in turn committed to hold us up in prayer before the One. And had a tender parting with the twins, whom they have surrendered to our care.
In the end, knowing the ultimate goal of the followers of Hamashiach will be the port that is next on our ship's schedule, has made it easy to decide what to do.
Elijah is returning to the city and the ship to bring news to the Purser about his family. I will remain here with Mary and the children for four days. If the Purser can join us before that time, that would be excellent. Otherwise we will begin a trek overland to the next port and plan to meet everyone there.
Elijah and I passed a few moments together, each committing the other to the love and care of our Master.
We bid God speed and confirmation crowned us.


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Tide have 7 players enter portal; 2 for Dawgs
Alabama had seven players enter the NCAA's transfer portal a day after the team's loss in the College Football Playoff National Championship, including reserve tight end Jahleel Billingsley.
The group of Alabama players appeared in the portal late Tuesday afternoon. Billingsley, a junior from Chicago, had 35 receptions for 543 yards and six touchdowns the past two seasons. He was a preseason All-SEC selection but played behind Cameron Latu at the end of the season. After recording a total nine receptions for 152 yards and a touchdown in wins over Southern Miss and Florida, Billingsley had seven receptions the rest of the season.
Other Alabama transfer portal entrants included linebacker Drew Sanders, who had 24 tackles, one sack, four quarterback hurries and two pass breakups. Reserve quarterback Paul Tyson, the great-grandson of former Alabama coach Paul "Bear" Bryant, also entered the portal, as did offensive lineman Tommy Brown, who started one game this season.
Georgia, which beat Alabama 33-18 in the championship game, had two players enter the portal on Tuesday: defensive back Ameer Speed, who started three games and recorded 13 tackles this season; and wide receiver Jaylen Johnson, who had three receptions for 29 yards in 14 games.
Cautious Chiefs prepping as if Mahomes will play
Sabalenka into first Australian Open quarterfinal
Penguins goalie Jarry, defenseman Petry back
Despite ankle injury, Mahomes leads Chiefs to win
Isaac returns in Magic win, feels 'relief and peace'

redpajama_set_name: RedPajamaCommonCrawl


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"I don't know," he said. "In terms of the hoopla around it, I keep my head down; try not to pay too much attention to it. I'm honestly so busy with other things, working on Boardwalk (the HBO series Boardwalk Empire), I've got a play this summer, I've got a family; I just don't have time to sit around and get too excited about any of it. I've got to keep working."
And work Shannon has, with several films hitting the theaters lately. He still has time for stage work with a production of Sam Shepard's Bluegrass-based drama Simpatico at Shannon's A Red Orchid Theatre in Chicago in July and August.
Shannon's unimpressed veneer should not be mistaken for disdain. This is not an art-house snob who deigned to do a potential blockbuster.
In fact, Shannon says one of the reasons he wanted to do the movie was to work with director Zack Snyder, known for his work on Watchmen , 300 and Sucker Punch .
"He's got a phenomenal visual sensibility that I really appreciate," Shannon says. "I think the movie's going to look incredible, at the very least," Shannon said. "And I think the rest of it will be pretty strong too."
The superhero franchise film is unfamiliar territory, but Shannon is in a familiar role: the bad guy. In fact, an article in the current issue of Rolling Stone calls Shannon "Hollywood's go-to villain," citing his recent performances as a contract killer in The Iceman and a disgraced government agent in Boardwalk Empire. He earned an Oscar nomination for best supporting actor in 2009 for his role in Revolutionary Road, as a troubled man in with searing insights in the lives of the lead couple, played by Leonardo DiCaprio and Kate Winslet. The winner that year in his category was the late Heath Ledger for playing The Joker in The Dark Knight, one of the rare times a superhero film has taken home a major Oscar.
As Zod in Man of Steel, Shannon plays one of Superman's fellow Kryptonians, a man who thinks they should unite in an invasion of Earth.
Shannon said making the movie was fun, but "it's still a lot of hard work, training and staging fights and whatnot. So it's serious fun."
If Man of Steel lives up to box office expectations, it should raise Shannon's profile even more.
Related stories from Lexington Herald Leader
movies-news-reviews
'Man of Steel': Superman without the warmth and wit
Michael Shannon plays Superman's nemesis, General Zod, center, in Man of Steel, directed by Zack Snyder. Shannon got his acting start with Actors Guild of Lexington. He has become, according to Rolling Stone magazine, "Hollywood's go-to villain."
Shannon and his girlfriend, actress Kate Arrington, attended Man of Steel's world premiere on Monday in New York. Evan Agostini/Invision/AP
Photo slideshow: Day 1 at the Forecastle Festival
By JOHN CARUCCI Associated Press
Pam Grier's role as the owner of a hardware store in rural Nebraska in the ABC sitcom "Bless This Mess" isn't a complete departure.
MORE MOVIE NEWS & REVIEWS
Movie News & Reviews
With 'The Lion King' and 'The Mandalorian,' director Jon Favreau cements his status as a Disney MVP
Movie review: Lynn Shelton's politically barbed 'Sword of Trust,' starring Marc Maron, takes on Civil War truthers
Movie review: Roddy Doyle's 'Rosie' is a heartbreaking tale of an Irish family's resilience
Movie review: David Crosby hoping for a little more time in elegiac doc 'Remember My Name'
Movie review: 'Ray & Liz' is a bleak, beautiful memoir of an unhappy English childhood


Founded in 1977 by entrepreneur Vittorio Moretti, Bellavista is nowadays one of the most refined and elegant classic method sparkling wines of Italy; synonymous with timeless Italian lifestyle and class.
Bellavista winery owns almost 200 hectares of vineyards, divided in 147 parcels, throughout 10 municipalities in the Franciacorta area.
The Estate not only comprises the vineyards, but also L'Albereta Relais & Châteaux, a gourmet restaurant and a renowned spa.
Each vineyard is harvested and fermented separately, then elevated in small French oak barriques, to enhance the natural expression of the different terroirs.
All activities are performed manually to guarantee the utmost attention to detail; from hand picking to manual remuage.
The name "Bellavista", in Italian "beautuful view", derives from the spectacular property position, on a high hill overlooking Lake Iseo.
In 1981 Vittorio Moretti met the talented oenologist Mattia Vezzola, whose work was paramount to the definition of Bellavista style and in 2007 Vezzola was named "Oenologist of the year" by Gambero Rosso – Slow Food.
At the winery, every action aims at respecting nature and achieving elegance, enjoyability and longevity, distinctive yet recognizable in every harvest.


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A-Mei Returns To Singapore For Concert In The New ...
A-Mei Returns To Singapore For Concert In The New Year
Tuesday September 15 2015
Taiwan's queen of Mando-pop, A-Mei or Chang Hui Mei, will be ringing in the new year with a concert in Singapore, at the National Stadium on January 9th
Featuring a stage set-up costing US$3.5 million (S$4.96 million), the singer will perform as her alter egos — aMEI and Amit — in what has been hailed as her best concert show in recent years. Organisers said fans can expect to have the best of both worlds, with an aural and visual feast with specially curated songs which that will run the gamut of aMEI's classic ballads, catchy dance tracks and AMIT's expressive rock jams. Tickets are available through SportsHub.
Laneway Announces Singapore Line-Up
Friday September 11 2015
Singapore Sports Hub
Wed Jun 18 2014 - Sun Jan 10 2016
A-Mei
1 Stadium Drive, Kallang


Is Good News Really Good News?
Lots of positive earnings surprises and two ways to interpret Moody's credit data.
Maybe, just maybe, the economy is bottoming.
For one day, at least, senior executives and investors can be hopeful after three high-profile companies reported quarterly results that offered reasons for optimism.
The biggest news: Microsoft said after the markets closed on Wednesday that fiscal fourth-quarter revenues would come in at $6.5 billion to $6.6 billion, up from previous guidance of $6.3 billion to $6.5 billion.
The markets reacted ebulliantly to the news.
Of course, it also announced that it will record a first-ever investment loss of $2.6 billion.
Motorola, meanwhile, also offered a bit of encouraging news, if you are in the mood to be an optimist.
Sure, it reported a second-quarter operating loss of $232 million, or 11 cents a share, excluding special items. And, yes, it compares with an operating profit of $551 million, or 25 cents a share, in the same period last year.
And, yes, Motorola warned in April that earnings would come in a few pennies worse than the first-quarter loss of 9 cents per share.
However, analysts then predicted, on average, that the quarterly loss would amount to 12 cents per share. Presto! Motorola beat expectations.
What's more, handset sales came in at $2.5 billion. Estimates called for $2.2 billion to $2.3 billion. Handset orders, which reflect the future, amounted to $2.9 billion versus an estimate of $2.8 billion.
And, although its semiconductor loss far exceeded estimates, the company predicted double-digit growth for next year.
Sure, the one-time profitable Internet company—not a small feat–wound up reporting pro forma earnings of one penny per share. The consensus forecast called for a breakeven quarter.
Alas, its net loss was $48.5 million, which includes all restructuring and acquisition costs. Last year, it reported net earnings of $53.3 million.
Wal-Mart Stores Inc. said that June same-store sales surged by 6.9 percent, much higher than Wal-Mart's and Wall Street's forecast of just 3 to 5 percent.
Paper products company Willamette Industries Inc. said that second- quarter earnings rose about 6 percent, to $70.8 million, or 64 cents per diluted share, two pennies better than the consensus forecast.
Enron Corp., the energy marketing and trading giant, said second- quarter earnings jumped nearly 40 percent, beating the average Wall Street estimate by 3 cents.
First Union said its operating earnings came in three pennies per share ahead of expectations even though they fell by 9 percent.
Harris Interactive Inc., the polling company, said that it expects its fiscal fourth-quarter loss to be smaller than analysts' forecasts due to growth in health-care research and to cost-cutting.
Personal care product maker Helen of Troy Limited said that stronger demand for its products boosted its fiscal first quarter results and also raised its earnings and sales estimates for the full year.
There are also two ways to interpret Moody's recent report on credit quality.
The credit rating agency said that the credit strength of U.S. corporations fell sharply in the second quarter of 2001 as rating downgrades outpaced upgrades by a factor of three-to-one.
This was the 13th consecutive quarter of credit deterioration, especially in the industrial sector.
And, Moody's warned of decidedly negative trends over the next three months.
Even so, Moody's said downgrades for U.S. companies fell to 152 in the quarter ended June 30 from 176 in the first quarter and 153 in fourth- quarter 2000.
And Moody's upgraded the bonds of 58 companies during the second quarter, seven more than the 51 in the first quarter and 23 percent more than the 47 in the fourth quarter of 2000.
However, the net result of these actions is that downgrades far exceeded upgrades.
Altogether, $148.2 billion of bonds were downgraded last quarter, compared with $259.6 billion in the first quarter, and $373.5 billion in fourth-quarter 2000. These are also clearly encouraging numbers.
And, the amount of debt upgraded totaled $80.4 billion last quarter, $28.8 billion in the first quarter, and $257.6 billion in the last period of 2000.
Meanwhile, a total of 52 companies were put on review for possible downgrade last quarter, compared with 39 reviews for possible upgrade. This is an improvement from the 66 reviews for downgrade and 29 reviews for upgrade initiated during the first quarter, and an even more dramatic turn from the 91 downgrade reviews and 27 upgrade reviews during fourth-quarter 2000. When a company's rating is put on review, the review generally concludes within three months, Moody's noted.
AOL Time Warner Inc. cut about 30 people in its interactive marketing group, according to Reuters.
A jury award against American Home Products Corp. regarding alleged health damage from the "fen-phen" diet drug cocktail has been reduced substantially, to $9.2 million from $56.5 million.
Alliance Capital Management LP filed a shelf registration to sell up to $600 million in debt securities and warrants.


Infective endocarditis (IE) most often results from bacterial infection, but infections caused by fungi, Coxiella, or Chlamydia are not rare. Infection usually involves heart valves not always previously known to be abnormal, in particular a bicuspid aortic valve, mitral valve prolapse, or (rarely) a septal defect or ventricular aneurysm. Coarctation of the aorta, patent ductus arteriosus, aneurysms, or arteriovenous shunts may be the site of infective endarteritis. Prosthetic valves may be involved, and infection at the site of implantation of foreign material including devices poses a particularly difficult problem.


