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In Terence Tao's notes page 1, cited below, he mentions that it is easy to see that

$\lim_{|x| \to \infty} xHf(x) = \frac{1}{\pi}\int f$

where $f$ is a Schwartz function and $H$ is the Hilbert transform. I'm not really seeing this. In particular, it seems to require exchanging the limits on the $\varepsilon$ and $|x|$, which I haven't been able to justify.

Any help would be greatly appreciated!

Tao: http://www.math.ucla.edu/~tao/247a.1.06f/notes4.pdf

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Let $f \in \mathcal{S}(\mathbb{R})$ be a Schwartz function. We start by showing that $\lim_{|x|\to \infty} H f (x) = 0$ holds. The only proof I know at the moment relies on the relationship $H f = - \mathrm{i} \mathcal{F}^{-1} [\mathcal{F}(f) \operatorname{sgn}]$ between the Hilbert and the Fourier transform. Since $\mathcal{F}(f) \operatorname{sgn} \in L^1 (\mathbb{R})$, the Riemann-Lebesgue lemma implies $H f \in C_0 (\mathbb{R})$. In particular, $H f$ vanishes at infinity.

The issues with the limit $\varepsilon \to 0^+$ can be resolved by writing the Hilbert transform without it. For $x \in \mathbb{R}$ we have \begin{align} \pi H f (x) &= \lim_{\varepsilon \to 0^+} \int \limits_{\mathbb{R} \setminus [-\varepsilon,\varepsilon]} \frac{f(x-t)}{t} \, \mathrm{d} t \stackrel{t \to -t}{=} - \lim_{\varepsilon \to 0^+} \int \limits_{\mathbb{R} \setminus [-\varepsilon,\varepsilon]} \frac{f(x+t)}{t} \, \mathrm{d} t \\ &= \frac{1}{2} \lim_{\varepsilon \to 0^+} \int \limits_{\mathbb{R} \setminus [-\varepsilon,\varepsilon]} \frac{f(x-t) - f(x+t)}{t} \, \mathrm{d} t = \int \limits_\mathbb{R} \frac{f(x-t) - f(x+t)}{2t} \, \mathrm{d} t \, , \end{align} so the Hilbert transform is essentially the integral of the central difference quotient.

Using this representation and the definition $g(x) = x f(x)$ for $x \in \mathbb{R}$, we can compute \begin{align} \pi x H f(x) - \int \limits_\mathbb{R} f(t) \, \mathrm{d} t &= \int \limits_\mathbb{R} \left[\frac{x f(x-t) - x f(x+t)}{2t} - \frac{1}{2} f(x-t) - \frac{1}{2} f(x+t)\right] \, \mathrm{d} t \\ &= \int \limits_\mathbb{R} \frac{g(x-t) - g(x+t)}{2t} \, \mathrm{d} t = \pi H g (x) \stackrel{|x| \to \infty}{\longrightarrow} 0 \, , \end{align} where the final limit follows from $g \in \mathcal{S}(\mathbb{R})$ and the first result.

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