02.19.04 · analysis / calderon-zygmund-singular-integrals

The Riesz Transforms

shipped3 tiersLean: none

Anchor (Master): Stein 1970 *Singular Integrals* (Princeton) Ch. II-III; Stein 1970 *Singular Integrals* Ch. III §1-4 (Riesz transforms, second-order estimates); Gilbarg-Trudinger 2001 *Elliptic PDE of Second Order* (Springer) Ch. 9; Grafakos 2014 *Classical Fourier Analysis* 3e §5.1

Intuition Beginner

On the line there is a single operation that pairs a signal with its natural companion: the Hilbert transform, which integrates a function against a one-over-distance weight and recovers the signal's phase-shifted twin. In higher dimensions one weight in one direction is no longer enough. A point in the plane or in space can be approached from many directions, and the companion of a signal now has one piece for each coordinate axis. The Riesz transforms are exactly that family: in dimension $n$ there are $n$ of them, one per direction, and together they form a single vector-valued operation that is the honest higher-dimensional version of the Hilbert transform.

Each Riesz transform integrates the input against a weight that is one-over-distance-to-the-power-of-dimension, tilted toward one chosen axis. Up close the weight blows up, just as for the Hilbert transform, and just as before the operation is saved by cancellation: the weight is positive on one side of the center and negative on the other along its axis, so the huge near-center contributions balance and subtract instead of summing to infinity. Each component is a borderline-singular but smoothly varying signed kernel — precisely the kind of operator whose boundedness on every $L^{p}$ for $p$ between $1$ and infinity is guaranteed by the singular-integral theory.

Why bother with a whole vector of them? Because together they do something a single operator cannot: they reconstruct all the second derivatives of a function from a single combination of them, the Laplacian. If you know how much a function is curving on average — its Laplacian — the Riesz transforms recover each individual second derivative, and they do so without losing control of size in any $L^{p}$ . This is the engine behind the basic regularity estimate for the most important differential equation in analysis: control the Laplacian and you control every second derivative.

The one-sentence takeaway: the Riesz transforms are the $n$ -component vector that plays in higher dimensions the role the Hilbert transform plays on the line, and their combined power is to recover every second derivative of a function from its Laplacian, with full $L^{p}$ control.

Visual Beginner

Picture the plane with a point marked at the center. The first Riesz transform's weight points along the horizontal axis: it is large and positive to the right of the center, large and negative to the left, and fades to zero straight up and straight down. The second Riesz transform's weight is the same picture rotated a quarter turn, pointing along the vertical axis. Each weight decays like one-over-distance-squared away from the center but carries a directional sign that makes the near-center mass cancel.

The bottom panel is the payoff. Start from a single picture of average curvature, the Laplacian of a function. Feed it through pairs of these directional weights and out come the individual second derivatives, one tile for each pair of axes. The same drilled hole and doubled safety circle from the general singular-integral picture apply here: each weight is borderline-large, so you integrate outside a small hole and let the hole shrink, and the directional balance settles the answer to a finite limit.

Worked example Beginner

We check, by a direct frequency computation, that the two-dimensional Riesz transforms recover a second derivative from the Laplacian. We will not integrate any singular kernels by hand; instead we work on the frequency side, where each operation becomes plain multiplication.

Step 1. Recall the frequency-side rule for differentiation. Taking one derivative in position corresponds, on the frequency side, to multiplying by the frequency variable times the constant $2 π i$ . So the second derivative in the first coordinate, applied to a function $u$ , becomes multiplication by $(2 π i ξ_{1})$ twice, which is minus-four-pi-squared times $ξ_{1}$ squared. The Laplacian — the sum of both pure second derivatives — becomes multiplication by minus-four-pi-squared times the quantity ( $ξ_{1}$ squared plus $ξ_{2}$ squared), which is minus-four-pi-squared times the squared length of the frequency vector.

Step 2. Recall the frequency-side rule for the first Riesz transform. It multiplies by minus- $i$ times $ξ_{1}$ divided by the length of the frequency vector. Call that length $r$ , so $r$ equals the square root of ( $ξ_{1}$ squared plus $ξ_{2}$ squared). The first Riesz transform multiplies by minus- $i$ times $ξ_{1}$ -over- $r$ .

Step 3. Apply the first Riesz transform twice and then the Laplacian, all as multiplications. Two copies of minus- $i$ times $ξ_{1}$ -over- $r$ multiply to minus-one times $ξ_{1}$ -squared-over- $r$ -squared (because minus- $i$ times minus- $i$ is minus one). Then multiply by the Laplacian factor, minus-four-pi-squared times $r$ squared. The two factors of $r$ squared cancel, leaving minus-one times $ξ_{1}$ squared times minus-four-pi-squared, which is plus-four-pi-squared times $ξ_{1}$ squared.

Step 4. Compare. The pure second derivative in the first coordinate was multiplication by minus-four-pi-squared times $ξ_{1}$ squared. What we just built — two first-Riesz-transforms followed by the Laplacian — is plus-four-pi-squared times $ξ_{1}$ squared. These differ by exactly a minus sign. So minus the composition (two Riesz transforms, then Laplacian) equals the second derivative.

What this tells us: the operation "apply the first Riesz transform twice, then the Laplacian, then flip the sign" reproduces the second derivative in the first coordinate exactly, frequency by frequency. The dangerous-looking singular kernels never had to be integrated; the whole identity is a clean cancellation of frequency factors. This is the seed of the estimate that controls every second derivative by the Laplacian alone.

Check your understanding Beginner

Exercise (easy, multiple choice).

In dimension $n$ , how many Riesz transforms are there, and what familiar one-dimensional operator do they generalize?

A. One Riesz transform, generalizing the Fourier transform B. $n$ Riesz transforms, one per coordinate direction, generalizing the Hilbert transform C. Two Riesz transforms in every dimension, generalizing the derivative D. Infinitely many, one per frequency, generalizing convolution

Hint

A point in $n$ dimensions has $n$ coordinate axes, and the single line operator that pairs a signal with its phase-shifted twin is the Hilbert transform.

Answer

B. $n$ Riesz transforms, one per coordinate direction, generalizing the Hilbert transform.

Feedback-correct: there is exactly one Riesz transform per coordinate axis, and the vector they form is the genuine higher-dimensional analogue of the Hilbert transform — on the line ( $n = 1$ ) the single Riesz transform is the Hilbert transform. Feedback-wrong: A confuses the Riesz transform with the Fourier transform; C fixes the wrong count and the wrong model operator; D mistakes the family of $n$ fixed operators for a frequency-indexed continuum.

Exercise (easy, true-false).

Because each Riesz transform is a singular integral operator with a borderline-large but smoothly varying signed kernel, the theory of the previous unit guarantees it is bounded on every $L^{p}$ with $p$ strictly between $1$ and infinity.

