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text:
Q: Query and select data from the test2 and test3 tables from MySQL? I am looking to get the data from the test2 and test3 tables to show up on the webpage, but I keep seeing a blank screen.
$query = "INSERT INTO test2,test3 VALUES(' ','$title','$stdate','$endate',' ',' ',' ',' ',' ','$fileName','$fileSize','$fileType',' ','$desc',' ','$postuser')";

} } }
else {
$query = "INSERT INTO test2,test3 VALUES(' ','$title','$stdate','$endate',' ',' ',' ',' ',' ','$fileName','$fileSize','$fileType',' ','$desc',' ','$postuser')";

}
mysql_query($query) or die('Error, query failed.         '. mysql_error());

}

$query4=" SELECT * FROM test2,test3";

A: Your approach feels inherently wrong, but nevertheless, here is corrected code:
$query = "INSERT INTO test2 VALUES(' ','$title','$stdate','$endate',' ',' ',' ',' ',' ','$fileName','$fileSize','$fileType',' ','$desc',' ','$postuser')";

} } }
else {
$query = "INSERT INTO test3 VALUES(' ','$title','$stdate','$endate',' ',' ',' ',' ',' ','$fileName','$fileSize','$fileType',' ','$desc',' ','$postuser')";

}
mysql_query($query) or die('Error, query failed.         '. mysql_error());

}

$query4="SELECT * FROM test2 UNION ALL SELECT * FROM test3";

A: How about this one
$dh = mysql_connect('localhost','root','') or die(mysql_error());

$db1 = mysql_select_db('test2',$dh);
mysql_query("SELECT * FROM tableFromTest2",$db1);

$db2 = mysql_select_db('test3',$dh);
mysql_query("SELECT * FROM tableFromTest3",$db2);

mysql_close($dh);

or
mysql_connect('localhost','root','') or die(mysql_error());

mysql_query("USE test2");
mysql_query("SELECT * FROM tableFromTest2");

mysql_query("USE test3");
mysql_query("SELECT * FROM tableFromTest3");

mysql_close($dh);

A: Looks like you want a simple union:
SELECT * FROM test2
UNION
SELECT * FROM test3

assuming the table structure of test2 and test3 are similar (as they are in your sample code).
Of course since you're inserting it twice (once into each table), you might want to get it back twice - so use UNION ALL. Also you can play with ORDER BY and DISTINCT Optimization.

redpajama_set_name: RedPajamaStackExchange


text:
The Secret to Veteran Success in Corporate America
by Mike Schindler | Aug 18, 2021 | In the News
What if what you learned in the service is the secret to your success in corporate America?
Bruce Evans, Army Veteran and VP of Franchise Development for the nationally renowned food chains Capriotti's and Wing Zone, shares how his military experience – and what specifically from his military service – opened the doors to his climb in corporate America.
Be sure to listen closely – because the secret to your success is revealed in this latest episode of The Military Wire. And there just might be an opportunity for you…with Bruce's team.
Bruce didn't have it easy. He faced some hardships. But there was a defining moment in his service that changed his life. And that moment has been one of the keys to his success.
Now driven by his innate sense of urgency and refusal to remain stagnant, Bruce has been at the forefront of the Capriotti's evolution for close to a decade now. He's helping to provide guidance to the brands network of 500+ nationwide franchisees, integrate modern tech advances (AI, delivery systems, app development) into the company's daily operations, and ensure that the culture and values that make Capriotti's the fan-favorite chain it is today remain consistent in every corner of the country.
And he's looking for great Veterans who are willing to lean in to discovering how to bring their skills to the franchise world. (He gives his phone number in the interview.)
The Military Wire, hosted by Mike Schindler, interviews some of America's most elite men and women who have served this country in hopes that you the listener will gain an "ah-ha" moment that will help you move your life forward. Mike is a US Navy Veteran, an award-winning author, and has been featured on Yahoo Finance, the Boston Globe, NY Post, USA Today, Q13 Fox, and numerous other media outlets discussing why today's Veterans are truly one of America's Greatest Assets. Mike is the author of two highly endorsed books, "Operation Military Family: How Military Couples are Fighting to Preserve their Marriages," and "U.S. Veterans in the Workforce: Why the 7% are America's Greatest Asset."

redpajama_set_name: RedPajamaCommonCrawl


It seems Homeschoolers Anonymous has made an increasingly large splash in the homeschooling world.
Prominent Christian homeschool leader Kevin Swanson himself felt the need to address the group in a recent broadcast on his Generations with Vision radio show. He gave it the title "Apostate Homeschoolers." If you click the link to listen, the section on Homeschoolers Anonymous starts at 5:00 and goes until 10:40, when Swanson moves on to the Boy Scouts.
What does Swanson blame for the growth of the "homeschool apostates" and their increased networking and online activism? NCFCA homeschool speech and debate. Oh yes. NCFCA was started by Christian homeschool leaders to equip a generation of homeschooled children to be culture warriors, fighting against the godless secularists and working to establish a Christian America. But apparently, according to Swanson, it's gone awry, and too many of its homeschool participants have left God's Truth for the faulty world of man's intellect and reason.
In other words, Swanson has stumbled upon the very real truth that indoctrination fails when you teach children how to think instead of what to think.
But if ensuring that your young people retain your beliefs requires teaching them what to think without ever teaching them how to think, the problem is with your beliefs, not with the fact that certain of your young people figure out how to think and then walk away. That this is the response of the Christian homeschooling world—that perhaps teaching kids how to think was a bad idea—then what they have to offer is very sad indeed.
Look how scary we are, with all of our researching and talking and thinking and socializing!
love the pictures at the end.
The part about Swanson's broadcast that bothered me the most was frankly his venom towards us. His tone is way beyond righteous indignation, he's moved into the realm of throwing ad hom's at people who formerly thought of his family as friends. The fact that he so gleefully burns bridges in order to scare people into buying his books and agreeing with his personal theology is about as pharisaical as it gets.
I'm gonna listen to this again, but take a shot every time he said humility, humble themselves, or Almighty God. It'll be the fastest I've ever gotten wasted!!
You might need an ambulance.
I listened to Kevin's broadcast and was unimpressed. He never even touched on the abuse that is exposed in our stories. For someone so long-winded, he says surprisingly little. He does manager to fit in quite a few ad hominen attacks and plugs for his new book.
Probably the biggest head scratcher was when he accused us of thinking we're smart because we've learned some "big words." LOL! Ah well, leaders like Kevin are going the way of the dinosaur, and good riddance.
The big words bit is particularly funny considering how flamboyant certain reconstructionist and quiverful authors tend to be. I don't think anyone can compete with the sheer breadth of useless vocabulary these guys use.


text:
Misheard Song Lyrics -> Artist -> M -> The Maranatha Singers
Misheard song lyrics (also called mondegreens) occur when people misunderstand the lyrics in a song. These are NOT intentional rephrasing of lyrics, which is called parody. This page contains all the misheard lyrics for The Maranatha Singers that have been submitted to this site and the old collection from inthe80s started in 1996. For more information about the misheard lyrics available on this site, please read our FAQ.
There are 6 misheard song lyrics for The Maranatha Singers on amIright currently.
"He Has Made Me Glad"
He has made me black.
He has made me glad.
He has made me flab.
He has made me flat.
He is Amy Grant
He has made me glad
He is Baby Grand.
He is maybe flat.
Praise 16 Power of Your Love


Bonaventura C, Johnson. So human an animal. Every year, depending upon geographic location and intensity of El Nio events, crops suffer from too little water and wither on the spot, or are lost to severe flooding, hailstorms..
Hast du eine Frage zu Essay schreiben. Beispiele, Merkmale und Aufbau. Antwort von cheasy3 - 21:25, gucke dir das mal an, dass müsste dir helfen verstoß melden, hast Du eine eigene Frage an unsere. Es geht dabei..
back to the 3 you "already had. Or they "vanish the one" (i.e., just ignore and forget about it) so that they add 777 to 888 and get 555. . My crowning glory moment was when I left a 25-page research paper for an AP History class to get done the night before it was due, giving myself a few hours to write perfection and got. Even though most adults can say those letters in order, just as they and children can say the names of numbers in order, it is very difficult, unless one practices a lot, to be able to group things in sets of "n" or to multiply. Let them do additions and subtractions on paper, checking their answers and their manipulations with different color (group value) poker chips. Unfortunately, too many teachers teach like that manager manages. Many conceptually distinct ideas occur together naturally in practice. Once children have gained facility with counting, and with counting by groups, especially groups of 10's and perhaps 100's, and 1000's (i.e., knowing that when you group things by 100's and 1000's that the series go "100, 200, 300. Whereas if you do teach subtractions from 11 through 18, you give them the option of using any or all three methods.
I take "place-value" to be about how and why columns represent what they do and how they relate to each other, not just knowing what they are english essay on christmas named. For example, numbers written in Roman numerals are pronounced the same as numbers in Arabic numerals. Although children should not be taught arithmetic only mechanically, there are some mechanical skills that can be relatively easily learned by children, and that are important or necessary for seeing more "interesting" number relationships. Column representations of groups are more difficult to comprehend than color representations, and I suspect that is (1) because they depend on location relative to other numerals which have to be (remembered to be) looked for and then examined, rather than on just one inherent. Be sure to give the child lots of encouragment and support during this difficult period.
Comprehension strategy instruction helps students become purposeful, active readers who are in control of their own reading comprehension. I tell my students that the first step in writing a research paper is to admit you have a research paper. Write up a schedule with a series of milestones to accomplish by a specific date (e.g. Find 10 sources by September 20, finish preliminary research by October 15 and keep. Handedness Research Institute Left.


JobTrain often hosts a variety of career fairs to help past students and members of the community get connected with employers who are seeking new talent.
Provide budding cooks with a reality check about how becoming a chef is more than just cooking well.
An incredibly received presentation at the National Trainers' Exchange in May 2017 led to, Art Taylor, JobTrain's Chief Program Officer, being asked to do an encore by UCLA's Labor Occupation Safety & Healthcare Program (LOSH).
JobTrain transforms lives and communities in Silicon Valley. We help the Valley's most in need reclaim their lives from poverty and unemployment by preparing them for successful, sustainable careers in high-demand and emerging fields.


Not negating the beauty of their renditions, Monica Dogra and Shibani Dandekar's joint performance at a recently concluded gala was also appreciated because they could replicate the camaraderie they share behind the cameras, on stage. "The industry keeps bringing us together," they say of their multiple joint ventures, which has now led to a six-city tour titled Monica-Shibani.
"We've got venues locked in Navi Mumbai, Pune, Delhi and Punjab, and will also head to two cities in the South. We're zeroing in on the dates. We've been working on our playlist for about a year," Dandekar tells mid-day, adding that repeated requests to collaborate from fans on social media was also reason behind their decision to embark on this tour. "There was sufficient curiosity about what we could offer as one," she says.
On the lines of their recent performance - which saw them belt out indie pop numbers along with celebrated tracks of American artistes like Beyonce - Dogra says their set for the tour will also see a mix of Bollywood chartbusters intertwined with pop songs. "When we performed together the last time, we realised that a mixed set of this nature had never been presented for the Indian audience. Also, it's about how we can present it."
Their pairing, Dogra says, will be as enjoyable for audiences as it is for them. "We're fairly similar, yet distinct. Shibani listens to a lot of pop music and chart-toppers. I think one of her best qualities is that she can step into the room and pick out a track that will become everybody's favourite. That's one thing I've not been the best at."
Dogra adds that collaborating with another musician helps her explore different facets to herself. "As an independent artiste, you get stuck in a particular identity. But when you collaborate, you find parts of yourself that you wouldn't know unless you worked with the other person. Shibani makes me sexier, I make her edgier."


