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The MA Visual Communication course is about communicating small and big data in visual ways – graphic design, visual communications, infographics.
2019 entry
2019 entry
This specialist programme reflects the dynamic contemporary practices within the creative industries. It stimulates you to create challenging solutions to real-world design issues to create innovators and leaders in the field of visual communication. You identify the key transferable skills to help you create or respond to career opportunities or undertake further research.
You consider branding, graphic design, illustration, marketing, user experience, new media and future technologies – exploring and visually representing important information. You explore scenario building, fore sighting and future proofing as important factors in establishing the direction of your work and potential developments in design, communication and cultural industries.
This course prepares you, as a creative individual, for professional practice in developing new business ideas, products, systems and artefacts. And it provides a stimulating environment to support high-level enquiry into emerging and future aspects of creative practice, through individual and collaborative action. You benefit from existing work that staff are doing around interactive technology and social design.
The programme begins with a group research project, sharing information and references from diverse sources. Collecting and analysing information from a theme of common interest helps to develop your awareness of the subject from multiple perspectives. Stage 1 involves developing professional skills, ideas, research, project work and the opportunity for co-working, partnerships and collaborations. The enterprise potential of your interests is evaluated and innovative outputs are proposed.
Stage 2 allows time to prepare, research and develop project proposals, culminating in a feasibility study for a negotiated major project. Preparation for this major work includes extended reading, visual and experiential references, critical thinking, and collecting and analysing information.
Stage 3 enables you to direct and display your major project work, supported by regular tutorial contact and studio interaction. You show future ambitions and plans for the project including how it may be distributed or realised beyond the University.
Critical Thinking and Theoretical Contexts
This module explores and interrogates key concepts and critical thinking in the 21st century. It promotes debate on developing themes within culture and the creative arts in theory and practice.
Through lectures you explore critical thinking and cultural philosophical approaches. Key issues are explored through talking and debating in seminars.
There are two assessments; a 20 minute presentation focusing on cultural theory in relation to your design practice and field of specialism, and a critical report that recognises design practice with respect to cultural and critical theory.
You learn a number of research methods and practices in order to carry out primary design research to identify solutions to real-world problems.
You cover investigations into practical design applications, strategic planning and theoretical speculations.
Through seminars you debate and discuss case studies and the practical application of research methods.
You are assessed on your research plan (20%) and a critical analysis of research methods appropriate to your own design practice, identifying how these are implemented within a design problem solving context (80%).
Enterprising Leadership and Project Management
You learn through seminars and student-led group work, with guest speakers providing real-world perspectives and expertise. You are assessed by a group portfolio, demonstrating innovation, inventiveness, an entrepreneurial mindset, leadership, creativity and project management. The portfolio contains a product development report and a group presentation detailing the process and findings.
MA Visual Communication Major Project
Post Digital Futures
Specialist Skills
Visual Communication and Creative Direction
Modules offered may vary.
At MA level it is vital that you take an active role in structuring your own learning, and engage with the relevant methods and underpinning theories of your discipline.
Tutorials, seminars and workshops enable you to apply key learning principles to your day-to-day interactions. Individual support, provided by a personal tutor, is an integral feature of the learning and teaching strategy.
Research is also an intrinsic part of your study. You need to find and make sense of a wide variety of information from books, newspapers, journals, magazines, websites, archives and many other sources, then analyse and discuss your findings to inform the creative process. Lectures and briefings introduce topics and impart key aspects of disciplinary knowledge, usually to larger groups.
You develop your practical and professional skills with hands-on experience, informed by subject knowledge and critical understanding. Practical workshops introduce specific skills, followed by independent learning, project work, tutorials and critiques.
Critical reflection is key to all successful problem solving and is essential to the design process. You are expected to test and assess your solutions against design criteria which you develop in the light of your research.
We use a variety of assessment methods throughout the modules, as specified in the module handbooks. These are primarily in-course assessments, where you submit work during the module rather than sit timed exams at the end.
The modules are generally project based and primarily assessed through appraising a portfolio of work, often accompanied by a verbal presentation. Design work is largely developmental – you are assessed on the process you use to achieve your solutions as well as the result, so it is essential that you provide clear evidence of your development work.
When you graduate you have the opportunity to go on to a range of design-related employment, develop new enterprise propositions or receive project funding to take your ideas to market. Further study at doctoral level is also an option.
You must have a second class (2.2) honours degree or higher in an arts, design or other creative discipline. We also consider relevant experience or equivalent qualifications. Applicants are normally interviewed and required to present a portfolio of work.
For additional information please see the entry requirements in our admissions section
International applicants can find out what qualifications they need by visiting Your Country
For this particular course, you may need to undertake an occupational health or work-based risk assessment check. If you have a disability, specific learning difficulty, mental health condition, autism spectrum condition, sensory impairment or medical condition that may require reasonable adjustments during an external placement, in the university or in a clinical practice area, this must be declared as part of the enrolment process. If you are unsure you can contact the relevant admissions or course tutor for guidance.
* Subject to University approval
2019 entry
2019 entry


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I wanted to break up the typical posts with a little bit of silly information about myself. I thought a 5 things you didn't know about me type of post would break the ice! 1. I'm afraid of aliens - as a child, I would stay awake hours in bed imagining that aliens would land … Continue reading 5 things you didn't know about me
Category: me
Don't call this a diet
At the end of March/beginning of April, I went through some personal issues that turned my own image of myself upside down - one being the spontaneous eruption of (itchy!) tiny red dots all over my body (also known as this, that I'd never even heard of it before!) which impeded my motivation to blog, … Continue reading Don't call this a diet
A literary take on me
Four years ago when I wrote this post, I shared that I had finally begun working on Master's degree. It has been a very long journey (mostly because I have had to go to school part time and work full time) and it always took a toll on the blog. It was always me juggling … Continue reading A literary take on me
My heart in San Francisco
One year ago I packed up my car, put my cat in the passenger seat and drove away from Las Vegas (with the boyfriend in tow behind me driving a huge Uhaul). Twelve hours later, I was opening the door to what would be my new little apartment in San Francisco. I consider this a … Continue reading My heart in San Francisco

subset: commoncrawl


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"And I am sure of this, that he who began a good work in you will bring it to completion at the day of Jesus Christ" (Philippians 1:6).
I received a little more this morning in line with the thoughts I shared last night in Tell Me More about the Jesus You Serve. If you haven't already, please click on the link, and read the post, and consider it alongside this one.
I see our works as evidence of our faith, which is why I believe James declares that faith without works is dead (2:14-26). Related themes seen in Romans and Galatians, for example, Paul declares to the Ephesians that they had been saved by grace, which is a gift of God, not from themselves, so that no one can boast. Paul goes on to call them God's handiwork and that good works were prepared in advance for them to do (2:8-10). Our actions testify to our sanctification. It is God who justifies; his love transforms us into the likeness of his son. And there's so much more to Jesus than what we might be accustomed to reading on the side of a coffee mug, or finding in a feel-good, inspirational meme. There was a reason Jesus was nailed to a cross. There was a reason he caused his critics to gnash their teeth. There is a reason the world hated him then and continues to hate him today. Brothers and sisters, we must cast off the old, and embrace the whole person of Jesus Christ, or we embrace none of him.
I suspect the notion that "once saved always saved" allows continuing in sin is a lot like the controversy Paul was addressing in Romans 3. Paul's response in verse 8 leaves no doubt about his position on abusing God's abounding mercy: "Why not say—as some slanderously claim that we say—'Let us do evil that good may result?' Their condemnation is just!" "Do not be conformed to this world, but be transformed by the renewal of your mind, that by testing you may discern what is the will of God, what is good and acceptable and perfect" (Romans 12:2). I don't understand claiming to cherish love, while reviling repentance. Are they not one and the same? The message God has put on my heart is one of love, but not the fallen, distorted love the flesh revels.
Now don't become too discouraged if your conversion to Christ doesn't necessarily eliminate all your personal struggles immediately. Trust me, I've been there and done that. Your bad habits didn't form overnight, and they probably won't be broken any sooner. Our experiences effect us, and it takes time to relearn better ways. But put your faith in the Good Shepherd; he is more than up for the challenge. God plays a long game. "And I am sure of this, that he who began a good work in you will bring it to completion at the day of Jesus Christ" (Philippians 1:6). Therefore. let us not be content to remain as lost sheep meandering in the field, but rather, desire to grow and mature, because brothers and sisters we've been found! Pray without ceasing, because you will be tested, and yes, you're going to stumble, but keep pressing toward the prize. For it is in pressing toward the prize that you will be revealed! After all, Jesus doesn't just save souls, though that would be enough. Jesus changes lives!