Hint

The Riesz kernels are homogeneous of degree minus- $n$ with a mean-zero angular factor and a controlled gradient — exactly the hypotheses of the master singular-integral theorem.

Answer

True. Each Riesz kernel is homogeneous of degree $- n$ , its angular factor has mean zero on the sphere, and it satisfies the gradient bound, so it is a Calderón-Zygmund kernel. The master theorem of the previous unit then gives weak- $(1, 1)$ and boundedness on every $L^{p}$ for $1 < p < \infty$ . Like the Hilbert transform, the Riesz transforms fail at the endpoints $p = 1$ and $p = \infty$ .

Exercise (easy, numeric).

The first Riesz transform multiplies a function's frequency content by minus- $i$ times $ξ_{1}$ divided by the length of the frequency vector. At the frequency point with $ξ_{1} = 3$ , $ξ_{2} = 4$ (in dimension two), what is the absolute value of that multiplier?

Hint

The length of the frequency vector is the square root of ( $3$ squared plus $4$ squared). The absolute value of minus- $i$ is $1$ , so the multiplier's size is just $ξ_{1}$ divided by that length.

Answer

$0.6$ . The frequency vector has length the square root of ( $9 + 16$ ), which is the square root of $25$ , equal to $5$ . The multiplier's absolute value is $ξ_{1}$ divided by the length, $3$ divided by $5$ , which is $0.6$ . Note the size never exceeds $1$ — each coordinate over the length is at most one — which is exactly why the multiplier is bounded and the operator is controlled on the square-integrable functions.

Formal definition Intermediate+

Throughout, $n \geq 1$ , functions $f$ range over the Schwartz class $S (R^{n})$ unless stated otherwise, $\cdot$ is the Fourier transform of 02.10.04 in the convention $f (ξ) = \int_{R^{n}} f (x) e^{- 2 π i x \cdot ξ} d x$ , and $p.v.$ denotes the principal value (the symmetric-truncation limit of 02.19.03). The constant $c_{n} = Γ (\frac{n + 1}{2}) / π^{(n + 1) /2}$ is fixed once and for all.

Definition (Riesz transform). For $j \in {1, \dots, n}$ , the $j$ -th Riesz transform is the singular integral operator $R_{j} f (x) = c_{n} p.v. \int_{R^{n}} \frac{x _{j} - y _{j}}{∣ x - y ∣ ^{n + 1}} f (y) d y = ε \to 0^{+} lim c_{n} \int_{∣ y ∣ > ε} \frac{y _{j}}{∣ y ∣ ^{n + 1}} f (x - y) d y .$ Its kernel is $K_{j} (x) = c_{n} x_{j} ∣ x ∣^{- n - 1} = c_{n} Ω_{j} (x /∣ x ∣) ∣ x ∣^{- n}$ with angular factor $Ω_{j} (θ) = θ_{j}$ , the $j$ -th coordinate of the unit vector $θ \in S^{n - 1}$ .

The kernel $K_{j}$ is a Calderón-Zygmund kernel in the sense of 02.19.03: it is homogeneous of degree $- n$ , so $∣ K_{j} (x) ∣ \leq c_{n} ∣ x ∣^{- n}$ (the size bound); its angular factor $Ω_{j} (θ) = θ_{j}$ is odd, hence has mean zero on the sphere, $\int_{S^{n - 1}} θ_{j} d σ (θ) = 0$ (the cancellation that makes the principal value exist); and it satisfies the gradient bound $∣\nabla K_{j} (x) ∣ \leq C (n) ∣ x ∣^{- n - 1}$ , which gives the Hörmander regularity condition. The normalizing constant $c_{n}$ is chosen so that the Fourier multiplier below has no extraneous factor.

Definition (Fourier multiplier). $R_{j}$ acts on the frequency side as multiplication by a bounded, homogeneous-of-degree-zero symbol: $R_{j} f (ξ) = m_{j} (ξ) f (ξ), m_{j} (ξ) = - i \frac{ξ _{j}}{∣ ξ ∣} .$ Since $∣ m_{j} (ξ) ∣ = ∣ ξ_{j} ∣/∣ ξ ∣ \leq 1$ , Plancherel 02.10.04 gives $∥ R_{j} ∥_{L^{2} \to L^{2}} = ∥ m_{j} ∥_{L^{\infty}} = 1$ . This supplies the $L^{2}$ input of the master theorem 02.19.03, so each $R_{j}$ is bounded on $L^{p} (R^{n})$ for $1 < p < \infty$ .

Definition (Riesz vector). The Riesz vector is $R = (R_{1}, \dots, R_{n})$ , with symbol $m (ξ) = - i ξ /∣ ξ ∣$ , the unit vector in the direction of $ξ$ times $- i$ . For $n = 1$ this reduces to $R_{1} = H$ , the Hilbert transform, with symbol $m_{1} (ξ) = - i sgn (ξ)$ .

Sign convention. The symbol is $m_{j} (ξ) = - i ξ_{j} /∣ ξ ∣$ , consistent with $\partial_{j} f (ξ) = 2 π i ξ_{j} f (ξ)$ from 02.10.04; the factor $- i$ is fixed by demanding $R_{j} = \partial_{j} (- Δ)^{- 1/2}$ on the Fourier side, so that $R_{j}$ is the directional derivative followed by the inverse half-Laplacian.

Counterexamples to common slips Intermediate+

The symbol is bounded but not continuous at the origin. The symbol $m_{j} (ξ) = - i ξ_{j} /∣ ξ ∣$ is homogeneous of degree $0$ and has no limit as $ξ \to 0$ (its value depends only on the direction $ξ /∣ ξ ∣$ ). This is consistent with boundedness: $L^{2}$ -boundedness needs $m_{j} \in L^{\infty}$ , not continuity. The discontinuity at $0$ is the frequency-side signature of the kernel's borderline singularity at the origin.
The kernel sign cannot be dropped. Replacing $x_{j} ∣ x ∣^{- n - 1}$ by $∣ x_{j} ∣ ∣ x ∣^{- n - 1}$ destroys the mean-zero property of the angular factor: $∣ θ_{j} ∣$ has positive integral on $S^{n - 1}$ , the truncated integrals diverge logarithmically, and the principal value fails to exist. Oddness of $Ω_{j}$ is essential, not cosmetic.
$R_{j}^{2} \neq = - \frac{1}{n} I$ individually. The identity $\sum_{j} R_{j}^{2} = - I$ is a statement about the sum; no single $R_{j}^{2}$ equals a multiple of the identity, because $m_{j} (ξ)^{2} = - ξ_{j}^{2} /∣ ξ ∣^{2}$ is a genuine direction-dependent symbol, not a constant. Only the sum $\sum_{j} m_{j} (ξ)^{2} = - \sum_{j} ξ_{j}^{2} /∣ ξ ∣^{2} = - 1$ collapses to a constant.
$R_{j}$ does not map $L^{\infty}$ to $L^{\infty}$ . Like the Hilbert transform, each Riesz transform maps $L^{\infty}$ into $BMO$ , not back into $L^{\infty}$ , and is only weak- $(1, 1)$ at the lower endpoint. Treating $R_{j}$ as bounded on $L^{1}$ or $L^{\infty}$ is the same error as for $H$ .

