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Second Strang estimate with a bounded extension to the comparison space

Statement

Assume Countable Choice. Let V be a real or complex Hilbert space, a:V×V→K a bounded coercive sesquilinear form (bilinear over R), and F a bounded functional with the test-linearity required by a. Let Vh be a finite-dimensional Hilbert space with norm ∥⋅∥h, not necessarily a subspace of V. Suppose the algebraic sum Xh=V+Vh is equipped with a norm extending these norms, and a~h:Xh×Vh→K is a bounded sesquilinear extension of a discrete form ah with bound Mh. Over C, both forms are linear in the first argument and conjugate-linear in the second; over R, they are bilinear. Assume ah is coercive on Vh with constant αh>0, and let Fh be bounded on Vh with the matching test-linearity. The exact and discrete equations have unique solutions u and uh, and

∥u−uh∥h≤(1+Mh/αh)inf⁡vh∈Vh∥u−vh∥h+δhnc(u)/αh,

where δhnc(u)=sup⁡wh∈Vh∖{0}∣Fh(wh)−a~h(u,wh)∣/∥wh∥h when Vh≠{0}, and δhnc(u):=0 when Vh={0}. The estimate requires the bounded extension and does not assert that every nonconforming, penalty, or discontinuous Galerkin form has one.

For a family of these problems, suppose Mh≤M‾<∞ and αh≥α‾>0 uniformly. Writing Eh=inf⁡vh∈Vh∥u−vh∥h, the uniform bound is ∥u−uh∥h≤(1+M‾/α‾)Eh+δhnc(u)/α‾. Thus uniform control of Eh and δhnc(u) gives a uniform error budget; if both tend to zero, the errors tend to zero, and if both are O(hp) for a fixed p>0, the error is O(hp). No rate is inferred without those approximation, residual, and stability bounds.

Facts & Assumptions

Given: Countable Choice; the exact Hilbert problem on V; the nonconforming space Vh and common comparison norm on Xh; a bounded extension a~h; discrete coercivity αh>0; and a bounded discrete load.

[F1]

The Lax--Milgram theorem under The Axiom of Countable Choice (ACω): the exact bounded coercive Hilbert problem has a unique solution.

[F2]

By hypothesis, a~h is sesquilinear with the stated slot conventions (bilinear over R). By definition, a~h agrees with ah on Vh×Vh and ∣a~h(z,wh)∣≤Mh∥z∥h∥wh∥h on Xh×Vh (Variational-crime settings and consistency defects).

[A1]

Re⁡ah(vh,vh)≥αh∥vh∥h2 for every vh∈Vh.

[A2]

δhnc(u) is the consistency residual at the exact solution, as defined in Variational-crime settings and consistency defects.

Proof

1.1F1given

Apply [F1] on V to obtain the unique exact solution u.

1.2F1A1given

The finite-dimensional Hilbert problem on Vh has bounded coercive form ah and bounded load Fh, so the same Lax–Milgram theorem [F1] gives a unique uh∈Vh.

2.1A1step 1.2choose

Fix vh∈Vh and set eh:=uh−vh. Then eh∈Vh, and [A1] gives αh∥eh∥h2≤Re⁡ah(eh,eh)=Re⁡(Fh(eh)−a~h(vh,eh)).

3.1F2A1A2step 2.1algebra

By the discrete equation and extension identity, ah(eh,eh)=a~h(u−vh,eh)+Fh(eh)−a~h(u,eh). The extension bound [F2] and residual definition [A2] therefore give αh∥eh∥h2≤(Mh∥u−vh∥h+δhnc(u))∥eh∥h.

4.1step 3.1givenalgebra∎

If eh=0 there is nothing to divide by; otherwise divide step 3.1 by αh∥eh∥h. The triangle inequality gives ∥u−uh∥h≤(1+Mh/αh)∥u−vh∥h+δhnc(u)/αh. Taking the infimum over vh proves the per-mesh bound. For the stated family, replace Mh/αh by M‾/α‾ and 1/αh by 1/α‾. The resulting constants are independent of h, yielding the stated uniform, convergence and rate consequences when the two explicit error terms have the corresponding control.

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