To view all posts, click/press the link here to visit the Amazing Tangled Grace main page.
Please sign up to follow my blog and receive notifications of new posts by email. God bless!


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11 June 2014
The new Consumer Contracts Act provides a number of strict information requirements in relation to the consumer's right of cancellation of 14 days. If the consumer is not provided with the relevant information, this may have serious effect on the agreement.
If a company concludes an agreement on the Internet with a consumer concerning goods, subscriptions (e.g magazines) or the like, the company must provide information on the right of cancellation:
If the company does not meet these requirements, the consumer will have a right of cancellation of 12 months (and not just 14 days).
According to the new Act, the company will also be subject to a fine if providing insufficient information on the right of cancellation to the consumer.
This also applies to the delivery of digital content (e.g. VOD or SVOD), but in this connection, it is important to note that it is possible, according to some specific conditions, to waive the consumer's right to cancel such agreements.
The new Act also stipulates a number of detailed conditions as to how the order process must be, and how the consumer is now entitled to cancel the purchase of goods that have already been used.
Finally, the new Act carries on the rule that continuing subscriptions may only bind the consumer for five months, after which the consumer may terminate the subscription at one month's notice. An important change is that failure to observe the above will also result in fines.
Nina Henningsen
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You have to kiss many frogs before you met prince charming is a phrase I apply, blame my thirst for getting the best dad for my kids. After the ton of disappointment from the facebook guy, I felt like asking for lollipops I had given Julia although my ethic instincts pulled me back. It was time to get a better strategy in search of someone better. While awaiting the K.C.S.E. results, I enrolled for some short courses; I hear it works a great deal in avoiding instances of replacing the house girl at home.
I would get up every morning and head to school enthusiastically with the urge of coming across him. This was suggested by my dressing code, I had now divorced sneakers, bandanas and refrained from stealing my little sister's dresses replacing them with shorts and heels. In fact, I missed lunch and trekked to school, just to save in efforts of spicing up my wardrobe. From my research, it was crystal clear that men are attracted to ladies who dress in a sexy way; by all means I had to strive for such a look.
In class, I made sure to walk in when everyone was settled; this way my heels would work miracles in catching their attention. If this failed to work, I would fake a call and walk outside pretending how urgent it was. Indeed, the heels did wonders as the tutor got to know my name. For your information, he is a tall dark guy who seemed wealthy. The looks did not bother me, my target was his wealth.
During his lessons, I would build castles in the air on our family, `Mr. and Mrs. Omondi's family or if you like Omosh's family. My life would change for the better, getting free rides to school and therefore no worries of acquiring `maskwembe'. With the good reputation of the Luo community, he would spoil me like nonsense and this would get me out of the `mtumba' club to shopping in Mr. Price. I would try my best to answer questions in class even if it meant doing irrelevant answers, what mattered most is that my presence was recognized.
On his 1st day, he scribbled his digits on the board and said that he liked it when students notify him if one could not make it to class; otherwise, it broke his heart when he came to class only to find quarter of students absent. At times, I would miss classes intentionally as that would pave way for him getting my texts. The text would flow like `morning sir Omondi, it was my sincere wish to attend the 11am class but due to unavoidable circumstances sadly I will not be present. Yours sincerely, Mumbi'. This message would be sent after a hundred and one reviews, God knows how anxious I got waiting for the reply; he never disappointed. `Okay Mumbi, enjoy your day. Thanks for the update'.
The next day, it would be my pleasure of sharing with school buddies how Omondi texted me. This motivated me to shift 100%energy in his subject, doing extra work and taking for marking, the primary school way. Something told me this was an added bonga point to the existing points of my sexy shorts, classy mtumba heels not to leave out my great participation in class.
On Friday after my visit to the library, it was time to get home for lack of a plot. All that was in my head was Math on how I would trek to save for a boutique heel. Something made me shift from my line of thoughts, I had spotted my target in a hotel and there and then made up my mind to spend the sh. 50 I was planning to save on a smokie, it was the only thing that matched my cash to the menu. I almost hugged him on the realization that he was all alone, no lady in the picture.
Me: hey teacher, you also visit my favorite hotel? Wow! Anyway, I came to grab a dish here and spotted you.
Omondi: very good madam, I am honored.
Just then, a lady approached the table and Omondi was quick to speak, sweetheart (I almost fell due to happiness, although this lasted 15seconds) meet one of my students Mumbi, she came to buy a dish and decided to say hi. The sweetheart said that it was a pleasure meeting me, wishing me a nice weekend.
Why is the world so unfair? I thought my sexy shorts and mtumba heels would win him.
Find out whether the project of `boutique heels' pushed through in the next episode.


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I know this Valentine's Day we agreed to not do anything for each other. Not flowers, not candy, not cards -just a nice date night with our daughter by our side. But I ended up breaking that and got you a little something something anyway.
While I'm thinking about it, I have broken many "promises" over the last two years alone.
I had wls, my mom died, I petitioned for custody of my sister, you left for korea, lillie was diagnosed with scoliosis, I had a hypoglycemic incident then a panic attack, you came home, and we moved while trying to sell our house… these aren't excuses. Just facts. Big fat obstacles life has thrown at us in the last two years. There were things inbetween those obstacles that probably pissed me off more than anything else because it was easy to target you with my frustrations than face everything head on by myself for a good 391 days out of all of it.
So, out of anything I could give you yesterday, today, or tomorrow, it's this.
You have been there. Maybe not in the way I expect from you, but you are always there even when you're not physically there with us. When I was 325 lbs, and your coworkers were asking you why you were even with me, you said "because I love her. you don't leave people you love because of one tiny flaw." Do you know how good that felt? To know that you thought the giant issue I had with myself was so minuscule to you and that you actually told people that you loved me despite how much I hated myself?
When my mom passed away and I got the phone call in the middle of the Wendy's drive through, with my sister screaming "WHAT?! NO!", lillie asking if her grammy was ok, and me crying uncontrollably in the front seat, you stayed calm. You told the lady at the window that we needed to cancel our order and drove us all the way back to TN, over 6 hours of driving nonstop practically, without complaint. I don't know if I ever actually thanked you for that. For being there while I figured out how to plan for my mother's service. For basically letting me fall apart without actually falling apart because you were there.
Then shortly after mom passed, I don't even know if I completely talked with you about it or not, but I probably just told you it was happening that we were going to petition for custody of my sister and you drove us to the court house and we all filled out the papers. You were willing to take on another child, who wasn't either of ours biologically, but you went with it.
Six months later you had to leave. And I bitched about it for a good month before you had to board that plane. All the things I'd have to do by myself, all the things I would be solely responsible for, double the work load on me without you. Yet you let me. You still told me you loved me.
When Lillie was diagnosed with scoliosis, you cried. I didn't expect you to cry. I didn't think it would have affected you the way that it did. And it amazed me. How much you love our daughter and want the best for her.
When I had the incident in the car driving with lillie and my blood sugar dropped so low that I almost got into a car accident and the next day when I had a panic attack thinking about what could have happened, you told me it would be ok. That everything is ok.
You finally came home last November and things have been stressful since. Reintegration is hard, even still as we try to continue settling into our new home, while selling/renting out our old home. And night after night, you're still here to tell me that you love me.
There have been days I accused you of awful things, because that was what I was taught to do when I get mad. The men in my life before you either failed me as fathers or weren't allowed to be apart of my life. And it's not an excuse to be upset with you when my issues frustrate me or my mind overthinks about all of the past.
I have failed as a wife, and even as you read this, I know you'll tell me I haven't. But I have. I have not even come close to showing or expressing how much I love you nearly as much as you continue to do even when you're at your wits end with me. I have often told you that you'll be better off without me in your life because I don't value or think of myself as a productive member of the family (mostly when I have a terrible day). You still tell me you love me and you don't want anyone else. When every tdy or deployment has come up, I complain and bitch about it like no one would believe. I HATE being without you. Absolutely hate it. And you still tell me that you'd rather be home as well, but it's your job. And now, as we try to figure out how to live with each other again after being away for so long, you still love me. I could not be more thankful for you. For not giving up on me. For not letting me walk away when I'm frustrated with myself. For encouraging me to be whatever I want to be in life. For being a good dad. For being a loving, brave, strong, and handsome husband.
No one's marriage is perfect.
We are not perfect. Life is not perfect. But it's always worth it. One day, we will look back at all this craziness and laugh. We will ask ourselves how we did it and I'll always tell you "because you kept your promise to always love me" and that is all I'll need.
