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. [theorem: Involution Property] For every $f \in L^p(\mathbb{R}, \mathcal{L}^1)$ with $1 < p < \infty$: \begin{align*} H(Hf) = H^2 f = -f \qquad \mathcal{L}^1\text{-almost everywhere}. \end{align*} Equivalently, $H^{-1} = -H$: the Hilbert transform is its own inverse up to a sign. [/theorem] The identity $H^2 = -\operatorname{Id}$ is most easily proved via the Fourier multiplier characterisation (see below): the Fourier multiplier of $H$ is $m(\xi) = -i\operatorname{sgn}(\xi)$, so $H^2$ has multiplier $m(\xi)^2 = (-i\operatorname{sgn}(\xi))^2 = -\operatorname{sgn}(\xi)^2 = -1$ for $\xi \neq 0$. The involution property means that $H$ is a "quarter-turn" in function space: applying $H$ four times returns to the original function ($H^4 = \operatorname{Id}$). This is analogous to multiplication by $i$ in $\mathbb{C}$, which satisfies $i^4 = 1$. Indeed, $H$ plays the role of multiplication by $-i$ on positive frequencies and by $+i$ on negative frequencies, as the Fourier characterisation makes precise. Another fundamental property is the relationship between $H$ and derivatives. Since the kernel of $H$ is $1/(\pi x)$ and the derivative of $\ln|x|$ is $1/x$ (in the [distributional](/page/Distribution) sense), one expects $H$ to commute with [differentiation](/page/Derivative). [theorem: Commutation With Derivatives] Let $f \in W^{1,p}(\mathbb{R})$ with $1 < p < \infty$ (the [Sobolev space](/page/Sobolev%20Space) of $L^p$ functions with $L^p$ [weak derivative](/page/Weak%20Derivative)). Then $Hf \in W^{1,p}(\mathbb{R})$ and \begin{align*} \frac{d}{dx}(Hf) = H\left(\frac{df}{dx}\right) \qquad \mathcal{L}^1\text{-almost everywhere}. \end{align*} [/theorem] This follows from the Fourier characterisation: differentiation multiplies the Fourier transform by $i\xi$, and the Hilbert transform multiplies by $-i\operatorname{sgn}(\xi)$. These multipliers commute because $(-i\operatorname{sgn}(\xi))(i\xi) = (i\xi)(-i\operatorname{sgn}(\xi))$. ## The Fourier Multiplier Characterisation The deepest insight into the Hilbert transform comes from its action on the Fourier transform. In the frequency domain, the Hilbert transform is remarkably simple: it multiplies each frequency component by $-i\operatorname{sgn}(\xi)$, which amounts to a phase shift of $-\pi/2$ for positive frequencies and $+\pi/2$ for negative frequencies. We use the Fourier transform convention \begin{align*} \hat{f}(\xi) = \int_\mathbb{R} f(x) e^{-ix\xi} \, d\mathcal{L}^1(x), \qquad f(x) = \frac{1}{2\pi} \int_\mathbb{R} \hat{f}(\xi) e^{ix\xi} \, d\mathcal{L}^1(\xi), \end{align*} so that Plancherel's theorem reads $\|f\|_{L^2}^2 = \frac{1}{2\pi}\|\hat{f}\|_{L^2}^2$. To compute the Fourier multiplier of $H$, we need the distributional Fourier transform of the kernel $K(x) = 1/(\pi x)$. This is a standard but delicate computation: $K$ is not in $L^1$ (it decays too slowly at infinity) and is not in $L^2$ (it is not square-integrable), so the Fourier transform must be interpreted in the sense of [tempered distributions](/page/Tempered%20Distributions). [theorem: Fourier Multiplier Of The Hilbert Transform] For every $f \in L^2(\mathbb{R}, \mathcal{L}^1)$: \begin{align*} \widehat{Hf}(\xi) = -i\operatorname{sgn}(\xi) \cdot \hat{f}(\xi) \qquad \text{for } \mathcal{L}^1\text{-a.e. } \xi \in \mathbb{R}, \end{align*} where $\operatorname{sgn}(\xi) = \begin{cases} +1 & \text{if } \xi > 0, \\ 0 & \text{if } \xi = 0, \\ -1 & \text{if } \xi < 0. \end{cases}$ Equivalently, the Hilbert transform is a Fourier multiplier operator with symbol $m(\xi) = -i\operatorname{sgn}(\xi)$. [/theorem] The proof has two steps. First, one computes the distributional Fourier transform of $\operatorname{p.v.