LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-10-02

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The Riesz transform is the principal value of its kernel, with the matching constant

Statement

Assume Countable Choice, use the $e^{- 2 π i x ξ}$ convention, and let $K_{j} (x) = c_{n} x_{j} / ∣ x ∣^{n + 1}$ with $c_{n} = Γ ((n + 1) / 2) / π^{(n + 1) / 2}$ be the Riesz kernel of Riesz transforms on Euclidean space. Then for every Schwartz function $f \in S (R^{n})$ :

the truncated integrals $\int_{∣ y ∣ > ε} K_{j} (y) f (x - y) d y$ converge as $ε ↓ 0$ for every $x \in R^{n}$ , with a limit that is continuous in $x$ ; and
that continuous function is a representative of the $L^{2}$ class $R_{j} f$ , whose Fourier multiplier is $- i ξ_{j} / ∣ ξ ∣$ .

Existence of the principal value is asserted only for Schwartz $f$ , pointwise in $x$ ; no almost-everywhere convergence for general $L^{2}$ or $L^{p}$ inputs is claimed.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $1 \leq j \leq n$ , the Riesz kernel $K_{j}$ , the symbol $m_{j} (ξ) = - i ξ_{j} / ∣ ξ ∣$ for $ξ \neq 0$ with $m_{j} (0) = 0$ , and the operator $R_{j} = F_{2}^{- 1} M_{m_{j}} F_{2}$ on $L^{2} (R^{n})$ .

[F1]

The Riesz kernel is $K_{j} (x) = c_{n} x_{j} / ∣ x ∣^{n + 1}$ with $0 < c_{n} < \infty$ , smooth and odd on $R^{n} ∖ {0}$ , and $∣ K_{j} (x) ∣ \leq c_{n} ∣ x ∣^{- n}$ ; the operator $R_{j}$ is the bounded $L^{2}$ operator with symbol $m_{j}$ . Riesz transforms on Euclidean space

[F2]

For $S (T) : = \int_{0}^{T} \frac{\sin u}{u} d u$ , one has $S (T) \to \frac{π}{2}$ , $∣ S (T) ∣ \leq 3$ for all $T \geq 0$ , and $∣ \int_{A}^{B} \frac{\sin u}{u} d u ∣ \leq 2 / A$ when $1 \leq A < B$ . Thus $∣ \int_{A}^{B} \frac{\sin u}{u} d u ∣ \leq 6$ for all $0 < A < B$ : use $∣ S (B) ∣ + ∣ S (A) ∣ \leq 6$ if $A < 1$ , and the tail bound if $A \geq 1$ . The sine integral under Countable Choice: uniform bounds and the value pi/2

[F3]

Polar coordinates: $\int_{R^{n}} h d λ_{n} = \int_{0}^{\infty} \int_{S^{n - 1}} h (r ω) r^{n - 1} d σ (ω) d r$ for nonnegative Borel $h$ and, by splitting real and imaginary parts into their positive and negative parts, for integrable complex Borel $h$ , and the finite Borel measure $σ$ is uniquely determined by this property. Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma

[F4]

$V_{m} (1) = π^{m / 2} / Γ (m / 2 + 1)$ for $m \geq 1$ , and volumes scale as $V_{m} (ρ) = V_{m} (1) ρ^{m}$ . The closed form for the volume of the unit $n$ -ball

[F5]

Fubini for L^1 functions on a sigma-finite product. Fubini's theorem for L^1 functions on a sigma-finite product

[F6]

Dominated convergence. Dominated convergence

[F7]

$(u * φ) (x) = ⟨ u_{y}, φ (x - y) ⟩$ defines the tempered convolution for $u \in S^{'}$ and Schwartz $φ$ . Convolution of a tempered distribution with a schwartz function

[F8]

$F (u * φ) = (F u) (F φ)$ for $u \in S^{'} (R^{n})$ and Schwartz $φ$ . Fourier transform converts allowed tempered convolutions to products

[F9]

Fourier transformation is a topological automorphism of $S^{'} (R^{n})$ , hence injective. Fourier transform is a topological automorphism of tempered distributions

[F10]

