# Math Help - Hilbert Transform

1. ## Hilbert Transform

Hi.

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|) 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'Hôpital's rule to conclude the limit is finite (is there an easier way to show this limit is finite?). Hence the integral exists.

Continuity.


\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 $B_x$ 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.