text:
We're not going to make you jump through hoops either—we know it's a timely decision to make and one that you have to be confident in. We keep our comparison service quick and easy to use by allowing you to find the right provider in your local area. Simply enter your zip code, view and compare the plans and deals across a range of providers, and then select the one that best suits your needs. We also provide all the contact details that you'll need in order to call and set up your service.
Customer service is just as Important as channel selection when it comes to shopping for a TV service. Not only do you want your TV provider to offer reliability, but you also want to see that your provider has a proven track record of keeping people happy. CableTV.com can help! We give you the data you need to find the cable company in Dallas, TX with the highest consumer ratings. By following our recommendation, you can minimize the chances of having a bad customer experience with your chosen provider.
To some executives, no company offers a more egregious example of how the value of sports has spiraled out of control than Time Warner Cable. In 2013 the cable company, now owned by Charter Communications Inc., agreed to pay an average $334 million a year to broadcast Los Angeles Dodgers games for the next 25 years on its cable channel, SportsNet LA. That's roughly eight times what Fox reportedly paid in the previous Dodgers deal. To cover the cost, Time Warner Cable initially charged almost $5 per month per subscriber, making it one of the most expensive in the bundle.
Choosing a TV is a big decision, and one that should be based on your specific needs. Most TVs on the market today are smart TVs and have software geared toward streaming Netflix and other platforms. There is a school of thought that it's actually better to not have a smart TV because eventually the software updates stop rolling out. So long as you have enough HDMI ports on your set, you can use any popular streaming device like a Roku or Amazon Fire TV.
We mentioned before: YouTube TV is not YouTube Red. Red is more like an advanced, commercial-free version of regular ol' YouTube. The YouTube TV option costs more and has a lot of catching up to do to rival the other live TV streaming services. It's in limited cities, has limited channels, and is available on limited devices (Chromecast, Android, iOS, LG and Samsung TVs, Xbox One, and browsers on PCs). It doesn't even bundle in YouTube Red as part of the service—you'd have to spend another $10 a month on Red to keep ads out of your regular YouTube streaming. Premium channel add-ons include $11-a-month Showtime, $5-a-month Shudder, and $7-per-month Sundance. The big plus: unlimited storage in the cloud-based DVR option.
The abbreviation CATV is often used for cable television. It originally stood for Community Access Television or Community Antenna Television, from cable television's origins in 1948. In areas where over-the-air TV reception was limited by distance from transmitters or mountainous terrain, large "community antennas" were constructed, and cable was run from them to individual homes. The origins of cable broadcasting for radio are even older as radio programming was distributed by cable in some European cities as far back as 1924.[citation needed]
The second change is our budget. Our cable and internet package was roughly $90 per month, and this was about 7 or 8 years ago. Prices have since risen quite a bit in most markets. That was too expensive for what we got out of it (the base TV package, with no premium channels, was over $50 a month after the introductory price expired). We only rarely watched TV, and then it was usually limited to a few channels. We cut the cable TV service and elected to pay for a faster internet service. We pay a lot for Internet, but we are paying for a service we use.
Lots of people choose satellite TV because you can choose from a wider variety of shows and options than you can with cable TV. Again, plans with a lot of channels, features, and variety are more likely to be expensive. However, there are some cases in which you might really want a specific channel that a cable provider won't have. For example, if you are from New York but you live in California, it might be easier to get Giants games with a satellite TV service than with a cable service.
After Charter Communications recently acquired Time Warner Cable and Bright House Networks, it rebranded as Spectrum and has worked to simplify its services. Unfortunately, that means you'll only have one option if you just want cable TV. That one TV-only plan is called TV Select. It comes with at least 125 channels and about 46% of the most popular channels. Notable absences include E! and the Travel Channel, but otherwise you'll be getting a fair number of major networks.
Amazon Prime Video has a pretty huge collection of newer movies, and more importantly, exclusive shows that are really well done. It's now $119 per year, and comes with an assortment of other perks like free shipping, a one million song library and access to free Kindle books. But for me, the value comes with Amazon's homegrown movies and series. Many, if not all, are shot in 4K HDR (Ultra HD) so it's nice to have that option without having to pay more for it (like Netflix).
The only reasons not to get an inexpensive antenna are because you don't want to fuss between different inputs, or you can't find a suitable window or another spot in your home. Other than that, it's the best way to ensure you get all your local channels, which many streaming services lack in some form. Read our comparison of the best antennas available right now.
Hulu ($7.99/mo., $11.99/mo.): Hulu's original content isn't as copious or as impressive as Netflix's, but it did just win the first ever "Best Dramatic Series" Emmy for a subscription streaming service, courtesy of "The Handmaid's Tale." Hulu is also becoming more and more of a boon to TV buffs, thanks to a growing library of classic older shows, as well as some current ABC, NBC, and Fox series. (In fact, one of Hulu's main selling-points for cord-cutters is it has deals to allow subscribers to watch the most recent episode or episodes of much of those networks' programming.) The lower price tier includes commercial breaks. The higher tier kills the ads. Hulu also has an option to add live TV (starting at $39.99/mo.), covered further down.
Streaming live TV is an effective cable TV alternative, but it does have some drawbacks. While it doesn't have contracts or equipment costs, it requires a strong internet connection. You'll want to get an internet download speed of 50 Mbps or more to avoid buffering. But before you sign up, just make sure all this doesn't end up costing more than a cable and internet bundle would.
This type of service is also used to circumvent sports network blackouts or simply to mask your identity online from would-be identity thieves. Of course, check with your content provider's terms of service to make sure you are not breaking any end-user agreements. To learn the differences between a Smart DNS and VPN check out my post on VPN vs Smart DNS.
The second factor you should consider is definitely the terms and length of your contract with the provider. Thoroughly go through the contract agreements and make sure that you are not locked in a long-term contract. Take note that the provider may increase the cable TV price after a year or so. Therefore, signing up for a long-term contract will prevent you from switching to a different provider when this happens.
For the base price, you get on-demand stuff from almost all of the networks (but not The CW) and even get them live in some markets. There are lots of basic cable stations (minus Viacom-owned stations like SyFy and Comedy Central). Each new tier of service adds more channels, going up to $44.99 to add some sports programming, and $54.99 for 90 channels, ending with $74.99—that Ultra package has those 90 channels plus paid cable services HBO and Showtime (but no Starz) for a little less than adding them separately.
Equipment, taxes and Cable One broadcast TV surcharge based on market (varies by location) and Cable One sports programming surcharge of $5.00 are in addition to the monthly rate. These surcharges are designed to partially offset the rapidly escalating rates Cable One is charged for carriage of TV stations and sports programming. Please see https://support.cableone.net/hc/en-us/articles/115012028888-Cable-TV-Surcharges- for more information.
Here's where it all begins: You're sick of paying an exorbitant cable or satellite bill, and you have a strong sense that if you just limited your spending to a few streaming subscription services, you would be much more satisfied with your home entertainment experience. So let's say that you already have a good TV, a speedy internet connection and a set-top box. (If you don't, we'll get to that later.) Who gets your money?
Netflix.com: There is a monthly cost associated with Netflix , but the added convenience of keeping the movies and TV shows as long as you want, and being able to stream a selection of movies and TV shows over the computer through one of the options above, make it worth it. $8.99 for the basic package, plus you can get a 1 month netflix free trial!
Showtime has made itself an add-on with just about any service that offers the option. It's available through Hulu, Amazon Prime, and CBS All Access, plus the live TV streaming services (below). Or use the apps on Apple TV, iOS, Android, Roku, and Xbox One. The price to get Showtime those services is generally a couple buck lower per month, a $24 a year savings.
When we went through the ordering process for Spectrum and got all the way to the check-out, we realized we never got to choose which DVR we wanted. Spectrum offers both Motorola and Cisco DVRs (neither of which stand out from competitor DVRs like the Genie or Hopper 3), and as far as we can tell, you get whichever one Spectrum decides to send you.
All the cable-replacement services offer some type of free trial period, so you can try before you buy. Because most require a credit card number, you'll have to keep track of when the trial period expires and cancel if you don't want to continue the service. And remember that package details can change often, so check the latest offers before signing up.
One thing is for sure: A number of Tampa Bay residents have moved away from cable, and many are interested in exploring that option. The Tampa Bay Times asked readers to reach out about their experience with cutting the cord, and received around 100 responses. The most-cited reasons for making the leap were dissatisfaction with a cable provider and cost.
For supporters of teams outside your local area, some sport-specific streaming options might also be attractive. Each major sports league offers some sort of online viewing option for somewhere in the neighborhood of $110 a year, with the caveat that local games are blacked out. (NFL fans can pay only $69.99 to watch any team they like, but must make do with replays.)
As part of this switch over, we installed an over-the-air antenna. This allows us to pick up a large number of local channels. We live close enough to the Des Moines metro area that we can pick up strong signals from a lot of stations in Des Moines. (It's worth noting that this is aided by the fact that there's a repeater tower within five miles of our house, which means the signals are strong.) This includes the major broadcast networks – ABC, CBS, Fox, NBC, PBS – and a large number of additional channels of varying quality, including an all-children's channel and a channel that seems to be nothing but a loop of the weather forecast.
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Coaxial cables are capable of bi-directional carriage of signals as well as the transmission of large amounts of data. Cable television signals use only a portion of the bandwidth available over coaxial lines. This leaves plenty of space available for other digital services such as cable internet, cable telephony and wireless services, using both unlicensed and licensed spectrum. Broadband internet access is achieved over coaxial cable by using cable modems to convert the network data into a type of digital signal that can be transferred over coaxial cable. One problem with some cable systems is the older amplifiers placed along the cable routes are unidirectional thus in order to allow for uploading of data the customer would need to use an analog telephone modem to provide for the upstream connection. This limited the upstream speed to 31.2k and prevented the always-on convenience broadband internet typically provides. Many large cable systems have upgraded or are upgrading their equipment to allow for bi-directional signals, thus allowing for greater upload speed and always-on convenience, though these upgrades are expensive.
Verizon FiOS TV: Verizon's FiOS fiber television service is available in a number of states along the eastern seaboard and in Oregon and California. The company shook up the pay TV industry in 2015 by launching its custom build-your-own package, which allows customers to choose themed channel packs, similar to Dish's Flex Pack. You can see what channels they offer through our channel guide to Verizon FiOS. You can also bundle FiOS TV with FiOS high-speed internet.
What you get: The sports-centric Fubo now offers about 75 channels for the base $45-a-month package, or more than 90 channels with the new $50-a-month Fubo Extra plan. With Fubo you get a mix of live and on-demand channels from broadcast networks (CBS, Fox, and NBC in most markets), cable channels (A&E, Bravo, FX, SyFy, USA), and sports networks (BeIn Sports, FS1, Golf Channel, NBA TV). You also now get Turner channels, plus sports programming such as the MLB on TBS and the NBA on TNT. Also in the mix: a robust roster of regional sports networks—including those from NBC, Fox, and Yes—for local-team action, including MLB and NHL games. The service comes with a free cloud DVR that lets you store 30 hours of shows, movies, and games.
Apple TV has direct access to some streaming services using the remote control, and for others you must use an Apple device to cast the service to the TV. In addition, neither Chromecast nor Apple TV provides native access to the Amazon streaming service, so if you are an Amazon Prime member using either of these devices you are missing out on the valuable benefits from your Prime membership.
Where Mediacom really suffers is its customer service. It consistently ranks at the bottom, a worrisome practice in an industry with an already poor reputation. Consumer Reports readers gave it 58 out of 100, ACSI gave it 56 out of 100. If you choose Mediacom as your cable provider, keep a keen eye on your billing statements and confirm any deals your promised.
However, please note that I'm not advocating doing ALL of these! If you did, you'd wind up spending more than you were paying for cable. That would be dumb. We currently subscribe to Sling TV, Netflix, and Amazon Prime. We are switching over to DIRECTV NOW starting next month because we have one AT&T phone and can use the data from that phone to watch for free.
At the outset, cable systems only served smaller communities without television stations of their own, and which could not easily receive signals from stations in cities because of distance or hilly terrain. In Canada, however, communities with their own signals were fertile cable markets, as viewers wanted to receive American signals. Rarely, as in the college town of Alfred, New York, U.S. cable systems retransmitted Canadian channels.
More channels and movies, plus plenty of extras and premium options. That's how we roll at Midco®, so you can command your own personal cable TV entertainment experience. You'll enjoy hundreds of channels, and smooth mobile streaming with TV Everywhere and mobile ON Demand. There's a Midco cable package for everyone – and you can combine cable TV with Midco Internet and home phone to save even more.
Hi Judy, it depends on where you live, as to the options available to you. To seek free options first put your zip into this site to see how many stations are within 60 miles of your location: https://www.antennasdirect.com/transmitter-locator.html. There are about 18 towers within 60 miles of me. Each town broadcasts multiple channels depending on the network they are connected to. With my 18 towers, I get 68 channels. If you can get enough channels in your area, it may be worth investing in a very good rooftop digital antenna. I have a Clear2 which cost me about $60 bucks, but their are even better ones out there. Depending on where your towers are located, you may want a very strong multi-directional antenna. That's going to be the best option for free OTA. I ran my antenna directly into where my cable connected to house, so all my lines in the house received the signal. I also purchased a $15 in-line signal booster. You can also find those on amazon or google them. Also, are you sure hughes net is only provider for your area? make sure you check every option to see an unlimited internet provider as unlimited internet and streaming device with Terrarium TV is ultimate cord cutter dream. What is your zip code? I'll check any additional options for you.
Always take a moment on a new TV to switch off any special features that are meant to reduce motion-blurring. (These go by different names depending on the manufacturer, but the setting typically has the word "motion" in it, as in Sony's "Motionflow" or LG's "TruMotion.") These settings tend to make movies and even most modern television shows look more flat and artificial, in what's often called "the soap opera effect."
PlayStation Vue is a streaming service similar to Sling TV, offering channels like AMC, TBS, Syfy and much more. However, this service starts at $39.99 per month. Check out this comparison of Sling TV Channels vs. PS Vue for the difference in each services channel lineup. You don't need a PlayStation gaming console to use the service either. While the service is available on PlayStation 3 and PlayStation 4, you can also use the service with Roku, Amazon Fire TV, Fire TV Stick, Google Chromecast, Android, iPad, and iPhone. Check out my review of PlayStation Vue for more details.
Way back when, cable and satellite TV were initially sold to consumers as an add-on: Get all of your local channels, along with uncut movies, more televisions programs and additional sports. But pulling the plug on cable doesn't necessarily mean you can just go back to broadcast networks only. Modern homes and apartments are no longer wired-up with antennas the way they were 30 years ago. Indoor HDTV antennas are available at prices ranging from under $20 to move than $150, but the quality of the signal and the number of channels that come in clearly vary depending on external factors, like whether you live near mountains or skyscrapers.
Maybe having a streaming subscription that delivers all of the content over the internet is your "cut the cord" solution. There are definitely some advantages to going with one of these services over traditional cable beyond just taking the actual cable that runs into your house out of the equation. And chief among these advantages is the lack of a contract, so you can turn them on one month and turn them off the next.


