text:
Interview: Joey King Chats About The Kissing Booth 3
Entertainment, Movies
Listen Here Missy Ep 2: Making New Friends, Friendship Icks and Toxic Friends
The Kissing Booth 3 wraps up its movie trilogy on Netflix with the release of the final movie.
Its star, and Executive Producer, Joey King chats about The Kissing Booth 3, how the movies changed her life, her favourite moments and saying goodbye to Elle…
Looking back on this journey from the beginning, what was it about the first Kissing Booth movie that you were drawn to?
My parents' generation had The Breakfast Club. We didn't have anything like this for my generation that was fun, that was lighthearted, that was just a really good time. Our lead girl, Elle, is not a nerd, and she's not popular. She's not an outcast. She's not overly social. She's just a girl. And I felt like her voice was just so in tune with mine, and I felt so connected to her immediately. I loved that she's just who she is, and she is different because she's just being who she is!
What's your favourite memory from doing the first movie?
One of the craziest things was I had to start my journey by learning those dance routines with Joel, and we had so much fun together. Other than throwing down at the occasional Bar
Mitzvah, I didn't have any professional dance training. And going through that process of learning those dances with Joel…oh my gosh! We would both get so frustrated with ourselves, but we had so much fun. We bonded so much in our rehearsals.
When did you realise that The Kissing Booth had become a real phenomenon?
I believed in the movie from the get go, but at the same time, it was just an independent film that didn't have a lot of money behind it and we didn't have a lot of time. We just kind of did it. We were all very proud of ourselves, but there was no game plan for marketing nor did we know if it would have a big life on Netflix. We were all just happy that we had a good time making it.
I remember the day that the trailer came out, I was really excited and made social posts about it. But then within about two days, it all of a sudden started to rack up views. We began to feel like, "Oh my God, people care about this. People want to see this." The trailer views just kept getting higher and higher and higher… and then when the movie came out, it exploded.
What do you think people were responding to?
I think so much of it can be attributed to the writing. Our director and writer, Vince Marcello, is brilliant and adapted the book so well. The dialogue is so funny and so real. I love how we don't take ourselves too seriously and I think audiences resonate with that, especially since there really hadn't been a movie like this in quite some time. Just playful stuff.
What has the fan support been like?
It's honestly unbelievable. Before the movie came out I had, just as a reference point, around 615,000 followers on Instagram. And now my numbers are much different than that. [She now has 18.2 million followers.] That's a very obvious showing of the fan support. I've gone on to do various other projects as well and it's amazing how fans have supported me along the way. It's so great how so many of them have been just excited to see what all of us do. It's really wonderful to see how it impacted them so much that they love all of us.
One of the big themes of these movies is the male-female friendship at their center between Elle and Lee. What do you think of that element?
Beth Reekles, who is the author of the books, was just so spot on with how those kinds of friendships are. One of my favorite things about these movies is that Elle and Lee's friendship never amounts to anything else, iIt never leads to that romantic zone. It's truly just a friendship. It makes me really happy because they are the most supportive people in each other's lives. And knowing that they'll always have each other is so beautiful. None of that ever feels flirtatious. I think that Joel captures that so well, Beth captured that so well, and our director Vince captured that so well.
A post shared by Joey King (@joeyking)
Your character also has this dreamy romance with Lee's brother Noah. What kind of feedback have you gotten about that from fans?
A lot of people do obsess about it and they're very much invested in the romance of this movie, which is amazing. It's a huge part of these films. It's so beautiful, it's so sweet. It's young, teen love
What I also think is really exciting and interesting about it is we got really nervous about how the audience would react when we introduced Taylor Zakhar Perez as Marco in the second movie. But, I have to say I was also excited because I thought Taylor just killed it with that role. The fans were so accepting of him. I realized he was such a great guy and I was rooting for him, too. Seeing and following the fan reactions was a blast. They still loved Noah, but also loved Marco and felt so conflicted. I credit that to the writing – Beth and Vince did such a good job of making Marco truly the nicest guy that you couldn't help but love him.
What was it like for you to step into an executive producer role on the second movie?
It was very, very exciting. I've been really interested in producing for a long time now. I've been an actor for almost 18 years and started out loving being in this business. It's always made me want to learn more and be as involved as I can be and this new role was a big learning curve. Vince was always very collaborative with me on the first film. Once I became an EP, it was nice to have even more honest conversations with one another. We had lots of talks about the final cut together and our marketing vision. I feel really excited because I always have a lot of ideas and a lot of opinions, but now I'm actually able to express them.
I also grew up playing Elle. I've been in this character's shoes for five years now so being able to have a say in a character that I love so much — and I am the demographic of the audience— it is exciting to have an effect on how the public gets to view that.
What was it like being a kind of leader on the set?
I always say my motto is: Don't take yourself too seriously, get the work done and be as professional as possible. If you're not having fun, why are you there?
We absolutely had fun which I believe is another reason why the films resonate with audiences, because we all love each other.
You had to wear wigs as Elle in the second and third movies because you were just coming off filming another role that required you to shave your head. Did Elle's hair and wardrobe help you to get into character?
The minute I put that wig on and put on my cute Elle summery clothes, I just felt so much like her. I felt so happy being in her shoes. I personally don't have a tailored-to-my-body summer
closet, but Elle does, so whenever I put on her clothes, I instantly felt more confident than whatever I came to work in. I even have my PJs tailored to my body [as Elle].
You mentioned those classic teen movies like The Breakfast Club. And you had the queen of those '80s John Hughes films, Molly Ringwald, on hand as Lee and Noah's mom. What was it like to have her as a mother figure on set?
Molly is so much fun. I cannot even put it into words how grateful I am that she agreed to be part of these films. She really took care of all of us and also knew how to throw down with us. She gave me lots of advice. Like: If you're not having fun, what are you doing? And if you're not being respectful as a professional also, what are you doing?
It's also cool because she brought her kids out with her. She just takes on that awesome mom role where she is such a great example for her kids.
In the third installment, Elle and Lee have a summer bucket list from when they were younger that they commit to completing during the summer before they go off to college. This includes skydiving, a flash mob, and lots of other activities that seemed like they must have been so fun to film. What was your favourite part?
My absolute favourite, which is only in the movie for like a few seconds, was sumo wrestling in the big suits. I have a whole four-minute-long video of us filming that. I have never in my life laughed harder. Every moment of filming with Joel is just always stupid fun.
Feel Good Movies To Watch RN
What is your favourite memory from the third movie overall?
I think my favourite memory from doing the second and third movies was becoming friends with and getting to know Taylor. He is such an important part of my life now, I couldn't imagine life without him. And even on top of that was the fact that we brought every single person back from the first movie, so we got to see everyone again. I love all of them. I feel so lucky to have gotten to know these people and gotten to come back and see all of them again. We had the privilege to all see each other again because of our fans. And we don't take that lightly.
Elle is dealing with some heavier stuff in this one: making big decisions about college, figuring out her relationships with Lee and Noah, trying to earn money, facing the fact that her dad is having his own new relationship, looking after her brother. How did it feel to make that transition?
Our characters all grew up a little bit. I liked that we gave our characters a chance to become adults in a way that naturally happens as you learn through mistakes you make. You use what happens to figure it out and fix it. But I think in this movie there was a sense of becoming more mature and we got to do a lot of crazy cool stuff. I mean, we had a bucket list montage. We're pretty lucky that we got to take these characters and bring them to bigger and better and higher places.
Without giving away too much, we can say that the movie's ending gives us a sense of where Elle is headed in her future, and it's a poignant and empowering conclusion for her. What did you think of her wrap-up?
Elle has to learn to make decisions for herself. I mean Elle is like me, Elle is at heart a big-time people pleaser and that's wonderful, but sometimes it pays to be selfish. I think she learned that you have to make decisions for yourself in order to be truly happy. I've always loved this character's dedication to her relationships. But I also love when she decides she should think for herself and no one else.
A post shared by The Kissing Booth (@thekissingboothnetflix)
Which character's journey is your favourite besides your own?
Oh man, my favourite, I have to say, is Lee. He's such a good friend. A lot of times, he isn't the best friend, but he always takes accountability and he loves Elle. He loves his friends and his family and his girlfriend just so ferociously. And I think that it shows in his wrap-up, too, that he has just become an even more wonderful version of himself. Every moment spent with Joel playing Lee, I'm in awe of how amazing he is at bringing that character to life. I watch him all day and he brings so much heart and humor and life to that character.
How did these movies change your life?
I am so proud to have worked on this trilogy and will be excited to show my kids when they're in their teens. I made amazing friends. Then there's the fan base, the way that the movie resonated with people across the world. It connects me with so many people with whom I would have never been connected. I was in an airport one time and this man, who was around 39, stopped me in the airport and said, "I love your movies!" It's so wonderful how wide an audience it reached. I believed in it from the very first time I read that script and knowing the world felt the same way that I did is just priceless.
The Kissing Booth 3 will be released 11 August only on Netflix.
Read More: QUIZ: Do You Belong In 'The Kissing Booth' Or 'To All The Boys I've Loved Before'?
Read More: Quiz: How Well Do You Remember 'The Kissing Booth 2'?
Read More: A Timeline of Joey King & Jacob Elordi's Relationship
Netflix The Kissing Booth
30 Halloween Movies To Watch


If I have current earnings for the SP500 of $160, should trailing or forward P/E be used to calculate the SP500's price?
What is the difference in using one over the other?
The answer depends on what the future price for the Index will be used for.
Trailing Earnings, or Trailing P/E is used as a consistent and quite reliable metric for Index performance relative to its long-term historical trend, and as such, it is the measure to use when a more quantitative metric is required.
Forward Earnings or Forward P/E is a forecast, an educated and informed estimation. As such, it is useful when attempting to asses Risk. However, Forward P/E depends on factors that are generally not under anyone's control, such as future interest rates, future commodity prices, regulatory norms, geo-political forces, etc. As Master Yoda would say: "The future, easy to see it is; but the future is always changing."
For the purposes of investment strategy, both must be used. Historical P/E will be used to ascertain where current Index price is relative to its historical trend, and Forward P/E will be used to ascertain Risk -- at the same time.
Some traders happen to like to look at the forward P/E.
Some traders happen to like to look at the trailing P/E.
Many traders have utterly no interest in the P/E, and have never looked at it for any reason.
Most basically, your sentence says something like "can I calculate the price using..."
You can't, in any way, "calculate" the price of a stock. It's just what someone happens to pay last in a trade.
Not the answer you're looking for? Browse other questions tagged stock-analysis stock-markets market-indexes price-earnings-ratio or ask your own question.


