## The Poisson Integral

So I’m sure I’m losing readers/upsetting some readers with my recent mumbo-jumbo, and I shall continue in that vein. I will answer a long standing question: What is the Poisson Integral? Before I begin, I will not be doing this in all its glory.

So let’s clear up what I consider to be a misnomer. This should not be called the Poisson integral, but rather the Poisson integral representation. Let U be the unit disc in the complex plane. Let T be the unit circle. Then if A is a vector space of continuous complex functions on the closed unit disc $\overline{U}$, A contains all polynomials and $\sup_{z\in U}|f(z)|=\sup_{z\in T}|f(z)|$ for every $f\in A$, then the Poisson integral representation

$\displaystyle f(z)=\frac{1}{2\pi}\int_{-\pi}^\pi \frac{1-r^2}{1-2r\cos(\theta -t) +r^2}f(e^{it})dt$ where $z=re^{it}$ is valid for every $f\in A$ and every $z\in U$.

OK. Let’s break this down. It isn’t so bad. (Why is this in my real analysis textbook?!?) So first note that A contains all polynomials. It may not contain anything extra than that, and that is OK. If we throw extra stuff in, then we need to make sure it is still a vector space. That next condition can be stated in a different way. It is sort of a “maximum modulus” type condition. We want $\|f\|_U=\|f\|_T$, where we use the typical norm for an operator on a Banach space ($\|f\|_K=\sup\{\|f(z)\| : z\in K, \|z\|\leq 1\}$).

Now to get the integral representation, fix a z in U. Now be the Riesz Representation Theorem we have that there is a Borel measure, $\mu_z$, such that $f(z)=\int_T fd\mu_z$. Since z is complex we can write it as $z=re^{i\theta}$ for some |r|<1. Let $u_n(w)=w^n$, then we have the that the collection $span\{u_n\}_n$ is dense in A by Stone-Weierstrass, so let’s see what we can get out of the integral representation for this collection.

$u_n(z)=z^n=r^ne^{in\theta}=\int_T u_nd\mu_z$ for n=0,1,2,… and since $u_{-n}=\overline{u_n}$ on T, we have $r^{|n|}e^{in\theta}=\int_T u_nd\mu_z$ now for $n=0,\pm 1, \pm 2, \ldots$.

Now define $\displaystyle P_r(\theta -t)=\sum_{-\infty}^\infty r^{|n|}e^{in(\theta-t)}$ where t is real. But by the Dominated Convergence Theorem we have that $\displaystyle r^{|n|}e^{in\theta}=\frac{1}{2\pi}\int_{-\pi}^\pi P_r(\theta-t)e^{int}dt$, but here we have constructed what we want $\int_T fd\mu_z=\frac{1}{2\pi}\int_{-\pi}^\pi f(e^{it})P_r(\theta - t)dt$.

But you say this doesn’t look the same as you said. Well, let’s deconstruct $P_r(\theta-t)$ which is called the Poisson kernel. Since the series is symmetrical about zero, we can break it into $\displaystyle \sum_{-\infty}^\infty r^{|n|}e^{in(\theta-t)}=1+2\sum_{n=1}^\infty (ze^{-it})^n=\frac{e^{it}+z}{e^{it}-z}=\frac{1-r^2+2ir\sin(\theta - t)}{|1-ze^{-it}|^2}$. But we only care about the real part (why?), so $\displaystyle P_r(\theta-t)=\frac{1-r^2}{1-2r\cos(\theta - t)+r^2}$, i.e. the Poisson integral in the form first stated!

## Language Folly

I know this isn’t going to make much sense to anyone, but if you want to say:
あなたの男の子がかわいいです。

Then you probably shouldn’t say:
あなたの男の子が嫌いです。

They may not look close, but it is the difference between kawaii and kirai (or your boy is cute vs I hate your boy).

## Fourier Series Theorem

I had many things that I wanted to talk about, but when I read this theorem, it was so shocking that I just had to post it. Now from a general intuition standpoint, you might think this theorem to be quite natural. But remember, most of us have been trained to think Fourier series are extremely nicely behaved. In fact, if I had read the wikipedia article when I was learning Fourier series years ago, I wouldn’t be surprised as much since under Divergence, it tells us this stuff.

So secretly we always are working with $L^2(T)$, or square-integrable $2\pi$-periodic functions. When we think in this way, we have the Fourier series of f at x given by the partial sums $s_n(f; x)=\frac{1}{2\pi}\int_{-\pi}^\pi f(t)D_n(x-t)dt$ where $D_n(t)=\sum_{k=-n}^n e^{ikt}$. It turns out quite simply that in $L^2$-norm the partial sums converge to f quite quickly. This nice convergence tricks us into thinking that we will have nice convergence all the time.

So if we switch to just continuous $2\pi$-periodic functions (which is a dense subset of $L^2(T)$), do we get something as simple as point-wise convergence (it would surely be too much to ask for uniform convergence)? Well, from a common measure theory theorem, since the sequence converges in $L^2$-norm we have a subsequence converging point-wise almost everywhere. But this leaves much room for error. How much error you ask?

It turns out that there is is a set $E\subset C(T)$ which is a dense $G_\delta$ in C(T) which has the following property: For each $f\in E$, the set $Q_f=\{x: s^*(f; x)=\infty\}$ is a dense $G_\delta$ in $\mathbb{R}$. Note that $s^*(f;x)=sup_n |s_n(f;x)|$.

This is pretty rough considering it means in non-technical terms that continuous functions are completely filled with functions for which the points where the Fourier series behaves badly is almost everywhere. Also, note that a “dense $G_\delta$” is uncountable (nice little topological proof if you want to try it), so this isn’t like some minimally dense set we’re talking about.

I was going to prove the theorem, but now I don’t think I will because, there may be at best one person that has made it this far. If interested drop a comment and I’ll gladly add the proof here, though. I just need to make sure there is an interest before going forth with it.

We should note that it doesn’t take much to correct the problems stated here. If we just make sure that our function is Lipschitz of some order, then we have a convergent Fourier series.

## The Fuss over Gravity

So a nice discussion on the extra spatial dimensions post has led me to elaborate on what the fuss over gravity is. I have a wide range of audience, and most are probably in the philosophy school due to Philosophers’ Carnival. I think it is sort of important to know the controversy in physics right now, especially as a philosopher pondering the nature of the universe, and with CERN nearing its completion.

It is true that this fuss is over gravity, but it isn’t in an extra-dimensional sense. The four fundamental forces are the strong force, electromagnetic force, weak force, and gravity. The first three fit perfectly together through quantum mechanics. Then gravity has its own theory: general relativity. I actually listed them in order of strength. Gravity is by far the weakest and the strong force is the strongest (hence the name).

The problem is to get gravity into a quantum mechanics theory. The problem has to do with the uncertainty principle. Essentially, gravity is nonrenormalizable. This is terribly challenging to explain, but when you attempt to correct for small errors, those errors have to go to zero, but with gravity the errors build up and actually go to infinity. In a sense gravity “blows up.” Now if we try the other way, to put the really small into general relativity, we get what are known as singularities. This may be possible to overcome, but the postulates of quantum mechanics cannot be carried over to curved space-time (which is the whole point of GR).

So instead of trying to put one into the other people are trying completely new things and trying to get QM and GR as special cases of a more general theory. There are many attempts to do this. There is loop quantum gravity, topological quantum field theory, string theory, etc. It turns out that so far string theory is the only one that can get gravity into it successfully. It also turns out that string theory demands extra dimensions.

Disclaimer: Before some crazy physicist yells at me that other theories have fit gravity into their picture, I am saying this from my own knowledge. I know that there are many many different attempts at this, it is to my knowledge that none have successfully incorporated gravity without severe compromises somewhere else.

## The House of Mirth

I lied to you again. It turns out that Gravity’s Rainbow has been delayed by a week. I started looking up influences and references in it, and it turns out I have a lot of reading to do before starting the novel. Second, the book club I joined meets this Sat, and I haven’t even started the book. The book is Edith Wharton’s The House of Mirth.

OK. Since I’m interested in post-modernism, I would never dream of reading Wharton on my own. She writes social realism, which seems to be a wasted genre. Realism definitely has its pluses, since it points out faults with society, but to be considered literary enough for a book club is beyond me. When your main concern is to get things right, how can you include the millions of other aspects of a work of literature that actually requires interpretation and hence would make good discussion.

I started it, and it is quite enjoyable. At the same time it is highly frustrating. Wharton’s intent is to point out many dreadful aspects of society (of her time), but unintentionally she is pointing out something much bigger in my mind. The clearest theme (and I’m quite sure it is unintentional) that comes across is how unreflective western society is. These people are going through their entire lives and not once asking themselves any of what I consider the “big questions.” They don’t even realize that there could be something to think and talk about outside of their shallow thoughts.

I could be wrong about whether Wharton wanted this to come across or not, but I don’t think so. Her themes (as realism dictates) are quite obvious. Since she is portraying realistically, though, the theme I just discussed inevitably comes across as well to someone concerned with that aspect of society.

This just reminds me how far art has come in such a short period of time. This was only about 20 years before Faulkner, but Faulkner is at such a higher level. Other than theme, I can’t really think of good things to talk about in Wharton. If we were to discuss Faulkner, we would have to start at such a more fundamental level. What is the plot of the novel? There would be disagreement. What are all these crazy devices being used? Why? Look at all these different levels and patterns emerging, etc.

When we move even further along the literary timeline we get even crazier things like: Character A is thinking about Character B. Suddenly the narrative shifts to B, and travels via analepsis back in time to her life with Character C. Character C takes up the narrative and analeptically shifts the focus back in time farther to an event that shaped his life; then the focus returns to C’s “present.” The prose is then recycled back to its starting point with Character A, who is currently inhabiting the reader’s “real-time” present.

I just don’t understand why a book club would pick a novel that is so one-dimensional with so many other novels out there that have extreme interpretations worth discussing.

## Property of the Plane

I was reading the blog topological musings, and the category talk reminded me of this neat thing I posted on AoPS awhile ago.

By using the Mayer-Vietoris sequence in cohomology, we determine whether or not $\mathbb{R}^2$ can be written as the union of two open connected sets, U and V, such that $U\cap V$ is disconnected.

Well, it seems that we are concerned with $H^0$ stuff, since that tells us the number of connected components.

The sequence gives $0\to H^0(\mathbb{R}^2)\to H^0(U)\oplus H^0(V)\to H^0(U\cap V)\to 0$, a nice short exact sequence in which we mostly know they are connected, so

$0\to \mathbb{R}\to \mathbb{R}\oplus \mathbb{R}\to H^0(U\cap V)\to 0$, basically the question reduces to: Is it possible for the second to last term to have dimension greater than 1, but the sequence to remain exact?

Well, no, since the exactness tells us the dimension of $H^0(U\cap V)$ is the dimension of $\frac{\mathbb{R}\oplus\mathbb{R}}{\mathbb{R}}\cong \mathbb{R}$.

So we have a nifty result: the plane cannot be broken into two connected parts in which the intersection of these two parts is disconnected.

Challenge for the readers: This seems extremely simple and obvious. Find a proof that does not require advanced techniques such as cohomology.

## When Art is Meaningless

I recently read Kafka on the Shore by Haruki Murakami. Murakami brings up many fantastic themes in a very complicated plot, but simple writing style. I would like to focus on a very often overlooked function of art.

Page 393:

“Writing things was important, wasn’t it?” Nakata asked.
“Yes, it was. The process of writing was important. Even though the finished product is completely meaningless.”

This quote reminds me that most of the time artists create to express themselves; to explore something that is bothering them, maybe. When interpreting a work, we often forget about this. This brings to mind two things. First, maybe we don’t have any right to judge a work of art. The work could be completely meaningless to everyone except the artist, but it did its job. It helped the artist through something. Who are we to judge whether it is good or not? Second, if we are to make judgments, write papers, critique, interpret, etc, then we really should take into consideration that the work could be meaningless to us.

This seems to have some interesting ramifications in analyzing math as an art form. When the lay person looks at a truly beautiful proof, all they get is something that is meaningless. The mathematician (eh erm, Andrew Wiles), may have been tormented by the concept for years. The end product that gets spewed out is something that the mathematician “needed” to do. It is a personal thing that doesn’t need to make sense to others.

This post is sort of interesting, since now that I finished this novel I am moving on to what I am calling the “Gravity’s Rainbow Challenge.” I will attempt to read (and probably write here about what I read) all of Gravity’s Rainbow in a two month period. This work is often considered one of the most challenging novels of all time. I need to bookmark this post to remind myself that despite thousands of Ph.D. thesis and many full length books being written on how to interpret this novel, the novel could be something that the author needed to get out and is not supposed to be something that outsiders get.

In other news. I gave up on that proof from the last post. It is too difficult, and was sort of a spur of the moment thing that I don’t have enough interest in to finish.