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

=== For the mild form you need to make sense of ===

∫0tGt−sP∇ ⁣⋅(u⊗u)(s) ds,\int_0^t G_{t-s}P\nabla\!\cdot( u\otimes u)(s)\,ds,∫0tGt−sP∇⋅(u⊗u)(s)ds,
where the integrand is a map [0,t]→E[0,t]\to E[0,t]→E. Below are two robust, standardly justified routes (one analytic, one algebraic) you can adopt inside your extended framework — with explicit statements you will need to prove or assume.

==== 1. Equip EEE (or at least the subspace in which all Picard iterates live) with a Banach-norm ∥⋅∥X\|\cdot\|_X∥⋅∥X. For example, choose an XXX like L2L^2L2, HkH^kHk, or CαC^\alphaCα if you embed classical functions there; if EEE is larger you still need a topology that makes the heat semigroup and Leray projector continuous operators Gt,P:  X→XG_t,P:\;X\to XGt,P:X→X. ====
# Show the integrand s↦Gt−sP∇ ⁣⋅(u⊗u)(s)s\mapsto G_{t-s}P\nabla\!\cdot( u\otimes u)(s)s↦Gt−sP∇⋅(u⊗u)(s) is strongly continuous (i.e. continuous in the ∥⋅∥X\|\cdot\|_X∥⋅∥X-norm) on [0,t][0,t][0,t] or at least on (0,t](0,t](0,t] with an integrable singularity at 000. For Picard iterates this is standard because each iterate is smooth for s>0s>0s>0 and heat-kernel estimates give control near 000.
# Then define the integral as the Bochner integral in XXX. Bochner integrals are limits of Riemann sums in the Banach norm and satisfy linearity, additivity, and the usual fundamental theorem relationships when the integrand is continuous. This gives a well-defined vector-valued integral in XXX.
# Remarks: Bochner integration also gives estimates (norm bounds) which you can use in contraction/Grönwall arguments. It avoids having to construct antiderivatives in EEE.

What you must assume/prove: existence of a Banach topology on the target space that makes the mapping continuous and the operators bounded; and that the extended multiplication u⊗uu\otimes uu⊗u maps the function class into that Banach space.

==== If you insist on defining the definite integral solely via values of antiderivatives and your extended linear limit LLL, you must: ====
# Define continuity of f:[0,t]→Ef:[0,t]\to Ef:[0,t]→E in terms of the limit LLL: i.e. fff is continuous at s0s_0s0 if for every sequence sn→s0s_n\to s_0sn→s0 (classically) we have Ln→∞f(sn)=f(s0)L_{n\to\infty} f(s_n)=f(s_0)Ln→∞f(sn)=f(s0). But that presupposes a way to take limits of sequences with values in EEE — you must make that precise.
# Construct a primitive F(x):=Ln→∞∑k=1nf(ξk)ΔxkF(x) := L_{n\to\infty} \sum_{k=1}^{n} f(\xi_k)\Delta x_kF(x):=Ln→∞∑k=1nf(ξk)Δxk (Riemann sums) and prove F′(x)=f(x)F'(x)=f(x)F′(x)=f(x) in the sense of your extended derivative. This again requires that LLL commutes with the relevant operations and that uniform continuity estimates hold to pass the limiting operation through subtraction/division.
# This route is essentially reproducing the classical proof of the FTC but with LLL replacing the classical limit. For it to succeed you need stronger and more explicit axioms: LLL must preserve uniform convergence and commute with multiplication by scalar step functions, etc. These are nontrivial axioms and are exactly the hidden extra assumptions you were trying to avoid.

Why this route is brittle: you again need control of sequences of Riemann sums and uniformity; the “antiderivative exists because the integrand is continuous” claim becomes equivalent to proving that your extension of limits has the same uniform-convergence properties as the classical limit — a major extra hypothesis.