Connective constant: Difference between revisions

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correct polynomial -- the one given previously was the polynomial for the critical point, not the connective constant
 
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In [[mathematics]], '''singular integral operators of convolution type''' are the [[singular integral operator]]s that arise on '''R'''<sup>''n''</sup> and '''T'''<sup>''n''</sup> through convolution by distributions; equivalently they are the singular integral operators that commute with translations. The classical examples in [[harmonic analysis]] are the [[harmonic conjugate|harmonic conjugation operator]] on the circle, the [[Hilbert transform]] on the circle and the real line, the [[Beurling transform]] in the complex plane and the [[Riesz transform]]s in Euclidean space. The continuity of these operators on ''L''<sup>2</sup> is evident because the [[Fourier transform]] converts them into [[Fourier multiplier|multiplication operators]]. Continuity on ''L<sup>p</sup>'' spaces was first established by [[Marcel Riesz]]. The classical techniques include the use of [[Poisson integral]]s, [[interpolation theory]] and the [[Hardy–Littlewood maximal function]]. For more general operators, fundamental new techniques, introduced by [[Alberto Calderón]] and [[Antoni Zygmund]] in 1952, were developed by a number of authors to give general criteria for continuity on ''L<sup>p</sup>'' spaces. This article explains the theory for the classical operators and sketches the subsequent general theory.
Four or five years ago, a reader of some of my columns bought the domain name jamesaltucher.com and gave it to me as a birthday gift. It was a total surprise to me. I didn't even know the reader. I hope one day we meet.<br>Two years ago a friend of mine, Tim Sykes, insisted I had to have a blog. He set it up for me. He even wrote the "About Me". I didn't want a blog. I had nothing to say. But about 6 or 7 months ago I decided I wanted to take this blog seriously. I kept putting off changing the "About Me" which was no longer really about me and maybe never was.<br>A few weeks ago I did a chapter in one of the books in Seth Godin's "The Domino Project". The book is out and called "No Idling". Mohit Pawar organized it (here's Mohit's blog) and sent me a bunch of questions recently. It's intended to be an interview on his blog but I hope Mohit forgives me because I want to use it as my new "About Me" also.<br>1. You are a trader, investor, writer, and entrepreneur? Which of these roles you enjoy the most and why?<br>When I first moved to New York City in 1994 I wanted to be everything to everyone. I had spent the six years prior to that writing a bunch of unpublished novels and unpublished short stories. I must've sent out 100s of stories to literary journals. I got form rejections from every publisher, journal, and agent I sent my novels and stories to.<br>Now, in 1994, everything was possible. The money was in NYC. Media was here. I lived in my 10�10 room and pulled suits out of a garbage bag every morning but it didn't matter...the internet was revving up and I knew how to build a website. One of the few in the city. My sister warned me though: nobody here is your friend. Everybody wants something<br>
 
And I wanted something. I wanted the fleeting feelings of success, for the first time ever, in order to feel better about myself. I wanted a girl next to me. I wanted to build and sell companies and finally prove to everyone I was the smartest. I wanted to do a TV show. I wanted to write books<br>
==L<sup>2</sup> theory==
But everything involved having a master. Clients. Employers. Investors. Publishers. The market (the deadliest master of all). Employees. I was a slave to everyone for so many years. And the more shackles I had on, the lonelier I got<br>
 
(Me in the Fortress of Solitude<br>
===Hilbert transform on the circle===
Much of the time, even when I had those moments of success, I didn't know how to turn it into a better life. I felt ugly and then later, I felt stupid when I would let the success dribble away down the sink<br>
{{See also|Harmonic conjugate}}
I love writing because every now and then that ugliness turns into honesty. When I write, I'm only a slave to myself. When I do all of those other things you ask about, I'm a slave to everyone else<br>
The theory for ''L''<sup>2</sup> functions is particularly simple on the circle.<ref>{{harvnb|Torchinsky|2004|pp=65–66}}</ref><ref>{{harvnb|Bell|1992|pp=14–15}}</ref> If ''f'' ∈ ''L''<sup>2</sup>('''T'''), then it has a Fourier series expansion
Some links<br>
 
33 Unusual Tips to Being a Better Write<br>
:<math>f(\theta)=\sum_{n \in \mathbf{Z}} a_n e^{in\theta}.</math>
"The Tooth<br>
 
(one of my favorite posts on my blog<br><br>
[[Hardy space]] H<sup>2</sup>('''T''') consists of the functions for which the negative coefficients vanish, ''a<sub>n</sub>'' = 0 for ''n'' < 0. These are precisely the square-integrable functions that arise as boundary values of holomorphic functions in the open unit disk. Indeed ''f'' is the boundary value of the function
2. What inspires you to get up and start working/writing every day<br>
 
The other day I had breakfast with a fascinating guy who had just sold a piece of his fund of funds. He told me what "fracking" was and how the US was going to be a major  [http://www.pcs-systems.co.uk/Images/celinebag.aspx http://www.pcs-systems.co.uk/Images/celinebag.aspx] oil player again. We spoke for two hours about a wide range of topics, including what happens when we can finally implant a google chip in our brains<br>
:<math>F(z)=\sum_{n\ge 0} a_n z^n,</math>
After that I had to go onto NPR because I firmly believe that in one important respect we are degenerating as a country - we are graduating a generation of indentured servants who will spend 50 years or more paying down their student debt rather than starting companies and curing cancer. So maybe I made a difference<br>
 
Then I had lunch with a guy I hadn't seen in ten years. In those ten years he had gone to jail and now I was finally taking the time to forgive him for something he never did to me. I felt bad I hadn't helped him when he was at his low point. Then I came home and watched my kid play clarinet at her school. Then I read until I fell asleep. Today I did nothing but write. Both days inspired me<br>
in the sense that the functions
It also inspires me that I'm being asked these questions. Whenever anyone asks me to do anything I'm infinitely grateful. Why me? I feel lucky. I like it when someone cares what I think. I'll write and do things as long as anyone cares. I honestly probably wouldn't write if nobody cared. I don't have enough humility for that, I'm ashamed to admit<br><br>
 
3. Your new book "How to be the luckiest person alive" has just come out. What is it about<br>
:<math>f_r(\theta)=F(re^{i\theta}),</math>
When I was a kid I thought I needed certain things: a [http://search.huffingtonpost.com/search?q=college+education&s_it=header_form_v1 college education] from a great school, a great home, a lot of money, someone who would love me with ease. I wanted people to think I was smart. I wanted people to think I was even special. And as I grew older more and more goals got added to the list: a high chess rating, a published book, perfect weather, good friends,  respect in various fields, etc. I lied to myself that I needed these things to be happy. The world was going to work hard to give me these things, I thought. But it turned out the world owed me no favors<br>
 
And gradually, over time, I lost everything I had ever gained. Several times.  I've paced at night so many times wondering what the hell was I going to do next or trying not to care. The book is about regaining your sanity, regaining your happiness, finding luck in all the little pockets of life that people forget about. It's about turning away from the religion you've been hypnotized into believing into the religion you can find inside yourself every moment of the day<br><br>
defined by the restriction of ''F'' to the concentric circles |''z''| = ''r'', satisfy
[Note: in a few days I'm going to do a post on self-publishing and also how to get the ebook for free. The link above is to the paperback. Kindle should be ready soon also.<br>
 
Related link: Why I Write Books Even Though I've Lost Money On Every Book I've Ever Writte<br>
:<math>\|f_r-f\|_2 \rightarrow 0.</math>
4. Is it possible to accelerate success? If yes, how<br><br><br>
 
Yes, and it's the only way I know actually to achieve success. Its by following the Daily Practice I outline in this post:<br>
The orthogonal projection ''P'' of ''L''<sup>2</sup>('''T''') onto H<sup>2</sup>('''T''') is called the '''Szegő projection'''. It is a bounded operator on ''L''<sup>2</sup>('''T''') with [[operator norm]] 1. By Cauchy's theorem
It's the only way I know to exercise every muscle from the inside of you to the outside of you. I firmly believe that happiness starts with that practice<br>
 
5. You say that discipline, persistence and psychology are important if one has to achieve success. How can one work on improving "psychology" part<br>
:<math>F(z)= {1\over 2\pi i} \int_{|\zeta|=1}  {f(\zeta)\over \zeta -z} \,d\zeta={1\over 2\pi} \int_{-\pi}^{\pi} {f(\theta) \over 1-e^{-i\theta}z} \, d\theta.</math>
Success doesn't really mean anything. People want to be happy in a harsh and unforgiving world. It's very difficult. We're so lucky most of us live in countries without major wars. Our kids aren't getting killed by random gunfire. We all have cell phones. We all can communicate with each other on the Internet. We have Google to catalog every piece of information in history!  We are so amazingly lucky already<br>
 
How can it be I was so lucky to be born into such a body? In New York City of all places? Just by being born in such a way on this planet was an amazing success<br>
Thus
So what else is there? The fact is that most of us, including me, have a hard time being happy with such ready-made success. We quickly adapt and want so much more out of life. It's not wars or disease that kill us. It's the minor inconveniences that add up in life. It's the times we feel slighted or betrayed. Or even slightly betrayed. Or overcharged. Or we miss a train. Or it's raining today. Or the dishwasher doesn't work. Or the supermarket doesn't have the food we like. We forget how good the snow tasted when we were kids. Now we want gourmet food at every meal<br>
 
Taking a step back, doing the Daily Practice I outline in the question above. For me, the results of that bring me happiness. That's success. Today. And hopefully tomorrow<br>
:<math>\displaystyle{F(re^{i\varphi})={1\over 2\pi} \int_{-\pi}^{\pi} {f(\varphi-\theta) \over 1-re^{i\theta}} \, d\theta.}</math>
6. You advocate not sending kids to college. What if kids grow up and then blame their parents about not letting them get a college education<br>
 
I went to one of my kid's music recitals yesterday. She was happy to see me. I hugged her afterwards. She played "the star wars theme" on the clarinet. I wish I could've played that for my parents. My other daughter has a dance recital in a few weeks. I tried to give her tips but she laughed at me. I was quite the breakdancer in my youth. The nerdiest breakdancer on the planet. I want to be present for them. To love them. To let them always know that in their own dark moments, they know I will listen to them. I love them. Even when they cry and don't always agree with me. Even when they laugh at me because sometimes I act like a clown<br>
When ''r'' = 1, the integrand on the right hand side has a singularity at θ = 0. The '''truncated Hilbert transform''' is defined by
Later, if they want to blame me for anything at all then I will still love them. That's my "what if"<br>
 
Two posts<br>
:<math>\displaystyle{H_\varepsilon f(\varphi) = {i\over \pi} \int_{\varepsilon\le |\theta| \le  \pi} {f(\varphi-\theta) \over 1-e^{i\theta}} \, d\theta={1\over \pi} \int_{|\zeta-e^{i\varphi}|\ge \delta} {f(\zeta)\over \zeta-e^{i\varphi}}\, d\zeta,}</math>
I want my daughters to be lesbian<br>
 
Advice I want to give my daughter<br><br><br>
where δ = |1 – ''e''<sup>''i''ε</sup>|. Since it is defined as convolution with a bounded function, it is a bounded operator on L<sup>2</sup>('''T'''). Now
7. Four of your favorite posts from The Altucher Confidential<br>
 
As soon as I publish a post I get scared to death. Is it good? Will people re-tweet? Will one part of the audience of this blog like it at the expense of another part of the audience. Will I get Facebook Likes? I have to stop clinging to these things but you also need to respect the audience. I don't know. It's a little bit confusing to me. I don't have the confidence of a real writer yet<br>
:<math>\displaystyle{H_\varepsilon{1}={i\over\pi}\int_\varepsilon^\pi 2 \Re (1-e^{i\theta})^{-1} \, d\theta ={i\over\pi}\int_\varepsilon^\pi 1 \, d\theta = i - {i \varepsilon\over \pi}.}</math>
Here are four of my favorites<br>
 
How I screwed Yasser Arafat out of $2mm (and lost another $100mm in the process<br>
If ''f'' is a polynomial in ''z'' then
It's Your Fault<br>
 
I'm Guilty of Torturing Wome<br>
:<math>\displaystyle{H_\varepsilon f(z) - {i(1-\varepsilon)\over \pi} f(z)={1\over \pi i} \int_{|\zeta -z|\ge \delta} {f(\zeta)-f(z)\over \zeta -z} \, d\zeta.}</math>
The Girl Whose Name Was a Curs<br>
 
Although these three are favorites I really don't post anything unless it's my favorite of that moment<br>
By Cauchy's theorem the right hand side tends to 0 uniformly as ε, and hence δ, tends to 0. So
8. 3 must-read books for aspiring entrepreneurs<br>
 
The key in an entrepreneur book: you want to learn business. You want to learn how to honestly communicate with your customers. You want to stand out<br>
:<math>\displaystyle{H_\varepsilon f \rightarrow if}</math>
The Essays of Warren Buffett by Lawrence Cunningha<br>
 
"The Thank you Economy" by Gary Vaynerchu<br>
uniformly for polynomials. On the other hand if ''u''(''z'') = ''z'' it is immediate that
"Purple cow" by Seth Godi<br>
 
9. I love your writing, so do so many others out there. Who are your favorite writers<br>
:<math>\displaystyle{\overline{H_\varepsilon f} = - u^{-1} H_\varepsilon( u \overline{f}).}</math>
"Jesus's Son" by Denis Johnson is the best collection of short stories ever written. I'm afraid I really don't like his novels though<br>
 
"Tangents" by M. Prado. A beautiful series of graphic stories about relationships<br>
Thus if ''f'' is a polynomial in ''z''<sup>−1</sup> without constant term
Other writers: Miranda July, Ariel Leve, Mary Gaitskill, Charles Bukowski, Celine, Sam Lipsyte, William Vollmann, Raymond Carver. Arthur Nersesian. Stephen Dubner<br><br>
 
(Bukowski<br><br><br><br><br><br><br><br><br>
:<math> \displaystyle{H_\varepsilon f \rightarrow -i f}</math> uniformly.
Many writers are only really good storytellers. Most writers come out of a cardboard factory MFA system and lack a real voice. A real voice is where every word exposes ten levels of hypocrisy in the world and brings us all the way back to see reality. The writers above have their own voices, their own pains, and their unique ways of expressing those pains. Some of them are funny. Some a little more dark. I wish I could write 1/10 as good as any of them<br><br>
 
10. You are a prolific writer. Do you have any hacks that help you write a lot in little time<br>
Define the '''Hilbert transform''' on the circle by
Coffee, plus everything else coffee does for you first thing in the morning<br>
 
Only write about things you either love or hate. But if you hate something, try to find a tiny gem buried in the bag of dirt so you can reach in when nobody is looking and put that gem in your pocket. Stealing a diamond in all the shit around us and then giving it away for free via writing is a nice little hack, Being fearless precisely when you are most scared is the best hack<br><br>
:<math>\displaystyle{H=i(2P-I).}</math>
11. I totally get and love your idea about bleeding as a writer, appreciate if you share more with the readers of this blog<br>
 
Most people worry about what other people think of them. Most people worry about their health. Most people are at a crossroads and don't know how to take the next step and which road to take it on. Everyone is in a perpetual state of 'where do I put my foot next'. Nobody, including me, can avoid that<br>
Thus if ''f'' is a trigonometric polynomial
You and I both need to wash our faces in the morning, brush our teeth, shower, shit, eat, fight the weather, fight the colds that want to attack us if we're not ready. Fight loneliness or learn how to love and appreciate the people who want to love you back. And learn how to forgive and love the people who are even more stupid and cruel than we are. We're afraid to tell each other these things because they are all both disgusting and true<br>
 
You and I both have the same color blood. If I cut my wrist open you can see the color of my blood. You look at it and see that it's the same color as yours. We have something in common. It doesn't have to be shameful. It's just red. Now we're friends. No matter whom you are or where you are from. I didn't have to lie to you to get you to be my friend<br>
:<math>\displaystyle{H_\varepsilon f \rightarrow Hf}</math> uniformly.
Related Links<br>
 
How to be a Psychic in Ten Easy Lesson<br>
It follows that if ''f'' is any L<sup>2</sup> function
My New Year's Resolution in 199<br><br><br>
 
12. What is your advice for young entrepreneurs<br>
:<math>\displaystyle{H_\varepsilon f \rightarrow Hf}</math> in the L<sup>2</sup> norm.
Only build something you really want to use yourself. There's got to be one thing you are completely desperate for and no matter where you look you can't find it. Nobody has invented it yet. So there you go - you invent it. If there's other people like you, you have a business. Else. You fail. Then do it again. Until it works. One day it will<br>
 
Follow these 100 Rules<br>
This is an immediate consequence of the result for trigonometric polynomials once it is established that the operators ''H''<sub>ε</sub> are uniformly bounded in [[operator norm]]. But on [–π,π]
The 100 Rules for Being a Good Entrepreneur<br>
 
And, in particular this<br>
:<math>\displaystyle{(1-e^{i\theta})^{-1}= [(1-e^{i\theta})^{-1} -i\theta^{-1}] +i\theta^{-1}.}</math>
The Easiest Way to Succeed as an Entrepreneu<br>
 
In my just released book I have more chapters on my experiences as an entrepreneur<br>
The first term is bounded on the whole of [–π,π], so it suffices to show that the convolution operators ''S''<sub>ε</sub> defined by
13. I advocate the concept of working at a job while building your business. You have of course lived it. Now as you look back, what is your take on this? Is it possible to make it work while sailing on two boats<br><br>
 
Your boss wants everything out of you. He wants you to work 80 hours a week. He wants to look good taking credit for your work. He wants your infinite loyalty. So you need something back<br>
:<math>\displaystyle{S_\varepsilon f(\varphi) =\int_{\varepsilon \le |\theta|\le \pi} f(\varphi-\theta)\theta^{-1}\,d\theta}</math>
Exploit your employer. It's the best way to get good experience, clients, contacts. It's a legal way to steal. It's a fast way to be an entrepreneur because you see what large companies with infinite money are willing to pay for. If you can provide that, you make millions. It's how many great businesses have started and will always start. It's how every exit I've had started<br>
 
14. Who is a "person with true moral fiber"? In current times are there any role models who are people with true moral fiber<br><br><br>
are uniformly bounded. With respect to the orthonormal basis ''e''<sup>''in''θ</sup> convolution operators are diagonal and their operator norms are given by taking the supremum of the moduli of the Fourier coefficients. Direct computation shows that these all have the form
I don't really know the answer. I think I know a few people like that. I hope I'm someone like that. And I pray to god the people I'm invested in are like that and my family is like that<br>
 
I find most people to be largely mean and stupid, a vile combination. It's not that I'm pessimistic or [http://Www.ehow.com/search.html?s=cynical cynical]. I'm very much an optimist. It's just reality. Open the newspaper or turn on the TV and watch these people<br>
:<math>\displaystyle{{1\over \pi}\left |\int_a^b {\sin t \over t}\, dt\right|}</math>
Moral fiber atrophies more quickly than any muscle on the body. An exercise I do every morning is to promise myself that "I'm going to save a life today" and then leave it in the hands of the Universe to direct me how I can best do that. Through that little exercise plus the Daily Practice described above I hope to keep regenerating that fiber<br><br>
 
15.  Your message to the readers of this blog<br>
with 0 < ''a'' < ''b''. These integrals are well-known to be uniformly bounded.
Skip dinner. But follow me on Twitter.<br><br><br><br>
 
Read more posts on The Altucher Confidential �
It also follows that, for  a continuous function ''f'' on the circle, ''H''<sub>ε</sub>''f'' converges uniformly to ''Hf'', so in particular pointwise. The pointwise limit is a [[Cauchy principal value]], written
More from The Altucher Confidentia<br>
 
Life is Like a Game. Here�s How You Master ANY Gam<br><br>
:<math>\displaystyle{Hf= \mathrm{P.V.}\,{1\over \pi} \int {f(\zeta)\over \zeta-e^{i\varphi}}\, d\zeta.}</math>
Step By Step Guide to Make $10 Million And Then Totally Blow <br><br>
 
Can You Do One Page a Day?
If ''f'' is just in L<sup>2</sup> then ''H''<sub>ε</sub>''f'' converges to ''Hf'' pointwise almost everywhere.  In fact define the [[Poisson integral|Poisson operators]] on L<sup>2</sup> functions by
 
:<math>T_r \left (\sum a_n e^{in\theta} \right )=\sum r^{|n|} a_n e^{in\theta},</math>
 
for ''r'' < 1. Since these operators are diagonal, it is easy to see that ''T<sub>r</sub>f'' tends to ''f'' in L<sup>2</sup> as ''r'' increases to 1. Moreover, as Lebesgue proved, ''T<sub>r</sub>f''  also tends pointwise to ''f'' at each [[Lebesgue point]] of ''f''. On the other hand, it is also known that ''T<sub>r</sub>Hf'' – ''H''<sub>1 – ''r''</sub> ''f'' tends to zero at each Lebesgue point of ''f''. Hence ''H''<sub>1 – ''r''</sub> ''f'' tends pointwise to ''f'' on the common Lebesgue points of ''f'' and ''Hf'' and therefore almost everywhere.<ref>{{harvnb|Krantz|1999}}</ref><ref>{{harvnb|Torchinsky|1986}}</ref><ref>{{harvnb|Stein|Rami|2005|pp=112–114}}</ref>
 
Results of this kind on pointwise convergence are proved more generally below for ''L<sup>p</sup>'' functions using the Poisson operators and the Hardy–Littlewood maximal function of ''f''.
 
The Hilbert transform has a natural compatibility with orientation-preserving diffeomorphisms of the circle.<ref>See:
*{{harvnb|Mikhlin|Prössdorf|1986}}
*{{harvnb|Segal|1981}}
*{{harvnb|Pressley|Segal|1986}}</ref> Thus if ''H'' is a diffeomorphism of the circle with
 
:<math>H(e^{i\theta})=e^{ih(\theta)},\,\,\, h(\theta+2\pi)=h(\theta)+2\pi,</math>
 
then the operators
 
:<math>H_\varepsilon^h f(e^{i\varphi})=\frac{1}{\pi}\int_{|e^{ih(\theta)} -e^{ih(\varphi)}|\ge \varepsilon} \frac{f(e^{i\theta})}{e^{i\theta}-e^{i\varphi}}e^{i\theta}\, d\theta,</math>
 
are uniformly bounded and tend in the strong operator topology to ''H''. Morevoer if ''Vf''(''z'') = ''f''(''H''(''z'')), then ''VHV''<sup>−1</sup> – ''H''  is an operator with smooth kernel, so a [[Hilbert–Schmidt operator]].
 
In fact if ''G'' is the inverse of ''H'' with corresponding function ''g''(θ), then
 
:<math>(VH^h_\varepsilon V^{-1}- H_\varepsilon) f(e^{i\varphi}) = {1\over \pi}\int_{|e^{i\theta}-e^{i\varphi}| \ge \varepsilon}\left[{g^\prime(\theta) e^{ig(\theta)} \over e^{ig(\theta)} - e^{ig(\varphi)}} - {e^{i\theta} \over e^{i\theta} - e^{i\varphi} }\right]\,f(e^{i\theta})\, d\theta.</math>
 
Since the kernel on the right hand side is smooth on '''T''' × '''T''', it follows that the operators on the right hand side are uniformly bounded and hence so too are the operators ''H''<sub>ε</sub><sup>''h''</sup>. To see that they tend strongly to ''H'', it suffices to check this on trigonometric polynomials. In that case
 
:<math>H^h_\varepsilon f(\zeta)={1\over \pi i} \int_{|H(z)-H(\zeta)|\ge \varepsilon} \frac{f(z)}{z -\zeta} dz= {1\over \pi i} \int_{|H(z)-H(\zeta)|\ge \varepsilon} {f(z)-f(\zeta)\over z -\zeta}\, dz + \frac{f(\zeta)}{\pi i} \int_{|H(z)-H(\zeta)|\ge \varepsilon} {dz\over z -\zeta}.</math>
 
In the first integral the integrand is a trigonometric polynomial in ''z'' and ζ and so the integral is a trigonometric polynomial in ζ. It tends in ''L''<sup>2</sup> to the trigonometric polynomial
 
:<math>{1\over \pi i} \int {f(z)-f(\zeta)\over z -\zeta}\, dz.</math>
 
The integral in the second term can be calculated by the [[principle of the argument]]. It tends in L<sup>2</sup> to the constant function 1, so that
 
:<math>\lim_{\varepsilon\to 0} H_\varepsilon^h f(\zeta) = f(\zeta) + {1\over \pi i} \int {f(z)-f(\zeta)\over z -\zeta}\, dz,</math>
 
where the limit is in L<sup>2</sup>. On the other hand the right hand side is imdependent of the diffeomorphism. Since for the identity diffeomorphism, the left hand side equals ''Hf'', it too equals ''Hf'' (this can also be checked directly if ''f'' is a trigonometric polynomial). Finally, letting ε → 0,
 
:<math>(VH V^{-1}- H) f(e^{i\varphi}) = \frac{1}{\pi} \int \left[{g^\prime(\theta) e^{ig(\theta)} \over e^{ig(\theta)} - e^{ig(\varphi)}} - {e^{i\theta} \over e^{i\theta} - e^{i\varphi}}\right]\,f(e^{i\theta})\, d\theta.</math>
 
The direct method of evaluating Fourier coefficients to prove the uniform boundedness of the operator ''H''<sup>ε</sup> does not generalize directly to ''L<sup>p</sup>'' spaces with 1 < ''p'' < ∞. Instead a direct comparison of ''H''<sup>ε</sup>''f'' with the [[Poisson integral]] of the Hilbert transform is used classically to prove this. If ''f'' has Fourier series
 
:<math>f(e^{i\theta})=\sum_{n\in \mathbf{Z}} a_n e^{in\theta},</math>
 
its Poisson integral is defined by
 
:<math>\displaystyle{P_rf(e^{i\theta})=\sum_{n\in \mathbf{Z}} a_n r^{|n|} e^{in\theta}={1\over 2\pi}\int_0^{2\pi} {(1-r^2)f(e^{i\theta})\over 1-2r\cos\theta + r^2}\,d\theta =K_r\star f(e^{i\theta}),}</math>
 
where the [[Poisson kernel]] ''K''<sub>''r''</sub> is given by
 
:<math>\displaystyle{K_r(e^{i\theta})=\sum_{n\in \mathbf{Z}} r^{|n|}e^{in\theta} ={1-r^2\over 1 - 2r\cos\theta + r^2}.}</math>
 
In ''f'' is in L<sup>''p''</sup>('''T''') then the operators ''P''<sub>''r''</sub> satisfy
 
:<math>\displaystyle{\|P_rf - f\|_p\rightarrow 0.}</math>
 
In fact the ''K''<sub>''r''</sub> are positive so
 
:<math>\displaystyle{\|K_r\|_1 ={1\over 2\pi} \int_0^{2\pi} K_r(e^{i\theta})\, d\theta =1.}</math>
 
Thus the operators ''P<sub>r</sub>'' have operator norm bounded by 1 on ''L<sup>p</sup>''. The convergence statement above follows by continuity from the result for trigonometric polynomials, where it is an immediate consequence of the formula for the Fourier coefficients of ''K''<sub>''r''</sub>.
 
The uniform boundedness of the operator norm of ''H''<sub>ε</sub> follows because ''HP<sub>r</sub>''  − ''H''<sub>1−''r''</sub> is given as convolution by the function ψ<sub>''r''</sub>, where<ref>{{harvnb|Garnett|2007|p=102}}</ref>
 
:<math>\begin{align}
\psi_r(e^{i\theta}) &=1+\frac{1-r}{1+ r} \cot \left(\tfrac{\theta}{2} \right ) K_r(e^{i\theta}) \\
&\le 1+ \frac{1-r}{1+r} \cot \left (\tfrac{1-r}{2} \right ) K_r(e^{i\theta})
\end{align}</math>
 
for  1 − ''r'' ≤ |θ| ≤ π, and, for |θ| < 1 − ''r'',
 
:<math>\displaystyle{\psi_r(e^{i\theta})=1+ {2r\sin \theta\over 1-2r\cos\theta +r^2}.}</math>
 
These estimates show that the ''L''<sup>1</sup> norms ∫ |ψ<sub>''r''</sub>| are uniformly bounded. Since ''H'' is a bounded operator, it follows that the operators ''H''<sub>ε</sub> are uniformly bounded in operator norm on ''L''<sup>2</sup>('''T'''). The same argument can be used on ''L<sup>p</sup>''('''T''') once it is known that that the Hilbert transform ''H'' is bounded in operator norm on ''L<sup>p</sup>''('''T''').
 
===Hilbert transform on the real line===
{{See also|Hilbert transform}}
As in the case of the circle, the theory for L<sup>2</sup> functions is particularly easy to develop. In fact, as observed by Rosenblum and Devinatz, the two Hilbert transforms can be related using the Cayley transform.<ref>See:
*{{harvnb|Devinatz|1967}}
*{{harvnb|Rosenblum|Rovnyak|1997}}
*{{harvnb|Rosenblum|Rovnyak|1994}}
*{{harvnb|Nikolski|1986}}</ref>
 
The '''Hilbert transform''' ''H''<sub>'''R'''</sub> on L<sup>2</sup>('''R''') is defined by
 
:<math>\widehat{H_{\mathbf{R}} f} = \left (i\chi_{[0,\infty)} -i\chi_{(-\infty,0]} \right ) \widehat{f},</math>
 
where the [[Fourier transform]] is given by
 
:<math>\displaystyle{\widehat{f}(t)={1\over \sqrt{2\pi}}\int_{-\infty}^\infty f(x) e^{-itx} \, dx.}</math>
 
Define the Hardy space H<sup>2</sup>('''R''') to be the closed subspace of L<sup>2</sup>('''R''') consisting of functions for which the Fourier transform vanishes on the negative part of the real axis. Its orthogonal complement is given by functions for which the Fourier transform vanishes on the positive part of the real axis. It is the complex conjugate of H<sup>2</sup>('''R'''). If ''P''<sub>'''R'''</sub> is the orthogonal projection onto H<sup>2</sup>('''R'''), then
 
:<math>\displaystyle{H_{\mathbf{R}}=i(2P_{\mathbf{R}}-I).}</math>
 
The Cayley transform
 
:<math>\displaystyle{C(x)={x-i\over x+i}}</math>
 
carries the extended real line onto the circle, sending the point at ∞ to 1, and the upper halfplane onto the unit disk.
 
Define the unitary operator from L<sup>2</sup>('''T''') onto L<sup>2</sup>('''R''') by
 
:<math>\displaystyle{Uf(x)=\pi^{-1/2} (x+i)^{-1} f(C(x)).}</math>
 
This operator carries the Hardy space of the circle H<sup>2</sup>('''T''') onto H<sup>2</sup>('''R''').  In fact for |''w''| < 1, the linear span of the functions
 
:<math>f_w(z)= \frac{1}{1-w z}</math>
 
is dense in H<sup>2</sup>('''T'''). Moreover
 
:<math>Uf_w(x)= \frac{1}{\sqrt{\pi}} \frac{1}{(1-w)(x-\overline{z})}</math>
 
where
 
:<math>\displaystyle{z=C^{-1}(\overline{w}).}</math>
 
On the other hand, for ''z'' ∈ '''H''', the linear span of the functions
 
:<math>\displaystyle{g_z(t)=e^{itz}\chi_{[0,\infty)}(t)}</math>
 
is dense in L<sup>2</sup>((0,∞)). By the [[Fourier inversion formula]], they are the Fourier transforms of
 
:<math>\displaystyle{h_z(x)=\widehat{g_z}(-x)={i\over \sqrt{2\pi}} (x + z)^{-1},}</math>
 
so the linear span of these functions is dense in H<sup>2</sup>('''R'''). Since ''U'' carries the ''f''<sub>w</sub>'s onto multiples of the ''h''<sub>z</sub>'s, it follows that ''U'' carries H<sup>2</sup>('''T''') onto H<sup>2</sup>('''R'''). Thus
 
:<math>\displaystyle{UH_{\mathbf{T}} U^* = H_{\mathbf{R}}.}</math>
 
In {{harvtxt|Nikolski|1986}}, part of the L<sup>2</sup> theory on the real line and the upper halfplane is developed by transferring the results from the circle and the unit disk. The natural replacements for concentric circles in the disk are lines parallel to the real axis in '''H'''. Under the Cayley transform these correspond to circles in the disk that are tangent to the unit circle at the point one. The behaviour of functions in H<sup>2</sup>('''T''') on these circles is part of the theory of [[Carleson measure]]s. The theory of singular integrals, however, can be developed more easily by working directly on '''R'''.
 
H<sup>2</sup>('''R''') consists exactly of L<sup>2</sup> functions ''f'' that arise of boundary values of holomorphic functions on '''H''' in the following sense:<ref>{{harvnb|Stein|Shakarchi|2005|pp=213–221}}</ref> ''f''  is in H<sup>2</sup> provided that there is a holomorphic function ''F''(''z'') on '''H''' such that the functions ''f<sub>y</sub>''(''x'') = ''f''(''x'' + ''iy'') for ''y'' > 0 are in L<sup>2</sup> and ''f<sub>y</sub>'' tends to ''f'' in L<sup>2</sup> as ''y'' → 0. In this case ''F'' is necessarily unique and given by [[Cauchy's integral formula]]:
 
:<math>\displaystyle{F(z)={1\over 2\pi i}  \int_{-\infty}^\infty {f(s)\over s-z}\, ds.}</math>
 
In fact, identifying H<sup>2</sup> with L<sup>2</sup>(0,∞) via the Fourier transform, for ''y'' > 0 multiplication by ''e''<sup>–''yt''</sup> on L<sup>2</sup>(0,∞) induces a contraction semigroup ''V''<sub>''y''</sub> on H<sup>2</sup>. Hence for ''f'' in L<sup>2</sub>
 
:<math>\displaystyle{{1\over 2\pi i} \int_{-\infty}^\infty {f(s)\over s-z}\, ds ={1\over \sqrt{2\pi}} \int_{-\infty}^\infty f(s) \widehat{g_z}(s) \, ds  = {1\over \sqrt{2\pi}} \int_{-\infty}^\infty \widehat{f}(s) g_z(s) \, ds= V_yPf(x).}</math>
 
If ''f'' is in H<sup>2</sup>, ''F''(''z'') is holomorphic for Im ''z'' > 0, since the family of L<sup>2</sup> functions ''g<sub>z</sub>'' depends holomorphically on ''z''.  Moreover ''f''<sub>''y''</sub> = ''V<sub>y</sub>f'' tends to ''f'' in ''H''<sup>2</sup> since this is true for the Fourier transforms. Conversely if such an ''F'' exists, by Cauchy's integral theorem and the above identity applied to ''f''<sub>''y''</sub>
 
:<math>\displaystyle{f_{y+t}=V_t Pf_y}</math>
 
for ''t'' > 0. Letting ''t'' tend to ''0'', it follows that ''Pf<sub>y</sub>'' = ''f<sub>y</sub>'', so that ''f<sub>y</sub>'' lies in H<sup>2</sup>. But then so too does the limit ''f''. Since
 
:<math>\displaystyle{V_t f_y=f_{y+t}=V_y f_t,}</math>
 
uniqueness of ''F'' follows from
 
:<math> \displaystyle{f_t=\lim_{y\rightarrow 0} f_{y+t}=\lim_{y\rightarrow 0} V_t f_y = V_t f.}</math>
 
For ''f'' in L<sup>2</sup>, the '''truncated Hilbert transforms''' are defined by
 
:<math>\begin{align}
H_{\varepsilon,R} f(x) &={1\over \pi}\int_{\varepsilon \le |y-x|\le R} {f(y)\over x-y} \, dy ={1\over \pi}\int_{\varepsilon \le |y|\le R} {f(x-y)\over y}\, dy \\
H_{\varepsilon} f(x) &={1\over \pi}\int_{ |y-x|\ge \varepsilon} {f(y)\over x-y} \, dy ={1\over \pi} \int_{ |y|\ge \varepsilon} {f(x-y)\over y}\, dy.
\end{align}</math>
 
The operators ''H''<sub>ε,''R''</sub> are convolutions by bounded functions of compact support, so their operator norms are given by the uniform norm of their Fourier transforms. As before the absolute values have the form
 
:<math>\displaystyle{{1\over \sqrt{2\pi}}\left|\int_a^b { 2 \sin t \over t} \, dt\right|.}</math>
 
with 0 < ''a'' < ''b'', so the operators ''H''<sub>ε,''R''</sub> are uniformly bounded in operator norm. Since ''H''<sub>ε,''R''</sub>''f'' tends to ''H''<sub>ε</sub>''f'' in ''L''<sup>2</sup> for ''f'' with compact support, and hence for arbitrary ''f'', the operators ''H''<sub>ε</sub> are also uniformly bounded in operator norm.
 
To prove that ''H''<sub>ε</sub> ''f'' tends to ''Hf'' as ε tends to zero, it suffices to check this on a dense set of functions. On the other hand,
 
:<math>\displaystyle{\overline{H_\varepsilon f} = -  H_\varepsilon( \overline{f}),}</math>
 
so it suffices to prove that ''H''<sub>ε</sub>''f'' tends to ''if'' for a dense set of functions in H<sup>2</sup>('''R'''), for example the Fourier transforms of smooth functions ''g'' with compact support in (0,∞). But the Fourier transform ''f'' extends to an entire function  ''F'' on '''C''', which is bounded on Im(''z'') ≥ 0. The same is true of the derivatives of ''g''. Up to a scalar these correspond to multiplying ''F''(''z'') by powers of ''z''. Thus ''F'' satisfies a [[Paley-Wiener theorem|Payley-Wiener estimate]] for Im(''z'') ≥ 0:<ref>{{harvnb|Hörmander|1990}}</ref>
 
:<math>\displaystyle{|F^{(m)}(z)|\le K_{N,m} (1+|z|)^{-N}}</math>
 
for any ''m'', ''N'' ≥ 0. In particular, the integral defining ''H''<sub>ε</sub>''f''(''x'') can be computed by taking a standard semicircle contour centered on ''x''. It consists of a large semicircle with radius ''R'' and a small circle radius ε with the two portions of the real axis between them. By Cauchy's theorem, the integral round the contour is zero. The integral round the large contour tends to zero by the Paley-Wiener estimate. The integral on the real axis is the limit sought. It is therefore given as minus the limit on the small semicircular contour. But this is the limit of
 
:<math>\displaystyle{{1\over \pi} \int_{\Gamma} {F(z)\over z-x} \, dz.}</math>
 
Where Γ is the small semicircular contour, oriented anticlockwise. By the usual techniques of contour integration, this limit equals ''if''(''x'').<ref>{{harvnb|Titchmarsh|1939|102–105}}</ref> In this case, it is easy to check that the convergence is dominated in L<sup>2</sup> since
 
:<math>H_\varepsilon f(x)=\frac{1}{\pi}\int_{|y-x|\ge \varepsilon} \frac{f(y)-f(x)}{y-x}\,dy=\frac{1}{\pi} \int_{|y-x|\ge \varepsilon} \int_0^1 f^\prime(x+t(y-x))\,dt\, dy</math>
 
so that convergence is dominated by
 
:<math>G(x)=\frac{1}{2\pi} \int_0^1\int_{-\infty}^\infty |f^\prime(x+ty)|\,dy</math>
 
which is in L<sup>2</sup> by the Paley-Wiener estimate.
 
It follows that for ''f'' on ''L''<sup>2</sup>('''R''')
 
:<math>\displaystyle{H_\varepsilon f \rightarrow H f.}</math>
 
This can also be deduced directly because, after passing to Fourier transforms, ''H''<sub>ε</sub> and ''H'' become multiplication operators by uniformly bounded functions. The multipliers for ''H''<sub>ε</sub> tend pointwise almost everywhere to the multiplier for ''H'', so the statement above follows from the [[dominated convergence theorem]] applied to the Fourier transforms.
 
As for the Hilbert transform on the circle, ''H''<sub>ε</sub>''f'' tends to ''Hf'' pointwise almost everywhere if ''f'' is an L<sup>2</sup> function. In fact, define the [[Poisson integral|Poisson operators]] on L<sup>2</sup> functions by
 
:<math>T_y f(x)=\int_{-\infty}^\infty P_y(x-t)f(t)\, dt,</math>
 
where the Poisson kernel is given by
 
:<math>P_y(x)=\frac{y}{\pi( x^2 +y^2)}.</math>
 
for ''y'' > 0. Its Fourier transform is
 
:<math>\displaystyle{\widehat{P_y}(t)=e^{-y|t|},}</math>
 
from which it is easy to see that ''T<sub>y</sub>f'' tends to ''f'' in L<sup>2</sup> as ''y'' increases to 0. Moreover, as Lebesgue proved, 
''T<sub>y</sub>f''  also tends pointwise to ''f'' at each [[Lebesgue point]] of ''f''. On the other hand, it is also known that ''T<sub>y</sub>Hf'' – ''H<sub>y</sub>f'' tends to zero at each Lebesgue point of ''f''. Hence ''H''<sub>ε</sub>''f'' tends pointwise to ''f'' on the common Lebesgue points of ''f'' and ''Hf'' and therefore almost everywhere.<ref>See:
*{{harvnb|Krantz|1999}}
*{{harvnb|Torchinsky|1986}}
*{{harvnb|Duoandikoetxea|2001|pp=49–51}}</ref><ref>{{harvnb|Stein|Shakarchi|2005|pp=112–114}}</ref>  The absolute values of the functions ''T<sub>y</sub>f'' − ''f'' and ''T<sub>y</sub>Hf'' – ''H<sub>y</sub>f'' can be bounded pointwise by multiples of the maximal function of ''f''.<ref>{{harvnb|Stein|Weiss|1971}}</ref>
 
As for the Hilbert transform on the circle, the uniform boundedness of the operator norms of ''H''<sub>ε</sub> follows from that of the ''T''<sub>ε</sub> if ''H'' is known to be bounded, since ''HT''<sub>ε</sub> − ''H''<sub>ε</sub> is the convolution operator by the function
 
:<math>g_\varepsilon(x) = \begin{cases}
\frac{x}{\pi( x^2 +\varepsilon^2)} & |x|\le \varepsilon \\
\frac{x}{\pi( x^2 +\varepsilon^2)} -\frac{1}{\pi x} & |x| >\varepsilon
\end{cases}</math>
 
The L<sup>1</sup> norms of these functions are uniformly bounded.
 
===Riesz transforms in the complex plane===
{{Main|Riesz transform}}
The complex Riesz transforms ''R'' and ''R''* in the complex plane are the unitary operators on L<sup>2</sup>('''C''') defined as multiplication by ''z''/|''z''| and its conjugate on the Fourier transform of an L<sup>2</sup> function ''f'':
 
:<math>\displaystyle{\widehat{Rf}(z)={\overline{z}\over |z|} \widehat{f}(z),\,\,\, \widehat{R^*f}(z)={z\over |z|} \widehat{f}(z).}</math>
 
Idenitifying '''C''' with '''R'''<sup>2</sup>, ''R'' and ''R''* are given by
 
:<math>\displaystyle{R=-iR_1 + R_2,\,\,\, R^*=-iR_1 - R_2,}</math>
 
where ''R''<sub>1</sub> and ''R''<sub>2</sub> are the Riesz transforms on '''R'''<sup>2</sup> defined below.
 
On  L<sup>2</sup>('''C'''), the operator ''R'' and its integer powers are unitary. They can also be expressed as singular integral operators:<ref>{{harvnb|Astala|Ivaniecz|Martin|2009|pp=101–102}}</ref>
 
:<math>\displaystyle{R^kf(w)=\lim_{\varepsilon\rightarrow 0} \int_{|z-w|\ge \varepsilon} M_k(w-z)f(z)\,dx\, dy,}</math>
 
where
 
:<math>\displaystyle{M_k(z)={k\over 2\pi i^k} {z^k \over |z|^{k+2}} \,\,\,\, (k\ge 1), \,\,\,\, M_{-k}(z) =\overline{M_k(z)}.}</math>
 
Defining the truncated higher Riesz transforms as
 
:<math>\displaystyle{R^{(k)}_\varepsilon f(w)=\int_{|z-w|\ge \varepsilon} M_k(w-z)f(z)\,dx\, dy,}</math>
 
these operators can be shown to be uniformly bounded in operator norm. For odd powers this can be deduced by the method of rotation of Calderón and Zygmund, described below.<ref>{{harvnb|Grafakos|2005}}</ref> If the operators are known to be bounded in operator norm it can also be deduced using the Poisson operators.<ref>{{harvnb|Stein|Weiss|1971}}</ref>
 
The Poisson operators ''T''<sub>''s''</sub> on '''R'''<sup>2</sup> are defined for ''s'' > 0 by
 
:<math>\displaystyle{T_sf(x) ={1\over 2\pi}\int_{\mathbf{R}^2} {s f(x)\over (|x-t|^2 + s^2)^{3/2}}\, dt.}</math>
 
They are given by convolution with the functions
 
:<math>\displaystyle{P_s(x)= {s\over 2\pi(|x|^2 + s^2)^{3/2} }.}</math>
 
''P''<sub>''s''</sub> is the Fourier transform of the function ''e''<sup>− ''s''|''x''|</sup>, so under the Fourier transform they correspond to multiplication by these functions and form a contraction semigroup on L<sup>2</sup>('''R'''<sup>2</sup>).  Since ''P''<sub>''y''</sub> is positive and integrable with integral 1, the operators ''T''<sub>''s''</sub> also define a contraction semigroup on each L<sup>''p''</sup> space with 1 < ''p'' <  ∞.
 
The higher Riesz transforms of the Poisson kernel can be computed:
 
:<math>\displaystyle{R^kP_s(z)={k\over 2\pi i^k} {z^k \over (|z|^2 +s^2)^{k/2+1}}}</math>
 
for ''k'' ≥ 1 and the complex conjugate for − ''k''. Indeed the right hand side is a harmonic function ''F''(''x'',''y'',''s'') of three variable and for such functions<ref>{{harvnb|Stein|Weiss|1971|p=51}}</ref>
 
:<math>\displaystyle{T_{s_1} F(x,y,s_2) = F(x,y,s_1+s_2).}</math>
 
As before the operators
 
:<math>\displaystyle{T_\varepsilon R^k - R^{(k)}_\varepsilon}</math>
 
are given by convolution with integrable functions and have uniformly bounded operator norms. Since the Riesz transforms are unitary on L<sup>2</sup>('''C'''), the uniform boundedness of the truncated Riesz transforms implies that they converge in the strong operator topology to the corresponding Riesz transforms.
 
The uniform boundedness of the difference between the transform and the truncated transform can also be seen for odd ''k'' using the Calderón-Zygmund method of rotation.<ref>{{harvnb|Grafakos|2008}}</ref><ref>{{harvnb|Stein|Weiss|1971|pp=222–223}}</ref> The group '''T''' acts by rotation on functions on '''C''' via
 
:<math>\displaystyle{U_\theta f(z)=f(e^{i\theta}z).}</math>
 
This defines a unitary representation on L<sup>2</sup>('''C''') and the unitary operators ''R''<sub>θ</sub> commute with the Fourier transform. If ''A'' is a bounded operator on L<sup>2</sup>('''R''') then it defines a bounded operator ''A''<sup>(1)</sup> on
L<sup>2</sup>('''C''') simply by making ''A'' act on the first coordinate. With the identification L<sup>2</sup>('''R'''<sup>2</sup>) = L<sup>2</sup>('''R''') ⊗ L<sup>2</sup>('''R'''), ''A''<sup>(1)</sup> = ''A'' ⊗ ''I''. If φ is a continuous function on the circle then a new operator can be defined by
 
:<math>\displaystyle{B ={1\over 2\pi} \int_0^{2\pi} \varphi(\theta) U_\theta A^{(1)} U_\theta^* \, d\theta.}</math>
 
This definition is understood in the sense that
 
:<math>\displaystyle{(Bf,g) ={1\over 2\pi} \int_0^{2\pi} \varphi(\theta) (U_\theta A^{(1)} U_\theta^*f,g) \, d\theta}</math>
 
for any ''f'', ''g'' in L<sup>2</sup>('''C'''). It follows that
 
:<math>\displaystyle{\| B \| \le {1\over 2\pi} \int_0^{2\pi} |\varphi(\theta)| \cdot\|A\|\, d\theta.}</math>
 
Taking ''A'' to be the Hilbert transform ''H'' on ''L''<sup>2</sup>('''R''') or its truncation ''H''<sub>ε</sub>, it follows that
 
:<math>\begin{align}
R &={1\over 2\pi} \int_0^{2\pi} e^{-i\theta} U_\theta H^{(1)} U_\theta^* \, d\theta,\\
R_\varepsilon &={1\over 2\pi} \int_0^{2\pi} e^{-i\theta} U_\theta H^{(1)}_\varepsilon U_\theta^* \, d\theta.
\end{align}</math>
 
Taking adjoints gives a similar formuls for ''R*'' and its truncation.  This gives a second way to verify estimate the norms of ''R'', ''R''* and their truncations. It has the advantage of being applicable also for ''L<sup>p</sup>'' spaces.
 
The Poisson operators can also be used to show that the truncated higher Riesz transforms of a function tend to the higher Riesz transform at the common Lebesgue points of the function and its transform. Indeed (''R<sup>k</sup>T''<sub>ε</sub> − ''R''<sup>(''k'')</sup><sub>ε</sub>)''f'' → 0 at each Lebesgue point of ''f''; while (''R<sup>k</sup>'' − ''R<sup>k</sup>T''<sub>ε</sub>)''f'' → 0 at each Lebesgue point of ''R<sup>k</sup>f''.<ref>{{harvnb|Stein|Weiss|1971}}</ref>
 
===Beurling transform in the complex plane===
{{see also|Beltrami equation}}
Since
 
:<math>{\overline{z}\over z}= \left({\overline{z}\over |z|}\right)^2,</math>
 
the Beurling transform ''T'' on ''L''<sup>2</sup> is the unitary operator equal to ''R''<sup>2</sup>. This relation has been used classically in {{harvtxt|Vekua|1962}} and {{harvtxt|Ahlfors|1966}} to establish the continuity properties of ''T'' on ''L<sup>p</sup>'' spaces. The results on the Riesz transform and its powers show that ''T'' is  the limit in the strong operator topology of the truncated operators
 
:<math>T_\varepsilon f(w)=-\frac{1}{\pi}\iint_{|z-w|\ge \varepsilon} \frac{f(z)}{(w-z)^2} dxdy.</math>
 
Accordingly ''Tf'' can be written as a Cauchy principal value integral:
 
:<math>Tf(w)=-\frac{1}{\pi} P.V. \iint \frac{f(z)}{(w-z)^2} dxdy=-\frac{1}{\pi}\lim_{\varepsilon \to 0}\iint_{|z-w|\ge \varepsilon} \frac{f(z)}{(w-z)^2} dxdy.</math>
 
From the description of ''T'' and ''T''* on Fourier transforms, it follows that if ''f'' is smooth of compact support
 
:<math>\begin{align}
T(\partial_{\overline{z}} f) &=\partial_z Tf, \\
T^*(\partial_z f) =\partial_{\overline{z}} T^*f.
\end{align}</math>
 
Like the Hilbert transform in one dimension, the Beurling transform has a compatibility with conformal changes of coordinate. Let Ω be a bounded region in '''C''' with smooth boundary ∂Ω and let φ be a univalent holomorphic map of the [[unit disk]] ''D'' onto Ω extending to a smooth diffeomorphism of the circle onto ∂Ω. If χ<sub>Ω</sub> is the [[indicator function|characteristic function]] of  Ω, the operator can χ<sub>Ω</sub>''T''χ<sub>Ω</sub> defines an operaror ''T''(Ω) on L<sup>2</sup>(Ω). Through the conformal map φ, it induces an operator, also denoted ''T''(Ω), on L<sup>2</sup>(''D'') which can be compared with ''T''(''D''). The same is true of the truncations ''T''<sub>ε</sub>(Ω) and ''T''<sub>ε</sub>(''D'').
 