I can't promise you that I won't get frustrated still. I can't promise you that I still won't fail as a wife. But, I can promise and will work on being better. Showing you how much you do matter to me. Focusing on being a better team member in our relationship. I know it won't happen overnight. I'm pretty sure that I'll always be a work in progress. But you knew that already and you've never made me feel bad about the surreal amount of baggage that I still carry with me all these years. I will be here for you when you need me. I do support all that you want to do even though it frustrates me to no end that you have to be gone a lot. It's all worth it. Every single bit of this crazy life. The good, the bad, the ugly, and the occasional wondrous parts of it, because we have you.


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Bitcoin explained like youre five
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|Soapier at 28.04.2018 в 03:53
Lovely ass, but the title has me befuddled. If the boyfriend is "sleeping" wouldn't he be woke up by the sound of ass cheeks being parted and her moaning? I'm a heavy sleeper but the sound and smell of sex wake me right outta bed lol
|Nicolas at 04.05.2018 в 05:40
i love it when guys think scenes in porn are actual relatable real life scenarios that they just havent had the fortune of finding out there them selves yet. Thats why porn is classified as FANTASY. Nobody is interesting or hot enouigh to keep the rapt attention of a group of guys over and over and ovefr again. Most faggots cant even manage having just one guy dedicated to them, but sure go ahead and reach for the stars and go get your group...lol
|Deafening at 05.05.2018 в 13:07
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|Cetorhinus at 25.05.2018 в 07:14
they should at least try to make this realistic. No muslim woman would walk around with her hair showing under her hijab, tattoos and bare feet. Fuckin stupid costume
|Cercis at 31.05.2018 в 21:20
|Greenlee at 01.06.2018 в 06:11
who she


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Billions and billions of clothing items are produced every year by the fashion industry to clothe the world. Most people love keeping up with the fashion trends, not giving any thought to the processes that go into producing the clothes. They do not know that the fashion industry is having a terrible impact on our fragile planet. Untreated toxic wastewater with lead, mercury and other toxins is dumped directly from textile factories into rivers. They are in fact one of the major polluters in the world, after the oil industry.
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Studying Anime Japanese - Basic Techniques A quick summary of this article: Watch lots of anime. Her teaching and research interests cover aspects of information and communications technologies in everyday life. Huge black tits com. Even if we exclude the limit cases-the sexual sadists and abusers-the Campaign can be seen as raising a major concern: that we are creating a technology for purposes based on unjust and abusive modes of human interaction that will perpetuate those practices.
What disturbs this columnist about the gate-crashers is their consummate self-absorption. Kira nerys nude. So I placed my order and sat impatiently waiting for the time to come when they would be shipped out and I could enjoy the creepy treasures. Boston General Boston's first album ranks among the fastest-selling debuts in rock history.
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If you have the ability to schedule a hazardous waste pickup for that old polish, do it. Due to a whole host of problems, she's condemned to Earth and made to reincarnate endlessly, always living an absolutely miserable life.
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It is difficult to imagine the tragic heroines, however, as having such practical considerations in mind. Finally, not all of the female longings and desires expressed in Harlequins are regressive.
When arguing against the necessity of baptism, they say these verses refer to Spirit baptism. Oedipus had come as a prince from Corinth, and saved Thebes by solving the riddle of the deadly Sphinx that menaced the city. So instead of going into denial and then give sexual predators a perfect excuse to hide and keep it a secret, isn't it common sense to educate children about their body so they know what's natural and what's abuse.
Articulate and pleasant, Levi--shown in the middle photo on the film's official website--responds to many of Fischer's questions, expressing enthusiasm for attending the camp so that he can become a more dedicated member of "God's army. For the record people can see through you and the ones that can't are just like you.
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As for the coordination, I really cannot emphasis enough that you do not have to organise the perfect occasion for everyone.
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Untitled - Rebecca Silus
Inspired by the Hennepin Avenue Bridge in Minneapolis, Silus uses landscape to explore concepts of history, time and memory.
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Jon Rahm let the other 17 players know he meant business last Friday down there at the Hero World Challenge hosted by none other than Tiger Woods.
All he did was shoot 63 and put everyone else on notice and he said after chewing up the Albany Golf Club that the cool trophy that depicts a tiger and the world, would look really good in his trophy case.
No, Jon Rahm does not lack for confidence.
He went into the Sunday finale tied at 13-under par with some pretty good competition — Ryder Cuppers Tony Finau and former Open champion Henrik Stenson.
Rahm set the tone early with four front-nine birdies. He basically doubled-up on Finau and Stenson, both were two-under but it was becoming apparent that it would be difficult at best to keep up with Rahm, who was playing like the Kansas City Chiefs offense.
It became no contest over that last nine with Rahm playing bogey-free while Finau and Stennie had some issues.
By day's end it wasn't even close. Rahm's 65 got him to 20-under, four better than runner-up Finau, five clear of third-place Justin Rose and six better than Stenson, who tied down a solo fourth.
That's the way Rahm can rock and roll. Keep in mind this kid just turned 24 and his confidence suitcase is packed really, really tight.
This was his first time at Tiger's event. Tiger saw Rahm up close and personal the final day of the Ryder Cup in France, got dusted by Rahmbo in the singles.
Still, Rahm was gutsy enough to ask the host for a little help earlier in the week.
"He was putting next to me and I asked him, because I'm not the best putter on Bermuda grass and I know it's kind of funny to say that now," Rahm recounted. "I said, 'Hey, Tiger can you please teach me how to putt on Bermuda?'" Woods laughed and offered some passing advice, "He told me it's all about feel," Rahm recalled.
When Tiger handed the Tiger Trophy to Rahm on Sunday, he couldn't resist:
'That's a pretty good performance for somebody who can't putt on Bermuda greens," said Woods to Rahm.
Rahm credited his clubs:
"I did tell him, 'My man, you should look at my stats because it was all ball-striking. I think I made one putt outside 10 feet, and I didn't miss a shot the last three days, it's as simple as that. But I don't think he believed me really."
Yeah, Tiger knows best. But we all know that when Rahm gets on a roll, he's beyond good, world-class good.
It was here last year that after he won, everyone was talking about Rickie Fowler using the victory as a launching pad for a great season. Didn't happen and Rickie went winless.
You get the feeling that won't happen with Jon Rahm. Rahm's fiery, emotional and he basically crushed a world-class field down there at that beautiful Tavistock-built luxury project.
He's in the same boat as everyone else, however. They see the host, they see 14 major championships and 80 PGA Tour wins.
Then they see the 125-foot yacht in the Albany marina and know that they're all chasing what probably cannot be caught.


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When Did They Repaint WalMart?
Apparently, after so many years of the subtly pleasing Earth-tone tan color schemes, WalMart is shaking things up with a new, brighter color pallet. It's something we've all slowly noticed over the last couple of years, a pop of orange here, a little off-yellow there. It's the facelift the retail giant needed to 'class' it up a little bit.
All the same, you can definitely tell someone didn't put a lot of thought into these colors. The blue is ok, if a little on the bright side, as if it isn't quite deep enough... But that mass of hyper-white edging the design out doesn't seem like a color anyone put any actual thought into. If you've lived in SWOK long, you already know where this is going...
Not today, and most likely not until next Spring, that pretty bright-white will almost have a yellow/orange glow to it. Not in a pleasing, subtle way... but in a 'wow that's dingy' way.
Why? We live in a sea of red dirt. A uniquely identified sticky clay that, well, sticks to everything. And it's everywhere around us most of the year. Floating in the air we breath, mocking us as we wash our cars... and now, setting sights on this bright, new behemoth out on the West side of Lawton.
Don't get me wrong, it doesn't make me upset... as if I were attached to what everyone still calls 'The New WalMart.' It just makes you wonder if these were the colors they had overstocked in the paint department.
Keep in mind, this wasn't a local decision that took place. A bad round of ideas at a local watering hole... This was a corporate decision straight out of Arkansas. I honestly can't tell if I just didn't notice sooner, or if it was done very quickly in between grocery trips. Either way, I don't like it.