I am writing this post from the 12th annual Offshore Alert conference in Miami Beach. The big topic this year? FATCA and offshore tax evasion. While the entire world gears up to enforce the new Foreign Account Tax Compliance Act (FATCA), experts and Senate staffers are finding many holes in the new legislation.
For those not already aware, FATCA is legislation passed by Congress in 2010 aimed at combatting tax evasion using foreign financial accounts. FATCA requires both individuals and banks to disclose offshore accounts belonging to U.S. taxpayers no matter where they reside.
Since the IRS's 2009 amnesty program (the Offshore Voluntary Disclosure Initiative), 43,000 taxpayers have come forward to report their foreign accounts. Unfortunately, the IRS believes that millions more still are not compliant. That is important because the penalties for failure to report a foreign bank account can be the greater of $100,000 or 50% of the highest historical balance of any unreported account. In willful cases, taxpayers could face prison.
FATCA will soon require offshore financial institutions – that includes banks, brokerage firms and insurance companies located outside the United States – to review their account base and report those account holders with apparent ties to the United States. On paper, the new legislative regime seems very effective. Fortunately or unfortunately – depending on your view point, there are many holes in FATCA.
One of the most glaring deficiencies concerns paper records. Banks are required to review and perform due diligence on electronic data. A few smaller banks and those in less developed countries still maintain paper records. Are those banks required to perform the same due diligence on paper records? That question has yet to be answered.
Perhaps is the biggest enforcement challenge is nominee accounts. The true tax evaders have learned the safest way to avoid detection is to open accounts in the name of another person or entity. We have heard one taxpayer who opened an account in Switzerland under the name of Seychelles trust. The Seychelles trust was in turn owned by an international business company based in Hong Kong.
Nominee accounts have been effective at thwarting Uncle Sam, yet that too is fraught with peril. The IRS considers nominee accounts as an affirmative act of tax evasion. If one is caught hiding money in the name of a fictitious entity expect criminal prosecution.
It will take years to fix the loopholes and creative tax evaders will always find or create a new loophole. Even without FATCA, however, the Justice Department and IRS have proven themselves very adept at finding offshore accounts. Time is running out for taxpayers and hard decisions must be made. Comply and face potential penalties or spend the next several years looking over your shoulder. That choice is a no brainer to us yet thousands of people still sit on the fence hoping that FATCA will go away. Unfortunately, it appears that most world powers are considering their own FATCA model.
Tax transparency is here to stay. For now, there are still many of cracks in the program.
Mahany & Ertl – America's Tax Lawyers. IRS services provided worldwide.


Zarif told CNN Chief National Security Correspondent Jim Sciutto that terminology used by the White House to describe the agreement differed from the text agreed to by Iran and the other countries in the talks — the United States, Britain, France, Russia, China and Germany.
"The White House version both underplays the concessions and overplays Iranian commitments" under the agreement that took effect Monday, Zarif said in Davos, Switzerland, where he was attending the World Economic Forum.
As part of the accord, Iran was required to dilute its stockpile of uranium that had been enriched to 20%, well above the 5% level needed for power generation but still below the level for developing a nuclear weapon.
In addition, the deal mandated that Iran halt all enrichment above 5% and "dismantle the technical connections required to enrich above 5%," according to a White House fact sheet issued in November after the initial agreement was reached.
Zarif accused the Obama administration of creating a false impression with such language.
"You don't need to over-emphasize it," Zarif said of the White House language. A separate summary sent out by the White House last week did not use the word dismantle.
Iranian and U.S. officials have tried to sell the nuclear agreement to domestic opponents in their respective countries who could scuttle it.
Iranian officials have called the interim pact a victory and said it failed to halt the nation's nuclear development program, while U.S. officials say the agreement essentially froze Iran's nuclear program and rolled back some capabilities.
Zarif noted the political pressure facing both sides, which includes a push in Congress for more sanctions against Iran that Tehran warns would destroy any chance for success in talks on a long-range nuclear agreement intended to prevent development of an Iranian nuclear weapon.
"All of us are facing difficulties and oppositions and concerns and misgivings," he said, noting he had been summoned Wednesday to Iran's parliament to answer questions.
Zarif and Rouhani traveled to Switzerland for annual gathering of world political and business leaders in Davos as a new round of Syrian talks started in Montreux before moving to Geneva.
Iran, a major backer of Syrian President Bashar al-Assad, was invited to the Syrian talks by U.N. Secretary General Ban Ki-moon, then disinvited under pressure from the United States because Tehran refused to endorse conditions in a previous agreement setting up the talks.


text:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">

<TITLE>
SchemaStringEnumEntry (XMLBeans 2.6.0 Documentation)
</TITLE>

<META NAME="date" CONTENT="2012-07-23">

<LINK REL ="stylesheet" TYPE="text/css" HREF="../../../stylesheet.css" TITLE="Style">

      Overview
      Package
    Class
      Tree
      Deprecated
      Index
      Help

 PREV CLASS
 NEXT CLASS

  FRAMES
 NO FRAMES

  All Classes

  SUMMARY: NESTED | FIELD | CONSTR | METHOD

DETAIL: FIELD | CONSTR | METHOD

org.apache.xmlbeans

Interface SchemaStringEnumEntry

public interface SchemaStringEnumEntry

Describes a code-generated string enumeration

See Also:
SchemaType.getStringEnumEntries()

Method Summary

 String
getEnumName()

          A capitalized name to use for the enumeration constant name

 int
getIntValue()

          An integer to represent this choice in the enumeration

 String
getString()

          The enumerated string value

Method Detail

getString

String getString()

The enumerated string value

getIntValue

int getIntValue()

An integer to represent this choice in the enumeration

getEnumName

String getEnumName()

A capitalized name to use for the enumeration constant name

      Overview
      Package
    Class
      Tree
      Deprecated
      Index
      Help

 PREV CLASS
 NEXT CLASS

  FRAMES
 NO FRAMES

  All Classes

  SUMMARY: NESTED | FIELD | CONSTR | METHOD

DETAIL: FIELD | CONSTR | METHOD

redpajama_set_name: RedPajamaGithub


Chale Wote Street Art Festival begins today August 15, 2016. The much anticipated festival which is dubbed the "Spirit Robot", will bring lovers of everything art from all walks of life together to have this great experience.
Last week, marked my 1-month stay here in Ghana. Watch my Vlog below to find out how it has been so far!


Oh, how we LOVE tools, but HATE goodbyes!
Whether you're into side hustling or you're a full-time consultant, trainer or freelancer, it's important to know the shortcuts that actually matter. There are so many tools, such an abundance of resources for making the best use of your time and our guest, Zach Sexton gives us some straight-up wisdom about tools he recommends that make the biggest difference in productivity.
A little about Zach, he loves to teach and share exciting tools, systems and resources that he's learned about over the years. He teaches what he needs to learn the most. As a former disorganized, all-over-the-place person, after he found systems and tools to keep himself on track he realized how much more effective he could be. Now Zach's made a career of sharing those productive ideas and concepts with his clients.
Brittany and Nick talk to Zack about his productivity journey, from his time as a disorganized person to a productivity expert and he really gives us amazing tips on what kind of tools and processes he used to help him to plan things better and not have to rely on keeping everything in his head. We even get a chance to talk about his experience with agile and how to use story points to plan your work more effectively. We also learned how Zack made the transition from full time employee to solopreneur and what that was like!
Lastly, Brittney and Nick make a special announcement you may not like, but definitely need to hear...enjoy Snappy Nation!
Connect with Zach on Twitter or at ZacharySexton.com.
So many people waste time going back-and-forth trying to figure out when to schedule an appointment. If this is you, you'll want to get in on Zach's challenge!
"Set up a scheduler, like Youcanbook.me or Calendly.com"
Put it in the footer of your email, and it allows people to book times with you automatically. Start making more connections with people without that ping-pong back-and-forth email exchange. Learn more about scheduler applications with Zach's free "Automate Your Booking" course.


Have you got your treasures on a reel to reel audio tape. Maybe your grandparents used reel to reel audio tapes to record "letters" to send back to the old country. I've seen many reel to reel audio recordings with precious audio of relatives that have long since passed away. Do you have reels of tape like these?
Reel-to-reel audio tape or open reel audio tape is a form of magnetic audio recording media that is held on an open reel rather than being contained in a cassette. These tapes were popular from around the mid 1950′s into the mid 1980′s. Most reel-to-reel tapes have either one or two stereo tracks recorded on them. This means that the tapes typically have four tracks (a left and right for each stereo pair). One stereo track was recorded on the tape and when the end of the tape was reached, the tape was flipped around and the second stereo track was recorded. This results in the two stereo tracks being recorded in opposite directions.


Tarion Home Warranty. For information in regards to New home construction and the applicable warranties.
Canada Mortgage and Housing Corporation. For information on mortgages, other government sponsored housing programs as well as a comprehensive catalog of publications and reports (some free) available to the public.
The Ottawa Construction Association. To help you find a reputable contractor for most work you might need to do around the house.
Homestars For an alternative peer review site on various trades and services.
Ottawa Real Estate Board. For help finding a realtor.
City of Ottawa. For too many things….such as bi-laws and more.
Ottawa Carleton District School Board.
Conseil des écoles publiques de l'Est de l'Ontario.
Conseil des écoles catholique de centre-est.
OC Transpo. For local transit.
© 2019 Foundation & Up Home Inspection Ltd.


text:
FactoryGirl.define do
  factory :user do
    sequence(:email)      { Faker::Internet.email }
    sequence(:username)   { Faker::Internet.user_name }
    password              '1234567890'
    password_confirmation '1234567890'
  end
end


Mayfield Garden, one of the world's largest privately owned cool climate gardens, needed a new website to showcase their spectacular views and features to visitors and event co-ordinators. They needed the new website to promote their events and enable visitors to buy tickets online.
We got in touch with our floral side to custom code, design and integrate a wordpress site that shows off all the gardens best assets. The site includes a gallery of images and a press section so that visitors can stay up to date with all the gardens latest events.
The client has responded very positively to the site that we created for them. They feel it really represents the garden's personality and best assets. The new site has attracted a large increase in visits and enquiries and the online ticketing system has created an increase in ticket sales.


Our goal at Gordon Powers Electrician is to be the most efficient Castlecrag electricians by providing the best value for your money. Our Castlecrag electricians are friendly, offer a superior level of customer service and will not leave you until you are 100% happy with the work carried out.
We have over 10 years of experience fixing all sorts of electrical issues in Castlecrag and all of its surrounding suburbs. There's simply no job that is too big, small or too complex for our electricians in Castlecrag. With our level 2 electricians and emergency electricians in Castlecrag, we have you covered at all times.
If you're located in Castlecrag, call us today on 02 9199 7480 for a level 2 electrician, emergency electrician, commercial electrician, smoke alarm installation, switchboard upgrade or installation, hot water repairs, CCTV installation and more. Gordon Powers can handle any electrical emergency!
Is your outdoor lighting and smoke detectors not working as intended? A local Castlecrag Electrician can keep you safe from fire outbreak and unwanted entrance. These are vital electrical components that need to be working properly as this can save your life and your assets. With the help of Gordon Powers Electrical, you will have constant assurance that your place is secured. We will repair everything so you just wait for the perfect completion of the work.
Gordon Powers Electrical guarantees that you can reach your friendly Castlecrag Electrician anytime you need help. We even operate during the holidays for your confidence that you have a Castlecrag Electrician who can reach your doorsteps with no issues. We have incomparable and expert experience in dealing with power supplies, lighting, fault finding and many more.
For any general electrical concerns, call Gordon Powers Electrical or email us at info@gordonpowers.com.au, subject line: Electrician Castlecrag. Expect that the hassle of any electrical problem will never bother you anymore!
Your local certified and skilled Gordon Powers Level 2 Electrician Castlecrag is just a dial away. Give us a call now and get excellent services. Our friendly customer service is waiting for you.
For your Switchboard/Fusebox Repairs and Upgrades Castlecrag, call us now. We will you the best solution to your Switchboard problems. Our friendly team will be with you right away.


text:
Q: How do I use bundle exec in Aptana Studio 3.0.9 to run the debugger? Aptana seems to start in the sh shell. So I'm trying to switch to zsh, and execute commands to start a rails app.
I've changed my debug executable so that it runs a unix executable with the following contents:
#!/usr/bin/env ruby
exec("zsh | (echo $SHELL && [[ -s '$HOME/.rvm/scripts/rvm' ]] && source '$HOME/.rvm/scripts/rvm' && rvm use 1.8.7@tinderbox && (bundle exec script/server -p3001 &))")

So, when I click my debug executable from the debug menu on the toolbar, the debug perspective appears, and it shows that
myApp [Ruby Application]
    Ruby
    /usr/bin/ruby

is running and hasn't been terminated.
but when I go to my browser and type in 0.0.0.0:3001, I get "Oops! Google Chrome could not connect to localhost:3001" - which means that the server sin't running.
I have the output of the executable set to a file on my desktop here is what happens every run:
 Fast Debugger (ruby-debug-ide 0.4.16, ruby-debug-base 0.10.4) listens on 127.0.0.1:55306
/bin/zsh

*

*Note: I get the same results with bash as with zsh

What is going on? and how do I get my command to work?

