Americans today live in an unpredictable economic atmosphere. During this time of uncertainty, it is vitally important that corporate entities keep their assets stable to prevent their companies from going under. Lawsuits against corporate management have consistently increased over time leaving the companies that keep hard-working Americans employed at risk of losing everything they have. To prevent this from happening, Dean & Draper Insurance Agency recommends that Texas companies protect their upper management and therefore their entire businesses with officers and directors liability insurance.
Directors and officers (D & O) liability insurance is essentially executive insurance that offers protection to a company against lawsuits related to management decisions. A Texas directors and officers insurance policy goes beyond what is offered with general corporate liability insurance in that it responds to accusations of mismanagement of a company or its finances. D & O liability insurance does not cover claims related to bodily injury. Instead, it can be used to protect management during lawsuits filed against a company. For example, a directors and officers policy offers coverage against claims of financial mishandling by even one member of management and can also offer protection in lawsuits related to discrimination or termination. As corporate lawsuits can be incredibly expensive, a directors and officers liability insurance policy can quite literally determine whether or not a Texas business stays afloat.
Don't let one disgruntled employee or one regretful act of mismanagement cost you your Texas business. The commercial insurance brokers at Dean & Draper will work with you and your company to assure you have the correct coverage with your D & O liability insurance plan.


text:
Walmart To Layoff 700–800 at Arkansas HQ
Retail giant Walmart will cut 700 to 800 jobs at its Arkansas headquarters.The job hits will affect the retailer's real estate, apparel, and health and wellness offices, according to Associated Press reports. The company's pharmacy, optical and in-store h
Retail giant Walmart will cut 700 to 800 jobs at its Arkansas headquarters.
The job hits will affect the retailer's real estate, apparel, and health and wellness offices, according to Associated Press reports. The company's pharmacy, optical and in-store heath clinic will now be operated under one division and Walmart will be moving some apparel positions from the Bentonville, Ark., office to New York.
Despite the cuts, Walmart says it still plans to hire thousands of new employees for its stores and clubs this year. In 2008, the discounter opened up 166 new stores and this year plans to open no more than 140.

redpajama_set_name: RedPajamaCommonCrawl


Microsoft has revealed the new limited edition Forza Motorsport 6 Xbox One game console. It features a blue livery with a stripe over the matte section with a special reflective logo on the side of the console celebrating the 10 year anniversary of Turn 10 Studios' Forza Motorsport series.
You can check out the brief promo video that's featured on the Xbox news wire.
The console comes with a digital copy of Forza Motorsport 6, a 1TB hard drive for extra storage, a special Forza 6 10 year anniversary DLC car pack (yes, they're already tempting you with DLC before the game is even out) and a specialized, blue livery controller.
The interesting thing about the controller is that it not only contains the new headphone jack so you have additional options when it comes to the listening experience you can gather through your wireless controller, but it also has some specially designed palm grips on the back of the controller.
As explained in the video, the grips are both a cosmetic and a utilitarian addition to the controller, adding a necessary bit of anti-slippage during your gaming sessions. This is an interesting addition to the controller because KontrolFreek recently announced that they're getting into the controller grip business. Annexing their offering of analog stick extenders, KontrolFreek added palm grips to their line-up considering that Microsoft seems to want to bud into the analog extender game with their new Elite Wireless Controller, which carries the hefty price tag of $150.
But getting back to the limited edition Xbox One for Forza 6... one of the other key changes to the console is that it has an engine sound for the start-up when you hit the power button. It also has an engine powering down sound when you turn it off.
I don't know how well the livery and new sound effects will help the console handle 1080p at 60fps but Turn 10 putting their efforts into play with the new game to not only increase the on-screen AI count but also attempt to improve the graphics is risky business. I imagine they'll have to optimize the crap out of this new Forza to avoid the pitfalls that Forza 5 ran into where audience members on the side of the track were 2D cardboard cutouts and the lighting was all pre-baked to help keep the game running as smooth as possible.
It's likely that the pre-baked shadows will make a return to the game because real-time dynamic lighting will murder the in-game performance, just look at Project CARS for an example of trying to really push the software technology beyond the limits of the hardware. It's a beautiful game but the Xbox One's GPU didn't really like agreeing with what Slightly Mad Studios was trying to do.
We'll find out just how well Turn 10 will manage Forza 6 when it drops on September 15th in North America and September 18th in Europe.
The controller will be a standalone accessory as well, launching alongside the game and the limited edition console. The console will be available for $399.99.


text:
Macrobathra eudesma is a moth in the family Cosmopterigidae. It was described by Oswald Bertram Lower in 1900. It is found in Australia.

References

Macrobathra
Moths described in 1900

redpajama_set_name: RedPajamaWikipedia


text:
Zulu's storied history symbolizes Mardi Gras for African Americans worldwide
From the Mardi Gras Indians to the Flambeaux carriers to the skeleton crews, the contribution of African Americans to...
Posted: Feb 13, 2018 9:03 PM
From the Mardi Gras Indians to the Flambeaux carriers to the skeleton crews, the contribution of African Americans to traditional New Orleans Mardi Gras is evident. But nothing compares to the flash and flair of Zulu.
"It symbolizes Mardi Gras for the African American community across the world," said Zulu President Naaman Stewart. "Everybody that comes to Mardi Gras wants a Zulu coconut, and they'll do just about anything to get it."
The Zulu Social Aid and Pleasure Club is known as much for its elaborate coconuts as it is for community service and bringing people together. Members are also responsible for what could be the largest formal African American affair in the country -- the Zulu Ball.
Zulu's roots trace back to the days of segregation, when black people weren't allowed to be in other Mardi Gras krewes.
The masks of Zulu are strikingly similar to the "blackface makeup" once popularized by the minstrel shows of old. In fact, Louis Armstrong faced backlash from the Civil Rights community when he wore the makeup as King of Zulu in 1949.
But to understand the Zulu makeup, you have to know the history.
In the early 1900s, social aid and pleasure clubs were a popular mechanism for improving day-to-day life in the African American community. One of these clubs, the Tramps, went to the Pythian Temple in 1909 to see a play titled "There Never Was and Never Will Be a King Like Me."
The play was based loosely off of the Zulu ethnic group in Africa. When the Tramps walked out, they were no longer tramps, but Zulu warriors instead.
When they formed the Zulu parade krewe, the city required all krewe members to wear masks. Zulu members had limited resources -- and theatrical inspiration. Thus, the Zulu tradition was born.
"Some people assume that it was blackface make up, when what it was was ash," Stewart said.
From the coconuts to the balls to lending a helping hand behind the scenes, the Zulu organization has revolutionized Mardi Gras participation for the New Orleans African American community.
Zulu rolls at 8 a.m. Mardi Gras day along the Uptown route.
Spike Lee talks about his Mardi Gras ride with Zulu
Mardi Gras Fast Facts
Couple gets married on Mardi Gras float
LSU professor creates biodegradable Mardi Gras beads
Six Flags debuts Mardi Gras Hangover ride
Singer grants 9-year-old cancer patient's wish at Mardi Gras
Group accuses Airbnb of price gouging for Mardi Gras 2018
Grandfather builds custom wagon floats every Mardi Gras
STEM students create lego robot to clean Mardi Gras debris


A welcome return to the darkness for Changes with his latest tune - De-Solving Ego.
Twisted and moody with industrial stabs that drive the tune alone over a solid groove that still manages to pull the urge to dance out of the mayhem.
Hints of melody bring a nice balance to the track which shows that in music, as in life, there is beauty in chaos.
'De-Solving Ego' is available now from Beatport and iTunes.


MVLS Records is a record label and publishing company. Our aim is to serve quality artists and bring their music to a global audience. We will achieve this using our growing network of licensees and distribution partners. The genres of our artists range from Pop and Rock through to Contemporary Gospel music.
MVLS Records executive team have over 25 years experience in the music entertainment business. Away from developing and promoting our current repertoire, part of the MVLS vision is to continue to provide an alternative option to musicians, artistes, writers and composers seeking to further their careers in the world of recording and publishing.
We are a UK based independent music publisher. Our primary focus is to sign, develop and promote songwriters' compositions and give them the platform to work with performing artists in all styles and genres of music, as well as equip them with the skills to write jingles and library music for radio, TV / film, advertising, as well as all forms of multimedia (including the gaming industry) globally.
MVLS Publishing is split into two main areas: Artist and Repertoire (A&R) and Administration (which comprises of sub-publishing, licensing and exploitation in the copyright content) internationally.
MVLS Publishing has access to a skilled in-house team of songsmiths, writers and producers who are all very much involved on a creative level from the initial conception of the idea through to the final delivery of the song content. We have an East London headquarters and studio where all these activities are able to take place. MVLS Publishing takes pride in creating the perfect environment for writers to achieve their best work whilst remaining undisturbed.
We are keen to speak to gifted and unpublished artists, musicians, songwriters and composers who are looking to partner with an innovative music publisher. Whatever your songwriting and /or composing circumstances are, talking to us may prove beneficial to you.
Solicited submissions should be sent to the address below. However, please note that we receive a significant amount of material. Please allow 2 weeks for us to respond to you.. Should you wish any material to be returned to you, please ensure to enclose a stamped addressed envelope and we will do our best to ensure this is done.
Under the MVLS umbrella is a music production company with a state of the art Pro-tools HD recording studio based in East London. We house our own roster of producers, writers, musicians/singers and composers to cater for clients looking for music in the arena of TV, film, jingles and artist, singers.


text:
Home/ ACU News/ Academic News/ Dalton receives Fulbright award to teach in Malaysia
Dalton receives Fulbright award to teach in Malaysia
Academic News / ACU News
Mackenzie Dalton
Photo by Scott Delony
Applying for a Fulbright Scholar Award, much less getting it, wasn't on Mackenzie Dalton's mind when she enrolled at ACU as a freshman.
Then she met Dr. Jason Morris, dean of the university's Honors College. He mentioned the opportunity and convinced her to apply. Mackenzie, who is from Little Rock, Arkansas, is an accounting major and English minor.
She has been accepted at Baylor Law School, but that will have to be postponed for a little while. One thing Mackenzie always wanted to do was teach English abroad. Morris provided the opportunity she was looking for. She applied for the Fulbright Scholar Award, received it, and now is planning to head to Malaysia as an English Teaching Assistant.
"It worked out," she said.
Everything seems to be working out for Mackenzie – and for ACU. Mackenzie is one of two ACU students who were named recipients of the Fulbright Scholar Award this year. Angela Jirik will be an English Teaching Assistant in Colombia. This year's award winners join a list of other ACU students and faculty honored with Fulbright scholarships and grants dating to the 1950s.
ACU is included in the list of U.S. colleges and universities that produced the most 2018-2019 Fulbright U.S. students. ACU is one of only three Texas institutions to receive the designation.
Mackenzie will graduate May 11 and then will spend some time with her family in Little Rock, including a brother who is a senior in high school.
"I'll head home to see him graduate," she said.
Then she will have a long wait before leaving for Malaysia, sometime in January. Some of the time will be spent on a road trip with her family, camping out and visiting several national parks in the West.
Mackenzie also wants to get a job or internship, preferably related to law, the career she wants to pursue after finishing her stint as a Fulbright Scholar. As of now, Mackenzie doesn't know exactly where she will be assigned in Malaysia, but she knows she will be somewhere in the country for 10 months.
She will be following in the footsteps of another ACU Fulbright honoree, Lindsie Lawson, who received the award in 2018 and left for Malaysia in January 2019. Her term will be up in November, so the two ACU graduates won't bump into each other in Malaysia. But Mackenzie is getting some tips from Lindsie as she waits to go abroad.
Since arriving at ACU, Mackenzie has discovered a world of opportunity, thanks to people like Morris and other faculty and staff. She spent part of the summer of 2017 in Leipzig, Germany, studying business under Dr. Jonathan Stewart. She served an internship in the summer of 2018 and now is preparing for Malaysia.
Once she gets home from the Malaysia trip, Mackenzie will be on the road to Waco, where she will enroll at Baylor Law School for the spring 2021 semester. Mackenzie is leaning toward immigration law after earning her degree – if not immigration law, then definitely a field with a social justice emphasis.
Mackenzie also understands business well and is drawn to that field, too. If she does go into business law, Mackenzie will find some way to use that skill to help others. She went on a spring break mission trip to the Dominican Republic in 2018 and found it to be rewarding.
Wherever life takes her, Mackenzie will maintain her ties to ACU.
"I have made some very, very good friends," she said.


The human brain is a magical thing and it's not wired up in the most straightforward way. Here's a trick you can perform on yourself, described by the great Stephen Fry on the BBC series QI. You could do it standing up, but seated is probably easier. Also a good idea to not be driving a car.
Toronto-born Mahdi Gilbert is, as far as I know, unique in the world of magic. He was born without hands and feet but created his own system to perform the kind of magic which would, under ordinary circumstances, be called "sleight-of-hand".
The trick you can see performed is a modern classic known as "Oil and Water"
Mahdi's work is, in part, inspired by another differently abled magician from Argentina, René Lavand. Lavand lost his right arm in an accident at an early age.
He also worked primarily with cards. Most of his performances revolved around storytelling and even the occasional poetry reading. He described his own work as "Lentidigitation" which was the opposite of "Prestidigitation", roughly translated as "Fast Finger Action".
If you watch it below, you'll repeatedly hear him say "No Se Pueder Hacer Más Lento" which translates to "It can't be done any slower."
The UK-based recreational math channel Numberphile is still on a magic spree. Here they explain a card trick made popular by the Scientific American columnist (and pretty much hero to nerds everywhere) Martin Gardner.
It's a little long for casual internet time-wasting the payoff at the end was entirely worth it.
I would totally try this in my next show if I thought for a second anyone in a Canadian audience would know what the word "polity" meant.
Bayes' Theorem, discovered by the Reverend Thomas Bayes in the eighteenth century. When you have something you're not sure of, it's the calculation you perform to update your belief when you encounter new evidence. It's essentially the mathematical underpinning of the scientific method and it's an incredibly valuable thing to understand.
All that coolness aside, this is still the most badass titles for a statistics lesson!
Add magic to your next event!


