text:
Evening links for December 3rd
By Zester	— 2 months ago
https://www.scmp.com/topics/china-property
https://www.scmp.com/business/banking-finance/article/3158254/debt-laden-developer-china-aoyuan-tumbles-after-failing
https://abc7.com/freeway-fatalities-deadly-crashes-traffic-accidents/11294645/
https://www.thegatewaypundit.com/2021/12/forbes-changes-headline-admits-mrna-vaccines-alter-dna-change-back/
https://www.theepochtimes.com/books-including-adult-sex-with-minors-back-in-school-library_4137486.html
https://www.marketwatch.https://www.scmp.com/topics/china-property
https://www.marketwatch.com/story/u-s-could-run-out-of-cash-as-soon-as-dec-21-experts-say-11638536433
https://finance.yahoo.com/news/oil-prices-have-tanked-so-hard-traders-are-assuming-planes-wont-fly-for-3-months-goldman-sachs-114609776.html
https://gizmodo.com/nso-groups-spyware-may-have-been-used-to-target-u-s-st-1848157598
https://www.theepochtimes.com/14-arrested-in-smash-and-grab-robberies-all-are-out-of-custody_4137750.html
https://www.thezimbabwean.co/2021/12/zimbabwe-not-eligible-for-debt-relief-until-after-paying-world-bank-debt-imf/
https://www.theepochtimes.com/nevada-becomes-first-state-to-impose-surcharge-on-unvaccinated-workers_4136987.html
com/story/u-s-could-run-out-of-cash-as-soon-as-dec-21-experts-say-11638536433
Daily Links for January 17th
By Zester	—	1 day ago
https://almaalomah.me/2022/01/16/576744/
https://www.dinardaily.net/t103167-the-central-bank-of-iraq-confirms-that-it-will-not-grant-a-license-to-any-encrypted-digital-currency
https://www.theepochtimes.com/mkt_morningbrief/court-quashes-familys-quest-to-try-ivermectin-as-recovered-patients-tout-effectiveness-of-alternative-treatments_4205837.html
https://bestnewshere.com/hackers-crack-pfizer-moderna-pfizer-servers-all-vaccine-death-data-now-public-from-all-companies-thanks-to-hackers/
https://www.cureus.com/articles/82162-ivermectin-prophylaxis-used-for-covid-19-a-citywide-prospective-observational-study-of-223128-subjects-using-propensity-score-matching
https://www.youtube.com/watch?v=omRU82LY1rE Katie Halper: Biden Lied AGAIN About Being Arrested & Only Right-Wing Media Covered It
https://www.theepochtimes.com/mkt_morningbrief/waste-of-time-to-keep-vaccinating-people-ex-head-of-uk-vaccine-taskforce_4216302.html
https://www.theepochtimes.com/mkt_morningbrief/fbi-identifies-texas-synagogue-hostage-taker-as-british-national-malik-faisal-akram_4216562.html
https://www.independent.co.uk/world/russian-invasion-ukraine-inevitable-b1993814.html
https://theintercept.com/2022/01/16/uae-2016-election-trump-clinton-george-nader/
https://www.theepochtimes.com/goodbye-pandemic-hello-endemic_4216339.html
Evening Links for September 17th
By Zester	—	4 months ago
https://www.theepochtimes.com/mkt_breakingnews/democrat-proposal-seeks-to-end-tax-break-on-exchange-traded-funds_4002385.html
https://t.me/John_McAfee_Channel/54 Anything to it?
https://www.ingersolllockwood.com/exercise-your-rights/
https://www.perild.com/2021/09/17/in-france-thousands-of-covid-unvaccinated-doctors-lost-their-jobs/
https://www.theepochtimes.com/mkt_breakingnews/red-states-lead-in-employment-blue-states-get-props-for-sharpness-of-rebound-report_4002575.html
https://www.msn.com/en-us/health/medical/fda-panel-votes-against-pfizers-booster-shot/ar-AAOyJMd
https://gab.com/SonOfEnos/posts/106937804329528145 The story needs shared.
https://www.instagram.com/tv/CT5sNajlcwj/?utm_medium=copy_link JP on freedom
https://articles.mercola.com/sites/articles/archive/2021/09/17/natural-immunity-vaccine-passports.aspx
https://www.theblaze.com/news/french-military-kills-isis-leader-who-ambushed-us-soldiers
https://www.bloombergquint.com/politics/biden-warns-states-cities-of-fallout-from-looming-debt-default
https://www.bitchute.com/video/gigUyK3yLtMU/ Funeral director
Daily Links for December 8th
By Zester	—	1 month ago
https://en.vietnamplus.vn/reference-exchange-rate-up-50-vnd/216718.vnp
https://youtu.be/5c0wpCufAUk Downing Street staff shown joking in leaked video about Christmas party they later denied | ITV News
https://youtu.be/mqnbuS53lE0 'Where was the moral compass in Downing Street as staff laughed about Christmas party?' | ITV News
https://www.manchestereveningnews.co.uk/news/greater-manchester-news/new-plan-b-plus-covid-22396872
https://www.theepochtimes.com/mkt_morningbrief/maxwell-accusers-offer-barrage-of-detail-for-alleged-trafficking-scheme_4142374.html
https://www.thegatewaypundit.com/2021/12/breaking-j6-attorney-undercover-law-enforcement-officer-passed-weapons-j6-protesters-get-arrested-clear-entrapment-video/
https://www.theepochtimes.com/mkt_morningbrief/nothing-short-of-human-rights-violations-greene-blasts-treatment-of-jan-6-political-prisoners_4143767.html
https://www.theepochtimes.com/mkt_morningbrief/amazons-cloud-computing-services-struck-by-major-outage_4143357.html
https://www.naturalnews.com/2021-12-07-decoded-the-globalist-plan-to-destroy-bitcoin-and-the-defi-crypto-ecosystem.html
https://youtu.be/ecjuPcn_Btk Commercial Real Estate Crash And Housing Bubble Burst Loom In China As Evergrande Collapses


Trying to get this blog accomplished at this time of year, is becoming more and more difficult. I am still checking the agendas and am trying to double check most agendas before leaving the classroom. Therefore, I have not been too worried about the blog and I'm very sorry for not keeping up on it.
Every student now has an independent book to read for December. They should be coming home and bringing it back to school the next day. Please ensure it's in the backpack and stays there when your child is not reading it, so that none get left at home.
This weekend the students are to make a point of reading at least 20 minutes...preferably 40, depending on ability.
Measurement, multiplication and division are all being connected over the next few weeks with our gingerbread houses. I am going to try it out this weekend with the following recipe http://www.simplyrecipes.com/recipes/how_to_make_a_gingerbread_house/print/ If you have something better and have tips you can share with me PLEASE DO! A letter is coming home tonight with a list of possible items you can send in. Please check off the item and send in on Monday. A quick email to me would be VERY helpful for me, in order for me to organize and purchase supplies over the weekend. I will buy a big back of flour at Costco, my big concern is the molasses and shortening (in replacement of the butter). Some eggs would be great too, but I understand that's difficult to send in; however, if you feel like dropping by and helping out, that would be great too. :-) The more adult hands I have the better. I am going to break the kids up into 7 groups of 3. Everyone will have photos taken with their finished house and will bring a part of it home.
Please understand, I have never done this before with 4th graders. Any ideas or problems that may arise, I would be most grateful for your advice and concern. I am explaining the project to them on Monday, along with looking at the recipe. I would like to make the dough on Tuesday, but will have to bake in batches, becasue we only have 2 ovens. I could probably bake some at home that night too. We will slowly put it together over the course of the week. I am looking at decorating it by Friday. Therefore, if you could send in the candy by Friday, that would be great.


text:
U.S. ban on Chinese telecom equipment moves forward amid concerns
Jenna McLaughlin
National Security and Investigations Reporter
Yahoo News April 17, 2019
WASHINGTON — Amid ongoing international debate about how to confront the national security threats from 5G, a congressionally mandated ban on U.S. government business deals with several Chinese telecom companies, including Huawei and ZTE, is on track to be implemented by late summer, according to the Pentagon.
The Pentagon, the General Services Administration and NASA — members of the government's Federal Acquisition Regulatory Council — are working to implement the congressional legislation by modifying their procurement regulations, according to the White House Office of Information and Regulatory Affairs. The new rule, which will apply to all executive agencies once complete, "is needed to protect U.S. networks against cyber activities conducted through Chinese Government-supported telecommunications equipment and services," according to a published description of the change, mandated by the 2019 defense spending bill.
"The review is in coordination and we expect the rule to be implemented by August 2019," wrote Lt. Col. Mike Andrews, a spokesperson for the Pentagon, in an email to Yahoo News.
However, Huawei is challenging the ban in federal court in Plano, Texas, the location of its U.S. customer service center. It's unclear whether or not the lawsuit could affect the implementation of the 2019 defense authorization bill.
Photo illustration: Yahoo News; photos: AP(3), Getty Images(2).
The Department of Justice declined to comment.
The proposed ban on government procurement of Chinese telecom equipment demonstrates how the Trump administration has struggled to find a way to fight what it argues is the national security threat posed by Chinese companies involved in next-generation communications networks, while also dealing with economic realities that make a complete ban unlikely.
The Trump administration had reportedly considered going further than the defense bill, using an executive order to ban the use of Huawei products for development of 5G, the next generation of high-speed communications networks. However, the administration appears more recently to be backing off from that plan after resistance from U.S. allies and from small and rural companies that fear the costs of cutting out Huawei.
FCC Chairman Ajit Pai, right, looks on as President Trump discusses the deployment of 5G technology in the United States on April 12. (Photo: Evan Vucci/AP)
Even as that larger ban languishes, the 2019 defense authorization bill, signed into law last August, would be a first step toward eliminating procurement of Huawei products within the federal government. The bill includes a section that restricts federal agencies, including the Pentagon, from purchasing or utilizing "covered telecommunications equipment" sold by Chinese companies that qualify as "government-supported."
While top national security and intelligence officials have not provided any hard evidence to prove Huawei has spied for Beijing, supporters of the ban argue that the company is required by Chinese law to comply with any government orders.
The bill covers Chinese companies — namely Huawei and ZTE — that sell communications devices and technology necessary for future 5G networks.
The Federal Acquisition Regulatory Council, which is responsible for implementing the ban, has some latitude in interpreting exactly what Congress meant when passing the 2019 defense authorization, according to Robert Huffman, an attorney with Akin Gump who focuses on government defense contracts and supply chain compliance issues. "The rule will have to interpret" key provisions within the law, including what a "substantial or essential component" is, to determine if a piece of Huawei technology would qualify for restriction, he explained.
"If the terms are construed broadly, it would mean contractors using any Huawei etc. equipment or services anywhere in the world would be barred from federal contracts, period," he wrote in an email to Yahoo News. "This is the position industry is so concerned about...Most do not believe this will come to pass, as it's just too disruptive."
Contractors and federal purchasers will need to pay much closer attention to the security of the supply chain, potentially including products manufactured by Huawei and sold by other companies, according to Huffman.
Huawei's mobile phones are displayed at a telecoms service shop in Hong Kong, Friday, March 29, 2019. (Photo: Kin Cheung/AP)
The prohibition also applies to surveillance and video recording companies in national security contexts, including Hytera Communications Corp., Hangzhou Technology Co. and Dahua Technology Co. For example, if a camera sold by one of those three companies was in use for surveillance of a government facility or for a public safety purpose, that specific use would no longer be allowed. The bill also banned purchase of Hikvision cameras for U.S. government agencies.
In the United Kingdom, British politicians have already raised the alarm about the use of surveillance cameras made by a Chinese firm for surveillance in government facilities, but they have not taken formal legal action to remove the already existing cameras.
This new U.S. prohibition mirrors a similar one included in the 2018 defense spending bill, making it illegal for the government to use or purchasing software from Kaspersky, a Russian cybersecurity and antivirus company. Kaspersky has been unsuccessfully trying to reverse that ban by revealing its source code for outside review and moving its lab from Russia to Switzerland. Two lawsuits brought by Kaspersky Lab alleging that the government's ban was unconstitutional and brought undue harm were dismissed in May 2018.
The new restrictions on Chinese technology companies, amid an ongoing trade war with China, have encountered more international debate than the similar provision barring Kaspersky. Officials in the United Kingdom appear open to continuing to work with Huawei, and focusing on mitigating the risk of Chinese government access to data. British intelligence agencies have conducted a review of Huawei's security and concluded there are serious vulnerabilities, but it's unclear whether those security risks will prompt the government to pursue a ban.
U.S. officials studying Huawei and 5G suggest the company is not going away anytime soon. According to one U.S. government consultant, who asked for anonymity because they were not authorized to speak to the press, Huawei is so deeply involved in developing the international standards for 5G that even its competitors, Nokia and Ericsson, will be beholden to the Chinese company to some extent. "Even if there is an outright U.S. ban… Huawei is still going to have a huge role to play," the expert told Yahoo News.
A man walks inside a Huawei's customer service center in Hong Kong, Friday, March 29, 2019. (Photo: Kin Cheung/AP)
In the meantime, Huawei appears determined to fight back; the company sued the U.S. government over the ban in federal court, arguing that the provision is unconstitutional because it targets the company without providing due process. On Monday, Huawei and the U.S. government agreed to a schedule of legal proceedings through September 2019, unless the case is dismissed by the judge beforehand.
Huawei did not respond immediately to a request for comment.
But even as that legal case proceeds, the U.S. government is going forward with its rule change, which would effectively block the company from federal contracts.
Congress appeared to be especially concerned by products sold by the telecommunications companies that can interact with or inspect user data, according to the bill. If the technology is incapable of routing or redirecting user traffic, or surveilling it, there may be an exception for government use.
While the new rule will restrict Huawei's reach within the United States, it also gives the secretary of defense wide latitude to work with the director of national intelligence and the head of the FBI to expand the restrictions to other companies owned or controlled by a "covered foreign country" — a power that may create uncertainty for U.S. business people and lawyers who work with firms abroad.
Huffman, the attorney at Akin Gump, expects Huawei's lawsuit to go the way of Kaspersky's. "That lawsuit will fail on the merits," he wrote.
However, even if it is not immediately thrown out, Huffman predicted, it would not affect the new rule without a decision on the merits.
"The only thing (other than new legislation) that would prevent or delay the rule from taking effect is if the Trump administration decides to 'trade it away' as part of its ongoing trade negotiations with China," he said.
Read more from Yahoo News:
AOC: Trump is 'absolutely' trying to incite violence against Omar
Trump confirms possible plan to send immigrants to sanctuary cities
Bernie Sanders supports letting jailed felons vote in elections
Buttigieg suggests Pence offered him empty praise when he came out
AOC on 'Game of Thrones' finale: It'd be 'badass' if it ended in democracy
Pentagon nominee regrets Turkey's 'drift' from West
Ericsson warns on negative margin impact, shares fall 7%
So...Did the Pentagon Use Ticks for Biological Warfare?
The Air Force's F-22 Raptors Won't Meet the Pentagon's 80-Percent Readiness Goal
Italy to drop emergency legislation on 5G golden powers

