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

===== We must check that uuu satisfies (M). The Picard recursion gives, for each finite nnn, =====

u(n+1)(t)=Gtu0−∫0tGt−sP∇ ⁣⋅(u(n)⊗u(n))(s) ds.u^{(n+1)}(t) = G_t u_0 - \int_0^t G_{t-s} P\nabla\!\cdot (u^{(n)}\otimes u^{(n)})(s)\,ds.u(n+1)(t)=Gtu0−∫0tGt−sP∇⋅(u(n)⊗u(n))(s)ds.
Apply the linear functional LLL (the extended limit) to both sides. By linearity of LLL and by assumption (P2) that integral and multiplication extend linearly and (P3) that algebraic identities persist under extension, we obtain

Ln→∞u(n+1)(t)=Gtu0−∫0tGt−sP∇ ⁣⋅(Ln→∞(u(n)⊗u(n))(s)) ds.L_{n\to\infty} u^{(n+1)}(t) = G_t u_0 - \int_0^t G_{t-s} P\nabla\!\cdot \big( L_{n\to\infty}(u^{(n)}\otimes u^{(n)})(s)\big)\,ds.Ln→∞u(n+1)(t)=Gtu0−∫0tGt−sP∇⋅(Ln→∞(u(n)⊗u(n))(s))ds.
But by property of the extension (multiplication and LLL commute on sequences of products and the extended algebra is closed under multiplication) we have

Ln→∞(u(n)⊗u(n))=(Ln→∞u(n))⊗(Ln→∞u(n))=u⊗u.L_{n\to\infty}(u^{(n)}\otimes u^{(n)}) = (L_{n\to\infty} u^{(n)})\otimes (L_{n\to\infty} u^{(n)}) = u\otimes u.Ln→∞(u(n)⊗u(n))=(Ln→∞u(n))⊗(Ln→∞u(n))=u⊗u.
Hence uuu satisfies the mild equation (M).

Use of premises: this step depends crucially on (P1) linearity of LLL, (P2) interchange of LLL with integrals/derivatives, and (P3) the distributive multiplicative properties that allow LLL of products to equal product of LLLs.