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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.