Submitted by More Better on Tue, 11/08/2005 - 11:57.
Submitted by Sudhir Kade on Thu, 01/12/2006 - 18:11.
Few people realize how many of us have dealt with issues of mental health at some point in our lives - it is no wonder people are often encouraged to be happy or not to worry. This topic of mental health has often been considered taboo because few people want to associate with mental illness, or being 'crazy'. Yet statistics show a majority of people on earth will grapple with some form of mental illness or disorder during their lives - every family on earth can relate to someone in their clan who has suffers from a mental disorder. The fear and anxiety (a mental disorder in its own right) around assoicaition with mental illness causes many mental illnesses to go undiagnosed and often, untreated. Among the most common of these is depression - anxiety disorders and eating disorders are not far behind.
As a former pre-med who dating a clinical psychology intern for a long time (and have a father and brother who are physicians), I learned a great deal about different mental conditions and a common framework for the diagnosis of mental conditions - the DSM !V. Yet despite all there was to learn I realized mental health and the proper treatment of mental disorders is still a very inexact science. I believe this is because all people are so unique and different that they cannot be exactly pigeonholed into specifc categories and pegged with exact disorders all the time. Pracitioners do the best that they can, however, and many medicines have been developed that have been proven to provide significant relief for millions of people. A persistent problem for many sufferers is denial - few people want to succumb to the thought that they may need medication for the rest of their lives. Still others try medication but are so put off by side effects suffered they discontinue treatment and cope in other ways - typically with illegal drugs or alcohol.
The self medication path is often chosen because it removes the sufferer of a mental condition from the stigma society (unfortunately) places on mental illness. Unfortunately these self-medicating paths can result in probelms with drug and alcohol addiction that leave the sufferer worse off than before.
Many feel that therapy plays just as important a role as medication in treating mental illness - science has proven that therapy, indeed has resulted in breakthrough results for many patients. Typically a combined approach is best, because it can help prevent or catch problems like discontinuation of medication and allows the patient a critical outlet for expression under promised confidentiality.
I'm not psychological or psychiatric expert but these are many concepts i've learned in the past and have thoughts about now. One of my biggest concerns, putting this in the context of social change and community developement is that we need to raise awareness around issues of mental health and help people overcome fears and anxieties around admission and treatment of a problem. This region offers some of the highest ranked and highly regarded mental health programs and services in the nation, if not the world. To successfully create a better quality of life for all people health issues need to be addressed - and mental health is no exception. The difficulty is that mental health problems are not easy to identify or are always immediately apparent - and as alluded to earlier, fear of association with mental illness and the association of mental illness with 'craziness' or 'insanity' keeps many individuals from admitting to a problem or coping with the problem through illicit means.
What's the answer? Many people could speak more expertly than I, but I believe that educating people as to how common these disorders are, teaching them not to stereotype or stigmatize those who admit to problems and get needed help for them, and encouraging everyone to participate in some form of therapy to experience its benefits could have positive outcomes. We could see a great improvement in a great number of people- on the whole creating a much happier, healthier, and productive populace.
I'd love to hear more thoughts from true experts - and just wanted to share some thoughts i've long had regarding mental health.
Submitted by MShafarenko on Mon, 01/23/2006 - 16:04.
As a recent psychology graduate from Case, and a clinically-diagnosed sufferer of panic disorder, I find this topic near and dear to me. Yet, in all my years of thinking about my condition and the condition of many psychologically-ill people, I have come to a different conclusion about mental health. Specifically, it addresses Norm's point on taking demographics into account.
I see our society suffering from a social disorder known as non-acceptance. I feel we live in and around a society of people who are unwilling to accept each other's differences, new ideas, change, and inclusion. The foremost value of honesty is replaced with complacency and ingenuine "niceness." We are prone to tell someone we like their dress in fear of hurting their feelings, instead of telling them our honest opinion that the dress is hideous and it would be in their best interest to change. With that, we have created a culture of distrust and fear of others not like us - because while they say one thing to our face, they say something completely different behind our backs.
This society with these cultures breeds a lot of isolation in people who yearn to be a part of a larger whole, to feel a sense of belonging, to be engaged in an effort to change the world into a better place. I think America has the highest prevelance of psychological disorders - now, some might say that that is because we are one of the more liberal countries in the world that no longer throws the mentally ill into asylums and recognizes these illnesses for what they are. I agree with that - but also say that we also have "civil rights" which aren't really TRUE but given at face value so those fighting for them will shut up. I do not believe American companies are inclusive of minorities or those with disabilities - they are merely forced to comply with a government law known as Equal Employment and Opportunity, so they tabulate their minority/disability quotas and feel that they have done their part to rid of racial/sexual prejudices in our society. Wrong - the prejudice and discrimination still exists on a very covert level, much like George Bush issuing executive power to spy on American citizens.
The larger point at hand is that I do not believe that all the people who have mental/psychological illness are predisposed to it and that a large part of these disorders have to do with the values of our society. Changing those is a matter of time (say 50+ years, if ever). Only then will we see our ideals materialize into our everyday lives, where humans use alternative energy for their consumptive purposes, where humans seek to help other humans through the struggle of life, where TRUE (not MTV or other POP) artists/musicians are given the resources and attention to create and share their talents and visions. The long-enduring fight for these ideals is what gets me going every day, while a silent struggle with the present reality is ever present.
Submitted by Sudhir Kade on Mon, 01/23/2006 - 17:11.
MIke, I really enjoyed your commentary here. You are exactly right when you speak of social disorder on a grand scale. You speak to the way so many groups and subgroups take exclusionary viewpoints and make prejudicial decisions. Many times this comes down to groupthink and the fact that many members of society instinctively ridicule or mock those different than us. This could be applied to race or religion, but i'll specifically address behavior. If someone sparks a conversation like others have never heard before or someone brings up points never made before these can be instinctively interpreted as odd or erratic behavior. I participate in coursework through the Gestalt Institute of Cleveland and it has been a very important aspect to be able to speak authentically from the heart, coming from a place of love. Say a phrase like that, or better yet - practice this often : and be prepared to face the consequences - in this case, perhaps being labeled as an eccentric. Yet everything in my heart tells me this is the right way we need to shift as a society - to transcend self-embeddedness and strive toward egolessness.
So I really like your comment alluding to society's definition of what normal is - it just might be that the great visionaries are simply ahead of their time and context - and that which they speak of amounts to that to which the masses cannot relate. I cannot tell you how many times I've proposed the advocacy of sustainability and it has just fallen on deaf ears.
One can be so easily labeled a tree-hugging fantatic by advocating greening measures to a conservative party. That same individual could be embraced with open arms and championed a hero by a liberal group.
You speak to values of society and the necessity for these to change via paradigm shift, This brings directly to mind the need to inculcate class cirriculum at early stages on up with the new set of values. For that it often either takes the monumental task of convincing school boards to advocate the new paradigm - or perhaps the creation of charter schools which bypass the bureaucratic process and 'just do it'. Two such (local) charter programs include the School for Intergenerational Learning and the Entrepreneurship Academy.
Take solace, Mike, in the fact that I join you in your fight, as do so many other agents for positive change. As we continue to build communities using technology and polarize them toward meaningful action we just might make a significant difference in the short and long term.
Submitted by Norm Roulet on Fri, 01/13/2006 - 15:49.
Wow that goes to the foundation of human being in the world, today. I'd be interested in seeing more data on levels of mental disorders by disorder and other demographics, with trends over time. Whatever we can better address we should. In NEO, this time of year, SAD (Seasonal Affective Disorder) is a big issue not found in other regions - we should be experts in dealing with that here.


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Here you can know the Top 10 richest cities of the world, Reported by human resources firm Mercer Consulting, the cost of living in North American, Asian and African cities has been rising this past year, despite the global slowdown.
About 214 cities worldwide and used New York City as the benchmark. However, no city is the all-rounder. For instance, Tokyo took the lead at the ranking of indexes of "economic vitality" and "research ability".TOP 10 RICHEST CITIES IN THE WORLD
Here are the List of Top 10 Richest cities in the world. Have a look at all of them below:
TOKYO (JAPAN)
Tokyo is the capital of Japan, The largest Metropolitan area in the world, and it is the home of the Japanese Imperial Family. Tokyo city still holds its position as richest city in spite of stagnant Japanese economy of 'the lost decade'. With a GDP of $1479 Billion.
NEW YORK (U.S.A)
As we all know the New York City is the most popular city of U.S.A with a population of over 8.1 million. New York is headquarter for all the major banks and financial institutions of the world. GDP in 2008 was $1406 Billion, New York Stock Exchange is world's largest stock exchange by market capitalization.
LOS ANGELES (U.S.A)
Los Angeles is the most populous city in the U.S state of California and the second most populous city in the United States, after New York City, with a population at the 2010 United States Census of 3,792,621. With GDP of $592 Los Angeles is a world centre of business, international trade, entertainment, culture, media, fashion, science, technology, and education.
CHICAGO (U.S.A)
Chicago is the third most populous city in U.S.A city has approximately 2.7 million residents, Chicago is on no 4th in the list of Top 10 Richest Cities with the GDP of $574 Billion.
LONDON (ENGLAND)
London is the capital city of England, one of the beautiful cities of the world. London is a leading global city, with strengths in the arts, commerce, education, entertainment, fashion, finance, healthcare, media, professional services, research and development, tourism and transport all contributing to its prominence. With the GDP of $565 Billion it is Europe's richest and most vibrant city.
Paris is the capital city of France and also the largest city of France, Paris has the has a population of about 2,200,000. With the GDP of €552.7 billion (US$813.4 billion) Paris is on no 6 in Top Ten Richest Cities in World.
OSAKA (JAPAN)
Osaka is a city in the Kansai region of Japan's main island of Honshu, Electronic giants Panasonic, Sharp, Sanyo have their headquarters in this city. GDP was $417 Billion in 2008.
Mexico City is one of the most important financial and cultural centers in North America. The city is located 2200 m above the sea level. Mexico City population of roughly 8,605,239 inhabitants (2000 data) and a metro population of roughly 17,809,471 (2000 data) with the GDP of $390 Billion is more than entire GDP of many Latin American countries.
PHILADELPHIA (U.S.A)
Is the largest city in the Commonwealth of Pennsylvania and the fifth-most-populous city in the United States. Major economic activities of city is oil refining and Financial Services. It has GDP of $388 Dollar.
SÃN PAULO ( BRAZIL )
San Paulo is the largest in the Brazil and the world's 7th largest city by population, San Paulo is main commercial city of Brazil. GDP of $388 Billion is almost same as Philadelphia.
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Everyone wants the best care possible for the seniors in their life once they're no longer able to live completely independently. It's always a good idea to look into what options are available as far as what provider to choose, but also what level of care is necessary. Some senior living apartments only provide a place to live and access to activities while others, that are assisted living, provide meals three times a day as well as assistance with medication, dressing, bathroom, and more. In this post, we will look at things you should consider when choosing a senior living provider.
Location is important. When possible it's a good idea for the residents to live somewhere where friends and family can come by and visit often. Aesthetics are also important. Your loved one should feel comfortable in their new home. How is the food? Is it something nutritious as well as something they will enjoy eating? Pay attention to staff. Are they friendly and happy?
Talk to residents who are already living there and their families. Ask them what they like and don't like about being there. Talk to staff, nurses, and volunteers to get a feel for what a typical day is like for seniors in their care.
This is a big one. It is so important your loved ones stay engaged. Whether it be senior fitness, wellness programs, salon & spas, or music.
At Heritage Community of Kalamazoo, we know how difficult it can be to make the transition to a senior living community. We do everything we can to enrich the lives of our residents and provide their families with peace of mind knowing they're in good hands. Please do not hesitate to call, stop by, or message us with any questions you may have.