A: type
zsh -l

and in new zsh shell:
rvm 1.8.7 do bundle exec script/server -p3001

or with rvm wrapper:
generate wrapper for bundle:
rvm wrapper 1.8.7 run bundle

it will generate run_bundle, find it:
which run_bundle

and use it in aptana:
/path/to/run_bundle exec script/server -p3001

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text:
\section{Introduction}

One of the most fundamental questions in quantum information theory concerns the ability to send information over a noisy quantum communication channel~\cite{Holevo06,c1,c2,c3,c4,c5,c6,c7,c8,c9,Has09,Yard08,Brandao,Fuk10,Bra11}.
Unlike classical channels, quantum channels exhibits an unintuitive phenomenon in which the optimal rate to transmit reliably classical or quantum information is not additive under taking tensor products of two (or more) quantum channels~\cite{Has09,Yard08}. The question of whether the product state classical capacity (i.e. Holevo capacity) is additive or not was an open problem for more than a decade and was shown by Shor~\cite{c6} to be equivalent to three other additivity conjectures; namely, the additivity of entanglement of formation, the strong super-additivity of entanglement of formation, and the additivity of the minimum entropy output of a quantum channel.

The discovery that all these quantities are not additive~\cite{Has09} left with it key problems in the field wide open. One such problem is how much entanglement between input signal states is needed to violate additivity. A partial answer for that was given in recent work~\cite{GF,FGA}, where it was shown that the minimum entropy output of a quantum channel is locally additive. This result indicated that while entanglement is needed to violate additivity, arbitrarily small amount of entanglement will not be sufficient. Here we extend this result by showing that the minimum Renyi output entropies with parameter $\alpha$ greater than 1 are also locally additive. However, unlike the von-Neumann entropy ($\alpha=1$), for $\alpha>1$ the Renyi entropies are given in terms of the log of the $\alpha$-norms (also known as $\ell_p$-norms, where for notational convenience we rename $p=\alpha$ presently) which are multiplicative under tensor product. Therefore, in order to prove local multiplicativity of the output $\alpha$-norms of a quantum channel, it is not possible to use exactly the same techniques as those used in the case $\alpha=1$ since the latter relied heavily on the additive nature of the von-Neumann entropy.

This paper is organized as follows. In Section~\ref{preli} we provide the precise definition of local additivity, including notations and preliminaries that will be used in the subsequent sections. Section~\ref{sec:main} is devoted to the statement and proof of the main result. The proof of the main result is based on 3 substantial lemmas that will be proved in Sections~\ref{sec:der},~\ref{sec:dir}, and~\ref{sec:add}. Finally, in Section~\ref{sec:concluding} we end with a few concluding remarks.

\section{Notations and Preliminaries}\label{preli}

Quantum channels are described in terms of completely-positive trace preserving linear maps (CPT maps).
A CPT map $\mathcal{N}: H_{d_{\rm in}}\to H_{d_{\rm out}}$ takes the set of $d_{\rm in}\times d_{\rm in}$ Hermitian matrices
$H_{d_{\rm in}}$ to a subset of the set of all $d_{\rm out}\times d_{\rm out}$ Hermitian matrices
$H_{d_{\rm out}}$. Any finite dimensional quantum channel can be characterized in terms of a unitary embedding followed by a partial trace (the Stinespring dilation theorem): for any CPT map $\mathcal{N}$ there exists an ancillary space of Hermitian matrices $H_{E}$ such that
$$
\mathcal{N}(\rho)=\mathop{\mathrm{Tr}}\nolimits_{E}\left[U(\rho\otimes |0\rangle_{E}\langle 0|) U^{\dag}\right]
$$
where $\rho\in H_{d_{\rm in}}$ and $U$ is a unitary matrix mapping states $|\psi\rangle|0\rangle_E$ with
$|\psi\rangle\in H_{d_{\text{in}}}$ to
$H_{d_{\text{out}}}\otimes H_{E}$.

For $\alpha\ge0$, the minimum $\alpha$-Renyi entropy output of a quantum channel $\mathcal{N}$ is defined by
\begin{equation}\label{minim}
S_{\alpha}^{\min}(\mathcal{N})\equiv\min_{\rho\in H_{d_{\text{in}},+,1}}S_\alpha\left(\mathcal{N}(\rho)\right)\;,
\end{equation}
where $H_{d_{\text{in}},+,1}\subset H_{d_{\text{in}}}$ is the set of all $d_{\rm in}\times d_{\rm in}$ positive semi-definite matrices with trace $=1$ (i.e. density matrices), and $$S_\alpha(\rho)\equiv\frac{1}{1-\alpha}\log\mathop{\mathrm{Tr}}\nolimits(\rho^\alpha)$$ is the $\alpha$-Renyi entropy with $0\leq \alpha\leq \infty$, where for $\alpha=0,1,\infty$ the Renyi entropies are defined in terms of the limits. For $0\leq \alpha\leq 1$ the Renyi entropies are concave in $\rho$, and therefore it follows that the minimization can be taken over all rank one matrices $\rho=|\psi\rangle\langle\psi|$ in $H_{d_{\text{in}},+,1}$. While for $\alpha>1$ the Renyi entropy is not concave in general (only Schur concave),  we can still take the minimization over rank-1 matrices since the $\alpha$-norm $\|\rho\|_\alpha=\left[\mathop{\mathrm{Tr}}\nolimits(\rho^\alpha)\right]^{1/\alpha}$ is convex for $\alpha\geq 1$.  To see why, note that for $\rho=\sum_jp_j|\psi_j\rangle\langle\psi_j|$, we have $\|\mathcal{N}(\rho)\|_\alpha\leq\sum_jp_j \|\mathcal{N}(|\psi_j\rangle\langle\psi_j|)\|_\alpha$, so that
\begin{align*}
S_{\alpha}\left(\mathcal{N}(\rho)\right)& =\frac{\alpha}{1-\alpha}\log\|\mathcal{N}(\rho)\|_\alpha
\geq \frac{\alpha}{1-\alpha}\log\sum_jp_j \|\mathcal{N}(|\psi_j\rangle\langle\psi_j|)\|_\alpha\\
& \geq\frac{\alpha}{1-\alpha}\log\max_{j}\|\mathcal{N}(|\psi_j\rangle\langle\psi_j|)\|_\alpha
=\min_jS_\alpha\left(\mathcal{N}(|\psi_j\rangle\langle\psi_j|)\right)\;.
\end{align*}
Therefore, for all $0\leq\alpha\leq\infty$ the minimum in~\eqref{minim} can be taken over all rank 1 matrices in $H_{d_{\text{in}},+,1}$.

For any such rank one density matrix $\rho=|\psi\rangle\langle\psi|$ we can define a bipartite pure state
$|\Psi\rangle=U|\psi\rangle|0\rangle_{E}$ in the bipartite subspace $\mathcal{K}\equiv \{|\Psi\rangle\big|\;|\psi\rangle\in H_{d_{\rm in}}\}$. Therefore, the minimum
Renyi entropy output of the channel $\mathcal{N}$ can be expressed in terms of the Renyi entanglement of the bipartite subspace  $\mathcal{K}$ defined by
$$
E_\alpha(\mathcal{K})\equiv\min_{|\phi\rangle\in\mathcal{K}\;,\;\|\phi\|=1}E_\alpha(|\phi\rangle)\;,
$$
where $E_\alpha(|\phi\rangle)\equiv S_\alpha\left(\mathop{\mathrm{Tr}}\nolimits_{E}(|\phi\rangle\langle\phi|)\right)$ is the Renyi entropy of entanglement. In~\cite{GN} it was pointed out that $E_\alpha(\mathcal{K})=0$ unless $\dim\mathcal{K}\leq (d_{\rm out}-1)(\dim H_{E}-1)$. This claim follows directly from the fact that the number of (bipartite) states in an unextendible product basis is at least $d_{\rm out}+\dim H_{E}-1$, cf.\ \cite{Ben99}.

With these notations, the non-additivity of the minimum Renyi entropy output of a quantum channel is equivalent to
the existence of two subspaces $\mathcal{K}_1\subset\mathbb{C}^{n_1}\otimes\mathbb{C}^{m_1}$ and $\mathcal{K}_2\subset\mathbb{C}^{n_2}\otimes\mathbb{C}^{m_2}$ such that
$$
E_\alpha(\mathcal{K}_1\otimes\mathcal{K}_1)<E_\alpha(\mathcal{K}_1)+E_\alpha(\mathcal{K}_2)\;.
$$

\subsection*{Local Minimum/Maximum}\label{local}

Let $\mathcal{K}\subset\mathbb{C}^{n}\otimes\mathbb{C}^{m}$ be a subspace of bipartite entangled states.
Since the bipartite Hilbert space $\mathbb{C}^{n}\otimes\mathbb{C}^{m}$ is isomorphic to the Hilbert space
of all $n\times m$ complex matrices $\mathbb{C}^{n\times m}$, we can view any bipartite state
$|\psi\rangle^{AB}=\sum_{i,j}x_{ij}|i\rangle|j\rangle$ in $\mathcal{K}$ as an $n\times m$ matrix $x$.
The reduced density matrix of $|\psi\rangle^{AB}$ is then given by $\rho_r\equiv\mathop{\mathrm{Tr}}\nolimits_{B}|\psi\rangle^{AB}\langle\psi|=xx^*$,
and the $\alpha$-Renyi entropy of entanglement of $|\psi\rangle^{AB}$ is given by
\begin{equation}\label{entropy}
E_\alpha(x)\equiv\frac{1}{1-\alpha}\log\mathop{\mathrm{Tr}}\nolimits[(xx^*)^\alpha]\;.
\end{equation}
In our notations, instead of using a dagger, we use $x^*$ to denote the hermitian conjugate of the matrix $x$.

Since the $\log$ function is continuous and monotonic, instead of showing that $E_\alpha$ is locally additive for $\alpha>1$, we will show that
$$
Q_\alpha(x)=\mathop{\mathrm{Tr}}\nolimits[(xx^*)^\alpha]
$$
is locally multiplicative.

If $x\in\mathcal{K}$ is a local minimum of $E_\alpha$ in $\mathcal{K}$ (i.e. $x\in\mathcal{K}$ is a local \emph{maximum} of $Q_\alpha$ in $\mathcal{K}$), then there exists a neighbourhood of $x$ in $\mathcal{K}$
such that $x$ is the minimum in that neighbourhood. Any state in a neighbourhood of $x$ can be written as
$ax+by$, where $a,b\in\mathbb{C}$ and $y\in\mathcal{K}$ is a matrix orthogonal to $x$; i.e. $\mathop{\mathrm{Tr}}\nolimits(xy^*)=0$.
We also assume that the state is normalized so that $|a|^2+|b|^2=1$.
Now, since the function $E_\alpha(x)$ (or $Q_\alpha(x)$) is independent of a global phase, we can assume that $a$ is a positive \emph{real} number.
We can also assume that $b$ is real since we can absorb its phase into $y$ (adding a phase to $y$ will not change its orthogonality to $x$). Thus, any normalized state in a neighbourhood of $x$ can be written as
$$
\frac{x+ty}{\sqrt{1+t^2}}\;\;\text{with}\;\;\mathop{\mathrm{Tr}}\nolimits(xy^*)=0\;,
$$
where $t\equiv b/a$ is a small real number and $y$ is normalized (i.e. $\mathop{\mathrm{Tr}}\nolimits(yy^*)=1$).

\begin{definition}\label{maindef}
$\;$\\
\textbf{(a)} A matrix $x\in\mathcal{K}$ is said to be a critical point of $Q_\alpha(x)$ in $\mathcal{K}$ if
$$
D_{y}Q_\alpha(x)\equiv \frac{d}{dt}Q_\alpha\left(\frac{x+ty}{\sqrt{1+t^2}}\right)\Big|_{t=0}=0\;\;\;\forall\;y\in x^{\perp}
$$
where the notation $D_{y}Q_\alpha(x)$ indicates that we are taking the directional derivative of $Q_\alpha$ in the direction of $y$,
and $x^\perp\subset\mathcal{K}$ denotes the subspace of all the matrices $y$ in $\mathcal{K}$ for which
$\mathop{\mathrm{Tr}}\nolimits(xy^*)=0$.\\
\textbf{(b)}  A matrix $x\in\mathcal{K}$ is said to be a non-degenerate local maximum of $Q_\alpha(x)$ in $\mathcal{K}$ if it is critical and
$$
D_{y}^{2}Q_\alpha(x)\equiv \frac{d^2}{dt^2}Q_\alpha\left(\frac{x+ty}{\sqrt{1+t^2}}\right)\Big|_{t=0}<0\;\;\;\forall\;y\in x^{\perp}.
$$
(This is a maximum for $Q_\alpha$, which gives a minimum for $E_\alpha$, since $\frac{1}{1-\alpha}<0$.) Moreover, a critical $x\in\mathcal{K}$ is said to be degenerate if there exists at least one direction $y$ such that
$D_{y}^{2}Q_\alpha(x)=0$.
\end{definition}

To be clear: {\em local additivity} of $E_\alpha$ is the statement that if $x^A$ and $x^B$ are local minima for $E_\alpha$ in two subspaces $\mathcal{K}^A$ and $\mathcal{K}^B$, then $x^A\otimes x^B$ is a local minimum for $E_\alpha$ in $\mathcal{K}^A\otimes\mathcal{K}^B$.  We make this completely precise (also considering a subtle point of degenerate local minima) in Theorem \ref{thm.main} below.

In our calculations we will assume that $x$ is diagonal (or equivalently, the bipartite state $x$ that represents is given in its
Schmidt form). This assumption results in no loss of generality, due to
the singular value decomposition theorem; namely, we can always find unitary matrices $u\in\mathbb{C}^{n\times n}$ and $v\in\mathbb{C}^{m\times m}$ such that $uxv$ is an $n\times m$ diagonal matrix with non-negative real numbers (the singular values of $x$) on the diagonal. Since $E_\alpha(x)=E_\alpha(uxv)$, we can assume without loss of generality that $x$ is a diagonal matrix.

