Another project to celebrate the new Millennium, the O2 – originally known as the Millennium Dome – is the largest domed structure in the world.
Occupying a prominent site at the northern tip of the Greenwich Peninsula (on the south bank of the River Thames), the building is 365 metres in diameter, 50 metres high at its highest point and features 12 100 metre high masts which hold up the Teflon-coated glass fibre dome using some 45 miles of steel cable.
It was designed by Sir Richard Rogers and his firm with construction commencing in 1997 after the project, conceived by the Conservative Government, was endorsed (and expanded) by the new Blair Labour Government.
The Dome was officially opened on 31st December, 1999, at a 'New Millennium Spectacular' attended by, among others, members of the Royal Family and government.
Throughout the following year the venue hosted what was known as the 'Millennium Experience', a celebration of the beginning of the new millennium which attracted more than six million visitors (a big figure but apparently only half of what was projected initially).
A controversial project since its very inception due to its cost and speculation about its use after the year 2000, the facility was largely devoid of life for several years after the year 2000 (apart from a few major events) but the site, which was eventually sold to consortium Meridian Delta Ltd which turned to then consortium member the Anschutz Entertainment Group (AEG) to oversee the transformation of the facility into a sports and entertainment complex.
The building, rebranded O2 following a deal with the telco company of the same name, reopened in 2007 (the first band to play there in a public show was Bon Jovi and more than 600 bands had played there as of last year) and has since been used for a range of events including music concerts and sports, the latter including some of the indoor sports played at the 2012 Olympic Games including gymnastics and basketball.
As well as a more than 20,000 seat sports arena, the O2 now features music venues, a cinema, bars, restaurants and shops as well as an exhibition space. There's also a 90 minute climb over the top of the Dome for the adventurous.
More than 60 million people visited the O2 since it opened in 2007.


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text:
3x3 success story Ulaanbaatar will make history at Jeddah Final
ULAANBAATAR (Mongolia) - At the core of one of 3x3's biggest success stories, Ulaanbaatar MMC Energy always appeared destined to make the World Tour's biggest stage.
And the Mongolian team will fulfil their wish when they make their historic debut at the Jeddah Final on December 18-19. They are the first team from Mongolia to ever make the season decider.
It was a nervous wait for them during the recently concluded Doha Masters after finishing 11th on the World Tour standings following 3x3's return in Hungary. The top 10 teams had already qualified for Jeddah but Ulaanbaatar MMC Energy - who didn't play in Doha - had to rely on the right results to ensure they received a ticket.
Fortunately for them and 12th ranked Amsterdam, none of the possible contenders made the required semi-finals ensuring Ulaanbaatar MMC Energy ticked another impressive box for 3x3's Cinderella story Mongolia.
With a population of just three million, the North East Asian country is a stellar No.7 on the combined 3x3 rankings and the women's team last year directly qualified for the Tokyo Olympics.
Mongolia is traditionally known for producing athletes for individual events, especially combat sports but the women's 3x3 team made history by becoming the country's first sports team to qualify for the Olympics.
Mongolia's men's team won the FIBA 3x3 Asia Cup 2017 on home soil and were runners-up the following two years.
The continual success of Ulaanbaatar MMC Energy on the World Tour over the past three years has been important for the growth of 3x3 in Mongolia. They are led by the enforcer Delgernyam Davaasambuu, who is built like a bull, and the athletic Dulguun Enkhbat.
They made their World Tour debut at the Utsunomiya Masters 2017 and made a stunning impression with victory over No.1 seed Ljubljana in the opener en route to a surprising semi-final finish.
It was their only appearance on the World Tour that season but they qualified for three Masters in 2018 highlighted by a quarter-final finish in Chengdu. Ulaanbaatar MMC Energy stepped it up last year and memorably broke through on the pro circuit when they stunned the field to win their first-ever 3x3 Challenger with victory in Jeju.
Davaasambuu played bully ball throughout to finish with 39 points in 5 games in an MVP performance that proved he is in the elite 3x3 category.
The triumph came just after their semi-final finish at the Los Angeles Masters but a lack of participation on the World Tour proved costly as Ulaanbaatar MMC Energy finished the season 14th on the standings to agonizingly miss out on the Utsunomiya Final.
They finally made it this time around but their form in Debrecen will cause some alarms after failing to make an impact in their 2 Masters appearances though they lost a tight quarter-final against NY Harlem at the Hungary Masters.
Of most concern was the form slump of Davaasambuu, who scored just 12 points overall and committed almost as many turnovers (9). He will need to recapture his past brilliance for the Mongolians to do what they traditionally do best and cause upsets in Jeddah.
Much of the scoring burden in Hungary fell on Tsenguunbayar Gotov, Mongolia's No.5 ranked player who handled the responsibility superbly.
But Ulaanbaatar MMC Energy, who boast 4 of Mongolia's top 5 ranked players, will need a more even spread to go deep in Saudi Arabia. They will also have to rediscover their physicality and passion - characteristics that make them normally so hard to beat.
But don't write them off just yet.
After all, these Mongolians have a special ability to produce their best against all the odds.
Nauris Miezis named FIBA 3x3 World Tour 2020 MVP
Wed, Dec 30, 2020
Most memorable stats of FIBA 3x3 World Tour 2020
Assembling the FIBA 3x3 World Tour 2020 Dream Team of the Season
Mon, Dec 28, 2020
Calm Miezis is 3x3's ultimate finisher in the clutch
Why 3x3's new champs Riga won the World Tour Final
Assembling the FIBA 3x3 World Tour Jeddah Final 2020 Dream Team
7 things we learnt from an epic FIBA 3x3 World Tour Final
Riga win first-ever FIBA 3x3 World Tour title at Jeddah Final 2020 on unbelievable Tissot Buzzer Beater
Miezis named FIBA 3x3 World Tour Jeddah Final MVP

redpajama_set_name: RedPajamaCommonCrawl


text:
import asyncio
import glob
import logging

import serial

logger = logging.getLogger(__name__)

BAUDRATE = 460800
# BAUDRATE = 230400
# BAUDRATE = 115200

SERIAL_DEVICE_PATTERN = '/dev/*.usbmodem*'

globals = {
    'serial_f': None,
    'video_buffer': None,
    'video_frame': 0
}

class SerialPortError(Exception): pass

def open_serial():
    for dev in glob.glob(SERIAL_DEVICE_PATTERN):
        try:
            s = serial.Serial(port=dev, baudrate=BAUDRATE, timeout=2)
            logging.info("Opened serial port {}".format(dev))
            return s
        except (serial.SerialException, OSError):
            logging.warn("Couldn't open serial port {}".format(dev))

    raise SerialPortError

def reset_to_top(serial_f):
    arduino_status = 1  # 0 indicates ready for entire frame
    try:
        while arduino_status != 0:
            if serial_f.inWaiting() > 0:
                arduino_status = ord(serial_f.read())
            else:
                serial_f.write(bytes((0,)))
                logger.info('byte')

    except Exception as e:
        logger.exception(e)
        init(globals['video_buffer'])

def write_serial():
    serial_f = globals['serial_f']
    video_buffer = globals['video_buffer']
    reset_to_top(serial_f)
    serial_f.write(video_buffer.uint8)

def init(video_buffer):
    loop = asyncio.get_event_loop()
    globals['video_buffer'] = video_buffer
    try:
        globals['serial_f'] = open_serial()
        loop.add_reader(globals['serial_f'].fileno(), write_serial)
    except SerialPortError:
        logger.critical("\n\nCOULD NOT OPEN SERIAL PORT\n\n")

redpajama_set_name: RedPajamaGithub


text:
'We weren't surprised': Students react with a weary shrug to city's plans to close Bronx school where boy was fatally stabbed
By Patrick Wall Dec 19, 2017, 6:04pm EST
Patrick Wall
Some students at the Urban Assembly School for Wildlife Conservation in the Bronx heard the news from a superintendent, who went from classroom to classroom informing them that their school would be closed after this school year. Others learned secondhand from friends, or from letters they were told to bring home.
But if it came as a surprise that their school — which serves fewer than 500 students in grades 6 to 12 — was among 14 troubled schools that the New York City education department announced Monday it intends to shutter, the decision nonetheless made sense to many students: One of their peers had fatally stabbed one boy and seriously wounded another during history class this September. The violence shocked the city and school community, yet it did not seem totally out of place at a school where student clashes were commonplace and instability was the order of the day.
"People were happy that the school is closing because the school is not really a great school," said Sarah Vega, a 12th-grader, during an interview Monday with four schoolmates that was arranged by the group, Teens Take Charge. "We weren't surprised," added Chloe Oliveras.
The school has been on the decline in recent years. Its state test scores have sunk further and further below the citywide average, with just 12 percent of students passing the English tests and 5 percent passing math last school year. (The high school's 75 percent four-year graduation rate is on par with the city average, and 8 points above the Bronx rate.)
At the same time, it has churned through several leaders and, since last academic year, lost over 100 students — including about 45 who received department permission to transfer in the wake of the stabbing. In addition, just five students listed the school as their first choice when applying to high schools for next year.
In an email to reporters Monday, the education department's press secretary, Toya Holness, listed steps the department had taken to stabilize the school, like boosting staff training, installing a new principal, and dispatching officials, including schools Chancellor Carmen Fariña, to visit.
"Despite these additional interventions, there continues to be instability for students and staff," Holness wrote, "and the Chancellor has determined that Wildlife students will be better served at another school."
The school has been open since 2007, but students said a feeling of disorder began to take hold under Astrid Jacobo, who was principal from December 2015 until she was removed in October. In a survey last academic year, just 36 percent of teachers said they considered the principal an effective manager. (Jacobo did not respond to an email seeking comment.)
Meanwhile, only 55 percent of students said they felt safe in the school's hallways and cafeteria, and just 22 percent said most students treat each other with respect. On Monday, students said fights were common, as were requests for "safety transfers" to other schools.
"It was like no discipline whatsoever," Chloe said. Added Sarah: "I don't feel safe in that school."
Wildlife Conservation had started to implement "restorative justice" practices, which push students to reflect on their misbehavior and try to repair any harm they've caused. Part of that effort involved a "justice panel" of students who would interview their classmates about their actions, try to figure out what caused them, and come up with "sanctions," which might include a written reflection or an apology letter to a teacher.
"Some kids take it seriously, some kids don't," said Sarah, who is on the panel. "Usually they don't listen to us because obviously they're not going to listen to someone who's so close to their age."
On Monday, Sarah added, the panel heard from a girl who had flown into a rage after a classmate flung a piece of paper at her; during her outburst, she threatened to throw a chair at her teacher.
The incident she described was eerily reminiscent of the one on Sept. 27, when 18-year-old Abel Cedeno reportedly snapped during history class after being pelted with bits of paper and pencils, before stabbing 15-year-old Matthew McCree to death and seriously wounding Ariane Laboy, 16. Cedeno, who has pleaded not guilty to manslaughter charges, later said that other students had taunted him with racist and homophobic slurs.
After the attack, the city sent in grief counselors and beefed up security, which included adding metal detectors. But, the students said, things still didn't feel right.
They had to walk past the classroom where their friend was killed — the "crime scene," as Chloe put it. Later, students objected when the administration wanted to reopen the room, so instead it was converted into an office, the students said. At one point, feeling unheard and out of sorts after the stabbing, some students held a silent sit-in in the hallway.
"Everyone was crying and throwing tantrums," in the days after the stabbing, said Miarah Cabassa, also in 12th grade. "We all was hurt."
In a statement, Frank Giaimo, the acting principal who replaced Jacobo, said he has focused on making sure all students and staff feel safe and welcomed at the school.
"Over the last several weeks, I've spent time getting to know the community and together, we will continue our work to provide students with high-quality instruction, support teachers and partner with families," he said.
Separately, the education department has promised to work individually with students at the closing schools to ensure they are offered spots at higher-performing schools next year. (The normal high-school application period ended Dec. 1.)
Seniors who are preparing to graduate from the School for Wildlife Conservation will never get that second chance to attend a safe, orderly high school. But for some younger students like Jaidan Oliveras, a 9th-grader who learned of the city's plans from his sister, Chloe, it felt like a reprieve.
"I was just happy because I'll get to transfer because I don't want to be at that school," said Jaidan, who has attended the school since the 6th grade. "At first, I actually enjoyed it. But at the end, I started getting tired of it."


