The Hilbert transform is the prototypical singular integral operator on the real line. Given a function $f : \mathbb{R} \to \mathbb{C}$, the Hilbert transform $Hf$ is defined — at least formally — by the [convolution](/page/Convolution)

\begin{align*} Hf(x) = \frac{1}{\pi} \int_{\mathbb{R}} \frac{f(t)}{x - t} \, d\mathcal{L}^1(t). \end{align*}

The integral does not converge absolutely (the kernel $1/(x - t)$ is not locally [integrable](/page/Integral) on $\mathbb{R}$), so the definition requires a principal-value prescription: one excises a symmetric interval around the singularity $t = x$ and takes a [limit](/page/Limit). Despite this delicate definition, the Hilbert transform is a bounded operator on $L^p(\mathbb{R})$ for all $1 < p < \infty$ — the celebrated theorem of Marcel Riesz (1928) — and it fails to be bounded precisely at the endpoints $p = 1$ and $p = \infty$.

The Hilbert transform appears throughout analysis under different guises. In complex analysis, it is the [boundary](/page/Boundary) operator that sends the real part of a holomorphic function in the upper half-plane to its imaginary part: it is the "harmonic conjugation" operator. In Fourier analysis, it acts as the multiplier $\xi \mapsto -i\operatorname{sgn}(\xi)$, converting cosines into sines and vice versa — the simplest nontrivial Fourier multiplier that is bounded but not given by a bounded function in the time domain. In signal processing, it shifts the phase of every frequency component by $\pm \pi/2$, producing the "analytic signal" from a real-valued signal. And in the theory of singular integrals, it serves as the model case against which all Calderón–Zygmund operators are measured.

## Motivation

[motivation] ### The Harmonic Conjugation Problem Let $u : \mathbb{R}^2_+ \to \mathbb{R}$ be a harmonic function on the upper half-plane $\mathbb{R}^2_+ = \{(x, y) : y > 0\}$ with boundary values $u(x, 0) = f(x)$. The Cauchy–Riemann equations guarantee the existence of a *harmonic conjugate* $v : \mathbb{R}^2_+ \to \mathbb{R}$ such that $u + iv$ is holomorphic. The conjugate $v$ is unique up to a constant (fixed by requiring $v \to 0$ at infinity), and the question is: given $f$, what are the boundary values $v(x, 0)$? The answer is given by the Poisson integral. The unique bounded harmonic function on $\mathbb{R}^2_+$ with boundary values $f$ is the Poisson integral $u(x, y) = P_y * f(x)$, where $P_y(x) = \frac{1}{\pi} \cdot \frac{y}{x^2 + y^2}$ is the Poisson kernel. Its harmonic conjugate is $v(x, y) = Q_y * f(x)$, where $Q_y(x) = \frac{1}{\pi} \cdot \frac{x}{x^2 + y^2}$ is the conjugate Poisson kernel. As $y \to 0^+$, the Poisson kernel $P_y$ converges to a delta mass (recovering $f$), while the conjugate Poisson kernel $Q_y$ converges — in the principal-value sense — to the kernel $1/(\pi x)$. The boundary values of $v$ are therefore \begin{align*} v(x, 0) = \lim_{y \to 0^+} Q_y * f(x) = \frac{1}{\pi} \operatorname{p.v.} \int_{\mathbb{R}} \frac{f(t)}{x - t} \, d\mathcal{L}^1(t) = Hf(x). \end{align*} The Hilbert transform is thus the operator that maps boundary values of the real part to boundary values of the imaginary part of a holomorphic function on the upper half-plane. ### Why Principal Values? The kernel $K(x) = 1/(\pi x)$ is odd ($K(-x) = -K(x)$) and decays like $1/|x|$ at infinity. On $\mathbb{R}$, the function $|x|^{-1}$ is not locally integrable: $\int_{-1}^{1} |x|^{-1} \, d\mathcal{L}^1(x) = +\infty$. (This is in contrast to the two-dimensional kernel $1/|z|$ used in the [Cauchy transform](/page/Cauchy%20Transform), which *is* locally integrable with respect to $\mathcal{L}^2$ because the area element $r \, dr$ compensates the $1/r$ singularity.) The convolution $K * f$ therefore requires a cancellation argument: the positive and negative parts of $K$ near the origin are infinite individually, but their symmetric cancellation produces a finite limit. This is the principal value. [/motivation]

## Definition

The definition of the Hilbert transform requires specifying the principal-value limit precisely. The key point is that the excision must be *symmetric* — removing $(x - \epsilon, x + \epsilon)$ rather than $(x - \epsilon, x + \delta)$ with $\epsilon \neq \delta$ — because the cancellation of the odd kernel depends on this symmetry.