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Percussion Ensemble: "World Percussion Bash"
Wednesday, May 27, 2015 - 7:30pm
Meany Hall—Studio Theatre
$15 ($10 students/seniors), Notecard
Join us for a true "World Percussion Bash" in this year-end concert featuring The UW Percussion Ensemble (directed by Thomas Collier), The UW Gamelan Ensemble (directed by Christina Sunardi), The UW Harry Partch Ensemble (directed by Charles Corey) and the UW Steel Drum Ensemble (directed by Shannon Dudley), presenting music from a variety of musical traditions.
UW Percussion Ensemble
Tom Collier, director
Gainsborough by Thomas Gauger
Morning of the World by William Maloof
CONDUCTOR BIO
Shannon Dudley, Steel Drum Ensemble
Shannon Dudley, associate professor of Ethnomusicology, holds a PhD from the University of California at Berkeley. He teaches courses that include music of Latin America and the Caribbean, American popular music, Music and Community, Comparative Musicianship and Analysis, and graduate seminars in Ethnomusicology. He also directs the UW steeelband.
Dudley has conducted research in Trinidad and Tobago, focusing on the history and music of steelbands. More recent research projects include the music of El Gran Combo, and salsa music in Puerto Rico generally, as well as Latino contributions to American popular music. His theoretical interests include nationalism, transculturation, and participatory music practices.
His publications include Carnival Music in Trinidad (Oxford University Press, 2004), as well as Music From Behind the Bridge (Oxford University Press, 2008), a history of Trinidad steelband music, and numerous other articles on Caribbean music, including and "Judging by the Beat: Calypso vs. Soca," Ethnomusicology (1996), and "El Gran Combo, Cortijo, and the Musical Geography of Cangrejos/Santurce, Puerto Rico," Journal of Caribbean Studies (2008).
Dudley is one of the curators (along with his wife, Marisol Berríos-Miranda, and Michelle Habell-Pallan) for American Sabor: Latinos in U.S. Popular Music, a bilingual museum exhibit that opened at the Experience Music Project in Seattle in 2008. American Sabor was exhibited in museums in several U.S. cities, culminating in a 3-month run at the International Gallery of the Smithsonian Institution in summer 2011. He is also guest curator for a smaller version of the exhibit prepared by the Smithsonian Institution Travelling Exhibit Service (SITES), and for their website (http://www.americansabor.org), and is currently working on a book manuscript.
In Seattle Dudley performs on steel pan with several local bands, and participates in the Seattle Fandango Project (SFP), a community music group that practices son jarocho. He has helped to bring visiting artists from Mexico to the University of Washington (including Son de Madera and Laura Rebolloso) and to promote active collaboration between community arts activists and university individuals and programs.
Thomas Collier, Percussion Ensemble
Director of percussion studies at the University of Washington School of Music since 1980, associate professor Tom Collier has performed and recorded with many important classical, jazz, and popular artists, in addition to recording and performing with his own jazz group. He is a veteran of more than 50 years in music — his first public appearance was at age five, on xylophone, and his first professional performances were made as a nine-year-old marimba virtuoso.
Collier has appeared in concert with many important jazz and popular artists, including Eddie Daniels, Roger Kellaway, Frank Zappa, Emil Richards, Laurendo Almeida, Buddy DeFranco, Diane Schurr, Peggy Lee, Herb Ellis, Ernestine Anderson, Natalie Cole, Mannheim Steamroller, The Beach Boys, Della Reese, and many more. Although his primary focus in recent years has been jazz, Collier has continued to perform occasionally as a featured mallet soloist with the Seattle Symphony, Spokane Symphony, Bellevue Philharmonic, Northwest Chamber Orchestra, and the Denver Symphony.
He has made commercial and educational recordings with his own jazz group for various jazz labels, including Inner City (New York), Music Minus One (New York), TC Records (Seattle), Nebula (Baton Rouge), and Studio 4 (Los Angeles). Collier has also made recordings with many internationally known jazz and popular artists, including Ernie Watts, Don Grusin, Bud Shank, Alex Acuna, Bobby Shew, Laurendo Almeida, Barbra Streisand, Ry Cooder, Nilsson, Howard Roberts, and others. In addition, he has appeared on many Hollywood film soundtracks under the direction of composers such as John Williams, Henry Mancini, Elmer Bernstein, Lalo Schifrin, Jerry Goldsmith, and Oliver Nelson.
Collier has also established a reputation as a jazz/percussion composer, with many of his compositions for jazz percussion ensemble published by Studio 4 Productions and distributed by Alfred Music. He has won 15 consecutive ASCAP Popular Panel Awards for his jazz and percussion compositions.
Collier has recorded several educational albums for Music Minus One and Studio 4 Productions, as well as presenting more than 300 jazz concerts in public schools around Washington for the Arts in Education Program, Washington State Arts Commission. The National Association of Jazz Educators presented him with an "Outstanding Service to Jazz Education" award in 1980.
A UW alumnus, Collier graduated from the School of Music in 1971 with a BA/BM in percussion performance.
Shannon Dudley
Tom Collier
Christina Sunardi
UW Steelband


Best prank ever? These women decided to scare strangers in the park...with donuts! The most surprising part of this prank is how few people wanted to partake in the treat.
As we approach the end of the school year, it's time to start planning the best senior pranks. Here are a few pranks for inspiration.
Here's another donut prank. This one is less scary, but a lot more disgusting. Do you think you'd fall for a donut covered in baby powder?
In case you really want to make people distrust you, here are few more food-related pranks you can try out. We're definitely going to do the Pringles one.


JTBC program "iKON Picnic" has come to an end.
This travel variety program showed iKON members as high school students of "Yang's High School" going on a field trip chaperoned by Jung Hyung Don, Jo Se Ho, and Kim Shin Young. The show came to an end after six episodes that made viewers conjure up fun memories from their school days and showed different sides from iKON's powerful on-stage persona.
Through a phone call on the conclusion of the program, Yang Hyun Suk revealed future plans and his affection for the group. "iKON will release an album soon. The variety program was fun, and I hope their album does well too. I don't express it, but I really value and love them," said Yang Hyun Suk. Stay tuned for more updates on iKON's January comeback!


I'm too busy this week to remind readers that I told you so... instead, I'll simply point out a link to an oldie, but goodie from early July of 2008. Enjoy.
And in case you aren't panicked enough... here is a link to an article about what happens to charities when the bank they use goes bust. Don't think this isn't happening already.
If you haven't reviewed FDIC rules on your nonprofit's (and affiliate's) bank account you are gmabling with your donor's money. Next week might be too late.
This is a good Q&A from a treasurer for a local Habitat for Humanity group on FDIC rules regarding their affiliate accounts.
It's been six weeks since Bruce E. Ivins, the alledged anthrax suspect, killed himself. However, his cremated remains are still stored at a funeral home in Maryland awaiting bizarre probate proceedings.
In a will he wrote last year, a few months before the Federal Bureau of Investigation focused the anthrax letters investigation on him, Dr. Ivins wrote of his wish to be cremated and have his ashes scattered. But fearing that his wife, Diane, and their two children might not honor the request, he came up with a novel way to enforce his demand: threatening to make a bequest to an organization he knew his wife opposed, Planned Parenthood.
"If my remains are not cremated and my ashes are not scattered or spread on the ground, I give to Planned Parenthood of Maryland" $50,000, Dr. Ivins wrote in the will. Court records value the estate at $143,000.
Ms. Ivins is a former president of Frederick County Right to Life, according to F.B.I. records. Bruce Ivins played keyboards at a Catholic church in Frederick and described himself in e-mail messages as "pro-life," but he was not an anti-abortion activist, said his lawyer, Paul F. Kemp. Ms. Ivins declined to comment, Mr. Kemp said.
I'm not sure if I've ever heard of a donor who uses his/her will to encourage action by threatening to make a donation to a group that the surviving heirs hated.
A doctor with McCain's medical charity who supplied her with prescriptions for the drugs lost his license and never practiced again. The charity, the American Voluntary Medical Team, eventually had to be closed in the wake of the controversy. Her husband was forced to admit publicly that he was absent much of the time she was having problems and was not aware of them.
"So many lives were damaged by this," said Jeanette Johnson, whose husband, John Max Johnson, surrendered his medical license. "A lot of good people. Doctors who volunteered their time. My husband. I cannot begin to tell you how painful it was. We moved far away to start over."
McCain's addiction also embroiled her with one of her charity's former employees, Tom Gosinski, who reported her drug use to the DEA and provided prosecutors with a contemporaneous journal that detailed the effects of her drug problems.
But if you still think my outrage is over the line.. please... keep emailing me.
If you don't know how the new Schedule G from the newly revised IRS Form 990 will effect your organization's fundraising reporting, you can now listen to a podcast with Geoffrey Peters, president of Creative Direct Response, who recently provided very useful information on how to get your organization ready to fill out the schedule.
Geoffrey Peters, a fundraiser and lawyer, is a noted legal expert on the regulation of nonprofits and their fundraisers. Download his complete bio.
Go to the Foundation Center to listen to the podcast.
A lot of people are being hurt by the collapse of Freddie Mac and Fannie Mae. As a fundraiser, I can't help but focus on the philanthropic implications. During a press conference describing the takeover, the government said it "will review the charitable activities."
By its account, Freddie Mac, along with its foundation, has invested more than $348 million in the community to date. Last year its annual Hoops for the Homeless campaign, a celebrity-studded basketball tournament to fight family homelessness, raised $900,000 for six local nonprofits that included Hannah House and So Others Might Eat.
Last year Fannie Mae dissolved its foundation -- an organization that has put more than $1 billion into education, affordable housing, education and economic development programs since 1979. The company said its philanthropic activities would be handled in-house, and it continued to give to local organization and initiatives. Among its 2007 programs, the company pledged $10 million to improve infrastructure in D.C. schools and $1 million in grants to help revitalize D.C. neighborhoods. Its annual Help the Homeless walkathon raised more than $7 million last year to support 175 local homeless service providers, such as Reston Interfaith, last year."
Some of you may remember that the embattled mortgage giants came under fire this past summer for making sizeable contributions to Jesse Jackson's Rainbow/PUSH Coalition and Citizenship Education Fund Annual Conference.
Combined these guys also gave almost $2 million in contributions to politicians during the 2006 election cycle.
I mean... Freddie Mac was recognized by the Washington Business Journal and Greater D.C. Cares as the top corporate philanthropist in the Washington region.
Do you understand how big this is?
A fellow fundraiser once told me that the only news worse than hearing that a major donor has been indicted is that a major donor has declared bankruptcy. Trust me. I have experience with this one... it's not always easy making a bankrupt corporate donor make good on a charitable pledge.
Alaska Gov. Sarah Palin will appear at around 30 fund-raising events in the next two months leading up to Election Day – about one every two days on average, officials with Sen. John McCain's presidential campaign said today.
Yet, no announcements on whether or not she ever plans to answers questions during a press conference.
Did she really steal drugs from her own charity?
Cindy McCain looked so nice in her speech at the GOP convention tonight.
It's hard to believe they featured the story of how she started her own nonprofit called the American Voluntary Medical Team, but they left out the part of the story where she was investigated by the DEA for stealing drugs to feed her addiction to prescription painkillers.
I believe she wore red that day. She granted semi-exclusive interviews to one TV station and three daily newspaper reporters in Arizona, tearfully recalling her addiction, which came about after painful back and knee problems and was exacerbated by the stress of the Keating Five banking scandal that had ensnared her husband. To make matters worse, McCain admitted, she had stolen the drugs from the American Voluntary Medical Team, her own charity, and had been investigated by the Drug Enforcement Administration.
"She was blonde and beautiful. A rich man's daughter who became a politically powerful man's wife. She had it all, including an insidious addiction to drugs that sapped the beauty from her life like a spider on a butterfly."
Hmmm. But that's old news, right?
You gotta hand it to Jerry Lewis.
The "freewheeling 82-year-old showman" co-hosted the 22-hour telethon yesterday that raised a record $65 million for the Muscular Dystrophy Association's annual Labor Day event.
Despite the sluggish economy, the 43rd annual fundraising blitz was $1.2 million more than last year's total. Even more impressive... the host also urged viewers to contribute to the Salvation Army's emergency disaster services because of the threats from Hurricane Gustav.


On July 6, President Shavkat Mirziyoyev, at the invitation of the President of Kazakhstan Nursultan Nazarbayev, arrived in Astana.
The First Deputy Prime Minister of the Republic of Kazakhstan Askar Mamin and other officials met the President of Uzbekistan.
As previously reported, Shavkat Mirziyoyev will attend the event timed to Astana's 20th anniversary.


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Message People Nearby With Spray
Written by Diogo Costa
In the past, whenpeople wanted to keep in touch with people they already knew, they would just call or text them. However, nowadays things are changing – people want to discover and communicate with people they do not know and, more than that, with people that are located around them at any given moment.
As usual, the world of tech is trying to follow the wishes of users and produce ways of satisfying those wishes. Recently, a new app launched promising to connect users that were previously unconnected, helping to discover and build a local community anywhere at any time – let's meet Spray.
Announced as the world's first app that lets you message everyone nearby without knowing their names, emails, phone numbers or social media accounts, Spray is a free messaging app available for Android and iOS. It allows users to instantly connect to everyone nearby to freely share information, ask questions, offer support, get involved, or even meet new people.
In a sense, Spray can be thought of as a "barrier breaker", providing a platform for strangers to meet. It protects privacy and assures anonymity at all times, with users doing their shares publicly and responding privately, but Spray is always sent from a real person – users are required to login with Facebook, Twitter or their phone number.
As one might expect, Spray allows users to share common messages, but also other content like images and files, that can be sent to anyone within range. Despite being a very new app, Spray managed to (with zero marketing) reach an achievement of 80,000 messages sent and 30,000 users worldwide, already helping festivals, cities, and brick and mortars by giving them the power of a local messenger.
It has also managed to get positive reviews, including some powerful people in the tech world. Mitch Lowe, co-founder of Netflix and Redbox, said that this app brings "a very innovative and twisted approach to connecting people nearby, that has not been implemented before. Thanks to Spray, the world will be better, safer and an easier place for everyone".
So, with all this positive impact, it should come as no surprise if we hear more about Spray in the near future.
Diogo Costa Writer
25 y/o, born and living in Portugal. Majored in Biology, but tech and computers were always a passion. Wrote for sites like Windows.Appstorm and MakeTechEasier.
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Mitchell, P. (2017) 'Literary geography and the digital: The emergence of neogeography.' In Tally, R.T. Jr. (ed) The Routledge Handbook of Literature and Space. London: Routledge, pp. 85-94.
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Ridanpää, J. (2007) 'Laughing at northernness: postcolonialism and metafictive irony in the imaginative geography'. Social & Cultural Geography, 8(6), pp. 907-928.
Ridanpää, J. (2010a) 'Metafictive geography.' Culture, Theory & Critique, 51(1), pp. 47-63.
Ridanpää, J. (2010b) 'A Masculinist Northern Wilderness and the Emancipatory Potential of Literary Irony.' Gender, Place & Culture, 17(3), pp. 319-335.
Ridanpää, J. (2014) 'Politics of literary humour and contested narrative identity (of a region with no identity)' Cultural Geographies, 21(4), pp. 711-726.
Travis, C. (2017) 'Joycean Chronotopography: Homer, Dante, Ulysses.' In Tally, R.T. Jr. (ed) The Routledge Handbook of Literature and Space. London: Routledge, pp. 323-336.
Waugh, P. (1985) Metafiction. The Theory and Practice of Self-Conscious Fiction. New York: Methuen.
Williams, R. (1973) The Country and the City. London: Chatto and Windus.