redpajama_set_name: RedPajamaCommonCrawl


text:
[![Build
Status](https://travis-ci.org/tstone2077/gitshelve.svg?branch=develop)](https://travis-ci.org/tstone2077/gitshelve)
[![Coverage Status](https://coveralls.io/repos/tstone2077/gitshelve/badge.svg?branch=develop)](https://coveralls.io/r/tstone2077/gitshelve?branch=feature%2Fdevelop)

Originally branched from https://github.com/duplys/git-issues

gitshelve is very useful standalone code inside of the git-issues.  This
repo is to separate it out into its own project.  That being the case,
the following changes were made:

1. original history has been retained [by using git filter-branch and
a simple command
line](http://git.661346.n2.nabble.com/Remove-all-files-except-a-few-files-using-filter-branch-td7567155.html).
2. code has been updated to support python 2 and 3 (2.7 and 3.2 tested)
3. unit tests have been written to cover 93% of the code.

Please be sure to view the LICENSE file for copying, modifying, or
distributing.

redpajama_set_name: RedPajamaGithub


An easy salt dough recipe you can use to make this sparkly salt dough candle holder craft.
Do you remember the salt dough candle holders we made for Diwali? They look just a good decked out in Christmas colours.
The salt dough recipe is the same but this time they're painted with red acrylic paint with lashings of red glitter for added sparkle.
A row of these on the mantle piece or in the centre of the Christmas dinner table look so festive.
So pretty! Love all the sparkle.
Thank you for linking to Kids Get Crafty.
These are so cute! I have a 25 Days of Handmade Christmas series running on my blog and I think this would be a great thing to share with my readers! If you're interested, I would just need the HTML for a guest post. I'm off to check out the tutorial now.
These are adorable….how long did they take to cook? We did salt dough ornaments and they took over 6 hrs to bake! Thanks for sharing at Thrifty Thursday!!
That's win – win Dawn! Hope you enjoy making them.
This would be a great gift idea – it's kid-friendly but will get used too!


California transplants will forever bemoan the fact that New York has no good Mexican food, which, if we're using California as a point of comparison is mostly still pretty true.One of the few exceptions, however, is El Atoradero, which evolved from a Bronx bodega to a full-fledged sit-down restaurant in Prospect Heights.Part of the allure of this gourmet take on your favorite Italian-American casserole is obviously the impressive tally in the title, which sort of begs the question: Are there really a full 100 layers of pasta, sauce and cheese inside? Each pasta sheet is razor-thin, making a proper fact-check nearly impossible.One thing is certain: There are an awful lot of layers, and the entire build-out is a thing of beauty -- a bolognese-stained monument to the lofty culinary status that Italian cuisine has finally achieved in NYC and rightfully deserves.It's called the Bee Sting, and it injects your tongue with all the right flavors: savory tomato, salty sopressata, sweet honey, and spicy chili.
Chef Waxman cuts Bell & Evans birds in half, seasons, and grills them before roasting in the oven. The herb-based salsa verde makes the dish pop New Yorkers have strong opinions about where to find the best bagel, but the only way to really judge a bagel is to try it in its natural form -- without any cream cheese or toasting. The bagel is dense but not bloated, and has a nice crunch that gives way to a pillowy and slightly salty interior.The square grandma slice has creamy melted cheese with pungent and robust red sauce interspersed.It tastes like the ideal dream of what pizza used to taste like "when you were a kid." The city needs more Best Pizza and less cheap pizza.Marco [Canora]'s gnocchi is pure perfection from texture to flavor.It is as simple as can be but warms the soul every time I eat it and inspires to keep pushing me to make the perfect pasta.


Clean, friendly staff and great organization. And on a 97 degree day, the entire unit is cooled so moving was a breeze!
the manager was fantastic. I really like the facility and I feel very secure about keeping my belongings there.
Excellent service, great facility. All exceeded expectations!
This storage facility is clean, secure, and easy to access. The staff is friendly and explains everything. The storage unit had high ceilings and was clean. I'd definitely recommend this facility.


The clients had bought the house next door and sectioned off part of the extensive garden for their own use. They wanted a design that would integrate the two gardens and make sense of the new L-shape, different levels and the existing swimming pool and hardwood deck. The garden is on the edge of Epping Forest, with mature beech, birch and oak trees all around, so it was important that the new design sit comfortably in the landscape and with the modern extension to the house.
The strong, geometric design centres on a magnificent, multi-stemmed Griselinia littoralis, which had been concealed behind large shrubs and fencing and whose twisted trunk becomes the focal point of the stepped porcelain pathways from the house terrace up to the pool. The tree was reshaped to form a natural arch, providing a vista across the garden, with an avenue of flowering Prunus calleryana 'Chanticleer', framed by clipped beech hedging, leading the eye towards the sundial sculpture and a large, informal perennial border at the end of the garden. A circular bench anchors the mature Swedish birch tree and provides a point of shady tranquillity.


text:
Our Beautiful Cervantes
The Town of Cervantes was named after an American whaling ship the "Cervantes", which ran aground on the Cervantes islands on the 29th June 1844. The Crew was fishing when a gale blew up. Before the vessel could make sail to weather out the gale at sea, it was driven on to a sand-bar.
The crew made it to shore, and three of them arrived in Perth on 6th July to report the loss. Captain Gibson had said that the vessel suffered only minor damage; however its keel was broken. Due to the long distance from the ship to Fremantle, which was the nearest place for repairs to be carried out, it was decided that the vessel was to be sold.
The wreckage was sold for 155 pounds. The purchaser, a Mr Wickstead, was contemplating forming a whale station, as the crew had reported an abundance of seals in the area where the wreckage had occurred.
This never eventuated and the ship was never recovered. 'Cervantes' Whaling Ship still lies in 2 meters of water. Pieces of old timber, probably from this wreck, are sometimes found on the beaches near Cervantes.
The Cervantes Cultural Committee
A committee formed in 2004 to promote the Arts & Culture in our region. We are a diverse group of farmers, fisherman, govt employees, retirees, living in Cervantes & supporting our local communities. Our major objective is to increase arts participation & appreciation in the community. Monthly meetings are held to organise Art/craft based workshops promoting the interests of our regional communities, and for the purpose of selecting the artists/craftspeople and workshop tutors who will be the principal attractions for the now biennial festival.
The Committee Members
Alanna Smith
Trish Sudholz
Ivan McLay
Murray Ford
Erica Ford
Ann Preitz
Patsy Briotti
Yvonne Jenkin
Evelyn Hawkins
Marilyn Williams
Jan Ackland
Jenny Seymour
Marilyn Gazeley


Another way to classify waves is to compare what the particles in the medium are doing to the direction the wave is traveling.
Transverse Waves - the particles in the medium oscillate in a direction perpendicular to the way the wave is traveling. A good example is a wave on a string.
Longitudinal Waves - the particles in the medium oscillate along the same direction as the way the wave is traveling. Sound waves are longitudinal waves.


text:
\section{Introduction}

In this note we are interested in giving an explicit representation for a particular class of curvature depending integral functionals defined on compact manifolds without boundary. We restrict our analysis to $C^{2,1}$ regular manifolds with non-negative mean curvature that can be identified as boundaries of mean-convex domains. More precisely, for a mean-convex domain $\Omega \subset \mathbb{R}^n$, denoted by $k(x)$ the mean curvature of $\partial \Omega$, we are interested in recovering the value of boundary integrals of the type
\begin{equation}\label{integral}
\int_{\partial \Omega} f(k(x)) \, d\mathcal{H}^{n-1}
\end{equation}
in terms of the behavior of $f$ inside $\Omega$ when $f$ satisfies suitable regularity assumptions. It turns out that, if $f$ is a differentiable non-increasing function (cf. Theorem \ref{mainthm} for the precise assumptions), \eqref{integral} can be expressed as the sum of volume integrals plus a defect measure $\delta^f$ concentrated on the ridge set of $\Omega$ as follows
\begin{equation}\label{formula}
\int_{\partial \Omega} f(k(x)) \,d \mathcal{H}^{n-1} =  \int_{\Omega} k(x) f(k(x))\,dx - \int_{\Omega} f'(k(x))|D^{2}d(x)|^{2} \, dx + \delta^{f}(\Sigma)
\end{equation}
where with $d$ and $\Sigma$ we have denoted the distance function from $\partial \Omega$ and its singular set respectively.

