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A bounded form defect preserves discrete coercivity

Statement

Let V be a Hilbert space and let Vh⊆V be a subspace. Suppose the exact form a is coercive with constant α>0 in the norm ∥⋅∥V. Let ah be a discrete form on Vh and suppose

∣ah(vh,vh)−a(vh,vh)∣≤ηh∥vh∥V2(vh∈Vh)

for some ηh≥0. Then ah is coercive with constant at least α−ηh whenever ηh<α. In particular, if ηh≤α/2, then its coercivity constant is at least α/2.

Facts & Assumptions

Given: A Hilbert space V, a subspace Vh, a coercive form a with constant α>0, a discrete form ah, and the diagonal defect bound with constant ηh.

[F1]

Bounded, coercive and symmetric sesquilinear forms: coercivity means Re⁡a(v,v)≥α∥v∥V2; symmetry is not required.

[A1]

For every vh∈Vh, ∣ah(vh,vh)−a(vh,vh)∣≤ηh∥vh∥V2.

Proof

1.1givenF1A1algebra

Fix vh∈Vh. By coercivity [F1] and the defect hypothesis [A1], Re⁡ah(vh,vh)=Re⁡a(vh,vh)+Re⁡(ah(vh,vh)−a(vh,vh))≥(α−ηh)∥vh∥V2.

2.1step 1.1algebra∎

If ηh<α, then α−ηh>0, so the inequality in step 1.1 is precisely coercivity with that constant. If ηh≤α/2, then α−ηh≥α/2.

Depends on

Used by

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