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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.

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    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.2012-05-18
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    @Andrew I found this multiplier formula in literature as well, but I don't think we're supposed to use it since nothing similar was ever mentioned.2012-05-19

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