text:
page 18: Wait another rotation ~ 4/12/2012
Scon-o: "You OKIE?"
Dana: "I feel like my head got kicked in the kidney."
Dana: "Where are we?"
Scon-o: "cefe^3"
Dana: "Well, I'm glad that SPOOLED okay.."
Dana: "Oh my."
Dana: "Is she HERE?"
UNSPOOL
Scon-o: "We'll question th' Ipsoid^3"
Dana: "How is THAT an Ipsoid?"
Scon-o: "O, weit-- now it's the cefe^3"
Scon-o: "We reely need the ipsoid more then the cefe- we'll weit"
Dana: "Wait for WHUT?"
Scon-o: "Just enother 'ROTEYTION!'"
Dana: "In what way is that a ROTATION?"
Scon-o: "Well- remember- we hev no reezin to be here..this is where MY sub-beristeh would go-- so whut we see is..."
Dana: "Is-- is WHAT?"
Scon-o: "We are onlee here in~thru the PERCEPTION of my sub-beristeh-- otherwise, why would we think this plece should BE?"
Dana: "That doesn't make sense"
Scon-o: "It can't, not really."
You somehow transmuted the title from "Nerva's Rex" to "Her Severe." :D Looking forward to the next mutation.
I love the partitioning/obscuring of the frames in Frames 1 & 2. Also how Scon-o is considerate enough to ask if Dana is okay. :) I find Scon-o and Dana both likable. I liked Dana from the time she called Nerva "Joan's friend" back on Page 7. Come to think of it, that was also a frame where she was asking Nerva if she was okay, too.
I read the sequence between Frames 6, 7, & 8 and can't tell if I'm not getting something implied or not. Did the yarn not spool? Or is Dana just boggling at the dizziness of how far they've come down? It wasn't enough to keep me from going on, but it was enough to make me go back and reread it several times.
Nice cefe^3!!! I like how it looks lathe-turned in all its renditions so that "roteytion" makes total sense.
That's really cool to know that this is done in ballpoint; somehow it changes my reading of it. I'd thought it was pencil up till now. Yes, you get quite a lot of gradation from your ballpoint!
Thanks so much for all your replies. It's really interesting to read your impressions of your own work!
I like the frame fragmentation/partitioning as well. Back on Transbunny somewhere, there's a place where two frames are shattered and overlay on eachother...though describing it that way makes it sound cooler than it probably is.
I'm glad that you find the characters likable -- I think the characters are presented implicitly rather than explicitly, but if they're still likable, then they're doing their job (they make a better window than a friend).
Yeah, the 6-8 isn't so great; in retrospect, it would have been nice to swap the order of 6 and 7. But yes, your read was correct: the "oh my" was about how the tether now extends a great distance up.
The rotating was...well, I hope next week will further illustrate the intent (I don't remember if it does, we'll find out at the same time).

redpajama_set_name: RedPajamaCommonCrawl


Even in an increasingly electronic world, businesses and non-profits still do a significant amount of mailings via the United States Postal Service (USPS.) For bulk mailings--advertisements, brochures, reminder postcards, mass letter mailings--the USPS offers two options: stamped and metered mail. Which option is best for your mailings?
When most think of mailing an item, their first thought is of placing a stamp on the item and dropping it in a mailbox. Stamped mail is perfect for the individual mailing a few pieces at a time.
If mailing as a part of a business or non-profit, buying stamps and applying them to each piece of mail might be too daunting when faced with thousands of pieces. The time and manpower costs might significantly increase the mailings' budget.
If stamped mail is preferred for a large mailing, businesses and groups are encouraged to apply for a pre-cancelled stamp permit. Pre-cancelled stamps allow mail to skip the usual cancellation process, which legitimises the stamp. Applying for a precanceled stamp permit is free. Once received, permit holders just show their permit at any post office to purchase pre-cancelled stamps. Pre-cancelled stamps have a short time frame in which they can be used--sometimes as short as one day.
Metered mail is widely used by businesses and non-profits who have bulk mailings. The cancellation and postage information is printed directly onto the piece of mail or a sticker by a postage meter.
Postage meters are available for purchase by five authorised manufacturers in the US: Pitney Bowes, Data-Pac Mailing Systems, FP Mailing Solutions, Hasler Inc. and Neopost. In addition to purchasing the meter and accessories, a business or non-profit must apply for a metered mail permit.
Using a meter lets a business or non-profit automate its bulk mailings. The postage meter will keep track of the amount of items you are mailing, and the cost can be deducted from a bank account you specify. If you are mailing many items of different weights, some meters come in conjunction with a postage scale, and you would then manually set the meter to the amount equal to the weight of the mailing.
Metered mail is then either picked up by the USPS or can be dropped in a metered mail specified mailbox.
Both stamped and metered mail have a place within a business or non-profit setting. Selecting a postage option will depend on an organisation's frequency and size of mailings. Is the organisation's mailings regulated to one holiday appeal letter and the occasional brochure sent to an individual upon request? The cost of leasing or purchasing a postage meter and the time invested in researching and setting up that option would not be appropriate for the organisation.
However, if mailing is a major part of the organisation's business--sending hundreds or thousands of the same piece at once--a postage meter will save time and effort. It will also give the business an opportunity to have postage meter service and supplies delivered, a significant time and resource saver.
Individuals mailing items do not have the volume to consider metered mail as a legitimate option. The cost of leasing the meter is not cost-effective for an individual. First-class stamped mail is the best option for individuals--no matter how long your annual holiday letter list is.
Given the USPS's recent moves to shutter thousands of post offices across the country, it will become increasingly inconvenient for individuals and businesses to handle mailings through the post office. This might cause more organisations to move toward metered mail. The USPS has an online option for purchasing large quantities of stamps, which will keep stamped mail a viable option for some.
Kat Hasenauer is a freelance sportswriter focusing on hockey and lacrosse. She is a regular contributor to several sports websites, and is well known for her informative Twitter feed. She also maintains her own sports blog, "...On Being a Sports Girl." A student affairs professional by day, she holds degrees from School of the Arts (Rochester, NY), Binghamton University and Boston University.


text:
Este retrato de Monsieur Bertin es un cuadro de Dominique Ingres. Además de las obras orientalistas por la que es conocido principalmente Ingres, cultivó el retrato, como en este óleo que mide 116 centímetros de alto y 96 de ancho. Actualmente se conserva en el Museo del Louvre de París.

Historia
Louis-François Bertin era el director del Journal des Débats, un rico empresario editorial en la época de Luis Felipe. Ingres realiza en este cuadro el arquetipo del retrato burgués que influirá profundamente a los pintores académicos como Léon Bonnat, pero también a pintores modernos como Degas y Picasso. Es una de sus obras más célebres.

Encargado y adquirido por Bertin en 1832, fue legado a su hija Louise Bertin después de su muerte. Ella a su vez se lo lega a su sobrina Marie Bertin esposa de Jules Bapst director del Journal des débats. La última propietaria Cécile Bapst, su sobrina, vendió el retrato al Louvre en 1897.

Análisis
El cuadro tiene un formato mediano, de 116 cm por 96 cm. Sobre un fondo marrón iluminado desde la derecha del cuadro, se destaca la silueta masiva de Bertin. Está retratado en tres cuartos sobre una silla cuyo brazo se ve.

Ingres lo representa de manera poco habitual, que en la época llegó a considerarse vulgar: con las manos, con los dedos abiertos, apoyadas en los muslos. No obstante, es esa postura decidida la que proporciona la clave de la personalidad del retratado, más que el rostro en sí.

La gama cromática es limitada: el negro de las vestiduras y el marrón oscuro del fondo. Viste una chaqueta y unos pantalones negros, con un chaleco marrón en satén sobre una camisa blanca cuyo cuello supera, y de un corbata del mismo color. Se percibe también un par de catalejos que sale del bolsillo derecho de los pantalones. De este entorno oscuro, destaca su rostro. Es un hombre envejecido, que tiene por entonces 66 años, ya que nació en 1766. Tiene el cabello gris claro casi blanco. El rostro presenta una expresión dura y observa directamente al espectador.

Puede verse, en la parte baja de la pared una decoración con un motivo griego, y el reflejo de una ventana sobre el brazo de la silla. En lo alto a la derecha está escrito L.F. BERTIN. En la esquina superior izquierda aparece la firma y fecha del cuadro: J.INGRES PINXIT 1832 .

Referencias
 Carrassat, P.F.R., Maestros de la pintura, Spes Editorial, S.L., 2005. ISBN 84-8332-597-7
 Hans Naef, Die Bildniszeichnungen von J.-A.-D. Ingres, Benteli Verlag Bern, 1979, tomo III, pág. 114-135.

Referencias

Retratos de Dominique Ingres
Cuadros de 1832
Cuadros de Francia de los años 1830
Francia en 1832
Pinturas del Museo del Louvre
Cuadros del Romanticismo

redpajama_set_name: RedPajamaWikipedia


Kaua'i RSVP, which is part of the Agency on Elderly Affairs, announced last week the kick-off of a collection drive to benefit Kaua'i Economic Opportunity's homeless emergency shelter.
The products being collected are body soap, deodorant, laundry detergent, razors, rubber slippers, shampoo, toothbrushes, toothpaste and towels.
The items can be dropped off Monday to Friday, from 8 a.m. to 4 p.m. at the Agency on Elderly Affairs/Kaua'i RSVP. The office is at the Līhu'e Civic Center, 4444 Rice St., Suite 330. The deadline is Sept. 7.
Additionally, donations will be accepted on two Saturdays, Aug. 13 and Aug. 27, from 9:30 a.m. to 12:30 p.m., at Kmart and Times Supermarket in Līhu'e.
The National Day of Service and Remembrance was established in 2002 as an annual tradition of engaging in charitable service on Sept. 11.


text:
Fickle Friends - You Are Someone Else
by Brett Dickinson
It's been a long wait for a Fickle Friends album. The five-piece from Brighton formed in 2013, and with singles dating back to 2014, it's been a long wait for the fans who have been with them from the beginning (sadly I cannot count myself in that number). Having shared stages with bands such as the Killers, Halsey and HAIM, they've used the time well, seen a lot, played a lot and won a lot of fans already as well as gaining a vast array of musical influences. The wait for the most part seems worth it - the band is remarkably sure of its sound for a first album, and it sounds very polished with smooth hooks, bright synths and a well-crafted combination of catchiness and meaning.
The album's start is spot on. The opening single Wake Me Up starts the album strongly providing it with a bright, bouncy and breezy tone to draw you in. The second and third songs Glue and Swim are arguably their best known songs, both having gained attention as singles. The songs are very synth heavy, but it is a sound that works for them, complementing the confident voice of lead singer Natassja Shiner. Swim in particular builds on the summery, bouncy feel of Wake Me Up. However song four, Bite, darkens the tone just with the addition of a very noticeable beat pulsing beneath the synths, suiting the depiction of a toxic relationship well. It retains some of the bounce, however, with an element of 80s dance music to it, different to the synth-pop of the first three songs.
The next songs Lovesick and Say No More are more guitar driven than the ones before, possibly with more similarities to the alt-pop, indie side of their influences. Yet it is a style they can carry off (and one that suits my musical preferences). Sadly Heartbroken, the next song, fails to straddle the line between catchy and hook-laden, with the high-pitched synth and relaxed vocals crossing the line into cheesy and slightly irritating. However the fact this is song 8 before they get the balance wrong is almost a compliment, yet the drop-off from Say No More to Heartbroken does blemish this section of the album for me.
The two minutes of In My Head serves almost as an interlude in the album, with a much slower, entrancing sound, allowing a focus on Shiner's voice and the lyrics (both of which could be far too easily overlooked by a casual listen). It's a really intriguing, personal piece and is going to divide opinion amongst their fans as it is such a different sound. Personally, I think it serves a beautiful, haunting interlude and would really like them to explore this side of their sound some more at some point.
The second part of the album continues to establish Fickle Friends' credentials and synth-pop sound, with songs such as Hello Hello and She which would be worthy of special mention on a lesser album but are just another song on this one. Rotation struggles initially as the mood change from In My Head is quite severe, but it recovers enough to be a solid, if unexceptional track. Paris is a slower, yet still danceable number for the first half allowing attention to focus on the emotive lyrics - "by the time you change your mind I might have left" - before a drop mid-song which picks up the tempo and showcases the lower ranges of Shiner's voice. Crossing continents, Brooklyn from which the album title You are Someone Else comes is very similar to Swim, synth heavy, catchy hooks, every aspect ready-made for a festival. Midnight is a song I can't quite decide on: I think I quite like it, it's a bit darker, a bit more haunting ("trouble always follows me") yet its just a bit different to the rest of the album and I'm not quite sure why.
The album overall is a very strong showing as a whole, rather than just some good singles combined with filler (which I find that this type of album can tend towards) and, with the sad exception of Heartbroken, there is not a bad song on the album. However, it would also probably be true that none of the songs are comparatively outstanding (it has no Mr. Brightside-type instantly memorable, album-defining song). Yet in spite of that minor failing it is a very strong, pop album especially for a debut album: the (incredibly long) time to release a first album seems to have paid off.