posted by Howard B. Owens in news, fire, byron.
An outbuilding, with a car nearby, is reported on fire at 6478 Freeman Road, Byron.
Byron fire, South Byron fire, and Stafford fire dispatched.
The vehicle has caught fire.
UPDATE 3:22 p.m.: A second alarm is requested. Mercy EMS for firefighter support also requested.
posted by Howard B. Owens in byron, news.
I don't know enough about vintage cars to tell you the make or year of this vehicle but driving through Byron this morning I found it sitting outside a building in the hamlet.
June Rouse was sorry to see it go -- the planter she has lovingly cultivated for 10 years around her mailbox on Lyman Road, Byron.
She learned just a day ahead of time that a new water line being installed on Lyman was going right through the side of the road her mailbox is on.
She tried to save what she could with a tractor, digging up big frozen clumps of earth with her bulbs in the dirt. She hopes she can replant everything this spring.
Photo submitted by June Rouse.
posted by Billie Owens in byron, bergen, Byron-Bergen High School, news, musical, Annie.
Byron-Bergen High School presents the spring musical "Annie," opening at 7 p.m. Thursday, March 7.
This rags-to-riches tale features the music and acting talents of: Molly Belknap, Jack Benstead, Sarah Bleiler, Justine Bloom, Shelby Bridge, Stephanie Buell, Fiona Burke, Dominick Butz, Caleb Calhoun, Hailey Canfield, Caris Carlson, Hannah Catalino, Emily Chaback, Destiny Colon, Sadie Cook, Jacey Donahue, Jeremy Donahue, Serenity Donahue, Kendan Dressler, Josh Fleming, Sara Fraser, Aurora Hiscutt, Jason Hoehn, Kelly Ireland, Sage Johnson, Callista Kinkelaar, Aiden Kulikowski, Naomi Mathias, Courtney Pakusch, Libby Piper, Alexandria Schuck, Chloe Shuskey, Deacon Smith, Isabelle Stevens, Nicole Stone, Alayna Streeter, Hannah VanSkiver, and Lexi Vurraro.
Erin Parnapy, Coltin Henry, and Suzanne Scholand provide stage and set management.
Direction and musical direction is by Laurence Tallman. Special Byron-Bergen staff cameos by Karen Tischer and Peter Spence.
There will also be performances at 7 p.m. on Friday and Saturday, March 8 and 9 in the Jr/Sr High School Auditorium. Tickets are $10 for adults and $8 for students and senior citizens. Tickets are available in advance in the High School Main Office during school hours, or at the door.
This show is presented with special arrangements with, and all authorized performance materials are supplied by Music Theater International (MTI).
Photos courtesy of Gretchen Spittler.
posted by Billie Owens in weather, byron, south byron, news.
Seven utility poles, possibly with power lines attached, are down in a field on Tripp Road near Townline Road in Byron. Byron and South Byron fire departments are called to investigate.
UPDATE 1:27 p.m.: National Grid crews are called to the scene.
Two people have been transported, including one by Mercy Flight, from the site of an accident involving a manure spreader and a van in the area of 6385 North Bergen Road, Byron.
The county's Crash Management Team is investigating the accident.
Byron and South Byron volunteer fire departments are on scene. Bergen's ambulance also responded.
One person was reportedly entrapped in a vehicle following the accident and extrication was required.
UPDATE Midnight: A work van with two male occupants rear-ended a manure spreader on North Bergen Road that was returning to the farm after working a nearby field.
The spreader was equipped with reflectors, said Byron Chief Robert Mruczek.
The driver and the passenger in the van were transported to Strong Memorial Hospital, one by Mercy Flight.
A short extrication was required for one of the patients, Mruczek said.
The driver of the farm equipment was transported to UMMC for evaluation.
The manure spreader was barely damaged.
"It's a strong piece of machinery," Mruczek said. "It's very solid."
The Sheriff's Office is investigating the accident and the names of those involved have not yet been released.
Information and photos via Alecia Kaus/Video News Service.
posted by Howard B. Owens in weather, byron, news.
Nancy Clark shared these photos with us of an ice jam and ice overflow on Black Creek near Walkers Corner Road in Byron from this morning.
A one-vehicle rollover accident is reported on Bank Street Road near Cockram Road, Byron.
Byron fire and South Byron fire dispatched.
UPDATE 11:51 a.m.: Road conditions are not good, according to a first responder. Elba fire requeted to the scene to assist.
UPDATE 11:53 a.m.: The vehicle is off the road. All responding units from Byron and South Byron are parked at the side of the road. The road is not blocked. Elba can stand down. Mercy EMS on scene.
A mother and her adult son lost their home at 6123 Cook Road, Byron, on Monday after a fire erupted inside the residence at 7:10 p.m.
Byron and South Byron responded to the fire with mutual aid from Bergen, Stafford, Elba, and departments in Orleans and Monroe counties. The City's Fast Team also responded.
Byron Chief Robert Mruczek said the house was fully involved when he arrived on scene and he immediately called for a second alarm.
The firefighting effort was hampered by the cold and the wind. With slick roads and fire hydrants buried in snow, it was difficult to get water to the scene, Mruczek said. A metal roof also prevented firefighters from opening vents.
"We were trying to go to the inside to knock it down as fast as we possibly could," Mruczek said. "That was hampered due to the weather conditions for our water supply, so we started to attack from the outside and then the wind kicked up. The wind did not help us."
The weather conditions made it tough on the volunteers who responded to the fire, Mruczek. A school bus was set up to provide a rehab station. Firefighters are rotated from fighting the fire to rehab about every 20 or 30 minutes so they stay fresh, stay hydrated, and don't risk hypothermia.
"They work hard," Mruczek said. "You know, they do their thing. They're on the hose line. They're trying to get into the building and you've got to keep on rotating in and out because it's tiresome in the cold. You freeze up, the water freezes on you and you've got to exchange."
Both mother and son escaped the house and were physically unharmed. They are being cared for by the Salvation Army. They had no pets.
"It's unfortunate," Mruczek said. "It's sad. With the winter weather, these things happen and you try to do the best you can and you hope for the best. Fortunately, nobody got hurt. That's the good thing. Everybody's OK."
posted by Billie Owens in fire, news, byron, south byron.
The fire at 6123 Cook Road in Byron, which was first reported shortly after 7 o'clock last night, has rekindled. Byron and South Byron fire departments are responding.
"It's going fairly good," says a first responder at the scene.
UPDATE 5:48 a.m.: A pumper and a tanker from Elba are called to provide mutual aid.


Efficacy of pneumatic compression and low-level laser therapy in the treatment of postmastectomy lymphoedema: a randomized controlled trial.
Giuseppe Portale, Mario Costantini, +5 authors F.A.C.S. Giovanni Zaninotto M.D.
Randomized clinical trial of intermittent pneumatic compression and low molecular weight heparin in trauma.
Intermittent pneumatic compression improves venous ulcer healing.
Treatment of achalasia with pneumatic dilatations.


text:
Web Games Day
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From victory over the «push» and the return
From Hemeimeem, Putin ordered a partial withdrawal of his country's
troops deployed in Syria "towards its permanent bases", to
concentrate its presence there along with certain points where it performs
specific tasks, including monitoring the areas of "de-escalation Zones ".
Hours after Putin's words, the Russian Defence Ministry confirmed that it had
already begun to implement the order.
It is necessary to read the Russian move in the context of efforts to
launch the path of "political settlement" in Syria, which was clearly
presented in Putin's speech yesterday. Also, the declaration of victory over
the «push» and the return of part of the Russian forces will be within Russia
in the interest of the president candidate for a second term. With Moscow's
broad criticism of the "illegal" US military presence on Syrian
territory, its recent decision will constitute an additional bargaining chip in
negotiations with American "partners." The withdrawal of troops is
unlikely to affect the strength of the Russian presence in Syria. The bases of Hemeimeem
and Tartous will remain an important center of operations, while some military
police, the Reconciliation Center and military advisers will maintain their
presence in various Syrian regions. The nature of the developments in the (http://www.al-akhbar.com, 2017)«de-escalation agreements
and the progress of the Geneva, Astana and Sochi conferences will have a
decisive role in determining the size and type of such a presence.
Putin, from the air base, praised the results of
counterterrorism operations in Syria and reiterated his readiness to work with
Turkey and Iran to "restore peaceful life and establish a political
process … and prepare for the Syrian national dialogue conference."
Putin thanked the soldiers of his forces for their role in "protecting
their country" during their participation in operations in Syria,
"helping the people of Syria to maintain their state, and defeat the
terrorists." He added that "Syria has been preserved as an
independent and sovereign state, and the refugees are returning to their homes
and favorable conditions have been created for reaching a political settlement
within the framework of the United Nations." The Russian Center for Reconciliation
will continue its work in accordance with international agreements, "If
the terrorists return to activity again, we will carry out unprecedented
strikes against them." For his part, Assad expressed his gratitude «on
behalf of the Syrian people … the role played by the Russian armed forces».
"The coming generations that will read about this war will not
differentiate between a Syrian martyr and a Russian martyr," he said.
"Your visit is an opportunity to discuss the second phase of counter-terrorism
and at the same time the political process in Syria." (https://sputniknews.com,
SDF and the coordination with the Iraqi forces
In light of these developments, the celebration
of the "Syrian Democratic Forces" witnessed a meeting with a number
of Iraqi army officers on the Syrian-Iraqi border north of the Euphrates River,
talking about a "joint center" for coordination between the two
parties, which called for an official Iraqi denial from the Joint Operations Command.
Common. "The consultative meeting, which was proposed by the Iraqi side,
discussed how best to deal with the infiltration of terrorist factions across
the border," said Mustafa Bali, director of the media center at SDF,
referring to an understanding of "a mechanism for exchange of information
and joint coordination to monitor border security" . He went so far as to
consider the meeting «an implicit recognition of one of the most important
neighboring countries legitimacy SDF, as a military force liberated northern
Syria from terrorism». On the other hand, the spokesman for the Iraqi
Operations Command, Brigadier General Yahiya Rasool, said that "there is
absolutely no agreement with SDF to establish a joint coordination center …
It was not a meeting to discuss the issue of securing the border."
Battles continue in the last bastions of «ISIS»
The Syrian army and its allies continued their
military operations against the last "hot spots" on the Al-Mayadin
al-Bukamal road. The army units were able to advance towards the town of
Salhiya in the countryside of al-Bukamal, and control of a number of oil wells
and towns.
The forces are approaching more to control the
entire road between the two cities and isolating "ISIS" in a large
desert enclave. The army launched attacks to secure the area between the town
of Hamaimh and al-Bukamal in the depth of the southeast desert of Deir al-Zour.
In parallel, the Syrian Democratic Forces
continue their attack on the Da'ash sites in the eastern and north-eastern
countryside of Deir al-Zour, where they took control of the villages of Abu
Hamam and Suwaidan and reached the outskirts of Alkishkieh.
The liberation of all areas of the South Euphrates
Less than 20 days after the liberation of the
border town of al-Bukamal, the Syrian army and its allies completed the
liberation of the entire Euphrates basin from the control of ISIS. The
organization is trapped in a wide desert enclave extending west to the
outskirts of the city of Sokhna. The restoration of the towns of the Euphrates
valley is no less important than the liberation of al-Bukamal. These towns are
the road between Deir al-Zour and the Iraqi border, and securing them will
enable the revitalization of the existing al-Bukamal crossing. Controlling them
also means expelling the organization from its last point of control, which
includes residential communities, so that it will have a desert area that could
be a starting point for security operations against the surrounding towns and
cities. T2 is expected to be the main target for these operations, given its
remote and isolated location in the Badia (http://www.plenglish.com/index.php?o=rn=20263=syrian-troops-adanced-to-al-bukamal-city-in-deir-ezzor,
(https://southfront.org/syrian-war-report-november-20-2017-government-troops-liberated-al-bukamal-from-isis/,
This field achievement came through the
liberation of the last towns between the axis of the army activity north of
al-Bukamal and southeast of the fields. Due to the importance of these sites to
«urging», It was costly to recover it. A large number of members of the army and its
allies were martyred during the fierce fighting in the vicinity of the towns of
Qouriya, Ashara and elsewhere.
In parallel with the official declaration of the
army on the liberation of the entire Euphrates basin, Moscow went out to
declare a defeat of «ISIS». President Vladimir Putin said in parallel with the
confirmation of his candidacy for a second presidential term, the end of the
military operations on the banks of the Euphrates with a "complete defeat
of terrorists", despite the existence of some "enclaves" of the
organization. It seems that Putin repeated what he said during his meeting with
his Syrian counterpart, Bashar al-Assad, on the need to "move to the next
stage", the political process, noting the efforts to hold the conference
«national dialogue» in Sochi. He referred to the ongoing work on a "new
constitution" for Syria, after which "the transition to presidential
and parliamentary elections" would take place.
In between the Syrian and Turkish governments was
IntroductionKaliningrad the current situation in Kaliningrad and what
Syrian the opposition using arms to defend themselves
us this us army school game. There are
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Open concept living dining and kitchen with stainless steel appliances breakfast bar & pantry on main floor with half bath. 2nd floor offers 2 spacious bedrooms and a full bath between them (second bedroom does not have a window). Master bdrm has a vaulted ceiling and walk in closet. Upstairs laundry. NEW CARPET THRU OUT! Patio by front entry for your outdoor living. Access the underground parking from your kitchen (2 parking spaces included). Great location! Live near restaurants parks schools bike trails and just minutes from Epic.