text:
Article Hindi
It is a weird kind of disappointment: Sara Ali Khan on Kedarnath not being released in Uttarakhand
The Kedarnath actress says that the state has given her so much and the least she could have given them back is her film.
Actress Sara Ali Khan, who made her debut with Abhishek Kapoor's Kedarnath, is disappointed over the film not being released in the state of Uttarakhand.
Fearing violence, the district magistrates of seven districts — Dehradun, Haridwar, Nainital, Udham Singh Nagar, Pauri, Tehri and Almora — had decided to not let the film be released. Khan has been appreciated for her maiden film and the actress is disappointed that the people of Uttarakhand will not be able to see the film.
"If I am being honest, yes I am. It is a weird kind of disappointment that it hasn't come from anything else other than the fact that Uttarakhand gave me a lot. Those 45 days in Kedarnath have given me a lot, and the least, the bare minimum that I could have done is given them back Kedarnath," replied Khan when we asked about the film not being released in the north Indian state.
Directed by Abhishek Kapoor, Kedarnath is a love story of a Hindu Brahmin girl and a Muslim porter, set in the backdrop of the devastating 2013 Kedarnath floods. There were apprehensions around the film, and one politician demanded a ban on the film, accusing it of promoting love jihad, a term coined by right-wing extremist Hindu groups for marriages between Muslim men and Hindu women.
After watching the trailer, there was concern expressed over the film suggesting that the floods were perhaps a curse, as the fictional character Mukku's love is not accepted.
However, after watching the film, there was nothing in the film that could hurt any sentiments. Director Kapoor has observed great sensibility while dealing with a sensitive issue and not disrespecting the victims of the floods. Neither Kapoor's characters nor his plot could be deemed offensive.
"It is interesting that you bring this up because that was one of my favourite things about this film. My character is so relatable. She is this high energy girl, she goes through this hardship in life but the setting is so new. Her family dynamics is so new, her world is so new. I think it is really beautiful when there is something familiar in a new setting. Because you have the novelty of an experience as well, but you also have the connection of familiarity which is what I think resonated with Kedarnath," said Khan.
Religion is a sensitive issue and seldom do we get to see inter-faith love stories in Hindi cinema. Would Kedarnath's good show bust the adverse perception of inter-faith love stories?
"I don't think anything is a formula. The only thing that I follow is conviction. I don't know if anything is safe or unsafe. The reason why Kedarnath has resonated is because the entire team working with this film has conviction in what we are saying. In the face of larger things in life, you don't think about being Hindu, Muslim, rich, poor. All you think about is being together," said the actress.
Khan has Simmba coming up on 28 December, and she has also been rumoured to feature alongside Tiger Shroff in Baaghi 3. She clarified that the Baaghi 3 rumour wasn't true.
From the time she was was rumoured to emulate her parents, actors Saif Ali Khan Khan and Amrita Singh, there's always been speculation that her mother Amrita takes an active interest in her daughter's choice of films. Khan said that the while her mother gives her advice, eventually she lets her make the final decision.


text:
A low-cost horizontal current bipolar transistor (HCBT) technology for the BiCMOS integration with FinFETs
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A Low Gate Turn-OFF Impedance Driver for Suppressing Crosstalk of SiC MOSFET Based on Different Discrete Packages
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Alternatives to Coscheduling a Network of Workstations
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Network of Workstations
Message Passing Interface
Aluminum oxide free-standing thin films to enable nitrogen edge soft X-ray scattering
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X ray scattering
X rays
Always-On Speech Recognition Using TrueNorth, a Reconfigurable, Neurosynaptic Processor
Tsai, W. Y., Barch, D. R., Cassidy, A. S., Debole, M. V., Andreopoulos, A., Jackson, B. L., Flickner, M. D., Arthur, J. V., Modha, D. S., Sampson, J. & Narayanan, V., Jun 1 2017, In : IEEE Transactions on Computers. 66, 6, p. 996-1007 12 p., 7750640.
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A Matrix-Based Approach to Global Locality Optimization
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A maximum entropy framework for semisupervised and active learning with unknown and label-scarce classes
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A metamaterial-enabled design enhancing decades-old short backfire antenna technology for space applications
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A method for the mass production of ICF targets
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A micro-vibration measurement method based on isostrain cantilever and cascade self-demodulated FBG pair
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A mobility-prediction-based relay deployment framework for conserving power in MANETs
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A model of network neutrality with usage-based prices
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A model of the spread of randomly scanning Internet worms that saturate access links
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A Multi-Cycle Switching Technique for Efficient Ultrasonic Wireless Power Delivery
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A multi-objective memetic optimization ap-proach to the circular antenna array de-sign problem
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A Multitask Grocery Assist System for the Visually Impaired: Smart glasses, gloves, and shopping carts provide auditory and tactile feedback
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An active damping method to reduce output impedance of converters in voltage source cascaded system
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An Adaptive Acoustic Echo Canceler for Hands-Free Teleconferencing
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An adaptive power-conserving service discipline for bluetooth (APCB) wireless networks
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An algorithm and a stability theory for downdating the ULV decomposition
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An alternate design paradigm for low-power, low-cost, testable hybrid systems using scaled LTPS TFTs
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An Alternative Approach to Call/Connection Control in Broadband Switching Systems
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Analysis and design of broad-band single-layer rectangular U-slot microstrip patch antennas
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Analysis and numerical modeling of a 20W microwave electrothermal thruster
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Cavity resonators
Analysis and optimization of gyrokinetic toroidal simulations on homogenous and heterogenous platforms
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Analysis for Crosstalk of SiC MOSFET with Different Packages in a Phase-Leg Configuration and a Low Gate Turn-Off Impedance Driver
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Analysis of chamber simulations of long collecting probes in high-speed dense plasmas
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Analysis of DIBL Effect and Negative Resistance Performance for NCFET Based on a Compact SPICE Model
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Analysis of error recovery schemes for networks on chips
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Analytical Determination of Conduction and Switching Power Losses in Flying-Capacitor-Based Active Neutral-Point-Clamped Multilevel Converter
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Analytical determination of conduction power loss and investigation of switching power loss for modified flying capacitor multicell converters
Khoshkbar-Sadigh, A., Dargahi, V. & Corzine, K., Feb 10 2016, In : IET Power Electronics. 9, 2, p. 175-187 13 p.
Analytical Expressions for the Mutual Coupling of Loop Antennas Valid from the RF to Optical Regimes
Nagar, J., Lu, B. Q., Pantoja, M. F. & Werner, D. H., Dec 2017, In : IEEE Transactions on Antennas and Propagation. 65, 12, p. 6889-6903 15 p., 8046065.
Analytical model and performance analysis of a network interface card
Cherukuri, N., Kandiraju, G. B., Gautam, N. & Sivasubramaniam, A., Jan 1 2004, In : International Journal of Modelling and Simulation. 24, 3, p. 179-189 11 p.
Queueing networks
Model Analysis
Analytical Model
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Analytical modelling of point-to-point CCC-HVDC revisited
Kaur, J. & Chaudhuri, N. R., Jan 1 2019, In : IET Generation, Transmission and Distribution. 13, 6, p. 916-926 11 p.
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Liang, M., Zheng, Q., Li, Y. & Ba, T., Jan 10 2017, In : Diangong Jishu Xuebao/Transactions of China Electrotechnical Society. 32, 1, p. 148-158 11 p.
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Analytical surrogate model for the aberrations of an arbitrary GRIN lens
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40 minutes before the race I was all ready to go with my transition laid out so I thought I would top up my tyres with air. I saw the mechanic put on the valve of the pump and then the next second he had taken the wheel valves apart. My deep section front wheels has a couple of valve extenders which apparently intrigued him so he thought he would have a look. I couldn't get the threads to re-align, and he continued to take more parts off the wheel. I couldn't get the threads to go back in and we couldn't get much air in without it so he handed my a spare wheel, complete with reflectors and spoke-e-dokes for the race. Brilliant. The language barrier hindered my efforts to explain that the wheel was fine and just needed un-buggering. I found an English speaker just before we ran to the start and I hope that in the 40mins that the swim would take that they would be able to fix it and when I ran to T1 my bike would have the wheel back on. Alas, it did not.
I knew what I had to do, separate what was in my control and what was not and get on with it. This really wasn't easy. I was in China, with a bike I had never ridden (but was very grateful to have!), and without my race wheels, I was now riding against some top athletes with a less than ideal set up. At least I was more fortunate than Erich who made one last adjustment to his wetsuit sleeve and wripped a huge gash in the arm like the incredible hulk. Just as we were being called through to the start line he was wrapping up his arm in electrical tape.
The swim was cold, for the Americans and Australians, warm for the British, though they had threatened to cut down the distance and indeed did so for the age group athletes. It was a 4 lap swim with a beach exit each lap. My goggles were full of water for the first two laps which meant I was a nuicanse to myself and the group. So I just hung out at the back of the lead group which was going at a pretty comfortable pace with only Kieran Doe getting away.
30m after the mount line was a little 10% hill which was fun. A sweeping coastal path was the first part of the route which was very enjoyable but needed great handling skills to maximise speed, which I did not have. The was also a 2km hill mid way round and some super fast descents - a real bike riders course, really enjoyable but I was really struggling to find any kind of rhythm and was slow.
My running legs felt good on the 4 lap out and back course. I was keeping pace with the best runners for 2 laps, and even though I didn't feel it, the times show that I slowed on the last lap. By this time I was still chasing hard to run down Lee who had blown after two laps in an effort to get in the prize money. But I was too slow and was 200m short of catching him.
I really enjoyed the race. Great venue, challenging course and just a lot of fun. My swim and run were descent enough, the bike split was what let me down. The new bike was running great, the random wheels less so, but in the end I just didn't have it in the legs, simple as that.


text:
Q: How can we show that the limit point $x$ of this sequence $x_n\in B_n$ is contained in $\bigcap_nB_n$? Let

*

*$(E,d)$ be a locally compact second-countable metric space;

*$\tau$ denote the topology on $E$ generated by $d$;

*$(K_n)_{n\in\mathbb N}\subseteq E$ such that $K_n$ is $\tau$-compact, $K_n\subseteq K_{n+1}^\circ$ and $d(\partial K_n,\partial K_{n+1})>0$ for all $n\in\mathbb N$;

*$\Delta\not\in E$ and $\tau^\ast$ denote the Alexandroff one-point extension of $(E,d)$ by $\Delta$; i.e. $$\tau^\ast=\tau\cup\{E^\ast\setminus B:B\subseteq E\text{ is }\tau\text{-compact}\}$$ ($\tau^\ast$ is metrizable);

*$x:[0,\infty)\to E^\ast$ and $$\zeta:=\inf\underbrace{\{t\ge0:x(t)=\Delta\}}_{=:\:I}.$$
Assume $$\forall t\ge0:x(t)\in E\Rightarrow(\forall N\in\tau:x(t)\in N\Rightarrow\exists\delta>0:x([t,t+\delta)\subseteq N)\tag1.$$ Let $$\sigma_n:=\inf\underbrace{\{t\ge0:x(t)\not\in K_n^\circ\}}_{=:\:I_n}$$ and $$\tau_n:=\inf\underbrace{\{t\ge0:x(t)\not\in K_n\}}_{=:\:J_n}$$ for $n\in\mathbb N$ and $$x(\infty):=\Delta.$$

Since $(\tau_n)_{n\in\mathbb N}$ is nondecreasing, $$\tau_n\xrightarrow{n\to\infty}\sup_{n\in\mathbb N}\tau_n=:\tau\tag2.$$ How can we show that $\tau=\zeta$?

I was able to show the following:

*

*If $n\in\mathbb N$, then $$\tau_n\le\sigma_{n+1}\le\tau_n+1\tag3.$$

*If $n\in\mathbb N$ with $\tau_n<\infty$, then $$\tau_n\le\zeta\tag4.$$

*If $n\in\mathbb N$, then $$x(\sigma_{n+1})\not\in K_n\tag5.$$

*Either $\zeta\in I$ or $I=\emptyset$.

By $(3)$, it must necessarily hold $$\sigma_n\xrightarrow{n\to\infty}\tau\tag6.$$ But now I begin to struggle ... Maybe we should proceed by showing that $x^\ast(\tau)=\Delta$. If $\tau=\infty$, then this is trivial by definition. So, assume $\tau<\infty$. Then $\tau_n<\infty$ for all $n\in\mathbb N$ and hence $$\tau\le\zeta\tag7$$ by $(4)$. If $\tau=\zeta$, then $x(\tau)=\Delta$ by the last bullet point above.

