Hilbert transform

Remarks

Example: for $f(x) = \frac{1}{1 + x^2}$, one has $Hf(x) = \frac{x}{1 + x^2}$. Check: $F(z) = \frac{i}{z + i}$ is holomorphic on the upper half-plane and tends to $0$ at infinity. On $\R$ it equals $\frac{1}{1 + x^2} + i \frac{x}{1 + x^2}$, so its imaginary part is the harmonic conjugate of its real part.

Non-example: the integral is singular at $y = x$, so the principal value is needed. The integral $\int_\R \frac{f(y)}{x - y}\, dy$ diverges absolutely at every $x$ near which $f$ is continuous and nonzero.

On the Fourier transform side, $\widehat{Hf}(\xi) = -i \sgn(\xi)\, \hat f(\xi)$ for $f \in L^2(\R)$. Hence $H$ is an isometry of $L^2(\R)$ with $H^2 = -\id$. For $1 < p < \infty$, $H$ is a bounded operator on $L^p(\R)$, a theorem of Marcel Riesz; for $p = 1$ it is only of weak type $(1, 1)$.