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For today, there are several Minecraft editions for different devices. You can play the desktop version on your computer, play on your smartphone or console. Strangely enough, developers very often share beta versions that contain a lot of new content but not all players know how to get these beta versions.
First, look at the edition you are using. They are very different and can only be used on certain devices. Secondly, follow the instructions below to download the Minecraft beta version. That's it, there's nothing complicated here.
The development team regularly publishes pre-release versions for the Java version, which are also known as snapshots. You can at any time become a beta tester and try out a snapshot. To do this, you need to find and enable "Enable Snapshots" in the launcher settings. Then select the latest snapshot and click play.
In fact, the pocket version (now called without a prefix) is the most popular and therefore the developers very often please the players with updates. According to statistics, a new beta version appears every week after the pre-release. To not miss the moment and enjoy new features you can install the beta version if you have a licensed copy of the game.
1. Go to the official beta test page - Google Play.
2. Click on the button to become a tester.
3. The game will automatically update to the latest beta version. If this does not happen then reinstall the game.
0. If you want to return a stable version of MCPE, then leave the testing program on the official GooglePlay page.
Xbox Insider Hub allows you to switch the active version of Minecraft, which will be downloaded from the Store.
1. Open the Store app.
2. Install Xbox Insider Hub and open it.
3. Select Insider content > Minecraft Beta > Join.
0. You can help developers if you tell them about the bugs you found.


KONAMI is celebrating both DDR and pop'n music's 20th anniversary with a crossover event. Both DDR and pop'n music get a brand new song named ANNIVERSARY ∴∵∴ ←↓↑→ by BEMANI Sound Team "U1 overground". DDR A gets シュレーディンガーの猫 by Cait Sith, and pop'n music peace gets Come to Life by ARM (IOSYS) feat. Nicole Curry. The event is live as of 11/15/2018, 10:00 AM JST. More info about the event can be found on the DDR 20th Anniversary site.


text:
Sweden Yeah Yeah Yeah: Caesars and The Lonely Boys
Caesars, The Lonely Boys
With just ten million people, Sweden definitely punches above its weight in international popular culture. I mean, sometimes it seems like every second person there must be a crime fiction writer or member of a rock combo. Maybe both. Today I feature just two fab Swedish bands that deserve more scrutiny from discerning international poprock fans: Caesars and The Lonely Boys.
First up: Caesars, aka Caesar's Palace, aka The Twelve Caesars. Confused? They started with Caesar's Palace for the first few albums but worried about problems with a certain US hotel chain and so shortened it, while the 12 was added for tours around Scandinavia, for reasons that are unclear. Well, suffice to say, everything can be found under the Caesars brand now. Altogether the band released five albums between 1998 and 2008 but the first two are not generally available, though some of their songs appear on the 2003 compilation 39 Minutes of Bliss (in an Otherwise Meaningless World). The group's breakout record was 2002's Love for the Streets and it's not hard to hear why, the album is chock-a-bloc full of hummable wonders that tap into a wide range of sixties and indie eighties motifs. From the anthemic "Over 'for It Started" to the tripping good-time feel of "Candy Kane" to the more country-ish Stonesy feel of "Cheap Glue" the LP slips by effortlessly in easygoing party mode. The overall strength of the record is typically overshadowed by its underground monster organ-heavy hit "Jerk It Out" but I urge listeners – don't stop there! It's a cool tune, for sure, but there is so much more to love from this band. Both 2005's Paper Tigers and 2008's Strawberry Weed are solid albums, full of melodic treats, like "We Got to Leave" from the former and "Stuck with You" from the latter.
https://poprockrecord.files.wordpress.com/2020/05/11-we-got-to-leave.m4aWe Got To Leave
Next, a band created to provide a soundtrack to a book about a fictional 1965 band. In 1995 one half of Swedish duo Roxette agreed to put together a group and songs for Swedish author Mats Olsson's 1995 novel The Lonely Boys. The results are 1965 fabulous! Per Gessle and his ragtag band of veterans from the Swedish music scene essentially become The Lonely Boys. They easily nail the sixties vibe as well as better known retro efforts like That Thing You Do, which came out a year later. The self-titled debut cranks through all the classic Beatles/Stones sounds but there is also a solid dollop of 1979 on tracks like "Lonely Boys" (with that slightly sped up Beatles sound). Then there's "I'm Not Like You" which draws more on the Kinks and Who while "Keep the Radio On" flashes more Merseybeat. The only cover here is the Jagger/Richards rarity "So Much In Love" but "Flowers on the Moon" also sounds pretty late 1969 Stonesy. And then there's the killer jangle on "Genius Gone Wrong." For the uninitiated, The Lonely Boys will an unavoidable full-record listening session.
https://poprockrecord.files.wordpress.com/2020/05/01-lonely-boys-2006-remastered-version-1.m4aLonely Boyshttps://poprockrecord.files.wordpress.com/2020/05/03-keep-the-radio-on-this-is-the-perfect-song-2006-remastered-version-1.m4aKeep the Radio Onhttps://poprockrecord.files.wordpress.com/2020/05/09-flowers-on-the-moon-1.m4aFlowers on the Moon
Seems Sweden used to be just famous for massages and a pretty good welfare state. Well, they needn't stop there. Get caught up on these Swedish pleasures online.


Nick Coleman is a recognized leader in Cyber security. He is Global Head, Cyber Security Intelligence Services at IBM. Previously he was The National Reviewer of Security at the UK Government and Authored the 'Coleman Report' published in the UK Houses of Parliament. He is an appointed Advisor to the Executive Director of the EU Security Agency (ENISA). He is an Honorary Professor at Lancaster University's GCHQ Accredited Centre of Teaching and Research Excellence. He is a Fellow of the Institution of Engineering and Technology and a Fellow of the BCS. He holds an MBA with Distinction from Manchester Business School.
The intersection of artificial intelligence, cybersecurity and EU policy will continue to generate increased interest — and debate — for years to come.
The EU's evolving regulatory environment on cybersecurity and data protection offers both possibilities and challenges for IBM and its clients.


Van Halen backstage pass Laminated For Unlawful Carnal Knowledge!
Van Halen backstage pass Laminated pass Balance North America 1995 !
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Van Halen backstage pass Laminated tour pass Summer '95 White pass!


This really struck me, since Nimoy rose to fame as Star Trek premiered at the pivotal mid-Sixties Uranus-Pluto conjunction, and he passed at the current Uranus-Pluto square. Astoundingly, it turns out he was also born under a Uranus-Pluto square! No wonder he identified so strongly with the role of Spock, the lone alien aboard the Starship Enterprise. And, at Nimoy's birth on March 26, 1931 the Sun, Mercury and North Node were conjunct the ultimate outsider, Uranus, in pioneering Aries. Clearly destined to "boldly go where no man has gone before," he will live forever as the pointy-eared Vulcan who helped ground the volatile Captain Kirk. With transiting Uranus conjunct Nimoy's natal Uranus while Jupiter trined it at his death, one hopes it was a relatively quick and painless passing.
Nimoy shed his mortal coil just prior to next week's Virgo Full Moon (March 5, 10:05 a.m. PST). The Virgo Moon (rational earth) opposing the Pisces Sun (emotional water) perfectly symbolizes the logical, even-keeled Vulcan faced with integrating his emotional, human side.
The final pass of the unruly-but-liberating Uranus-Pluto square peaks mid-March, so we could all use a little extra grounding this month. Do some yoga, work in the garden or volunteer your services near this Full Moon. Staying close to the Earth will help balance your head and heart, so you can truly live long and prosper!
← Aquarius New Moon: Wood Sheeps and Helpful Peeps!