Now, please assume that $$x(\sigma_n)\xrightarrow{n\to\infty}x(\tau)\tag8.$$ Noting that $$\{\Delta\}=E^\ast\setminus E=E^\ast\setminus\bigcup_{n\in\mathbb N}K_n=\bigcap_{n\in\mathbb N}(E^\ast\setminus K_n)\tag9,$$ is this enough to conclude?

redpajama_set_name: RedPajamaStackExchange


text:
'Supergirl' Season 2 Soundtrack Details
Posted: October 9, 2017 by filmmusicreporter in TV Music Albums
Tags: Blake Neely, score, Soundtrack, Supergirl, The CW
WaterTower Music and La-La Land Records will release the second soundtrack album for the CW drama Supergirl. The album features selections of the original music from the show's second season composed by Blake Neely (The Flash, Arrow, Riverdale, Legends of Tomorrow, Blindspot, Everwood). The soundtrack will be released digitally tomorrow. A CD version is now available for pre-order on Amazon. The labels have previously released a soundtrack album featuring Neely's music from the series' first season. Supergirl is developed by Greg Berlanti, Ali Adler & Sarah Schechter based on the DC Comic characters and stars Melissa Benoist, Calista Flockhart, Mehcad Brooks, David Harewood and Chyler Leigh. The show is returning for its third season tonight on the CW. Visit the official show website for updates.
Here's the track list of the album:
1. Superman (2:17)
2. Famous Cousins Working Together (2:01)
3. First Day as a Reporter (2:14)
4. Goodbye to Cat (2:05)
5. Fighting Fire with Fire (2:45)
6. Cage Fighting Aliens (2:22)
7. Mon-El's First Day as an Intern (1:19)
8. Always a Sidekick (2:17)
9. Roulette / Thing from Another Planet (2:46)
10. Guardian (2:37)
11. Alex Tells Maggie (1:32)
12. Virus Threat (3:07)
13. Sense About Her (2:46)
14. Star-Crossed Martians (3:14)
15. Mind Meld (3:15)
16. The Luthors (3:09)
17. Mxyzptlk (2:45)
18. Duel with Mxyzpttlk (3:04)
19. Cadmus (1:39)
20. All I Thought About (2:53)
21. Saving Her Sister and the Ark (2:48)
22. Mon-El's True History (3:58)
23. Why Are You Doing This? (2:58)
24. Almost Losing Alex (2:01)
25. Jimmy Bonds with Marcus (2:31)
26. Daxamites Invade (1:52)
27. Forcing a Marriage (1:54)
28. Fighting Superman (2:47)
29. Releasing the Lead / Mon-El Says Goodbye (3:35)
30. I Am So Proud of You (1:49)
31. Thirty Five Years Ago (1:23)
Cindylover1969 says:
Andrew Kreisberg, not Sarah Schechter.


If you like foods that melt, then stock up on various kinds of cheese. There's mozzarella, cheddar, Swiss and Monterey Jack. You can get cheese in block form, shredded or sliced. There are numerous uses for cheese in the home. Top off a casserole, add shredded cheese to a taco or enjoy a grilled cheese sandwich with multiple types of cheese on it. Find coupons on Giving Assistant to enjoy cheese from Kraft and other companies at reduced prices. A promo code can also be used to get cheese in a bulk order so that you have plenty in your refrigerator.


Nacel's student exchange programs in Brazil offer the unique opportunity to live with a local family and study at a local high school. Like just another Brazilian teenager!
If you have an intermediate level of Portuguese and want to improve it as much as you can, come and participate in this program. Live an unforgettable experience and discover the educational system of South America's largest country.
What should I expect from this High school exchange in Brazil?
The student can study one semester or one academic year at one High school located in Greater São Paulo, Brazil.
Students will live with Brazilian host families who will welcome them as if they were a new member of the family.
To make it easier for you to adapt to Brazil, you will benefit from a three-month Portuguese course.
As soon as you arrive in the country, you will attend an orientation session from a local coordinator who will provide you with assistance and support throughout your stay. We will keep your parents informed of your adventures in the country.
Although, graduating from a high school in Brazil is not possible, you can still participate in this experience. International students from all over the world study in Brazil to immerse themselves in the lifestyle of Brazilians and to get their world-famous accent!
Sign up for this authentic exchange program in Brazil with Nacel!
Please request more information through the contact form.
Saying good bye to my parents was difficult; I mean I was not going to see them in a long time and I kept wondering how was it going to be to get up in another house and say "good morning" to a family that wasn't my family. That really scared me a lot. When I cross the gate of the plane, I looked into myself and notice that I was going to experience a huge change. I was not going to be a teenager anymore that relies on his parent´s to get everything that I need and also most importantly, I realized that I needed to start to make my own life. When I arrived to Brazil something strange happened. I was no longer afraid, it was like something changed in me and made me feel that everything was going to be fine. When I first met my Brazilian father I knew that this experience was going to be incredible and when I met my Brazilian mother I felt happy and secure. Suddenly I felt that a needed to know this lovely country from the top to the bottom and did my thing. Brazil it's a country full of contradictions, on one hand it's a growing economy with a first class growing economy and population; on the other hand they have also people that lives in extreme poverty. That reminded me of Mexico a lot, and also made me thought, what it´s going wrong in the world? I mean, how aren't people realizing what it´s going on? They live like they're in a bubble that cover´s them from the reality that their living? I got very angry with myself, because I had to go to another country to realized that Mexico it´s in a very bad situation too. I guess that the big difference about Brazil and Mexico it´s that Brazil wants to make a change and the ones that are searching for it are the young ones. I saw how the teenager´s wanted the best for their future and how they were searching for ways to improve the economy, health and most importantly the education. They are doing it and Brazil it´s changing because of them. I really didn't know why I chose Brazil to make my academic exchange, but something inside me told me that Brazil had a lot to show me and teach me. My feeling weren't wrong at all. Now I have tons of new ideas and a new sense of change and I don't know how, but I would like to share these with my family, friends and the entire world. So now I know why I chose Brazil, and I couldn't be happier with my choice. I really would have loved to stay a little bit more and without a doubt I would love to go back, mainly to visit my Brazilian family and also my new friends that I really love and also consider them part of my family and why not to dance a little bit of samba.


text:
Winding Through The Milky Way
An album of new songs coming from the heart of the Earth.
The opening song reaching into our primordial center and pulling forward into the conscious, the memories of our ancestors to wake us to our destiny.
Winding Through the Milky Way marks the culmination of an ongoing dream. Harjo found her songs first through poetry, and published seven award winning books of poetry, including How We Became Human, New and Selected Poems. She picked up a saxophone in the late 80's and put together the band Joy Harjo and Poetic Justice. With the band she released her first CD Letter from the End of the Twentieth Century , now a classic. Two solo CD's followed: Native Joy for Real, and She Had Some Horses. Among Harjo's honors are the Eagle Spirit Achievement Award for overall contributions in the arts, the New Mexico Governor's Award for Excellence in the Arts and the Lifetime Achievement Award from the Native Writers Circle of the Americas. She performs internationally solo and with her new band. She premiered a reading of her one-woman show (with original music), Wings of Night Sky, Wings of Morning Light at the Public Theater in NYC in 2007. She co-wrote the signature film of the National Museum of the American Indian, A Thousand Roads. Harjo writes a column "Comings and Goings" for her tribal newspaper, the Muscogee Nation News. She lives mostly in Honolulu, Hawai'i where she is a member of the Hui Nalu Canoe Club, and in Albuquerque, New Mexico. Produced by Larry Mitchell.
Fast Horse Recordings 2008, Mekko Productions Inc., 2009
If you'd like an authographed CD, order through PayPal.
If purchasing through PayPal — CDs are $14.99 plus $3 shipping and handling (U.S. orders only). A PayPal account is not required to pay through PayPal. All Major Credit cards accepted.If you'd like your CD autographed, please put it in the notes when processing your payment through PayPal or send us an email requesting your CD by autographed.
*For larger or international orders, Contact for shipping and handling prices.
Cover Art by Joy Harjo
Layout and Design: Teri Hitt
Opening Song for the Maker
Rabbit Is Up to Tricks
This Is My Heart
No Huli
Flute Loop One
Flute Loop Two
Witchi Tai To
Goin' Home


text:
\section{ \bf Introduction and main results}
Let $\sigma : \Re \rightarrow \Re $ be a continuous function and $B$
be an $\Re$-valued standard Brownian motion on a complete filtered
probability space $(\Omega, {\mathcal F},\{{\mathcal F}_t\}_{t\in
[0,1]}, {\bf P})$ satisfying the usual conditions. For $x\geq  0 $,
we consider the following   stochastic differential equation on
$\Re_+=[0, +\infty )$ with reflecting boundary condition:
\begin{eqnarray}
X_t(x)=x +\int^t_0 \sigma (X_s(x))\circ dB_s +L^x_t, \ \forall \ t
\in [0,1],
\end{eqnarray}
where $\circ $ denotes the Stratonovich integral. A
 pair $ (X_t(x), L_t^x, t\in [0,1])$ is called a solution to equation (1.1)
if \\
(i)$ X_0(x)=x$, $ X_t(x)\geq 0 $ for $t\in [0,1]$,\\
(ii) $X_t(x)$, $ L_t^x $ are continuous and adapted to $ \{{\mathcal
F}_t\}_{t\in [0,1]}  $,\\
(iii) $L_t^x  $ is non-decreasing with $ L_0^x=0 $ and
\begin{eqnarray}
\int^t_0 \chi_{\{ X_s(x)=0\}}dL_s^x=L_t^x,
\end{eqnarray}
(iv) $ (X_t(x), L_t^x)$ satisfies Eq.(1.1) almost surely for every
$t\geq 0$.

 There now
exists a considerable body of literature devoted to the study of
reflected stochastic differential equations(see
\cite{s3,s4,s5,s6,s7,s9, s8} and references therein ). It is
well-known that Eq.(1.1) has a unique solution for any given initial
value $x\geq 0$ if $\sigma $ and its derivative are Lipschitz
continuous
functions.\\
\vskip 0.3cm

Now consider the following question: Does there still  exist  a pair
$ (X_t, L_t, t\in [0,1])$ to solve Eq.(1.1) if the initial value is
an arbitrary non-negative random variable $Z$ which may depends on
the whole Brownian paths ? \vskip 0.3cm The answer is not
immediately clear because one needs to deal with anticipating
stochastic integration. The main purpose of the present paper is to
give an affirmative answer to the above
 question. More precisely, our main result is the following
\begin{Them}
Assume that the function $\sigma $, its derivatives $ \sigma' $ and
$ \sigma''  $ are Lipschitz continuous, and $Z$ is a nonnegative
random variable. Then there is a pair $ (X_t(Z), L_t^Z, t\in [0,1])$
that solves the following anticipating reflected  SDE,
\begin{eqnarray}
X_t(Z)=Z +\int^t_0 \sigma (X_s(Z))\circ dB_s +L^Z_t, \ \forall \ t
\in [0,1],
\end{eqnarray}
and satisfies \\
(i)$ X_0(Z)=Z$, $ X_t(Z)\geq 0 $ for $t\in [0,1]$,\\
(ii) $X_t(Z)$, $ L^Z_t $ are continuous,\\
 (iii) $L^Z_t  $ is
non-decreasing with $ L^Z_0=0 $ and
\begin{eqnarray}
\int^t_0 \chi_{\{ X_s(Z)=0\}}dL^Z_s=L^Z_t.
\end{eqnarray}
\end{Them}
Where the stochastic integral in (1.3) is interpreted as
anticipating Stratonovich integral. Let us now recall the
definition. For any $t\in [0,1]$, let $\pi$ denote an arbitrary
partition of the interval $[0,t]$ of the form: $\pi = \{0=t_0< t_1<
\cdots < t_n=t \} $. Let $ \|\pi \|=\sup\limits_{0\leq k \leq
n-1}\{(t_{k+1}-t_k ) \}$ denote the norm of $\pi $. For a stochastic
process $ f=\{ f_s, \ s\in [0,1] \}$, we define its Riemann sums
$S_\pi(f, t) $ by
\begin{eqnarray}
S_\pi(f, t)=\sum^{n-1}_{k=0} \frac{1}{t_{k+1}-t_k}\bigg (
\int^{t_{k+1}}_{t_k} f_s ds  \bigg )( B_{t_{k+1}}-B_{t_k}).
\end{eqnarray}
We have the following
\begin{Def}
We say that a stochastic  process $ f=\{ f_s, \ s\in [0,1] \}$ such
that $ \int^1_0 \chi_{\{s\leq t \}} |f_s | ds < +\infty $ a.s. is
Stratonovich integrable if the family  $S_\pi(f, t) $ converges in
probability as $\|\pi \|\rightarrow  0 $. In this case the limit
will be called the  Stratonovich integral of the  process $ f $ on
$[0,1]$ and will be denoted by $ \int^t_0 f_s\circ dB_s $.
\end{Def}
Let us now describe our approach. To prove Theorem 1.1, the natural
idea is to replace $x$ in (1.1) by the initial random variable $Z$
and prove that the pair $(X_t(Z), L_t^Z)$ satisfies the anticipating
SDE. To achieve this, the key is to establish the following
 substitution formula
\begin{equation}
\int^t_0 \sigma (X_s(x))\circ dB_s\big |_{x=Z}=\int^t_0 \sigma
(X_s(Z))\circ dB_s
\end{equation}
for all $t\in [0,1]$.\\
To obtain (1.6), it seems that we can not apply the existing
substitution formula in the literature (see [9],[10]) because the
regularity of the solution $X_t(x)$ of (1.1) with respect to the
initial value $x$ is not good enough to satisfy the required
hypothesis. Instead, we prove (1.6) by showing
 the  uniform convergence (w.r.t. x) of
the corresponding Riemann Sums  $S_\pi(t,x)$ (Theorem 5.1 in section
5 below ). The Garsia, Rodemich and Rumsey's Lemma and   moments
estimates for one-point and two-point motions(Theorem 4.1, 4.2 in
section 4 below ) will play an important role. For the proof of
(1.4), we need to study the continuity of the random field $\int^t_0
l(X_s(x))dL_s^x$ for any continuous function $l(y)$  on $(0, \infty
)$ with compact support.

