TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck pass

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Babuska-Brezzi solvability by kernel coercivity and inf-sup

Statement

Assume Countable Choice. Let $V, Q$ be real Hilbert spaces, and let $a : V \times V \to R$ and $b : V \times Q \to R$ be bounded bilinear forms. If $Q = {0}$ , the theorem reduces to Lax–Milgram on $V$ with $Z = V$ and $p = 0$ . Otherwise, if $Q \neq {0}$ , assume $V \neq {0}$ . Define $B : V \to Q$ by $(B v, q)_{Q} = b (v, q)$ and set $β : = \inf_{q \neq 0} \sup_{v \neq 0} b (v, q) / (∥ v ∥_{V} ∥ q ∥_{Q}) > 0$ . Suppose $Z : = \ker B$ is closed and $a (z, z) \geq α ∥ z ∥_{V}^{2}$ on $Z$ , with $α > 0$ . For every $F \in V^{'}$ and $G \in Q^{'}$ , there is a unique $(u, p) \in V \times Q$ satisfying $a (u, v) + b (v, p) = F (v), b (u, q) = G (q),$ for all $v \in V, q \in Q$ . If $M_{a} = ∥ a ∥$ , then one stability bound is $∥ u ∥_{V} \leq β^{- 1} ∥ G ∥ + α^{- 1} (∥ F ∥ + M_{a} β^{- 1} ∥ G ∥), ∥ p ∥_{Q} \leq β^{- 1} (∥ F ∥ + M_{a} ∥ u ∥_{V}) .$

Facts & Assumptions

Given: Countable Choice, real Hilbert spaces, bounded $a, b$ , $β > 0$ , and coercivity of $a$ on $Z$ .

[A1]

Countable Choice is required by the Lax–Milgram step on the closed constraint kernel. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

The inf-sup bound yields a bounded linear right inverse $R : Q \to V$ with $B R = I$ and $∥ R ∥ \leq β^{- 1}$ . (Hilbert inf-sup gives a bounded constraint right inverse)

[F2]

Lax–Milgram gives a unique solution on the Hilbert subspace $Z$ for a bounded coercive form, and its norm is controlled by the right-hand side divided by $α$ . (The Lax--Milgram theorem, The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F3]

Riesz representation identifies bounded functionals on $V$ and $Q$ with inner products against unique vectors. (Riesz representation for Hilbert spaces)

Proof

1.1A1F1F3

If $Q = {0}$ , then $G = 0$ , $Z = V$ , and Lax–Milgram on $V$ gives the unique solution $(u, 0)$ with $∥ u ∥ \leq α^{- 1} ∥ F ∥$ ; so suppose $Q \neq {0}$ . By [F3], write $G (q) = (g, q)_{Q}$ for a unique $g \in Q$ with $∥ g ∥ = ∥ G ∥$ . Let $u_{g} = R g$ using [F1]. Then $B u_{g} = g$ , $∥ u_{g} ∥ \leq β^{- 1} ∥ G ∥$ , and the constraint equation will hold for $u = u_{g} + z$ with any $z \in Z$ .

2.1A1F2step 1.1

On the closed Hilbert subspace $Z$ , solve $a (z, w) = F (w) - a (u_{g}, w)$ for all $w \in Z$ by [F2]. The restricted form is bounded and coercive by hypothesis, and the functional norm is at most $∥ F ∥ + M_{a} ∥ u_{g} ∥$ . Therefore $∥ z ∥ \leq α^{- 1} (∥ F ∥ + M_{a} β^{- 1} ∥ G ∥)$ and $u = u_{g} + z$ has the stated bound. The equation $a (u, w) = F (w)$ holds for every $w \in Z$ .

3.1A1F1givenstep 1.1step 2.1

Define $ℓ (v) = F (v) - a (u, v)$ . It vanishes on $Z$ by step 2.1. Set $Λ (q) = ℓ (R q)$ . This is bounded by $∥ R ∥ ∥ ℓ ∥ ∥ q ∥$ , and for every $v$ , $v - R B v \in Z$ , so $ℓ (v) = ℓ (R B v) = Λ (B v)$ .

4.1A1F1F3step 3.1step 1.1

By [F3], there is a unique $p \in Q$ with $Λ (q) = (q, p)_{Q}$ . Hence $ℓ (v) = (B v, p)_{Q} = b (v, p)$ , giving the first mixed equation; step 1.1 gives the second. Also $∥ p ∥ \leq β^{- 1} ∥ ℓ ∥ \leq β^{- 1} (∥ F ∥ + M_{a} ∥ u ∥)$ .

5.1A1F1givenF3∎

For uniqueness, subtract two solutions. Their velocity difference $w$ lies in $Z$ ; testing the first difference equation with $w$ gives $a (w, w) = 0$ , so coercivity gives $w = 0$ . The pressure difference then pairs to zero with every $B v$ ; since [F1] makes $B$ onto, it pairs to zero with all $Q$ , hence is zero.

Depends on

Used by

Dependency tree · two levels

25 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Jean-Luc Guermond, MATH 661 Fall 2025, Chapters 53–54 (complete course notes) (standard reference, not scraped)
Daniele Boffi, Franco Brezzi, Michel Fortin, Mixed Finite Element Methods and Applications (complete textbook) (standard reference, not scraped)