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Girl Out of Water, by Laura Silverman
Anise can't imagine living anywhere but with her Dad in their oceanfront house in Santa Cruz, surfing every day, and spending time with her friends. But things are changing -- one last summer before her friends start to go away to college and beyond. So, Anise is committed to making the most of this summer.
But then her aunt is in a bad accident and Dad announces that they are going to have to spend the summer at her aunt's house - in Nebraska - to take care of her cousins while her aunt recovers. There's definitely no surfing in Nebraska! And worse, no friends either! She'll miss the whole summer! Will her friends even remember her when she gets back?
Nebraska, in the end, isn't all that bad. She spends the summer caring for her cousins and develops a stronger bond with them. And she meets a handsome one-armed skateboarder named Lincoln with whom she grows close. But as the summer comes to close, she finds herself torn between her two worlds -- the comfort of the old and the new world she's recently discovered.
The best way to describe my disquiet with this book is to point out that the culminating sequences of the story are told in rushed flashback, as if even the author knew that they would be so dull that she didn't want to spend time with them! But then, how interesting can it be to have your climax be having your protagonist admit that she's been terrible to her friends all summer, by ignoring them? Or, in other words, this Santa Cruz world of Anise's was so uninteresting to the author that she barely mentions it for over half of the book.
And the human growth story (about a young women discovering that absence does not have to mean forgetting is simply slipped out there for our approval). The subplots provide plenty of additional frustrations, including Lincoln's alleged disability, which is played down so far that it serves no actual purpose in the story (Hey! There's this guy and he only has one arm, but you'd never know it because he does everything and we never call it out for any reason). I understand Silverman's desire to paint disabled people as "normal" but this is really about invisibility. Anises's cousin and her anger at her mother's injury or Anise's half-hearted searching for her mother also hung out there as unresolved.
In sum, it was light, fluffy, and readable, but really fell apart story-wise for me.
Windfall, by Jennifer E. Smith
Alex, Approximately, by Jenn Bennett
The Warden's Daughter, by Jerry Spinelli
Defy the Stars, by Claudia Gray
Hidden, by Miriam Halahmy
Radio Silence, by Alice Oseman
It Started With Goodbye, by Christina June


= APPLICATION RECEIVED = Perfectly and quietly nestled on the elevated ground floor at the north facing front of this ideally located double bricked complex boasting a fresh, contemporary and luxurious renovation, common manicured gardens upon entry and a short walk to Dulwich Hill Shopping Precinct or Lewisham Train Station. This outstanding apartment is the perfect infusion of stylish design, location and affordability.
= Two generous sundrenched north facing double bedrooms with mirrored built in wardrobes.


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In light of COVID-19 mandates, MT Sunrise Rotary will be hosting virtual meetings until further notice. Our next meeting will be 7:00 AM Friday (3/5).
Our program this week will feature Patrick Marchese, Chairman Mequon Parks and Open Spaces and Mike Gies, Mequon Forestry Superintendent. Patrick and Mike will be discussing Mequon's Park & Open Space long range plans for Rotary Park & along the Riverwalk in vicinity of the Gateway, with emphasis on improvements to be made this year. (Scroll down for bio)
Bruce Carter (3/5)
Danila Danesi (3/12)
My grandfather once told me that there were two kinds of people: those who do the work and those who take the credit. He told me to try to be in the first group; there was much less competition. – Indira Gandhi
Member Spotlight - Bruce Carter
Hi I'm Bruce Carter! My family re-located to Mequon-Thiensville from the Kansas City area just about 4 years ago. I'm a CPA but work in operations. I received a degree in Finance & Banking from the University of Missouri. I'm Vice President of Home Health Operations for Advocate Aurora Post Acute Division and spend my time commuting between Milwaukee and Chicago. Recently my wife and I established Advanced Business Solutions, a business consulting and tax services LLC, which we fondly call our "second act".
I've been married for 33 years to my best friend, Beth. Have 4 fantastic children: Bridget, who resides in Houston, Texas with her husband Jarrod, and our first grand child, George. Twins: William a 1st year law student at Ohio State and Olivia – a Theater Major, improv specialist, and looking to become a script writer, who got a little delayed moving to Chicago with the pandemic and is working as a Patient Service Rep for Aurora Urgent Clinic downtown Milwaukee. Andrew, is a Freshman at Kansas State University working on a Professional Sales Management degree and a member of Delta Sig fraternity. Finally, we have our 6 year old Golden Retriever, June.
Most important lesson learned in business is to over communicate and add value. I joined Rotary when we re-located at the recommendation of a good friend, who is a member of Rotary in Leawood, KS. I value service over self. Most memorable Rotary moment, is the pleasure meeting each one of you and look forward to creating those memorable moments!
Aziz Abu Sarah - Meaningful Conversations
In case you were unable to catch Aziz Abu Sarah's talk on 2/26 or just want to see it again you're welcome to watch the recording of our meeting by clicking the image above.
Peanuts by Charles Schulz
A Candlelight Walk in the Woods - Photo Gallery
The Perfect Place
21-year-old to open bike shop along Interurban Trail in Mequon
By Lisa Curtis - Special to the News Graphic
MEQUON — Biking enthusiast Sam Wolf knew that he wanted to be his own boss someday. He just didn't realize how soon that day would come.
While many of his peers are in college or simply trying to figure out a career path, the 21-year old Wolf has signed a lease in one of the coveted Foxtown retail spaces.
Wolf will open Trailside Recreation bike store this month, catering to both casual and seasoned bicyclists. His shop at 11090 Weston Drive, Suite 300, is directly adjacent to the Ozaukee Interurban Trail.
It was his uncle, a developer who worked on Foxtown, who told him one night at dinner that he had the "perfect place" for a bike shop.
"We went out there and looked at it, and I said, 'yeah, this is the perfect spot,'" Wolf said. "We got started on it within that week."
Trailside Recreation will offer rentals and the sale of bicycles and eBikes, bicycle equipment and apparel.
Wolf, who lives in Grafton and grew up in Cedar Grove, worked for Erik's bike shop in Grafton for about five years. He said he tends to work on bikes more than he rides, but loves to spend time on the Pleasant Valley Nature Park mountain bike trails.
How we got there: WHO African region is certified free of wild poliovirus
In a challenging time, Rotary members and partners have shown drive in overcoming challenges
By Diana Schoberg Image credit: Andrew Esiebo
It was May 1997, and Sam Okudzeto, then a member of the African Regional PolioPlus Committee from Ghana, was flying to Sierra Leone for what he anticipated would be a routine annual meeting about polio eradication in Africa. But when his plane touched down and he looked out the window, he saw that soldiers, guns drawn, had flooded the tarmac.
As Okudzeto made his way to passport control, he asked an airport official what was going on. "Listen carefully," he was told. "You can hear the guns." There had been a coup that morning.
"All we heard was boom, boom, boom," Okudzeto recalls. "Then I realized that the aircraft that had brought us had gone."
With no taxis running, Okudzeto and the other meeting participants who had been on the plane walked to a nearby hotel. "We all went to our rooms and put our luggage down and then went to the restaurant," he recalls. "I asked for the biggest and juiciest sole fish I had ever had in my life, because it might be my last supper." (Luckily it wasn't, and four days later, Okudzeto and the others caught a helicopter out of the country.)
"There is an enemy in life — and it's fear," he says now. "For those who are not afraid, it's amazing what you can do. It's fantastic to see the result now: Africa is [wild] polio-free."
Craig Burnett - 2020-21 District Governor
District 6270 Update
Raised in suburban Detroit, Craig has made his home in a variety of places in the Midwestern U.S., settling in Oshkosh 29 years ago, and he hopes the Badger fans among you will forgive him for his degrees from Michigan State University.
He joined the Rotary Club of Oshkosh Southwest in 1999 and has served in a variety of board and committee roles since then, including the presidency in 2008-09. He served as Assistant Governor for the 5 Oshkosh and Fond du Lac Rotary clubs from 2015 - 2018.
Over his decades in Oshkosh, he has also served in a wide range of non-Rotary volunteer leadership roles within the community. At present, he serves as a volunteer conflict mediator for the Winnebago Conflict Resolution Center, is on the board of directors of his community's 137 year-old performing arts theater The Grand Oshkosh, and is a co-producer of the TEDxOshkosh conference.
Craig is the owner of the Oshkosh-based Web consulting and development firm Clareo Consulting, LLC, and has been married to his wife Heidi, a veterinarian, for 24 amazing years. In the spare moments of down-time he can find, he loves to do anything outdoors, especially if it involves wilderness spaces without electricity where modern technology does not work.
Lucia Francis - MT Sunrise Rotary


text:
Q: finishActivity doesn't recognize in Fragment I previously start speech recognition service from MainActivity by startActivityForResult and put onActivityResult also in the main. I also stop the service by finishActivity in the MainActivity. All work O.K.
Now I pass the active of the service to the Fragment and also the onActivityResult in the Fragment. It get there the replay. But when i put the finishActivity in the Fragment it mark it in red and didn't recognize it.
It is look it need activity so I add before getActivity.finishActivity(THE REQUEST CODE) and it run. But it is look that it doesn't close the service any more.
Where could be the problem?
Bar.

A: Try using this.stopSelf() if in the same service
Or inside your onActivityResult try calling stopService(intent)
Refer
alternative to finish() method for Service class? To kill it dead
for the same

A: See startActivityForResult() from a Fragment and finishing child Activity, doesn't call onActivityResult() in Fragment. The problem is the fragment's startActivityForResult() is remapping your request code, but the getActivity().finishActivity() is not mapping back. You need to start the activity from MainActivity instead of the fragment.

redpajama_set_name: RedPajamaStackExchange


If you want products that are safe for the Guaifenesin Protocol, if you have sensitivities or allergies to salicylates, gluten or soy or if you simply want fragrance free products these products are for you! You can feel secure in knowing that I have a vested interest in always keeping these products safe. Why waste time and energy reading labels when it's very possible to make a mistake? Companies do change their ingredients quite often but many times keep their packaging exactly the same. I have seen many people make those mistakes therefore making the Guaifenesin Protocol ineffective. "Knowing" that all of the products you use on a daily basis are Salicylate Free has to be your #1 priority to be successful in this endeavor. If you have allergies or sensitivities to salicylates these products will help to make your life easier and worry free as well.
"I am deeply grateful to Andrea Rose. We were unable to find products in the beginning that were and would remain salicylate free. She answered my call for help and created a variety that are now cherished by many women and men. I am able to tell all of our patients how safe they are with her products AND IT WILL NEVER CHANGE.
Andrea has willingly continued her direct contact with people with fibromyalgia and has helped them solve an untold number of problems. She has certainly extended herself well beyond anything I expected. She remains inspirational."
Having suffered with Fibromyalgia for many years I know first hand the devastation this disease can do. I feel so very lucky to have found the Guaifenesin Protocol. It reversed this disease and allowed me to regain my life!
This protocol is not easy but the alternative wasn't acceptable to me. I just bit the bullet and did what I had to do and I am here to tell you that it works!
The most difficult part of the protocol was finding products that did not contain any ingredients that contained plant oils or extracts since they are Salicylates and according to Dr St Amand and years of his experience with patients they will block the effects of the guaifenesin so that the protocol simply will not work.
That meant reading every label of every product that I needed to use on a daily basis from toothpaste, hair care, skin care and make up. It was extremely frustrating and scary since I wanted to get well and was terribly afraid I would make a mistake. I also wanted to help Dr St Amand in some way since this protocol was saving my life! When I voiced my concerns he thought for a moment and then said "Why not start your own cosmetic company? Do you know how that would help me?" Dr St Amand knew that all too often patients were making mistakes when they searched for safe products. That in turn would sabotage their progress and success and subsequently the protocol did not work for them.
I instinctively knew at that moment that this was a need that had to be filled! Within 6 months my company was born!
Remember, if you are on the Guaifenesin Protocol, these products are just as important as the correct dose of guai.
There are no gimmicks here. Simply high quality products developed to keep you safe and to make your lives easier. My products are salicylate free, fragrance free, gluten free, soy free, animal cruelty free and most products are also paraben free.
My company was designed to help those of you on the Guaifenesin Protocol and for those of you who cannot tolerate salicylates. Since I have a vested interest you can always be sure that all products you purchase from me will be safe.
Best of health to you always!