\vskip 0.3cm This paper is organized as follows. Section 2 is to
study the  regularity of  the solution $ ( X_t(x), L^x_t ) $ of
Eq.(1.1). In Section 3 we prove the continuity of functional of
local times. Section 4  is to study moments estimates for one-point
and two-point motions. In Section 5 we prove  the uniform
convergence (w.r.t.x) of the Riemann Sums
      $S_\pi(t,x)$. The proof of Theorem 1.1 will be completed in Section 6.

\setcounter{equation}{0}
\section{\bf Regularity of the solution $ ( X_t(x), L^x_t ) $ of Eq.(1.1)}
We first recall the deterministic Skorohod  problem(see\cite{s1}).
\begin{Def}
Let $ y\in \big\{f\in C( [0,1]\longrightarrow \Re ), \ f(0)\geq 0
\big \}$. We will  say that a pair $(x, k) $ of functions on $[0,1]$
is a solution of the Skorohod  problem  associated with $y$ if \\
(i) $x_t=y_t+ k_t$, $t\in [0,1]$,\\
(ii) $x_t\geq 0 $, $t\in [0,1]$,\\
(iii) $k$ is increasing, continuous, $k(0)=0 $ and satisfies
\begin{eqnarray}
k_t=\int^t_0 \chi_{\{x_s=0\}}dk_s.
\end{eqnarray}
In this case, the function $k$ is given by
\begin{eqnarray}
k_t=-\inf_{s\leq t}\{ (y_s \wedge 0)\}.
\end{eqnarray}
\end{Def}

\begin{Prop}
Assume that the function $ \sigma $ satisfies the same conditions as
in Theorem 1.1, and  $ ( X_t(x), L^x_t ) $  is a solution of
Eq.(1.1). Then there is a constant $c$ such that
\begin{eqnarray}
{\bf E}\big \{\sup_{t\leq 1}|X_t(x)-X_t(y)|^{p}  \big \} \leq \exp\{
cp^2 +cp\}|x-y|^p
\end{eqnarray}
for any $x, y \in \Re_+$ and $p\geq 1$.
\end{Prop}
{\bf Proof.}\ By H\"{o}lder inequality, we need only to prove
Proposition 2.1 for $p\geq 4$. \  Let $ a(x):= \frac{1}{2}(\sigma
\sigma')(x) $ for any $x\in \Re $, we have
\begin{eqnarray}
&& X_t(x)=x + \int^t_0 \sigma (X_s(x)) dB_s + \int^t_0 a(X_s(x)) ds
+L^x_t,\ t\in [0,1],\nonumber\\ \\
&& X_t(y)=y + \int^t_0 \sigma (X_s(y)) dB_s + \int^t_0 a(X_s(y)) ds
+L^y_t,\ t\in [0,1].\nonumber\\
\end{eqnarray}
By the reflection principle (2.2),
\begin{eqnarray}
&& L^x_t=-\inf_{s\leq t}\big \{\big (x + \int^t_0 \sigma (X_s(x))
dB_s + \int^t_0 a(X_s(x)) ds      \big )\wedge 0      \big \},\ t\in
[0,1], \nonumber\\
\\
&& L^y_t=-\inf_{s\leq t}\big \{\big (y + \int^t_0 \sigma (X_s(y))
dB_s + \int^t_0 a(X_s(y)) ds      \big )\wedge 0      \big \},\ t\in
[0,1].\nonumber\\
\end{eqnarray}
Thus,
\begin{eqnarray}
 |X_t(x)-X_t(y)|& \leq &  |x-y| + \sup_{s\leq t}\big |\int^t_0 [ \sigma
(X_s(x))- \sigma (X_s(y)) ]dB_s\big | + | L^x_t-L^y_t|\nonumber\\
&\leq & 2 |x-y| +2 \sup_{s\leq t}\big |\int^t_0 [ \sigma (X_s(x))-
\sigma (X_s(y)) ]dB_s\big |\nonumber\\
&&\quad  +2\int^t_0 | a (X_s(x))- a (X_s(y)) |ds.
\end{eqnarray}
Set $ Y_t(x, y)=\sup_{s\leq t}\{|X_s(x)-X_s(y)|\}$,
$\psi_t(x,y)=\|Y_t(x, y)\|_p $ for any $ x, y \in \Re_+ $ and $p\geq
4$. By Burkh\"o{}lder (see \cite{s12}) and H\"{o}lder inequalities,
\begin{eqnarray}
&& \big \|\sup_{s\leq t}\big |\int^t_0 [ \sigma (X_s(x))- \sigma
(X_s(y)) ]dB_s \big | \big \|_p\nonumber\\
&& \leq c p^{\frac{1}{2}}\| \big ( \int^t_0[ \sigma (X_s(x))- \sigma
(X_s(y)) ]^2ds      \big )^{\frac{1}{2}} \|_p\nonumber\\
&& \leq c p^{\frac{1}{2}}( \int^t_0 \psi_s(x,y) ^2ds      \big
)^{\frac{1}{2}},
\end{eqnarray}
where we have used  Lipschitz continuity of $ \sigma $.\\
Similarly,
\begin{eqnarray}
 \big \|\int^t_0 | a (X_s(x))- a (X_s(y)) |ds\big \|_p
  \leq c ( \int^t_0 \psi_s(x,y) ^2ds      \big
)^{\frac{1}{2}}.
\end{eqnarray}
Using (2.8), (2.9) and (2.10),we get that
\begin{eqnarray}
\psi_t(x,y) ^2\leq 12 |x-y|^2 + (12c^2p +3c^2)\int^t_0 \psi_s(x,y)
^2ds .
\end{eqnarray}
It follows from  Gronwall's lemma  and (2.11) that
\begin{eqnarray*}
{\bf E}\big \{\sup_{t\leq 1}|X_t(x)-X_t(y)|^{p}  \big \} \leq \exp\{
12cp^2 +3c^2p+3p\}|x-y|^p.
\end{eqnarray*}
Thus we complete the proof. \quad $\Box $

In view of  (2.6) and (2.7), the following is a direct consequence
of Proposition 2.1.
\begin{Prop}
Assume that the function $ \sigma $ satisfies the same conditions as
in Theorem 1.1, and $ ( X_t(x), L^x_t ) $  is a solution of
Eq.(1.1). Then there is a constant $c$ such that
\begin{eqnarray}
{\bf E}\big \{\sup_{t\leq 1}|L^x_t-L^y_t|^{p}  \big \} \leq \exp\{
cp^2 +cp\}|x-y|^p
\end{eqnarray}
for any $x, y \in \Re_+$ and $p\geq 1$.
\end{Prop}
Similar arguments lead to  the following result.
\begin{Prop}
Assume that the function $ \sigma $ satisfies the same conditions as
in Theorem 1.1, $ ( X_t(x), L^x_t ) $  is a solution of Eq.(1.1).
Then there is a constant $c$ such that
\begin{eqnarray}
&&{\bf E}\big \{\sup_{0\leq t\leq 1}|X_t(x)|^{p}  \big \} \leq
\exp\{
cp^2 +cp\}(1+x)^p,\\
&&{\bf E}\big \{\sup_{0\leq t\leq 1}|L^x_t|^{p}  \big \} \leq \exp\{
cp^2 +cp\}(1+x)^p
\end{eqnarray}
for any $x \in \Re_+$ and $p\geq 1$.
\end{Prop}
Next, we  study the  regularity of the solution $ ( X_t(x), L^x_t )
$ of Eq.(1.1) w.r.t. $t$.
\begin{Prop}
Assume that the function $ \sigma $ satisfies the same conditions as
in Theorem 1.1, $ ( X_t(x), L^x_t ) $  is a solution of Eq.(1.1).
Then,  for any $ R>0 $ and $p\geq 1 $,  there exist constants $C(p,
R)$ such that, for $s, t \in [0, 1]$,
\begin{eqnarray}
&&\sup_{0\leq x \leq R}{\bf E}\big\{ |X_t(x)-X_s(x)|^{2p}
\big \} \leq C(p, R)|t-s|^p,\\
&&\sup_{0\leq x \leq R}{\bf E}\big\{ |L^x_t-L^x_s|^{2p} \big \} \leq
C(p, R)|t-s|^p.
\end{eqnarray}
\end{Prop}
{\bf Proof.} For $ 0\leq s\leq t \leq 1 $, let $f(t)\equiv -x
-\int^t_0 \sigma (X_s(x)) dB_s - \int^t_0 a(X_s(x)) ds$. By
reflection principle,
$$ L_t^x=\sup_{0\leq u \leq t}\{f(u)\vee 0    \}       .$$
Noting that
\begin{equation}
L^x_t-L^x_s \leq \sup_{s\leq u \leq t}\{| f(u) - f(s)| \},
\end{equation}
 by Burkh\"o{}lder (see \cite{s12}) and H\"{o}lder inequalities, we have
\begin{eqnarray}
{\bf E}\{|L^x_t-L^x_s|^{2p}\}& \leq & c(p) {\bf E} \{\sup_{s\leq u
\leq t}|\int^u_s \sigma (X_v(x)) dB_v          |^{2p} \}\nonumber\\
&&+ c(p) {\bf E} \{\sup_{s\leq u \leq t}|\int^u_s a(X_v(x)) dv
|^{2p} \}\nonumber\\
&\leq & c(p) {\bf E} \big (\int^t_s \sigma (X_v(x))^2 dv \big
)^p\nonumber\\
&&  +c(p){\bf E} \big ( |\int^t_s a(X_v(x)) dv | \big
)^{2p}\nonumber\\
&\leq & c(p)( 1+R^{2p})( |t-s|^p + |t-s|^{2p})\nonumber\\
&\leq & c(p, R )|t-s|^p,
\end{eqnarray}
and this implies (2.16).
 Since
$$ | X_t(x) - X_s(x) | \leq |\int^u_s \sigma (X_v(x)) dB_v    |
+  |\int^t_s a(X_v(x)) dv |  + | L^x_t-L^x_s     |, $$ the
inequality (2.15) follows from (2.16), Burkh\"o{}lder (see
\cite{s12}) and H\"{o}lder inequalities. Thus we complete the proof
of Proposition 2.4. \quad $\Box$