$⟨ F u, φ ⟩ = ⟨ u, F φ ⟩$ with no conjugate on the right-hand side. Fourier transform of a tempered distribution

[F11]

For a continuous curve $h : [a, b] \to R^{2}$ differentiable on $(a, b)$ , the bound $∣ h^{'} (t) ∣ \leq D$ implies $∣ h (b) - h (a) ∣ \leq D (b - a)$ . Identify $C$ with $R^{2}$ when applying this inequality. The mean value inequality: if $f : [a, b] \to R^{m}$ is continuous and differentiable on $(a, b)$ with $∥ f^{'} ∥_{2} \leq M$ , then $∥ f (b) - f (a) ∥_{2} \leq M (b - a)$

[F12]

The real one-variable chain rule applies to compositions of real scalar functions; below it is applied separately to the real and imaginary parts of each coordinate section of $f$ . The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$

[F13]

Schwartz seminorms: for every integer $N \geq 0$ there is a finite constant $C_{N} (φ)$ with $∣ φ (x) ∣ \leq C_{N} (φ) (1 + ∣ x ∣)^{- N}$ . Schwartz space and its seminorms

[F14]

Linear change of variables for Lebesgue measure. A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not

[F15]

The Gamma function satisfies $Γ (1) = 1$ . The real Gamma functional equation $Γ (s + 1) = s Γ (s)$

Proof

technique · direct

1.1F1F6F11F12F13

Fix $f \in S (R^{n})$ and $x \in R^{n}$ . Write $g_{x} (y) : = f (x - y) - f (x)$ , and put $D_{k} : = \sup_{z} ∣ \partial_{k} f (z) ∣ < \infty$ for $1 \leq k \leq n$ by [F13], and set $M : = \sum_{k = 1}^{n} D_{k}$ . Join $x$ to $x - y$ by the $n$ coordinate segments with successive endpoints $z^{(k)} = x - \sum_{ℓ = 1}^{k} y_{ℓ} e_{ℓ}$ , where $z^{(0)} = x$ . On the $k$ -th segment, the real one-variable chain rule [F12] on both components gives $(d / d t) f (z^{(k - 1)} - t y_{k} e_{k}) = - y_{k} \partial_{k} f (z^{(k - 1)} - t y_{k} e_{k})$ for $0 < t < 1$ ; this follows from the definition of the coordinate partial derivative and is valid also when $y_{k} = 0$ , when the curve is constant. Applying [F11] to this complex curve viewed in $R^{2}$ bounds its increment by $D_{k} ∣ y_{k} ∣$ . Telescoping gives $∣ g_{x} (y) ∣ \leq \sum_{k} D_{k} ∣ y_{k} ∣ \leq M ∣ y ∣$ for every $y$ . Hence on $∣ y ∣ < 1$ the bound $∣ K_{j} (y) g_{x} (y) ∣ \leq c_{n} M ∣ y ∣^{1 - n}$ is integrable in $n$ dimensions, while $1 + ∣ y ∣ \leq (1 + ∣ x ∣) (1 + ∣ x - y ∣)$ and the Schwartz bound [F13] with $N = n + 2$ give $∣ f (x - y) ∣ \leq C_{n + 2} (f) (1 + ∣ x ∣)^{n + 2} (1 + ∣ y ∣)^{- n - 2}$ . Thus on $∣ y ∣ > 1$ , $∣ K_{j} (y) f (x - y) ∣ \leq c_{n} C_{n + 2} (f) (1 + ∣ x ∣)^{n + 2} ∣ y ∣^{- n - 2}$ is integrable. Since $\int_{ε < ∣ y ∣ < 1} K_{j} (y) f (x) d y = 0$ by oddness of $K_{j}$ and symmetry of the annulus, $\int_{∣ y ∣ > ε} K_{j} (y) f (x - y) d y = \int_{ε < ∣ y ∣ < 1} K_{j} (y) g_{x} (y) d y + \int_{∣ y ∣ > 1} K_{j} (y) f (x - y) d y$ , and $ε ↓ 0$ in the first term yields the absolutely convergent limit $J (x) : = \int_{∣ y ∣ < 1} K_{j} (y) g_{x} (y) d y + \int_{∣ y ∣ > 1} K_{j} (y) f (x - y) d y$ . For $x_{k} \to x$ , the sequence $(x_{k})$ is bounded, so the tail constants $(1 + ∣ x_{k} ∣)^{n + 2}$ have a common finite bound. This and the common small-ball bound $c_{n} M ∣ y ∣^{1 - n}$ supply integrable dominators for [F6]; continuity of $f$ gives pointwise convergence in both integrals, hence $J (x_{k}) \to J (x)$ .