\section{Main Results}\label{sec:main}

In this section we state and prove the main result of this paper.
The proof is based on 3 lemmas that will be proved in 3 subsequent sections.

\begin{theorem} \label{thm.main}
Let $\mathcal{K}^{A}$ and  $\mathcal{K}^{B}$ be two subspaces of $n_1\times m_1$ and $n_2\times m_2$ complex matrices, respectively. Let $x^A$ and $x^B$ be two normalized complex matrices in $\mathcal{K}^A$ and $\mathcal{K}^B$, respectively. Then, for $\alpha>1$:\\

{\rm \bf{(a)}} If $x^A$ and $x^B$ are non-degenerate local minima of $E_\alpha$ in $\mathcal{K}^{A}$ and  $\mathcal{K}^{B}$, respectively, then $x^A\otimes x^B$ is a non-degenerate local minimum of $E_\alpha$ in $\mathcal{K}^{A}\otimes\mathcal{K}^{B}$.\\

{\rm \bf{(b)}} If $x^A$ and $x^B$ are local minima of $E_\alpha$ in $\mathcal{K}^{A}$ and  $\mathcal{K}^{B}$, with at least one of them being non-degenerate, then $x^A\otimes x^B$ is a local minimum of $E_\alpha$ in $\mathcal{K}^{A}\otimes\mathcal{K}^{B}$.
\end{theorem}

Implicit in the above theorem is the fact that if $x^A$ and $x^B$ are critical points of $E_\alpha$ in $\mathcal{K}^{A}$
and $\mathcal{K}^{B}$, respectively, then $x^A\otimes x^B$ is a critical point of $E_\alpha$
in $\mathcal{K}^{A}\otimes\mathcal{K}^{B}$. This fact was observed in~\cite{AIM08} (see also~\cite{Maxim}),
and was later stated in~\cite{FGA}.
It follows from the linearity in $y$ of the condition given in Eq.~(\ref{critical}) (see the next section) for critical points. We will therefore focus in this section on the higher order directional derivatives of $E_\alpha$ (or equivalently of $Q_\alpha$).

For the proof of Theorem \ref{thm.main}, we can assume without loss of generality that $n_1=m_1$, $n_2=m_2$, by padding the matrices with extra rows/columns of $0$s.
From the singular valued decomposition (see the argument below Definition~\ref{maindef}) we can assume
without loss of generality that $x^A=\mathop{{\rm diag}}\nolimits\{\sqrt{p_1},\ldots,\sqrt{p_{n_1}}\} $
and $x^B=\mathop{{\rm diag}}\nolimits\{\sqrt{q_1},\ldots,\sqrt{q_{n_2}}\}$,
where $p_j$ and $q_k$ are non-negative and $\sum_{j=1}^{n_1}p_i=\sum_{k=1}^{n_2}q_{j}=1$.

We first assume that both $x^A$ and $x^B$ are non-degenerate local maxima of $Q_\alpha$.
We need to show that $D_{y}^{2}Q_\alpha(x)<0$ for all $y\in x^\perp$, where $x\equiv x^A\otimes x^B$.
The most general
$y\in \left(x^A\otimes x^B\right)^\perp$ can be written as
\begin{equation}\label{generalform}
y=c_1x^A\otimes y^{B}+c_2y^{A}\otimes x^B+c_3y'\;,
\end{equation}
where $y^{A}\in (x^A)^\perp$, $y^{B}\in (x^B)^\perp$, and $y'\in\left(x^A\right)^\perp\otimes \left(x^B\right)^\perp$
are all normalized. The numbers $c_j$ can be chosen to be real because we can absorb their phases into $y^{A}$, $y^{B}$, and $y'$. They also
satisfy $c_{1}^{2}+c_{2}^{2}+c_{3}^{2}=1$, so that $y$ is normalized.

\begin{lemma}\label{directions}  If $x^A$ and $x^B$ are critical points, then
\begin{equation}
D_{y}^{2}Q_\alpha(x)=c_{1}^{2}D_{x^A\otimes y^{B}}^{2}Q_\alpha(x)+c_{2}^{2}D_{y^{A}\otimes x^B}^{2}Q_\alpha(x)+c_{3}^{2}D_{y'}^{2}Q_\alpha(x).
\end{equation}
\end{lemma}

It is therefore enough to consider the three directions $x^A\otimes y^{B}$, $y^{A}\otimes x^B$ and $y'$ separately. We will show that in each of this directions, the second order derivatives $D_{x^A\otimes y^{B}}^{2}Q_\alpha(x)$, $D_{y^{A}\otimes x^B}^{2}Q_\alpha(x)$, and $D_{y'}^{2}Q_\alpha(x)$ are all negative, so that $D_{y}^{2}Q_\alpha(x)<0$.

Consider first the simple case where $y=x^A\otimes y^{B}$. Here we have
\begin{align}\label{simple}
Q_\alpha\left(\frac{x+ty}{\sqrt{1+t^2}}\right)& =Q_\alpha\left(x^A\otimes \frac{x^B+ty^{B}}{\sqrt{1+t^2}}\right)
=Q_\alpha\left(x^A\right)Q_\alpha\left(\frac{x^B+ty^{B}}{\sqrt{1+t^2}}\right)\;.
\end{align}
Since $x^B$ is a non-degenerate local maximum, and $Q_\alpha(x^A)>0$, we must have $D_{y}^{2}Q_\alpha(x)<0$ for $y=x^A\otimes y^{B}$. The case $y=y^{A}\otimes x^B$ is similar.

Consider now the case in which $y\in \left(x^A\right)^\perp\otimes \left(x^B\right)^\perp$.
To prove that $D_{y}^{2}Q_\alpha(x)<0$ in this case, we will use the following explicit computation.

\begin{lemma}\label{derivatives}
Let $x\in\mathcal{K}\subset\mathbb{C}^{n\times n}$ and $y\in x^{\perp}$ both normalized.
Denote the eigenvalues of $\rho\equiv xx^*$ by $\{p_j\}_{j=1}^{n}$, and decompose the complex matrix
$y=w+iz$ such that the $n\times n$ matrices $w$ and $z$ are both Hermitian. Then
\begin{equation}\label{deriv}
\frac{1}{2\alpha}D_{y}^{2}Q_\alpha(x)=-\mathop{\mathrm{Tr}}\nolimits\left[\rho^\alpha\right]+\mathop{\mathrm{Tr}}\nolimits\left[w\Phi_{\rho}^{-}(w)+z\Phi_{\rho}^{+}(z)\right]\;,
\end{equation}
where $\Phi^{\pm}_{\rho}$ are self-adjoint linear maps acting on the Hilbert space of $n\times n$ complex matrices (equipped with the Hilbert-Schmidt inner product), defined by the following Hadamard product:
\begin{equation} \label{e.phi.simple}
\left[\Phi^{\pm}_{\rho}(y)\right]_{jk} \equiv \phi_{jk}^{\pm}\,y_{jk}\;\;\;\;\text{where}\;\;\;\;\phi_{jk}^\pm \equiv \frac{p_j^{\alpha-1/2}\pm p_k^{\alpha-1/2}}{p_j^{1/2}\pm p_k^{1/2}}.
\end{equation}
Furthermore, if $p_j=p_k$ then $\phi_{jk}^{+}= p_{j}^{\alpha-1}$ and $\phi_{jk}^{-}=(2\alpha-1)p_{j}^{\alpha-1}$.
\end{lemma}
Note that the second order derivative is well behaved even if some $p_j$ are zero.
Now, denote by $\rho^{A}\equiv x^Ax^{A*}$ and $\rho^{B}\equiv x^Bx^{B*}$ the two density matrices associated with the two local maxima. From the lemma above it follows that
$D_{y}^{2}Q_\alpha(x^A\otimes x^B)<0$ for all $y\in \left(x^A\right)^\perp\otimes \left(x^B\right)^\perp$ if and only if
\begin{equation}\label{main}
\mathop{\mathrm{Tr}}\nolimits\left[w\Phi_{\rho^A\otimes\rho^B}^{-}(w)+z\Phi_{\rho^A\otimes\rho^B}^{+}(z)\right]< \mathrm{Tr\;}[(\rho^A)^\alpha]\mathrm{Tr\;}[(\rho^B)^\alpha] \;\;\;\;,\;\;\;\forall\;y=w+iz\in\left(x^A\right)^\perp\otimes \left(x^B\right)^\perp\;.
\end{equation}
Since $x^{A}$ and $x^B$ are local maxima of $Q_\alpha$ in their respective subspaces, it follows that  both $D_{y^{A}}^{2}Q_\alpha(x^{A})<0$ and $D_{iy^{A}}^{2}Q_\alpha(x^{A})<0$, since both $y^{A}$ and $iy^{A}$ belong to $\left(x^{A}\right)^\perp$. In particular,
\begin{align}\label{complex}
0 &>\frac{1}{4\alpha}\left(D_{y^{A}}^{2}Q_\alpha(x^A)+D_{iy^{A}}^{2}Q_\alpha(x^A)\right)\nonumber\\
 &=-\mathrm{Tr\;}[(\rho^A)^\alpha]+\frac{1}{2}\mathop{\mathrm{Tr}}\nolimits\left[w^{A}\left(\Phi_{\rho^A}^{-}+\Phi_{\rho^A}^{+}\right)(w^{A})+z^{A}\left(\Phi_{\rho^A}^{-}+\Phi_{\rho^A}^{+}\right)(z^{A})\right]\nonumber\\
 &=-\mathrm{Tr\;}[(\rho^A)^\alpha]+\mathop{\mathrm{Tr}}\nolimits\left[y^{A*}\left(\Phi_{\rho^A}^{-}+\Phi_{\rho^A}^{+}\right)(y^{A})\right]
\end{align}
where the last equality follows from the self-adjointness of $\Phi_{\rho^A}^{-}+\Phi_{\rho^A}^{+}$ with respect to the Hilbert-Schmidt inner product, and the decomposition $y^{A}=w^{A}+iz^{A}$ where $w^A$ and $z^A$ are Hermitian. We therefore arrive at the following inequalities:
\begin{equation} \label{assumption}
\begin{aligned}
\mathop{\mathrm{Tr}}\nolimits\left[y^{A*}\left(\Phi_{\rho^A}^{-}+\Phi_{\rho^A}^{+}\right)(y^{A})\right]&<\mathrm{Tr\;}[(\rho^A)^\alpha]\;\;\;\;\;\forall\;y^A\in\left(x^{A}\right)^\perp\\
\mathop{\mathrm{Tr}}\nolimits\left[y^{B*}\left(\Phi_{\rho^B}^{-}+\Phi_{\rho^B}^{+}\right)(y^{B})\right]&<\mathrm{Tr\;}[(\rho^B)^\alpha]\;\;\;\;\;\forall\;y^B\in\left(x^{B}\right)^\perp.
\end{aligned}
\end{equation}

In the final step towards the proof of the theorem, we will be using the following key operator estimate.

\begin{lemma} \label{additivity}
Let $\Psi$ be the self-adjoint linear operator acting on $\mathcal{K}^A\otimes\mathcal{K}^B$, given by the following Hadamard product:
\begin{equation}
[\Psi(y)]_{jk,\ell m} = \psi_{jk,\ell m}\,y_{jk,\ell m}
\end{equation}
where
\begin{equation}
\psi_{jk,\ell m}=(\phi_{j\ell}^{+A}+\phi_{j\ell}^{-A})(\phi_{km}^{+B}+\phi_{km}^{-B})=\frac{p_j^\alpha-p_\ell^\alpha}{p_j-p_\ell}\frac{q_k^\alpha-q_m^\alpha}{q_k-q_m}
\end{equation}
and $\{p_j\}$ and $\{q_k\}$ are the eigenvalues of $\rho^A$ and $\rho^B$.
Then
\begin{equation}\label{thm}
\Phi_{\rho^{A}\otimes\rho^{B}}^{\pm}\leq \Psi\;.
\end{equation}
\end{lemma}