 \setcounter{equation}{0}
\section{ \bf Continuity of functionals of local times }
Let $ ( X_t(x), L^x_t ) $  be a solution of Eq.(1.1). Because of
Propositions 2.1-2.2 and Proposition 2.4, we may assume that
$X_t(x), L_t^x$ are jointly continuous in $(t,x)$. Let  $ l(y) $ be
a continuous function  on $(0, \infty )$ with compact support. Put
$F(t, x)=\int^t_0 l(X_s(x))dL^x_s$ for $x\in \Re_+$. We have the
following
\begin{Prop}
Assume that the function $ \sigma $ satisfies the same conditions as
in Theorem 1.1, and $ ( X_t(x), L^x_t ) $  is a solution of
Eq.(1.1). Then the function $F(t,x)$ is jointly continuous in $(t,
x)$.
\end{Prop}
{\bf Proof.} \ Since
\begin{eqnarray}
|F(t,x)-F(s,x)|\leq \sup_{y>0}\{|l(y)|\}|L^x_t-L^x_s|,
\end{eqnarray}
the function $F(t,x)$ is continuous in $t$ uniformly with respect to
$x$ in any compact set by Proposition 2.4 and Kolmogorov's
continuity criterion( see Theorem 1.4.1 in \cite{s20} ). Thus, it
suffices to show the continuity of $ F(t,x)$ w.r.t. $x$ for any
fixed $t$. Let $x_n,x\in \Re_+ $ with  $ x_n \longrightarrow x $ as
$ n\longrightarrow +\infty $. By Propositions 2.1-2.2, and
Kolmogorov's continuity criterion( see Theorem 1.4.1 in \cite{s20}
), we have
\begin{eqnarray}
L^{x_n}_t\longrightarrow L^x_t, \quad  X_t(x_n)\longrightarrow
X_t(x),
\end{eqnarray}
uniformly in $t$, as $n\longrightarrow +\infty $. Therefore, there
exists a constant $C\geq 1$ such that for all  $n\geq 1$
\begin{eqnarray}
 L^{x_n}_t \leq C + L^x_1.
 \end{eqnarray}
Since the function $l(x)$ is bounded  and continuous, by (3.2) and
(3.3),
\begin{eqnarray}
| \int^t_0 [ l(X_s(x_n))-l(X_s(x))]dL^{x_n}_s|\longrightarrow 0
\end{eqnarray}
as $n\longrightarrow +\infty$. Because $L^{x_n}_t $ and $ L^{x}_t$
are increasing and continuous, by (3.2), the  sequence of finite
measures $ dL^{x_n}_t $ on $[0,1]$ converges weakly to the finite
measure $ dL^{x}_t $ on $[0,1]$. Therefore, for bounded continuous
function $ l(X_s(x))$ on $[0,1]$, we have
\begin{eqnarray}
\lim_{n\rightarrow \infty }\int^t_0 l( X_s(x))dL^{x_n}_s=\int^t_0 l(
X_s(x))dL^{x}_s.
\end{eqnarray}
The proof of Proposition 3.1 follows from (3.4) and (3.5).\quad
$\Box $

\setcounter{equation}{0}
\section{{\bf  Moments estimates for
one-point and two-point motions}}

For any $R >0 $ and  $ x \in [0, R] $, let $ ( X_t(x), L^x_t ) $
 be a solution of Eq.(1.1).We define  $ S_\pi (t,x) $ and $I(t,x) $
by\begin{eqnarray*}
S_\pi (t,x)&:=&S_\pi(\sigma (X_{\cdot} (x)), t),\\
I(t, x)&:=& \int^t_0\sigma (X_s (x))\circ dB_s
        =\int^t_0\sigma (X_s (x))dB_s +\frac{1}{2}
        \int^t_0 (  \sigma     \sigma' )(X_s (x)) ds.
\end{eqnarray*}
Write
\begin{eqnarray}
&&S_\pi(t,x)\nonumber\\
&=&\sum^{n-1}_{k=0} \frac{1}{t_{k+1}-t_k}\bigg (
\int^{t_{k+1}}_{t_k} \sigma ( X_s(x) ) ds  \bigg )(
B_{t_{k+1}}-B_{t_k})\nonumber\\
&=& \sum^{n-1}_{k=0}\sigma ( X_{t_k}(x) )(
B_{t_{k+1}}-B_{t_k})\nonumber\\
&+&\sum^{n-1}_{k=0} \frac{1}{t_{k+1}-t_k}\bigg (
\int^{t_{k+1}}_{t_k} (\sigma ( X_s(x) )-\sigma ( X_{t_k}(x) )) ds
\bigg )( B_{t_{k+1}}-B_{t_k}).\nonumber\\
\end{eqnarray}
By Ito's formula, for $s\geq t_k$,
\begin{eqnarray}
\sigma ( X_s(x) )&-&\sigma ( X_{t_k}(x) )\nonumber\\
&=&\int_{t_k}^s\sigma^{\prime} ( X_u(x) )\sigma ( X_u(x) )
dB_u+ \int_{t_k}^s\sigma^{\prime} ( X_u(x) ) dL_u^x \nonumber\\
&+&\frac{1}{2}\int_{t_k}^s(\sigma^{\prime})^2( X_u(x) ) \sigma (
X_u(x) ) du+\frac{1}{2}\int_{t_k}^s \sigma^{\prime\prime}( X_u(x)
)\sigma^2 ( X_u(x)) du.\nonumber\\
\end{eqnarray}
Thus we can write $S_\pi(t,x)-I(t,x )$ as follows:
\begin{equation}
S_\pi(t,x)-I(t,x )=A_{1\pi}+A_{2\pi}+A_{3\pi}+A_{4\pi},
\end{equation}
where
\begin{eqnarray*}
A_{1\pi}(x)&:=&\sum^{n-1}_{i=0} \sigma
(X_{t_i}( x))( B_{t_{i+1}}-B_{t_i})-\int^1_0 \sigma(X_s(x))dB_s,\\
A_{2\pi}(x)&:= &\sum^{n-1}_{i=0}\frac{1}{t_{i+1}-t_i} \int_{t_i}^{t_{i+1}}ds
\big (\int_{t_i}^{s}\sigma'(X_u(x))\sigma(X_u(x))dB_u\big )(B_{t_{i+1}}-B_{t_i})\\
&&- \frac{1}{2}\int_{0}^{t}\sigma'(X_s(x))\sigma(X_s(x))ds,\\
A_{3\pi}(x)&:=& \sum^{n-1}_{i=0}\frac{1}{t_{i+1}-t_i}
\int_{t_i}^{t_{i+1}}ds \big ( \int_{t_k}^s[(\sigma^{\prime})^2(
X_u(x) )
\sigma ( X_u(x) )\nonumber\\
&&+\sigma^{\prime\prime}( X_u(x) )\sigma^2 ( X_u(x)) ]du\big ) \times (B_{t_{i+1}}-B_{t_i}),\\
A_{4\pi}(x)&:=& \sum^{n-1}_{i=0} \frac{1}{t_{i+1}-t_i}
\int_{t_i}^{t_{i+1}}ds\big (
 \int_{t_k}^s\sigma^{\prime}( X_u(x) )dL_u^x\big )\times (B_{t_{i+1}}-B_{t_i}).
\end{eqnarray*}

\begin{proposition}
Assume that the function $ \sigma $ satisfies the same conditions as
in Theorem 1.1, $ ( X_t(x), L^x_t ) $  is a solution of Eq.(1.1).
Then  for any $ p\geq 2$ and $R>0$ there exist  constants $C(p, R )
$
 such that
\begin{equation}
\sup_{ x\in [0,  R]}{\bf E} \big \{|S_\pi(t,x)-I(t,x ) |^{2p}\big
\}\leq C(p, R ) \|\pi\|^{\frac{1}{2}p}.
\end{equation}
\end{proposition}

{\bf Proof.} In the sequel, we will use $c(p)$ to denote a generic
constant which depends only on $p$ and whose value may be different
from line to line. By Burkholder-Davis-Gundy inequalities, we have
\begin{eqnarray}
\{{\bf E} \{ |A_{1\pi}(x)|^{2p}\}\}^{\frac{1}{p}}&\leq & c(p) \bigg
[
{\bf E} \bigg ( \sum^{n-1}_{i=0}\int_{t_i}^{t_{i+1}}(\sigma(X_s(x))-
\sigma(X_{t_i}(x))^2ds\bigg )^p \bigg ]^{\frac{1}{p}  }\nonumber\\
&\leq & c(p)\sum^{n-1}_{i=0}\bigg \{ {\bf E}
\{\vert\int_{t_i}^{t_{i+1}}(\sigma(X_s(x))-\sigma(X_{t_i}(x))^2ds\vert^{p}
\} \bigg
\}^{\frac{1}{p}}\nonumber\\
&\leq&c(p)\sum^{n-1}_{i=0} \{\vert\int_{t_i}^{t_{i+1}}({\bf E}[(\sigma(X_s(x))
-\sigma(X_{t_i}(x))^{2p}])^{\frac{1}{p}}ds\nonumber \\
&\leq & c(p)\|\pi\|,
\end{eqnarray}
where we have used Proposition 2.4. For $p\geq 1$, by H\"{o}lder
inequality,
\begin{eqnarray}
&&{\bf E} \{ |A_{3\pi}(x)|^{2p}\}\nonumber\\
&\leq &{\bf E} \{ \bigg (\sum^{n-1}_{i=0}(\frac{1}{t_{i+1}-t_i}
 \int_{t_i}^{t_{i+1}}ds \int_{t_i}^s[(\sigma^{\prime})^2( X_u(x) )
 \sigma ( X_u(x) )\nonumber\\
 &&+\sigma^{\prime\prime}( X_u(x) )\sigma^2 ( X_u(x)) ]du)^2\bigg )^p
 \times \big (\sum^{n-1}_{i=0}(B_{t_{i+1}}-B_{t_i})^2\big )^p\}\nonumber\\
&\leq & c(p)\|\pi\|^p\sup_{\pi}{\bf E} \{\big
(\sum^{n-1}_{i=0}(B_{t_{i+1}}
-B_{t_i})^2\big )^p\}\nonumber\\
&\leq& c(p)\|\pi\|^p .
\end{eqnarray}
\begin{eqnarray}
{\bf E} \{ |A_{4\pi}(x)|^{2p}\}
&\leq &{\bf E} \{ \big (\sum^{n-1}_{i=0}(\frac{1}{t_{i+1}-t_i} \int_{t_i}^{t_{i+1}}ds
 \int_{t_i}^s\sigma^{\prime}( X_u(x) )dL_u^x)^2\big )^p\nonumber\\
&&\times \big (\sum^{n-1}_{i=0}(B_{t_{i+1}}-B_{t_i})^2\big )^p\}\nonumber\\
&\leq & c(p){\bf E} \{ \big (\sum^{n-1}_{i=0}(L_{t_{i+1}}^x-L_{t_i}^x)^2\big )^p\times
 \big (\sum^{n-1}_{i=0}(B_{t_{i+1}}-B_{t_i})^2\big )^p\}\nonumber\\
&\leq & c(p){\bf E} \{ \big (\sup_i(L_{t_{i+1}}^x-L_{t_i}^x)L_1^x\big )^p\times
\big (\sum^{n-1}_{i=0}(B_{t_{i+1}}-B_{t_i})^2\big )^p\}\nonumber\\
&\leq & c(p)({\bf E}\{\big (\sup_i(L_{t_{i+1}}^x-L_{t_i}^x)\big
)^{3p}\})^{\frac{1}{3}}
 ({\bf E}\{\big (L_{1}^x\big )^{3p}\})^{\frac{1}{3}}\nonumber\\
&&\times {\bf E} \{\big (\sum^{n-1}_{i=0}(B_{t_{i+1}}-B_{t_i})^2\big )^p\}\nonumber\\
&\leq& c(p)\|\pi\|^{\frac{p}{2}},
\end{eqnarray}
where Proposition 2.4 and Kolmogorov's continuity criterion( see
Theorem 1.4.1 in \cite{s20} ) were used in the last inequality.
Using Fubini Theorem,  $A_{2\pi}$ can be further written as
\begin{equation}
A_{2\pi}(x)=A_{2\pi}^{(1)}(x)+A_{2\pi}^{(2)}(x)+A_{2\pi}^{(3)}(x),
\end{equation}
where
\begin{eqnarray*}
A_{2\pi}^{(1)}(x)&:=& \sum^{n-1}_{i=0}\frac{1}{t_{i+1}-t_i}
\int_{t_i}^{t_{i+1}}(t_{i+1}-u) \bigg (\sigma^{\prime}( X_u(x))
\sigma ( X_u(x) )\\
&&-\sigma^{\prime}( X_{t_i}(x))\sigma ( X_{t_i}(x) \bigg )du,\\
A_{2\pi}^{(2)}(x)&:=&
-\frac{1}{2}\sum^{n-1}_{i=0}\int_{t_i}^{t_{i+1}} \big
(\sigma^{\prime}( X_u(x))\sigma ( X_u(x) )-\sigma^{\prime}(
X_{t_i}(x))
\sigma ( X_{t_i}(x) \big )du,\\
A_{2\pi}^{(3)}(x)&:= &\sum^{n-1}_{i=0}\{ \frac{1}{t_{i+1}-t_i} \big
(\int_{t_i}^{t_{i+1}}(t_{i+1}-u)\sigma'(X_u(x))\sigma(X_u(x))
dB_u\big )(B_{t_{i+1}}-B_{t_i})\\
&&-\frac{1}{t_{i+1}-t_i} \big (\int_{t_i}^{t_{i+1}}(t_{i+1}-u)
\sigma'(X_u(x))\sigma(X_u(x))du\big )\}.
\end{eqnarray*}
Since $\sigma^{\prime}\sigma$ is Lipschitz continuous, it follows
that
\begin{eqnarray}
\{{\bf E} \{ |A_{2\pi}^{(1)}(x)|^{p}\}\}^{\frac{1}{p}}&\leq &
\sum^{n-1}_{i=0}\int_{t_i}^{t_{i+1}} \big \{{\bf E}
\{\|\sigma^{\prime}( X_u(x))
\sigma ( X_u(x) )\nonumber\\
&&-\sigma^{\prime}( X_{t_i}(x))\sigma ( X_{t_i}(x) \|^p\}
\big \}^{\frac{1}{p}}du\nonumber\\
&\leq & c\sum^{n-1}_{i=0}\int_{t_i}^{t_{i+1}} \{{\bf E}
\{\| X_u(x)-X_{t_i}(x) \|^p\}\}^{\frac{1}{p}}du\nonumber\\
&\leq & c\|\pi\|^{\frac{1}{2}}.
\end{eqnarray}
Similar arguments lead to
\begin{equation}
\{{\bf E} \{ |A_{2\pi}^{(2)}(x)|^{p}\}\}^{\frac{1}{p}}\leq
c\|\pi\|^{\frac{1}{2}}.
\end{equation}
Noting that $A_{2\pi}^{(3)}$ is a  martingale. Using
Burkholder-Davis-Gundy inequalities, we obtain that
\begin{eqnarray}
&&\{{\bf E} \{ |A_{2\pi}^{(3)}(x)|^{2p}\}\}^{\frac{1}{p}}\nonumber\\
&\leq & c(p) \bigg [ {\bf E} \bigg ( \sum^{n-1}_{i=0}\{
\frac{1}{t_{i+1}-t_i} \big
(\int_{t_i}^{t_{i+1}}(t_{i+1}-u)\sigma'(X_u(x))\sigma(X_u(x))dB_u\big
)
(B_{t_{i+1}}-B_{t_i})\nonumber\\
&&-\frac{1}{t_{i+1}-t_i} \big (\int_{t_i}^{t_{i+1}}(t_{i+1}-u)
\sigma'(X_u(x))\sigma(X_u(x))du\big )\}^2\bigg )^p \bigg
]^{\frac{1}{p}}
\nonumber\\
&\leq& c(p)\sum^{n-1}_{i=0}\bigg [ {\bf E}\{ \frac{1}{t_{i+1}-t_i}
\big (\int_{t_i}^{t_{i+1}}(t_{i+1}-u)
\sigma'(X_u(x))\sigma(X_u(x))dB_u\big )(B_{t_{i+1}}-B_{t_i})\nonumber\\
&&-\frac{1}{t_{i+1}-t_i} \big (\int_{t_i}^{t_{i+1}}(t_{i+1}-u)
\sigma'(X_u(x))\sigma(X_u(x))du\big )\}^{2p} \bigg ]^{\frac{1}{p}}\nonumber\\
&\leq& c(p)\sum^{n-1}_{i=0}\bigg \{ \big ({\bf E}\{ \big
|\frac{1}{t_{i+1}-t_i} \int_{t_i}^{t_{i+1}}
(t_{i+1}-u)\sigma'(X_u(x))\sigma(X_u(x))dB_u\big )(B_{t_{i+1}}
-B_{t_i})\|^{2p}\big )^{\frac{1}{p}}\nonumber\\
&&+\big ({\bf E}\{ \big |\frac{1}{t_{i+1}-t_i} \big
(\int_{t_i}^{t_{i+1}}(t_{i+1}-u)\sigma'(X_u(x))
\sigma(X_u(x))du\big |\}^{2p}\big )^{\frac{1}{p}}\bigg \}\nonumber\\
&\leq & c(p)\|\pi\|,
\end{eqnarray}
where H\"o{}lder inequality was used for the last inequality.
Combining the estimates for $A_{2\pi}^{(1)}$, $A_{2\pi}^{(2)}$ and
$A_{2\pi}^{(3)}$  together, we deduce that
\begin{equation}
\{{\bf E} \{ |A_{2\pi}(x)|^{2p}\}\}^{\frac{1}{p}}\leq
c\|\pi\|^{\frac{1}{2}}.
\end{equation}
Now (4.4) follows from (4.5), (4.6), (4.7) and (4.12). The proof is
complete. \quad $\Box$ \vskip 0.4cm

