---
tag: 2KJV7187NR
kind: definition
name: Finite Hilbert transform
msc: [44A15, "45E05"]
defines: 5JYBXH9BP6
uses: [0SYR10HD4R, SASP14MA1Z, NSCEBXM7ZM]
sources:
  - url: https://en.wikipedia.org/wiki/Hilbert_transform
---

## Definition

Let $1 < p < \infty$ and let $f \in L^p(-1, 1)$, a function in the [[$L^p$ space|SASP14MA1Z]] of the interval.
The *finite Hilbert transform* of $f$ is the function $T f$ on $(-1, 1)$ given by
$$(T f)(x) = \frac{1}{\pi}\, \mathrm{PV}\!\int_{-1}^{1} \frac{f(t)}{t - x}\, dt = \frac{1}{\pi} \lim_{\varepsilon \to 0^+} \int_{\set{t \in (-1, 1) : \abs{t - x} > \varepsilon}} \frac{f(t)}{t - x}\, dt,$$
the [[Cauchy principal value|0SYR10HD4R]] integral.
The limit exists for almost every $x \in (-1, 1)$.

## Remarks

If $\tilde f$ is $f$ on $(-1, 1)$ and $0$ elsewhere, then $T f = -H\tilde f$ on $(-1, 1)$, where $H$ is the [[Hilbert transform|NSCEBXM7ZM]] with kernel $\frac{1}{\pi (x - t)}$.
By the theorem of M. Riesz on $H$, the operator $T$ is bounded on $L^p(-1, 1)$ for $1 < p < \infty$.
Some authors use the kernel $\frac{1}{x - t}$ instead, which changes the sign of $T$.

Example: $f(t) = \sqrt{1 - t^2}$ gives $T f(x) = -x$.
Check: for $\abs{z} > 1$, $\int_{-1}^{1} \frac{\sqrt{1 - t^2}}{t - z}\, dt = \pi\bigl(\sqrt{z^2 - 1} - z\bigr)$ with the branch of the root that behaves like $z$ at infinity.
Its boundary values from above and below on $(-1, 1)$ are $\pi\bigl(\pm i\sqrt{1 - x^2} - x\bigr)$, and the principal value is their average $-\pi x$.

Non-example of injectivity: $g(t) = (1 - t^2)^{-1/2}$ lies in $L^p(-1, 1)$ for $1 < p < 2$ and $T g = 0$ on $(-1, 1)$.
So $T$ is not injective on these spaces.
For $p \ge 2$ the function $g$ is not in $L^p(-1, 1)$.