text:
Four Ebola-response workers killed in attacks, forcing withdrawal from critical DRC region
By Helen Branswell @HelenBranswell
FABRICE COFFRINI/AFP/Getty Images
Four people working for the Ebola containment effort in the Democratic Republic of the Congo were killed overnight Thursday in attacks on two locations where response teams were operating. Another six people were injured in the violence, forcing the World Health Organization and partner agencies to withdraw staff from a strategically critical location, a senior WHO official said.
The attack was the deadliest yet during a 17-month outbreak in which response workers have been frequently targeted for violence, said Dr. Mike Ryan, executive director of the WHO's emergencies program, speaking with reporters from Geneva.
WHO Director-General Tedros Adhanom Ghebreyesus condemned the attacks.
"These constant attacks must stop," Tedros, as he is known, said in a video the WHO distributed on social media. "We call on everyone who has a role to play to end this cycle of violence."
Three of the deaths occurred in an ambush of an accommodations camp housing 189 response workers at a place called Biakato Mines. In a separate attack at a coordination office in Mangina, a policeman guarding response workers was killed.
Two Ebola treatments yield 'substantial decrease' in mortality, landmark trial shows
"We are heartbroken that they died as they worked to save others," Ryan said.
To date, he said, there have been 386 attacks on response workers, with at least seven deaths and 77 injuries. Many other times, response teams have been caught in the crossfire between the multiple rebel groups that operate in the Ebola-affected area and the government and United Nations. peacekeeping troops in the region.
The latest violence dashes hopes that the outbreak might be in its final throes.
Months of intense effort to try to stop transmission of Ebola in the provinces of North Kivu and Ituri were paying off. There have been fewer cases to date in November than any month since the outbreak began. For perhaps the first time, the response teams had a good handle on where the virus was spreading, with 98% of cases occurring in two chains of transmission. Containment efforts were working.
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"We were in a position where we were really starting to get on top of this virus," Ryan said. "Ebola was retreating. And now it is likely to resurge. … When responders aren't in the community, the virus spreads undetected and uninterrupted."
This is the second largest Ebola outbreak ever, with more than 3,300 cases and 2,200 deaths. Though this one is almost 10 times smaller than the West African outbreak of 2014-2016, Ryan and others have called the North Kivu-Ituri outbreak the most difficult ever, occurring as it is in a region that is home to upwards of 8 million people and that has been mired in conflict for more than two decades.
Despite use of a highly effective vaccine and drugs that have been proven in this outbreak to reduce deaths, the response workers have struggled to gain the acceptance and assistance of people in affected communities.
Community resistance remains a problem at Lwemba, a health zone near Biakato Mines that is currently the place where most transmission is occurring. Even before the violence, response teams had encountered difficulties in Lwemba. The camp at Biakato Mines housed people who had been working in Lwemba. "Now without our teams on the ground, we have even less access," Ryan said.
Violence earlier in the outbreak fueled spread of the Ebola virus, with a massive surge of cases in March, April, and May after multiple late February attacks on response operations, including two treatment centers that were razed.
Ryan was clearly bracing for the prospect that this pattern is about to be repeated. "We've seen the impact previously of long-term interruptions of operations," he said, adding that it's impossible to predict at this point how soon full response operations will be able to resume.
Helen Branswell
Senior Writer, Infectious Disease
Helen Branswell covers issues broadly related to infectious diseases, including outbreaks, preparedness, research, and vaccine development.
helen.branswell@statnews.com
@HelenBranswell
infectious diseaase
Why are they doing this? Has anyone interviewed them?
Rachele Raanan says:
How many American troops have been active in the DRC? How many of those has President Trump withdrawn?
craig cornish says:
As President Trump withdraws our military from around the world signaling our unwillingness to send our troops "overseas," expect greater power and control worldwide of armed bandits, criminals, and nations.
By Damian Garde and Lev Facher
FDA approves an Ebola vaccine, long in development, for the first time
By Helen Branswell
FDA approves Amarin fish-oil-derived drug to prevent heart attacks and strokes
By Matthew Herper
FDA changes course on Sarepta's once-rejected muscular dystrophy drug
By Damian Garde
New data show Bluebird, Bristol CAR-T drug could be pioneering myeloma therapy. Competitors may be close behind
By Adam Feuerstein and Damian Garde


The new BenQ W5000 home theater projector starts shipping this month. This 1080p resolution, single-chip DLP projector, delivers an extremely sharp, high contrast image. It is priced at $4,995, which is somewhat higher than a number of competitive 1080p projectors on the market at this time. However, it comes through with a dazzling picture that justifies the price.
Excellent contrast and black level. Don't let the 10,000:1 contrast rating fool you. That is a relatively low number in today's competitive market where contrast specifications range from 15,000:1 up to 50,000:1. But in any given scene, the real contrast you experience on the W5000 is in many instances higher than that of competing models with much higher contrast ratings.
Extremely sharp image. The W5000 is more razor sharp than any 1080p product we've seen yet. This is particularly apparent in close-up facial images, where every imperfection in the skin is resolved in extraordinary detail. The projector's sharpness may be attributed in part to exceptionally high optical precision in the lens.
Three-dimensional image. The high native contrast and exceptional sharpness combine to produce a visual depth, or apparent three-dimensionality, that is well above average as compared to other 1080p models on the market today.
Bright in Brilliant Color mode. There are two options for Brilliant Color, on or off, and the setting you choose has a large impact on light output. When it is on, the ANSI lumen output in Cinema mode ranges from 350 lumens to 600 lumens depending upon where you set the manual iris. Thus it is brighter than average when compared to other home theater projectors. However, turning Brilliant Color off reduces average lumen output by about about 32% in the highlights and 50% in mid-tones. The advantage of turning Brilliant Color off is that it makes digital noise levels much less apparent in scenes that are particularly susceptible to noise. This is a key trade-off, and the user will make a decision to run with Brilliant Color either on or off depending on screen size and gain, ambient light, and the viewer's sensitivity to noise artifacts.


Smart Trim Forskolin is a simple to use Coelus Forskolin formula that helps to prevent obesity and overweight issues to manage healthy formula. You don't need to incline toward slimming down or practices in view of its multi advantage of forskolin which is found in the tuberous underlying foundations of Coleus. There are dreamlike different advantages of this weight reduction supplement however it serves for the most part for fat consuming procedure and keeping the body fit as a fiddle and sizes. By taking out additional body weight and controlling fat strands it destroys a few passing causing illnesses normally. At the point when the body begins putting away fat than you require then it begins creating strange wellbeing conditions inside in your body. The fixings are single yet compelling yet there are a few different substitutes to help fat weight reduction framework without harming your body in any negative way. This is a dietary pill supplement which should be gone up against normal routine to render regular weight reduction. So what's you work here is to just take the pills since it likewise accompanies craving suppressant recipe to control a few realities of gorging or passionate dietary issues.
The item offers out with the little name of the producer and the creation unit is arranged in Langley Virginia U.S.A. The creators guarantee much about this forskolin weight reduction supplement as due to its normal fixings it advances totally common weight misfortune procedure to keep up a sound constitution. They have likewise given data about Ingredients and splendidly weight administration arrangement. Presently in the event that you are as yet disparaging your hefty condition then you are propelling yourself towards grave perils. By dealing with your body weight and keeping fit as a fiddle you can really keep these medical issues just you are fat doesn't mean you can never dispose of those greasy sections. I know for the vast majority of the general population weight reduction have never been a simple decision to set aside a few minutes taking procedure and moderate fat consuming procedure which barely give you palatable outcomes. Conventional techniques incorporate running, works out, appropriate eating regimen, no fat eating low carbs and numerous all the more however let me approach you are these strategies helpful for today's present-day way of life? The greater part of us would likely say a major NO! on account of time taking procedure and we scarcely give whenever to our physical make-up upkeep on a typical day by day schedule.
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Yet, here we are acquainting a superior alternative with being fit & alluring without much concern. Who doesn't need to look fit as a fiddle & hot? So it's the very wish of everybody except the arrangements or choices are not what anybody could take after in view of exorbitant consumption, intrusive supplement, and low eating regimen sustenance. It looks essentially like an advanced way of life so is it conceivable to look attractive furthermore, fit as a fiddle in moment way? I am extremely upbeat to reveal to you that now it's totally conceivable with Smart Trim Forskolin a weight reduction framework as of late presented after the arrangement of inquires about. The thing what made it so unique is its unadulterated common Ingredients and forskolin. Let's discover more about this forskolin weight reduction supplement.
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We interview Mr Tan Cheng Siong, recipient of the President's Design Award 2012, to find out more about Pandan Valley Condominium, one of the first projects of its kind in Singapore.
Q: What motivated the development and design of Pandan Valley Condominium?
TCS: Pandan Valley is, similar to Pearl Bank Apartments, a part of the government's effort to beat the land shortage problem. At the time there were only two types of property: the first typology was landed property various estates, in the style of settlements in America and Australia. The second was HDB apartments. These two distinct typologies catered almost exclusively to the higher and lower income groups respectively.
Then, the government proposed to stop further development of landed property sometime in the 1970s, and this upset the people in that demographic, who now had to look to apartment living. But in those days, apartments were not appealing; they came in small point blocks, and there wasn't space for community and family. This is when the idea for building Condominiums began, so that community spaces could be organised into one project, something that residents could own and share.
The challenge was to entice landed-property customers to high rise living, by offering more than what a terrace house can provide. While the terrace house had land, yards, and multiple storeys, Condominiums offered swimming pools, shops, basketball courts, playgrounds and landscaping, which terrace houses could not get. Also as a strategy to bridge the gap, the units had to be big, almost comparable to that of terrace houses, ranging from 150-250 square metres, a good size for families.
Q: As one of the first developments of its type, what was the public reception like?
TCS: At the time of its release, Singaporeans were not interested. They still wanted their terrace houses, and made many excuses not to buy in to the Condominium concept. The initial few months were difficult, with slow sales despite newspaper announcements and articles. Even the faculty at NUS were resistant, a market we thought we could tap on because of its proximity.
But overnight, with the sudden economic boom, Pandan Valley became popular. I was told that now it's very popular for expatriate rental, and some visiting professors at NUS who were assigned housing there were very impressed, asking me why I didn't do more of the same. That gave me a lot of satisfaction.
Since then, I've seen condominiums built on a small pocket of land – those are not condominiums, those are just apartments with a swimming pool!


text:
Sephora launches Addison Rae's Item Beauty and Hyram Yarbro's Selfless
by Adink January 10, 2023
Influential brands Item Beauty and Selfless by Hyram are being pulled from Sephora.
Addison Rae launched Item Beauty in 2020 and Hyram Yarbro launched Selfless in 2021.
Industry sources told Insider that influencer-founded beauty brands have seen mixed results over the years.
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It appears that the beauty products of two popular influencers – Addison Rae and Hyram Yarbro – may no longer be available in shopping malls in America.
An insider has confirmed from a well-informed industry source that Sephora will be pulling Item Beauty Addison Rae and Selfless by Hyram Hyrama Yarbro from its shelves. While Item products are still available for purchase on the beauty giant's website, Selfless products are nowhere to be found.
Representatives for Rae and Yarbro did not respond to Insider's requests for comment prior to publication. Sephora also did not respond to an Insider's request for comment.
Item Beauty was launched in August 2020 by Rae, who gained internet fame by recreating viral dances on TikTok. Her brand positions itself as a "clean, science-backed" line, free of chemicals found in traditional cosmetics.
While Item originally launched as a direct-to-consumer brand – selling to customers via an online store – it entered into an exclusive partnership with Sephora in August 2021.
Yarbro, a beauty vlogger whose skincare teaching videos have earned him millions of followers on YouTube and TikTok, launched his Selfless by Hyram skincare line directly at Sephora in June 2021.
Selfless advertises itself as gentler than other skincare lines on the market, offering products with lower concentrations of chemicals such as retinol and salicylic acid than competitors.
Rae and Yarbro's fame suggested that their brands were poised to succeed with their target audience – the millions of fans who followed them on TikTok, YouTube and Instagram. However, a well-informed source in the beauty industry said Sephora customers did not connect with either brand, according to correspondence seen by Insider.
"It's sort of a fallen star," said Shannon Coyne, co-founder of Asbury and Coyne Advisors, noting that the brand's growth has slowed significantly in recent months.
"You lose TikTok and Instagram and you're out of business"
Hyram Yarbro has millions of followers on social media, but his brand isn't doing so well at Sephora.
That companies like Item and Selfless are struggling is no shock to beauty connoisseurs. Over the past three years, Insider has discovered that as many as 25 celebrities and influencers have launched beauty or skincare brands, and only the best of them have been able to climb to the top.
The news of Item and Selfless comes just a week after Morphe Cosmetics, a beauty chain known for its collaborations with influencers, announced it would close all of its stores in the US.
"Celebrities rely on their surname, but there has to be more than just a name, because they fall in and out of favor," Coyne said.
Item products had been heavily discounted at Sephora for weeks, signaling that the beauty chain was likely looking to get rid of stock.
Moreover, Rae herself has been relatively silent about the brand on social media. She last posted an entry about the brand on her personal Instagram feed on September 29, 2022.
"Sephora is a retailer, not a manufacturer," Jeffrey Ten, a beauty industry veteran and president of Global Indie Brand Development, told Insider. "They're selling you space. They don't build your brand. You are responsible for building your brand."
Both brands also saw a drop in social media followers. Rank has gone from 355,000 Instagram followers in November 2021 to 334,000 today, according to Social Blade data. And Selfless has grown from 196,000 Instagram followers in November 2021 to 170,000 today, according to Social Blade.
"You lose TikTok and Instagram and you're out of business," Ten said.
TikTok stars like Rae have struggled to form lasting bonds with their fans. While it's a good platform for discovery, there's a limited ability to sell in-app products, agents and developers told Insider.
Celebrity endorsed brands include bright spots like Selena Gomez's Rare Beauty
That's not to say that all celebrity-founded beauty brands fared poorly.
Dae Hair, founded by Instagram influencer Amber Fillerup Clark, is touted by investors and analysts as an example of a brand that resonates well with customers.
In December, the brand announced an $8 million Series A deal led by Verity Venture Partners – a rare bright spot in a time when many investors have pulled out of new deals in the consumer space.
Similarly, singer Selena Gomez's Rare Beauty has been cited by several industry sources as another great example of celebrity-backed beauty success.
The brand launched at Sephora in September 2020 and has seen steady growth. Rare Beauty website traffic increased over 70% between November 2021 and November 2022, according to data provided to Insider by analytics platform Likeweb. One industry expert said the brand is likely doing well at Sephora as well.
Whether a beauty brand will stand the test of time ultimately seems to depend on how much personal capital a celebrity or influencer is willing to invest in it.
"Celebrities and influencers with their embedded communities can be great brand boosters and early adopters, but just being a talent-driven brand today is not enough," one investor focused on beauty startups told Insider.
"Consumers are extremely astute and well-informed and the market is highly competitive; In order to truly resonate with the consumer, it is imperative that brands are able to stand on their own in terms of product quality, effectiveness, value and brand positioning," they added. .
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redpajama_set_name: RedPajamaCommonCrawl