With this lemma at hand, we are ready to prove inequality~\eqref{main}, and thus Theorem \ref{thm.main}(a).
First, observe that
\begin{equation}\label{beg}
\mathop{\mathrm{Tr}}\nolimits\left[w\Phi_{\rho^A\otimes\rho^B}^{-}(w)+z\Phi_{\rho^A\otimes\rho^B}^{+}(z)\right]
\leq\mathop{\mathrm{Tr}}\nolimits\left[w\Psi(w)+z\Psi(z)\right]=\mathop{\mathrm{Tr}}\nolimits\left[y^*\Psi(y)\right].
\end{equation}
Now, note that any
$y\in \left(x^A\right)^\perp\otimes \left(x^B\right)^\perp$ has a Schmidt decomposition given by
 \begin{equation}\label{yprime}
 y=\sum_{i}r_i y^{A}_{i}\otimes y^{B}_{i}\;,
 \end{equation}
 where
 \begin{equation}\label{ort}
 \mathop{\mathrm{Tr}}\nolimits[y^{A}_{i}y^{A*}_{i'}] = \mathop{\mathrm{Tr}}\nolimits[y^{B}_{i}y^{B*}_{i'}]=\delta_{ii'}\;,
 \end{equation}
 and $r_i$ are real non-negative numbers such that $\sum_{i}r_{i}^{2}=1$.
Therefore, substituting~\eqref{yprime} into the right-hand-side of~\eqref{beg} gives
\begin{align*}
\mathop{\mathrm{Tr}}\nolimits\left[y^*\Psi(y)\right] & =\sum_{i,i'}r_{i}r_{i'}\mathop{\mathrm{Tr}}\nolimits\left[y^{A*}_{i'}\otimes y^{B*}_{i'}\Psi(y^{A}_{i}\otimes y^{B}_{i})\right]\\
& =\sum_{i,i'}r_ir_{i'}\left[\sum_{j,\ell}\left(y^{A*}_{i'}\right)_{j\ell}\left(y^{A}_{i}\right)_{\ell j}\left(\phi_{j\ell}^{+A}+\phi_{j\ell}^{-A}\right)\sum_{k,m}\left(y^{B*}_{i'}\right)_{km}\left(y^{B}_{i}\right)_{m k}\left(\phi_{km}^{+B}+\phi_{km}^{-B}\right)\right]\\
&=\sum_{i,i'}r_ir_{i'}\mathop{\mathrm{Tr}}\nolimits\left[y_{i'}^{A*}\left(\Phi_{\rho^A}^{-}+\Phi_{\rho^A}^{+}\right)(y_{i}^{A})\right]
\mathop{\mathrm{Tr}}\nolimits\left[y_{i'}^{B*}\left(\Phi_{\rho^B}^{-}+\Phi_{\rho^B}^{+}\right)(y_{i}^{B})\right].
\end{align*}
Now, denote
\begin{align}
& A_{i'i}\equiv\mathop{\mathrm{Tr}}\nolimits\left[y_{i'}^{A*}\left(\Phi_{\rho^A}^{-}+\Phi_{\rho^A}^{+}\right)(y_{i}^{A})\right],\nonumber\\
& B_{i'i}\equiv\mathop{\mathrm{Tr}}\nolimits\left[y_{i'}^{B*}\left(\Phi_{\rho^B}^{-}+\Phi_{\rho^B}^{+}\right)(y_{i}^{B})\right].
\end{align}
Since the operators $\Phi_{\rho^A}^{-}+\Phi_{\rho^A}^{+}$ and $\Phi_{\rho^B}^{-}+\Phi_{\rho^B}^{+}$ are self-adjoint, the matrices $A$ and $B$ are Hermitian.  Inequality \eqref{assumption} then gives the following upper bounds.
\begin{lemma} \label{lem.A<Tr} The Hermitian matrices $A,B$ above satisfy
\[ 0\leq A < \mathrm{Tr\;}[(\rho^A)^\alpha] \qquad \text{and} \qquad 0\leq B < \mathrm{Tr\;}[(\rho^B)^\alpha]. \]
\end{lemma}

\begin{proof} Fix a unit vector $v = [v_i]$; the claim is that $0\leq \langle v,Av\rangle < \mathrm{Tr\;}[(\rho^A)^\alpha]$ (and similarly for $B$ and $\rho^B$).  Note that
\begin{align*} \langle v,Av\rangle = \sum_{i,i'} \overline{v_{i'}} v_i A_{i'i} &= \sum_{i,i'} \overline{v_{i'}} v_i \mathop{\mathrm{Tr}}\nolimits\left[y_{i'}^{A*}\left(\Phi_{\rho^A}^{-}+\Phi_{\rho^A}^{+}\right)(y_{i}^{A})\right] \\
&= \mathrm{Tr\;}\left[\sum_{i,i'} (v_{i'}y^A_{i'})^\ast \left(\Phi_{\rho^A}^{-}+\Phi_{\rho^A}^{+}\right)(v_iy_i^A)\right] = \mathrm{Tr\;}\left[y^{A\ast}\left(\Phi_{\rho^A}^{-}+\Phi_{\rho^A}^{+}\right)(y^A)\right],
\end{align*}
where $y^A=\sum_i v_iy_i$.  From \eqref{ort}, it follows that $y^A$ is also a unit vector; it is in the subspace $(x^A)^\perp$ (being a linear combination of $y_i$ that are in this subspace). Note that the RHS of the equation above is non-negative since $(p_{j}^{\alpha}-p_{k}^{\alpha})/(p_j-p_k)\geq 0$ for all $j,k$. The rest of the statement of the lemma now follows by \eqref{assumption} (the case for $B$ and $\rho^B$ is similar). \end{proof}

Now, diagonalize $A$ and $B$: $A=U^*D^AU$ and $B=V^*D^BV$ where $U$ and $V$ are unitary matrices and $D^A$ and $D^B$ are diagonal matrices.  Lemma \ref{lem.A<Tr} shows that the diagonal entries of $D^A$ are all less than $\mathrm{Tr\;}[(\rho^A)^\alpha]$, and the diagonal entries of $D^B$ are all less than $\mathrm{Tr\;}[(\rho^B)^\alpha]$. With this in mind, we get
\begin{align*}
\mathop{\mathrm{Tr}}\nolimits\left[y^*\Psi(y)\right] =\sum_{i,i'}r_{i}r_{i'}A_{i'i}B_{i'i}
&=\sum_{i,i'}r_{i}r_{i'}\sum_{k}U^{*}_{i'k}D^{A}_{kk}U_{ki}\sum_jV^{*}_{i'j}D^{B}_{jj}V_{ji} \\
&= \sum_{j,k} D^A_{kk}D^B_{jj} \sum_{i,i'} r_ir_{i'} U^\ast_{i'k}U_{ki}V^\ast_{i'j}V_{ji}.
 \end{align*}
Note that the internal sum can be written as
\[ \sum_{i,i'} r_ir_{i'} U^\ast_{i'k}U_{ki}V^\ast_{i'j}V_{ji} = \sum_i r_i U_{ki}V_{ji} \, \sum_{i'} r_{i'} U_{i'k}^\ast V_{i'j}^\ast = \left|\sum_{i}r_iU_{ki}V_{ji}\right|^2.  \]
In particular, the internal sum is positive.  Hence, using Lemma \ref{lem.A<Tr}, we have
\begin{equation} \label{e.penult} \mathrm{Tr\;}[y^\ast\Psi(y)] =\sum_{k,j}D^{A}_{kk}D^{B}_{jj}\left|\sum_{i}r_iU_{ki}V_{ji}\right|^2 < \mathrm{Tr\;}[(\rho^A)^\alpha]\mathrm{Tr\;}[(\rho^B)^\alpha]\sum_{j,k}\left|\sum_{i}r_iU_{ki}V_{ji}\right|^2. \end{equation}
Finally, we may again expand this sum.  Using the unitarity of $U$ and $V$, we find
\[ \sum_{j,k}\left|\sum_{i}r_iU_{ki}V_{ji}\right|^2 = \sum_{i,i',j,k}r_{i}r_{i'}U^{*}_{i'k}U_{ki}V^{*}_{i'j}V_{ji} = \sum_{i,i'}r_{i}r_{i'}\delta_{i'i}\delta_{i'i} = \sum_i r_i^2 = 1.  \]
Combining this with \eqref{e.penult} yields $\mathrm{Tr\;}[y^\ast\Psi(y)] < \mathrm{Tr\;}[(\rho^A)^\alpha]\mathrm{Tr\;}[(\rho^B)^\alpha]$.  This, combined with \eqref{beg}, proves \eqref{main}, concluding the proof of part~(a) of the theorem.

To prove part (b), we assume w.l.o.g. that $x^A$ is a strict (i.e. non-degenerate) local maximum of $Q_\alpha$, and $x^B$ is a degenerate local maximum of $Q_\alpha$. We therefore have
\begin{equation} \begin{aligned}
& D_{y^{A}}^{2}Q_\alpha(x^A)<0\;\;\;\;\;\;\forall\;\;y^A\in(x^A)^\perp\\
& D_{y^{B}}^{2}Q_\alpha(x^B)\leq 0\;\;\;\;\;\;\forall\;\;y^B\in(x^B)^\perp.
\end{aligned}\end{equation}

First note that from~\eqref{simple} it follows that $x^A\otimes x^B$ is a \emph{degenerate} local maximum of $Q_\alpha$ (recall $\alpha>1$) in any direction of the form $y=x^A\otimes y^{B}$, with $y^B\in(x^B)^\perp$.
Similar arguments shows that $x^A\otimes x^B$ is a \emph{non-degenerate} local maximum of $Q_\alpha$ in any direction of the form $y=y^A\otimes x^B$ where $y^A\in(x^A)^\perp$. It is therefore left to consider directions $y\in \left(x^A\right)^{\perp}\otimes \left(x^B\right)^\perp$. In these directions $x^A\otimes x^B$ is a \emph{non-degenerate} local maximum of $Q_\alpha$ since $D_{y}^{2}Q_\alpha(x^A\otimes x^B)<0$ for $y\in \left(x^A\right)^{\perp}\otimes \left(x^B\right)^\perp$. To see this, note that~\eqref{e.penult} still holds with strict inequality since the first equation of~\eqref{assumption} still holds, while the second equation of~\eqref{assumption} holds with $\leq$ sign. This concludes the proof of part (b) of the theorem.

\section{Proof lemma~\ref{derivatives}}\label{sec:der}

\subsection{Linearization}

Let $x,y\in\mathcal{K}\subset\mathbb{C}^{n\times n}$.
We define two Hermitian matrices $X$ and $Y$ in $\mathbb{C}^{2n\times 2n}$
corresponding to $x$ and $y$, respectively:
\begin{equation}\label{lin}
 X=\left[\begin{array}{cc} 0 & x\\ x^{*} &0 \end{array}\right]\;\text{and}\;
Y=\left[\begin{array}{cc} 0 & y \\mathbf{y}^{*}&0 \end{array}\right]\;,
 \end{equation}
 where $0$ stands for the $n\times n$ zero matrix. Note that  $\text{Tr}(X^2)=\text{Tr}(Y^2)=2$ if $\text{Tr}(xx^*)=\text{Tr}(yy^*)=1$.

 We will also denote by $\mathcal{L}\subset\mathbb{C}^{2n\times 2n}$
 the linearization space corresponding to $\mathcal{K}$; that is,
 $$
 \mathcal{L}:=\{Y\in\mathbb{C}^{2n\times 2n}\;|\;y\in\mathcal{K}\}\;,
 $$
 where $Y$ corresponds to $y$ as in Eq.~(\ref{lin}).
 Note that for $Y_1,Y_2\in\mathcal{L}$ and for $r_1,r_2\in\mathbb{R}$, the matrix $r_1Y_1+r_2Y_2$ is also in
 $\mathcal{L}$; that is, $\mathcal{L}$ is a vector space over the \emph{real} numbers.

 Define
 \begin{equation}
 G_\alpha(X)=\text{Tr}\left[X^{2\alpha}\right]\;.
 \end{equation}
 Then a key observation is that
 \begin{equation}
 G_\alpha(X)=2Q_\alpha(x)\;.
 \end{equation}
 Thus, in order to calculate the directional derivatives of $Q_\alpha$ in $\mathcal{K}$ , we will first find the derivatives of $G_\alpha$ in $\mathcal{L}$, and then translate these calculations back to the space $\mathcal{K}$.

 Consider the point $(x+ty)/\sqrt{1+t^2}$ (here $0<t\in\mathbb{R}$), in a neighbourhood of $x$.
 Recall from our discussion earlier that $\mathop{\mathrm{Tr}}\nolimits(xy^*)=0$ and assume the normalization $\text{Tr}(xx^*)=\text{Tr}(yy^*)=1$. This point is mapped to $(X+tY)/\sqrt{1+t^2}$ in the space $\mathcal{L}$.
 Moreover, the condition $\mathop{\mathrm{Tr}}\nolimits(xy^*)=0$ is equivalent to the conditions $\mathop{\mathrm{Tr}}\nolimits(XY)=\mathop{\mathrm{Tr}}\nolimits(X\tilde{Y})=0$, where
 \begin{equation}
\tilde{Y}:=\left[\begin{array}{cc} 0 & iy \\-iy^{*}& 0\end{array}\right]\;.
 \end{equation}
Now, note that
 \begin{align}\label{approx}
 G_\alpha\left(\frac{X+tY}{\sqrt{1+t^2}}\right)=\frac{1}{(1+t^2)^\alpha}G_\alpha(X+tY)&=G_\alpha(X+tY)-\alpha t^2 G_\alpha(X+tY)+O(t^4).
\end{align}
Recall that $x=\text{diag}\{p_1,...,p_n\}$.
Hence, the eigenvalues of the matrix $X$ are $\{\lambda_j\}_{j=1,...,2n}$,
where $\lambda_j=\sqrt{p_j}$ for $j=1,...,n$, $\lambda_j=-\sqrt{p_{j-n}}$ for $j=n+1,...,2n$.
We will be working with a basis in which $X=\text{diag}\{\lambda_1,...,\lambda_{2n}\}$. In this basis $Y$ does not have the form
given in Eq.(\ref{lin}). To understand the form of $Y$, we now discuss the diagonalization of $X$.

Recall that the matrix $x\in\mathbb{C}^{n\times n}$ is diagonal
(the singular value decomposition theorem).
In order to make $X$ diagonal we conjugate it with the following generalization of the Hadamard matrix:
$$
\text{diag}\{\lambda_1,...,\lambda_{2n}\}=UXU\;\text{and}\;U=\frac{1}{\sqrt{2}}\left[\begin{array}{cc} I & I\\ I &-I
\end{array}\right]
$$
where $I$ is the $n\times n$ identity matrix.
Under this change of basis, $Y$ takes the form
\begin{equation}
Y=\left[\begin{array}{cc} w & -iz\\ iz &-w\end{array}\right]\;,
\label{forms}
\end{equation}
where $w$ and $z$ are the $n\times n$  Hermitian matrices defined by $y=w+iz$, or equivalently:
$$
w\equiv \frac{1}{2}\left(y+y^{*}\right)\;\text{and}\;z=\frac{1}{2i}(y-y^*)\;.
$$

\subsection{Taylor Expansion}

In~\cite{GF} the following analytic matrix Taylor expansion was given.
\begin{theorem}[\cite{GF}, Theorem 2] \label{taylor} Let $A,B\in\mathbb{C}^{n\times n}$, and suppose $A=\mathrm{diag}(\lambda_1,\ldots,\lambda_n)$.  If $f\colon\mathbb{R}\to\mathbb{C}$ is real analytic on an open neighborhood of the eigenvalues $\{\lambda_1,\ldots,\lambda_n\}$ of $A$, then
 \begin{align}\label{polcase1}
 \mathop{\mathrm{Tr}}\nolimits [f(A+tB)]=\mathop{\mathrm{Tr}}\nolimits [f(A)]+
 t\sum_{j=1}^{n}f'(\lambda_j)[B]_{jj}+t^2\sum_{j,k=1}^n \frac{f'(\lambda_j)-f'(\lambda_k)}{2(\lambda_j-\lambda_k)}[B]_{jk}[B]_{kj}+O(t^3)\;.
 \end{align}
 \end{theorem}
\begin{remark}
The expansion above can be naturally generalized to higher than the second order, but second order suffices for our present purposes.  The second order coefficient involves a mixture of a genuine derivative and a difference quotient of $f$.  The expression must be interpreted by taking limits when there is division by $0$; for example, when $\lambda_j=\lambda_k$, the coefficient is
$$
\frac{f'(\lambda_j)-f'(\lambda_k)}{2(\lambda_j-\lambda_k)}\equiv\frac{1}{2}f''(\lambda_j)\;\;\;\text{for}\;\;\lambda_j=\lambda_k
$$
agreeing with the normal Taylor expansion when $n=1$.
\end{remark}

We would like to extend this Taylor expansion beyond the analytic category.  To give meaning to the expression $f(A+tB)$, we henceforth assume $A,B$ are Hermitian, and then interpret this through functional (spectral) calculus.  We would like to apply \eqref{polcase1} to the function $f(x) = g_\alpha(x) \equiv x^{2\alpha}$, which is $C^2$ (for $\alpha>1$) but not analytic; we will see that a version of \eqref{polcase1} holds, but the error term will not generally be $O(t^3)$ but instead $O(t^{2\alpha})$ for $1<\alpha<3/2$.

