## 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.