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

=== Below is a compact, ordered chain of statements (assumptions ⇒ lemmas ⇒ conclusions). I separate what is standard and provable from what is missing / conditional so you can see exactly which extra claim would turn the argument into a proof of “no blowup”. ===

==== - Let u0∈C∞u_0\in C^\inftyu0∈C∞ be divergence-free initial data on R3\mathbb{R}^3R3 (or T3\mathbb{T}^3T3). ====
* Fix s>32s>\tfrac32s>23; set X:=Hs(R3)X:=H^s(\mathbb{R}^3)X:=Hs(R3).
'' u(t)u(t)u(t) denotes a putative solution; T∗T^''T∗ its maximal time of existence in the HsH^sHs-class.

==== 1. Local existence (Picard in C([0,T];X)C([0,T];X)C([0,T];X)). If u0∈Hsu_0\in H^su0∈Hs with s>3/2s>3/2s>3/2, there exists T>0T>0T>0 and a unique mild solution u∈C([0,T];Hs)u\in C([0,T];H^s)u∈C([0,T];Hs). (Reference: standard semigroup/Picard theory.) ====
# Smoothing for positive time. For any t>0t>0t>0, u(t)∈C∞u(t)\in C^\inftyu(t)∈C∞ (instantaneous spatial smoothing from the heat semigroup). (Consequence of smoothing estimates for etΔe^{t\Delta}etΔ.)
# Blowup alternative (maximality property). There exists a maximal T∗∈(0,∞]T^''\in(0,\infty]T∗∈(0,∞] such that the solution extends on [0,T∗)[0,T^'')[0,T∗) with: if T∗<∞ then lim⁡t↑T∗∥u(t)∥Hs=+∞.\text{if }T^''<\infty\text{ then }\lim_{t\uparrow T^''}\|u(t)\|_{H^s} = +\infty.if T∗<∞ then t↑T∗lim∥u(t)∥Hs=+∞. (Standard ODE/continuation principle in Banach spaces.)
# Energy equality / inequality for sufficiently regular solutions. If uuu is smooth enough (classical), then for all ttt before blowup 12ddt∥u(t)∥L22+ν∥∇u(t)∥L22=0,\tfrac12\frac{d}{dt}\|u(t)\|_{L^2}^2 + \nu\|\nabla u(t)\|_{L^2}^2 = 0,21dtd∥u(t)∥L22+ν∥∇u(t)∥L22=0, yielding ∥u(t)∥L2≤∥u0∥L2\|u(t)\|_{L^2}\le\|u_0\|_{L^2}∥u(t)∥L2≤∥u0∥L2 for all ttt. For weak limits one gets the energy inequality. (This controls the L2L^2L2-norm globally but not higher norms.)

==== To conclude no finite-time blowup (i.e., T∗=∞T^*=\inftyT∗=∞ and uuu is global classical), one of the following must be true: ====

Path A — Direct a priori control of the HsH^sHs norm
* (P-A1) There is a provable a-priori estimate: ∀t≥0,  ∥u(t)∥Hs≤F(∥u0∥Hs,t)\forall t\ge0,\; \|u(t)\|_{H^s}\le F(\|u_0\|_{H^s},t)∀t≥0,∥u(t)∥Hs≤F(∥u0∥Hs,t) with FFF finite for all finite times.
'' From (P-A1) + blowup alternative ⇒ T∗=∞T^''=\inftyT∗=∞. (This is exactly the missing global a-priori bound — the Clay core.)

Path B — Control a critical norm that implies no blowup (known conditional criteria)
'' (P-B1) Show a known conditional criterion holds globally, e.g. Beale–Kato–Majda (BKM): if ∫0T∗∥ω(⋅,t)∥L∞ dt<∞\int_0^{T^''}\|\omega(\cdot,t)\|_{L^\infty}\,dt <\infty∫0T∗∥ω(⋅,t)∥L∞dt<∞ then no blowup occurs (i.e. HsH^sHs-norm cannot blow up).
* So if you can prove the BKM quantity is finite for all time, then no blowup. (This reduces the Clay problem to controlling a critical/endpoint norm.)