1.2F1F3

For $0 < ε < R$ and $ξ \neq 0$ put $Λ_{ε, R} (ξ) : = \int_{ε < ∣ y ∣ < R} K_{j} (y) e^{- 2 π i y \cdot ξ} d y$ . The cosine part of the integrand is odd in $y$ , so it integrates to zero on the symmetric annulus, and $K_{j} (y) = c_{n} y_{j} / ∣ y ∣^{n + 1}$ gives $Λ_{ε, R} (ξ) = - i c_{n} \int_{ε < ∣ y ∣ < R} \frac{y_{j}}{∣ y ∣^{n + 1}} \sin (2 π y \cdot ξ) d y$ . Polar coordinates [F3] turn this into $Λ_{ε, R} (ξ) = - i c_{n} \int_{ε}^{R} \frac{d r}{r} \int_{S^{n - 1}} ω_{j} \sin (2 π r ξ \cdot ω) d σ (ω)$ .

1.3F3F14

For $ξ \neq 0$ one has $\int_{S^{n - 1}} sgn (ξ \cdot ω) ω_{j} d σ (ω) = A_{n} \frac{ξ_{j}}{∣ ξ ∣}$ with $A_{n} : = \int_{S^{n - 1}} ∣ ω_{1} ∣ d σ$ : by [F3] the measure $σ$ is invariant under the orthogonal map $ω \mapsto R ω$ , so substituting $ω = R^{T} u$ for an orthogonal map with $R (ξ / ∣ ξ ∣) = e_{1}$ (take $R = I$ if $v : = ξ / ∣ ξ ∣ = e_{1}$ , and otherwise take $R = I - 2 w w^{T} / ∣ w ∣^{2}$ with $w = v - e_{1}$ ) and reflecting $u_{k} \mapsto - u_{k}$ for $k \neq 1$ (which preserves $sgn (u_{1}) u_{1}$ and kills the other components by oddness) leaves only $A_{n} (R^{T} e_{1})_{j} = A_{n} ξ_{j} / ∣ ξ ∣$ .

1.4F3F4F5F14F15

$A_{n} = 2 π^{(n - 1) / 2} / Γ ((n + 1) / 2)$ : compute $C : = \int_{B_{n}} ∣ x_{1} ∣ d x$ twice. Polar coordinates [F3] give $C = \int_{0}^{1} r^{n} d r \cdot A_{n} = A_{n} / (n + 1)$ ; for $n \geq 2$ , slicing at $x_{1} = t$ gives, by [F5], [F14] and [F4], $C = \int_{- 1}^{1} ∣ t ∣ V_{n - 1} (1) (1 - t^{2})^{(n - 1) / 2} d t = 2 V_{n - 1} (1) / (n + 1) = 2 π^{(n - 1) / 2} / ((n + 1) Γ ((n + 1) / 2))$ , hence $A_{n} = (n + 1) C = 2 V_{n - 1} (1) = 2 π^{(n - 1) / 2} / Γ ((n + 1) / 2)$ . For $n = 1$ the sphere is $S^{0} = {- 1, 1}$ : the defining identity of [F3], applied to functions supported in the annulus $1 < ∣ x ∣ < 2$ , shows that the measure $σ$ is the counting measure $δ_{- 1} + δ_{1}$ , so $A_{1} = \int_{S^{0}} ∣ ω_{1} ∣ d σ = 1 + 1 = 2$ , while $2 π^{0} / Γ (1) = 2$ by $Γ (1) = 1$ of [F15]. Hence $A_{n} = 2 π^{(n - 1) / 2} / Γ ((n + 1) / 2)$ for every $n \geq 1$ , and $c_{n} A_{n} π / 2 = 1$ by cancellation of $Γ ((n + 1) / 2)$ and $π^{(n + 1) / 2}$ .