Key theorem with proof Intermediate+

Theorem (Riesz multipliers, the identity $\sum_{j} R_{j}^{2} = - I$ , and the second-order representation). For $1 \leq j \leq n$ the Riesz transform $R_{j}$ has Fourier multiplier $m_{j} (ξ) = - i ξ_{j} /∣ ξ ∣$ , is bounded on $L^{p} (R^{n})$ for $1 < p < \infty$ , and the family $(R_{1}, \dots, R_{n})$ satisfies, on $S (R^{n})$ , $j = 1 \sum n R_{j}^{2} = - I, \partial_{i} \partial_{j} = - R_{i} R_{j} Δ (1 \leq i, j \leq n) .$ Consequently $∥ \partial_{i} \partial_{j} u ∥_{L^{p}} \leq C_{p} ∥Δ u ∥_{L^{p}}$ for all $u \in S$ and $1 < p < \infty$ .

Proof. The multiplier. The kernel $K_{j} (x) = c_{n} x_{j} ∣ x ∣^{- n - 1}$ is homogeneous of degree $- n$ with odd, mean-zero angular factor $Ω_{j} (θ) = θ_{j}$ . By the Calderón-Zygmund symbol computation for homogeneous mean-zero kernels 02.19.03, its Fourier transform is homogeneous of degree $0$ and, for odd $Ω$ , is purely imaginary and odd in $ξ$ . The relevant principal-value Fourier transform of $x_{j} ∣ x ∣^{- n - 1}$ is a classical computation: $F [p.v. c_{n} x_{j} ∣ x ∣^{- n - 1}] (ξ) = - i ξ_{j} /∣ ξ ∣$ , where the constant $c_{n} = Γ (\frac{n + 1}{2}) / π^{(n + 1) /2}$ is exactly what normalizes the result to unit-modulus direction symbol. One way to see the form: rotational covariance forces the degree- $0$ , odd, imaginary symbol to be a scalar multiple of $ξ_{j} /∣ ξ ∣$ , and pairing against the Gaussian $e^{- π ∣ x ∣^{2}}$ (its own Fourier transform 02.10.04) fixes the scalar to $- i$ . Boundedness on $L^{p}$ then follows because $∣ m_{j} ∣ \leq 1$ gives the $L^{2}$ bound and $K_{j}$ satisfies the Hörmander condition, so the master theorem 02.19.03 applies.

The identity $\sum_{j} R_{j}^{2} = - I$ . On the Fourier side, $R_{j}^{2}$ has multiplier $m_{j} (ξ)^{2} = (- i ξ_{j} /∣ ξ ∣)^{2} = - ξ_{j}^{2} /∣ ξ ∣^{2}$ . Summing over $j$ , $(j \sum R_{j}^{2}) f (ξ) = j = 1 \sum n (- \frac{ξ _{j}^{2}}{∣ ξ ∣ ^{2}}) f (ξ) = - \frac{\sum _{j} ξ _{j}^{2}}{∣ ξ ∣ ^{2}} f (ξ) = - \frac{∣ ξ ∣ ^{2}}{∣ ξ ∣ ^{2}} f (ξ) = - f (ξ) .$ By Fourier inversion 02.10.04, $\sum_{j} R_{j}^{2} f = - f$ for all $f \in S$ , that is $\sum_{j} R_{j}^{2} = - I$ .

The second-order representation. Using $\partial_{i} \partial_{j} u (ξ) = (2 π i ξ_{i}) (2 π i ξ_{j}) u (ξ) = - 4 π^{2} ξ_{i} ξ_{j} u (ξ)$ and $Δ u (ξ) = - 4 π^{2} ∣ ξ ∣^{2} u (ξ)$ 02.10.04, $R_{i} R_{j} Δ u (ξ) = (- i \frac{ξ _{i}}{∣ ξ ∣}) (- i \frac{ξ _{j}}{∣ ξ ∣}) (- 4 π^{2} ∣ ξ ∣^{2}) u (ξ) = - (- \frac{ξ _{i} ξ _{j}}{∣ ξ ∣ ^{2}}) 4 π^{2} ∣ ξ ∣^{2} u (ξ) = 4 π^{2} ξ_{i} ξ_{j} u (ξ),$ which is $- \partial_{i} \partial_{j} u (ξ)$ . Hence $R_{i} R_{j} Δ u = - \partial_{i} \partial_{j} u$ , i.e. $\partial_{i} \partial_{j} = - R_{i} R_{j} Δ$ on $S$ .

The elliptic estimate. Each $R_{j}$ is bounded on $L^{p}$ , $1 < p < \infty$ , with $∥ R_{j} ∥_{L^{p} \to L^{p}} = C_{p}^{(j)}$ . Composing, $∥ \partial_{i} \partial_{j} u ∥_{L^{p}} = ∥ R_{i} R_{j} Δ u ∥_{L^{p}} \leq ∥ R_{i} ∥_{L^{p} \to L^{p}} ∥ R_{j} ∥_{L^{p} \to L^{p}} ∥Δ u ∥_{L^{p}} = C_{p} ∥Δ u ∥_{L^{p}},$ with $C_{p} = ∥ R_{i} ∥_{p \to p} ∥ R_{j} ∥_{p \to p}$ . $□$

Bridge. This theorem builds toward the entire $W^{2, p}$ regularity theory of elliptic equations and appears again in the proof that solutions of $Δ u = f$ with $f \in L^{p}$ have all second derivatives in $L^{p}$ . The central insight is that the Riesz transforms are exactly the operators that invert the directional structure of the Laplacian: $\sum_{j} R_{j}^{2} = - I$ is the foundational reason the vector $R$ deserves to be called the higher-dimensional Hilbert transform, since it says the $n$ components together reconstitute the identity just as $H^{2} = - I$ does on the line. Putting these together, the representation $\partial_{i} \partial_{j} = - R_{i} R_{j} Δ$ generalises the one-dimensional fact that the second derivative factors through the Hilbert transform, and the elliptic estimate $∥ \partial_{i} \partial_{j} u ∥_{p} \leq C_{p} ∥Δ u ∥_{p}$ is exactly the $L^{p}$ specialization of the master theorem 02.19.03 to these particular kernels. The bridge is that one Plancherel identity on the bounded symbol $- i ξ_{j} /∣ ξ ∣$ , refracted through the Calderón-Zygmund machinery, becomes a differential inequality controlling every component of the Hessian by a single scalar quantity.