#TuesdayTips Do you know what the..
Do you know what the Number 1 reason people don't set goals is? It's because they are afraid of failure. Sometimes clients say, 'I don't know what I want?' – what they often mean is, 'I'm not clear that I can achieve what I really want so I'll pretend I don't know.' At other times, they genuinely don't know because they've suppressed their dreaming for so long that they're not aware of their real thoughts and desires.
Perhaps you fail to achieve your goals as often as you succeed and you wonder why that is. Today I'd like to explore the link between achieving your goals and giving your unconscious mind clear direction.
In Neuro-Linguistic Programming (NLP) one of the key areas we help people deal with, is letting go negative emotions – the core emotions of anger, fear, guilt, shame, and sadness. It only takes a few minutes, and usually clients will feel an immediate sense of relief which is fantastic. The negative emotions are gone, but your unconscious mind needs clear direction on how to think. If you don't provide that direction, then your unconscious mind will quickly fall back into the old habits of thinking using those negative patterns. So, you've dealt with the problem – but now you have some re-training to do as well.
It's the same with S.M.A.R.T. goals and your unconscious mind. You've created your S.M.A.R.T. goal – it is specific, measurable, as-if-now, realistic, and has a time frame – you've visualised it and made it as visceral as possible but you need to communicate that goal clearly to your unconscious mind so that it will help you achieve it. Your unconscious mind really wants to please you – but you need to tell it how it can do that and you need to make sure the instruction is clearly embedded.
When you create your S.M.A.R.T. goals and visualise them, that's step #1. The rest of the process involves reminding your unconscious mind of those goals – that's where vision boards, pictures, and regular review of those goals comes in. It may sound like hard work, but the truth is that once your unconscious mind knows what it is aiming for, it will support and help you at every turn. You'll find yourself taking the needed decisions and completing your actions much more easily and effortlessly than ever before. It almost removes the need for self-discipline because your unconscious mind is working for you day-in, day-out, helping you to accomplish everything you ever dreamed of doing.
Think about that! Think about having your feet on a path towards the goals you really want, and moving forwards almost by instinct – easily and relentlessly. That's what happens when you form your goals well, and then harness the power of your unconscious mind to draw you towards them.
Failure to achieve these goals isn't really an option – because you only fail when you are conflicted about your goals. This time, you've set your goals carefully using the S.M.A.R.T. method, and you've given clear directions to your unconscious mind via your visualisation, and frequent reminders. All of a sudden you'll find yourself smashing goal after goal as you work on them using a combination of deliberate action and effortless motivation.
Meta Description: How to achieve your goals every time without conflict and stress by harnessing the power of your unconscious mind.
Your Unconscious Mind Is Controlling Your Life – Does That Scare You?


text:
Feature Properties
Real Reports
Caribbean Directory
Design Den
Editor's Obsessions
The Lust List
My Island
1411 results found for ""
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Summer 2011 Issue Credits
www.reallifecaribbean.com Vol 6, issue #3 Summer 2011 PUBLISHER mc2 Ltd Melanie Roddam melanie@reallife.ky EDITORIAL Editor Kyle Fulton kyle@reallife.ky Writer Juliet Austin juliet@reallife.ky SALES & MARKETING Sales Director Melanie Ro...
Contributors Bios - Summer 2011
SANJAY M. AMIN A Director for BCQS Group International, Sanjay first moved to the Caribbean in 1997 to open the BCQS office in the British Virgin Islands, then picked up sticks and headed to Barbados in 2006 to open their office, finally hitti...
Creating a Green Future
An often-quoted definition of sustainable development is development that "meets the needs of the present without compromising the ability of future generations' needs." Global warming has already affected the Caribbean with temperatures in the regio...
All Systems Go - Hurricane Preparedness
Knowledge Fortune favours the prepared mind Know the risk factors from storm surge, flooding and wind for your area and gauge your home's vulnerability. Identify official public shelters and designated Emergency Medical Centres. Know when to take a...
Artist Lynn Parotti
Water: omnipresent; ebbing, flowing, rippling, ever changing; an integral element of the Caribbean psyche. Growing up in The Bahamas, cradled by the evocative Caribbean Sea, artist, Lynn Parotti's internal compass is intimately and intricately draw...
24 Hours in Grenada
A land of spices, exotic flowers and rare fruits, Grenada's culture is all embracing enthusiasm expressed through music, cooking, dance and sport - a nation of welcoming people awaiting your arrival. Spicy Splendor Oliver's Restaurant, located on t...
reallifecaribbean on Instagram
Mr. Jamaica: Gordon 'Butch' Stewart
Sir Royston Hopkin of the Spice Island Beach Resort
Bella Verde, Cayman Islands
Embrace the Outdoors
TRIO ARCHITECTURE MOVES TO NEW OFFICE IN CAMANA BAY


The day will include speakers such as Mick Keogh, ACCC deputy chair and John Gladigau launching Australia's first farmer-owned mutual.
It will also be an opportunity to hear how a group of growers are transitioning to a whole new industry; what a berry collective learned from its first export experience; how a forestry collective scored a deal for its secondary timbers; how a newly formed Victorian milk co-op scored a major processing deal; the NSW garlic co-op that sold-out its first harvest at double market price; what a turf growers' co-op overcame to now represent 10% of its national industry; the fishing co-op trialling a new online auction system and an ethical merino co-op that is launching a blockchain-backed 'green' guarantee.
The day-long, no-cost event includes great networking opportunities and all refreshments. Book through the link below.


For intermediate Koha ILS users who wish to quickly generate a single tab MARC framework.
Last Thursday, Ashish Kumar Barik, librarian at our new client-partner Midnapore City College filed a support ticket asking for a custom single sheet MARC21 framework or what is more commonly referred to by LIS professionals as a "worksheet". He wrote that he wanted the following tags 000, 003, 005, 008, 020, 040, 041, 044, 082, 100, 245, 250, 260, 300, 440, 490, 500, 504, 650, 700, 942 and that the sub-tags/fields should be set as in the default marc framework shipped with Koha. We promised him his new framework. Being new to this side of Koha, he of course had missed out two key fields without which his system would be rendered practically useless i.e. the two local use tags – 952 and 999. Koha uses 952 to handle holdings (item) information and 999 is purely an internal tag used to track the bibliographic records.
Now anyone who has ever setup a new MARC framework knows that it can be a laborious and time consuming task. Further, there are chances of introducing inadvertent human errors that may lead to error or bad data when used as a part of the framework. As a result, at L2C2 Technologies we have developed several well defined strategies to manage custom marc frameworks for our clients. In today's blog, we are going to share the simplest of the techniques we use in cases like this. The outcome of this exercise is a 100% error free marc framework generated in less than 2 minutes.
LEGAL DISCLAIMER: The next steps involve directly accessing and making changes in the the Koha database. So use these instructions at your own risk, if you face any data loss, corruption or system errors we are not responsible.
We used a regex capable editor like Notepad++ to quote the fields mentioned by Ashish, so that 000, 003, 005, 008, 020, 040, 041, 044, 082, 100, 245, 250, 260, 300, 440, 490, 500, 504, 650, 700, 942 became '000', '003', '005', '008', '020', '040', '041', '044', '082', '100', '245', '250', '260', '300', '440', '490', '500', '504', '650', '700', '942'. And while we did that, we also added the following fields missing in his list i.e. '952', '999'.
Next we copied the default framework into MCC1 as its base, since that is what Ashish had wanted. At this point, the MCC1 framework is exactly same as the default framework of Koha.
Next we fired up the MySQL console and logged in with the user id and passwd from MCC's koha-conf.xml, and chose Ashish's database in this case koha_mcc for the next steps.
MySQL client told us 152 rows were affected.
This time MySQL reported that 3416 rows were updated.
MySQL reported 341 rows were affected.
Coming back to MCC's Koha staff client, we did the most important thing i.e. running MARC Bibliographic framework test. The test came out clean without any error.
That's it! MCC's custom MARC framework is ready for use. Click on the image below and then zoom in to see the details up close.
The launch of NECOPAC the first volunteer-run, freely available, public z39.50 service in North Eastern India.
On the auspicious occasion of Rongali Bihu (for the Assamese) and Poila Boishakh (for the Bengali), on behalf of the NECOPAC team, I'm happy to announce the start of NECOPAC z39.50 service. This is the first freely available, public z39.50 (copy cataloging) service in North Eastern India. As on date there are 92,333 bibliographic records in the database, all of which are volunteer contributed. We are expecting more records to be contributed soon.
The germ of the idea of NECOPAC started about 2 years back during a chance meeting of a group of young like-minded library professionals and technology specialists in Guwahati, Assam.
The north eastern Indian is home to 8 states – Assam, Manipur, Tripura, Meghalaya, Mizoram, Nagaland, Arunachal Pradesh and Sikkim. The region boasts of great cultural heritage and diversityr. Libraries are an important means of preserving and disseminating this rich heritage, traditional knowledge, literature and other creative output.
Freely available copy cataloging of Indian publications using z39.50 service, to this day, remains largely a distant dream for most Indian library professionals. The scenario in the North East is even more difficult when it comes to publications in the local languages from this part of India.
The NECOPAC z39.50 service is a volunteer-run collaborative attempt at bridging this service gap. It is the first freely available, public z39.50 service in the entire North Eastern India. For us to grow bigger and be able to serve more LIS professionals in the NE, we need your active support and contributions i.e. a copy of your bibliographic records.
DISCLAIMER : The bibliographic records are provided on AS-IS basis by this service and the NECOPAC team does not vouch for the correctness or accuracy of these records.


Suits take on a new dimension for next summer. After becoming an emblematic piece of the feminine wardrobe, they appear in all their splendour, through a tuxedo or a more casual version for every day.For the summer months, we especially love the fluidity of materials that are suppler, lighter and more comfortable for the strong heat. Note the wide suits that can easily be worn with trainers, a style already adopted by the singers from the group, Haim.From the catwalk to the city, Giambattista Valli reinterpreted an iconic suit.The couturier delivers his own version of the trend that has invaded podiums: the white suit is proposed to suit everyone's styles and has never looked so elegant, likewise with checks, becoming an essential in wardrobes.The colours used for suit trousers are very striking. We see them in pastel tones, with powder pink being especially present as well as in more casual and natural colours with earthy, camel and brown tones. Until they explode. The sun brings colour, and suit trousers play with various colours including pink, orange and green that can influence a more positive attitude.Finally, amongst new trends, we see stripes and polka dots appear for evenings and for a new type of tuxedo, brocade plays with shimmers in a suit that brings a crazy presence that was well adopted by Carla Bruni who showed for Dolce & Gabbana.Interviews with the Haim sisters: I love suiting, so you can tell, I love the suits, so all the suits that she showed were amazing, beautiful.Interiew with Giambattista Valli:The first suit was really symbolic for Yoko Ono as well as its silhouette and spirit and the very powerful but simple idea behind it.


Shipra Jain is an Indian-based science student who graduated from University of Delhi with a Bachelor's degree in Physics in 2009. She received a Master's degree in Physics from Indian Institute of Technology (Guwahati) in the year 2011 with a specialization in experimental material physics. After that she migrated to atmospheric physics and started working... Follow the author @	and view more articles.


Founded in 2007 by Gosia Piatek, Kowtow is a label committed to creating positive change to only use renewable and sustainable fibres and ethical manufacturing to deliver collections that are utilitarian, minimal and carry an understated sense of femininity. All Kowtow garments are certified by non-profit internationally recognised organisations. Independent audits and accreditation bodies ensure to remain true and supporting fair wages, no child labor, gender equality & grower community.


