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The Hilbert transform of an $H^{1}$ atom is integrable

Example

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ and let $a$ be a $(1, \infty, 0)$ -atom supported in a compact cube $Q \subseteq R^{n}$ . Let $T$ be the Hilbert transform when $n = 1$ (The Hilbert transform is an L2 isometry and squares to minus the identity) and the vector $(R_{1}, \dots, R_{n})$ of Riesz transforms when $n \geq 2$ (Riesz transforms on Euclidean space); each component is read as its $L^{2}$ operator, with norm $B \leq 1$ (Riesz transforms are L2 contractions and square to minus the identity in sum), and has an odd kernel with a first-difference bound ( $δ = 1$ ) whose constants depend only on $n$ (Riesz kernel size, difference and spherical-cancellation bounds). Then every component $T a$ lies in $L^{1} (R^{n})$ and $∥ T a ∥_{L^{1}} \leq C_{n} (A_{1} + A_{2}^{'} + A_{3} + B),$ where $A_{1}, A_{2}^{'}, A_{3}, B$ are the kernel size, Holder, cancellation and $L^{2}$ constants of the component. Write $ℓ = ℓ (Q)$ and let $Q^{†}$ be the concentric cube of side length $2 \sqrt{n} ℓ$ . The near/far split is explicit: $\int_{Q^{†}} ∣ T a ∣ \leq ∣ Q^{†} ∣^{1 / 2} ∥ T a ∥_{L^{2}} \leq (2 \sqrt{n})^{n / 2} B, \int_{(Q^{†})^{c}} ∣ T a ∣ \leq C_{n, δ} A_{2}^{'} ∥ a ∥_{L^{1}} \leq C_{n, δ} A_{2}^{'},$ the far estimate using only the zeroth moment of $a$ and the Holder bound for the kernel.

Facts & Assumptions

Given: Countable Choice and $n \geq 1$ , a $(1, \infty, 0)$ -atom $a$ supported in a compact cube $Q$ with centre $c_{Q}$ , and a component operator $T$ as in the example.

[L1]

$a \in L_{c}^{\infty}$ , $∥ a ∥_{L^{\infty}} \leq ∣ Q ∣^{- 1}$ , $∥ a ∥_{L^{1}} \leq 1$ and $\int a = 0$ ( $H^{p}$ atoms with a prescribed moment order, Axis-parallel rectangles in $R^{m}$ and their volume).

[F1]

The Hilbert transform is an $L^{2}$ isometry and is skew-adjoint; its action on Schwartz functions is the principal-value integral with $k (x) = 1 / (π x)$ (The Hilbert transform is an L2 isometry and squares to minus the identity, The Hilbert transform is skew-adjoint on L2, The Hilbert transform is the tempered convolution with pv(1/(pi x)) and has signum Fourier multiplier, Truncated Hilbert transform and principal value). Each $R_{j}$ is an $L^{2}$ contraction with purely imaginary Fourier symbol $- i ξ_{j} / ∣ ξ ∣$ , and its Schwartz action is the principal-value integral with $k (x) = c_{n} x_{j} / ∣ x ∣^{n + 1}$ (Riesz transforms are L2 contractions and square to minus the identity in sum, Riesz transforms on Euclidean space, The Riesz transform is the principal value of its kernel, with the matching constant). Plancherel preserves the inner product (Plancherel theorem). The Riesz kernels obey the size, first-difference and spherical-cancellation bounds of Riesz kernel size, difference and spherical-cancellation bounds; the kernel constants use Calderón–Zygmund kernels and their associated operators.

[F2]

The atom belongs to $H^{1}$ with finite norm by Atoms have uniformly bounded $H^{p}$ quasi-norm and uniformly bounded test pairings.

[F3]

A cube of side length $ℓ$ has measure $ℓ^{n}$ ; its concentric cube of side length $2 \sqrt{n} ℓ$ has measure $(2 \sqrt{n})^{n} ℓ^{n}$ , and every point $y$ of the original cube satisfies $∣ y - c_{Q} ∣ \leq \sqrt{n} ℓ / 2$ (Axis-parallel rectangles in $R^{m}$ and their volume, A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[F4]

Fubini applies to integrable functions, Tonelli to nonnegative functions, and polar coordinates give $\int_{∣ z ∣ \geq R} ∣ z ∣^{- n - 1} d z = σ (S^{n - 1}) / R$ for $R > 0$ (Fubini's theorem for L^1 functions on a sigma-finite product, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma). Equality of regular distributions implies equality almost everywhere for locally integrable functions on an open set (Locally integrable functions embed in distributions).