It's Sometimes Legal to Park in a Red Zone in SF? Whaaaaaa??
Editor's Note: Welcome to the SF Parking Hack of the Week, wherein we'll hook you up with one awesome, game-changing SF parking hack every Friday courtesy of the guy who wrote the book on them (no, like, really, he wrote a book about them).
First (this is important): the part of the curb on either side of a driveway, or the "curb cut" as SFMTA calls it, is legally considered to be part of THE DRIVEWAY, not the curb. There were 29,272 citations issued last year for blocking a driveway. Many of these citations were given for having part of the bumper of the car jutting into the "curb cut" space. Logic leads us to believe that the curb cut is part of the curb, not part of the driveway, but illogically/legally, it is indeed part of the driveway.
Also (this is important-er): you can't park in the red-painted area on the curb which extends the curb cut for another 3ft on either side of the driveway -- narrow driveway owners pay the city handsomely for this paint job to make it easier for them to get in and out.
That said (this is important-est!): if the red-painted curb does not have an official DPT, MTA, or SFMTA mark on it, it's not a legal red zone (haaaaack alert!). Only red zones painted by the city with a DPT, SFMTA, or MTA stencil are enforceable, meaning all those spots in the Marina that you just barely couldn't fit in because of the little bit of red after the curb cut? If it doesn't have the stencil on it, feel free to get all up in it.
David LaBua is a leader in the sustainable urban mobility movement, author of Finding the Sweet Spot - The Insider's Guide to Parking in SF, and the founder of VoicePark. Have an SF parking question? Hit him up at sanfrancisco@thrillist.com.


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Draw for M20 EHF EURO 2022 held in Gondomar
EHF / Courtney Gahan18 February 2022, 13:00
The 16 nations set to contest the M20 EHF EURO 2022 in Portugal on 7–17 July learned their preliminary round opponents on Friday afternoon as the draw took place in Gondomar.
The competition was forced into cancellation in 2020 due to the Covid-19 pandemic, making the 2022 edition the first in four years. Slovenia were the last U20 champions crowned after beating France in the final in 2018, while Germany picked up the bronze medal ahead of 2022 hosts Portugal in fourth.
However, the generation that will vie for the 2022 title did have the chance to don their national jerseys for a younger age category championship, as a one-off M19 EHF EURO was held in 2021, with Germany taking the title.
Therefore considered the European current title holders in this generation, Germany were seeded into pot 1 for the draw for the 13th edition of the event. Their preliminary round opponents will be Iceland, Italy and Serbia in group D.
The silver medallists from the M19 EHF EURO 2021, Croatia, will take on Sweden, France and Montenegro in group C.
Spain, who clinched bronze at the M19 EHF EURO 2021, will meet Portugal, Norway and Poland in group A, while Slovenia — fourth place at the 2021 event — will start in group B. Their opponents will be Denmark, Hungary and Faroe Islands.
From the preliminary round, the top two teams will progress to the main round, while the remainder will continue in the intermediate round. The semi-finalists will then be the two top-ranked teams in each of the two main round groups.
A number of special guests attended or assisted with the draw, including EHF President Michael Wiederer, Portuguese Handball Federation President Miguel Laranjeiro, and representatives of the hosting municipalities.
Portugal national team members from the past, present and future also had a role in the draw, with the likes of legendary Carlos Resende, Rui Silva, Fabio Magalhaes and Martim Costa and participating.
"I love the handball today and I am very pleased to see that we have a good generation. I sincerely hope that this event will be a good opportunity for them to grow — not just to win, but to grow as players," said Carlos Resende, while Fabio Magalhaes commented:
"I'm very pleased to be here, seeing the growth of these guys. I have the opportunity to train with them in the club and the national team, and they are doing a good job.
"They are going to achieve great things in the future. They will have to start now in their own home, playing this important competition, and make the best job they can."
Denmark are the record winners of the M20 EHF EURO, with four titles and six medals overall. Germany have won three times, Spain twice and Slovenia, Poland and Yugoslavia each once.
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Younger Age Category
Croatia, Poland book last tickets to under-21 worlds
REVIEW: The two teams join the 12 European nations already qualified for the World Championship in Germany and Greece
read more 3 weeks ago
Two places for U21 World Championship to be booked
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Younger Age Category analysis provides insight into game dev…
NEWS: Analyses of the seven Younger Age Category events to have taken place in 2022 have been published
'Hand in Hand': Costa brothers approach the top together
NEWS: In a 25-minute film, Sporting's future stars Francisco and Martim Costa give exclusive insights in their lives
Eight "Respect Your Talent" alumni who can shine in EHF EURO…
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text:
\section{Introduction}
\label{xxsec1}

Let $\mathcal{A}$ be an associative $\ast$-algebra over the complex field $\Bbb{C}$.
For any $A, B\in \mathcal{A}$, one can denote a ``new product" of $A$ and $B$ by
$A\diamond B=AB+BA^\ast$, and this new product $\diamond$ is usually said to be \textit{$\ast$-Jordan product}.
Such kind of product based on Jordan bracket naturally
appears in relation with the so-called Jordan $\ast$-derivations and plays an important
role in the problem of representability of quadratic functionals by sesqui-linear
functionals on left-modules over $\ast$-algebras (see \cite{Molnar2, MolnarSemrl, Semrl1}).
The product is workable for us to characterize ideals, see  \cite{BresarFosner, Molnar1, Molnar3}.
Especial attention has been paid to understanding mappings which preserve the product
$AB+BA^\ast$ between $\ast$-algebras, see \cite{DaiLu, HuoZhengLiu, LiLu, LiLuFang1, ZhaoLi2}.

The question of to what extent the multiplicative structure of an algebra determines its
additive structure has been considered by many researchers  over the past decades.
In particular, they have investigated under which conditions bijective mappings
between algebras preserving the multiplicative structure necessarily preserve the
additive structure as well. The most fundamental result in this direction is due
to W. S. Martindale III \cite{Martindale} who  proved that every bijective
multiplicative mapping from a prime ring containing a nontrivial idempotent
onto an arbitrary ring is necessarily additive. Later, a number of authors
considered the Jordan-type product or Lie-type product and proved that,
on certain associative algebras or rings, bijective mappings which preserve
any of those products are automatically additive, see
\cite{BaiDu, BresarFosner, CuiLi, DaiLu, HuoZhengLiu, LiLu, LiLuFang1, Molnar4, Molnar5, ZhaoLi2}.

An additive mapping $\delta: \mathcal{A} \longrightarrow \mathcal{A}$ is called an
\textit{additive derivation} if $\delta(AB)=\delta(A)B+A\delta (B)$ for all  $A,B \in \mathcal{A}$.
Furthermore, $\delta$ is said to be an \textit{additive $\ast$-derivation} provided that $\delta$
is an additive derivation and satisfies $\delta(A^{\ast})=\delta(A)^\ast$ for all $A\in \mathcal{A}$.
Let $\delta: \mathcal{A}\longrightarrow \mathcal{A}$ be a mapping
(without the additivity assumption). We say that $\delta$ is a
\textit{nonlinear $\ast$-Jordan derivation} if
$$
\delta(A \diamond B)=\delta(A) \diamond B+A \diamond \delta(B),
$$
holds true for all $A,B\in \mathcal{A}$. Similarly, a mapping $\delta:
\mathcal{A}\longrightarrow \mathcal{A}$ is called a \textit{nonlinear $\ast$-Jordan triple derivation} if it satisfies the condition
$$
\delta(A \diamond B \diamond C)=\delta(A) \diamond B \diamond C+A \diamond \delta(B)\diamond C+A \diamond B \diamond \delta(C)
$$
for all $A, B, C\in \mathcal{A}$, where $A \diamond B \diamond C=(A \diamond B) \diamond C$ .
We should be aware that $\diamond$ is not necessarily associative.

Given the consideration of $\ast$-Jordan derivations and $\ast$-Jordan triple derivations, we can
further develop them in one natural way. Suppose that $n\geq 2$ is a fixed positive
integer. Let us see a sequence of polynomials with $\ast$
$$
\begin{aligned}
p_1(x_1)&=x_1,\\
p_2(x_1,x_2)&=x_1 \diamond x_2=x_1x_2+x_2x_1^\ast,\\
p_3(x_1,x_2,x_3)&=p_2(x_1,x_2) \diamond x_3=(x_1 \diamond x_2) \diamond x_3,\\
p_4(x_1,x_2,x_3,x_4)&=p_3(x_1,x_2,x_3) \diamond x_4=((x_1 \diamond x_2) \diamond x_3) \diamond x_4,\\
\cdots &\cdots,\\
p_n(x_1,x_2,\cdots,x_n)&=p_{n-1}(x_1,x_2,\cdots,x_{n-1}) \diamond x_n\\
&=\underbrace{(\cdots ((}_{n-2}x_1 \diamond x_2)\diamond x_3)\diamond \cdots \diamond x_{n-1}) \diamond x_n.
\end{aligned}
$$
Accordingly, a \textit{nonlinear $\ast$-Jordan $n$-derivation} is a mapping $\delta:
\mathcal{A} \longrightarrow \mathcal{A}$ satisfying the condition
$$
\delta(p_n(A_1, A_2,\cdots, A_n))=\sum_{k=1}^n
p_n(A_1,\cdots, A_{k-1}, \delta(A_k), A_{k+1},\cdots, A_n)
$$
for all $A_1,A_2,\cdots, A_n\in \mathcal{A}$. This notion makes the best use of the definition of Lie-type derivations
and that of $\ast$-Lie-type derivations,  see \cite{FosnerWeiXiao, Lin1, Lin2}.
By the definition, it is clear that every $\ast$-Jordan derivation is a
$\ast$-Jordan 2-derivation and each $\ast$-Jordan triple derivation is a $\ast$-Jordan 3-derivation.
One can easily check that each nonlinear $\ast$-Jordan derivation on $\mathcal{A}$ is
a nonlinear $\ast$-Jordan triple derivation. But, we don't know whether the converse statement is true.
$\ast$-Jordan 2-derivations, $\ast$-Jordan 3-derivations and $\ast$-Jordan $n$-derivations
are collectively referred to as \textit{$\ast$-Jordan-type derivations}.
$\ast$-Jordan-type derivations on operator algebras have been studied by
several authors. Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B(H)}$ be the algebra of all
bounded linear operators on $\mathcal{H}$.  Li et al in \cite{LiLuFang2} showed that if $\mathcal{A}\subseteq \mathcal{B(H)}$ is a von Neumann
algebra without central summands of type $I_1$, then $\delta: \mathcal{A}\longrightarrow \mathcal{B(H)}$ is a
nonlinear $\ast$-Jordan derivation if and only if $\delta$ is an additive $\ast$-derivation.
More recently, this result is extended to the case of nonlinear $\ast$-Jordan triple derivations by Zhao and Li \cite{ZhaoLi1}.
Taghavi et al \cite{TaghaviRohiDarvish} and Zhang \cite{Zhang} independently investigate
$\ast$-Jordan derivations on factor von Neumann algebras, respectively. It turns out that
each nonlinear $\ast$-Jordan derivation on a factor von Neumann algebra is an additive $\ast$-derivation.

Inspired by the afore-mentioned works, we will concentrate on giving a description of nonlinear
$\ast$-Jordan-type derivations on von Neumann algebras. The organization of this paper is as follows.
We recall and collect some indispensable facts with respect to von Neumann algebras in the
second section \ref{xxsec2}. The third Section \ref{xxsec3} is devoted to our main result
Theorem \ref{xxsec3.1} and its proof. The main theorem states that every nonlinear
$\ast$-Jordan-type derivation on a von Neumann algebra without central summands of type $I_1$
is an additive $\ast$-derivation. Similar statements are also given for factor von Neumann algebras
and standard operator algebras without proofs. Some potential topics for the future research are presented in the last Section \ref{xxsec4}.

\section{Preliminaries}
\label{xxsec2}

Throughout this paper, $\mathcal{H}$ denotes a complex Hilbert space and $\mathcal{B}(\mathcal{H})$
is the algebra of all bounded linear operators on $\mathcal{H}$. A \textit{von Neumann algebra}
$\mathcal{A}$ is weakly closed, self-adjoint algebra of operators on $\mathcal{H}$ containing
the identity operator $I$. The set $\mathcal{Z}(\mathcal{A}) =\{ S \in \mathcal{A}: ST=TS \ \ \text{for all}\ \ T\in \mathcal{A}\}$
is called the \textit{centre} of $\mathcal{A}$.
A projection $P$ is called a \textit{central abelian projection} if $P \in \mathcal{Z}(\mathcal{A})$
and $P\mathcal{A}P$ is abelian. Recall that the \textit{central carrier} of $A$, denoted by $\overline{A}$, is
the smallest central projection $P$ satisfying the condition $PA=A$. It is straightforward to
check that the central carrier of $A$ is the projection onto the closed subspace spanned by
$\{ BA(x):B \in \mathcal{A}, x\in \mathcal{H} \}$. If $A$ is self-adjoint, then the \textit{core}
of $A$, denoted by $\underline{A}$, is $\text{sup}\{ S\in \mathcal{Z}(\mathcal{A}): S=S^{\ast}, S \leq A\}$. If $P$
is a projection, it is clear that $\underline{P}$ is the largest central projection $Q$ with $Q\leq P$.
A projection $P$ is said to be \textit{core-free} if $\underline{P}=0$. It is not difficult to see
that $\underline{P}=0$ if and only if $\overline{I-P}=I$.

