Mac and cheese gets hot hot hot in this buffalo cheddar mac and cheese with garlic croutons and broccoli.
When MAC AND CHEESE MANIA WEEK just so happens to coincide with your birthday week?
Basically the definition of kismet.
Because we know that calories don't count on your birthday week, and calories also don't count on holidays (and what is MAC AND CHEESE MANIA WEEK if not a 120 hour holiday?).
And the last time I checked, a minus minus a minus equals a minus.
So the way I see it is that I'll be BURNING calories this week by eating mac and cheese.
So now you're probably like what is MAC AND CHEESE MANIA WEEK and how can I get involved/justify eating carbs and cheese every day for a week straight.
Well, it's a blogger-made holiday hosted by Rachel and Brandy and sponsored by Door to Door Organics and OXO, that celebrates everyone's favorite comfort food! A whole slew of fabulous bloggers (including yours truly) have come together to create every possible variety of mac and cheese that your hearts could desire.
There will also be great prizes up for grabs, including Door to Door Organics Gift Cards, OXO Gift Baskets, Zak Designs Prize Pack, a year's supply of Cabot cheese, a Vermont Creamery gift basket, a risotto kit from Marx Foods, Barilla Pasta, Anolon Advanced Bronze 4.5 Qt. Tapered Stockpot, Rachael Ray Stoneware Casseroval, Wusthof CLASSIC Studio Block Set, Microplane Graters and Kitchen Tools, a SavorX Starter Block with spices and Le Cordon Bleu by Swissmar oven to table roasters. Wow. That was a mouthful.
Head on over to Rachel's and Brandy's blogs to find out more about how to win these fabulous items and to see what other great recipes have been cooked up for this fun event!
But first, let's discuss this mac and cheese and why you need it in your week. I stumbled across a spicy (no, but REALLY SPICY) buffalo sauce-infused cheddar cheese at Whole Foods last week (and it was on SALE, no less!) and just knew that I needed it in mac and cheese form with little bites of broccoli florets scattered throughout. For health.
I was originally hellbent on making my own breadcrumbs for a breadcrumb crust until I got bitten by a lazy (post 11 mile run) bug and didn't feel like breaking out my food processor. So croutons it was! These are extra special because they are tossed with a roasted garlic-infused olive oil, but if you don't have such a thing hanging around your cabinets, regular olive oil will do!
The resulting mac and cheese was MEGA creamy, MEGA cheesy, and BIG on spice. The.Boy and I sweated through every delicious bite.
And went back for seconds.
Because, birthday/holiday week negative calories. Gotta take advantage while we can.
Toss bread cubes with garlic-infused olive oil. Spread out on a parchment-lined baking sheet. Sprinkle with salt and black pepper. Toast in the oven for 10-15 minutes, or until crispy on the outside.
Bring a pot of salted water to a boil. Add pasta and cook until according to package instructions, adding in the broccoli florets during the last 2 minutes of cooking. Strain.
Meanwhile, Bring 1 cup of milk to a simmer. Whisk together the remaining ½ cup milk and 2 tbsp flour. Pour into the simmering milk and whisk until warm and slightly thickened. Stir in the cheese, simmering until melted.
Toss the cheese sauce with the pasta and broccoli. Serve topped with garlicky croutons.
Spicy & cheesy? Sounds heavenly.
This looks so good! And with all that broccoli it is definitely healthy.
Happy birthday week! You have fun eating all the calories in the world. 😀 And I love that pasta shape. Fun!
Buffalo cheddar cheese?!?!?!?! Ok yes. and obviously in this form!
Yes to this. A thousand times yes. This will be my dinner tonight.
I've never heard of buffalo cheddar cheese – bet my boys would gobble this right up!
This sounds amazing. I may have to make something cheesy myself now to overcome the cravings!
Happy birthday week Joanne! I hope you have a great one!
What a perfect, easy recipe! I've never heard of buffalo infused cheese. Gonna make it a mission to find it now. I always need a good excuse to go to Whole Foods, this is definitely one!
Yummm! I bet the buffalo cheese works wonders here. I can definitely make an Inspiralized version of this!
First off, a big HAPPY BIRTHDAY to you! And I definitely think mac and cheese is an awesome way to celebrate (among other things, of course). Hope you have a wonderful bday week and that this is a wonderful year for you.
Who doesn't love mac and cheese? We Love the ingredients in this recipe! Have a wonderful Birthday too!
Haaaaaaaaaappy birthday week!! I'm so excited for mac n cheese week, I can barely stand it! Those pasta noodles! OMG, those pasta noodles! Never have I ever seen noodles like that and they look like they trap in so much cheesy goodness in their open-face glory. I've never tried buffalo mac and now it's time to get my buffalo mac ON!
The best week ever! I love that you kept the broccoli florets in it 🙂 What a fun idea!
I'm stunned you didn't put some squash in there – THAT would have made it the best birthday dinner! Also? I WANT ALL THE MAC AND CHEESE. Lol. Love yours!
Buffalo cheddar deliciousness is what this is! … and the croutons! I die.
I definitely need this mac and cheese in my life! It sounds/looks amazing!
Happy Birthday Week! (Which is totally a legitimate thing.) I'd like this spicy mac and cheese on my birthday, especially with those crisp croutons. Enjoy this gorgeous cake and live it up on your birthday!
Happy Birthday Joanne, mac n cheese is the perfect way to celebrate birthday week for sure and this looks awesome. Love the broccoli bits.
That looks mega delicious! Love the add of garlic croutons.
I love that you added some buffalo flair to your mac n cheese. I love a little heat and this sounds delicious.
Hot-diggity-dog – happy Week-Long Birthday Celebrations Joanne!
Mac n cheese is my daughters all time fav and she will love these week long mac n cheese fest!
I can definitely handle a week long holiday that involves eating mac and cheese! Love the spiciness!
This sounds so delicious! I have been on a Buffalo kick lately… I'm surprised at myself that I didn't try to do a buffalo mac! Ha! Well you've done it for me. So now I'm off to the store to find that amazing cheese! Also, HAPPY BIRTHDAY!
give me mac and cheese any time (even though I really shouldn't eat all that cheese). I generally pass croutons by – pick them out of my salad, actually, but these look really good! Whole wheat, no?
This is seriously my idea of heaven you know how much I love spicy foods! Dang kids are cramping my style 😉 I put broccoli in my M&C too!
An 11 mile run justifies eating Mac and Cheese for an entire year. I need to get in on the celebrations!
What a great way to celebrate your birthday week. It looks TASTY!!!
We had mac and cheese the past 2 weeks. I'm either way out of the loop, or totally ahead of the game.
I would NEVER call you lazy, Joanne! But the croutons were a great solution.
love the pasta shape, love the broccoli, love the flavors. very nice take on mac and cheese!
I love buffalo flavored cheese! And I love mac and cheese more. What a great way to start your birthday week!
Happy Birthday to you!! I hope it's fantastic!
That looks SO GOOD! I might modify this and use some Cabot Bacon Cheddar I have lurking in my fridge.
I would love to have this for lunch right about now. So delicious!
mike would most definitely add hot sauce to this 🙂 what a great way to start celebrating your birthday week!
BEST BIRTHDAY WEEK EVER! This reminds me of the velveeta broccoli mac we had as kids (yes, I was the weird kid who liked the broccoli one), except 1,000x classier and more flavorful!
Happy Birthday! Buffalo mac and cheese is a fantastic idea. That kick of spice sounds like a great way to change things up.
You cannot go wrong with broccoli and cheddar, why not put it in mac n' cheese form?
What a fun twist on Mac and Cheese! SO yummy – I'm all about anything buffalo chicken! This is such a fun link up!
I love the added spice and the croutons. Well done, Joanne!
Happy Birthday indeed! I need to find some of this cheddar, I have not seen buffalo cheddar before!
what a fun recipe! looks amazing, joanne.
Mac and cheese with CROUTONS?! I need like five bowls!
Happy birthday week! I must try this! Buffalo cheddar? Yum yum..!!
I could have this Mac and cheese dish any time, definitely a delish comfort food.
Love the look of that pasta!
This is stunning, Joanne! I want it for diiiiiinner asap!
Happy B-day week Joanne!!! Hope it's wonderful! I'm so glad you posted the crouton recipe 'cause I have half a loaf of semi-stale multi-grain that needs to get used up. Do you have any idea what kind of pasta that is? I've never seen anything like it. Love the idea of using broccoli with the mac and cheese. Again, Happiest of Birthdays sweetie!
such a simple and great looking mac and cheese!
Happy birthday, Joanne!! This looks amazing!!! Yes to anything buffalo!
I love mac n cheese and this dish sounds like a winner to me!
Happy Birthday week!! I hope it is a great one,. I could eat this mac and cheese. I like this, it not your everyday mac and cheese.
I love the shape of this macaroni—helps to hold onto more cheese!
Happy birthday Joanne and you couldn't find a better way to celebrate than with this delicious pasta dish!
Woohoo! I always love birthday week- I eat whatever I want and say well, of course I have to celebrate, it would be wrong not to. Cheese is always an amazing celebration and this mac and cheese is a great birthday-week-celebration-meal-starter. Woo, that's a mouthful!
I'm totally digging this mac n cheese. All my favourite things in one dish.
You just inspired my love affair with buffalo-flavored-everything to continue! This sounds and looks crazy good! I have to get my hands on some of that cheese!
A whole week of Mac and Cheese sounds like a great week to me! Happy Birthday!
Croutons on mac and cheese!?! I love it!!! Thanks for helping us kick off mac and cheese mania with this amazing recipe!
I feel this recipe was totally made for me. Buffalo cheddar cheese and delicious croutons….yes please! Also, I'm totally going to find that buffalo cheese. I need it in my life!


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How to talk to children about Russia's attack on Ukraine
News about Russia's invasion of Ukraine has dominated headlines for nearly a week. The images seen can be difficult for adults, and even harder for children. Darcy Walker Krause, the Executive Director of the Uplift Center for Greiving Children, joins to offer tips on speaking with children about the ongoing issue.
Weather Authority: Thursday, 5 a.m. forecast
FOX 29's Sue Serio has your Thursday morning weather forecast.
If you're looking for a place to grab some cookies with your favorite Eagles players' faces on them, this Berwyn bakery has you covered.
Weather Authority: 10 p.m. Wednesday forecast
FOX 29 Weather Authority 10 p.m. Wednesday forecast.
Police: Officer shot during West Philadelphia traffic stop expected to recover, 3 suspects arrested
Police Commissioner Danielle Outlaw said an officer who was critically injured when a traffic stop escalated to a shooting Wednesday in West Philadelphia is expected to recover. Three suspects were later arrested in connection to the shooting and their car was recovered.
Eagles fans flocking to Arizona for long Super Bowl weekend
FOX 29's Shaynah Ferreira speaks with Eagles fans at Philadelphia International Airport as they make their way to Arizona for Super Bowl LVII.
Eagles fans drive to Super Bowl LVII in their RV
Mike, Alex, and Shiba catch up with Elizabeth and Harry Reeves - two very dedicated Eagles fans who drove to the game in their RV!
Philadelphia brewery offering a family-friend place to watch the Eagles this Sunday
If you can't make it to Phoenix to cheer on the Eagles, there are plenty of places to watch back home! Wissahickon Brewing Company is getting ready to host the whole family for a Super Bowl watch party.
Over 500 students cheer on the Eagles at Bethel Springs Elementary School
Team spirit is at an all-time high at Bethel Springs Elementary School as students and staff gather for an Eagles pep rally!
You Decide: Philadelphia's Next Mayor
Philadelphia is in the middle of what may be a historic season. With so many candidates in the running for the city's next mayor, it's important to have all the information to make educated decisions.
Stakeholders, Philly Police Commissioner discuss gun violence and resources
Philadelphia leaders came together to tackle the gun violence crisis and the overall message is police and community leaders are working together to prevent gun violence and resources are available for survivors.
Dr. Martin Luther King Jr.: The Dream, Then and Now
It's been 60 years since the "I Have a Dream Speech," and 55 years since Martin Luther King Jr. was assassinated. To celebrate MLK Day, FOX 29 talks with local people who had their own personal experiences with Dr. King.
Documentary 'Trigger' works to address gun violence in Philadelphia
Jamie Brunson, Gabriel Bryant and Ramon Cruz Jr. join "Good Day Philadelphia" to discuss the documentary "Trigger."
Video: $20,000 reward offered for suspect with 'distinctive limp' in deadly Frankford shooting

redpajama_set_name: RedPajamaCommonCrawl


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You are here: Home > Things To See & Do > Castlefield Gallery
Castlefield Gallery
Type:Gallery
2 Hewitt Street
Knott Mill
M15 4GB
Nearest Tram Stop
Founded in 1984, the gallery is at the forefront of supporting emerging artists and new art practice in the UK.
Beyond displaying a diverse range of contemporary art across up to seven temporary exhibitions per year and presenting offsite projects, the gallery runs New Art Spaces; pop-up project spaces for artists and creatives across Greater Manchester, and CG Associates, a professional development scheme for artists, writers and independent curators.
Ticket Tariff
Adult Free
Note: Prices are a guide only and may change on a daily basis.
By Metrolink Tram:To Deansgate/Castlefield.By Bus:The No.2 Metroshuttle, the free city centre bus, takes a circular route from Victoria Train Station. Nearest stop to the gallery is on Whitworth Street West by Deansgate Train Station.By Train:To Deansgate Train Station.Alternatively, to Manchester Piccadilly, Manchester Oxford Road or Manchester Victoria Train Stations. Take the Metrolink from Victoria or Piccadilly (Altrincham or Eccles direction) to Deansgate/Castlefield. We are within walking distance from Oxford Road train station – continue down Whitworth Street West until you reach Deansgate Train Station.
By Car:Manchester is ringed by the M60, linking to the UK's extensive motorway network. Follow signs to the city centre and park near Deansgate Train Station. Nearest all-day car parking is on Chester Road and Little Peter Street in Knott Mill. Short-term street parking is available on Deansgate and Whitworth Street West.
sidmep2018
Hidden gem
Saturday, 21st April 2018
Not one to miss! Perfect sized gallery to pop in and experience the most exciting contemporary art in Manchester Read full review
l0uise20100
Amazing Gallery
Thursday, 4th October 2018
Great art. Great exhibitions. Great gallery. I love it. A lot of good artists are exhibiting there.. Read full review
Stay284801
Mixture!
Small non imposing gallery. A few excellent exhibits and a few not in my opinion. Courteous and helpful staff. Thoughtfully laid out. Good exposure for contemporary artists and their work. Read full review
Imported Facilities
Guided Tours Available for Groups
Guided Tours Compulsory for Groups
Dates and Opening Times
Monday - Tuesday Closed*
Wednesday - Sunday 13:00 - 18:00*
Dates and times marked with a '*' are provisional and may be subject to change
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An incomparable venue in the city.
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Showcasing internationally renowned and Designated collections of fine art, decorative…
Friday, 24th January 2020 - Saturday, 28th March 2020
Thumbs UpThumbs Up, a group exhibition that will share work by artists Michael Beard, Frances Disley, Harry Meadley, and Leslie Thompson.


