# Hilbert transform

of a function $f$

The improper integral

\begin{equation}\label{eq:1} g\left(x\right) = \frac{1}{\pi} \int_{0}^{\infty}\frac{f\left(x+t\right)-f\left(x-t\right)}{t}\mathrm{d}t. \end{equation}

If $f \in L ( - \infty ,\ \infty )$, the function $g$ exists for almost-all values of $x$. If $f \in L _{p} (- \infty ,\ \infty )$, $p \in (1,\ \infty )$, the function $g$ also belongs to $L _{p} (- \infty ,\ \infty )$ and the inversion formula

$$\tag{2} f (x) \ = \ - { \frac{1} \pi } \int\limits _{0} ^ \infty \frac{g (x + t) - g (x - t)}{t} \ dt$$

is valid almost-everywhere. Here

$$\tag{3} \int\limits _ {- \infty} ^ \infty | g (x) | ^{2} \ dx \ \leq \ M _{p} \int\limits _ {- \infty} ^ \infty | f (x) | ^{p} \ dx,$$

where the constant $M _{p}$ depends only on $p$.

Formulas (1) and (2) are equivalent to the formulas

$$\tag{4} g (x) \ = \ { \frac{1} \pi } \int\limits _ {- \infty} ^ \infty \frac{f (t)}{t - x} \ dt,$$

$$\tag{5} f (x) \ = \ { \frac{1} \pi } \int\limits _ {- \infty} ^ \infty \frac{g (t)}{t - x} \ dt,$$

in which the integrals are understood in the sense of the principal value.

The integral

$$\tag{6} g (x) \ = \ { \frac{1}{2 \pi} } \int\limits _{0} ^ {2 \pi} f (t) \ \mathop{\rm cotan}\nolimits \ \frac{t - x}{2} \ dt,$$

understood in the sense of its principal value, is also called the Hilbert transform of $f$. This integral is often called the Hilbert singular integral. In the theory of Fourier series the function $g$ defined by (6) is said to be conjugate with $f$.

If $f \in L (0,\ 2 \pi )$, $g$ exists almost-everywhere, and if $f$ satisfies a Lipschitz condition of order $\alpha \in (0,\ 1)$, $g$ exists for any $x$ and satisfies the same condition. If $f \in L _{p} (0,\ 2 \pi )$, $p \in (1,\ \infty )$, then $g$ has the same property, and an inequality analogous to (3) in which the integrals are taken over the interval $(0,\ 2 \pi )$ is valid. Thus, the integral operators generated by the Hilbert transform are bounded (linear) operators on the respective spaces $L _{p}$.

If $f$ satisfies a Lipschitz condition, or if $f \in L _{p} (0,\ 2 \pi )$, and also

$$\int\limits _{0} ^ {2 \pi} g (x) \ dx \ = \ 0,$$

the following inversion formula is valid:

$$\tag{7} f (x) \ = \ - { \frac{1}{2 \pi} } \int\limits _{0} ^ {2 \pi} g (t) \ \mathop{\rm cotan}\nolimits \ \frac{t - x}{2} \ dt,$$

and

$$\int\limits _{0} ^ {2 \pi} f (x) \ dx \ = \ 0.$$

In the class of functions which satisfy a Lipschitz condition, equation (7) is valid everywhere, and in the class of functions with integrable $p$- th power, it is valid almost-everywhere.

Each one of the pairs of formulas written above, such as (4) or (5), may be considered as an integral equation of the first kind, and the second formula yields the solution of this equation.

If the functions $\mathop{\rm cotan}\nolimits \{ (t - x)/2 \}$ and ${1/(t - x)}$ are considered as kernels of integral operators, they are often referred to as the Hilbert kernel and as the Cauchy kernel. In the case of the unit circle, there exists a simple relationship between these kernels:

$$\frac{d \tau}{t - \xi} \ = \ { \frac{1}{2} } \left ( \mathop{\rm cotan}\nolimits \ { \frac{t - x}{2} } + i \right ) \ dt,$$

where $\xi = e ^{ix}$, $\tau = e ^{it}$.

How to Cite This Entry:
Hilbert transform. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Hilbert_transform&oldid=44364
This article was adapted from an original article by B.V. Khvedelidze (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article