text:
Vaping can lead to smoking regardless of what students use in their e-cigarettes. National Institute on Drug Abuse.
The use of e-cigarettes among teenagers has grown immensely in recent years. Science News for Students.
Vaping Trend Sweeps Across U.S. Campuses
Published on February 26, 2018 in Features by Brooks Lalley
A viral craze is sweeping across high school and college campuses across the United States. Unlike the harmless middle school trends of yesteryear, like Webkinz and Silly Bandz, this "epidemic" originates from basic lifestyle choices. However, like alcoholism and drug abuse, vaping is a toxic underground trend that has detrimental effects.
The vaping trend has taken campuses by storm and is the cause of serious health risks. According to the National Center on Addiction and Substance Abuse (NCASA), "vaping is the act of inhaling and exhaling the aerosol, often referred to as vapor, which is produced by an e-cigarette or similar device." The devices themselves can resemble a flash drive or pen, making it easier to hide on school campuses. Part of the appeal among teenagers is the belief that "e-cigarettes are less harmful than other tobacco products" as well as the ability for these devices to use a variety of flavors, from mango to mint (SurgeonGeneral.gov).
Teens are more likely to use e-cigarettes than cigarettes. National Institute on Drug Abuse.
Sophomore Mark Baisier said, "From what I know, I think vaping has become so popular because nicotine gives people the same effect as cigarettes or dip. It has become more popular due to the fact that it does not require the tobacco flavor that many find unappealing. It is also very easy for someone to conceal a vape with them because they can be so small and don't leave a smell."
However, the vapor itself is what raises a red flag on health concerns. While vaping might seem like a nontoxic alternative to smoking cigarettes, there remain hidden dangers. "The e-liquid in vaporizer products usually contains a propylene glycol or vegetable glycerin-based liquid with nicotine, flavoring and other chemicals and metals, but not tobacco. Some people use these devices to vape THC, the chemical responsible for most of marijuana's mind-altering effects, or even synthetic drugs like flakka, instead of nicotine," according to NCASA.
Many of these chemicals have been linked to cancer, as well as respiratory and heart disease. Because e-cigarettes were introduced on the market nearly a decade ago and have only recently caught national attention, many of the health effects of vaping have not yet been analyzed.
As a result, parents and school administrators are largely unaware of this fad and have yet to establish firm rules on the restriction of these substances. Wesleyan School, however, has included a section about vaping in its handbook; it reads, "The possession, use or distribution of tobacco or any nicotine products of any form on campus or at off-campus school functions are prohibited. Examples include, but are not limited to, chewing tobacco, cigarettes, e-cigarettes and vaping devices." This policy was put in place to protect students from harmful substances and maintain the integrity of the drug-free Wesleyan community.
Assistant Dean of Students Mary Stephenson said, "Nicotine, which is a part of each of these products, is an addictive, habit-forming substance that is particularly harmful to the adolescent brain. Because the adolescent brain is not fully developed, it is especially vulnerable to the harmful effects of nicotine. I'm not a scientist, but there are a number of studies that show the damaging effects nicotine has on the developing function and capacity of the adolescent brain. As such, we want to clearly discourage the use of such products, which is why they are not allowed to be used by anyone on our campus (adults or students). We want to make a clear stand in word and by example that tobacco products are not part of a healthy lifestyle and encourage our students to live tobacco free."
"Many vape devices have the capability to vaporize substances other than nicotine, such as opiates, variants of marijuana, and other illegal drugs, not to mention legal products that are not meant for human consumption or inhaling. In order to be consistent with our drug and alcohol policy, we feel it is imperative to not just be a tobacco-free campus, but also a vape-free campus. These devices have no place on our campus," said Stephenson.
Despite the negative effects of vaping, arguments have been made in favor of the new fad. Sophomore respondent, who requested anonymity said, "Vaping can have positive effects on the community as a whole, because I personally know adults who have stopped smoking through vaping. However, since vaping has only been around for a short period of time, we have no idea whether it is a healthier alternative or not. I personally believe that it is a healthier alternative and will have positive benefits down the road on smokers and 'would have been' smokers, but at the end of the day…only time will tell."
Wesleyan parent Suzanne Baisier said, "Although I believe that the choice to vape or not to vape should be made by the student and his/her parents, I appreciate the 'vape free' school that we are a part of, because I feel that it puts less 'peer pressure' on my kids to vape. If the student body were allowed to do so on campus, it is my belief that kids who otherwise would not be likely to vape, might give it a try. The fewer opportunities for negative peer pressure on campus, the better off our student body will be as a whole."
Tags: drugse-cigaretteshealthillegalNational Center on Addiction and Substance Abusenicotinetobaccovaping
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text:
Travel News: March 8th 2021
March 9, 2021 by Treska - Travel Smart with Koloman, posted in News
Thailand to reduce quarantine period for vaccinated travellers
According to Reuters, Thailand will from next month reduce its mandatory quarantine from 14 to seven days for foreigners arriving in the country who have been vaccinated against the coronavirus, its Health Minister said on Monday.
Vaccinations must be administered within three months of the travel period and visitors will still be required to show negative COVID-19 test results within three days of their departure, Anutin Charnvirankul told a news conference.
Those not yet inoculated but with coronavirus-free certificates would be quarantined for 10 days, he said.
"Foreigners travelling to Thailand with vaccination certificates in accordance with the requirements of each brands, will need to quarantine for only seven days," Anutin said, referring to the doses needed to be effective.
Thailand's flight limits, its strict entry requirements and mandatory quarantine for all arrivals have been central to its success in limiting the spread of the virus to just over 26,000 cases and 85 deaths. For the rest of the article please click here and continue reading on Reuters.
Source: Trully Classy
covid-19 Pandemic tames Airbnb in Europe's tourist hotspots
According to the latest updates from Reuters, the COVID-19 pandemic has achieved what many mayors across Europe have tried and failed to do: wipe out tens of thousands of Airbnbs from city centres and so help lower rental costs for locals, in some places by as much as 15%.
While Europe's cities have long welcomed tourists, critics say the surge of properties listed on short-lettings site Airbnb in recent years had priced many locals out of their own housing markets, turning historic neighbourhoods into soulless spaces.
Property management companies and landlords contacted by Reuters in cities including Lisbon, Barcelona, Prague, and Venice said the collapse of tourism in the pandemic meant some hosts had now replaced holidaymakers with mid- to long-term tenants, moved in themselves, or given up properties altogether.
Data from holiday rental analytics firm AirDNA showed the number of Airbnb listings with at least one night booked or available in the past month in Europe's 50 largest cities plunged 21.9% year-on-year in 2020.
Airbnb says it has…for the rest of the article please click here and continue reading on Reuters.
Source: Airbnb Newsroom
traveling by train to croatia? – possible thanks to regio jet
Last summer when the Covid-19 did get weaker and borders were re-opening again, the Czech based carrier Regio Jet did announce new routes and travel possibilities from Czech republic to Croatia. This year they came back with new routes and announcements:
Source: Regiojet
The train connection between Czech capital Prague main station (Praha hl. n) to Rijeka and Split will start on the 28th of May. In the first few weeks they will operate 3 times/week, starting the Summer Holiday (1st of July until the 31st of August) they will operate daily. Tickets are already available from 22.9 EUR From Prague to Rijeka (one way), however those are not the only routes available. Thanks to the great cooperation with the bus companies in Crotia, you can access another 56 locations from Rijeka and Split as visible on the map below:
For further details and buying the tickets, please click here and access RegioJet.
Vietnam could welcome vaccinated tourist from abroad already in july
Accouting to Lan Houng from e.vnexpress.net, Tourism experts proposed Vietnam's tourism industry prepare to welcome vaccinated foreign visitors back to the country from the third quarter in a similar vein to regional peers.
Nguyen Huu Tho, chairman of Vietnam Tourism Association (VTA), said Vietnam has been one of the most successful countries in containing Covid-19 outbreaks, creating favorable conditions for the tourism industry to open up to international markets since the beginning of the third quarter.
He cited Thailand, which plans to open its doors to foreign tourists already vaccinated against the coronavirus from July 1, and Singapore, which is ready to do the same.
"Vietnam should prepare to receive international visitors from July to help the heavily-hit tourism industry recover as 95 percent of inbound travel firms have suspended operation. Promoting domestic tourism could bring in revenue but cannot save the tourism industry. The roadmap to opening up the international market is the only solution," he added. For further details, please click here and continue reading on e.vnexpress.net.
Source: e.vnexpress.net
Vaccinated travelers to Madeira can skip COVID-19 testing
According to Andrea Smith from Lonely Planet, Visitors must prove their status through an official vaccination certificate that includes the date of the two doses taken and the immunization period for each vaccine. The requirement can also be waived for those who can provide proof of recovery from the virus in the previous 90 days.
To qualify, documentation in English must be submitted to the Madeirasafe.com app before the trip is taken. The Madeira Promotion Bureau has described the new development as a "vote of confidence in this new phase the world is going through."
Passengers who aren't vaccinated but meet the country's entry requirements can still visit, but they are required to have a negative PCR test carried out within 72 hours prior to boarding. If they arrive without it, they can take the test on arrival at the airport, and are required to quarantine at their destination until they receive the result. Those who don't take the test upon arrival are required to self-isolate for a period of 14 days where they are staying. Find out more about Madeira's entry requirements here.
For further details about Madeira, please click here and continue reading on Lonely Planet
Skip quarantine in new york – currently only for domestic travelers
According to Andrea Smith from Lonely Planet,
New York state is lifting quarantine and COVID-19 testing restrictions on vaccinated domestic travelers. The caveat is that the new ruling applies within 90 days of the traveler's second inoculation, due to the lack of data around how long vaccine-related immunity will last.
New York governor, Andrew N. Cuomo, said that travelers will still have to undergo testing and quarantine if the 90-day period since vaccination took place has elapsed. The rules for New York at present are that travelers must provide a negative COVID-19 viral test taken within three calendar days of travel to the city and undergo a mandatory 10-day quarantine.
The same applies to domestic travelers who haven't been fully vaccinated yet. The relaxing of restrictions does not apply to international travelers over two years, who must also provide a negative COVID-19 viral test taken within three days of travel and continue to quarantine, even if they've been vaccinated.
Cuomo also said that beginning 2 April, event, arts and entertainment venues can reopen at 33% capacity, up to 100 people indoors and up to 200 people outdoors. If all attendees present proof of negative test prior to entry, capacity can increase…for the rest of the article please click here and read the full text on Lonely Planet.
Previous postTravel News: March 7th 2021
Next postTravel News: March 9th 2021


While researching the cause of beehive demise across the country, Javier stumbles onto a breakthrough—a virus carried by bugs that insecticides can't eradicate. The discovery explains a string of illnesses and deaths, and it means a vaccine is possible. But instead of his achievement being celebrated, his bosses try to kill him.
After evading capture, Javier meets Liz, a widow who works at a homeless shelter. Together, they become fugitives, and in searching for a vaccine, they discover an intricate chain of secrets that lead to the most powerful entities in the country.
November 2, 2017 in Uncategorized.
Drake and the Fliers has a wonderful new review!


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Add some action to your coding loop with a blinking LED light.
Can you complete the challenge? Does this challenge give you any more ideas to try out? Once you've updated the code, make sure to hit the Upload button!
Find the delay value where the LED is blining so fast that you can't actually see it turn off.
As you program more, you'll get better at imagining which line of code is running as you watch your program work on Code Rocket. You might even start to think of the program more as a 'circle' than a single sheet of code.
1. If you let the example code run, how many times will the code loop?


Ready to find a better energy deal in Muttaburra, 4732, Queensland?
Muttaburra is a suburb located in the Barcaldine Region, Queensland, Australia. The population of Muttaburra is approximately 88 people. Best place to be in Muttaburra is Caravan Park. Some suburbs around Muttaburra are Tablederry, Cornish Creek, Sardine, Bangall, Ibis, Morella, Aramac, Upper Cornish Creek and Pelican Creek.
Why Muttaburra should compare energy prices using Econnex?


text:
\section{ Introduction and Main results} \label{sec:intro}

\subsection{Introduction}
The purpose of this note is to present a \emph{quantitative} two weight theorem for the Hardy-Littlewood maximal operator  when the underlying space is a space of homogeneous type $\mathcal{S}$ (SHT in the sequel), endowed with a quasimetric $\rho$ and a doubling measure $\mu$ (see Section \ref{sec:SHT} for the precise definitions).
We briefly recall some background on this problem in the \emph{euclidean} or \emph{classical} setting,  when we are working in $\mathbb{R}^n$ and we consider Lebesgue measure and euclidean metric. We also assume that in this classical setting all the  maximal operators involved and $A_p$ classes of weights are defined over cubes. Let $M$ stand for the usual uncentered Hardy-Littlewood maximal operator:
\begin{equation*}
Mf(x) = \sup_{Q\ni x } \frac{1}{|Q|}\int_{Q} |f|\,dx.
\end{equation*}
The problem of characterizing the pair of weights for which the maximal operator is bounded  between weighted Lebesgue spaces was solved by Sawyer \cite{sawyer82b}:
To be more precise, if $1<p<\infty$ we define for any pair of weights $w,\sigma$, the (two weight) norm,
\begin{equation}  \label{MainQuestion}
\|M(\cdot \sigma)\|_{L^p(w)}:= \sup_{f\in L^p(\sigma)}  \frac{ \|M(f \sigma) \|_{L^p(w)}}{ \|f\|_{L^p(\sigma)} }
\end{equation}
then Sawyer showed that  $\|M(\cdot \sigma)\|_{L^p(w)}$ is finite if and and only if
\begin{equation*}
 \sup_Q \frac{\int_Q (M(\chi_Q \sigma )^p \ wdx}{ \sigma(Q)}<\infty,
\end{equation*}
where the supremum is taken over all the cubes in $\mathbb{R}^n$.
A quantitative precise version of this result is the following: if we define
\begin{equation*}
 [w,\sigma]_{S_p}:= \left(\frac1{\sigma(Q)} \,\int_Q M(\sigma\chi_Q)^pw\ dx\right)^{1/p}.
\end{equation*}
then
\begin{equation}\label{eq:moen}
 \|M(\cdot \sigma)\|_{L^p(w)} \sim p'[w,\sigma]_{S_p},
\end{equation}
where $\frac{1}{p}+\frac{1}{p'}=1$. This result is due to K. Moen and can be found in \cite{Moen:Two-weight}.