 Next result is the moment estimates for
the two point motions.
\begin{proposition}
Assume that the function $ \sigma $ satisfies the same conditions as
in Theorem 1.1, $ ( X_t(x), L^x_t ) $  is a solution of Eq.(1.1).
Then  for any $ p\geq 2$ and $R>0$ there exists  constants $C(p, R )
$ and $\beta$, independent of the partition $\pi$,
 such that
\begin{equation}
{\bf E} \big \{\sup_{ t\in [0,  1]}|S_\pi(t,x)-S_\pi(t,y) |^{p}\big
\}\leq C(p, R ) |x-y|^{\frac{1}{2}p},
\end{equation}
for all $x, y\in [0, R]$.
\end{proposition}
{\bf Proof.} Similarly as (4.3), write
\begin{equation}
S_\pi(t,x)-S_\pi(t,y)
=A_{1\pi}(x,y)+A_{2\pi}(x,y)+A_{3\pi}(x,y)+A_{4\pi}(x,y),
\end{equation}
where
\begin{eqnarray*}
A_{1\pi}(x,y)&:=&\sum^{n-1}_{i=0}\big ( \sigma (X_{t_i}( x))- \sigma
(X_{t_i}( y))\big )( B_{t_{i+1}}-B_{t_i}),\\
A_{2\pi}(x,y)&:= &\sum^{n-1}_{i=0}\big \{\frac{1}{t_{i+1}-t_i}
\bigg ( \int_{t_i}^{t_{i+1}}ds \int_{t_i}^{s}\big (\sigma'(X_u(x))\sigma(X_u(x))\\
&&-\sigma'(X_u(y))\sigma(X_u(y))\big )dB_u\bigg )\times(B_{t_{i+1}}-B_{t_i})\big \},\\
A_{3\pi}(x,y)&:=& \sum^{n-1}_{i=0}\bigg \{\frac{1}{t_{i+1}-t_i}
 \int_{t_i}^{t_{i+1}}ds\big \{ \int_{t_k}^s[(\sigma^{\prime})^2( X_u(x) )
 \sigma ( X_u(x) )\\
 &&+\sigma^{\prime\prime}( X_u(x) )\sigma^2 ( X_u(x))-(\sigma^{\prime})^2( X_u(y) )
 \sigma ( X_u(y) )\\
 &&-
\sigma^{\prime\prime}( X_u(y) )\sigma^2 ( X_u(y)) ]du\big\}
\times (B_{t_{i+1}}-B_{t_i})\bigg \},\\
A_{4\pi}(x,y)&:=& \sum^{n-1}_{i=0}\frac{1}{t_{i+1}-t_i}
 \int_{t_i}^{t_{i+1}}ds\big\{ \int_{t_i}^s\sigma^{\prime}( X_u(x) )dL_u^x
 \\
 &&-\int_{t_i}^s\sigma^{\prime}( X_u(y) )dL_u^y\big \}(B_{t_{i+1}}-B_{t_i})\\
&=& \sigma^{\prime}(0)\sum^{n-1}_{i=0}\frac{1}{t_{i+1}-t_i}
\int_{t_i}^{t_{i+1}}ds\big\{
\int_{t_i}^sdL_u^x-\int_{t_i}^sdL_u^y\big \} (B_{t_{i+1}}-B_{t_i}).
\end{eqnarray*}
By Burkholder-Davis-Gundy inequalities, the Lipschitz continuity of
$\sigma$
 and (2.3), it follows easily that
\begin{equation}
{\bf E} \big \{\sup_{ t\in [0,  1]}|A_{1\pi}(x,y) |^{p}\big \}\leq
C(p, R ) |x-y|^{p},
\end{equation}
By virtue of (2.3) and the Lipschitz continuity of $\sigma'\sigma$,
\begin{eqnarray}
&&\bigg ({\bf E} \big \{\sup_{ t\in [0,  1]}|A_{2\pi}(x,y)
|^{p}\big \}\bigg )^{\frac{1}{p}}\nonumber\\
&\leq &\sum^{n-1}_{i=0}\big \{\frac{1}{t_{i+1}-t_i}
\int_{t_i}^{t_{i+1}}ds\bigg ( {\bf E}
\big \{\vert \int_{t_i}^{s}\big (\sigma'(X_u(x))\sigma(X_u(x))\nonumber\\
&&-\sigma'(X_u(y))\sigma(X_u(y))\big )dB_u\bigg )
\times(B_{t_{i+1}}-B_{t_i})\vert^p\big \}\bigg )^{\frac{1}{p}}\big\}\nonumber\\
&\leq &\sum^{n-1}_{i=0}\big \{\frac{1}{t_{i+1}-t_i}
\int_{t_i}^{t_{i+1}}ds\bigg ( {\bf E} \big \{\vert
\int_{t_i}^{s}\big (\sigma'(X_u(x))\sigma(X_u(x))\nonumber\\
&&-\sigma'(X_u(y))\sigma(X_u(y))\big )dB_u\vert^{2p}\bigg
)^{\frac{1}{2p}}\bigg ( {\bf E} \big \{\vert (B_{t_{i+1}}
-B_{t_i})\vert^{2p}\big \}\bigg )^{\frac{1}{2p}}\big\}\nonumber\\
&\leq &\sum^{n-1}_{i=0}\big \{\frac{1}{t_{i+1}-t_i}
\int_{t_i}^{t_{i+1}}ds\bigg (\int_{t_i}^{s} \big \{{\bf E}[\vert
X_u(x))-X_u(y)\vert^{2p}]
\big\}^{\frac{1}{p}}du\bigg )^{\frac{1}{2}}\nonumber\\
&&\times (t_{i+1}-t_i)^{\frac{1}{2}} \big\}\nonumber\\
 &\leq & c(p)|x-y|.
\end{eqnarray}
By a similar, but simpler argument, we also found that
\begin{equation}
\bigg ({\bf E} \big \{\sup_{ t\in [0,  1]}|A_{3\pi}(x,y) |^{p}\big
\}\bigg )^{\frac{1}{p}}\leq c(p)|x-y|.
\end{equation}
Observe that
\begin{eqnarray}
(A_{4\pi}(x,y))^2&\leq &c\sum^{n-1}_{i=0}\frac{1}{t_{i+1}-t_i}
\int_{t_i}^{t_{i+1}}ds\big ( L_s^x-L_{t_i}^x-L_s^y+L_{t_i}^x\big )^2\nonumber\\
&&\times \sum^{n-1}_{i=0}(B_{t_{i+1}}-B_{t_i})^2\nonumber\\
&\leq &c(\sup_{0\leq s\leq 1}\vert L_s^x-L_s^y\vert)\sum^{n-1}_{i=0}
\frac{1}{t_{i+1}-t_i} \int_{t_i}^{t_{i+1}}\big (\vert  L_s^x-L_{t_i}^x\vert
 +\vert L_s^y-L_{t_i}^x\vert \big )ds\nonumber\\
&&\times \sum^{n-1}_{i=0}(B_{t_{i+1}}-B_{t_i})^2\nonumber\\
&\leq &c(\sup_{0\leq s\leq 1}\vert L_s^x-L_s^y\vert)\sum^{n-1}_{i=0}
\big (\vert  L_{t_{i+1}}^x-L_{t_i}^x\vert +\vert L_{t_{i+1}}^y-L_{t_i}^x\vert \big )\nonumber\\
&&\times \sum^{n-1}_{i=0}(B_{t_{i+1}}-B_{t_i})^2.
\end{eqnarray}
It follows that
\begin{eqnarray}
&&{\bf E} \big \{\sup_{ t\in [0,  1]}|A_{4\pi}(x,y)
|^{2p}\big \}\nonumber\\
&\leq &c\big ({\bf E} \big \{(\sup_{0\leq s\leq 1} \vert
L_s^x-L_s^y\vert^{3p}\big \}\big )^{\frac{1}{3}}
\big ({\bf E} \big \{(  L_{1}^x+L_{1}^y)^{3p}\big \})^{\frac{1}{3}}\nonumber\\
&&\times \big ({\bf E} \big \{\big (\sum^{n-1}_{i=0}
(B_{t_{i+1}}-B_{t_i})^2)\big )^{3p}\big
\}\big )^{\frac{1}{3}}\nonumber\\
&\leq & c(p, R) |x-y|^{\frac{1}{2}p}.
\end{eqnarray}
Putting together above estimates (4.15)-(4.17) and (4.19), we arrive
at (4.13). \quad $\Box$