2.1step 1.2step 1.3step 1.4F2F6

In 1.2 let $R = 1 / ε$ and $ε ↓ 0$ . For each fixed $ω \in S^{n - 1}$ with $ξ \cdot ω \neq 0$ , the substitution $u = 2 π r (ξ \cdot ω)$ (with orientation, [F2]) gives $\int_{ε}^{1 / ε} \frac{\sin (2 π r ξ \cdot ω)}{r} d r \to \frac{π}{2} sgn (ξ \cdot ω)$ ; when $ξ \cdot ω = 0$ the integral is zero. In all cases [F2] bounds its absolute value by $6$ , uniformly in $ε$ and $ω$ . Since the sphere has finite measure, [F6] on $S^{n - 1}$ gives $\lim_{ε ↓ 0} Λ_{ε, 1 / ε} (ξ) = - i c_{n} \int_{S^{n - 1}} ω_{j} \frac{π}{2} sgn (ξ \cdot ω) d σ (ω) = - i \frac{c_{n} π}{2} A_{n} \frac{ξ_{j}}{∣ ξ ∣} = - i \frac{ξ_{j}}{∣ ξ ∣}$ by 1.3 and the constant identity of 1.4.

3.1step 1.1step 2.1F5F6F10

Define the tempered distribution $W_{j}$ by the symmetric principal-value pairing $⟨ W_{j}, φ ⟩ : = \lim_{ε ↓ 0} \int_{ε < ∣ y ∣ < 1 / ε} K_{j} (y) φ (y) d y$ for $φ \in S$ ; the two-piece bound of 1.1 shows the limit exists, is finite, and is Schwartz-continuous. By [F10], $⟨ F W_{j}, φ ⟩ = ⟨ W_{j}, \hat{φ} ⟩ = \lim_{ε ↓ 0} \int_{ε < ∣ y ∣ < 1 / ε} K_{j} (y) \hat{φ} (y) d y$ ; the double integrand is absolutely integrable since $\int_{ε < ∣ y ∣ < 1 / ε} ∣ K_{j} (y) ∣ d y \int_{R^{n}} ∣ φ (ξ) ∣ d ξ < \infty$ , so [F5] applies, giving $\int_{ε < ∣ y ∣ < 1 / ε} K_{j} (y) \hat{φ} (y) d y = \int_{R^{n}} φ (ξ) Λ_{ε, 1 / ε} (ξ) d ξ$ . By 2.1 the bracket converges to $- i ξ_{j} / ∣ ξ ∣$ pointwise off the null set ${ξ = 0}$ , and by the uniform bound of 2.1 it is dominated by a constant times $∣ φ (ξ) ∣$ ; [F6] therefore yields $⟨ F W_{j}, φ ⟩ = \int_{R^{n}} φ (ξ) (- i ξ_{j} / ∣ ξ ∣) d ξ$ , i.e. $F W_{j} = m_{j}$ as tempered distributions.

4.1step 1.1step 3.1F1F7F8F9∎

By [F8] and 3.1, $F (W_{j} * f) = (F W_{j}) (F f) = m_{j} \hat{f}$ ; by [F1] the $L^{2}$ class $R_{j} f$ has Fourier transform $m_{j} \hat{f}$ as well, so the two tempered distributions agree and [F9] gives $W_{j} * f = R_{j} f$ . By [F7] and the definition of $W_{j}$ in 3.1, the value $(W_{j} * f) (x) = ⟨ W_{j}, f (x - \cdot) ⟩$ is exactly the limit $J (x)$ of 1.1; the continuity in 1.1 therefore makes $J$ a continuous representative of the $L^{2}$ class $R_{j} f$ , and the truncated integrals $\int_{∣ y ∣ > ε} K_{j} (y) f (x - y) d y$ converge to it at every $x$ .

Depends on

Used by

Riesz kernel size, difference and spherical-cancellation bounds Lemma

Dependency tree · two levels

122 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Loukas Grafakos, Classical Fourier Analysis, third edition (standard reference, not scraped)
Richard S. Laugesen, Harmonic Analysis Lecture Notes (standard reference, not scraped)