Inversion of the Hilbert transform

Remarks

For $f \in L^2(\R)$ the Fourier transform of $Hf$ is $-i \sgn(\xi)\, \hat f(\xi)$. Applying this twice multiplies $\hat f$ by $(-i \sgn \xi)^2 = -1$ for $\xi \ne 0$, which gives $H^2 f = -f$ on $L^2$. For general $p$, M. Riesz's theorem says that $H$ is bounded on $L^p$, so $H^2$ is bounded too, and the identity extends from the dense subspace $L^2 \cap L^p$ by continuity. The hypothesis $p > 1$ is needed: for $f \in L^1(\R)$ the function $Hf$ need not lie in $L^1$. For example, the indicator function of $[0, 1]$ has $Hf(x) = \frac{1}{\pi} \log \abs{\frac{x}{x - 1}}$, which behaves like $\frac{1}{\pi x}$ as $\abs{x} \to \infty$ and is not integrable.

Example: for $f(x) = \frac{1}{1 + x^2}$ one has $Hf(x) = \frac{x}{1 + x^2}$ and then $H\big[\frac{x}{1 + x^2}\big] = -\frac{1}{1 + x^2}$.

Going deeper

For a student

The inversion formula says that $H$ is, up to sign, its own inverse; on $L^2$ it is a unitary operator with $H^* = -H$. It explains the symmetry between the two halves of a Hilbert transform pair. An example is the pair of real and imaginary parts of the boundary values of a function in the Hardy space of the upper half-plane. Together with the boundedness on $L^p$ for $1 < p < \infty$, it is the starting point of the theory of singular integral operators.