Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass
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.

A Dirichlet Laplacian inverse is uniformly bounded in the maximum norm

Example

For h=1/N, N≥2, let AN act on vectors (v1,…,vN−1) with v0=vN=0 by

(ANv)i=2vi−vi−1−vi+1h2,1≤i<N.

With the vector maximum norm ∥v∥∞=max⁡1≤i<N∣vi∣, the inverse satisfies ∥AN−1∥∞→∞≤1/8 for every N≥2.

Facts & Assumptions

Given: N≥2, h=1/N, v0=vN=0, and the displayed matrix action of AN.

[A1]

The grid vector norm is ∥v∥∞=max⁡1≤i<N∣vi∣, and the matrix uses zero endpoint values.

Verification

1.1givenA1algebra

Set xi=ih and ϕi=xi(1−xi)/2 for 0≤i≤N. Then ϕ0=ϕN=0, −δh2ϕi=1, so ANϕ=1, and 0≤ϕi≤1/8 because x(1−x)≤1/4 on [0,1].

1.2givenA1algebra

If w0=wN=0 and ANw≥0 at every interior index, a negative interior minimum wk<0 would give (ANw)k=(2wk−wk−1−wk+1)/h2≤0. Equality is then forced, so both neighboring values equal wk; repeating this along the finite grid reaches a zero endpoint, a contradiction. Therefore wi≥0 everywhere.

2.1givenA1step 1.2algebra

The kernel of AN is zero: if ANv=0 and v has a positive interior maximum, the equation forces both neighboring values to equal that maximum, which propagates to a zero endpoint; applying the same argument to −v rules out a negative minimum. Thus v=0, and the square finite matrix is invertible.

3.1step 1.1step 1.2step 2.1A1algebra∎

Let g be any right-hand side, let v=AN−1g, and put M=∥g∥∞. If M=0, then v=0. If M>0, AN(Mϕ+v)=M1+g≥0 and AN(Mϕ−v)=M1−g≥0; step 1.2 makes both grid functions nonnegative. Hence ∣vi∣≤Mϕi≤M/8 for each i, proving ∥AN−1g∥∞≤∥g∥∞/8.

Used by

Nothing in the library uses this result yet.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.