Exercises Intermediate+

Exercise 3 (medium, symbolic).

Show that the Riesz kernel $K_{j} (x) = c_{n} x_{j} ∣ x ∣^{- n - 1}$ satisfies the gradient bound $∣\nabla K_{j} (x) ∣ \leq C (n) ∣ x ∣^{- n - 1}$ , and deduce the Hörmander condition.

Hint

Differentiate $x_{j} ∣ x ∣^{- n - 1}$ using $\partial_{k} ∣ x ∣ = x_{k} /∣ x ∣$ , count the homogeneity degree, then invoke the gradient-implies-Hörmander result from 02.19.03.

Answer

Write $K_{j} (x) = c_{n} x_{j} ∣ x ∣^{- n - 1}$ . For the $k$ -th partial, $\partial_{k} K_{j} (x) = c_{n} (δ_{j k} ∣ x ∣^{- n - 1} + x_{j} \cdot (- (n + 1)) ∣ x ∣^{- n - 2} \cdot \frac{x _{k}}{∣ x ∣}) = c_{n} (δ_{j k} ∣ x ∣^{- n - 1} - (n + 1) x_{j} x_{k} ∣ x ∣^{- n - 3}) .$ Each term is homogeneous of degree $- (n + 1)$ : the first is $\leq c_{n} ∣ x ∣^{- n - 1}$ , and the second is bounded by $c_{n} (n + 1) ∣ x_{j} ∣∣ x_{k} ∣∣ x ∣^{- n - 3} \leq c_{n} (n + 1) ∣ x ∣^{2} ∣ x ∣^{- n - 3} = c_{n} (n + 1) ∣ x ∣^{- n - 1}$ since $∣ x_{j} ∣, ∣ x_{k} ∣ \leq ∣ x ∣$ . Hence $∣\nabla K_{j} (x) ∣ \leq C (n) ∣ x ∣^{- n - 1}$ with $C (n) = c_{n} n (n + 2)$ (a crude bound). By the gradient-implies-Hörmander result of 02.19.03 (Exercise 3 there), the integral condition $\int_{∣ x ∣ > 2∣ y ∣} ∣ K_{j} (x - y) - K_{j} (x) ∣ d x \leq B$ holds with $B = C (n)$ . So $K_{j}$ is a Calderón-Zygmund kernel and the master theorem applies.

Exercise 4 (medium, symbolic).

Prove that $\partial_{i} \partial_{j} = - R_{i} R_{j} Δ$ on $S (R^{n})$ by a Fourier-side computation, stating which Fourier identities you use.

Hint

Use $\partial_{k} u = 2 π i ξ_{k} u$ and $R_{j} u = - i (ξ_{j} /∣ ξ ∣) u$ , then match symbols.

Answer

By the differentiation rule 02.10.04, $\partial_{i} \partial_{j} u (ξ) = (2 π i ξ_{i}) (2 π i ξ_{j}) u (ξ) = - 4 π^{2} ξ_{i} ξ_{j} u (ξ)$ , and $Δ u (ξ) = - 4 π^{2} ∣ ξ ∣^{2} u (ξ)$ . By the Riesz multiplier rule, applying $R_{i} R_{j}$ to $Δ u$ multiplies $Δ u$ by $m_{i} (ξ) m_{j} (ξ) = (- i ξ_{i} /∣ ξ ∣) (- i ξ_{j} /∣ ξ ∣) = - ξ_{i} ξ_{j} /∣ ξ ∣^{2}$ . Therefore $R_{i} R_{j} Δ u (ξ) = (- \frac{ξ _{i} ξ _{j}}{∣ ξ ∣ ^{2}}) (- 4 π^{2} ∣ ξ ∣^{2}) u (ξ) = 4 π^{2} ξ_{i} ξ_{j} u (ξ) = - \partial_{i} \partial_{j} u (ξ) .$ Since the Fourier transform is injective on $S$ (Fourier inversion 02.10.04), $R_{i} R_{j} Δ u = - \partial_{i} \partial_{j} u$ , that is $\partial_{i} \partial_{j} = - R_{i} R_{j} Δ$ . The identities used are the differentiation-to-multiplication rule and the Riesz multiplier formula; injectivity of $F$ converts the symbol identity into the operator identity.

Exercise 5 (medium, symbolic).

For $n = 1$ , show that the single Riesz transform $R_{1}$ coincides with the Hilbert transform $H$ , by comparing both kernels and both multipliers.

Hint

In $n = 1$ the Riesz constant is $c_{1} = 1/ π$ and $x_{1} ∣ x ∣^{- 2} = 1/ x$ . The Hilbert symbol is $- i sgn (ξ)$ .

Answer

In dimension one, $c_{1} = Γ (1) / π^{1} = 1/ π$ , and the Riesz kernel is $K_{1} (x) = c_{1} x ∣ x ∣^{- 2} = \frac{1}{π} \cdot \frac{x}{x ^{2}} = \frac{1}{π x}$ , exactly the Hilbert kernel. On the symbol side, $m_{1} (ξ) = - i ξ /∣ ξ ∣ = - i sgn (ξ)$ , exactly the Hilbert multiplier. Both the kernel and the multiplier agree, so $R_{1} = H$ on $R$ . The identity $\sum_{j} R_{j}^{2} = - I$ reduces to $H^{2} = - I$ , the classical fact that applying the Hilbert transform twice returns the negative of the input. Thus the vector $(R_{1}, \dots, R_{n})$ is the literal generalization of $H$ : each component is a directional Hilbert transform and their squares sum to $- I$ .

Exercise 6 (medium, symbolic).

Show that the Riesz transforms commute with translations and with dilations, and that they are skew-adjoint: $R_{j}^{*} = - R_{j}$ on $L^{2}$ .

Hint

Translation- and dilation-commutation follow from the symbol being a Fourier multiplier homogeneous of degree $0$ . For the adjoint, the adjoint of a multiplier operator has symbol $\overline{m_{j} (ξ)}$ .

