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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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    @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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$Hf$ 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.)