FORGET Santa Claus, '80s R&B superstar Bobby Brown is coming to town.
THERE will be no beating around the bush during the upcoming 'Straight Talk, No Chaser' event featuring an all-male panel who will engage an all-female audience in raw, uncensored conversations about what really goes on in the hearts and minds of men.
A Bahamian produced show about movies and Caribbean cinema culture will have its international debut next month.
THE LOVE she has for Bahamian culture and the home-grown rake n' scrape and Junkanoo beats adds the spice she needs to keep pushing her music skills and talent.
THE young members of the Dance Bahamas School danced their way into the hearts of their audiences last weekend in a holiday performance which was just a taste of what is to come from the company in 2015.
AS the face of Cable Bahamas, local music sensation Angelique Sabrina played a pivotal part in ensuring that the grand opening of the company's newest flagship store in the Mall At Marathon was sensational.
IF YOU are excited about Bahamas Junkanoo Carnival in May 2015, but have not yet decided on a group to hit the streets and dance with during the Road March, you may just want to take a closer look at the Xtasy Band.
TEEN singing sensation Angelique Sabrina is topping off a successful year with the release of her first album, "One For The People".
IT IS a feeling of pure bliss for singer Cambrell Poitier to have been selected as one of the semi finalists for the 2015 Bahamas Junkanoo Carnival CD Compilation.
IN A concert to benefit talented music majors at the College of the Bahamas, providing them with full scholarships, the Nassau Renaissance Singers (NRS) are excited to present their annual "Music For Christmas" holiday extravaganza.


text:
\section{Introduction}\label{intr}

Physical systems involving long-range interactions, long-time
memories, or fractal structures can hardly be treated within the
traditional background of statistical physics. The Tsallis entropies have been widely
adopted in this direction \cite{gmt}. As a rule, stationary states
of such systems are described by one-parametric extensions of the
Zipf-Mandelbrot power-law distribution. Generalized entropies have also
found use as alternate measures of an informational content. For
instance, the entropic uncertainty principle has been expressed in
terms of both the R\'{e}nyi \cite{zpv08} and the Tsallis entropies
\cite{majer,rast104}. Studies of generalized entropies allow to
fit some properties of the standard entropy. The connection
between strong subadditivity of the von Neumann entropy and the
Wigner-Yanase-Dyson conjecture is a remarkable example (see
\cite{nielsen,wehrl} and references therein).

The relative entropy, or Kullback-Leibler divergence \cite{KL51},
is frequently used as a measure of statistical distinguishability.
Csiszar's $f$-divergence \cite{ics67,ics08} and Petz's
quasi-entropies \cite{petz86,petz10} are famous generalizations of
the Kullback-Leibler measure to the classical and quantum cases,
respectively. In both the classical and quantum regimes,
properties of the relative entropy are the subject of active
research. So, the development of a standard background to
generalized entropies consists an important issue. In the
present paper, we examine continuity properties of the Tsallis
relative entropy. The obtained bounds are expressed in terms of
the trace distance between two probability distributions or
density operators. In this regard, our bounds characterize a
continuity property in the sense of Fannes \cite{AE05}.

The paper is organized as follows. In Section \ref{dfnt}, the main
definitions are given. One consequence of the monotonicity of the
quantum $f$-divergence is considered in Section \ref{acof}. A
family of lower continuity bounds on the Tsallis relative entropy
of order $\alpha\in(0;1)$ is derived in Section \ref{inpt}. In
their essence, these inequalities are one-parametric extensions of
the Pinsker inequality. The case of R\'{e}nyi relative entropy is
considered as well. In Section \ref{upbn}, upper continuity bounds
on the Tsallis relative $\alpha$-entropy of two probability
distributions in terms of the minimal probability in its second
argument are obtained. Fano type upper bounds on the conditional
Tsallis entropy are derived for all $\alpha>0$ in Section
\ref{ngfn}. As is shown, these bounds lead to generalized
inequalities of Fannes type.

\section{Definitions and notation}\label{dfnt}

In the classical regime, we will consider probability
distributions over the finite index set $\mcx$ of cardinality $N$. The
trace distance between probability distributions $P=\{p(x)\}$ and
$Q=\{q(x)\}$ are then defined as
\begin{equation}
D(P,Q):=\frac{1}{2}{\>}\sum\nolimits_{x\in\mcx}|p(x)-q(x)|
\ . \label{pdpq}
\end{equation}
Let $\mathcal{L}(\hh)$ be the space of linear operators on
finite-dimensional Hilbert space $\hh$. We also use the notations
$\lsp(\hh)$ to denote the positive semidefinite operators. For any
operator $\ax$, we put $|\ax|\in\lsp(\hh)$ as a unique positive
square root of $\ax^{*}\ax\geq{\mathbf{0}}$. The trace norm
$\|\ax\|_1:=\tr|\ax|$ and the trace distance, defined as
\begin{equation}
\rd(\ax,\ay):=\frac{1}{2}{\>}\|\ax-\ay\|_{1}\equiv\frac{1}{2}{\>}\tr|\ax-\ay|
\ , \label{rdrs}
\end{equation}
are widely used in both the mathematical physics and quantum
information theory. Using the Ky Fan norms, the partitioned
versions of the above measures can be adopted properly
\cite{rast1023,rast103}. By $\kr(\ax)$ and $\sop(\ax)$ we denote
the kernel and the support of operator $\ax$. Eigenvalues of the
operator $\ax$ form the multi-set $\spc(\ax)$. For two operators
$\ax$ and $\ay$ on $\hh$, we define the Hilbert--Schmidt inner
product by
\begin{equation}
\langle\ax{\,},\ay\rangle_{\rm{hs}}
:={\rm{tr}}(\ax^{*}\ay)
\ . \label{hsdef}
\end{equation}

For positive $\alpha\not=1$, the Tsallis $\alpha$-entropy of
probability distribution $P=\{p(x)\}$ is defined by \cite{tsallis}
\begin{equation}
H_{\alpha}(P):=\frac{1}{1-\alpha}{\,}\left(\sum\nolimits_{x\in\mcx} p(x)^{\alpha}
- 1 \right)\equiv
-\sum\nolimits_{x\in\mcx} p(x)^{\alpha}\ln_{\alpha}p(x)
\ , \label{tsaent}
\end{equation}
where $\ln_{\alpha}z\equiv\bigl(z^{1-\alpha}-1\bigr)/(1-\alpha)$
is the $\alpha$-logarithm. The maximal value $\ln_{\alpha}N$ is reached for the uniform distribution, when $p(x)=1/N$ for all $x\in\mcx$. The Shannon entropy
$H_{1}(P)=-\sum_{x}p(x)\ln{p}(x)$ is obtained in the limit
$\alpha\to1$. By $h_{\alpha}(u)$ we denote the binary Tsallis
$\alpha$-entropy, that is
\begin{equation}
h_{\alpha}(u):= H_{\alpha}\bigl(\{u,1-u\}\bigr)=
-u^{\alpha}\ln_{\alpha}u-(1-u)^{\alpha}\ln_{\alpha}(1-u)
{\ }\qquad (u\in[0;1])
\ . \label{hbin}
\end{equation}
This function is concave, since its second derivative is negative.
It is clear that $h_{\alpha}(u)=h_{\alpha}(1-u)$. Another
important one-parametric generalization is the R\'{e}nyi entropy
(see, e.g., \cite{ics08,petz08} and references therein)
\begin{equation}
R_{\alpha}(P):=\frac{1}{1-\alpha}{\ }\ln\left(\sum\nolimits_{x\in\mcx} p(x)^{\alpha}\right)
\ . \label{rpdf}
\end{equation}
The entropies (\ref{tsaent}) and (\ref{rpdf}) are connected by the
equality
$(1-\alpha)R_{\alpha}(P)=\ln\bigl[1+(1-\alpha)H_{\alpha}(P)\bigr]$.
The quantum analogs of these entropies are respectively defined as
\begin{align}
{\rm{H}}_{\alpha}(\rho)&:=\frac{1}{1-\alpha}{\>}\Bigl(\tr(\rho^{\alpha})-1\Bigr)
\ , \label{qtsdf}\\
{\rm{R}}_{\alpha}(\rho)&:=\frac{1}{1-\alpha}{\ }\ln\bigl[\tr(\rho^{\alpha})\bigr]
\ . \label{qrndf}
\end{align}
Subadditivity of the quantum Tsallis entropy (\ref{qtsdf}) for
$\alpha>1$ has been conjectured by Raggio \cite{raggio} and later
proved by Audenaert \cite{auden07}. This result has been extended
to some of so-called unified entropies \cite{rastjst}. The
subadditivity property was generally believed to be true for the
Wigner-Yanase entropy, until counterexamples were given
\cite{hansen07,seiringer07}. Meantime, if the bipartite state is
pure then it is sufficient for the subadditivity. Other sufficient
conditions for subadditivity of the Wigner-Yanase entropy are
obtained in \cite{CH10}.

The standard relative entropy of $P=\{p(x)\}$ to $Q=\{q(x)\}$ is
defined as $H_1(P||Q)=-\sum_{x} p(x)\ln\bigl[q(x)/p(x)\bigr]$
\cite{ics08}. For density operators $\rho$ and $\sigma$, the
quantum relative entropy is expressed as \cite{petz08}
\begin{equation}
\rh_1(\rho||\sigma):=\tr(\rho\ln\rho-\rho\ln\sigma)
\ . \label{relan}
\end{equation}
In the classical regime, the Tsallis relative $\alpha$-entropy is
introduced by \cite{borland}
\begin{equation}
H_{\alpha}(P||Q):=-\sum\nolimits_{x\in\mcx}p(x){\,}\ln_{\alpha}\frac{q(x)}{p(x)}=
\frac{1}{1-\alpha}{\,}\left(1-\sum\nolimits_{x\in\mcx}p(x)^{\alpha}q(x)^{1-\alpha} \right)
\ . \label{srtdef}
\end{equation}
Basic properties of this measure are discussed in
\cite{borland,fky04}. The R\'{e}nyi relative entropy is defined as \cite{ics08}
\begin{equation}
R_{\alpha}(P||Q):=\frac{-1}{1-\alpha}{\ }\ln\left(
\sum\nolimits_{x\in\mcx}p(x)^{\alpha}q(x)^{1-\alpha}\right)
\ . \label{rrdef}
\end{equation}
It is convenient to extend the definition (\ref{srtdef}) to any
positive-valued functions $A$ and $B$ on the finite set $\mcx$.
For given set $A=\{a(x)\}$, we put the index subset
$\mcx_A=\{x:{\,}a(x)\neq0\}$ and its complement $\tom_A$. For
$\alpha>1$, the ''Tsallis relative $\alpha$-entropy'' of
$A=\{a(x)\}$ to $B=\{b(x)\}$ is defined as
\begin{equation}
H_{\alpha}(A||B):=
\left\{
\begin{array}{ll}
\frac{1}{\alpha-1}\Bigl(\sum\nolimits_{x\in\mcx_A}
a(x)^{\alpha}b(x)^{1-\alpha}-\sum\nolimits_{x\in\mcx_A} a(x)\Bigr)\>,
& {\ }{\ } \tom_B\subset\tom_A \ , \\
+\infty\ , & {\ }{\ } {\rm{otherwise}} \ .
\end{array}
\right.
\label{geendf}
\end{equation}
Omitting the second entry, we obtain the definition for
$0<\alpha<1$. For any positive scalar $\lambda$, we have
\begin{equation}
H_{\alpha}(\lambda{A}||\lambda{B})=\lambda{\,}H_{\alpha}(A||B)
\ , \label{homg}
\end{equation}
i.e. it is a homogeneous function of degree one. For
$\alpha\in(0;1)$ and density operators $\rho$ and $\sigma$, we
define the Tsallis relative entropy as
\begin{equation}
\rh_{\alpha}(\rho||\sigma):=\frac{1}{1-\alpha}
{\ }\Bigl(1-\tr\left(\rho^{\alpha}\sigma^{1-\alpha}\right)\Bigr)
\ . \label{ahadef}
\end{equation}
For $\alpha>1$, the right-hand side of (\ref{ahadef}) is
well-defined whenever $\kr(\sigma)\in\kr(\rho)$. In the singular
case, when the term $\kr(\sigma)\cap\sop(\rho)\neq\emptyset$
occurs, the right-hand side of (\ref{ahadef}) is dealt similar to
the standard relative entropy (\ref{relan}). Namely, relative
entropies are defined to be $+\infty$. Extending (\ref{rrdef}) to
the quantum case, we define
\begin{equation}
{\rm{R}}_{\alpha}(\rho||\sigma):=\frac{-1}{1-\alpha}{\ }\ln\bigl[\tr(\rho^{\alpha}\sigma^{1-\alpha})\bigr]
\ . \label{radef}
\end{equation}
For these entropies, we have the equality
\begin{equation}
(\alpha-1){\rm{R}}_{\alpha}(\rho||\sigma)=\ln\bigl[1+(\alpha-1)\rh_{\alpha}(\rho||\sigma)\bigr]
\ , \label{alrha}
\end{equation}
and the same relation in classical setting. For $\alpha>1$ and
$\am,\bn\in{\mathcal{L}}_{+}(\hh)$, we also introduce
\begin{equation}
\rh_{\alpha}(\am||\bn):=
\left\{
\begin{array}{ll}
\frac{1}{\alpha-1}{\,}\Bigl(\tr(\am^{\alpha}\bn^{1-\alpha})-\tr(\am)\Bigr)\ ,
& {\ }{\ } \kr(\bn)\subset\kr(\am) \ , \\
+\infty\ , & {\ }{\ } {\rm{otherwise}} \ .
\end{array}
\right.
\label{qendf}
\end{equation}