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text:
\section{Introduction}
\par Let $(M^n,g) (n\geq3)$ be an $n$-dimensional Riemannian manifold with the Riemannian curvature tensor $Rm=\{R_{ijkl}\}$, the Weyl curvature tensor $W=\{W_{ijkl}\}$, the Ricci tensor $Ric=\{R_{ij}\}$ and the scalar curvature $R$. For any manifold of dimension $n\geq4$, the Bach tensor, introduced by Bach \cite{B}, is defined as \begin{eqnarray}
B_{ij}\equiv\frac{1}{n-3}\nabla^k\nabla^lW_{ikjl}+\frac{1}{n-2}R^{kl}W_{ikjl}.
\end{eqnarray}
Here and hereafter the Einstein convention of summing over the repeated indices will be adopted. In \cite{KL}, Korzynski and Lewandowski proved that the Bach tensor can be identified with the Yang-mills current of the  Cartan  normal conformal connection.
Recall that a metric $g$ is called Bach-flat if the Bach tensor vanishes. It is easy to see that $B_{ij}=0$ if $(M^n,g)$ is either locally conformally flat, or an Einstein manifold. In the case of $n=4$, $g$ is Bach-flat if and only if it is a critical metric of the functional (see \cite{{Be},{GN}}) $$W:g\mapsto\int_{M}{|W_g|}^2dV_g.$$

Now we introduce the definition of the Yamabe constant. Given a complete Riemannian $n$-manifold $(M^n,g)$ of dimension $n\geq3$, the Yamabe constant $Y(M,[g])$ ($[g]$ is the conformal class of $g$) is defined as
$$Y(M,[g])\equiv\inf\limits_{\tilde{g}\in[g]}\frac{\int_M{R_{\tilde{g}}}dV_{\tilde{g}}}{\left(\int_MdV_{\tilde{g}}\right)^{\frac{n-2}{n}}}=\inf\limits_{0\neq u\in C^\infty_0(M^n)}\frac{\int_{M}\left(\frac{4(n-1)}{n-2}|\nabla u|^2+Ru^2\right)dV_g}{\left(\int_{M}|u|^\frac{2n}{n-2}dV_g\right)^\frac{n-2}{n}}.$$
The important works of Aubin \cite{A}, Schoen \cite{S}, Trudinger \cite{T} and Yamabe \cite{Y} showed that  for compact manifolds the infimum in the above is always achieved. There are noncompact complete Riemannian manifolds of negative scalar curvature with positive Yamabe constant. For example, any simply connected complete locally conformally flat manifold has positive Yamabe constant (see \cite{SY} ). Furthermore, for compact manifolds, $Y(M,[g])$ is determined by the sign of the scalar curvature $R$ (see \cite{A} ), and for noncompact manifolds, $Y(M,[g])$ is always positive if $R$ vanishes (see \cite{D}).

The curvature pinching phenomenon plays an important role in global differential geometry. Some isolation theorems of Weyl curvature tensor of positive Einstein manifolds are given in \cite{{HV},{IS},{Si}}, when its $L^{\frac{n}{2}}$-norm is small. Recently, two rigidity theorems of Weyl curvature tensor of positive Einstein manifolds are given
in \cite{{C},{FX2},{FX3}}, which improve results due to \cite{{HV},{IS},{Si}}.  The first author and Xiao have studied compact manifolds with harmonic curvature to obtain some rigidity results in \cite{{F},{FX}}. Here when a  Riemannian manifold satisfies $\delta Rm= \{\nabla^lR_{ijkl}\}=0$, we call it a manifold with harmonic curvature.
Bach-flat manifolds have been studied by many authors. For any complete Bach-flat manifold, Kim \cite{K} has studied their rigidity phenomena  and derived that a complete  Bach-flat 4-manifold $M^4$ with nonnegative constant scalar curvature and the positive Yamabe constant is an Einstein manifold if the $L^2$-norm of the trace-free Riemannian curvature tensor $\mathring{Rm}$ is small enough. Later, Chu \cite{Ch} improved Kim's result and showed that $M^4$ is in fact a space of constant curvature under the same assumptions.  Chu and Feng \cite{CF} proved the rigidity result for  $n$-dimensional Bach-flat manifolds with constant scalar curvature and positive Yamabe constant. For a compact Bach-flat manifold $M^4$ with the positive Yamabe constant, Chang et al. \cite{CQY} proved that $M^4$ is conformal equivalent to the standard four-sphere provided that the $L^2$-norm of the Weyl curvature tensor $W$ is small enough, and also showed that there is only finite diffeomorphism class with a bounded $L^2$-norm of $W$.

Now, we are interested in $L^p$ pinching problems for  compact Bach-flat manifolds with positive constant scalar curvature. In this paper, under some $L^p$ pinching conditions,  we show that the  compact Bach-flat manifold with positive constant scalar curvature is spherical space form or Einstein manifold. More precisely, we have the following theorems:
\begin{theorem}
Let $(M^n,g) (n\geq4)$ be an $n$-dimensional  compact Bach-flat Riemannian manifold with positive constant scalar curvature. For $p\geq\frac{n}{2}$, if
$$\left(\int_{M}|\mathring{Rm}|^pdV_g\right)^{\frac{1}{p}}<\varepsilon(n)Y(M,[g])^\frac{n}{2p}R^{1-\frac{n}{2p}},$$
where $\varepsilon(n)$ is a constant depending only on $n$, i.e.,
\[
\varepsilon(n)=\left\{
\begin{array}{cc}
\frac{n-2}{4(n-1)\left(C(n)+(n-2)\sqrt{\frac{n-2}{2(n-1)}}\right)}, &\mbox{if $n=4,5$ and $p=\frac{n}{2}$,}\\
\left[\frac{(n-2)(2p-n)}{n(6-n)}\right]^{\frac{n}{2p}}\frac{(6-n)p}{2(n-1)(2p-n)\left(C(n)+(n-2)\sqrt{\frac{n-2}{2(n-1)}}\right)}, &\mbox{if $n=4,5$ and $\frac{n}{2}<p<\frac{2n}{n-2}$,}\\
\frac{1}{(n-1)\left(C(n)+(n-2)\sqrt{\frac{n-2}{2(n-1)}}\right)}, &\mbox{if $n\geq6$ and $p\geq\frac{n}{2}$ or if $n=4,5$ and $p\geq\frac{2n}{n-2}$,}
\end{array}\right.
\]
and $C(n)$ is defined in Lemma 2.1,
then $(M^n,g)$ is isometric to a quotient of the round $\mathbb{S}^n$.
\end{theorem}
\begin{corollary}
Let $(M^n,g) (n\geq4)$ be an $n$-dimensional  compact Bach-flat Riemannian manifold with positive constant scalar curvature. If
$$\left(\int_{M}|\mathring{Rm}|^{\frac n2}dV_g\right)^{\frac{2}{n}}<\varepsilon(n)Y(M,[g]),$$
where
\[
\varepsilon(n)=\left\{
\begin{array}{cc}
\frac{n-2}{4(n-1)\left(C(n)+(n-2)\sqrt{\frac{n-2}{2(n-1)}}\right)}, &\mbox{if $n=4,5$,}\\
\frac{1}{(n-1)\left(C(n)+(n-2)\sqrt{\frac{n-2}{2(n-1)}}\right)}, &\mbox{if $n\geq6$,}
\end{array}\right.
\]
then $(M^n,g)$ is isometric to a quotient of the round $\mathbb{S}^n$.
\end{corollary}
\begin{remark}
The above $L^p$-pinching condition in Theorem $1.1$ is invariant under any homothety. $L^{\frac n2}$ trace-free Riemannian curvature pinching theorems have been shown by Kim \cite{K}, Chu \cite{C}, and Chu and Feng \cite{CF}, in which the pinching constant  are not explicit, respectively.  Theorem $1.1$ extends the $L^{p}$ trace-free Riemannian curvature pinching theorems given by \cite{{C},{CF},{K}} in power $p=\frac n2$ to  $p\geq\frac n2$.
\end{remark}
\begin{theorem}
Let $(M^n,g) (n\geq4)$ be an $n$-dimensional  compact Bach-flat Riemannian manifold with positive constant scalar curvature. If
\begin{eqnarray}
\left(\int_{M}\left|W+\frac{\sqrt{n}}{2\sqrt{2}(n-2)}\mathring{Ric}\circledwedge g\right|^{\frac{n}{2}}dV_g\right)^{\frac{2}{n}}<C_1(n)Y(M,[g]),
\end{eqnarray}
where
\[
C_1(n)=\left\{
\begin{array}{cc}
\sqrt{\frac{n-2}{32(n-1)}}, &\mbox{if $n=4,5$,}\\
\frac{1}{\sqrt{2(n-2)(n-1)}}, &\mbox{if $n\geq6$,}
\end{array}\right.
\]
then $(M^n,g)$ is an Einstein manifold. In particular, for $n=4,5$, then $(M^n,g)$ is isometric to a quotient of the round $\mathbb{S}^n$; for $n\geq6$, if the pinching constant in (1.2) is weakened to $\frac{2Y(M,[g])}{nC_2(n)}$, where $C_2(n)$ is defined in Lemma 2.1 of \cite{FX3}, then $(M^n,g)$ is isometric to a quotient of the round $\mathbb{S}^n$.
\end{theorem}

\begin{remark}
When $n\geq6$, the inequality (1.2) of this theorem is optimal. The critical case is given by the following example. If $(\mathbb{S}^1(t)\times\mathbb{S}^{n-1},g_t)$ is the product of the circle of radius $t$ with $\mathbb{S}^{n-1}$, and if $g_t$ is the standard product metric normalized such that $Vol(g_t)=1$, we have $W=0$, $g_t$ is a Yamabe metric for small $t$ (see \cite{S2}), and $\left(\int_{M}|\mathring{Ric}|^{\frac{n}{2}}dV_g\right)^{\frac{2}{n}}=\frac{Y(M,[g])}{\sqrt{n(n-1)}}$, which is the critical case of the inequality (1.2) in Theorem $1.4$. We know that $(\mathbb{S}^1(t)\times\mathbb{S}^{n-1},g_t)$ is not Einstein.
\end{remark}
\begin{corollary}
Let $(M^4,g)$ be a $4$-dimensional  compact Bach-flat Riemannian manifold with positive constant scalar curvature. If
 \begin{eqnarray}
 \int_{M}|W|^2dV_g+\frac{5}{4}\int_{M}|\mathring{Ric}|^2dV_g\leq\frac{1}{48}\int_{M}R^2dV_g,
 \end{eqnarray}
then $(M^4,g)$ is isometric to a quotient of the round $\mathbb{S}^4$.
\end{corollary}
\begin{remark}
By the Chern-Gauss-Bonnet formula,
the pinching condition (1.3) in Corollary $1.6$ is equivalent to the following
\begin{eqnarray}
\int_{M}|W|^2dV_g+\frac{2}{39}\int_{M}R^2dV_g\leq\frac{160}{13}\pi^2\chi(M),
\end{eqnarray}
where $\chi(M)$ is the Euler-Poincar\'{e} characteristic of $M$.
\end{remark}
\begin{theorem}
Let $(M^n,g)$ be an $n$-dimensional compact Bach-flat Riemannian manifold with positive constant scalar curvature. If
\begin{eqnarray}
|W|^2+\frac{{n}}{2(n-2)}|\mathring{Ric}|^2\leq\frac{1}{{2(n-2)(n-1)}}R^2,
\end{eqnarray}
then $(M^n,g)$ is isometric to either an Einstein manifold or a quotient of $\mathbb{S}^1\times \mathbb{S}^{n-1}$ with the product metric.
\end{theorem}
\begin{corollary}
Let $(M^n,g)$ be an $n=4$ or $5$-dimensional compact Bach-flat Riemannian manifold with positive constant scalar curvature. If
 $$
 |W|^2+\frac{{n}}{2(n-2)}|\mathring{Ric}|^2\leq\frac{1}{{2(n-2)(n-1)}}R^2,
$$
then $(M^n,g)$ is isometric to either a quotient of the round $\mathbb{S}^n$ or a quotient of $\mathbb{S}^1\times \mathbb{S}^{n-1}$ with the product metric.
\end{corollary}

