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Dyadic pieces have annular Fourier support and uniformly bounded rescaled kernels

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). With the fixed partition and notation of The inhomogeneous dyadic frequency partition and its Littlewood-Paley operators:

for $j \geq 1$ the kernel $K_{j} = F^{- 1} φ_{j}$ lies in $S (R^{n})$ and satisfies $K_{j} (x) = 2^{(j - 1) n} K_{1} (2^{j - 1} x) (x \in R^{n})$ and $\int_{R^{n}} K_{j} = 0$ ; also ${\tilde{K}}_{j} \in S$ , with ${\tilde{K}}_{j} = K_{j - 1} + K_{j} + K_{j + 1}$ (a finite sum of Schwartz kernels; no scaling law for the companions);
for every $N \geq 0$ there are constants $C_{N}, C_{N}^{'} < \infty$ depending only on $n, ψ, N$ with $∣ K_{j} (x) ∣ \leq C_{N} 2^{j n} (1 + 2^{j} ∣ x ∣)^{- N}, ∣ {\tilde{K}}_{j} (x) ∣ \leq C_{N}^{'} 2^{j n} (1 + 2^{j} ∣ x ∣)^{- N}$ for all $j \geq 1$ and $x \in R^{n}$ , while $K_{0}, {\tilde{K}}_{0} \in S$ ;
consequently $∥ K_{j} ∥_{1} \leq C$ and $∥ {\tilde{K}}_{j} ∥_{1} \leq C$ uniformly in $j$ , and for $f \in L^{p} (R^{n}; C)$ , $1 \leq p \leq \infty$ , $∥ f * K_{j} ∥_{p} \leq C ∥ f ∥_{p}, ∥ f * {\tilde{K}}_{j} ∥_{p} \leq C ∥ f ∥_{p}$ uniformly in $j \geq 0$ , that is $∥ Δ_{j} f ∥_{p} \leq C ∥ f ∥_{p}$ and $∥ {\tilde{Δ}}_{j} f ∥_{p} \leq C ∥ f ∥_{p}$ .

Facts & Assumptions

Given: the fixed partition $(φ_{j})$ of Existence of a smooth inhomogeneous dyadic frequency partition with its companion sequence $({\tilde{φ}}_{j})$ and the kernels $K_{j} = F^{- 1} φ_{j}$ , ${\tilde{K}}_{j} = F^{- 1} {\tilde{φ}}_{j}$ of The inhomogeneous dyadic frequency partition and its Littlewood-Paley operators; an integer $N \geq 0$ .

[F1]

The symbols satisfy $φ_{0} = ψ$ , $φ_{j} (ξ) = ψ (2^{- j} ξ) - ψ (2^{- (j - 1)} ξ)$ for $j \geq 1$ and ${\tilde{φ}}_{j} = φ_{j - 1} + φ_{j} + φ_{j + 1}$ with $φ_{- 1} = 0$ ; $ψ$ is Schwartz, $ψ (0) = 1$ and $ψ = 0$ for $∣ ξ ∣ \geq 2$ ; and $φ_{j} (ξ) = φ_{1} (2^{- (j - 1)} ξ)$ for $j \geq 1$ (Existence of a smooth inhomogeneous dyadic frequency partition, The inhomogeneous dyadic frequency partition and its Littlewood-Paley operators).

[F2]

$K_{j}, {\tilde{K}}_{j} \in S (R^{n})$ , $F K_{j} = φ_{j}$ and $F {\tilde{K}}_{j} = {\tilde{φ}}_{j}$ (Fourier transform acts continuously on Schwartz space); for $h \in S (R^{n})$ the inversion formula $h (x) = \int_{R^{n}} \hat{h} (ξ) e^{2 π i x \cdot ξ} d ξ$ holds with an absolutely convergent integral (Fourier inversion on Schwartz space), and for $L^{1}$ functions the distributional transform is the regular distribution of the integral transform (Fourier transform agrees with l one and plancherel transforms), so the integral transform of the integrable function $K_{j}$ equals $φ_{j}$ pointwise.