 The following result can be proved similarly
as Proposition 4.2.
\begin{proposition}
Assume that the function $ \sigma $ satisfies the same conditions as
in Theorem 1.1, $ ( X_t(x), L^x_t ) $  is a solution of Eq.(1.1).
Then  for any $ p\geq 2$ and $R>0$ there exists  a constant $C(p, R
)$,
 such that
\begin{equation}
{\bf E} \big \{\sup_{ t\in [0,  1]}|I(t,x)-I(t,y) |^{p}\big \}\leq
C(p, R ) |x-y|^{p}
\end{equation}
for all $x, y\in [0, R]$.
\end{proposition}
\setcounter{equation}{0}
\section{{\bf Uniform convergence of the Riemann sums}}
\begin{theorem}
Assume that the function $ \sigma $ satisfies the same conditions as
in Theorem 1.1, $ ( X_t(x), L^x_t ) $  is a solution of Eq.(1.1).
Then  for any $ p\geq 2$ and $R>0$,
\begin{equation}
\lim_{\|\pi\|\rightarrow 0}{\bf E} \big \{\sup_{ x\in [0,
R]}|S_\pi(t,x)-I(t,x ) |^{2p}\big \}=0.
\end{equation}
\end{theorem}
{\bf Proof}. By Propositions 4.2 and 4.3, it follows from
Garsia-Rodemich and Rumsey's Lemma (cf.\cite{s14})that there exists
a constant $\beta_0>0$, independent of $\pi$, such that for $x, y\in
[0,R]$,
\begin{equation}
|S_\pi(t,x)-S_\pi(t,y)|\leq K_{\pi}(\omega)|x-y|^{\beta_0},
\end{equation}
\begin{equation}
|I(t,x)-I(t,y)|\leq K(\omega)|x-y|^{\beta_0},
\end{equation}
where $ K_{\pi}(\omega), K(\omega)$ are random variables that
satisfy
\begin{equation}
\sup_{\pi}{\bf E}\{|K_{\pi}|^p\}<\infty,\quad {\bf
E}\{|K|^p\}<\infty,
\end{equation}
for all $p\geq 1$. This is possible because the constant in
Propositions 4.2 is independent of $\pi$. Thus, given any
$\varepsilon>0$, there is $\delta>0$ such that
\begin{equation}
{\bf E}\{\sup_{x\in [0,R]}\sup_{y: |y-x|\leq \delta}
|S_\pi(t,y)-S_\pi(t,x)|^p\}\leq \varepsilon,
\end{equation}
\begin{equation}
{\bf E}\{\sup_{x\in [0,R]}\sup_{y: |y-x|\leq
\delta}|I(t,y)-I(t,x)|^p\}\leq \varepsilon.
\end{equation}
On the other hand, for any  $\delta>0$, we can find $x_1, ..., x_m$
such that
$$ [0,R]\subset \cup_{i=1}^m B(x_i, \delta).$$
Consequently,
\begin{eqnarray}
\sup_{ x\in [0,  R]}|S_\pi(t,x)&-&I(t,x )
|^{p}\nonumber\\
&\leq & \sup_{x_i}\sup_{y\in B(x_i,\delta)}|S_\pi(t,y)-I(t,y )|^{p}\nonumber\\
&\leq& c(p)\sup_{x_i}\sup_{y\in
B(x_i,\delta)}|S_\pi(t,y)-S_\pi(t,x_i)|^{p}
\nonumber\\
&&+c(p)\sum_{i=1}^m|S_\pi(t,x_i)-I(t,x_i)|^{p}\nonumber\\
&&+c(p)\sup_{x_i}\sup_{y\in B(x_i,\delta)}|I(t,y)-I(t,x_i)|^{2p}.
\end{eqnarray}
By virtue of (5.5) and (5.6), this implies that
\begin{eqnarray}
{\bf E}\{\sup_{ x\in [0,  R]}|S_\pi(t,x)&-&I(t,x )
|^{p}\}\nonumber\\
&\leq& 2c(p)\varepsilon+c(p){\bf
E}\{\sum_{i=1}^m|S_\pi(t,x_i)-I(t,x_i)|^p\}.
\end{eqnarray}
Let first $\|\pi\|\rightarrow 0$ and then $\varepsilon\rightarrow 0$
to get (5.1) by Proposition 4.1 and (5.8). Thus we complete the
proof. \quad $\Box$

\setcounter{equation}{0}
\section{ \bf Proof of Theorem 1.1 }
We will prove that $(X_t:=X_t(Z), L_t^Z)$ solves the anticipating
reflected SDE (1.3). Let $l(y)$ be any given continuous function on
$(0,\infty)$ with compact support. Since $ ( X_t(x), L^x_t ) $ is a
solution of Eq.(1.1),  $ F_t(x)=\int_0^tl(X_s(x))dL_s^x=0 $ for all
$\omega$ in  some measurable set  $\Omega_{t,x}$ with ${\bf P}
(\Omega_{t,x} )=1$. By continuity of $ F_t(x) $ proved in
Proposition 3.1,  we have
\begin{eqnarray*}
{\bf P}\big ( \omega\ : F_t(x, \omega)=0 \quad \mbox{for all
$(t,x)\in [0,1]\times [0, +\infty )$} \big )=1.
\end{eqnarray*}
Hence
\begin{eqnarray}
{\bf P}\big ( \omega\ : F_t(Z, \omega)=0 \quad \mbox{for all $t\in
[0,1]$} \big )=1,
\end{eqnarray}
for any non-negative random variable $Z$.  This implies that
$$L_t^Z=\int_0^t\chi_{\{X_s(Z)=0\}}dL_s^Z$$
Next we prove (1.6). By Theorem 5.1, the following  holds almost
surely on $\{\omega; Z(\omega)\leq M\}$
$$ \int^t_0 \sigma (X_s(x))\circ dB_s\big |_{x=Z}\chi_{\{\omega; Z(\omega)\leq M\}}$$
$$=\lim_{\|\pi\|\rightarrow 0}S_\pi(t,x)\big |_{x=Z}\chi_{\{\omega; Z(\omega)\leq M\}}$$
$$= \int^t_0 \sigma (X_s(Z))\circ dB_s\chi_{\{\omega; Z(\omega)\leq M\}}.$$
Letting $M\rightarrow \infty$ we obtain the substitution formula
(1.6), and therefore prove the Theorem. \quad \quad $\Box$
\vskip 6pt

{\bf Acknowledgements.} This work is supported by NSFC and SRF for
ROCS, SEM. The author would like to thank both for their generous
financial support.

redpajama_set_name: RedPajamaArXiv


> The Amish make solid oak comic book boxes?
> THE AMISH make solid oak COMIC BOOK BOXES!?
> SOLID OAK COMIC BOOK BOXES!!!?
drawers specifically designed for easy access to your entire collection.
he's willing to ship--lucky for us, he's willing to deliver to Toledo.
> Comic Con? They'd clean up!
were there, they were selling them for $1,000 each (they're well worth it).


Store Credit Cards: Which ones are the best? Store Credit Cards: Which ones are the best?
Store credit cards, also referred to as retail credit cards are credit cards issued by retail stores. Mostly these cards allow for purchases from the issuer retail store and a few other stores. Some though can be used in more stores. These cards come with so many perks and offers. As such, they can be a good way to save up on purchases on your retail.
However, before applying for that card you need to be cautious. Always ask yourself whether you need the card. Also, store credit cards tend to have high-interest rates on balances accrued. If you are not the type to pay up on balances keep away.
Personal Factors to consider before taking up a retail card?
Despite what you might be told, there are no free credit cards. Therefore you have to ask yourself whether you need one. If you are taking up a card to build up on your credit history retail credit cards may be your thing. Nonetheless, if you are taking up a credit card for shopping sprees, you are better off with a bank backed credit card. They are more convenient.
Holding up on a huge number of credit cards is never an advisable thing to do. It leaves you prone to forgetting to pay up on some of the card balances. Additionally, having idle credit cards on your credit file does not mirror well among credit issuers.
Taking up a credit card may seem like a sweet thing until one has to start paying up. If you are always will running late on insurance premiums and utility bills then credit cards are not for you.
If that retail credit card is being sold to you as a way of saving on purchases consider looking at how much you are likely to save. If you are spending more on interest payments than you are saving you might be better off without that card in tow.
The interest charged on that card determines the cost of holding that card. Store credit cards generally charge higher interest than banks. In the United States, for instance, they charge interests of between 27% and 30%. As a rule of thumb, if the interest charged is higher by more than 10% the rate charged by banks, it may be time to turn your back and say bye.
Certain store issued credit cards can only be used within their stores and a chosen few stores. On the other hand, other retail cards allow for use of the cards in a variety of places. Cards that are Visa, American Express or MasterCard enabled are more suitable since they can be used in many places across the whole world.
Most retail credit cards carry a lower purchase limit than bank issued credit cards. This may inhibit your ability to purchase certain items on credit. If you are likely to use credit cards to purchase high-value items then you should consider using cards issued directly by banks. Otherwise, if you are a good customer you can always try to negotiate a review on your limits.
Retail store managers know that you are likely to have another credit card tucked in somewhere else. To entice you to use their cards they will give you offers. One common offer is the cash back offer where for each purchase made on the credit cards you earn points. These points can be redeemed by purchasing another item in the store. Sounds free? Most retail card issuers have this program, so why not jump in on it. While at it, choose a card with the best offer.
Certain card issues have rewards schemes. Some offer rewards on reaching a particular level of purchases. Others offer rewards on card application anniversaries. Consider the rewards on offer and how redeemable they are. As regards redemption, some cards will allow your rewards to remain redeemable within a period of up to 24 months while others will only give you a few weeks to redeem your rewards. Also look at whether the reward can be converted into cash (not available on most cards) or you have to spend the reward in the store.
A good number of stores of have other special offers tied to their credit cards. Some may give retail card holder's priority purchase rights on certain items. Additionally holding the card might allow you to benefit from discounts that are exclusively given to credit cardholders. Other retailers, such as Amazon, offer competitive if not free offers for shipping and packaging.
Do all store credit card issuers report your spending history to CRBs? No, not all cards. If you are looking to improve on your credit history you should ask if there are report submissions to CRBs. The more the number of CRBs receiving the report the better.
• Retail credit cards are easier to apply and gain access to as compared to bank credit cards.
• Generally, have expensive interest rates as compared to bank credits.
• Most retail credit card issuers are never out rightly forthcoming on any hidden charges and fees.
Are retail credit cards worth taking?
It all depends on your needs and spending habits. Retail credit cards are a good form of credit history build up but they are quite expensive. If you are able to spend more and pay on time, you can apply for that card. However, if you have a poor credit card history, pocket your hands and walk away. Otherwise, shop around for the best offers, because there is always a better offer somewhere.
« Ever wondered about the HSN credit card? Read this to learn everything!
I used to have terrible credit. When I was 18, I got a defaulted student loan, along with a ton of unpaid medical bills that were reporting negatives to my credit report. I decided one day I was going to fix my credit and become more financially responsible. Through the soft pull credit card method, I was able to get approved for credit cards that I normally wouldn't be able to.
I decided that I would take my information from building my score from 450 to 750 and put it on a blog. This is the result! Feel free to contact me with any details or if you need help. Just post a comment and I'll respond ASAP.
Ever wondered about the HSN credit card? Read this to learn everything!
© 2019 Shopping Cart Trick.


In the category of Talons, the Hannie created by Betty London knows how to play the game! The brand has opted for a leather upper in pink that will show off your look. It has a carefully designed synthetic outer sole. Perfect for showing off your feminine style.