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A reply to Fred's OPE-L 4992. He wrote: "your interpretation explicitly abstracts from market prices. Even if changing market prices were another possible cause of changes in prices of production, this would not explain why your prices of production change." No Fred, this isn't so, and that's why the Chapter 6 evidence matters, as does your invention of an evil twin profit rate to immunize your interpretation against it. First, you confuse an interpretation with a specific illustration of the interpretation under particular circumstances. The TSSI does not abstract from market prices. Quite the contrary. I will be happy to provide references if you need them. Second, the specific illustration (in Kliman and McGlone's 1988 paper) does not abstract from market prices either. What we wrote, and what you quoted, is that it abstracts from market price *oscillations*. That is, as the CONTEXT in which the quote appears makes clear, we depicted only the movement of values and production prices in our periods 1 through 14. We did not depict the movement of market prices around the prices of production during that time-span. For instance, in period 1, prices of production are assumed to reign, and they are assumed to reign in period 2 as well. But between periods 1 and 2, i.e., in period 1.1, 1.2, 1.3, etc. (keep in mind that the length of a period is arbitrary), market prices can deviate from prices of production. We chose not to depict such situations; that is all. There is, however, a different sense -- a sense quite pertinent to the present discussion -- in which we did not abstract from market prices. One way of putting this is that we did not abstract from a change in market prices *before* period 1 commences. Keep in mind that market prices are just prices at which things are actually bought and sold. Thus prices of production are particular market prices, and "value prices" are other particular market prices. Now, what we depicted was a situation in which, prior to period 1, the output prices were one set of market prices -- "value prices" -- and in period 1, the output prices were a different set of market prices -- production prices. Hence, market prices changed between period 0, let's call it, and period 1. Those changes in market prices caused the general rate of profit, values, and production prices to change and to keep changing. Marx nowhere wrote, to my knowledge, that the effects of the initial change in market prices all had to be immediate, that there couldn't be lagged effects. As Allin notes, that is all that is going on in the illustration in question. Ciao, Andrew ("Drewk") Kliman Dept. of Social Sciences Pace University Pleasantville, NY 10570 USA phone: (914) 773-3968 fax: (914) 773-3951 Home: 60 W. 76th St. #4E New York, NY 10023 USA "The practice of philosophy is itself theoretical. It is the critique that measures the individual existence by the essence, the particular reality by the Idea."
This archive was generated by hypermail 2b30 : Thu Mar 01 2001 - 14:01:39 EST


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Spurred by soaring demand for its X5 sport-utility, BMW will likely expand its only U.S. manufacturing plant in Spartanburg, S.C., BMW's worldwide production chief said Monday.
Since establishing the plant at Spartanburg in 1996 to build the X3 roadster, BMW has seen its US sales grow from 50,000 to 200,000 units a year.
Norbert Reithofer, member of BMW's management board, worldwide production, said a large part of that success was down to the decision to locate a plant in the United States. That demonstrated BMW's commitment to the market and sales improved as a direct result.
"Plants are not unimportant for markets," said Reithofer at the Automotive News Europe Congress in Gothenburg, Sweden, on Monday. "The U.S.A. is very important and we will think to expand at Spartanburg."
BMW also is building new plants in Goodwood, England, and Leipzig, Germany. Asked where else BMW might decide to establish new production facilities, Reithofer said: "We have some ideas. We will probably look toward Asia. We already have a CKD [complete knockdown] plant in Thailand."
Reithofer contrasted the state-of-the-art plant BMW is building in Leipzig with the comparatively low-tech Goodwood facility, where it will make a new generation of Rolls-Royces.
"The new home of Rolls-Royce only has two robots," he said. "High-quality wood and leather work will be carried out by hand, following Rolls-Royce tradition. This just proves there can be no single model for a factory of the future."
In contrast to Goodwood, the sleek Leipzig plant won't be dedicated to the production of a single model.
Reithofer said it was essential that BMW -- and other carmakers -- continue to help struggling suppliers such as Germany's Peguform.
"We have to learn to manage our networks and keep them alive," he said. "The auto industry cannot be allowed to have weak suppliers. We have to think more as OEMs about how to manage medium-size suppliers with 3,000 to 10,000 employees."
A slide Reithofer showed during his presentation appeared to indicate plans for a sport-utility version of the Mini.
In response to a question from the Congress audience about Mini brand extensions, Reithofer said: "We have a diesel coming next, then a convertible. After that, we may have to think about the concepts you mention."
Reithofer isn't worried about the future BMW 1 series competing with the Mini.
"The Mini is front-wheel drive and is in a different league," he said.
BMW is a family owned business and Reithofer was asked if the German carmaker might consider a merger in the future.
"We are not interested in a merger," he said. "But cooperations are always important."


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Multipliers of Lorentz spaces
- Ashlyn Bryant
- 3 months ago
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1 Multipliers of Lorentz spaces By Yann-Kunn CHEN Hang-Chin LAI* 1. Introduction \mathfrak{i} \Gamma Let G be locally compact abelian groups in Pontrjagin duality \lambda respectively with Haar measures \eta such that the Plancherel theorem holds. In this paper we investigate the multipliers on Lorentz spaces L(p, q)(g), 1\leq p\leq\infty, 1\leq q\leq\infty. For convenience, we review briefy in section 2 what we need on the fundamental theory of L(p, q) spaces. Many propositions concerning {}^{t}l(p, q) spaces one can refer to Hunt [2], Blozinsky [1], Yap [8] [9]. In this paper, it is essentially to investigate the multipliers of L(p, q) in which the identity \mathfrak{m}(l^{1}(g), F(G))\cong F(G) will be true for the cases of F(G)_{-}^{-}L\}p, q)(g) or F(G)=A(p, q)(g) defined in Yap [8]. It follows that the identity \mathfrak{m}(l^{1}(g), A^{p}(G))\cong A^{p}(G) for 1\leq p\leq 2 in Lai [4 ; Proposition 5.2] will be a consequence in this paper. Further we would give an an\delta swer for the question risen in [4]. 2. Preliminaries on L(p, q) spaces DEFINITIONS. Let f be a measurable function defined on a measure space (X, \mu). We assume that the functions f are finite valued almost everywhere for y>0, \mu\{x\in X; f(x) >y\}<\infty The distribution function of f is defined by \lambda_{f}(y)=\mu\{x\in X; f(x) >y\}, y>01 The (nonnegative) rearrangment of f is defined by ]^{*}(t)= \inf\{y>0 ; \lambda_{f}(y)\leq t\}=\sup\{y>0 ; \lambda_{f}(y)>t\}, t>01 The average function of f is defined by f^{**}(t)= \frac{1}{t}\int_{0}^{t}f^{*}(s)ds, t>0. *) The second author was partially supported by the Mathematics Research Center, Taiwan, Republic of China.
2 248 Y. K. Chen H. C. Lai Note that \lambda_{f}(\cdot), f^{*}(\cdot), f^{**}(\cdot) are nonincreasing right continuous functions on (0, \infty) (cf. Hunt [2]). The Lorentz space denoted by L(p, q)(x, \mu), for brevity by L(p, q), is defined to be the collection of a1!f such that where f_{(p,q)}^{*}<\infty f_{(p,q)}^{*}= ( \frac{q}{p}\int_{0}^{\infty}[t^{1/p}f^{*}(t)]^{q}\frac{dt}{t})^{1/q}, 0<p<\infty, 0<q<\infty \backslash \sup_{t>0}t^{1/p}f^{*}(t), 0<p\leq\infty, q=\infty It follows from Hunt [2 ; p. 253] that f_{(p,p)}^{*}=f_{p} if 0<q_{1}\leq q_{2} \leq\infty f_{(p,q_{2})}^{*}\leqf_{p,q_{1})}^{*}\backslash, 0<p<\infty, then holds so. L(p, q_{1})\subset L(p, q_{2}) Evidently L(p,p)=L^{p} algebraically, in particular, if, then L(p,p) 1\leq p\leq\infty is isometrically isomorphic to where L(p,p) takes L^{p} _{(p,p)}^{*} as its norm. \lambda^{-} Now we consider to be a locally compact Hausdorff space \mu is a positive Borel measure, then it can be shown that: every function in L(p, q)(x, \mu) is locally integrable if only if any one of the cases p= 1=q ; p=\infty=q or 1<p<\infty, 1\leq q\leq\infty holds. Throughout we will assume that X is locally compact Hausdorff space with positive Borel measure \mu, that function f in L(p, q) is locally integrable. It is known that the functional endows a topology in L(p, q) _{(p,q)}^{*} such that L(p, q) is a topological vector space, the limit f_{n}arrow f in this topology means that (see Hunt [2; p. 257]). f_{n}-f_{(p,q)}^{*}arrow 0 We also intr0- duce (see Hunt [2]) the following function. \sup_{\mu(e)\geq t}\frac{1}{\mu(e)}\int_{e} f(x) d\mu, t\leq\mu(x) E\subset X f^{\star\star}(t)= \frac{1}{t}\int_{x} f(x) d\mu, t>\mu(x), f_{(p,q)}^{\star}=f^{\star\star}_{(p,q)}^{*} f_{(p,q)}=f^{**}_{(p,p)}^{*_{1}} In [2; p. 258], Hunt proved that L(p, q) is a Banach space under the _{(p,q)}^{\star} norm for 1<p<\infty, 1\leq q\leq\infty. Moreover f_{(p,q)}^{*} \leqf_{(p,q)}^{\star}\leqf^{*}_{(p,q\rangle}^{\star}\leq\frac{p}{p-1}f_{(p,q)}^{*}). It follows immediately that PROPOSITION If 1<p<\infty, 1\leq q\leq\infty _{(p.q)}^{\star} _{(p,q)}^{*} then are equivalmt norms on L(p, q) hence L(p, q) is a Banach space under _{(p.q)}^{\star} _{(p,q)} _{(p,q)}^{*} the noms or which induces the same topology as does.