However, it is still an open problem to find a characterization more closely related to the $A_p$ condition of Muckenhoupt which is easier to use in applications.  Indeed, recall that the two weight $A_p$ condition:
\begin{equation*}
\sup_Q\left(\Xint-_Q w\ dx\right)\left(\Xint-_Q v^{-\frac1{p-1}}\ dx\right)^{p-1}<\infty
\end{equation*}
is necessary for the boundedness of $M$ from $L^p(v)$ into $L^{p}(w)$ (which is clearly equivalent, setting $\sigma=v^{1-p'}$, to the two weight problem), but it is not sufficient. Therefore, the general idea is to strengthen the $A_p$ condition to make it sufficient. The first result on this direction is due to Neugebauer \cite{Neugebauer}, proving that, for any $r>1$, it is sufficient to consider the following ``power bump'' for the $A_p$ condition:
\begin{equation}\label{Neug}
\sup_Q\left(\Xint-_Q w^{r}\ dx\right)^\frac{1}{r}\left(\Xint-_Q v^{-\frac{r}{p-1}}\ dx\right)^{(p-1)r}<\infty
\end{equation}

Later, the first author improved this result in \cite{perez95} by considering a different approach which allows to consider much larger classes weights. The new idea is to replace \emph{only} the average norm associated
to the weight $v^{-\frac1{p-1}}$  in \eqref{Neug} by an ``stronger'' norm which is often called a ``bump". This norm is defined
in terms of an appropriate Banach function $X$ space satisfying certain special property. This property is related to the $L^p$ boundedness of a natural maximal function related to the space. More precisely, for a given Banach function space $X$, the local $X$-average of a measurable function $f$ associated to the cube $Q$ is defined as
\begin{equation*}
 \|f\|_{X,Q}=\left\|\tau_{\ell(Q)}(f\chi_Q)\right\|_X,
\end{equation*}
where $\tau_\delta$ is the dilation operator $\tau_\delta f(x)=f(\delta x)$, $\delta>0$ and $\ell(Q)$ stands for the sidelength of the cube $Q$. The natural maximal operator associated to the space $X$ is defined as
\begin{equation*}
M_{X}f(x)= \sup_{Q:x\in Q} \|f\|_{X,Q}
\end{equation*}
and the key property is that the maximal operator $M_{X'}$ is bounded on  $L^p(\mathbb{R}^n)$
where $X'$ is the associate space to $X$ (see \eqref{bnessX'} below).

As a corollary of our main result, Theorem \ref{thm:main},  we will give a quantitative version of the main result from \cite{perez95} regarding sufficient conditions for the two weight inequality to hold:

\begin{theorem}\label{thm:perez-bump}
Let $w$ and $\sigma$ be a pair of weights that satisfies the condition
\begin{equation}\label{keySufficientCondition}
\sup \left(\Xint-_Q w\ dx\right) \|\sigma^{1/p'}\|^p_{X,Q} <\infty.
\end{equation}
Suppose, in addition, that the maximal operator associated to the associate  space is bounded on $L^p(\mathbb{R}^n)$:
\begin{equation}\label{bnessX'}
 M_{X'}: L^p(\mathbb{R}^n)\to L^p(\mathbb{R}^n).
\end{equation}
Then there is a finite positive constant $C$ such that:
\begin{equation*}
\|M(\cdot \sigma)\|_{L^p(w)} \leq C.
\end{equation*}

\end{theorem}

In this note we give a different  result of this type with the hope that it may lead to different, possible better, conditions for the two weight problem for Singular Integral Operators.

Most of the interesting examples are obtained when $X$ is an Orlicz  space $L_\Phi $ defined in term of the Young function $\Phi$ (see Section \ref{sec:SHT} for the precise definitions). In this case,  the local average with respect to $\Phi$ over a cube $Q$ is
\begin{equation*}
 \|f\|_{\Phi,Q} =\|f\|_{\Phi,Q,\mu}= \inf\left\{\lambda >0:
 \frac{1}{\mu(Q)}\int_{Q} \Phi\left(\frac{ |f|}{ \lambda } \right)
dx \le 1\right\}
\end{equation*}
where $\mu$ is here the Lebesgue measure. The  corresponding maximal function is
\begin{equation}\label{eq:maximaltype}
M_{\Phi}f(x)= \sup_{Q:x\in Q} \|f\|_{\Phi,Q}.
\end{equation}

Related to condition \eqref{keySufficientCondition} we introduce here the following quantities.

\begin{definition}\label{def:Ap-multiple}
Let $(\mathcal{S},d\mu)$ be a SHT. Given a ball $B\subset \mathcal{S}$, a Young function $\Phi$ and two weights $w$ and $\sigma$, we define the quantity
\begin{equation}\label{eq:A_p-local}
 A_p(w,\sigma,B,\Phi):=\left( \Xint-_{B} w\, d\mu\right)\|\sigma^{1/p'}\|^p_{\Phi,B}
\end{equation}
and we say that a pair of weights belong to the $A_{p,\Phi}$ class if
\begin{equation*}
[w,\sigma,\Phi]_{A_p}:=\sup_B A_p(w,\sigma,B,\Phi) <\infty,
\end{equation*}
where the $\sup$ is taken over all balls in the space. In the particular case of $\Phi(t)=t^{p'}$, this condition  corresponds to the classical $A_p$ condition and we use the notation
\begin{equation*}
 [w,\sigma]_{A_p}:=\sup_B\left(\Xint-_{B} w\ d\mu\right)\left(\Xint-_{B} \sigma\ d\mu\right)^{p-1}.
\end{equation*}

\end{definition}

We define now a generalization of the Fuji-Wilson constant of a $A_{\infty}$ weight $\sigma$ as introduced  in \cite{HP}  by means of a Young function $\Phi$:
\begin{equation*}
[\sigma,\Phi]_{W_p}:=\sup_B\frac{1}{\sigma(B)}\int_B M_{\Phi}\left(\sigma^{1/p}\chi_B\right)^p\ d\mu
\end{equation*}

Note that the particular choice of $\Phi_p(t):=t^p$ reduces to the $A_\infty$ constant ( see Definition \eqref{eq:Ainfty} from Section \ref{sec:SHT}):
\begin{equation}\label{eq:WpPhi-p--Ainfty}
[\sigma,\Phi_p]_{W_p}=\sup_B\frac{1}{\sigma(B)}\int_B M\left(\sigma\chi_B\right)\ d\mu =[\sigma]_{A_\infty}.
\end{equation}

\subsection{Main results}
Our main purpose in the present note is to address the problem mentioned above within the context of spaces of homogeneous type. In this context, the Hardy--Littlewood maximal operator $M$ is defined over balls:
\begin{equation}\label{eq:maximal-SHT}
Mf(x) = \sup_{B\ni x } \frac{1}{\mu(B)}\int_{B} |f|\,d\mu.
\end{equation}

The Orlicz type maximal operators are defined also with balls and with respect to the measure $\mu$ in the natural way.

Our main result is the following theorem.
\begin{theorem} \label{thm:main}
Let  $1 < p < \infty$ and let $\Phi$ be any Young function with conjugate function $\bar\Phi$. Then, for any pair of weights $w,\sigma$,
there exists a structural constant $C>0$ such that the (two weight) norm defined in \eqref{MainQuestion} satisfies
\begin{equation}\label{eq:main}
\|M(\cdot \sigma)\|_{L^p(w)} \leq C p'\left( [w,\sigma,\Phi]_{A_p}[\sigma,\bar\Phi]_{W_p}\right)^{1/p},
\end{equation}
\end{theorem}

We emphasize that \eqref{eq:main}, which is even new in the usual context of Euclidean Spaces, fits into the spirit of the $A_p-A_{\infty}$ theorem derived in \cite{HP} and \cite{HPR1}. The main point here is that we have a two weight result with a better condition and with a  proof that avoids completely the use of the sharp quantitative reverse H\"older inequality  for $A_{\infty}$ weights proved in these papers. This property is, of course, of independent interest but it is not used in our results.

From this Theorem, we derive several corollaries. First, we have a direct proof of the two weight result derived in \cite{HP} using the $[w]_{A_{\infty}}$ constant of Fujii-Wilson \eqref{eq:Ainfty}.

\begin{corollary}\label{cor:mixed-two-weight}
Under the same hypothesis of Theorem \ref{thm:main}, we have that there exists a structural constant $C>0$ such that
\begin{equation}\label{eq:mixed-two}
\|M(\cdot \sigma)\|_{L^p(w)} \leq Cp'\left([w,\sigma]_{A_p}[\sigma]_{A_\infty}\right)^{1/p}.
\end{equation}
\end{corollary}

Note that the result in Theorem \ref{thm:main} involves two suprema like in Corollary \ref{cor:mixed-two-weight}.  It would interesting to find out if there is a version of this result involving only one supremum. There is some evidence that it could be the case, see for example \cite{HP}, Theorem 4.3. See also the recent work \cite{LM}.

As a second consequence of  Theorem \ref{thm:main}, we have the announced quantitative version of Theorem \ref{thm:perez-bump}:

\begin{corollary}\label{cor:precise-bump}
Under the same hypothesis of Theorem \ref{thm:main}, we have that there exists a structural constant $C>0$ such that
\begin{equation*}
\|M(\cdot \sigma)\|_{L^p(w)} \leq C p' [w,\sigma,\Phi]_{A_p}^{1/p} \|M_{\bar\Phi}\|_{L^p(\mathbb{R}^n)}
\end{equation*}
\end{corollary}
We remark that this approach produces a non-optimal dependence on $p$, since we have to pay with one $p'$ for using Sawyer's theorem. However, the ideas from the proof of Theorem \ref{thm:main} can be used to derive a direct proof of Corollary \ref{cor:precise-bump} without the $p'$ factor. We include the proof in the appendix.

Finally, for the one weight problem, we recover the known mixed bound.
\begin{corollary}\label{cor:mixed-one-weight}
For any $A_p$ weight $w$ the following mixed bound holds:
\begin{equation*}
\|M\|_{L^p(w)} \leq  C p' \left([w]_{A_p}[\sigma]_{A_\infty}\right)^{1/p}
\end{equation*}
where $C$ is an structural constant and as usual $\sigma=w^{1-p'}$ is the dual weight.
\end{corollary}

\begin{remark}
To be able to extend the proofs to this general scenario, we need to use (and prove) suitable versions of classical tools on this subject, such as Calder\'on--Zygmund decompositions. We remark that in previous works (\cite{PW-JFA}, \cite{SW}) most of the results are proved under the assumption that the space has non-empty annuli. The main consequence of this property is that in that case the measure $\mu$ enjoys a reverse doubling property, which is crucial in the proof of Calder\'on--Zygmund type lemmas. However, this assumption implies, for instance, that the space has infinite measure and no atoms (i.e. points with positive measure) and therefore constraints the family of spaces under study.  Recently, some of those results were proven without this hypothesis, see for example \cite{Pradolini-Salinas}. Here we choose to work without the annuli property and therefore we need to adapt the proofs from \cite{PW-JFA}. Hence, we will need to consider separately the cases when the space has finite or infinite measure. An important and useful result on this matter is the following:
\end{remark}

\begin{lemma}[\cite{vili}]\label{lem:bounded-finite}
Let  $(\mathcal S,\rho,\mu)$ be a space of homogeneous type. Then $\mathcal{S}$ is  bounded if and only if $\mu(\mathcal S)<\infty$.
\end{lemma}

\subsection{Outline}
The article is organized as follows. In Section \ref{sec:prelim} we summarize some basic needed results on spaces of homogeneous type and Orlicz spaces. We also include a Calder\'on--Zygmund type decomposition lemma. In Section \ref{sec:proofs} we present the proofs of our results. Finally, we include in Section  \ref{sec:appendix} an Appendix with a direct proof of a slightly better result than Corollary \ref{cor:mixed-two-weight}.