It's a modest start to much larger plans.
After months of waiting and no shortage of hype, Tesla's first Model 3 is finally rolling off the production line. Elon Musk has revealed that the initial production unit is ready (not pictured as of this writing) pending a "final checkout." Musk will get the first car, as he has with earlier models, but he didn't call dibs -- while Ira Ehrenpreis was the first to plunk down a full deposit, he gave that spot to Musk as a birthday present. It's safe to say that either would be part of an exclusive club given that a mere 30 Model 3s will reach buyers by the end of July.
It's not certain just how much has changed on the Model 3 since it was unveiled back in 2016. You likely won't get the full scoop on that until a handover party scheduled for July 28th. However, Musk has already hinted that there shouldn't be any dramatic changes versus the release candidate prototypes that appeared in March. The past few months have largely been spent tweaking and testing components to ensure the Model 3 is ready for the road.
It's a largely symbolic announcement when most Model 3 pre-order customers won't even get their vehicles until 2018. Production is only slated to ramp up to 20,000 cars per month by December, which hardly puts a dent in the hundreds of thousands of reservations made since last spring. With that said, you really are looking at the start of Tesla's next chapter. Musk and crew can now say they're producing EVs aimed at the mainstream (albeit the higher end), not just a subset of the luxury car crowd.
Has Tesla Model S and Model X Demand Peaked?
Tesla co-founder and chief executive Elon Musk has shared the first images of ... Mr Musk tweeted a picture captioned "First Production Model 3" on Saturday.
Tesla has been working on a limited production run of the Model 3 electric car. The first 30 new cars are expected to be delivered to customers on July 28.
Elon Musk gets first Tesla Model 3 as his 46th birthday gift; everything you need to ... CEO Elon Musk has been gifted the first production model of Tesla Model 3.
Musk tweeted earlier that the first production model of the vehicle was being constructed and undergoing a final check before posting two photos of the car.
Before unveiling pictures of the historic auto invention, Elon Musk tweeted, "Production unit 1 of Model 3 is now built and going through final checkout. Pics soon.
Musk tweets first photos of Tesla's mass market electric Model 3 ... co-founder Elon Musk tweeted on July 9, 2017, a photo of the first production Tesla Model 3.
The Model 3 is an incredibly important vehicle to the company, as its the first 'affordably priced' car from Tesla and the first to be produced in high volumes.
The very first production-spec Tesla Model 3 has rolled out of the company's ... Production unit 1 of Model 3 is now built and going through final checkout.


When you work on your own and do not intend on partnering up with anyone or hiring anyone, you only need to worry about your personal wants and needs when it comes to work. It gives you a great deal of freedom in regard to the choices that you make, so you can pick an office space in Atlanta that truly has everything you need. If you have a home office and a lack of productivity as a result, you should start looking at office spaces with the intention to lease one as it is a dependable way to solve your problem.
Whether your business can benefit from collaborating with other professionals or not, the fact that working in an office with other professionals allows you to have this option is quite beneficial. It is easier to make business connections when you are working with them in the same building.
If it is peace and quiet that is keeping you from working productively, you can easily find office space that provides you with complete silence for your business. It is a matter of making it a top priority and not choosing an office space that has you working alongside other professionals in the same room.
By working in an office, you will naturally be in a productive environment. It is not uncommon for this to spark your productivity because seeing others working hard and being successful is not fun when you are on the sidelines slacking. These professionals will motivate you to work just as hard as them.
If you have any questions about office spaces, contact us at any time.


text:
Ботанічний сад університету Берна або Бернський ботанічний сад () — ботанічний сад у місті Берн (кантон Берн, Швейцарія).  Ботанічний сад є членом Міжнародної ради ботанічних садів з охорони рослин (BGCI) і має код BERN.

Ботанічний сад Бернського університету (BOGA), заснований 1859 року, є барвистим оазисом у центрі Берна. Колекція саду налічує близько 5500 видів рослин з усього світу і використовується в сфері освіти, науки і культури. Головна спеціалізація ботанічного саду — альпійські рослини.

Рослинні колекції саду

Альпінарій
Альпінарій пропонує відвідувачам коротку подорож по Піренеям, Альпам, Карпатам, Кавказу, Гімалаям і Скелястим горам. Тут росте близько 1500 різних видів, це найбільш багата і цікава частина ботанічного саду, його «серце».

Альпінарій був створений в п'ятдесяті роки минулого століття з великих блоків вапняку. Ця частина саду виглядає дуже натурально через крутий схил і розташування каменів на ньому.

Корисні, декоративні та лікарські рослини
 Сад півоній,
 Сад лікарських трав,
 Садові культури,
 Дикий сад.

Географічні регіони
 Австралія,
 Нова Зеландія,
 Сукуленти відкритого ґрунту Нового Світу,
 Північноамериканські водно-болотні угіддя,
 Центральноєвропейські деревні рослини,
 Азія,
 Азійські водно-болотні угіддя,
 Середземномор'я,
 Південна Африка,
 Американські водні рослини.

Оранжереї
 Оранжерея Середземноморських рослин,
 Оранжерея орхідей,
 Пальмова оранжерея,
 Оранжерея папороті,
 Оранжерея сукулентів.

Дендрарій
Протягом десятиліть в ботанічному саду створена чудова колекція дерев і чагарників (близько 1000 видів). Через дуже сприятливу кліматичну ситуацію в саду (південний схил пагорба), тут, на висоті 530 метрів над рівнем моря, добре себе почувають навіть ніжні деревні рослини. Деревні рослини можна знайти на всій території саду.

Кімнатні рослини
У літні місяці дубові ванни з кімнатними рослинами розташовані в семи різних місцях ботанічного саду.

Екологічні зони
 Зернове поле,
 Ставок,
 Швейцарські водні та болотні рослини,
 Сухий луг,
 Сад метеликів і гусениць,
 Грот,
 Будинок комах,
 Бджолиний вулик.

Галерея

Посилання
 Офійційний сайт ботанічного саду
 BGCI — Botanischer Garten der Universitat Bern

Берн
Природоохоронні об'єкти, засновані 1859
Берн


Made by hand in the USA with up to 8 layers of papier-mâché for durability.
Covered with 100% imported linen.
A cotton wadding layer under the linen facilitates easy pinning to the form.
Cast Iron black Automatic Horseshoe Base (with adjustable height) included with all dress forms.
Please allow up to 10 weeks for delivery.
10% discount for students. Just fax, e-mail or mail us proof of enrollment in a fashion related university and we will send you a coupon code to use on your order.


text:
\section{Introduction and Preliminaries}

\quad The fixed point theory is an essential tool not only in the field of nonlinear analysis but also in almost all applicable branches of importance.  After Banach's contraction principle (\cite{Ban}), over the course of time there were multiple generalizations  have been made for example, Geraghty's Theorem (\cite{G}). Moreover in \cite{Kho} a $k$-dimensional extension of the Geraghty's result is also provided. Never the less, Kannan and Fisher both have given two independent type of contraction operators (see \cite{Kan}, \cite{F}) which are completely independent of the Banach's contraction. Also, in a recent paper by F. Fogh, S. Behnamian and F. Pashaie (see, \cite{Kg}) introduced the concept of Kannan-Geraghty contraction.

In this paper, We extend the $k$-dimensional extension of the Geraghty's Theorem stated in \cite{Kho} to infinite dimension. We also introduce and develop a new notion of $H_k$ contraction for Geraghty-Kannan type operator from \cite{Kg} to provide both $k$ dimensional extension and infinite dimensional extension. We also provide a same family of $k$-dimensional extension for Fisher type contractions and also develop similar infinite dimensional extension in this case. As application of the theorems proved in this paper we define the notion of two independent type of infinite integral equations, namely infinite dimensional Fredholm equation and Uryshon type integral equation and, provide existential proof of their unique solution.

Next, we discuss some of the preliminaries which will be needed later for proving our main theorems.

\begin{theorem}\label{2.1} (\cite{G}) Let $(X, d)$ be a complete metric space and $T : X \to X$ be a
mapping. If $T$ satisfies the following inequality:
\begin{align*}
d(T x, T y) \leq \beta(d(x, y)) d(x, y)
\end{align*}
 where $\beta : [0, \infty) \to [0, 1)$ is a function which satisfies the condition;
 \begin{align*}
\lim\limits_{n\to \infty}\beta(t_n) = 1 ~implies \lim\limits_{n\to \infty}t_n = 0
\end{align*}

Then $T$ has a unique fixed point $u\in X$ and ${T^nx}$ converges to $u$ for each $x \in X.$

\end{theorem}
We denote by $\mathbb{G}$ a set of all functions $\beta$ given in Theorem \ref{2.1}.

\begin{theorem}\label{2.2} (\cite{Kan}) Let $(X, d)$ be a complete metric space and $T : X \to X$ be a
mapping. If $T$ satisfies the following inequality:
\begin{align*}
d(T x, T y) \leq c(d(Tx, x)+ d(Ty, y))
\end{align*}
 where $c\in (0, 1/2)$,

Then $T$ has a unique fixed point $u \in X $.
\end{theorem}

\begin{theorem}\label{2.3} (\cite{F}) Let $(X, d)$ be a complete metric space and $T : X \to X$ be a
mapping. If $T$ satisfies the following inequality:
\begin{align*}
d(T x, T y) \leq c( d(Tx, y)+ d(Ty, x) )
\end{align*}
 where $c\in (0, 1/2)$, $x,y\in X$.

Then $T$ has a unique fixed point $u\in X $.
\end{theorem}

\begin{definition}\label{2.4} (Kannan-Geraghty Map) \cite{Kg} For any complete metric space (X,d) and self map $T:X\to X$, $T$ is defined to be a Kannan-Geraghty self map if and only if it satisfies the following:
		$$d(Tx , Ty)\leq\frac{\beta(d(x,y))}{2}~(d(Tx,x)+d(Ty,y))$$
where, $ \beta \in  \mathbb{G}$, $x,y\in X$.
\end{definition}

\begin{theorem}\label{2.5} (\cite{Kg}) Let $(X, d)$ be a complete metric space and $T : X \to X$ be a
mapping. If $T$ is a Kannan-Geraghty self map on $(X, d)$, then $T$ has a unique fixed point $u \in X $ and for any $x_0 \in X$, $T^nx_0$ converges to $u$.
\end{theorem}

\begin{theorem}\label{2.6}(\cite{Kho}) Let $(X, d)$ be a complete metric space and $k \in \mathbb{N} $ let, $T : X^k \to X$
be such that, for all $u_1,u_2, . . . , u_{k+1}\in X$, we have the following inequality,

\begin{align*}
d(T (u_1, . . . , u_k), T (u_2, . . . , u_{k+1}))&\leq
M((u_1, . . . , u_k), (u_2, . . . , u_{k+1})))M((u_1, . . . , u_k), (u_2,. . . ,  u_{k+1}))
\end{align*}
where, $ \beta \in  \mathbb{G}$ and $M : X^k \times X^k \to [0,\infty)$ is as.\\

$M((u_1, . . . , u_k), (u_2,. . . ,  u_{k+1}))
= max\left\lbrace d(u_k, u_{k+1}),~d(u_k, T (u_1, . . . , u_k)),~ d( u_{k+1}, T (u_2,  . . . ,  u_{k+1}))\right\rbrace$.