[definition: Hilbert Transform On The Real Line] Let $f \in L^p(\mathbb{R}, \mathcal{L}^1)$ with $1 \le p < \infty$, or more generally let $f$ be a locally integrable function on $\mathbb{R}$ with sufficient decay at infinity. The **Hilbert transform** of $f$ is \begin{align*} Hf(x) := \frac{1}{\pi} \operatorname{p.v.} \int_{\mathbb{R}} \frac{f(t)}{x - t} \, d\mathcal{L}^1(t) = \frac{1}{\pi} \lim_{\epsilon \to 0^+} \int_{|x - t| > \epsilon} \frac{f(t)}{x - t} \, d\mathcal{L}^1(t). \end{align*} [/definition]

The principal-value limit exists for $\mathcal{L}^1$-almost every $x$ when $f \in L^p$ with $1 \le p < \infty$. For Schwartz-class [functions](/page/Function) $f \in \mathcal{S}(\mathbb{R})$, the limit exists everywhere and defines a smooth function. The existence for general $L^p$ functions is a nontrivial result that follows from the $L^p$-boundedness theory (for $p > 1$) or from the Calderón–Zygmund decomposition (for $p = 1$, giving existence almost everywhere via the weak-type $(1,1)$ bound).

An equivalent formulation that is sometimes more convenient replaces the symmetric excision with a regularisation. Combining the contributions from $t > x$ and $t < x$ symmetrically:

\begin{align*} Hf(x) = \frac{1}{\pi} \lim_{\epsilon \to 0^+} \int_\epsilon^\infty \frac{f(x - t) - f(x + t)}{t} \, d\mathcal{L}^1(t). \end{align*}

This formula makes the cancellation explicit: the numerator $f(x - t) - f(x + t)$ vanishes at $t = 0$ (at least when $f$ is continuous), so the integrand has a removable singularity.

The Hilbert transform also acts naturally on periodic functions, where it is called the conjugate function. On the unit circle, the role of the kernel $1/(\pi x)$ is played by the cotangent.

[definition: Conjugate Function On The Circle] Let $f : \mathbb{T} \to \mathbb{C}$ be an integrable function on the unit circle $\mathbb{T} = \mathbb{R}/(2\pi\mathbb{Z})$. The **conjugate function** (or **periodic Hilbert transform**) of $f$ is \begin{align*} \tilde{f}(\theta) := \frac{1}{2\pi} \operatorname{p.v.} \int_{-\pi}^{\pi} f(\theta - t) \cot\left(\frac{t}{2}\right) \, d\mathcal{L}^1(t). \end{align*} [/definition]

The cotangent kernel $\cot(t/2)/(2\pi)$ is the periodic analogue of $1/(\pi t)$: near $t = 0$, $\cot(t/2) \approx 2/t$, so $\frac{1}{2\pi}\cot(t/2) \approx \frac{1}{\pi t}$, recovering the Hilbert transform kernel. The conjugate function satisfies the same $L^p$-boundedness properties as the Hilbert transform on $\mathbb{R}$.

## Basic Algebraic Properties

The Hilbert transform has several structural properties that follow directly from the symmetry of the kernel $K(x) = 1/(\pi x)$. These properties constrain the operator and play a crucial role in applications.

The kernel $K$ is odd, real-valued, and homogeneous of degree $-1$: $K(\lambda x) = \lambda^{-1} K(x)$ for $\lambda > 0$. These symmetries propagate to the operator. The oddness means that $H$ applied to an even function yields an odd function and vice versa. The homogeneity implies that $H$ commutes with dilations: if $f_\lambda(x) = f(\lambda x)$ for $\lambda > 0$, then $H(f_\lambda) = (Hf)_\lambda$. Similarly, $H$ commutes with translations: $H(\tau_a f) = \tau_a(Hf)$, where $\tau_a f(x) = f(x - a)$.

The most important algebraic property is the *involution* identity: applying the Hilbert transform twice gives the negative identity.