Our initial motivation for this study come from the understanding of a useful consequence of the area formula and Fubini's theorem, which is sometimes referred to as Heintze-Karcher's inequality (see \cite{HeiKar78-000} and  \cite[Theorem 6.16]{MonRos05-000}). In our setting it provides an upper bound of the measure of $\Omega$ in terms of the boundary integral of $1/k$. Indeed, it states that for any regular strictly mean-convex domain $\Omega\subset \mathbb{R}^{n}$ we have
\begin{equation}\label{HK-ineq}
\cal L^n (\Omega) \leq \frac{1}{n}\int_{\partial \Omega} \frac{1}{k(x)}\, d \cal{H}^{n-1},
\end{equation}
where $\cal L^n$ denotes the $n$-dimensional Lebesgue measure. The case of equality in \eqref{HK-ineq} is particularly interesting, as is known that equality occurs if and only if $\Omega$ is a ball and this property, combined with Minkowski integral formula, gives an elegant proof of the classical Alexandrov's theorem that identifies the sphere as the unique compact connected surface with constant mean curvature (see.  \cite[Theorem 6.17]{MonRos05-000}). This point of view has been recently used in \cite{He:2009p175} to prove an anisotropic version of Alexandrov's theorem for compact embedded hypersurfaces with constant anisotropic mean curvature.

Our goal was to obtain a sharp estimate for the error term in \eqref{HK-ineq},
and this led us to the study of the boundary integral
\begin{equation*}\label{1suf}
\int_{\partial \Omega} f(k(x)) \, d \cal{H}^{n-1}.
\end{equation*}

An application of \eqref{formula} with $f(t)=1/t$ allows us to write
\[
\int_{\partial \Omega} \frac{1}{k(x)} \,d \mathcal{H}^{n-1} - n \cal{L} (\Omega)= \int_{\Omega} \frac{|D^{2}d(x)|^{2}-(n-1)k^2(x)}{k^2(x)} \, dx + \delta^{f}(\Sigma) \geq 0
\]
filling the gap in \eqref{HK-ineq} by explicitly expressing the error term as a sum of a volume integral and a defect measure in the spirit of  \cite{Aviles:1996p212}.

The idea behind the proof of the main result can be easily explained by a formal application of the divergence theorem. Indeed, recalling that, denoted by $\nu(x)$ the unit inner normal to $\partial \Omega$ at a point $x$, we have $\nu(x)=\nabla d(x)$, the divergence theorem should give us
\[
\int_{\partial \Omega} f(k) d \cal{H}^{n-1} = \int_{\partial \Omega}f(k)\nabla d \cdot \nu \,d \cal{H}^{n-1} \approx - \int_\Omega \operatorname{div} \left( f(k)\nabla d \right) .
\]
The meaning of the last integral in the previous formula has to be clarified, since the integrand term a priori is just defined as a distribution. This is a key step of the proof of \eqref{formula} in Theorem \ref{mainthm}, whose main ingredient is indeed the identification of $-\operatorname{div}\left( f(k)\nabla d \right)$ as a non-negative Radon measure which is absolutely continuous with respect to the $(n-1)$-dimensional Hausdorff measure. The formula then follows once we identify the densities of its Lebesgue decomposition.

The paper is structured as follows. In the next section we recall some preliminary results on the regularity of the distance function and on functionals of measures. In Section \ref{sec-meas} we prove that $-\operatorname{div}\left( f(k)\nabla d \right)$ is indeed a non-negative Radon measure in a neighborhood of $\Omega$. The proof of the main result is presented in Section \ref{sec-main}. In the last section the anisotropic version of the representation formula \eqref{formula} is briefly discussed.
\section{Preliminaries}

We start recalling some classical regularity results concerning the distance function from the boundary of a domain that have been proved in  \cite{Li:2005p201,Li:2005p47}. Let $\Omega$ be a domain in $\mathbb{R}^n$. We define the singular set $\Sigma$ as the complement of the open set $G$ defined as the largest open subset of $\Omega$ such that for every $x\in G$ there is a unique closest point on $\partial \Omega$ from $x$. The following regularity result holds true.
\begin{theorem}[Li-Nirenberg] \label{lini-01}
Suppose that $\partial \Omega$ is of class $C^{\ell,\alpha}$, with $\ell\geq 2$ and $0<\alpha\leq 1$, then the distance function belongs to $C^{\ell,\alpha}(G\cup \partial \Omega)$.
\end{theorem}
Let us comment on the properties of the singular set $\Sigma = \Omega\setminus G$ of the distance function to the boundary $\partial \Omega$ (sometimes called also {\it ridge} of $\Omega$ or {\it medial axes}). It is known that $\Sigma$ is always a connected set and has finite $(n-1)$-dimensional Hausdorff measure, in particular $\Sigma$ is a $(n-1)$-rectifiable set of finite perimeter. Another useful property is resumed here. Consider $y\in \partial \Omega$, the \emph{cut point} of $y$, denoted by $m(y)$ is defined as the point when, moving along the inner normal from $y$, the set $\Sigma$ is hit for the first time.
\begin{theorem}\label{lini-02}
Let $\Omega$ be of class $C^{2,1}$. Then from every point $y\in \partial \Omega$, the length $s(y)$ of the segment joining $y$ and its cut point $m(y)$ is Lipshitz continuous in $y$.
\end{theorem}
As pointed out in \cite[Remark 1.2]{Li:2005p201} the regularty condition $C^{2,1}$ is sharp for the validity of the previous result.

\begin{definition}[mean-convex domain]\label{meanconvex} Let $\Omega$ be a bounded domain in $\mathbb{R}^n$ with boundary $\partial \Omega$ of class $C^{1,1}$ . We let $\nu$ be the interior unit normal to $\partial \Omega$ and $\mathbf{H}_{\partial \Omega}$ the mean curvature vector of $\partial \Omega$. We say that $\Omega$ is mean-convex if $\mathbf{H}_{\partial \Omega}$ is pointing inside $\Omega$ at every point, i.e. $\mathbf{H}_{\partial \Omega} \cdot \nu \geq 0$.
\end{definition}

In the sequel we will assume that $\Omega$ is a mean-convex domain. Since the boundary of $\Omega$ is supposed to be of class $C^2$ this is equivalent to require that the mean curvature $k(y)$ is non-negative for any $y\in\partial \Omega$.

Denote by $d(x)$ the distance function form $\partial \Omega$. We will write, with a slight abuse of notation, $k(x)=-\Delta d(x)$ and $\nu(x)=\nabla d(x)$. One of the steps needed for the proof of the identity \eqref{formula}  is the fact that the distribution $S=-\operatorname{div} (f(k)\nabla d)$ is a non-negative Radon measure if $\Omega$ is mean-convex. Let $S$ be a real distribution, i.e. $S(\phi)$ is real for any real valued $\phi\in \mathcal{D}$, we say that $S$ is non-negative if $S(\phi)\geq 0$ for any $\phi \in \mathcal{D}$ with $\phi \geq 0$. We will make use of the following well known result (see \cite[Th\'eor\`eme V]{Schwartz:un}).
\begin{theorem}\label{Measure}
A non-negative distribution $S$ can be identified as a continuous linear form on $\mathcal{D}^k$ equipped with the topology induced by $C^k$ ($k\geq 0$). Moreover $S$ is identified with a non-negative Radon measure $\mu_S$ through the equality
\[
S(\phi) = \int \phi\, d\mu_S.
\]
\end{theorem}

Here we recall some notations on functionals of measures, we refer to \cite{Demengel:1984cx} and \cite{Goffman:1964dj} for further details. In order to understand the behavior of continuous functions at infinity (needed to deal with functions of Radon measures) we will use the following notion. Let $f:\mathbb{R}^m\to \mathbb{R}\cup \{\infty\}$ with $f(0)< +\infty$, we define its \emph{recession function} as
\[
f^\infty(p):=\lim_{t\nearrow \infty} \frac{f(t p)}{t}.
\]

The recession function can be used to give a meaning to functions depending on pairs of Radon measures. Let $ f:\mathbb{R}^m \to [0,\infty]$ be continuous and such that $f^\infty$ is well defined. The recession function is positively homogeneous of degree $1$, it is finite along the direction of (at most) linear growth of $f$ and is infinite along the direction of superlinear growth. Moreover it is clear that if $\|f\|_\infty < \infty$ then $f^\infty=0$ identically. Let $\mu$ and $\lambda$ be respectively an $\mathbb{R}^m$-valued and a positive measure in $\Omega \subset \mathbb{R}^n$, we can define the measure
\[
G(\mu,\lambda) := f\left(\frac{\mu}{\lambda}\right) \lambda + f^\infty\left(\frac{\mu^s}{|\mu^s|}\right) |\mu^s|\,,
\]
i.e. for a measurable set $E$
\[
G(\mu,\lambda)(E) := \int_{E }f\left(\frac{\mu}{\lambda}(x)\right) d \lambda(x) + f^\infty\left(\frac{\mu^s}{|\mu^s|}(x)\right) |\mu^s|(x),
\]
where $\mu^s$ is the singular part of $\mu$ with respect to $\lambda$ and $ \frac{\mu}{\lambda}$ denotes the Radon-Nikodym derivative of $ \mu $ with respect to $ \lambda $.
When $\lambda$ is the Lebesgue measure we will indicate the measure $G(\mu,\lambda)$ with $f(\mu)$. It is worth to observe that if $f$ is a convex and lower semicontinuous function, then the recession function is always well defined and the functional $G$ turns out to be lower semicontinuous with respect to the weak-$*$ convergence of measures (cf. \cite[Theorem 2.34]{AmbFusPal00-000}).

\begin{lemma}
Let $f$ be a real-valued bounded continuous function in $(0,+\infty)$ and let $\mu$ be a positive Radon measure. Then $f(\mu)$ is bounded, i.e. $f(\mu)$ is absolutely continuous with respect to $\mathcal{L}^n$ and its density is an $L^\infty$ function.
\end{lemma}
\begin{proof}
Using the previous notation we have
\[
f(\mu) := f\left(\frac{\mu}{\mathcal{L}^n}\right) \mathcal{L}^n + f^\infty\left(\frac{\mu^s}{|\mu^s|}\right) |\mu^s|.
\]
Now we observe that $f^\infty$ is identically zero by the boundedness assumption on $f$. This imply the desired result since we can write
\[
f(\mu) := f\left(\frac{\mu}{\mathcal{L}^n}\right) \mathcal{L}^n
\]
and $\| f\left(\frac{\mu}{\mathcal{L}^n}\right)\|_\infty \leq \| f\|_\infty $ .
\end{proof}

