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Thread: Hilbert Transform

  1. #1
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    Hilbert Transform

    Hi.

    The Hilbert transform is given by

    $\displaystyle Hf(x) := p.v.\frac1\pi \int_{-\infty}^\infty \frac{f(t)}{x-t}\ dt$

    I would like
    (a) if $\displaystyle f\in C_c^0(\Bbb R)$ (is continuous and compactly supported), then $\displaystyle Hf$ exists and is continuous, and
    (b)$\displaystyle H^2 = -I$.

    I have a tentative proof of the first; I really need help on the second.

    Proof of (a).

    Existence of Integral.

    Let f be continuous and compactly supported. By changing variables, etc., we can consider only the existence of

    $\displaystyle p.v.\int_{-\infty}^\infty \frac{f(t)}{t}\ dt$.

    Let $\displaystyle g(t) := f(t) - f(0)$ (so that $\displaystyle g(0) = 0$) to obtain

    $\displaystyle p.v.\int_{-\infty}^\infty \frac{g(t)}{t}\ dt + \underbrace{p.v.\int_{-\infty}^\infty \frac{f(0)}{t}\ dt}_{= 0}$.

    Claim. If $\displaystyle g \in C^0([0,d])$ and $\displaystyle g(0) = 0$, then $\displaystyle p.v.\int_0^d \frac{g(t)}t\ dt$ exists.

    From this, we can conclude that the principal value integral exists near the pole ($\displaystyle \int_{|x-t|<d}$) and from compact support, that it exists away from the pole ($\displaystyle \int_{|x-t|>d}$) and existence follows.

    Proof of Claim.
    The integrand is bounded for $\displaystyle t>0$; it is also bounded as $\displaystyle t\to 0+$ because $\displaystyle |g| \leq G$ for some $\displaystyle G\in C^1(\Bbb R)$ so that we have $\displaystyle \lim_{t\to 0+}\frac{|g(t)|}t \leq \lim_{t\to 0} \frac{G(t)}t$ and we can apply l'HŰpital's rule to conclude the limit is finite (is there an easier way to show this limit is finite?). Hence the integral exists.


    Continuity.

    $\displaystyle
    \begin{aligned}
    \left|\ p.v.\int_{-\infty}^\infty \frac{f(t)}{x-t}\ dt - p.v. \int_{-\infty}^\infty \frac{f(t)}{z-t}\ dt\ \right| &= \left|\ p.v.\int_{-\infty}^\infty \frac{(x-z)f(t)}{(x-t)(z-t)}\ dt\ \right|\\
    &= |x - z|\left|\ p.v.\int_{-\infty}^\infty \frac{f(t)}{(x-t)(z-t)}\ dt\ \right|
    \end{aligned}
    $

    Let $\displaystyle B_x$ and $\displaystyle B_z$ be disjoint balls containing x and z, respectively, so the last integral is bounded iff $\displaystyle p.v.\int_{B_x} \frac{f(t)}{(x-t)(z-t)}\ dt + p.v.\int_{B_z} \frac{f(t)}{(x-t)(z-t)}\ dt$ is bounded. The integrals are similar so we consider only the first. Setting $\displaystyle g(t) := f(t)/(z-t)$, we see g is continuous on $\displaystyle B_x$ and the integral in question is $\displaystyle Hg(x)$, so we have that it is finite and remains finite as $\displaystyle |x-z|\to 0$, so that we have continuity (this argument is rough; will touch up later; also, better arguments appreciated!)


    I know this is sloppy; sorry. Is this correct? and can anyone point me in the right direction on (b)? Thanks a whole lot—I really appreciate it. _(_ _)_

    (p.s. is there an \inline{...} command? Integrals look kind of ugly in the middle of lines of text... thanks)

    Edit. I proved (b) for all simple functions by direct computation on an indicator function and then used linearity. Since simple functions are dense in (say) $\displaystyle L^1(\Bbb R)$, I should have (b) for all of $\displaystyle L^1$, correct? Thanks.
    Last edited by sleepingcat; Feb 14th 2009 at 02:52 PM.
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