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?
Note, since this is a homework question, please don't be overly explicit in your answers.