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Harmonic Analysis: The Hilbert Transform September 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)}
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 paper August 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.