\section{A consequence of monotonicity of the $f$-divergence}\label{acof}

In this section, we prove one result which will be used for
obtaining quantum bounds of Pinsker type. The Tsallis relative
entropy (\ref{srtdef}) is closely related to the Csisz\'{a}r
$f$-divergence \cite{ics67}. Let $z\mapsto{f}(z)$ be a convex
function on $z\in[0;+\infty)$ with $f(1)=0$. The Csisz\'{a}r
$f$-divergence of $P=\{p(x)\}$ from $Q=\{q(x)\}$ is defined as
\cite{ics67,ics08}
\begin{equation}
S_{f}(P||Q):=\sum_{x\in\mcx} q(x){\>}f{\!}\left(\frac{p(x)}{q(x)}\right)
\ . \label{cfdf}
\end{equation}
Taking $f_{\alpha}(z)=(\alpha-1)^{-1}z^{\alpha}$ with positive
$\alpha\neq1$ and adding corresponding constant, the formula
(\ref{cfdf}) leads to the right-hand side of (\ref{srtdef}). The
definition (\ref{cfdf}) can generally be used without the
normalization condition.

In the following, we use the convention that powers of a positive
semidefinite operator are only taken on its support. Namely, by
$\am^{-1}$ and $\am^{0}$ we respectively denote the generalized
inverse of $\am$ and the projection onto its support. A quantum
counterpart of Csisz\'{a}r's $f$-divergence is introduced as
follows \cite{hmp10}. For an operator
$\am\in\mathcal{L}_{+}(\hh)$, let $\llv_{\am}$ and $\rrv_{\am}$
denote the left and the right multiplications by $\am$,
respectively, defined as
\begin{equation}
\llv_{\am}:{\>} \ax\mapsto\am\ax\ , \qquad
\rrv_{\am}:{\>} \ax\mapsto\ax\am \ , \qquad \ax\in\mathcal{L}(\hh)
\ . \label{lrmults}
\end{equation}
Left and right multiplications commute with each other, namely
$\llv_{\am}\rrv_{\bn}=\rrv_{\bn}\llv_{\am}$ for
$\am,\bn\in\mathcal{L}_{+}(\hh)$. Let $z\mapsto{f}(z)$ be a
continuous function on $z\in[0;+\infty)$. Taking the set
$\bigl\{ab^{-1}:{\>}a\in\spc(\am),{\>}b\in\spc(\bn)\bigr\}$, we
write \cite{hmp10}
\begin{equation}
f(\llv_{\am}\rrv_{\bn^{-1}}):=\sum_{a\in\spc(\am)}{\,}\sum_{b\in\spc(\bn)}
f(ab^{-1}){\,}\llv_{\np_{a}}{\,}\rrv_{\nq_{b}}
\ , \label{fabpq}
\end{equation}
where the formulas $\am=\sum_{a}a{\,}\np_{a}$ and
$\bn=\sum_{b}b{\,}\nq_{b}$ express the spectral decompositions of
$\am$ and $\bn$, respectively. If $\kr(\bn)\subset\kr(\am)$, then
the $f$-divergence of $\am$ with respect to $\bn$ is defined as
\cite{hmp10}
\begin{equation}
\rs_{f}(\am||\bn):=\bigl\langle\bn^{1/2},f(\llv_{\am}\rrv_{\bn^{-1}}){\,}\bn^{1/2}\bigr\rangle_{\rm{hs}}
\ . \label{qsfdef}
\end{equation}
Let $\pen$ be the identity operator. In general case, the quantum
$f$-divergence is defined by the formula
\begin{equation}
\rs_{f}(\am||\bn):=\underset{\veps\searrow{0}}{\lim}{\>}\rs_{f}(\am||\bn+\veps\pen)
\ . \label{qsfdep}
\end{equation}
Basic properties of the quantity (\ref{qsfdef}) are discussed in
the paper \cite{hmp10}. Using the function
$f_{\alpha}(z)=(\alpha-1)^{-1}z^{\alpha}$, for
$\kr(\bn)\subset\kr(\am)$ we obtain
\begin{equation}
\rs_{\alpha}(\am||\bn)=\frac{1}{\alpha-1}{\ }
\bigl\langle\bn^{1/2},(\llv_{\am}\rrv_{\bn^{-1}})^{\alpha}{\,}\bn^{1/2}\bigr\rangle_{\rm{hs}}
=\frac{1}{\alpha-1}{\ }\tr(\am^{\alpha}\bn^{1-\alpha})
\ . \label{qsfdaf}
\end{equation}
Adding the term $(1-\alpha)^{-1}\tr(\am)$ in the right-hand side
of (\ref{qsfdaf}), we have $\rh_{\alpha}(\am||\bn)$ in view of
(\ref{qendf}). One of the most important properties of relative
entropies is their monotonicity under the action of
trace-preserving completely positive (TPCP) maps \cite{uhlmann77}. For a general discussion of a role of stochastic maps in quantum theory, see the paper \cite{busch99}.
Many fundamental results of quantum information theory are closely
related to the monotonicity of the standard relative entropy
\cite{JR10,nielsen,vedral02}. General conditions for the monotonicity
of the quantum $f$-divergence are obtained in \cite{hmp10}. If the
map $\varPhi$ is TPCP-map and the function $f$ is operator convex
on $[0;+\infty)$ then
\begin{equation}
\rs_{f}\bigl(\varPhi(\am)\big|\big|{\,}\varPhi(\bn)\bigr)\leq\rs_{f}(\am||\bn)
\ . \label{fdmon}
\end{equation}
Note that the inequality (\ref{fdmon}) has generally been
established in \cite{hmp10} under weaker conditions on the maps.
From the monotonicity (\ref{fdmon}) we can derive a simple upper
bounds on the quantum $f$-divergence in terms of classical one.

\begin{Thm}\label{conmon}
Let $\am,\bn\in\lsp(\hh)$, and let $\ip_{\pm}$ be projectors on
the eigenspaces corresponding to positive and negative eigenvalues
of the difference $(\am-\bn)$. There holds
\begin{equation}
\rs_{f}(\am||\bn)\geq{S}_{f}\bigl(\{u'_{\pm}\}\big|\big|\{v'_{\pm}\}\bigr)
\ , \label{sfuf}
\end{equation}
where $u'_{\pm}=\tr(\ip_{\pm}\am)$ and $v'_{\pm}=\tr(\ip_{\pm}\bn)$.
\end{Thm}

{\bf Proof.} Using the Jordan decomposition of Hermitian operator
\begin{equation}
\am-\bn=\sum\nolimits_{\mu>0}\mu{\,}|\mu\rangle\langle\mu|-
 \sum\nolimits_{\nu>0}\nu{\,}|\nu\rangle\langle\nu|
\ , \label{jorde}
\end{equation}
we define projectors $\ip_{+}=\sum_{\mu}|\mu\rangle\langle\mu|$ and
$\ip_{-}=\sum_{\nu}|\nu\rangle\langle\nu|$. When the difference
$(\am-\bn)$ has zero eigenvalues, corresponding eigenvectors
should be included to the orthonormal sets $\{|\mu\rangle\}$ and
$\{|\nu\rangle\}$ anyhow; then $\ip_{+}+\ip_{-}=\pen$.
Consider the TPCP-map $\psv$ defined as
\begin{equation}
\am\mapsto\psv(\am)=\sum\nolimits_{\mu}|\mu\rangle\langle\mu|\am|\mu\rangle\langle\mu|+
\sum\nolimits_{\nu}|\nu\rangle\langle\nu|\am|\nu\rangle\langle\nu|=
\sum\nolimits_{\mu} u_{\mu}|\mu\rangle\langle\mu|
+\sum\nolimits_{\nu} u_\nu|\nu\rangle\langle\nu|
\ , \label{qogr}
\end{equation}
where probabilities $u_{\mu}=\langle\mu|\am|\mu\rangle$ and
$u_{\nu}=\langle\nu|\am|\nu\rangle$. Putting
$v_{\mu}=\langle\mu|\bn|\mu\rangle$ and
$v_{\nu}=\langle\nu|\bn|\nu\rangle$, we further write
\begin{equation}
\psv(\bn)=\sum\nolimits_{\mu} v_{\mu}|\mu\rangle\langle\mu|
+\sum\nolimits_{\nu} v_{\nu}|\nu\rangle\langle\nu|
\ . \label{qogs}
\end{equation}
So the outputs $\psv(\am)$ and $\psv(\bn)$ are diagonal in the
same basis. Combining this fact with the inequality (\ref{fdmon}),
we obtain
\begin{equation}
\rs_{f}(\am||\bn)\geq\rs_{f}\bigl(\psv(\am)\big|\big|{\,}\psv(\bn)\bigr)=
S_{f}\bigl(\{u_{\mu},u_{\nu}\}\big|\big|\{v_{\mu},v_{\nu}\}\bigr)
\ . \label{moncr}
\end{equation}
The final step is to use the monotonicity in classical regime. Let
us put the 2-by-$d$ transition probability matrix
\begin{equation}
\tm=
\begin{pmatrix}
 1 & \cdots & 1 & 0 & \cdots& 0 \\
 0 & \cdots & 0 & 1 & \cdots& 1 \\
\end{pmatrix}
\ , \label{tt2}
\end{equation}
in which the units of the first row act on $\mu$-components, the
units of the second row act on $\nu$-components. This matrix maps
the ordered sets $\{u_{\mu},u_{\nu}\}$ and $\{v_{\mu},v_{\nu}\}$ to $\{u'_{+},u'_{-}\}$
and $\{v'_{+},v'_{-}\}$, respectively. It is clear that
\begin{equation}
u'_{+}=\sum\nolimits_{\mu}u_{\mu}=\tr(\ip_{+}\am)
\ , \qquad u'_{-}=\sum\nolimits_{\nu}u_{\nu}=\tr(\ip_{-}\am)
\ , \label{umun}
\end{equation}
and analogously for $v'_{\pm}$ with $\bn$ instead of $\am$. By the
monotonicity, the right-hand side of (\ref{moncr}) is larger than
or equal to the right-hand side of (\ref{sfuf}). $\blacksquare$

The function $z\mapsto{z}^{\alpha}$ is operator concave on
$\lsp(\hh)$ for $0<\alpha<1$ (see, e.g., theorem 4.2.3 in
\cite{bhatia07}). So the function
$f_{\alpha}(z)=(\alpha-1)^{-1}z^{\alpha}$ is operator convex for
$0<\alpha<1$. Combining this with the inequality (\ref{sfuf}), we
then obtain
\begin{equation}
\rs_{\alpha}(\am||\bn)\geq{S}_{\alpha}\bigl(\{u'_{\pm}\}\big|\big|\{v'_{\pm}\}\bigr)
\ . \label{pinab}
\end{equation}
For density operators, we have $u'_{\pm}=\tr(\ip_{\pm}\rho)$ and
$v'_{\pm}=\tr(\ip_{\pm}\sigma)$. Up to a notation, the result
(\ref{pinab}) with density operators was presented in
\cite{rast10fn} (see theorem IV.1 therein). Writing the relation
\begin{equation}
\|\am-\bn\|_{1}=|u'_{+}-v'_{+}|+|u'_{-}-v'_{-}|
\ , \label{abtrn}
\end{equation}
we should estimate the right-hand of (\ref{sfuf}) from below in
terms of the distance $\|\am-\bn\|_{1}$.