To round off the proof of our main theorem, we need to give some necessary lemmas.

 \begin{lemma}\label{xxsec2.1}{\rm \cite[Lemma 4]{Miers}}
Let $\mathcal{A}$ be a von Neumann without central summands of type $I_1$.
Then each nonzero central projection in $\mathcal{A}$ is the central carrier of
a core-free projection in $\mathcal{A}$.
\end{lemma}

\begin{lemma} \label{xxsec2.2}{\rm \cite[Lemma 2.2]{LiLuFang2}}
Let $\mathcal{A}$ be a von Neumann algebra on a Hilbert space $\mathcal{H}$.
Let  $A \in \mathcal{B}(\mathcal{H})$ and $P \in \mathcal{A}$ is a projection with
$\overline{P}=I$. If $ABP=0$ for all $B\in \mathcal{A}$, then $A=0$.

\end{lemma}

\begin{lemma}\label{xxsec2.3}
Let $\mathcal{A}$ be a von Neumann algebra without central summands of type $I_1$.
For any $A\in \mathcal{A}$ and for any positive integer $n\geq 2$, we have
$$
p_n\left (A, \frac{1}{2}I, \cdots, \frac{1}{2}I \right)= \frac{1}{2}\left(A+A^{\ast} \right)
$$
and
$$
p_n\left (\frac{1}{2}I, \cdots, \frac{1}{2}I,A \right)= A.
$$
 \end{lemma}

\begin{proof}
By a recursive calculation, we know that
$$
\begin{aligned}
p_n\left(A, \frac{1}{2}I, \cdots, \frac{1}{2}I \right )&=p_{n-1}\left(\frac{1}{2}(A+A^{\ast}), \frac{1}{2}I, \cdots, \frac{1}{2}I\right )\\
&=p_{n-2}\left(\frac{1}{2} \left( A+A^{\ast} \right), \frac{1}{2}I, \cdots, \frac{1}{2}I \right )\\
&=p_{n-3}\left(\frac{1}{2} \left( A+A^{\ast} \right), \frac{1}{2}I, \cdots, \frac{1}{2}I \right )\\
&=\cdots\\
&=\frac{1}{2}\left ( A+A^{\ast} \right ).
\end{aligned}
$$
Similarly, we also have
$$
\begin{aligned}
p_n\left (\frac{1}{2}I, \cdots, \frac{1}{2}I,A \right)&=p_{n-1}\left (\frac{1}{2}I, \cdots, \frac{1}{2}I,A \right)\\
&=p_{n-2}\left (\frac{1}{2}I, \cdots, \frac{1}{2}I,A \right)\\
&=\cdots\\
 &=A.
\end{aligned}
$$
\end{proof}

Let $\mathcal{A}$ be a von Neumann algebra without central summands of type $I_1$.
By Lemma \ref{xxsec2.1}, we know that there exists a nonzero central projection
$P$ such that $\underline{P}=0$ and $\overline{P}=I$. For the convenience of discussion,
let us set $P_1=P$, $P_2=I-P$. We write $\mathcal{A}_{ij}=P_i\mathcal{A}P_j$.
Thus one gets $\mathcal{A}=\sum_{i,j=1}^2 \mathcal{A}_{ij}$. We denote the imaginary
unit by ${\rm i}$.

\begin{lemma}\label{xxsec2.4}
Let $\mathcal{A}$ be a von Neumann algebra without central summands of type $I_1$.
For any  $A\in \mathcal{A}$ with $A=\sum_{i,j =1}^2 A_{ij}$ and $A_{ij}\in \mathcal{A}_{ij}$, we have
\begin{enumerate}
\item[(a)] $P_1 \diamond A=0$ implies that $A_{12}=A_{21}=A_{11}=0$.
\item[(b)] $P_2 \diamond A=0$ implies that $A_{12}=A_{21}=A_{22}=0$.
\item[(c)] $(P_2-P_1)\diamond A=0$ implies that $A_{11}=A_{22}=0$.
\end{enumerate}
\end{lemma}

\begin{proof}
Let us first prove the assertion (a).
We have
$$
\begin{aligned}
0&=P_1 \diamond A=P_1A+AP_1^{\ast}=P_1A+AP_1\\
&=A_{12}+A_{21}+2A_{11},
\end{aligned}
$$
which leads to $A_{12}=A_{21}=A_{11}=0$.

The other two assertions can be achieved by an analogous manner.
\end{proof}

\section{Main Theorem and Its Proof}
\label{xxsec3}

We are in a position to give the main theorem of this article which can be stated
as follows.

 \begin{theorem} \label{xxsec3.1}
Let $\mathcal{A}$ be a von Neumann algebra without central summands of
type $I_1$. Then a mapping $\delta: \mathcal{A} \longrightarrow  \mathcal{B(H)}$ satisfies the rule
$$
\delta(p_n(A_1, A_2,\cdots, A_n))=\sum_{k=1}^n
p_n(A_1,\cdots, A_{k-1}, \delta(A_k), A_{k+1},\cdots, A_n)
$$
for all $A_1, A_2,\cdots, A_n \in \mathcal{A}$ if and only if $\delta$ is an additive $\ast$-derivation.
 \end{theorem}

\begin{proof}
The proof of this theorem can be realized via a series of claims.

\vspace{2mm}

\noindent {\bf Claim 1.} $\delta(0)=0$.
$$
\delta(0)=\delta(p_n(0,0,\cdots,0))=\sum_{k=1}^{n}\overset{\ \ \ \  k}{p_n(0,\cdots,\delta(0),\cdots,0)}=0.
$$

\noindent {\bf Claim 2.}  For any $A \in \mathcal{A}$, we have $\delta \left( \frac{1}{2}I \right) \diamond A=0$.
\vspace{2mm}

Note that  the fact $\frac{1}{2}I=p_n\left(\frac{1}{2}I,\cdots,\frac{1}{2}I \right)$.  By Lemma \ref{xxsec2.3} it follows that
$$
\begin{aligned}
\delta \left( \frac{1}{2}I \right)=&\delta\left (p_n\left(\frac{1}{2}I,\cdots,\frac{1}{2}I \right) \right)\\
=&\sum_{k=1}^{n}\overset{\ \ \ \  k}{p_n\left(\frac{1}{2}I,\cdots,\frac{1}{2}I,\delta \left( \frac{1}{2}I \right),\frac{1}{2}I,\cdots,\frac{1}{2}I \right)}\\
=&\frac{n-1}{2}\left ( \delta \left( \frac{1}{2}I \right) +\delta \left( \frac{1}{2}I \right)^{\ast}\right)+\delta \left( \frac{1}{2}I \right).
\end{aligned}
$$
This gives
$$
\delta \left( \frac{1}{2}I \right) +\delta \left( \frac{1}{2}I \right)^{\ast}=0.
$$
That is,
$$
\delta \left( \frac{1}{2}I \right) \diamond \frac{1}{2}I=0. \eqno(3.1)
$$
Using the relation (3.1), we get
$$
\begin{aligned}
\delta(A)=&\delta\left (p_n\left(\frac{1}{2}I,\cdots,\frac{1}{2}I,A \right)\right)\\
=&\sum_{k=1}^{n-1}\overset{\ \ k}{p_n\left(\frac{1}{2}I,\cdots,\frac{1}{2}I,\delta \left( \frac{1}{2}I \right),\frac{1}{2}I,\cdots,\frac{1}{2}I,A \right)}+p_n\left(\frac{1}{2}I,\cdots,\frac{1}{2}I,\delta(A) \right)\\
=&\delta \left( \frac{1}{2}I \right)A +A\delta \left( \frac{1}{2}I \right)^{\ast}+\delta(A).
\end{aligned}
$$
for all $A \in \mathcal{A}$. Thus we obtain
$$
\delta \left( \frac{1}{2}I \right)A +A\delta \left( \frac{1}{2}I \right)^{\ast}=0.
$$
That is,
$$
\delta \left( \frac{1}{2}I \right) \diamond A=0.
$$

\noindent{\bf Claim 3.} For any $A_{ll}\in \mathcal{A}_{ll}, B_{ij}\in \mathcal{A}_{ij}\, (i,j, l=1,2, i \neq j)$, we have
$$
\delta(A_{ll}+B_{ij})=\delta(A_{ll})+\delta(B_{ij}).
$$
We only need to prove the case of $i=l=1,j=2$, and the proofs of the other cases are rather similar
and are omitted here. Let us write
$$
M=\delta(A_{11}+B_{12})-\delta(A_{11})-\delta(B_{12}).
$$
It is sufficient for us to show that $M=0$. Since
$$
p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2,A_{11} \right)=0
$$
and
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I, P_2,A_{11}+ B_{12} \right)= p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I,  P_2,B_{12} \right),
$$
we by {\bf Claim 2} have
$$
\begin{aligned}
&p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, \delta\left(P_2\right),A_{11}+ B_{12}\right)+ p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2, \delta\left(A_{11}+ B_{12}\right) \right)\\
=&\delta\left( p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2, A_{11}+ B_{12} \right)\right)\\
=&\delta\left( p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, P_2,A_{11} \right)\right)+\delta\left( p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2, B_{12} \right)\right)\\
=&p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,\delta\left(P_2\right), A_{11}\right)+ p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2, \delta\left(A_{11}\right) \right)\\
&+p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, \delta\left(P_2\right),B_{12} \right)+p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, P_2,\delta\left(B_{12}\right) \right).
\end{aligned}
$$
We therefore get
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I,P_2,M \right)=0.
$$
In light of Lemma \ref{xxsec2.3}, we obtain
$$
p_2\left ( P_2, M \right)=P_2 \diamond M=0.
$$
It follows from Lemma \ref{xxsec2.4} that
$$
M_{12}=M_{21}=M_{22}=0.
$$
Notice that
$$
p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2-P_1, B_{12} \right)= ( P_2-P_1)\diamond B_{12}=0
$$
and
$$
p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2-P_1, A_{11}+ B_{12} \right)= p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, P_2-P_1,A_{11}\right).
$$
By {\bf Claim 2}, we observe that
$$
\begin{aligned}
&p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, \delta(P_2-P_1),A_{11}+ B_{12} \right)\\
&+ p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I, P_2-P_1, \delta\left(A_{11}+ B_{12}\right) \right)\\
=&\delta\left( p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, P_2-P_1,A_{11}+ B_{12}\right)\right)\\
=&\delta\left( p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, P_2-P_1,B_{12} \right)\right)+\delta\left( p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I,  P_2-P_1,A_{11} \right)\right)\\
=&p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,\delta(P_2-P_1), B_{12} \right)+p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I, P_2-P_1, \delta\left(B_{12}\right) \right)\\
&+p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I, \delta(P_2-P_1), A_{11}  \right)+ p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, P_2-P_1, \delta\left(A_{11} \right) \right).
\end{aligned}
$$
Thus we arrive at
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I, P_2-P_1, M \right)=0.
$$
Taking into account Lemma \ref{xxsec2.3}, we get
$$
p_2\left (  P_2-P_1, M \right)=(P_2-P_1)\diamond M=0.
$$
Applying Lemma \ref{xxsec2.4} yields that
$$
M_{11}=0.
$$
We therefore have $M=0$. That is,
$$
\delta(A_{11}+B_{12})=\delta(A_{11})+\delta(B_{12}).
$$
The other cases can be verified by an analogous manner.

