L² boundedness of the Hilbert transform

Step 1 (find the multiplier via a regularization). The kernel $1/(\pi x)$ is not integrable, so approximate it: for $\epsilon>0$ let $k_\epsilon(x) = \frac{1}{\pi}\frac{x}{x^2+\epsilon^2}$, an odd, integrable approximation that converges to the principal-value kernel as $\epsilon\to 0$. A direct (contour or tables) computation gives $\hat k_\epsilon(\xi) = -i\,\operatorname{sgn}(\xi)\, e^{-2\pi\epsilon|\xi|}$.

Step 2 (pass to the limit). As $\epsilon \to 0^+$, $e^{-2\pi\epsilon|\xi|} \to 1$ for every fixed $\xi \ne 0$, and $k_\epsilon * f \to Hf$ for nice $f$. Passing to the limit inside the convolution theorem gives the multiplier identity $\widehat{Hf}(\xi) = -i\,\operatorname{sgn}(\xi)\, \hat f(\xi)$: the Hilbert transform acts on the frequency side by multiplication by $-i\,\operatorname{sgn}(\xi)$.

Step 3 (apply Plancherel). Since $|-i\,\operatorname{sgn}(\xi)| = 1$ for every $\xi \ne 0$ (a single point has measure zero and does not affect the integral), $|\widehat{Hf}(\xi)|^2 = |\hat f(\xi)|^2$ for a.e. $\xi$. Integrating and invoking Plancherel's theorem on both sides gives $\|Hf\|_{L^2} = \|f\|_{L^2}$: the Hilbert transform is an isometry on $L^2(\mathbb{R})$, in particular bounded.