\section{preliminaries}\label{sec:prelim}

In this section we first summarize some basic aspects regarding spaces of homogeneous type and Orlicz spaces. Then, we include a Calder\'on--Zygmund (C--Z) decomposition lemma adapted to our purposes.

\subsection{Spaces of homogeneous type}\label{sec:SHT}
A quasimetric $d$ on a set $\mathcal{S}$ is a function $d:{\mathcal S} \times
{\mathcal S} \rightarrow [0,\infty)$ which satisfies
\begin{enumerate}
 \item $d(x,y)=0$ if and only if $x=y$;
\item $d(x,y)=d(y,x)$ for all $x,y$;
 \item there exists a finite constant $\kappa \ge 1$ such that, for all $x,y,z \in \mathcal{S}$,
\begin{equation*}
d(x,y)\le \kappa (d(x,z)+d(z,y)).
\end{equation*}
\end{enumerate}

Given $x \in \mathcal{S}$ and $r > 0$,  we define the ball with center $x$ and radius $r$, $B(x,r) := \{y \in {\mathcal{S}} :d(x,y) < r\}$ and we denote its radius $r$ by $r(B)$ and its center $x$ by $x_B$.
A space of homogeneous type $({\mathcal{S}},d,\mu)$ is a set $\mathcal{S}$ endowed with a quasimetric $d$ and a doubling  nonnegative Borel measure $\mu$ such that
\begin{equation}\label{eq:doubling}
 \mu(B(x,2r)) \le C\mu(B(x,r))
\end{equation}

Let $C_\mu$ be the smallest constant satisfying \eqref{eq:doubling}. Then $D_\mu = \log_2 C_\mu$ is called the doubling order of $\mu$. It follows that
\begin{equation}
\frac{\mu(B)}{\mu(\tilde{B})} \le
C^{2+\log_2\kappa}_{\mu}\left(\frac{r(B)}{r(\tilde{B})}\right)^{D_\mu} \;\mbox{for all
balls}\; \tilde{B} \subset B.
\end{equation}
In particular for $\lambda>1$ and $B$ a  ball, we have that
\begin{equation}\label{eq:doublingDIL}
 \mu(\lambda B) \le (2\lambda)^{D_\mu}   \mu(B).
\end{equation}
Here, as usual, $\lambda B$ stands for  the dilation of a ball $B(x,\lambda r)$ with $\lambda>0$.
Throughout this paper, we will say that a constant $c=c(\kappa,\mu)>0$ is a \emph{structural constant} if it  depends only on the quasimetric constant $\kappa$ and the doubling constant $C_\mu$.

An elementary but important property of the quasimetric is the following. Suppose that we have two balls $B_1=B(x_1,r_1)$ and $B_2=B(x_2,r_2)$ with non empty intersection. Then,
\begin{equation}\label{eq:engulfing}
 r_1\le r_2 \Longrightarrow B_1\subset \kappa(2\kappa+1)B_2.
\end{equation}
This is usually known as the ``engulfing'' property and follows directly from the quasitriangular property of the quasimetric.

In a general space of homogeneous type, the balls $ B(x,r)$ are not necessarily open, but by a theorem of
Macias and Segovia \cite{MS}, there is a continuous quasimetric
$d'$ which is equivalent to $d$ (i.e., there are positive
constants $c_{1}$ and $c_{2}$ such that $c_{1}d'(x,y)\le d(x,y)
\le c_{2}d'(x,y)$ for all $x,y \in \mathcal{S}$) for which every ball is
open. We always assume that the quasimetric $d$ is continuous and
that balls are open.

We will adopt the usual notation: if $\nu$ is a measure and $E$ is a measurable set, $\nu(E)$ denotes the $\nu$-measure of $E$. Also, if $f$ is a measurable function on $(\mathcal S,d,\mu)$ and $E$ is a measurable set, we will use the notation $f(E):=\int_E f(x)\ d\mu$.
We also will denote the $\mu$-average of $f$ over a ball $B$ as $f_{B} = \Xint-_B f d\mu$.
We recall that a weight $w$ (any non negative measurable function) satisfies the $A_p$ condition for $1<p<\infty$ if
\begin{equation*}
   [w]_{A_p}:=\sup_B\left(\Xint-_B w\ d\mu\right)\left(\Xint-_B w^{-\frac{1}{p-1}}\ d\mu\right)^{p-1},
\end{equation*}
where the supremum is taken over all the balls in $\mathcal{S}$. The $A_{\infty}$ class is defined in the natural way by $A_{\infty}:=\bigcup_{p>1}A_p$

This class of weights can also be characterized by means of an appropriate constant. In fact, there are various different definitions of this constant, all of them equivalent in the sense that  they define the same class of weights. Perhaps the more classical and known definition is the following  due  to Hru\v{s}\v{c}ev
\cite{Hruscev} (see also \cite{GCRdF}):
\begin{equation*}
[w]^{exp}_{A_\infty}:=\sup_B \left(\Xint-_{B} w\,d\mu\right) \exp \left(\Xint-_{B} \log w^{-1}\,d\mu  \right).
\end{equation*}
However, in \cite{HP} the authors use a ``new'' $A_\infty$ constant (which was originally introduced implicitly  by Fujii in \cite{Fujii} and later by Wilson in \cite{Wilson:87}), which seems to be better suited. For any $w\in A_\infty$, we define
 \begin{equation}\label{eq:Ainfty}
     [w]_{A_\infty}:= [w]^{W}_{A_\infty}:=\sup_B\frac{1}{w(B)}\int_B M(w\chi_B )\ d\mu,
 \end{equation}
where $M$ is the usual Hardy--Littlewood maximal operator. When the underlying space is $\mathbb{R}^d$, it is easy to see that $[w]_{A_\infty}\le c [w]^{exp}_{A_\infty}$ for some structural $c>0$.  In fact, it is shown in \cite{HP} that there are examples  showing that $[w]_{A_\infty}$ is much smaller than $[w]^{exp}_{A_\infty}$
The same line of ideas yields the inequality in this wider scenario. See the recent work of Beznosova and Reznikov \cite{BR} for a comprehensive and thorough study of these different $A_\infty$ constants.
We also refer the reader to the forthcoming work of Duoandikoetxea, Martin-Reyes and Ombrosi  \cite{DMRO} for a discussion regarding different definitions of $A_\infty$ classes.

\subsection{Orlicz spaces}\label{sec:Orlicz}
We recall here some basic definitions and facts about Orlicz spaces.

A function $\Phi:[0,\infty) \rightarrow [0,\infty)$ is called a Young function if it is continuous, convex, increasing and satisfies  $\Phi(0)=0$ and $\Phi(t) \rightarrow \infty$ as $t \rightarrow \infty$.  For Orlicz spaces, we are usually only concerned about the behaviour of Young functions for $t$ large.
The space $L_{\Phi}$ is a Banach function space with the Luxemburg norm
\[
\|f\|_{\Phi} =\|f\|_{\Phi,\mu} =\inf\left\{\lambda >0: \int_{\mathcal{S}}
\Phi( \frac{ |f|}{\lambda }) \, d\mu \le 1 \right\}.
\]
Each Young function $\Phi$ has an associated complementary Young function $\bar{\Phi}$ satisfying
\begin{equation*}
t\le \Phi^{-1}(t)\bar{\Phi}^{-1}(t) \le 2t \label{propiedad}
\end{equation*}
for all $t>0$. The function $\bar{\Phi}$ is called the conjugate
of $\Phi$, and the  space $L_{\bar{\Phi}}$ is called the conjugate
space of $L_{\Phi}$. For example, if $\Phi(t) = t^p$ for $1 < p <
\infty$, then $\bar{\Phi}(t) = t^{p'}, p' = p/(p-1)$, and the
conjugate space of $L^p(\mu)$ is $L^{p'}(\mu)$.

A very important property of Orlicz spaces is the generalized
H\"older inequality
\begin{equation}\label{eq:HOLDERglobal}
\int_{\mathcal{S}} |fg|\, d\mu \le 2 \|f\|_{\Phi}\|g\|_{\bar{\Phi}}.
\end{equation}
Now we introduce local versions of Luxemburg norms. If $\Phi$ is a Young function, let
\begin{equation*}
\|f\|_{\Phi,B} =\|f\|_{\Phi,B,\mu}= \inf\left\{\lambda >0:
\frac{1}{\mu(B)}\int_{B} \Phi\left(\frac{ |f|}{ \lambda }\right)  \,
d\mu \le 1\right\}.
\end{equation*}
Furthermore, the local version of the generalized H\"older inequality (\ref{eq:HOLDERglobal}) is
\begin{equation}\label{eq:HOLDERlocal}
\frac{1}{\mu(B)}\int_{B}fg\, d\mu \le 2 \|f\|_{\Phi,B}\|g\|_{\bar{\Phi},B}.
\end{equation}
Recall the definition of the maximal type operators $M_\Phi$ from \eqref{eq:maximaltype}:
\begin{equation}\label{eq:maximaltype-SHT}
M_{\Phi}f(x)= \sup_{B:x\in B} \|f\|_{\Phi,B}.
\end{equation}
An important fact related to this sort of operator is that its boundedness is related to the so called $B_p$ condition. For any  positive function  $\Phi$ (not necessarily a Young function), we have that
\begin{equation*}
\|M_{\Phi}\|^p_{L^{p}(\mathcal{S})}\,   \leq c_{\mu,\kappa}\, \alpha_{p}(\Phi),
\end{equation*}
where $\alpha_{p}(\Phi)$ is the following tail condition
\begin{equation}\label{eq:Phi-p}
\alpha_{p}(\Phi)= \,\int_{1}^{\infty} \frac{\Phi(t)} { t^p }
\frac{dt}{t} < \infty.
\end{equation}
It is worth noting that in the recent article \cite{LL} the authors define the appropriate analogue of the $B_p$ condition in order to characterize the boundedness of the \emph{strong} Orlicz-type maximal function  defined over rectangles both in the linear and multilinear cases. Recent developments and improvements can also be found in \cite{Masty-Perez}, where the authors addressed the problem of studying the maximal operator between Banach function spaces.