\section{Pinsker type inequalities for $\alpha\in(0;1)$}\label{inpt}

Studies of distinguishability measures and relations between them
consist an actual issue of quantum information theory. The Pinsker
inequality \cite{ics67} and its quantum analog expressed as
\cite{hots81}
\begin{equation}
\rh_1(\rho||\sigma)\geq2{\,}\rd(\rho,\sigma)^2
\ , \label{qpins}
\end{equation}
are well-known results of such a kind. Various lower and upper
bounds on the relative entropy (\ref{relan}) were obtained in Ref.
\cite{AE05}. As given in terms of difference distances, these
bounds characterize a continuity property in the sense of Fannes
\cite{AE05}. Recall that Fannes' inequality bounds from above a
potential change of the von Neumann entropy in terms of trace norm
distance \cite{fannes}. Fannes' inequality has been extended to
the Tsallis $q$-entropy \cite{yanagi,zhang} and its partial sums
\cite{rast1023}. The authors of the paper \cite{still90} proved
the inequalities
\begin{equation}
\rh_{\alpha}(\rho||\sigma)\leq\rh_1(\rho||\sigma)\leq\rh_{\beta}(\rho||\sigma)
\ , \label{p1qin}
\end{equation}
where $0\leq\alpha<1$ and $1<\beta\leq2$. So, for $1<\beta\leq2$
the relative entropy $\rh_{\beta}(\rho||\sigma)$ is bounded from
below by the right-hand side of Eq. (\ref{qpins}). More detailed
lower bounds on the relative entropy (\ref{relan}) are presented
in Ref. \cite{AE05}. By (\ref{p1qin}), these lower bounds are all
valid for $\rh_{\beta}(\rho||\sigma)$ with $1<\beta\leq2$.

Let $\ip_{+}$ be a projector on the eigenspace corresponding to
positive eigenvalues of the difference $(\rho-\sigma)$. For
normalized density operators, the inequality (\ref{pinab})
together with the definitions (\ref{srtdef}) and (\ref{ahadef})
leads to the bound
\begin{equation}
\rh_{\alpha}(\rho||\sigma)\geq{H}_{\alpha}\bigl(\{u,1-u\}\big|\big|\{v,1-v\}\bigr)
\ , \label{piniq}
\end{equation}
where we write $u=\tr(\ip_{+}\rho)$ and $v=\tr(\ip_{+}\sigma)$ for
brevity. Denoting $t=|u-v|$, we also have $\|\rho-\sigma\|_{1}=2t$
and $\rd(\rho,\sigma)=t$. In the paper \cite{rast10fn}, for
$u,v\in[0;1]$ we have proved the inequality (see lemma IV.2
therein)
\begin{equation}
\sqrt{uv}+\sqrt{(1-u)(1-v)}\leq\sqrt{1-t^2}
\ , \label{tih12}
\end{equation}
whence
${H}_{1/2}\bigl(\{u,1-u\}\big|\big|\{v,1-v\}\bigr)\geq2\left(1-\sqrt{1-t^2}\right)$
and, therefore, $\rh_{1/2}(\rho||\sigma)\geq\rd(\rho,\sigma)^2$.
We shall now estimate from below the Tsallis relative entropy
(\ref{qendf}) for arbitrary $\alpha\in(0;1)$.

\begin{Lem}\label{conal}
Let $u,v\in[0;1]$ and $g(t)=1-\sqrt{1-t^2}$. For
$\alpha\in[0;1/2]$ and $t=|u-v|$, there holds
\begin{equation}
u^{\alpha}v^{1-\alpha}+(1-u)^{\alpha}(1-v)^{1-\alpha}\leq1-2\alpha{g}(t)
\ . \label{alcon}
\end{equation}
\end{Lem}

{\bf Proof.} For fixed $u$ and $v$, we define the function
\begin{equation}
\Phi_{uv}(\alpha)=u^{\alpha}v^{1-\alpha}+(1-u)^{\alpha}(1-v)^{1-\alpha}+2\alpha{g}(t)-1
\ . \label{alcon1}
\end{equation}
The claim (\ref{alcon}) is equivalent to the inequality
$\Phi_{uv}(\alpha)\leq0$ for all $\alpha\in[0;1/2]$. First, we
have $\Phi_{uv}(0)=0$ obviously; second, $\Phi_{uv}(1/2)\leq0$ in
view of the relation (\ref{tih12}). Third, $\Phi_{uv}(\alpha)$ is
a convex function of the parameter $\alpha$. Indeed, for
$u,v\neq0,1$ we write down
\begin{equation}
\frac{\partial^2\Phi_{uv}}{\partial\alpha^2}=u^{\alpha}v^{1-\alpha}\left(\ln\frac{u}{v}\right)^2+
(1-u)^{\alpha}(1-v)^{1-\alpha}\left(\ln\frac{1-u}{1-v}\right)^2\geq0
\ . \label{parph}
\end{equation}
If a convex function is negative at the end points of some
interval, it is negative in this interval everywhere.
$\blacksquare$

Combining this with the statement of Lemma \ref{conal} gives a
lower continuity estimate on the Tsallis relative $\alpha$-entropy
for $\alpha\in(0;1)$. We formulate for two positive operators with
equal traces.

\begin{Thm}\label{bbpn}
Let $\am,\bn\in\lsp(\hh)$, $\tr(\am)=\tr(\bn)=\theta$,
$\rd(\am,\bn)=\tau$ and $g(t)=1-\sqrt{1-t^2}$. For all
$\alpha\in(0;1)$, there holds
\begin{equation}
\rh_{\alpha}(\am||\bn)\geq\varkappa_{\alpha}{\,}\theta{\,}{g}\bigl(\tau/\theta\bigr)
\ , \label{rhavf}
\end{equation}
where the factor $\varkappa_{\alpha}=2\alpha(1-\alpha)^{-1}$ for
$0<\alpha\leq1/2$ and $\varkappa_{\alpha}=2$ for
$1/2\leq\alpha<1$.
\end{Thm}

{\bf Proof.} Using the theorem precondition and (\ref{abtrn}), we
have $u'_{+}+u'_{-}=v'_{+}+v'_{-}=\theta$ and
$\|\am-\bn\|_{1}=2(u'_{+}-v'_{+})$, whence $\tau=u'_{+}-v'_{+}$.
It then follows from (\ref{pinab}), that
\begin{equation}
(1-\alpha){\,}\rs_{\alpha}(\am||\bn)\geq(1-\alpha){\,}S_{\alpha}\bigl(\{u'_{\pm}\}\big|\big|\{v'_{\pm}\}\bigr)
=-\theta\left[u^{\alpha}v^{1-\alpha}+(1-u)^{\alpha}(1-v)^{1-\alpha}\right]
\ , \label{abpin}
\end{equation}
where $u=u'_{+}/\theta$, $v=v'_{+}/\theta$, and $u-v=\tau/\theta$.
Since the map (\ref{qogr}) is trace-preserving,
$\tr\bigl(\varPsi(\am)\bigr)=\theta$. Combining this with
(\ref{qendf}) and (\ref{abpin}) leads to
\begin{equation}
(1-\alpha){\,}\rh_{\alpha}(\am||\bn)\geq
\theta\left[1-u^{\alpha}v^{1-\alpha}-(1-u)^{\alpha}(1-v)^{1-\alpha}\right]
\ , \label{al12}
\end{equation}
Due to the inequality (\ref{alcon}), for $0<\alpha\leq1/2$ the
right-hand side of (\ref{al12}) is not less than
$2\alpha{\,}\theta{\,}g(\tau/\theta)$. Hence the claim
(\ref{rhavf}) with $\varkappa_{\alpha}=2\alpha(1-\alpha)^{-1}$
follows. For $1/2\leq\alpha<1$, we put $\beta=1-\alpha$ and also
write
\begin{equation}
\beta{\,}\rh_{\alpha}(\rho||\sigma)\geq
\theta\left[1-u^{1-\beta}v^{\beta}-(1-u)^{1-\beta}(1-v)^{\beta}\right]\geq2\beta{\,}\theta{\,}g(\tau/\theta)
\ . \label{al21}
\end{equation}
Hence the claim (\ref{rhavf}) with $\varkappa_{\alpha}=2$ follows.
$\blacksquare$

For probability distributions, the lower continuity bound
(\ref{rhavf}) is rewritten with the classical trace distance
$\tau=D(P,Q)$. Expanding the function $g(\tau/\theta)$ into power series,
we obtain a family of lower bounds of the Pinsker type. Due to the
binomial theorem, we actually have
\begin{equation}
\rh_{\alpha}(\am||\bn)\geq\varkappa_{\alpha}\sum_{n=1}^{\infty}
\binom{1/2}{n}(-1)^{n+1}\frac{\tau^{2n}}{\theta^{2n-1}}
\ . \label{exbn}
\end{equation}
For all $n$, the coefficient $\binom{1/2}{n}(-1)^{n+1}$ is
positive. So, any partial sum of the series  (\ref{exbn}) provides
a lower continuity bound. In particular, for normalized density
operators we have a quadratic bound
\begin{equation}
\rh_{\alpha}(\rho||\sigma)\geq\frac{\varkappa_{\alpha}}{2}{\>}\rd(\rho,\sigma)^2
\ . \label{pquad}
\end{equation}
For $1/2\leq\alpha<1$, the multiplier $\varkappa_{\alpha}/2=1$,
whereas for $\alpha=1$ we have the quadratic bound (\ref{qpins})
with the multiplier two. This is evidence for that the series
(\ref{exbn}) does not provide an expansion with the best constants
at powers of the trace distance. For the standard relative entropy
$H_1(P||Q)$, these constants have been the subject of long-time
research (for details, see \cite{FHT03} and references therein).
It would be interesting to develop methods for finding the best
constants in Pinsker type inequalities for the Tsallis relative
$\alpha$-entropy. For instance, we could try to extend those ways
that are known for the case $\alpha=1$. Nevertheless, the results
(\ref{rhavf}) and (\ref{exbn}) resolve the question in a general
sense, since they provide non-trivial lower continuity bounds on
the Tsallis relative $\alpha$-entropy. For $\alpha\in(0;1)$, we
also combine (\ref{alrha}) with (\ref{rhavf}) and hence obtain
\begin{equation}
{\rm{R}}_{\alpha}(\rho||\sigma)\geq\frac{1}{\alpha-1}{\ }
\ln\bigl[1-(1-\alpha)\varkappa_{\alpha}g(\tau)\bigr]\geq\varkappa_{\alpha}g(\tau)
\ , \label{hravf}
\end{equation}
where $\tau=\rd(\rho,\sigma)$. Indeed, the function
$(\alpha-1)^{-1}\ln\bigl[1-(1-\alpha){\,}x\bigr]$ increases with
$x\in\bigl[0;\frac{1}{1-\alpha}\bigr)$ and $-\ln(1-\xi)\geq\xi$
for $\xi\in[0;1)$. The inequality (\ref{hravf}) can be regarded as
the bound of Pinsker type on the R\'{e}nyi relative entropy for
$\alpha\in(0;1)$. In particular, we have a version of
(\ref{pquad}) with ${\rm{R}}_{\alpha}(\rho||\sigma)$ instead of
$\rh_{\alpha}(\rho||\sigma)$. Thus, we have obtained a family of
lower continuity bounds in terms of the trace distance on both the
relative entropies (\ref{ahadef}) and (\ref{radef}).