3 Multipliers of Lorentz spaces 249 _{(p,q)} PROOF. It is immediately that is a norm in L(p, q). By previous inequalities, it is sufficient to show that f_{(p,q)}^{*} \leqf_{(p,q)}\leq\frac{1}{p-1}f_{(p,q)}^{*} Indeed, if q\neq\infty since f^{*}\leq f^{**}=(f^{*})^{**})=f^{**})^{*}, it follows that f_{(p,q)}^{*} \leqf_{(p,q)}=(\frac{q}{p}\int_{0}^{\infty}[t^{1/p}f^{**}(t)]^{q}\frac{dt}{t})^{1/q} =( \frac{q}{p}\int_{0}^{\infty}[\int_{0}^{t}f^{*}(x)dx]^{q}t^{(q/p)-1-q}dt)^{1/q} \leq\frac{p}{p-1}(\frac{q}{p}\int_{0}^{\infty}f^{*}(x)^{q}x^{t/pq)-1}dx)^{1/q} (Hardy inequality) = \frac{p}{p-1}f_{(p,q)}^{*} If q=\infty, we have f_{(p,\infty)}^{*}= \sup_{t>0}t^{1/p}f^{*}(.t)\leq\sup_{t>0}t^{1/p}f^{**}(t)=f_{(p,\infty)} = \sup_{t>0}t^{(1_{j} p)-1\int_{0}^{t}f^{*}(s)ds\leq\sup_{t>0}}t^{(1\prime_{p)-1\int_{0}^{t}s^{-1/p}f_{(p.\infty)}^{*}ds}} = \frac{p}{p-1}f_{(p,\infty)}^{*}, this shows f_{(p.\infty)}^{*}. \leqf_{(p.\infty)}\leq\frac{p}{p-1}f_{(p,\infty)}^{*} Q.E.D. From now on we shall consider that the space L(p, q) endows the _{(p,q)} norm for 1<p<\infty, 1\leq q\leq\infty. In the cases p=1=q p=\infty=q, are already the complete norms of L(1,1) L(\infty, _{(1,1)}^{*} _{(\infty,\infty)}^{*} \infty), thus for the discussion of L(1,1) L(\infty, \infty), we mean that they have complete norms respectively. By Hunt [2 ; pp. _{(1,1)}^{*} _{(\infty,\infty)}^{*} ], we have PROPOSITION (i) For any one of the cases p=1=q;p=\infty=q or 1<p<\infty 1\leq q\leq\infty, the space L(p, q) is a Banach space with respect to the _{(p,q)} norm. (ii) The conjugate space L(p, 1) is L(p, \infty) where 1/p+1/p =1. The conjugate space L(p, q), 1<p<\infty, 1<q<\infty, is L(p, q ) where 1/q+1/q =1 hence they are reflexive. The dual pair is of the form \langle f, g\rangle=\int fgd\mu, f\in L(p, q), g\in L(p, q ).
4 be 250 Y. K. Chen H. C. Lai 3. Multiplierson L(p, q)(g) spaces Let A be a commutative normed algebra Ba (two sided) A-module normed linear space. We denote by \mathfrak{m}(a, B) the set of all bounded linear mappin.g. T:Aarrow B such that T(ab)=a(Tb) for any a, b\in A1 Each element is called the multiplier T\in \mathfrak{m}(a, B) \otimes f A to B. Under the usual convolution, the space L^{p}(G) for 1\leq p\leq\infty A^{p}(G) for 1\leq p\leq\infty are -module it was known that L^{1} \mathfrak{m}(l^{1}(g), L^{1}(G))\cong M(G) \mathfrak{m}(l^{1}(g), L^{p}(G))\cong L^{p}(G) 1<p<\infty \mathfrak{m}(l^{1}(g), A^{p}(G))\cong A^{p}(G) 1\leq p\leq 2 where G is a locally compact abelian group, M(G) denotes the bounded regular measures A^{p}(G) is, with the norm f^{p}= \max(f_{1}, \hat{f}_{p}), a Banach algebra of all functions f \hat{f} in L^{1}(G) whose Fourier transforms being to L^{p}(\Gamma). Blozinsky ] 1] proved that the usual convolution * on simple function space S_{0}(G) can be uniquely extended to L^{1}*L(p, q) where every function in L(p, q) is locally integrable hence L(p, q)(g) is an -module Banach L^{1} space with respect to. _{(p,q)} Let A(p, q)(g) be the subspace of L^{1}(G) with Fourier transforms in L(p, q)(\gamma) provided each function in L(p, q)(\gamma) is locally integrable. For every fea(p, q)(g), we supply a norm in A(p, q) by f_{a(p,q)}= \max(f_{1}, \hat{f}_{(p,q)}) This norm is equivalent to the sum norm. In particular, if f_{1}+\hat{f}_{(p,q)} p=q, then A(p, q)=a^{p}. In [8], Yap showed that A(p, q)(1<p<\infty, 1\leq q\leq\infty) is a Segal algebra with respect to the sum norm, so it is also a Segal algebra with respect to our given norm. In this section we investigate that whether the following identities hold \mathfrak{m}(l^{1}(g), L(p, q)(g))\cong L(p, q)(g) \mathfrak{m}(l^{1}(g), A(p, q)(g))\cong A(p, q)(g) [ For convenient, we state some lemmas which are probably not all new. LEMMA Let \lambda Haar measure of G. Thm (i)\lambda_{f_{\delta}}=\lambda,, (ii) f_{s}^{*}=
5 y\epsilon\sigma_{\alpha} Multipliers of Lorentz spaces 251 f^{*}, f_{s}^{**}=f^{**}, f_{s}_{(p.q)}^{*}=f_{(p,q)}^{*} where f_{s}(x)=f(x-s). f_{s}_{(p,q)}=f_{(p,q)} PROOF. (i) Since \{x\in G; f_{s}(x) >t\}=\langle x\in G ; f(x-s) >t\}=\{y\in G; f(y) \lambda_{f_{g}}(t)=\lambda_{f}(t) >t\}+s, we have. (ii) This is a consequence of (i). Q.E.D. sarrow f_{s} LEMMA For every f\in L(p, q)(g), 1<p<\infty, 1\leq q<\infty, the mapping of G into L(p, q)(g) is continuous. PROOF. Since simple functions are dense in L(p, q)(g), it is sufficient to show that for any simple function f, sarrow f_{s} is continuous. Let f= \sum_{l\approx 1}^{n}k_{i} \chi_{b_{i}}, then f_{s}= \sum_{i\underline{-}1}^{n}k_{i}\chi_{e_{i}+s}. Now (f_{s}-f)^{**} \leq\sum_{i=1}^{n} k (\chi_{e_{i}+s}-\chi_{e_{t}})^{**} where \Delta 1 x\in(e_{i}+s)\delta E_{i} \chi_{e_{i}+s}-\chi_{e_{i}} (x)=\{ 0 otherwise, the character- \chi_{e} denotes the symmetric difference of sets is istic function on E. We have \lambda[e_{i}+s)\delta E_{i}], t<1 \lambda(\chi_{e_{i}+\epsilon}-\chi_{e_{t}}.)