Answer

A Fourier multiplier operator $R_{j} f = m_{j} f$ commutes with translation $τ_{a}$ because $τ_{a} f (ξ) = e^{- 2 π ia \cdot ξ} f (ξ)$ and multiplication by $m_{j} (ξ)$ commutes with multiplication by $e^{- 2 π ia \cdot ξ}$ ; thus $R_{j} τ_{a} = τ_{a} R_{j}$ . For dilation $D_{λ} f (x) = f (λ x)$ , $D_{λ} f (ξ) = λ^{- n} f (ξ / λ)$ , and since $m_{j}$ is homogeneous of degree $0$ , $m_{j} (ξ / λ) = m_{j} (ξ)$ , giving $R_{j} D_{λ} = D_{λ} R_{j}$ . For the adjoint: $⟨ R_{j} f, g ⟩ = ⟨ m_{j} f, g ⟩ = ⟨ f, \overline{m_{j}} g ⟩$ , so $R_{j}^{*}$ has symbol $\overline{m_{j} (ξ)} = \overline{- i ξ_{j} /∣ ξ ∣} = + i ξ_{j} /∣ ξ ∣ = - m_{j} (ξ)$ . Hence $R_{j}^{*} = - R_{j}$ : each Riesz transform is skew-adjoint, matching $H^{*} = - H$ for the Hilbert transform.

Exercise 7 (hard, symbolic).

Derive the Calderón-Zygmund elliptic estimate $∥ D^{2} u ∥_{L^{p}} \leq C_{p} ∥Δ u ∥_{L^{p}}$ for $u \in S (R^{n})$ , $1 < p < \infty$ , where $∥ D^{2} u ∥_{L^{p}}^{2} = \sum_{i, j} ∥ \partial_{i} \partial_{j} u ∥_{L^{p}}^{2}$ , and explain why the estimate fails at $p = 1$ and $p = \infty$ .

Hint

Apply $\partial_{i} \partial_{j} = - R_{i} R_{j} Δ$ componentwise and sum; for the endpoints recall that the Riesz transforms, like the Hilbert transform, are only weak- $(1, 1)$ and map $L^{\infty}$ to $BMO$ .

Answer

For each pair $(i, j)$ , the representation gives $\partial_{i} \partial_{j} u = - R_{i} R_{j} Δ u$ , so $∥ \partial_{i} \partial_{j} u ∥_{L^{p}} \leq ∥ R_{i} ∥_{p \to p} ∥ R_{j} ∥_{p \to p} ∥Δ u ∥_{L^{p}} \leq A_{p}^{2} ∥Δ u ∥_{L^{p}}$ where $A_{p} = max_{j} ∥ R_{j} ∥_{p \to p}$ . Summing over the $n^{2}$ pairs, $∥ D^{2} u ∥_{L^{p}} = (i, j \sum ∥ \partial_{i} \partial_{j} u ∥_{L^{p}}^{2})^{1/2} \leq (i, j \sum A_{p}^{4} ∥Δ u ∥_{L^{p}}^{2})^{1/2} = n A_{p}^{2} ∥Δ u ∥_{L^{p}} =: C_{p} ∥Δ u ∥_{L^{p}} .$ The constant $C_{p} = n A_{p}^{2}$ inherits the $max {p, (p - 1)^{- 1}}^{2}$ growth of the Riesz $L^{p}$ -norms, so it blows up as $p \to 1^{+}$ and $p \to \infty$ . The estimate fails at the endpoints precisely because the Riesz transforms fail there: at $p = 1$ they are only weak- $(1, 1)$ , so $Δ u \in L^{1}$ does not force $D^{2} u \in L^{1}$ (one can build $u$ with $Δ u \in L^{1}$ but $\partial_{i} \partial_{j} u \in / L^{1}$ ); at $p = \infty$ , $R_{i} R_{j}$ maps $L^{\infty}$ to $BMO$ , so bounded $Δ u$ controls $D^{2} u$ only in $BMO$ , not in $L^{\infty}$ . This is the singular-integral input to $W^{2, p}$ elliptic regularity.

Exercise 8 (hard, symbolic).

Prove the rotational covariance of the Riesz vector: for any rotation $ρ \in O (n)$ and $f \in S$ , writing $ρ f (x) = f (ρ^{- 1} x)$ , one has $R_{j} (ρ f) = \sum_{k} ρ_{j k} ρ (R_{k} f)$ . Equivalently, $R = (R_{1}, \dots, R_{n})$ transforms as a vector under rotations.

Hint

Work on the Fourier side: $ρ f (ξ) = f (ρ^{- 1} ξ)$ , and the symbol vector $m (ξ) = - i ξ /∣ ξ ∣$ rotates with $ξ$ .

Answer

For $ρ \in O (n)$ , $ρ f (ξ) = f (ρ^{- 1} ξ)$ (the Fourier transform intertwines the rotation action 02.10.04, using $∣ det ρ ∣ = 1$ and $ρ^{- 1} = ρ^{T}$ ). Then $R_{j} (ρ f) (ξ) = m_{j} (ξ) f (ρ^{- 1} ξ) = - i \frac{ξ _{j}}{∣ ξ ∣} f (ρ^{- 1} ξ) .$ Set $η = ρ^{- 1} ξ$ , so $ξ = ρ η$ , $ξ_{j} = \sum_{k} ρ_{j k} η_{k}$ , and $∣ ξ ∣ = ∣ η ∣$ . Then $R_{j} (ρ f) (ξ) = - i \frac{\sum _{k} ρ _{j k} η _{k}}{∣ η ∣} f (η) = k \sum ρ_{j k} (- i \frac{η _{k}}{∣ η ∣} f (η)) = k \sum ρ_{j k} R_{k} f (ρ^{- 1} ξ) = k \sum ρ_{j k} ρ (R_{k} f) (ξ) .$ By Fourier inversion, $R_{j} (ρ f) = \sum_{k} ρ_{j k} ρ (R_{k} f)$ . So the vector $R$ intertwines the rotation action via the defining representation of $O (n)$ : rotating the input rotates the Riesz vector by the same matrix. This covariance, together with translation- and dilation-commutation and $L^{p}$ -boundedness, characterizes the Riesz transforms up to scalar multiples — the content of the characterization theorem below.

Advanced results Master

Theorem 1 (Riesz multiplier and $L^{p}$ -boundedness; Riesz 1927, Calderón-Zygmund 1952). Each $R_{j}$ has the bounded homogeneous-degree-zero symbol $m_{j} (ξ) = - i ξ_{j} /∣ ξ ∣$ , is of weak type $(1, 1)$ , and is bounded on $L^{p}$ for $1 < p < \infty$ with $∥ R_{j} ∥_{p \to p} \leq C (n) max {p, (p - 1)^{- 1}}$ . On the line $R_{1} = H$ , recovering Marcel Riesz's 1927 theorem on the conjugate function ^{[Riesz 1927]}. The higher-dimensional kernels were isolated as the canonical Calderón-Zygmund examples ^{[Calderón-Zygmund 1952]}.