Next month's 2018 BET Hip Hop Awards look to be a Drake-dominated affair. The network today (Sept. 12) revealed the slate of award nominees, led by Drizzy with 11 nominations and Cardi B with 10.
Both rappers' nominations include Best Hip Hop Video, MVP of the Year, and Album of the Year for Scorpion and Invasion of Privacy, respectively. J. Cole's KOD, Migos' Culture II and The Carters' Everything Is Love are also competing for Album of the Year.
Childish Gambino, for his part, pulled six nominations, while Travis Scott and Kendrick Lamar each earned five. All three are nominated in the Lyricist of the Year category.
This year's Best New Hip Hop Artist category, won last year by Cardi, includes Blocboy JB, Juice Wrld, Lil Baby, Rich the Kid and the late XXXTentacion. Both Cardi and X were also nominated in the same category for the 2018 American Music Awards, which was released today.
The show, hosted on Oct. 6 at The Fillmore Miami Beach at Jackie Gleason Theater in Miami Beach, Fla., will be broadcast on BET Networks on Oct. 16. The show's performance list has not yet been announced.
Last year's program featured Cardi performing "Bodak Yellow," Eminem dissing Donald Trump, and of course, BET's famous cyphers. See the nominations in full below.
21 Savage & Offset & Metro Boomin - "Ric Flair Drip"
Dej Loaf & Leon Bridges - "Liberated"


Sharp knives gave Nathanial Zimet his edge in winning "Chopped"
Chef Nathanial Zimet of Boucherie said he knew he had a chance of winning the Food Network's "Chopped" competition show when he saw the knives the other chefs brought to New York.
"You had to bring seven knives," Zimet said the day after the rest of the world learned he won $10,000 and the title Chopped Champ. "When we got there, we were told to open our knives and give them to (the producers.) I was looking at everybody's knives and that's when I really felt confident.
"My assumption is what we do is a profession. You care for the tools of your trade," Zimet said. "I don't know how you could effectively compete if you're not serious about your tools."
Zimet used his knives to expert advantage. The judges commented on his butchering of the Scottish red leg partridge, one of the surprise ingredients sprung on the chefs in the entree round.
And he made sausage of some of it and wrapped it in a kale leaf, serving the meatball with a sauce of melted peanut butter taffy (which Zimet had never eaten) and red curry paste, over a hash of sweet potato and kale.
Zimet said last week that his preparation for the show consisted of sharpening his knives (and he's opening another restaurant next year in the Marigny neighborhood.) He watched the show, which features chefs preparing the ingredients of mystery baskets in short time periods, a few times before he went to New York.
"I told myself I wiould have some recipes in my head. But to be quite honest, I didn't do that. I figured, in the nature of true competition, to just show 'em what I got," Zimet said. "I tried to really focus on utilizing completely the ingredients given to me. I thought that was my best bet."
Zimet and his crew knocked off early Tuesday night and went to nearby Cooter Brown's to watch the show. A friend made him a cake to celebrate the win. Melissa Samuels of Melissa's Fine Pastries also made a replica of his purple food truck, Que Crawl, for his big benefit party, when the community rallied to raise funds after he was shot in April 2011.
He competed only 11 months after the shooting.
The first round of competition challenged the chefs to create an appetizer from brook trout, rhubarb, dried strawberries and lemon verbena. Zimet used the head of the fish to make a fumet with the dried strawberries and rhubarb. Judge Alex Guarnaschelli said of his sauce, "I just want to drink it, but the fish doesn't belong there."
Right after that, the judges asked him about the shooting. After he showed how the bullets passed through his arm and torso, Guarnaschelli said, "Your sauce tastes better." All three judges expressed amazement that he was even there.
Personal chef Courtney Renn of Atlanta was eliminated after the appetizer round for failing to plate all four items. All three remaining contestants overcooked the partridge in the entree round, but Jimmy Tessier, chef de cuisine at Bacio, in the Tropicana Las Vegas Hotel and Resort, overcooked it the worst and was chopped.
That left Zimet in the dessert round with Akiko Thurnauer, a New York chef/owner of a Japanese restaurant named Family Recipe. The mystery ingredients were dulce de leche, pineapple, sesame seed buns and apricot paste.
Zimet grabbed a rolling pin and repurposed the sesame seed bun into a dumpling wrapper, around a filling of apricot paste, dulce de leche and mascarpone. It looked delicious, pinched together in a crown shape.
"I've done a lot of things where I take bread and roll it out to make, like, a pasta," Zimet said. "I was comfortable with all the ingredients. I thank my training for that." A 36-year-old native of North Carolina, he attended Le Cordon Bleu in London and Sydney, Australia.
Zimet grilled the pineapple and chopped it into a salad with mint to top the dumplings, which started to ease open as he cooked them in butter on top of the stove. This led to the best line of the night.
He decided to whip cream, by hand, with 90 seconds left. Guarnaschelli fretted that he could be using a mixer to save time.
"Are we not professionals?" Zimet said, indignant at the very idea.
His niece asked him the same thing one Thanksgiving. His reply: "'Machine, shoo'! I don't allow things like that in my kitchen. That's laziness. I don't believe in making pasta by machine, either...pasta is better if made by hand. Everything's better if you touch it."
Zimet said the entire experience was a good time. His business partner, James Denio, went with him to New York and they ate at Madison Park and Per Se.
"I felt strong all the way through," he said. And he looked it.
As for the $10,000 prize, the newest of New Orleans' six "Chopped" champs said he is using it to buy appliances for his new house.
"I'm super stoked," Zimet said of his new home, the first he's ever owned. "It's eight blocks from the restaurant, so I can be even closer to that place which I never leave."


As technology advances, so do LEDs and the result is a more visually striking and exceptionally durable bulb. The new Opticore™ LED bulb features SMD technology like that used in flat panel LED TVs to create consistent illumination from the base to the tip and eliminate the light directionality that occurs in standard LED bulbs. In addition to advanced technology, Opticore™ LED bulbs are constructed with polycarbonate which is 200 times stronger than glass and the bulbs are infused with color to create a rich hue that won't fade, chip or crack. Opticore™ LED bulbs are available in a variety of popular color options as well as faceted and smooth lenses to create dynamic light displays like you've never seen before!


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Greedy News
Smack Talk & Slander
Photo Recon
googlehostedservice.html
Drop in trigger jobs for 870 Shotguns
Shooting Illustrated has a great story on a new Timney drop in sear for the Remington 870 shotgun. Well known for their excellent trigger kits for rifles like the Ruger 10/22 or the AR, this Timney solution will be a welcome option for the 870. While the 870 is very easy to break down, and even upgrade yourself, many do not want to mess with the actual triggers. Timney now makes even this task rather simple.
By 3eyedSamurai at 01/16/2010 - 18:51
15% off at OPSGEAR
OPSGear has a bunch of items already on discount, check out their Clearence section for up to 50% off. But Greedy Gun Runner was just told that with this discount code below you can get up to an additional 15% off!
Greedy Sales
3eyedSamurai's blog
New Military Run Flat Tires
Following the successful completion of a 21-month collaborative agreement, the U.S. Army Research Laboratory awarded a four-year contract to Resilient Technologies to continue the development and testing of its unique, patent pending non-pneumatic (airless) tire (NPT) for the High Mobility Multipurpose Wheeled Vehicle (HMMWV or Humvee). The NPT will not go flat if shot or hit by shrapnel from a roadside bomb. This groundbreaking product will safeguard U.S. military personnel by providing continuous mobility during hostile events.
Hi Res HQ Photos of Military Vehicles and Aircraft
You have to check out this amazing site for HiRes/HQ images of weapons, aircraft and vehicles.They are very well organized and easy to browse through. We have used them to create our very own wallpapers. Enjoy!
Jets, Helos, Props
http://www.primeportal.net/the_airstrip.htm
Now that's a cake
We recently threw a party for one of our Greedy Gun Runner's and of course a regular cake just wouldn't do. Here is a pic of our Claymore Cake.
Summer Glau from the Terminator TV show uses KJW Airsoft Glocks
She just did the new promo photo shoots for the TV show, and I saw she is using KJW Airsoft Glocks, even for a nationwide print campaign. I'm going to capitalize on this by writing an article about it. Most of the photos make it too hard to tell, but being me, I found Hi-Res images that show all the trades. She is also using an M16 in one new poster from the same shoot, and I can't tell the brand, but I would expect that too to be airsoft.
Call of Duty WaW Der Riese (the giant) leaked?
Believe us when we say this sounds a little weird. One of our GGRs left his XBOX on last night while still in a World at War zombie map and this morning it asks him to update his WaW game. After the update his XBOX has lists of players for the new Der Riese zombie map! The leader board shows up with the developer's gamer tags even.
Airtronic produces an RPG-7 for the US Military (with minor design inputs from the original)
The most famous individual anti-armor weapon in history is now in U.S. hands. Airtronic USA is announcing their production of the RPG –7 rocket-propelled grenade launcher along with a complete complement of ammunition – all manufactured in United States facilities. The Airtronic USA product is a combination of the simplicity of action and operation of the original RPG-7 design matched with American manufacturing precision and technology.
MultiCam sells excess fabric to ShamWow
CRYE has joined with German Based company, ShamWow, in an effort to cut 2nd quarter losses. The decision has been made to cut rolls of literal miles of Multicam fabric, to be sold to German company ShamWow.
It is expected this is an attempt to recoup some of the costs of CRYE's fabrication loses. Be on the lookout for MiltiWow, the new super absorbent towel, to hit the shelves in June. It is being marketed by the German company as the Towel that you can never misplace, making reference to the pattern that can be easily seen in front of nearly any surrounding found in nature on this planet.
By Bishop at 05/18/2009 - 19:14
Rumors & Smack Talk
Multicam Loses Proposed UN Contract
Another smashing blow to Multicam as it has once again lost hope of a large contract. Our GGR reporter met with Mulicam CEO, who we interviewed last week while he sipped on a buttery nipple at a local pub called the Pompas Rose. The CEO said he "had hoped the UN would recognize the marginal improvement." But even at paintball players pricing, the UN couldn't be coerced to see the negligible improvement as worthwhile. It seems they will be staying with their UN blue rather than downgrading to Multicam.


Dr. Faria Asad is a Dermatologist from Lahore Dr. Faria Asad is a certified Dermatologist and his qualification is MBBS. Dr. Faria Asad is a well-reputed Dermatologist in Lahore. Dr. Faria Asad is affiliated with different hospitals and you can find his affiliation with these hospitals here.
Dr. Faria Asad Address and Contacts Dr. Faria Asad is a Dermatologist based in Lahore. Dr. Faria Asad phone number is 0300 065 435, (042) 35167251 and she is sitting at the Dental Clinic in different timeframes.
Dr. Faria Asad is one of the most patient-friendly Dermatologist and his appointment fee is around Rs. 500 to 1000. You can find Dr. Faria Asad in different hospitals from 8 to 9 pm. You can find the complete list and information about Dr. Faria Asad's timings, clinics, and hospitals.
If you are looking for a Dermatologist in Lahore then you can try Dr. Faria Asad as a Dermatologist You can find the patient reviews and public opinions about Dr. Faria Asad here. You can also compare Dr. Faria Asad with other Dermatologists here with the compare button above.