The ALLIANCE has four meeting rooms and four hot desk available for hire by our members.
The ALLIANCE has 4 meeting rooms that are available for hire between 9 am and 5 pm, Monday to Friday (excluding office closures).
Hot desks are available for use by our members between 9.30 am and 4 pm, Monday to Friday. Members should contact the ALLIANCE on 0141 404 0231 or email roombookings@alliance-scotland.org.uk to book a hot desk.
If you wish to make use of the facilities and are not already a member, please see our membership section for information on how to become a member organisation or individual supporter of the ALLIANCE, or contact us at membership@alliance-scotland.org.uk. If you complete this booking form and are not already a member, the ALLIANCE will use the information provided to contact you to establish whether you want to join as a member.


The coal industry came to Kent in 1912 and Betteshanger Colliery started production in 1927 and was the last pit in Kent, closing in 1989. Rugby was first played at Betteshanger (The Colliers) in 1936, and was disbanded in 1939 at the outbreak of World War II. The club reformed in season 1948-49 when the Yorkshire and Lancashire rugby league enthusiasts combined with the Welsh exiles. The Colliers were highly respected throughout their 57-year history winning county 7's championship twice, as well as Kent 2, Kent 1 and London Div.4 SE.
Following closure of Betteshanger Colliery and the RM barracks both clubs found themselves competing for an ever-decreasing pool of players and in 2004 the two clubs engaged in discussion about amalgamation. Although they had very different backgrounds and histories the joining of the clubs was the right decision and in 2006 Deal and Betteshanger Lions RFC was born, and has proved very successful.
The Club is extremely proud that it is one of the very few sides at Level 7 that play predominantly home grown talent blended with a few players that have moved into the area. The club endeavours to field three regular senior sides. The club also has a thriving mini's and junior section where we coach all children from 6 to 17years. The coaches work hard to ensure the clubs ethos of teamwork, respect, enjoyment and discipline, are values to be aspired to at all times. Such is our commitment to promote the RFU values, Deal and Betteshanger was one of the first Rugby Clubs in East Kent to be awarded the coveted "Seal of Approval".


Taking over the family business isn't the easiest task, especially in music. But Arwen Lewis is fearlessly following in father's footsteps in more ways than you'd expect. The singer/songwriter is set to release her debut album soon, and today, we're excited to premiere a song from it. In case you aren't aware, Lewis' father, Peter, played in Moby Grape, and she is taking a bold step in covering his band's hard rocking "Omaha" in addition to a number of additional tracks by the San Francisco rock legends. Not that she finds it too daunting or anything like that.
"This song always made me feel like someone cared about me when I listened to it, and in the little time that I spent around Skip, I got the sense that he wanted people to know that he cared about them," Arwen says. "'Omaha' is a part of this record because of that message, this music has been made to be shared with everyone who wants to experience it."
"Skip had more energy than anyone I'd ever seen on stage, and people loved him," Peter adds. "When he wasn't performing he would fall into trances unexpectedly sometimes. At least the way he explained it to me, in the middle of one of these he had an epiphany about the band being together forever and wrote the song."
Arwen Lewis' debut Arwen will be released on October 2 via Omad Records.


text:
import datetime
import logging
import csv
import os
import time

import yaml

import dbmodels

# Set logging level
logging.basicConfig(level = logging.ERROR)

# Read config
with open("/etc/politraf/config.yaml", 'r') as stream:
    try:
        config = (yaml.safe_load(stream))
        OTX_KEY = config['otx_key']
        TRAF_FILE = config['traf_file']
        url = config['db_url']
        name = config['username']
        passw = config['password']
    except yaml.YAMLError as e:
        logging.error(e)
    logging.info("Config is OK")

# Init clickhouse
try:
    db = dbmodels.Database('politraf', db_url=url, username=name, password=passw, readonly=False, autocreate=True)
except Exception as e:
    logging.error(e)

class TReceiver():

    def __init__(self, traf_file):
        try:
            self.gs = traf_file
        except Exception as e:
            logging.error(e)

    def write_traf(self):
        start = time.time()
        logging.info("Processing traffic ...")
        with open(self.gs, newline='') as csvfile:
            spamreader = csv.reader(csvfile, delimiter=',', quotechar='|')
            os.remove(self.gs)
            for event in spamreader:
                try:
                    today = event[0]
                    timestamp = event[1]
                    protocol = event[2]
                    src_addr = event[3]
                    src_port = event[4]
                    dst_addr = event[5]
                    dst_port = event[6]
                    qry_name = event[7]
                    db.insert([dbmodels.CONNStats_buffer(event_date=today, timestamp=timestamp, protocol=protocol, src_addr=src_addr, src_port=src_port, dst_addr=dst_addr, dst_port=dst_port, qry_name=qry_name)])

                except Exception as e:
                    logging.error(e)
        #os.remove(self.gs)
        end = time.time()
        time_to_end = end - start
        print('Time to load traffic ', time_to_end)

def main():

    # Create a receiver
    t_receiver = TReceiver(traf_file=TRAF_FILE)

    # Write IOC files
    t_receiver.write_traf()

if __name__ == '__main__':
    main()

redpajama_set_name: RedPajamaGithub


Last weekend, protesters led by the People's Alliance for Democracy (PAD) vowed to mobilize hundreds of thousands of people in the streets to push for the ouster of Thailand's elected government. What happened during the past few days? What did the protesters accomplish?
Monday – Parliament building was occupied.
Thursday – Don Mueang Airport takeover. Bangkok's air travel completely paralyzed.
Thousands are stranded in the airports. Cab drivers couldn't enter the premises. Thailand's Prime Minister has already declared a State of Emergency in the airports. There are coup rumors; PAD wants the military to intervene to end the political crisis.
The provocative actions of PAD have angered many residents and visitors. Even mainstream media in Thailand, which used to be sympathetic to PAD, are now criticizing the group.
Thailand's tourism industry is now suffering. Some Thai residents are even advising their friends not to visit Thailand until next year.
"Unbelievable. I don't think my hubby will be able to fly to Bangkok and then further south this weekend… Seems really hard to take a bus or the train as well! This is just stupid.
"Isn't the airport supposed to be like the most secured place in any country? And yet, these mustards can just easily barge in.
"What if they were a group of armed militants instead of aunties with clappers?
"Thanks to the PAD's latest airport-paralyzing move, my friend's cousin who was in a brutal car accident in Surat-thani can't be flown back to Bangkok for proper medical care and instead lies in a coma in an inept hospital down south with a fractured skull, broken ribs and broken arm.
PAD believes the Prime Minister is a puppet of ousted leader Thaksin Shinawatra. On the other hand, PAD is accused of having close ties to some sections of the elite and military. PAD is popular among the urban middle class.
"What is really astonishing to me is how the foreign embassies in Bangkok have remained silent over this crisis, except for worthless travel advisories.
It all started with the military coup in 2006, then the military sided with PAD and the elites who saw Thaksin as a blocking stone to their money-making opportunities. They thought that with Thaksin removed, they could make Thailand their own kingdom. Now they knew the coup did not help at all. The military knew it was a mistake but just continued to let this mistake become bigger. I believe if Thaksin is allowed back to govern, he can solve Thailand's present problem. There will be bloodshed, but it is inevitable. Unless Thailand wants these lawless, military-backed terrorist-like PAD protesters to call the shots whenever they do like a particular government, it is now time for the King and the ordinary people of Thailand to step in and assist Thaksin to come in and help resolve this growing issue.
May God bless the King and the people of Thailand.
It all started with the military coup in 2006, then the military sided with PAD and the elites who saw Thaksin as a blocking stone to their money-making opportunities. They thought that with Thaksin removed, they could make Thailand their own kingdom. Now they knew the coup did not help at all. The military knew it was a mistake but just continued to let this mistake become bigger. I believe if Thaksin is allowed back to govern, he can solve Thailand's present problem. There will be bloodshed, but it is inevitable. Unless Thailand wants these lawless, military-backed terrorist-like PAD protesters to call the shots whenever they do not like a particular government, it is now time for the King and the ordinary people of Thailand to step in and assist Thaksin to come in and help resolve this growing issue.
This is an SMS I got in the middle of the night from our reporter at the Airport. I think it says it all.


text:
Scarlett Johansson as Black Widow is inspired by the #MeToo movement
Scarlett Johansson, the American actress, seems inspired by #MeToo for women, giving them the strength to stand up and say what they want.
The tale of her next Marvel cinematic-universe super heroin picture, as reported by the 36-year-old actress Scarlett Johansson, began to take form in 2017 when a few women confronted one of the most prominent film producers in Hollywood.
Scarlett Johansson remarked that the authors quickly compared the leading character of the Black Widow's film with other real-life women who speak up following abuses from the leadership.
"We must remark on the amazing women's movement supporting other women, and come across these shared trauma experiences by truly supporting and coming out," she said in an interview with Yahoo.
Read Also: The 'Black Widow' of Scarlett Johansson is offered via streaming service for $30
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Walls360 created a collection of custom promotional graphics for BEGSONLAND BRANDING + MARKETING in LAS VEGAS.
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text:
\section{Introduction}
Let $\Omega$  be a locally Lipschitz domain in ${\rm I\hspace{-0.2em}R}^{2}$  and define $Nf = \nabla \cdot Tf = {\rm div}\left(Tf\right),$  where  $f\in C^{2}(\Omega)$  and
$Tf= \frac{\nabla f}{\sqrt{1+\left|\nabla f\right|^{2}}}.$
Let $H\in C^{1}(\overline{\Omega})$  satisfy the condition
\begin{equation}
\label{(16.60)}
\left|\int_{\Omega} H\eta \ dx\right| \le \frac{1}{2} \int_{\Omega} |D\eta|\ dx \ \ \ \ {\rm for \ all \ } \eta\in C^{1}_{0}(\Omega)
\end{equation}
(e.g. \cite[(16.60)]{GT}).
Here and throughout the paper, we adopt the sign convention that the curvature of $\Omega$  is nonnegative when $\Omega$  is convex.
Consider the Dirichlet problem
\begin{eqnarray}
\label{eq:D}
Nf & = & 2H  \mbox{  \ in \ } \Omega    \\
f & = & \phi  \mbox{ \ on \ } \partial \Omega.
 \label{bc:D}
\end{eqnarray}
We wish to understand the boundary behavior of a solution of (\ref{eq:D})-(\ref{bc:D}).