(t)=\{ 0 t\geq 1 This shows that Hence (\chi_{e_{i}+s}-\chi_{e_{i}})^{*}(t)=/1 0 t<\lambda((e_{i}+s)\delta E_{i}) t\geq\lambda((e_{i}+s)\delta E_{i}). \chi_{e_{i}+s}-\chi_{fi}_{(p,q)}^{*}=[\lambda((e_{i}+s)\delta E_{i})]^{1/p}arrow 0(sarrow 0). f_{s}-f_{(p,q)}^{*}arrow 0(sarrow 0) f_{s}-f_{(p,q)}arrow 0(sarrow 0) Q.E.D. LEMMA There is an approximate i&ntity \{a_{\alpha}\}ofl^{1}(g) such that a_{\alpha}_{1}=1 f*a_{\alpha}arrow f for every f\in L(p, q), 1<p<\infty, 1\leq q<\infty. It holds dso for f\in L(p, q), 1<p<\infty, 1<q<\infty. PROOF. For a proof of simple function follows immediately from \{U_{\alpha}\} Lemma Indeed, let be a decreasing neighborhood system at the origin in G, a_{\alpha} for each \alpha, we assume that is non-negative continuous U_{\alpha} function with support in such that. If \int_{g}a_{\alpha}(x)d\lambda(x)=1 g is a simple function, then g*a_{\alpha}-g_{(p,q)} \leq\int_{g}g_{y}-g_{(p,q)}a_{\alpha}(y)d\lambda(y) \leq\supg_{y}-g_{(p,q)}arrow.0 Lemma 3. 2)
6 is 252 Y. K. Chen H. C. Lai the limit being taken over the net of \alpha. As the simple fnctuions are dense in L(p, q), thus for any fel(p, q), there exists a simple function \epsilon>0 g such that f-g_{(p,q)}<\epsilon f*a_{\alpha}-f_{(p,q)}\leqf*a_{\alpha}-g*a_{\alpha}_{(p,q)} +g*a_{\alpha}-g_{(p,q)} +g-f_{(p,q)} \lim_{\alpha} <2\epsilon+g*a_{\alpha}-g_{(p,q)} sup f*a_{\alpha}-f_{(p,q)}\leq 2_{61} Since \epsilon arbitraty, we have f*a_{a}arrow f for every f\in L(p, q). Q.E.D. One of the main results is the following THEOREM The multiplier space \mathfrak{m}(l^{1}(g), L(p, q)(g)) is isomtric isomorphic to L(p, q)(g) for 1<p<\infty, 1<q<\infty. Moreover for any T\in \mathfrak{m}(l^{1}(g), L(p, q)(g)), there is a unique f in L(p, q)(g) such that Ta=a*f for every a\in L^{1}(G). PROOF. By Proposition 2. 2, we see that the dual pair is of the form \langle f, g\rangle=\int_{g}f(x)g(x)d\lambda(x), f\in L(p, q ), gel(p, q). Moreo\backslash ^{\vee}er for every \mathfrak{g}\in L (p, q ) there is a unique g\in L(p, q) such that \mathfrak{g}(f)=\int_{g}fgd\lambda for every f\in L(p, q ). Now if a\in L^{1}(G), f\in L(p, q ), we define, where a is the a\mathfrak{g}f=\tilde{a}*f reflexive function of a. Then it follows from Blozinsky [1; Theorem 2. 9] that L(p, q ) is an -module under L^{1} \Phi, naturaly if for every a\in L^{1}(G), \mathfrak{g}\in L (p, q ), we define a\otimes \mathfrak{g}(f)=\mathfrak{g}(a\mathfrak{g}f) for f\in L(p, q ) [ \otimes Then L (p, q ) is a!sol^{1}(g)-module under. Since there is an approximate identity in L^{1}(G) such that \{a_{\alpha}\} a_{\alpha}_{1}=1 a_{\alpha}\mathfrak{g}f-f_{(\mathcal{d},q )}=a_{\alpha}*\tilde{f}-\tilde{f}_{(p,q )}arrow 0 it follows from Liu Rooij [6; Lemma 2. 8] that hence \mathfrak{m}(l^{1}(g), L (p q )(G)\cong L (p, q )(G), \mathfrak{m}(l^{1}(g), L(p, q)(g))\cong L(p, q)(g). 1<p, q<\infty t We have to claim that the group algebra L^{\grave{i}}(G) acting on L (p, q ) corresponds to acting on L(p, q). That is for any a\in L^{1}(G), a\otimes \mathfrak{g} in
7 Multipliers of Lorentz spaces 253 L (p, q ) corresponds to a*g in L(p, q) whenever \mathfrak{g}\in L (p, q ) corresponds to g\in L(p, q). Indeed, for every f\in L(p, q ), a \otimes \mathfrak{g}(f)=\mathfrak{g}(a\oplus f)=\int_{g}(\tilde{a}*f)g d \lambda=\int_{g}f(a*g)d\lambda. Hence for each T \in \mathfrak{m}(l^{1}(g), L (p, q )(G)), there is a unique \mathfrak{g}\in L (p, q ) such that T a=a\otimes \mathfrak{g} for every. a\in L^{1}(G) This implies that for every T\in \mathfrak{m}(l^{1}(g), L(p, q)(g)) there is a unique g\in L (p, q) such that Ta=a*g for every. a\in L^{1}(G) Thus the theorem is proved. Q.E.D. In order to characterize the multipliers \mathfrak{m}(l^{1}(g), A(p, q)(g)), we define a space M(p, q)(g) as follows M(p, q)(g)=\{\mu\in M(G);\hat{\mu}\in L(p, q)(\gamma), 1\leq p\leq\infty, 1\leq q\leq\infty\} \hat{\mu} where is the Fourier Stieltjes transform of a bounded regular measure \mu in M(G). For every \mu\in M(p, q)(g) we supply a norm by \mu_{m(p,q)}=\max\{\mu, \hat{\mu}_{(p,q)}\}, 1\leq p, q\leq\infty Denote by M(p, q)(g) the space of M(p, q)(g) with the norm \mu_{u^{p}} In particular if p=q, we denote by M^{p}(G)=M(p,p)(G) with the norm = \max\{\mu, \hat{\mu}_{p}\}, \mu\in M^{p}(G). Note that M(\infty, \infty)(g)=m(g). _{M(p,q)}. THEOREM The space M(p, q)(g)(1\leq p\leq\infty, 1\leq q\leq\infty) is a commutative Banach algebra. PROOF. Evidently M(p, q)(g) is a normed linear space. Suppose that \{\mu_{n}\} \{\mu_{n}^{\nwarrow}\} is a Cauchy sequence in M(p, q)(g)\subset M(G). Then is also a Cauchy sequence in L(p, q)(\gamma). By the completeness of M(G) L(p, q) (\Gamma), there are \mu\in M(G) h\in L(p, q)(\gamma) such that \mu_{n}-\mu_{m(g)}arrow 0 \hat{\mu}_{n}-h_{(p,q)}arrow 0( \{\dot{\grave{\mu}}_{n_{o}}\} It follows from Yap [8; Lemma 2. 2] that there is a subsequence \{\mu_{n}\} which converges pointwise almost everywhere to h. Thus of \acute{\dot{\mu}}_{n}-\hat{\mu}_{\infty}\leq\mu_{n}-\mu_{m(g)}arrow 0 implies \hat{\mu}=h., this shows that M(p, q)(g) is complete with respect to the norm Further, for any \mu_{1}, \mu_{2} in M(p, q)(g), we have _{M(p,q)}.