\subsection{Calder\'on--Zygmund decomposition for spaces of homogeneous type}

\

The following lemma is a classical result in the theory, regarding a decomposition of a generic level set of the Hardy--Littlewood maximal function $M$. Some variants can be found in \cite{AimarPAMS} for $M$ and in \cite{AimarTAMS} for the centered maximal function $M^c$ . In this latter case, the proof is straightforward. We include here a detailed proof for the general case of $M$ where some extra subtleties are needed.

\begin{lemma}[Calder\'on--Zygmund decomposition]\label{lem:stoppingtime}
Let $B$ be a fixed ball and let $f$ be a bounded nonnegative measurable function. Let $M$ be the usual non centered Hardy--Littlewood maximal function. Define the set $\Omega_{\lambda}$ as
\begin{equation}\label{eq:omegalambda}
\Omega_\lambda = \{x \in  B:  Mf(x) >\lambda\},
\end{equation}
Let $\lambda>0$ be such that $\lambda\ge \Xint-_B f\ d\mu$. If $\Omega_{\lambda}$ is non-empty, then given $\eta > 1$, there exists a countable family
$\{B_i\}$ of pairwise disjoint balls such that, for $\theta=4\kappa^2+\kappa$,
\begin{itemize}
\item[i)] $\displaystyle \cup_{i} B_{i}\subset \Omega_{\lambda} \subset
\cup_{i} \theta B_{i}$,
\item[ii)] For all $i$,
\begin{equation*}
\lambda <\frac{1}{\mu(B_i)} \int_{B_i} f d\mu.
\end{equation*}
\item[iii)] If $B$ is any ball such that $B_i\subset B$ for some $i$ and $r(B)\ge \eta r(B_i)$, we have that
\begin{equation}\label{eq:ballmaximal1}
\frac{1}{\mu(\eta B)} \int_{\eta B} f d\mu  \le \lambda.
\end{equation}
\end{itemize}
\end{lemma}

\begin{proof}
Define, for each $x\in\Omega_\lambda$, the following set:
 \begin{equation*}
 \mathcal{R}^\lambda_x=\left\{r>0: \Xint-_{B} f\ d\mu >\lambda, x\in B=B(y,r)\right\},
 \end{equation*}
which is clearly non-empty. The key here is to prove that $\mathcal{R}^\lambda_x$ is bounded. If the whole space is bounded, there is nothing to prove. In the case of unbounded spaces, we argue as follows.
Since the space is of infinite measure (recall Lemma \ref{lem:bounded-finite}), and clearly $S=\bigcup_{r>0} B(x,r)$,
we have that $\mu(B(x,r))$ goes to $+\infty$ when $r\to\infty$ for any $x\in \mathcal{S}$. Therefore, for $K=\kappa(2\kappa+1)$, we can choose $r_1$ such that the ball $B_1=B(x,r_1)$ satisfies the inequality
\begin{equation*}
\mu(B_1)\ge \frac{2(2K)^{D_\mu}\|f\|_{L^1}}{\lambda}
\end{equation*}
Suppose now that $\sup\mathcal{R}^\lambda_x=+\infty$. Then we can choose a ball $B_2=B(y,r_2)$ for some $y$ such that $x\in B_2$, $\Xint-_{B_2}f\ d\mu>\lambda$ and $r_2>r_1$. Now, by the engulfing property \eqref{eq:engulfing} we obtain that $B_1\subset KB_2$. The doubling condition \eqref{eq:doublingDIL} yields
\begin{equation*}
 \mu(B_1)\le \mu(KB_2)\le (2k)^{D_\mu}\mu(B_2)
\end{equation*}
Then we obtain that
\begin{equation*}
\frac{2\|f\|_{L^1}}{\lambda}\le \mu(B_2)
<  \frac{\|f\|_{L^1}}{\lambda}
\end{equation*}
which is a contradiction. We conclude that, in any case, for any $x\in \Omega_\lambda$, we have that $\sup \mathcal{R}^\lambda_x<\infty$.

Now fix $\eta>1$. If $x \in \Omega_\lambda$, there is a ball $B_{x}$ containing $x$, whose radius $r(B_x)$ satisfies $\frac{\sup \mathcal{R}^\lambda_x}{\eta} < r(B_x)\leq \sup \mathcal{R}^\lambda_x$, and for which $\Xint-_{B_x} f \ d\mu > \lambda$. Thus the ball $B_x$ satisfies ii) and iii). Also
note that $\Omega_\lambda = \bigcup_{x\in \Omega_{\lambda}} B_x$.
Picking a Vitali type subcover of $\{B_{x}\}_{x\in
\Omega_{\lambda}}$ as in \cite{SW}, Lemma 3.3, we obtain a
family of pairwise disjoint balls $\{B_{i}\} \subset
\{B_{x}\}_{x\in \Omega_{\lambda}}$ satisfying i).  Therefore
$\{B_i\}$ satisfies i), ii) and iii).
\end{proof}

We will need another important lemma, in order to handle simultaneously decompositions of level sets at different scales.

\begin{lemma}\label{lem:disjointing}
Let $B$ be a ball and let $f$ be a bounded nonnegative measurable function. Let also $a \gg 1$ and, for each integer $k$ such that $a^k>\Xint-_B f\ d\mu$, we define $\Omega_{k}$ as
\begin{equation}\label{eq:Omega-k}
\Omega _{k} = \left\{x\in B: Mf(x) >a^{k}  \right\},
\end{equation}
Let $\{E_i^k\}_{i,k}$ be defined by $E_i^k=B_i^k\setminus \Omega_{k+1}$, where the family of balls $\{B_i^k\}_{i,k}$ is obtained by applying Lemma \ref{lem:stoppingtime} to each $\Omega_k$.
Then, for $\theta=4\kappa^2+\kappa$ as in the previous Lemma and $\eta=\kappa^2(4\kappa+3)$, the following inequality holds:
\begin{equation}\label{eq:Bik vs Eik}
\mu(B_i^k\cap \Omega_{k+1})< \frac{(4\theta\eta)^{D_\mu}}{a}\mu(B_i^k).
\end{equation}
Consequently, for sufficiently large $a$, we can obtain that
\begin{equation}\label{eq:Bik vs Eik one half}
\mu(B_i^k) \le 2\mu(E_i^k).
\end{equation}
\end{lemma}

\begin{proof}
To prove the claim, we apply Lemma \ref{lem:stoppingtime} with $\eta=\kappa^2(4\kappa+3)$. Then, by part i), we have that, for $\theta=4\kappa^2+\kappa$
 \begin{equation*}
\Omega_{k+1}\subset \bigcup_m\theta B^{k+1}_m
\end{equation*}
and then
\begin{equation}\label{eq:decompBik-k+1}
\mu(B_{i}^{k}  \cap \Omega_{k+1} )\le \sum_{m} \mu( B_{i}^{k} \cap
\theta B_{m}^{k+1} ).
\end{equation}
Suppose now that $B_i^k\cap \theta B_m^{k+1}\neq \emptyset$. We claim that $r(B_{m}^{k+1})\le r(B_{i}^{k})$. Suppose the contrary, namely $r(B_{m}^{k+1})> r(B_{i}^{k})$. Then, by property \eqref{eq:engulfing}, we can see that $B_{i}^{k} \subset \kappa^2(4\kappa+3) B_{m}^{k+1}=\eta B_{m}^{k+1}$.
For $B=\eta B_{m}^{k+1}$, part iii) from Lemma \ref{lem:stoppingtime} gives us that the average satisfies
\begin{equation}\label{eq:avg-etaBmk+1}
 \frac{1}{\mu(B)}\int_B f\ d\mu\le a^k.
\end{equation}
Now, by the properties of the family $\{B_m^{k+1}\}_m$ and the doubling condition of $\mu$, we have that, for $a>(2\eta)^{D_\mu}$,
\begin{equation}\label{eq:ak}
\frac{1}{\mu(\eta B_{m}^{k+1})}\int_{\eta B_{m}^{k+1}} f\ d\mu>\frac{a^{k+1}}{ (2\eta)^{D_\mu}}>a^k.
\end{equation}
This last inequality contradicts \eqref{eq:avg-etaBmk+1}. Then, whenever $B_i^k\cap \theta B_m^{k+1}\neq \emptyset$, we have that
$r(B_{m}^{k+1})\le r(B_{i}^{k})$ and from that it follows that $ B_{m}^{k+1}\subset \eta B_{i}^{k}$. The sum \eqref{eq:decompBik-k+1} now becomes
\begin{eqnarray*}
\mu(B_{i}^{k}  \cap \Omega_{k+1} )& \le & \sum_{m:B_{m}^{k+1}\subset \eta B_{i}^{k}} \mu( B_{j}^{k} \cap \theta B_{m}^{k+1} )\\
& \le &  (2\theta)^{D_\mu} \sum_{m:B_{m}^{k+1}\subset \eta B_{i}^{k}} \mu(B_{m}^{k+1} )\\
&\le & \frac{(2\theta)^{D_\mu}}{a^{k+1}}\int_{\eta B_i^k}f\ d\mu
\end{eqnarray*}
since the sets $\{B_m^{k+1}\}_m$ are pairwise disjoint. Finally, by part iii) of Lemma \ref{lem:stoppingtime}, we obtain
\begin{equation*}
 \mu(B_{i}^{k}  \cap \Omega_{k+1})\le \frac{(4\theta\eta)^{D_\mu}}{a}\mu(B_i^k),
\end{equation*}
which is inequality \eqref{eq:Bik vs Eik}.
\end{proof}

\section{Proofs of the main results}\label{sec:proofs}

We present here the proof or our main results. Our starting point is a version of the sharp two weight inequality \eqref{eq:moen} valid for SHT from \cite{kairema:twoweight}:

\begin{theorem}[\cite{kairema:twoweight}]\label{thm:kairema}
Let $(\mathcal{S},\rho,\mu)$ a SHT. Then the H--L maximal operator $M$ defined by \eqref{eq:maximal-SHT} satisfies the bound
\begin{equation}\label{eq:kairema}
 \left\|M(f\sigma)\right\|_{L^p(w)}\le C p'[w,\sigma]_{S_p}\|f\|_{L^p(\sigma)},
\end{equation}
where $[w,\sigma]_{S_p}$ is the Sawyer's condition with respect to balls:
\begin{equation}
[w,\sigma]_{S_p}:=\sup_B  \left(\frac1{\sigma(B)} \int_B M(\sigma\chi_B)^pw\ d\mu\right)^{1/p}.
\end{equation}
\end{theorem}

We now present the proof of the main result.
\begin{proof}[Proof of Theorem \ref{thm:main}]
By Theorem \ref{thm:kairema}, we only need to prove that
\begin{equation*}
 [w,\sigma]_{S_p}\le C [w,\sigma,\Phi]^{1/p}_{A_p}[\sigma,\bar\Phi]^{1/p}_{W_p}
\end{equation*}
for some constant $C$, for any Young function $\Phi$, for any $1<p<\infty$. Let $B$ be a fixed ball $B$ and consider the sets $\Omega_k$ from \eqref{eq:Omega-k} for the function $\sigma\chi_B$ for any $k\in \mathbb{Z}$. We remark here that in order to apply a C--Z decomposition of these sets, we need the level of the decomposition to be larger that the average over the ball. We proceed as follows. Take any $a>1$ and consider $k_0\in\mathbb{Z}$ such that
\begin{equation}\label{eq:small-average}
a^{k_0-1}< \Xint-_B \sigma\ d\mu \le a^{k_0}.
\end{equation}
Now, let $A$ be the set of the small values of the maximal function:
\begin{equation*}
 A=\left\{x\in B: M(\sigma\chi_B)\le a\Xint-_B \sigma\ d\mu\right\}.
\end{equation*}

For any $x\in B\setminus A$, we have that
\begin{equation*}
 M(\sigma\chi_B)(x)> a\Xint-_B \sigma\ d\mu>a^{k_0}\ge\Xint-_B \sigma\ d\mu.
\end{equation*}

Therefore,
\begin{eqnarray*}
\int_B M(\sigma\chi_B)^p w \ d\mu  & =  & \int_A M(\sigma\chi_B)^p w \ d\mu  +\int_{B\setminus A} M(\sigma\chi_B)^p w \ d\mu  \\
& \le & a^p w(B) \left(\Xint-_{B} \sigma\ d\mu\right)^p + \sum_{k\ge k_0} \int_{ \Omega_{k}\setminus \Omega_{k+1}} M(\sigma\chi_B)^p w\ d\mu\\
&= & I + II
\end{eqnarray*}

The first term $I$ can be bounded easily. By the general H\"older inequality \eqref{eq:HOLDERlocal}, we obtain
\begin{eqnarray*}
 I & \le & 2a^p\left(\Xint-_B w\ d\mu \right)\|\sigma^{1/p'}\|^p_{\Phi,B}\|\sigma^{1/p}\|^p_{\bar\Phi,B}\ \mu(B)\\
&\le & 2[w,\sigma,\Phi]_{A_p}\int_B M_{\bar\Phi}(\sigma^{1/p}\chi_B)^p\ d\mu
\end{eqnarray*}