Then there exists a point $u \in X$ such that $T (u, u, . . . , u) = u.$
\end{theorem}

We denote by
$\mathbb{N}$ (resp. $\mathbb{N}_0$) the set of positive (nonnegative) integers.  The aim of the next section is to generalize and extend Theorem \ref{2.1}, Theorem \ref{2.5} and \ref{2.6} as well as the k-dimensional extension of the result given in \cite{G} to infinite-dimension. We denote the infinite tuples of points $(x_1, x_2, ...)$ by $(x_k)_{i=1} ^{\infty}$ and the infinite tuples $(x_1, x_2, ...x_{k-1}, x_{k},x_{k},...)$ with the $k-$th point repeated by $(x_{j,\hat{k}})_{j=1} ^\infty.$

\section{Main results}

\quad In this section we present the main results of the paper. First, let us introduce some new notions concerning the infinite-dimensions of the points. Further let us consider, in all the cases, that $\left(X,d\right)$ is a $complete$ $metric$ $space$ and $k\in\mathbb{N}$.

\begin{definition}\label{d3.1} (Extended $M_k$ function) For any $(u_{i})_{i=1}^{\infty},(v_{i})_{i=1}^{\infty}\in\prod_{i=1}^{\infty} X$ and  $T:\prod_{i=1}^{\infty} X\to X$, we define the extended function $M_{k}:\prod_{l=1}^\infty X \times \prod_{l=1}^\infty X\rightarrow X$ as follows,

$$M_{k}\left((u_{i})_{i=1}^{\infty},(v_{i})_{i=1}^{\infty}\right)=\left\{\begin{array}{ll} {\max\left\lbrace\sup\limits_{l\geqslant k} d\left(u_{l},v_{l}\right),\sup\limits_{l\geqslant k} d\left(u_{l},T(u_{i,\hat{l}})_{i=1}^{\infty}\right), \sup\limits_{l\geqslant k} d\left(v_{l},T(v_{i,\hat{l}})_{i=1}^{\infty}\right)\right\rbrace},\\ \hspace{20pt}{\text{ if all supremum exist}}\\
{\max\left\lbrace d\left(u_{k},v_{k}\right), d\left(u_{k},T(u_{i,\hat{k}})_{i=1}^{\infty}\right), d\left(v_{k},T(v_{i,\hat{k}})_{i=1}^{\infty}\right)\right\rbrace},\\ \hspace{20pt} {\text{ if one fails to exists}}\\
\end{array}\right.$$

\end{definition}

\begin{example}
 For $k=1\text{, }X=[0,1]$ the operator $T:\prod_{i=1}{\infty}X\to X$ defined as $T((u_{i})_{i=1}^{\infty})=u_k$, $M_{k}\left((u_{i})_{i=1}^{\infty},(v_{i})_{i=1}^{\infty}\right)= \abs{u_k-v_k}=\abs{u_1-v_1}$
\end{example}

\begin{definition}\label{d3.2} ($H_k$ contraction) An operator $T:\prod_{l=1}^\infty X\rightarrow X$ is called a $H_{k}$ $contraction$ if and only if it satisfies the following inequality,
\begin{equation}\label{hc}
		d\left(T\left((u_{i,\widehat{k}})_{i=1}^{\infty}\right), T\left((v_{i,\widehat{k}})_{i=1}^{\infty}\right)\right) \leqslant\beta\left(M_{k}\left((u_{i,\widehat{k}})_{i=1}^{\infty},(v_{i,\widehat{k}})_{i=1}^{\infty}\right)\right)M_{k}\left((u_{i,\widehat{k}})_{i=1}^{\infty},(v_{i,\widehat{k}})_{i=1}^{\infty}\right),
\end{equation}
for all $u_1,~u_2,~\ldots,~u_{k},~ v_1,~ v_2, \ldots,~ v_k\in X$, where $\beta\in\mathbb{G}$.
\end{definition}

\begin{example}
 For example the operator $T:\prod_{i=1}^{\infty}X\to X$ where $X=[0,1]$, defined by $T((x_i)_{i=1}^{\infty}):=cx_k$ is a $H_k$ contraction where $k\in\mathbb{N}$ be any fixed number and $c\in(0,1)$.
\end{example}

\begin{definition}\label{d3.3} (Kannan-Geraghty $H_k$ contraction) An operator $T:\prod_{l=1}^{k} X\rightarrow X$ is called a Kannan-Geraghty contraction of dimension $k$ if and only if it satisfies the following inequality
\begin{equation*}
  d\left(T\left((u_{i})_{i=1}^{k}\right), T\left((v_{i})_{i=1}^{k}\right)\right) \leqslant\frac{\beta(d(u_k,v_{k}))}{2}(d(T\left((u_{i})_{i=1}^{k}\right), u_k)+ d(T\left((v_{i})_{i=1}^{k}\right), v_{k})),
\end{equation*}
for all $u_1, u_2,\ldots,u_{k}, v_1, v_2, \ldots, v_k\in X$, where $\beta\in\mathbb{G}$.
\end{definition}

\begin{definition}\label{d3.4} (Extended Kannan-Geraghty $H_k$ contraction) An operator $T:\prod_{l=1}^{\infty} X\rightarrow X$ is called a Extended Kannan-Geraghty $H_k$ contraction if and only if it satisfies the following inequality
\begin{equation*}
		d\left(T\left((u_{i,\widehat{k}})_{i=1}^{\infty}\right),T\left((v_{i,\widehat{k}})_{i=1}^{\infty}\right)\right) \leqslant\frac{\beta\left(d(u_k,v_{k})\right)}{2}(d(T\left((u_{i,\widehat{k}})_{i=1}^{\infty}\right),u_k)+d(T\left((v_{i,\widehat{k}})_{i=1}^{\infty}\right),v_{k})),
\end{equation*}
for all $u_1,~ u_2,\ldots,~u_{k}, v_1,~ v_2, \ldots,~ v_k\in X$, where $\beta\in\mathbb{G}$.

\end{definition}

\begin{definition}\label{d3.5} (Fisher-Geraghty $H_k$ contraction)  An operator $T:\prod_{l=1}^{k} X\rightarrow X$ is called a Fisher-Geraghty $H_k$ contraction of dimension $k$ if and only if it satisfies the following inequality,
\begin{equation*}
	d\left(T\left((u_{i})_{i=1}^{k}\right), T\left((v_{i})_{i=1}^{k}\right)\right) \leqslant\frac{\beta(d(u_k,v_k))}{2}(d(T\left((u_{i})_{i=1}^{k}\right),v_{k})+d(T\left((v_{i})_{i=1}^{k}\right),u_{k})),
\end{equation*}
for all $u_1,~ u_2,~\ldots,~u_{k},~ v_1,~ v_2,~ \ldots,~ v_k\in X$, where $\beta\in\mathbb{G}$.
\end{definition}

\begin{definition}\label{d3.6} (Extended Fisher-Geraghty $H_k$ contraction)  An operator $T:\prod_{l=1}^{\infty} X\rightarrow X$ is called a Extended Kannan-Geraghty $H_k$ contraction if and only if it satisfies the following inequality
\begin{equation*}
d\left(T\left((u_{i,\widehat{k}})_{i=1}^{\infty}\right),T\left((v_{i,\widehat{k}})_{i=1}^{\infty}\right)\right) \leqslant\frac{\beta\left(d(u_k,v_{k})\right)}{2}(d(T\left((u_{i,\widehat{k}})_{i=1}^{\infty}\right),v_{k})+d(T\left((v_{i,\widehat{k}})_{i=1}^{\infty}\right),u_{k})),
\end{equation*}
for all $u_1,~ u_2,~\ldots,~u_{k},~ v_1,~ v_2,~ \ldots,~ v_k\in X$, where $\beta\in\mathbb{G}$.
\end{definition}

\begin{example}
 As an example to Definition \ref{d3.3} and Definition \ref{d3.5} we can consider for $X=[0,1]$, $T:\prod_{i=1}^{k}X\to~X$ defined as $T((x_i)_{i=1}^{k})=cx_k$ for some $c\in(0,\frac{1}{2})$.
\end{example}

\begin{example}
 As an example to Definition \ref{d3.4} and Definition \ref{d3.6} we can consider for $X=[0,1]$, $T:\prod_{i=1}^{\infty} X\to~X$ defined as $T((x_i)_{i=1}^{\infty})=cx_k$ for some $c\in(0,\frac{1}{2})$.
\end{example}
\noindent Let $T:\prod_{i=1}^{k}X\to~X$ be any operator. Let us choose $x_1, x_2, \ldots, x_k\in X$ Now we define the following:
\begin{definition}\label{d3.7}($k$-Picard sequence with respect to the operator $T$)
The $k$-Picard sequence with respect to the operator $T$ based on the base point set $\lbrace x_1, x_2, \ldots, x_k\rbrace$ is definded as  $x_{n+k}:=T((x_{n+i-1})_{i=1}^{k})$, for all $n\geq k$.
\end{definition}

\begin{example}
 If we fix $k=1$ then the base point set is singleton and the $1$- Picard sequence with respect to $T$ based on $\lbrace x_0\rbrace$ is basically the Picard sequence of $T$ based on the base point $\lbrace x_0\rbrace$ defined by $x_n:= T(x_{n-1}) ~\forall n\geqslant 1$ for some $x_0\in X$.
\end{example}

\noindent Let $T:\prod_{i=1}^{\infty}X\to~X$ be any operator. Let us choose $x_1, x_2, \ldots, x_k\in X$ Now we define the following:

\begin{definition}\label{d3.8}(Infinite $k$-Picard sequence with respect to the operator $T$)
The $k$-Picard sequence with respect to the operator $T$ based on the base point set $\lbrace x_1, x_2, \ldots, x_k\rbrace$ is definded as  $x_{n+k}:=T((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty})$, for all $n\geq k$.
\end{definition}

\begin{example}
If we fix $k=1$ then the base point set is singleton and the infinite $1$- Picard sequence with respect to $T$ based on the base point $\lbrace x_0\rbrace$ is basically the sequence  defined by $x_n:= T((x_{n-1})_{n=1}^{\infty}) ~\forall n\geqslant 1$ for some $x_0\in X$.
\end{example}

Let us give our first main fixed point result which is a generalization of Banach contraction principle with respect to the infinite-dimensional notion introduced in our paper.
\begin{theorem}\label{3.1}
Let $\left(X,d\right)$ be a complete metric space. $T:\prod_{l=1}^\infty X \rightarrow X$ is a $ H_{k}$ contraction for some $k\in\mathbb{N}$ . Then there exists $u\in X$ such that $T((u)_{i=1}^{\infty})=u$ and the infinite $k$-Picard sequence for $T$ converges to $u$.
\end{theorem}
\begin{proof}
		Let $x_1,x_2\ldots,x_k\in X$. Then infinite $k$-Picard sequence is defined as follows:
		\newline
		$\forall~n\in\mathbb{N} $ we define,
		\begin{equation}\label{s}
		x_{n+k}:=T\left((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty}\right)
		\end{equation}
		\newline

			We claim that $\left\lbrace d\left(x_{n+k},x_{n+k+1}\right)\right\rbrace_{n\in\mathbb{N}}$ is convergent.

		From Definition (\ref{d3.2})
		\begin{flushleft}
			$d\left(x_{n+k+1},x_{n+k+2}\right)=d\left(T\left((x_{n+i,\widehat{n+k}})_{i=1}^{\infty}\right), T\left((x_{n+i+1,\widehat{n+1+k}})_{i=1}^{\infty}\right)\right)$

			$\leqslant\beta\left(M_{k}\left((x_{n+i,\widehat{n+k}})_{i=1}^{\infty}, (x_{n+1+i,\widehat{n+1+k}})_{i=1}^{\infty}\right)\right)M_{k}\left((x_{n+i,\widehat{n+k}})_{i=1}^{\infty}, (x_{n+1+i,\widehat{n+1+k}})_{i=1}^{\infty}\right).$

		\end{flushleft}
		\begin{flushleft}
			Let
$M_{k}\left((x_{n+i,\widehat{n+k}})_{i=1}^{\infty},(x_{n+i+1,\widehat{n+1+k}})_{i=1}^{\infty}\right)=\max\left\lbrace\sup\limits_{l\geqslant k} d\left(u_{l},v_{l}\right),\sup\limits_{l\geqslant k} d\left(u_{l},T(u_{i,\widehat{l}})_{i=1}^{\infty}\right),\sup\limits_{l\geqslant k} d\left(v_{l},T(v_{i,\widehat{l}})_{i=1}^{\infty}\right)\right\rbrace$
		\end{flushleft}
		where
		\begin{gather}
		u_{l}:=x_{n+l},\: \forall\: 1\leqslant l\leqslant k
		\\
		u_{l}:=x_{n+k},\: \forall\: l> k
		\\
		v_{l}:=x_{n+l+1},\: \forall\: 1\leqslant l\leqslant k
		\\
		u_{l}:=x_{n+k+1},\: \forall\: l> k.
		\end{gather}
		But
		\begin{flushleft}
			$\sup\limits_{l\geqslant k} d\left(u_{l},v_{l}\right)=d\left(x_{n+k},x_{n+k+1}\right)$,\\
			$\sup\limits_{l\geqslant k} d\left(u_{l},T\left((u_{i,\widehat{l}})_{i=1}^{\infty}\right)\right)=d\left(x_{n+k},T\left((x_{n+i,\widehat{n+k}})_{i=1}^{\infty}\right)\right)$,\\
			$\sup\limits_{l\geqslant k} d\left(v_{l},T\left((v_{i,\widehat{l}})_{i=1}^{\infty}\right)\right)=d\left(x_{n+k+1},T\left((x_{n+i+1,\widehat{n+k+1}})_{i=1}^{\infty}\right)\right)$.
		\end{flushleft}
		\begin{flushleft}
			$\implies M_{k}\left((x_{n+i,\widehat{n+k}})_{i=1}^{\infty},(x_{n+i+1,\widehat{n+k+1}})_{i=1}^{\infty}\right)$\\$=\max\left\lbrace d\left(x_{n+k},x_{n+k+1}\right), d\left(x_{n+k},T\left((x_{n+i,\widehat{n+k}})_{i=1}^{\infty}\right)\right),d\left(x_{n+k+1},T\left((x_{n+1+i,\widehat{n+k+1}})_{i=1}^{\infty}\right)\right)\right\rbrace$
			$\implies M_{k}\left((x_{n+i,\widehat{n+k}})_{i=1}^{\infty},(x_{n+1+i,\widehat{n+k+1}})_{i=1}^{\infty}\right)=\max\left\lbrace d\left(x_{n+k},x_{n+k+1}\right), d\left(x_{n+k},x_{n+k+1}\right),d\left(x_{n+k+1},x_{n+k+2}\right)\right\rbrace$
			$=\max\left\lbrace d\left(x_{n+k},x_{n+k+1}\right),d\left(x_{n+k+1},x_{n+k+2}\right)\right\rbrace$.
		\end{flushleft}

		If $M_{k}\left((x_{n+i,\widehat{n+k}})_{i=1}^{\infty},(x_{n+i+1,\widehat{n+k+1}})_{i=1}^{\infty}\right)=d\left(x_{n+k+1},x_{n+k+2}\right)$ then, from Definition (\ref{d3.2}), we get

				\begin{flushleft}
			$d\left(x_{n+k+1},x_{n+k+2}\right)=d\left(T\left((x_{n+i,\widehat{n+k}})_{i=1}^{\infty}\right),T\left((x_{n+i+1,\widehat{n+k+1}})_{i=1}^{\infty}\right)\right)$
			\\
			$\leqslant\beta\left(M_{k}\left((x_{n+i,\widehat{n+k}})_{i=1}^{\infty},(x_{n+i+1,\widehat{n+k+1}})_{i=1}^{\infty}\right)\right)M_{k}\left((x_{n+i,\widehat{n+k}})_{i=1}^{\infty},(x_{n+i+1,\widehat{n+k+1}})_{i=1}^{\infty}\right)$
			\\
			$=\beta\left(d\left(x_{n+k+1},x_{n+k+2}\right)\right)d\left(x_{n+k+1},x_{n+k+2}\right)$
			\\
			$<d\left(x_{n+k+1},x_{n+k+2}\right).$
		\end{flushleft}

Then we get a contradiction.