[F3]

For $n \geq 1$ , if $h \in S (R^{n})$ then $h \in L^{1} (R^{n})$ (Schwartz derivatives are integrable), and for every $N \geq 0$ the quantity $\sup_{u \in R^{n}} (1 + ∣ u ∣)^{N} ∣ h (u) ∣$ is finite and bounded by a finite sum of the seminorms $p_{α 0} (h)$ , because $(1 + ∣ u ∣)^{N} \leq C_{N} \sum_{∣ α ∣ \leq N} ∣ u^{α} ∣$ (Schwartz space and its seminorms, $C^{k}$ maps and multi-index derivative notation in Euclidean space).

[F4]

Young's inequality: for $1 \leq p \leq \infty$ , $f \in L^{p}$ and $g \in L^{1}$ the convolution $f * g$ is defined almost everywhere and $∥ f * g ∥_{p} \leq ∥ f ∥_{p} ∥ g ∥_{1}$ (Young's convolution inequality under Countable Choice).

Proof

technique · direct

1.1F1algebra

Rescaling law for the pieces. For $j \geq 1$ and every $ξ$ , $φ_{j} (ξ) = ψ (2^{- j} ξ) - ψ (2^{- (j - 1)} ξ) = φ_{1} (2^{- (j - 1)} ξ)$ , because $φ_{1} (2^{- (j - 1)} ξ) = ψ (2^{- 1} 2^{- (j - 1)} ξ) - ψ (2^{- (j - 1)} ξ)$ , and more generally $φ_{k + j - 1} (ξ) = φ_{k} (2^{- (j - 1)} ξ)$ for every $k \geq 1$ by the same computation with $k$ in place of $1$ . The companions are sums, not rescalings: ${\tilde{φ}}_{j} = φ_{j - 1} + φ_{j} + φ_{j + 1}$ , hence ${\tilde{K}}_{j} = K_{j - 1} + K_{j} + K_{j + 1}$ by linearity of the inverse transform, and no rescaling law is claimed for them.

2.1F1F2step 1.1algebra

Rescaling law for the kernels. Since $K_{1} = F^{- 1} φ_{1} \in S$ and $K_{j} = F^{- 1} φ_{j} = F^{- 1} (φ_{1} (2^{- (j - 1)} \cdot))$ by step 1.1, the inversion formula applied at $x$ gives $K_{j} (x) = \int φ_{1} (2^{- (j - 1)} ξ) e^{2 π i x \cdot ξ} d ξ$ ; substituting $ξ = 2^{j - 1} η$ , $d ξ = 2^{(j - 1) n} d η$ , this becomes $K_{j} (x) = 2^{(j - 1) n} \int φ_{1} (η) e^{2 π i (2^{j - 1} x) \cdot η} d η = 2^{(j - 1) n} K_{1} (2^{j - 1} x)$ . For the companions, step 1.1 gives ${\tilde{K}}_{j} = K_{j - 1} + K_{j} + K_{j + 1}$ (the inverse transform is linear and each $K_{i}$ lies in $S$ ); a finite sum of Schwartz kernels again lies in $S$ .

3.1F1F2step 2.1algebra

Mean zero of the high-frequency kernels. For $j \geq 1$ the integral ${\hat{K}}_{j} (0) = \int_{R^{n}} K_{j} (x) d x$ is the value at the origin of the integral transform of $K_{j}$ , which by [F2] equals $φ_{j} (0)$ ; since $ψ (0) = 1$ by [F1], $φ_{j} (0) = ψ (0) - ψ (0) = 0$ .