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Covid concerns aside, Tigers ready to tip off season in Sioux Falls
by: Mike Ceide
PORTLAND, OREGON – NOVEMBER 12: D.J. Jeffries #0 of the Memphis Tigers drives to the basket on Francis Okoro #33 of the Oregon Ducks during the first half of the game at Moda Center on November 12, 2019 in Portland, Oregon. (Photo by Steve Dykes/Getty Images)
SIOUX FALLS, S.D. — With the college basketball season tipping off on Wednesday, Wichita State is.the latest team to pull out of the Tigers' season opening event, the Crossover Classic.
But unlike Duke, Ohio State, Utah, Dayton, Texas A&M or Creighton, the Shockers actually made it all the way to Sioux Falls Monday night before positive covid tests forced Wichita State to withdraw.
No such problems for Penny Hardaway and the Tigers, who touched down in South Dakota on Monday and passed their covid tests so their game with Saint Mary's Wednesday…is on.
For all the covid issues in South Dakota, of which there are many, Hardaway doesn't seem overly concerned and actually excited to get a chance to play while many others are not.
"Now that you see the system and how they have everyone kind of sectioned off in their own little space, their own area. Everyone is wearing masks so no. The tournament's done a great job with how they've set it up so we're not walking around on pins and needles," Hardaway said. "Nine months ago, canceling conference tournaments. Canceling the NCAA Tournament and not knowing if we were going to have a season. To be here right now. To be able to play a game, especially with tournaments around the country being canceled because of COVID situations, we're definitely fortunate to be here."
But not everyone made the trip for the Tigers.
Hardaway announcing that newcomer Ahmad Rand is not with the team in Sioux Falls because of contact tracing.
Tigers and Saint Mary's tip off the season Wednesday at 1 p.m.
KNOXVILLE, Tenn. (WATE) -- This week, Jan. 14-21, the Pat Summitt Foundation and the Southeastern Conference will honor the former Lady Vols basketball coach with the 10th We Back Pat Week.
SEC member institutions will offer support to the Foundation during each of their 18 women's basketball home games in 14 cities through a variety of efforts including promotional public address announcements and video endorsements.
Grizzlies' Morant upgraded to questionable vs. T'Wolves
by Mike Ceide / Jan 14, 2021
MEMPHIS -- Rarely has an injury report brought so much excitement to Grizzly fans everywhere.
Just over two weeks after severely spraining his ankle against the Brooklyn Nets, Ja Morant is listed as questionable,upgraded from out, for Friday night's game against the Minnesota Timberwolves. Now there's no guarantee Morant will play but there is a chance after the Grizzlies' star guard has missed the team's last nine games.


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Back in 2015, Variety reported the actress had exited the film due to "creative differences" with writer Nicole Holofcener.
Even Oscar winners get fired. Julianne Moore opened up about getting axed from the Oscar-nominated film "Can You Ever Forgive Me?" during Thursday's episode of "Watch What Happens Live With Andy Cohen."
Back in 2015, Variety reported the actress had exited the film due to "creative differences" with writer Nicole Holofcener. However, the "Still Alice" star decided to set the record straight during the late-night show after a caller asked her about the decision to leave.
"I didn't leave that movie. I was fired," she said.
When asked about Holofcener's reasoning for the termination, Moore replied, "I think she didn't like what I was doing."
"We had just been kind of rehearsing and doing pre-production and stuff, and I think that her idea of where the character was was different than my idea of [where] the character was," she said. "And so, she fired me."
Moore then admitted she hadn't seen the movie yet because "it's still kind of painful." However, she indicated she had no bad blood with Melissa McCarthy, who starred in the film and received an Oscar nomination for her role.
"I love Melissa McCarthy. I worship her. I think she's fantastic," she said. "So, I'm sure she's great."
After fellow guest Diane von Furstenberg asked her how it felt to be fired, Moore acknowledged "it's pretty bad."
"The only other time I was fired was when I was working at a yogurt stand when I was 15," she said. "So yeah, it felt bad."
"Can You Ever Forgive Me?" was up for three Academy Awards. Not only was McCarthy nominated for Best Actress in a Leading Role but Richard E. Grant was also up for Best Actor in a Supporting Role. In addition, the movie received a nod for Best Adapted Screenplay.
Of course, Moore has been recognized by The Academy, as well. In fact, she's been nominated for an Oscar five times and took home the trophy in 2015 for "Still Alice."
(E! and Bravo are both part of NBCUniversal).


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text:
\section*{\Cref*{sec:pf:model} (p. \pageref*{sec:pf:model})}
\label{omit:pf:model}
\addcontentsline{toc}{section}{Section \ref*{sec:pf:model} (p. \pageref*{sec:pf:model})}

The two utility possibility frontiers may be written as
\begin{align*}
	F^0(u)
	&= u - \lambda \phi^{-1}(u)
	\\
	\text{and} \quad
	F^1(u)
	&= u + \lambda \max_{L \geq 0} \left\{
	w L - \phi^{-1} \left( u + \kappa(L) \right)
	\right\} .
\end{align*}

\begin{proof}[Proof of \Cref*{lemma:UI_assns}]
	%
	Clearly $F^0,F^1$ are well-defined and continuous on $\left[ 0, \infty \right)$.
	It remains to show that they are strictly concave with peaks $u^0,u^1$ satisfying $u^1 < u^0$, and that the gap $F^1 - F^0$ is strictly decreasing.

	\begin{namedthm}[Claim.]
		%
		\label{claim:Lstar_interior}
		%
		For each $u>0$, the maximisation problem in the expression for $F^1(u)$
		has a unique solution $L^\star(u) \in (0,\infty)$.
		Furthermore, $\lim_{u \to \infty} L^\star(u) = 0$.
		%
	\end{namedthm}

	\begin{proof}[Proof]%
		\renewcommand{\qedsymbol}{$\square$}
		%
		Fix $u>0$, and write $f : \mathbf{R}_+ \to \mathbf{R}$ for the objective. Its (right-hand) derivative is
		%
		\begin{equation*}
			f'(L)
			= w - \frac{ \kappa'(L) }
			{ \phi'\left( \phi^{-1} \left( u + \kappa(L) \right) \right) } .
		\end{equation*}
		%
		Clearly $f'$ is strictly decreasing, so there is at most one maximiser.
		We have $f'(0) = w > 0$, whereas $f'(L) < 0$ for any large enough $L>0$ since
		%
		\begin{equation*}
			\lim_{L \to \infty} f'(L)
			\leq w - \lim_{L \to \infty} \frac{ \kappa'(L) }
			{ \phi'\left( \phi^{-1} \left( \kappa(L) \right) \right) }
			= - \infty ,
		\end{equation*}
		%
		where the equality holds since
		the numerator is bounded away from zero as $L \to \infty$
		while the denominator vanishes.
		Thus the unique maximiser $L^\star(u)$ of $f$ is interior, and therefore satisfies the first-order condition $f'\left( L^\star(u) \right) = 0$.
		As $u \to \infty$ in the first-order condition, the denominator in the fraction vanishes since $\lim_{C \to \infty} \phi'(C) = 0$,
		requiring the numerator also to vanish,
		which since $\kappa' > 0$ on $(0,\infty)$ demands that $L^\star(u) \to 0$.
		%
	\end{proof}%
	\renewcommand{\qedsymbol}{$\blacksquare$}

	$F^0$ is strictly concave since $\phi$ is.
	Its (right-hand) derivative
	%
	\begin{equation*}
		F^{0\prime}(u)
		= 1 - \frac{ \lambda }{ \phi'\left( \phi^{-1}(u) \right) }
	\end{equation*}
	%
	is strictly positive at $u=0$ since $\lim_{C \to 0} \phi'(C) = \infty$,
	and is strictly negative for $u$ large enough since
	$\phi^{-1}(u) \to \infty$ as $u \to \infty$
	and $\lim_{C \to \infty} \phi'(C) = 0$.
	Thus $F^0$ is uniquely maximised at $u^0 \in (0,\infty)$ satisfying the first-order condition
	%
	\begin{equation*}
		\phi'\left( \phi^{-1}\left( u^0 \right) \right) = \lambda .
	\end{equation*}

	$F^1$ is also strictly concave: for $u \neq u'$ in $[0,\infty)$ and $\eta \in (0,1)$,
	we have
	%
	\begin{multline*}
		F^1( \eta u + (1-\eta) u' )
		\\
		\begin{aligned}
			&= \eta u + (1-\eta) u'
			+ \lambda \max_{L \geq 0} \left\{
			w L - \phi^{-1} \left( \eta u + (1-\eta) u' + \kappa(L) \right)
			\right\}
			\\
			&> \eta u + (1-\eta) u'
			\\
			&\quad
			+ \lambda \max_{L \geq 0} \left\{
			w L
			- \eta \phi^{-1} \left( u + \kappa(L) \right)
			- (1-\eta) \phi^{-1} \left( u' + \kappa(L) \right)
			\right\}
			\\
			&\geq \eta \left[ u
			+ \lambda \max_{L \geq 0} \left\{
			w L
			- \phi^{-1} \left( u + \kappa(L) \right)
			\right\}
			\right]
			\\
			&\quad
			+ (1-\eta) \left[ u'
			+ \lambda \max_{L \geq 0} \left\{
			w L
			- \phi^{-1} \left( u' + \kappa(L) \right)
			\right\}
			\right]
			\\
			&= \eta F^1(u) + (1-\eta) F^1(u') ,
		\end{aligned}
	\end{multline*}
	%
	where the strict inequality holds since $-\phi^{-1}$ is strictly concave.
	By the envelope theorem, we have
	%
	\begin{equation*}
		F^{1\prime}(u)
		= 1 - \frac{ \lambda }
		{ \phi'\left( \phi^{-1}\left(
		u + \kappa\left( L^\star(u) \right)
		\right) \right) } .
	\end{equation*}
	%
	This expression is strictly negative for $u$ large enough
	because the denominator in the fraction vanishes as $u \to \infty$
	since $L^\star(u) \to 0$ by the \hyperref[claim:Lstar_interior]{claim}.
	Thus $F^1$ has a unique maximiser $u^1 \in [0,\infty)$,
	which satisfies the first-order condition
	%
	\begin{equation*}
		\phi'\left( \phi^{-1} \left(
		u^1 + \kappa\left( L^\star\left(u^1\right) \right)
		\right) \right)
		\leq \lambda ,
		\quad \text{with equality if $u^1>0$.}
	\end{equation*}

	To show that $u^0 > u^1$, consider two cases.
	If $u^1 = 0$, then $u^0 > 0 = u^1$.
	If instead $u^1 > 0$, then the first-order conditions for $u^0$ and $u^1$ together yield
	%
	\begin{equation*}
		\phi'\left( \phi^{-1}\left( u^0 \right) \right)
		= \lambda
		= \phi'\left( \phi^{-1} \left(
		u^1 + \kappa\left( L^\star\left(u^1\right) \right)
		\right) \right) .
	\end{equation*}
	%
	Thus
	%
	\begin{equation*}
		u^0 = u^1 + \kappa\left( L^\star\left(u^1\right) \right) > u^1
	\end{equation*}
	%
	since $L^\star\left( u^1 \right) > 0$ by the \hyperref[claim:Lstar_interior]{claim}
	and $\kappa > 0$ on $(0,\infty)$.

	It remains only to show that $F^1 - F^0$ is strictly decreasing, so that $u^\star = 0$.
	It suffices to show that $F^{1\prime} < F^{0\prime}$ on $(0,\infty)$.
	So fix any $u>0$.
	Since $L^\star(u) > 0$ by the \hyperref[claim:Lstar_interior]{claim},
	and $\kappa > 0$ on $(0,\infty)$, we have
	%
	\begin{equation*}
		F^{0\prime}(u)
		= 1 - \frac{ \lambda }{ \phi'\left( \phi^{-1}(u) \right) }
		>  1 - \frac{ \lambda }
		{ \phi'\left( \phi^{-1}\left(
		u + \kappa\left( L^\star(u) \right)
		\right) \right) }
		= F^{1\prime}(u) .
		\qedhere
	\end{equation*}
	%
\end{proof}

\section*{\Cref*{suppl:ext:drop_F1-F0_strict} (p. \pageref*{suppl:ext:drop_F1-F0_strict})}
\label{omit:ext:drop_F1-F0_strict}
\addcontentsline{toc}{section}{Supplemental appendix \ref*{suppl:ext:drop_F1-F0_strict} (p. \pageref*{suppl:ext:drop_F1-F0_strict})}

To formalise and prove the trichotomy asserted in \cref*{footnote:trichotomy} (p. \pageref*{footnote:trichotomy}),
consider frontiers $F^0,F^1$ satisfying all of our model assumptions,
except for the requirement that $u^\star$ be a strict local maximum of $F^1-F^0$.

\begin{definition}
	%
	\label{definition:saddle}
	%
	$\widebar{u} \in (0,\infty)$ is a \emph{saddle point} of a continuous function $\psi$ defined on $[0,\infty)$ iff both
	%
	\begin{enumerate}[label=(\alph*)]

		\item \label{bullet:saddle:deriv}
		for all $\varepsilon > 0$, there are $u,u' \in (0,\infty)$ with $\widebar{u}-\varepsilon < u < \widebar{u} < u' < \widebar{u}+\varepsilon$ such that $\abs*{ \psi(u')-\psi(u) } / (u'-u) < \varepsilon$,%
			\footnote{In case $\psi$ is differentiable at $\widebar{u}$,
			condition \ref{bullet:saddle:deriv} reduces to the requirement that $\psi'\left(\widebar{u}\right)=0$.
			Our definition thus extends the usual definition of a saddle point, which applies only to differentiable functions.}
		and

		\item \label{bullet:saddle:maxmin}
		$\widebar{u}$ is neither a local maximum nor a local minimum of $\psi$.

