TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck pendingverified 2026-10-23 (gpt-6.1-sol)

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.

Pressure recovery for periodic incompressible Sobolev velocity

Statement

Let $Q_{d} = R^{d} / (2 π Z)^{d}$ have normalized measure $d μ = (2 π)^{- d} d x$ . For $α = (α_{1}, \dots, α_{d}) \in N_{0}^{d}$ , write $∣ α ∣ = \sum_{j} α_{j}$ and $\partial^{α} = \partial_{1}^{α_{1}} \dots \partial_{d}^{α_{d}}$ . Give $C^{\infty} (Q_{d}; R)$ the Fr'echet topology generated by $p_{m} (ϕ) = \max_{∣ α ∣ \leq m} ∥ \partial^{α} ϕ ∥_{\infty}$ , for $m \in N_{0}$ , where the derivatives are those of the smooth periodic lift to $R^{d}$ . A periodic distribution is a continuous real-linear functional on this test space, extended complex-linearly to complex-valued tests. For such a distribution $S$ , define $\hat{S} (k) = ⟨ S, e^{- i k \cdot x} ⟩$ for $e_{k} (x) = e^{i k \cdot x}$ and define its mean to be $\hat{S} (0) = ⟨ S, 1 ⟩$ . For an integrable function $g$ , the pairing is $⟨ g, ϕ ⟩ = \int_{Q_{d}} g ϕ d μ$ . Define distributional derivatives by $⟨ \partial_{j} S, ϕ ⟩ = - ⟨ S, \partial_{j} ϕ ⟩$ . For a smooth scalar test, $\nabla ϕ = (\partial_{1} ϕ, \dots, \partial_{d} ϕ)$ and $Δ ϕ = \sum_{j} \partial_{j}^{2} ϕ$ ; for a smooth vector field $v$ , $div v = \sum_{j} \partial_{j} v_{j}$ , $(v \otimes v)_{i j} = v_{i} v_{j}$ , and $(v \cdot \nabla) v_{i} = \sum_{j} v_{j} \partial_{j} v_{i}$ . In particular, $div u = 0$ distributionally means $- \int_{Q_{d}} u \cdot \nabla ϕ d μ = 0$ for every real test function $ϕ$ . Equality of periodic distributions means equality on every such test function. Assume Countable Choice. Let $d \in {1, 2, 3}$ , let $s > d / 2 + 1$ , and let $u \in H^{s} (Q_{d}; R^{d})$ satisfy $div u = 0$ in this sense. Then $u \otimes u \in H^{s}$ , the mean-zero scalar $π = Q (u \otimes u)$ belongs to $H^{s}$ , and in $H^{s - 1}$ , $P (u \cdot \nabla u) = u \cdot \nabla u + \nabla π .$ Among mean-zero periodic distributions $\tilde{π}$ for which this identity holds as an equality of periodic distributions, this pressure is unique. The formula is a torus pressure recovery result and asserts no wall pressure regularity.

Statement references

Terence Tao, 254A Notes 1: Local well-posedness of the Navier–Stokes equations (2018).

Erick Schulz, Introduction to the Navier–Stokes Equations, McGill MATH 595 notes (2014).

Facts & Assumptions

Given: The hypotheses and conventions stated in the formulation.

[F1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): countable families of nonempty sets admit a choice function.

[F2]

Periodic fluid Fourier Sobolev spaces: $H^{s}$ is the completion in $\sum_{k} w_{k}^{2 s} ∣ \hat{q} (k) ∣^{2}$ with $w_{k} = (1 + ∣ k ∣^{2})^{1 / 2}$ on the normalized periodic torus; for $s \geq 0$ each element has a unique $L^{2}$ representative.

[F3]

Periodic Fourier embeddings and products: when $σ > d / 2$ , multiplication is continuous $H^{σ} \times H^{σ} \to H^{σ}$ , and when $σ > 1 + d / 2$ , $H^{σ}$ embeds in $C^{1}$ .

