PropositionStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pendingverified 2026-10-23 (gpt-6.1-sol)

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Gauge-invariant pressure sensor weights

Statement

Assume Countable Choice. Let $d \in {2, 3}$ and let $D \subset R^{d}$ be a nonempty bounded connected Lipschitz open set, with $V = H_{0}^{1} (D; R^{d})$ and $Q = L_{0}^{2} (D)$ . Take $g \in L^{2} (D; R)$ and define $S_{g} (q) : = \int_{D} g q d x$ for $q \in Q$ ; this is a bounded pressure sensor. Put $\overset{ˉ}{g} : = ∣ D ∣^{- 1} \int_{D} g d x$ . On the normalized pressure space, $S_{g} = S_{g - \overset{ˉ}{g}}$ . Whenever $p$ denotes an unnormalized pressure representative in a weak pressure term or sensor reading below, take $p \in L^{2} (D; R)$ ; a normalized pressure is denoted $q \in Q \subset L^{2} (D; R)$ . The weak Stokes pressure term $- \int_{D} p div v d x$ is unchanged by $p \mapsto p + c$ for $v \in V$ and $c \in R$ . The unnormalized sensor reading $\int_{D} g p d x$ is unchanged under every pressure shift $p \mapsto p + c$ if and only if $\int_{D} g d x = 0$ , equivalently $g \in Q$ .

Statement references

Guermond, MATH 661, Chapter 53, Remark 53.4 and §53.2.1, pressure constant and weak pressure term.

Boffi–Brezzi–Fortin, Mixed Finite Element Methods and Applications, §8.2.1, Remark 8.2.1, pp. 462–463, pressure modulo constants and zero-mean representative.

Facts & Assumptions

Given: Countable Choice; a nonempty bounded connected Lipschitz domain $D \subset R^{d}$ , $d \in {2, 3}$ ; the zero-boundary velocity space $V$ ; the mean-zero pressure space $Q$ ; an $L^{2}$ sensor weight $g$ ; and, for the unnormalized assertions, a pressure representative $p \in L^{2} (D; R)$ .

[A1]

Countable Choice is assumed by the Stokes-space and zero-boundary divergence suppliers. Stokes velocity and pressure spaces and weak forms Zero mean of the divergence of a zero-boundary field

[F1]

Stokes velocity and pressure spaces and weak forms: $V = H_{0}^{1} (D; R^{d})$ , $Q = L_{0}^{2} (D)$ , and for $q \in Q$ the weak pressure form is $b (v, q) = - \int_{D} q div v d x$ .

[F2]

Zero mean of the divergence of a zero-boundary field: every $v \in H_{0}^{1} (D; R^{d})$ has $div v \in L^{2} (D)$ and $\int_{D} div v d x = 0$ .

[F3]

Cauchy-Schwarz inequality for $L^{2}$ : the integral pairing of two $L^{2} (D)$ functions is finite and has absolute value at most the product of their $L^{2}$ norms.

[F4]

Bounded Stokes pressure sensors: a bounded pressure sensor is a bounded real-linear functional on $Q$ with its operator norm.

Proof

1.1givenF1F3F4algebra

Since $D$ is nonempty and open it contains a ball, while boundedness gives finite measure; thus $0 < ∣ D ∣ < \infty$ and $1 \in L^{2} (D)$ . [F3] therefore makes $\int_{D} g d x$ finite. For $q \in Q$ , [F3] gives $∣ S_{g} (q) ∣ \leq ∥ g ∥_{2} ∥ q ∥_{2}$ , so $S_{g} \in Q^{'}$ . Also $g - \overset{ˉ}{g} \in Q$ and $S_{g} (q) - S_{g - \overset{ˉ}{g}} (q) = \overset{ˉ}{g} \int_{D} q d x = 0$ ; hence the normalized-gauge sensor depends only on the mean-zero part of $g$ .

1.2A1F2F3algebra

Under the Countable Choice assumption [A1], apply [F2]: every $v \in V$ has $div v \in L^{2} (D; R)$ and $\int_{D} div v d x = 0$ . Since $D$ has finite measure, $p + c \in L^{2} (D; R)$ and [F3] makes both pressure pairings finite. Their difference is $- c \int_{D} div v d x = 0$ , so adding a constant leaves the weak pressure term unchanged.

2.1givenF1F3algebra∎

For any $p \in L^{2} (D; R)$ and $c \in R$ , boundedness of $D$ gives $p + c \in L^{2} (D; R)$ , and [F3] makes both sensor pairings finite. Their difference is $\int_{D} g (p + c) d x - \int_{D} g p d x = c \int_{D} g d x$ . If $\int_{D} g = 0$ , every shift leaves the reading unchanged; conversely, invariance for every shift, in particular $c = 1$ , forces $\int_{D} g = 0$ . By the definition of $Q$ , this is equivalent to $g \in Q$ .

Gauge-invariant pressure sensor weights

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources