## Harmonic Analysis: The Hilbert TransformSeptember 30, 2010

Posted by Sarah in Uncategorized.
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I’ve really been enjoying my harmonic analysis class and I thought I’d write up some recent notes.

The Poisson kernel, $P_r(\theta)$, is defined as
$\sum r^|k| e^{i k \theta} = 1 + 2 Re(\frac{z}{1-z})$
$= Re(1 + 2 \frac{z}{1-z}) = Re(\frac{1+z}{1-z}.$
This is the real part of a holomorphic function. Now, what are the properties of that holomorphic function?
We define the Herglotz kernel as
$f * H_r(\theta) = f * P_r + i f * q_r$
where $q_r(\theta) = Im(\frac{1+z}{1-z})$.
As $r \to 1, f * P_r(\theta) \to f$ (this is why the Poisson kernel solves the Dirichlet problem in the disc.) Similarly, as $r \to 1, i f * q_r(\theta) \to \tilde{f}$, some unknown function. This is the conjugate function of f, and convolution with $q_r$ is known as the Hilbert transform.

Now, we can write $q_r(\theta)$ explicitly as
$i q_r = \sum sgn k r^|k| e^{i k \theta}$
similar to the Poisson kernel, except for the sign function. Summing this series,
$q_r(\theta) = \frac{i 2 r \sin \theta}{1 - 2 r cos \theta + r^2}$ while
$P_r(\theta) = \frac{1 - r^2}{1 - 2 r cos \theta + r^2}.$
This conjugate function of $f$ need not be bounded, even if $f$ is. In other words, the Hilbert transform is not bounded in the $L^\infty$ norm. It’s not bounded in the $L^1$ norm either. But it’s clearly bounded in the $L^2$ norm.

The Hilbert transform can be shown to be bounded for all $L^p, 1 < p < 2$, by first showing that it satisfies a weak-type inequality and then showing that all linear operators satisfying a weak-type inequality are bounded in $L^p, 1 < p < 2$ (this is called the Marcinkiewicz Interpolation Theorem.) I might type that up another time.

If we let $F = f + i \tilde{f}$, then $F$ is sort of an envelope for $f$, larger in absolute value and smoother, and having the property that it stretches and shrinks with the function.

## What is the Selberg Trace Formula?August 19, 2010

Posted by Sarah in Uncategorized.
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Ngo Bau Chau was one of the winners of the Fields Medal for 2010, so I thought I’d try to understand what it is he studied. The laudation gives a clear explanation of his contribution to the Langlands program. Ngo’s proof applies the Selberg trace formula, and since that’s accessible, I thought I’d try to understand that.

If $G$ is a Lie group and $\Gamma$ is a cofinite discrete group, we want to understand the operator

$R(f) = \int_\Gamma f(y) R(y) dy$
on $L^2(G/\Gamma)$.

$R(f)$ is an integral operator, with kernel
$K(f) = \sum_{\gamma \in \Gamma} f(x^{-1} \gamma y)$
And now we can express the trace of the operator as
$\int_G \sum_{\gamma \in \Gamma} f(x^{-1} \gamma x) dx$
The result is
$\sum_{[\gamma]} \int_{\Gamma_\gamma / G} f(x^{-1} \gamma x) dx = \sum_{\pi} tr(\pi(f))$
where $[\gamma]$ are the conjugacy classes of $\gamma$, $\Gamma_\gamma$ is the centralizer of $\gamma$ in $\Gamma$, and $\pi$ are the automorphic representations.

When $\Gamma = \mathbb{Z}$ and $G = \mathbb{R}$, the Selberg Trace Formula is the Poisson Summation Formula

$\sum_{n = -\infty}^\infty f(t + nT) = \frac{1}{T}\sum_{k = -\infty}^\infty \hat{f}(k/T) exp(2 \pi i k/T t).$

The Poisson Summation formula can be seen to be a trace formula as follows:
The eigenvectors of the positive Laplacian on the unit circle are $m^2$ for integer $m$, and the eigenfunctions are $\phi_m = (2\pi)^{1/2} e^{i mx}.$
Consider the linear operator
$(Lf)(x) = \int_0^{2\pi} k(x, y) f(y) dy$
with kernel
$k(x, y) = \sum h(m)\phi_m(x) \bar{\phi_m(y)}$
Then
$L \phi_m = h(m) \phi_m$ and so the Poisson Summation Formula says
$Tr L = \sum_m h(m) = \sum \int h(\rho) e^{2 \pi i n \rho} d\rho$

The Selberg Trace Formula is proven in a similar manner to the proof of the Poisson Summation Theorem. If $R$ is the regular representation of $G$ on $L^2(\Gamma/G)$
$[R(g) \phi](x) = \phi(xg), g\in G, x \in \Gamma / G$
then we can write R as
$R(f)\phi(x) = \int_G f(x^{-1}g) \phi(g) dg$
fixing a Haar measure on $G$. Then, by splitting the integral
$R(f) \phi(x) = \int_{\Gamma / G} \sum_{\gamma \in \Gamma} f(x^{-1} \gamma g)\phi(g) dg = \int_{\Gamma / G} K_f(x, y) \phi(y) dy$.

We now take the trace of this operator in two different ways:
on one side,
$tr R = \int_{\Gamma / G} K_f(x, x) dx$
We break the sum over $\gamma$ into conjugacy classes of $\gamma$. Each conjugacy class contributes
$\int_{\Gamma / G} \sum_{\delta \in \Gamma_\gamma / \Gamma}f(x^{-1}\delta^{-1} \gamma \delta x) dx = \int_{\Gamma / G} f(x^{-1} \gamma x) dx$
On the other side, we can compute the trace alternatively by quoting a result that says $L^2(\Gamma / G)$ decomposes into a direct sum of irreducible representations of G:
$\sum_{\pi} tr(\pi(f))$.

This completes the proof. Conjugacy classes and irreducible representations are dual — in the same way that integers (in the Fourier series) and functions $e^{2 \pi i nx}$ are dual in the Poisson summation formula.

So what can you do with the trace formula? One thing is to let $\Gamma$ be the fundamental group of a Riemann surface, and describe the spectrum of differential operators such as the Laplace-Beltrami operator using geometric data like the lengths of geodesics. (Way back in sophomore year I was trying to learn about this from Peter Buser’s book.) The Selberg Trace Formula is also useful in analyzing the Riemann zeta function.

See these notes for a bunch of background on the Selberg Trace Formula.

## Meyer/Basarab paperAugust 18, 2010

Posted by Sarah in Uncategorized.
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this is what I’m reading through right now: “A variant on the compressed sensing of Emmanuel Candes,” by Yves Meyer and Matei Basarab. I’m not done with it, but it involves reconstructing a function by sub-Nyquist sampling of its Fourier coefficients, using an $\ell^1$-minimization technique. It’s a generalization of a theorem by Terence Tao about functions on finite fields. Looks to be interesting.