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'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.

$\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)

I provedEdit.(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.