Verification

technique · direct

1.1F1F4algebra

Kernel bounds. For the Hilbert kernel and $∣ h ∣ \leq ∣ z ∣ / 2$ , $∣ k (z - h) - k (z) ∣ = ∣ h ∣ / (π ∣ z ∣ ∣ z - h ∣) \leq (2 / π) ∣ h ∣ ∣ z ∣^{- 2}$ . Its annular size constant is $A_{1} = 2 \log 2 / π$ and its cancellation constant is $A_{3} = 0$ by oddness. For Riesz kernels [F1] gives $A_{2}^{'} = c_{n} 2^{n + 1} (3 n + 4)$ ; polar coordinates give $A_{1} \leq c_{n} σ (S^{n - 1}) \log 2$ , and spherical cancellation gives $A_{3} = 0$ . Thus in either case $∣ k (z - h) - k (z) ∣ \leq A_{2}^{'} ∣ h ∣ ∣ z ∣^{- n - 1}$ for $∣ h ∣ \leq ∣ z ∣ / 2$ , with constants depending only on $n$ .

1.2L1F1F4algebra

Off-support representation. Each $T$ is skew-adjoint: this is [F1] for the Hilbert transform, and follows for Riesz transforms from Plancherel and $\overline{m_{j}} = - m_{j}$ . Put $Ω = R^{n} ∖ Q$ and $g (x) = \int_{Q} k (x - y) a (y) d y$ on $Ω$ . For $φ \in C_{c}^{\infty} (Ω)$ the supports of $a$ and $φ$ have positive distance, so $T φ (y) = \int k (y - x) φ (x) d x$ on $Q$ by the Schwartz principal-value formulas [F1], and the double integral is absolutely integrable. Skew-adjointness, Fubini and the real odd kernel give $⟨ T a, φ ⟩ = - ⟨ a, T φ ⟩ = \int_{Ω} g (x) \overline{φ (x)} d x$ . The function $g$ is locally bounded on $Ω$ , since the kernel is bounded on each compact set separated from $Q$ and $a \in L^{1}$ ; also $T a \in L^{2} \subset L_{l o c}^{1}$ . Therefore the injectivity of regular distributions gives $T a (x) = g (x)$ almost everywhere on $Ω$ .

1.3L1F1F3algebra

Near estimate. Write $ℓ = ℓ (Q)$ and let $Q^{†}$ be the concentric cube of side length $2 \sqrt{n} ℓ$ . By [F3], $∣ Q^{†} ∣ = (2 \sqrt{n})^{n} ∣ Q ∣$ . Since $a \in L^{2}$ , Cauchy-Schwarz and the $L^{2}$ bound give $\int_{Q^{†}} ∣ T a ∣ \leq ∣ Q^{†} ∣^{1 / 2} ∥ T a ∥_{L^{2}} \leq ∣ Q^{†} ∣^{1 / 2} B ∥ a ∥_{L^{2}} \leq (2 \sqrt{n})^{n / 2} B$ , because $∥ a ∥_{L^{2}} \leq ∥ a ∥_{\infty} ∣ Q ∣^{1 / 2} \leq ∣ Q ∣^{- 1 / 2}$ .

2.1step 1.1step 1.2L1F3F4algebra

Far estimate. For $y \in Q$ one has $∣ y - c_{Q} ∣ \leq \sqrt{n} ℓ / 2$ , while $x \notin Q^{†}$ gives $∣ x - c_{Q} ∣ \geq \sqrt{n} ℓ$ . By 1.2 and $\int a = 0$ , $T a (x) = \int_{Q} [k (x - y) - k (x - c_{Q})] a (y) d y$ almost everywhere there. For $h = y - c_{Q} \neq 0$ , step 1.1 and polar coordinates give $\int_{∣ z ∣ \geq 2 ∣ h ∣} ∣ k (z - h) - k (z) ∣ d z \leq A_{2}^{'} ∣ h ∣ σ (S^{n - 1}) / (2 ∣ h ∣) = σ (S^{n - 1}) A_{2}^{'} / 2$ ; for $h = 0$ the difference is identically zero. Tonelli consequently gives $\int_{(Q^{†})^{c}} ∣ T a ∣ \leq \int_{Q} ∣ a (y) ∣ \int_{∣ z ∣ \geq 2 ∣ y - c_{Q} ∣} ∣ k (z - (y - c_{Q})) - k (z) ∣ d z d y \leq σ (S^{n - 1}) A_{2}^{'} ∥ a ∥_{1} / 2 \leq σ (S^{n - 1}) A_{2}^{'} / 2$ .

3.1step 1.3step 2.1F2algebra∎

Conclusion. Steps 1.3 and 2.1 give $∥ T a ∥_{1} \leq (2 \sqrt{n})^{n / 2} B + σ (S^{n - 1}) A_{2}^{'} / 2 \leq C_{n} (A_{1} + A_{2}^{'} + A_{3} + B)$ for every component. The atom is in $H^{1}$ by [F2], and these estimates prove directly that its $L^{2}$ transform is integrable, without requiring smoothness of the atom.

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Sources

Mark Williams, Notes on Harmonic Analysis (January 11, 2022) (standard reference, not scraped)
Li-An Daniel Wang, Multiplier Theorems on Anisotropic Hardy Spaces (PhD dissertation, University of Oregon, 2012) (standard reference, not scraped)

The Hilbert transform of an H1H1 atom is integrable