			Hence $M_{k}\left((x_{n+i,\widehat{n+k}})_{i=1}^{\infty},(x_{n+i+1,\widehat{n+k+1}})_{i=1}^{\infty}\right)=d\left(x_{n+k},x_{n+k+1}\right).$
		\begin{flushleft}
			$\implies d\left(x_{n+k+1},x_{n+k+2}\right)\leqslant\beta\left(d\left(x_{n+k},x_{n+k+1}\right)\right)d\left(x_{n+k},x_{n+k+1}\right)<d\left(x_{n+k},x_{n+k+1}\right)$
		\end{flushleft}
		$\implies \left\lbrace x_{n+k}\right\rbrace_{n\in\mathbb{N}}$ is a strictly decreasing sequence.\\

	Using Monotone Convergence Theorem, for some $(r\geqslant 0)\in \mathbb{R}$, we get
		\begin{equation}\label{l}
		\lim_{n\to\infty}d( x_{n+k},x_{n+k+1})=r.\\
		\end{equation}

		We claim $\lim\limits_{n\to\infty}d( x_{n+k},x_{n+k+1})=0$.

		Suppose the contrary and assuming that $r>0$ we get
		\begin{flushleft}
			$$d\left(x_{n+k+1},x_{n+k+2}\right)\leqslant\beta\left(d\left(x_{n+k},x_{n+k+1}\right)\right)d\left(x_{n+k},x_{n+k+1}\right)$$ $$\implies\frac{d\left(x_{n+k+1},x_{n+k+2}\right)}{d\left(x_{n+k},x_{n+k+1}\right)}\leqslant\beta\left(d\left(x_{n+k},x_{n+k+1}\right)\right).$$
		\par This implies $\lim\limits_{n\to\infty}\beta\left(d\left(x_{n+k},x_{n+k+1}\right)\right)\geqslant 1.$
		\par Since
			$\beta\in\mathbb{G}\text{ implies}\lim\limits_{n\to\infty}\beta\left(d\left(x_{n+k},x_{n+k+1}\right)\right)\leqslant 1$
			$\implies\lim\limits_{n\to\infty}\beta\left(d\left(x_{n+k},x_{n+k+1}\right)\right)=1$
we obtain
			\begin{equation}
			\lim\limits_{n\to\infty}d\left(x_{n+k},x_{n+k+1}\right)=0.\label{8}
			\end{equation}
		\end{flushleft}

		We now claim that $\left\lbrace x_{n+k}\right\rbrace_{n\in\mathbb{N}}$ is Cauchy and we prove it by contradiction. If we suppose the contrary $\exists~\varepsilon>0$ such that we can find some subsequences $\left\lbrace x_{\left(m\right)+k}\right\rbrace_{p\in\mathbb{N}}$, $\left\lbrace x_{n\left(p\right)+k}\right\rbrace_{p\in\mathbb{N}}$ with $m\left(p\right)>n\left(p\right)>p$ such that for every $p$ we have
			\begin{equation}
			d\left(x_{m\left(p\right)+k},x_{n\left(p\right)+k}\right)\geqslant\varepsilon\label{9}
			\end{equation}

			Moreover, corresponding to each $n(p)$ we can choose least of such $m\left(p\right)$ satisfying (\ref{9}). Then
			\begin{equation}\label{10}
			d\left(x_{m\left(p\right)+k-1},x_{n\left(p\right)+k}\right)<\varepsilon.
			\end{equation}

			From (\ref{8}),(\ref{10}) and using triangle inequality we get
			\begin{equation}
			d\left(x_{m\left(p\right)+k-1},x_{n\left(p\right)+k-1}\right)\leqslant d\left(x_{m\left(p\right)+k-1},x_{n\left(p\right)+k}\right)+d\left(x_{n\left(p\right)+k-1},x_{n\left(p\right)+k}\right)\label{11}
			\end{equation}
			\begin{center}
				$<\varepsilon+d\left(x_{n\left(p\right)+k-1},x_{n\left(p\right)+k}\right)$
			\end{center}
			and
			\begin{equation}\label{12}
			\varepsilon \leqslant d\left(x_{n\left(p\right)+k},x_{m\left(p\right)+k}\right)\\
			~\leqslant d\left(x_{n\left(p\right)+k},x_{n\left(p\right)+k-1}\right)+d\left(x_{n\left(p\right)+k-1},x_{m\left(p\right)+k-1}\right)+d\left(x_{m\left(p\right)+k-1},x_{m\left(p\right)+k}\right).
			\end{equation}

			If $p\to\infty$ in (\ref{11}) and using (\ref{12}) we get
			\begin{equation}\label{13}
			\lim_{p\to\infty}d\left(x_{m\left(p\right)+k-1},x_{n\left(p\right)+k-1}\right)=\varepsilon.
			\end{equation}

		\begin{flushleft}
			On the other hand, if
			$M_{k}\left((x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=1}^{\infty},(x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty}\right)$ \\$=\max\left\lbrace\sup\limits_{l\geqslant k-1} d\left(u_{l},v_{l}\right),\sup\limits_{l\geqslant k-1} d\left(u_{l},T\left((u_{i,\widehat{l}})_{i=1}^{\infty}\right)\right),\sup\limits_{l\geqslant k-1} d\left(v_{l},T\left((v_{i,\widehat{l}})_{i=1}^{\infty}\right)\right)\right\rbrace$
			where
			\begin{gather}
			u_{l}:=x_{n\left(p\right)+l-1},\: \forall\: 1\leqslant l\leqslant k
			\\
			u_{l}:=x_{n\left(p\right)+k-1},\: \forall\: l> k
			\\
			v_{l}:=x_{m\left(p\right)+l-1},\: \forall\: 1\leqslant l\leqslant k
			\\
			u_{l}:=x_{m\left(p\right)+k-1},\: \forall\: l> k
			\end{gather}
			$\implies M_{k}\left((x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=1}^{\infty},(x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty}\right)$

			$=\max\lbrace d\left(x_{n(p)+k-1},x_{m(p)+k-1}\right),  d\left(x_{n(p)+k-1},T\left((x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=1}^{\infty}\right)\right)$,\\
			$\hspace{6cm} d\left(x_{m\left(p\right)+k-1},T\left((x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty})\right)\right)\rbrace$\\
			$=\max\lbrace d\left(x_{n(p)+k-1},x_{m(p)+k-1}\right), d(x_{n(p)+k-1},x_{n(p)+k}), d(x_{m(p)+k-1},x_{m(p)+k})\rbrace $.
		\end{flushleft}
		Using (\ref{8}) and (\ref{13}) we get
		\begin{equation}
		\lim_{p\to\infty}M_{k}\left((x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=1}^{\infty},(x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty}\right)=\varepsilon.\label{18}
		\end{equation}
		By (\ref{hc}) and (\ref{18}) we get
		\begin{align}\label{e8}
&\varepsilon\leq d\left(x_{n\left(p\right)+k},x_{m\left(p\right)+k}\right)=d\left(T(x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=1}^{\infty},T(x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty})\right)\nonumber\\ &\leq\beta\left(M_{k}\left((x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=1}^{\infty},(x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty}\right)\right)\\
&\ \ \ \ \ M_{k}\left((x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=0}^{\infty},(x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty}\right)\nonumber\\
&<M_{k}\left((x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=1}^{\infty},(x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty}\right).\nonumber
			\end{align}

			If we suppose
			    $$\sup\limits_{l\geqslant k} d\left(u_{l},v_{l}\right)=d\left(x_{n(p)+k-1},x_{m(p)+k-1}\right)$$
				$$\sup\limits_{l\geqslant k} d\left(u_{l},T(u_{i,\widehat{l}})_{i=1}^{\infty}\right)=d\left(x_{n(p)+k-1},T\left((x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=1}^{\infty}\right)\right)$$
				$$\sup\limits_{l\geqslant k} d\left(v_{l},T(v_{i,\widehat{l}})_{i=1}^{\infty}\right)=d\left(x_{m(p)+k-1},T\left((x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty}\right)\right)$$
\begin{align*}
			&\implies M_{k}\left((x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=1}^{\infty},(x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty}\right)\\&=\max\lbrace d\left(x_{n(p)+k},x_{m(p)+k}\right),d\left(x_{n(p)+k-1}, T\left((x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=1}^{\infty}\right)\right),\\ &d\left(x_{m\left(p\right)+k-1},T\left((x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty})\right)\right)\rbrace
			\\&=\max\lbrace d\left(x_{n(p)+k}, x_{m(p)+k}\right), d(x_{n(p)+k-1}, x_{n(p)+k}), d(x_{m(p)+k-1}, x_{m(p)+k})\rbrace
	\end{align*}
		\begin{flushleft}
			Letting $p\to\infty$ and using (\ref{hc}), (\ref{e8}), (\ref{8}) and (\ref{18}) we get\\ $\lim\limits_{p\to\infty}\beta\left(M_{k}\left((x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=1}^{\infty},(x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty}\right)\right)$\\$\lim\limits_{p\to\infty}M_{k}\left((x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=1}^{\infty},(x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty}\right)\geqslant\varepsilon$.\\

			From (\ref{18})		$$\beta\left(M_{k}\left((x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=1}^{\infty},(x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty}\right)\right)\geqslant 1$$
			Since $\beta\in\mathbb{G}$.
			$\implies \lim\limits_{p\to\infty}\beta\left(M_{k}\left((x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=1}^{\infty},(x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty}\right)\right)= 1$
		$\implies\lim\limits_{p\to\infty}M_{k}\left((x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=1}^{\infty},(x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty}\right)=0$
		\end{flushleft}
		We get a contradiction to (\ref{18}).
		Hence $\{x_{n+k}\}_{n\in\mathbb{N}}$ is a Cauchy sequence.

		Since $\left(X,d\right)$ is complete there exists $u\in X$ such that
		\begin{equation}\label{e9}
		\lim\limits_{n\to\infty}x_{n+k}=u.
		\end{equation}

		Then we claim that $d\left(T\left((u)_{i=1}^{\infty}\right),u\right)=0$.

		\begin{flushleft}
			If we suppose contrary, then $d\left(u,T\left((u)_{i=1}^{\infty}\right)\right)>0$,
			\begin{align*}
			M\left((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty},\left((u)_{i=1}^{\infty}\right)\right)&=\max\left\lbrace d(x_{n+k-1},u),~d\left(x_{n+k-1},T(x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty}\right),d\left(u,T\left((u)_{i=1}^{\infty}\right)\right)\right\rbrace\\
		&=\max\left\lbrace d\left(x_{n+k-1},u\right),d\left(x_{n+k},x_{n+k-1}\right),d\left(u,T((u)_{i=1}^{\infty})\right)\right\rbrace.\\
			\end{align*}

		\end{flushleft}
		Letting $n\to\infty$ and using (\ref{8}), (\ref{e9}) we get	$$\lim\limits_{n\to\infty}M\left((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty},(u)_{i=1}^{\infty}\right)=d\left(u,T((u)_{i=1}^{\infty})\right)\neq 0.$$
		\begin{flushleft}
			Then we have
			\begin{align*}
d\left(u,T\left((u)_{i=1}^{\infty}\right)\right)&\leqslant d\left(x_{n+k},u\right)+d\left(x_{n+k},T((u)_{i=1}^{\infty})\right)\\ &=d\left(x_{n+k},u\right)+d\left(T\left((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty}\right),T((u)_{i=1}^{\infty})\right)
			\\ & \leqslant d\left(x_{n+k},u\right)+\beta\left(M\left(
			(x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty},(u)_{i=1}^{\infty}\right)\right)M\left(
			(x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty},(u)_{i=1}^{\infty}\right)
			\\
			& \leqslant d\left(x_{n+k},u\right)+\beta\left(d\left(u,T((u)_{i=1}^{\infty})\right)\right)d\left(u,T((u)_{i=1}^{\infty})\right).
			\end{align*}
			\end{flushleft}

		Letting $n\to\infty$ and using (\ref{e9}) we get $\lim_{n\to\infty}\beta\left(d\left(u,T((u)_{i=1}^{\infty})\right)\right)\geqslant 1$.

Since $\beta\in\mathbb{G}\implies\lim\limits_{n\to\infty}\beta\left(d\left(u,T((u)_{i=1}^{\infty})\right)\right)=1.$

Then we have $\lim\limits_{n\to\infty} d\left(u,T((u)_{i=1}^{\infty})\right)=0$, which implies $d\left(u,T((u)_{i=1}^{\infty})\right)=0$. So we get a contradiction.

