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Thread: Hilbert Transform

  1. #1
    May 2008

    Hilbert Transform


    The Hilbert transform is given by

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

    I would like
    (a) if f\in C_c^0(\Bbb R) (is continuous and compactly supported), then Hf exists and is continuous, and
    (b) 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

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

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

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

    Claim. If g \in C^0([0,d]) and g(0) = 0, then 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 ( \int_{|x-t|<d}) and from compact support, that it exists away from the pole ( \int_{|x-t|>d}) and existence follows.

    Proof of Claim.
    The integrand is bounded for t>0; it is also bounded as t\to 0+ because |g| \leq G for some G\in C^1(\Bbb R) so that we have \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.


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

    Let B_x and B_z be disjoint balls containing x and z, respectively, so the last integral is bounded iff 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 g(t) := f(t)/(z-t), we see g is continuous on B_x and the integral in question is Hg(x), so we have that it is finite and remains finite as |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) L^1(\Bbb R), I should have (b) for all of L^1, correct? Thanks.
    Last edited by sleepingcat; Feb 14th 2009 at 03:52 PM.
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