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How statement and proof provenance work

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  • 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.
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

An explicit finite model separates ideal and practical test vectors

Statement

Let U=V=R2 with Euclidean inner product, A=(1101), b(x,y)=(Ax)⋅y, Uh=span⁡(e2), practical test space Vr=span⁡(e1), and load F(y)=e1⋅y. Then Te2=(1,1), Tre2=e1, and ideal and practical trial coefficients are 1/2 and 1.

Facts & Assumptions

Given: The stated finite-dimensional trial/test data.

[F1]

In this finite model, define Tw∈V by (Tw,y)V=b(w,y) for every y∈V; existence and uniqueness follow from finite-dimensional Riesz representation. (Finite-dimensional Riesz representation: every functional is uniquely v↦⟨v,w⟩)

[F2]

Define Trw∈Vr by (Trw,y)V=b(w,y) for every y∈Vr; finite-dimensional Riesz representation again gives existence and uniqueness. (Finite-dimensional Riesz representation: every functional is uniquely v↦⟨v,w⟩)

[F3]

Finite-dimensional Riesz systems have unique representatives. (Finite-dimensional Riesz representation: every functional is uniquely v↦⟨v,w⟩)

Proof

1.1F1F2givenalgebra

Ae2=(1,1), so (Ae2,y)V=b(e2,y) for every y∈V, and [F1] gives Te2=(1,1). For every y=te1∈Vr, (e1,y)V=t=b(e2,y), so [F2] gives Tre2=e1.

2.1F1F2F3step 1.1algebra∎

For ideal coefficient c, the equation is b(ce2,Te2)=F(Te2), hence 2c=1 and c=1/2. For practical tests it is b(ce2,e1)=F(e1), hence c=1. Uniqueness follows from the positive Gram scalars 2 and 1 in [F3].

Remarks

The Euclidean Riesz maps defined here give a finite-dimensional analogue of Ideal Poisson DPG trial-to-test Riesz map and Exact local finite Riesz solves for practical DPG tests.

Depends on

Used by

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Dependency tree · two levels

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Sources