Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge 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.

The symmetric shifted solution operator is positive and self-adjoint

Statement

Assume Countable Choice. In the symmetric case of The L2 operator associated with a symmetric elliptic form, with Ω open and μ≥β, the shifted solution operator Kμ of The shifted elliptic solution operator, regarded on L2(Ω), is self-adjoint and positive: (Kμf,g)L2=(f,Kμg)L2,(Kμf,f)L2=aμ(Kμf,Kμf) ≥ α∥Kμf∥H012 ≥ 0 for all f,g∈L2(Ω), with (Kμf,f)L2=0 if and only if f=0; in particular Kμ is injective. Moreover a(Kμf,v)=(f−μKμf,v)L2 for every v∈H01(Ω), so Kμf∈D(L) and L(Kμf)=f−μKμf. Symmetry of Kμ is verified from the form; no self-adjointness of the differential expression is assumed.

Facts & Assumptions

Given: Countable Choice; the symmetric divergence-form case with form a and μ≥β; the shifted solution operator Kμ and the shifted form aμ=a+μ(⋅,⋅)L2; f,g∈L2(Ω).

[F1]

Defining identity and coercivity: aμ(Kμh,v)=(h,v)L2 for all v∈H01(Ω) and all h∈L2(Ω), and aμ(u,u)≥α∥u∥H012 with α=θ/2 for u∈H01(Ω) (The shifted elliptic solution operator, A sufficiently large shift is coercive, Bounded, coercive and symmetric sesquilinear forms).

[F2]

Symmetry: a(u,v)=a(v,u)‾ and (v,u)L2=(u,v)L2‾, so aμ is symmetric as well; in particular aμ(u,u) is real (The L2 operator associated with a symmetric elliptic form, Bounded, coercive and symmetric sesquilinear forms, The formal adjoint and the adjoint weak Dirichlet problem).

[F3]

Density: H01(Ω) is dense in L2(Ω) (Smooth compactly supported functions of an open set are dense in L2).

[F4]

Hilbert adjoints and positivity: an operator T on a Hilbert space is self-adjoint when (Tf,g)=(f,Tg) for all f,g, and positive when (Tf,f)≥0 (The Hilbert-space adjoint of a bounded operator, Self-adjoint, positive, unitary and normal operators).

[F5]

The operator L and its domain: u∈D(L) with Lu=h means u∈H01(Ω) and a(u,v)=(h,v)L2 for all v∈H01(Ω) (The L2 operator associated with a symmetric elliptic form, The associated elliptic operator is densely defined, symmetric and lower bounded).

Proof

technique · direct
1.1F1F2F4givenalgebra

Self-adjointness. By [F1] applied to g and [F2], and then to f, (Kμf,g)L2=(g,Kμf)L2‾=aμ(Kμg,Kμf)‾=aμ(Kμf,Kμg)=(f,Kμg)L2, for all f,g∈L2(Ω); hence Kμ is self-adjoint by [F4].

2.1F1F3F4step 1.1givenalgebra

Positivity and injectivity. Taking g=f in the computation of step 1.1 and using [F1], (Kμf,f)L2=aμ(Kμf,Kμf)≥α∥Kμf∥H012≥0. If (Kμf,f)L2=0, then α∥Kμf∥H012≤0, so Kμf=0; then for every v∈H01(Ω) the defining identity gives (f,v)L2=aμ(Kμf,v)=0, and density of H01(Ω) in L2(Ω) ([F3]) gives f=0. Conversely f=0 gives Kμf=0 and hence (Kμf,f)L2=0; thus Kμ is positive and injective.

3.1F1F5givenalgebra∎

Range description. For f∈L2(Ω) and every v∈H01(Ω), a(Kμf,v)=aμ(Kμf,v)−μ(Kμf,v)L2=(f−μKμf,v)L2, because the datum f−μKμf lies in L2(Ω). By [F5] this says Kμf∈D(L) and L(Kμf)=f−μKμf.

Depends on

Used by

Dependency tree · two levels

66 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