}(1/x)$. The distribution $\operatorname{p.v.}(1/x)$ acts on a test function $\varphi$ by $\langle \operatorname{p.v.}(1/x), \varphi \rangle = \lim_{\epsilon \to 0^+}\int_{|x|>\epsilon} \varphi(x)/x \, d\mathcal{L}^1(x)$. Its Fourier transform is $\widehat{\operatorname{p.v.}(1/x)}(\xi) = -i\pi\operatorname{sgn}(\xi)$. (This can be verified by computing the Fourier transform of the regularised kernel $x/(x^2 + \epsilon^2)$ — which has Fourier transform $-i\pi e^{-\epsilon|\xi|}\operatorname{sgn}(\xi)$ — and taking $\epsilon \to 0^+$.) Since $Hf = \frac{1}{\pi}\operatorname{p.v.}(1/x) * f$, the convolution theorem gives $\widehat{Hf}(\xi) = \frac{1}{\pi} \cdot (-i\pi\operatorname{sgn}(\xi)) \cdot \hat{f}(\xi) = -i\operatorname{sgn}(\xi)\hat{f}(\xi)$. Since $|m(\xi)| = |\!-\!i\operatorname{sgn}(\xi)| = 1$ for all $\xi \neq 0$, Plancherel's theorem immediately gives $\|Hf\|_{L^2} = \|f\|_{L^2}$: the Hilbert transform is an $L^2$-isometry. This is the easiest case of the $L^p$-boundedness theorem and serves as the starting point for the interpolation arguments that establish boundedness for other values of $p$. The Fourier multiplier characterisation also makes the involution $H^2 = -\operatorname{Id}$ transparent: $m(\xi)^2 = (-i\operatorname{sgn}(\xi))^2 = -1$ for $\xi \neq 0$, so $\widehat{H^2 f}(\xi) = -\hat{f}(\xi)$. [example: Hilbert Transform Of A Cosine] Let $f(x) = \cos(\omega x)$ for some $\omega > 0$. Formally (in the distributional sense), $\hat{f}(\xi) = \pi(\delta(\xi - \omega) + \delta(\xi + \omega))$, so \begin{align*} \widehat{Hf}(\xi) = -i\operatorname{sgn}(\xi) \cdot \pi(\delta(\xi - \omega) + \delta(\xi + \omega)) = \pi(-i\delta(\xi - \omega) + i\delta(\xi + \omega)). \end{align*} Inverting the Fourier transform: \begin{align*} Hf(x) = \frac{1}{2\pi}\left(-i\pi e^{i\omega x} + i\pi e^{-i\omega x}\right) = \frac{-i e^{i\omega x} + ie^{-i\omega x}}{2} = \frac{e^{i\omega x} - e^{-i\omega x}}{2i} = \sin(\omega x). \end{align*} The Hilbert transform converts cosine into sine — a phase shift of $\pi/2$. Similarly, $H(\sin(\omega x)) = -\cos(\omega x)$: sine maps to negative cosine, a further $\pi/2$ shift. After two applications, $H^2(\cos(\omega x)) = H(\sin(\omega x)) = -\cos(\omega x)$, confirming $H^2 = -\operatorname{Id}$. [/example] [example: Hilbert Transform Of The Poisson Kernel] Let $f(x) = P_y(x) = \frac{1}{\pi} \cdot \frac{y}{x^2 + y^2}$ for fixed $y > 0$. This is the Poisson kernel — a positive, even, smooth function that integrates to $1$. Its Fourier transform is $\hat{f}(\xi) = e^{-y|\xi|}$. The Hilbert transform has Fourier transform $\widehat{Hf}(\xi) = -i\operatorname{sgn}(\xi) e^{-y|\xi|}$. Inverting: \begin{align*} Hf(x) &= \frac{1}{2\pi}\int_\mathbb{R} (-i\operatorname{sgn}(\xi)) e^{-y|\xi|} e^{ix\xi} \, d\mathcal{L}^1(\xi) \\ &= \frac{-i}{2\pi}\int_0^\infty e^{-y\xi} e^{ix\xi} \, d\xi + \frac{i}{2\pi}\int_{-\infty}^0 e^{y\xi} e^{ix\xi} \, d\xi \\ &= \frac{-i}{2\pi} \cdot \frac{1}{y - ix} + \frac{i}{2\pi} \cdot \frac{1}{y + ix} \\ &= \frac{-i(y + ix) + i(y - ix)}{2\pi(y^2 + x^2)} = \frac{2x}{2\pi(x^2 + y^2)} = \frac{1}{\pi} \cdot \frac{x}{x^2 + y^2}. \end{align*} This is the conjugate Poisson kernel $Q_y(x)$. The result confirms the motivating discussion: the Poisson kernel and conjugate Poisson kernel are Hilbert transform pairs: \begin{align*} H\left(\frac{y}{\pi(x^2 + y^2)}\right) = \frac{x}{\pi(x^2 + y^2)}. \end{align*} The Poisson kernel is even and the conjugate Poisson kernel is odd, consistent with the Hilbert transform mapping even functions to odd functions. [/example] ## $L^p$ Boundedness: The Theorem of M. Riesz The central result in the theory of the Hilbert transform is the $L^p$-boundedness theorem, proved by Marcel Riesz in 1928. It asserts that the Hilbert transform is a bounded linear operator on $L^p(\mathbb{R})$ for every $1 < p < \infty$, and it fails to be bounded at the endpoints $p = 1$ and $p = \infty$. This theorem was a landmark achievement — it was the first result establishing the boundedness of a singular integral operator with a non-integrable kernel, and it opened the door to the Calderón–Zygmund theory of singular integrals in higher dimensions. The $L^2$ case is immediate from the Fourier multiplier characterisation (since the multiplier has modulus $1$), but the extension to $p \neq 2$ requires fundamentally new ideas. Riesz's original proof used complex methods (subharmonicity of $|F|^p$ for holomorphic $F$ with suitable modifications); later proofs by Calderón and Zygmund introduced real-variable techniques (decompositions, maximal functions, interpolation) that generalised to far wider settings. [theorem: $L^p$ Boundedness of the Hilbert Transform] For every $1 < p < \infty$, there exists a constant $C_p > 0$ such that \begin{align*} \|Hf\|_{L^p(\mathbb{R})} \le C_p \|f\|_{L^p(\mathbb{R})} \qquad \text{for all } f \in L^p(\mathbb{R}, \mathcal{L}^1). \end{align*} The best constant satisfies $C_p = \cot(\pi/(2p^*))$ where $p^* = \max(p, p/(p-1))$, with equality achieved by suitable [test functions](/page/Test%20Function) (Pichorides, 1972). For $p = 2$, the sharp constant is $C_2 = 1$ (the Hilbert transform is an isometry on $L^2$). [/theorem] The sharp constant $C_p = \cot(\pi/(2p^*))$ blows up as $p \to 1^+$ and as $p \to \infty$: $C_p \sim \frac{1}{\pi} \cdot \frac{1}{p - 1}$ as $p \to 1^+$ and $C_p \sim \frac{p}{\pi}$ as $p \to \infty$. This blow-up is not an artefact of the proof — the Hilbert transform is genuinely unbounded on $L^1$ and $L^\infty$. The failure at the endpoints is qualitatively different in the two cases. At $p = 1$, the Hilbert transform maps $L^1$ functions to functions that are "barely" non-integrable — they satisfy a weak-type estimate instead. [theorem: Weak Type One One Bound] There exists an absolute constant $C > 0$ such that for every $f \in L^1(\mathbb{R}, \mathcal{L}^1)$ and every $\lambda > 0$: \begin{align*} \mathcal{L}^1\left(\{x \in \mathbb{R} : |Hf(x)| > \lambda\}\right) \le \frac{C}{\lambda} \|f\|_{L^1(\mathbb{R})}. \end{align*} That is, $H$ is of weak type $(1, 1)$. [/theorem] The weak-type bound says that the distribution function of $|Hf|$ decays at least as fast as $1/\lambda$, which is the borderline rate for $L^1$ membership (a function is in $L^1$ if and only if $\int_0^\infty \mathcal{L}^1(\{|f| > \lambda\}) \, d\lambda < \infty$, and the integral of $1/\lambda$ diverges). The Hilbert transform of an $L^1$ function is thus "just barely" not in $L^1$ — it lies in the weak $L^1$ space $L^{1,\infty}$. At $p = \infty$, the failure is more severe: $H$ does not even map $L^\infty$ into $L^\infty$. The problem is that the Hilbert transform of a bounded function can grow logarithmically. [example: Hilbert Transform Of An Indicator Function] Let $f = \mathbb{1}_{[a,b]}$ for $-\infty < a < b < \infty$. This is the simplest nontrivial example and already exhibits the endpoint failures. The Hilbert transform is computed by direct integration. For $x \notin \{a, b\}$: \begin{align*} Hf(x) = \frac{1}{\pi} \operatorname{p.v.} \int_a^b \frac{dt}{x - t}. \end{align*} When $x \notin [a,b]$, the integral is not singular: \begin{align*} Hf(x) = \frac{1}{\pi} \int_a^b \frac{dt}{x - t} = \frac{1}{\pi}\left[-\ln|x - t|\right]_{t=a}^{t=b} = \frac{1}{\pi} \ln\frac{|x - a|}{|x - b|}. \end{align*} When $x \in (a, b)$, the principal value gives (splitting at $x$ and using symmetry of the excision): \begin{align*} Hf(x) &= \frac{1}{\pi} \lim_{\epsilon \to 0^+}\left(\int_a^{x-\epsilon}\frac{dt}{x - t} + \int_{x+\epsilon}^b \frac{dt}{x-t}\right) \\ &= \frac{1}{\pi}\lim_{\epsilon \to 0^+}\left(\ln(x - a) - \ln\epsilon + \ln\epsilon - \ln(b - x)\right) = \frac{1}{\pi}\ln\frac{x - a}{b - x}. \end{align*} The $\ln\epsilon$ terms cancel — this is the crucial cancellation that makes the principal value finite. Both cases combine into: \begin{align*} H(\mathbb{1}_{[a,b]})(x) = \frac{1}{\pi}\ln\left|\frac{x - a}{x - b}\right| \qquad \text{for } x \notin \{a, b\}. \end{align*} **Behaviour near the endpoints.** As $x \to a^+$ (entering the interval from the left endpoint): $|x - a| \to 0$ and $|x - b| \to b - a > 0$, so $Hf(x) \to -\infty$ logarithmically. As $x \to b^-$ (approaching the right endpoint from inside): $|x - a| \to b - a$ and $|x - b| \to 0$, so $Hf(x) \to +\infty$. **Endpoint failures.** The function $Hf$ has logarithmic singularities at $x = a$ and $x = b$. Near $x = a$: \begin{align*} |Hf(x)| \approx \frac{1}{\pi}|\ln|x - a|| \qquad \text{as } x \to a. \end{align*} Since $\int_0^1 |\ln t| \, dt = 1 < \infty$, the singularity is integrable — but $Hf$ also decays like $1/x$ at infinity (not fast enough for $L^1$). Specifically, for $|x| \gg b$: \begin{align*} Hf(x) = \frac{1}{\pi}\ln\left(1 - \frac{b-a}{x - b}\right) \approx \frac{-(b-a)}{\pi(x - b)} \sim \frac{a - b}{\pi x}, \end{align*} so $Hf \sim (a-b)/(\pi x)$. The tail $O(1/x)$ is not integrable on $\mathbb{R}$, confirming $Hf \notin L^1(\mathbb{R})$ even though $f \in L^1$. This is the failure of strong-type $(1,1)$. The fact that $f = \mathbb{1}_{[a,b]} \in L^\infty$ but $Hf$ has logarithmic blow-up at $a$ and $b$ also demonstrates the failure at $p = \infty$: $Hf \notin L^\infty$. [/example] [example: Hilbert Transform Of A Gaussian] Let $f(x) = e^{-x^2}$. This is a Schwartz function, so $Hf$ is also smooth. The Fourier transform of $f$ is $\hat{f}(\xi) = \sqrt{\pi} e^{-\xi^2/4}$, so \begin{align*} \widehat{Hf}(\xi) = -i\operatorname{sgn}(\xi) \sqrt{\pi} e^{-\xi^2/4}. \end{align*} The inverse Fourier transform does not have a closed-form expression in terms of elementary functions. However, $Hf$ can be expressed in terms of the *Dawson function* $D(x) := e^{-x^2}\int_0^x e^{t^2} \, d\mathcal{L}^1(t)$: \begin{align*} H(e^{-x^2})(x) = \frac{2}{\sqrt{\pi}} D(x) = \frac{2}{\sqrt{\pi}} e^{-x^2} \int_0^x e^{t^2} \, d\mathcal{L}^1(t). \end{align*} This can be verified by computing the principal-value integral directly. Write: \begin{align*} H(e^{-x^2})(x) = \frac{1}{\pi}\operatorname{p.v.