\section{The measure $-\operatorname{div} (f(k)\nabla d)$}\label{sec-meas}

In this section we prove that $-\operatorname{div} (f(k)\nabla d)$ is a non-negative Radon measure. The first step is to prove that $k:=-\Delta d$ is a positive Radon measure. We will use the \emph{normal coordinates} to represent points in $\Omega\setminus \Sigma$. For $z \in \Omega \setminus \Sigma$ we can write
$z = x + t \nu_x$ for a unique $x \in \partial \Omega$ and $0 < t < s(x)$, where $s(x) = \langle m(x) - x,
\nu_x\rangle$ and $\nu_x$ is the unit inner normal to $\partial \Omega$ at $x$. We recall that by Theorem \ref{lini-02} the function $s(x)$ is Lipschitz continuous. Moreover from Theorem \ref{lini-01} we know that $-\Delta d(z)$ is well-defined in $ G \cup \partial \Omega $. We can explicitly compute it in terms of the principal curvatures $k_r(x)$ of $\partial \Omega$ in the point $x$, we have indeed (see for instance \cite[Section 14.6, Lemma 14.17]{Gilbarg:2001p77})
\begin{equation}\label{dlaplacian}
-\Delta d(z) = \sum_{r=1}^{n-1} \frac{k_r(x)}{1- t k_r(x)}\, .
\end{equation}

\begin{lemma}Let $ \Omega $ be a mean-convex domain with boundary of class $ C^{2,1} $. Then $ -\Delta d $ is a non-negative Radon measure. If $ k(x) \geq c > 0 $ for any $x\in \partial \Omega$, then $-\Delta d$ is a positive Radon measure uniformly bounded from below, in particular $-\Delta d\geq c $ in distributional sense.
\end{lemma}

\begin{proof}
Consider $S:=-\Delta d$ which is well-defined as a distribution on $\Omega$ since $\|\nabla d\|_\infty\leq 1$. Let $\varphi \geq 0$ be a test function. We use \eqref{dlaplacian} to infer, by explicit calculations, that if $\varphi$ is supported in $\Omega\setminus \Sigma$ we have $S(\varphi)\geq c$.
Let $\theta$ be a cut-off function such that $\theta = 1$ on $[0, 1]$, $\theta = 0$ on $[2, \infty)$ and $\theta$ is non-increasing $C^1$ function. Define for $\varepsilon > 0$
\[
\psi_\varepsilon(x,t)=\theta \left(\frac{s(x) - t}{\varepsilon}\right)\;\; \textrm{ with }\;\;\varepsilon < \inf_{x\in \partial \Omega} \frac{s(x)}{2}.
\]
We note that, by definition,
\begin{equation}\label{positive}
\langle \nabla \psi_\varepsilon, \nabla d\rangle = \frac{\partial  \psi_\varepsilon}{\partial t}  \geq 0.
\end{equation}
We use this cut-off to write $S(\varphi) = S(\psi_\varepsilon \varphi) + S((1 - \psi_\varepsilon) \varphi)$. As noted before, since $(1 - \psi_\varepsilon) \varphi$ is supported in $\Omega\setminus \Sigma$ we have $S((1 - \psi_\varepsilon) \varphi)\geq c$. For the other term we write, using \eqref{positive}
\[
S(\psi_\varepsilon \varphi)= \int_{\Omega} \varphi \, \langle \nabla \psi_\varepsilon , \nabla d \rangle + \int_{\Omega} \psi_\varepsilon \, \langle \nabla \varphi, \nabla d \rangle \geq  \int_{\Omega} \psi_\varepsilon \, \langle \nabla \varphi, \nabla d \rangle.
\]
Here and hereafter we often suppress $dx$ unless confusion occurs.
Since the last term tends to zero for $\varepsilon$ going to zero we get $S(\varphi)\geq 0$. The claim follows from the Theorem \ref{Measure}.

\end{proof}

\begin{rem}\label{bound}From the previous two lemmata we easily infer that if $f$ is bounded and if $\Omega$ is mean-convex, then the distribution $f(k)\nabla d$ can be identified with an essentially bounded map in $\Omega$.
\end{rem}

Using the same idea of the previous lemma we are able now to prove the following.

\begin{theorem}\label{thm-measurediv} Let $\Omega$ be a mean-convex domain with boundary of class $C^{2,1}$ and let $f$ be a non-negative and non-increasing function of class $C^1$ in a neighborhood of the interval $[\min_{\partial\Omega}k(x), \infty)$, then the distribution \[S=-\operatorname{div} [f(k)\nabla d]\] is a non-negative Radon measure in a sufficiently small neighborhood of $\overline{\Omega}$.
\end{theorem}
\begin{proof} We start by observing that the regularity of $\partial \Omega$ allows us to extend in a $C^{2,1}$ way the distance function in a $\delta$-neighborhood of $\overline{\Omega}$ defined by
\[
\Omega_\delta:= \{x\in\mathbb{R}^n \;:\; d(x, \Omega) < \delta \}.
\]
By the regularity assumptions on $f$, we can choose $\delta$ sufficiently small in order to have $f(k(y))\leq c < \infty$ for any $y\in \Omega_\delta\setminus \Omega$. We will continue to denote by $d(x)$ the extended distance function from $\partial \Omega$ defined in $\Omega_\delta$.
Now we observe that if $\varphi$ is a test function supported in $\Omega_\delta \setminus \Sigma$ with $\varphi \geq 0$, then $S(\varphi) \geq 0$. This follows by direct calculations. Indeed, since in $\Omega_\delta \setminus \Sigma$ the distance function is $C^{2,1}$, we can differentiate two times the equality $|\nabla d |=1$ to get (using Einstein notation of summated indices)
\[
\partial_{lji}d \,\partial_{i}d + \partial_{ji}\, \partial_{li}d = 0,
\]
that implies, for $l=j$
\begin{equation}\label{secondform}
\langle\nabla d, \nabla k \rangle = - (\nabla d \cdot \nabla) \operatorname{div} \nabla d =  |D^{2}d|^{2}
\end{equation}
and consequently
\[
\operatorname{div}(f(k) \nabla d) = f'(k)\langle \nabla d , \nabla k \rangle - f(k) k \leq 0.
\]
Hence the last inequality follows as in the previous proof, observing that by \eqref{dlaplacian} the function $k$ is uniformly bounded by a positive constant in any compact set far from $\Sigma$.

For a general $\varphi$ we can use the same cut-off procedure used in the proof of the previous lemma. We can therefore write as before
\[
S(\varphi) = S(\psi_\varepsilon \varphi) + S((1 - \psi_\varepsilon) \varphi).
\]
We observed already that the second term is positive due to the aforementioned explicit calculation. It remains to bound the quantity
\[
S(\psi_\varepsilon \varphi)= \int_{\Omega_\delta} \varphi \,f(k) \, \langle \nabla \psi_\varepsilon , \nabla d \rangle + \int_{\Omega_\delta}  \psi_\varepsilon \, \langle \nabla \varphi, f(k) \nabla d \rangle \geq  \int_{\Omega_\delta} \psi_\varepsilon \, \langle \nabla \varphi, f(k) \nabla d \rangle,
\]
where the last inequality is a consequence of \eqref{positive} and of the non-negativity of the function $f$. Finally we observe that Remark \ref{bound}  and the definition of $\psi_\varepsilon$ allow us to infer that
\[
\lim_{\varepsilon \to 0} \int_{\Omega_\delta} \psi_\varepsilon \, \langle \nabla \varphi, f(k) \nabla d \rangle = 0
\]
concluding the proof by Theorem \ref{Measure}.
\end{proof}

We observe now that the measure $\mu:=-\operatorname{div}[f(k) \nabla d] $ is absolutely continuous with respect to $\mathcal{H}^{n-1}$. To this aim it is sufficient to bound the upper $(n-1)$-density of the measure $\mu$  (see \cite[Theorem 2.56]{AmbFusPal00-000})
\[
\Theta_{n-1}(\mu,x) = \limsup_{\rho\to 0} \frac{\mu(B_\rho(x))}{\rho^{n-1}}
\]
for any $x\in \Omega$, where $B_\rho(x)$ denotes the closed ball of radius $\rho$ centered at $x$. This can be achieved using a smoothing argument and integration by parts. Indeed we recall that by Remark \ref{bound} $f(k)\nabla d$ is in $L^\infty(\Omega)$. Then if we consider a family of standard mollifiers $\rho_\varepsilon$ we have (up to subsequences) $|\rho_\varepsilon * f(k)\nabla d | \leq C $ for some $C< \infty$. Moreover, by standard properties of convolutions of measures (see \cite[Theorem 2.2]{AmbFusPal00-000}) we also have that $-\operatorname{div}(f(k) \nabla d)  * \rho_\varepsilon$ is a $C^\infty$ function that locally weak-$*$ converges to $-\operatorname{div}(f(k) \nabla d) $ in the sense of measures. Let $x \in \Omega$, we use the lower semicontinuity of the total variation with respect to weak-$*$ convergence of measures to ensure that (for sufficiently small $\rho$) we have
\[
\mu(B_\rho(x)) \leq \lim_{\varepsilon\to 0} \int_{B_\rho(x)}  -\operatorname{div}(f(k) \nabla d)  * \rho_\varepsilon \, dx = -\int_{\partial B_\rho(x)} \langle f(k) \nabla d * \rho_\varepsilon , \nu \rangle \,d \mathcal{H}^{n-1} \leq C \omega_n \rho^{n-1}
\]
which proves the claim.

\section{Main result}\label{sec-main}
In this section we will precisely state and prove the main result of the paper. To this aim, we will need to know more precise properties of the ridge set. In particular, it will be useful to understand its {\it stratified} structure. The singular set of the distance function is characterized by the property that if $y\in \Sigma$ then there exist at least two points in $\partial \Omega$ in which the value $d(x):=d(x,\partial \Omega)$ of the distance function is attained. Let us consider the set $\Sigma_0\subset \Sigma$ of the points where the distance function form $\partial \Omega$ is attained exactly  in two points. Namely we set
\[
\Sigma_0:= \left\{  x\in \Sigma \;| \begin{array}{c} \text{there exists a unique pair} \, (y_1(x),y_2(x)) \in \partial \Omega \times \partial \Omega \\ \text{such that} \,y_1(x)\not=y_2(x)\,,\; d(x)=|x-y_1|=|x-y_2| \end{array} \right\}.
\]
For a fixed $x\in \Sigma_0$ let us denote by $\nu_1$ and $\nu_2$ the inner normal directions to $\partial \Omega$ at $y_1(x)$ and $y_2(x)$ respectively. We observe that, due to the expression \eqref{dlaplacian}, when moving along $\nu_i$ from $y_i$ toward $x\in \Sigma_0$, we can identify two limit values for the extended curvature function $k=-\Delta d$, namely for $j\in\{1,2\}$,
\[
k^j(x)=\lim_{t\to s(y_j)} \sum_{r=1}^{n-1} \frac{k_r(y_j)}{1- t k_r(y_j)}.
\]

Our main result can be stated as follows.

\begin{theorem}\label{mainthm}
Let $\Omega$ be a mean-convex domain with boundary of class $C^{2,1}$ and let $f$ be a non-negative and non-increasing  function of class $C^1$ in a neighborhood of the interval $[\min_{\partial\Omega}k(x), \infty)$. Then the following formula holds
 \begin{equation}\label{general-iso}
\int_{\partial \Omega} f(k(x)) \,d \mathcal{H}^{n-1} =  \int_{\Omega} k(x) f(k(x))\,dx - \int_{\Omega} f'(k(x))|D^{2}d(x)|^{2} \, dx + \delta^{f}(\Sigma),
 \end{equation}
 where
  \[
 \delta^f(\Sigma)=\int_{\Sigma_{0}} \frac{|\nu_1(x)-\nu_2(x)|}{\sqrt{2}}\big[f(k^1(x))+f(k^2(x))\big] \,d \mathcal{H}^{n-1}(x).
 \]
\end{theorem}

Once we know that the distribution $S=-\operatorname{div}(f(k)\nabla d)$ is a measure on $\Omega_\delta$ which is absolutely continuous with respect to the $(n-1)$-dimensional Hausdorff measure, a key step toward the proof of the main result is to identify the densities of its absolutely continuous part and of its singular part. The next lemma provides the desired identification.

\begin{lemma}\label{lemma-density} Let $f$ be as in Theorem \ref{mainthm}, then for the measure $-\operatorname{div}(f(k)\nabla d)$ the following decomposition holds
\begin{equation}\label{represent}
\begin{split}
-\operatorname{div}(f(k)\nabla d)  = &  \big(f(k) k -f'(k)|D^2 d|^2  \big) d\mathcal{L}^n \\
 & +  \left(\frac{|\nu_1(x)-\nu_2(x)|}{\sqrt{2}}\big[f(k^1(x))+f(k^2(x))\big] \right) \, \mathcal{H}^{n-1}|_{\Sigma_0}  \;.
\end{split}
\end{equation}