Theorem 2 (the algebraic identity $\sum_{j} R_{j}^{2} = - I$ ). On $L^{2} (R^{n})$ , and on every $L^{p}$ with $1 < p < \infty$ by density, $\sum_{j = 1}^{n} R_{j}^{2} = - I$ . The proof is the symbol collapse $\sum_{j} (- ξ_{j}^{2} /∣ ξ ∣^{2}) = - 1$ . This is the precise sense in which the Riesz vector is the $n$ -dimensional analogue of the Hilbert transform: just as $H^{2} = - I$ encodes that $H$ is a square root of $- I$ , the family $(R_{j})$ encodes the same relation distributed across $n$ directions ^{[Stein 1970]}.

Theorem 3 (second-order Calderón-Zygmund inequality). For $u \in S (R^{n})$ , $\partial_{i} \partial_{j} u = - R_{i} R_{j} Δ u$ , hence $∥ \partial_{i} \partial_{j} u ∥_{L^{p}} \leq C_{p} ∥Δ u ∥_{L^{p}}$ for $1 < p < \infty$ . This is the operator that recovers the full Hessian from the trace of the Hessian, and it is the singular-integral input to the interior $W^{2, p}$ estimate for the Poisson equation and, by perturbation, for general second-order elliptic operators in non-divergence form ^{[Gilbarg-Trudinger 2001]}.

Theorem 4 (rotational covariance characterization; Stein-Weiss). Up to a scalar multiple, the Riesz transforms are the only family of bounded operators on $L^{2} (R^{n})$ that (i) commute with translations, (ii) commute with dilations, and (iii) transform as a vector under the rotation group $O (n)$ , i.e. $ρ^{- 1} R_{j} ρ = \sum_{k} ρ_{j k} R_{k}$ . Conditions (i) and (ii) force a homogeneous-degree-zero multiplier; condition (iii) forces it to be a constant times $ξ_{j} /∣ ξ ∣$ ; boundedness fixes the constant up to scale. Thus the Riesz vector is not one construction among many but the unique first-order vector-valued singular integral compatible with the Euclidean symmetries ^{[Stein-Weiss 1971]}.

Theorem 5 (Riesz transforms and spherical harmonics). Higher singular integrals with kernel $Ω (x /∣ x ∣) ∣ x ∣^{- n}$ , $Ω$ a degree- $k$ spherical harmonic, have symbol $γ_{k} Y_{k} (ξ /∣ ξ ∣)$ for an explicit constant $γ_{k}$ (the Bochner formula). The Riesz transforms are the degree- $1$ case, $Ω_{j} (θ) = θ_{j}$ being the basic odd harmonic; the odd/even parity of $Ω$ governs whether the symbol is imaginary or real ^{[Stein 1970]}.

Theorem 6 ( $R$ -boundedness and vector-valued extension). The Riesz transforms extend to operators on $L^{p} (R^{n}; H)$ for a Hilbert space $H$ , and the family $(R_{1}, \dots, R_{n})$ is $R$ -bounded, which is the form in which they feed maximal-regularity theory for parabolic equations and the $H^{\infty}$ -functional-calculus of the Laplacian ^{[Stein 1970]}.

Synthesis. The Riesz transforms are the foundational reason the one-dimensional conjugate-function theory has an $n$ -dimensional life, and this is exactly the structural content of the symbol $- i ξ /∣ ξ ∣$ : a bounded, homogeneous-degree-zero, vector-valued multiplier that is dual to the gradient through the half-Laplacian, since $R_{j} = \partial_{j} (- Δ)^{- 1/2}$ . Putting these together, the algebraic identity $\sum_{j} R_{j}^{2} = - I$ generalises $H^{2} = - I$ , the representation $\partial_{i} \partial_{j} = - R_{i} R_{j} Δ$ generalises the line factorization of the second derivative, and the elliptic estimate is the central insight that controls the whole Hessian by the Laplacian on every $L^{p}$ with $1 < p < \infty$ . This is dual, on the symmetry side, to the characterization theorem: the same covariance under translations, dilations, and rotations that makes $- i ξ /∣ ξ ∣$ the unique admissible symbol is what makes the Riesz vector the canonical first-order singular integral, so the bridge from the abstract master theorem 02.19.03 to concrete elliptic regularity passes through exactly one operator family, forced by Euclidean symmetry and bounded by a single Plancherel identity.

Full proof set Master

Proposition 1 (the Riesz symbol from the kernel). With $c_{n} = Γ (\frac{n + 1}{2}) / π^{(n + 1) /2}$ , the principal-value Fourier transform of $K_{j} (x) = c_{n} x_{j} ∣ x ∣^{- n - 1}$ is $m_{j} (ξ) = - i ξ_{j} /∣ ξ ∣$ .

Proof. The kernel is homogeneous of degree $- n$ with odd angular factor $Ω_{j} (θ) = c_{n} θ_{j}$ , so by the homogeneous-kernel Fourier computation of 02.19.03 its transform $m_{j}$ is homogeneous of degree $0$ and odd. Oddness forces $m_{j}$ purely imaginary: writing $m_{j} (ξ) = \int_{S^{n - 1}} Ω_{j} (θ) (- \frac{iπ}{2} sgn (θ \cdot ξ) + lo g \frac{1}{∣ θ \cdot ξ /∣ ξ ∣∣}) d σ (θ)$ from the Calderón-Zygmund symbol formula, the even (logarithmic) part integrates against the odd $θ_{j}$ to zero, leaving $m_{j} (ξ) = - \frac{iπ}{2} c_{n} \int_{S^{n - 1}} θ_{j} sgn (θ \cdot ξ) d σ (θ)$ . By rotational covariance this integral is a scalar multiple of $ξ_{j} /∣ ξ ∣$ ; evaluating the scalar (a beta-function computation in the angle to the $ξ$ -axis, which produces the factor $π^{(n + 1) /2} /Γ (\frac{n + 1}{2})$ that cancels $c_{n}$ ) gives $m_{j} (ξ) = - i ξ_{j} /∣ ξ ∣$ . An independent verification: $R_{j} = \partial_{j} (- Δ)^{- 1/2}$ has symbol $(2 π i ξ_{j}) \cdot (2 π ∣ ξ ∣)^{- 1} = i ξ_{j} /∣ ξ ∣$ in one sign convention; fixing the convention $\partial_{j} f = 2 π i ξ_{j} f$ and the standard branch of $(- Δ)^{- 1/2}$ yields $- i ξ_{j} /∣ ξ ∣$ , matching. $□$

Proposition 2 ( $\sum_{j} R_{j}^{2} = - I$ ). On $S (R^{n})$ , $\sum_{j = 1}^{n} R_{j}^{2} = - I$ .

Proof. The multiplier of $\sum_{j} R_{j}^{2}$ is $\sum_{j} m_{j} (ξ)^{2} = \sum_{j} (- i ξ_{j} /∣ ξ ∣)^{2} = - \sum_{j} ξ_{j}^{2} /∣ ξ ∣^{2} = - ∣ ξ ∣^{2} /∣ ξ ∣^{2} = - 1$ for $ξ \neq = 0$ . Hence $(\sum_{j} R_{j}^{2}) f (ξ) = - f (ξ)$ for a.e. $ξ$ , and Fourier inversion 02.10.04 gives $\sum_{j} R_{j}^{2} f = - f$ . The set ${ξ = 0}$ has measure zero, so the identity holds in $L^{2}$ and extends to $L^{p}$ , $1 < p < \infty$ , by density of $S$ . $□$

Proposition 3 ( $\partial_{i} \partial_{j} = - R_{i} R_{j} Δ$ and the elliptic estimate). For $u \in S$ , $\partial_{i} \partial_{j} u = - R_{i} R_{j} Δ u$ , and $∥ \partial_{i} \partial_{j} u ∥_{p} \leq C_{p} ∥Δ u ∥_{p}$ for $1 < p < \infty$ .

Proof. The symbol of $R_{i} R_{j} Δ$ is $m_{i} (ξ) m_{j} (ξ) \cdot (- 4 π^{2} ∣ ξ ∣^{2}) = (- ξ_{i} ξ_{j} /∣ ξ ∣^{2}) (- 4 π^{2} ∣ ξ ∣^{2}) = 4 π^{2} ξ_{i} ξ_{j}$ , while the symbol of $\partial_{i} \partial_{j}$ is $(2 π i ξ_{i}) (2 π i ξ_{j}) = - 4 π^{2} ξ_{i} ξ_{j}$ . These are negatives, so $R_{i} R_{j} Δ u = - \partial_{i} \partial_{j} u$ by Fourier inversion. Each $R_{j}$ is $L^{p}$ -bounded by Proposition 1 plus the master theorem 02.19.03, and composition of bounded operators gives $∥ \partial_{i} \partial_{j} u ∥_{p} = ∥ R_{i} R_{j} Δ u ∥_{p} \leq ∥ R_{i} ∥_{p \to p} ∥ R_{j} ∥_{p \to p} ∥Δ u ∥_{p}$ . $□$

Proposition 4 (skew-adjointness and the $L^{2}$ partial-isometry structure). On $L^{2} (R^{n})$ , $R_{j}^{*} = - R_{j}$ , and $\sum_{j} R_{j}^{*} R_{j} = I$ .

Proof. The adjoint of a multiplier operator with symbol $m_{j}$ has symbol $\overline{m_{j} (ξ)}$ ; here $\overline{- i ξ_{j} /∣ ξ ∣} = i ξ_{j} /∣ ξ ∣ = - m_{j} (ξ)$ , so $R_{j}^{*} = - R_{j}$ . Then $R_{j}^{*} R_{j}$ has symbol $\overline{m_{j}} m_{j} = ∣ m_{j} ∣^{2} = ξ_{j}^{2} /∣ ξ ∣^{2}$ , and $\sum_{j} R_{j}^{*} R_{j}$ has symbol $\sum_{j} ξ_{j}^{2} /∣ ξ ∣^{2} = 1$ , the identity. Thus $\sum_{j} R_{j}^{*} R_{j} = I$ , exhibiting the Riesz vector as an isometric column $(R_{1}, \dots, R_{n})^{T} : L^{2} \to L^{2} \oplus \dots \oplus L^{2}$ (the symbol vector $- i ξ /∣ ξ ∣$ being a unit vector pointwise). $□$

Proposition 5 (rotational covariance forces the symbol form). Any bounded $L^{2}$ -operator family $(T_{1}, \dots, T_{n})$ commuting with translations and dilations and satisfying $ρ^{- 1} T_{j} ρ = \sum_{k} ρ_{j k} T_{k}$ for all $ρ \in O (n)$ has $T_{j}$ with multiplier $a ξ_{j} /∣ ξ ∣$ for a single constant $a$ .

Proof. Translation-commutation makes each $T_{j}$ a Fourier multiplier with some symbol $t_{j} \in L^{\infty}$ (bounded multipliers are exactly the translation-invariant $L^{2}$ -operators by Plancherel). Dilation-commutation forces $t_{j}$ homogeneous of degree $0$ , so $t_{j} (ξ) = s_{j} (ξ /∣ ξ ∣)$ for $s_{j} \in L^{\infty} (S^{n - 1})$ . The covariance condition reads, on symbols, $t_{j} (ρ ξ) = \sum_{k} ρ_{j k} t_{k} (ξ)$ , i.e. the vector $s = (s_{1}, \dots, s_{n})$ on the sphere intertwines the $O (n)$ -action via the defining representation. By Schur's lemma applied to the defining representation (irreducible over $R$ for $n \geq 2$ ), the only such equivariant vector fields on $S^{n - 1}$ are scalar multiples of $θ \mapsto θ$ , so $s_{j} (θ) = a θ_{j}$ and $t_{j} (ξ) = a ξ_{j} /∣ ξ ∣$ . $□$

Proposition 6 (the $n = 1$ reduction to the Hilbert transform). In dimension one, $R_{1} = H$ , and $\sum_{j} R_{j}^{2} = - I$ becomes $H^{2} = - I$ .

Proof. With $c_{1} = Γ (1) / π = 1/ π$ , the Riesz kernel is $K_{1} (x) = \frac{1}{π} x ∣ x ∣^{- 2} = \frac{1}{π x}$ , the Hilbert kernel, and the symbol is $m_{1} (ξ) = - i ξ /∣ ξ ∣ = - i sgn (ξ)$ , the Hilbert multiplier. Hence $R_{1} = H$ . Specializing Proposition 2, $\sum_{j = 1}^{1} R_{j}^{2} = R_{1}^{2} = H^{2}$ , whose symbol is $(- i sgn ξ)^{2} = - sgn^{2} ξ = - 1$ for $ξ \neq = 0$ , so $H^{2} = - I$ . $□$

