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Second Strang estimate with a bounded extension to the comparison space
Statement
Assume Countable Choice. Let be a real or complex Hilbert space, a bounded coercive sesquilinear form (bilinear over ), and a bounded functional with the test-linearity required by . Let be a finite-dimensional Hilbert space with norm , not necessarily a subspace of . Suppose the algebraic sum is equipped with a norm extending these norms, and is a bounded sesquilinear extension of a discrete form with bound . Over , both forms are linear in the first argument and conjugate-linear in the second; over , they are bilinear. Assume is coercive on with constant , and let be bounded on with the matching test-linearity. The exact and discrete equations have unique solutions and , and
where when , and when . 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 and uniformly. Writing , the uniform bound is Thus uniform control of and gives a uniform error budget; if both tend to zero, the errors tend to zero, and if both are for a fixed , the error is . No rate is inferred without those approximation, residual, and stability bounds.
Facts & Assumptions
Given: Countable Choice; the exact Hilbert problem on ; the nonconforming space and common comparison norm on ; a bounded extension ; discrete coercivity ; and a bounded discrete load.
The Lax--Milgram theorem under The Axiom of Countable Choice (): the exact bounded coercive Hilbert problem has a unique solution.
By hypothesis, is sesquilinear with the stated slot conventions (bilinear over ). By definition, agrees with on and on (Variational-crime settings and consistency defects).
for every .
is the consistency residual at the exact solution, as defined in Variational-crime settings and consistency defects.
Proof
Apply [F1] on to obtain the unique exact solution .
The finite-dimensional Hilbert problem on has bounded coercive form and bounded load , so the same Lax–Milgram theorem [F1] gives a unique .
Fix and set . Then , and [A1] gives .
By the discrete equation and extension identity, . The extension bound [F2] and residual definition [A2] therefore give .
If there is nothing to divide by; otherwise divide step 3.1 by . The triangle inequality gives . Taking the infimum over proves the per-mesh bound. For the stated family, replace by and by . The resulting constants are independent of , yielding the stated uniform, convergence and rate consequences when the two explicit error terms have the corresponding control.
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Sources
- Jean-Luc Guermond, MATH 661 Fall 2025, Part VI, Chapter 27, Error analysis with variational crimes (standard reference, not scraped)
- Joachim Schöberl, Numerical Methods for Partial Differential Equations, Chapter 25, Non-conforming Finite Element Methods (standard reference, not scraped)