\vspace{2mm}

\noindent{\bf Claim 4.} For any $B_{12}\in \mathcal{A}_{12}, C_{21}\in \mathcal{A}_{21}$, we have
$$
\delta(B_{12}+C_{21})=\delta(B_{12})+\delta(C_{21}).
$$
We only need to show that
$$
M=\delta(B_{12}+C_{21})-\delta(B_{12})-\delta(C_{21})=0.
$$
Since
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I,P_2-P_1, B_{12} \right)= (P_2-P_1)\diamond B_{12}=0
$$
and
$$
p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2-P_1,C_{21} \right)= p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2-P_1,B_{12}+C_{21} \right)=0,
$$
we obtain
$$
\begin{aligned}
&p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,\delta(P_2-P_1), B_{12}+C_{21} \right)\\
&+ p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2-P_1, \delta\left(B_{12}+C_{21}\right) \right)\\
=&\delta\left( p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2-P_1, B_{12}+C_{21} \right)\right)\\
=&\delta\left( p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, P_2-P_1,B_{12} \right)\right)+\delta\left( p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2-P_1, C_{21} \right)\right)\\
=&p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, \delta(P_2-P_1),B_{12} \right)+p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2-P_1,\delta\left(B_{12}\right) \right)\\
&+p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, \delta(P_2-P_1), C_{21} \right)+ p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2-P_1, \delta\left(C_{21}\right) \right).
\end{aligned}
$$
Hence, we have
$$
p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, P_2-P_1,M \right)=0.
$$
Applying Lemma \ref{xxsec2.3} gives
$$
p_2\left ( P_2-P_1, M \right)=(P_2-P_1)\diamond M=0.
$$
By Lemma \ref{xxsec2.4}, we know that
$$
M_{11}=M_{22}=0.
$$
Note that the facts
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I,B_{12},P_1 \right)=0
$$
and
$$
p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, B_{12}+C_{21},P_1\right)=p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,C_{21},P_1 \right).
$$
Using similar computations as the above, we get
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I,M,P_1 \right)=0.
$$
In view of Lemma \ref{xxsec2.3} and the fact $M_{11}=M_{22}=0$, one can see that
$$
M_{21}=0.
$$
On the other hand, we should remark that
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I, B_{12}+C_{21},P_2\right)=p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I,B_{12},P_2 \right)
$$
and
$$
p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,C_{21},P_2 \right)=0.
$$
Using similar arguments as the above, one can get $M_{12}=0$.
\vspace{2mm}

\noindent{\bf Claim 5.} For all $A_{11}\in \mathcal{A}_{11}, D_{22}\in \mathcal{A}_{22}$, we have
$$
\delta(A_{11}+D_{22})=\delta(A_{11})+\delta(D_{22}).
$$
It is sufficient to prove that
$$
M=\delta(A_{11}+D_{22})-\delta(A_{11})-\delta(D_{22})=0.
$$
Since
$$
p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2, A_{11} \right)=0
$$
and
$$
p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2,A_{11}+ D_{22} \right)= p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I,P_2, D_{22} \right),
$$
we konw that
$$
\begin{aligned}
&p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I, \delta\left(P_2\right), A_{11}+ D_{22}\right)+ p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2, \delta\left(A_{11}+ D_{22}\right) \right)\\
=&\delta\left( p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, P_2,A_{11}+ D_{22}\right)\right)\\
=&\delta\left( p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, P_2,D_{22}\right)\right)+\delta\left( p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,P_2, A_{11} \right)\right)\\
=&p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,\delta\left(P_2\right),D_{22} \right)+p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I, P_2,\delta\left(D_{22}\right) \right)\\
&+p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I, \delta\left(P_2\right), A_{11}\right)+ p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I, P_2, \delta\left(A_{11}\right) \right).
\end{aligned}
$$
Thus we obtain
$$
p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, P_2,M \right)=0.
$$
By invoking of Lemma \ref{xxsec2.3}, we arrive at
$$
p_2\left ( P_2, M \right)=P_2 \diamond M=0.
$$
It follows from Lemma \ref{xxsec2.4} that
$$
M_{12}=M_{21}=M_{22}=0.
$$
We should remark that
$$
p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, P_1,D_{22} \right)= P_1\diamond D_{22}=0
$$
and that
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I, P_1,A_{11}+D_{22} \right)=p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I, P_1,A_{11}\right).
$$
Using similar discussions as the above, one can get
$$
M_{11}=0.
$$
Hence we conclude that $M=0$. That is,
$$
\delta(A_{11}+D_{22})=\delta(A_{11})+\delta(D_{22}).
$$

\noindent{\bf Claim 6.} For any $A_{11}\in \mathcal{A}_{11}, B_{12}\in \mathcal{A}_{12},C_{21}\in \mathcal{A}_{21}$ and $D_{22}\in \mathcal{A}_{22}$, we have
\begin{enumerate}
\item[(a)] $\delta(A_{11}+B_{12}+C_{21})=\delta(A_{11})+\delta(B_{12})+\delta(C_{21})$,

\item[(b)]   $\delta(B_{12}+C_{21}+D_{22})=\delta(B_{12})+\delta(C_{21})+\delta(D_{22})$.
\end{enumerate}

Let us first prove the result (a). For convenience, let us set
$$
M= \delta(A_{11}+B_{12}+C_{21})-\delta(A_{11})-\delta(B_{12})-\delta(C_{21}).
$$
We shall prove that $M=0$. In view of the facts
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_2, A_{11} \right)=0
$$
and
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_2, A_{11}+B_{12}+C_{21} \right)=p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_2, B_{12}+C_{21} \right),
$$
we by {\bf Claim 4} get
$$
\begin{aligned}
&p_n \left ( \frac{1}{2}I,\cdots, \frac{1}{2}I , \delta\left(P_2\right),A_{11}+B_{12}+C_{21} \right)\\
&+p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_2, \delta\left(A_{11}+B_{12}+C_{21} \right)\right)\\
=&\delta \left(p_n \left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_2, A_{11}+B_{12}+C_{21} \right)\right)\\
=&\delta \left(p_n \left ( \frac{1}{2}I,\cdots, \frac{1}{2}I , P_2,B_{12}+C_{21} \right)\right)+\delta \left(p_n \left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_2, A_{11} \right)\right)\\
=&p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,\delta\left(P_2\right), B_{12}  +C_{21}\right)+p_n\left(\frac{1}{2}I,\cdots, \frac{1}{2}I , P_2, \delta\left(B_{12} \right)+\delta\left(C_{21} \right)\right)\\
&+p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,\delta\left(P_2\right), A_{11} \right)+p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_2, \delta\left(A_{11} \right)\right)
\end{aligned}
$$
By Lemma \ref{xxsec2.3} we know that
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_2, M\right)=P_2\diamond M=0.
$$
It follows from Lemma \ref{xxsec2.4} that
$$
M_{12}=M_{21}=M_{22}=0.
$$
In order to show $M_{11}=0$, we should note that
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_2-P_1, C_{21} \right)=0
$$
and that
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_2-P_1, A_{11}+B_{12}+C_{21} \right)=p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_2-P_1, A_{11}+B_{12} \right).
$$
Using {\bf Claim 3}, we see that
$$
\begin{aligned}
&p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, \delta(P_2-P_1),A_{11}+B_{12}+C_{21} \right)\\
&+ p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I, P_2-P_1,\delta\left(A_{11}+B_{12}+C_{21}\right) \right)\\
=&\delta\left( p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I,P_2-P_1, A_{11}+B_{12}+C_{21} \right)\right)\\
=&\delta\left( p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I, P_2-P_1,A_{11}+B_{12} \right)\right)+\delta\left( p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, P_2-P_1,C_{21} \right)\right)\\
=&p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I,\delta(P_2-P_1), A_{11}+B_{12} \right)+p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I, P_2-P_1, \delta\left(A_{11}\right) +\delta\left(B_{12}\right)\right)\\
&+p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I,\delta(P_2-P_1), C_{21} \right)+ p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I, P_2-P_1,\delta\left(C_{21}\right) \right).
\end{aligned}
$$
This implies that
 $$
0=p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_2-P_1, M\right)=(P_2-P_1)\diamond M.
$$
According to Lemma \ref{xxsec2.4}, we know that $M_{11}=0$. Thus we arrive at
$$
\delta(A_{11}+B_{12}+C_{21})=\delta(A_{11})+\delta(B_{12})+\delta(C_{21}) .
$$
Considering the relations
$$
\delta\left(p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_1,B_{12}+C_{21} +D_{22} \right)\right)
$$
and
$$
\delta\left(p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_2-P_1,B_{12}+C_{21} +D_{22} \right)\right),
$$
together with the previous calculations, we assert that
$$
\delta(B_{12}+C_{21}+D_{22})=\delta(B_{12})+\delta(C_{21})+\delta(D_{22}).
$$

\noindent{\bf Claim 7.} For any $A_{11}\in \mathcal{A}_{11}, B_{12}\in \mathcal{A}_{12},C_{21}\in \mathcal{A}_{21}$ and $D_{22}\in \mathcal{A}_{22}$, we have
$$
 \delta(A_{11}+B_{12}+C_{21}+D_{22})=\delta(A_{11})+\delta(B_{12})+\delta(C_{21})+\delta(D_{22}).
$$

We only need to prove that
$$
M= \delta(A_{11}+B_{12}+C_{21}+D_{22})-\delta(A_{11})-\delta(B_{12})-\delta(C_{21})-\delta(D_{22})=0.
$$
Note the facts that
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_1, D_{22}\right)=0
$$
and
$$p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_1, A_{11}+B_{12}+C_{21}+D_{22}\right)=p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_1, A_{11}+B_{12}+C_{21}\right).
$$
Applying {\bf Claim 6} (a) yields that
$$
\begin{aligned}
 &p_n \left ( \frac{1}{2}I,\cdots, \frac{1}{2}I , \delta\left(P_1\right), A_{11}+B_{12}+C_{21}+D_{22}\right)\\
 &+ p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_1, \delta\left(A_{11}+B_{12}+C_{21}+D_{22}\right) \right) \\
 =&  \delta\left(p_n \left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_1, A_{11}+B_{12}+C_{21}+D_{22}\right)\right) \\
 =&  \delta\left(p_n \left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_1, A_{11}+B_{12}+C_{21}\right)\right)+ \delta\left(p_n \left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_1, D_{22}\right) \right) \\
 =&p_n \left ( \frac{1}{2}I,\cdots, \frac{1}{2}I , \delta\left(P_1\right), A_{11}+B_{12}+C_{21}\right)+p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I , \delta\left(P_1\right),D_{22}\right)\\
 &+ p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I , P_1, \delta\left(A_{11}+B_{12}+C_{21}\right) \right) + p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I ,  P_1, \delta\left(D_{22}\right) \right) \\
=&p_n \left ( \frac{1}{2}I,\cdots, \frac{1}{2}I , \delta\left(P_1\right), A_{11}+B_{12}+C_{21}+D_{22}\right)\\
&+ p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I , P_1, \delta\left(A_{11}\right)+\delta\left(B_{12}\right)+\delta\left(C_{21}\right)+\left(D_{22}\right) \right).
\end{aligned}
$$
Thus we obtain
$$
 0=p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_1,  M \right)= P_1 \diamond M.
$$
So $M_{12}=M_{21}=M_{11}=0$ by Lemma \ref{xxsec2.4}.
\vspace{2mm}

Similarly, using the relations
$$
p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_2, A_{11}\right)=0
$$
and
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_2, A_{11}+B_{12}+C_{21}+D_{22}\right)=p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_2, B_{12}+C_{21}+D_{22}\right),
$$
one can get $M_{22}=0$.  The proof of this claim is completed.
\vspace{2mm}

\noindent{\bf Claim 8.} For any $A_{ij},B_{ij}\in \mathcal{A}_{ij}\, (i,j=1,2)$, we have
$$
\delta(A_{ij}+B_{ij})=\delta(A_{ij})+\delta(B_{ij}).
$$

\noindent{\bf Case 1:}  $i \neq j$.