	\end{enumerate}
	%
\end{definition}

\begin{namedthm}[Claim.]
	%
	If $u^\star>0$, then either
	(i) $u^\star$ is a local maximum of $F^1-F^0$,
	(ii) $u^\star$ is a saddle point of $F^1-F^0$, or
	(iii) both $F^0$ and $F^1$ are kinked at $u^\star$.
	%
\end{namedthm}

Assuming that $u^\star$ is a strict local maximum of $F^1-F^0$
thus amounts to ruling out the saddle-point case (ii),
the (pathological) mutual-kink case (iii),
and the case in which $F^1-F^0$ is constant on a left-neighbourhood of $u^\star$.

\begin{proof}
	%
	Assume that $u^\star>0$ and that neither (i) nor (ii) holds;
	we shall deduce (iii).
	By its definition, $u^\star$ cannot be a local minimum of $F^1-F^0$.
	Then by definition of `saddle point',
	there must be an $\varepsilon>0$ such that
	%
	\begin{equation*}
		\frac{ \abs*{ \left(F^1-F^0\right)(u') - \left(F^1-F^0\right)(u) } }
		{u'-u}
		\geq \varepsilon
	\end{equation*}
	%
	for all $u,u' \in (0,\infty)$ satisfying $u^\star-\varepsilon < u < u^\star < u' < u^\star+\varepsilon$.
	Since $F^1-F^0$ is strictly decreasing on $\left[u^\star,u^0\right]$,
	the expression in the numerator must be strictly negative for $u$ sufficiently close to $u^\star$,
	and thus $F^{1+}(u^\star) \leq F^{0+}(u^\star) - \varepsilon < F^{0+}(u^\star)$ and (similarly) $F^{1-}(u^\star) < F^{0-}(u^\star)$.
	By definition of $u^\star$, the frontiers $F^0,F^1$ share a supergradient $\eta \in \mathbf{R}$ at $u^\star$,
	so that $F^{j+}(u^\star) \leq \eta \leq F^{j-}(u^\star)$ for both $j \in \{0,1\}$.
	Then
	%
	\begin{equation*}
		F^{1+}(u^\star) < F^{0+}(u^\star)
		\leq \eta
		\leq F^{1-}(u^\star) < F^{0-}(u^\star) ,
	\end{equation*}
	%
	implying that both $F^0$ and $F^1$ are kinked at $u^\star$.
	%
\end{proof}

\section*{\Cref*{suppl:ext:random_F1} (p. \pageref*{suppl:ext:random_F1})}
\label{omit:ext:random_F1}
\addcontentsline{toc}{section}{Supplemental appendix \ref*{suppl:ext:random_F1} (p. \pageref*{suppl:ext:random_F1})}

For $u \in [0,\infty)$,
write $M_u$ for the space of all maps $\widehat{X} : \mathcal{F} \to [0,\infty)$ that satisfy
$\mathbf{E}\bigl( \widehat{X}(\boldsymbol{F}) \bigr) = u$.

\begin{proof}[Proof of \Cref*{lemma:randomF1_assns}]
	%
	For concavity, take $u,u^\dag \in [0,\infty)$ and $\lambda \in (0,1)$.
	Let $\widehat{X} \in M_u$ and $\widehat{X}^\dag \in M_{u^\dag}$ be maximisers:
	%
	\begin{equation*}
		\mathbf{E}\left( \boldsymbol{F}\left( \widehat{X}( \boldsymbol{F} ) \right) \right)
		= F^1(u)
		\quad \text{and} \quad
		\mathbf{E}\left( \boldsymbol{F}\left( \widehat{X}^\dag( \boldsymbol{F} ) \right) \right)
		= F^1\left( u^\dag \right) .
	\end{equation*}
	%
	Then $\lambda \widehat{X} + (1-\lambda) \widehat{X}^\dag$ belongs to $M_{\lambda u + (1-\lambda) u^\dag}$, so that
	%
	\begin{multline*}
		F^1\left( \lambda u + (1-\lambda) u^\dag \right)
		\\
		\begin{aligned}
			&\geq \mathbf{E}\left(
			\boldsymbol{F}\left(
			\lambda \widehat{X}( \boldsymbol{F} )
			+ (1-\lambda) \widehat{X}^\dag( \boldsymbol{F} )
			\right)
			\right)
			&&
			\\
			&\geq \mathbf{E}\left(
			\lambda \boldsymbol{F}\left( \widehat{X}( \boldsymbol{F} ) \right)
			+ (1-\lambda) \boldsymbol{F}\left( \widehat{X}^\dag( \boldsymbol{F} )\ \right)
			\right)
			&& \text{since $\boldsymbol{F}$ is concave a.s.}
			\\
			&= \lambda F^1(u) + (1-\lambda) F^1\left( u^\dag \right) .
			&&
		\end{aligned}
	\end{multline*}

	To see that $F^1$ attains a maximum at $u^1$, observe that $F^1$ is bounded above by $\mathbf{E}\left( \boldsymbol{F}\left( U^1( \boldsymbol{F} ) \right) \right)$,
	and that it attains this value at $u^1$.
	The peak is unique because for $u \neq u^1$, any $\widehat{X} \in M_u$ must have $\widehat{X} \neq U^1$ on a non-null set, and thus
	%
	\begin{equation*}
		\boldsymbol{F}\left( \widehat{X}(\boldsymbol{F}) \right)
		\leq \mathrel{(<)} \boldsymbol{F}\left( U^1(\boldsymbol{F}) \right)
		\quad \text{a.s. (with positive probability).}
	\end{equation*}

	For upper semi-continuity, observe that since $F^1$ is concave,
	it is continuous on the interior of its effective domain, which is a convex set that contains $u^1$.
	$F^1$ is trivially continuous off the closure of its effective domain, where it is constant and equal to $-\infty$.
	It remains only to establish that
	%
	\begin{equation*}
		\limsup_{u' \to u} F^1(u') \leq F^1(u)
	\end{equation*}
	%
	for $u$ on the boundary of the effective domain (there are at most two such $u$s).
	It suffices to show for an arbitrary decreasing sequence $(u_n)_{n \in \mathbf{N}}$ in the interior of the effective domain converging to some $u \in \left[ 0, u^1 \right]$ that
	%
	\begin{equation*}
		\lim_{n \to \infty} F^1(u_n) \leq F^1(u)
	\end{equation*}
	%
	(where the limit exists since $\left( F^1(u_n) \right)_{n \in \mathbf{N}}$ is eventually monotone),
	and similarly for increasing sequences converging to $u \geq u^1$.
	We show the former, omitting the analogous argument for the latter.

	For each $n \in \mathbf{N}$, let $\widehat{X}_n \in M_{u_n}$ be a maximiser at $u_n$:
	%
	\begin{equation*}
		\mathbf{E}\left( \boldsymbol{F}\left( \widehat{X}_n( \boldsymbol{F} ) \right) \right)
		= F^1(u_n) .
	\end{equation*}
	%
	Since $\bigcup_{ u' \in \left[ 0, u^1 \right] } M_{u'}$ is compact (because bounded),
	the sequence $\bigl( \widehat{X}_n \bigr)_{n \in \mathbf{N}}$ admits a convergent subsequence $\bigl( \widehat{X}_{n_k} \bigr)_{k \in \mathbf{N}}$, whose limit we denote by $\widehat{X}$.
	We have
	%
	\begin{align*}
		F^1(u)
		&\geq \mathbf{E}\left( \boldsymbol{F}\left( \widehat{X}( \boldsymbol{F} ) \right) \right)
		&& \text{since $\widehat{X} \in M_u$}
		\\
		&\geq \mathbf{E}\left( \lim_{k \to \infty} \boldsymbol{F}\left( \widehat{X}_{n_k}( \boldsymbol{F} ) \right) \right)
		&& \text{since $\boldsymbol{F}$ is upper semi-continuous a.s.}
		\\
		&= \lim_{k \to \infty}
		\mathbf{E}\left( \boldsymbol{F}\left( \widehat{X}_{n_k}( \boldsymbol{F} ) \right) \right)
		&& \text{by bounded convergence}
		\\
		&= \lim_{k \to \infty} F^1\left( u_{n_k} \right)
		&&
		\\
		&= \lim_{n \to \infty} F^1\left( u_n \right)
		&& \text{since $\left( F^1(u_n) \right)_{n \in \mathbf{N}}$ is convergent.}
		\qedhere
	\end{align*}
	%
\end{proof}

\section*{\Cref*{suppl:no_delay} (p. \pageref*{suppl:no_delay})}
\label{omit:no_delay}
\addcontentsline{toc}{section}{Supplemental appendix \ref*{suppl:no_delay} (p. \pageref*{suppl:no_delay})}

\begin{proof}[Proof of \Cref*{observation:no_delay}, first part]
	%
	Let $\left( x^0, X^1 \right)$ be an IC mechanism with $X^1_t \neq \max\left\{ X^0_t, u^1 \right\}$ for some period $t \in \mathbf{R}_+$.
	Consider the alternative mechanism $\bigl( x^0, X^{1\dag} \bigr)$,
	where $X^{1\dag}_t \coloneqq \max\left\{ X^0_t, u^1 \right\}$
	and $X^{1\dag}_s \coloneqq X^1_s$ for $s \neq t$.
	This mechanism is clearly IC,
	and the principal earns the same payoff if the breakthrough occurs at $\tau \neq t$.
	She earns a strictly higher payoff if the breakthrough occurs at $t$:
	%
	\begin{itemize}

		\item If $X^0_t > u^1$,
		then $X^1_t > X^0_t$ since $X^1_t \geq X^0_t$ by IC, and
		%
		\begin{equation*}
			X^1_t \neq \max\left\{ X^0_t, u^1 \right\} = X^0_t
			\quad \text{by hypothesis.}
		\end{equation*}
		%
		Thus $u^1 < X^0_t = X^{1\dag}_t < X^1_t$, so that
		%
		\begin{equation*}
			F^1\left( X^{1\dag}_t \right) > F^1\left( X^1_t \right)
			\quad \text{since $F^1$ is strictly decreasing on $\left[ u^1, \infty \right)$.}
		\end{equation*}

		\item If $X^0_t \leq u^1$, then $X^{1\dag}_t = u^1 = \max\left\{ X^0_t, u^1 \right\} \neq X^1_t$, so that
		%
		\begin{equation*}
			F^1\left( X^{1\dag} \right) > F^1\left( X^1_t \right)
			\quad \text{since $F^1$ is uniquely maximised at $u^1$.}
			\qedhere
		\end{equation*}

	\end{itemize}
	%
\end{proof}

\begin{proof}[Proof of \Cref*{observation:no_delay}, second part]
	%
	Since $F^0$ is affine,
	the principal's payoff from a mechanism $\left( x^0, X^1 \right)$
	under a distribution $G$ with $G(0) = 0$ may be written%
		\footnote{Adapt the payoff-rewriting exercise in the sketch proof of \Cref*{theorem:deadline} (p. \pageref*{eq:affine_payoff}).}
	%
	\begin{equation*}
		\Pi_G\left( x^0, X^1 \right)
		= \mathbf{E}_G\left(
		F^0\left( X_0^0 \right)
		+ e^{-r\tau} \phi\left( X^0_\tau \right)
		\right) ,
	\end{equation*}
	%
	where $\phi : [0,\infty) \to \mathbf{R}$ is defined by
	$\phi(u) \coloneqq F^1\left( \max\left\{ u, u^1 \right\} \right) - F^0(u)$.
	By inspection, $\phi$ is strictly decreasing on $\left[ 0, u^0 \right]$.

	Take any mechanism $\left( x^0, X^1 \right)$, assuming without loss of generality that $x^0 \leq u^0$ and that $X^1 = \max\left\{ X^0, u^1 \right\}$,
	and suppose that $\left( x^0, X^1 \right)$ does not satisfy
	%
	\begin{equation}
		x^0_t =
		\begin{cases}
			u^0		& \text{for $t \leq T$} \\
			0		& \text{for $t > T$}
		\end{cases}
		\quad \text{for some deadline $T \in [0,\infty]$;}
		\label{eq:no_delay_dl}
	\end{equation}
	%
	we shall show that it is dominated.
	Consider the mechanism $\bigl( x^{0\dag}, X^{1\dag} \bigr)$ defined by
	%
	\begin{equation*}
		x^{0\dag}_t
		\coloneqq
		\begin{cases}
			u^0		& \text{for $t \leq T$} \\
			0		& \text{for $t > T$}
		\end{cases}
	\end{equation*}
	%
	and $X^{1\dag} \coloneqq \max\bigl\{ X^{0\dag}, u^1 \bigr\}$,
	where $T$ is such that $X^{0\dag}_0 = X^0_0$.
	It is easy to see that $X^{0\dag} \leq X^0$,%
		\footnote{Adapt the argument for the \hyperref[claim:dl_leq_origjoin]{claim} in the proof of \Cref*{theorem:deadline} (\cref*{app:pf_theorem_deadline}, p. \pageref*{claim:dl_leq_origjoin}).}
	and there must be periods in which $X^{0\dag} \neq X^0$ since $x^0$ does not have the form \eqref{eq:no_delay_dl}.
	Since $\phi$ is strictly decreasing, it follows that
	%
	\begin{equation*}
		\Pi_G\left( x^{0\dag}, X^{1\dag} \right)
		\geq \Pi_G\left( x^0, X^1 \right)
		\quad \text{for any distribution $G$ with $G(0)=0$,}
	\end{equation*}
	%
	and the inequality is strict for any distribution $G$ that assigns positive probability to those times at which $X^{0\dag} \neq X^0$.
	%
\end{proof}

\section*{\Cref*{suppl:Euler_lemma_pf} (p. \pageref*{suppl:Euler_lemma_pf})}
\label{omit:Euler_lemma_pf}
\addcontentsline{toc}{section}{Supplemental appendix \ref*{suppl:Euler_lemma_pf} (p. \pageref*{suppl:Euler_lemma_pf})}

To prove \Cref*{lemma:euler_integrable_supergradient} and the Gateaux lemma, we shall use the following standard integration-by-parts result:%
	\footnote{See e.g. Theorem 18.4 in \textcite[p. 236]{Billingsley1995}.}

\begin{namedthm}[IBP lemma.]
	%
	\label{lemma:int_parts}
	%
	Let $\nu$ be a finite measure on $\mathbf{R}_+$, and let $L$ be a $\nu$-integrable function $\mathbf{R}_+ \to \mathbf{R}$ satisfying $L(t) = L(0) + \int_0^t l$ for some (Lebesgue-)integrable $l : \mathbf{R}_+ \to \mathbf{R}$.
	Then
	%
	\begin{equation*}
		\int_{[0,T]} L \mathrm{d} \nu = L(T) \nu([0,T]) - \int_0^T \nu([0,t]) l(t) \mathrm{d} t
		\quad \text{for every $T \in \mathbf{R}_+$.}
	\end{equation*}
	%
\end{namedthm}