\end{lemma}

Before proving the Lemma \ref{lemma-density} we state two preliminary lemmata that will be useful throughout the proof. The first one is an orthogonality result of linear algebra.

\begin{lemma}\label{lemma-linear}
Let $e_i$ for $i \in \{0, 1, 2\}$ be three different unit normals in $\mathbb{R}^n$. Then $e_i - e_0$ for  $i = 1, 2$ are linearly independent.
\end{lemma}

\begin{proof}
Assume that $c_1$ and $c_2$ in $\mathbb{R}^n$ satisfy $c_1 (e_1 - e_0) + c_2 (e_2 - e_1) = 0$.
Thus $c_1 e_1 + c_2 e_2 = (c_1 + c_2) e_1$. If $c_1 + c_2$ is not zero, then $e_0$, $e_1$, $e_2$ lie on the same line which is
impossible since $e_i$ has length one. If $c_1 + c_2 = 0$, then $c_1 = 0$, $c_2
= 0$ since $e_1$ and $e_2$ are linearly independent. We thus conclude $c_1 =
c_2 =0$.
\end{proof}

The second one gives us some informations on the Jacobian of the Lipschitz map $m$ at regular points which are inverse images of conjugate points.

\begin{lemma}\label{jacob} Let $x \in \partial\Omega$ be such that $m$ and $s$ are differentiable at  $x$. Assume that
\[
\lim_{t \to s(x)} \sum_{r=1}^{n-1} \frac{k_r(x)}{1- t k_r(x)}= \infty\;,
\]
then the Jacobian of $m$ at $x$ is zero.
\end{lemma}

\begin{proof}

By assumption we know that there exist $r\in \{1,\dots n-1\}$ such that $s(x) = 1/k_r(x)$ and by definition we have
\[
m(x) = x+s(x) \nu(x),
\]
where $\nu(x)$ is the inner unit normal to $\partial \Omega$. Let $\tau_r(x)$ be the unit tangent vector at $x$ such that it is the eigenvector of the Weigarten map corresponding to the eigenvalue $k_r(x)$. Differentiating along the direction $\tau_r$ we get
\[
D_{\tau_r} m(x) = D_{\tau_r} s(x) \nu(x) + \big(1 - s(x)k_r(x)\big) \tau_r(x) =D_{\tau_r} s(x) \nu(s) .
\]

Assume that $D_{\tau_r} s$ is not zero at $x$. By the implicit function theorem we observe that $\Sigma$ is tangent to $\nu$ at $m(x)$. One often assumes $C^1$ regularity to get a $C^1$ implicit function. However, if one is interested in differentiability of the implicit function at $x$, the (Fr\'echet) differentiability of $m$ at $x$ is enough. Thus the set $\Sigma$ is tangent to the ray from $x$ to $m(x)$. This contradicts the fact that $s(x)$ is Lipschitz. Thus we conclude that $D_{\tau_r} m(x) = 0$. This implies that the Jacobian of $m$ at $x$ is zero.
\end{proof}

\begin{proof}[Proof of Lemma \ref{lemma-density}]
The proof will be divided in several steps.

\bigskip
{\sc{Step 1 :}}  First we justify the expression of $\delta^f(\Sigma)$ analyzing the local representation of $-\operatorname{div}(f(k)\nabla d)$ near $\Sigma_0$. Let $U\subset \Omega$ be an open set such that $U\cap \Sigma_0 \not = \emptyset$ and there exist two connected, mutually disjoint and relatively open sets $\mathcal{N}_1$ and $\mathcal{N}_2$ on $\partial \Omega$ with the property that
\begin{equation}\label{surface}
U\cap \Sigma_0 = \{ x\in U \;| \; d(x,\mathcal{N}_1)-d(x,\mathcal{N}_2) = 0 \}.
\end{equation}
We use the notation $d_i(x)=d(x,\mathcal{N}_i)$ for $i=\{1,2\}$. Suppose moreover that $k\leq M <\infty$ in $U$. Under this condition, by the regularity of $\partial \Omega$ we can write $U=U_1\cup U_2 \cup (\Sigma_0\cap U)$, where $\Sigma_0\cap U$ is a $C^{2,1}$ hypersurface and
\[
U_1:= \{ x\in U \;|\; d_1(x)< d_2(x) \} \;\;,\;\;\;\; U_2:=\{x\in U \;|\;  d_1(x)> d_2(x) \}.
\]
Let $\phi$ be a test function supported in $U$, we have
\[
\begin{split}
\langle -\operatorname{div}(f(k)\nabla d), \phi \rangle =  & \int_{U}  f(k)\nabla d \cdot \nabla \phi \, dx  =  \int_{U_1}  f(k)\nabla d \cdot \nabla \phi \, dx +  \int_{U_2}   f(k)\nabla d \cdot \nabla \phi \, dx \\
= & \int_{U_1}  f(-\Delta d_1)\nabla d_1 \cdot \nabla \phi \, dx +  \int_{U_2}   f(-\Delta d_2)\nabla d_2 \cdot \nabla \phi \, dx \\
= & - \int_{U_1}  f'(-\Delta d_1)|D^2 d_1|^2  \phi \, dx -  \int_{U_2}   f'(-\Delta d_2)|D^2 d_2|^2  \phi \, dx \\
& -  \int_{U_1}  f(-\Delta d_1) \Delta d_1  \phi \, dx -  \int_{U_2}   f(-\Delta d_2) \Delta d_2   \phi \, dx \\
& + \int _{\partial U_1} \phi f(-\Delta d_1)\nabla d_1 \cdot \nu_1 \, d\mathcal{H}^{n-1}  + \int _{\partial U_2} \phi f(-\Delta d_2)\nabla d_2 \cdot \nu_2 \, d\mathcal{H}^{n-1} \\
= & - \int_{U}  f'(k)|D^2 d|^2  \phi \, dx + \int_{U} f(k) k\, \phi \,dx  \\
 & + \int_{\Sigma_0}  \frac{|\nu_1(x)-\nu_2(x)|}{\sqrt{2}}[f(k^1(x))+f(k^2(x))] \,\phi \,d \mathcal{H}^{n-1}(x) .
\end{split}
\]
We note explicitly that, in order to justify the previous calculations, we can extend up to the boundary, in a canonical way, all the quantities we are interested in, when considered separately on $U_1$ and $U_2$. In the last equality of the previous formula we have used \eqref{dlaplacian}, \eqref{secondform} and the fact that we can explicitly calculate the expression of the normal to the boundary $\partial U_i$ in the portion that lives on $\Sigma_0$ (which is the only one that gives a contribution to the integral since $\phi$ is supported in $U$). More precisely in view of \eqref{surface}, we have
\[
\nu_1 = \frac{\nabla d_1-\nabla d_2}{|\nabla d_1-\nabla d_2|}=-\nu_2\,,
\]
from which it follows that
\[
\nabla d_i \cdot \nu_i = \frac{1-\nabla d_1 \cdot \nabla d_2}{\sqrt{2} \sqrt{1- \nabla d_1 \cdot \nabla d_2}} = \frac{|\nabla d_1-\nabla d_2|}{\sqrt{2}}\,.
\]
Note that along a geodesic line the gradient of the distance function points always in the same direction.

\bigskip

\noindent {\sc{Step 2 :}}  We give now an estimate of the size of the subset of $\Sigma \setminus \Sigma_0$ where the curvature remains bounded. For a given $z\in\Sigma$ we define
\[
\kappa(z) := \sup\left \{ \lim_{t\to s(y)} \sum_{r=1}^{n-1} \frac{k_r(y)}{1- t k_r(y)}  \; | \; y \in \partial \Omega \;\text{ with } \; d(z)=|z-y| \right\}.
\]
We are interested in the points of $\Sigma$ where $\kappa$ is bounded. For $M>0$ define
\[
G_M := \left\{ z\in \Sigma \; | \begin{array}{l} \kappa(z)\leq M \; \text{and there are  } y_1(z),y_2(z),y_3(z) \in \partial \Omega \\ \text{such that } d(z)=|z-y_i(z)| \; \text{for any } i\in\{1,2,3\} \end{array}\right  \}.
\]
We claim that $\cal H^{n-1}(G_M) =0 $. The proof of the claim is based on the observation that for a fixed $z\in G_M$ and for any collection of neighborhoods $\cal N_i$ of $y_i(z)$ in $\partial \Omega$ such that $y_i(z) \not \in \cal N_j$ if $j\not=i$, we have that $z$ is contained in the intersection of the two hypersurfaces given by
\[
\Gamma_1 := \big\{x \in \Omega \;:\; d(x,\cal N_1)=d(x,\cal N_3) \big\} \;\; \text{ and }\;\;\; \Gamma_2 := \big\{x \in \Omega \;:\; d(x,\cal N_2)=d(x,\cal N_3) \big\} .
\]
It is easy to check, as in {\sc{Step 1}} that, for $i\in \{1,2\}$, the normal vector of $\Gamma_i$ is parallel to $\nabla d(x,\cal N_i)-\nabla d(x,\cal N_3)$. Lemma \ref{lemma-linear} ensures us that $\Gamma_1$ and $\Gamma_2$ are transversal and thus $\cal H^{n-2}(\Gamma_1 \cap \Gamma_2) < \infty$.
The claim will be proved once we know that we can reduce to consider just countably many hypersurfaces of the type $\Gamma_i$ to cover the set $G_M$.

We start by defining for any $h\in \mathbb{N}$ the set
\[
G^h_M := \left\{ z\in G_M \; | \begin{array}{l} \text{there exist } \; y_1,y_2,y_3 \in \partial \Omega  \text{ with } d(z)=|z-y_i|  \\  \text{and } |y_i-y_j|>\frac{1}{h} \text{ for any }\,i,j\in \{1,2,3\} \end{array}\right \}.
\]
Choose $\rho_h >0$ such that for any $y\in \cal N_h(y_i(z))=B_{\rho_h}(y_i)\cap \partial \Omega$ we have, independently of $z \in G^h_M$,
\[
\sum_{r=1}^{n-1} \frac{k_r(y)}{1- s(y) k_r(y)} \leq M+1.
\]
Since the set
\[
L := \left\{ y\in\partial \Omega \;:\; \sum_{r=1}^{n-1} \frac{k_r(y)}{1- s(y) k_r(y)} \leq M+1 \right \}
\]
is compact, we may find a finite collection of points $\{y_i\}_{i\in I}$ with $y_i \not \in \cal N_h(y_j)$ for $i\not = j$ and
\[
L\subset \bigcup_{i\in I}\cal N_h(y_i).
\]
For any pair of points $y_i, y_j$ with $i,j \in I$ we can consider the hypersurfaces
\[
\Gamma_{i,j}  := \big\{x \in \Omega \;:\; d(x,\cal N_h(y_i))=d(x,\cal N_h(y_j)) \big\}
\]
As before, we have that $\Gamma_{i,l}$ and $\Gamma_{j,l}$ are transversal. Then the set $G^h_M$ is contained in the finite union of the intersections of the type $\Gamma_{j,l}\cap \Gamma_{i,l}$ for $i,j,l \in I$. The claim follows by noticing that $G_M \subseteq \bigcup_{h\in \mathbb{N}} G_M^h$.