Path C — Show approximations converge strongly enough to pass equality and bootstrap regularity
* (P-C1) Construct approximations u(n)u^{(n)}u(n) (Picard/Galerkin) satisfying exact energy equality and higher uniform bounds.
* (P-C2) Prove strong convergence u(n)→uu^{(n)}\to uu(n)→u in the norm(s) that control the nonlinearities (e.g. strong in L∞(0,T;Hs)L^\infty(0,T;H^s)L∞(0,T;Hs) and in L2(0,T;Hs+1)L^2(0,T;H^{s+1})L2(0,T;Hs+1)).
'' Then energy equality and higher estimates pass to the limit, giving higher uniform control and hence T∗=∞T^''=\inftyT∗=∞. (This is another way of producing the a-priori control in Path A; but proving P-C2 is as hard as the Clay problem.)

==== - Gap 1 — Passing limits through nonlinearities/integrals. Strong convergence in the necessary norms must be proven; an algebraic extended-limit LLL without continuity does not justify interchanging limit and integral or passing nonlinear quantities to the limit. Consequence: you cannot deduce the energy equality or higher bounds for the limit unless you have strong convergence or continuity properties of the limit operator. ====
* Gap 2 — Energy controls only L2L^2L2. Energy equality gives dissipation but only controls the L2L^2L2-norm and the time integral of ∥∇u∥L22\|\nabla u\|_{L^2}^2∥∇u∥L22. Those are insufficient to bound supercritical norms (like HsH^sHs, s>1s>1s>1), so blowup in HsH^sHs remains possible.
* Gap 3 — Critical norm control is the real obstacle. All successful criteria reduce the problem to proving boundedness of a scaling-critical quantity (e.g. vorticity L∞L^\inftyL∞ integral, Serrin norms, or scaling-invariant Besov norms). Those are nontrivial and are not produced by the above energy bounds.

==== One of these (any single item suffices logically): ====
# Global HsH^sHs a-priori bound: for given smooth u0u_0u0, sup⁡t≥0∥u(t)∥Hs<∞\sup_{t\ge0}\|u(t)\|_{H^s}<\inftysupt≥0∥u(t)∥Hs<∞. (Direct.)
# Critical norm bound (BKM/Serrin): ∫0∞∥ω(⋅,t)∥L∞ dt<∞\int_0^\infty \|\omega(\cdot,t)\|_{L^\infty}\,dt<\infty∫0∞∥ω(⋅,t)∥L∞dt<∞ or Serrin condition holds globally.
# Strong compactness of approximants: Picard/Galerkin approximants converge strongly in the precise norms that allow passing the energy equality and higher estimates to the limit.
# Well-behaved extended-limit operator LLL: if you work in a generalized algebra, prove LLL is continuous on the Bochner space, preserves classicality, and commutes with nonlinear operations and integrals (this is a heavy foundational requirement).

==== 1. u0∈Hs, s>3/2u_0\in H^s,\ s>3/2u0∈Hs, s>3/2 ⇒ local mild solution u∈C([0,T];Hs)u\in C([0,T];H^s)u∈C([0,T];Hs). ====
# - (one of 1–4 above) ⇒ uniform control of ∥u(t)∥Hs\|u(t)\|_{H^s}∥u(t)∥Hs for all ttt.
# By the blowup alternative, uniform control ⇒ T∗=∞T^*=\inftyT∗=∞.
# Smoothing for t>0t>0t>0 ⇒ u(⋅,t)∈C∞u(\cdot,t)\in C^\inftyu(⋅,t)∈C∞ for every t>0t>0t>0.
# Therefore global classical solution — i.e. no finite-time blowup.

If you want, I can now produce a one-page checklist showing exactly which estimates (with explicit norms) you need to prove at each step for Path A or Path B, or else draft a short proof-of-concept showing how to obtain the Picard local existence with explicit constants so you can see where the a-priori bound must enter. Which follow-up do you prefer?