Our approach is to study the function $F(t)= \mathrm{Tr\;}[f(A+tB)]$ as an ordinary calculus function, and apply the single-variable version of Taylor's theorem to it.  We use the Kato--Rellich theorem (cf.\ \cite[p.\ 122]{Kato} see also~\cite{Friedland}) on eigenvalue perturbation, which says the following.

\begin{theorem}[Kato, Rellich] \label{t.Rellich} Let $(a,b)\subset\mathbb{R}$ be an open interval, and let $M\colon(a,b)\to\mathbb{C}^{n\times n}$ be a Hermitian matrix-valued real analytic function.  Then there is a real analytic function $U\colon(a,b)\to U(n,\mathbb{C})$ (the $n\times n$ unitary matrices) and $n$ real analytic functions $\mu_1,\ldots,\mu_n\colon(a,b)\to\mathbb{R}$ such that
\[ M(t) = U(t)^\ast \mathrm{diag}(\mu_1(t),\ldots,\mu_n(t))U(t). \]
\end{theorem}
Here, we call a matrix-valued function real analytic if all of its entries are standard $\mathbb{C}$-valued real analytic functions.  The Kato--Rellich theorem thus states the remarkable fact that the eigenvalues and eigenvectors of a Hermitian matrix $M(t)$ all depend analytically on the parameter $t$, provided the entries of $M(t)$ depend analytically on $t$.

Specializing to the case that $M(t) = A+tB$ where $A=\mathrm{diag}(\lambda_1,\ldots,\lambda_n)$, the Taylor series of the eigenvalue $\mu_j(t)$ was computed in \cite[Appendix A]{GF}; the result is
\begin{equation} \label{e.mu.Rellich} \mu_j(t) = \lambda_j + t[B]_{jj} + t^2\sum_{k\colon\lambda_k\ne\lambda_j}\frac{1}{\lambda_j-\lambda_k}|[B]_{jk}|^2 + O(t^3). \end{equation}
With this in hand, it is now straightforward to generalize Theorem \ref{taylor} to the present case, where $f$ is generally only a little smoother than $C^2$.

\begin{theorem} \label{t.Taylor.2} Let $(a,b)$ be an open interval in $\mathbb{R}$, let $0<\epsilon<1$, and let $f\in C^{2+\epsilon}(a,b)$, meaning that $f\in C^2(a,b)$ and $f''$ is $\epsilon$-H\"older continuous on $(a,b)$.  Let $A,B\in\mathbb{C}^{n\times n}$ be Hermitian matrices, with $A = \mathrm{diag}(\lambda_1,\ldots,\lambda_n)$.  Then the function $F(t) = \mathrm{Tr\;}[f(A+tB)]$ is also in $C^{2+\epsilon}(a,b)$, and
\begin{equation} \label{polcase2}
F(t) = \mathop{\mathrm{Tr}}\nolimits [f(A)]+ t\sum_{j=1}^{n}f'(\lambda_j)[B]_{jj}+t^2\sum_{j,k=1}^n \frac{f'(\lambda_j)-f'(\lambda_k)}{2(\lambda_j-\lambda_k)}|[B]_{jk}|^2+O(t^{2+\epsilon}).
\end{equation}
\end{theorem}

\begin{proof} By Theorem \ref{t.Rellich}, the matrix $A+tB$ can be diagonalized as $A+tB = U(t)^\ast\mathrm{diag}(\mu_1(t),\ldots,\mu_n(t))U(t)$ where $\mu_1,\ldots,\mu_n$ are analytic in $t$.  By functional (spectral) calculus, $f(A+tB)$ is defined to be
\[ f(A+tB) = U(t)^\ast\mathrm{diag}(f(\mu_1(t)),\ldots,f(\mu_n(t)))U(t) \]
and so
\[ F(t) = \mathrm{Tr\;}[f(A+tB)] = \sum_{j=1}^n f(\mu_j(t)). \]
Since the functions $\mu_j$ are analytic and $f\in C^{2+\epsilon}(a,b)$, it follows that the functions $f\circ\mu_j$ are in $C^{2+\epsilon}(a,b)$, and therefore so is their sum $F$.  Therefore, from Taylor's theorem
\[ F(t) = F(0) + tF'(0)+\frac12t^2F''(0) + O(t^{2+\epsilon}). \]
Indeed, to see this, use Taylor's theorem to first order with mean value remainder: $F(t) = F(0) + tF'(0) + \frac12t^2 F''(\xi)$ for some $\xi\in[0,t]$; but since $F''$ is $\epsilon$-H\"older continuous, $|F''(\xi)-F''(0)| \le C|\xi-0|^\epsilon \le Ct^\epsilon$ for some constant $C$.  So $\frac12t^2F''(\xi) = \frac12t^2F''(0) + O(t^{2+\epsilon})$ as required.

It remains only to compute the coefficients $F(0)$, $F'(0)$, and $F''(0)$, which we can now easily do using \eqref{e.mu.Rellich}. First we have
\[ F(0) = \sum_{j=1}^n f(\mu_j(0)) = \sum_{j=1}^n f(\lambda_j) = \mathrm{Tr\;}[f(A)]. \]
Next, applying the chain rule,
\[ F'(0) = \sum_{j=1}^n f'(\mu_j(0))\mu_j'(0) = \sum_{j=1}^n f'(\lambda_j)[B]_{jj}. \]
Differentiating one more time, we have $\frac{d^2}{dt^2}f(\mu_j(t)) = f''(\mu_j(t))(\mu_j'(t))^2 + f'(\mu_j(t))\mu_j''(t)$.  Setting $t=0$ and summing gives
\[ F''(0) = \sum_{j=1}^n \left[f''(\lambda_j)([B]_{jj})^2 + 2f'(\lambda_j)\sum_{k\colon j\ne k}\frac{1}{\lambda_j-\lambda_k}|[B]_{jk}|^2\right]. \]
To see this has the form given in \eqref{polcase2}, denote by $\delta(j,k) = 0$ if $\lambda_j=\lambda_k$, while $\delta(j,k)=1$ if $\lambda_j\ne\lambda_k$.  Then
\begin{align*} F''(0) = \sum_{j=1}^n f''(\lambda_j)([B]_{jj})^2 + 2\sum_{j,k=1}^n f'(\lambda_j)\frac{\delta(j,k)}{\lambda_j-\lambda_k}|[B]_{jk}|^2. \end{align*}
Break up the second sum into its two copies, and in the second one reverse the roles of $j$ and $k$:
\begin{align*} 2\sum_{j,k=1}^n f'(\lambda_j)\frac{\delta(j,k)}{\lambda_j-\lambda_k}|[B]_{jk}|^2 &= \sum_{j,k=1}^n f'(\lambda_j)\frac{\delta(j,k)}{\lambda_j-\lambda_k}|[B]_{jk}|^2 + \sum_{j,k=1}^n f'(\lambda_k)\frac{\delta(j,k)}{\lambda_k-\lambda_j}|[B]_{kj}|^2 \\
&= \sum_{j,k=1}^n \delta(j,k) \frac{f'(\lambda_j)-f'(\lambda_k)}{\lambda_j-\lambda_k}|[B]_{jk}|^2
\end{align*}
where in the second line we have used the fact that $B$ is Hermitian, so $|[B]_{jk}| = |[B]_{kj}|$.  Now, as noted above, if $\lambda_j=\lambda_k$, we interpret the difference quotient to mean $\frac{f'(\lambda_j)-f'(\lambda_k)}{\lambda_j-\lambda_k} \equiv f''(\lambda_j)$.  As such, if there is some $j\ne k$ with $\lambda_j=\lambda_k$, then this term will appear twice in the sum (once at the index $(j,k)$, and again at the index $(k,j)$), with opposite signs; hence, $\delta(j,k)$ is automatically accounted for whenever $j\ne k$, and so we have
\[ F''(0) = \sum_{j=1}^n f''(\lambda_j)([B]_{jj})^2 + \sum_{1\le j\ne k\le n} \frac{f'(\lambda_j)-f'(\lambda_k)}{\lambda_j-\lambda_k}|[B]_{jk}|^2. \]
The first sum is precisely the missing diagonal terms from the second sum, owing to the fact that $B$ is Hermitian and so $([B]_{jj})^2 = |[B]_{jj}|^2$.  This completes the proof.  \end{proof}

\begin{remark} It is worth noting that an alternate proof is possible, avoiding the Kato-Rellich theorem, using Fourier analysis and the useful identity that, for any $C^1$ function $h$, the difference quotient can be written in the form
\[ \frac{h(x)-h(y)}{x-y} = \int_0^1 h'(sx+(1-s)y)\,ds. \]
This allows one to quickly recover the second order Taylor expansion of $\mathrm{Tr\;}[f(X+tY)]$ given in \eqref{polcase2} for functions $f$ that are slightly smoother than $C^2$ (functions $f$ that are Fourier transforms of complex measures with finite absolute second moment).  This is an unnecessary technical restriction, but suffices to deal with the desired function $f(x) = x^{2\alpha}$ after a smooth cutoff, and it gives a little more motivation to explain {\em why} the mixed difference quotient derivative appears in the Taylor expansion.  For more details, see \cite[Section 3]{KNPS}.
\end{remark}

We now use Theorem \ref{t.Taylor.2} to compute the Taylor expansion of the function $G_\alpha(X+tY)$ in a neighbourhood of $t=0$, and use it to prove Lemma~\ref{derivatives}.

\subsection{The first and second derivatives of $Q_\alpha(x)$}

We summarize the statement of Lemma~\ref{derivatives} in the following theorem, where we also include the first order derivative.

\begin{theorem}
Let $\rho\equiv x^2=\mathop{{\rm diag}}\nolimits\{p_1,\ldots,p_{n}\}$. Decompose $y=w+iz$ with $w,z$ Hermitian. Then,
\begin{align}
&D_{y}^{1}Q_\alpha(x) =\alpha\mathop{\mathrm{Tr}}\nolimits(w x^{2\alpha-1})\nonumber\\
&D_{y}^{2}Q_\alpha(x)=2\alpha\left[-\mathop{\mathrm{Tr}}\nolimits\left(\rho^{\alpha}\right)+\mathop{\mathrm{Tr}}\nolimits\left(w\Phi_{\rho}^{-}(w)+z\Phi_{\rho}^{+}(z)\right)
\right]
\label{se}
\end{align}
where $\Phi^\pm_\rho$ are defined in \eqref{e.phi.simple}.

\end{theorem}

\begin{remark}
The condition for $x\in\mathcal{K}$ to be critical is $D_{y}^{1}Q_\alpha(x)=0$ which is equivalent to $\mathop{\mathrm{Tr}}\nolimits[(y^*+y)x^{2\alpha-1}]=0$ for all $y\in\mathcal{K}$ such that $\mathop{\mathrm{Tr}}\nolimits(xy^*)=0$. Moreover, if $x$ is critical then we also have
$D_{iy}^{1}Q_\alpha(x)=0$ for all $y\in x^\perp\subset\mathcal{K}$.
Hence, if $x$ is critical we must have
\begin{equation}\label{critical}
\mathop{\mathrm{Tr}}\nolimits(y^*x^{2\alpha-1})=\mathop{\mathrm{Tr}}\nolimits(yx^{2\alpha-1})=0
\end{equation}
for all $y\in x^\perp\subset\mathcal{K}$.
\end{remark}
\begin{proof} Since $\alpha>1$, the function $g_\alpha(x) = x^{2\alpha}$ is $C^{2+\epsilon}(\mathbb{R})$ for any $\epsilon\le 2(\alpha-1)$.  (We think of $g_\alpha(x) = (x^2)^\alpha$, so $g'(x) = 2\alpha (x^2)^{\alpha-1/2}$ and $g''(x) = 2\alpha(2\alpha-1)(x^2)^{\alpha-1}$.)  Therefore, we may apply Theorem \ref{t.Taylor.2} to the function $f=g_\alpha$, with $A=X$ and $B=Y$.  Applying the expansion~\eqref{polcase2} to $F(t) = G_\alpha(X+tY) = \mathrm{Tr\;}[g_\alpha(X+tY)]$ gives
\begin{align}\label{tay}
 G_\alpha(X+tY)=G_\alpha(X)+
 t\sum_{j=1}^{2n}g'_\alpha(\lambda_j)Y_{jj}+t^2\sum_{j,k=1}^{2n} \frac{g'_\alpha(\lambda_j)-g'_\alpha(\lambda_k)}{2(\lambda_j-\lambda_k)}\left|Y_{jk}\right|^2+O(t^{2+\varepsilon}).
 \end{align}