\bigskip
\noindent {\sc{Step 3 :}} We analyze now the measure of the set
\[
K:=\{z \in \Sigma \;:\; \kappa(z) = +\infty \}.
\]
We claim that $\cal H^{n-1}(K)=0$. Since $K$ is a subset of the image through the Lipschitz map $m$ of the set
\[
A_\infty:=\left\{ x\in \partial \Omega \; : \; \lim_{t \to s(x)} \sum_{r=1}^{n-1} \frac{k_r(x)}{1- t k_r(x)}= \infty  \right\},
\]
it is sufficient to prove that
\begin{equation}\label{zeromeas}
\cal H^{n-1}\big(m(A_\infty)\big)=0.
\end{equation}
We first observe that at $\cal H^{n-1}$-a.e. $x$ in $A_\infty$ the mappings $m$ and $s$ are differentiable by the Rademacher's theorem. Then the
equality \eqref{zeromeas} follows from Lemma \ref{jacob} and from the area formula for Lipschitz maps (cf. \cite[Theorem 2.71]{AmbFusPal00-000} or \cite[Theorem 3.2.3]{Fed69-000}).

\bigskip
\noindent {\sc{Step 4 :}} From {\sc{Step 3}}, since the measure $\mu:=-\operatorname{div}(f(k)\nabla d)$ is absolutely continuous with respect to $\cal H^{n-1}$, we have that the set $K$ is negligible with respect to $\mu$. It is then enough to prove \eqref{represent} on the set $\Omega_\delta \setminus K$. The claim will follow by continuity of measures along increasing sequences of sets once we have proved that \eqref{represent} holds on the set
\[
W_M:= \{ x\in  \Omega_\delta \,:\, \kappa(x) < M\},
\]
for any $M>0$. To this aim we observe that, since $z\in \Sigma_0$ is a conjugate point only if $\kappa(z)=\infty$,  by \cite[Theorem 1] {Li:2005p47}, we can find an open covering of $\Sigma_0 \cap  W_M$ given by $\{U_i\}_{i\in I}$, with $U_i$ satisfying \eqref{surface} as in {\sc Step 1}. We can then consider  the open covering of $W_M\setminus G_M$ given by $U_0 \cup \bigcup_{i\in I}U_i$ where $U_0$ is such that
 $\partial \Omega \subset U_0$ and $U_0 \cap \Sigma = \emptyset$.

If $\phi$ is a test function supported on $U_0$, we can apply the integration by parts to obtain
\begin{equation}\label{U1}
\langle -\operatorname{div}(f(k)\nabla d), \phi \rangle = - \int_{U_0}  f'(k(x))|D^2 d(x)|^2  \phi(x) \, dx + \int_{U_1} f(k(x)) k(x)\, \phi(x) \,dx.
\end{equation}
Finally, using a partition of unity subordinate to the cover $U_0 \cup \bigcup_{i\in I}U_i$, from \eqref{U1} and the {\sc Step 1}, we have the claim.
\end{proof}

We are finally in position to prove the main theorem.

\begin{proof}[Proof of Theorem \ref{mainthm}]
Let us consider the characteristic function of $\Omega$, $\chi_\Omega$ and approximate it using a standard mollifier with $\chi^\varepsilon_\Omega =\rho_\varepsilon * \chi_\Omega \in C^\infty_{0}(\Omega_{2\varepsilon})$. We explicitly observe that $\nabla \chi^\varepsilon_\Omega=0$ in $\Omega_{-2\varepsilon}$ since $\chi^\varepsilon_\Omega(x) = 1$ for any $x\in \Omega_{-2\varepsilon}$. We can then compute
\begin{equation}\label{limit-01}
\begin{split}
\lim_{\varepsilon\to 0}  \langle -\operatorname{div}(f(k)\nabla d),  \chi^\varepsilon_\Omega \rangle  = & \lim_{\varepsilon\to 0}  \int_{(\partial \Omega)_\varepsilon} f(k)\nabla d \cdot \nabla \chi^\varepsilon_\Omega \, dx \\
 = & \lim_{\varepsilon \to 0} \int_{(\partial \Omega)_\delta} f(k) \nabla d \cdot d D(\chi^\varepsilon_{\Omega}) \\
= & \int_{(\partial \Omega)_\delta} f(k) \nabla d \cdot d D(\chi_{\Omega}) 	\\
= & \int_{\partial \Omega} f(k) \, d \mathcal{H}^{n-1},
\end{split}
\end{equation}
where $(A)_\delta$ denotes the $\delta$-neighborhood of $A$. To conclude the proof it is sufficient to observe that from \eqref{represent} we also have
\[
\begin{split}
\lim_{\varepsilon\to 0}  \langle -\operatorname{div}(f(k)\nabla d),  \chi^\varepsilon_\Omega \rangle  = & \lim_{\varepsilon\to 0}  \int_{\Omega_\varepsilon}  \big(f(k) k -f'(k)|D^2 d|^2 \big)  \, dx \\
  & + \int_{\Sigma_{0}} \frac{|\nu_1(x)-\nu_2(x)|}{\sqrt{2}}[f(k^1(x))+f(k^2(x))] \,d \mathcal{H}^{n-1}(x) \\
= &  \int_{\Omega}  \big( f(k(x)) k(x)-f'(k(x))|D^2 d(x)|^2 \big)  \, dx \\
  & + \int_{\Sigma_{0}} \frac{|\nu_1(x)-\nu_2(x)|}{\sqrt{2}}[f(k^1(x))+f(k^2(x))] \,d \mathcal{H}^{n-1}(x).
\end{split}
\]
\end{proof}

\section{The anisotropic case}

In this section we briefly describe how our approach can be also used in the anisotropic setting to provide a representation for boundary integrals involving anisotropic curvatures. We restrict ourself here to a formal discussion implicitly assuming all the regularity properties and the necessary assumptions needed to perform the calculations, without mentioning them.

We start recalling some useful notations (cf. for instance \cite{Bellettini:2009p223}). Let $\varphi^0$ be a one-homogeneous convex function that will be our anisotropy and let $\varphi$ be its dual function. The anisotropic distance function of a
point $x$ from a set $C$ is
defined by
\[
d_\varphi(x,C)=\inf_{y\in C}\varphi(y-x).
\]
For a given domain $\Omega$ we consider the signed anisotropic distance function from
$\partial \Omega$, defined as
\[
d_\varphi(x)=d_\varphi(x,\Omega)-d_\varphi(x,\mathbb{R}^n\setminus \Omega).
\]
We denote by $n_\varphi$ the Cahn-Hoffman vector and $k_\varphi$ the
anisotropic mean curvature. More precisely, defined
\[
T_{\varphi^0} := \frac{1}{2} \nabla \big( ( \varphi^0)^2 \big),
\]
we set
\[
n^*_{\varphi}=\frac{n(x)}{\varphi^0(n(x))} \;\; ,\;\;\; n_\varphi = T_{\varphi^0} (n^*_{\varphi}).
\]
We have, near $\partial \Omega$,
\[
n^*_{\varphi} = \nabla d_\varphi \;,\;\;
\langle n^*_{\varphi} , n_\varphi \rangle = 1 \;, \;\;
k_\varphi=-\operatorname{div} n_\varphi.
\]

As in the isotropic case, a formal application of the Gauss theorem leads to
\[
\begin{split}
\int_{\partial \Omega} f(k_{\varphi})\varphi^{0}(n) \, d\cal H^{n-1} & = \int_{\partial \Omega}
f(k_{\varphi})\varphi^{0}(n)\langle n_\varphi^*,n_\varphi
\rangle \, d\cal H^{n-1}  \\
 & =\int_{\partial \Omega}  f(k_{\varphi})\,n_\varphi \cdot n  \, d\cal H^{n-1}  \approx
-\int_\Omega \operatorname{div}
\left(
f(k_{\varphi})\,n_\varphi
\right).
\end{split}
\]
The main issue, as in the previous case, is to give a representation of the measure $- \operatorname{div}
\left(
f(k_{\varphi})\,n_\varphi
\right)$. Let us observe that, outside of the singular set of $d_\varphi$ we have
\[
\operatorname{div}\left(f(k_{\varphi})\,n_\varphi \right) = f'(k_{\varphi}) \nabla k_{\varphi} \cdot n_\varphi -  f(k_{\varphi})k_{\varphi} \,;
\]
moreover, the term $ \nabla k_{\varphi} \cdot n_\varphi$ can be expressed in terms of the squared
norm of the $\varphi$-Weingarten operator. To this aim for instance we can consider a proper test function $\phi$ and compute
\begin{equation*}
  \begin{split}
  \int_\Omega (n_\varphi \cdot \nabla) \operatorname{div}n_\varphi \phi  &
  = - \int_\Omega \partial_j  (n_\varphi^i \phi)  \partial_i (n^j_\varphi) \\
  & = -
  \int_\Omega  \partial_j  (n_\varphi^i)  \partial_i (n^j_\varphi) \phi -
  \int_\Omega n_\varphi^i \partial_i (n^j_\varphi)  \partial_j \phi \\
 & = \int_\Omega \| \nabla n_\varphi\|^2 \phi,
\end{split}
\end{equation*}
where $\nabla n_\varphi$ denotes the Jacobian matrix of $n_\varphi$ and $\|\nabla n_\varphi\|^2$ denotes the square of its Euclidean norm, i.e. the sum of the squares of the components of the matrix $\nabla n_\varphi$. In the last equality we have used the symmetry of the Jacobian of the Cahn-Hoffman vector and the Euler formula for homogeneous
functions. We explicitly note that $\| \nabla n_\varphi\|^2$ is the squared norm of the anisotropic Weingarten operator that correspond to $\| D^2 d\|^2$ in the isotropic setting (cf. Lemma 3.2 and Remark 3.3 in \cite{BelFra00-000}).

Proceeding similarly as in the previous section, using $d_\varphi$ instead of $d$, we may derive the following representation formula
\[
\int_{\partial \Omega} f(k_{\phi})\varphi^{0}(n) d\cal H^{n-1}= \int_\Omega f'(k_{\varphi}) \| \nabla n_\varphi\|^2dx - \int_\Omega f(k_{\varphi})k_{\varphi}\,dx + \delta_\varphi^f(\Sigma_\varphi),
\]
where
\[
\delta_\varphi^f(\Sigma_\varphi) = \int_{\Sigma_0^\varphi}\big( f(k_{\varphi}^1(x))n_\varphi(y_1)- f(k_{\varphi}^2(x))n_\varphi(y_2) \big) \frac{n^*_\varphi(y_1)-n^*_\varphi(y_2)}{|n^*_\varphi(y_1)-n^*_\varphi(y_2)|} d\cal H^{n-1}.
\]

\section{Acknowledgments}
This work was initiated while the second author visited
the Department of Mathematics of Hokkaido University during 2010. Its
hospitality is gratefully acknowledged. This work was completed when the
second author visited the School of Engineering Science of Osaka
University during 2012/13 under Marie Curie project IRSES-2009-247486. The work of the
first author has been partly supported by the Japan Society for the
Promotion of Science (JSPS) through grants for scientific research Kiban
(S) 21224001 and Kiban (A) 23244015.