Now, for the second term $II$, we first note that

\begin{eqnarray*}
\int_{B\setminus A} M(\sigma\chi_B)^p w \ d\mu & = & \sum_{k\ge k_0} \int_{ \Omega_{k}\setminus \Omega_{k+1}} M(\sigma\chi_B)^p w\ d\mu\\
& \le & a^p\sum_{k\ge k_0} a^{kp} w(\Omega_{k})
\end{eqnarray*}

By the choice of $k_0$, we can apply Lemma \ref{lem:stoppingtime} to perform a C--Z decomposition at all levels $k\ge k_0$ and obtain a family of balls $\{B^k_i\}_{i,k}$ with the properties listed in that lemma. Then,

\begin{eqnarray*}
\int_{B\setminus A} M(\sigma\chi_B)^p w \ d\mu &\le & a^{p} \sum_{k,i} \left(\Xint-_{B_i^k}\sigma\chi_B\ d\mu \right)^{p} w(\theta B_i^k)\\
&\le & a^{p} \sum_{k,i} \left(\frac{\mu(\theta B_i^k)}{\mu(B_i^k)}\Xint-_{\theta B_i^k}\sigma^\frac{1}{p}\sigma^\frac{1}{p'}\chi_B\ d\mu \right)^{p} w(\theta B_i^k)
 \end{eqnarray*}
We now proceed as before, using  the local generalized Holder inequality \eqref{eq:HOLDERlocal} and  the doubling property \eqref{eq:doublingDIL} of the measure (twice). Then we obtain
\begin{equation*}
 \int_{B\setminus A} M(\sigma\chi_B)^p w  d\mu \le 2a^{p} (2\theta)^{(p+1)D_\mu}[w,\sigma,\Phi]_{A_p}\sum_{k,i}\left\| \sigma^\frac{1}{p}\chi_B\right\|^p_{\bar{\Phi},\theta B_i^k}\mu(B_i^k)
\end{equation*}
The key here is to use Lemma \ref{lem:disjointing} to pass from the family $\{B_i^k\}$ to the pairwise disjoint family $\{E_i^k\}$. Then, for $a\ge 2(4\theta\eta)^{D_\mu}$, we can bound the last sum as follows
\begin{eqnarray*}
\sum_{k,i}\left\| \sigma^\frac{1}{p}\chi_B\right\|^p_{\bar{\Phi},\theta B_i^k}\mu(B_i^k)& \le &  2 \sum_{k,i}\left\| \sigma^\frac{1}{p}\chi_B\right\|^p_{\bar{\Phi},\theta B_i^k}\mu(E_i^k)\\
&\le&  2 \sum_{k,i} \int_{E_i^k}M_{\bar{\Phi}}(\sigma^\frac{1}{p}\chi_B)^p\ d\mu\\
&\le&  2  \int_B M_{\bar{\Phi}}(\sigma^\frac{1}{p}\chi_B)^p\ d\mu
\end{eqnarray*}
since the sets $ \{ E_{k,j}\}$ are pairwise disjoint. Collecting all previous estimates and dividing by $\sigma(B)$, we obtain the desired estimate
\begin{equation*}
 [w,\sigma]^p_{S_p}\le 4 a^{p} (2\theta)^{(p+1)D_\mu}[w,\sigma,\Phi]_{A_p} [\sigma,\bar\Phi]_{W_p},
\end{equation*}
and the proof of Theorem \ref{thm:main} is complete.
\end{proof}

It remains to prove Corollary \ref{cor:mixed-two-weight}. To that end, we need to consider the special case of $\Phi(t)=t^{p'}$.

\begin{proof}[Proof of Corollary \ref{cor:mixed-two-weight}]
Considering then  $\Phi(t)=t^{p'}$, the quantity \eqref{eq:A_p-local} is
\begin{eqnarray*}
A_p(w,\sigma,B,\Phi) & = &\left( \Xint-_{B} w\, d\mu\right)\|\sigma^{1/p'}\|^p_{\Phi,B} \\
& = & \left( \Xint-_{B} w(y)\, d\mu\right) \left( \Xint-_{B} \sigma \, d\mu\right)^{p-1}.
\end{eqnarray*}
In addition, we have from \eqref{eq:WpPhi-p--Ainfty} that $[\sigma,\overline{\Phi_{p'}}]_{W_p}=[\sigma,\Phi_p]_{W_p}=[\sigma]_{A_\infty}$  and therefore we obtain \eqref{eq:mixed-two}.
\end{proof}

For the proof of Corollary \ref{cor:precise-bump}, we simply use the boundedness of $M_{\bar\Phi}$  on $L^p(\mu)$,
\begin{equation*}
 [\sigma,\bar\Phi]_{W_p}:=\sup_B\frac{1}{\sigma(B)}\int_B M_{\bar\Phi}\left(\sigma^{1/p}\chi_B\right)^p\ d\mu
 \leq \|M_{\bar\Phi}\|^p_{L^p}.
\end{equation*}

The proof of Corollary \ref{cor:mixed-one-weight} is trivial.

\section{Appendix}\label{sec:appendix}

We include here a direct proof of version of Corollary \ref{cor:precise-bump} which is better in terms of the dependence on $p$. Precisely, we have the following Proposition.

\begin{proposition}\label{pro:precise-bump-sharp-p}
 Let  $1 < p < \infty$. For any pair of weights $w,\sigma$ and any Young function $\Phi$,
there exists a structural constant $C>0$ such that
\begin{equation*}
\|M (f\sigma)\|_{L^p(w)}\leq C [w,\sigma,\Phi]^{1/p}_{A_p} \|M_{\bar\Phi}\|_{L^p}\|f\|_{L^{p}(\sigma)}
\end{equation*}
\end{proposition}

\begin{proof}[Proof of Proposition  \ref{pro:precise-bump-sharp-p}]
By density it is enough to prove the inequality for each nonnegative bounded function with compact support  $f$. We first consider the case of unbounded $S$. In this case we have $\Xint-_S f\sigma\ d\mu=0$. Therefore, instead of the sets from sets from \eqref{eq:Omega-k}, we consider
\begin{equation*}\label{eq:Omega-k-global}
\Omega _{k} = \left\{x\in \mathcal{S}: M(f\sigma)(x) >a^{k}  \right\},
\end{equation*}
for any $a>1$ and any $k\in \mathbb{Z}$. Then, we can write

\begin{equation*}
\int_{\mathcal S} M(f\sigma)^p w \ d\mu = \sum_{k} \int_{ \Omega_{k}\setminus \Omega_{k+1}} M(f\sigma)^p w\ d\mu
\end{equation*}
Then, following the same line of ideas as in the proof of Theorem \ref{thm:main}, we obtain
\begin{equation*}
 \int_{\mathcal S} M(f\sigma)^p w  d\mu   \le 2a^{p} (2\theta)^{(p+1)D_\mu}[w,\sigma,\Phi]_{A_p} \sum_{k,i}\left\| f\sigma^\frac{1}{p}\right\|^p_{\bar{\Phi},\theta B_i^k}\mu(B_i^k)
\end{equation*}
By Lemma \ref{lem:disjointing} we can replace the family $\{B_i^k\}$ by the pairwise disjoint family $\{E_i^k\}$ to obtain the desired estimate:
\begin{equation*}
 \int_{\mathcal S} M(f\sigma)^p w \ d\mu \le 4a^{p} (2\theta)^{(p+1)D_\mu}[w,\sigma,\Phi]_{A_p}  \|M_{\bar{\Phi}}\|_{L^p}^p\int_{S}f^p\sigma\ d\mu.
\end{equation*}
In the bounded case, the whole space is a ball and we can write $\mathcal S=B(x,R)$ for any $x$ and some $R>0$. The problem here is to deal with the small values of $\lambda$, since we cannot apply Lemma \ref{lem:disjointing} for $a^k\le \Xint-_S f\sigma\ d\mu$.  We then take any $a>1$ and consider $k_0\in\mathbb{Z}$ to verify \eqref{eq:small-average}:
\begin{equation*}
a^{k_0-1}< \Xint-_S f\sigma\ d\mu \le a^{k_0}
\end{equation*}
and argue as in the proof of Theorem \ref{thm:main}.
\end{proof}

Now, from this last proposition, we can derive another proof of the mixed bound \eqref{eq:mixed-two} from Corollary \ref{cor:mixed-two-weight}. The disadvantage of this approach with respect to the previous one is that we need a deep property of $A_\infty$ weights: the sharp Reverse H\"older Inequality. In the whole generality of SHT, we only know a \emph{weak} version of this result from the recent paper \cite{HPR1}:
\begin{theorem}[Sharp  weak Reverse H\"older Inequality, \cite{HPR1}]\label{thm:SharpRHI}
Let $w\in A_\infty$. Define the exponent $r(w)=1+\frac{1}{\tau_{\kappa\mu}[w]_{A_{\infty}}}$,
where $\tau_{\kappa\mu}$ is an structural constant.
Then,
\begin{equation*}
  \left(\Xint-_B w^{r(w)}\ d\mu\right)^{1/r(w)}\leq 2(4\kappa)^{D_\mu}\Xint-_{2\kappa B} w\ d\mu,
\end{equation*}
where $B$ is any ball in $\mathcal S$.
\end{theorem}
The other ingredient for the alternative proof of Corollary \ref{cor:mixed-two-weight} is  the known estimate for the operator norm for $M$. For any $1<q<\infty$, we have that $\|M\|^q_{L^q}\sim q'$.

\begin{proof}[Another proof of Corollary \ref{cor:mixed-two-weight}]
Consider the particular choice of $\Phi(t)=t^{p'r}$ for $r>1$. Then  quantity \eqref{eq:A_p-local} is

\begin{equation*}
A_p(w,\sigma,B,\Phi) =\left( \Xint-_{B} w(y)\, d\mu\right) \left( \Xint-_{B} \sigma^r\, d\mu\right)^{p/rp'}
\end{equation*}
If we choose $r$ from the sharp weak reverse H\"older property (Theorem \ref{thm:SharpRHI}), we obtain that
\begin{eqnarray*}
A_p(w,\sigma,B,\Phi) & = &
\left( \Xint-_{B} w\ d\mu\right)\left(2(4\kappa)^{D_\mu}\Xint-_{2\kappa B} \sigma\ d\mu\right)^{p-1}\\
&\le& 2^{p-1}(4\kappa)^{pD_\mu}\left( \Xint-_{2\kappa B} w\ d\mu\right)\left(\Xint-_{2\kappa B} \sigma\ d\mu\right)^{p-1}\\
&\le & 2^{p-1}(4\kappa)^{pD_\mu}[w,\sigma]_{A_p}
\end{eqnarray*}
And therefore the proof of Proposition \ref{pro:precise-bump-sharp-p} gives
\begin{equation*}
\|M (f\sigma)\|_{L^p(w)} \leq C [w,\sigma]_{A_p}^{1/p}  \|M_{\bar{\Phi}}\|_{L^{p}(\mathcal{S},d\mu)}  \, \|f\|_{L^{p}(\sigma)}.
\end{equation*}
We conclude with the proof by computing $\|M_{\bar \Phi}\|_{L^p}$ for $\Phi(t)=t^{p'r}$. We use \eqref{eq:Phi-p}, and then we obtain that $\|M_{\bar \Phi}\|^p_{L^p}\le c r'p'$.
But, by the choice of $r$, it follows that $r'\sim [\sigma]_{A_\infty}$ and we obtain \eqref{eq:mixed-two}.
\end{proof}

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