Let ''U''<sub>ε</sub> be the disk |''z'' − ''w''| < ε and ''V''<sup>ε</sub> the region |φ(''z'') − φ(''w'')|  < ε. On ''L''<sup>2</sup>(''D'')
 
:<math>\begin{align}
T_\varepsilon(\Omega)f(w) &= -\frac{1}{\pi} \iint_{D\backslash V_\varepsilon} \left [{\varphi^\prime(w)\varphi^\prime(z) \over (\varphi(z)-\varphi(w))^2}f(z)\right ]dxdy,\\
T_\varepsilon(D)f(w) &=-{1\over \pi} \iint_{D\backslash U_\varepsilon} {f(z) \over (z-w)^2}dxdy,
\end{align}</math>
 
and the operator norms of these truncated operators are uniformly bounded. On the other hand if
 
:<math>T^\prime_\varepsilon(D)f(w) = -{1\over \pi} \iint_{D\backslash V_\varepsilon} \frac{f(z)}{(z-w)^2} dxdy,</math>
 
then the difference between this operator and ''T''<sub>ε</sub>(Ω) is a truncated operator with smooth kernel ''K''(''w'',''z''):
 
:<math>K(w,z)=-{1\over \pi} \left[{\varphi^\prime(w)\varphi^\prime(z)\over (\varphi(z)-\varphi(w))^2} -{1\over (z-w)^2}\right].</math>
 
So the operators ''T′''<sub>ε</sub>(''D'') must also have uniformly bounded operator norms. To see that their difference tends to 0 in the strong operator topology, it is enough to check this for ''f'' smooth of compact support in ''D''. By Green's theorem<ref>{{harvnb|Astala|Iwaniecz|Martin|2009|pp=93–95}}</ref>
 
:<math>\left (T_\varepsilon(D)-T^\prime_\varepsilon(D) \right )f(w)= \frac{1}{\pi}\iint_{U_\varepsilon} {\partial_zf(z)\over z-w}dxdy-{1\over \pi}\iint_{V_\varepsilon} {\partial_zf(z)\over z-w}dxdy+{1\over 2\pi i}\int_{\partial U_\varepsilon} \frac{f(z)}{z-w}d\overline{z}-\frac{1}{2\pi i}\int_{\partial V_\varepsilon} {f(z)\over z-w}\, d\overline{z}.</math>
 
All four terms on the right hand side tend to 0. Hence the difference ''T''(Ω) − ''T''(''D'') is the [[Hilbert–Schmidt operator]] with kernel ''K''.
 
For pointwise convergence there is simple argument due to {{harvtxt|Mateu|Verdera|2006}} showing that the truncated integrals converge to ''Tf'' precisely at its Lebesgue points, that is almost everywhere.<ref>{{harvnb|Astala|Iwaniecz|Martin|2009|pp=97–98}}</ref> In fact ''T'' has the following symmetry property for ''f'', ''g'' ∈ ''L''<sup>2</sup>('''C''')
 
:<math>\iint (Tf) g = -{1\over \pi}\lim \int_{|z-w|\ge \varepsilon} \frac{f(w)g(z)}{(w-z)^2} =\iint f (Tg).</math>
 
On the other hand if χ is the [[characteristic function]] of the disk ''D''(''z'',ε) with centre ''z'' and radius ε, then
 
:<math>T\chi(w) = -\varepsilon^2 \frac{1-\chi(w)}{(w-z)^2}.</math>
 
Hence
 
:<math>T_\varepsilon(f)(z)={1\over \pi\varepsilon^2}\iint f (T\chi)= {1\over \pi\varepsilon^2}\iint (Tf)\chi = \mathbf{Av}_{D(z,\varepsilon)}\, Tf.</math>
 
By the [[Lebesgue differentiation theorem]], the right hand side converges to ''Tf'' at the Lebesgue points of ''Tf''.
 
===Riesz transforms in higher dimensions===
{{Main|Riesz transform}}
For ''f'' in the Schwartz space of '''R'''<sup>''n''</sup>, the ''j''th '''Riesz transform''' is defined by
 
:<math>R_j f(x) =c_n\lim_{\varepsilon \to 0} \int_{|y|\ge \varepsilon} f(x-y){y_j\over |y|^{n+1}}dy= \frac{c_n}{n-1}\int \partial_j f(x-y){1\over |y|^{n-1}} dy,</math>
 
where
 
:<math>c_n=\Gamma\left(\tfrac{n+1}{2}\right)\pi^{-\frac{n+1}{2}}.</math>
 
Under the Fourier transform:
 
:<math>\widehat{R_j f}(t)={it_j\over |t|}\widehat{f}(t).</math>
 
Thus ''R<sub>j</sub>'' corresponds to the operator ∂<sub>''j''</sub>Δ<sup>−1/2</sup>, where Δ = −∂<sub>1</sub><sup>2</sup> − ... −∂<sub>''n''</sub><sup>2</sup> denotes the Laplacian on '''R'''<sup>''n''</sup>. By definition ''R<sub>j</sub>'' is a bounded and skew-adjoint operator for the ''L''<sup>2</sup> norm and
 
:<math>R_1^2 + \cdots + R_n^2 = -I.</math>
 
The corresponding truncated operators
 
:<math>R_{j,\varepsilon} f(x) =c_n\int_{|y|\ge \varepsilon} f(x-y){y_j\over |y|^{n+1}} dy</math>
 
are uniformly bounded in the operator norm. This can either be proved directly or can be established by the '''Calderón−Zygmund method of rotations''' for the group SO(''n'').<ref>{{harvnb|Grafokos|2008|pp=272–274}}</ref> This expresses the operators ''R<sub>j</sub>'' and their truncations in terms of the Hilbert transforms in one dimension and its truncations. In fact if ''G'' = SO(''n'') with normalised Haar measure and ''H''<sup>(1)</sup> is the Hilbert transform in the first coordinate, then
 
:<math>\begin{align}
R_j &=\int_G \varphi(g) gH^{(1)}g^{-1} \, dg, \\
R_{j,\varepsilon} &=\int_G \varphi(g) gH_\varepsilon^{(1)} g^{-1} \, dg, \\
R_{j,\varepsilon,R} &=\int_G \varphi(g) gH_{\varepsilon,R}^{(1)} g^{-1} \, dg.
\end{align}</math>
 
where φ(''g'') is the (1,''j'') matrix coefficient of ''g''.
 
In particular for ''f'' ∈ ''L''<sup>2</sup>, ''R''<sub>''j'',ε</sub>''f'' → ''R<sub>j</sub>f'' in ''L''<sup>2</sup>. Moreover ''R''<sub>''j'',ε</sub>''f'' tends to ''R<sub>j</sub>'' almost everywhere. This can be proved exactly as for the Hilbert transform by using the Poisson operators defined on ''L''<sup>2</sup>('''R'''<sup>''n''</sup>) when '''R'''<sup>''n''</sup> is regarded as the boundary of a halfspace in '''R'''<sup>''n''+1</sup>. Alternatively it can be proved directly from the result for the Hilbert transform on '''R''' using the expression of ''R<sub>j</sub>'' as an integral over ''G''.<ref>{{harvnb|Grafakos|2008}}</ref><ref>{{harvnb|Stein|Weiss|1971|pp=222–223, 236–237}}</ref>
 
The Poisson operators ''T<sub>y</sub>'' on '''R'''<sup>''n''</sup> are defined for ''y'' > 0 by<ref>{{harvnb|Stein|Weiss|1971}}</ref>
 
:<math>T_yf(x) =c_n\int_{\mathbf{R}^n} \frac{y f(x)}{\left (|x-t|^2 + y^2 \right )^{\frac{n+1}{2}}} dt.</math>
 
They are given by convolution with the functions
 
:<math>P_y(x)=c_n \frac{y}{\left (|x|^2 + y^2 \right )^{\frac{n+1}{2}}}.</math>
 
''P''<sub>''y''</sub> is the Fourier transform of the function ''e''<sup>−''y''|''x''|</sup>, so under the Fourier transform they correspond to multiplication by these functions and form a contraction semigroup on L<sup>2</sup>('''R'''<sup>''n''</sup>).  Since ''P<sub>y</sub>'' is positive and integrable with integral 1, the operators ''T''<sub>''y''</sub> also define a contraction semigroup on each ''L<sup>p</sup>'' space with 1 < ''p'' <  ∞.
 
The Riesz transforms of the Poisson kernel can be computed
 
:<math>R_j P_\varepsilon(x)= c_n \frac{x_j}{\left(|x|^2 + \varepsilon^2 \right)^{\frac{n+1}{2}}}.</math>
 
The operator ''R<sub>j</sub>T''<sub>ε</sub> is given by convolution with this function. It can be checked directly that the operators ''R<sub>j</sub>T''<sub>ε</sub> − ''R''<sub>''j'',ε</sub> are given by convolution with functions uniformly bounded in ''L''<sup>1</sup> norm. The operator norm of the difference is therefore uniformly bounded. We have (''R<sub>j</sub>T''<sub>ε</sub> − ''R''<sub>''j'',ε</sub>)''f'' → 0 at each Lebesgue point of ''f''; while (''R<sub>j</sub>'' − ''R<sub>j</sub>T''<sub>ε</sub>)''f'' → 0 at each Lebesgue point of ''R<sub>j</sub>f''. So ''R''<sub>''j'',ε</sub>''f'' → ''R<sub>j</sub>f'' on the common Lebesgue points of ''f'' and ''R<sub>j</sub>f''.
 
==L<sup>p</sup> theory==
 
===Elementary proofs of M. Riesz theorem===
The theorem of [[Marcel Riesz]] asserts that singular integral operators that are continuous for the L<sup>2</sup> norm are also continuous in the L<sup>''p''</sup> norm for 1 < ''p'' < ∞ and that the operator norms vary continuously with ''p''.
*'''Bochner's proof for Hilbert transform on the circle.'''<ref>{{harvnb|Grafakos|2005|p=215−216}}</ref> Once it is established that the operator norms of the Hilbert transform on L<sup>''p''</sup>('''T''') are bounded for even integers, it follows from the [[Riesz–Thorin interpolation theorem]] and duality that they are bounded for all ''p'' with 1 < ''p'' < ∞ and that the norms vary continuously with ''p''. Moreover the arguments with the Poisson integral can be applied to show that the truncated Hilbert transforms ''H''<sub>ε</sub> are uniformly bounded in operator norm and converge in the strong operator topology to ''H''.
 
:It is enough to prove the bound for real trigonometric polynomials without constant term:
 
::<math>\displaystyle{f(e^{i\theta}) =\sum_{m=1}^N a_m e^{im\theta} + a_{-m} e^{-im\theta},\,\,\,\,\,\,\,a_{-m}=\overline{a_m}.}</math>
 
:Since ''f'' + ''iHf'' is a polynomial in ''e''<sup>''i'' θ</sup> without constant term
 
::<math>\displaystyle{{1\over 2\pi}\int_0^{2\pi} (f+iHf)^{2n} \, d\theta = 0.}</math>
 
:Hence, taking the real part and using [[Hölder's inequality]]:
 
::<math>\displaystyle{\|Hf\|_{2n}^{2n} \le \sum_{k=0}^{n-1} {2n\choose 2k} |((Hf)^{2k},f^{2n-2k})|\le  \sum_{k=0}^{n-1} {2n\choose 2k} \|Hf\|_{2n}^{2k}\cdot\|f\|_{2n}^{2n-2k}.}</math>
 
:So the M. Riesz theorem follows by induction for ''p'' an even integer and hence for all ''p'' with 1 < ''p'' < ∞.
 
*'''Cotlar's proof for Hilbert transform on the line.'''<ref>{{harvnb|Grafakos|2005|p=255−257}}</ref> Once it is established that the operator norms of the Hilbert transform on L<sup>''p''</sup>('''R''') are bounded when ''p'' is a power of 2, it follows from the [[Riesz–Thorin interpolation theorem]] and duality that they are bounded for all ''p'' with 1 < ''p'' < ∞ and that the norms vary continuously with ''p''. Moreover the arguments with the Poisson integral can be applied to show that the truncated Hilbert transforms ''H''<sub>ε</sub> are uniformly bounded in operator norm and converge in the strong operator topology to ''H''.
 
:It is enough to prove the bound when ''f'' is a Schwartz function. In that case the following identity of Cotlar holds:
 
::<math>\displaystyle{(Hf)^2= f^2 +2H(fH(f)).}</math> 
:In fact, write ''f'' = ''f''<sub>+</sub> + ''f''<sub>-</sub> according to the ±''i'' eigenspaces of ''H''. Since ''f'' ± ''iHf'' extend to holomorphic functions in the upper and lower half plane, so too do their squares. Hence
 
::<math>\displaystyle{f^2 -(Hf)^2=(f_+ + f_-)^2 + (f_+-f_-)^2 =2(f_+^2 + f_-^2)=-2iH(f_+^2 -f_-^2)=-2H(f(Hf)).}</math>
 
:(Cotlar's identity can also be verified directly by taking Fourier transforms.)
 
:Hence, assuming the M. Riesz theorem for ''p'' = 2<sup>''n''</sup>,
 
::<math>\displaystyle{\|Hf\|^2_{2^{n+1}}=\|(Hf)^2\|_{2^n} \le \|f^2\|_{2^n} + 2\|H(fH(f))\|_{2^n}\le \|f\|_{2^{n+1}}^2 + 2\|H\|_{2^n}\|f\|_{2^{n+1}}\|Hf\|_{2^{n+1}}.}</math>
 
:Since
 
::<math>\displaystyle{R^2 > 1+2 \|H\|_{2^n} R}</math>
 
:for ''R'' sufficiently large, the M. Riesz theorem must also hold for ''p'' = 2<sup>''n'' + 1</sup>.
 
:Exactly the same method works for the Hilbert transform on the circle.<ref>{{harvnb|Gohberg|Krupnik|1992|pp=19–20}}</ref> The same identity of Cotlar is easily verified on trigonometric polynomials ''f'' by writing them as the sum of the terms with non-negative and negative exponents, i.e. the ±''i'' eigenfunctions of ''H''. The L<sup>''p''</sup> bounds can therefore be established when ''p'' is a power of 2 and follow in general by interpolation and duality.
 
*'''Calderón–Zygmund method of rotation.''' The method of rotation for Riesz transforms and their truncations applies equally well on L<sup>''p''</sup> spaces for 1 < ''p'' < ∞. Thus these operators can be expressed in terms of the Hilbert transform on '''R''' and its truncations. The integration of the functions Φ from the group '''T''' or SO(''n'') into the space of operators on L<sup>''p''</sup> is taken in the weak sense:
 
::<math>\displaystyle{(\int_G \Phi(x)\, dx\, f,g) =\int_G (\Phi(x)f,g)\, dx}</math>
 
:where ''f'' lies in L<sup>''p''</sub> and ''g'' lies in the [[dual space]] L<sup>''q''</sup> with
 
::<math>\displaystyle{{1\over p} + {1\over q} = 1.}</math>
 
:It follows that Riesz transforms are bounded on L<sup>''p''</sup> and that the differences with their truncations are also uniformly bounded. The continuity of the L<sup>''p''</sup> norms of a fixed Riesz transform is a consequence of the [[Riesz–Thorin interpolation theorem]].
 
===Pointwise convergence===
{{see also|Fatou's theorem}}
The proofs of pointwise convergence for Hilbert and Riesz transforms rely on the [[Lebesgue differentiation theorem]], which can be proved using the [[Hardy-Littlewood maximal function]].<ref>See:
*{{harvnb|Stein|Weiss|1971|pp=12–13}}
*{{harvnb|Torchinsky|2004}}</ref>  The techniques for the simplest and best known case, namely the Hilbert transform on the circle, are a prototype for all the other transforms. This case is explained in detail here.
 
Let ''f'' be  in L<sup>p</sup>('''T''') for ''p'' > 1. The Lebesgue differentiation theorem states that
 
:<math>\displaystyle{A(\varepsilon)= {1\over 2\varepsilon}\int_{x-\varepsilon}^{x+\varepsilon} |f(t)-f(x)| \, dt \to 0}</math>
 
for almost all ''x'' in '''T'''.<ref>{{harvnb|Torchinsky|2005|pp=41–42}}</ref><ref>{{harvnb|Katznelson|2004|pp=10–21}}</ref><ref>{{harvnb|Stein|Shakarchi|112-114}}</ref> The points at which this holds are called the '''Lebesgue points''' of ''f''. Using this theorem it follows that if ''f'' is an integrable function on the circle, the Poisson integral ''T<sub>r</sub>f''  tends pointwise to ''f'' at each [[Lebesgue point]] of ''f''. In fact, for ''x'' fixed, ''A''(ε) is  a continuous function on [0,π]. Continuity at 0 follows because ''x'' is a Lebesgue point and elsewhere because, if ''h'' is an integrable function, the integral of |h| on intervals of decreasing length tends to 0 by [[Hölder's inequality]].
 
Letting ''r'' = 1 − ε, the difference can be estimated by two integrals:
 
:<math>2\pi|T_{r}f(x) - f(x)|=\int_{0}^{2\pi} |(f(x-y)-f(x))P_r(y)|\, dy\le \int_{|y|\le \varepsilon} + \int_{|y|\ge \varepsilon}.</math>
 
The Poisson kernel has two important properties for ε small
 
:<math>\begin{align}
\sup_{y\in [-\varepsilon,\varepsilon]} |P_{1-\varepsilon}(y)| &\le \varepsilon^{-1}. \\
\sup_{y\notin (-\varepsilon,\varepsilon)} |P_{1-\varepsilon}(y)| &\to 0.
\end{align}</math>
 
The first integral is bounded by ''A''(ε) by the first inequality so tends to zero as ε goes to 0; the second integral  tends to 0 by the second inequality.
 
The same reasoning can be used to show that ''T''<sub>1 − ε</sub>''Hf'' – ''H''<sub>ε</sub>''f'' tends to zero at each Lebesgue point of ''f''.<ref>{{harvnb|Garnett|2007|pp=102–103}}</ref> In fact the operator ''T''<sub>1 − ε</sub>''Hf'' has kernel ''Q<sub>r</sub>'' + ''i'', where the conjugate Poisson kernel ''Q''<sub>''r''</sub> is defined by
 
:<math>\displaystyle{Q_r(\theta)={2r\sin \theta\over 1 -2r \cos\theta + r^2}.}</math>
 
Hence
 
:<math>\displaystyle{2\pi|T_{1-\varepsilon} H f(x) - H_\varepsilon f(x)|\le \int_{|y|\le \varepsilon} | f(x-y)-f(x)|\cdot|Q_r(y)|\, dy + \int_{|y|\ge \varepsilon} |f(x-y)-f(x)|\cdot |Q_1(y)-Q_r(y)|\,dy.}</math>
 
The conjugate Poisson kernel has two important properties for ε small
 
:<math>\begin{align}
\sup_{y\in [-\varepsilon,\varepsilon]} |Q_{1-\varepsilon}(y)| &\le \varepsilon^{-1}.\\
\sup_{y\notin (-\varepsilon,\varepsilon)} |Q_1(y)-Q_{1-\varepsilon}(y)| &\to 0.
\end{align}</math>
 
Exactly the same reasoning as before shows that the two integrals tend to 0 as ε → 0.
 
Combining these two limit formulas it follows that ''H''<sub>ε</sub>''f'' tends pointwise to ''f'' on the common Lebesgue points of ''f'' and ''Hf'' and therefore almost everywhere.<ref>{{harvnb|Krantz|1999}}</ref><ref>{{harvnb|Torchinsky|1986}}</ref><ref>{{harvnb|Stein|Shakarchi|2005|pp=112–114}}</ref>
 
===Maximal functions===
{{see also|Hardy–Littlewood maximal function}}
Much of the ''L<sup>p</sup>'' theory has been developed using maximal functions and maximal transforms. This approach has the advantage that it also extends to L<sup>1</sup> spaces in an appropriate "weak" sense and gives refined estimates in ''L<sup>p</sup>'' spaces for ''p'' > 1. These finer estimates form an important part of the techniques involved in [[Lennart Carleson]]'s solution in 1966 of [[Lusin's conjecture]] that the Fourier series of L<sup>2</sup> functions converge almost everywhere.<ref>{{harvnb|Arias de Reyna|2002}}</ref> In the more rudimentary forms of this approach, the L<sup>2</sup> theory is given less precedence: instead there is more emphasis on the L<sup>1</sup> theory, in particular its measure-theoretic and probabilistic aspects; results for other ''L<sup>p</sup>'' spaces are deduced by a form of [[Marcinkiewicz interpolation theorem|interpolation]] between L<sup>1</sup> and L<sup>∞</sup> spaces. The approach is described in numerous textbooks, including the classics {{harvtxt|Zygmund|1977}} and {{harvtxt|Katznelson|2004}}. Katznelson's account is followed here for the particular case of the Hilbert transform of functions in L<sup>1</sup>('''T'''), the case not covered by the development above.<ref>See also:
*{{harvnb|Torchinsky|2005}}
*{{harvnb|Grafakos|2008}}
*{{harvnb|Krantz|1999}}</ref>
 
If ''f'' is an L<sup>1</sup> function on the circle its maximal function is defined by<ref>{{harvnb|Krantz|1999|p=71}}</ref>
 
:<math>\displaystyle{f^*(t)=\sup_{0<h\le \pi}  {1\over 2h} \int_{t-h}^{t+h} |f(s)|\, ds.}</math>
 
''f''* is finite almost everywhere and is of weak L<sup>1</sup> type. In fact for λ > 0 if
 
:<math>\displaystyle{E_f(\lambda)=\{x:\,|f(x)| > \lambda\}, \,\, f_\lambda =\chi_{E(\lambda)} f,}</math>
 
then<ref>{{harvnb|Katznelson|2004|pp=74–75}}</ref>
 
:<math>m(E_{f^*}(\lambda))\le {8\over \lambda}\int_{E_f(\lambda)} |f|\le {8\|f\|_1\over \lambda},</math>
 
where ''m'' denotes Lebesgue measure.
 
The Hardy−Littlewood inequality above leads to a proof that almost every point ''x'' of '''T''' is a [[Lebesgue point]] of an integrable function ''f'', so that
 
:<math>\displaystyle{\lim_{h\to 0} \frac{\int^{x+h}_{x-h}|f(t)-f(x)|\, dt}{2h} \to 0.}</math>
 
In fact let
 
:<math>\omega(f)(x)=\limsup_{h\to 0} \frac{\int^{x+h}_{x-h}|f(t)-f(x)|\, dt}{2h} \le f^*(x) +|f(x)|.</math>
 
If ''g'' is continuous, then the ω(''g'') =0, so that  ω(''f'' − ''g'') = ω(''f''). On the other hand ''f'' can be approximated arbitrarily closely in L<sup>1</sup> by continuous ''g''. Then, using [[Chebychev's inequality]],
 
:<math>m\{x: \, \omega(f)(x)> \lambda\} =m\{x: \, \omega(f-g)(x)> \lambda\} \le m\{x: \, (f-g)^*(x)> \lambda\} + m\{x: \, |f(x)-g(x)|> \lambda\} \le C\lambda^{-1}\|f-g\|_1.</math>
 
The right hand side can be made arbitrarily small, so that ω(''f'') = 0 almost everywhere.
 
The Poisson integrals of an L<sup>1</sup> function ''f'' satisfy<ref>{{harvnb|Katznelson|2004|p=76}}</ref>
 
:<math>\displaystyle{|T_rf|\le f^*.}</math>
 
It follows that ''T''<sub>''r''</sub> ''f'' tends to ''f'' pointwise almost everywhere. In fact let
 
:<math>\displaystyle{\Omega(f)=\limsup_{r\rightarrow 1} |T_rf -f|.}</math>
 
If ''g'' is continuous, then the difference tends to zero everywhere, so Ω(''f'' − ''g'') = Ω(''f''). On the other hand ''f'' can be approximated arbitrarily closely in L<sup>1</sup> by continuous ''g''. Then, using [[Chebychev's inequality]],
 
:<math>m\{x: \, \Omega(f)(x)> \lambda\} =m\{x: \, \Omega(f-g)(x)> \lambda\} \le m\{x: \, (f-g)^*(x)> \lambda\} + m\{x: \, |f(x)-g(x)|> \lambda\} \le C\lambda^{-1}\|f-g\|_1.</math>
 
The right hand side can be made arbitrarily small, so that Ω(''f'') = 0 almost everywhere. A more refined argument shows that convergence occurs at each Lebesgue point of ''f''.
 
If ''f'' is integrable the conjugate Poisson integrals are defined and given by convolution by the kernel ''Q''<sub>''r''</sub>. This defines ''Hf'' inside |''z''| < 1. To show that ''Hf'' has a radial limit for almost all angles,<ref>{{harvnb|Katznelson|2004|p=64}}</ref> consider
 
:<math>\displaystyle{F(z)=\exp -f(z) -iHf(z),}</math>
 
where ''f''(''z'') denotes the extension of ''f'' by Poisson integral. ''F'' is holomorphic in the unit disk with |''F''(''z'')| ≤ 1. The restriction of ''F'' to a countable family of concentric circles gives a sequence of functions in L<sup>∞</sup>('''T''') which has a weak ''g'' limit in L<sup>∞</sup>('''T''') with Poisson integral ''F''. By the L<sup>2</sup> results, ''g'' is the radial limit for almost all angles of ''F''. It follows that ''Hf''(''z'') has a radial limit almost everywhere. This is taken as the definition of ''Hf'' on '''T''', so that ''T''<sub>''r''</sub>''H'' f tends pointwise to ''H'' almost everywhere.  The function ''Hf'' is of weak L<sup>1</sup> type.<ref>{{harvnb|Katznelson|2004|p=66}}</ref>
 
The inequality used above to prove pointwise convergence for L<sup>''p''</sup> function with 1 < ''p'' < ∞ make sense for L<sup>1</sup> functions by invoking the maximal function. The inequality becomes
 
:<math>\displaystyle{|H_\varepsilon f - T_{1-\varepsilon}Hf|\le 4f^*.}</math>
 
Let
 
:<math>\displaystyle{\omega(f)=\limsup_{\varepsilon\rightarrow 0} |H_\varepsilon f - T_{1-\varepsilon}Hf|.}</math>
 
If ''g'' is smooth, then the difference tends to zero everywhere, so ω(''f'' − ''g'') = ω(''f''). On the other hand ''f'' can be approximated arbitrarily closely in L<sup>1</sup> by smooth ''g''. Then
 
:<math>m\{x: \, \omega(f)(x)> \lambda\} =m\{x: \, \omega(f-g)(x)> \lambda\} \le m\{x: \, 4(f-g)^*(x)> \lambda\} \le C\lambda^{-1}\|f-g\|_1.</math>
 
The right hand side can be made arbitrarily small, so that ω(''f'') = 0 almost everywhere. Thus the difference for ''f'' tends to zero almost everywhere. A more refined argument can be given<ref>{{harvnb|Katznelson|2004|pp=78–79}}</ref> to show that, as in case of  L<sup>''p''</sup>, the difference tends to zero at all Lebesgue points of ''f''. In combination with the result for the conjugate Poisson integral, it follows that, if ''f'' is in L<sup>1</sup>('''T'''), then ''H''<sub>ε</sub>''f'' converges to ''Hf'' almost everywhere, a theorem originally proved by Privalov in 1919.
 
==General theory==
{{harvtxt|Calderón|Zygmund|1952}} introduced general techniques for studying singular integral operators of convolution type. In Fourier transform the operators are given by multiplication operators. These will yield bounded operators on L<sup>2</sup> if the corresponding multiplier function is bounded. To prove boundedness on L<sup>''p''</sup> spaces, Calderón and Zygmund introduced a method of decomposing L<sup>1</sup> functions, generalising the [[rising sun lemma]] of [[F. Riesz]]. This method showed that the operator defined a continuous operator from L<sup>1</sup> to the space of functionc of weak L<sup>1</sup>. The [[Marcinkiewicz interpolation theorem]] and duality then implies that the singular integral operator is bounded on all L<sup>''p''</sup> for 1 < ''p'' < ∞. A simple version of this theory is described below for operators on '''R'''. As {{harvtxt|de Leeuw|1965}} showed, results on '''R''' can be deduced from corresponding results for '''T''' by restricting the multiplier to the integers, or equivalently periodizing the kernel of the operator. Corresponding results for the circle were originally established by Marcinkiewicz in 1939. These results generalize to '''R'''<sup>''n''</sup> and '''T'''<sup>''n''</sup>. They provide an alternative method for showing that the Riesz transforms, the higher Riesz transforms and in particular the Beurling transform define bounded operators on L<sup>''p''</sup> spaces.<ref>See:
*{{harvnb|Hörmander|1990}}
*{{harvnb|Torchinsky|2005}}
*{{harvnb|Grafakos|2008}}
*{{harvnb|Stein|1970}}
*{{harvnb|Stein|Weiss|1971|pp=257–267}}</ref>
 
===Calderón-Zygmund decomposition===
{{see also|Rising sun lemma|Calderón–Zygmund lemma}}
Let ''f'' be a non-negative integrable or continuous function on [''a'',''b'']. Let ''I'' = (''a'',''b''). For any open subinterval ''J'' of [''a'',''b''], let ''f''<sub>''J''</sub> denote the average of |''f''| over ''J''. Let α be a positive constant greater than ''f''<sub>''I''</sub>. Divide ''I'' into two equal intervals (omitting the midpoint). One of these intervals must satisfy ''f''<sub>''J''</sub> < α since their sum is 2''f''<sub>''I''</sub> so less than 2α.  Otherwise the interval will satisfy α ≤ ''f''<sub>''J''</sub> < 2α. Discard such intervals and repeat the halving process with the remaining interval, discarding intervals using the same criterion. This can be continued indefinitely. The discarded intervals are disjoinr and their union is an open set Ω. For points ''x'' in the complement, they lie in a nested set of intervals with lengths decreasing to 0 and on each of which the average of ''f'' is bounded by α. If ''f'' is continuous these averages tend to |''f''(''x'')|. If ''f'' is only integrable this is only true almost everywhere, for it is true at the [[Lebesgue point]]s of ''f'' by the [[Lebesgue differentiation theorem]].  Thus ''f'' satisfies |''f''(x)| ≤ α almost everywhere on  Ω<sup>''c''</sup>, the complement of Ω. Let ''J''<sub>''n''</sub> be the set of discarded intervals and define the "good" function ''g'' by
 
:<math>\displaystyle{g(x)= \mathbf{Av}_{J_n}(f)\,\,\, (x\in J_n), \,\,\,\,\,g(x) = f(x)\,\,\, (x\in \Omega^c).}</math>
 
By construction |''g''(''x'')| ≤ 2α almost everywhere and
 
:<math>\displaystyle{\|g\|_1 \le \|f\|_1.}</math>
 
Combining these two inequalities gives
 
:<math>\displaystyle{\|g\|_p^p \le (2\alpha)^{p-1}\|f\|_1.}</math>
 
Define the "bad" function ''b'' by ''b'' = ''f'' − ''g''. Thus ''b'' is 0 off Ω and equal to ''f'' minus its average on ''J''<sub>''n''</sub>. So the average of ''b'' on ''J''<sub>''n''</sub> is zero and
 
:<math>\displaystyle{\|b\|_1 \le 2\|f\|_1.}</math>
 
Moreover since |''b''| ≥ α on Ω
 
:<math>\displaystyle{m(\Omega)\le \alpha^{-1} \|f\|_1.}</math>
 
The decomposition
 
:<math>\displaystyle{f(x) =g(x) + b(x)}</math>
 
is called the '''Calderón–Zygmund decomposition'''.<ref>{{harvnb|Torchinsky|2005|pp=74–76,84–85}}</ref>
 
===Multiplier theorem===
{{see also|Multiplier (Fourier analysis)|Marcinkiewicz interpolation theorem}}
Let ''K''(''x'') be a kernel defined on '''R'''\{0} such that
 
:<math>W(f)=\lim_{\varepsilon\to 0}\int_{|x|\ge \varepsilon} K(x)f(x)\,dx</math>
 
exists as a [[tempered distribution]] for ''f'' a [[Schwartz function]]. Suppose that the Fourier transform of ''T'' is bounded, so that convolution by ''W'' defines a bounded operator ''T'' on L<sup>2</sup>('''R'''). Then if ''K'' satisfies '''Hörmander's condition'''
 
:<math>A=\sup_{y\ne 0} \int_{|x|\ge 2|y|}|K(x-y) - K(x)|\, dx <\infty,</math>
 
then ''T'' defines a bounded operator on L<sup>''p''</sup> for 1 < ''p'' < ∞ and a continuous operator from L<sup>1</sup> into functions of weak type L<sup>1</sup>.<ref>{{harvnb|Grafakos|2008|pp=290–293}}</ref>
 
In fact by the Marcinkiewicz interpolation argument and duality, it suffices to check that if ''f'' is smooth of compact support then
 
:<math>m\{x:\, |Tf(x)| \ge 2\lambda\} \le (2A+4\|T\|)\cdot \lambda^{-1} \|f\|_1.</math>
 
Take a Calderón−Zygmund decomposition of ''f'' as above
 
:<math>f(x)=g(x)+b(x)</math>
 
with intervals ''J''<sub>''n''</sub> and with α = λμ, where  μ > 0. Then
 
:<math>m\{x:\, |Tf(x)| \ge 2\lambda\} \le m\{x:\, |Tg(x)| \ge \lambda\} + m\{x:\, |Tb(x)| \ge \lambda\}.</math>
 
The term for ''g'' can be estimated using Chebychev's inequality:
 
:<math>m\{x:\, |Tg(x)| \ge 2\lambda\} \le \lambda^{-2} \|Tg\|_2^2 \le \lambda^{-2}\|T\|^2 \|g\|_2^2 \le 2\lambda^{-1}\mu \|T\|^2 \|f\|_1.</math>
 
If ''J''* is defined to be the interval with the same centre as ''J'' but twice the length, the term for ''b'' can be broken up into two parts:
 
:<math>m\{x:\, |Tb(x)| \ge \lambda\}\le m\{x:\, x\notin \cup J_n^*,\,\,\, |Tb(x)| \ge \lambda\} +m(\cup J_n^*).</math>
 
The second term is easy to estimate:
 
:<math>m(\cup J_n^*)\le \sum m(J_n^*)=2\sum m(J_n) \le 2\lambda^{-1} \mu^{-1} \|f\|_1.</math>
 
To estimate the first term note that
 
:<math>b=\sum b_n, \qquad b_n =(f - \mathbf{Av}_{J_n}(f))\chi_{J_n}.</math>
 
Thus by Chebychev's inequality:
 
:<math>m\{x:\, x\notin \cup J_m^*,\,\,\, |Tb(x)| \ge \lambda\}\le \lambda^{-1}\int_{(\cup J_m^*)^c} |Tb(x)|\, dx \le \lambda^{-1} \sum_n\int_{( J_n^*)^c} |Tb_n(x)|\, dx.</math>
 
By construction the integral of ''b<sub>n</sub>'' over ''J<sub>n</sub>'' is zero. Thus, if ''y<sub>n</sub>'' is the midpoint of ''J<sub>n</sub>'', then by Hörmander's condition:
 
:<math>\int_{( J_n^*)^c} |Tb_n(x)|\, dx= \int_{( J_n^*)^c} \left|\int_{J_n} (K(x-y)-K(x-y_n))b_n(y)\, dy\right|\, dx \le \int_{J_n} |b_n(y)| \int_{( J_n^*)^c} |K(x-y)-K(x-y_n)|\, dxdy \le A\|b_n\|_1.</math>
 
Hence
 
:<math>m \left \{x:\, x\notin \cup J_m^*, |Tb(x)| \ge \lambda \right \} \le \lambda^{-1} A \|b\|_1 \le 2 A\lambda^{-1}\|f\|_1.</math>
 
Combining the three estimates gives
 
:<math>m\{x:\, |Tf(x)| \ge \lambda\} \le \left (2\mu\|T\|^2 +2\mu^{-1} + 2A \right )\lambda^{-1}\|f\|_1.</math>
 
The constant is minimized by taking
 
:<math>\mu=\|T\|^{-1}.</math>
 
The Markinciewicz interpolation argument extends the bounds to any L<sup>''p''</sup> with 1 < ''p'' < 2 as follows.<ref>{{harvnb|Hörmander|1990|p=245}}</ref> Given ''a'' > 0, write
 
:<math>f=f_a + f^a,</math>
 
where ''f''<sub>''a''</sub> = ''f'' if |''f''| < ''a'' and 0 otherwise and ''f''<sup>''a''</sup> = ''f'' if |''f''| ≥ ''a'' and 0 otherwise. Then by Chebychev's inequality and the weak type L<sup>1</sup> inequality above
 
:<math>m\{x:\, |Tf(x)| > a\} \le m \left \{x:\, |Tf_a(x)| > \tfrac{a}{2} \right \}+ m \left \{x:\, |Tf^a(x)| > \tfrac{a}{2} \right \}\le 4a^{-2}\|T\|^2 \|f_a\|_2^2 +C a^{-1}\|f^a\|_1.</math>
 
Hence
 
:<math>\begin{align}
\|Tf\|_p^p &= p\int_0^\infty a^{p-1} m\{x:\, |Tf(x)|> a\} \, da \\
&\le p \int_0^\infty a^{p-1} \left ( 4a^{-2}\|T\|^2 \|f_a\|_2^2 +C a^{-1}\|f^a\|_1 \right ) da \\
&=4\|T\|^2 \iint_{|f(x)|<a} |f(x)|^2 a^{p-3}\,dx\, da + 2C\iint_{|f(x)|\ge a} |f(x)| a^{p-2}\,dx\, da \\
&\le \left (4\|T\|^2(2-p)^{-1} + C (p-1)^{-1} \right ) \int |f|^p \\
&=C_p \|f\|_p^p.
\end{align}</math>
 
By duality
 
:<math>\|Tf\|_q \le C_p \|f\|_q.</math>
 
Continuity of the norms can be shown by a more refined argument<ref>{{harvnb|Torchinsky|2005|pp=87–91}}</ref> or follows from the [[Riesz–Thorin interpolation theorem]].
 
==Notes==
{{reflist|colwidth=30em}}
 
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*{{citation|last=Rosenblum|first= Marvin|last2= Rovnyak|first2= James|title=
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*{{citation|title=Introduction to Fourier Analysis on Euclidean Spaces|first=Elías M. |last=Stein|first2= Guido L.|last2= Weiss|publisher=Princeton University Press|year= 1971|
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[[Category:Operator theory]]
[[Category:Harmonic analysis]]
[[Category:Singular integrals]]

Latest revision as of 16:58, 5 November 2014

Four or five years ago, a reader of some of my columns bought the domain name jamesaltucher.com and gave it to me as a birthday gift. It was a total surprise to me. I didn't even know the reader. I hope one day we meet.
Two years ago a friend of mine, Tim Sykes, insisted I had to have a blog. He set it up for me. He even wrote the "About Me". I didn't want a blog. I had nothing to say. But about 6 or 7 months ago I decided I wanted to take this blog seriously. I kept putting off changing the "About Me" which was no longer really about me and maybe never was.
A few weeks ago I did a chapter in one of the books in Seth Godin's "The Domino Project". The book is out and called "No Idling". Mohit Pawar organized it (here's Mohit's blog) and sent me a bunch of questions recently. It's intended to be an interview on his blog but I hope Mohit forgives me because I want to use it as my new "About Me" also.
1. You are a trader, investor, writer, and entrepreneur? Which of these roles you enjoy the most and why?
When I first moved to New York City in 1994 I wanted to be everything to everyone. I had spent the six years prior to that writing a bunch of unpublished novels and unpublished short stories. I must've sent out 100s of stories to literary journals. I got form rejections from every publisher, journal, and agent I sent my novels and stories to.
Now, in 1994, everything was possible. The money was in NYC. Media was here. I lived in my 10�10 room and pulled suits out of a garbage bag every morning but it didn't matter...the internet was revving up and I knew how to build a website. One of the few in the city. My sister warned me though: nobody here is your friend. Everybody wants something
And I wanted something. I wanted the fleeting feelings of success, for the first time ever, in order to feel better about myself. I wanted a girl next to me. I wanted to build and sell companies and finally prove to everyone I was the smartest. I wanted to do a TV show. I wanted to write books
But everything involved having a master. Clients. Employers. Investors. Publishers. The market (the deadliest master of all). Employees. I was a slave to everyone for so many years. And the more shackles I had on, the lonelier I got
(Me in the Fortress of Solitude
Much of the time, even when I had those moments of success, I didn't know how to turn it into a better life. I felt ugly and then later, I felt stupid when I would let the success dribble away down the sink
I love writing because every now and then that ugliness turns into honesty. When I write, I'm only a slave to myself. When I do all of those other things you ask about, I'm a slave to everyone else
Some links
33 Unusual Tips to Being a Better Write
"The Tooth
(one of my favorite posts on my blog

2. What inspires you to get up and start working/writing every day
The other day I had breakfast with a fascinating guy who had just sold a piece of his fund of funds. He told me what "fracking" was and how the US was going to be a major http://www.pcs-systems.co.uk/Images/celinebag.aspx oil player again. We spoke for two hours about a wide range of topics, including what happens when we can finally implant a google chip in our brains
After that I had to go onto NPR because I firmly believe that in one important respect we are degenerating as a country - we are graduating a generation of indentured servants who will spend 50 years or more paying down their student debt rather than starting companies and curing cancer. So maybe I made a difference
Then I had lunch with a guy I hadn't seen in ten years. In those ten years he had gone to jail and now I was finally taking the time to forgive him for something he never did to me. I felt bad I hadn't helped him when he was at his low point. Then I came home and watched my kid play clarinet at her school. Then I read until I fell asleep. Today I did nothing but write. Both days inspired me
It also inspires me that I'm being asked these questions. Whenever anyone asks me to do anything I'm infinitely grateful. Why me? I feel lucky. I like it when someone cares what I think. I'll write and do things as long as anyone cares. I honestly probably wouldn't write if nobody cared. I don't have enough humility for that, I'm ashamed to admit

3. Your new book "How to be the luckiest person alive" has just come out. What is it about
When I was a kid I thought I needed certain things: a college education from a great school, a great home, a lot of money, someone who would love me with ease. I wanted people to think I was smart. I wanted people to think I was even special. And as I grew older more and more goals got added to the list: a high chess rating, a published book, perfect weather, good friends, respect in various fields, etc. I lied to myself that I needed these things to be happy. The world was going to work hard to give me these things, I thought. But it turned out the world owed me no favors
And gradually, over time, I lost everything I had ever gained. Several times. I've paced at night so many times wondering what the hell was I going to do next or trying not to care. The book is about regaining your sanity, regaining your happiness, finding luck in all the little pockets of life that people forget about. It's about turning away from the religion you've been hypnotized into believing into the religion you can find inside yourself every moment of the day

[Note: in a few days I'm going to do a post on self-publishing and also how to get the ebook for free. The link above is to the paperback. Kindle should be ready soon also.
Related link: Why I Write Books Even Though I've Lost Money On Every Book I've Ever Writte
4. Is it possible to accelerate success? If yes, how


Yes, and it's the only way I know actually to achieve success. Its by following the Daily Practice I outline in this post:
It's the only way I know to exercise every muscle from the inside of you to the outside of you. I firmly believe that happiness starts with that practice
5. You say that discipline, persistence and psychology are important if one has to achieve success. How can one work on improving "psychology" part
Success doesn't really mean anything. People want to be happy in a harsh and unforgiving world. It's very difficult. We're so lucky most of us live in countries without major wars. Our kids aren't getting killed by random gunfire. We all have cell phones. We all can communicate with each other on the Internet. We have Google to catalog every piece of information in history! We are so amazingly lucky already
How can it be I was so lucky to be born into such a body? In New York City of all places? Just by being born in such a way on this planet was an amazing success
So what else is there? The fact is that most of us, including me, have a hard time being happy with such ready-made success. We quickly adapt and want so much more out of life. It's not wars or disease that kill us. It's the minor inconveniences that add up in life. It's the times we feel slighted or betrayed. Or even slightly betrayed. Or overcharged. Or we miss a train. Or it's raining today. Or the dishwasher doesn't work. Or the supermarket doesn't have the food we like. We forget how good the snow tasted when we were kids. Now we want gourmet food at every meal
Taking a step back, doing the Daily Practice I outline in the question above. For me, the results of that bring me happiness. That's success. Today. And hopefully tomorrow
6. You advocate not sending kids to college. What if kids grow up and then blame their parents about not letting them get a college education
I went to one of my kid's music recitals yesterday. She was happy to see me. I hugged her afterwards. She played "the star wars theme" on the clarinet. I wish I could've played that for my parents. My other daughter has a dance recital in a few weeks. I tried to give her tips but she laughed at me. I was quite the breakdancer in my youth. The nerdiest breakdancer on the planet. I want to be present for them. To love them. To let them always know that in their own dark moments, they know I will listen to them. I love them. Even when they cry and don't always agree with me. Even when they laugh at me because sometimes I act like a clown
Later, if they want to blame me for anything at all then I will still love them. That's my "what if"
Two posts
I want my daughters to be lesbian
Advice I want to give my daughter


7. Four of your favorite posts from The Altucher Confidential
As soon as I publish a post I get scared to death. Is it good? Will people re-tweet? Will one part of the audience of this blog like it at the expense of another part of the audience. Will I get Facebook Likes? I have to stop clinging to these things but you also need to respect the audience. I don't know. It's a little bit confusing to me. I don't have the confidence of a real writer yet
Here are four of my favorites
How I screwed Yasser Arafat out of $2mm (and lost another $100mm in the process
It's Your Fault
I'm Guilty of Torturing Wome
The Girl Whose Name Was a Curs
Although these three are favorites I really don't post anything unless it's my favorite of that moment
8. 3 must-read books for aspiring entrepreneurs
The key in an entrepreneur book: you want to learn business. You want to learn how to honestly communicate with your customers. You want to stand out
The Essays of Warren Buffett by Lawrence Cunningha
"The Thank you Economy" by Gary Vaynerchu
"Purple cow" by Seth Godi
9. I love your writing, so do so many others out there. Who are your favorite writers
"Jesus's Son" by Denis Johnson is the best collection of short stories ever written. I'm afraid I really don't like his novels though
"Tangents" by M. Prado. A beautiful series of graphic stories about relationships
Other writers: Miranda July, Ariel Leve, Mary Gaitskill, Charles Bukowski, Celine, Sam Lipsyte, William Vollmann, Raymond Carver. Arthur Nersesian. Stephen Dubner

(Bukowski








Many writers are only really good storytellers. Most writers come out of a cardboard factory MFA system and lack a real voice. A real voice is where every word exposes ten levels of hypocrisy in the world and brings us all the way back to see reality. The writers above have their own voices, their own pains, and their unique ways of expressing those pains. Some of them are funny. Some a little more dark. I wish I could write 1/10 as good as any of them

10. You are a prolific writer. Do you have any hacks that help you write a lot in little time
Coffee, plus everything else coffee does for you first thing in the morning
Only write about things you either love or hate. But if you hate something, try to find a tiny gem buried in the bag of dirt so you can reach in when nobody is looking and put that gem in your pocket. Stealing a diamond in all the shit around us and then giving it away for free via writing is a nice little hack, Being fearless precisely when you are most scared is the best hack

11. I totally get and love your idea about bleeding as a writer, appreciate if you share more with the readers of this blog
Most people worry about what other people think of them. Most people worry about their health. Most people are at a crossroads and don't know how to take the next step and which road to take it on. Everyone is in a perpetual state of 'where do I put my foot next'. Nobody, including me, can avoid that
You and I both need to wash our faces in the morning, brush our teeth, shower, shit, eat, fight the weather, fight the colds that want to attack us if we're not ready. Fight loneliness or learn how to love and appreciate the people who want to love you back. And learn how to forgive and love the people who are even more stupid and cruel than we are. We're afraid to tell each other these things because they are all both disgusting and true
You and I both have the same color blood. If I cut my wrist open you can see the color of my blood. You look at it and see that it's the same color as yours. We have something in common. It doesn't have to be shameful. It's just red. Now we're friends. No matter whom you are or where you are from. I didn't have to lie to you to get you to be my friend
Related Links
How to be a Psychic in Ten Easy Lesson
My New Year's Resolution in 199


12. What is your advice for young entrepreneurs
Only build something you really want to use yourself. There's got to be one thing you are completely desperate for and no matter where you look you can't find it. Nobody has invented it yet. So there you go - you invent it. If there's other people like you, you have a business. Else. You fail. Then do it again. Until it works. One day it will
Follow these 100 Rules
The 100 Rules for Being a Good Entrepreneur
And, in particular this
The Easiest Way to Succeed as an Entrepreneu
In my just released book I have more chapters on my experiences as an entrepreneur
13. I advocate the concept of working at a job while building your business. You have of course lived it. Now as you look back, what is your take on this? Is it possible to make it work while sailing on two boats

Your boss wants everything out of you. He wants you to work 80 hours a week. He wants to look good taking credit for your work. He wants your infinite loyalty. So you need something back
Exploit your employer. It's the best way to get good experience, clients, contacts. It's a legal way to steal. It's a fast way to be an entrepreneur because you see what large companies with infinite money are willing to pay for. If you can provide that, you make millions. It's how many great businesses have started and will always start. It's how every exit I've had started
14. Who is a "person with true moral fiber"? In current times are there any role models who are people with true moral fiber


I don't really know the answer. I think I know a few people like that. I hope I'm someone like that. And I pray to god the people I'm invested in are like that and my family is like that
I find most people to be largely mean and stupid, a vile combination. It's not that I'm pessimistic or cynical. I'm very much an optimist. It's just reality. Open the newspaper or turn on the TV and watch these people
Moral fiber atrophies more quickly than any muscle on the body. An exercise I do every morning is to promise myself that "I'm going to save a life today" and then leave it in the hands of the Universe to direct me how I can best do that. Through that little exercise plus the Daily Practice described above I hope to keep regenerating that fiber

15. Your message to the readers of this blog
Skip dinner. But follow me on Twitter.



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