Hence, $d\left(u,T((u)_{i=1}^{\infty})\right)=0\implies T((u)_{i=1}^{\infty}))=u$.
\end{proof}

\begin{remark} Theorem \ref{3.1} is a proper generalization to the Theorem \ref{2.6} since in case of the simplest operator on $\prod_{i=1}^{\infty}X\to X$ the contraction condition of Theorem \ref{2.6} is not applicable but on the other hand the $H_k$ contraction (see Definition \ref{d3.2} ) is easily applicable for the infinite case. Also if we restrict the operator to any finite $k$ dimension through an easy calculation it is obvious that it is an equivalent statement of Theorem \ref{2.6}.

\end{remark}

\begin{theorem}\label{3.2}.
Let $\left(X,d\right)$ be a complete metric space and $T:\prod_{l=1}^{\infty} X \rightarrow X$ be an extended Kannan-Geraghty $H_k$ contraction for some $k\in\mathbb{N}$. Then $\exists u\in X$ such that $T((u)_{i=1}^{\infty})=u$ and for any $x_1, \ldots, x_k \in X$, the infinite $k$-Picard sequence converges to $u$.\\
\end{theorem}
\begin{proof}
  Let $x_1, \ldots, x_k\in X$. For all $n\in\mathbb{N}$ we define the infinite $k$-Picard sequence as follows
		\begin{equation}\label{s}
		x_{n+k}:=T\left((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty}\right).
		\end{equation}

		We claim that $\lim_{n\to\infty}d(x_{n+k},x_{n+k+1})=0$. Then we have
		\begin{flushleft}			$d\left(x_{n+k+1},x_{n+k+2}\right)=d\left(T\left((x_{n+i,\widehat{n+k}})_{i=1}^{\infty}\right),T\left((x_{n+i+1,\widehat{n+1+k}})_{i=1}^{\infty}\right)\right)$
			\\		$\leqslant\frac{\beta\left(d(x_{n+k},x_{n+k+1})\right)}{2}(d(T\left((x_{n+i,\widehat{n+k}})_{i=1}^{\infty}\right),x_{n+k})+d(T\left((x_{n+i+1,\widehat{n+k+1}})_{i=1}^{\infty}\right),x_{n+k+1}))$
			\\
			$<~\frac{1}{2}(d(x_{n+k},x_{n+k+1})+d(x_{n+k+1},x_{n+k+2}))\implies d(x_{n+k+1},x_{n+k+2})<d(x_{n+k},x_{n+k+1})$.
		\end{flushleft}

Then our sequence  is a monotone decreasing sequence which is bounded below. Then there exists $r\geq~0$ such that
\begin{equation}\label{1}
  \lim\limits_{n\to~0}d(x_{n+k},x_{n+k+1})=r.
\end{equation}

We claim that $r=0$.

If we consider the contrary and suppose $r>0$ we have
\begin{flushleft}
  $\lim\limits_{n\to\infty}d(x_{n+k+1},x_{n+k+2})\leqslant\frac{\lim\limits_{n\to\infty}\beta\left(d(x_{n+k},x_{n+k+1})\right)}{2}(\lim\limits_{n\to\infty}d(x_{n+k},x_{n+k+1})+\lim\limits_{n\to\infty}d(x_{n+k+1},x_{n+k+2}))$

  $\implies\frac{2\lim\limits_{n\to\infty}d(x_{n+k+1},x_{n+k+2})}{\lim\limits_{n\to\infty}d(x_{n+k},x_{n+k+1})+\lim\limits_{n\to\infty}d(x_{n+k+1},x_{n+k+2})}\leqslant~\lim\limits_{n\to\infty}\beta\left(d(x_{n+k},x_{n+k+1})\right)$.

  \par From (\ref{1}) we get

  $\frac{2r}{2r}\leqslant~\beta\left(d(x_{n+k},x_{n+k+1})\right)\implies\lim\limits_{n\to\infty}\beta\left(d(x_{n+k},x_{n+k+1})\right)\geqslant~1$.
\end{flushleft}

Since $\beta\in\mathbb{G}$ then $\lim\limits_{n\to\infty}\beta\left(d(x_{n+k},x_{n+k+1})\right)\leqslant~1$.

Using the well known "Sandwich Theorem" we obtain
\begin{equation}
  \lim\limits_{n\to\infty}\beta\left(d(x_{n+k},x_{n+k+1})\right)=1\implies\lim\limits_{n\to\infty}d(x_{n+k},x_{n+k+1})=0.
\end{equation}

\begin{flushleft}

 Further, we have

   $d(x_{n+k},x_{m+k})=d(T((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty}),T((x_{m+i-1,\widehat{m+k-1}})_{i=1}^{\infty}))$
  \\
  $\leqslant~\frac{\beta\left(d(x_{n+k-1},x_{m+k-1})\right)}{2}(d(T((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty}),x_{n+k-1})+d(T((x_{m+i-1,\widehat{m+k-1}})_{i=1}^{\infty}),x_{m+k-1}))$
  \\
  $<\frac{1}{2}(d(x_{n+k-1},x_{n+k})+d(x_{m+k-1},x_{m+k}))$.

  \par For sufficiently large enough $n,m\in\mathbb{N}$
  \[d(x_{n+k},x_{m+k})<\varepsilon\]
  for a fixed $\varepsilon>0$.

  Then the sequence $\{x_{n+k}\}_{n\in\mathbb{N}}$ is a Cauchy sequence.

  Since $(X,d)$ is a complete metric space, there exists a $u\in~X$ such that,
  \begin{equation}\label{2}
   \lim\limits_{n\to\infty}x_{n+k}=u.
  \end{equation}

  \par We claim that $T((u)_{i=1}^{\infty})=u$.

\par If we suppose the contrary we have $d(u,T((u)_{i=1}^{\infty}))>0$.

  Then, by (\ref{1}) and (\ref{2}), for an arbitrary $\varepsilon>0$ and a sufficiently large $n$ we get
  \begin{align*}
d\left(u,T\left((u)_{i=1}^{\infty}\right)\right)&\leqslant d\left(x_{n+k},u\right)+d(x_{n+k},T((u)_{i=1}^{\infty}))\\ &=d\left(x_{n+k},u\right)+d\left(T\left((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty}\right),T((u)_{i=1}^{\infty})\right)
			\\ & \leqslant d\left(x_{n+k},u\right)+\frac{\beta\left(d(u,x_{n+k-1})\right)}{2}(d(T((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty}),x_{n+k-1})+d(T((u)_{i=1}^{\infty}),u))
			\\
			& \leqslant \frac{1}{2}~d\left(u,T((u)_{i=1}^{\infty})\right)+\frac{\varepsilon}{2}+\frac{\varepsilon}{2}
 \leqslant \frac{1}{2}~d(u,T((u)_{i=1}^{\infty})+\varepsilon
\\
			\hspace{-25pt} &  \implies\frac{1}{2}~d\left(u,T((u)_{i=1}^{\infty})\right)\leqslant\varepsilon. ~\text{ This is a contradiction.}
  \end{align*}

  Hence, $d(T((u)_{i=1}^{\infty}),u)=0\implies~T((u)_{i=1}^{\infty})=u.$

  Then the conclusion follows.
\end{flushleft}
\end{proof}

Next, we will provide a new result for multivalued proper extension of Theorem \ref{2.5} which is also a generalization of Kannan (Theorem \ref{2.3}) as a result of Theorem \ref{3.2}.

\begin{corollary}\label{3.3}
Let $(X,d)$ be a complete metric space and $T$ be a Kannan-Geraghty $H_k$ contraction. Then T has a fixed point and every $k$-Picard sequence for $T$ converges to $u$.
\end{corollary}

\begin{proof}
Let us choose $x_0, \ldots, x_{k}\in~X$. We define a $k$-Picard sequence by
\begin{equation}
  x_{n+k}:=T(x_n,\ldots,x_{n+k-1}).
\end{equation}

If we follow same steps as in the proof of Theorem \ref{3.2} we get the required fixed point.
\end{proof}

\begin{remark}\label{3.4} For $k=1$ we get Theorem \ref{2.5} in \cite{Kg} which proves that Corollary \ref{3.3} is a proper generalization of Theorem \ref{2.5}.
\end{remark}
\begin{theorem}\label{3.5}
Let $\left(X,d\right)$ be a complete metric space and $T:\prod_{l=1}^{\infty} X \rightarrow X$ be an extended Fisher-Geraghty $H_k$ contraction for some $ k\in\mathbb{N}$. Then there exists $u\in X$ such that $T((u)_{i=1}^{\infty})=u$ and for any $x_1, x_2, \ldots x_k$ the infinite $k$-Picard sequence converges to $u$.
\end{theorem}

\begin{proof}
  Let $x_1, \ldots, x_k\in X$.  For all $n\in\mathbb{N} $ we define infinite $k$- Picard sequence by
		\begin{equation}\label{s}
		x_{n+k}:=T\left((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty}\right).
		\end{equation}

		We claim $\lim\limits_{n\to\infty}d(x_{n+k},x_{n+k+1})=0$.

Then we have
		\begin{flushleft}
			$d\left(x_{n+k+1},x_{n+k+2}\right)=d\left(T\left((x_{n+i,\widehat{n+k}})_{i=1}^{\infty}\right),T\left((x_{n+i+1,\widehat{n+1+k}})_{i=1}^{\infty}\right)\right)$
			\\
			$\leqslant\frac{\beta\left(d(x_{n+k},x_{n+k+1})\right)}{2}(d(T\left((x_{n+i,\widehat{n+k}})_{i=1}^{\infty}\right),x_{n+k+1})+d(T\left((x_{n+i,\widehat{n+k+1}})_{i=1}^{\infty}\right),x_{n+k}))$
			\\
			$<~\frac{1}{2}(d(x_{n+k+1},x_{n+k+1})+d(x_{n+k},x_{n+k+2}))$
			\\
			$<~\frac{1}{2}(d(x_{n+k},x_{n+k+1})+d(x_{n+k+1},x_{n+k+2}))\implies d(x_{n+k+1},x_{n+k+2})<d(x_{n+k},x_{n+k+1})$
		\end{flushleft}

Hence our sequence is a monotone decreasing sequence bounded below. This implies thrat there exsists $r\geq~0$ such that
\begin{equation}\label{3}
  \lim\limits_{n\to~0}d(x_{n+k},x_{n+k+1})=r.
\end{equation}

We claim that $r=0$.

If we suppose the contrary and $r>0$ we get
\begin{flushleft}
  $\lim\limits_{n\to\infty}d(x_{n+k+1},x_{n+k+2})$

  $\leqslant\frac{\lim\limits_{n\to\infty}\beta\left(d(x_{n+k},x_{n+k+1})\right)}{2}(\lim\limits_{n\to\infty}d(x_{n+k},x_{n+k+1})+\lim\limits_{n\to\infty}d(x_{n+k+1},x_{n+k+2}))$

  $\implies\frac{2\lim\limits_{n\to\infty}d(x_{n+k+1},x_{n+k+2})}{\lim\limits_{n\to\infty}d(x_{n+k},x_{n+k+1})+\lim\limits_{n\to\infty}d(x_{n+k+1},x_{n+k+2})}\leqslant~\lim\limits_{n\to\infty}\beta\left(d(x_{n+k},x_{n+k+1})\right)$

  Using (\ref{3}) we obtain

  $\frac{2r}{2r}\leqslant~\lim\limits_{n\to\infty}\beta\left(d(x_{n+k},x_{n+k+1})\right)\implies\lim\limits_{n\to\infty}\beta\left(d(x_{n+k},x_{n+k+1})\right)\geqslant~1 $
\end{flushleft}

\par Since $\beta\in\mathbb{G}$ we have	$\lim\limits_{n\to\infty}\beta\left(d(x_{n+k},x_{n+k+1})\right)\leqslant 1$.
\par From the well know in the related literature, Sandwich Theorem we get
\begin{equation}\label{l}
 \lim\limits_{n\to\infty}\beta\left(d(x_{n+k},x_{n+k+1})\right)=1 \implies\lim\limits_{n\to\infty}d(x_{n+k},x_{n+k+1})=0.
\end{equation}

\begin{flushleft}

 We claim that
			$\left\lbrace x_{n+k}\right\rbrace_{n\in\mathbb{N}}$ is a Cauchy and we want to prove this by contradiction. Then, using the contrary there exists $\varepsilon>0$ such that we can find subsequences $\left\lbrace x_{m\left(p\right)+k}\right\rbrace_{p\in\mathbb{N}}$,$\left\lbrace x_{n\left(p\right)+k}\right\rbrace_{p\in\mathbb{N}}$ with $m\left(p\right)>n\left(p\right)>p$ such that for every $p$ we have
			\begin{equation}\label{e2}
			d\left(x_{m\left(p\right)+k},x_{n\left(p\right)+k}\right)\geqslant\varepsilon.
			\end{equation}

\par Moreover, correseponding to each $n(p)$ we can choose least of such $m\left(p\right)$ satisfying (\ref{e2}) so that,
			\begin{equation}\label{e3}
			d\left(x_{m\left(p\right)+k-1},x_{n\left(p\right)+k}\right)<\varepsilon.
			\end{equation}

			Using (\ref{e2}), (\ref{e3}) and the triangle inequality we get
			\begin{equation}\label{e4}
			d\left(x_{m\left(p\right)+k-1},x_{n\left(p\right)+k-1}\right)\leqslant d\left(x_{m\left(p\right)+k-1},x_{n\left(p\right)+k}\right)+d\left(x_{n\left(p\right)+k-1},x_{n\left(p\right)+k}\right)
			\end{equation}
			\begin{center}
				$<\varepsilon+d\left(x_{n\left(p\right)+k-1},x_{n\left(p\right)+k}\right)$
			\end{center}
			and
			\begin{align*}\label{e5}
			\varepsilon &\leqslant d\left(x_{n\left(p\right)+k},x_{m\left(p\right)+k}\right)\\
			&\leqslant d\left(x_{n\left(p\right)+k},x_{n\left(p\right)+k-1}\right)+d\left(x_{n\left(p\right)+k-1},x_{m\left(p\right)+k-1}\right)+d\left(x_{m\left(p\right)+k-1},x_{m\left(p\right)+k}\right).
			\end{align*}

			Letting $p\to\infty$ in (\ref{e4}) and (\ref{l}) we get
			\begin{equation}\label{e6}
			\lim\limits_{p\to\infty}d\left(x_{m\left(p\right)+k-1},x_{n\left(p\right)+k-1}\right)=\varepsilon.
			\end{equation}

\begin{flushleft}
    $d(x_{n(p)+k},x_{m(p)+k})=d(T((x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=1}^{\infty}),T((x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty}))$
    \\
    $\leqslant\frac{\beta(d(x_{n(p)+k-1},x_{m(p)+k-1}))}{2}(d(T((x_{n(p)+i-1,\widehat{n(p)+k-1}})_{i=1}^{\infty}),x_{m(p)+k-1})$
    \\
    $+d(T((x_{m(p)+i-1,\widehat{m(p)+k-1}})_{i=1}^{\infty}),x_{n(p)+k-1}))$\
    \\$<\frac{1}{2}(d(x_{n(p)+k},x_{m(p)+k-1})+d(x_{m(p)+k},x_{n(p)+k-1}))$
    \\$\leqslant\frac{1}{2}(\varepsilon+d(x_{m(p)+k},x_{m(p)+k-1})+d(x_{m(p)+k-1},x_{n(p)+k})+d(x_{n(p)+k},x_{n(p)+k-1}))$
    \\$\leqslant\frac{1}{2}(2\varepsilon+d(x_{m(p)+k},x_{m(p)+k-1})+d(x_{n(p)+k},x_{n(p)+k-1}))$
\end{flushleft}

    Then from (\ref{e6}) and (\ref{l}) we get $\varepsilon<\varepsilon$. So this is a contradiction.