}\int_\mathbb{R} \frac{e^{-t^2}}{x - t} \, d\mathcal{L}^1(t) = \frac{1}{\pi}\operatorname{p.v.}\int_\mathbb{R} \frac{e^{-(x-s)^2}}{s} \, d\mathcal{L}^1(s) \end{align*} (substituting $s = x - t$). Expanding $e^{-(x-s)^2} = e^{-x^2} e^{2xs} e^{-s^2}$ and using the fact that the principal value of $\int e^{-s^2}/s \, ds = 0$ (by oddness), the computation reduces to evaluating the imaginary part of $\int_\mathbb{R} e^{-s^2 + 2ixs}/s \, ds$, which yields the Dawson function. The Dawson function $D(x)$ is odd (consistent with $Hf$ being odd when $f$ is even), satisfies $D(x) \sim 1/(2x)$ as $x \to \infty$ (so $Hf(x) \sim 1/(\sqrt{\pi} x)$, the expected $O(1/x)$ decay), and has a global maximum of approximately $D(0.924) \approx 0.541$. The Hilbert transform of the Gaussian is thus a smooth odd function that rises from $0$ at the origin, reaches a peak at $x \approx 0.924$, and decays like $1/x$ at infinity. [/example] ## Connection to Hardy Spaces and Analytic Functions The Hilbert transform is intimately connected to the theory of Hardy spaces — the spaces of holomorphic functions on the upper half-plane with controlled boundary behaviour. This connection provides both the original motivation for studying $H$ and the most powerful tools for analysing it. A holomorphic function $F : \mathbb{R}^2_+ \to \mathbb{C}$ on the upper half-plane belongs to the **Hardy space** $H^p(\mathbb{R}^2_+)$ (for $0 < p \le \infty$) if \begin{align*} \|F\|_{H^p}^p := \sup_{y > 0} \int_\mathbb{R} |F(x + iy)|^p \, d\mathcal{L}^1(x) < \infty. \end{align*} For $1 \le p \le \infty$, every $F \in H^p(\mathbb{R}^2_+)$ has non-tangential boundary values $F(x) := \lim_{y \to 0^+} F(x + iy)$ existing $\mathcal{L}^1$-almost everywhere, and the boundary function belongs to $L^p(\mathbb{R})$. The Hilbert transform characterises which $L^p$ functions arise as boundary values. Writing $F = u + iv$ with $u, v$ real-valued, the Cauchy–Riemann equations force $v = Hu$ on the boundary (up to a constant), where $H$ is the Hilbert transform. Conversely, given $f \in L^p(\mathbb{R})$ with $1 < p < \infty$, the function $F(x + iy) = P_y * f(x) + iQ_y * f(x)$ belongs to $H^p(\mathbb{R}^2_+)$ with boundary values $f + iHf$. The projection from $L^2(\mathbb{R})$ onto the boundary values of $H^2(\mathbb{R}^2_+)$ is the **Szegő projection** (or **Riesz projection**), which in the Fourier domain simply restricts to positive frequencies. It is related to the Hilbert transform by a simple algebraic identity. [theorem: Riesz Projection And The Hilbert Transform] Define the Riesz projection $P_+ : L^2(\mathbb{R}, \mathcal{L}^1) \to L^2(\mathbb{R}, \mathcal{L}^1)$ by $\widehat{P_+ f}(\xi) = \mathbb{1}_{\xi > 0} \hat{f}(\xi)$ (restriction to positive frequencies). Then: \begin{align*} P_+ = \frac{1}{2}(\operatorname{Id} + iH). \end{align*} Equivalently, $P_+ f = \frac{1}{2}(f + iHf)$ for $f \in L^2(\mathbb{R})$. The range of $P_+$ consists of those $L^2$ functions whose [Fourier transforms](/page/Fourier%20Transform) are supported on $[0, \infty)$ — the boundary values of $H^2$ functions. [/theorem] The formula $P_+ = \frac{1}{2}(\operatorname{Id} + iH)$ is verified on the Fourier side: $\frac{1}{2}(1 + i(-i\operatorname{sgn}(\xi))) = \frac{1}{2}(1 + \operatorname{sgn}(\xi)) = \mathbb{1}_{\xi > 0}$ for $\xi \neq 0$. The complementary projection $P_- = \frac{1}{2}(\operatorname{Id} - iH)$ restricts to negative frequencies. This decomposition has a signal-processing interpretation: given a real-valued signal $f$, the *analytic signal* associated to $f$ is $f_a = f + iHf = 2P_+ f$. The analytic signal has no negative-frequency content and can be written as $f_a(x) = A(x) e^{i\phi(x)}$, where $A(x) = |f_a(x)|$ is the *instantaneous amplitude* (envelope) and $\phi(x) = \arg(f_a(x))$ is the *instantaneous phase*. The instantaneous frequency is $\phi'(x)/(2\pi)$. [example: Analytic Signal Of A Modulated Cosine] Consider the amplitude-modulated signal $f(x) = A(x)\cos(\omega_0 x)$ where $A(x) > 0$ varies slowly relative to the carrier frequency $\omega_0 > 0$. If $A$ is bandlimited to frequencies much smaller than $\omega_0$, then $Hf(x) \approx A(x)\sin(\omega_0 x)$ (the Hilbert transform shifts the carrier by $\pi/2$ while approximately preserving the envelope). The analytic signal is: \begin{align*} f_a(x) = f(x) + iHf(x) \approx A(x)(\cos(\omega_0 x) + i\sin(\omega_0 x)) = A(x)e^{i\omega_0 x}. \end{align*} The instantaneous amplitude is $|f_a(x)| = A(x)$, recovering the envelope. The instantaneous phase is $\phi(x) = \omega_0 x$, and the instantaneous frequency is $\omega_0/(2\pi)$. This is the basis of envelope detection in communications and the Hilbert–Huang transform in signal analysis. The approximation $Hf \approx A(x)\sin(\omega_0 x)$ is exact when $A$ is constant (pure cosine) and becomes increasingly accurate as the bandwidth of $A$ shrinks relative to $\omega_0$ — this is the *Bedrosian theorem*, which states that $H(fg) = fHg$ whenever the Fourier support of $f$ is contained in $[-\Omega, \Omega]$ and the Fourier support of $g$ is contained in $\mathbb{R} \setminus (-\Omega, \Omega)$ (i.e., the spectra are disjoint). [/example] ## The Hilbert Transform and Singular Integrals The Hilbert transform is the simplest example of a Calderón–Zygmund singular integral operator, and the techniques developed for its study — principal-value truncations, maximal operators, Calderón–Zygmund decompositions — form the template for the general theory. In higher dimensions $\mathbb{R}^n$ with $n \ge 2$, the Hilbert transform is replaced by the *Riesz transforms* $R_j f = c_n \operatorname{p.v.}(x_j/|x|^{n+1}) * f$, which are the individual components of the "vector-valued Hilbert transform" and satisfy the same $L^p$-boundedness properties. A Calderón–Zygmund operator on $\mathbb{R}^n$ is a bounded linear operator $T : L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$ whose distributional kernel $K(x, y)$ satisfies — away from the diagonal — the size condition $|K(x,y)| \le C/|x - y|^n$ and the regularity condition $|\nabla_x K(x,y)| + |\nabla_y K(x,y)| \le C/|x - y|^{n+1}$. The Hilbert transform kernel $1/(\pi(x - t))$ in one dimension satisfies these with $n = 1$. The Calderón–Zygmund theorem then asserts that *every* operator satisfying these conditions is bounded on $L^p$ for $1 < p < \infty$ and of weak type $(1,1)$ — a vast generalisation of the Riesz theorem. The maximal Hilbert transform provides quantitative control on the principal-value convergence. Define: \begin{align*} H^* f(x) := \sup_{\epsilon > 0} \left|\int_{|x-t|>\epsilon} \frac{f(t)}{\pi(x - t)} \, d\mathcal{L}^1(t)\right|. \end{align*} The maximal Hilbert transform satisfies the same $L^p$-bounds as $H$ itself: $\|H^* f\|_{L^p} \le C_p' \|f\|_{L^p}$ for $1 < p < \infty$ and the same weak-type $(1,1)$ estimate. The strong-type bound implies, in particular, that the principal-value limit defining $Hf$ exists almost everywhere — the maximal function controls the oscillation of the truncated operators. ## Problems [problem] (a) Use the Fourier multiplier characterisation to prove that for $f, g \in L^2(\mathbb{R}, \mathcal{L}^1)$: \begin{align*} \int_\mathbb{R} Hf(x) \cdot g(x) \, d\mathcal{L}^1(x) = -\int_\mathbb{R} f(x) \cdot Hg(x) \, d\mathcal{L}^1(x). \end{align*} That is, $H$ is skew-adjoint on $L^2(\mathbb{R})$: $H^* = -H$. (b) Deduce that $\int_\mathbb{R} f(x) \cdot Hf(x) \, d\mathcal{L}^1(x) = 0$ for every real-valued $f \in L^2(\mathbb{R})$. Interpret this geometrically. [/problem] [solution] **Part (a).** By Plancherel's theorem (with the convention $\int fg \, d\mathcal{L}^1 = \frac{1}{2\pi}\int \hat{f}\overline{\hat{g}} \, d\mathcal{L}^1$): \begin{align*} \int_\mathbb{R} Hf \cdot \bar{g} \, d\mathcal{L}^1 = \frac{1}{2\pi}\int_\mathbb{R} \widehat{Hf}(\xi) \overline{\hat{g}(\xi)} \, d\mathcal{L}^1(\xi) = \frac{1}{2\pi}\int_\mathbb{R} (-i\operatorname{sgn}(\xi))\hat{f}(\xi)\overline{\hat{g}(\xi)} \, d\mathcal{L}^1(\xi). \end{align*} On the other hand: \begin{align*} \int_\mathbb{R} f \cdot \overline{Hg} \, d\mathcal{L}^1 &= \frac{1}{2\pi}\int_\mathbb{R} \hat{f}(\xi) \overline{\widehat{Hg}(\xi)} \, d\mathcal{L}^1(\xi) = \frac{1}{2\pi}\int_\mathbb{R} \hat{f}(\xi) \overline{(-i\operatorname{sgn}(\xi))\hat{g}(\xi)} \, d\mathcal{L}^1(\xi) \\ &= \frac{1}{2\pi}\int_\mathbb{R} (i\operatorname{sgn}(\xi))\hat{f}(\xi)\overline{\hat{g}(\xi)} \, d\mathcal{L}^1(\xi), \end{align*} using $\overline{-i\operatorname{sgn}(\xi)} = i\operatorname{sgn}(\xi)$ (since $\operatorname{sgn}(\xi)$ is real). Comparing the two expressions: \begin{align*} \int_\mathbb{R} Hf \cdot \bar{g} \, d\mathcal{L}^1 = -\int_\mathbb{R} f \cdot \overline{Hg} \, d\mathcal{L}^1. \end{align*} When $f$ and $g$ are real-valued, $\bar{g} = g$ and $\overline{Hg} = Hg$ (since $H$ maps real functions to real functions), giving $\int Hf \cdot g = -\int f \cdot Hg$. **Part (b).** Setting $g = f$ (real-valued) in part (a): \begin{align*} \int_\mathbb{R} Hf \cdot f \, d\mathcal{L}^1 = -\int_\mathbb{R} f \cdot Hf \, d\mathcal{L}^1. \end{align*} Both sides are the same number (just reordering the product), so $\int f \cdot Hf = -\int f \cdot Hf$, which forces $\int f \cdot Hf \, d\mathcal{L}^1 = 0$. Geometrically, this says that $f$ and $Hf$ are orthogonal in $L^2(\mathbb{R})$. Together with $\|Hf\|_{L^2} = \|f\|_{L^2}$ (the isometry property), this means the map $f \mapsto (f, Hf)$ embeds $L^2(\mathbb{R})$ into a subspace where $f$ and $Hf$ are orthogonal vectors of equal length. In signal processing language, a signal and its Hilbert transform carry "independent" information — they form the real and imaginary parts of the analytic signal. [/solution] [problem] Compute the Hilbert transform of $f(x) = \frac{1}{1 + x^2}$ directly from the principal-value integral (without using Fourier transforms) and verify the result using the Fourier multiplier characterisation. [/problem] [solution] **Step 1: Direct computation via principal value.