\section{Upper continuity bounds for $\alpha>1$}\label{upbn}

One of basic features of the standard relative entropy is its
unboundedness. The relative $\alpha$-entropy enjoys the
same for $\alpha>1$. So we may ask a behaviour of the functional
$H_{\alpha}(P||Q)$ as the minimal probability in $Q$ goes to zero.
Of course, in the quantum case this question is more difficult
due to the non-commutativity. For the standard relative entropy, such an upper
bound was presented in \cite{BR2}, and more bounds were given in \cite{AE05}. For the
quantum relative $\alpha$-entropy of order $\alpha>1$, upper
continuity bounds in terms of the minimal eigenvalue of its second
entry were obtained in \cite{rast1009}. It turns out that in the
commutative case these bounds can be sharpened significantly. Our
derivation will mainly based on the joint convexity. Namely, for each positive
$\alpha\neq1$ the quantity (\ref{geendf}) enjoys
\begin{equation}
H_{\alpha}\left(\theta{A}^{(1)}+(1-\theta)A^{(2)}\bigm|\bigm|\theta{B}^{(1)}+(1-\theta)B^{(2)}\right)
\leq\theta H_{\alpha}(A^{(1)}||B^{(1)})+(1-\theta)H_{\alpha}(A^{(2)}||B^{(2)})
\label{tab12}
\end{equation}
for all $0\leq\theta\leq1$. This relation can be proved by means
of so-called ''generalized log-sum inequality'' (see formula (16)
in \cite{borland}). The properties (\ref{homg}) and (\ref{tab12})
lead to the following statement.

\begin{Lem}\label{baah}
Let $A$, $B$, $C$ be three sets of positive numbers, and $\am$,
$\bn$, and $\cn$ three positive operators. There holds
\begin{align}
& H_{\alpha}(A+C||B+C)\leq{H}_{\alpha}(A||B) & (0\leq\alpha<\infty)
\ , \label{haabc}\\
& \rh_{\alpha}(\am+\cn||\bn+\cn)\leq\rh_{\alpha}(\am||\bn) & (0\leq\alpha\leq2)
\ . \label{haab2}
\end{align}
\end{Lem}

{\bf Proof.} Using (\ref{homg}) and (\ref{tab12}), we merely write
\begin{equation}
H_{\alpha}(A+C||B+C)=2{\,}H_{\alpha}\bigl((A+C)/2\bigm|\bigm|(B+C)/2\bigr)\leq
H_{\alpha}(A||B)+H_{\alpha}(C||C)=H_{\alpha}(A||B)
\label{haab1}
\end{equation}
in view of obvious $H_{\alpha}(C||C)=0$. The quantum relative
entropy (\ref{qendf}) also enjoys both the homogeneity of degree
one and the joint convexity, but the latter only for
$0\leq\alpha\leq2$ (see, e.g., the review \cite{JR10}). Rewriting
the above arguments with the quantum relative $\alpha$-entropy
instead of the classical one, we have arrived at the claim
(\ref{haab2}). $\blacksquare$

For the standard relative entropy (\ref{relan}), the relation
(\ref{haab2}) was proved in \cite{AE05}. The inequality
(\ref{haabc}) can be utilized for obtaining an upper bound on
$H_{\alpha}(P||Q)$ in terms of the trace distance $D(P,Q)$ and the
minimal probability
\begin{equation}
q_0:=\min\{q_j:{\>}j\in\Omega_P\}
\ . \label{qmdf}
\end{equation}
Here we apply that any sum in $H_{\alpha}(P||Q)$ is effectively
restricted to the index subset $\Omega_P$. Defining the set
$\Delta=P-Q$ with elements $\delta_j=p_j-q_j$, we put another set
$\Bar{Q}$ with positive elements
\begin{equation}
\Bar{q}_j:=\max\{q_0,-\delta_j\}
\ . \label{bqdf}
\end{equation}
Writing $Q=\Bar{Q}+(Q-\Bar{Q})$ and using the property
(\ref{haabc}) with $C=Q-\Bar{Q}$, we obtain
\begin{equation}
H_{\alpha}(P||Q)=H_{\alpha}\left(\Delta+\Bar{Q}+(Q-\Bar{Q})\bigm|\bigm|\Bar{Q}+(Q-\Bar{Q})\right)\leq
H_{\alpha}(\Delta+\Bar{Q}{\,}||\Bar{Q})
\ . \label{baq}
\end{equation}
Use of (\ref{haabc}) is correct here due to positivity of both
$\Delta+\Bar{Q}$ and $Q-\Bar{Q}$: we clearly have
$\delta_j+\max\{q_0,-\delta_j\}\geq0$ and
$q_j-\max\{q_0,q_j-p_j\}\geq0$. The maximization of the right-hand
side of (\ref{baq}) is under the conditions $\sum_j\delta_j=0$ and
$\sum_j|\delta_j|=2{\,}D(P,Q)$. We separately consider the two
cases, $D(P,Q)\leq{q_0}$ and $q_0<D(P,Q)\leq1-q_0$.

\begin{Thm}\label{thal1}
Let $q_0$ be defined by (\ref{qmdf}),
$\tom_Q\subset\tom_P$ and $\tau=D(P,Q)$.
For $\alpha>1$, the Tsallis relative $\alpha$-entropy is bounded
from above as
\begin{align}
& H_{\alpha}(P||Q)\leq\frac{1}{\alpha-1}
{\,}\Bigl(
(q_0+\tau)^{\alpha}q_0^{1-\alpha}+(q_0-\tau)^{\alpha}q_0^{1-\alpha}-2q_0
\Bigr)
& (\tau\leq q_0)
\ , \label{obon1}\\
& H_{\alpha}(P||Q)\leq\frac{1}{\alpha-1}
{\,}\Bigl(
(q_0+\tau)^{\alpha}q_0^{1-\alpha}-(q_0+\tau)
\Bigr)
 & (q_0<\tau\leq1-q_0)
\ . \label{obon2}
\end{align}
\end{Thm}

{\bf Proof.} In the case $\tau\leq q_0$, we have
$\max(q_0,-\delta_j)=q_0$ for all $j$. By $x_i$ and $(-y_j)$ we
respectively denote positive and negative elements of the set
$\Delta=\{\delta_j\}$. The conditions $\sum_j\delta_j=0$ and
$\sum_j|\delta_j|=2\tau$ are rewritten as
\begin{equation}
\sum\nolimits_{i\in\omega_x}x_i=\sum\nolimits_{j\in\omega_y}y_j=\tau
\ , \label{xycond}
\end{equation}
where $\omega_x$ and $\omega_y$ are corresponding subsets of the
set $\Omega_P$ of cardinality $n$. The right-hand side of
(\ref{baq}) is represented as the function
\begin{equation}
F(x_i,y_j)=\frac{1}{\alpha-1}{\,}\left(
\sum\nolimits_{i\in\omega_x}(q_0+x_i)^{\alpha}q_0^{1-\alpha}+
\sum\nolimits_{j\in\omega_y}(q_0-y_j)^{\alpha}q_0^{1-\alpha}
-n q_0
\right)
\ . \label{ffun}
\end{equation}
This function should be maximized under the conditions $0\leq
x_i$, $0\leq y_j$ and (\ref{xycond}), which define a simplex.
Recall that the global maximum of a convex function relative to a
convex set is reached at one of the extreme points of that set
\cite{rockaf}. Hence the maximum of $F(x_i,y_j)$ is equal to the
right-hand side of (\ref{obon1}) and reached, when one of the
$x_i$'s and one of the $y_j$'s are equal to $\tau$ and other are
all zero.

In the case $q_0<\tau$, we add negative elements $(-z_k)$ such
that $q_0<z_k$ and
$\sum_{j\in\omega_y}y_j+\sum_{k\in\omega_z}z_k=\tau$. Before
maximization, we further simplify the right-hand side of
(\ref{baq}). Putting the sets $\Delta_x=\{x_i\}$,
$\Delta_y=\{y_j\}$ and $\Delta_z=\{z_k\}$, we write
\begin{equation}
\Delta=\Delta_x-\Delta_y-\Delta_z
\ , \qquad
\Bar{Q}=q_0(I_x+I_y)+\Delta_z
\ , \qquad
\Delta+\Bar{Q}=q_0(I_x+I_y)+\Delta_x-\Delta_y
\ . \label{dbdb}
\end{equation}
Here $I_x$ denote the indicator of the set $\omega_x$ taking the
value one for $j\in\omega_x$ and zero for $j\not\in\omega_x$. Note
that if the set $\Omega_Z$ does not intersect with both the
$\Omega_A$ and $\Omega_B$ then
$H_{\alpha}(A||B+Z)=H_{\alpha}(A||B)$. Using this fact twice and
the inequality (\ref{haabc}) again with positive
$C=q_0I_y-\Delta_y$, we rewrite the right-hand side of (\ref{baq})
in a form
\begin{align}
H_{\alpha}\left(q_0(I_x+I_y)+\Delta_x-\Delta_y\bigm|\bigm| q_0(I_x+I_y)+\Delta_z\right)
&=H_{\alpha}\left(q_0I_x+\Delta_x+C\bigm|\bigm| q_0I_x+\Delta_y+C\right)
\nonumber\\
&\leq{H}_{\alpha}\left(q_0I_x+\Delta_x\bigm|\bigm| q_0I_x+\Delta_y\right)
={H}_{\alpha}\left(q_0I_x+\Delta_x\bigm|\bigm| q_0I_x\right)
\ . \label{zcag}
\end{align}
The latter is expressed as
$G(x_i)=(\alpha-1)^{-1}\left(\sum_{i\in\omega_x}(q_0+x_i)^{\alpha}q_0^{1-\alpha}-(n_xq_0+\tau)\right)$,
where $n_x$ is cardinality of the $\omega_x$. Under the conditions
$0\leq x_i$ and $\sum_ix_i=\tau$, the maximum of $G(x_i)$ is equal
to the right-hand side of (\ref{obon2}) and reached, when one of
the $x_i$'s is $\tau$ and other are all zero. $\blacksquare$

The upper bounds (\ref{obon1}) and (\ref{obon2}) have a behavior
$q_0^{1-\alpha}$ with respect to the minimal probability $q_0$.
For the quantum relative entropy $\rh_{\alpha}(\rho||\sigma)$,
upper continuity bounds with a similar dependence on the minimal
eigenvalue of $\sigma$ were obtained in our previous paper
\cite{rast1009}. The bounds (\ref{obon1}) and (\ref{obon2}) are
stronger, but their proof is quite restricted to the commutative
case. The principal point is that positivity of diagonal elements
of a matrix do not imply positivity of matrix itself (except for
the case of diagonal matrices). So, the proof of Theorem
\ref{thal1} is purely classical in character. On the other hand,
weaker bounds in the paper \cite{rast1009} have been proved just
for the quantum case. Note that the inequalities (\ref{obon1}) and
(\ref{obon2}) can be rewritten in terms of $\alpha$-logarithm as
\begin{align}
& H_{\alpha}(P||Q)\leq -(q_0+\tau_0){\>}\ln_{\alpha}\left(\frac{q_0}{q_0+\tau}\right)
-(q_0-\tau){\>}\ln_{\alpha}\left(\frac{q_0}{q_0-\tau}\right)
& (\tau\leq{q}_0)
\ , \label{lobon1}\\
& H_{\alpha}(P||Q)\leq -(q_0+\tau){\>}\ln_{\alpha}\left(\frac{q_0}{q_0+\tau}\right)
& (b_0<\tau\leq1-q_0)
\ . \label{lobon2}
\end{align}
Using (\ref{alrha}) in the classical regime, we see that the
bounds (\ref{lobon1}) and (\ref{lobon2}) remain valid with
$R_{\alpha}(P||Q)$ instead of $H_{\alpha}(P||Q)$. This claim
follows from the points that $\ln\{1+(\alpha-1){\,}x\}$ increases
with positive $x$ and $\ln(1+\xi)\leq\xi$ for $\xi\geq0$. The
relations (\ref{lobon1}) and (\ref{lobon2}) are
$\alpha$-parametric extensions of the upper bounds obtained in
Ref. \cite{AE05} for the standard relative entropy. In effect, the
upper continuity bounds (\ref{obon1}) and (\ref{obon2}) could be
used for classical systems which are well described within
non-extensive thermostatistics.

