Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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.

Hermitian Galerkin solutions are best in the energy norm

Statement

Assume the hypotheses and notation of Céa quasi-optimality for coercive conforming Galerkin problems, and suppose additionally that a is Hermitian: a(v,w)=a(w,v)‾. Then ∥v∥a:=a(v,v) is the energy norm and the Galerkin solution is its best approximation: ∥u−uh∥a=inf⁡vh∈Vh∥u−vh∥a.

Facts & Assumptions

Given: The bounded coercive Galerkin setting of the cited Céa theorem and the additional Hermitian identity for a.

[F1]

The Céa theorem supplies unique continuous and discrete solutions (Céa quasi-optimality for coercive conforming Galerkin problems).

[F2]

The error u−uh is orthogonal to Vh in the form a (Galerkin orthogonality for conforming approximations).

[F3]

A bounded coercive form is Hermitian exactly when a(v,w)=a(w,v)‾; coercivity gives Re⁡a(v,v)≥α∥v∥V2 (Bounded, coercive and symmetric sesquilinear forms).

Proof

1.1F3given

Hermitian symmetry makes a(v,v) real, and coercivity makes it positive for every nonzero v. Thus a is an inner product and ∥⋅∥a is its induced norm.

1.2F2F3given

Put e=u−uh and, for arbitrary vh∈Vh, put ηh=uh−vh. By [F2], a(e,ηh)=0; Hermitian symmetry gives a(ηh,e)=a(e,ηh)‾=0.

2.1F2F3step 1.2algebra

Since u−vh=e+ηh, sesquilinearity and step 1.2 give ∥u−vh∥a2=a(e+ηh,e+ηh)=∥e∥a2+∥ηh∥a2≥∥e∥a2. Therefore ∥u−uh∥a≤∥u−vh∥a for every vh∈Vh.

3.1F1step 2.1algebra∎

Taking the infimum in step 2.1 gives one inequality. The choice vh=uh gives the reverse inequality, so equality holds.

Depends on

Used by

Dependency tree · two levels

17 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