** \begin{align*} Hf(x) = \frac{1}{\pi}\operatorname{p.v.}\int_\mathbb{R} \frac{1}{(1 + t^2)(x - t)} \, d\mathcal{L}^1(t). \end{align*} Decompose by partial fractions: \begin{align*} \frac{1}{(1 + t^2)(x - t)} = \frac{1}{1 + x^2}\left(\frac{x - t}{1 + t^2} + \frac{1}{x - t}\right) = \frac{1}{1+x^2}\left(\frac{x}{1+t^2} - \frac{t}{1+t^2} + \frac{1}{x-t}\right). \end{align*} Actually, let us use a cleaner partial fraction. Write $\frac{1}{(1+t^2)(x-t)} = \frac{A + Bt}{1+t^2} + \frac{C}{x - t}$. Multiplying through: $1 = (A + Bt)(x - t) + C(1 + t^2)$. Setting $t = x$: $1 = C(1 + x^2)$, so $C = 1/(1+x^2)$. Comparing the $t^2$ coefficient: $0 = -B + C$, so $B = C = 1/(1+x^2)$. Comparing the constant: $1 = Ax + C$, so $A = (1 - C)/x = x/(1 + x^2)$ (for $x \neq 0$; the case $x = 0$ can be handled by [continuity](/page/Continuity)). Therefore: \begin{align*} \frac{1}{(1+t^2)(x-t)} = \frac{1}{1+x^2}\left(\frac{x + t}{1+t^2} + \frac{1}{x-t}\right). \end{align*} Now integrate: \begin{align*} Hf(x) = \frac{1}{\pi(1+x^2)}\left[\int_\mathbb{R}\frac{x + t}{1+t^2}\,d\mathcal{L}^1(t) + \operatorname{p.v.}\int_\mathbb{R}\frac{dt}{x - t}\right]. \end{align*} The first integral: $\int_\mathbb{R}\frac{x}{1+t^2}\,dt = x\pi$ and $\int_\mathbb{R}\frac{t}{1+t^2}\,dt = 0$ (odd integrand). So the first integral equals $x\pi$. The second integral: $\operatorname{p.v.}\int_\mathbb{R}\frac{dt}{x-t}$. Substituting $s = x - t$: $\operatorname{p.v.}\int_\mathbb{R}\frac{ds}{s} = 0$ (the principal value of $1/s$ against Lebesgue measure is zero by oddness). Therefore: \begin{align*} Hf(x) = \frac{1}{\pi(1 + x^2)} \cdot x\pi = \frac{x}{1 + x^2} = \frac{-x}{1+x^2} \cdot (-1). \end{align*} Wait — let us recheck signs. We have: \begin{align*} Hf(x) = \frac{1}{\pi(1+x^2)}(x\pi + 0) = \frac{x}{1+x^2}. \end{align*} Recalling $f(x) = 1/(1+x^2) = \pi P_1(x)$ (the Poisson kernel at height $y = 1$, up to the factor $\pi$), we expected $Hf = \pi Q_1(x) = \pi \cdot \frac{x}{\pi(x^2+1)} = \frac{x}{x^2+1}$. This is consistent: \begin{align*} H\left(\frac{1}{1+x^2}\right)(x) = \frac{x}{1+x^2}. \end{align*} **Step 2: Verification via Fourier multiplier.** The Fourier transform of $f(x) = 1/(1+x^2)$ is $\hat{f}(\xi) = \pi e^{-|\xi|}$. Therefore: \begin{align*} \widehat{Hf}(\xi) = -i\operatorname{sgn}(\xi)\cdot \pi e^{-|\xi|}. \end{align*} The inverse Fourier transform of $-i\operatorname{sgn}(\xi)e^{-|\xi|}$ was computed in the Poisson kernel example earlier in this page: \begin{align*} \frac{1}{2\pi}\int_\mathbb{R} (-i\operatorname{sgn}(\xi))\pi e^{-|\xi|} e^{ix\xi}\,d\mathcal{L}^1(\xi) = \frac{1}{2}\int_\mathbb{R} (-i\operatorname{sgn}(\xi)) e^{-|\xi|+ix\xi}\,d\mathcal{L}^1(\xi) = \frac{x}{1+x^2}. \end{align*} (The last equality follows from the computation in the Poisson kernel example with $y = 1$.) Both methods give the same answer, confirming the result. [/solution] ## References - Duoandikoetxea, J., *Fourier Analysis* (2001). - Grafakos, L., *Classical Fourier Analysis* (2014). - King, F. W., *Hilbert Transforms*, Vols. 1–2 (2009). - Muscalu, C. and Schlag, W., *Classical and Multilinear Harmonic Analysis*, Vol. I (2013). - Stein, E. M., *Singular Integrals and Differentiability Properties of Functions* (1970). - Stein, E. M. and Weiss, G., *Introduction to Fourier Analysis on Euclidean Spaces* (1971).