Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generated
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.

BMO averages on nested cubes grow at most logarithmically

Statement

There is Cn<∞ such that for every b∈BMO(Rn) and all cubes Q⊆R, writing ℓ(Q) for the side length, ∣bQ−bR∣≤Cn(1+log⁡2(ℓ(R)/ℓ(Q)))∥b∥BMO.

Facts & Assumptions

Given: b∈BMO(Rn) and cubes Q⊆R with side lengths ℓ(Q)≤ℓ(R), together with the mean bE=∣E∣−1∫Eb, the seminorm and the cube and volume conventions of BMO seminorm and the quotient by constants and Axis-parallel rectangles in Rm and their volume.

[F1]

For every cube E, ∣E∣−1∫E∣b−bE∣≤∥b∥BMO, and for cubes E⊆F one has ∣bE−bF∣≤∣E∣−1∫E∣b−bF∣≤(∣F∣/∣E∣)∥b∥BMO (BMO seminorm and the quotient by constants).

[F2]

A cube is a nondegenerate axis-parallel cube; the concentric cube with side length 2kℓ(Q) has volume 2kn∣Q∣, and a cube whose side length is at most that of R and which is contained in R has volume at most ∣R∣ (Axis-parallel rectangles in Rm and their volume).

Proof

technique · direct
1.1F2algebra

If Q=R the claim is trivial, so assume Q⊊R. Let m≥0 be the least integer such that 2mℓ(Q)≥2ℓ(R), and let R′ be the cube concentric with Q of side length 2mℓ(Q). The centres of Q and R both lie in R, so their coordinatewise distance is at most ℓ(R)/2; every point of R is therefore within coordinate distance ℓ(R) of the centre of Q, and R⊆R′. Since Q⊊R, ℓ(Q)<ℓ(R), so m≥1. Minimality gives 2m−1ℓ(Q)<2ℓ(R) and hence 2mℓ(Q)<4ℓ(R). Therefore ∣R′∣=2mn∣Q∣<4n∣R∣.

2.1step 1.1F1

Let Q=Q0⊆Q1⊆⋯⊆Qm=R′ be the concentric cubes of side lengths 2kℓ(Q). Since ∣Qk+1∣=2n∣Qk∣, [F1] gives ∣bQk−bQk+1∣≤∣Qk∣−1∫Qk∣b−bQk+1∣≤2n∥b∥BMO for each k<m, and the triangle inequality over the m steps gives ∣bQ−bR′∣≤m2n∥b∥BMO.

2.2step 1.1F1

Since R⊆R′, [F1] and step 1.1 give ∣bR′−bR∣≤∣R∣−1∫R∣b−bR′∣≤(∣R′∣/∣R∣)∥b∥BMO≤4n∥b∥BMO.

3.1step 2.1step 2.2algebra∎

Combining steps 2.1 and 2.2, ∣bQ−bR∣≤(2nm+4n)∥b∥BMO. From 2mℓ(Q)<4ℓ(R) we get m<2+log⁡2(ℓ(R)/ℓ(Q)), hence m≤2+log⁡2(ℓ(R)/ℓ(Q))≤2(1+log⁡2(ℓ(R)/ℓ(Q))). Therefore ∣bQ−bR∣≤(2n+1+4n)(1+log⁡2(ℓ(R)/ℓ(Q)))∥b∥BMO, which is the claim with Cn=2n+1+4n.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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