In his lecture notes on the Hilbert transform, Tao claims without proof that if $f$ is in the Schwartz class, then $$ \lim_{|x|\to\infty}xHf(x)=\frac{1}{\pi}\int_{\bf R}f\tag{*} $$ where $H$ is the Hilbert transform. Intuitively, the $x$ on the left might "cancel out" the denominator in the definition of $Hf$. A naive attempt of change of variables yields: $$ x\int_{|t|>\epsilon}\frac{f(x-t)}{t}\ dt =\int_{\epsilon+x}^\infty\frac{xf(u)}{x-u}\ du +\int_{-\infty}^{x-\epsilon}\frac{xf(u)}{x-u}\ du. $$
If "$\lim_{|x|\to\infty}$" can be put inside the integral sign, then we are done. But I don't know how to justify it since I don't have a uniform estimate for $|\frac{x}{x-u}-1|$.
Would anyone elaborate how ($*$) is done?
Here is an excerpt of the notes:
Thanks to Daniel Fischer's comment, I have an answer.
Note that $$ \pi Hf(x) = \int_{|y-x| > 1} \frac{f(y)}{x-y}\ dy + \lim_{\epsilon\to 0}\int_{\epsilon\leq|y-x| \leq 1} \frac{f(y)-f(x)}{x-y}\ dy\tag{2} $$ This can be show by observing that convolution commutes: $f*g=g*f$ and that $$ \int_{\epsilon<|y-x| \leq 1} \frac{f(x)}{x-y}\ dy=0 $$ by "symmetry".
Since we have the pointwise limit: $$ \lim_{|x|\to\infty}1_{|x-y|>1}(y)\frac{f(y)}{x-y}=0, $$ dominated convergence shows that the first term on the RHS of (2) is gone as $|x|\to\infty$. On the other hand, the mean value theorem implies that the absolute value of the second term can be controlled by $$ 2|f'(\xi_x)|,\quad \xi_x\in (x-1,x+1), $$ which goes to $0$ as $|x|\to\infty$ since $f$ is in the Schwartz space $\mathcal{S}({\bf R})$. We have thus shown that
Now note that1 $$ xHf(x)=\frac{1}{\pi}\int_{\bf R}f+Hg(x), $$ where $g(x)=xf(x)\in\mathcal{S}({\bf R})$. We are done.
1 Observe that $$ x\int_{|t|>\epsilon}\frac{f(x-t)}{t}\ dt =\int_{|u-x|>\epsilon}\frac{x-u}{x-u}\cdot f(u)\ du +\int_{|u-x|>\epsilon}\frac{u\cdot f(u)}{x-u}\ du. $$