Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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.

Cheeger sweep and layer cake

Statement

For a finite d-regular graph on n2 vertices and a nonnegative f supported on at most n/2 vertices, use the unnormalized inner product f,g0=vf(v)g(v) and energy E(f)=f,(IM)f0. Then hf021du<vAuvf(u)2f(v)22E(f)f02. Also γ2h, and there exists a nonzero nonnegative function g, supported on at most n/2 vertices, with E(g)γg02: namely, the positive part of a suitable sign of a nonzero mean-zero μ2 eigenvector. These are the indicator and positive-part conclusions of the preceding lemma in the unnormalized inner product.

Facts & Assumptions

Given: the objects and hypotheses in the statement above.

[F1]

Let n2 and use normalized edge expansion h and algebraic gap γ=1μ2. Then γ2h. Moreover some sign of a nonzero mean-zero μ2 eigenvector has positive part f0 supported on at most n/2 vertices and satisfying f,(IM)fγf2. (Cheeger indicator and positive part energy).

[F2]

For vectors u,v in a real or complex inner product space, u,vuv. Equality holds if and only if u and v are linearly dependent, including the case in which either vector is zero. (Cauchy–Schwarz: u,vuv, with equality exactly for linearly dependent vectors).

Proof

1.1

Order coordinates f1fn0 and let a=suppfn/2. Put fa+1=0. Each difference of squared values telescopes across initial segments, so the middle numerator equals i=1a(fi2fi+12)cut({1,,i}). Each cut is at least dhi, and i=1ai(fi2fi+12)=ifi2. For a=0 all sums are zero.

givenalgebra
2.1

Factor f(u)2f(v)2=f(u)f(v)f(u)+f(v) and apply Cauchy–Schwarz with weights Auv for u<v. The first squared sum is dE(f); the second is at most 2u<vAuv(f(u)2+f(v)2)2df02. Dividing by d proves the upper estimate. Loop terms vanish in the difference sum and only reduce the nonloop degree sum.

F2step 1.1
3.1

Multiplying all normalized inner products by n leaves the preceding lemma's inequalities unchanged; thus its positive-part and indicator estimates have exactly the stated unnormalized form. No division by E(f) or f0 was used above, so zero functions and zero energy are included.

F1step 2.1

Depends on

Used by

Dependency tree · two levels

9 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