8 254 Y. K. Chen H. C. Lai \mu_{1^{*}}\mu_{2}_{m(p,q)}=\max\{\mu_{1}*\mu_{2}_{m(g)}, \hat{\mu}_{1}\cdot\hat{\mu}_{2}_{(p,q)}\} \leq\max\{\mu_{1}\mu_{2}, \hat{\mu}_{1}\hat{\mu}_{2}_{(p,q)}\} \leq\mu_{1}_{m(p,q)}\mu_{2}_{m(p,q)}1 Hence M(p, q)(g) is a commutative Banach algebra. Q.E.D. It is immediately that A(p, q) is an ideal of M(p, q). The question arises that whether the space A(p, q) is a proper ideal of M(p, or not. q), Precisely we have the following THEOREM (i) If 1\leq q\leq p\leq 2, then M(p, q)=a(p, q) (ii) If p>2, 1\leq q\leq\infty, then M(p, q)\neq\supset A(p, q). PROOF. (i) For 1\leq q\leq p\leq 2, \mu\in M(p, q), we have \mu\in L(p, q)\subset L(p, p). This implies \mu\in M(p,p). It follows from Liu Rooij [6; Lemma 2. 3] that \mu is absolutely continuous, so that \mu\in A(p, q). For (ii), we consider G=T \Gamma=Z. Wiener Wintner [7] proved that there is a nonnegative singular measure \mu on T such that \mu(n)= 0(n^{-1 2+}.) for any. Thus for \epsilon>0 p>2,1\leq q\leq\infty, we choose a singular measure having the property of [7] showed. Then there are constants C_{1}>C_{2}>0 n_{0} such that so C_{1}, n <n_{0} \hat{\mu}(n) \leq f(n)=\{ C_{2} n ^{-1/2+\epsilon}, n \geq n_{0} f^{*}(t)=\{ C_{1}, t<2n_{0} C_{2}m^{-1/2+\epsilon}, 2m\leq t<2m+2, m\geq n_{0} f_{(p,q)}^{*}= \{\frac{q}{p}\int_{0}^{2n_{0}}c_{1}^{a_{x^{q/p-1}}}dx+\sum_{n=n_{0}}^{\infty}(c_{2}n^{-1./2+\epsilon})^{q}..((2n+2)^{q/p_{\sim}}-(2n)^{q/p})\}^{1/q}, = \{K_{1}+K_{2}\sum_{n=n_{0}}n^{-q^{1}2+\in q}(n+1)^{q/p}-n^{q/p})\}^{1/q}\infty where K_{1}, K_{2} are constants. If p>2,1\leq q<\infty, we have - \frac{q}{2}+\epsilon q+\frac{q}{p}-1 <-1 provided \epsilon(<\frac{p-2}{2pq}) is sufficient small, since n^{-q/2+8q}\cdot((n+1)^{q//}p-n^{q_{l}p)=0(n^{-q/2+\epsilon q+q p-1)}}, the series \sum_{n=p_{0}} converges. Hence f_{(p,q)}^{*}<\infty \mu_{(p,q)}<\infty proves \mu\in
9 have has Multipliers of Lorentz spaces 255 M(p, q) but \mu\not\in A(p, q). If p>2, q=\infty, we have f_{(p,q)}^{*}= \sup_{t>0}t^{1/p}f^{*}(t)<\infty. Hence \mu\in M(p, \infty), \mu\not\in A(p_{ },\infty). Q.E.D. REMARK. We do not know what happen for the case 1<p<q\leq 2. The following lemma is useful the proof follows immediately from the properties of Segal algebras. \{e_{\alpha}\} Lemma There is an approximate idmtity of A(p, q)(g), 1< p<\infty, 1\leq q<\infty, which is a bounded approximate identity of L^{1}(G) such e_{\alpha}_{1}\leq 1 \acute{e}_{\alpha} that compact support for all \alpha. We have seen in Lemma 3. 3 that there is an approximate identity of L(p, q)(g) which is the bounded approximate identity of. L_{1}(G) One can choose the approximate identity like as Lemma 3. 7 that the Fourier transforms have compact supports. For convenient, we state it as following \{e_{\alpha}\} L^{1}(\Gamma) LEMMA There is a bounded approximate i&ntity of \hat{e}_{\alpha} such that has cmpact support in G e_{\alpha}*f-f_{(p,q)}arrow 0 for any f\in L(p, q)(\gamma), 1<p<\infty, 1\leq q<\infty LEMMA (i) The space A^{1}(G) is contained in A(p, q)(g) for 1\leq p<\infty, 1\leq q\leq\infty. (ii) The Fourier transforms \hat{a^{1}(g}) A\hat{(p,q)}(G) are dmse in L(p, q) for (\Gamma) 1\leq p<\infty, 1\leq q<\infty. PROOF. Suppose that f=\hat{g}, g\in A^{1}G, then f\in L^{1}(\Gamma) \int_{0}^{\infty}f^{*}dt= \int_{\gamma} f d\eta<\infty. If p\geq q, we have \int_{0}^{\infty}x^{q^{j}p-1} f^{*}(x)dx\leqf\int_{0}^{1}x^{q/p-1}dx+\int_{1}^{\infty}x^{q/p-1}f^{*}(x)^{q}dx \leq\frac{q}{p}f_{\infty}+\int_{1}^{\infty}f^{*}(x)^{q}dx <\infty : so fel\{p,q) (\Gamma). If p<q, then fel\{p, p) \subset L(p, q). This shows that g\ni A(p, q) \hat{a^{1}(g} ) \subset A\acute{(p}\hat{q)(},G ) \subset L(p, q)(\gamma). It is sufficient to show that \hat{a^{1}(g}) is dense in L(p, q)(\gamma). \{e_{\alpha}\} Let is L^{1}(\Gamma) be an approximate identity \hat{e}_{\alpha} of L(p, q)(\gamma) such that compact support. Then for every f\in L(p, q) (\Gamma) any, \epsilon>0 \alpha_{0} there is such that e_{\alpha_{0}}*f-f^{(p,q)}<\epsilon l Since the simple functions are dense in L(p, q)(\gamma), thus for the given e> 0, there exists a simple function g such that f-g_{(p,q)}.<\epsilon,
10 256 Y. K. Chen H. C. Lai e_{\alpha_{0}}*g-f_{(p,q)}\leqe_{\alpha_{0}}_{1}g-f_{(p,q)}+e_{\alpha_{0}}*f-f_{(p,q)} <(C+1)\epsilon where e_{\alpha_{0}}_{1}\leq \acute{\acute{e}}_{\alpha_{0}}*g=\hat{e}_{a_{0}}\hat{g}\in C. But C_{c}(G)\subset L^{1}(G), we This shows that \hat{a^{1}(g} ) is dense in L(p, q)(\gamma). Q.E.D. We use the symboles appear in Liu Rooij [6]. 3. 9, Theorem 2. 4, [6 ; Lemma 2. 8], we obtain see that e_{\alpha_{0}}*g\in A^{1}(G). Thus by Lemma LEMMA Let H be the closures of \{(\tilde{f}, -\hat{f});f\in A^{1}(G)\} in C_{0}(G)\cross L(p, q )(\Gamma) Then J=\{(\mu, h) ; \mu\in M(G), h\in L(p, q)(\gamma), \int_{g}\tilde{f}d\mu=\int_{\gamma}\hat{f}hd\eta, f\in A^{1}(G)\} \{C_{0}(G)V_{H}L(p, q )(\Gamma)\} \cong M(G)\Lambda_{J}L(p, q)(\gamma), 1<p<\infty, 1<q<\infty If p=1, then q=1, the same result holds for C_{0}(\Gamma) \infty)(\gamma)=l^{\infty}(\gamma). in place of L(\infty, LEMMA For 1<p<\infty, 1<q<\infty or p=1=q, then M(G)\Lambda_{J}L(p, q)(\gamma)\equiv M(p, q)(g) (\mu,\acute{\grave{\mu}}) where any element \mu in M(p, q)(g) is regarded as the pair in L(p, q)(\gamma) so that M(p, q) is embedded in M(G)\cross L(p, q)(\gamma). M(G)\cross PROOF. If (\mu, h)\in J, then it follows from \hat{a^{1}(}g ) dense in L(F, q )(\Gamma) for \int_{\gamma}\hat{f}\hat{\mu}d\eta=\int_{\gamma}\hat{f}hd\eta f\in A^{1}(G) h=\hat{\mu} that. (In detail see Liu Rooij [6].) Q.E.D. It is clear that M(p, q)(g) is an L^{1}(G) -module under convolution thus we have following theorem. THEOREM For 1<p<\infty, 1<q<\infty or p=1=q, the multiplier algebra \mathfrak{m}(l^{1}(g), M(p, q)(g)) is isometricdly isomorphic to M(p, q)(g). Moreover for every multiplier T of L^{1}(G) into M(p, q)(g) can be expressed as the form PROOF. T(a)=a*\mu for some \mu\in M(p, q)(g) any a\in L^{1}(G). From Lemma Lemma 3. 11, we see that \{C_{0}(G)V_{H}L(p, q )(\Gamma)\} \cong M(p, q)(g). Using this identity [6; Lemma 2. 8], the theorem follows immediately. To this end, we have only to show that C_{0}(G)V_{H}L(p, q )(\Gamma) L^{1}\langle G)- is module under certain operation there is an approximate identity of
