Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated
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The square function of a low-frequency-localised function

Example

Assume Countable Choice (The Axiom of Countable Choice (ACω)). Let f∈S(Rn) have Fourier transform supported in {∣ξ∣≤1}. Then Δjf=0 for every j≥1 and Δ0f=f, so Sf=∣Δ0f∣=∣f∣ and hence ∥Sf∥p=∥f∥p for every 1≤p<∞. The example records that in the region where exactly one piece of the partition is nonzero the square function degenerates to the modulus of that piece, so for nonzero functions in this class the strict-range coefficients satisfy cp≤1≤Cp.

Verification

Given: Countable Choice and f∈S(Rn) with supp⁡f^⊂{∣ξ∣≤1}.

[L1] The partition satisfies φ0=ψ with ψ=1 on {∣ξ∣≤1}, and for j≥1 the piece φj vanishes on {∣ξ∣≤2j−1} (Existence of a smooth inhomogeneous dyadic frequency partition, The inhomogeneous dyadic frequency partition and its Littlewood-Paley operators).

[L2] For f∈S one has Δjf^=φjf^, and Fourier inversion gives g=F−1(g^) for Schwartz g (The inhomogeneous dyadic frequency partition and its Littlewood-Paley operators, Fourier inversion on Schwartz space); the square function is Sf=(∑j≥0∣Δjf∣2)1/2 (The Littlewood-Paley square function).

1.1L1givenalgebra

The pieces on the low-frequency ball. Since supp⁡f^⊂{∣ξ∣≤1} and ψ=1 there, φ0f^=ψf^=f^. For j≥1, supp⁡f^ is contained in the vanishing region {∣ξ∣≤2j−1} of φj (because 2j−1≥1), so φjf^=0.

2.1L2step 1.1algebra∎

The square function. By step 1.1 and [L2], Δ0f=F−1(f^)=f and Δjf=F−1(0)=0 for every j≥1; hence only the j=0 term of the square function survives, Sf=∣Δ0f∣=∣f∣, and therefore ∥Sf∥p=∥∣f∣∥p=∥f∥p for every 1≤p<∞. For a nonzero function in this class, cp∥f∥p≤∥Sf∥p≤Cp∥f∥p therefore forces cp≤1≤Cp.

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