Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)
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.

Coercivity makes a small form step a strict contraction

Statement

Assume Countable Choice, used through A bounded form is represented by a unique bounded operator. Let H≠{0} be a real or complex Hilbert space and let a be a bounded coercive sesquilinear form on H with constants M>0,α>0 (so necessarily α≤M); let A be its operator. Put qρ:=1−2ρα+ρ2M2 for 0<ρ<2α/M2. Then 0≤qρ<1, and for every w∈H the map Tρ,w(u):=u−ρ(Au−w) is a strict contraction of H with constant qρ: ∥Tρ,w(u)−Tρ,w(v)∥≤qρ∥u−v∥ for all u,v. In particular I−ρA is a strict contraction with the same constant. The estimate is the only place where the coercivity constant and the bound enter the contraction argument.

Facts & Assumptions

Given: Countable Choice; a Hilbert space H≠{0} with inner product linear in the first argument; a bounded coercive sesquilinear form a with constants M>0,α>0; its operator A, a(u,v)=(Au,v); and a real ρ with 0<ρ<2α/M2.

[F1]

A is linear with a(u,v)=(Au,v) for all u,v; Re⁡a(u,u)=Re⁡(Au,u)≥α∥u∥2 and ∥Au∥≤M∥u∥ (Bounded, coercive and symmetric sesquilinear forms, A bounded form is represented by a unique bounded operator, A coercive form operator is bounded below).

[F2]

Inner-product norm expansion: ∥z−ρAz∥2=∥z∥2−2ρRe⁡(Az,z)+ρ2∥Az∥2, and Re⁡(Az,z)≤∣(Az,z)∣≤∥Az∥ ∥z∥ by Cauchy--Schwarz; the induced length is a norm with the triangle inequality (Cauchy–Schwarz: ∣⟨x,y⟩∣≤∥x∥ ∥y∥, with equality exactly for dependent pairs, The induced length is a norm, Real and imaginary parts, complex conjugation, and modulus).

[F3]

A map S:H→H is a strict contraction with constant q when ∥S(u)−S(v)∥≤q∥u−v∥ for all u,v and 0≤q<1 (Lipschitz map, α-Hölder map for rational 0<α≤1, and contraction).

[F4]

Nonnegative square roots: for t≥0 there is a unique s≥0 with s2=t, denoted t (Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}). If 0≤x≤y, then x>y would imply x−y=(x−y)(x+y)>0, a contradiction; hence x≤y.

Proof

1.1F1algebra

On a nonzero Hilbert space the named constants satisfy α≤M: choosing u≠0 and dividing u by its norm, α≤Re⁡a(u,u)≤∣a(u,u)∣≤M∥u∥2=M, The identity 1−2ρα+ρ2M2=M2(ρ−α/M2)2+1−α2/M2 makes the radicand nonnegative; it may vanish when α=M and ρ=α/M2.

1.2F1F2algebra

Contraction estimate: for z∈H and 0<ρ<2α/M2 the expansion of [F2] together with [F1] gives ∥z−ρAz∥2=∥z∥2−2ρRe⁡(Az,z)+ρ2∥Az∥2≤(1−2ρα+ρ2M2)∥z∥2=qρ2∥z∥2, so ∥(I−ρA)z∥≤qρ∥z∥; taking z=u−v and using Tρ,w(u)−Tρ,w(v)=(I−ρA)(u−v) gives the contraction estimate for every w.

2.1F3F4step 1.1step 1.2algebra∎

The constant lies in [0,1): the radicand is a quadratic in ρ with minimum 1−α2/M2 at ρ=α/M2, which is nonnegative because α≤M by step 1.1; at the endpoints ρ=0 and ρ=2α/M2 it equals 1, and for 0<ρ<2α/M2 either directly 1−2ρα+ρ2M2<1 or by the strict minimum unless ρ=α/M2 and α=M, in which case the radicand vanishes and qρ=0; in every case 0≤qρ<1. Hence I−ρA, and with it Tρ,w, is a strict contraction with constant qρ.

Depends on

Used by

Dependency tree · two levels

43 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