\section{Notes on Fano and Fannes inequalities}\label{ngfn}

In this section, we will obtain upper bounds on the conditional
Tsallis $\alpha$-entropy for all $\alpha>0$. It is convenient to
change the notation slightly as follows. Let $X$ and $Y$ be
discrete random variables with probabilities $\{p_{X}(x)\}$ and
$\{p_{Y}(y)\}$, each supported on the $N$-point set $\mcx$. By
$p_{XY}(x,y)$ and $p_{X|Y}(x|y)$ we respectively denote the joint
and conditional probabilities. The joint $\alpha$-entropy and the
conditional $\alpha$-entropy are respectively defined as
\begin{align}
 & H_{\alpha}(X,Y):=\frac{1}{1-\alpha}\left(\sum\nolimits_{x,y}p_{XY}(x,y)^\alpha-1\right)
\ , \label{joind}\\
 & H_{\alpha}(X|Y):=\sum\nolimits_{y}p_{Y}(y)^{\alpha}H_{\alpha}(X|y)
\ , \label{cond0}
\end{align}
where
$H_{\alpha}(X|y)=(1-\alpha)^{-1}\bigl(\sum\nolimits_{x}p_{X|Y}(x|y)^\alpha-1\bigr)$.
Rewriting
$H_{\alpha}(X|y)=-\sum_{x}p_{X|Y}(x|y)^{\alpha}\ln_{\alpha}p_{X|Y}(x|y)$,
we further obtain
\begin{equation}
H_{\alpha}(X|Y)=-\sum\nolimits_{x,y}p_{XY}(x,y)^{\alpha}\ln_{\alpha}p_{X|Y}(x|y)
\ , \label{rwha}
\end{equation}
due to the relation $p_{Y}(y){\,}p_{X|Y}(x|y)=p_{XY}(x,y)$. We will
follow the original scheme of derivation (see the classical text
\cite{fano}, section 6.2). The probability of error is expressed
as
\begin{equation}
P_e=\sum\nolimits_{y}p_{Y}(y){\,}q(e|y)
\ , \qquad
q(e|y)=1-p_{X|Y}(y|y)=\sum\nolimits_{x\neq{y}}p_{X|Y}(x|y)
\ . \label{pep}
\end{equation}

\begin{Lem}\label{metfan}
For all $\alpha\in(0;\infty)$, there holds
\begin{equation}
H_{\alpha}(X|Y)\leq\sum\nolimits_{y}p_{Y}(y)^{\alpha}h_{\alpha}\bigl(q(e|y)\bigr)+
\ln_{\alpha}(N-1)\sum\nolimits_{y}p_{Y}(y)^{\alpha}q(e|y)^{\alpha}
\ . \label{resub}
\end{equation}
\end{Lem}

{\bf Proof.} Using the expression for $p(e|y)$ and the
definition (\ref{hbin}), we write
\begin{align}
H_{\alpha}(X|y)&=-p_{X|Y}(y|y)^{\alpha}\ln_{\alpha}p_{X|Y}(y|y)-
\sum\nolimits_{x\neq{y}}p_{X|Y}(x|y)^{\alpha}\ln_{\alpha}p_{X|Y}(x|y)
\nonumber\\
&=h_{\alpha}\bigl(q(e|y)\bigr)+q(e|y)^{\alpha}\ln_{\alpha}q(e|y)-
\sum\nolimits_{x\neq{y}}p_{X|Y}(x|y)^{\alpha}\ln_{\alpha}p_{X|Y}(x|y)
\ . \label{conw}
\end{align}
Due to $q(e|y)=\sum_{x\neq{y}}p_{X|Y}(x|y)$ and the properties of
$\alpha$-logarithm, the second and third terms in the right-hand
side of (\ref{conw}) are combined as
\begin{align}
&-\sum\nolimits_{x\neq{y}}
p_{X|Y}(x|y)^{\alpha}\ln_{\alpha}p_{X|Y}(x|y)+q(e|y){\,}q(e|y)^{\alpha-1}\ln_{\alpha}q(e|y)\nonumber\\
&=-\sum\nolimits_{x\neq{y}} p_{X|Y}(x|y)^{\alpha}\Bigl(\ln_{\alpha}p_{X|Y}(x|y)-p_{X|Y}(x|y)^{1-\alpha}q(e|y)^{\alpha-1}\ln_{\alpha}q(e|y)\Bigr)
\nonumber\\
&=-\sum\nolimits_{x\neq{y}} p_{X|Y}(x|y)^{\alpha}\left(\ln_{\alpha}p_{X|Y}(x|y)+p_{X|Y}(x|y)^{1-\alpha}\ln_{\alpha}\frac{1}{q(e|y)}{\>}\right)
\label{onx}\\
&=-q(e|y)^{\alpha}\sum\nolimits_{x\neq{y}} \frac{p_{X|Y}(x|y)^{\alpha}}{q(e|y)^{\alpha}}
{\ }\ln_{\alpha}\frac{p_{X|Y}(x|y)}{q(e|y)}\leq
q(e|y)^{\alpha}\ln_{\alpha}(N-1)
\ . \label{stn1}
\end{align}
Here we used the identities
$\ln_{\alpha}(1/\xi)=-\xi^{\alpha-1}\ln_{\alpha}\xi$ (right before
(\ref{onx})) and
$\ln_{\alpha}(\xi{z})=\ln_{\alpha}\xi+\xi^{1-\alpha}\ln_{\alpha}z$
(right before (\ref{stn1})). Substituting (\ref{stn1}) in
(\ref{conw}) and further in (\ref{cond0}), we obtain
(\ref{resub}). $\blacksquare$

\begin{Thm}\label{tbow}
Let random variables $X$ and $Y$ take values on the same finite
set with cardinality $N$. For given value of the error probability
$P_e$, the conditional entropy $H_{\alpha}(X|Y)$ is bounded from
above as
\begin{align}
H_{\alpha}(X|Y) &\leq\frac{P_e^{{\,}\alpha}-\alpha{P}_e}{1-\alpha}+P_e^{{\,}\alpha}\ln_{\alpha}\bigl[N(N-1)\bigr]
 & (0<\alpha<1)
\ , \label{fanin0}\\
H_{\alpha}(X|Y) &\leq h_{\alpha}(P_e)+P_e^{{\,}\alpha}\ln_{\alpha}(N-1)
 & (1<\alpha<\infty)
\ . \label{fanin1}
\end{align}
\end{Thm}

{\bf Proof.} For $\alpha\in(0;1)$, we use the expression
$h_{\alpha}(u)=(1-\alpha)^{-1}[u^{\alpha}+(1-u)^{\alpha}-1]\leq(1-\alpha)^{-1}(u^{\alpha}-\alpha{u})$,
which follows from (\ref{hbin}) and the inequality
\begin{equation}
1-(1-u)^{\alpha}=\int\nolimits_0^u\alpha(1-t)^{\alpha-1}dt
\geq\int\nolimits_0^u\alpha{\,}dt=\alpha{u}
\ . \label{inu11}
\end{equation}
By these relations and $\xi_{y}=p_{Y}(y){\,}q(e|y)$, the first sum in
the right-hand side of (\ref{resub}) is no larger than
\begin{equation}
\frac{1}{1-\alpha}{\>}\sum\nolimits_{y} p_{Y}(y)^{\alpha}\bigl[q(e|y)^{\alpha}-\alpha{q}(e|y)\bigr]
\leq\frac{1}{1-\alpha}\left(\sum\nolimits_{y} \xi_{y}^{\alpha}-\alpha{P}_e\right)
\ , \label{xypal}
\end{equation}
in view of
$\sum_{y}p_{Y}(y)^{\alpha}q(e|y)\geq\sum_{y}p_{Y}(y){\,}q(e|y)=P_e$.
Using the H\"older inequality, we also obtain
\begin{equation}
\max\left\{\sum\nolimits_{y=1}^{N} \xi_y^{\alpha}:{\ }0\leq\xi_y\leq1,{\ }
\sum\nolimits_{y=1}^{N}\xi_{y}=P_e\right\}=N^{1-\alpha}P_e^{{\,}\alpha}
\ , \label{funx}
\end{equation}
which is reached for $\xi_{y}=P_e/N$. So the term
$(1-\alpha)^{-1}\left(N^{1-\alpha}P_e^{{\,}\alpha}-\alpha{P}_e\right)$
is an upper bound for the right-hand side of (\ref{xypal}).
Combining this with the product of $\ln_{\alpha}(N-1)$ and
(\ref{funx}) finally gives (\ref{fanin0}).

Using $p_{Y}(y)^{\alpha}\leq{p}_{Y}(y)$ for $\alpha>1$ and Jensen's
inequality for the concave function (\ref{hbin}), we have
\begin{equation}
\sum\nolimits_{y}p_{Y}(y)^{\alpha}h_{\alpha}\bigl(q(e|y)\bigr)\leq
\sum\nolimits_{y}p_{Y}(y){\,}h_{\alpha}\bigl(q(e|y)\bigr)\leq
h_{\alpha}\left(\sum\nolimits_{y}p_{Y}(y){\,}q(e|y)\right)=h_{\alpha}(P_e)
\ . \label{hconc}
\end{equation}
For $\alpha>1$, there holds $\sum\nolimits_{y}
p_{Y}(y)^{\alpha}q(e|y)^{\alpha}\leq\left(\sum\nolimits_{y}
p_{Y}(y){\,}q(e|y)\right)^{\alpha}=P_e^{{\,}\alpha}$ (this can be
rewritten as $\|\centerdot\|_{\alpha}\leq\|\centerdot\|_1$ in
terms of the vector norms). By these two points, the inequality
(\ref{resub}) leads to (\ref{fanin1}). $\blacksquare$

For $\alpha>1$, the inequality (\ref{fanin1}) with $P_e$ instead
of $P_e^{\alpha}$ was presented in \cite{sf06}. So we obtain an
improvement of the known result. The inequality (\ref{fanin0}) for
$0<\alpha<1$ is a new bound. By construction, the bound
(\ref{fanin0}) is not sharp. Nevertheless, this inequality is
sufficiently exact for small values of $P_e$. In any event, both
the bounds of Theorem \ref{tbow} show that $P_e\to0$ implies
$H_{\alpha}(X|Y)\to0$. On the other hand, if $H_{\alpha}(X|Y)$ is
large then the probability of making an error in inference must be
large as well. In this regard, the essence of our inequalities
concurs with a typical use of the standard Fano inequality in
terms of the Shannon entropies.