\subsection{Proof of Lemma \ref*{lemma:euler_integrable_supergradient} (p. \pageref*{lemma:euler_integrable_supergradient})}
\label{omit:Euler_lemma_pf:euler_integrable_supergradient}

Define $\Phi : \mathbf{R}_+ \to [0,\infty]$ by
\begin{equation*}
	\Phi(t) \coloneqq r \int_0^t e^{-rs} \phi^0(s) \mathrm{d} s .
\end{equation*}
We first show that $\mathbf{E}_G(\Phi(\tau))<\infty$,
so that $\Phi$ is $G$-integrable.
Note that
\begin{align*}
	\mathbf{E}_G\left( r \int_0^\tau \frac{e^{-rt}}{1-G(t)} \mathrm{d} t \right)
	&= \lim_{T \to \infty} \int_{[0,T]}
	\left( r \int_0^s \frac{e^{-rt}}{1-G(t)} \mathrm{d} t \right)
	G(\mathrm{d} s)
	\\
	&= \lim_{T \to \infty} r \int_0^T e^{-rt}\frac{G(T)-G(t)}{1-G(t)}\mathrm{d} t
	= r \int_0^\infty e^{-rt} \mathrm{d} t
	= 1 ,
\end{align*}
where the first and third equalities hold by monotone convergence, and the second follows from the \hyperref[lemma:int_parts]{IBP lemma}
with $\nu$ the measure associated with $G$
and
\begin{equation*}
	l(t) \coloneqq
	\begin{cases}
		r e^{-rt} / [1-G(t)]	& \text{for $t < T$} \\
		0						& \text{for $t \geq T$}
	\end{cases}
	\qquad \text{if $G(T)<1$}
\end{equation*}
and $l(t) \coloneqq r e^{-rt}$ if $G(T)=1$.
Thus since $(x,X)$ satisfies the Euler equation with $\phi^0,\phi^1$,
(\ref*{eq:euler}) yields
\begin{align*}
	\mathbf{E}_G\left( \Phi(\tau) \right)
	&= \mathbf{E}_G\left( r \int_0^\tau e^{-rt}
	\frac{-\int_{[0,t]}\phi^1 \mathrm{d} G}{1-G(t)} \mathrm{d} t\right)
	\\
	&\leq \mathbf{E}_G\left( r \int_0^\tau e^{-rt}
	\frac{\int_{\mathbf{R}_+}\abs*{\phi^1} \mathrm{d} G}{1-G(t)} \mathrm{d} t\right)
	= \mathbf{E}_G\left(\abs*{\phi^1(\tau)}\right)
	< \infty ,
\end{align*}
where the final inequality holds since $\phi^1$ is $G$-integrable.

To show that $x$ belongs to $\mathcal{X}_G$,
fix a $u \in \left(0,u^0\right)$;
we shall prove that the maps $\psi^0_{x,u}$ and $\psi^1_{X,u}$ (defined on p. \pageref*{eq:euler_psi01_defn}) are $G$-integrable.
For the former, let $\mathcal{T}$ be the set of $t \in \mathbf{R}_+$ at which $x_t \geq u$.
The map $t \mapsto F^{0\prime}(x_t,u)$ is bounded on $\mathcal{T}$
(by zero below and by $F^{0+}(u)$ above) since $x \leq u^0$ and $F^0$ is increasing and concave,
so it suffices to show that
\begin{equation*}
	\varphi(t)
	\coloneqq r \int_0^t e^{-rs}
	F^{0\prime}\left(x_s,u\right) \boldsymbol{1}_{\mathbf{R}_+ \setminus \mathcal{T}}(s) \mathrm{d} s
\end{equation*}
is $G$-integrable.
Since $(x,X)$ satisfies the Euler equation, $\phi^0(t)$ is a supergradient of $F^0$ at $x_t$ for a.e. $t \in \mathbf{R}_+$ such that $G(t) < 1$.
Thus since $x < u$ on $\mathbf{R}_+ \setminus \mathcal{T}$ and $F^0$ is concave, we have
\begin{equation*}
	\phi^0(t)
	\geq F^{0+}(x_t)
	= F^{0\prime}(x_t,u)
	\quad \text{for a.e. $t \in \mathbf{R}_+ \setminus \mathcal{T}$ with $G(t)<1$,}
\end{equation*}
so that
\begin{equation*}
	\varphi(t)
	\leq r \int_0^t e^{-rs} \phi^0(s) \boldsymbol{1}_{\mathbf{R}_+ \setminus \mathcal{T}}(s) \mathrm{d} s
	\leq \Phi(t)
	\quad \text{for all $t \in \mathbf{R}_+$ with $G(t)<1$.}
\end{equation*}
Since $\varphi$ and $\Phi$ are continuous
and the former is non-negative,
it follows that $0 \leq \varphi \leq \Phi$ on the support of $G$,
so that $\varphi$ is $G$-integrable since $\Phi$ is.

It remains to show that $\psi^1_{X,u}$ is $G$-integrable.
Choose $\varepsilon \in \left(0,u \wedge \left[u^0 - u \right]\right)$ so that $\varepsilon < u^1 \wedge \left(u^0 - u^1 \right)$ if $u^1 > 0$,
and let
\begin{equation*}
	\mathcal{T}'
	\coloneqq
	\begin{cases}
		\left\{t \in \mathbf{R}_+ : X_t < u + \varepsilon\right\}
		& \text{if $u^1 = 0$}
		\\
		\left\{t \in \mathbf{R}_+ : \left(u \wedge u^1\right)-\varepsilon < X_t < \left(u \vee u^1\right) + \varepsilon\right\}
		& \text{if $u^1 > 0$.}
	\end{cases}
\end{equation*}
We will show that $t \mapsto \psi^1_{X,u}(t) = e^{-rt} F^{1\prime}(X_t,u)$
is bounded on $\mathcal{T}'$
and $G$-integrable on $\mathbf{R}_+ \setminus \mathcal{T}'$.
The former follows from the fact that
\begin{equation*}
	\text{$F^1$ is Lipschitz continuous on}\quad
	\begin{cases}
		\left[0,u+\varepsilon\right]
		& \text{if $u^1=0$} \\
		\left[ \left(u \wedge u^1\right)-\varepsilon,
		\left(u \vee u^1\right) + \varepsilon\right]
		& \text{if $u^1>0$,}
	\end{cases}
\end{equation*}
so that $t \mapsto F^{1\prime}(X_t,u)$ is bounded on $\mathcal{T}'$.
For the latter,
note that (by definition of $\mathcal{T}'$,)
every $t \in \mathbf{R}_+ \setminus \mathcal{T}'$ has
either $X_t < u, u^1$ or $X_t > u,u^1$.
Since $(x,X)$ satisfies the Euler equation, $\phi^1(t)$ is a supergradient of $F^1$ at $X_t$ for $G$-a.e. $t \in \mathbf{R}_+$.
Thus for $G$-a.e. $t \in \mathbf{R}_+ \setminus \mathcal{T}'$, we have
\begin{align*}
	\text{either}\quad
	F^{1\prime}(X_t,u) = F^{1+}(X_t)
	\quad &\text{and} \quad
	0 < F^{1+}(X_t) \leq \phi^1(t)
	\\
	\text{or}\quad
	F^{1\prime}(X_t,u) = F^{1-}(X_t)
	\quad &\text{and} \quad
	0 > F^{1-}(X_t) \geq \phi^1(t)
\end{align*}
by the concavity of $F^0$,
so that
\begin{equation*}
	\abs*{\psi^1_{X,u}(t)}
	\leq \abs*{F^{1\prime}(X_t,u)}
	\leq \abs*{\phi^1(t)}
	\quad \text{for $G$-a.e. $t \in \mathbf{R}_+ \setminus \mathcal{T}'$,}
\end{equation*}
which implies that $\psi^1_{X,u}$ is $G$-integrable on $\mathbf{R}_+ \setminus \mathcal {T}'$
since $\phi^1$ is.
\qed

\subsection{Proof of the claim in the proof of Lemma \ref*{lemma:integrable_supergradient} (p. \pageref*{claim:integrable_supergradient})}
\label{omit:Euler_lemma_pf:integrable_supergradient}

\renewcommand{\qedsymbol}{$\square$}
For $\alpha \in (0,1)$, define $h^0_\alpha, h^1_\alpha : \mathbf{R}_+ \to (-\infty,\infty]$ by
\begin{align*}
	h^0_\alpha(t)
	&\coloneqq r e^{-rt}
	\frac{ F^0\left( x_t+\alpha \left[ u - x_t \right] \right)
	- F^0( x_t ) }
	{ \alpha }
	\\
	\text{and}\quad
	h^1_\alpha(t)
	&\coloneqq e^{-rt}
	\frac{ F^1\left( X_t + \alpha \left[u-X_t\right] \right) - F^1(X_t) }
	{\alpha} .
\end{align*}
Clearly both are measurable for each $\alpha \in (0,1)$.
By definition, we have
\begin{equation*}
	\mathrm{D} \pi_G\bigl( x, x^\dag-x \bigr)
	= \lim_{\alpha \downarrow 0} \left[
	\mathbf{E}_G\left( \int_0^\tau h^0_\alpha \right)
	+ \mathbf{E}_G\left( h^1_\alpha(\tau) \right)
	\right] ;
\end{equation*}
we must show that the limit exists and satisfies the asserted inequality.

As $\alpha \downarrow 0$,
$h^0_\alpha$ and $h^1_\alpha$
converge pointwise
to
$h^0 : \mathbf{R}_+ \to (-\infty,\infty]$
and $h^1 : \mathbf{R}_+ \to [-\infty,\infty]$, respectively, given by
\begin{equation*}
	h^0(t)
	\coloneqq r e^{-rt} F^{0\prime}\left( x_t, u \right)
	\left( u - x_t \right)
	\quad \text{and} \quad
	h^1(t)
	\coloneqq e^{-rt} F^{1\prime}\left( X_t, u \right)
	\left( u - X_t \right) .
\end{equation*}
We shall show that
\begin{enumerate}[label=(\alph*)]

	\item \label{bullet:integrable_supergradient_a}
	$t \mapsto \int_0^t h^0 \boldsymbol{1}_{\mathcal{T}}$
	and $h^1 \boldsymbol{1}_{\mathcal{T}'}$ are $G$-integrable,

	\item \label{bullet:integrable_supergradient_b}
	$\mathbf{E}_G\left( \int_0^\tau h^0 \right)$ and $\mathbf{E}_G\left( h^1(\tau) \right)$ exist and are $>-\infty$, and

	\item \label{bullet:integrable_supergradient_c}
	$\mathbf{E}_G\left( \int_0^\tau h^0_\alpha \right)
	\to \mathbf{E}_G\left( \int_0^\tau h^0 \right)$
	and $\mathbf{E}_G\left( h^1_\alpha(\tau) \right)
	\to \mathbf{E}_G\left( h^1(\tau) \right)$
	as $\alpha \downarrow 0$.

\end{enumerate}
This suffices since then
$C \coloneqq
  \mathbf{E}_G\left( \int_0^\tau h^0 \boldsymbol{1}_{\mathcal{T}} \right)
+ \mathbf{E}_G\left( h^1(\tau) \boldsymbol{1}_{\mathcal{T}'}(\tau) \right)$ is finite,
\begin{align*}
	\mathrm{D} \pi_G\bigl( x, x^\dag-x \bigr)
	&= \mathbf{E}_G\left( \int_0^\tau h^0 \right)
	+ \mathbf{E}_G\left( h^1(\tau) \right)
	\\
	&= \mathbf{E}_G\left( \int_0^\tau h^0 \boldsymbol{1}_{\mathbf{R}_+ \setminus \mathcal{T}} \right)
	+ \mathbf{E}_G\left( h^1(\tau) \boldsymbol{1}_{\mathbf{R}_+ \setminus \mathcal{T}'}(\tau) \right)
	+ C ,
\end{align*}
and moreover any $t \in \mathbf{R}_+ \setminus \mathcal{T}$ has
(by definition of $\mathcal{T}$)
\begin{equation*}
	u-x_t \geq \varepsilon
	\quad \text{and thus} \quad
	h^0(t) \geq r e^{-rt} F^{0\prime}(x_t,u) \varepsilon
	= r e^{-rt} \abs*{ F^{0\prime}(x_t,u) } \varepsilon ,
\end{equation*}
while any $t \in \mathbf{R}_+ \setminus \mathcal{T}'$ has
\begin{align*}
	\text{either}\quad
	u-X_t \geq \varepsilon'
	\quad &\text{and} \quad
	X_t < u^1
	\quad \text{(hence $F^{1\prime}\left(X_t,u\right) > 0$)}
	\\
	\text{or (if $u^1 > 0$)}\quad
	u-X_t \leq \varepsilon'
	\quad &\text{and} \quad
	X_t > u^1
	\quad \text{(hence $F^{1\prime}\left(X_t,u\right) < 0$)}
\end{align*}
by definition of $\mathcal{T}'$,
so that $h^1(t) \geq e^{-rt}\abs*{F^{1\prime}\left(X_t,u\right)} \varepsilon'$.