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


Welcome to Fittsburgh! How Joe and Anthony Vennare are Shaping up Pittsburgh to be a Better and Healthier City.
Being healthy in Pittsburgh has never been so easy, thanks to Anthony and Joe Vennare (yup, you guessed it they're bros). But not those kind of bro's, like the related kind). They launched Fittsburgh, a site about healthy eating and living here in Pittsburgh just this past May, and it is already taking off! Pittsburghers are clearly in dire need of change and variety for ways to move and eat well here in the 'Steel City.' The Vennare brothers make this possible by providing a resource to find locally grown food and all unique (and most importantly FUN) ways to workout! Pittsburgh is an amazing smorgasbord for all sorts of amazing food and fitness styles that thanks to Joe and Anthony we are no longer blind to! I recently had the pleasure of chatting with Anthony, one half of this dynamic duo, about fitness, food and doing what you love.
It seems that with each passing year we get less and less music with meaning, music that has soul, music with a message. THANK GOODNESS these guys came along! MKTO consist of Tony Oller and Malcolm Kelley. Two former Nickelodeon stars that decided to break out, grow up and put music out there to inspire and make people feel good.
In their self titled debut album they touch on issues of being bullied to showing some respect to women. Much like their single 'Classic' that's blowing up right are these guys style. They may be younger but they have a cool old school vibe that I really dig. Not since Macklemore's 'Same Love' have I heard and really felt such passion emanating through my speakers. I think I really like these guys so much because there here to uplift and inspire people to dream and dream BIG. This is something that has drastically gone away in a day and age of fast pace living and even faster money. Do yourself a favor and check these guys out, you won't be sorry! Check out my favorite's 'Classic' and 'American Dream' after the jump!


text:
GM's record profit sharing to boost Michigan economy as EV push intensifies
Red Wings' Frans Nielsen relishing return to Nassau Coliseum
Ted Kulfan
Uniondale, N.Y. — When the regular season schedule comes out in the summer, most players don't pay attention.
They'll live it when the time comes, and anyway, it becomes a blur anyway.
But for Frans Nielsen, when the schedule came out in June, he couldn't wait. Nielsen wanted to know whether the Red Wings would be playing at the refurbished Nassau Coliseum when the Wings visited the New York Islanders.
"I'm not going to lie, I did look at the schedule when it came out to see if we were going to play at the Coliseum or not," said Nielsen, who played for the Islanders — at the Coliseum — for 10 seasons before joining the Red Wings. "It's definitely going to be different than playing in Brooklyn.
"I'm excited, I never thought I was going to play there again."
More:Red Wings' Mike Green out 3-5 weeks; Filip Hronek recalled from G.R.
More:Red Wings tweak struggling power play
More:Red Wings to focus on limiting opponents' chances
The Islanders are splitting their home schedule between the Coliseum and the Barclays Center in Brooklyn. They're expected to open a new arena near Belmont Park on 2021.
This will be the Red Wings' first game at Nassau Coliseum since March 2015.
Nielsen scored his first goal in the arena, experienced his first NHL playoffs there, and was consistently amazed by the support and passion of Islanders fans.
"A lot of great memories. It means a lot to me, that place," Nielsen said. "It's fun to go back. It's awesome that some of these young kids in here (the Red Wings' locker room) get to play a game there.
"It's very similar to the Joe (Louis Arena). It's really loud. Now that they (the Islanders) came back again, the fans are a little more excited there, I've heard, and it's like a playoff atmosphere every night when they're playing there."
Nielsen went back to the comparison of Joe Louis Arena, where maybe the rink didn't have the creature comforts of the newer arenas around North America, but had a special place in the heart of the home fans.
"The Joe was old, too, but it's special to everyone in here," Nielsen said. "It was just the same (with Nassau Coliseum). There is history, they won four Cups and you still see those guys. When you talk to fans around there, that place meant so much to those people in the heyday when they were winning.
"It was old and it is old. But it's just like the Joe. It's a pretty cool place."
When the Wings and Islanders met last week, the game turned unexpectedly physical, especially after Niklas Kronwall crushed the Islanders' Anders Lee with an open-ice hit that forced Lee briefly out of the game to get stitches.
Dylan Larkin got into an altercation with Josh Bailey, after Bailey nailed Kronwall in the corner.
Some people feel the edginess could carry over into Saturday's game, although Blashill wasn't so sure.
"Every game has a life of its own," Blashill said. "We have to go win the hockey game. We can't worry about anything else. If they want to worry about extra-curricular stuff, then let them worry about it."
Blashill was more concerned about Friday's loss against Ottawa.
"We had an opportunity to win at home and we didn't win," Blashill said. "We have to go find a way to win."
ted.kulfan@detroitnews.com


text:
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We are born to strive. And to thrive. But strive towards what, exactly?
Innately gifted to succeed against adversity, we rise to the challenge again and again. We pick ourselves up, dust ourselves off, and weigh back in. Our biggest challenge is picking the right challenge.
Back in the day, when most people lived and died within 20 miles of where they were born, the challenges were straightforward: get enough food and fuel to keep body and soul together, and raise a family so you'd have someone to look after you in old age. If there was enough time and energy left over, have some fun. Career choices were limited. Communication involved people within walking or riding distance, or those you could reach by letter (with a delay of several weeks or months).
Fast forward to today: instantly accessible, we are awash in information and choices. We have choice overload; there is no end of what's available to buy, eat and do. We work in a rapidly changing job market within a highly complex, global economy where many occupations are being commoditized. To thrive, we have to be savvy, discerning and self-managing. And if our biggest challenge is picking the right challenge, we have endless distractions to keep us from landing on the right one.
Human brains have adapted over millennia, yet they still retain some core insecurities – those that kept us safe in less developed surroundings. As Dan Baker theorizes in "What Happy People Know", all humans carry around a deep uneasiness on two topics: "I don't have enough" and "I'm not good enough". Those two fears drive a lot of our choices, consciously or not. We think we are making rational decisions, but we are really just responding that age old, very deeply rooted concern about getting enough to eat and being accepted by a group, because for humans, safety comes in numbers; we are wired to seek social acceptance.
We are drawn like magnets to endless options, yet have difficulty processing it all, so we live by accident, following the law of unintended effects. We drive the extra few miles to get to the grocery store with aisles and aisles of choices, but buy the same things over and over, and then often fail to turn them into meals when we get home. It's an odd paradox, that of choice. The human brain is both attracted to choice and paralyzed by too many options. In the Harvard Business School Article "More Isn't Always Better," Barry Schwarz cites the 2000 study by psychologists Sheena Iyengar and Mark Lepper in which shoppers at an upscale food market were offered $1 off any jam they sampled. On one day, 24 jams were offered and on another day, only 6 jams were put on display. The large display attracted far more attention – people swarmed to it like bees, yet the smaller display resulted in 10x the number of purchases.
Why is that? Because making choices is one of the most exhausting things the brain can do. It's why we avoid it like the plague, as David Rock tells us in Your Brain at Work. Prioritizing involves several parts: understanding a new idea, making a decision, recalling items from memory, and inhibiting competing calls for our attention, all at once. No wonder we put off making decisions! They call for a ton of mental bandwidth and when we can't muster sufficient energy or capacity, we do nothing, or fall back on a habit or familiar non-choice.
Without a clearly defined purpose for our lives, without having thought through "what challenge am I best suited to overcome?", it is easy to fall into downward spirals, focusing on the wrong priorities, or no priorities at all – living by accident instead of living on purpose. Worse yet, looking for outside reasons for our own inner turmoil. Meanwhile, the problem is between our ears, not our neighbour's badly behaved dog or our unreasonable boss.
It's ironic that in developed economies, where many basic needs are readily met, we get ourselves so mired down, and make things so much more complicated than they need to be. We live life as a problem to be solved, rather an an experience to be enjoyed. Many of us are not experiencing greater contentment. Quite the reverse.
So what's the answer? If we are overwhelmed with options – too much of a good thing – what do we need to do? Two things: be discerning and be intentional. Our only true limitation is time. Each one of us has only 24 hours in a day, no matter how much we try to cram into it.
Today, our ultimate challenge is knowing what to leave behind, what to shut out, and what to ignore. The quality of our lives is more dependent on what we take out of them, rather than what we add to them; what we remove ourselves from, not what we involve ourselves in. FOMO drives a lot of behaviour, with unintended effects.
What exactly do we want to accomplish?
If all else fails and the world goes to hell in a handcart, what would we want to retain?
When our lives are over, what do we want to be remembered for?
Answer those questions. We need to determine what we are especially good at, what makes our hearts sing, why we are here. Align our activities and relationships with that. Limit our commitments to those things that move us in the direction of our purpose. Live life on purpose, with purpose.
This post is the first in a new eight part series offered by Authentic Impact. We hope you enjoy it!