If $\partial\Omega$  is smooth and the curvature of $\partial\Omega$  is greater than or equal to $2|H|$  at each point of $\partial\Omega,$
then this problem is well-posed.
If $\phi\in C^{0}(\partial\Omega),$  then there exists a unique solution of (\ref{eq:D})-(\ref{bc:D}) in $C^{2}(\Omega)\cap C^{0}(\overline{\Omega})$
and $f=\phi$  on $\partial\Omega$  (\cite[Theorem 16.10]{GT}).
On the other hand, one can choose a distinguished point ${\mathcal O}\in\partial\Omega$  and use the ``gliding hump'' construction as in \cite{Lan:89}
and \cite[Theorem 3]{LS1}, in conjunction with \cite[Theorem 2]{NoraKirk1},
to prove that there exist $\phi\in L^{\infty}(\partial\Omega)$  such that the (unique) variational solution $f$  of (\ref{eq:D})-(\ref{bc:D}) is
in $C^{2}(\Omega)$  but $f$  is discontinuous at ${\mathcal O}$  and none of the radial limits of $f$  at ${\mathcal O}$  exist.

When the appropriate geometric conditions (e.g. convexity of the domain in the case of the minimal surface equation on ${\rm I\hspace{-0.2em}R}^{2}$) are not satisfied, then
the Dirichlet problem is ill-posed and a classical solution of (\ref{eq:D})-(\ref{bc:D}) may not exist (see, for example, \cite{JS1968} and \cite[\S 406]{NitscheBook}).
Interest in determining sufficient conditions for the existence of classical solutions of (\ref{eq:D})-(\ref{bc:D}) is long standing and
one method is to impose ``smallness'' conditions on the Dirichlet boundary data $\phi.$
When $H\equiv 0$  in (\ref{eq:D}), an early result is A. Korn's classic 1909 paper (\cite{Korn}); J.C.C. Nitsche discusses some of the history
of this problem for the minimal surface equation in \cite[\S 285 \& \S 412]{NitscheBook}.
G. Williams (\cite{Williams1984,Williams1985}), C.P. Lau (\cite{Lau}), K. Hayasida and M. Nakatani (\cite{HayasidaNakatani}),
M. Bergner (\cite{Bergner}), J. Ripoll and F. Tomi (\cite{RT}) and
many others have investigated limiting $\phi$  in order to prove a classical solution exists.

Rather than imposing a ``smallness'' condition on the Dirichlet data $\phi$  and trying to obtain a classical solution of (\ref{eq:D})-(\ref{bc:D}),
we wish to impose a (local) condition on the curvature of the domain, place no restrictions on the Dirichlet data $\phi\in L^{\infty}(\partial\Omega)$  and
prove that the variational solution $f$  extends to be continuous at a point $\mathcal{O}$  of $\partial\Omega$  (or on an open subset $\Gamma$  of $\partial\Omega$)
in the sense that $f\in C^{0}\left(\Omega\cup \{\mathcal{O}\}\right)$  (or $f\in C^{0}\left(\Omega\cup \Gamma\right)$)
when $\phi$  is continuous at $\mathcal{O}$  (or on $\Gamma$).
In general, no classical solution may exist and the variational solution is the best approximation to a classical solution.

We shall assume that $\partial\Omega$  is smooth, ${\mathcal O}\in\partial\Omega$  is a distinguished point, the curvature $\Lambda$  of $\partial\Omega$  satisfies
\begin{equation}
\label{Condition1}
\Lambda({\bf x}) < -2|H({\bf x})| \ \ \ \ {\rm for} \ {\bf x}\in\partial\Omega, \ |{\bf x}-{\mathcal O}|<\delta
\end{equation}
for some $\delta>0$  and  $\phi\in L^{\infty}(\partial\Omega).$
In Theorem \ref{Thm1}, we prove that the (unique) variational solution $f\in C^{2}(\Omega)$  of (\ref{eq:D})-(\ref{bc:D}) is
continuous at ${\mathcal O}$  if $\phi$  is continuous at ${\mathcal O}.$
In Theorem \ref{Thm2}, we assume $H\equiv 0,$  relax (\ref{Condition1}) slightly and obtain the same result as in Theorem \ref{Thm1}.
In Theorem \ref{Thm3}, we prove that the radial limits $Rf(\cdot)$  of $f$  exist at ${\mathcal O}$   if $\phi$  is discontinuous at ${\mathcal O}$
(even if $\phi$  does not have one-sided limits at ${\mathcal O}$).

The idea of the proof is to describe the graph of $f$  parametrically in isothermal coordinates, prove that it is uniformly continuous on its (open)
parameter domain and therefore extends uniquely to a continuous function on the closure of the parameter domain.
In the case of Theorems \ref{Thm1} and \ref{Thm2}, this is equivalent to proving that $f$  is uniformly continuous on the intersection of $\Omega$  and
an open neighborhood $U$  of ${\mathcal O}$  and therefore $f$  extends uniquely to a continuous function on $\overline{\Omega\cap U}.$
Of course, just as in \cite[Theorem 4.2]{Bour}, \cite{Lin} and \cite{Simon},  this extension of $f$  needs not equal $\phi.$
Differences between our results and those, for example, in the papers by Bourni, Lin and Simon are the additional requirements imposed on the domain or the boundary
data; \cite{Bour}  requires the graph of $\phi$  to be a specific type of limit of the graphs of $C^{1,\alpha}$  functions,
\cite{Lin}  requires $\phi$  to be Lipschitz and \cite{Simon} requires $H\equiv 0,$  $\partial\Omega$  to be $C^{4}$  and  $\phi$  to be Lipschitz.
In Example \ref{Example1}, we present an illustration of the value of symmetry and prove that $f$  can be continuous at ${\mathcal O}$  even if $\phi$
is not; this illustration uses the same domain as that mentioned in \cite{Simon}.

In the remaining case where $-2|H({\mathcal O})|\le \Lambda({\mathcal O})<2|H({\mathcal O})|,$  the behavior of the variational solution at
${\mathcal O}$  is unknown.  The Dirichlet problem (\ref{eq:D})-(\ref{bc:D})
does not have a classical solution in $C^{2}(\Omega)\cap C^{0}(\overline{\Omega})$  for all $\phi\in C^{\infty}(\partial\Omega)$  (\cite[Corollary 14.13]{GT})
and the standard gliding hump argument does not work in this situation.
On the other hand, not all of the comparison functions needed here are available and so the conclusions of Theorems \ref{Thm1} and \ref{Thm2} may not hold.
Moonies (\cite{FinnMoon,Liang92}, see also \cite{Liang96}) are not bounded below but do illustrate some properties of this case,
with $H\equiv 1,$  $\Lambda\equiv -2$  on one component of the boundary and $\Lambda\equiv \frac{1}{R},$  $\frac{1}{2}<R<1,$  on the other component.
As in \cite{JL}, these cases illustrate the strong differences between uniformly elliptic (genre zero) and prescribed mean curvature (genre two) equations
(see also Remark \ref{Remark2});
for Laplace's equation in ${\rm I\hspace{-0.2em}R}^{2},$  for example, boundary curvature would play no role in the solvability of Dirichlet problems and
the gliding hump construction could always be used, exactly as in the first case above.
\vspace{3mm}

\begin{thm}
\label{Thm1}
Suppose $\Omega$  is a locally Lipschitz  domain in ${\rm I\hspace{-0.2em}R}^{2},$  $\Gamma$  is a $C^{2,\lambda}$  open subset of $\partial\Omega$  for some $\lambda\in (0,1),$
the  curvature $\Lambda({\bf x})$   of $\Gamma$  at ${\bf x}$  is less than $-2|H({\bf x})|$  for ${\bf x}\in\Gamma.$
Suppose $\phi\in L^{\infty}(\partial\Omega),$  ${\bf y}\in \Gamma,$
either $f$  is symmetric with respect to a line through ${\bf y}$  or $\phi$  is continuous at ${\bf y},$  and $f\in BV(\Omega)$  minimizes
\begin{equation}
\label{AA}
J(u)=\int_{\Omega} \sqrt{1+|Du|^{2}}\ d{\bf x} + \int_{\Omega}  2Hu \ d{\bf x} + \int_{\partial\Omega} |u-\phi|\ ds
\end{equation}
for $u\in BV(\Omega).$  Then $f\in C^{0}(\Omega\cup \{ {\bf y}\}).$
If $\phi\in C^{0}(\Gamma),$  then $f\in C^{0}(\Omega\cup\Gamma).$
\end{thm}
\vspace{3mm}

\noindent When $H\equiv 0,$  the strict inequality (\ref{Condition1}) can be relaxed.

\begin{thm}
\label{Thm2}
Suppose $\Omega$  is a locally Lipschitz  domain in ${\rm I\hspace{-0.2em}R}^{2},$  $\Gamma$  is a $C^{2,\lambda}$  open subset of $\partial\Omega$
for some $\lambda\in (0,1),$  and the  curvature $\Lambda$  of $\Gamma$  is nonpositive and vanishes, at most, at a finite number of points of $\Gamma.$
For each point ${\bf x}_{0}\in \Gamma$  at which $\Lambda({\bf x}_{0})=0,$  suppose there exist $C>0$  and $\delta>0$  such that
\begin{equation}
\label{c-bound}
|\Lambda({\bf x})|\ge C|{\bf x}-{\bf x}_{0}|^{\lambda} \ \ {\rm for} \ \  {\bf x}\in \Gamma, \ |{\bf x}-{\bf x}_{0}|<\delta.
\end{equation}
Suppose $\phi\in L^{\infty}(\partial\Omega),$  ${\bf y}\in \Gamma,$
either $f$  is symmetric with respect to a line through ${\bf y}$  or $\phi$  is continuous at ${\bf y},$  and $f\in BV(\Omega)$  minimizes
\begin{equation}
\label{JJ}
J(u)=\int_{\Omega} \sqrt{1+|Du|^{2}}\ d{\bf x} + \int_{\partial\Omega} |u-\phi| \ ds
\end{equation}
for $u\in BV(\Omega).$  Then $f\in C^{0}(\Omega\cup \{ {\bf y}\}).$
If $\phi\in C^{0}(\Gamma),$  then $f\in C^{0}(\Omega\cup\Gamma).$
\end{thm}
\vspace{3mm}

\begin{example}
\label{Example1}
Let $\Omega=\{(x,y)\in {\rm I\hspace{-0.2em}R}^{2} : 1< (x+1)^{2}+y^{2}<\cosh^{2}(1)\}$  and $\phi(x,y)=\sin\left(\frac{\pi}{x^2+y^2}\right)$  for $(x,y)\neq (0,0)$
(see Figure \ref{Cat} for a rough illustration of the graph of $\phi$).  Set ${\mathcal O}=(0,0)$  and   $H\equiv 0.$
Let $f\in C^{2}(\Omega)$  minimize (\ref{JJ}) over $BV(\Omega).$
Then Theorem \ref{Thm1} (with ${\bf y}={\mathcal O}$) implies $f\in C^{0}\left(\overline{\Omega}\right),$  even though $\phi$  has no limit at ${\mathcal O}.$
\end{example}

\begin{figure}[htb]
\centerline{
\includegraphics[width=1.5in]{Trace1.pdf}}
\caption{$\Omega$  and part of the graph of $\phi$  \label{Cat}}
\end{figure}

\noindent We shall prove the following theorem on the existence and behavior of the radial limits of variational solutions of (\ref{eq:D})-(\ref{bc:D})
and use this to prove Theorems \ref{Thm1} and \ref{Thm2}.
At a point ${\bf y}\in\partial\Omega,$  we let $\alpha({\bf y})$  and $\beta({\bf y})$  be the angles which the tangent rays to $\partial\Omega$  at
${\bf y}$  make with the positive $x-$axis such that
\[
\{ {\bf y}+r(\cos\theta,\sin\theta) : 0<r<\epsilon(\theta), \alpha({\bf y})<\theta<\beta({\bf y}) \} \subset \Omega\cap B_{\delta}({\bf y})
\]
for some $\delta>0$  and some function $\epsilon(\cdot):(\alpha({\bf y}),\beta({\bf y}))\to (0,\delta).$
If $\partial\Omega$  is smooth at ${\bf y},$  $\beta({\bf y})=\alpha({\bf y})+\pi.$
For $\theta\in (\alpha({\bf y}),\beta({\bf y})),$  $Rf(\theta,{\bf y})=\lim_{r\downarrow 0}f({\bf y}+r(\cos\theta,\sin\theta))$  if this limit exists.
Also $Rf(\alpha({\bf y}),{\bf y})=\lim_{\Gamma_{1}({\bf y})\ni {\bf x}\to {\bf y}} f({\bf x})$  if this limit exists and
$Rf(\beta({\bf y}),{\bf y})=\lim_{\Gamma_{2}({\bf y})\ni {\bf x}\to {\bf y}} f({\bf x})$  if this limit exists,
where $\partial \Omega \cap B_{\delta}({\bf y})\setminus \{{\bf y}\}$ consists of disjoint, open arcs  $\Gamma_{1}({\bf y})$  and $\Gamma_{2}({\bf y})$
whose tangent rays approach the rays $\theta= \alpha({\bf y})$  and $\theta= \beta({\bf y})$ respectively, as the point ${\bf y}$ is approached.
\vspace{5mm}

\begin{thm}
\label{Thm3}
Suppose $\Omega$  is a locally Lipschitz  domain in ${\rm I\hspace{-0.2em}R}^{2},$  $\Gamma$  is a $C^{2,\lambda}$  open subset of $\partial\Omega$
for some $\lambda\in (0,1).$  Suppose either (a) $H\equiv 0$  in a neighborhood of $\Gamma,$
the curvature $\Lambda$  of $\Gamma$  is nonpositive and vanishes at only a finite number of points of $\Gamma,$  at each of which (\ref{c-bound}) holds,
or  (b) the  curvature $\Lambda({\bf x})$   of $\Gamma$  at ${\bf x}$  is less than $-2|H({\bf x})|$  for ${\bf x}\in\Gamma.$
Let $f\in C^{2}(\Omega)\cap L^{\infty}(\Omega)$  satisfy $Nf=2H$  in $\Omega.$
Suppose that ${\bf y}\in\partial\Omega.$     Then the limits
\[
\lim_{\Gamma_{1}({\bf y})\ni {\bf x}\to {\bf y}} f({\bf x})=z_{1} \ {\rm and} \
\lim_{\Gamma_{2}({\bf y})\ni {\bf x}\to {\bf y}} f({\bf x})=z_{2}
\]
exist, $Rf(\theta,{\bf y})$  exists for each $\theta\in [\alpha({\bf y}),\beta({\bf y})],$
$Rf(\cdot,{\bf y})\in C^{0}([\alpha({\bf y}),\beta({\bf y})]),$  and $Rf(\cdot,{\bf y})$ behaves in one of the following ways:

\noindent (i) $Rf(\cdot,{\bf y})=z_{1}$  is a constant function  and $f$  is continuous at ${\bf y}.$

\noindent (ii) There exist $\alpha_{1}$ and $\alpha_{2}$ so that $\alpha({\bf y}) \leq \alpha_{1}
< \alpha_{2} \leq \beta({\bf y}),$  $Rf=z_{1}$  on $[\alpha({\bf y}), \alpha_{1}],$  $Rf=z_{2}$  on $[\alpha_{2}, \beta({\bf y})]$
and $Rf$ is strictly increasing (if $z_{1}<z_{2}$) or strictly decreasing  (if $z_{1}>z_{2}$)  on $[\alpha_{1}, \alpha_{2}].$
\end{thm}

\section{Proofs}

Let  $Q$  be the operator on $C^{2}(\Omega)$  given by
\begin{equation}
\label{Q}
Qf({\bf x}) \myeq Nf({\bf x}) - 2H({\bf x}), \ \ \ \ {\bf x}\in\Omega.
\end{equation}
Let $\nu$  be the exterior unit normal to $\partial\Omega,$  defined almost everywhere on $\partial\Omega.$
We assume that for almost every ${\bf y}\in\partial\Omega,$  there is a continuous extension $\hat\nu$  of $\nu$  to a neighborhood of ${\bf y}.$

\begin{definition}
\label{Bernstein+}
Given a locally Lipschitz domain $\Omega,$ a \underline{\bf upper Bernstein pair} $\left(U^{+},\psi^{+}\right)$  for a curve $\Gamma\subset \partial\Omega$
and a function $H$  in (\ref{Q}) is a $C^{1}$  domain $U^{+}$  and a function $\psi^{+}\in C^{2}(U^{+})\cap C^{0}\left(\overline{U^{+}}\right)$  such that
$\Gamma\subset\partial U^{+},$  $\nu$  is the exterior unit normal to $\partial U^{+}$  at each point of $\Gamma$
(i.e. $U^{+}$  and $\Omega$  lie on the same side of $\Gamma$), $Q\psi^{+}\le 0$  in $U^{+},$  and $T\psi^{+}\cdot\nu=1$  almost everywhere on an open subset
of $\partial U^{+}$  containing $\overline{\Gamma}$  in the same sense as in  \cite{CF:74a}; that is, for almost every ${\bf y}\in\Gamma,$
\begin{equation}
\label{zero}
\lim_{U^{+}\ni {\bf x}\to {\bf y}} \frac{\nabla \psi^{+}({\bf x})\cdot \hat\nu({\bf x})}{\sqrt{1+|\nabla \psi^{+}({\bf x})|^{2}}} = 1.
\end{equation}
\end{definition}

\begin{definition}
\label{Bernstein-}
Given a domain $\Omega$  as above, a \underline {\bf lower Bernstein pair} $\left(U^{-},\psi^{-}\right)$  for a curve $\Gamma\subset \partial\Omega$
and a function $H$  in (\ref{Q}) is a $C^{1}$  domain $U^{-}$  and a function $\psi^{-}\in C^{2}(U^{-})\cap C^{0}\left(\overline{U^{-}}\right)$
such that $\Gamma\subset\partial U^{-},$  $\nu$  is the exterior unit normal to $\partial U^{-}$  at each point of $\Gamma$
(i.e. $U^{-}$  and $\Omega$  lie on the same side of $\Gamma$),
$Q\psi^{-}\ge 0$  in $U^{-},$  and $T\psi^{-}\cdot\nu=-1$  almost everywhere on an open subset of $\partial U^{-}$  containing $\overline{\Gamma}$
in the same sense as in  \cite{CF:74a}.
\end{definition}

The existence of Bernstein pairs is established in Lemmas \ref{Prop2} and \ref{Prop3}.

\begin{lem}
\label{Prop1}
Let $a<b,$  $\lambda\in (0,1),$  $\psi\in C^{2,\lambda}([a,b])$  and
$
\Gamma=\{(x,\psi(x))\in {\rm I\hspace{-0.2em}R}^{2} : x\in [a,b]\}
$
such that
$\psi'(x)< 0$  for $x\in [a,b],$  $\psi''(x)<0$  for $x\in [a,b]\setminus J,$    there exist $C_{1}>0$  and $\epsilon_{1}>0$  such that
if $\bar x\in J$  and $|x-\bar x|<\epsilon_{1},$  then $\psi''(x)\le -C_{1}|x-\bar x|^{\lambda},$  where $J$  is a finite subset of $(a,b).$
Then there exists an open set $U\subset{\rm I\hspace{-0.2em}R}^{2}$  with  $\Gamma\subset \partial U$  and a function
$h\in C^{2}\left(U\right)\cap C^{0}\left(\overline{U}\right)$  such that  $\partial U$  is a closed, $C^{2,\lambda}$  curve,
$\Gamma$  lies below $U$  in ${\rm I\hspace{-0.2em}R}^{2}$  (i.e. the exterior unit normal $\nu=(\nu_{1}(x),\nu_{2}(x))$  to $\partial U$ satisfies $\nu_{2}(x)<0$
for $a\le x\le b$),   $Nh=0$  in $U$  and $Th\cdot\nu=1$  almost everywhere on an open subset of $\partial U^{+}$  containing $\overline{\Gamma}$
(i.e. (\ref{zero}) holds).
\end{lem}
\vspace{3mm}

\noindent {\bf Proof:} We may assume that $a,b>0.$
There exists $c>b$  and $k\in C^{2,\lambda}([-c,c])$  with $k(-x)=k(x)$  for $x\in [0,c]$  such that $k(x)=-\psi(x)$  for $x\in [a,b],$
$k''(x)>0$  for $x\in [-c,c]\setminus J,$  where $J$  is a finite set, $k''(0)>0,$  and the set
\[
K=\{(x,k(x))\in {\rm I\hspace{-0.2em}R}^{2} : x\in [-c,c]\}
\]
is strictly concave (i.e. $tk(x_{1})+(1-t)k(x_{2})>k\left(tx_{1}+(1-t)x_{2}\right)$  for each $t\in (0,1)$  and
$x_{1},x_{2}\in [-c,c]$  with $x_{1}\neq x_{2}$).
From \cite[pp.1063-5]{EL1986B}, we can construct a domain $\Omega(K,l)$ such that $K\subset\partial\Omega(K,l)$  and
$\Omega(K,l)$  lies below $K$  (i.e. the outward unit normal to  $\Omega(K,l)$  at $(x,k(x))$  is
$\nu(x)=\frac{\left(-k'(x),1\right)}{\sqrt{1+\left(k'(x)\right)^{2}}};$  see \cite[Figure 4]{EL1986B})
and a function $F^{+}\in C^{2}\left(\Omega(K,l)\right)\cap C^{0}\left(\overline{\Omega(K,l)}\right)$  such that the function
\[
\mu({\bf x})\myeq \frac{\left(\nabla F^{+}({\bf x}),-1\right)}{\sqrt{1+|\nabla F^{+}({\bf x})|^{2}}}, \ \ \ {\bf x}\in\Omega(K,l),
\]
extends continuously to a function on $\Omega(K,l)\cup K$  and $\mu(x,k(x))\cdot \nu(x)=1$  for $x\in [-c,c].$
Now let $V$  be an open subset of $\Omega$  with $C^{2,\lambda}$  boundary such that $\{(x,-\psi(x)) : x\in [a,b]\}\subset \partial V$  and
$\partial V\cap \left(\partial\Omega(K,l)\setminus K\right)=\emptyset$  and then let
$U=\{(x,-y) : (x,y)\in V\}$  and $h(x,y)=F^{+}(x,-y)$  for $(x,y)\in\overline{U}.$  \qed

\begin{lem}
\label{Prop2}
Suppose $\Omega$  is a locally Lipschitz  domain in ${\rm I\hspace{-0.2em}R}^{2},$  $\Gamma$  is a $C^{2,\lambda}$  open subset of $\partial\Omega$  for some $\lambda\in (0,1),$
and the  curvature $\Lambda$  of $\Gamma$  is either negative or nonpositive, vanishes only at finite number of points of $\Gamma$  and (\ref{c-bound})
holds at each such point.  Let ${\bf y}\in\Gamma.$
Then there exist $\delta>0$  and upper and lower Bernstein pairs $\left(U^{\pm},\psi^{\pm}\right)$  for $(\Gamma\cap B_{\delta}({\bf y}),0).$
\end{lem}
\vspace{3mm}

\noindent {\bf Proof:}
Let $\Omega\subset{\rm I\hspace{-0.2em}R}^{2}$  be an open set,  $\Gamma\subset\partial\Omega$  be a $C^{2,\lambda}$  curve and ${\bf y}\in \Gamma$
be a point at which we wish to have upper and lower Bernstein pairs for $H\equiv 0.$
Notice that (\ref{c-bound}) implies that the ``curvature'' condition in Lemma \ref{Prop1} (i.e. after rotating $\Omega,$  the condition that if
$\Lambda(\bar x,\psi(\bar x))=0,$  then $\psi''(x)\le -C_{1}|x-\bar x|^{\lambda}$  when $x$  is near $\bar x$)  is satisfied.
Choose a neighborood $V$  of ${\bf y}$  and a rigid motion $\zeta:{\rm I\hspace{-0.2em}R}^{2}\to{\rm I\hspace{-0.2em}R}^{2}$  such that $\Sigma\myeq V\cap \partial\Omega \subset\Gamma$
and the curve $\zeta\left(\Sigma\right)$  satisfies the hypotheses of Lemma \ref{Prop1}.
Let $U$  and $h$  be as given in the conclusion of Lemma \ref{Prop1}.
Then $\left(\zeta^{-1}(U),h\circ\zeta\right)$  will be an upper Bernstein pair for $\Sigma$  and $H\equiv 0$  and
$\left(\zeta^{-1}(U),-h\circ\zeta\right)$  will be a lower Bernstein pair for $\Sigma$  and $H\equiv 0.$  \qed
\vspace{3mm}

\begin{lem}
\label{Prop3}
Suppose $\Omega$  is a $C^{2,\lambda}$  domain in ${\rm I\hspace{-0.2em}R}^{2}$  for some $\lambda\in (0,1),$  ${\bf y}\in\partial\Omega$  and
$\Lambda({\bf y})< -2|H({\bf y})|,$  where $\Lambda({\bf y})$  denotes the curvature of $\partial\Omega$  at ${\bf y}.$
Then there exist $\delta>0$  and upper and lower Bernstein pairs $\left(U^{\pm},\psi^{\pm}\right)$  for $(\Gamma,H),$   where
$\Gamma=B_{\delta}({\bf y})\cap {\partial}\Omega.$
\end{lem}
\vspace{3mm}