Recalling that $\lambda_j=\sqrt{p_j}$ for $j=1,...,n$, $\lambda_j=-\sqrt{p_{j-n}}$ for $j=n+1,...,2n$ gives
\begin{equation}
\sum_{j=1}^{2n}g'_\alpha(\lambda_j)Y_{jj}=2\alpha\sum_{j=1}^{n}p^{\alpha-1/2}_{j}w_{jj}=2\alpha\mathop{\mathrm{Tr}}\nolimits\left(x^{2\alpha-1}w\right)
\end{equation}
where we have used the form~\eqref{forms} of $Y$.
Similarly, for the second terms in~\eqref{tay} we get
\begin{align}
\sum_{j,k=1}^{2n} \frac{g'_\alpha(\lambda_j)-g'_\alpha(\lambda_k)}{2(\lambda_j-\lambda_k)}\left|Y_{jk}\right|^2
=
\sum_{j,k=1}^{n} \frac{g'_\alpha(\sqrt{p_j})-g'_\alpha(\sqrt{p_k})}{\sqrt{p_j}-\sqrt{p_k}}\left|w_{jk}\right|^2+\sum_{j,k=1}^{n} \frac{g'_\alpha(\sqrt{p_j})+g'_\alpha(\sqrt{p_k})}{\sqrt{p_j}+\sqrt{p_k}}\left|z_{jk}\right|^2
\end{align}
where we have used $g_\alpha(-\sqrt{p_k})=-g_\alpha(\sqrt{p_k})$.
Substituting $g'_\alpha(\sqrt{p_j})=2\alpha p_{j}^{\alpha-1/2}$
gives
\begin{equation}
\sum_{j,k=1}^{2n} \frac{g'_\alpha(\lambda_j)-g'_\alpha(\lambda_k)}{2(\lambda_j-\lambda_k)}\left|Y_{jk}\right|^2
=2\alpha\mathop{\mathrm{Tr}}\nolimits\left(w\Phi_{\rho}^{-}(w)+z\Phi_{\rho}^{+}(z)\right).
\end{equation}
Now, combining \eqref{approx} with \eqref{polcase2} to $0$th order, we have

\begin{align*} G_\alpha\left(\frac{X+tY}{\sqrt{1+t^2}}\right)=\frac{1}{(1+t^2)^\alpha}G_\alpha(X+tY)&=G_\alpha(X+tY)-\alpha t^2 G_\alpha(X+tY)+O(t^4) \\
&= G_\alpha(X+tY) - \alpha t^2 G_\alpha(X) + O(t^3)
\end{align*}
and combining this with the full force of \eqref{polcase2} yields
\begin{equation}
 G_\alpha\left(\frac{X+tY}{\sqrt{1+t^2}}\right)=G_\alpha(X)+t2\alpha\mathop{\mathrm{Tr}}\nolimits\left(x^{2\alpha-1}w\right)+\alpha t^2\left[-G_\alpha(X)
 +2\mathop{\mathrm{Tr}}\nolimits\left(w\Phi_{\rho}^{-}(w)+z\Phi_{\rho}^{+}(z)\right)\right]+O(t^{2+\varepsilon}).
 \end{equation}
Finally, since $Q_\alpha(x)=\frac12 G_\alpha(X)$, we conclude
 \begin{equation}
 Q_\alpha\left(\frac{X+tY}{\sqrt{1+t^2}}\right)=Q_\alpha(x)+t\alpha\mathop{\mathrm{Tr}}\nolimits\left(x^{2\alpha-1}w\right)+\alpha t^2\left[-Q_\alpha(x)
 +\mathop{\mathrm{Tr}}\nolimits\left(w\Phi_{\rho}^{-}(w)+z\Phi_{\rho}^{+}(z)\right)\right]+O(t^{2+\varepsilon}).
 \end{equation}
 This completes the proof.  \end{proof}

\section{Proof of Lemma~\ref{directions}}\label{sec:dir}
Consider the general form of $y$ given in~\eqref{generalform}. Writing $y=w+iz$ with $w,z$ Hermitian, we therefore have
\begin{align}
w & =c_1x^A\otimes w^B+c_2w^A\otimes x^B+c_3w'\nonumber\\
z & =c_1x^A\otimes z^B+c_2\omega^A\otimes x^B+c_3z'
\end{align}
where we decomposed $y^{A}=w^A+iz^A$, $y^B=w^B+iz^B$, and $y'=w'+iz'$.
Now substituting these into the formula~\eqref{deriv} of Lemma~\ref{derivatives} for the second directional derivative of $Q_\alpha$ gives
\begin{align}\label{crossterms}
\frac{1}{2\alpha}D_{y}^{2}Q_\alpha(x^A\otimes x^B)  =-\mathop{\mathrm{Tr}}\nolimits\left[\left(\rho^A\otimes\rho^B\right)^\alpha\right]
&+c_{1}^{2}\mathop{\mathrm{Tr}}\nolimits\left[x^A\otimes w^B\Phi_{\rho^A\otimes\rho^B}^{-}(x^A\otimes w^B)+x^A\otimes z^B\Phi_{\rho^A\otimes\rho^B}^{+}(x^A\otimes z^B)\right]\nonumber\\
&+c_{2}^{2}\mathop{\mathrm{Tr}}\nolimits\left[w^A\otimes x^B\Phi_{\rho^A\otimes\rho^B}^{-}(w^A\otimes x^B)+z^A\otimes x^B\Phi_{\rho^A\otimes\rho^B}^{+}(z^A\otimes x^B)\right]\nonumber\\
&+c_{3}^{2}\mathop{\mathrm{Tr}}\nolimits\left[w'\Phi_{\rho^A\otimes\rho^B}^{-}(w')+z'\Phi_{\rho^A\otimes\rho^B}^{+}(z')\right]+\text{cross terms}\nonumber\\
& =c_{1}^{2}D_{x^A\otimes y^{B}}^{2}Q_\alpha(x)+c_{2}^{2}D_{y^{A}\otimes x^B}^{2}Q_\alpha(x)+c_{3}^{2}D_{y'}^{2}Q_\alpha(x)+\text{cross terms}
\end{align}
where in the last equality we have used the normalization $c_{1}^{2}+c_{2}^{2}+c_{3}^{2}=1$. The cross terms are all the elements that have two distinct terms to the right and left of $\Phi_{\rho^A\otimes\rho^B}^{\pm}$. We now show that all these terms are zero if $x^A$ and $x^B$ are critical points.

First, recall that w.l.o.g.\ we assume that both $x^A$ and $x^B$ are square diagonal matrices. Moreover, since they are critical points we get from~\eqref{critical} that
\begin{equation}\label{cri}
\mathop{\mathrm{Tr}}\nolimits\left[w^A(x^A)^{2\alpha-1}\right]=\mathop{\mathrm{Tr}}\nolimits\left[z^A(x^A)^{2\alpha-1}\right]=0
\end{equation}
and
\begin{equation}
\mathop{\mathrm{Tr}}\nolimits\left[w^B(x^B)^{2\alpha-1}\right]=\mathop{\mathrm{Tr}}\nolimits\left[z^B(x^B)^{2\alpha-1}\right]=0\;.
\end{equation}

Next, note that
\begin{equation}
\left[\Phi_{\rho^A\otimes\rho^B}^{\pm}(y)\right]_{jk,\ell m}=\phi^{AB\pm}_{jk,\ell m}y_{jk,\ell m}\;\;\text{ where }\;\;
\phi^{AB\pm}_{jk,\ell m}\equiv\frac{(p_jq_k)^{\alpha-1/2}\pm (p_\ell q_m)^{\alpha-1/2}}{(p_jq_k)^{1/2}\pm (p_\ell q_m)^{1/2}}
\end{equation}
where $\{p_j\}$ and $\{q_k\}$ are the eigenvalues of the diagonal matrices $\rho^A\equiv\left(x^A\right)^2$ and $\rho^B=(x^B)^2$. Note also that $\psi^{AB\pm}_{jk,\ell k}=\phi^{A\pm}_{j\ell}q^{\alpha-1}_{k}$ and similarly $\phi^{AB\pm}_{jk,j m}=p_{j}^{\alpha-1}\phi^{B\pm}_{km}$. Therefore, since $x^A$ is diagonal we get
\begin{equation}
\Phi_{\rho^A\otimes\rho^B}^{\pm}(x^A\otimes w^B)=\left(x^A\right)^{2\alpha-1}\otimes\Phi_{\rho^B}^{\pm}(w^B),
\end{equation}
and similarly
\begin{equation}
\Phi_{\rho^A\otimes\rho^B}^{\pm}(w^A\otimes x^B)=\Phi_{\rho^A}^{\pm}(w^A)\otimes\left(x^B\right)^{2\alpha-1}\;.
\end{equation}
Therefore, computing the first cross terms,
\begin{align}
& \mathop{\mathrm{Tr}}\nolimits\left[x^A\otimes w^B\Phi_{\rho^A\otimes\rho^B}^{-}(w^A\otimes x^B)\right]=\mathop{\mathrm{Tr}}\nolimits\left[w^A\otimes x^B\Phi_{\rho^A\otimes\rho^B}^{-}(x^A\otimes w^B)\right]\nonumber\\
&=\mathop{\mathrm{Tr}}\nolimits\left[\left(w^A\otimes x^B\right)\left(\left(x^A\right)^{2\alpha-1}\otimes\Phi_{\rho^B}^{-}(w^B)\right)\right]
=\mathop{\mathrm{Tr}}\nolimits\left[w^A\left(x^A\right)^{2\alpha-1}\right]\mathop{\mathrm{Tr}}\nolimits\left[ x^B\Phi_{\rho^B}^{-}(w^B)\right]=0\;,
\end{align}
where the first equality follows from the fact that $\Phi_{\rho^A\otimes\rho^B}^{-}$ is self-adjoint, and the last one from~\eqref{cri}.
Next, the cross terms
\begin{equation}\label{cross}
\mathop{\mathrm{Tr}}\nolimits\left[x^A\otimes w^B\Phi_{\rho^A\otimes\rho^B}^{-}(w')\right]=\mathop{\mathrm{Tr}}\nolimits\left[w'\Phi_{\rho^A\otimes\rho^B}^{-}(x^A\otimes w^B)\right]=\mathop{\mathrm{Tr}}\nolimits\left[w'\left(\left(x^A\right)^{2\alpha-1}\otimes\Phi_{\rho^B}^{-}(w^B)\right)\right]\;.
\end{equation}
To see that this term is also zero, recall the expression for $y'$ in~\eqref{yprime}. It gives
\begin{equation}
w'=\frac{y'+y^{\prime*}}{2}=\frac{1}{2}\sum_{i}r_i \left(y^{A}_{i}\otimes y^{B}_{i}+y^{*A}_{i}\otimes y^{*B}_{i}\right)\;.
\end{equation}
Hence, the right hand side of~\eqref{cross} becomes
\begin{align}
&\mathop{\mathrm{Tr}}\nolimits\left[w'\left(\left(x^A\right)^{2\alpha-1}\otimes\Phi_{\rho^B}^{-}(w^B)\right)\right] =\frac{1}{2}\sum_{i}r_i \mathop{\mathrm{Tr}}\nolimits\left[\left(y^{A}_{i}\otimes y^{B}_{i}+y^{*A}_{i}\otimes y^{*B}_{i}\right)\left(\left(x^A\right)^{2\alpha-1}\otimes\Phi_{\rho^B}^{-}(w^B)\right)\right]\nonumber\\
&=
\frac{1}{2}\sum_{i}r_i\left(\mathop{\mathrm{Tr}}\nolimits\left[y^{A}_{i}\left(x^A\right)^{2\alpha-1}\right]\mathop{\mathrm{Tr}}\nolimits\left[y^{B}_{i}\Phi_{\rho^B}^{-}(w^B)\right]+
\mathop{\mathrm{Tr}}\nolimits\left[y^{*A}_{i}\left(x^A\right)^{2\alpha-1}\right]\mathop{\mathrm{Tr}}\nolimits\left[y^{*B}_{i}\Phi_{\rho^B}^{-}(w^B)\right]\right)=0\;,
\end{align}
where in the last equality we have used~\eqref{critical} for $y_{i}^{A}$ and $y_{i}^{*A}$. Using similar arguments for the final cross terms yields
\begin{equation}
\mathop{\mathrm{Tr}}\nolimits\left[w^A\otimes x^B\Phi_{\rho^A\otimes\rho^B}^{-}(w')\right]=\mathop{\mathrm{Tr}}\nolimits\left[w'\Phi_{\rho^A\otimes\rho^B}^{-}(w^A\otimes x^B)\right]=0\;.
\end{equation}
Therefore, we have shown that all the cross terms of $w$ in~\eqref{crossterms} are zero.
Using similar arguments it follows that all the cross terms of $z$ are also zero. This completes the proof of Lemma~\ref{directions}.

\section{Proof of Lemma~\ref{additivity}}\label{sec:add}
We will prove that
$\Phi_{\rho^{A}\otimes\rho^{B}}^{\pm}\leq \Psi$ by showing it for the components. That is, we will show that
$\phi_{jk,\ell m}^{\pm}\leq \psi_{jk,\ell m}$ for all indices $j,k,\ell,m$; it is straightforward to verify that this is equivalent to the operator inequality for Haadamard product operators. This componentwise inequality is equivalent to
\begin{equation} \label{a} \frac{(p_jq_k)^{\alpha-1/2}\pm (p_\ell q_m)^{\alpha-1/2}}{(p_jq_k)^{1/2}\pm (p_\ell q_m)^{1/2}} \le \frac{p_j^\alpha- p_\ell^\alpha}{p_j-p_\ell}\frac{q_k^\alpha-q_m^\alpha}{q_k-q_m}. \end{equation}

The following simple lemma shows that $\Phi_{\rho^{A}\otimes\rho^{B}}^{\pm}\leq \Psi$ if and only if
$\Phi_{\rho^{A}\otimes\rho^{B}}^{-}\leq \Psi$.

\begin{lemma} \label{lemma.diff.quotient} Let $r,s\ge 0$ and $\beta\ge1$.  Then
\begin{equation} \label{e.ineq.30} \frac{r^\beta-s^\beta}{r-s} \ge \frac{r^\beta + s^\beta}{r+s}\ge 0. \end{equation}
\end{lemma}
\begin{proof} First suppose $r\ne 0$ and $s\ne r$.  Dividing through by $r$ and setting $t=s/r$, the desired inequalities are
\begin{equation} \label{e.ineq.31} \frac{1-t^\beta}{1-t} \ge \frac{1+t^\beta}{1+t} \ge 0. \end{equation}
The second inequality is manifestly satisfied.  It is also easy to see that $\frac{1-t^\beta}{1-t}\ge 0$ (in fact whenever $\beta\ge 0$), simply by considering the two cases $t<1$ and $t>1$.  For the first inequality in \eqref{e.ineq.31}, we simplify
\begin{equation} \frac{1-t^\beta}{1-t} - \frac{1+t^\beta}{1+t} = \frac{2t}{1+t}\frac{1-t^{\beta-1}}{1-t} \end{equation}
and, by what we just showed, this is $\ge 0$ as well.