    Then
    \begin{equation}
     \{x_{n+k}\}_{n\in\mathbb{N}}~\text{ is a Cauchy sequence.}
    \end{equation}

  Since $(X,d)$ is complete there exists a $u\in~X$ such that
  \begin{equation}\label{4}
   \lim\limits_{n\to\infty}x_{n+k}=u.
  \end{equation}
 \par  We claim that $T((u)_{i=1}^{\infty})=u$.

 \par If we suppose the contrary we have $d(u,T((u)_{i=1}^{\infty}))>0$. Then, by (\ref{3}) and (\ref{4}) for arbitrary an $\varepsilon>0$ and a sufficiently large $n$ we get
  \begin{align*}
&d\left(u,T\left((u)_{i=1}^{\infty}\right)\right)\leqslant d\left(x_{n+k},u\right)+d(x_{n+k},T((u)_{i=1}^{\infty}))\\ &=d\left(x_{n+k},u\right)+d\left(T\left((x_{n+i-1,\widehat{n+k-1}})_{i=1}^{\infty}\right),T((u)_{i=1}^{\infty})\right)
			\\ & \leqslant d\left(x_{n+k},u\right)+\frac{\beta\left(d(u,x_{n+k-1})\right)}{2}(d(T((x_{n+i-1,\widehat{n+k-1}})_{i=1})^{\infty}),u)+d(T((u)_{i=1}^{\infty}),x_{n+k-1}))
			\\
			& \leqslant d\left(x_{n+k},u\right)+\frac{\beta\left(d(u,x_{n+k-1})\right)}{2}(d(T((x_{n+i-1,\widehat{n+k-1}})_{i=1})^{\infty}),u)+d(T((u)_{i=1}^{\infty}),u)+d(u,x_{n+k-1}))
			\\
			& \leqslant \frac{1}{2}~d\left(u,T((u)_{i=1}^{\infty})\right)+\frac{\varepsilon}{3}+\frac{\varepsilon}{6}
			\\
			& \leqslant \frac{1}{2}~d(u,T((u)_{i=1}^{\infty})+\frac{1}{2}\varepsilon\\
			\hspace{-25pt} &  \implies d\left(u,T((u)_{i=1}^{\infty})\right)\leqslant\varepsilon.
  \end{align*}

 \par Hence $d(T((u)_{i=1}^{\infty}),u)=0$ which implies $T((u)_{i=1}^{\infty})=u.$ Then the conclusion follows.

\end{flushleft}
\end{proof}

\begin{corollary}\label{3.6}
Let $(X,d)$ be a complete metric space and $T$ be a Fisher-Geraghty $H_k$ contraction. Then $T$  has a fixed point.
\end{corollary}
\begin{proof}

Choosen any $x_0, \ldots, x_{k}\in~X$ we define
\begin{equation}
  x_{n+k}:=T(x_n,\ldots,x_{n+k-1}).
\end{equation}
Using same steps as in the proof of Theorem \ref{3.5} we get the conclusion.
\end{proof}

\begin{remark}\label{3.7} For $k=1$ we get a new type of extension of Theorem \ref{2.3} proved in \cite{F} and for any $k$, Corollary \ref{3.6} also gives the multidimensional extension of the same theorem stated in \cite{F}.
\end{remark}
\begin{remark} The Banach Fixed Point Theorem(\cite{Ban}), Theorem \ref{2.1} , Theorem \ref{2.2} , Theorem \ref{2.3}, Theorem \ref{2.5} and Theorem \ref{2.6}  are all applicable only in complete metric space as well as the theorems described in \cite{Kho} but  Theorem \ref{3.1}, Theorem \ref{3.2}, Theorem \ref{3.5} which we have proved are talking about the space $\prod_{i=1}^{\infty}X$( where $X$ is a complete metric space) which is indeed metrizable but might not be complete.
\end{remark}

\section{Applications to integral equations}

\quad In \cite{n-dim} H Singh et al. introduced the notion of multi-dimensional Fredholm integral equation taking into account $n\in\mathbb{N}$ as dimension of the equation.
$$u(x_1,...,x_n)=f(x_1,...,x_n)+\underbrace{\int\limits_{0}^{1}\int\limits_{0}^{1}...\int\limits_{0}^{1}}_{n-times}K(x_1,...,x_n,y_1,...,y_n)u(y_1,...,y_n)dy_1...dy_n,$$
where $(x_1,...,x_n)\in D=\underbrace{([0,1]\times[0,1]\times...\times [0,1])}_{n-times}$ and $f(x_1,...,x_n), K(x_1,...,x_n,y_1,...,y_n)$ are known continuous functions defined on $D$ and $D^2$ respectively and $u(x_1,...,x_n)$ is a unknown function.

The Fredholm integral equations play an important role in modelling of physics phenomena described by two or three dimensions. In the same way, they have applications in astrophysics models thinking of the four dimensions of a neutron star or a black hole.

Thinking of this aspects, if we extend to infinity the dimension "n" of the previous multi-dimensional integral equation we introduce a new notion, {\it the infinite dimensional Fredholm integral equation}, as follows.

\begin{equation}\label{w1}
u\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)=f\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)+\int\limits_{0}^{1}...\int\limits_{0}^{1}...K\left( (t_{i,\widehat{k}})_{i=1}^{\infty},(s_{i,\widehat{k}})_{i=1}^{\infty}\right)u\left((s_{i,\widehat{k}})_{i=1}^{\infty}\right)ds_1...ds_n... .
\end{equation}

where $f\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right):[0,1]\to\mathbb{R}$ and $K:[0,1]\times\mathbb{R}\to\mathbb{R}$ are two known continuous functions and $u\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)$ is an unknown function.

Further, let us give our first application of the main results of this paper proving the existence of a solution of infinite dimensional Fredholm integral equation (\ref{w1}).

\begin{theorem}

Let $X=C([0,1],\mathbb{R})$  the set of real continuous functions on $[0,1]$ and let $d:X\times X\to\mathbb{R}_{+}$ given by
\begin{equation}\label{w2}
d\left((u_{i,\widehat{k}})_{i=1}^{\infty},T(u_{i,\widehat{k}})_{i=1}^{\infty}\right)=\sup\limits_{t\in[0,1]}\mid (u_{i,\widehat{k}})_{i=1}^{\infty}-T(u_{i,\widehat{k}})_{i=1}^{\infty}\mid.\end{equation}
Define $T:\prod\limits_{i=1}^{\infty}X\to X$ by
\begin{equation}\label{w3}
Tu\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)=f\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)+\int\limits_{0}^{1}...\int\limits_{0}^{1}...K\left( (t_{i,\widehat{k}})_{i=1}^{\infty},(s_{i,\widehat{k}})_{i=1}^{\infty}\right)u\left((s_{i,\widehat{k}})_{i=1}^{\infty}\right)ds_1...ds_n... .
\end{equation}

Assume the following holds:

\begin{enumerate}
  \item[$(i)$]$\left| u\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)-Tu\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)\right|\leq\frac{1}{2}\left| d\left(T\left((u_{i,\widehat{k}})_{i=1}^{\infty}\right), u_k\right)+d\left(T\left((u_{i+1,\widehat{k+1}})_{i=1}^{\infty}\right),u_{k+1}\right)\right|$;
  \item[$(ii)$] there exists a constant $\delta\in(0,1)$ such that $K\left((t_{i,\widehat{k}})_{i=1}^{\infty},(s_{i,\widehat{k}})_{i=1}^{\infty}\right)<\delta$;
  \item[$(iii)$]let $\beta:[0,\infty)\to[0,1)$ be defined as $\beta\left(z\right)=\frac{1}{\gamma}<1$ for every $\gamma>0$ and $z\in \mathbb{R}_{+}$.
\end{enumerate}

Then the infinite dimensional Fredholm integral equation (\ref{w1}) has a solution.

\end{theorem}

\begin{proof}

We can prove the existence of a solution of infinite dimensional Fredholm integral equation if we show that the operator $T$ defined by (\ref{w3}) has a fixed point.

We can easy remark that the space $X=(C[0,1],\mathbb{R})$ endowed with the metric $d$ defined by relation (\ref{w2}) form a complete metric space.
Then we shall show that all the hypothesis of Theorem \ref{3.2} are verified.

We have the following estimation
\begin{align*}
  &\left| Tu\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)-T^2u\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)\right| =|\int\limits_{0}^{1}...\int\limits_{0}^{1}...K\left( (t_{i,\widehat{k}})_{i=1}^{\infty},(s_{i,\widehat{k}})_{i=1}^{\infty}\right)u\left((s_{i,\widehat{k}})_{i=1}^{\infty}\right)ds_1...ds_n...\\
  & - \int\limits_{0}^{1}...\int\limits_{0}^{1}...K\left( (t_{i,\widehat{k}})_{i=1}^{\infty},(s_{i,\widehat{k}})_{i=1}^{\infty}\right)Tu\left((s_{i,\widehat{k}})_{i=1}^{\infty}\right)ds_1...ds_n...| \\
  & \leq \int\limits_{0}^{1}...\int\limits_{0}^{1}...K\left( (t_{i,\widehat{k}})_{i=1}^{\infty},(s_{i,\widehat{k}})_{i=1}^{\infty}\right)\left|u\left((s_{i,\widehat{k}})_{i=1}^{\infty}\right)-Tu\left((s_{i,\widehat{k}})_{i=1}^{\infty}\right)\right|ds_1...ds_n...\\
  & \leq \frac{\delta}{2\gamma}\left| d\left(T\left((u_{i,\widehat{k}})_{i=1}^{\infty}\right), u_k\right)+d\left(T\left((u_{i+1,\widehat{k+1}})_{i=1}^{\infty}\right),u_{k+1}\right)\right|\int\limits_{0}^{1}...\int\limits_{0}^{1}...ds_1...ds_n...\\
  &=\frac{\delta}{2\gamma}\left| d\left(T\left((u_{i,\widehat{k}})_{i=1}^{\infty}\right), u_k\right)+d\left(T\left((u_{i+1,\widehat{k+1}})_{i=1}^{\infty}\right),u_{k+1}\right)\right|.
\end{align*}

Taking supremum on both sides we get

$$\sup\limits_{t\in[0,1]}\left| Tu\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)-T^2u\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)\right|$$
$$\leq \sup\limits_{t\in[0,1]}\frac{\delta }{2\gamma}\left| d\left(T\left((u_{i,\widehat{k}})_{i=1}^{\infty}\right), u_k\right)+d\left(T\left((u_{i+1,\widehat{k+1}})_{i=1}^{\infty}\right),u_{k+1}\right)\right|.$$

Then, for $z=d(u_k,u_{k+1})$ and $\delta\in(0,1)$ we obtain

\begin{align*}
  d\left(Tu\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right),T^2u\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)\right)\\&\hspace{-70pt}\leq \frac{\delta\beta(d(u_k,u_{k+1}))}{2}\left(d\left(T\left((u_{i,\widehat{k}})_{i=1}^{\infty}\right), u_k\right)+d\left(T\left((u_{i+1,\widehat{k+1}})_{i=1}^{\infty}\right),u_{k+1}\right) \right)\\
  &\hspace{-70pt}\leq \frac{\beta(d(u_k,u_{k+1}))}{2}\left(d\left(T\left((u_{i,\widehat{k}})_{i=1}^{\infty}\right), u_k\right)+d\left(T\left((u_{i+1,\widehat{k+1}})_{i=1}^{\infty}\right),u_{k+1}\right) \right)
\end{align*}

In conclusion, all the hypothesis of  Theorem \ref{3.2} are accomplished. Then the operator $T$ has a fixed point, which means the infinite dimensional Fredholm integral equation (\ref{w1}) has a solution.
\end{proof}

The following application involve another type of integral equations, Urysohn type integral equations. We extend the known cases of this type of integral equations to {\it infinite dimensional Urysohn integral equation}.
\begin{equation}\label{urh}
  u\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)=f\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)+\int\limits_{a}^{b}...\int\limits_{a}^{b}... P\left( (t_{i,\widehat{k}})_{i=1}^{\infty},(s_{i,\widehat{k}})_{i=1}^{\infty},u\left((s_{i,\widehat{k}})_{i=1}^{\infty}\right)\right)ds_1...ds_n... .
\end{equation}
where $f\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right):[a,b]\to\mathbb{R}$ and $K:[a,b]\times[a,b]\times\mathbb{R}\to\mathbb{R}$ are two known continuous functions and $u\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)$ is an unknown function.