\noindent {\bf Proof:} We may assume $H\not\equiv 0$  in any neighborhood of ${\bf y}$  since Lemma \ref{Prop2} covers this case.
There exists $\delta_{1}>0$  such that  $\Lambda({\bf x})< -2|H({\bf x})|$  for each ${\bf x}\in\partial\Omega \cap B_{\delta_{1}}({\bf y}).$
There exists a $\delta_{2}\in (0,\delta_{1}/2)$  such that
\[
\Lambda_{0}\myeq \sup\{\Lambda({\bf x}) : {\bf x}\in\partial\Omega\cap B_{\delta_{2}}({\bf y})\} <
\inf\{-2|H({\bf x})| : {\bf x}\in\overline{\Omega}\cap B_{\delta_{2}}({\bf y})\} \myeq -2H_{0}.
\]
Let $z=h(\hat r),$    be a unduloid surface (see for example \cite{FinnMoon,LS1}) defined on the annulus
${\mathcal A}= {\mathcal A}({\bf q})\myeq \{ {\bf x}\in {\rm I\hspace{-0.2em}R}^{2} : r_{1}\le \hat r({\bf x})\le r_{2} \}$   with constant mean curvature $-H_{0}<0$  which
becomes vertical at $\hat r=r_{1},r_{2}$   (with $Th\cdot\nu=1$  on $\hat r({\bf x})=r_{1}$), where $\hat r=|{\bf x}-{\bf p}|,$
${\bf p}={\bf p}({\bf q})={\bf q}+r_{1}\nu({\bf q})$  for ${\bf q}\in \partial\Omega\cap B_{\delta_{2}}({\bf y}),$
$r_{1}=\frac{1-\sqrt{1+4c_{0}H_{0}}}{2H_{0}},$   $r_{2}=\frac{1+\sqrt{1+4c_{0}H_{0}}}{2H_{0}},$  and $c_{0}\in\left(-\frac{1}{4H_{0}},0\right)$  is arbitrary.
Now ${\mathcal A}({\bf q})$  touches $\partial\Omega$  at ${\bf q}$  and there exists $0<\delta_{3}<\delta_{2}/2$  such that
$\{ {\bf x}\in B_{\delta_{3}}({\bf y}) : |{\bf x}-{\bf p}({\bf q})|=r_{1}\} \subset \Omega\cup \{ {\bf q} \}$  and
$h'(r)\le 0$  if $r_{1}<r<r_{1}+\delta_{3}.$

Set $z_{1}=h(r_{1})-h(r_{2})$  and $z_{2}>z_{1}.$
Let $\psi^{+}$  be the variational solution of the Dirichlet problem $N\psi^{+}=-2H_{0}$  in $U=\Omega\cap B_{\delta_{3}}({\bf y}),$
$\psi^{+}=z_{2}$  on $\Gamma=B_{\delta_{3}}({\bf y})\cap\partial\Omega$  and $\psi^{+}=0$  on $\partial U\setminus\Gamma.$
By \cite[Theorem 5.1]{FinnBook}, we see that $\psi^{+}\le z_{1}<z_{2}$  on $\Gamma$  and so $T\psi^{+}\cdot\nu=1$  on $\Gamma;$
this follows, for example, from \cite{Bour,Lin}.
Now define $\psi^{-}=-\psi^{+}$  in $U.$  \qed
\vspace{3mm}

\begin{rem}
\label{Remark1}
Suppose $\Omega$  is a $C^{2,\lambda}$  domain in ${\rm I\hspace{-0.2em}R}^{2}$  for some $\lambda\in (0,1),$  ${\bf y}\in\partial\Omega$  and
$\Lambda({\bf y})< 2|H({\bf y})|,$  where $\Lambda({\bf y})$  denotes the curvature of $\partial\Omega$  at ${\bf y}.$
If $H$  is non-negative  in $U\cap\Omega$  for some neighborhood $U$  of ${\bf y},$  then the argument which establishes
\cite[Corollary 14.13]{GT}  and boundary regularity results (e.g. \cite{Bour,Lin}) imply that there exist $\delta>0$  and an upper
Bernstein pair $\left(U^{+},\psi^{+}\right)$  for $(\Gamma,H),$  where $\Gamma=B_{\delta}({\bf y})\cap {\partial}\Omega.$
If $H$  is non-positive  in $U\cap\Omega$  for some neighborhood $U$  of ${\bf y},$  then there exist $\delta>0$  and a lower
Bernstein pair $\left(U^{-},\psi^{-}\right)$  for $(\Gamma,H),$  where $\Gamma=B_{\delta}({\bf y})\cap {\partial}\Omega.$
\end{rem}
\vspace{3mm}

\noindent {\bf Proof of Theorem \ref{Thm3}:}
We note, as in \cite{LS1}, that the conclusion of Theorem \ref{Thm3} is a local one and so, for small $\delta>0,$  we can replace $\Omega$  by a $C^{2,\lambda}$
set $\Omega^{*}$  such that $\Omega\cap B_{\delta}({\bf y}) = \Omega^{*}\cap B_{\delta}({\bf y})$  and $\Omega^{*}\subset  B_{2\delta}({\bf y}).$
We may assume $\Omega$  is a bounded domain.  Set  $S_{0} = \{ ({\bf x},f({\bf x})) : {\bf x} \in \Omega \}.$
From the calculation on page 170 of \cite{LS1},  we see that the area of $S_{0}$  is finite; let $M_{0}$  denote this area.
For $\delta\in (0,1),$  set
\[
p(\delta) = \sqrt{\frac{8\pi M_{0}}{\ln\left(\frac{1}{\delta}\right)}}.
\]
Let $E= \{ (u,v) : u^{2}+v^{2}<1 \}.$
As in \cite{EL1986A,LS1}, there is a parametric description of the surface $S_{0},$
\begin{equation}
\label{PARAMETRIC}
Y(u,v) = (a(u,v),b(u,v),c(u,v)) \in C^{2}(E:{{\rm I\hspace{-0.2em}R}}^{3}),
\end{equation}
which has the following properties:

\noindent $\left(a_{1}\right)$  $Y$ is a diffeomorphism of $E$ onto $S_{0}$.

\noindent $\left(a_{2}\right)$
Set $G(u,v)=(a(u,v),b(u,v)),$  $(u,v)\in E.$   Then $G \in C^{0}(\overline{E} : {{\rm I\hspace{-0.2em}R}}^{2}).$

\noindent $\left(a_{3}\right)$
Set $\sigma({\bf y})=G^{-1}\left(\partial \Omega\setminus \{ {\bf y} \}\right);$
then $\sigma({\bf y})$ is a connected arc of $\partial E$  and $Y$ maps $\sigma({\bf y})$  onto $\partial \Omega\setminus \{ {\bf y} \}.$
We may assume the endpoints of $\sigma({\bf y})$  are ${\bf o}_{1}({\bf y})$  and ${\bf o}_{2}({\bf y}).$
(Note that ${\bf o}_{1}({\bf y})$  and ${\bf o}_{2}({\bf y})$  are not assumed to be distinct.)

\noindent $\left(a_{4}\right)$
$Y$ is conformal on $E$: $Y_{u} \cdot Y_{v} = 0, Y_{u}\cdot Y_{u} = Y_{v}\cdot Y_{v}$
on $E$.

\noindent $\left(a_{5}\right)$
$\triangle Y := Y_{uu} + Y_{vv} = H\left(Y\right)Y_{u} \times Y_{v}$  on $E$.
\vspace{2mm}

\noindent From Lemma \ref{Prop2}  when $H\equiv 0$  and Lemma \ref{Prop3}  when $H$  satisfies (\ref{(16.60)}), we see that
upper and lower Bernstein pairs $\left(U^{\pm},\psi^{\pm}\right)$  for $(\Gamma,H)$  exist.
Notice that for each $C\in{\rm I\hspace{-0.2em}R},$   $Q(\psi^{+}+C)=Q(\psi^{+})\le 0$  on $\Omega\cap U^{+}$
and $Q(\psi^{-}+C)=Q(\psi^{-})\ge 0$  on $\Omega\cap U^{-},$  and so
\begin{equation}
\label{Barrier+}
N(\psi^{+}+C)({\bf x}) \le 2H({\bf x},f({\bf x}))=Nf({\bf x}) \ \ {\rm for} \ \ {\bf x}\in\Omega\cap U^{+},
\end{equation}
and
\begin{equation}
\label{Barrier-}
N(\psi^{-}+C)({\bf x}) \ge 2H({\bf x},f({\bf x}))=Nf({\bf x}) \ \ {\rm for} \ \ {\bf x}\in\Omega\cap U^{-}.
\end{equation}
Let $q$  denote a modulus of continuity for $\psi^{+}$  and $\psi^{-}.$

Let $\zeta({\bf y})=\partial E\setminus\sigma({\bf y});$  then $G(\zeta({\bf y}))=\{{\bf y}\}$  and ${\bf o}_{1}({\bf y})$  and ${\bf o}_{2}({\bf y})$
are the endpoints of $\zeta({\bf y}).$
There exists a $\delta_{1}>0$  such that if ${\bf w}\in E$  and ${\rm dist}\left({\bf w}, \zeta({\bf y})\right)\le 2\delta_{1},$
then $G({\bf w})\in U^{+}\cap U^{-}.$
Now $T\psi^{\pm}\cdot \nu=\pm 1$  (in the sense of \cite{CF:74a}) almost everywhere on an open subset $\Upsilon^{\pm}$  of $\partial U^{\pm}$
which contains $\overline{\Gamma};$  there exists a $\delta_{2}>0$  such that
$\left(\partial U^{\pm} \setminus \Upsilon^{\pm}\right) \cap \{{\bf x}\in {\rm I\hspace{-0.2em}R}^{2} : |{\bf x}-{\bf y}|\le 2p(\delta_{2})\}=\emptyset.$
Set $\delta^{*}=\min\{\delta_{1},\delta_{2}\}$  and
\[
V^{*}= \{ {\bf w}\in E : {\rm dist}({\bf w},\zeta({\bf y}))<\delta^{*} \}.
\]
Notice if ${\bf w}\in V^{*},$  then $G({\bf w})\in U^{+}\cap U^{-}.$
\vspace{3mm}

\noindent {\bf Claim:}   $Y$  is uniformly continuous on $V^{*}$  and so extends to a continuous function on $\overline{V^{*}}.$
\vspace{3mm}

\noindent {\bf Pf:}  Let $\epsilon>0.$  Choose $\delta\in \left(0,\left(\delta^{*}\right)^{2}\right)$  such that  $p(\delta)+2q(p(\delta))<\epsilon.$
Let ${\bf w}_{1},{\bf w}_{2}\in V^{*}$  with $|{\bf w}_{1}-{\bf w}_{2}|<\delta;$  then
$G({\bf w}_{1}), G({\bf w}_{2})\in U^{+}_{1} \cap U^{-}.$
Set $C_{r}({\bf w}) = \{ {\bf u} \in E : |{\bf u} - {\bf w}| = r \}$  and  $B_{r}({\bf w}) = \{ {\bf u} \in E : |{\bf u} - {\bf w}| < r \}.$
From the Courant-Lebesgue Lemma (e.g. Lemma $3.1$ in \cite{Cour:50}), we see that there exists $\rho=\rho(\delta)\in \left(\delta,\sqrt{\delta}\right)$
such that the arclength $l_{\rho}({\bf w}_{1})$  of $Y(C_{\rho}({\bf w}_{1}))$  is less than $p(\delta).$
Notice that ${\bf w}_{2}\in B_{\rho(\delta)}({\bf w}_{1}).$   Let
\[
k(\delta)({\bf w}_{1})= \inf_{{\bf u}\in C_{\rho(\delta)}({\bf w}_{1})}c({\bf u}) = \inf_{ {\bf x}\in G(C_{\rho(\delta)}({\bf w}_{1})) } f({\bf x})
\]
and
\[
m(\delta)({\bf w}_{1})= \sup_{{\bf u}\in C_{\rho(\delta)}({\bf w}_{1})}c({\bf u}) = \sup_{ {\bf x}\in G(C_{\rho(\delta)}({\bf w}_{1})) } f({\bf x});
\]
then
\[
m(\delta)({\bf w}_{1})-k(\delta)({\bf w}_{1})\le l_{\rho} < p(\delta).
\]
Fix ${\bf x}_{0}\in C^{\prime}_{\rho(\delta)}({\bf w}_{1}).$
Set
\[
C^{+}=\inf_{{\bf x}\in U^{+}\cap C^{\prime}_{\rho(\delta)}({\bf w}_{1}) } \psi^{+}({\bf x}) \ \ {\rm and} \ \
C^{-}=\sup_{{\bf x}\in U^{-}\cap C^{\prime}_{\rho(\delta)}({\bf w}_{1}) } \psi^{-}({\bf x}).
\]
Then $\psi^{+}-C^{+}\ge 0$  on $U^{+}\cap C^{\prime}_{\rho(\delta)}({\bf w}_{1})$  and
$\psi^{-}-C^{-}\le 0$  on $U^{-}\cap C^{\prime}_{\rho(\delta)}({\bf w}_{1}).$
Therefore, for ${\bf x}\in U^{+}\cap U^{-}\cap C^{\prime}_{\rho(\delta)}({\bf w}_{1}),$  we have
\[
k(\delta)({\bf w}_{1})+\left(\psi^{-}({\bf x})-C^{-}\right) \le f({\bf x}) \le m(\delta)({\bf w}_{1})+\left(\psi^{+}({\bf x})-C^{+}\right).
\]
Set
\[
b^{+}({\bf x})= m(\delta)({\bf w}_{1})+\left(\psi^{+}({\bf x})-C^{+}\right) \ \ \ \
{\rm for} \ \ {\bf x}\in U^{+}\cap \overline{G\left(B_{\rho(\delta)}({\bf w}_{1})\right)}
\]
and
\[
b^{-}({\bf x})= k(\delta)({\bf w}_{1})+\left(\psi^{-}({\bf x})-C^{-}\right) \ \ \ \
{\rm for} \ \ {\bf x}\in  U^{-} \cap \overline{G\left(B_{\rho(\delta)}({\bf w}_{1})\right)}.
\]
Now $\rho(\delta)<\sqrt{\delta}<\delta^{*}\le\delta_{2};$  notice that if ${\bf w}\in \overline{B_{\rho(\delta)}({\bf w}_{1})},$  then
$|{\bf w}-{\bf w}_{1}|<\delta_{2}$  and  $|G({\bf w})-{\bf y}|<2p(\delta_{2})$  and thus if
${\bf x}\in \overline{G\left(B_{\rho(\delta)}({\bf w}_{1})\right)} \cap \partial U^{\pm},$  then ${\bf x}\in \Upsilon^{\pm}.$
From (\ref{Barrier+}) and (\ref{Barrier-}), the facts that $b^{-}\le f$  on $U^{-} \cap C^{\prime}_{\rho(\delta)}({\bf w}_{1})$
and $f\le b^{+}$  on $U^{+} \cap C^{\prime}_{\rho(\delta)}({\bf w}_{1})$  and the general comparison principle (\cite[Theorem 5.1]{FinnBook}),
we have
\begin{equation}
\label{uniformA}
b^{-}\le f\ {\rm on} \ U^{-} \cap \overline{G\left(B_{\rho(\delta)}({\bf w}_{1})\right)}
\end{equation}
and
\begin{equation}
\label{uniformB}
f\le b^{+}\ {\rm on} \ U^{+}\cap \overline{G\left(B_{\rho(\delta)}({\bf w}_{1})\right)}.
\end{equation}

