Editing Openai/6935580c-34e4-8001-9136-7bbdf1499790 (section)

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Do not rely on exotic limit extensions. Work in a concrete Banach/Sobolev space and prove all steps in that topology. The canonical program:
# Choose X=Hs(R3)X = H^s(\mathbb{R}^3)X=Hs(R3) with s>32s>\tfrac32s>23 (so HsH^sHs is an algebra and u↦P∇ ⁣⋅(u⊗u)u\mapsto P\nabla\!\cdot(u\otimes u)u↦P∇⋅(u⊗u) maps Hs→Hs−1H^s\to H^{s-1}Hs→Hs−1).
# Construct a local mild (strong) solution by Picard in C([0,T];Hs)C([0,T];H^s)C([0,T];Hs); show ∥u∥L∞(0,T;Hs)\|u\|_{L^\infty(0,T;H^s)}∥u∥L∞(0,T;Hs) controlled on the existence interval. (This is routine.)
# Derive a priori estimates for higher Sobolev norms; show that if these norms stay finite you can extend the solution. (This produces the standard blowup alternative.)
# To get global existence you must obtain global a priori bounds that prevent norm blowup. That is exactly the open hard part of Clay. So you can either: - Prove additional a priori bounds (this is essentially attempting the Clay proof), or - Prove conditional results: e.g., if a certain critical norm stays finite (Serrin-type condition, Beale–Kato–Majda criterion for vorticity), then the solution is global. These are known conditional theorems and are useful but do not solve Clay unconditionally.

Conclusion for Route 1: This path is rigorous and standard. It will produce clean conditional theorems or full solutions in special cases (small data, symmetry reductions), but it cannot be used to claim an unconditional global result without proving the missing a-priori bound — which is precisely the Clay challenge.

===== If you are committed to the approach using the extended linear limit LLL and total multiplication, then you must explicitly add and prove the additional properties that bridge the gap to classical functions. Concretely you must show: =====
# (C1) Classical-closure: the subspace C\mathcal{C}C of classical real-valued C∞C^\inftyC∞ functions is closed under the topology/operation in which you apply LLL. Equivalently, whenever a sequence of classical functions {fn}\{f_n\}{fn} is such that Ln→∞fnL_{n\to\infty} f_nLn→∞fn exists, then Ln→∞fnL_{n\to\infty} f_nLn→∞fn is a classical function (and coincides with the classical pointwise / uniform limit when that exists).
# (C2) Compatibility with nonlinearities: for sequences fn,gnf_n,g_nfn,gn of classical functions, L(fngn)=L(fn)⋅L(gn)L(f_n g_n) = L(f_n)\cdot L(g_n)L(fngn)=L(fn)⋅L(gn) whenever products make sense classically — and the same commuting property for derivatives and integrals (i.e., LLL commutes with differentiation and integration on the relevant sequences).
# (C3) Continuity properties: LLL preserves norm bounds needed for energy estimates: for example, if ∥fn∥Hs\|f_n\|_{H^s}∥fn∥Hs is uniformly bounded then L(fn)L(f_n)L(fn) lies in HsH^sHs and the same norm estimates transfer.

If you can construct such an LLL (or prove these properties for a known generalized function algebra), then the Picard-limit argument becomes legitimate and the limit will be classical. But a few remarks:
* This is nontrivial: many generalized-function frameworks (tempered distributions, Colombeau algebras, nonstandard analysis variants) either do not have all these properties or require careful reinterpretation of “classical function”. You must either construct LLL explicitly and prove (C1)–(C3), or show your axioms imply them.
* If you succeed, the result becomes conditional on these new axioms. The Clay Prize requires a proof in standard accepted mathematics; a proof resting on a new foundational axiom must include a convincing, peer-reviewed construction of that axiom system and why it is acceptable (consistent, conservative extension, etc.).

Conclusion for Route 2: Possible but heavy: it displaces the hard part of Clay onto the foundational task of building/justifying the extended algebra.