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WBR001 Is God really calling me—and to what?
August 26, 2015 at 7 pm Eastern
A vocation? Who, me? Yes, YOU! We are all called by God, and in this premiere webinar from A Nun's Life Ministry, you'll discover the wider meaning of vocational calling -- it's not just for religious life!
Presenter Sister Teresita Richards, SND, will provide tips, techniques, and resources for growing your vocation from right where you are. She will talk about our baptismal call as it relates to vocation, and will identify practical ways you can discover your own calling through prayer and reflection.
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Sister Teresita has been a Sister of Notre Dame for 33 years. She holds a Bachelor of Arts in Religious Studies from Lourdes University, and a Masters in Pastoral Ministry from St. John's University in Collegeville Minnesota.
Sister Teresita is currently the Vocation Director for the Toledo Province of the Sisters of Notre Dame. In the past she has served as Pastoral Director at three catholic high schools and as a Theology teacher. In her free time she enjoys cooking, baking and camping.
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text:
Encyclopedias almanacs transcripts and maps
Jacob, Testament of
JACOB, TESTAMENT OF , apocryphal work based on Genesis 49. Reference to an apocryphal book of Jacob is perhaps to be found in the Apostolical Constitutions 6:16 which speaks of "apocryphal books of… the three patriarchs." A testament of Jacob is to be found together with the testaments of Abraham and Isaac in Arabic, Ethiopic, and Coptic. This work is a patchwork of biblical verses, based substantially on Genesis 47:29–49:1. It was doubtless composed as a supplement to the extant testaments of Abraham and Isaac, both of which, and especially the former, contain important original material. This apocryphal work appears in a Paris Greek manuscript of the 12th century as a separate work under the title "Testament of Jacob." In Jerusalem, there is an Armenian manuscript 939, in which Gen. 47:27–50:26 precedes the Testaments of the 12 patriarchs, and under the same title as the preceding. The Testament was a common literary form in the period of the Second Temple, and doubtless the fact that Jacob is the one patriarch to whom the Bible assigns a last testament played a role in this development.
M.R. James, Testament of Abraham (1892), 6, 131–61; idem, The Lost Apocrypha of the Old Testament (1920), 18f.; G.H. Box, Testament of Abraham (1927), 55–89; Stone, in: Revue des Etudes Arméniennes, N.S., 5 (1968), 261–70.
[Michael E. Stone]
Encyclopaedia Judaica
"Jacob, Testament of ." Encyclopaedia Judaica. . Encyclopedia.com. 16 Jan. 2023 <https://www.encyclopedia.com>.
"Jacob, Testament of ." Encyclopaedia Judaica. . Encyclopedia.com. (January 16, 2023). https://www.encyclopedia.com/religion/encyclopedias-almanacs-transcripts-and-maps/jacob-testament
"Jacob, Testament of ." Encyclopaedia Judaica. . Retrieved January 16, 2023 from Encyclopedia.com: https://www.encyclopedia.com/religion/encyclopedias-almanacs-transcripts-and-maps/jacob-testament
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Psalms, Book of
Genesis Rabbah
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Jacob, Rosamund (1888–1960)
Jacob, Patriarch
Jacob, Naomi Ellington (1889–1964)
Jacob, Naomi Ellington
Jacob, Mr. ("Jacob of Simla") (ca. 1850-1921)
Jacob, Merle (Lynn)
Jacob, Maxime
Jacob, Max
Jacob, Joseph M.
Jacob, James R.
Jacob, Jack Frederick Raphael
Jacob, Heinrich Eduard
Jacob, Gunther (Wenceslaus)
Jacob, Gordon (Percival Septimus)
Jacob, Christian 1955-
Jacob, Blessing of
Jacob, Berthold
Jacob, Benjamin
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text:
Vitamin D supplementation found to reduce COVID-19-related mortality, says study
By Angela Betsaida B. Laguipo, BSNJan 6 2021
As the coronavirus disease 2019 (COVID-19) pandemic evolves, many agents have been tested to see if they can reduce the risk of severe disease and death among infected patients. One of the widely explored agents that have been shown to reduce disease progression is vitamin D.
Study: The impact of vitamin D supplementation on mortality rate and clinical outcomes of COVID-19 patients: A systematic review and meta-analysis. Image Credit: Anastasiia Chepinska / Shutterstock
Caused by the severe acute respiratory syndrome coronavirus 2 (SARS-CoV-2), COVID-19 is primarily a respiratory illness but is known to affect other parts of the body, from the gastrointestinal tract to the central nervous system. It can also overwhelm the immune system in severe cases, hence the scientific interest in whether immune-boosting supplementation can help fight off the infection.
Researchers at the Tabriz Health Services Management Research Center, Tabriz University of Medical Sciences, and the Shahid Beheshti University of Medical Sciences showed that prescribing vitamin D supplementation to COVID-19 patients seems to decrease the mortality rate, the severity of the disease, and serum levels of inflammatory markers.
The coronavirus pandemic has spread to 191 countries and regions, with over 86.47 million cases and 1.86 million deaths. With the global spread of the SARS-CoV-2 and rise in infection and mortality toll, efficient management of the illness has become an absolute priority.
Many scientists have since studied various agents to determine those that can prevent disease progression and COVID-19-related death. Several studies have suggested the positive impact of vitamin D on COVID-19 patients.
Vitamin D is a fat-soluble micronutrient that could facilitate the function of the immune system. In the past, studies have shown the potential link between vitamin D deficiency and severe outcomes in COVID-19 patients
This systematic review aims to evaluate the effects of vitamin D supplementation on clinical outcomes and mortality rate of COVID-19 patients.
Even before the current pandemic, the immunomodulatory role of vitamin D has been studied in the treatment of other upper respiratory infections. Vitamin D can help regulate the immune system, thanks to receptors found in various innate immune system cells. The vitamin can also suppress the adaptive immune response in the affected lung epithelial cells, hence, preventing the pro-inflammatory agents' harm on the affected sites.
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Lastly, vitamin D plays an important role in protecting against the direct damage of the inflammatory factors, which are released during viral diseases.
The study, published on the pre-print medRxiv* server, highlights the role of vitamin D supplementation in reducing the risk of severe illness and death due to COVID-19.
The team conducted a systematic review to arrive at the study findings, including research from PubMed, Web of Knowledge, Scopus, Embase, The Cochrane Library, and Ovid up to December 16, 2020.
The team has found that in three of the studies, the survival and mortality rate of the patients were studied, revealing that a significantly lower mortality rate among the intervention groups compared to the control groups. Moreover, two of the studies showed that the clinical outcomes based on the World Health Organization (WHO) Ordinal Scale for Clinical Improvement (OSCI) score for COVID-19, were lower in patients who received vitamin D supplementation.
Further, a study showed that there is a lower rate of intensive care unit (ICU) admission among those who received vitamin D, and another study reported a marked decrease in serum levels of fibrinogen, which increases in levels in any condition that causes inflammation or tissue damage.
Vitamin D supplementation seems to decrease the mortality rate, the severity of the disease, and the inflammatory markers' levels among the COVID-19 infected patients, leading to a better prognosis and increased survival… More studies should be conducted to determine the optimum dosage and route of vitamin D supplementation and further investigate the potential prophylactic effects."
To date, many countries report surging COVID-19 cases, including the United States and the United Kingdom, with over 21 million and 2.78 million cases, respectively. Since October 2020, the UK has experienced a second wave of the pandemic, with a new variant that is thought to be around 70% more infectious now circulating.
Knowing more about potential therapies to reduce the risk of death tied to COVID-19 is important as the world waits for the full deployment of COVID-19 vaccines.
*Important Notice
medRxiv publishes preliminary scientific reports that are not peer-reviewed and, therefore, should not be regarded as conclusive, guide clinical practice/health-related behavior, or treated as established information.
COVID-19 Dashboard by the Center for Systems Science and Engineering (CSSE) at Johns Hopkins University (JHU) - https://gisanddata.maps.arcgis.com/apps/opsdashboard/index.html#/bda7594740fd40299423467b48e9ecf6
Nikniaz, L., Akbarzadeh, M.A., Hosseinifard, H., and Hosseini, M.S. (2021). The impact of vitamin D supplementation on mortality rate and clinical outcomes of COVID-19 patients: A systematic review and meta-analysis. medRxiv. doi: https://doi.org/10.1101/2021.01.04.21249219, https://www.medrxiv.org/content/10.1101/2021.01.04.21249219v1
Posted in: Medical Science News | Medical Research News | Disease/Infection News | Healthcare News
Tags: Coronavirus, Coronavirus Disease COVID-19, Gastrointestinal Tract, Immune Response, Immune System, Immunomodulatory, Inflammation, Intensive Care, Mortality, Nervous System, Pandemic, Research, Respiratory, Respiratory Illness, SARS, SARS-CoV-2, Severe Acute Respiratory, Severe Acute Respiratory Syndrome, Syndrome, Vitamin D, Vitamin D Deficiency
Angela Betsaida B. Laguipo
Angela is a nurse by profession and a writer by heart. She graduated with honors (Cum Laude) for her Bachelor of Nursing degree at the University of Baguio, Philippines. She is currently completing her Master's Degree where she specialized in Maternal and Child Nursing and worked as a clinical instructor and educator in the School of Nursing at the University of Baguio.
Laguipo, Angela. (2021, January 06). Vitamin D supplementation found to reduce COVID-19-related mortality, says study. News-Medical. Retrieved on January 18, 2021 from https://www.news-medical.net/news/20210106/Vitamin-D-supplementation-found-to-reduce-COVID-19-related-mortality.aspx.
Laguipo, Angela. "Vitamin D supplementation found to reduce COVID-19-related mortality, says study". News-Medical. 18 January 2021. <https://www.news-medical.net/news/20210106/Vitamin-D-supplementation-found-to-reduce-COVID-19-related-mortality.aspx>.
Laguipo, Angela. "Vitamin D supplementation found to reduce COVID-19-related mortality, says study". News-Medical. https://www.news-medical.net/news/20210106/Vitamin-D-supplementation-found-to-reduce-COVID-19-related-mortality.aspx. (accessed January 18, 2021).
Laguipo, Angela. 2021. Vitamin D supplementation found to reduce COVID-19-related mortality, says study. News-Medical, viewed 18 January 2021, https://www.news-medical.net/news/20210106/Vitamin-D-supplementation-found-to-reduce-COVID-19-related-mortality.aspx.
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Just my experience with Vit. D and Covid 19. In Feb 2020 for a completely
Larry Jessip says:
Just my experience with Vit. D and Covid 19. In Feb 2020 for a completely unrelated problem my urology dr. did a test for my Vit. D levels and found them low. He proscribed 50,000 IU Vit. D2 pills at once a week for 2 months and once a mo. following. In April I had a case of Covid toe. Middle toe on each foot turned black. Looked like Gangrene actually. But then they came out of it. I think the Vit. D supplement stopped the Covid before it became severe.
Also in December one morning I woke up and stumbled into the kitchen for breakfast and realized after a few bites I had no sense of taste or smell. Everything was bland. So figuring it was a early Covid symptom I took one of the last D2 supplements I had. At lunchtime I was fine. I suspect myself that megadoses of Vit. D slow down or stop outright the virus,
Artificial Intelligence in Healthcare


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text:
Myadestes elisabeth o  ruiseñor , especie de ave endémica de Cuba, pertenece a la familia Turdidae del orden Passeriformes. Es la más melodiosa de las aves cubanas.

Nombres
Myadestes del griego, significa "que come moscas", elisabeth por ser dedicado a Elisabeth, y retrusus del latín "escondido". En inglés es llamado Cuban solitaire.

Distribución
Existe en bosques semicaducifolios o de pinos que tengan barrancos cársticos. Estos solo en las montañas de la Sierra de los Órganos y de la Sierra del Rosario en Pinar del Río en el occidente, y en las de la Sierra Maestra y del Macizo de Sagua-Baracoa en el oriente. Está ausente en las montañas de Guamuhaya en el centro de la Isla Cuba. En la Isla de la Juventud tenía una subespecie descrita en 1905, que no se ha vuelto a encontrar, por lo que se considera extinta.

Subespecies
M. e. elisabeth (Lembeye, 1850) de la Isla de Cuba.
M. e. retrusus Bangs & Zappey, 1905 de Isla de la Juventud, está extinta.

Descripción
Mide cerca de 19 cm de largo. El dorso es de color gris-oliváceo o gris-parduzco. Abajo desde la garganta a la base de la cola tiene color blanco-grisáceo que hacia el medio del vientre y de la garganta se vuelve blanco. Un anillo blanco rodea al ojo, y este es castaño. El pico es grisáceo. A los lados del pico tiene bandas como bigotes color castaño. Los extremos de las plumas exteriores de la cola tienen bordes blancos. Las patas son amarillas. Ambos sexos son similares. El inmaduro tiene el dorso más oliváceo y muy levemente punteado de castaño ventralmente. Se posa en las ramas para realizar su canto, que no cambia y no es penetrante. Se parece a la fricción de un dedo mojado en el borde de una copa de cristal. Se alimentan de frutos, semillas e insectos que busca en vuelo entre los árboles.

Nido
Anida entre mayo y julio en hoyos entre las rocas calcáreas o en huecos de árboles, cuyas aberturas suelen estar protegidas por las hojas de bromeliáceas. Dentro de estos huecos hace como una copa con paja, fibras y raíces que recubre con musgos. Pone dos o tres huevos blanco-verdosos con pequeños puntos negros o pardos.

Véase también
 Aves en Cuba

Referencias
 Blanco, P. (2001). El tenor de la montaña. Flora y Fauna 5(1):34-35. (Empresa Nacional para la Conservación de la Flora y la Fauna, Cuba).

 García, F. (1987). Las Aves de Cuba. Especies endémicas. Subespecies endémicas. Tomos I y II. Editorial Gente Nueva, La Habana. 207 pp.

 Garrido, O.H.; Kirkconnell, A. (2000). Birds of Cuba. Helm Field Guides, Londres. 253 pp.

elisabeth
Aves de Cuba
Fauna endémica de Cuba
Animales descritos en 1850
Taxones descritos por Juan Lembeye

redpajama_set_name: RedPajamaWikipedia


Another new release from today is the latest video from Upfront, taken from his upcoming double A-side. Look At This is produced by fellow Split Prophets member, Badhabitz and the video was made by regular collaborator, Chris Lucas. The single is set for release on 29 March, with pre-orders available now via the SP Bandcamp.
Crown Of Thornz is the latest leak from the new Billy Whizz EP, Always Finding Trouble. This time round the Bristol-based producer draws on more local talent from Verbal Highz representative, G00SE and his rapid fire lyricism. The track is accompanied by another solid video from regular collaborator, Chris Lucas. The EP is out today on AFT Raps.
Fresh cut from the impossible to pigeonhole Bristol crew, D£DW8, as they ready the release of their next EP. ROB£RT RIPL£Y is as looming and ominous as you would expect from the duo, made even more so by the fear and loathing, bad trip style video from Chris Lucas. The EP, AR£ W£ GOD$ Y£T?, is set for release on 8 February, with pre-orders and merch available via Bandcamp.
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Twizzy is due to return to the fore with a new album very soon and so to get us in the mood, yesterday saw the release of a new video from the project. Report is produced by fellow 3HB member, Billy Whizz and the highly entertaining video was made by Chris Lucas. The album, Old Dog No Tricks, is set for release on 10 September on AFT Raps.