\noindent Since the diameter of $G\left(B_{\rho(\delta)}({\bf w}_{1})\right)\le p(\delta),$  we have
$\left|\psi^{\pm}({\bf x})-C^{\pm}\right|\le q(p(\delta))$  for ${\bf x}\in U^{\pm}\cap G\left(B_{\rho(\delta)}({\bf w}_{1})\right).$
Thus, whenever ${\bf x}_{1},{\bf x}_{2}\in \overline{G\left(B_{\rho(\delta)}({\bf w}_{1})\right)},$  we have
${\bf x}_{1},{\bf x}_{2} \in U^{+}\cap U^{-}.$
Since $c({\bf w})=f\left(G({\bf w})\right),$  $G({\bf w}_{1})\in U^{+}\cap U^{-}$  and $G({\bf w}_{2})\in U^{+}\cap U^{-},$   we have
\[
b^{-}\left(G({\bf w}_{1})\right)-b^{+}\left(G({\bf w}_{2})\right) \le c({\bf w}_{1})-c({\bf w}_{2}) \le
b^{+}\left(G({\bf w}_{1})\right)-b^{-}\left(G({\bf w}_{2})\right)
\]
or
\[
-\left[m(\delta)({\bf w}_{1})-k(\delta)({\bf w}_{1}) +\left(\psi^{+}(G({\bf w}_{2}))-C^{+}\right) - \left(\psi^{-}(G({\bf w}_{1}))+C^{-}\right)\right]
\]
\[
\le c({\bf w}_{1})-c({\bf w}_{2}) \le
\]
\[
\left[m(\delta)({\bf w}_{1})-k(\delta)({\bf w}_{1}) +\left(\psi^{+}(G({\bf w}_{1}))-C^{+}\right) - \left(\psi^{-}(G({\bf w}_{2}))+C^{-}\right)\right].
\]
Since $|\psi^{\pm}(G({\bf w}))-C^{\pm}| \le q(p(\delta))$  for ${\bf w}\in B_{\rho(\delta)}({\bf w}_{1})\cap U^{\pm},$  we have
\[
|c({\bf w}_{1})-c({\bf w}_{2})| \le  p(\delta)+2q(p(\delta))<\epsilon.
\]
Thus $c$  is uniformly continuous on  $V^{*}$  and, since $G \in C^{0}(\overline{E} : {{\rm I\hspace{-0.2em}R}}^{2}),$  we see that  $Y$  is uniformly continuous on $V^{*}.$
Therefore $Y$  extends to a continuous function, still denote $Y,$  on $\overline{V^{*}}.$ \qed
\vspace{3mm}

Notice that
\[
\lim_{\Gamma_{1}\ni {\bf x}\to {\bf y}} f({\bf x})=\lim_{\partial E\ni {\bf w}\to {\bf o}_{1}({\bf y})} c({\bf w})=c({\bf o}_{1}({\bf y}))
\]
and
\[
\lim_{\Gamma_{2}\ni {\bf x}\to {\bf y}} f({\bf x})=\lim_{\partial E\ni {\bf w}\to {\bf o}_{2}({\bf y})} c({\bf w})=c({\bf o}_{2}({\bf y}))
\]
and so, with $z_{1}=c({\bf o}_{1}({\bf y}))$  and $z_{2}=c({\bf o}_{2}({\bf y})),$  we see that (\ref{Apple}) holds.

Then the limits
\begin{equation}
\label{Apple}
\lim_{\Gamma_{1}\ni {\bf x}\to {\bf y}} f({\bf x})=z_{1} \ {\rm and} \
\lim_{\Gamma_{2}\ni {\bf x}\to {\bf y}} f({\bf x})=z_{2}
\end{equation}
exist,

Now we  need to consider two cases:
\begin{itemize}
\item[$\left(A\right)$] ${\bf o}_{1}({\bf y})= {\bf o}_{2}({\bf y}).$
\item[$\left(B\right)$] ${\bf o}_{1}({\bf y})\neq {\bf o}_{2}({\bf y}).$
\end{itemize}
These correspond to Cases 5 and 3 respectively in Step 1 of the proof of \cite[Theorem 1]{LS1}.
\vspace{2mm}

\noindent {\bf Case (A):}  Suppose ${\bf o}_{1}({\bf y})= {\bf o}_{2}({\bf y});$  set ${\bf o}={\bf o}_{1}({\bf y})= {\bf o}_{2}({\bf y}).$
Then $f$  extends to a function in $C^{0}\left(\Omega \cup \{ {\bf y}\}\right)$  and case (i) of Theorem \ref{Thm3} holds.
\vspace{2mm}

\noindent  {\bf Pf:} Notice that $G$  is a bijection of $E\cup \{{\bf o}\}$  and $\Omega\cup \{{\bf y}\}.$  Thus we may define $f=c\circ G^{-1},$
so $f\left(G({\bf w})\right)=c({\bf w})$  for ${\bf w}\in E\cup \{{\bf o}\};$  this extends $f$  to a function defined on $\Omega\cup \{{\bf y}\}.$
Let $\{\delta_{i}\}$  be a decreasing sequence of positive numbers converging to zero and consider the sequence of open sets
$\{G(B_{\rho(i)}({\bf o}))\}$  in $\Omega,$  where $\rho(i)=\rho(\delta_{i}({\bf o})).$
Now ${\bf y}\notin G(C_{\rho(i)}({\bf o}))$  and so there exist $\sigma_{i}>0$  such that
\[
P(i)  =\{ {\bf x}\in\Omega : |{\bf x}-{\bf y}|<\sigma_{i}\} \subset G(B_{\rho(i)}({\bf o}))
\]
for each $i\in{\rm I\hspace{-0.2em}N}.$  Thus if  ${\bf x}\in P(i),$  we have $|f({\bf x}) - f({\bf y})|<p(\delta_{i})+2q(p(\delta_{i})).$
The continuity of $f$  at ${\bf y}$  follows from this.  \qed
\vspace{3mm}

\noindent {\bf Case (B):}  Suppose ${\bf o}_{1}({\bf y})\neq {\bf o}_{2}({\bf y}).$   Then case (ii) of Theorem \ref{Thm3} holds.
\vspace{2mm}

\noindent  {\bf Pf:}
As at the end of Step 1 of the proof of \cite[Theorem 1]{LS1}, we define $X:B\to{\rm I\hspace{-0.2em}R}^{3}$  by $X=Y\circ g$ and $K:B\to{\rm I\hspace{-0.2em}R}^{2}$
by $K=G\circ g,$  where $B=\{(u,v)\in{\rm I\hspace{-0.2em}R}^{2} : u^{2}+v^{2}<1, \ v>0\}$  and $g:\overline{B}\to \overline{E}$  is either a conformal or an indirectly
conformal (or anticonformal) map from $\overline{B}$  onto $\overline{E}$  such that  $g(1,0)= {\bf o}_{1}({\bf y}),$
$g(-1,0)= {\bf o}_{2}({\bf y})$  and $g(u,0)\in  {\bf o}_{1}({\bf y}){\bf o}_{2}({\bf y})$  for each $u\in [-1,1],$  where
${\bf ab}$  denotes the (appropriate) choice of arc in $\partial E$  with ${\bf a}$  and ${\bf b}$  as endpoints.

Notice that $K(u,0)={\bf y}$  for $u\in [-1,1].$  Set $x=a\circ g,$  $y=b\circ g$  and $z=c\circ g,$  so that $X(u,v)=(x(u,v),y(u,v),z(u,v))$
for $(u,v)\in B.$  Now, from Step 2 of the proof of \cite[Theorem 1]{LS1},
\[
X\in C^{0}\left(\overline{B}:{\rm I\hspace{-0.2em}R}^{3}\right)\cap C^{1,\iota}\left(B\cup\{(u,0):-1<u<1\}:{\rm I\hspace{-0.2em}R}^{3}\right)
\]
for some $\iota\in (0,1)$  and $X(u,0)=({\bf y},z(u,0))$  cannot be constant on any nondegenerate interval in $[-1,1].$
Define $\Theta(u)= {\rm arg}\left( x_{v}(u,0)+iy_{v}(u,0) \right).$  From equation \cite[(12)]{LS1}, we see that
\[
\alpha_{1}=\lim_{u\downarrow -1} \Theta(u) \ \ \ \ {\rm and} \ \ \ \  \alpha_{2}=\lim_{u\uparrow 1} \Theta(u);
\]
here $\alpha_{1}<\alpha_{2}.$
As in Steps 2-5 of the proof of \cite[Theorem 1]{LS1}, we see that $Rf(\theta)$  exists when $\theta\in \left(\alpha_{1},\alpha_{2}\right),$
\[
\overline{G^{-1}\left( L(\alpha_{2}) \right)} \cap \partial E = \{ {\bf o}_{1}({\bf y}) \} \    (\& \
\overline{K^{-1}\left( L(\alpha_{2}) \right)} \cap \partial B = \{ (1,0) \})   \  {\rm when}  \  \alpha_{2}<\beta({\bf y})
\]
\[
\overline{G^{-1}\left( L(\alpha_{1}) \right)} \cap \partial E = \{ {\bf o}_{2}({\bf y}) \} \ (\& \
\overline{K^{-1}\left( L(\alpha_{1}) \right)} \cap \partial B = \{ (-1,0) \})  \  {\rm when} \ \alpha_{1}>\alpha({\bf y})
\]
where  $L(\theta)= \{{\bf y}+(r\cos(\theta),r\sin(\theta))\in \Omega : 0<r<\delta^{*} \},$
and we see that  $Rf$ is strictly increasing or strictly decreasing on $(\alpha_{1}, \alpha_{2}).$
We may argue as in Case A to see that $f$  is uniformly continuous on
\[
\Omega^{+} =\{{\bf y}+(r\cos(\theta),r\sin(\theta))\in \Omega : 0<r<\delta, \alpha_{2}\le \theta<\beta({\bf y})+\epsilon\}
\]
and $f$  is uniformly continuous on
\[
\Omega^{-} =\{{\bf y}+(r\cos(\theta),r\sin(\theta))\in \Omega : 0<r<\delta, \alpha({\bf y})-\epsilon< \theta\le\alpha_{1}\}
\]
for some small $\epsilon>0$  and $\delta>0,$  since  $G$  is a bijection of $E\cup \{{\bf o}_{1}({\bf y})\}$  and $\Omega\cup \{{\bf y}\}$
and a bijection of $E\cup \{{\bf o}_{2}({\bf y})\}$  and $\Omega\cup \{{\bf y}\}.$
Theorem \ref{Thm3} then follows, as in \cite{NoraKirk1}, from Steps 2-5 of the proof of \cite[Theorem 1]{LS1}.
\qed
\vspace{3mm}

\noindent {\bf Proof of Theorem \ref{Thm2}:}
From Theorem \ref{Thm3}, we see that the radial limits $Rf(\theta,{\bf y})$  exist for each $\theta\in [\alpha({\bf y}),\beta({\bf y})].$
Set $z_{1}=Rf(\alpha({\bf y}),{\bf y}),$  $z_{2}=Rf(\beta({\bf y}),{\bf y})$  and $z_{3}=\phi({\bf y}).$
If $z_{1}=z_{2},$  then case (i) of Theorem \ref{Thm3} holds.
(If $f$  is symmetric with respect to a line through ${\bf y},$  then $z_{1}=z_{2}$  and we are done.)