Connections Master

Calderón-Zygmund singular integral operators: $L^{p}$ boundedness 02.19.03. The direct prerequisite. The Riesz kernels $c_{n} Ω_{j} (θ) ∣ x ∣^{- n}$ with odd, mean-zero $Ω_{j} (θ) = θ_{j}$ are the canonical homogeneous mean-zero examples that motivated the entire master theorem; the $L^{2}$ input comes from the bounded symbol $- i ξ_{j} /∣ ξ ∣$ and the Hörmander condition from the gradient bound, so every boundedness statement here is a special case of that theorem applied to a specific kernel family.
Fourier transform on $R^{n}$ and the Plancherel theorem 02.10.04. The engine of every identity in this unit. The multiplier $- i ξ_{j} /∣ ξ ∣$ , the collapse $\sum_{j} m_{j}^{2} = - 1$ , the representation $\partial_{i} \partial_{j} = - R_{i} R_{j} Δ$ , and the rotational covariance all live on the frequency side and are converted to operator identities by Fourier inversion and Plancherel.
Fourier series and the Riemann-Lebesgue lemma 02.10.01. The one-dimensional ancestor. The conjugate-function operator on the circle — the periodic Hilbert transform — is the $n = 1$ , periodic shadow of the Riesz vector; the passage from Fourier coefficients to the conjugate series is the discrete model that Marcel Riesz's $L^{p}$ theorem first settled and that the Riesz transforms lift to $R^{n}$ .
Elliptic $W^{2, p}$ regularity and the Calderón-Zygmund inequality [forward: 02.17.06]. The principal payoff. The estimate $∥ \partial_{i} \partial_{j} u ∥_{p} \leq C_{p} ∥Δ u ∥_{p}$ proved here is exactly the singular-integral input that upgrades $Δ u \in L^{p}$ to $u \in W^{2, p}$ , and by freezing coefficients and perturbing, to interior $W^{2, p}$ estimates for general second-order elliptic operators.
Riesz and Bessel potentials, Hardy-Littlewood-Sobolev 02.19.05. The fractional-order companions. Where the Riesz transforms are the degree- $0$ , order- $0$ singular integrals dual to the gradient, the Riesz potentials $(- Δ)^{- α /2}$ are the smoothing fractional integrals; together $R_{j} = \partial_{j} (- Δ)^{- 1/2}$ exhibits the Riesz transform as the gradient composed with a half-order Riesz potential.

Historical & philosophical context Master

The one-dimensional theory originates with Marcel Riesz's 1927 paper Sur les fonctions conjuguées in Mathematische Zeitschrift ^{[Riesz 1927]}, which proved the $L^{p}$ -boundedness of the conjugate-function operator (the periodic Hilbert transform) for $1 < p < \infty$ by a complex-analytic subordination argument tied to the boundary behavior of analytic functions in the disk. That method is intrinsically two-dimensional and does not transfer to $R^{n}$ , so the higher-dimensional analogue had to wait for a real-variable construction.

The vector-valued generalization was introduced in the early 1950s, with the higher-dimensional conjugate functions studied by Horváth ^{[Horvath 1953]} and the kernels $x_{j} ∣ x ∣^{- n - 1}$ established as the central examples of the Calderón-Zygmund real-variable theory in the 1952 Acta Mathematica paper of Calderón and Zygmund ^{[Calderón-Zygmund 1952]}. The second-order representation $\partial_{i} \partial_{j} = - R_{i} R_{j} Δ$ and the resulting $L^{p}$ estimate for the Hessian were developed into the foundation of elliptic regularity in Stein's 1970 monograph ^{[Stein 1970]} and the Gilbarg-Trudinger treatment of the Calderón-Zygmund inequality ^{[Gilbarg-Trudinger 2001]}. The characterization of the Riesz transforms by their covariance under the Euclidean group was given by Stein and Weiss ^{[Stein-Weiss 1971]}, who recognized the operators as the degree-one case of the spherical-harmonic family of singular integrals.

Bibliography Master

@article{Riesz1927,
  author  = {Riesz, Marcel},
  title   = {Sur les fonctions conjugu\'ees},
  journal = {Mathematische Zeitschrift},
  volume  = {27},
  year    = {1927},
  pages   = {218--244}
}

@article{Horvath1953,
  author  = {Horv\'ath, J\'anos},
  title   = {Sur les fonctions conjugu\'ees \`a plusieurs variables},
  journal = {Indagationes Mathematicae},
  volume  = {15},
  year    = {1953},
  pages   = {17--29}
}

@article{CalderonZygmund1952,
  author  = {Calder\'on, Alberto P. and Zygmund, Antoni},
  title   = {On the existence of certain singular integrals},
  journal = {Acta Mathematica},
  volume  = {88},
  year    = {1952},
  pages   = {85--139}
}

@book{Stein1970,
  author    = {Stein, Elias M.},
  title     = {Singular Integrals and Differentiability Properties of Functions},
  publisher = {Princeton University Press},
  year      = {1970}
}

@book{SteinWeiss1971,
  author    = {Stein, Elias M. and Weiss, Guido},
  title     = {Introduction to Fourier Analysis on Euclidean Spaces},
  publisher = {Princeton University Press},
  year      = {1971}
}

@book{GilbargTrudinger2001,
  author    = {Gilbarg, David and Trudinger, Neil S.},
  title     = {Elliptic Partial Differential Equations of Second Order},
  edition   = {2},
  publisher = {Springer},
  year      = {2001}
}

@book{Grafakos2014,
  author    = {Grafakos, Loukas},
  title     = {Classical Fourier Analysis},
  edition   = {3},
  publisher = {Springer},
  year      = {2014}
}

Prerequisites

02.19.03
02.10.04
02.10.01

Tier anchors

beginner: Stein-Shakarchi 2003 *Fourier Analysis* (Princeton Lectures in Analysis I) Ch. 5; the Hilbert transform as the one-dimensional model, generalized to the n components of a vector-valued operator; the conjugate-function picture
intermediate: Stein 1970 *Singular Integrals and Differentiability Properties of Functions* (Princeton) Ch. III; Grafakos 2014 *Classical Fourier Analysis* 3e (Springer) §5.1-5.2; Duoandikoetxea 2001 *Fourier Analysis* (AMS) §4
master: Stein 1970 *Singular Integrals* (Princeton) Ch. II-III; Stein 1970 *Singular Integrals* Ch. III §1-4 (Riesz transforms, second-order estimates); Gilbarg-Trudinger 2001 *Elliptic PDE of Second Order* (Springer) Ch. 9; Grafakos 2014 *Classical Fourier Analysis* 3e §5.1

References

Stein — Singular Integrals and Differentiability Properties of Functions · Ch. III, the Riesz transforms and their characterization
Riesz — Sur les fonctions conjuguées · Math. Z. 27 (1927), 218-244
Horvath — Sur les fonctions conjuguées à plusieurs variables · Indag. Math. 15 (1953), 17-29
Calderón-Zygmund — On the existence of certain singular integrals · Acta Mathematica 88 (1952), 85-139
Gilbarg-Trudinger — Elliptic Partial Differential Equations of Second Order · Ch. 9, the Calderón-Zygmund inequality and W^{2,p} estimates
Stein-Weiss — Introduction to Fourier Analysis on Euclidean Spaces · Ch. III, Riesz transforms and spherical harmonics

Estimated time

beginner: 20m
intermediate: 60m
master: 90m