One of Chanel's latest films for his Chance perfume, presents the young actress, Lily Newmark, walking alone in the city of Venice. And, this commercial was directed by Eva Michon.
The campaign is divided into four films, made in the same way, but highlighting different personalities like Selah Marley (Lauryn Hill's daughter), Belen Chavanne (Argentinian model and actress) and Angela Yuen (Chinese Actress).
Alone in Venice. Create your CHANCE.
Alone in Venice. Live your CHANCE.
Alone in Venice. Follow your CHANCE.


Your letter was amazing, considering you were sitting in a bathtub nearly asleep. I'm impressed. I hope these electronic devices don't shock you in such circumstances.
Because you asked I'll briefly describe my feelings about the recent Chagall and Magritte shows I saw while in New York City.
Chagall's use of color is mind blowing to me, and the brush strokes he uses are very spontaneous, like he's breathing the painting process (as opposed to Magritte's, which is almost minimal, controlled, and real thin, like a watercolor wash at times).
The imagery Marc C started using early in his life was from his dreams, or imagination, which at that time was very risky. He was brave. He painted bravely with his colors juxtapositioning the real-unreal, and he was extremely courageous depicting "life" in the Russian shetl of his youth as an imaginary flight of fancy. I'm sure he took a LOT of slack from his peers, rabbi, etc. in his little village for doing so. This is brave art.
Also, his use of Christ on the Cross was a real surprise, one that Carter and I both weren't expecting. I thought I knew about Chagall's work (in my mind he is a surrealist, always, and a colorist) but neither of us knew about his "hidden messages" conveyed through the use of the "icon" Jesus. There he is, the Jew of Jews (at least the Romans said so on the sign above his criminal's cross: INRI, king of Jews, mocking him). I believe, from seeing the way he used the crucifixion scene, over and over, even with "baby blue Jesus" in one, more abstracted work, that he felt Jesus epitomized the suffering of the Jews during the nazi regime. With the black and white shawl across Jesus's loins (I forget what the Jewish prayer shawl is called) he isn't depicting Jesus as an emblem of Christianity, but as the height of Jews being misunderstood, stigmatized, ostracized … and killed in the nazi world that Chagall knew all too well (he fled Europe to escape them, remember).
Speaking of Jews and nazis, I highly recommend the book i just finished, Edmund de Waal's "The Hare with Amber Eyes." It's a scholarly memoir, an inside view of what it was like to be Jewish … for about 100 years before and after the nazis. follows a family of bankers, from Vienna, who lost everything during the nazi occupation. The read was not crushingly sad so much as very informative, and shockingly real. A collection of Japanese netsukes is the theme, as the collection of these small sculptures passed from family member to family member from Paris, to Vienna, to Hong Kong, to the UK.
So funny, but I was standing (yes, it was very crowded, both shows!) staring at one of the early pieces and this funny little man, about 45, balding, working class, wearing a black T-shirt with motorcycle logos on it, comes up to me and says, so sweetly emboldened wanting to know about ahrt, I suppose) … "So … what's all this mean????" And thus began my humorously fueled rapport all throughout the next hour of viewing stunning Magrittes, describing what they meant to me to Johnny. He was all aglow, and soon "got it" and was able to pick out the different motifs (people represented by weird cut-outs reed-like paper, the rounded columns, his mistrust of words, the mix of dream and reality, etc.) so by the time Carter found us, the three of us continued amid the tidal wave of people, standing and discussing different works that stood out to each of us. We had a blast! Especially seeing the Light! go on in Johnny's eyes as he began to make sense of what before, to him, was incomprehensible hog-wash. Ahrt. Now he began to sense the power of what Art, good art, can convey.
Carter, my consort's favorite Magritte was the ham with an eyeball in the middle, very simple composition: a bottle of wine, a glass, a fork and knife. We all agreed it was "you are what you eat" … the way Carter is NOW living. So he was drawn to that one. Here's a bad shot of it from my iphone (yes, I illegally took photos, bad Lordflea!) But you can't really "see" the human eyeball in the middle of that hamsteak, can you? dern.
My fav?: well, the famous one where he paints a pipe and says: "Ceci ne c'est pas un pipe." Of course! This means how words are so fickle, and as a writer and artist, who has always had a hard time trying to decide WHICH of these two mediums I ought to trust to communicate the important stuff i have in my mind and heart, I can relate. But I love Magritte's MIND. The way he analyses life. It's all a big puzzle to him, and his paintings are like clue in a murder mystery, sending out little messages for his viewer to "get it" whatever "it" is that Magritte gets turned on by. And apparently, I like puzzles, and I like to try to figure out what Magritte's "message' it. Most of his works, I could, but only a few left me scratching my head.
My new friend Mick Stevens is a NYorker cartoonist, and I find trying to figure out his (and other NYorker toons) the same as looking at a Magritte. They're puzzles in images. Here's one of Mick's I've borrowed from his website.
The painting style of Magritte is pretty flat, like I said, but he does what he needs to do, paints fast, with little paint, no extra brush stroke and gets the hell out of painting as soon as he supplies the clues he wants to share about whatever subject, or dream, or news item he's focused on. Minimalism, for sure. He must be the Grandpappy of it, although REAL minimalism, I would say, didn't start for another thirty-forty years after Magritte's first use of minimal painting.


text:
Q: Do threads share time slot alloted to process Do threads belonging to the same process share the time alloted to the process or the time alloted to the process depends on the number of threads in the process....i.e a time slot is alloted to each thread by CPU...???

A: A general answer doesn't exist since it depends on the policy adopted by any specific OS.
Three common choices are:

*

*1 to 1, every thread has it's own schedulable entity

*N to 1, all threads of the same process aren't seen by the OS kernel, so just the whole process is scheduled (bad for multi-threading but good for legacy, they are somewhat similar to green threads)

*N to M, N threads are mapped to M schedulable entities (this is an hybrid approach that tries to be a compromise but it raises complexity of the scheduling)

More detailed explaination here.

A: For Windows, there's good documentation about how threads are selected to run here.  Thread priority is a combination of per-process and per-thread information.


text:
Q: How to have user input text to field, have text sent to database, and then displayed on a page? I am creating a website for a small personal project. Basically I need a text field where the user puts a quote, a submit button which sends it to a data base, and then all the quotes in the database displayed below that. I cannot figure out how to do this.

I have tried to use Javascript to display a quote after it is submitted, however everything disappears upon reloading the page, and it is not saved to a database.

A: You will need to use server side code to achieve this
Javascript is client side so data cannot be inserted to a database
These links should help you achieve what you need
https://www.w3schools.com/php/php_forms.asp
https://www.w3schools.com/php/php_mysql_insert.asp
https://www.w3schools.com/php/php_mysql_select.asp

A: You can do it on several ways:
Standard html form

*

*Set action form to php file. If you click submit then page will be refeshed.

*In php script you should veryfy if data are correct and sava data into database
or file

*Redirect user back to page with form. Bellow form you should fetch and display
rows from database.

Using AJAX - without refreshing page

*

*After click send button you can verify data in Javascript and send
them (via AJAX) to backend (php script)

*In backend verify if data are ok and save them to db/file. Backend should output JSON results. Message info ans status.

*In Javascript you should parse JSON result and make some actions like, show message to user.

*Next you can fetch (via ajax) list of all (or eg. last 10) rows from table in database

A: Try the following

first Set action form to php file (index.php).

<form method="Post" action="form.php">
<fieldset class="form-group">
    Username
    <input type="text" class="form-control" name="username" >
</fieldset>
<fieldset class="form-group">
    Lastname
    <input type="text" class="form-control" name="lastname" >
</fieldset>
<button type="button" name="submit" class="btn btn-primary">Submit</button>

form.php this will be second page

    <?php
 $server="localhost";
 $username ="root";
 $password ="";
 $database ="you_database";

 $connection =mysqli_connect($server ,$username ,$password,$database); //start connection

 if ($connection==TRUE) // test connection
  {
    echo "connected succefull";
 }

 if (isset($_POST['submit']))  //test submit name in form
 {
// declare parameter to initialized in text fild in form
// and those fild should appear in serve (apache mysql)
    $username =$_POST['username'];
    $lastname =$_POST['lastname'];

    $query ="INSERT INTO your_table  VALUES(NULL, '$username' ,'$lastname' ");
    $result =mysqli_query($connection ,$query);

    if ($result == TRUE) {

        echo "data insert succefull";
    }else{
        echo "failed to insert";
    }

 }
 mysqli_error($connection);

 ?>


ADVISORS Senior Consultant and trainer, Eng. Ziad El Dana has delivered a Project Management Professional Exam Preparation Course in partnership with the Order of Engineers in Beirut, Lebanon.
The course was conducted from March 18th, 2019 till April 5th, 2019 (8 Evening Sessions - MWF) and was delivered to a group of architects and engineers from multiple sector and engineering disciplines which made the interaction and discussions in class very beneficial.


Just who knew that plastic surgery was some thing treasured by the masses? The task is not a thing this is certainly liked, but the results are well worth money paid. The healing time has also reduced for the ages. It used to be that a lot of time ended up being demanded when you look at the recovery process. Immediately bruising and scarring is reduced significantly. It is because of many innovative improvements which have happened for the ages.
Cosmetic surgery is certainly a passion for people, lady and adolescents in the usa, however it's a thriving company across Asia and Europe, despite economic downturn.
Visual operation, consisting of a huge selection of aesthetic procedures and procedures, is growing in popularity all over the world, with more than 9 million medical and nonsurgical aesthetic methods done this past year in the United States alone.
Obtaining plastic surgery operation or plastic surgery treatment done in the western countries such as UK, USA, Australian Continent or Canada can are priced at a fortune, whereas the expense of plastic cosmetic surgery in lots of parts of asia might arrive as a pleasing wonder.
The high worldwide need, outrageously costly price of cosmetic processes in conjunction with painful extended hanging listings include key factors which have escalated the popularity of aesthetic treatment options in lower- and middle-income Asian and Latin and Central American countries such as for example India, Thailand, Mexico and Costa Rica.
To know extra about Mark Sofonio and website, please go to our very own site Rancho Mirage.
While Conway makes intercontinental headlines for their plastic cosmetic surgery trip, he is not the only one in the male search for looks enhancements. In accordance with the United states Society for visual Plastic Surgery (ASPS), significantly more than 900,000 people got cosmetic surgery procedures carried out in 2005. That seasons, the top five methods for men included: lipoplasty (liposuction), rhinoplasty (nose reshaping), blepharoplasty (aesthetic eyelid operation), male chest decrease and facelift.
Though people of most centuries are getting plastic cosmetic surgery, the majority is guys over 40. In a recent CBS Information document, doctors show two major causes men get plastic cosmetic surgery. Guys, they say, need a more youthful appearance to compete with young peers in the employees. And, they note, those who find themselves separated and matchmaking once again wish to generate on their own more inviting for any opposite gender. Whatever the reasons, here are some things to consider about these procedures before arranging the procedure.
Liposuction, the most popular methods among both women and men, occurs when surplus fat is taken away from below the facial skin by suction. But, getting rid of a great deal fat can put skin really loose to hang from the physique. This will leave anyone with a really unsightly look and will getting very challenging for any individual to control on a day-to-day foundation. Extra surgical procedure is frequently times needed to cut-off the extra skin. In many cases, guys do not need to bring this additional procedures because of their great skin elasticity, which allows her surface to adjust to the smaller sizes with general simplicity when compared with female, whose surface is the opposing.
This page was last edited on 3 February 2019, at 13:04.


IGNOU Offers Degree, Diploma and Certificate courses.
KKHSOU offers Degree, diploma and certificate courses through open and distance learning system.
Post Graduate Programme in Assamese, Political Science, Economics and Mathematics.
Bachelors' Programme under the Directorate of Distance Education,Dibrugarb University in all subjects.
For details: Dr. Jadumoni Das (Coordinator), Ph. 94351-51461.


Tim Holtz Idea-ology collection. Philosophy Tags.
A collection of 12 stamped metal tags that can be used as jewelry pieces or for adorning papercrafts.
Words include, "SIMPLIFY", "IMAGINE", "CHERISH", "BELIEVE", and lots more.


The Alameda County Health Pathway Partnership (ACHPP) is a consortium of pathway programs and organizations that aim to increase the diversity of the healthcare workforce by providing mentorship, academic enrichment, leadership development, and career exposure to disadvantaged and minority youth.
The Alameda County Health Care Services Agency has served as the facilitator of the consortium since 2007. Members are comprised of fifteen different organizations that provide short-term and long-term opportunities to youth in middle school through college and graduate students from underserved communities throughout Alameda County.