Now, if $s=r\ne 0$, we interpret the terms by taking the limit $s\to r$, which corresponds to $t\to 1$, and so \eqref{e.ineq.31} becomes $\beta\ge 1\ge 0$, which is true given the assumptions of the lemma.  Finally, if $r=0$, then \eqref{e.ineq.30} is the true statement $1\ge 1\ge 0$ if $s\ne 0$ (and similarly if $s=0$, evaluated by taking the limit $s\to0$).  This concludes the proof. \end{proof}

Applying the above lemma to \eqref{a}, with $r=\sqrt{p_jq_k}$ and $s=\sqrt{p_\ell q_m}$ shows that it is sufficient to prove:
\begin{equation}
\label{b} \frac{(p_jq_k)^{\alpha-1/2}- (p_\ell q_m)^{\alpha-1/2}}{(p_jq_k)^{1/2}- (p_\ell q_m)^{1/2}} \le \frac{p_j^\alpha- p_\ell^\alpha}{p_j-p_\ell}\frac{q_k^\alpha-q_m^\alpha}{q_k-q_m}.
\end{equation}
We first contend with some degenerate cases.  Suppose either $p_j=0$ or $q_k=0$; then the inequality reduces to
\begin{equation} \label{e.reduced.ineq} (p_\ell q_m)^{\alpha-1} \le p_\ell^{\alpha-1} \frac{q_k^\alpha-q_m^\alpha}{q_k-q_m}. \end{equation}
If $p_\ell=0$ this holds vacuously as $0\le 0$; otherwise we divide through by $p_\ell^{\alpha-1}$, giving
\begin{equation} \label{e.reduced.ineq.2} q_m^{\alpha-1} \le \frac{q_k^\alpha-q_m^\alpha}{q_k-q_m}. \end{equation}
It is easy to verify that this holds true for all $q_m\ge0$ and all $\alpha\ge 1$.  Thus, \eqref{a} holds true in these degenerate cases.  We therefore assume $p_j,q_k>0$.  Henceforth, let $s=\frac{p_\ell}{p_j}$ and $t=\frac{q_m}{q_k}$. Dividing both sides of \eqref{a} through by $(p_jq_k)^{\alpha-1/2}$, our final goal is to prove the following.

\begin{proposition} \label{c.strong.conj} For all $s,t\ge 0$ and $\alpha\ge 1$,
\begin{equation} \label{con} \frac{1-(st)^{\alpha-1/2}}{1-\sqrt{st}}\leq
\frac{1-s^\alpha}{1-s}\frac{1-t^\alpha}{1-t}. \end{equation}
The inequality is strict if $\alpha>1$ and at least one of $s,t$ is $>0$.
\end{proposition}

\begin{proof} Inequality \eqref{con} takes the form $f(st) \le g(s)g(t)$ where
\begin{equation} f(t) = \frac{1-t^{\alpha-1/2}}{1-t^{1/2}}, \qquad g(t) = \frac{1-t^\alpha}{1-t}. \end{equation}
First consider the case $s=t$.  We compute
\begin{equation} \label{e.g-f} g(t)^2 - f(t^2) = \frac{(1-t^\alpha)^2}{(1-t)^2} - \frac{1-t^{2\alpha-1}}{1-t} = \frac{(1-t^\alpha)^2 - (1-t)(1-t^{2\alpha-1})}{(1-t)^2}. \end{equation}
The numerator simplifies to
\begin{equation} \label{e.s=t} (1-2t^\alpha + t^{2\alpha}) - (1-t-t^{2\alpha-1}+t^{2\alpha}) = -2t^\alpha + t + t^{2\alpha-1} = t(t^{2(\alpha-1)} - 2t^{\alpha-1}+1) = t(t^{\alpha-1}-1)^2 \ge 0. \end{equation}
Thus, we know $f(t^2) \le g(t)^2$.  We would like to conclude that $f(st) \le g(s)g(t)$.  Let $r=\sqrt{st}$; what we just proved shows that
\begin{equation} \label{e.con2} f(st) = f(r^2) \le g(r)^2 = g(\sqrt{st})^2. \end{equation}
Therefore, to prove the desired inequality, it suffices to show that $g(\sqrt{st})^2 \le g(s)g(t)$.  Now, $g$ is a positive function of a positive variable, so we can define a new function $h(\xi) = \ln g(e^\xi)$, where $\xi\in\mathbb{R}$.  Then the requirement that $g(\sqrt{st})^2 \le g(s)g(t)$ becomes the statement that
\begin{equation} h\left(\frac12(\xi + \zeta)\right) \le \frac12\left(h(\xi)+h(\zeta)\right). \end{equation}
Therefore, the proof will be complete once we show that $h$ is convex.

To be explicit, the function $h$ is
\begin{equation} h(\xi) = \ln\frac{1-e^{\alpha \xi}}{1-e^\xi}. \end{equation}
The function is manifestly smooth for $\xi\ne 0$, and is continuous on $\mathbb{R}$ if we define its value at $0$ to be the limit $\ln\alpha$.  Note also that
\begin{equation} h(-\xi) = (1-\alpha)\xi + h(\xi). \end{equation}
Hence, it suffices to show that $h$ is convex on $(0,\infty)$.  On this domain, $h(\xi) = \ln(e^{\alpha\xi}-1)-\ln(e^\xi-1)$, and so
\begin{equation} h''(\xi) =  -\frac{\alpha^2 e^{\alpha\xi}}{(e^{\alpha\xi}-1)^2}+\frac{e^\xi}{(e^\xi-1)^2}. \end{equation}
Our goal is to show that $h''(\xi)\ge 0$ for all $\xi>0$.  Note that $h''(\xi) = \upsilon(1,\xi)-\upsilon(\alpha,\xi)$, where
\begin{equation} \upsilon(\alpha,\xi) = \frac{\alpha^2 e^{\alpha\xi}}{(e^{\alpha\xi}-1)^2}. \end{equation}
Hence, to show the desired conclusion that $h''(\xi)\ge 0$ for all $\xi>0$, it suffices to show that for each $\xi>0$ the function $\alpha\mapsto\upsilon(\alpha,\xi)$ is decreasing.  We compute the derivative
\begin{equation} \label{e.d/dalpha} \frac{\partial}{\partial\alpha}\upsilon(\alpha,\xi) = -\alpha \frac{e^{\alpha \xi}}{(e^{\alpha\xi}-1)^3}[(\alpha \xi-2)e^{\alpha \xi} + \alpha \xi+2]. \end{equation}
The factor $-\alpha \frac{e^{\alpha \xi}}{(e^{\alpha\xi}-1)^3}$ is $<0$.  The remaining factor takes the form $\chi(\alpha\xi)$, where
\begin{equation} \chi(u) = (u-2)e^u + u + 2. \end{equation}
Elementary calculus shows that $\chi$ is smooth, $\chi'(u) = (u-1)e^u + 1$, and $\chi''(u) = ue^u$.  In particular, $\chi(0) = \chi'(0) = \chi''(0)=0$.  Since $\chi''(u)>0$ for $u>0$, $\chi'$ is increasing on this domain, so $\chi'(u) > \chi'(0)=0$.  Thus $\chi$ is increasing, and since $\chi(0)=0$, $\chi(u)\ge 0$.  We conclude that $\frac{\partial}{\partial \alpha}\upsilon(\alpha,\xi) < 0$ for $\alpha,\xi>0$, as desired, thus proving \eqref{con}.

As to the strictness: \eqref{e.g-f} and \eqref{e.s=t} show that $f(t^2)<g(t)^2$ for $t>0$ (the case $s=t=1$ reduces \eqref{con} to $2\alpha-1 \le \alpha^2$ which is strict for $\alpha>1$).  Hence, we also get strictness in \eqref{con} whenever $st\ne 0$, thanks to \eqref{e.con2}.  If only one of $s,t$ is $0$ (say $s=0$ but $t\ne 0$), then \eqref{con} becomes $1 \le \frac{1-t^\alpha}{1-t}$, which is easily verified to be strict for $t>0$ and $\alpha>1$.  \end{proof}

\begin{remark} In fact, to conclude that the derivative in \eqref{e.d/dalpha} is $<0$ only needed $\alpha>0$.  But since $h_\alpha''(\xi) = \upsilon(1,\xi)-\upsilon(\alpha,\xi)$, this means that for $0<\alpha<1$, the function $h_\alpha$ is actually {\em concave}.  Hence our proof of \eqref{con} fails in this regime.  (This does not, however, mean that \eqref{con} is necessarily false there.) \end{remark}

\section{Concluding Remarks}\label{sec:concluding}

We have shown that the minimum Renyi entropy output of a quantum channel with Reyni parameter $\alpha>1$
is locally additive. This result extends the work of~\cite{GF} from $\alpha=1$ to $\alpha> 1$, and thereby demonstrates that local additivity holds for a large class of entropy functionals.  In~\cite{HW} Hayden and Winter showed that there are counterexamples for the global additivity conjecture for all Renyi entropies with $\alpha>1$. Hence, the current work complements their result by showing that these counter examples corresponds to a global effect of quantum channels, and cannot be a consequence of local properties of the channels involved.

In Appendix B of~\cite{FGA} (see also~\cite{GF}), it was shown that
both the local and global additivity conjectures are false for all Renyi entropies over the real numbers.
This in turn implies that a straightforward argument involving just directional derivatives
could not provide a proof of local additivity in the general complex case. Hence,
our method to prove local additivity strongly involved the complex structure. In particular, in~\eqref{complex} we use explicitly the assumption that $D_{y}^{2}Q_\alpha(x^A)<0$ in both directions $y=y^A$ and $y=iy^A$, where $i=\sqrt{-1}$.

While both proofs of local additivity for $\alpha=1$ and $\alpha>1$ use explicitly the complex structure, they exhibit key differences. The distinction follows from the fact that in the case $\alpha>1$ we essentially prove local multiplicativity of the $Q_\alpha$ functions, whereas in the von-Neumann case we prove local additivity. The later is somewhat more simple since certain cross-terms cancel out due to the additivity property of the von-Neumann case.  In particular, given $y=\sum_{i}r_iy_{i}^{A}\otimes y_{i}^{B}\in \left(x^A\right)^\perp\otimes \left(x^B\right)^\perp$, we had to show that
$D_{y}^{2}E_\alpha(x^A\otimes x^B)>0$ (or equivalently $D_{y}^{2}Q_\alpha(x^A\otimes x^B)<0$). In the $\alpha>1$ case, this was done using the fact that $D_{y^A}^{2}E_\alpha(x^A)>0$ and $D_{y^B}^{2}E_\alpha(x^B)>0$ for any
$y^A\in\text{span}\{y_{i}^{A}\}$ and $y^{B}\in\text{span}\{y_{i}^{B}\}$, respectively. On the other hand, in the case $\alpha=1$, all we needed to use is that $D_{y^{A}_{i}}^{2}E_\alpha(x^A)>0$ and $D_{y^{B}_{i}}^{2}E_\alpha(x^B)>0$ for all $i$.
This simplification was possible in the $\alpha=1$ case since the additive nature of the von-Neumann entropy led to the cancellation of the cross terms in the linear combination of $y=\sum_{i}r_iy_{i}^{A}\otimes y_{i}^{B}$. This cancellation does not occur in the $\alpha>1$ case, and instead we had to diagonalize the matrices $A$ and $B$ and use other arguments (see Lemma~5 and the arguments below it).

Another key difference between the $\alpha=1$ case and the $\alpha>1$ case is related to Lemma 4.
Lemma~4 in the limit $\alpha\to 1$ does \emph{not} reduce to the analogous lemma that was used in the $\alpha=1$ case. Again, the main reason for this is the multiplicativity versus additivity properties of the $\alpha>1$ case and the $\alpha=1$ case, respectively. In particular, Lemma~4 (or Proposition~\ref{c.strong.conj} that is used to prove Lemma 4) becomes trivial in the limit $\alpha\to 1$ and cannot be used to prove local additivity for the case $\alpha=1$.

In both cases of $\alpha=1$ (see~\cite{GF}) and $\alpha>1$ we had to assume that at least one of the two local minima is strict. The main reason for that is that otherwise it seems to be possible that  $D_{y}^{2}E_\alpha(x^A\otimes x^B)=0$ (rather than strictly positive) for some $y\in\left(x^A\right)^\perp\otimes \left(x^B\right)^\perp$. In order to study this case,
one will need to study third and fourth order directional derivatives which lead to very cumbersome expressions. It is therefore left open if local additivity holds in this case.

Finally, the case $\alpha<1$ was not studied in this paper since Lemma 4 fails to hold in this limit (in fact, in this case we need a similar lemma with the inequality {\em reversed}, since we are interested in local {\em minima} and not local maxima of $Q_\alpha$). Hence, the techniques used here can not be applied directly to this case, and we leave the study of this case for future work.\\

\subsection*{Acknowledgments}
We extend thanks to Mark Girard for many stimulating discussions on topics that are closely related to this work.

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