For this new type of Urysohn integral equation let us give the following result.

\begin{theorem}
Let $X=C([a,b],\mathbb{R})$  the set of real continuous functions on $[a,b]$ and let $d:X\times X\to\mathbb{R}_{+}$ given by
\begin{equation}\label{urh1}
d\left((u_{i,\widehat{k}})_{i=1}^{\infty},T(u_{i,\widehat{k}})_{i=1}^{\infty}\right)=\max\limits_{t\in[a,b]}\mid (u_{i,\widehat{k}})_{i=1}^{\infty}-T(u_{i,\widehat{k}})_{i=1}^{\infty}\mid\cdot e^{-sin^{-1}\alpha},\end{equation}
with $\alpha>0$.

Define $T:\prod\limits_{i=1}^{\infty}X\to X$ by
\begin{equation}\label{urh2}
Tu\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)=f\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)+\int\limits_{a}^{b}...\int\limits_{a}^{b}...P\left( (t_{i,\widehat{k}})_{i=1}^{\infty},(s_{i,\widehat{k}})_{i=1}^{\infty},u\left((s_{i,\widehat{k}})_{i=1}^{\infty}\right)\right)ds_1...ds_n... .
\end{equation}

Assume the following holds:

\begin{enumerate}
  \item[$(i)$]$\left| P\left((t_{i,\widehat{k}})_{i=1}^{\infty},(s_{i,\widehat{k}})_{i=1}^{\infty},u\left((s_{i,\widehat{k}})_{i=1}^{\infty}\right)\right)-P\left((t_{i,\widehat{k}})_{i=1}^{\infty},(s_{i,\widehat{k}})_{i=1}^{\infty},Tu\left((s_{i,\widehat{k}})_{i=1}^{\infty}\right)\right)\right|\\ \leq\frac{1}{\tau}\left| (u_{i,\widehat{k}})_{i=1}^{\infty}-T(u_{i,\widehat{k}})_{i=1}^{\infty}\right|$;
   \item[$(ii)$]let $\beta:[0,\infty)\to[0,1)$ be defined as $\beta\left(z\right)=\frac{1}{\tau}<1$ for every $\tau>0$ and $z\in X$.
\end{enumerate}

Then the infinite dimensional Urysohn integral equation (\ref{urh}) has a solution.

\end{theorem}

\begin{proof}

It is easy to check the space $X=(C[0,1],\mathbb{R})$ endowed with the metric $d$ defined by relation (\ref{urh1}) is a complete metric space.

To prove the existence of a solution of infinite dimensional Urysohn integral equation we shall show that the operator $T$ defined by (\ref{urh2}) has a fixed point.

We have the following estimation
\begin{align*}
  &\left| Tu\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)-T^2u\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)\right| =|\int\limits_{a}^{b}...\int\limits_{a}^{b}...P\left( (t_{i,\widehat{k}})_{i=1}^{\infty},(s_{i,\widehat{k}})_{i=1}^{\infty},u\left((s_{i,\widehat{k}})_{i=1}^{\infty}\right)\right)ds_1...ds_n...\\
  & - \int\limits_{a}^{b}...\int\limits_{a}^{b}...P\left( (t_{i,\widehat{k}})_{i=1}^{\infty},(s_{i,\widehat{k}})_{i=1}^{\infty},Tu\left((s_{i,\widehat{k}})_{i=1}^{\infty}\right)\right)ds_1...ds_n...| \\
  &\leq \int\limits_{a}^{b}...\int\limits_{a}^{b}...\left| P\left((t_{i,\widehat{k}})_{i=1}^{\infty},(s_{i,\widehat{k}})_{i=1}^{\infty},u\left((s_{i,\widehat{k}})_{i=1}^{\infty}\right)\right)-P\left((t_{i,\widehat{k}})_{i=1}^{\infty},(s_{i,\widehat{k}})_{i=1}^{\infty},Tu\left((s_{i,\widehat{k}})_{i=1}^{\infty}\right)\right)\right|ds_1...ds_n...\\
  & \leq \frac{1}{\tau}\int\limits_{a}^{b}...\int\limits_{a}^{b}... \left| (u_{i,\widehat{k}})_{i=1}^{\infty}-T(u_{i,\widehat{k}})_{i=1}^{\infty}\right|ds_1...ds_n...\\
  & \leq \frac{e^{sin^{-1}\alpha}}{\tau}\left|(u_{i,\widehat{k}})_{i=1}^{\infty}-T(u_{i,\widehat{k}})_{i=1}^{\infty}\right|e^{-sin^{-1}\alpha}\int\limits_{a}^{b}...\int\limits_{a}^{b}...ds_1...ds_n...\\
  &\leq \frac{e^{sin^{-1}\alpha}}{\tau}\left|(u_{i,\widehat{k}})_{i=1}^{\infty}-T(u_{i,\widehat{k}})_{i=1}^{\infty}\right|e^{-sin^{-1}\alpha}.
   \end{align*}

Applying maximum on both sides we get
$$\max\limits_{t\in[a,b]}\left| Tu\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)-T^2u\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)\right|e^{-sin^{-1}\alpha}\leq \frac{1}{\tau}\max\limits_{t\in[a,b]} \left|(u_{i,\widehat{k}})_{i=1}^{\infty}-T(u_{i,\widehat{k}})_{i=1}^{\infty}\right|e^{-sin^{-1}\alpha}.$$

For $z=M_{k}\left((u_{i,\widehat{k}})_{i=1}^{\infty},(u_{i+1,\widehat{k+1}})_{i=1}^{\infty}\right)$ we get
\begin{align*}
d\left(Tu\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right),T^2u\left((t_{i,\widehat{k}})_{i=1}^{\infty}\right)\right)&\leq \beta\left(M_{k}\left((u_{i,\widehat{k}})_{i=1}^{\infty},(u_{i+1,\widehat{k+1}})_{i=1}^{\infty}\right)\right)d\left((u_{i,\widehat{k}})_{i=1}^{\infty},T(u_{i,\widehat{k}})_{i=1}^{\infty}\right)\\
&\leq \beta\left(M_{k}\left((u_{i,\widehat{k}})_{i=1}^{\infty},(u_{i+1,\widehat{k+1}})_{i=1}^{\infty}\right)\right)M_{k}\left((u_{i,\widehat{k}})_{i=1}^{\infty},(u_{i+1,\widehat{k+1}})_{i=1}^{\infty}\right)
\end{align*}

Then all the conditions of Theorem \ref{3.1} are satisfied; result that the operator $T$ has a fixed point.

Then the infinite dimensional Urysohn type integral equation (\ref{urh}) has a solution.
\end{proof}

\section{Conclusion}

\quad Here in this article we have shown some metric combinatorial arguments that can be profitably applied to extend Geraghty's theorem, Kannan-Geraghty theorem and, Fisher's theorem as well as a particular extension for the infinite cases. This idea could also motivate to find a  similar infinite extension of the other contractions and its related operators to find fixed points or coincidence points. We also defined two new types of infinite integral equations which would encourage the study of the infinite integral equations and their solutions. From the discussion we have some interesting problems as follows.

\begin{enumerate}
 \item Can a similar concept be extended to the notion of coincidence points?

 \item We have discussed the fixed point theorem for the contraction from countable products of the complete metric space $X$ (with respect to uniform metric topology) to the space $X$. Can it further be extended to  an uncountable product of complete metric space which is not necessarily a metric space?
\end{enumerate}

\noindent \textbf{Acknowledgements} The  authors would like to thank Suprokash Hazra who has helped by his valuable comments in developing the key ideas of this paper.

redpajama_set_name: RedPajamaArXiv


text:
Clarkesworld Magazine
Storytelling Science Fiction Fantasy Horror Short Stories Speculative Fiction Audio Drama Sci-Fi / Fantasy Stories Clarkesworld Magazine
Creato da Clarkesworld Magazine, autore scoperto da Player FM e dalla nostra community - Il copyright è detenuto dall'editore, non da Player FM, e l'audio viene riprodotto direttamente dal suo server. Clicca sul pulsante Iscriviti per rimanere aggiornato su Player FM, o incolla l'URL del feed in un altra app per i podcast.
Clarkesworld Magazine « »
Catching the K Beast by Chen Qian (audio)
This episode features "Catching the K Beast" written by Chen Qian. Published in the April 2021 issue of Clarkesworld Magazine and read by Kate Baker. The text version of this story can be found at: https://clarkesworldmagazine.com/chen_04_21 Support us on Patreon at https://www.patreon.com/join/clarkesworld?
1273 episodi
Simile a Clarkesworld Magazine
LIGHTSPEED MAGAZINE - Science Fiction and Fantasy Story Podcast (Sci-Fi | Audiobook | Short Stories)
Adamant Press
Edited by bestselling anthologist John Joseph Adams, LIGHTSPEED is a Hugo Award-winning, critically-acclaimed digital magazine. In its pages, you'll find science fiction from near-future stories and sociological SF to far-future, star-spanning SF. Plus there's fantasy from epic sword-and-sorcery and contemporary urban tales to magical realism, science-fantasy, and folk tales. Each month, LIGHTSPEED brings you a mix of originals and reprints featuring a variety of authors, from the bestseller ...
Nightmare Magazine - Horror and Dark Fantasy Story Podcast (Audiobook | Short Stories)
Edited by bestselling, award-winning anthologist John Joseph Adams, NIGHTMARE is a digital magazine of horror and dark fantasy. In its pages, you will find all kinds of horror and dark fantasy, from zombie stories and haunted house tales, to visceral psychological horror. Every month NIGHTMARE will bring you a mix of original fiction and reprints, and featuring a variety of authors: from the bestsellers and award-winners you already know to the best new voices you haven't heard of yet. When ...
The Other Stories | Sci-Fi, Horror, Thriller, WTF Stories
Hawk & Cleaver | A Digital Story Studio bringing you the best new stories to watch, read, sniff, and absorb.
These aren't the stories your mother used to tell you ... no, these are The Other Stories. The Other Stories is a weekly short story podcast. A modern take on The Twilight Zone, Tales From The Crypt, or The Outer Limits. Sci-Fi, Horror, Thriller, WTF stories delivered right to your podcast feed every Monday morning. See acast.com/privacy for privacy and opt-out information.
The Orphans
The Light and Tragic Company
The Orphans is a cinematic sci-fi audio drama about survival in a harsh universe: castaways on a hostile world, A.I.s with unprecedented emotions, strangers who share faces, love and loss in a far-flung future. Each season explores a new vantage point in an ever-expanding and inter-connected galaxy!
Escape Artists, Inc
The Sound of Horror
Mission To Zyxx
An improvised science fiction sitcom following a team of ambassadors as they attempt to establish diplomatic relations with planets in the remote and chaotic Zyxx Quadrant… better known as the "ass end of space."
Night Vale Presents
Twice-monthly community updates for the small desert town of Night Vale, where every conspiracy theory is true. Turn on your radio and hide. Never listened before? It's an ongoing radio show. Start with the current episode, and you'll catch on in no time. Or, go right to Episode 1 if you wanna binge-listen.
Turn off the lights, get into bed and plug in your earphones. It's time for a creepy bedtime story. For the discerning horror fan, we cover the most chilling cases from around the world. From the paranormal to the supernatural, unsolved mysteries and strange deaths to cryptids, conspiracy theories and the most disturbing of true crimes, all told in a unique and creepy way. Join us every week for a new scary story. bedtime.stories@outlook.com
Adam Bash
SAYER is a narrative fiction podcast set on Earth's man-made second moon, Typhon. The eponymous SAYER is a highly advanced, self-aware AI created to help acclimate new residents to their new lives, and their new employment with Ærolith Dynamics. New episodes release every other week.
Fire on the Mound
William J. Meyer, Steve Rudolph, William Seegers
"The fire does not come from the mind." One thousand years ago... Mtilan, the Unborn King, pledged his life to Legion, the fallen spirit from beyond the veil of time. Through dark and forbidden magic the broken man gained both immortality and incredible power -- but lost his soul. Over a millennium later Pekra and his sister Lely race to find the mythical cottonwood tree and protect it from Mtilan and his master Legion -- but to succeed they must first befriend their father's killer -- a for ...


Super Micro News Impact on Nutanix?
Is the recent announcement about Super Micro a significant negative impact to Nutanix or will it not be an issue for them with their customers?
Can you please explain clearly what is HCI[HyperConverged Infrastructure] in terms of Nutanix?
Suppose in Nutanix,I am working on qualifying all components like SSDs,DIMMS,NIC,HDDs..on Nutanix nodes?Can you relate HCI to this?
Hi All, One of my CVM (03 hosts) suddenly restart and all VMs restart follow. What's happen with my system? Where can I look at the log? AOS: 5.5.4 Nutanix 20170830.156 NCC 3.5.3.1 Many thanks.
Hi... Multi-tenancy is the key common attribute of both public and private clouds. Does the current version of Nutanix Enterprise Cloud OS support multi-tenancy features? Where can I get more details about that? Thanks!
Hi All, I'm performing a Nutanix PoC with 4 nodes (4x Nutanix NX-T00-2NL3-G5 wit 512gb RAM each). Does anyone have any best practice for running a Nutanix proof of concept and any final PoC report template for to complete and certify? Thanks!