Note that
$$
\begin{aligned}
 p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_i+A_{ij}, P_j+B_{ij}\right) =( P_i+A_{ij}) \diamond (P_j+B_{ij}) \\
 =A_{ij}+B_{ij}+A_{ij}^{\ast}+B_{ij} A_{ij}^{\ast}.
\end{aligned}
$$
In light of {\bf Claim 6}, we know that
$$
 \begin{aligned}
\ \ \  &\delta\left(p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_i+A_{ij}, P_j+B_{ij}\right)\right) \\
 =&\delta(A_{ij}+B_{ij})+\delta\left(A_{ij}^{\ast}\right)+\delta\left( B_{ij} A_{ij}^{\ast}\right).
\end{aligned}\eqno(3.2)
$$
 On the other hand, we by {\bf Claim 3} and {\bf Claim 4} have
 $$
 \begin{aligned}
&\delta\left(p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_i+A_{ij}, P_j+B_{ij}\right)\right) \\
=&p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , \delta\left(P_i+A_{ij}\right), P_j+B_{ij}\right)+ p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_i+A_{ij}, \delta\left(P_j+B_{ij}\right) \right)\\
=&p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,\delta\left(P_i \right), P_j\right)+ p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , \delta\left(A_{ij}\right),P_j\right)\\
 &+p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,\delta\left(P_i\right), B_{ij}\right)+ p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I ,\delta\left(A_{ij}\right), B_{ij} \right)\\
&+p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_i, \delta\left(P_j\right)\right)+ p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_i, \delta\left(B_{ij}\right) \right)\\
&+p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I ,A_{ij},  \delta\left(P_j\right)\right)+ p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,A_{ij}, \delta\left(B_{ij}\right) \right)\\
=&\delta\left(p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_i, P_j\right)\right)+ \delta\left(p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I ,P_i, B_{ij}\right) \right)\\
&+\delta\left(p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I ,A_{ij},  P_j\right)\right)+ \delta\left(p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I ,A_{ij}, B_{ij}\right) \right)\\
=&\delta\left( P_i \diamond B_{ij}\right)+\delta\left(A_{ij} \diamond P_j\right)+ \delta\left(A_{ij}\diamond B_{ij}\right)\\
 =&\delta(B_{ij})+\delta\left(A_{ij}+A_{ij}^{\ast}\right)+\delta\left(B_{ij} A_{ij}^{\ast}\right)\\
 =&\delta\left(B_{ij}\right)+\delta\left(A_{ij}\right)+\delta\left(A_{ij}^{\ast}\right)+\delta\left(B_{ij} A_{ij}^{\ast}\right).
\end{aligned}\eqno(3.3)
$$
Compare (3.2) with (3.3) gives
$$
\delta\left(B_{ij}+A_{ij}\right)=\delta\left(B_{ij}\right)+\delta\left(A_{ij}\right).
$$

\noindent{\bf Case 2:}  $i=j$.
\vspace{2mm}

Let us set $M=\delta(A_{ii}+B_{ii})-\delta(A_{ii})-\delta(B_{ii})$. Let us take $l=1,2$, but $l \neq i$. Since
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_l, A_{ii}\right) =0
$$
and
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_l, B_{ii}\right) =p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_l, A_{ii}+B_{ii}\right)=0,
$$
we know that
 $$
 \begin{aligned}
& p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , \delta\left(P_l\right), A_{ii}+B_{ii}\right)+p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_l,\delta\left(A_{ii}+B_{ii}\right) \right) \\
=&\delta\left(p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_l, A_{ii}+B_{ii}\right)\right) \\
=&\delta\left(p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_l, A_{ii}\right)\right)+\delta\left(p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , P_l, B_{ii}\right)\right) \\
=& p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , \delta\left( P_l\right),A_{ii}\right)+p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I ,\delta\left( P_l\right), B_{ii}\right)\\
&+p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_l, \delta\left(A_{ii}\right) \right)+p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_l, \delta\left(B_{ii}\right) \right).
\end{aligned}
$$
Then we have
$$
p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_l, M \right)=P_l \diamond M=0.
$$
By invoking of Lemma \ref{xxsec2.4}, we arrive at $M_{li}=M_{il}=M_{ll}=0$.
\vspace{2mm}

The last step is to show that $M_{ii}=0$. Since
$$
\begin{aligned}
  p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_l,C_{li}, A_{ii}\right)=&C_{li}A_{ii}+A_{ii}C_{li}^{\ast},\\
  p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I ,P_l,C_{li}, B_{ii}\right)=&C_{li}B_{ii}+B_{ii}C_{li}^{\ast},\\
\end{aligned}
$$
and by {\bf Case 1} of this claim and {\bf Claim 4}, we have
$$
\begin{aligned}
&p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , \delta(P_l),C_{li},A_{ii}+B_{ii}\right)+p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I ,P_l,\delta(C_{li}), A_{ii}+B_{ii}\right) \\
  &+p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_l,C_{li}, \delta(A_{ii}+B_{ii})\right)  \\
  =&\delta\left( p_n\left( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_l,C_{li}, A_{ii}+B_{ii}\right) \right)\\
=&\delta\left( p_3\left(P_l,C_{li}, A_{ii}+B_{ii}\right) \right)\\
  =&\delta \left(  C_{li}(A_{ii}+B_{ii})+(A_{ii}+B{ii})C_{li}^{\ast}\right)\\
  =&\delta (C_{li}A_{ii}+A_{ii}C_{li}^{\ast})+\delta ( C_{li}B_{ii}+B{ii}C_{li}^{\ast})\\
  =&\delta\left( p_n\left( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_l,C_{li}, A_{ii}\right) \right)+\delta\left( p_n\left( \frac{1}{2}I,\cdots, \frac{1}{2}I , P_l,C_{li},B_{ii}\right) \right)\\
  =&p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I ,\delta(P_l),C_{li}, A_{ii}\right)+ p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I ,\delta(P_l),C_{li}, B_{ii}\right)\\
  &+p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I ,P_l,\delta(C_{li}), A_{ii}\right) +p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I ,P_l,\delta(C_{li}), B_{ii}\right)\\
&+p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_l,C_{li}, \delta(A_{ii})\right)  +p_n\left ( \frac{1}{2}I,\cdots, \frac{1}{2}I ,P_l,C_{li}, \delta(B_{ii})\right)  \\
\end{aligned}
$$
Thus we obtain
$$
\begin{aligned}
0=&p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I ,P_l, C_{li},M\right)\\
=& p_{3}\left ( P_l,C_{li},M\right) \\
=&C_{li}\diamond M.
\end{aligned}
$$
It follows that $C_{li}M_{ii}+M_{ii}C_{li}^{\ast}=0$. That is, $M_{ii}C_{li}^{\ast}=0$ for
all $C_{li} \in \mathcal{A}_{li}$. Note that $\overline{I-P}=I$. In light of Lemma \ref{xxsec2.2},
we conclude that $M_{ii}=0$.
\vspace{2mm}

As an immediate consequence of the previous Claims, we have

\vspace{2mm}

\noindent {\bf Claim 9.}  $\delta$ is an additive mapping.

\vspace{2mm}

Let us next show that $\delta$ is a $\ast$-derivation.

\vspace{2mm}

\noindent {\bf Claim 10.}  For any $A\in \mathcal{A}$, we have  $\delta(A^{\ast})=\delta(A)^{\ast}$.

\vspace{2mm}

In view of Lemma \ref{xxsec2.3}, we konw that
$$
p_n\left ( A, \frac{1}{2}I,\cdots, \frac{1}{2}I \right)=\frac{1}{2}(A+A^{\ast}).
$$
It follows that
$$
\begin{aligned}
\frac{1}{2}(\delta(A)+\delta(A^{\ast}))=&\delta\left(p_n\left ( A,  \frac{1}{2}I,\cdots, \frac{1}{2}I \right)\right)\\
=&p_n\left ( \delta\left(A\right),  \frac{1}{2}I,\cdots, \frac{1}{2}I \right)\\
=&\frac{1}{2}(\delta(A)+\delta(A)^{\ast}).
\end{aligned}
$$
Thus we get $\delta(A^{\ast})=\delta(A)^{\ast}$.

\vspace{2mm}

We next prove that $\delta$ is actually a derivation.

\vspace{2mm}

\noindent {\bf Claim 11.}  For any $A,B \in \mathcal{A}$, we have  $\delta(AB)=\delta(A)B+A\delta(B).$

Since
$$
p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I, A, B\right) =A\diamond B=AB+BA^{\ast},
$$
we obtain
$$
\begin{aligned}
\delta( AB+BA^{\ast})=&\delta\left( p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , A,B\right) \right)\\
=&p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , \delta\left( A\right),B\right) +p_n\left (\frac{1}{2}I,\cdots, \frac{1}{2}I , A,\delta\left( B\right) \right)\\
=&\delta(A)\diamond B+A\diamond \delta(B)\\
=&\delta(A)B+B\delta(A)^{\ast}+A\delta(B)+\delta(B)A^{\ast}.
 \end{aligned}
$$
It follows that
$$
\delta( AB)+\delta(BA^{\ast})=\delta(A)B+B\delta(A)^{\ast}+A\delta(B)+\delta(B)A^{\ast}.
\eqno(3.4)
$$
Replacing $A$ (resp. $B$) by $\text{i}A$ (resp. $\text{i}B$) in (3.4) and using {\bf Claim 9}, we arrive at
$$
\delta( AB)-\delta(BA^{\ast})=\delta(A)B-B\delta(A)^{\ast}+A\delta(B)-\delta(B)A^{\ast}.\eqno(3.5)
$$
Combining (3.4) with (3.5) gives
$$
\delta( AB)=\delta(A)B+A\delta(B).
$$
\end{proof}

By an analogous manner, we can prove

\begin{theorem}\label{xxsec3.2}
Let $\mathcal{B(H)}$ be the algebra of all bounded linear operators on a complex Hilbert space
$\mathcal{H}$ and $\mathcal{A}\subseteq \mathcal{B(H)}$ be a factor von Neumann algebra.
Then a mapping $\delta: \mathcal{A} \longrightarrow  \mathcal{A}$ satisfies the rule
$$
\delta(p_n(A_1, A_2,\cdots, A_n))=\sum_{k=1}^n
p_n(A_1,\cdots, A_{k-1}, \delta(A_k), A_{k+1},\cdots, A_n)
$$
for all $A_1, A_2,\cdots, A_n \in \mathcal{A}$ if and only if $\delta$ is an additive $\ast$-derivation.
\end{theorem}

$\mathcal{B}(\mathcal{H})$ denotes the algebra of all bounded linear
operators on a complex Hilbert space $\mathcal{H}$. Let us denote the subalgebra of
all bounded finite rank operators by $\mathcal{F}(\mathcal{H})\subseteq \mathcal{B}(\mathcal{H})$.
We call a subalgebra $\mathcal{A}$ of $\mathcal{B(H)}$ a \textit{standard operator algebra}
if it contains $\mathcal{F}(\mathcal{H})$. It should be remarked that a standard operator
algebra is not necessarily closed in the sense of weak operator topology. This is quite
different from von Neumann algebras which are always weakly closed.

From ring theoretic prespective, standard operator algebras and factor von Neumann
algebras are both prime, whereas von Neumann algebras are usually semiprime.
Recall that an algebra $\mathcal{A}$ is \textit{prime} if $A\mathcal{A}B=\{ 0\}$ impliess
either $A=0$ or $B=0$. An algebra is \textit{semiprime} if $A\mathcal{A}A=\{ 0\}$ impliess $A=0$.
Every standard operator algebra has the center $\Bbb{C}I$, which is also the center of  arbitrary
factor von Neumann algebra. An operator $P\in \mathcal{B}(\mathcal{H})$ is said to be
a \textit{projection} provided $P^{\ast}=P$ and $P^2=P$. Any operator $A \in \mathcal{B}(\mathcal{H})$
can be expressed as $A=\mathfrak{R}A+\text{i}\mathfrak{I}A$, where i is the
imaginary unit, $\mathfrak{R}A=\frac{A+A^{\ast}}{2}$ and $\mathfrak{I}A=\frac{A-A^{\ast}}{2\text{i}}$.
Note that both
 $\mathfrak{R}A$ and  $\mathfrak{I}A$ are self-adjoint.

Combining our current methods with the techniques of \cite{Lin1}, one can get

\begin{theorem}\label{xxsec3.3}
Let $\mathcal{H}$ be an infinite dimensional complex Hilbert space and $\mathcal{A}$ be a
standard operator algebra on $\mathcal{H}$ containing the identity operator $I$.
Suppose that $\mathcal{A}$ is closed under the adjoint operation.
Then a mapping $\delta: \mathcal{A} \longrightarrow  \mathcal{B(H)}$ satisfies the rule
$$
\delta(p_n(A_1, A_2,\cdots, A_n))=\sum_{k=1}^n
p_n(A_1,\cdots, A_{k-1}, \delta(A_k), A_{k+1},\cdots, A_n)
$$
for all $A_1, A_2,\cdots, A_n \in \mathcal{A}$ if and only if $\delta$ is an additive $\ast$-derivation.
\end{theorem}

We must point out that the technical routes and proving methods of Theorems \ref{xxsec3.2} and \ref{xxsec3.3}
are fairly similar to those of Theorem \ref{xxsec3.1}, and hence its proofs are omitted here for saving space.

\section{Related Topics for Future Research}
\label{xxsec4}

The main purpose of this article is to concentrate on studying nonlinear $\ast$-Jordan-type derivations
on operator algebras. The involved operator algebras are based on the algebra $\mathcal{B(H)}$
of all bounded linear operators on a complex Hilbert space $\mathcal{H}$, such as standard operator algebras,
factor von Neumann algebras, von Neumann algebras without central summands of type $I_1$.
Note that, unlike von Neumann algebras which are always weakly closed,
a standard operator algebra is not necessarily closed.  The current
work together with \cite{Jing, LiLuFang2, LiZhaoChen, Lin1, Lin2, TaghaviRohiDarvish, Zhang, YuZhang, ZhaoLi1}
indicates that it is feasible to investigate $\ast$-Jordan-type derivations and $\ast$-Lie-type
derivations on operator algebras under a unified framework---$\eta$-$\ast$-Jordan-type derivations.
We have good reasons to believe that characterizing $\eta$-$\ast$-Jordan-type
derivations on operator algebras is also of great interest. In the light of the motivation
and contents of this article, we would like to end this article by proposing  several open questions.

