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First Strang estimate for conforming perturbed Galerkin problems

Statement

Assume Countable Choice. Let V be a real or complex Hilbert space, a:V×V→K a bounded coercive form with boundedness constant M and coercivity constant α>0, and F:V→K a bounded functional linear in the real case and conjugate-linear in the complex test argument. Let Vh⊆V be finite-dimensional, and let ah:Vh×Vh→K and Fh:Vh→K be bounded, with Fh having the same test-linearity as F, and ah coercive with constant αh>0 in the inherited norm. Both a and ah are sesquilinear (bilinear over R), linear in their first argument and conjugate-linear in their second over C. There are unique exact and discrete solutions u and uh, and

∥u−uh∥V≤inf⁡vh∈Vh((1+M/αh)∥u−vh∥V+δa,h(vh)+δF,hαh),

where the defects are those of Variational-crime settings and consistency defects. For a family of meshes this is a uniform estimate whenever inf⁡hαh>0 and the displayed approximation and defect terms are uniformly controlled. The term δF,h retains source approximation and data oscillation.

Facts & Assumptions

Given: Countable Choice; V,a,F,Vh,ah,Fh with the Hilbert, boundedness, coercivity, and finite-dimensional hypotheses of the Statement.

[F1]

Bounded, coercive and symmetric sesquilinear forms: ∣a(v,w)∣≤M∥v∥V∥w∥V and Re⁡a(v,v)≥α∥v∥V2; the real bilinear case does not require symmetry.

[F2]

The Lax--Milgram theorem under The Axiom of Countable Choice (ACω): a bounded coercive form on a Hilbert space with a bounded functional has a unique solution to its variational equation.

[F3]

A finite-dimensional normed subspace is closed: a finite-dimensional subspace of a normed space is closed; therefore Vh with the inherited inner product is a Hilbert space.

[F4]

Variational-crime settings and consistency defects: δa,h(vh) and δF,h are the indicated suprema of the form and load defects against nonzero tests in Vh.

[A1]

ah is sesquilinear (bilinear over R), with the stated first- and second-argument conventions; it is bounded and Re⁡ah(vh,vh)≥αh∥vh∥V2 on Vh; Fh is bounded.

Proof

1.1F1F2given

The hypotheses on V,a,F are exactly those of [F2], so [F2] gives a unique u∈V satisfying a(u,v)=F(v) for every v∈V.

1.2F2F3A1given

By [F3], Vh is closed in V, hence complete in its inherited Hilbert norm. The restriction of ah is bounded and coercive by [A1], and Fh is bounded; applying [F2] on Vh gives a unique uh∈Vh with ah(uh,wh)=Fh(wh) for every wh∈Vh.

2.1A1step 1.2choose

Fix any vh∈Vh and put eh:=uh−vh. Discrete coercivity and the discrete equation give αh∥eh∥V2≤Re⁡ah(eh,eh)=Re⁡(Fh(eh)−ah(vh,eh)).

3.1F1F4givenstep 2.1algebra

Since eh∈Vh⊆V, the exact equation gives F(eh)=a(u,eh), and Fh(eh)−ah(vh,eh)=[Fh(eh)−F(eh)]+a(u−vh,eh)+[a(vh,eh)−ah(vh,eh)]. The boundedness of a [F1] and the defect definitions [F4] therefore imply αh∥eh∥V2≤(δF,h+M∥u−vh∥V+δa,h(vh))∥eh∥V.

4.1step 3.1algebra∎

If eh=0 the comparison is immediate; otherwise divide step 3.1 by αh∥eh∥V and use ∥u−uh∥V≤∥u−vh∥V+∥eh∥V. This proves the displayed bound for each vh, and taking the infimum proves the theorem.

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