Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-09-08 (Codex)
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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.

A smooth bump between concentric Euclidean balls

Statement

Let 0<r<R and n≥1. Then there exists a smooth function ρ:Rn→[0,1] such that ρ=1 on B‾r(0) and supp⁡(ρ)⊆BR(0).

Facts & Assumptions

Proof

technique · direct
1.1L1givenconstruct

Put s=(r+R)/2 and d=s2−r2>0, and define u(x)=(s2−∑j=1nxj2)/d and ρ(x)=σ(u(x)). The coordinate derivatives of u are ∂iu=−2xi/d, ∂j∂iu=−2δij/d, and zero at every higher order, so u is smooth. Here δij is 1 when i=j and 0 otherwise.

2.1F1L1step 1.1

Every iterated coordinate derivative of ρ is a finite sum of products of derivatives of u and factors σ(k)(u): induction follows by applying the chain and product rules in [L1] to each factor. These factors are continuous by [F1] and step 1.1, so ρ is smooth.

2.2F1step 1.1

If ∥x∥≤r, then u(x)≥1, so ρ(x)=1; if ∥x∥≥s, then u(x)≤0, so ρ(x)=0. Also 0≤ρ≤1 by [F1].

3.1F2step 1.1step 2.1step 2.2∎

The nonzero locus lies in Bs(0), whose closure is contained in the closed ball B‾s(0). Thus [F2] gives supp⁡ρ⊆B‾s(0)⊆BR(0), since s<R. Together with steps 2.1 and 2.2 this proves the claim.

Depends on

Used by

…and 17 more results.

Dependency tree · two levels

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