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Ideal ultraweak Poisson DPG is the energy best approximation

Statement

Assume Countable Choice and the real trial/test spaces and bilinear form of Ultraweak Poisson DPG spaces and plan-consistent bilinear form, with ∣b(w,v)∣≤5∥w∥U∥v∥V as in The ultraweak Poisson DPG bilinear form is bounded. The ideal trial-to-test map T of Ideal Poisson DPG trial-to-test Riesz map is well defined, bounded linear, and satisfies ∥T∥≤5. Let x∈U be any exact ultraweak solution and let Uh⊂U be finite dimensional. The ideal DPG equations with test space T(Uh) have a unique solution xh, which is the orthogonal projection of x in the induced energy inner product (w,z)E=(Tw,Tz)V. Consequently ∥x−xh∥E=inf⁡zh∈Uh∥x−zh∥E=∥B(x−xh)∥V′.

Facts & Assumptions

Given: Countable Choice, the real Hilbert trial/test spaces and bounded bilinear form, an exact solution x, and a finite-dimensional Uh.

[A1]

Countable Choice is declared for Hilbert Riesz maps. (The Axiom of Countable Choice (ACω))

[F1]

The trial operator has trivial kernel. (The ultraweak Poisson trial operator has trivial kernel)

[F2]

The ideal procedure defines Tw as the Riesz representative of b(w,⋅). (Ideal Poisson DPG trial-to-test Riesz map)

[F3]

Riesz represents bounded real functionals uniquely and identifies each dual norm with the norm of its representative. (Riesz representation for Hilbert spaces)

[F4]

The DPG bilinear form is bounded by ∣b(w,v)∣≤5∥w∥U∥v∥V. (The ultraweak Poisson DPG bilinear form is bounded)

Proof

1.1A1F1F2F3F4givenalgebra

For every w∈U, [F4] makes v↦b(w,v) a bounded functional on V, and [F3] gives the unique Tw in [F2], with ∥Tw∥V≤5∥w∥U. Bilinearity of b and uniqueness of the Riesz representative imply T(αw1+βw2)=αTw1+βTw2 for real scalars α,β, so T is bounded linear. By [F1], Tw=0 implies w=0; hence (w,z)E=(Tw,Tz)V is a positive-definite inner product. For every wh∈Uh, the ideal equation is (Txh,Twh)V=F(Twh)=(Tx,Twh)V, by [F2] and exactness.

2.1F1step 1.1algebra

Choose an orthonormal basis e1,…,eN of the finite-dimensional image T(Uh) and define P(Tx)=∑j(Tx,ej)Vej. The equations in step 1.1 identify Txh=P(Tx); Pythagoras proves it uniquely minimizes distance to that image. Injectivity of T gives the unique xh and the best-approximation identity.

3.1F2F3step 2.1algebra∎

The functional B(x−xh)=b(x−xh,⋅) has Riesz representative T(x−xh) by [F2]. Therefore [F3] gives ∥B(x−xh)∥V′=∥T(x−xh)∥V=∥x−xh∥E.

Depends on

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Nothing in the library uses this result yet.

Cited to discharge well-definedness by Ideal Poisson DPG trial-to-test Riesz map.

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