3.2F2F3step 2.1algebra

Pointwise decay. Fix $N \geq 0$ . By [F3] applied to $K_{1}$ there is $A_{N} < \infty$ , depending only on $n, ψ, N$ , with $∣ K_{1} (u) ∣ \leq A_{N} (1 + ∣ u ∣)^{- N}$ for all $u$ ; since $1 + 2^{j} ∣ x ∣ \leq 2 (1 + 2^{j - 1} ∣ x ∣)$ for $j \geq 1$ , step 2.1 gives, for $j \geq 1$ , $∣ K_{j} (x) ∣ = 2^{(j - 1) n} ∣ K_{1} (2^{j - 1} x) ∣ \leq A_{N} 2^{(j - 1) n} (1 + 2^{j - 1} ∣ x ∣)^{- N} \leq A_{N} 2^{N + n} 2^{j n} (1 + 2^{j} ∣ x ∣)^{- N}$ , so the asserted bound holds with $C_{N} : = A_{N} 2^{N + n}$ . For the companions, step 2.1 writes ${\tilde{K}}_{j} = K_{j - 1} + K_{j} + K_{j + 1}$ ; summing the bounds just proved for these three kernels, and for $j = 1$ also the Schwartz bound of $K_{0}$ from [F3], gives $∣ {\tilde{K}}_{j} (x) ∣ \leq C_{N}^{'} 2^{j n} (1 + 2^{j} ∣ x ∣)^{- N}$ after absorbing the fixed factors $3$ , $2^{N}$ and $2^{n}$ into $C_{N}^{'}$ . Since $K_{0} = F^{- 1} ψ$ and ${\tilde{K}}_{0} = F^{- 1} {\tilde{φ}}_{0}$ are inverse transforms of Schwartz functions, both lie in $S$ by [F2].

4.1F3step 3.2algebra

Uniform $L^{1}$ bounds. Take $N = n + 1$ in step 3.2 and substitute $u = 2^{j} x$ : for $j \geq 1$ , $\int ∣ K_{j} ∣ \leq C_{N} 2^{j n} \int (1 + 2^{j} ∣ x ∣)^{- N} d x = C_{N} \int (1 + ∣ u ∣)^{- N} d u = : C < \infty$ , and likewise $\int ∣ {\tilde{K}}_{j} ∣ \leq C_{N}^{'} \int (1 + ∣ u ∣)^{- N} d u \leq C$ after enlarging $C$ ; the integrals are finite: on $∣ u ∣ \leq 1$ the integrand is bounded and the ball lies in a finite-volume box; on $2^{k} < ∣ u ∣ \leq 2^{k + 1}$ its integral is at most $2^{- k (n + 1)} (2^{k + 2})^{n} = 2^{2 n - k}$ , and $\sum_{k \geq 0} 2^{- k} < \infty$ (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). For $j = 0$ , [F3] gives $∥ K_{0} ∥_{1} + ∥ {\tilde{K}}_{0} ∥_{1} < \infty$ , and these two constants are absorbed into $C$ .

5.1F4step 4.1algebra

Uniform $L^{p}$ bounds. For $f \in L^{p}$ , $1 \leq p \leq \infty$ , Young's inequality [F4] applied to $f$ and $K_{j} \in L^{1}$ gives $∥ f * K_{j} ∥_{p} \leq ∥ K_{j} ∥_{1} ∥ f ∥_{p} \leq C ∥ f ∥_{p}$ , and likewise for ${\tilde{K}}_{j}$ , uniformly in $j \geq 0$ .

6.1step 1.1step 2.1step 3.1step 3.2step 4.1step 5.1∎

Clauses 1, 2 and 3 are steps 2.1 with 3.1, step 3.2 and steps 4.1 with 5.1, respectively.

Depends on

Used by

Sharp frequency cutoffs have kernels that are not in L1 Counterexample
The Littlewood-Paley square function Definition
Dyadic pieces are uniformly Mihlin multipliers and uniformly Lp-bounded Lemma
L2 almost orthogonality of the dyadic pieces Lemma

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Sources

Loukas Grafakos, Classical Fourier Analysis, 3rd ed. (Springer GTM 249) (standard reference, not scraped)
Terence Tao, Math 247A Lecture Notes 4 (UCLA, Fall 2006) (standard reference, not scraped)
Mark Williams, Notes on Harmonic Analysis (January 11, 2022) (standard reference, not scraped)