[F4]

Periodic Leray and pressure maps: $P$ and $Q$ are the stated bounded Fourier multipliers and $\nabla Q F = - (I - P) div F$ .

[F5]

The Fourier basis and Parseval's identity on the finite torus: under Countable Choice, the characters $e^{2 π i k \cdot y}$ form a complete orthonormal basis of $L^{2} (R^{d} / Z^{d})$ , so Fourier synthesis recovers each $L^{2}$ function from its coefficients; the change of variables $x = 2 π y$ gives the same completeness for $e^{i k \cdot x}$ on $Q_{d}$ with its normalized measure.

Proof

Given: Countable Choice, $d \in {1, 2, 3}$ , $s > d / 2 + 1$ , and divergence-free $u \in H^{s}$ .

1.1givenF2F3F4

Since $s > d / 2$ , [F3] gives $u \otimes u \in H^{s}$ . The bounded map $Q$ in [F4] therefore gives a mean-zero $π = Q (u \otimes u) \in H^{s}$ . The coefficientwise derivative convention in [F4] gives $\partial_{j} u \in H^{s - 1}$ because $∣ k_{j} ∣ \leq w_{k}$ ; also $u \in H^{s - 1}$ since $w_{k} \geq 1$ . Since $s - 1 > d / 2$ , [F3] places each product $u_{j} \partial_{j} u_{i}$ and hence $u \cdot \nabla u$ in $H^{s - 1}$ . The embedding in [F3] gives a $C^{1}$ representative of $u$ , since $s > 1 + d / 2$ .

2.1givenF3F4step 1.1algebra

The $C^{1}$ representative from step 1.1 obeys the ordinary product rule. With the distributional derivative convention in the Statement, this is also the periodic distributional identity $div (u \otimes u)_{i} = \sum_{j} \partial_{j} (u_{i} u_{j}) = u \cdot \nabla u_{i} + u_{i} div u = u \cdot \nabla u_{i}$ . Applying the multiplier identity in [F4] yields $\nabla π = - (I - P) (u \cdot \nabla u)$ , hence the asserted decomposition.

3.1givenF1F2F5step 2.1algebra∎

If another mean-zero periodic distribution $\tilde{π}$ gives the same decomposition, then $\nabla S = 0$ for $S = \tilde{π} - π$ . The derivative convention implies $\hat{\partial_{j} S} (k) = i k_{j} \hat{S} (k)$ , so for each $k \neq 0$ the vector equation $i k \hat{S} (k) = 0$ gives $\hat{S} (k) = 0$ ; the zero mode also vanishes because $S$ has mean zero. To see that this forces $S = 0$ as a distribution, fix $ϕ \in C^{\infty} (Q_{d})$ . For every integer $N \geq 1$ , periodic integration by parts gives $w_{k}^{2 N} ∣ \hat{ϕ} (k) ∣ \leq ∥ (1 - Δ)^{N} ϕ ∥_{L^{1} (d μ)}$ . Choosing $N$ arbitrarily large, the lattice shell bound $# {k \in Z^{d} : n \leq ∣ k ∣ < n + 1} \leq C_{d} (n + 1)^{d - 1}$ follows by placing disjoint unit cubes at those lattice points: their union lies in an annulus of fixed enlarged width, whose volume is $O_{d} ((n + 1)^{d - 1})$ ; the $n = 0$ shell is finite. This bound shows that the Fourier series of $ϕ$ and each of its differentiated series converge absolutely and uniformly. Its sum is therefore a smooth function with the same Fourier coefficients as $ϕ$ ; by [F5] the two functions agree in $L^{2}$ , hence everywhere by continuity. Thus finite Fourier sums converge to $ϕ$ in every seminorm $p_{m}$ . The distribution $S$ vanishes on every character and hence on every finite Fourier sum; its continuity gives $⟨ S, ϕ ⟩ = 0$ . Since $ϕ$ was arbitrary, $S = 0$ , proving uniqueness among all mean-zero periodic distributions.

Pressure recovery for periodic incompressible Sobolev velocity

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources