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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.

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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 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.