To obtain \ref{bullet:integrable_supergradient_a}--\ref{bullet:integrable_supergradient_c} for $h^0$,
observe that $F^0$ is $K$-Lipschitz on $\left[ u-\varepsilon, u^0 \right]$ for some $K>0$ since it is concave and increasing on $\left[0,u^0\right]$.
Thus
\begin{equation*}
	\abs*{ h^0_\alpha(t) }
	\leq r e^{-rt} K \abs*{ x_t - u }
	\leq r e^{-rt} K u^0
	\quad \text{for all $\alpha \in (0,1)$ and $t \in \mathcal{T}$,}
\end{equation*}
which implies that
\begin{equation*}
	\abs*{ \int_0^t h^0_\alpha \boldsymbol{1}_{\mathcal{T}} }
	\leq \int_0^t \abs*{ h^0_\alpha } \boldsymbol{1}_{\mathcal{T}}
	\leq K u^0 r \int_0^t e^{-rt} \boldsymbol{1}_{\mathcal{T}}
	\leq K u^0
	\quad \text{for all $\alpha \in (0,1)$.}
\end{equation*}
Thus applying the dominated convergence theorem twice yields
\begin{equation*}
	\lim_{\alpha \downarrow 0}
	\mathbf{E}_G\left( \int_0^\tau h^0_\alpha \boldsymbol{1}_{\mathcal{T}} \right)
	= \mathbf{E}_G\left(
	\lim_{\alpha \downarrow 0}
	\int_0^\tau h^0_\alpha \boldsymbol{1}_{\mathcal{T}} \right)
	= \mathbf{E}_G\left( \int_0^\tau h^0 \boldsymbol{1}_{\mathcal{T}} \right)
	\in \mathbf{R} ,
\end{equation*}
giving us property \ref{bullet:integrable_supergradient_a}.

Meanwhile,
$h^0_\alpha \boldsymbol{1}_{\mathbf{R}_+ \setminus \mathcal{T}}$
is (pointwise) decreasing in $\alpha$ since $F^0$ is concave,
and is non-negative since $F^0$ is increasing
on $\left[ 0, u^0 \right]$ and $x \leq u < u^0$ on $\mathbf{R}_+ \setminus \mathcal{T}$.
Clearly the same is true of $t \mapsto \int_0^t h^0_\alpha \boldsymbol{1}_{\mathbf{R}_+ \setminus \mathcal{T}}$,
Thus applying the monotone convergence theorem twice yields
\begin{equation*}
	0
	\leq \lim_{\alpha \downarrow 0}
	\mathbf{E}_G\left( \int_0^\tau h^0_\alpha \boldsymbol{1}_{\mathbf{R}_+ \setminus \mathcal{T}} \right)
	= \mathbf{E}_G\left(
	\lim_{\alpha \downarrow 0}
	\int_0^\tau h^0_\alpha \boldsymbol{1}_{\mathbf{R}_+ \setminus \mathcal{T}} \right)
	= \mathbf{E}_G\left( \int_0^\tau h^0 \boldsymbol{1}_{\mathbf{R}_+ \setminus \mathcal{T}} \right) ,
\end{equation*}
where the rightmost term is well-defined.
Thus $h^0$ satisfies \ref{bullet:integrable_supergradient_b} and \ref{bullet:integrable_supergradient_c}.

To derive \ref{bullet:integrable_supergradient_a}--\ref{bullet:integrable_supergradient_c} for $h^1$,
observe that
\begin{equation*}
	\text{$F^1$ is Lipschitz continuous on}\quad
	\begin{cases}
		\left[0,u+\varepsilon'\right]
		& \text{if $u^1=0$} \\
		\left[ \left(u \wedge u^1\right)-\varepsilon',
		\left(u \vee u^1\right) + \varepsilon'\right]
		& \text{if $u^1>0$}
	\end{cases}
\end{equation*}
since it is concave and maximised at $u^0$.
Hence the family $\smash{\left( h^1_\alpha \right)_{\alpha \in (0,1)}}$ is uniformly bounded on $\mathcal{T}'$, so that
\begin{equation}
	\lim_{\alpha \downarrow 0}
	\mathbf{E}_G\left( h^1_\alpha(\tau) \boldsymbol{1}_{\mathcal{T}'}(\tau) \right)
	= \mathbf{E}_G\left( h^1(\tau) \boldsymbol{1}_{\mathcal{T}'}(\tau) \right)
	\in \mathbf{R}
	\label{eq:integrable_supergradient_1T}
\end{equation}
by bounded convergence.
Thus $h^1$ satisfies property \ref{bullet:integrable_supergradient_a}.

Define $\widebar{\alpha} \coloneqq \varepsilon' / \left( \varepsilon' + \abs*{u^1-u} \right)$.
For any $t \in \mathbf{R}_+ \setminus \mathcal{T}'$ we have
either $X_t \leq \left( u \wedge u^1 \right) - \varepsilon' < u$
or (if $u^1>0$) $X_t \geq \left( u \vee u^1 \right) + \varepsilon' > u$.
Thus for any $\alpha \in \left( 0, \widebar{\alpha} \right)$,
either
\begin{equation*}
	X_t
	< (1-\alpha) X_t + \alpha u
	\leq (1-\alpha) \left( u \wedge u^1 \right) - (1-\alpha) \varepsilon' + \alpha u
	< u^1
\end{equation*}
(where the last inequality holds by $\alpha<\widebar{\alpha}$)
or (similarly)
\begin{equation*}
	X_t
	> (1-\alpha) X_t + \alpha u
	\geq (1-\alpha) \left( u \vee u^1 \right) + (1-\alpha) \varepsilon' + \alpha u
	> u^1 .
\end{equation*}
In short, $(1-\alpha) X_t + \alpha u$ lies between $X_t$ and $u^1$
for any $t \in \mathbf{R}_+ \setminus \mathcal{T}'$ and $\alpha \in \left( 0, \widebar{\alpha} \right)$,
which since $F^1$ is concave and maximised at $u^1$ implies that
$\left( 0, \widebar{\alpha} \right) \ni \alpha \mapsto h^1_\alpha \boldsymbol{1}_{\mathbf{R}_+ \setminus \mathcal{T}'}$ is (pointwise) decreasing and non-negative.
Hence
\begin{equation*}
	0
	\leq \lim_{\alpha \downarrow 0}
	\mathbf{E}_G\left( h^1_\alpha(\tau) \boldsymbol{1}_{\mathbf{R}_+ \setminus \mathcal{T}'}(\tau) \right)
	= \mathbf{E}_G\left( h^1(\tau) \boldsymbol{1}_{\mathbf{R}_+ \setminus \mathcal{T}'}(\tau) \right)
\end{equation*}
by the monotone convergence theorem,
where the right-hand side is well-defined.
This together with \eqref{eq:integrable_supergradient_1T}
implies that $h^1$ satisfies \ref{bullet:integrable_supergradient_b} and \ref{bullet:integrable_supergradient_c}.
\qed
\renewcommand{\qedsymbol}{$\blacksquare$}

\subsection{Proof of the Gateaux lemma (p. \pageref*{lemma:euler_gateaux})}
\label{omit:Euler_lemma_pf:euler_gateaux}

For $\alpha \in (0,1)$, define $h^0_\alpha, h^1_\alpha : \mathbf{R}_+ \to (-\infty,\infty]$:
\begin{align*}
	h^0_\alpha(t)
	&\coloneqq r e^{-rt}
	\frac{ F^0\left( x_t+\alpha \left[ x^\dag_t - x_t \right] \right)
	- F^0( x_t ) }
	{ \alpha }
	\\
	\text{and}\quad
	h^1_\alpha(t)
	&\coloneqq e^{-rt}
	\frac{ F^1\left( X_t + \alpha \left[X^\dag_t-X_t\right] \right) - F^1(X_t) }
	{\alpha} .
\end{align*}
Clearly both are measurable for each $\alpha \in (0,1)$.
We have
\begin{equation*}
	\mathrm{D} \pi_G\bigl( x, x^\dag-x \bigr)
	= \lim_{\alpha \downarrow 0} \left[
	\mathbf{E}_G\left( \int_0^\tau h^0_\alpha \right)
	+ \mathbf{E}_G\left( h^1_\alpha(\tau) \right)
	\right]
\end{equation*}
by definition;
we must show that the limit exists and has the asserted value.

As $\alpha \downarrow 0$,
$h^0_\alpha$ and $h^1_\alpha$
converge pointwise
to
$h^0 : \mathbf{R}_+ \to (-\infty,\infty]$
and $h^1 : \mathbf{R}_+ \to [-\infty,\infty]$, respectively, given by
\begin{align*}
	h^0(t)
	&\coloneqq r e^{-rt} F^{0\prime}\left( x_t, x^\dag_t \right)
	\left( x^\dag_t - x_t \right)
	\\
	\text{and}\quad
	h^1(t)
	&\coloneqq e^{-rt} F^{1\prime}\left( X_t, X^\dag_t \right)
	\left( X^\dag_t - X_t \right) .
\end{align*}

\begin{namedthm}[Claim A.]
	%
	\label{claim:gateaux_integral_convergence}
	%
	$\mathrm{D}\pi_G\bigl(x,x^\dag-x\bigr)$ exists
	and is finite, equal to
	$\mathbf{E}_G\left( \int_0^\tau h^0 \right)
	+ \mathbf{E}_G\left( h^1(\tau) \right)$.
	%
\end{namedthm}

By \hyperref[claim:gateaux_integral_convergence]{claim A}, we have
\begin{multline*}
	\mathrm{D} \pi_G\left( x, x^\dag-x \right)
	= \mathbf{E}_G\left( \int_0^\tau h^0 \right)
	+ \mathbf{E}_G\left(h^1(\tau)\right)
	\\
	\begin{aligned}
		&= \mathbf{E}_G\left( r \int_0^\tau e^{-rs} \phi^0(s)
		\left[ x_s^\dag - x_s \right] \mathrm{d} s \right)
		+ \mathbf{E}_G\left( e^{-r\tau} \phi^1(\tau)
		\left[ X^\dag_\tau - X_\tau \right] \right)
		\\
		&\quad
		+ \mathbf{E}_G\left( r \int_0^\tau e^{-rt}
		\left[ F^{0\prime}\left( x_t, x^\dag_t \right) - \phi^0(t) \right]
		\left[ x_t - x^\dag_t \right] \mathrm{d} t \right)
		\\
		&\quad
		+ \mathbf{E}_G\left( e^{-r\tau}
		\left[ F^{1\prime}\left( X_\tau, X^\dag_\tau \right) - \phi^1(\tau) \right]
		\left[ X^\dag_\tau - X_\tau \right] \right) ,
	\end{aligned}
\end{multline*}
where the last equality holds since $t \mapsto r \int_0^t e^{-rs}\phi^0(s) \mathrm{d} s$ and $\phi^1$ are $G$-integrable
and $x,x^\dag,X,X^\dag$ are bounded.
The proof is completed by invoking the following claims to rewrite the first two terms in the desired form:

\begin{namedthm}[Claim B.]
	%
	\label{claim:gateaux_algebra_b}
	%
	It holds that
	%
	\begin{equation*}
		\mathbf{E}_G\left( r \int_0^\tau e^{-rs}\phi^0(s)
		\left[ x_s^\dag - x_s \right] \mathrm{d} s
		\right)
		= r \int_0^\infty e^{-rt} [1-G(t)]\phi^0(t)
		\left( x_t^\dag - x_t \right) \mathrm{d} t .
	\end{equation*}
	%
\end{namedthm}

\begin{namedthm}[Claim C.]
	%
	\label{claim:gateaux_algebra_c}
	%
	It holds that
	%
	\begin{equation*}
		\mathbf{E}_G\left(
		e^{-r\tau}\phi^1(\tau) \left[ X^\dag_\tau - X_\tau \right]
		\right)
		= r \int_0^\infty e^{-rt}
		\left( \int_{[0,t]} \phi^1 \mathrm{d} G \right)
		\left( x^\dag_t - x_t \right) \mathrm{d} t .
	\end{equation*}
	%
\end{namedthm}

\begin{proof}[Proof of {\hyperref[claim:gateaux_integral_convergence]{claim A}}]%
	\renewcommand{\qedsymbol}{$\square$}
	%
	We must show that $t \mapsto \int_0^t h^0$ and $h^1$
	are $G$-integrable and
	%
	\begin{equation*}
		\mathbf{E}_G\left( \int_0^\tau h^0_\alpha \right)
		\to \mathbf{E}_G\left( \int_0^\tau h^0 \right)
		\quad \text{and} \quad
		\mathbf{E}_G\left( h^1_\alpha(\tau) \right)
		\to \mathbf{E}_G\left( h^1(\tau) \right)
		\quad \text{as $\alpha \downarrow 0$.}
	\end{equation*}
	%
	Fix a $u \in \left(0,u^0\right)$.