11 Multipliers of Lorentz spaces 257 L^{1}(G) with the property as in [6; Lemma 2. 8]. For any \mu\in M(p, q)(g), there corresponds an element U_{\mu}\in[C_{0}(G)V_{H} L(p, q)(\gamma)] defined by U_{\mu}(f, g)= \int_{g}fd\mu+\int_{\gamma}g\acute{\grave{\mu}}d\eta, (f, g)\in C_{0}(G)V^{H}L(p, q )(\Gamma). \otimes We will define an operation for which C_{0}(G)V_{H}L(p, q )(\Gamma) is an L^{1}(G)- \otimes module. This operation induces an operation over its dual space which is defined by a\otimes U_{\mu}(f, g)=u_{\mu}(a\otimes(f, g)) a\otimes(f, g)=(\tilde{a}*f, dg) for any (f, g)\in C_{0}(G)V_{H}L(p, q )(\Gamma) a\in L^{1}(G). At first we have to show [C_{0}(G)V_{H}L(p, q )(\Gamma)] M(p, q)(g) are the same space acted by L^{1}(G) under the operations \otimes *respectively. That is holds a\otimes U_{\mu}=U_{a*\mu} for any a\in L^{1}(G) \mu\in M(p, q)(g). In fact for (f, g)\in C_{0}(G)V_{H}L(F, q ) (\Gamma), we have U_{a\mu}(f, g)= \int_{g}f(x)d(a*\mu)(x)+\int_{\gamma}d\hat{\mu}gd\eta = \int_{g}f(x)a*\mu(x)d\lambda(x)+\int_{\gamma}\hat{a}\hat{\mu}gd\eta (for a*\mu\in L^{1}(G)), a\otimes U_{\mu}(f, g)=u_{\mu}(a\otimes(f, g))=u_{\mu} (a-*f, \^ag) = \int_{g}\tilde{a}*f(x)d\lambda(x)+\int_{\gamma}\hat{a}\hat{\mu}gd\eta = \int_{g}f(y)a*\mu(y)d\lambda(y)+\int_{\gamma}\hat{a}\hat{\mu}gd\eta =U_{a*\mu}(f, g). Now we show that C_{0}(G)V_{H}L(p, q )(\Gamma) is L^{1}(G) \otimes -module under. In fact, if (\tilde{f}, -\hat{f})\in H, then it is clearly, a\otimes(\tilde{f}.,-\hat{f})=\tilde{a}^{*}\tilde{f} -\^a \hat{f}) \in H. Further for (/, g)\in C_{0}(G)\cross L(p, q )(\Gamma), we have a\otimes(f, g)= (, \^ag)ll a\wedge*f = \inf\{f _{\infty}+g _{(p,q )} ; (f, g )\cong ( a\sim*f, \^ag) mod H\} = \inf\{\tilde{a}*f+\tilde{h}_{\infty}+\hat{a}g-\hat{h}_{(pq )},, ; h\in A^{1}(G)\}. \leq\inf\{\tilde{a}*f+\overline{a*k}+\hat{a}g-a*k_{(pq )},, ; k\in A^{1}(G)\} since a*k\in L^{1}*A^{1}(G)\subset A^{1}(G), \leq\inf\{a_{1}f+\hslash_{\infty}+d_{\infty}g-\hat{k}_{(p,q )} ; k\in A^{1}(G)\} \leqa_{1} inf \{f _{\infty}+g _{(p,q )} ; (f, g )\cong(f, g) mod H\} =a_{1}(f, g)
12 258 Y. K. Chen H. C. Lai Hence a\otimes(f, g)\leqa_{1}(f, g) To complete the proof, it remains to show that there is a bounded approximate identity of L^{1}(G) with \{e_{\alpha}\} e_{\alpha}_{1}\leq 1 such that e_{\alpha}\otimes(f, g)-(f, g)arrow 0 for every (/, g)\in C_{0}(G)V_{H}L(F, q )(\Gamma). Let \{e_{\alpha}\} be a bounded approximate identity of L^{1}(G) e_{\alpha}_{1}\leq with \hat{e}_{\alpha} Fourier transform has compact support such that 1 the e_{\alpha}*f-f_{a(pq )}"arrow 0 for every f\in A(p, q )(G) (Lemma 3. 7.) 1 Since A(\hat{p,q )}(G) is dense in L(p, q )(\Gamma), thus for any g\in L(p, q )(\Gamma) \epsilon>0 there is h\in A(\hat{p,q )}(G) such that g-h_{(pq )}"<\epsilon, it follows that \acute{e}_{\alpha}g-g_{(pq\prime)}"\leqq\hat{e}_{\alpha}g-\hat{e}_{\alpha}h,p q ) +p_{\alpha}h-h_{(pq )}"+h-g_{(pq )} Since for \check{h}\in A(p, q )(G) with <2\epsilon+\hat{e}_{\alpha}h-h_{(pq )l} \check{h}=h\wedge, \epsilon \hat{e}_{\alpha}h-h_{(pq )}"= \hat{ e_{\alpha}*}\check{h}-h_{(p\prime,q\prime)}\leqe_{\alpha}*\hslash-\check{h}_{a(p\prime,q\prime)}arrow 0 is arbitrary, we have, \acute{e}_{\alpha}g-g_{(p\prime,q\prime)}arrow 01 Therefore for (f, g)\in C_{0}(G)V_{H}L(p, q )(\Gamma), we have e_{\alpha}\otimes(f, g)-(f, g)=(\tilde{e}_{\alpha}*f-f,\acute{e}_{\alpha}g-g \leq\tilde{e}_{\alpha}*f-f_{\infty}+\acute{e}_{\alpha}g-g_{(p\prime,q\prime)} arrow 0, the limit being taken over all \alpha. This proof is completed. Q.E.D. THEOREM For 1<p, q<\infty or p=1=q, the identity \mathfrak{m}(l^{1}(g), A(p, q)(g))\cong M(p, q)(g) holds. Furthermore, if T\in \mathfrak{m}(l^{1}(g), A(p, q)(g)), there is a unique \mu\in M(p, q)(g) such that Ta=a*\mu for any a\in L^{1}(G). PROOF. Evidently, for any T\in \mathfrak{m}(l^{1}(g), A(p, q)(g)), it follows that T\in \mathfrak{m}(l^{1}(g), M(p, q)(g)). Conversely for T\in \mathfrak{m}(l^{1}(g), M(p, q)(g)), it follows from Theorem that there is a unique \mu\in M(p, q)(g) such that Ta=a*\mu,\cdot
13 Multipliers of Lorentz spaces 259 holds for every a\in L^{1}(G). Thus a*\mu_{(p,q)}=\hat{a}\hat{\mu}_{(p,q)}\leqd_{\infty}\hat{\mu}_{(p,q)} \leqa_{1}\mu_{m(p,q)}<\infty, since a*\mu\in L^{1}(G), we see that a*\mu\in A(p, q)(g). That is T\in \mathfrak{m}(l^{1}(g), A(p, q)(g)). Hence by Theorem again, we obtain \mathfrak{m}(l^{1}(g), A(p, q)(g))\cong \mathfrak{m}(l^{1}(g), M(p, q)(g))\underline{\simeq}m(p, q)(g). Applying Theorem 3. 6 Theorem 3. 13, we have COROLLARY (i) If (ii) If 2<p, 1<q<\infty, then 1<q\leq p<2 or p=1=q, then \mathfrak{m}(l^{1}(g), A(p, q)(g))\cong A(p, q)(g). \mathfrak{m}(l^{1}(g), A(p, q)(g))\cong M(p, q)(g)\neq A\overline{\sim}(p, q)(g). Q.E.D. REMARK 1. If p=q, 1\leq p<\infty, then Theorem 3. 6 Theorem (or Corollary 3. 14) induce for \mathfrak{m}(l^{1}(g), A^{p}(G))\cong A^{p}(G) 1\leq p\leq 2 \mathfrak{m}(l^{1} (G), for A^{p}(G))\cong M^{p}(G)\neq\supset A^{p}(G) 2<p<\infty. This answers the question risen in Lai [4]. The classical multiplier problem in L^{1}(G) is a special case in our context. That is \mathfrak{m}(l^{1}(g), L^{1}(G))\cong \mathfrak{m}(l^{1}(g), M(G))\cong M(G). REMARK 2. From Corollary (ii), one sees that, in general, if S(G) is a proper Segal algebra, the multiplier algebra \mathfrak{m}(l^{1}(g), S(G)) needs not be isometrically isomorphic to S(G) itself. References Institute of Mathematics National Tsing Hua University Taiwan, Republic of China [1 ] A. P. BLOZINSKY: On a convolution theorem for L(p, q) spaces, Trans. Amer. Math. Soc, 164 (1972), A. P. BLOZINSKY: Convolution of L(p, q) functions, Proc. Amer. Math. Soc. 32 (1972), [2] R. A. HUNT: On L(p, q) spaces, Enseignement Math., 12 (1966), [3] H. C. LAI: On some propenies of A^{p}(G) -algebras, Proc. Japan Acad., 23 (1969), [4] H. C. LAI: On the multipliers of A^{p}(G) algebra, Tdhoku Math. J., 23 (1971),
14 260 Y. K. Chen ond H. C. Lai [5] R. LARSEN: An introduction to the Theory of Multipliers, Springer-Verlag Berlin Heidelberg [6] T. S. L1U A. C. M. Van ROOIJ: Sum intersections of normed linear spaces, Math. Nachr., 42 (1969), [7] N. WIENER A. WINTNER: Fourier-Stieltjes transforms singular infinite convolution, Amer. J. Math., 60 (1938), [8] L. Y. H. YAP: On two classes of subalgebras in L_{1}(G), Proc. Japan Acad., 48 (1972), [9] L. Y. H. YAP: Some remarks on convolution operators L(p,q) spaces, Duke Math. J., 36 (1969), [10] L. Y. H. YAP: On the impossibility of representing certain functions by convolutions, Math. Sc., 26 (1970), (Received January 15, 1974)


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