Suppose otherwise that $z_{1}\neq z_{2};$  we may assume that $z_{1}<z_{3}$  and $z_{1}<z_{2}.$
Then there exist $\alpha_{1}, \alpha_{2}\in [\alpha({\bf y}),\beta({\bf y})]$  with $\alpha_{1}<\alpha_{2}$  such that
\[
Rf(\theta,{\bf y}) \ \ {\rm is}
\left\{ \begin{array}{ccc}
{\rm constant}(=z_{1}) & {\rm for} & \alpha({\bf y})\le\theta\le\alpha_{1}\\
{\rm strictly \ increasing} & {\rm for} & \alpha_{1}\le\theta\le\alpha_{2}\\
{\rm constant}(=z_{2}) & {\rm for} & \alpha_{2}\le\theta\le\beta({\bf y}). \\
\end{array}
\right.
\]
From Theorem \ref{Thm3}, we see that $Rf(\theta,{\bf y})$  exists for each $y\in \Gamma$  and $\theta\in [\alpha({\bf y}),\beta({\bf y})]$
and $f$  is continuous on $\Omega\cup\Gamma\setminus \Upsilon$  for some countable subset $\Upsilon$  of $\Gamma.$
Let $z_{0}\in \left(z_{1},\min\{z_{2},z_{3}\}\right)$  and $\theta_{0}\in (\alpha_{1},\alpha_{2})$  satisfy $Rf(\theta_{0},{\bf y})=z_{0}.$
Let $C_{0}\subset\Omega$  be the $z_{0}-$level curve of $f$  which has ${\bf y}$  and a point ${\bf y}_{0}\in \partial\Omega\setminus \{{\bf y}\}$  as endpoints.
Let ${\bf y}_{1}\in \Gamma_{1}({\bf y})\cap\Gamma\setminus \Upsilon$  and ${\bf y}_{2}\in C_{0}$  such that the (open) line segment $L$  joining
${\bf y}_{1}$  and ${\bf y}_{2}$  is entirely contained in $\Omega.$
Let $M<\min\{z_{1},\inf_{L} f\}.$
Let $\Pi$  be the plane containing $({\bf y},z_{0})$  and $L\times \{ M\}$  and let $h$  be the affine function on ${\rm I\hspace{-0.2em}R}^{2}$  whose graph is $\Pi.$
Let $\Omega_{0}$  be the component of $\Omega\setminus \left(C_{0}\cup L\right)$  whose closure contains $B_{\delta}({\bf y})\cap \Gamma_{1}({\bf y})$
for small $\delta>0.$
By first choosing ${\bf y}_{2}$  sufficiently near ${\bf y}$  and then choosing ${\bf y}_{1}$  sufficiently near ${\bf y},$  we may assume that
${\bf y}$  is the furthest point on $C_{0}$  between ${\bf y}$  and ${\bf y}_{2}$  away from $L.$
Then $h=M<f$  on $L,$  $h\le z_{0}=f$  on the portion of $C_{0}$  between  ${\bf y}$  and ${\bf y}_{2},$
and $h({\bf y})=z_{0}>Rf(\theta,{\bf y})$  for each $\theta \in [\alpha({\bf y}),\theta_{0}).$
Thus $h\le f$  on $\Omega\cap\partial\Omega_{0}$  and $h>f$  on $B_{\delta}({\bf y})\cap \partial\Omega_{0}$  for some sufficiently small $\delta>0.$

Then there is a curve $C\subset \Omega_{0}$  on which $f=h$   whose endpoints are ${\bf y}_{3}$  and ${\bf y},$  for some
${\bf y}_{3}\in \Gamma_{1}({\bf y})$  between ${\bf y}_{1}$  and ${\bf y},$  such that $h>f$  in $\Omega_{1},$  where $\Omega_{1}\subset\Omega_{0}$
is the open set bounded by $C$  and the portion of $\Gamma_{1}({\bf y})$  between ${\bf y}$  and ${\bf y}_{3}.$
(In Figure \ref{FOUR_F}, on the left,  $\{\left({\bf x},h\left({\bf x}\right)\right): {\bf x}\in C\}$  is in red, $L$  is in dark blue, $C_{0}$  is in yellow, and
the light blue region is a portion of $\partial_{{\bf y}}^{1}\Omega\times{\rm I\hspace{-0.2em}R},$  and, on the right, $\Omega_{1}$  is in light green and
$\Gamma_{2}({\bf y})$  is in magenta.)
Now let $g\in C^{2}(\Omega)$  be defined by $g=f$  on $\Omega\setminus\overline{\Omega_{1}}$  and $g=h$  on $\Omega_{1}$  and observe that
$J(g)<J(f).$  (The functional $J(f)$  includes the area of the blue surface in Figure \ref{FOUR_FC}, which is a subset of $\partial\Omega\times{\rm I\hspace{-0.2em}R},$
and the area of the purple surface of this figure, which is the subset of the graph of $f$  over $\Omega_{1}$  while $J(g)$  does not include the areas
of the blue and purple surfaces and instead includes the area of the green surface on the left side of Figure \ref{FOUR_F},
which is the portion of the plane $\Pi$  over $\Omega_{1}.$)
This contradicts the fact that $f$  minimizes $J.$  Thus it must be the case that $z_{1}=z_{2},$  case (i) of Theorem \ref{Thm3} holds
and $f$  is continuous at ${\bf y}.$
(Notice that the set $\Omega_{1}=\{ {\bf x}\in\Omega_{0} : h({\bf x})>f({\bf x})\}$  could be more complex but the proof is unchanged; if
$V$  is an open set in $\Omega_{0}$  with $h<f$  in $V,$  $h=f$  on $\partial V\cap\Omega$  and $h\le f$  on $\partial V\cap \partial\Omega$
is an ``inclusion'' in $\Omega_{1},$  it does not matter; we still set $g=h$  on $\Omega_{1}$  and $g=f$  on $\Omega\setminus\Omega_{1}.$)
\qed

\begin{figure}[ht]
\centering
\includegraphics{Figure4A.pdf}
\caption{Side View of $\Pi\cap \left(\Omega\times{\rm I\hspace{-0.2em}R}\right)$ (left) and $\Omega_{1}$ (right)  \label{FOUR_F}}
\end{figure}

\begin{figure}[ht]
\centering
\includegraphics{Figure4C.pdf}
\caption{Graph of $f$  over $\Omega_{1}$  (purple) and part of $\partial\Omega\times{\rm I\hspace{-0.2em}R}$  (blue)  \label{FOUR_FC}}
\end{figure}

\vspace{3mm}

\noindent {\bf Proof of Theorem \ref{Thm1}:}
The proof of Theorem \ref{Thm1} is essentially the same as that of Theorem \ref{Thm2};
we replace $f$  with a function $g$  and obtain a contradiction by showing $J(g)<J(f),$  where
\[
J(u)=\int_{\Omega} \sqrt{1+|Du|^{2}} d{\bf x} + \int_{\Omega} \left( \int_{c}^{u({\bf x})} 2H({\bf x},t)  \ dt \right) d{\bf x}
+ \int_{\partial\Omega} |u-\phi| ds
\]
for $u\in BV(\Omega).$

As in the proof of Theorem \ref{Thm2}, we assume $z_{1}<z_{3},$  $z_{1}<z_{2},$  and  $Rf(\theta,{\bf y})$  is strictly increasing on
$(\alpha_{1},\alpha_{2})$  and constant on $[\alpha({\bf y}),\alpha_{1}]$  and $[\alpha_{2},\beta({\bf y})].$
Let $z_{0}\in \left(z_{1},\min\{z_{2},z_{3}\}\right)$  and $\theta_{0}\in (\alpha_{1},\alpha_{2})$  satisfy $Rf(\theta_{0},{\bf y})=z_{0}.$
Extend $H$  so $H\in C^{1}({\rm I\hspace{-0.2em}R}^{2}).$
Let $R>0$  be small enough that $2R|H({\bf x})|\le 1$  for  all ${\bf x}\in B_{2R}({\bf y}).$
Set $\theta_{b}=(\theta_{0}+\beta({\bf y}))/2$  and $T=\{ {\bf y}+r\left(\cos\theta_{b},\sin\theta_{b}\right) : r\in{\rm I\hspace{-0.2em}R}\},$  and let $C(R)$
be the circle of radius $R$  which passes through ${\bf y},$  is tangent at ${\bf y}$  to the line $T,$  and intersects $\Gamma_{1}({\bf y}).$
Let $V(R)$   be the open disk inside $C(R).$
(In Figure \ref{FIVE}, the blue arcs are part of $\partial\Omega,$  locally $\Omega$  lies below these blue arcs, the red rays represent $\theta=\theta_{0},$
and, on the right, $V(R)$  is the yellow disk,  $C(R)$  is the black circle, and $T$  is the green line.)
Notice that (\ref{eq:D})-(\ref{bc:D}) in the domain $V(R)$  is solvable for all $\phi\in C^{0}(C(R)).$

Let $h\in C^{2}\left(\overline{V(R)}\right)$  satisfy $Qh=0$  in $V(R),$  $h({\bf y})=z_{0},$
$h < f$  on $C(R)\cap\Omega$  (recall $Rf(\theta_{b},{\bf y})>Rf(\theta_{0},{\bf y})=z_{0}$),
and $h\le z_{0}$  on $C(R)\setminus \Omega.$
Set $U=\{ {\bf x}\in\Omega\cap V(R) : h({\bf x})>f({\bf x}) \}$  (see Figure \ref{FIVE}) and notice that
$\Gamma_{1}({\bf y})\cap B_{\delta}({\bf y})\subset \overline{U}\subset V(R)\cup \{{\bf y}\}$  for some $\delta>0.$
Now define $g\in C^{0}(\Omega)$  by $g=h$  on $U$  and $g=f$  on $\Omega\setminus U.$
Then $J(g)<J(f)$  as before in the proof of Theorem \ref{Thm2}.  \qed
\vspace{3mm}

\begin{figure}[ht]
\centering
\includegraphics{Figure5E.pdf}
\caption{$U$  (green region; left); $C(R)$  (black circle; right) \label{FIVE}}
\end{figure}

\noindent {\bf Proof of Example \ref{Example1}:}
By Theorem \ref{Thm1}, $f$  is continuous on $\Omega\cup \{(0,0)\}.$  Clearly $f$  is continuous at $(x,y)$  when $(x+1)^2+y^2=\cosh^{2}(1).$
By \cite{Simon}, $f$  is continuous at $(x,y)$  when $(x+1)^2+y^2=1$  and $(x,y)\neq(0,0).$
The parametrization (\ref{PARAMETRIC}) of the graph of $f$  (restricted to $\Omega\setminus \{(x,0):x<0\}$) satisfies $Y\in C^{0}(\overline{E}).$
Notice that $\zeta((0,0))=\{ {\bf o} \}$  (since $\beta((0,0))-\alpha((0,0))=\pi$  and $z_{1}=z_{2}$)  for some ${\bf o}\in \partial E.$
Suppose $G$  in $(a_{2})$  is not one-to-one.  Then there exists a nondegenerate arc $\zeta\subset\partial E$  such that $G(\zeta)=\{{\bf y}_{1}\}$  for some
${\bf y}_{1}\in \partial\Omega$  and therefore $f$  is not continuous at ${\bf y}_{1},$  which is a contradiction.
Thus $f=g\circ G^{-1}$  and so $f\in C^{0}\left(\overline{\Omega}\right).$
(The continuity of $G^{-1}$  follows, for example, from Lemma $3.1$ in \cite{Cour:50}.)
\qed

\begin{rem}
\label{Remark2}
The term ``order of non-uniformity'', used in \cite{Lau}, and the term genre, established in 1912 by Bernstein, adopted by Serrin (\cite[p. 425]{Ser:69})
and used in \cite{JL}, are numbers which measure the variation  from  uniform ellipticity of a quasilinear elliptic operator;
the genre and the ``order of non-uniformity'' for the minimal surface operator are both equal to two ($2$) while
the genre and the ``order of non-uniformity'' for uniformly elliptic operators are both equal to zero ($0$).
We are confused by the invention for a new phrase for an existing and named phenomenon.
\end{rem}

redpajama_set_name: RedPajamaArXiv


text:
Сен-Венсан () — коммуна в Италии, располагается в автономном регионе Валле-д'Аоста.

Население составляет 4846 человек (2008 г.), плотность населения составляет 233 чел./км². Занимает площадь 21 км². Почтовый индекс — 11027. Телефонный код — 0166.

Покровителем коммуны почитается  священномученик Викентий Сарагосский, празднование 22 января.

В городке находится известное казино, открытое в 1946 году.

