# Integrability of the Hilbert transform of a Schwartz function

Given a Schwartz function $f\colon\mathbb{R}\to\mathbb{R}$, define its Hilbert transform by $$(Hf)(x)=\frac{1}{\pi}\left(\int_{|t|\leq 1}\frac{f(x-t)-f(x)}{t}\,dt + \int_{|t|\geq 1}\frac{f(x-t)}{t}\,dt\right)$$ (the first integral is interpreted as an appropriate limit/principal value).

It can be shown that $Hf$ is continuous and that $\lim_{|x|\to\infty}x(Hf)(x)=\frac{1}{\pi}\int_{\mathbb{R}}f(t)\,dt=\frac{1}{\pi}\hat{f}(0)$. Using this, it is easy to prove that, if $Hf$ is absolutely integrable, $\hat{f}(0)$ must be $0$. I want to prove the converse.

So assume $\lim_{|x|\to\infty}x(Hf)(x)=0$. This by itself doesn't guarantee integrability of $Hf$, since $Hf$ might behave like $\frac{1}{x\log x}$ at infinity. A stronger decay condition for $Hf$ is needed, but I'm not sure where to get it from. The fact that $f$ is Schwartz should be important; does this imply $Hf$ decays faster than any polynomial?

I don't know if it is supposed to use the formula $F[Hf]=-i\,\mathrm{sign\,} \xi \hat{f}(\xi)\,$. Since $\hat{f}$ is smooth (from the Schwartz class), the possible slow decrease of $Hf$ is given by (and equal to $C_1/x$, where $C_1$ is proportional to) the jump of $F[Hf]$ at the origin. The next member of asymptotic should be like $C_2/x^2$, there $C_2$ depends on the jump of the first derivative of $F[Hf]$ at the origin etc. –  Andrew May 18 '12 at 9:29
$Hf$ does not have to decay faster than any polynomial; as @Andrew suggested, $1/x^2$ is the best we can expect.
The assumption $\int_{\mathbb R}f(t)\,dt=0$ implies that the antiderivative $F(t):=\int_{-\infty}^t f(s)\,ds$ is also integrable (in fact, it's in the Schwartz class too). This should allow you to integrate the second integral by parts (and the first one can be estimated with the MVT.)