# Hilbert Transform: limit of xHf(x)

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!

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.