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

$lim_x xHf(x) = frac1piint f$

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

Let $f in mathcalS(mathbbR)$ be a Schwartz function. We start by showing that $lim_x H f (x) = 0$ holds. The only proof I know at the moment relies on the relationship $H f = - mathrmi mathcalF^-1 [mathcalF(f) operatornamesgn]$ between the Hilbert and the Fourier transform. Since $mathcalF(f) operatornamesgn in L^1 (mathbbR)$, the Riemann-Lebesgue lemma implies $H f in C_0 (mathbbR)$. 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 $x in mathbbR$ we have
beginalign
pi H f (x) &= lim_varepsilon to 0^+ int limits_mathbbR setminus [-varepsilon,varepsilon] fracf(x-t)t , mathrmd t stackrelt to -t= - lim_varepsilon to 0^+ int limits_mathbbR setminus [-varepsilon,varepsilon] fracf(x+t)t , mathrmd t \
&= frac12 lim_varepsilon to 0^+ int limits_mathbbR setminus [-varepsilon,varepsilon] fracf(x-t) - f(x+t)t , mathrmd t = int limits_mathbbR fracf(x-t) - f(x+t)2t , mathrmd t , ,
endalign
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 $x in mathbbR$, we can compute
beginalign
pi x H f(x) - int limits_mathbbR f(t) , mathrmd t &= int limits_mathbbR left[fracx f(x-t) - x f(x+t)2t - frac12 f(x-t) - frac12 f(x+t)right] , mathrmd t \
&= int limits_mathbbR fracg(x-t) - g(x+t)2t , mathrmd t = pi H g (x) stackrelxlongrightarrow 0 , ,
endalign
where the final limit follows from $g in mathcalS(mathbbR)$ and the first result.