What is a bik file and how do I open a bik file?
A BIK file extension results from a proprietary video format developed by RAD Game Tools. BIK files are used for playing movie clips in video games on both PCs and console systems, allowing standard video to be highly compressed while optimizing video quality for the target platform. BIK files are often compressed from AVI, MOV or other common video file formats.


text:
PAIRENT PACT
Alcohol remains the #1 drug of choice for young people and alcohol related tragedies are the #1 cause of death for teenagers in the country. Underage drinking is NOT a harmless "rite of passage":
Ninety percent of alcohol consumed by underage drinkers is consumed while binge drinking, according to the Centers for Disease Control. Young people use methods such as "shotgunning" or chugging to consume large amounts of alcohol quickly. Other dangerous trends such as "slimming" (inserting vodka-soaked tampons into the vagina or anus) and "eyeball shots" (pouring vodka into the eye sockets) allow them to get drunk faster and with no trace of alcohol on their breath. This puts them at greater risk for potentially fatal alcohol poisoning and injuries.
Short term or moderate drinking impairs learning and memory far more in youth than adults. "Adolescent drinkers perform worse in school, are more likely to fall behind, and have an increased risk of social problems, depression suicidal thoughts and violence." (AMA 2002)
The earlier a young person begins drinking, the greater the likelihood they will become addicted to alcohol. Delaying the first use of alcohol is good prevention. (NIAAA)
These "designer drugs" are synthetic (human-made) drugs chemically related to cathinone, a stimulant found in the khat plant. Khat is a shrub grown in East Africa and southern Arabia. Bath salts were originally sold in mini-marts and smoke shops with names such as Ivory Wave, Purple Wave and Vanilla Sky. Contrary to their name, these are not items used in a bath. They are marketed as cheap substitutes for other stimulants such as methamphetamine and cocaine. Bath salts usually take the form of a white or brown crystal-like powder and are sold in small plastic or foil packages labeled "not for human consumption." Use of these drugs has been associated with adverse reactions including agitation, paranoia, hallucinations, chest pain and suicidality. On Sept. 7, 2011, the U.S. Drug Enforcement Administration (DEA) invoked its "emergency scheduling authority" to control three synthetic stimulants — mephedrone, MDPV, and methylone — commonly used in these "bath salts."
Whether in powder or smokeable rock (crack) form, cocaine is a powerful and highly addictive stimulant. Cocaine is sold in a white powder form that can be snorted or injected. Crack is a rock form of cocaine that is smokeable. Currently, cocaine is available in a very pure form. Synthetic forms of cocaine such as crystal methamphetamine (crystal, meth, crank) are also available.
Dextromethorphan is a semi synthetic opiod (similar to morphine) that is found in more than 120 over-the-counter medicines. These relatively inexpensive and easily available cough suppressants (such as Robotussin or Coricidin) when taken in heavy doses can produce hallucinations, loss of motor control, elevated blood pressure and irregular heartbeat. Overdoses can cause seizures, comas and death. Young people may refer to the experience as "Robotripping" or "Dexing".
ECSTASY (XTC)
This is a designer drug that is also known as MDMA. It was originally used as an aid in couples therapy since it produces feelings of love or intimacy. Ecstasy (also called, "e", "x", "molly" or "adam") has both stimulant and hallucinogenic properties; therefore, it has all the same risks associated with these classifications of drugs. Its additional effect of feelings of intimacy may lead adolescents to engage in sexual relationships they might otherwise avoid. Ecstasy is associated with the teen rave scene.
Energy products such as "Red Bull" can contain very high levels of caffeine that can cause adverse consequence including dehydration. Some studies have linked early use of energy drinks with later use of stimulant drugs such as cocaine.
Also called Ma Huang, Ephedra is a naturally occurring substance derived from plants. Its principal active ingredient is ephedrine, which when chemically synthesized is regulated as a stimulant drug. Ephedra, especially when taken with other stimulants, is associated with side effects such as heart attack, stroke, heart palpitations, psychiatric and upper gastrointestinal effects, tremors and insomnia. Because of these risks, the FDA has prohibited the sale of dietary supplements containing Ephedra.
There are various forms of hallucinogens available. In general hallucinogens have the effect of heightening and changing sensory input. Inanimate objects may appear to move or breath for instance, or someone may feel that they can see music or feel a color. LSD is often sold in blotter format – thin pieces of paper that have been saturated in liquid LSD and can be absorbed through mucous membranes such as the mouth or eye. Psilocybin Mushrooms (shrooms) are similar in appearance to regular mushrooms and can be eaten or made into a tea.
Like cocaine, there is currently a supply of very pure heroin available. Because of this purity, today's heroin can be snorted or smoked as well as injected. Heroin is a very powerful central nervous system depressant that can suppress breathing and very quickly cause a fatal overdose. It is physically addictive in every form andcan result in painful withdrawal symptoms. Synthetic forms of morphine (such as fentanyl) are also currently available.
These include easily available household products that have a smell. Users inhale the fumes from these products from the container itself, by putting the product on a rag, or by spraying the product into a plastic bag. The short term intoxicating effects can result in falls, heart rate abnormalities (sudden sniffing death), and hallucinations/ delusions. In addition, some deaths are caused by suffocation caused by the use of a plastic bag.
Kratom is derived from a leaf grown on trees in Southeast Asia, where local populations historically consumed it as an herbal remedy to common physical ailments and for energy. It acts as a stimulant when consumed in low doses but as a sedative when consumed in high doses. Some claim it has psychoactive properties and can counter opiate addictions. Kratom is legally sold in the US in certain smoke shops and online.
While not a new drug, marijuana presents some new issues for today's youth. These include the following facts:
Today's marijuana is significantly more powerful than the marijuana of the seventies.
The age of first use has declined from late adolescence to early adolescence.
Method of use has changed – instead of smoking marijuana cigarettes (joints), young people are now likely to be smoking marijuana cigars (blunts) and using devices.
Recent research indicates that marijuana use causes learning and memory deficits, and that it can be as addictive as other drugs.
Sometimes referred to as Angel Dust, PCP is a large animal tranquilizer that causes a feeling of "no pain, no presence, no problem". Under the influence of PCP, a person may be unable to experience pain and therefore appear "stronger" than they otherwise would. PCP is sometimes put on marijuana and smoked (a "dusted blunt").
PRESCRIPTION PAINKILLERS (Vicodin, Oxycontin)
Concern about the abuse of prescription painkillers has risen dramatically in the U.S. Of particular concern is the abuse of pain medications containing opiates, marketed under such brand names as Vicodin, OxyContin, Percocet, Demerol, and Darvon. The incidence of emergency department visits related to prescription painkiller abuse has more than doubled in recent years. Death by overdose is a large risk with many of these drugs. The suppression of breathing can be immediately fatal. Physical tolerance develops quickly with these pain killers, causing physical dependence and withdrawal side effects when the drugs are stopped. Other side effects include: nausea, vomiting, constipation and itching.
PRESCRIPTION SEDATIVES (Ambien, Seconal)
These sedatives are prescribed for insomnia. They are highly addictive and can lead to fatal overdoses, especially when consumed with alcohol. Use of these drugs can cause impairing effects such as loss of motor coordination, anxiety and confusion. In the withdrawal stage, people can suffer from tremors and seizures. Withdrawal is also potentially fatal.
PRESCRIPTION STIMULANTS (Ritalin, Atterol)
These drugs are prescribed to treat attention deficit disorder (ADD). Studies have shown that students with ADD who take these medications (as prescribed and monitored by a physician) are LESS likely to develop substance abuse problems than students with ADD who do not receive proper treatment. However, for students without ADD, these stimulant drugs have similar dangerous side effects and consequences as other stimulant drugs.
PRESCRIPTION TRANQUILIZERS (Valium, Xanax)
These benzodiazepines are prescribed to provide relief from anxiety. They cause drowsiness and are therefore very dangerous in combination with driving. They also cause problems with learning and memory. Drug tolerance develops quickly to these drugs—increasingly higher doses are needed to achieve the desired effect. There is a lengthy withdrawal period when use is discontinued. Overdose is a risk when these drugs are combined with other depressant drugs, including alcohol. Other side effects include lightheadedness, vertigo, poor muscle coordination and nightmares.
A perennial herb from Mexico is being sold legally in the United States and on the Internet with no federal regulations. Salvia can be smoked or chewed and produces hallucinogenic effects in less than a minute. The hallucinations can last up to a half hour. It is inexpensive, easy to get, and used more by adolescents and young adults than older adults.
This drug is actually ketamine, a small animal tranquilizer. In addition to its anesthetizing effects, it produces feelings of dissociation and reduced anxiety. Special K can be purchased on the street in a greenish powder form (often referred to as "bumps"). It is associated with the youth club scene.
This is the name of synthetic cannabis (marijuana) that is an herbal and chemical product that mimics the effect of marijuana. It is also sold under other names such as K2. Use of spice and other synthetic marijuana products has been associated with problem behaviors including the onset of irrational and psychotic-like behaviors. In 2010, the U.S. Drug Enforcement Agency used emergency powers to ban many synthetic cannabinoids. However, since the formula can be changed slightly to keep it legal, these products may still be available in some communities.
Anabolic steroids are drugs derived from the male hormone testosterone. The term "anabolic steroids" means muscle-building hormones. While they have approved medical uses, they are illegal to possess, distribute or use for non-medical reasons. They also have many side effects including trembling, severe acne, stunted growth, wide mood swings, uncontrolled anger and aggressiveness, feelings of invincibility and paranoia. Young people who use steroids usually use them to build muscles, not to get high.
Thank you! You will be subscribed to our email lists.
Thank You's
All Rights Reserved | © 2019 | Addiction Is Real, Inc.


God calls. We respond. A vocation is a response to a specific call that Jesus Christ places on our hearts. It is our individual way of living discipleship in the Church.
Our primary call, through the Sacrament of Baptism, is to be a follower (disciple) of Jesus Christ. It is to be a holy person. Within that first call to follow Christ, God calls us in specific ways to live a vocation to the married life, the single life, the consecrated religious life, or the ordained life.
Vocation to the Diocesan Priesthood is a life of serving God's people within the parishes and diocese by being a visible sign of the presence of Christ.
Vocation to the Single Life involves using the freedom of a single person to live as an example of great generosity and service to God and others.
By Christ Jesus crucified we have freedom from darkness and by His glorious resurrection hope unto eternal life!
Tell us about a great experience that you had during the summer of 2016.
This summer, I was blessed to spend a week and a half vacationing with my family in Rapid City, South Dakota. Though we see each other regularly, it was a fantastic experience to go somewhere and simply bond as a family. We were able to visit the presidents on Mt. Rushmore, horseback trails in Wyoming, appreciate God's handiwork in the Badlands and much more. Perhaps my favorite memory was when we were stopped by a herd of buffalo crossing the Black Hills Nature Trail and had them pass within arm's reach of our vehicle!
When did your commitment to the Catholic Faith grow deeper?
My commitment to the Catholic Faith deepened shortly after high school, emerging from a crisis of personal faith. Despite 18 years of weekly Mass participation, Christian music, one-on-one religious education lessons, and a positive, uplifting homeschool environment my parents had blessed me with, I was not sure I wanted to be Catholic any longer. Although I knew much of what the Catholic Church taught, I did not really understand why it taught these things. Furthermore, personal reasons also began to trigger a desire to potentially leave my childhood religion for a more accommodating Christian denomination.
So over the course of the following months, I set out to disprove the Catholic Faith, focusing primarily on doctrines and traditions that I either did not fully understand or disagreed with -- Purgatory, infant Baptism, and saintly intercession to name a few. I raised my objections with Catholics I knew and each time they were unable to answer my questions I counted towards disproving the Faith. Though I became more and more comfortable with the idea of leaving the Faith, it was the Eucharist that destroyed my entire strategy. Try as I might, I could not disprove it through reason, Scripture or Church history and slowly began to conclude that not only is the Eucharist the REAL DEAL, but that His Church upon Earth subsists in the Catholic Church!
Offer us a few words about one Gift of the Holy Spirit and how it has helped you.
The Gift of Fortitude has helped empower me to do things I normally would not do. It has given me the courage to honestly reflect upon my faith and life journey as well as give witness to others on how God has touched my life. Without this gift I probably would not have been able to give a witness talk for my parish or apply for seminary. Additionally, it made me more comfortable in discussing my beliefs and defending the Faith in public. Perhaps my fondest memories in the private sector will be my religious and political discussions with my co-workers and friends. Though some were uncomfortable at times, these honest discussions helped me understand and appreciate different points of view and hopefully vice versa.
Share an interesting fact about your high school years.
I starting earning college credit at age 16 by taking CLEP and DANTES exams and had accumulated over 90 credits by the time I graduated high school, which counted towards my accredited bachelor's from Thomas Edison State University.
Other than prayer, what is your fuel?
Water and strong, black coffee are staples in my everyday life, though on occasion nothing beats a good iced mocha!
As a priest, God willing, where would you most like to celebrate the Holy Mass?
I would like to celebrate Mass at St. Peter and Paul Parish in Hortonville as it is my home parish and enjoy how the entire parish sings during the liturgy.
St. Stephen for his vision and singleness of purpose towards the proclamation of the Gospel of the Lord Jesus Christ. And through his intercession, I too may learn simplicity of life and be filled with the same zeal for our Lord.