Let $\mathcal{A}$ be an associative $\ast$-algebra over the complex field $\mathbb{C}$
and $\eta$ be a non-zero scalar.  For any $A, B\in \mathcal{A}$, we can denote a ``new product" of $A$ and $B$ by
$A\diamond_{\eta} B=AB+\eta BA^\ast$. This new product $\diamond_{\eta}$ is
usually said to be \textit{$\eta$-$\ast$-Jordan product}.  Clearly, $1$-$\ast$-Jordan product $\diamond_1$ is the
so-called $\ast$-Jordan product, and $(-1)$-$\ast$-Jordan product $\diamond_{-1}$ is the so-called $\ast$-Lie product. Therefore,
it is reasonable to say that $\eta$-$\ast$-Jordan products organically unify
$\ast$-Jordan products with $\ast$-Lie products.  There are considerable works which are
devoted to the study of mappings preserving the $\eta$-$\ast$-Jordan product
between $\ast$-algebras, see \cite{CuiLi, DaiLu, HuoZhengLiu, LiLu, LiLuFang1, Molnar4, Molnar5, ZhaoLi2}
and the references therein.

Let $\delta: \mathcal{A}\longrightarrow \mathcal{A}$ be a mapping
(without the additivity assumption). We say that $\delta$ is a
\textit{nonlinear $\eta$-$\ast$-Jordan derivation} if
$$
\delta(A \diamond_{\eta} B)=\delta(A) \diamond_{\eta} B+A \diamond_{\eta} \delta(B),
$$
holds true for all $A,B\in \mathcal{A}$. Similarly, a mapping $\delta:
\mathcal{A}\longrightarrow \mathcal{A}$ is called a \textit{nonlinear $\eta$-$\ast$-Jordan triple derivation} if it satisfies the condition
$$
\delta(A \diamond_{\eta} B \diamond_{\eta} C)=\delta(A) \diamond_{\eta} B \diamond_{\eta} C+A \diamond_{\eta} \delta(B)\diamond_{\eta} C+A \diamond_{\eta} B \diamond_{\eta} \delta(C)
$$
for all $A, B, C\in \mathcal{A}$, where $A \diamond_{\eta} B \diamond_{\eta} C=(A \diamond_{\eta} B) \diamond_{\eta} C$ .
We should note that $\diamond$ is not necessarily associative.

Taking into account the definitions of $\eta$-$\ast$-Jordan derivations and $\eta$-$\ast$-Jordan
triple derivations, one can propose one much more common notion. Suppose that $n\geq 2$ is a fixed positive
integer. Let us see a sequence of polynomials with scalar $\eta$ and $\ast$
$$
\begin{aligned}
p_1(x_1)&=x_1,\\
p_2(x_1,x_2)&=x_1 \diamond_{\eta} x_2=x_1x_2+\eta x_2x_1^\ast,\\
p_3(x_1,x_2,x_3)&=p_2(x_1,x_2) \diamond_{\eta} x_3=(x_1 \diamond_{\eta} x_2) \diamond_{\eta} x_3,\\
p_4(x_1,x_2,x_3,x_4)&=p_3(x_1,x_2,x_3) \diamond_{\eta} x_4=((x_1 \diamond_{\eta} x_2) \diamond_{\eta} x_3) \diamond_{\eta} x_4,\\
\cdots &\cdots,\\
p_n(x_1,x_2,\cdots,x_n)&=p_{n-1}(x_1,x_2,\cdots,x_{n-1}) \diamond_{\eta} x_n\\
&=\underbrace{(\cdots ((}_{n-2}x_1 \diamond_{\eta} x_2)\diamond_{\eta} x_3)\diamond_{\eta} \cdots \diamond_{\eta} x_{n-1}) \diamond_{\eta} x_n.
\end{aligned}
$$
Accordingly, a \textit{nonlinear $\eta$-$\ast$-Jordan $n$-derivation} is a mapping $\delta:
\mathcal{A} \longrightarrow \mathcal{A}$ satisfying the condition
$$
\delta(p_n(A_1, A_2,\cdots, A_n))=\sum_{k=1}^n
p_n(A_1,\cdots, A_{k-1}, \delta(A_k), A_{k+1},\cdots, A_n)
$$
for all $A_1,A_2,\cdots, A_n\in \mathcal{A}$. This notion is motivated by the definition of $\ast$-Jordan-type derivations
and that of $\ast$-Lie-type derivations. Then each $\ast$-Jordan derivation is a
$1$-$\ast$-Jordan 2-derivation and every $\ast$-Jordan triple derivation is a $1$-$\ast$-Jordan 3-derivation.
Likewise, each $\ast$-Lie derivation is a
$(-1)$-$\ast$-Jordan 2-derivation and every $\ast$-Lie triple derivation is a $(-1)$-$\ast$-Jordan 3-derivation.
$\eta$-$\ast$-Jordan 2-derivations, $\eta$-$\ast$-Jordan 3-derivations and $\eta$-$\ast$-Jordan $n$-derivations
are collectively referred to as \textit{$\eta$-$\ast$-Jordan-type derivations}.
$\eta$-$\ast$-Jordan-type derivations on operator algebras are intensively studied by
several authors, \cite{Jing, LiLuFang2, LiZhaoChen, Lin1, Lin2, TaghaviRohiDarvish, Zhang, YuZhang, ZhaoLi1} .
A basic question in this line is to investigate whether each nonlinear $\eta$-$\ast$-Jordan-type derivation
on an operator algebra $\mathcal{A}$ with $\ast$ is an additive $\ast$-derivation.
In view of the current work and existing results in this direction, we propose several open questions.

\begin{question}\label{xxsec4.1}
Let $\mathcal{H}$ be an infinite dimensional complex Hilbert space and $\mathcal{A}$ be a
standard operator algebra on $\mathcal{H}$ containing the identity operator $I$.
Let $\eta$ be a non-zero scalar.
Suppose that $\mathcal{A}$ is closed under the adjoint operation.
A mapping $\delta: \mathcal{A} \longrightarrow  \mathcal{B(H)}$ satisfies the following condition:
$$
\delta(p_n(A_1, A_2,\cdots, A_n))=\sum_{k=1}^n
p_n(A_1,\cdots, A_{k-1}, \delta(A_k), A_{k+1},\cdots, A_n)
$$
for all $A_1, A_2,\cdots, A_n \in \mathcal{A}$. Is $\delta$ an additive $\ast$-derivation ? Does
the relation $\delta(\eta A)=\eta \delta(A)$ hold for any $A\in \mathcal{A}$ ?
\end{question}

\begin{question}\label{xxsec4.2}
Let $\mathcal{B(H)}$ be the algebra of all bounded linear operators on a complex Hilbert space
$\mathcal{H}$ and $\mathcal{A}\subseteq \mathcal{B(H)}$ be a factor von Neumann algebra. Suppose that $\eta$ is a non-zero scaler.
Let $\delta: \mathcal{A} \longrightarrow  \mathcal{B(H)}$ be a mapping such that
$$
\delta(p_n(A_1, A_2,\cdots, A_n))=\sum_{k=1}^n
p_n(A_1,\cdots, A_{k-1}, \delta(A_k), A_{k+1},\cdots, A_n)
$$
for all $A_1, A_2,\cdots, A_n \in \mathcal{A}$. Is $\delta$ an additive $\ast$-derivation ? Do we have
the relation $\delta(\eta A)=\eta \delta(A)$ for any $A\in \mathcal{A}$ ?
\end{question}

\begin{question}\label{xxsec4.3}
Let $\mathcal{B(H)}$ be the algebra of all bounded linear operators on a complex Hilbert space
$\mathcal{H}$ and $\mathcal{A}\subseteq \mathcal{B(H)}$ be a von Neumann algebra without central
summands of type $I_1$. Let $\eta$ be a non-zereo scalar. A mapping $\delta: \mathcal{A}\longrightarrow \mathcal{B(H)}$
satisfies the following conditions:
$$
\delta(p_n(A_1, A_2,\cdots, A_n))=\sum_{k=1}^n
p_n(A_1,\cdots, A_{k-1}, \delta(A_k), A_{k+1},\cdots, A_n)
$$
for all $A_1, A_2,\cdots, A_n \in \mathcal{A}$. Is $\delta$ an additive $\ast$-derivation ? Can we get
the relation $\delta(\eta A)=\eta \delta(A)$ for any $A\in \mathcal{A}$ ?
\end{question}

\bigskip

redpajama_set_name: RedPajamaArXiv


text:
Watch Mischievous Bear Snoop in Officer's Car Outside Courthouse
By Anabelle Doliner On 8/4/21 at 11:08 AM EDT
News Bears Wildlife Animals California
A hilarious video from the Sierra County Sheriff's Office in Downieville, California, shows an unusually furry intruder breaking into a correction officer's car.
The clip, featured above, was caught via security camera on June 6 but was released to the public on Sunday. The footage shows a bear approaching a car parked in the otherwise-empty Sierra County Courthouse lot.
The curious creature opens the sedan's back door and pokes its head in for a closer look. After examining the car's backseat, the bear decides to wander off, leaving the door ajar. The incident is said to have occurred at around 9:30 p.m.
"The next morning, when the officer got in her car to go home, she was puzzled to notice the door ajar light illuminated," the Sheriff's Office posted on Facebook. "She confirmed the rear passenger door was not fully latched; knowing the door had not been ajar when she went to work the previous evening, she took a look through the security footage and discovered the culprit."
"Thankfully, this bear found nothing of interest," the post added, along with the following advice: "Please don't forget to lock your doors and secure/remove anything that may attract our bears!"
Viewers couldn't get enough of the clip, which racked up thousands of views since it was posted earlier this week.
"This is something that happens somewhat frequently in Downieville," Sierra County Sheriff's Office Deputy Holly Bayly told Newsweek in a statement. "There have been a few incidents of a bear getting in the car or even stuck in the car. Since this one was captured on video, we thought it would be a great way to raise awareness."
According to Bayly, the officer actually had some snacks in her car, but the bear luckily left them untouched.
"When I remarked that it was fortunate she didn't keep snacks in the car, she [the officer] said, 'Well, there was a bag of diet chips on the back seat. I guess he wasn't interested in diet chips, though,'" Bayly said.
A hilarious video from the Sierra County Sheriff's Office in Downieville, California, shows an unusually furry intruder breaking into a correction officer's car. Above, Kamchatka brown bear is pictured at the Tierpark Hagenbeck zoo on June 28, 2021, in Hamburg, Germany. (Photo by MORRIS MAC MATZEN/AFP via Getty Images)
Often mischievous, bears don't shy away from aspects of the human world that interest them—especially if they suspect food is involved. According to the Get Bear Smart Society, bears "will inspect [odors], noises and objects to determine if they are edible or playable."
Last month, a bear became trapped in a car, causing serious damage to the vehicle. As documented in a viral video posted to TikTok, the Gatlinburg, Tennessee, tourist managed to scare the bear away with the help of his friends.
Days later, it was reported that two bears were spotted "checking showtimes" at an AMC Theatre in a Morristown, Tennessee, mall. "Check out our visitor this morning looking for showtimes of Space Jam at AMC College Square 12!" the mall joked on Facebook. "In all seriousness, please be cognizant of two bears in our area. Do not approach or feed the bears."
That warning is echoed by the National Park Service.
"Their behavior is sometimes unpredictable. Although rare, attacks on humans have occurred, inflicting serious injuries and death," the agency posted on its website. "Each bear and each experience is unique; there is no single strategy that will work in all situations and that guarantees safety."
It notes, however, that "most bear encounters end without injury."
Updated 08/04/2021, 4:10 p.m. ET: This story has been updated with comment from the Sierra County Sheriff's Office.