Демография
Динамика населения:

Администрация коммуны

Галерея

Примечания

Ссылки
 Национальный институт статистики
 Национальный институт статистики

redpajama_set_name: RedPajamaWikipedia


text:
Q: From Array to MySQL Not the best with MySQL, but I'm scraping data from websites with python, and storing the results in a mysql database. All the data I've collected have been placed into custom objects that I created (i.e each object is a car with a respective make, model, year, etc.) Now where I'm having an issue is that each car has a number of images associated with it, and it can range from zero to ten images. The question I have is how can I store these images in a MySQL table and connect it to its respective car.
This is the table I've created for each car so far (python code):
    command = ("""CREATE TABLE `%s`(
            `auctionHouse` varchar(255) NOT NULL,
            `auction` varchar(255) NOT NULL,
            `lot` varchar(255) NOT NULL,
            `year` int,
            `make` varchar(255) NOT NULL,
            `model` varchar(255),
            `description` text,
            `fullName` varchar(255) NOT NULL,
            `salePrice` varchar(255),
            `highPrice` int NOT NULL,
            PRIMARY KEY (`auction`,`lot`));
            """
           )%(auction)

Do I have to create another table for holding images and then when I query the car, use a join?

A: You do not have to create another table, but I would STRONGLY suggest you do.  You specified that you have a finite number of images per car, so you could have 10 columns in your car table.  But what happens when you decide to add more?
If you plan to store your images as Blobs, then most definitely don't create 10 columns; go ahead and create another table.
If you choose to create another table, I would suggest using Innodb and specify the foreign key constraint.
I would also suggest that you use a transaction when when you create the car entry, and associate the images to that car.
You need the id of the car you created to use it as the value in your image table to relate the image to the corresponding car.
http://www.w3schools.com/php/php_mysql_insert_lastid.asp
Then you can use a Join to get all the images related to the car, or you can query the images table by itself if you only need the images and you have the car Id.

redpajama_set_name: RedPajamaStackExchange


text:
Public School at Home Programs…
Alternative Learning Programs Overview ( Public School at Home )
Alternative Learning Programs (ALPs) are public school programs that do not require full-time on-site attendance, but do enroll the student as a full-time student. There are three basic models of ALPs. One type known as Parent Partnership Programs market heavily to homeschoolers and cause us the most concern. Many use homeschooling terminology in their names and in documentation. Most of these programs were also started by homeschooling parents, and some were state or local homeschooling leaders. This has caused much confusion in the homeschooling community.
The Office of the Superintendent of Public Instruction (OSPI) defines the status of a student according to who is in control of supervising the child's education. When a parent enrolls in an ALP, they must fill out a Student Learning Plan and report hours on-site and off-site. According to the OSPI, this places the student under the direct supervision of the public school and therefore makes him/her a public school student. ALPs have always allowed off-site independent study, but this should not be confused with homeschooling.
In the document titled, Washington State's Laws Regulating Home-based Instruction (published by the OSPI), question 11 (part three, page 17):
Are there any circumstances by which a school district may provide the same model of instruction that home-based students receive to students who are full-time public school students?
No. The definition of home-based instruction found in RCW 28A.225.010(4) does not meet direct supervision requirements for the allocation of state funds. Full-time equivalent student is defined in WAC 392-121-121. However, WAC 392-121-183 Finance, General Apportionment: Alternative Learning Experience Requirements authorizes school districts to adopt rules by which a school district may receive full-time enrollment for teaching/learning experiences conducted off-campus. Under such rules a student is enrolled as a public school student. Alternative school district programs must meet minimum program hours and direct supervision requirements to receive state funds.
The History of Alternative Learning Programs
In the early 1990's the State Legislature and the State Office of the Superintendent of Public Instruction (OSPI) began funding programs for alternative learning. The programs were authorized for grades 9 through 12 (RCW 28A150.220 1b) because it was hoped that they would help decrease high school drop-out rates. In 1995 the state superintendent's office passed a revision to the rules (WAC 392-121-183) governing the funding of these programs and made the programs available for grades K-12. This opened the door for some to exploit the funding of these programs. The WAC development team that worked on the revision in 1995 attempted to make it clear that these programs were available for public school students, not for students who were home-based or in a private school. This did not deter those who saw an opportunity to carry out their own agendas, whether that involved marketing the programs for personal profit, getting back their tax dollars, or growing the school district's revenue.
The programs were started in order to provide services to a variety of students, including those who are gifted, home-based, or at-risk. The programs were designed for less on-site hours than a traditional school, with the student doing work off-site or at home. In the case of a child who was not high school age, it was expected that the parents would supervise instruction at home. Some have tried to define this as homeschooling. There is no provision in the home-based instruction law or in any public school law to allow for government-funded homeschools.
There are only three legal ways to educate children in the state of Washington: home-based (homeschooling), private school, or public school. Just as there are many ways to homeschool and a variety of private schools, there are now many models of public school. The question is, who is in direct supervision of the child's education? The only model that allows the parents to be in direct supervision of their child's education is homeschooling.
RCW 28A.200.020 Home-based instruction — The state hereby recognizes that parents who are causing their children to receive instruction under RCW 28A.225.010 (4) shall be subject only to those minimum state laws and regulations which are necessary to insure that a sufficient basic educational opportunity is provided to the children receiving such instruction. Therefore, all decisions relating to philosophy or doctrine, selection of books, teaching materials, and curriculum, and methods, timing, and place in the provision or evaluation of home-based instruction shall be the responsibility of the parent except for matters specifically referred to in this chapter.
In the 2005 legislative session, the legislature passed a law (SB5828) that allowed for the expansion of alternative learning programs (ALPs). The new law allowed for digital learning programs, more commonly known as "virtual schools." As a result, the OSPI was required to write rules to govern these programs.
In addition, during the past couple of years the State Auditor's Office and the legislature (through the members of the Joint Legislative Audit and Review Committee) have been conducting an investigation into these programs due to their many abuses and lack of compliance, specifically within the Parent Partnership Programs (PPP).
Because of the new law for "virtual schools" and the investigation by the legislature and the State Auditor's Office, rules were written for all ALPs and OSPI developed "implementation guidelines" for these rules.
The new law/rules allow for part-time attendance in all ALPs, which means parents are no longer required to enroll in the program full-time, thereby losing their homeschool status and their freedoms under the home-based instruction law. Some school districts are not complying with this new law/rule.
CHN also worked to have a "full disclosure" form required by law so that parents will be informed of the difference between the program and homeschooling under the home-based instruction law RCW 28A.200. Many sites are not adhering to this legal requirement. This form is called the "Statement of Understanding" we have included a sample for your use.
If your program is not complying with these changes please e-mail us at chnow@msn.com and we will try to assist you.
There has been some confusion among parents who have children enrolled in these programs; much of the confusion can be attributed to the fact that many of the PPPs do not comply with existing laws and rules. Many PPPs do not make it clear to parents that 1) The school district is responsible for supervision of the child's education and 2) that all curriculum must be approved by the local school district and cannot be religious, even if the parent bought the material and is using it at home. If these hours are claimed for public funding then the teaching must be secular. This is not a change; this has always been the law/rules for alternative learning programs.
Links to sources:
Implementation Guideline from the Office of Superintendent of Public Instruction (OSPI) – includes new rules governing ALPs, implementation guidelines, and a chart of changes. http://www.k12.wa.us/AlternativeEd/ProgramImplementationGuidelines/default.aspx.


text:
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Gibraltar: nothing short of awesome
Many events take place in the iconic Casemates Square. / SUR
There can surely be nowhere else on earth which offers so much variety, excitement, adventure, history, nature and entertainment in such a small and beautiful space. Gibraltar is a true marvel, in many different ways
Debbie Bartlett |
1 July 201913:30
They say nice things come in small parcels and that certainly applies to Gibraltar, which may be only 6.7 square kilometres in size but is packed full of interesting places to visit and things to do, making it an ideal place to spend a day, a weekend or a longer break. In fact, the fun begins as soon as you arrive: unless you are a boat owner you either walk or drive across the border to this British Overseas Territory from La Línea de la Concepción in neighbouring Spain, or fly in to Gibraltar airport, marvelling at the view of the famous Rock as your aircraft approaches. Either way, you will then have to cross the airport runway to reach the town, by car, bus or on foot, which is a unique and somewhat surreal experience in itself.
Fascinating, from top to bottom
The great thing about Gibraltar is that it is all fascinating. However, that leads to a question for first-time visitors: where to go first?The city centre is a good starting point, especially the iconic Casemates Square, where many events take place throughout the year. Lined with cafés and restaurants, it is an ideal place to sit and plan the day ahead.
The skywalk was inaugurated by Mark Hamill. / SUR
Main Street, which is - rather unsurprisingly - the main street, is historic and charming with modern shops and facilities. By strolling along here you can take in two cathedrals, the King's Chapel, the Convent (home to the Governor of Gibraltar), 'No. 6', which is Gibraltar's equivalent of No. 10 Downing Street and, at the end of Main Street, the Trafalgar Cemetery, a key landmark in local history which commemorates the Battle of Trafalgar in 1805. If you carry on walking you will come to the lovely Alameda botanical gardens and the wildlife park, and en route you will see the cable car base station, in the car park.
One of the most rewarding and exciting experiences on Gibraltar is a trip to the Upper Rock Nature Reserve, especially if you take the six-minute cable car trip. At the top the three terraces have spectacular views of two continents, two bodies of water and three countries. The cable car runs every ten to fifteen minutes from 9.30am to 7.15pm, and you can buy a ticket for the Nature Reserve attractions from the base station or the souvenir shop at the top station. That gives you entry to the Great Siege Tunnels, the City Under Siege exhibition, the Moorish castle and the 100 Ton Gun.
The Nature Reserve ticket also includes entry to the Apes Den. Everybody has heard of Gibraltar's famous barbary macaques, and they are just as cute and furry as you expect. They are also extremely mischievous - hang on to hats, cameras, bags and any food you may have with you. These guys move fast! Despite them appearing to be bundles of fun and although they are used to people being around, it is important to remember that these apes are still wild animals. Treat them with respect, don't crowd them, and be aware of the warning sign: if they feel intimidated they will raise their eyebrows and purse their lips into a round shape. It means they need a bit of space, so just step away so they realise you're not a threat, and you'll all get on harmoniously.
The Gibraltar flag flies over the marina. / SUR
One of Gibraltar's newest attractions, and probably the most thrilling, is the Skywalk, which was inaugurated last year by Star Wars actor Mark Hamill. Entry to the Skywalk is also included in the Nature Reserve ticket. What is it? Well, imagine a walkway 340 metres directly above sea level, which is higher than the tallest point of the Shard building in London, with the floor and balustrade made of four layers of laminated glass. Yep, glass. So it looks as if there is nothing beneath your feet except the drop.
The Skywalk is 2.5 metres wide and projects to a maximum of 6.7 metres. Despite its apparent fragility it is in fact extremely strong. It could bear the weight of five elephants, or 340 people, although no more than 50 people are allowed onto it at the same time. Not to be missed for those who want the sensational experience of floating on air, while looking at truly spectacular views.
If that isn't enough for you and you want even more of an adrenalin rush, the Windsor Suspension Bridge at Royal Anglian Way will undoubtedly provide it. This is another new attraction in Gibraltar and is also included in the Nature Reserve ticket. It is only for the adventurous, stretching 71 metres across a 50-metre-deep gorge. Those who dare to keep their eyes open will be stunned by the views of the Strait, the bay and the city of Gibraltar.
If you don't like heights and you prefer to go downwards instead, you will love Lower St Michael's Cave, an incredibly beautiful natural cave with a lake. This must be booked well in advance, and is available for groups of between five and ten people. Quite fit people, we must add. The cave is fully illuminated, but the tour takes approximately three hours and there is some scrambling and minor climbing with ropes involved.
For those who love boats and the sea, Gibraltar in the summer will be a real treat. The modern marinas are atmospheric at any time of day (and check out the amazing Sunborn floating hotel in Ocean Village, while you're there) and there are environmentally responsible boat trips to spot dolphins and whales. Trips are also occasionally organised to give a close-up view from the sea of the Gorham's Cave complex, the last known home of the Neanderthals, which is now a Unesco World Heritage Site.
These are just a few of Gibraltar's many attractions, so if you're there for a few days and want to relax on a beach for a while you can do that too. This is Gibraltar, so even the beaches are unusual: Camp Bay, in the South District, was originally a quarry and is now a lido, while Little Bay has a small rocky beach under the cliffs. On the East side, Catalan Bay is a pretty Genoan-style village with a lovely cove, or you can choose the more secluded Sandy Bay or Eastern Beach, close to the airstrip, which is the largest in Gibraltar.
Whether you want to go up, down or around the Rock, hang out in the city centre or enjoy the view from the sea, Gibraltaltar is nothing short of awesome.
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How Do I Set My Rates?
Setting session rates is required before conducting your first session. You can do this by completing the following steps: My Network> My Account> Session Types. From here, you can input the session type description, rate, and duration in minutes. You can input as many session types as you like, and all of these are viewable to your consumers on your profile.


text:
package org.spongepowered.server.mixin.world;

import net.minecraft.server.MinecraftServer;
import net.minecraft.world.WorldType;
import org.spongepowered.asm.mixin.Mixin;
import org.spongepowered.common.interfaces.world.IMixinWorldType;

@Mixin(WorldType.class)
public abstract class MixinWorldType implements IMixinWorldType {

    @Override
    public int getSpawnFuzz() {
        return Math.max(5, MinecraftServer.getServer().getSpawnProtectionSize() - 6);
    }

}