	For $h^0$, define $\varphi : \mathbf{R}_+ \to [0,\infty]$ by
	%
	\begin{equation*}
		\varphi(t)
		\coloneqq r e^{-rt}
		\left[ F^{0\prime}(x_t,u) + F^{0\prime}\left(x_t^\dag,u\right) \right] .
	\end{equation*}
	%
	Recall the definition of $\psi^0_{x,u},\psi^0_{x^\dag,u}$ (p. \pageref*{eq:euler_psi01_defn}).
	Since $x$ and $x^\dag$ belong to $\mathcal{X}_G$,
	the map
	%
	\begin{equation*}
		t \mapsto \int_0^t \varphi = \psi^0_{x,u}(t) + \psi^0_{x^\dag,u}(t)
	\end{equation*}
	%
	is $G$-integrable,
	and thus $\varphi$ (being non-negative) is Lebesgue-integrable on $(0,t)$ for any $t \in \mathbf{R}_+$ with $G(t)<1$.%
		\footnote{If $\varphi$ were not integrable on $(0,t)$,
		then $\smash{\psi^0_{x,u}(s) + \psi^0_{x^\dag,u}(s) = \int_0^s \varphi \geq \int_0^t \varphi = \infty}$ for every $s \geq t$,
		which if $G(t)<1$ implies
		$\smash{\mathbf{E}_G( \psi^0_{x,u}(\tau) + \psi^0_{x^\dag,u}(\tau) ) = \infty}$,
		a contradiction.}
	Since $F^0$ is concave, we have
	%
	\begin{equation*}
		\abs*{ h^0_\alpha(t) }
		\leq \varphi(t) \abs*{ x_t - x^\dag_t }
		\leq \varphi(t) u^0
		\quad \text{for all $\alpha \in (0,1)$ and $t \in \mathbf{R}_+$,}
	\end{equation*}
	%
	and thus
	%
	\begin{equation*}
		\abs*{ \int_0^t h^0_\alpha }
		\leq \left[\psi^0_{x,u}(t) +\psi^0_{x^\dag,u}(t)\right] u^0
		\quad \text{for all $\alpha \in (0,1)$ and $t \in \mathbf{R}_+$.}
	\end{equation*}
	%
	Thus applying the dominated convergence theorem twice yields
	%
	\begin{equation*}
		\lim_{\alpha \downarrow 0}
		\mathbf{E}_G\left( \int_0^\tau h_\alpha^0 \right)
		= \mathbf{E}_G\left( \lim_{\alpha \downarrow 0}
		\int_0^\tau h_\alpha^0 \right)
		= \mathbf{E}_G\left(\int_0^\tau h^0 \right)
		\in \mathbf{R} .
	\end{equation*}

	Similarly, since $F^1$ is concave and $X,X^\dag$ take values in $\left[0,u^0\right]$, we have
	%
	\begin{align*}
		\abs*{h^1_\alpha(t)}
		&\leq \left( \abs*{\psi^1_{X,u}(t)}
		+ \abs*{\psi^1_{X^\dag,u}(t)} \right)
		\abs*{X^\dag_t- X_t}
		\\
		&\leq \left(\abs*{\psi^1_{X,u}(t)}+\abs*{\psi^1_{X^\dag,u}(t)}\right) u^0
		&& \text{for all $\alpha \in (0,1)$ and $t \in \mathbf{R}_+$.}
	\end{align*}
	%
	The right-hand side is $G$-integrable by since $x$ and $x^\dag$ belong to $\mathcal{X}_G$,
	so
	%
	\begin{equation*}
		\lim_{\alpha \downarrow 0}
		\mathbf{E}_G\left( h^1_\alpha(\tau) \right)
		= \mathbf{E}_G\left( h^1(\tau) \right)
		\in \mathbf{R}
	\end{equation*}
	%
	by dominated convergence.
	%
\end{proof}%
\renewcommand{\qedsymbol}{$\blacksquare$}

\begin{proof}[Proof of {\hyperref[claim:gateaux_algebra_b]{claim B}}]%
	\renewcommand{\qedsymbol}{$\square$}
	%
	Fix a $T \in \mathbf{R}_+$ such that $\lim_{t \uparrow T} G(t) < 1$, and note that $l_T : \mathbf{R}_+ \to \mathbf{R}$ given by
	%
	\begin{equation*}
		l_T(t)
		\coloneqq r e^{-rt} \phi^0(t) \left[ x_t^\dag - x_t \right]
		\boldsymbol{1}_{[0,T)}(t)
	\end{equation*}
	%
	is (Lebesgue-)integrable since $t \mapsto r \int_0^t e^{-rs}\phi^0(s) \mathrm{d} s$ is $G$-integrable and $x,x^\dag$ are bounded.
	We may therefore apply \hyperref[lemma:int_parts]{IBP lemma} (p. \pageref{lemma:int_parts}) to $l_T$ and the measure associated with $G$ to obtain
	%
	\begin{multline*}
		\int_{[0,T]} \left(
		r \int_0^t e^{-rs}\phi^0(s) \left[ x_s^\dag - x_s \right] \mathrm{d} s
		\right) G( \mathrm{d} t)
		\\
		= r \int_0^T [G(T)-G(t)]
		e^{-rt}\phi^0(t)
		\left[ x_t^\dag - x_t \right]
		\mathrm{d} t .
	\end{multline*}

	Define $\widebar{T} \coloneqq \sup\{ t \in \mathbf{R}_+: G(t) < 1 \}$.
	If $\widebar{T} < \infty$
	and $G$ has an atom at $\widebar{T}$,
	then we obtain the desired equation by setting $T \coloneqq \widebar{T}$.
	Suppose for the remainder that this is not the case.
	It suffices to show that
	$\chi : \mathbf{R}_+ \to [0,\infty]$ given by
	%
	\begin{equation*}
		\chi(t)
		\coloneqq
		\begin{cases}
			[1-G(t)] r e^{-rt} \phi^0(t)
			& \text{for $t < \widebar{T}$} \\
			0
			& \text{for $t \geq \widebar{T}$}
		\end{cases}
	\end{equation*}
	%
	satisfies $\int_0^\infty \chi < \infty$ (so that it is Lebesgue-integrable),
	for then letting $T \uparrow \widebar{T}$ above yields the desired result by the dominated convergence theorem
	since $t \mapsto r \int_0^t e^{-rs}\phi^0(s) \mathrm{d} s$ is $G$-integrable
	and $x,x^\dag$ are bounded.

	To show that $\int_0^\infty \chi$ is finite, note that applying the \hyperref[lemma:int_parts]{IBP lemma} as above to
	$x$ and $\widetilde{x}^\dag \coloneqq x + k$ (where $k \in \mathbf{R}$ is a constant)
	yields
	%
	\begin{equation*}
		\int_{[0,T]} \left(
		r \int_0^t e^{-rs}\phi^0(s) \mathrm{d} s
		\right) G(\mathrm{d} t)
		= r \int_0^T [G(T)-G(t)]
		e^{-rt}\phi^0(t)
		\mathrm{d} t
	\end{equation*}
	%
	for any $T \in \mathbf{R}_+$ with $G(T) < 1$.
	Letting $T \uparrow \widebar{T}$
	and using the dominated (monotone) convergence theorem on the left-hand (right-hand) side yields
	%
	\begin{equation*}
		\int_{\left[0,\widebar{T}\right)} \left(
		r \int_0^t e^{-rs}\phi^0(s) \mathrm{d} s
		\right) G(\mathrm{d} t)
		= \int_0^{\widebar{T}} \chi ,
	\end{equation*}
	%
	which since $G$ has no atom at $\widebar{T}$ implies that
	%
	\begin{equation*}
		\mathbf{E}_G\left( r \int_0^\tau e^{-rt} \phi^0(t) \mathrm{d} t \right)
		= \int_0^{\widebar{T}} \chi
		= \int_0^{\infty} \chi .
	\end{equation*}
	%
	Thus $\int_0^\infty \chi < \infty$
	since $t \mapsto r \int_0^t e^{-rs}\phi^0(s) \mathrm{d} s$ is $G$-integrable.
	%
\end{proof}%
\renewcommand{\qedsymbol}{$\blacksquare$}

\begin{proof}[Proof of {\hyperref[claim:gateaux_algebra_c]{claim C}}]%
	\renewcommand{\qedsymbol}{$\square$}
	%
	Define a measure $\nu_+$ on $\mathbf{R}_+$ by
	%
	\begin{equation*}
		\nu_+(A)
		\coloneqq \int_A \max\left\{\phi^1,0\right\} \mathrm{d} G
		\quad \text{for any measurable $A \subseteq \mathbf{R}_+$.}
	\end{equation*}
	%
	Since $\phi^1$ is $G$-integrable, $\nu_+$ is absolutely continuous with respect to (the measure associated with) $G$, with Radon--Nikod\'{y}m derivative $\max\{\phi^1,0\}$.
	Furthermore, the function $L : \mathbf{R}_+ \to \mathbf{R}$ defined by
	%
	\begin{equation*}
		L(t) \coloneqq e^{-rt}\left( X^\dag_t - X_t \right)
		\quad \text{for each $t \in \mathbf{R}_+$}
	\end{equation*}
	%
	is $\nu_+$-integrable since $X^\dag$ and $X$ are bounded.
	Thus for any $T \in \mathbf{R}_+$, we have
	%
	\begin{equation*}
		\int_{[0,T]} L \max\left\{\phi^1,0\right\} \mathrm{d} G
		= \int_{[0,T]} L \mathrm{d} \nu_+ .
	\end{equation*}
	%
	Furthermore, $L$ satisfies $L(t) = L(0) + \int_0^t l$, where $\smash{l(t) \coloneqq - e^{-rt} \bigl( x^\dag_t - x_t \bigr)}$.
	Thus for every $T \in \mathbf{R}_+$, the \hyperref[lemma:int_parts]{IBP lemma} yields
	%
	\begin{multline*}
		\int_{[0,T]} e^{-rt}\left(X^\dag_t -X_t\right) \nu_+(\mathrm{d} t)
		\\
		= e^{-rT} \left( X^\dag_T - X_T \right) \nu_+([0,T])
		+ r \int_0^T \nu_+([0,t])
		e^{-rt} \left( x^\dag_t - x_t \right) \mathrm{d} t .
	\end{multline*}
	%
	Applying the same argument to
	$\nu_-(A) \coloneqq - \int_A \min\left\{ \phi^1, 0 \right\} \mathrm{d} G$
	and subtracting yields, for each $T \in \mathbf{R}_+$,
	%
	\begin{multline*}
		\int_{[0,T]} e^{-rt}\left( X^\dag_t - X_t \right)
		\phi^1(t) G(\mathrm{d} t)
		\\
		= e^{-rT} \left( X^\dag_T - X_T \right) \int_{[0,T]}
		\phi^1 \mathrm{d} G
		+ r \int_0^T
		\left( \int_{[0,t]} \phi^1 \mathrm{d} G \right)
		e^{-rt} \left( x^\dag_t - x_t \right) \mathrm{d} t .
	\end{multline*}
	%
	Letting $T \to \infty$ yields the desired equation
	by boundedness of $x,x^\dag,X,X^\dag$
	and dominated convergence.
	%
\end{proof}%
\renewcommand{\qedsymbol}{$\blacksquare$}

With all three claims established,
the proof is complete.
\qed

\section*{\Cref*{suppl:construction:pf_lemma_approx_F} (p. \pageref*{suppl:construction:pf_lemma_approx_F})}
\label{omit:construction:pf_lemma_approx_F}
\addcontentsline{toc}{section}{Supplemental appendix \ref*{suppl:construction:pf_lemma_approx_F} (p. \pageref*{suppl:construction:pf_lemma_approx_F})}

Here's an example of a sequence of technologies
satisfying properties \ref*{bullet:thmC_pf_general:a}--\ref*{bullet:thmC_pf_general:d}.
Choose an $N \in \mathbf{N}$ such that $1/N < (u^0-u^1)/3$.
Given $n \geq N$, let $I_n \coloneqq [ 1/n , u^0 - 2/n ]$.
Fix $\gamma^n \in ( 0, 1/u^0n )$ and $\delta^n \in ( 0, 1/n )$.
For $j \in \{0,1\}$, define $f^j_n : I_n \rightarrow \mathbf{R}$ by
$\smash{f^j_n(u)
\coloneqq \frac{1}{\delta^n}\int_u^{u+\delta^n} F^{j+}
- \gamma^n u}$.
Given $\zeta^n \in (0,2/n)$, define $F^0_n,F^1_n : I_n \rightarrow \mathbf{R}$ by
\begin{align*}
	F^0_n(u)
	&\coloneqq F^0\left( u^\star + \frac{1}{n} \right)
	+ \int_{u^\star+1/n}^u f^0_n
	\\
	\text{and} \quad
	F^1_n(u)
	&\coloneqq F^1\left( u^\star + \frac{1}{n} \right)
	+ \int_{u^\star+1/n}^u f^1_n + \zeta^n .
\end{align*}
$F^0_n,F^1_n$ are strictly concave and differentiable, with (continuous) derivatives $f^0_n,f^1_n$ such that
\begin{equation}
	F^{j-}(u+1/n) - 1
	\leq f^j_n(u)
	\leq F^{j+}(u)
	\quad \text{for $j \in \{0,1\}$ and $u \in I_n$.}
	\label{eq:derivative_bounds}
	\tag{$\natural$}
\end{equation}
\iffalse
(For the left-hand side,
$f^j_n(u)
\geq F^{j+}\left( u + \delta^n \right) - \frac{1}{u^0 n} u^0
\geq F^{j+}\left( u + \delta^n \right) - 1$.)
\fi
As $n \to \infty$,
$f^j_n \to F^{j+}$ pointwise for both $j \in \{0,1\}$ since $F^{j+}$ is right-continuous,
so that $F^0_n \to F^0$ and $F^1_n - \zeta^n \to F^1$ pointwise by the bounded convergence theorem.
In fact, the convergence is uniform \parencite[][Theorem 10.8, p. 90]{Rockafellar1970}.
Thus for any $\varepsilon^n \in (0,1/n)$,
provided $\delta^n$ and $\gamma^n$ are sufficiently small,
we have that $F^0_n - F^0$ and $F^1_n - \zeta^n - F^1$ are bounded by $\varepsilon^n$,
so that choosing $\zeta^n > 2 \varepsilon^n$ ensures $F^1_n > F^0_n$ (as $F^1 \geq F^0$).
\iffalse
Because
\begin{equation*}
	F^1_n - F^0_n
	= \left( F^1 - F^0 \right)
	+ \left( F^1_n - F^1 - \zeta^n \right)
	+ \left( F^0 - F^0_n \right)
	+ \zeta^n
	\geq 0 - 2 \varepsilon^n + \zeta^n
	> 0 .
\end{equation*}
\fi

Since $F^1_n(1/n) > F^0_n(1/n)$, we may extend $F^0_n,F^1_n$ to $[0,u^0-2/n]$ while
preserving strict concavity, (continuous) differentiability and $F^1_n \geq F^0_n$
and ensuring that $F^1_n-F^0_n$ is strictly increasing on a neighbourhood of $0$
(which guarantees that $u^\star_n > 0$)
and that $F^{0\prime}_n,F^{1\prime}_n$ are bounded above on $(0,1/n]$ by $\max_{j \in \{0,1\}} F^{j+}(1/n) + 1$.
(The latter is possible because $F^{j\prime}_n(1/n) \leq F^{j+}(1/n)$ by \eqref{eq:derivative_bounds}.)
Further extend $F^1_n$ to all of $[0,\infty)$ by making it
differentiable and strictly concave on $\left[u^0-2/n,\infty\right)$, with derivative bounded below by $F^{1\prime}_n(u^0-2/n)-1/n$.
Since $F^1_n(u^0-2/n) > F^0_n(u^0-2/n)$, we may also extend $F^0_n$ to $[0,\infty)$ while
preserving strict concavity, (continuous) differentiability and $F^1_n \geq F^0_n$
and ensuring that $u^0_n \leq u^0$
and that $F^{0\prime}_n$ is bounded below by $F^{1-}(u^0-1/n)- 2$.
(The latter is possible since
$F^{1\prime}_n$ is bounded below by $F^{1\prime}_n(u^0-2/n)-1/n \geq F^{1-}(u^0-1/n)- 2$
by \eqref{eq:derivative_bounds}.)
If $u^\star_n$ is not a strict local maximum of $F^1_n-F^0_n$,
then we may make it so by decreasing $F^1_n$ pointwise on $[0,u^\star_n)$ by a small amount.

\begin{claim}
	%
	\label{claim:monster}
	%
	The sequence $(F^0_n,F^1_n)_{n=N}^\infty$ satisfies the model assumptions and properties \ref*{bullet:thmC_pf_general:a}--\ref*{bullet:thmC_pf_general:d}
	provided $\varepsilon^n \in (0,1/n)$ is chosen sufficiently small.
	%
\end{claim}

\begin{proof}
	%
	It is clear that $(F^0_n,F^1_n)_{n=N}^\infty$ satisfies property \ref*{bullet:thmC_pf_general:a},
	and that for each $n \in \mathbf{N}$, $F^0_n,F^1_n$ satisfy the all of the model assumptions except for $u^1_n < u^0_n$ (the conflict of interest).

