TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Critical Besov coefficient control implies a best m-term squared-error rate

Statement

Let $J \geq 3$ , $N = 2^{J}$ , $s > 0$ , and $p = (s + 1 / 2)^{- 1} \in (0, 2)$ . Let $U \in R^{N \times N}$ be orthogonal, $x \in R^{N}$ , and $c = U x$ . Regard $c$ as one coarse block $a \in R^{4}$ and detail blocks $d_{j} \in R^{2^{j}}$ for $2 \leq j \leq J - 1$ , with the norm of Finite wavelet-coefficient Besov quasi-norm on a dyadic multilevel transform. For $1 \leq m \leq N$ , hard retention of the $m$ largest coefficients has reconstruction error $∥ x - U^{T} H_{m} (c) ∥_{2} \leq \sqrt{\frac{p}{2 - p}} ∥ c ∥_{b_{p, p}^{s} (J)} m^{- s},$ where $c = U x$ and $U$ is the orthogonal analysis matrix.

Facts & Assumptions

Given: $s > 0$ , $p = (s + 1 / 2)^{- 1}$ , a finite orthogonal coefficient vector $c = U x$ , and $1 \leq m \leq N$ .

[F1]

Finite wavelet-coefficient Besov quasi-norm on a dyadic multilevel transform: at $p = (s + 1 / 2)^{- 1}$ the scale exponent $p (s + 1 / 2 - 1 / p)$ is zero, and the coefficient quasi-norm is the $ℓ^{p}$ quasi-norm.

[F2]

Hard m-term retention is optimal for an orthonormal coefficient basis: hard retention of the $m$ largest coefficients minimizes the orthogonal reconstruction error and its squared error is the sum of the discarded squares.

Proof

1.1givenF1choosealgebra

Since $s > 0$ , $p < 2$ , and [F1] gives $\sum_{n = 1}^{N} ∣ c_{n} ∣^{p} = ∥ c ∥_{b_{p, p}^{s} (J)}^{p}$ . Arrange the absolute coefficients in nonincreasing order $b_{1} \geq \dots \geq b_{N} \geq 0$ . For each $n$ , $n b_{n}^{p} \leq \sum_{r = 1}^{n} b_{r}^{p} \leq ∥ c ∥_{b_{p, p}^{s} (J)}^{p}$ , so $b_{n}^{2} \leq ∥ c ∥_{b_{p, p}^{s} (J)}^{2} n^{- 2 / p}$ .

2.1givenF2step 1.1algebra∎

By [F2], the squared best m-term error is $\sum_{n = m + 1}^{N} b_{n}^{2}$ . Because $2 / p > 1$ , the decreasing function $t^{- 2 / p}$ satisfies $\sum_{n = m + 1}^{N} n^{- 2 / p} \leq \int_{m}^{\infty} t^{- 2 / p} d t = \frac{p}{2 - p} m^{1 - 2 / p} .$ Combining this with step 1.1 and taking square roots gives a factor $m^{1 / 2 - 1 / p} = m^{- s}$ , since $1 / p = s + 1 / 2$ . When $m = N$ , the actual error is zero by retaining every coefficient, so the bound still holds.

Depends on

Used by

Smooth periodic samples have a best m-term D4 Besov-tail rate Corollary

Dependency tree · two levels

4 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

Stéphane Mallat, A Wavelet Tour of Signal Processing, 3rd ed., Chapter 9, §§9.2.2–9.2.3 (standard reference, not scraped)
Gilbert Strang and Kevin Amaratunga, Wavelets and Filter Banks, MIT OpenCourseWare 18.327, Handout 15 (standard reference, not scraped)