\section{Proof of Theorem 1.1}
Let $(M^n,g)(n\geq3)$ be an $n$-dimensional  complete Riemannian manifold with the metric $g=\{g_{ij}\}$. Denote by $Ric=\{R_{ij}\}$ and $R$ the Ricci tensor and the scalar curvature, respectively. It is well known that the Riemannian curvature tensor $Rm=\{R_{ijkl}\}$ of $M^n$ can be decomposed into three orthogonal components which have the same symmetries as
$Rm$ $$R_{ijkl}=W_{ijkl}+V_{ijkl}+U_{ijkl},$$ $$V_{ijkl}=\frac{1}{n-2}(\mathring{R}_{ik}g_{jl}-\mathring{R}_{il}g_{jk}+\mathring{R}_{jl}g_{ik}-\mathring{R}_{jk}g_{il}),$$ $$U_{ijkl}=\frac{R}{n(n-1)}(g_{ik}g_{jl}-g_{il}g_{jk}),$$
where $W=\{W_{ijkl}\}$, $V=\{V_{ijkl}\}$ and $U=\{U_{ijkl}\}$ denote the Weyl curvature tensor, the traceless Ricci part and the scalar curvature part, respectively, and $\mathring{Ric}=\{\mathring{R}_{ij}\}=\{R_{ij}-\frac{R}{n}g_{ij}\}$ is the trace-free Ricci tensor.
Denote by $\mathring{Rm}=\{\mathring{R}_{ijkl}\}=\{R_{ijkl}-U_{ijkl}\}$  the trace-free Riemannian curvature tensor.
 In local coordinates, the norm of a $(0,4)$-type tensor $T$ is defined as $$|T|^2=|T_{ijkl}|^2=g^{im}g^{jn}g^{ks}g^{lt}T_{ijkl}T_{mnst}.$$

The following equalities are easily obtained from the properties of Riemannian curvature tensor:
\begin{equation}
g^{ik}\mathring{R}_{ijkl}=\mathring{R}_{jl,}
\end{equation}
\begin{equation}
\mathring{R}_{ijkl}+\mathring{R}_{iljk}+\mathring{R}_{iklj}=0,
\end{equation}
\begin{equation}
\mathring{R}_{ijkl}=\mathring{R}_{klij}=-\mathring{R}_{jikl}=-\mathring{R}_{ijlk},
\end{equation}
\begin{equation}
|\mathring{Rm}|^2=|W|^2+|V|^2=|W|^2+\frac{4}{n-2}|\mathring{Ric}|^2.
\end{equation}
Moreover, under the assumption of constant scalar curvature, we get
\begin{equation}
\nabla_h\mathring{R}_{ijkl}+\nabla_l\mathring{R}_{ijhk}+\nabla_k\mathring{R}_{ijlh}=0,
\end{equation}
and
\begin{eqnarray}
\nabla^lW_{ijkl}&=&\nabla^lR_{ijkl}-\nabla^lV_{ijkl}-\nabla^lU_{ijkl}\nonumber\\
&=&\nabla^l\mathring{R}_{ijkl}-\nabla^lV_{ijkl}\nonumber\\
&=&\frac{n-3}{n-2}\left(\nabla_j\mathring{R}_{ik}-\nabla_i\mathring{R}_{jk}\right)\nonumber \\
&=&\frac{n-3}{n-2}\nabla^l\mathring{R}_{ijkl}.
\end{eqnarray}
Since $n\geq3$, from (2.4) we see that
\begin{eqnarray}
|\mathring{Ric}|^2\leq\frac{n-2}{4}|\mathring{Rm}|^2.
\end{eqnarray}

Let $\Lambda^2(M)$ and $\otimes^2(M)$  denote the space of skew symmetric $2$-tensors  and  $2$-tensors, respectively. It is easy to know that the dimension of $\Lambda^2(M)$ and $\otimes^2(M)$ is $\frac{n(n-1)}{2}$ and  $n^{2}$, respectively. Let $T=\{T_{ijkl}\}$ be a tensor with the same symmetries as the Riemannian curvature tensor. It defines a symmetric operator $T:\Lambda^2(M)\rightarrow\Lambda^2(M)$ by
$$(T\omega)_{kl}:=T_{ijkl}\omega_{ij},$$
with $\omega\in\Lambda^2(M)$. Similarly, it also defines a symmetric operator $T:\otimes^2(M)\rightarrow\otimes^2(M)$ by
$$(T\theta)_{kl}:=T_{kilj}\theta_{ij},$$
with $\theta\in\otimes^2(M)$.

In order to prove Theorem $1.1$, we need the following lemma:
\begin{lemma}
Let $(M^n,g) (n\geq3)$ be an n-dimensional Riemannian manifold with constant scalar curvature, then
\begin{eqnarray}
\mathring{R}^{ijkl}\Delta\mathring{R}_{ijkl}\geq
-C(n)|\mathring{Rm}|^3+2\mathring{R}^{ijkl}\nabla_l\nabla^m\mathring{R}_{ijkm}+A(n)R|\mathring{Rm}|^2,
\end{eqnarray}
where
\[
A(n)=\left\{
\begin{array}{cc}
\frac{1}{n-1}, &\mbox{if $R\geq0$,}\\ \frac{2}{n}, &\mbox{if $R<0$,}
\end{array}\right.
\]
and $C(n)=\frac{4(n^2-2)}{n\sqrt{n^2-1}}+\frac{n^2-n-4}{\sqrt{(n-2)n(n^2-1)}}+\sqrt{\frac{(n-2)(n-1)}{n}}$.
\end{lemma}
\begin{remark}
Although Lemma $2.1$ and the explicit coefficient of the term $|\mathring{Rm}|^3$ of (2.8) have been proved in \cite{Ch} and \cite{FX} respectively, for completeness, we also write it out.
\end{remark}
\begin{proof} To simplify the notations, we will compute at an arbitrarily chosen point $p\in M$ in normal coordinates centered at $p$ so that $g_{ij}=\delta_{ij}$. We obtain from (2.3) and (2.5) that
\begin{eqnarray}
\mathring{R}_{ijkl}\Delta\mathring{R}_{ijkl}&=&\mathring{R}_{ijkl}\nabla_m\nabla_m\mathring{R}_{ijkl}\nonumber \\ &=&2\mathring{R}_{ijkl}\nabla_m\nabla_l\mathring{R}_{ijkm}\nonumber \\ &=&2\mathring{R}_{ijkl}(\nabla_l\nabla_m\mathring{R}_{ijkm}+R_{hilm}\mathring{R}_{hjkm} +R_{hjlm}\mathring{R}_{ihkm}+R_{hklm}\mathring{R}_{ijhm}+R_{hmlm}\mathring{R}_{ijkh}),
\end{eqnarray}
where the Ricci identities are used in the last equality of (2.9). By the definition of trace-free Riemannian curvature tensor and (2.1), from (2.9) we get
\begin{eqnarray}
\mathring{R}_{ijkl}\Delta\mathring{R}_{ijkl}&=&2\mathring{R}_{ijkl}\nabla_l\nabla_m\mathring{R}_{ijkm}+2\mathring{R}_{ijkl}(R_{hilm}\mathring{R}_{hjkm}+R_{hjlm}\mathring{R}_{ihkm}+R_{hklm}\mathring{R}_{ijhm})+2R_{hl}\mathring{R}_{ijkl}\mathring{R}_{ijkh}\nonumber \\ &=&2\mathring{R}_{ijkl}\nabla_l\nabla_m\mathring{R}_{ijkm}+2\mathring{R}_{ijkl}(\mathring{R}_{hilm}\mathring{R}_{hjkm}+\mathring{R}_{hjlm}\mathring{R}_{ihkm}+\mathring{R}_{hklm}\mathring{R}_{ijhm})\nonumber\\
&&+2\mathring{R}_{ijkl}\mathring{R}_{ijkh}\mathring{R}_{hl}+\frac{2R}{n}|\mathring{Rm}|^2
+\frac{2R}{n(n-1)}\mathring{R}_{ijkl}(\mathring{R}_{ljki}+\mathring{R}_{ilkj}+\mathring{R}_{ijlk}+\mathring{R}_{jk}\delta_{li}-\mathring{R}_{ik}\delta_{lj})\nonumber\\
&=&2\mathring{R}_{ijkl}\nabla_l\nabla_m\mathring{R}_{ijkm}+2\mathring{R}_{ijkl}(\mathring{R}_{hilm}\mathring{R}_{hjkm}
+\mathring{R}_{hjlm}\mathring{R}_{ihkm}+\mathring{R}_{hklm}\mathring{R}_{ijhm})\nonumber\\
& &+2\mathring{R}_{ijkl}\mathring{R}_{ijkh}\mathring{R}_{hl}-\frac{4R}{n(n-1)}|\mathring{Ric}|^2+\frac{2R}{n}|\mathring{Rm}|^2\nonumber\\
&=&2\mathring{R}_{ijkl}\nabla_l\nabla_m\mathring{R}_{ijkm}-2\left(2\mathring{R}_{ijlk}\mathring{R}_{ihlm}\mathring{R}_{hjmk}+\frac{1}{2}\mathring{R}_{ijkl}\mathring{R}_{ijhm}\mathring{R}_{hmkl}\right)\nonumber\\
& &+2\mathring{R}_{ijkl}\mathring{R}_{ijkh}\mathring{R}_{hl}-\frac{4R}{n(n-1)}|\mathring{Ric}|^2+\frac{2R}{n}|\mathring{Rm}|^2.
\end{eqnarray}
\par We consider $\mathring{Rm}$ as a trace-free symmetric operator on $\Lambda^2(M)$ and $\otimes^2(M)$. By the algebraic inequalities $tr(T^3)\leq\frac{m-2}{\sqrt{m(m-1)}}|T|^3$ for trace-free symmetric $m$-matrices $T$ and $\lambda_i\leq\sqrt{\frac{m-1}{m}}|T|$ for the eigenvalues $\lambda_i$ of $T$ in \cite{H}, we get
\begin{eqnarray}
\left|2\mathring{R}_{ijlk}\mathring{R}_{ihlm}\mathring{R}_{hjmk}+\frac{1}{2}\mathring{R}_{ijkl}\mathring{R}_{ijhm}\mathring{R}_{hmkl}\right|
\leq2|\mathring{R}_{ijlk}\mathring{R}_{ihlm}\mathring{R}_{hjmk}|+\frac{1}{2}|\mathring{R}_{ijkl}\mathring{R}_{ijhm}\mathring{R}_{hmkl}|\nonumber\\
\leq\left(\frac{2(n^2-2)}{n{\sqrt{n^2-1}}}+\frac{n^2-n-4}{2\sqrt{(n-2)n(n^2-1)}}\right)|\mathring{Rm}|^3,
\end{eqnarray}
and
\begin{eqnarray}
|\mathring{R}_{ijkl}\mathring{R}_{ijkh}\mathring{R}_{hl}|\leq\sqrt{\frac{n-1}{n}}|\mathring{Ric}||\mathring{Rm}|^2.
\end{eqnarray}
Combining with (2.7), (2.10), (2.11) and (2.12), we get
\begin{eqnarray}
\mathring{R}_{ijkl}\Delta\mathring{R}_{ijkl}&\geq&
-\left(\sqrt{\frac{(n-1)(n-2)}{n}}+\frac{4(n^2-2)}{n\sqrt{n^2-1}}+\frac{n^2-n-4}{\sqrt{(n-2)n(n^2-1)}}\right)|\mathring{Rm}|^3\nonumber\\
& &+2\mathring{R}^{ijkl}\nabla_l\nabla^m\mathring{R}_{ijkm}+A(n)R|\mathring{Rm}|^2.
\end{eqnarray}
\end{proof}
\textbf{Proof of Theorem $1.1$.}
For simplicity of natation, we denote by $(\delta\mathring{Rm})_{ijk}=\nabla^l\mathring{R}_{ijkl}$ the divergence of the trace-free Riemannian curvature tensor and $u=|\mathring{Rm}|$. By the Kato inequality $|\nabla\mathring{Rm}|^2\geq|\nabla|\mathring{Rm}||^2$, we get
\begin{eqnarray}
\mathring{R}^{ijkl}\Delta\mathring{R}_{ijkl}
&\leq&\mathring{R}^{ijkl}\Delta\mathring{R}_{ijkl}+|\nabla\mathring{Rm}|^2-|\nabla|\mathring{Rm}||^2\nonumber\\
&=&\frac{1}{2}\Delta|\mathring{Rm}|^2-|\nabla|\mathring{Rm}||^2\nonumber\\
&=&|\mathring{Rm}|\Delta|\mathring{Rm}|=u\Delta u,
\end{eqnarray}
which together with Lemma $2.1$ and integrating on $M^n$ give
$$\int_{M}u\Delta udV_g\geq-C(n)\int_{M}u^3dV_g+2\int_{M}\mathring{R}^{ijkl}\nabla_l\nabla^m\mathring{R}_{ijkm}dV_g+\frac{R}{n-1}\int_{M}u^2dV_g.$$
Moreover, using the Stokes's theorem, we get
\begin{eqnarray}
\int_{M}|\nabla u|^2dV_g\leq C(n)\int_{M}u^3dV_g+2\int_{M}|\delta\mathring{Rm}|^2dV_g-\frac{R}{n-1}\int_{M}u^2dV_g.
\end{eqnarray}
Since $M^n$ is Bach-flat, we have
$$B_{ij}=\frac{1}{n-3}\nabla^k\nabla^lW_{ikjl}+\frac{1}{n-2}R^{kl}W_{ikjl}=0.$$
Multiplying the above equality by $\mathring{R}^{ij}$ and integrating on $M^n$ give
\begin{eqnarray}
0&=&\frac{1}{n-3}\int_{M}\mathring{R}^{ij}\nabla^k\nabla^lW_{ikjl}dV_g+\frac{1}{n-2}\int_{M}\mathring{R}^{ij}R^{kl}W_{ikjl}dV_g\nonumber\\
&=&-\frac{1}{n-2}\int_{M}\nabla^k\mathring{R}^{ij}\nabla^l\mathring{R}_{ikjl}dV_g+\frac{1}{n-2}\int_{M}\mathring{R}^{ij}\mathring{R}^{kl}W_{ikjl}dV_g\nonumber\\
&=&-\frac{1}{n-2}\int_{M}\frac{1}{2}\left(\nabla^k\mathring{R}^{ij}-\nabla^i\mathring{R}^{kj}\right)\nabla^l\mathring{R}_{ikjl}dV_g+\frac{1}{n-2}\int_{M}\mathring{R}^{ij}\mathring{R}^{kl}W_{ikjl}dV_g\nonumber\\
&=&-\frac{1}{2(n-2)}\int_{M}|\delta\mathring{Rm}|^2dV_g+\frac{1}{n-2}\int_{M}\mathring{R}^{ij}\mathring{R}^{kl}W_{ikjl}dV_g,
\end{eqnarray}
where (2.6) and the second Bianchi identities are used in the second line and the third line of (2.16) respectively. Using (2.7) and the Huisken inequality ( see Lemma 3.4 of  \cite{H})
$$|\mathring{R}^{ik}\mathring{R}^{jl}W_{ijkl}|\leq\sqrt{\frac{n-2}{2(n-1)}}|W||\mathring{Ric}|^2,$$
we have
\begin{eqnarray}
\int_{M}|\delta\mathring{Rm}|^2dV_g\leq\frac{n-2}{2}\sqrt{\frac{n-2}{2(n-1)}}\int_{M}u^3dV_g.
\end{eqnarray}
Combining with (2.15) and (2.17), we obtain
\begin{eqnarray}
\int_{M}|\nabla u|^2dV_g\leq E(n)\int_{M}u^3dV_g-\frac{R}{n-1}\int_{M}u^2dV_g,
\end{eqnarray}
where $E(n)=C(n)+\sqrt{\frac{(n-2)^3}{2(n-1)}}$. From (2.18), using Young's inequality and the H\"{o}lder inequality, we get
\begin{eqnarray}
\int_{M}|\nabla u|^2dV_g
&\leq&\frac{nE(n)}{2p}\epsilon^{-\frac{2p-n}{n}}\int_{M}u^{2+\frac{2p}{n}}dV_g+\left(\frac{(2p-n)\epsilon E(n)}{2p}-\frac{R}{n-1}\right)\int_{M}u^2dV_g\nonumber\\
&\leq&\frac{nE(n)}{2p}\epsilon^{-\frac{2p-n}{n}}\left(\int_{M}u^{\frac{2n}{n-2}}dV_g\right)^{\frac{n-2}{n}}\left(\int_{M}u^pdV_g\right)^{\frac{2}{n}}\\
& &+\left(\frac{(2p-n)\epsilon E(n)}{2p}-\frac{R}{n-1}\right)\int_{M}u^2dV_g\nonumber,
\end{eqnarray}
where $\epsilon$ is a positive constant. By the definition of Yamabe constant $Y(M,[g])$, we get
\begin{eqnarray}
\frac{n-2}{4(n-1)}Y(M,[g])\left(\int_{M}u^{\frac{2n}{n-2}}dV_g\right)^{\frac{n-2}{n}}
&\leq&\int_{M}|\nabla u|^2dV_g+\frac{(n-2)R}{4(n-1)}\int_{M}u^2dV_g\nonumber\\
&\leq&(1+\eta)\int_{M}|\nabla u|^2dV_g+\frac{(n-2)R}{4(n-1)}\int_{M}u^2dV_g,
\end{eqnarray}
where $\eta\geq0$ is a constant. Substituting (2.19) into (2.20), we conclude that
\begin{eqnarray}
\left\{\frac{n-2}{4(n-1)}Y(M,[g])-(1+\eta)\frac{nE(n)}{2p}\epsilon^{-\frac{2p-n}{n}}\left(\int_{M}u^pdV_g\right)^{\frac{2}{n}}\right\}\left(\int_{M}u^{\frac{2n}{n-2}}dV_g\right)^{\frac{n-2}{n}}\nonumber\\
\leq\left\{\frac{(n-2)R}{4(n-1)}+(1+\eta)\left(\frac{(2p-n)\epsilon E(n)}{2p}-\frac{R}{n-1}\right)\right\}\int_{M}u^2dV_g.
\end{eqnarray}

How we select $(\epsilon, \eta)$ to maximize $\left(\int_{M}u^pdV_g\right)^{\frac{2}{n}}$ in (2.21)  is equivalent to a problem of finding $(\epsilon, \eta)$ on a domain $\mathcal{D}$ which minimizes a function
$$F(\epsilon, \eta):=\frac{nE(n)}{2p}(1+\eta)\epsilon^{-\frac{2p-n}{n}}.$$
Here,  the domain $\mathcal{D}$ consists of points $(\epsilon, \eta)$ which satisfies inequalities
$$G(\epsilon, \eta):=\frac{(n-2)R}{4(n-1)}+(1+\eta)\left(\frac{(2p-n)\epsilon E(n)}{2p}-\frac{R}{n-1}\right)\leq0, \ \epsilon>0, \ \eta\geq0.$$
In the case of $p=\frac{n}{2}$, since we have \\
$$
F(\epsilon, \eta)=E(n)(1+\eta)  \ \
\begin{array}{cc}
&\mbox{and}
\end{array}
\ \ \ \
G(\epsilon, \eta)=\frac{R}{n-1}\left(\frac{n-6}{4}-\eta\right),
$$
we can set
$$
\eta=\left\{
\begin{array}{cc}
\frac{n-6}{4}, &\mbox{if $n\geq6$,}\\
0, &\mbox{if $n=4,5$}
\end{array}
\right.
, \ \
\begin{array}{cc}
&\mbox{$\epsilon =$ (any positive number).}
\end{array}
$$
In the case of $p>\frac{n}{2}$. In order to minimize $F(\epsilon, \eta)$ and $G(\epsilon, \eta)=0$, we can set
$$
\eta=\left\{
\begin{array}{cc}
\frac{(n-2)p-2n}{2n}, &\mbox{if $n=4,5$ and $p\geq\frac{2n}{n-2}$ or $n\geq6$}\\
0, &\mbox{if $n=4,5$ and $\frac n2<p<\frac{2n}{n-2}$}
\end{array}
\right.
,$$
$$
\epsilon =\left\{
\begin{array}{cc}
\frac{R}{(n-1)E(n)}, &\mbox{if $n=4,5$ and $p\geq\frac{2n}{n-2}$ or $n\geq6$}\\
\frac{p(6-n)R}{2(n-1)(2p-n)E(n)}, &\mbox{if $n=4,5$ and $\frac n2<p<\frac{2n}{n-2}$.}
\end{array}
\right.
$$
 In conclusion,
we can choose
$$\left(\int_{M}u^pdV_g\right)^{\frac{1}{p}}<\varepsilon(n)Y(M,[g])^\frac{n}{2p}R^{1-\frac{n}{2p}}$$
such that (2.21) implies $\left(\int_{M}u^{\frac{2n}{n-2}}dV_g\right)^{\frac{n-2}{n}}=0$, i.e., $\mathring{Rm}=0$. Hence $(M^n,g)$ is isometric to a quotient of the round $\mathbb{S}^n$.
\section{Proof of Theorem 1.4}
Now, we compute the Laplacian of $|\mathring{Ric}|^2$.
\begin{lemma}
Let $(M^n,g) (n\geq4)$ be a complete Bach-flat $n$-manifold with constant scalar curvature, then
\begin{eqnarray}
\Delta|\mathring{Ric}|^2=2|\nabla\mathring{Ric}|^2-4\mathring{R}_{ij}\mathring{R}_{kl}W_{ikjl}+\frac{2n}{n-2}\mathring{R}_{ij}\mathring{R}_{jk}\mathring{R}_{ik}
+\frac{2R}{n-1}|\mathring{Ric}|^2.
\end{eqnarray}
\end{lemma}
\begin{remark}
Although Lemma $3.1$ has been proved in \cite{CGY}, for completeness, we also write it out.\end{remark}
\begin{proof}We obtain from (2.3) and (2.5) that
\begin{eqnarray}
\Delta|\mathring{Ric}|^2&=&2|\nabla\mathring{Ric}|^2+2\mathring{R}_{ij}\nabla_k\nabla_k\mathring{R}_{ij}\nonumber \\
&=&2|\nabla\mathring{Ric}|^2+2\mathring{R}_{ij}\nabla_k(\nabla_j\mathring{R}_{ik}-\nabla_l\mathring{R}_{ilkj})\nonumber \\
&=&2|\nabla\mathring{Ric}|^2+2\mathring{R}_{ij}\nabla_k\nabla_j\mathring{R}_{ik}+2\mathring{R}_{ij}\nabla_k\nabla_l\mathring{R}_{ikjl}.
\end{eqnarray}
Since the scalar curvature is constant, by the Ricci identities, we get
\begin{eqnarray}
\mathring{R}_{ij}\nabla_k\nabla_j\mathring{R}_{ik}&=&\mathring{R}_{ij}(\nabla_j\nabla_k\mathring{R}_{ik}+\mathring{R}_{hk}R_{hijk}+\mathring{R}_{ih}R_{hkjk})\nonumber\\
&=&\mathring{R}_{ij}\mathring{R}_{hk}R_{hijk}+\mathring{R}_{ij}\mathring{R}_{ih}R_{hj}\nonumber\\
&=&\mathring{R}_{ij}\mathring{R}_{hk}[W_{hijk}+\frac{1}{n-2}(\mathring{R}_{ik}\delta_{hj}+\mathring{R}_{hj}\delta_{ik}-\mathring{R}_{ij}\delta_{hk}-\mathring{R}_{hk}\delta_{ij})\nonumber\\
& & +\frac{R}{n(n-1)}(\delta_{ik}\delta_{jh}-\delta_{ij}\delta_{hk})]+\mathring{R}_{ij}\mathring{R}_{ih}\mathring{R}_{hj}+\frac{R}{n}|\mathring{Ric}|^2\nonumber\\
&=&\mathring{R}_{ij}\mathring{R}_{hk}W_{hijk}+\frac{n}{n-2}\mathring{R}_{ij}\mathring{R}_{jk}\mathring{R}_{ik}+\frac{R}{n-1}|\mathring{Ric}|^2,
\end{eqnarray}
and
\begin{eqnarray*}
\mathring{R}_{ij}\nabla_k\nabla_l\mathring{R}_{ikjl}
&=&\mathring{R}_{ij}\nabla_k\nabla_lW_{ikjl}+\frac{1}{n-2}\mathring{R}_{ij}\nabla_k\nabla_l(\mathring{R}_{ij}\delta_{kl}+\mathring{R}_{kl}\delta_{ij}-\mathring{R}_{il}\delta_{jk}-\mathring{R}_{jk}\delta_{il})
\nonumber\\
&=&\mathring{R}_{ij}\nabla_k\nabla_lW_{ikjl}+\frac{1}{n-2}(\mathring{R}_{ij}\Delta\mathring{R}_{ij}-\mathring{R}_{ij}\nabla_j\nabla_l\mathring{R}_{il}-\mathring{R}_{ij}\nabla_k\nabla_i\mathring{R}_{kj})
\nonumber\\
&=&\mathring{R}_{ij}\nabla_k\nabla_lW_{ikjl}+\frac{1}{n-2}\left[\mathring{R}_{ij}\Delta\mathring{R}_{ij}-\mathring{R}_{ij}\nabla_k(\nabla_k\mathring{R}_{ij}+\nabla_l\mathring{R}_{jlki})\right]\nonumber\\
&=&\mathring{R}_{ij}\nabla_k\nabla_lW_{ikjl}+\frac{1}{n-2}\mathring{R}_{ij}\nabla_k\nabla_l\mathring{R}_{ikjl}.
\end{eqnarray*}
Since $M^n$ is Bach-flat, we get from the above equation that
\begin{eqnarray}
\mathring{R}_{ij}\nabla_k\nabla_l\mathring{R}_{ikjl}=-\mathring{R}_{ij}\mathring{R}_{kl}W_{ikjl}.
\end{eqnarray}
Combining with (3.2), (3.3) and (3.4), we obtain
$$\Delta|\mathring{Ric}|^2=2|\nabla\mathring{Ric}|^2-4\mathring{R}_{ij}\mathring{R}_{kl}W_{ikjl}+\frac{2n}{n-2}\mathring{R}_{ij}\mathring{R}_{jk}\mathring{R}_{ik}
+\frac{2R}{n-1}|\mathring{Ric}|^2.$$
This completes the proof of Lemma $3.1$.
\end{proof}
\begin{lemma}
On every $n$-dimensional Riemannian manifold, the following estimate holds
$$\left|-W_{ijkl}\mathring{R}_{ik}\mathring{R}_{jl}+\frac{n}{2(n-2)}\mathring{R}_{ij}\mathring{R}_{jk}\mathring{R}_{ik}\right|
\leq\sqrt{\frac{n-2}{2(n-1)}}|\mathring{Ric}|^2\left(|W|^2+\frac{n}{2(n-2)}|\mathring{Ric}|^2\right)^{\frac{1}{2}}.$$
\end{lemma}
\begin{remark}
We follow these proofs of Proposition 2.1 in \cite{C} and Lemma 4.7 in \cite{Bo} to prove this lemma. For completeness, we also write it out.
In general, according to the proof of Lemma $3.3$, we can obtain
$$\left|-W_{ijkl}\mathring{R}_{ik}\mathring{R}_{jl}+K\mathring{R}_{ij}\mathring{R}_{jk}\mathring{R}_{ik}\right|
\leq\sqrt{\frac{n-2}{2(n-1)}}|\mathring{Ric}|^2\left(|W|^2+\frac{2(n-2)K^2}{n}|\mathring{Ric}|^2\right)^{\frac{1}{2}},$$
where $K$ is a constant.
\end{remark}
\begin{proof}
First of all we have
$$(\mathring{Ric}\circledwedge g)_{ijkl}=\mathring{R}_{ik}g_{jl}-\mathring{R}_{il}g_{jk}+\mathring{R}_{jl}g_{ik}-\mathring{R}_{jk}g_{il},$$
$$(\mathring{Ric}\circledwedge \mathring{Ric})_{ijkl}=2(\mathring{R}_{ik}\mathring{R}_{jl}-\mathring{R}_{il}\mathring{R}_{jk}),$$
where $\circledwedge$ denotes the Kulkarni-Nomizu product.
An easy computation shows
$$W_{ijkl}\mathring{R}_{ik}\mathring{R}_{jl}=\frac{1}{4}W_{ijkl}(\mathring{Ric}\circledwedge \mathring{Ric})_{ijkl},$$
$$\mathring{R}_{ij}\mathring{R}_{jk}\mathring{R}_{ik}=-\frac{1}{8}(\mathring{Ric}\circledwedge g)_{ijkl}(\mathring{Ric}\circledwedge \mathring{Ric})_{ijkl}.$$
Hence we get the following identity
\begin{eqnarray}
-W_{ijkl}\mathring{R}_{ik}\mathring{R}_{jl}+\frac{n}{2(n-2)}\mathring{R}_{ij}\mathring{R}_{jk}\mathring{R}_{ik}=-\frac{1}{4}\left(W+\frac{n}{4(n-2)}\mathring{Ric}\circledwedge g\right)_{ijkl}(\mathring{Ric}\circledwedge \mathring{Ric})_{ijkl}.
\end{eqnarray}
Since $\mathring{Ric}\circledwedge \mathring{Ric}$ has the same symmetries of the Riemannian curvature tensor, it can be orthogonally decomposed as
$$\mathring{Ric}\circledwedge \mathring{Ric}=T+V'+U',$$
where $T$ is totally trace-free and
$$V'_{ijkl}=-\frac{2}{n-2}\left(\mathring{Ric}^2\circledwedge g\right)_{ijkl}+\frac{2}{n(n-2)}|\mathring{Ric}|^2(g\circledwedge g)_{ijkl},$$
$$U'_{ijkl}=-\frac{1}{n(n-1)}|\mathring{Ric}|^2(g\circledwedge g)_{ijkl},$$
where $\left(\mathring{Ric}^2\right)_{ik}=\mathring{R}_{ip}\mathring{R}_{kp}$. Taking the squared norm one obtains
$$|\mathring{Ric}\circledwedge \mathring{Ric}|^2=8|\mathring{Ric}|^4-8|\mathring{Ric}^2|^2,$$
$$|V'|^2=\frac{16}{n-2}|\mathring{Ric}^2|^2-\frac{16}{n(n-2)}|\mathring{Ric}|^4,$$
$$|U'|^2=\frac{8}{n(n-1)}|\mathring{Ric}|^4.$$
In particular, one has
$$|T|^2+\frac{n}{2}|V'|^2=|\mathring{Ric}\circledwedge \mathring{Ric}|^2+\frac{n-2}{2}|V'|^2-|U'|^2=\frac{8(n-2)}{n-1}|\mathring{Ric}|^4.$$
We now estimate the right hand side of (3.5). Using the fact that $W$ and $T$ are totally trace-free and the Cauchy-Schwarz inequality we obtain
\begin{eqnarray*}
\left|\left(W+\frac{n}{4(n-2)}\mathring{Ric}\circledwedge g\right)_{ijkl}(\mathring{Ric}\circledwedge \mathring{Ric})_{ijkl}\right|^2
&=&\left|\left(W+\frac{n}{4(n-2)}\mathring{Ric}\circledwedge g\right)_{ijkl}(T+V')_{ijkl}\right|^2\\
&=&\left|\left(W+\frac{\sqrt{2n}}{4(n-2)}\mathring{Ric}\circledwedge g\right)_{ijkl}\left(T+\sqrt{\frac{n}{2}}V'\right)_{ijkl}\right|^2\\
&\leq&\left|W+\frac{\sqrt{2n}}{4(n-2)}\mathring{Ric}\circledwedge g\right|^2\left(|T|^2+\frac{n}{2}|V'|^2\right)\\
&=&\frac{8(n-2)}{n-1}|\mathring{Ric}|^4\left(|W|^2+\frac{n}{2(n-2)}|\mathring{Ric}|^2\right).
\end{eqnarray*}
This estimate together with (3.5) concludes this proof.
\end{proof}

\textbf{Proof of Theorem $1.4$.} By Lemmas $3.1$ and $3.3$ and the Kato inequality $|\nabla\mathring{Ric}|^2\geq|\nabla|\mathring{Ric}||^2$, we get
\begin{eqnarray}
|\mathring{Ric}|\Delta|\mathring{Ric}|\geq
-\sqrt{\frac{2(n-2)}{n-1}}|\mathring{Ric}|^2\left(|W|^2+\frac{n}{2(n-2)}|\mathring{Ric}|^2\right)^{\frac{1}{2}}+\frac{R}{n-1}|\mathring{Ric}|^2.
\end{eqnarray}
Set $u=|\mathring{Ric}|$. By (3.6), we compute
\begin{eqnarray}
u^\alpha\Delta u^\alpha&=&u^\alpha\left(\alpha(\alpha-1)u^{\alpha-2}|\nabla u|^2+\alpha u^{\alpha-1}\Delta u\right)\nonumber \\
&=&\frac{\alpha-1}{\alpha}|\nabla u^\alpha|^2+\alpha u^{2\alpha-2}u\Delta u\nonumber \\
&\geq&\frac{\alpha-1}{\alpha}|\nabla u^\alpha|^2-\alpha\sqrt{\frac{2(n-2)}{n-1}}\left(|W|^2+\frac{n}{2(n-2)}u^2\right)^{\frac{1}{2}}u^{2\alpha}+\frac{\alpha R}{n-1}u^{2\alpha}.
\end{eqnarray}
Integrating (3.7) on $M^n$ and using Stoke's theorem, we have
\begin{eqnarray}
0\geq\left(2-\frac{1}{\alpha}\right)\int_{M}|\nabla u^\alpha|^2dV_g-\alpha\sqrt{\frac{2(n-2)}{n-1}}\int_{M}\left(|W|^2+\frac{n}{2(n-2)}u^2\right)^{\frac{1}{2}}u^{2\alpha}dV_g
+\frac{\alpha R}{n-1}\int_{M}u^{2\alpha}dV_g.
\end{eqnarray}
For $2-\frac{1}{\alpha}>0$, by the definition of Yamabe constant and H\"{o}ider inequality, we obtain from (3.8) that
\begin{eqnarray}
0&\geq&\left\{\left(2-\frac{1}{\alpha}\right)\frac{n-2}{4(n-1)}Y(M,[g])
-\alpha\sqrt{\frac{2(n-2)}{n-1}}\left(\int_{M}\left(|W|^2+\frac{n}{2(n-2)}u^2\right)^{\frac{n}{4}}dV_g\right)^{\frac{2}{n}}\right\}\left(\int_{M}u^{\frac{2n\alpha}{n-2}}dV_g\right)^{\frac{n-2}{n}}\nonumber \\
& &+\frac{4\alpha^2-2(n-2)\alpha+n-2}{4\alpha(n-1)}R\int_{M}u^{2\alpha}dV_g.
\end{eqnarray}
\textbf{Case 1.} when $n\geq6$, taking $\alpha=\frac{(n-2)\left(1+\sqrt{1-\frac{4}{n-2}}\right)}{4}$, from (3.9), we get
\begin{eqnarray}
0&\geq&\left\{\frac{Y(M,[g])}{\sqrt{2(n-1)(n-2)}}-\left(\int_{M}\left(|W|^2+\frac{n}{2(n-2)}u^2\right)^{\frac{n}{4}}dV_g\right)^{\frac{2}{n}}\right\}\left(\int_{M}u^{\frac{2n\alpha}{n-2}}dV_g\right)^{\frac{n-2}{n}}.
\end{eqnarray}
Since $W$ is totally trace-free, one has
$$\left|W+\frac{\sqrt{n}}{2\sqrt{2}(n-2)}\mathring{Ric}\circledwedge g\right|^2=|W|^2+\frac{n}{2(n-2)}|\mathring{Ric}|^2$$
and the pinching condition (1.2) implies that $M^n$ is Einstein.\\
\textbf{Case 2.} When $n=4,5$, we have $\frac{4\alpha^2-2(n-2)\alpha+n-2}{4\alpha(n-1)}>0$ for $\alpha>\frac{1}{2}$, and from (3.9), we get
\begin{eqnarray*}
0&\geq&\left\{\left(1-\left(1-\frac{1}{\alpha}\right)^2\right)\sqrt{\frac{n-2}{32(n-1)}}Y(M,[g])-\left(\int_{M}\left(|W|^2+\frac{n}{2(n-2)}u^2\right)^{\frac{n}{4}}dV_g\right)^{\frac{2}{n}}\right\}\left(\int_{M}u^{\frac{2n\alpha}{n-2}}dV_g\right)^{\frac{n-2}{n}}.
\end{eqnarray*}
Taking $\alpha=1$, we have
\begin{eqnarray}
0&\geq&\left\{\sqrt{\frac{n-2}{32(n-1)}}Y(M,[g])-\left(\int_{M}\left(|W|^2+\frac{n}{2(n-2)}u^2\right)^{\frac{n}{4}}dV_g\right)^{\frac{2}{n}}\right\}\left(\int_{M}u^{\frac{2n}{n-2}}dV_g\right)^{\frac{n-2}{n}}.
\end{eqnarray}
Thus the pinching condition (1.2) implies that $M^n$ is Einstein.
\par In particular, for $n=4,5$, the pinching condition (1.2) implies
\begin{eqnarray}
\left(\int_{M}|W|^{\frac{n}{2}}dV_g\right)^{\frac{2}{n}}<\sqrt{\frac{n-2}{32(n-1)}}Y(M,[g])<\frac{2Y(M,[g])}{C_2(n)n},
\end{eqnarray}
where the constant  \[
C_2(n)=\left\{
\begin{array}{cc}
\frac{\sqrt{6}}{2}, &\mbox{if $n=4$,}\\
\frac{8\sqrt{10}}{15}, &\mbox{if $n=5$,}\\
\frac{4(n^2-2)}{n\sqrt{n^2-1}}+\frac{n^2-n-4}{\sqrt{(n-2)(n-1)n(n+1)}},&\mbox{if $n\geq6$}
\end{array}\right.
\]
is defined in Lemma 2.1 of \cite{FX3}. By the rigidity result for positively curved Einstein manifolds (see Theorem 1.1 of \cite{FX3}), (3.12) implies that $M^n$ is isometric to a quotient of the round $\mathbb{S}^n$.
For $n\geq6$, we can choose $\alpha$ such that $\frac{4\alpha^2-2(n-2)\alpha+n-2}{4\alpha(n-1)}>0$ and
$$0\geq\left\{\frac{2Y(M,[g])}{C_2(n)n}-\left(\int_{M}\left(|W|^2+\frac{n}{2(n-2)}u^2\right)^{\frac{n}{4}}dV_g\right)^{\frac{2}{n}}\right\}\left(\int_{M}u^{\frac{2n\alpha}{n-2}}dV_g\right)^{\frac{n-2}{n}}.
$$
From Case 1, the pinching condition (1.2) implies that $M^n$ is Einstein. Hence, the pinching condition (1.2) implies
\begin{eqnarray}
\left(\int_{M}|W|^{\frac{n}{2}}dV_g\right)^{\frac{2}{n}}<\frac{2Y(M,[g])}{C_2(n)n}.
\end{eqnarray}
By the rigidity result for positively curved Einstein manifolds (see Theorem 1.1 of \cite{FX3}), (3.13) implies that $M^n$ is isometric to a quotient of the round $\mathbb{S}^n$.\\
\textbf{Proof of Corollary $1.6$.} To prove Corollary $1.6$, we need the following lemma which was proved by Gursky (see \cite{G}). For completeness, we also write it's proof out.
\begin{lemma}
Let $(M^4,g)$ be a complete 4-dimensional manifold, then the following estimate holds
$$\int_{M}R^2dV_g-12\int_{M}|\mathring{Ric}|^2dV_g\leq Y(M,[g])^2,$$
moreover, the inequality is strict unless $(M^4,g)$ is comformally Einstein.
\end{lemma}
\begin{proof}
By the Chern-Gauss-Bonnet formula (see the Equation 6.31 of \cite{Be})
\begin{eqnarray}
\int_{M}|W|^2dV_g-2\int_{M}|\mathring{Ric}|^2dV_g+\frac{1}{6}\int_{M}R^2dV_g=32\pi^2\chi(M),
\end{eqnarray}
and the conformal invariance of $\int_{M}|W|^2dV_g$, we find that $-2\int_{M}|\mathring{Ric}|^2dV_g+\frac{1}{6}\int_{M}R^2dV_g$ is also conformally invariant.
 Let $\tilde{g}\in[g]$ be a Yamabe metric. Then
\begin{eqnarray*}
Y(M,[g])^2&=&\frac{\left(\int_M{R_{\tilde{g}}}dV_{\tilde{g}}\right)^2}{\int_MdV_{\tilde{g}}}=\int_M{R}_{\tilde{g}}^2dV_{\tilde{g}}\\
&\geq&\int_M{R}_{\tilde{g}}^2dV_{\tilde{g}}-12\int_M|{\mathring{Ric}}_{\tilde{g}}|_{\tilde{g}}^2dV_{\tilde{g}}\\
&=&\int_MR^2dV_g-12\int_M|\mathring{Ric}|^2dV_g.
\end{eqnarray*}
 The equality case follows immediately.
\end{proof}