	For \ref*{bullet:thmC_pf_general:b} and the conflict of interest, since and $u^0$, $u^1$ and $u^\star$ are strict local maxima of (respectively) $F^0$, $F^1$ and $F^1-F^0$, we have
	%
	\begin{equation*}
		u^0 - \frac{1}{n} \leq u^0_n \leq u^0 ,
		\quad
		\abs*{u^1_n-u^1} \leq \frac{1}{n}
		\quad \text{and} \quad
		\abs*{u^\star_n-u^\star} \leq \frac{1}{n}
	\end{equation*}
	%
	for $\varepsilon^n > 0$ sufficiently small, as $F^0_n - F^0$ and $F^1_n - (F^1+\zeta^n)$ are bounded by $\varepsilon^n$ on $I_n$.
	In particular, $u^1_n < u^0_n$ (the conflict of interest) since $n \geq N$.
	Letting $n \to \infty$ yields \ref*{bullet:thmC_pf_general:b}.

	For the first part of \ref*{bullet:thmC_pf_general:c},
	fix a $u \in (0,u^0]$ at which $F^{1-}$ is finite.
	If $u=u^0$,
	then for each $n \geq N$,
	$F^{0\prime}_n, F^{1\prime}_n$ are bounded below by $F^{1-}(u^0-1/n) - 2$
	(by construction for $F^{0\prime}_n$, and by (\ref{eq:derivative_bounds}) for $F^{1\prime}_n$).
	Suppose instead that $u<u^0$.
	Choose a $u' \in (u,u^0)$, and let $N' \geq N$
	satisfy $1/N' < u < u^0 - 2/N'$ and $1/N' < u'-u$.
	Then $u$ belongs to $I_n$ for every $n \geq N'$,
	so that $F^{j\prime}_n(u) \geq F^{j-}(u+1/n) - 1 \geq F^{j-}(u') - 1$ by (\ref{eq:derivative_bounds}) for both $j \in \{0,1\}$.
	The set
	%
	\begin{equation*}
		\bigcup_{j \in \{0,1\}}
		\left\{ F^{j\prime}_n(u) : n \in \{1,\dots,N'\} \right\}
	\end{equation*}
	%
	is bounded since it is finite.
	Thus $(F^{0\prime}_n)_{n \in \mathbf{N}}$ and $(F^{1\prime}_n)_{n \in \mathbf{N}}$ are uniformly bounded below on $[0,u]$.

	The argument for the second part of \ref*{bullet:thmC_pf_general:c} is analogous:
	fix a $u \in [0,u^0)$ at which $F^{0+},F^{1+}$ are finite.
	If $u=0$,
	then $F^{0\prime}_n,F^{1\prime}_n$ are bounded above by $\max_{j \in \{0,1\}} F^{0+}_j(1/n)+1$ for all $n \geq N$
	by construction.
	If instead $u>0$,
	then $F^{j\prime}_n(u) \leq F^{j+}(u)$ for all $n \geq N$ such that $1/n < u < u^0 - 2/n$ by (\ref{eq:derivative_bounds}),
	and the set
	%
	\begin{equation*}
		\bigcup_{j \in \{0,1\}}
		\left\{ F^{j\prime}_n(u) :
		n \in \left\{1, \dots,
		\floor*{1/u} \vee
		\floor*{ \left. 2 \middle/ \left(u^0-u\right) \right. }
		\right\} \right\}
	\end{equation*}
	%
	is finite and thus bounded;
	so $(F^{0\prime}_n)_{n \in \mathbf{N}}$ and $(F^{1\prime}_n)_{n \in \mathbf{N}}$ are uniformly bounded above on $\left[u,u^0\right]$.

	For \ref*{bullet:thmC_pf_general:d}, fix $j \in \{0,1\}$ and a convergent sequence $(u_n)_{n=N}^\infty$ with limit $u \in [0,\infty)$ such that $0 < u_n \leq u^0_n$ for each $n \geq N$ and $\lim_{n \to \infty} F^{j\prime}_n(u_n)$ exists.
	Since $u^0_n \leq u^0$ for each $n \geq N$, we have $u \leq u^0$.
	From above,
	%
	\begin{equation}
		F^{j\prime}_n(u_n)
		= f^j_n(u_n)
		\;
		\begin{cases}
			\leq F^{j+}(u_n)
			& \text{if $u_n \geq 1/n$} \\
			\geq F^{j-}(u_n+1/n) - \gamma^n u_n
			& \text{if $u_n \leq u^0-2/n$}.
		\end{cases}
		\label{eq:monster_fn_dag}
		\tag{$\dag$}
	\end{equation}
	%
	Letting $n \to \infty$ yields
	%
	\begin{equation*}
		\lim_{n \to \infty} F^{j\prime}_n(u_n)
		\;
		\begin{cases}
			\leq F^{j-}(u)
			& \text{if $u>0$} \\
			\geq F^{j+}(u)
			& \text{if $u<u^0$.}
		\end{cases}
	\end{equation*}
	%
	If $u \in (0,u^0)$, then both cases are satisfied,
	so
	%
	\begin{equation*}
		F^{j+}(u) \leq \lim_{n \to \infty} F^{j\prime}_n(u_n) \leq F^{j+}(u) .
	\end{equation*}
	%
	If $u=0$, then the second case applies, so that
	$F^{j+}(0) \leq \lim_{n \to \infty} F^{j\prime}_n(u_n)$,
	which suffices.
	If $u=u^0$, then the first case holds, so it remains only to show that $F^{j+}( u^0 ) \leq \lim_{n \to \infty} F^{j\prime}_n(u_n)$.
	For $j=0$, we have
	$F^{0+}(u^0) \leq 0 \leq \lim_{n \to \infty} F^{0\prime}_n(u_n)$
	since $u_n \leq u^0_n$ for each $n \geq N$.
	For $j=1$, \eqref{eq:monster_fn_dag} implies
	%
	\begin{multline*}
		F^{1-}\left( \min\left\{ u_n + 1/n, u^0 - 1/n \right\} \right)
		- \gamma^n u_n - 1/n
		\quad
		\\
		\leq F^{1\prime}_n\left( \min\left\{ u_n, u^0 - 2/n \right\} \right) - 1/n
		\leq F^{1\prime}_n(u_n)
		\quad \text{for each $n \geq N$,}
	\end{multline*}
	%
	where the second inequality holds since $F^{1\prime}_n$ is bounded below by $F^{1\prime}_n(u^0-2/n)-1/n$. Letting $n \to \infty$ yields
	$F^{1+}( u^0 )
	\leq \lim_{n \to \infty} F^{1\prime}_n(u_n)$.
	%
\end{proof}

\section*{\Cref*{suppl:mcs:pf_lemma_euler_optimal} (p. \pageref*{suppl:mcs:pf_lemma_euler_optimal})}
\label{omit:mcs:pf_lemma_euler_optimal}
\addcontentsline{toc}{section}{Supplemental appendix \ref*{suppl:mcs:pf_lemma_euler_optimal} (p. \pageref*{suppl:mcs:pf_lemma_euler_optimal})}

\begin{proof}[Proof of \Cref*{observation:piG_str_conc}]
	%
	It suffices to show that $\pi_G$ is strictly concave.
	For distinct $x,x^\dag \in \mathcal{X}$ (i.e. $x \neq x^\dag$ on a non-null set) and any $\lambda \in (0,1)$, we have
	%
	\begin{multline*}
		\pi_G\left( \lambda x + (1-\lambda) x^\dag \right)
		\\
		\begin{aligned}
			&= \mathbf{E}_G\left( r \int_0^\tau e^{-rt}
			F^0\left( \lambda x_t + (1-\lambda) x^\dag_t \right) \mathrm{d} t
			+ e^{-r\tau} F^1\left( \lambda X_\tau + (1-\lambda) X^\dag_\tau \right)
			\right)
			\\
			&> \mathbf{E}_G\biggl( r \int_0^\tau e^{-rt}
			\left[ \lambda F^0(x_t)
			+ (1-\lambda) F^0\left( x^\dag_t \right) \right] \mathrm{d} t
			\\
			&\qquad\qquad\quad + e^{-r\tau} \left[ \lambda F^1(X_\tau)
			+ (1-\lambda) F^1\left( X^\dag_\tau \right) \right]
			\biggr)
			\\
			&= \lambda \pi_G(x) + (1-\lambda) \pi_G\left( x^\dag \right) ,
		\end{aligned}
	\end{multline*}
	%
	where the inequality holds since $F^1$ is concave,
	$F^0$ is strictly concave, $G$ has unbounded support and $x \neq x^\dag$ on a non-null set.
	%
\end{proof}

\printbibliography[heading=bibintoc]

\end{document}

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text:
Fire EMS, Firefighting, Health & Safety
Indian River County (FL) Provides Timeline for Firefighters' Temporary Housing
Assistant Chief for Emergency Medical Services Brian Burkeen said that the first modular housing unit for firefighters is expected to arrive in about two weeks, but that it will be late June or early July before workers can live in them, reports veronews.com.
Modulars were ordered to house firefighters and paramedics from Station 7 on Ranch Road near Interstate 95 and Station 1 on Old Dixie Highway in Vero. Station 1 is set to be gutted and renovated at a cost of more than $1.1 million. The County is considering demolishing Station 7 and purchasing or leasing property to build anew in a different location.
"We have had conversations with the manufacturer in regards to the modular buildings and the expected delivery date of the first one is June 9, 2015. This date of delivery may change given any scheduling conflicts from the builder," Burkeen said in response to questions from VeroNews.com.
A purchase order dated April 10 shows that two modular buildings were ordered from Modular Building Systems International in Clermont, at an all-in cost including setup and transport of just under $100,000 each. Each 24-foot by 60-foot commercial-grade modular building will house up to five people and will be rated for wind speeds of up to 170 miles per hour once secured properly on the site.
In the meantime, the Board of County Commissioners told the staff in late March to get firefighters out of those two stations as soon as possible to remove the first responders from potential environmental hazards, including mold and debris from rodent infestations.
Management personnel and the ladder company from Station 1 were moved to other stations in Vero for the duration of the renovations, leaving a skeleton crew of firefighters and paramedics at the Old Dixie station.
The Board initially approved $250,000 for the modulars, and last week commissioners added $200,000 to that in hopes of expediting the process.
The units are manufactured in Georgia, Burkeen said, and "will be delivered straight from the factory."
After the modulars arrive, they will need to be hooked up to utilities and to the county's emergency communications system.
"Applications for permit at Station 1 has been submitted to the City of Vero Beach and we are awaiting approval of that permit," Burkeen said.
The Emergency District staff is evaluating some options for the workers at Station 7, including an idea floated by Chairman Wesley Davis last week to house firefighters on the existing fire station property instead of moving operations twice.
"Once arrived and permits are obtained, we anticipate that they will be move-in ready within 21 to 30 days given there are no delays. We do not anticipate any delays in the process," Burkeen said.
On Tuesday, the Indian River Association of Fire Fighters Local 2201 President John O'Connor released a statement, plus more than 200 pages of investigative reports on four fire stations. O'Connor in that statement demanded firefighters be moved out of the four affected stations which tested positive for mold.
Firefighters claim that the water intrusion that resulted in the mold problems has been caused by the County's failure to properly maintain and repair leaks and other structural issues at the stations, some dating back to damage from the 2004 and 2005 hurricanes. Last fall, the County replaced the entire roof on Station 11 at Orchid.
Station 11 and Station 10 in Fellsmere are in the process of being repaired, but the decision was made by management for firefighters to shelter in place during the repairs, as work was scheduled to be completed in a shorter amount of time than it would take for modular buildings to arrive.
The union says more than 70 firefighters have filed workers' compensation claims for exposure to mold in the stations where they work and live while on shift. Of those, 14 so far have tested positive for mold-related toxins in their systems.
Read more of the story here http://bit.ly/1dyIq1r
Fireground Safety
Behind the Flames: Dealing with Electrical Injuries