The Garden Tour You've Always Wanted to go on!
Join us at Hatley Park National Historic Site to experience a complimentary garden tour lead by our garden professionals. Take a journey through the beautiful Italian, Japanese and Rose Gardens and visit our Glasshouse.
Look forward to seeing you over Garden Days weekend.


text:
using UnityEngine;
using System.Collections;
namespace GDGeek{
	public class VoxelFilterGlow : VoxelFilter {

		public override VoxelGeometry.MeshData filter(VoxelGeometry.MeshData data){
			VoxelGeometry.MeshData clone = (VoxelGeometry.MeshData)data.Clone ();
			for (int i = 0; i < clone.colors.Count; ++i) {

				Color c = clone.colors [i];
				Vector2 uv = Vector2.zero;

				uv.x =1.0f;

				clone.uvs [i] = uv;
			}
			return clone;
		}
	}
}

redpajama_set_name: RedPajamaGithub


Free download java game AquaLife from Bogee for mobil phone, 2005 year released. Free java games to your cell phone.
Sea world is full of mysteries and beautiful scenery! Food chain under the water does not differ from the ground. The big eat the small, small eat very small, grow up and eat the already large. In deep-sea world you have to be careful and wary of large hungry fish. From past the floating fish schools tear namely the one that you in the teeth. Also growth of your fish contribute to the pearls and various bonuses, came flowing on the surface of the water.
Download exciting java game from Bogee - AquaLife on your mobile phone and play. AquaLife - interesting free java game. You can download many mobile games on your cell phone absolutely free! On the site of a large list of screen sizes of the game AquaLife - 176x208, 240x320!


text:
Q: How to add a "Authorization=Bearer" header with Indy in Delphi? I'm trying to do a POST request using an access_token, and it works fine using POSTMAN, but when I try to do the same request on Delphi, I can't find a way to add the "Authorization=Bearer eyxxxxxx..." to the Request header, as POSTMAN does.
POSTMAN Request (working well):
POST /somepath HTTP/1.1
Host: someurl.com.br
Authorization: Bearer eyJhbGciOiJSUzI1NiJ9.....
Content-Type: application/json
(body content ommited)
Indy Request generated by Delphi, captured by HTTP Analyzer (always returning 401 Forbidden error, because the absence of "Authorization=Bearer" part):
POST /somepath HTTP/1.1
Host: someurl.com.br
Content-Type: application/json
(body content ommited)
I've tried to add the header using the code below, but the header part with the "Authorization=Bearer eyxxxxxx..." isn't generated on Request, returning the 401 Forbidden error.
FIdHTTP.Request.CustomHeaders.FoldLines := False;
FIdHTTP.Request.CustomHeaders.Add('Authorization=Bearer ' + txtToken.Text);

A: Just found the problem. I added the wrong separator between the "Authorization" and "Bearer" words.
Wrong:
FIdHTTP.Request.CustomHeaders.FoldLines := False;
FIdHTTP.Request.CustomHeaders.Add('Authorization=Bearer ' + txtToken.Text);

Correct:
FIdHTTP.Request.CustomHeaders.FoldLines := False;
FIdHTTP.Request.CustomHeaders.Add('Authorization:Bearer ' + txtToken.Text);

After replacing the '=' by ':', I received the expected response, like the one received by POSTMAN.


text:
Pertti Ilmari Purhonen (* 14. Februar 1942 in Helsinki; † 5. Februar 2011 in Porvoo) war ein finnischer Boxer.

Amateur
Purhonens größter Erfolg ist der Gewinn der Bronzemedaille im Weltergewicht bei den Olympischen Spielen 1964 in Tokio. Er verlor das Halbfinale gegen Ričardas Tamulis aus der Sowjetunion nach Punkten.

Bei Europameisterschaften war er weniger erfolgreich. Bei den Titelkämpfen 1961 in Belgrad verlor er im ersten Kampf gegen Richard McTaggart. 1963 in Moskau traf er im Viertelfinale auf Tamulis und war ihm auch in diesem Kampf unterlegen. Bei seiner Teilnahme an den Europameisterschaften 1967 in Rom scheiterte schon in der ersten Runde am Dänen Jørgen Hansen, der im Viertelfinale seinerseits Manfred Wolke unterlag.

Purhonen gewann allerdings acht finnische Meistertitel: 1960 und 1961 im Leichtgewicht, 1962 im Halbweltergewicht, 1963 und 1964 sowie 1966 bis 1968 im Weltergewicht.

Profi
Seine Profikarriere war sehr kurz und dauerte nur 2 Kämpfe, die er beide gewann. Im Februar 1969 besiegte er den Franzosen Gérard Hedin nach Punkten und im Dezember des gleichen Jahres Jack Johnson von den Niederländischen Antillen durch KO in der dritten Runde.

Spätes Leben
Purhonen litt seit etwa 2005 an der Alzheimer-Krankheit, an der er im Februar 2011 starb.

Weblinks

Einzelnachweise

Teilnehmer der Olympischen Sommerspiele 1964
Olympiateilnehmer (Finnland)
Boxer (Finnland)
Finne
Geboren 1942
Gestorben 2011
Mann

redpajama_set_name: RedPajamaWikipedia


text:
Yusuke Yachi (; born 2 January 1980) is a Japanese race walker.

He finished seventeenth at the 2007 World Championships in Osaka, in his major international debut race.

Achievements

References

1980 births
Living people
Japanese male racewalkers
20th-century Japanese people
21st-century Japanese people


The modern Travelodge Hotel Melbourne Southbank is perfectly located in the vicinity of Melbourne�s tourist attractions. Within proximity to the river, excellent cafes, the Aquarium, the Crown Casino, and the Royal Botanical Gardens, this is a great location for a vacation. Besides an impeccable location, the Southbank boasts the Travelodge standard of rooms which include all the necessary facilities guests might need. Guests can spend the evening networking around the art galleries and shopping centers of Melbourne or hit up some of the vibrant bars and restaurants in the area. Travelodge Hotel Melbourne Southbank features a great location and an affordable accommodation, making it a worthy stay!
Located right in the heart of Melbourne, this 3.5-star property is surrounded by exciting attractions that will give you a wonderful stay. The popular attractions include shopping precincts, theaters, galleries, bars, and restaurants. Just a stone's throw away from vibrant Bourke St. Mall, Federation Square, Southbank, Her Majesty's and Regent Theaters, Queen Victoria Markets, and Melbourne Central Shopping Complex, it is an excellent location for leisure travelers. They also feature fine rooms that are well appointed. Hotel facilities include valet parking, a luggage storage area, and concierge. Undoubtedly a great location, it is definitely worth your stay.
Excellent for business travelers, the Rydges on Swanston Hotel features large and modern meeting rooms, business facilities, and a hip restaurant great for entertaining. From the post-modern oriental looking restaurant to the sleek and comfortably stocked rooms, this hotel has a standard of elegance that emerges everywhere. Entertain clients with a world-class dining experience frequented by Melbourne residents, or treat yourself to a swim in the roof-top infinity pool boasting views of the city. With eleven conference facilities featuring the best in audio and visual equipment, fully catered services by highly trained receptionists, and natural lighting combined with soundproof wall paneling, your clients won�t be disappointed. After all is said and done, you�re smack dab in the center of Carlton, one of the coolest cosmopolitan centers in Melbourne for shopping and nightlife.
This contemporary hotel, located in the heart of Melbourne, offers excellent accommodation and a prime location for city exploration and business trips. This property is located across from Myer Department Store in Bourke Street Mall and is surrounded by lively cafes, fashion and accessory boutiques, as well as Melbourne's famous shopping arcades and major department stores such as David Jones and Zara. Guests can choose from a selection of different room types which feature amenities including a small refrigerator, in-room safe, air conditioning, ironing board and iron, tea and coffee maker, LCD TV with selected Foxtel channels, and an en-suite bathroom with shower. Guests enjoy 2 hours of free Wi-Fi access per 24 hour period. Take advantage of the hotel�s discount parking arrangement with Wilson�s Galleria located at the corner of Little Collins and Elizabeth Streets. Causeway Inn On The Mall Hotel offers comfort, affordability, and a friendly staff that is willing to go beyond the call of duty to make sure your stay is a pleasurable one.
With its convenient location close to Melbourne's city center, Oaks On Market Hotel is a perfect choice for travelers looking to explore all that the city has to offer. Guests staying here can enjoy access to many shopping, entertainment, and eating options. The guestrooms are spacious and well-appointed with a host of modern conveniences such as flat screen TVs, Internet access, and in-house movies. Delight in extra luxuries like rooms with stunning Yarra views, individual stereo systems, climate control, an indoor heated lap pool, fully equipped gymnasium, and sauna. Guests are welcome at the hotel's two licensed restaurants serving delicious food and refreshing drinks. All things considered, Oaks On Market Hotel is excellently located, well appointed, and reasonably priced.
Nestled in the heart of Essendon, Alexander Motor Inn & Apartments is an ideal spot from which to discover Melbourne. The city center is merely 9.Km away and the airport can be reached within 9 minutes. A well-kempt environment and its proximity to Sia Medical Centre Essendon, Eastbrooke Family Clinic Essendon, Golden Mile Heritage Trail give to this hotel a special charm.
Take advantage of a wealth of unrivaled services and amenities at this Melbourne hotel. Guests of the hotel can enjoy on-site features like free Wi-Fi in all rooms, daily housekeeping, laundromat, postal service, express check-in/check-out.
Experience high quality room facilities during your stay here. Some rooms include television LCD/plasma screen, internet access � wireless (complimentary), non smoking rooms, air conditioning, heating, provided to help guests recharge after a long day. Besides, the hotel's host of recreational offerings ensures you have plenty to do during your stay. Alexander Motor Inn & Apartments is your one-stop destination for quality hotel accommodations in Melbourne.


(KFVS) - The mild temperatures will continue into the evening as clouds begin to enter the area. Morning lows will be in the 60s.
Thursday will be partly cloudy and very breezy with a slight chance of storms late in the day across our western counties. Highs in the upper 70s to low 80s.
Winds Thursday will at times gust over 40 mph.
**FIRST ALERT-CHANCE OF STORMS** Our next focus is storms Thursday night into Friday morning. We will have several chances of storms to watch over the next week.
Storms will become more likely Thursday night as a cold front moves through the area. A few of these storms could be severe, with strong wind being the main threat.


Last week, we did a Q&A with Talent Acquisition expert Elaine Davidson on How to Select a New Applicant Tracking System. In the interview, Elaine also listed the top mistakes she's seen companies make when selecting an ATS. We decided to segment those out and include them here. Enjoy!
The company goes with what is trending or the platform that a direct competitor uses without really understanding what else is on the ATS market. There are new ATS platforms hitting the market all the time – and the founders of these companies are getting more and more savvy.
Clients often rush to make a decision rather than taking the extra few weeks or even months to learn the landscape and make an educated decision.
Assuming that the ATS vendor is a recruiting expert is a very common mistake and probably the most costly of all. The ATS vendor is an expert on their technology. Some of the ATS firms have in-house consultants that are part of the implementation team.
As a client, you need to understand what that person's background is – how long they have been in the space, and their level of understanding of Talent Acquisition at the executive level. Your consultant – whether in-house or hired by you as an independent 3rd party adviser, must be in a position to help you make bold decisions about your recruiting function, see what is possible, and think at least 3 steps ahead.
Many companies know the value of an applicant tracking system but they don't link it to the value of a CRM tool. The ATS is used once an individual becomes a candidate. Most ATS providers also offer a CRM tool that enables your recruiters to build pipelines, create Talent Communities, and ultimately fill positions more quickly. It is a mistake to not at least consider the CRM as part of your solution.
Operational Excellence and the future state of the recruiting process are not emphasized nearly enough in the selection of an ATS. It is imperative that you understand the unique process for your organization and you have the ATS system built accordingly.
Too many clients abandon their non-negotiables because they are told that the system "can't do that".It is a mistake to look at the ATS selection and implementation as a purely technical project. It's an organic process that should reflect the organization's personality. You should take full advantage of the creative space that begins with a white sheet of paper to design the vision you have for your team.
Lack of buy-in and communication from key stakeholders is a mistake from beginning to end. If you are selecting a new applicant tracking system, solicit the views of others beyond just Recruiters. What do Hiring Managers want to gain? What is their biggest pain point? What does your Executive team want to see in terms of reporting, dashboards, and metrics? Do people want self service? The point is to over-communicate and make this a company project that is focused on your Human Capital as opposed to just a recruiting tool.
I am a 3-time CEO, author and proud father of my preemie baby Maverick! We created Ongig to transform your job descriptions to boost quality candidates and diversity. Ongig supercharges your job pages through video, images, chat, social sharing, microsite creation and much more. Jobs can be more easily found through Artificial Intelligence-based job search and all pages are Mobile and SEO- optimized. Ongig also analyzes the text of your JDs and our pro copywriting team will rewrite your job descriptions. Early clients of Ongig include Salesforce, Yelp, GoDaddy, Verizon, Intel & Autodesk. Request a demo if you'd like to chat about partnering!