By Lemma $3.5$, we get
\begin{eqnarray}
\int_{M}|W|^2dV_g+\int_{M}|\mathring{Ric}|^2dV_g-\frac{Y(M,[g])^2}{48}\leq\int_{M}|W|^2dV_g+\frac{5}{4}\int_{M}|\mathring{Ric}|^2dV_g-\frac{1}{48}\int_{M}R^2dV_g.
\end{eqnarray}
Moreover, the inequality is strict unless $(M^4,g)$ is comformally Einstein. In the first case $``\,<"$, Theorem $1.4$ immediately implies  Corollary $1.6$; In the second case $``\,="$, $g$ is conformally Einstein. Since $g$ has constant scalar curvature, $g$ is Einstein from the proof of Obata Theorem (see Proposition 3.1 of \cite{LP}). By the rigidity result for positively curved Einstein manifolds (see Theorem 1.1 of \cite{FX3}), (3.15) implies that $M^4$ is isometric to a quotient of the round $\mathbb{S}^4$.
\section{Proof of Theorem $1.8$}
\noindent
\textbf{Proof of Theorem $1.8$.}  From (3.1), by Lemma $3.3$, we get
\begin{eqnarray}
\Delta|\mathring{Ric}|^2\geq2|\nabla\mathring{Ric}|^2+4\sqrt{\frac{n-2}{2(n-1)}}|\mathring{Ric}|^2\left\{\frac{R}{\sqrt{2(n-1)(n-2)}}-\left(|W|^2+\frac{n}{2(n-2)}|\mathring{Ric}|^2\right)^{\frac{1}{2}}\right\}.
\end{eqnarray}
Note that (1.5) is equivalent that the second of RHS of (4.1) is nonnegative.
By  the maximum principle, from (4.1) we get $\nabla\mathring{Ric}=0$. Since $M^n$ has positive constant scalar curvature, $M^n$ is a manifold with parallel Ricci tensor.
 Hence $M^n$ is a manifold with harmonic curvature. Using the same argument as in the proof of (3.1), we obtain a Weitzenb\"{o}ck formula (see (2.20) in \cite{F})
\begin{eqnarray}
\Delta|\mathring{Ric}|^2=2|\nabla\mathring{Ric}|^2-2\mathring{R}_{ij}\mathring{R}_{kl}W_{ikjl}+\frac{2n}{n-2}\mathring{R}_{ij}\mathring{R}_{jk}\mathring{R}_{ik}
+\frac{2R}{n-1}|\mathring{Ric}|^2.
\end{eqnarray}
By Remark $3.4$, we get
\begin{eqnarray}\left|-W_{ijkl}\mathring{R}_{ik}\mathring{R}_{jl}+\frac{n}{n-2}\mathring{R}_{ij}\mathring{R}_{jk}\mathring{R}_{ik}\right|
\leq\sqrt{\frac{n-2}{2(n-1)}}|\mathring{Ric}|^2\left(|W|^2+\frac{2n}{n-2}|\mathring{Ric}|^2\right)^{\frac{1}{2}}.\end{eqnarray}
Combing (4.2) with (4.3), we have
\begin{eqnarray}
\Delta|\mathring{Ric}|^2\geq2|\nabla\mathring{Ric}|^2+2|\mathring{Ric}|^2\left\{\frac{1}{n-1}R-\sqrt{\frac{n-2}{2(n-1)}}\left(|W|^2+\frac{2n}{n-2}|\mathring{Ric}|^2\right)^{\frac{1}{2}}\right\}.
\end{eqnarray}
\textbf{Case 1.} $W\neq 0$. By (4.4),
the inequality (1.5) implies
\begin{eqnarray}
0&\geq&2\sqrt{\frac{n-2}{2(n-1)}}|\mathring{Ric}|^2\left\{\sqrt{\frac{2}{(n-1)(n-2)}}R-\left(|W|^2+\frac{2n}{n-2}|\mathring{Ric}|^2\right)^{\frac{1}{2}}\right\}\nonumber\\
&\geq&
2\sqrt{\frac{n-2}{2(n-1)}}|\mathring{Ric}|^2\left\{\left(4|W|^2+\frac{2n}{n-2}|\mathring{Ric}|^2\right)^{\frac{1}{2}}-\left(|W|^2+\frac{2n}{n-2}|\mathring{Ric}|^2\right)^{\frac{1}{2}}\right\}.
\end{eqnarray}
By (4.5), we get $\mathring{Ric}=0$, i.e., $(M^n, {g})$ is  Einstein.\\
\textbf{Case 2.} $W=0$. Note that the inequality $n(n-1)|\mathring{Ric}|^2\leq R^2$ is equivalent to equation (1.5).  From (4.4),  we have
\begin{eqnarray}
0=\frac{2n}{n-2}\mathring{R}_{ij}\mathring{R}_{jk}\mathring{R}_{ik}
+\frac{2R}{n-1}|\mathring{Ric}|^2\geq\frac{2}{n-1}|\mathring{Ric}|^2\left({R}-\sqrt{(n-1)n}|\mathring{Ric}|\right)\geq0.
\end{eqnarray}
Hence
at every point, either
$\mathring{Ric}$ is null, i.e., $M^n$ is Eninstein, and by conformally flatness it has constant positive sectional curvature, or ${R}-\sqrt{(n-1)n}|\mathring{Ric}|=0$, according to the estimation of trace-free symmetric $2$-tensors, it has an eigenvalue of multiplicity
$(n-1)$ and another of multiplicity $1$. Since the Ricci tensor is parallel, by the de Rham decomposition Theorem, $M^n$ is covered isometrically by the product of Einstein manifolds. We have $R=\sqrt{(n-1)n}|\mathring{Ric}|.$ Since $M^n$ is conformally flat and has positive
scalar curvature, then the only possibility is that $M^n$ is covered isometrically by $\mathbb{S}^1\times \mathbb{S}^{n-1}$ with the product metric.
 So $(M^n,g)$ is isometric to either an Einstein manifold or a quotient of $\mathbb{S}^1\times \mathbb{S}^{n-1}$ with the product metric.\\
\textbf{Proof of Corollary $1.9$.} By Theorem $1.8$, we consider the case that $M^n$ is Einstein.  Using the same argument as in the proof of (3.1), we obtain a Weitzenb\"{o}ck formula for Einstein manifolds (see (5) in \cite{FX3})\begin{eqnarray}
\Delta|W|^2=2|\nabla W|^2-2C_3(n)|W|^3+\frac{4R}{n}|W|^2,
\end{eqnarray}
where $C_3(4)=\frac{\sqrt{6}}{2}$ and $C_3(5)=\frac{8\sqrt{10}}{15}$.
 From (4.7), the condition of Corollary $1.9$ implies that
$$
\Delta|W|^2=2|\nabla W|^2+\left(\frac{4R}{n}-2C_3(n)|W|\right)|W|^2\geq0.
$$
Hence $W=0$, i.e., $M^n$ is conformally flat. So $(M^n,g)$ is isometric to  a quotient of the round $\mathbb{S}^n$. This completes the proof of Corollary $1.9$.
\begin{remark}
Let $(M^n,g) (n\geq4)$ be an $n$-dimensional  compact Bach-flat Riemannian manifold with positive constant scalar curvature. If
$$
|W|^2+\frac{{n}}{2(n-2)}|\mathring{Ric}|^2<\frac{1}{{2(n-2)(n-1)}}R^2,
$$
then $M^n$ is an Einstein manifold.
\end{remark}
\textbf{Acknowledgments:} The authors thank the referee for his helpful suggestions.

redpajama_set_name: RedPajamaArXiv


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What to Do About Google Chrome's New SSL Policy
By Rachel Newcombe
In Ask Our Gurus, Web Design
What to Do About Google Chrome's New SSL Policy2017-09-132017-09-13https://www.envisionup.com/wp-content/uploads/2017/07/eup-logo.pngEnvisionUPhttps://www.envisionup.com/wp-content/uploads/2017/09/blog-google-ssl.jpg200px200px
Google is getting more security conscious all the time. When you use the Google Chrome browser, you've probably noticed that green padlock and the word "Secure" beside some website addresses:
Other websites – maybe even yours – show as not secure:
Starting in October, Google is upping the ante on security. It won't just be web pages with credit card or password forms; it will be all pages with forms, and every single page in Google Chrome's Incognito mode.
Why is Google Doing This?
Studies show that people don't see a lack of a secure sign as a warning. A lot of information gets shared on the Internet. Many users don't realize that the sites they are sharing their information on aren't as secure as others.
Google's plan to clearly label unsafe pages is another step towards making their browser safer and protecting the privacy of its users from outside sources, like hackers.
In January 2017, Chrome Began Flagging Certain HTTP Forms
In 2016, Chrome announced that starting in 2017, sites that did not have an SSL certificate would begin being marked as unsafe.
The process of identifying pages as not secure is a gradual one that started with the release of Chrome version 56.
HTTP pages that had a password or credit card field for a user to fill out were marked as unsafe in the address bar.
In October 2017, Warnings Will Appear on All Form and Incognito Pages
In October, Chrome's SSL policy will grow to include:
All HTTP pages where you enter data – even search boxes
All HTTP Incognito pages
Every page on the standard Google Chrome browser will load normally. But when a user begins filling out any kind of field, including name, phone number, or even search boxes on the website, a warning will appear in the address bar.
Eventually, all HTTP Pages Will Be Flagged as Not Secure
According to a recent post on the Chromium Forum, the goal is to eventually show 'Not Secure' on all HTTP pages across the browser. That means that even if you have no forms for users to fill in, your website could still be flagged as unsafe on the Chrome browser.
You're probably wondering why this matters to you. Your website seems to work just fine. That could be true, but it will matter to your online audience.
If they see the clear warning that your website is unsafe, your audience will be less likely to visit and interact with it.
Approximately 50% of Internet users use Google Chrome as their browser. With such a large percentage of your target audience using this browser, you don't want them to be deterred by a "not secure" warning.
You also don't want to damage your credibility by putting their personal information at risk due to lack of proper encryption.
That's why you should get an SSL certificate as soon as possible.
What's an SSL Certificate, and How Does It Affect My Website?
An SSL (or Secure Sockets Layer) certificate is what adds the 'S' to HTTPS in the domain search field in your browser. HTTPS signals that all data between your website and the user's browser is automatically encrypted and secure.
Think of it as a bridge between your website and Chrome. The information goes back and forth over the bridge. An SSL certificate adds an extra layer of support to this bridge, making sure it won't be damaged or tampered with. Without it, your bridge is more susceptible to hackers and other potential threats.
When you're using an http page, that signals to your audience that your 'bridge is at risk' – there's nothing encrypting the data.
An SSL certificate is attached to your server and your domain, and creates the encrypted passage between the web server and browser through a public and a private key.
Then users will see your domain name with https in front of it. It means that it is safe to pass information between the browser and the web page.
Talk to a Professional About Getting an SSL Certificate
The best approach to getting an SSL certificate is to talk to a professional. There's a lot that goes into the process of switching over your website pages, and you don't want to miss any important steps.
Rachel Newcombe
Rachel is a content marketer for EnvisionUP. She is Inbound Marketing and Google Analytics certified. She lives in Ottawa, and is a Nova Scotian at heart. She enjoys reading, movies both good and bad, and socializing with friends.
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