Uniform continuity is an important property of the von Neumann
entropy. The first result in this issue was given by Fannes
\cite{fannes}. The Tsallis entropy itself \cite{yanagi,zhang} and
its partial sums \cite{rast1023} also enjoy this property. Using
the classical Fano inequality, Fannes' bound has been sharpened
(see theorem 3.8 and its proof of Csisz\'{a}r in \cite{petz08}).
We shall now show that the Fano type inequalities (\ref{fanin0})
and (\ref{fanin1}) lead to the Fannes inequality in terms of
Tsallis entropies. Using properties of the $\alpha$-logarithm, the
joint entropy (\ref{joind}) can be re-expressed as \cite{sf06}
\begin{equation}
H_{\alpha}(X,Y)=H_{\alpha}(X)+H_{\alpha}(Y|X)=H_{\alpha}(Y)+H_{\alpha}(X|Y)
\ . \label{hmwh}
\end{equation}
Hence, in view of $H_{\alpha}(Y|X)\geq0$, the difference
$H_{\alpha}(X)-H_{\alpha}(Y)\leq{H}_{\alpha}(X|Y)$ is bounded from
above by the right-hand side of (\ref{fanin0}) for
$\alpha\in(0;1)$ and by the right-hand side of (\ref{fanin1}) for
$\alpha\in(1;+\infty)$. For given probability distributions
$\{p_{X}(x)\}$ and $\{p_{Y}(y)\}$, the joint probability mass
function $p_{XY}(x,y)$ can be built in such a way that
$P_e=D(X,Y)=(1/2)\sum_{x}\left|p_{X}(x)-p_{Y}(x)\right|$, -- this
follows from the coupling inequality (see, e.g., the book
\cite{lindvall}). Setting $\{p_{X}(x)\}=\spc(\rho)$ and
$\{p_{Y}(y)\}=\spc(\sigma)$, we then have
$D(X,Y)\leq\rd(\rho,\sigma)$ (see, e.g., lemma 11.1 in
\cite{petz08}). Assuming $N\geq2$, the right-hand side of
(\ref{fanin0}) increases with $P_e$ for all $0\leq{P}_e\leq1$, the
right-hand side of (\ref{fanin1}) increases with $P_e$ for all
$0\leq{P}_e\leq{N}/(N+1)$. Replacing $D(X,Y)$ with larger
$\rd(\rho,\sigma)$, we get the following result.

\begin{Thm}\label{ffnn}
Let $d$ be dimensionality of the Hilbert space and
$\tau=\rd(\rho,\sigma)$; then
\begin{align}
\bigl|\rh_{\alpha}(\rho)-\rh_{\alpha}(\sigma)\bigr| & \leq
\frac{\tau^{\alpha}-\alpha\tau}{1-\alpha}+\tau^{\alpha}\ln_{\alpha}\bigl[d(d-1)\bigr]
 & (0<\alpha<1 \ , \quad 0\leq\tau\leq1)
\ , \label{hmwh20}\\
\bigl|\rh_{\alpha}(\rho)-\rh_{\alpha}(\sigma)\bigr| & \leq
h_{\alpha}(\tau)+\tau^{\alpha}\ln_{\alpha}(d-1)
 & \left(1<\alpha \ , \quad 0\leq\tau\leq\frac{d}{d+1}\right)
\ . \label{hmwh21}
\end{align}
\end{Thm}

The inequality (\ref{hmwh21}), when $\alpha>1$, is just the
uniform estimate derived by a direct method in \cite{zhang}. This
inequality is the best bound known for the Tsallis entropies in
the parameter range $\alpha>1$. In the limit $\alpha\to1$, the
inequality (\ref{hmwh21}) reproduces the statement of theorem 3.8
in \cite{petz08}. For $\alpha\in(0;1)$, there exists another
inequality
\begin{equation}
\bigl|\rh_{\alpha}(\rho)-\rh_{\alpha}(\sigma)\bigr|
\leq\frac{(2\tau)^{\alpha}-2\tau}{1-\alpha}+(2\tau)^{\alpha}\ln_{\alpha}d
\ , \label{hmwh3}
\end{equation}
provided that $\|\rho-\sigma\|_1=2\tau\leq\alpha^{1/(1-\alpha)}$.
The bound (\ref{hmwh3}) was actually proved in the paper
\cite{yanagi} for all $\alpha\in[0;2]$, but the bound
(\ref{hmwh21}) is better for $\alpha\geq1$. Comparing our bound
(\ref{hmwh20}) with (\ref{hmwh3}), we see the following. In
general, the bound (\ref{hmwh20}) is weaker but covers all
acceptable values $\tau\in[0;1]$ of the trace distance. The scope
of (\ref{hmwh3}) is restricted to the range
$0\leq2\tau\leq\alpha^{1/(1-\alpha)}$. In low dimensions, however,
the bound (\ref{hmwh20}) can be better than the bound
(\ref{hmwh3}). Say, for $d=2$ and $\alpha=1/2$ the bound
(\ref{hmwh3}) holds for $0\leq\tau\leq1/8$. In this range, the
right-hand side of (\ref{hmwh3}) is larger than the right-hand
side of (\ref{hmwh20}). Moreover, for sufficiently small $\tau$
the difference between the two bounds is up to $40$ \%. Thus, the
bound (\ref{hmwh20}) has some practical interest, at least in the
primary qubit case.

redpajama_set_name: RedPajamaArXiv


text:
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First-Timer's Guide to Rome
The Eternal City dazzles with iconic Roman ruins, a wealth of art and architecture, and lively historic squares, but Rome can also overwhelm first-time visitors who want to see as many highlights as possible in a limited amount of time. Here are tips for taking in local treasures on your first trip to Rome.
Opt for Skip-the-Line Tours
Rome is home to some of the most visited monuments in the world, and waiting in long lines to enter highlights like the Colosseum and the Vatican Museums can steal precious hours from your day. Pre-booking skip-the-line entry is imperative if you want to maximize your time admiring the magnificent amphitheater where gladiators once fought, Michelangelo's Sistine Chapel, and other unforgettable sights.
Explore the City on Wheels
Rome is a sprawling city, and it can be a hike to get from one attraction to the next on foot. Hop aboard a Vespa, a vintage Fiat 500, an Ape Calessino tuk tuk, a bike, or even a Segway, and zip from square to square to take in the Trevi Fountain, Spanish Steps, and Pantheon. If you want to do some walking, a hop-on-hop-off bus tour is ideal.
Sightsee from Dawn to Dusk...and Beyond
Opt for early entry or night tours to stretch your day to the max, and enjoy the added benefits of no lines or crowds (and cooler temperatures, if you are visiting in the summer). Experience the peaceful atmosphere of the Vatican at dawn, the La Dolce Vita vibe of the historic center at dusk, and the timeless romance of the Colosseum by moonlight.
Explore the Neighborhoods
Rome offers more than world-famous landmarks and museums. This historic city is a patchwork of unique neighborhoods, each with a distinct character. Discover bohemian Trastevere, chic Monti, and the storied Jewish ghetto.
Stroll the Markets
Rome has a proud culinary tradition, and its outdoor food markets are centers of community life. Take a morning walk through stalls selling seasonal produce, artisan cheeses, cold cuts, and other specialties. Foodies can opt for a cooking experience that combines a market stroll with a hands-on lesson.
Get Out of Town
Rome's delights aren't confined to its urban center. Just outside the city walls, you can bike along the ancient Appian Way, head to the Castelli Romani for a wine tasting, or see where popes have summered for centuries at Castel Gandolfo.
Did you know?You can reserve your spot today and pay when you're ready.
You can reserve your spot today and pay when you're ready.Learn more
Fast Track - Vatican Tour with Museums, Sistine Chapel & Raphael rooms
Discover Vatican City including the Vatican Museums, St. Peter's Basilica, and the Sistine Chapel. Home to the world's largest private portfolio of artwork, the Vatican is a must-see for visitors to Rome, and the best way to explore it is with a guide. Learn about Raphael's Rooms, the Michelangelo frescoes of the Sistine Chapel, the 'La Pietà' sculpture in St. Peter's Basilica and more. Choose from several start times or upgrade to a small-group tour with just 12 people.
Jewish Ghetto and Navona Food Wine and Sightseeing Tour of Rome
Immerse yourself in the flavors of the 'Eternal City' on a half-day tour through Rome's iconic foods and wine. Visit a series of local restaurants in Campo Marzio and the Jewish Ghetto for tastings that range from fish and meat to pizza, pasta, and desserts served alongside Italian wines. Take in the city's historic sites as you go, with time to explore classic piazzas, ancient Roman sites, temples, and a synagogue. Finish your tour with inside tips on the best places to eat, drink and explore in Rome.
Rome by Vespa: Classic Rome Tour with Pick up
Zip through the bustling streets of Rome as the Romans do — on a Vespa! Hang on for the ride as your private driver and guide whisks you around on a vintage scooter for an exciting half-day tour of Eternal City highlights. Wind through traffic circles and mazelike streets, easily hopping off for a closer look at Imperial monuments and baroque piazzas. Learn about the history of Rome, from ancient emperors to the modern metropolis, and get an insider's glimpse at some of the city's best-kept secrets.
Rome by Night Tour Including Dinner
See the Eternal City lit up at night on this evening tour of Rome with dinner. Begin at a restaurant near the Trevi Fountain for a traditional Italian meal, then take a walk around the area—part of Rome's historical center—to enjoy the nighttime ambiance. Afterward, a panoramic coach tour takes you to see illuminated attractions, finishing at the Colosseum.
Full Course Cooking Class in Rome with Market Visit, Pasta & Dessert Making
Get the complete Roman experience and take a private cooking class during your holiday. This 4-hour excursion starts with shopping for your ingredients at the local market with your host. Then head to your class in a 17th-century palace and prepare a 4-course meal from start to finish. Sit down and enjoy the fruits of your labor and sample local wines and create beautiful memories of your time in Rome.
Hand sanitiser available to travellers and staff
Gear/equipment sanitised between use
Explore the landmarks of ancient Rome on a Segway tour with an experienced guide. On your own electric Segway, cruise by major Rome attractions like the Roman Forum, Colosseum, and Circus Maximus. Learn about the Eternal City's 2,000-year-old ruins while you zoom along the streets, and stop by the Mouth of Truth, Rome's legendary 'lie detector.' Limited to just eight people, this tour provides a small-group experience.
Rome Vintage Fiat 500 Self-Drive Tour by Convoy
Explore Rome as if you were in a classic Italian film on this sightseeing tour by vintage Fiat 500. Following a knowledgeable guide, you'll drive yourself through the Eternal City in your own Easter egg–colored Fiat, cruising down lanes that are often inaccessible to larger autos. See Roman landmarks like Circus Maximus, the Colosseum and the Arch of Constantine, and discover the Garden of the Oranges and other off-the-beaten-path sights, as well. Limited to just 10 people, this tour provides a small-group experience.
Frascati and Castelli Romani Private Wine Tour from Rome
Escape the bustle of Rome for a half-day private wine tour in the Frascati area. Visit vineyards and a wine cellar to learn about local production techniques and then taste a variety of the region's wines alongside traditional foods. Round-trip transportation by private car transfer is included.
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Located close to Little Chalfont station and shops is this extremely well presented three bedroom semi – detached family home refurbished within the last 3 years by the current owners with potential to extend into the loft subject to planning permission.
Steps leading to porch and front door to entrance hall with staircase to first floor and doors to kitchen/dining room and sitting room. Sitting room with bay window and feature fire and double doors to dining room. Kitchen/dining room with door to rear garden and door to conservatory. Upstairs there are two double bedrooms and 1 single bedroom currently being used as an office and a family bathroom.
In accordance to the Estate Agents Act 1979 we must inform perspective purchasers that the vendor of this property is an employee at Sewell & Gardner.


text:
Oakthorpe Park is a small park and a larger, informal and transiently used neighbourhood name denoting a small fraction of the south of the London Borough of Enfield in Palmers Green, north London, UK. It is situated next to the North Circular.

Rarely marked on maps, the Oakthorpe Park neighbourhood corresponds to part of the 2004 to date  Bowes ward and its nearest train or underground stations are at Wood Green, Palmers Green, Bounds Green and Bowes Park.  The A406 road North Circular Road passes east–west through the north of the neighbourhood.

Nearest areas
Wood Green
Bowes Park
Edmonton
White Hart Lane (ward)
New Southgate

References

Areas of London
Districts of the London Borough of Enfield

redpajama_set_name: RedPajamaWikipedia


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