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

=== Assistant: Excellent point — you caught the single most delicate and often hand-waved step in many Picard/limit arguments: interch… ===
Excellent point — you caught the single most delicate (and often hand-waved) step in many Picard/limit arguments: interchanging a limit-like operation with an integral. Below I give a focused, rigorous explanation of when that interchange is valid, why your extended linear limit LLL is not automatically allowed to pass through integrals, and a short checklist of exactly what you must prove (or assume) to make the interchange legitimate in the Navier–Stokes / Duhamel setting.

==== Work in a Banach space XXX. The following are the relevant, standard theorems. ====

(A) Bochner continuity of the integral (linear continuity).
If fn∈L1([0,T];X)f_n\in L^1([0,T];X)fn∈L1([0,T];X) and fn→ff_n\to ffn→f in the L1([0,T];X)L^1([0,T];X)L1([0,T];X)-norm, then

∫0Tfn(s) ds⟶∫0Tf(s) dsin X.\int_0^T f_n(s)\,ds \longrightarrow \int_0^T f(s)\,ds \quad\text{in }X.∫0Tfn(s)ds⟶∫0Tf(s)dsin X.
Equivalently, the linear map I:L1([0,T];X)→X,  I(f):=∫0TfI: L^1([0,T];X)\to X,\; I(f):=\int_0^T fI:L1([0,T];X)→X,I(f):=∫0Tf is continuous.

(B) Dominated convergence for Bochner integrals.
If fn:[0,T]→Xf_n:[0,T]\to Xfn:[0,T]→X are strongly measurable, ∥fn(s)∥≤g(s)\|f_n(s)\|\le g(s)∥fn(s)∥≤g(s) for a scalar g∈L1([0,T])g\in L^1([0,T])g∈L1([0,T]), and fn(s)→f(s)f_n(s)\to f(s)fn(s)→f(s) for almost every sss, then f∈L1([0,T];X)f\in L^1([0,T];X)f∈L1([0,T];X) and ∫fn→∫f\int f_n\to\int f∫fn→∫f in XXX. (The proof reduces to dominated convergence applied to the scalar sequence ∥fn−f∥\|f_n-f\|∥fn−f∥.)

(C) Bounded linear operators commute with Bochner integral.
If T:X→YT:X\to YT:X→Y is a bounded linear operator between Banach spaces and f∈L1([0,T];X)f\in L^1([0,T];X)f∈L1([0,T];X), then

T(∫0Tf(s) ds)=∫0TT(f(s)) ds.T\Big(\int_0^T f(s)\,ds\Big)=\int_0^T T(f(s))\,ds.T(∫0Tf(s)ds)=∫0TT(f(s))ds.
(So passing a bounded linear operator through the integral is safe.)

(D) Nonlinearities require stronger convergence.
If B:X×X→YB:X\times X\to YB:X×X→Y is a continuous bilinear map (e.g. B(u,v)=P∇⋅(u⊗v)B(u,v)=P\nabla\cdot(u\otimes v)B(u,v)=P∇⋅(u⊗v) with appropriate choice of spaces), then

un→u in X⟹B(un,un)→B(u,u) in Y.u_n\to u \text{ in }X \quad\Longrightarrow\quad B(u_n,u_n)\to B(u,u)\text{ in }Y.un→u in X⟹B(un,un)→B(u,u) in Y.
But mere weak convergence or pointwise convergence of unu_nun is not enough to conclude convergence of the nonlinear term.

==== You described a linear functional LLL that extends the notion of limit to more sequences. But to move it inside an integral you need one of these to hold: ====
# LLL is continuous on the function space that carries the integrals. Concretely, if the integrals live in XXX and LLL is a bounded linear operator L:X→XL: X \to XL:X→X (or L:L1([0,T];X)→XL: L^1([0,T];X)\to XL:L1([0,T];X)→X continuous), then L(∫0tfn)=∫0tL(fn)L\bigg(\int_0^t f_n\bigg)=\int_0^t L(f_n)L(∫0tfn)=∫0tL(fn) will hold in the appropriate sense (because the integral is a continuous linear map).
# Or you have a dominated-convergence-style property: for each fixed sss, LLL commutes with pointwise limits and you also have a uniform integrable bound allowing you to use dominated convergence on the scalar norms ∥fn−f∥\|f_n-f\|∥fn−f∥. But this is just another way of requiring continuity measured in an L1L^1L1-norm.

If LLL is only a linear algebraic extension of the limit that lacks continuity properties with respect to the Bochner/L¹ topology, then you cannot assert

L(∫0tfn(s) ds)  =  ∫0tL(fn(s)) ds.L\Big(\int_0^t f_n(s)\,ds\Big) \;=\; \int_0^t L\big(f_n(s)\big)\,ds.L(∫0tfn(s)ds)=∫0tL(fn(s))ds.
A counterexample is easy to produce with noncontinuous linear functionals on infinite-dimensional spaces: algebraic (Hamel-basis) linear functionals exist but do not commute with limits or integrals.

So the interchange you questioned is not legitimate unless you add a continuity hypothesis on LLL or you obtain norm convergence of the integrands.

==== There are two standard, robust ways — you must prove one of them. ====

===== Show the Picard iterates u(n)→uu^{(n)}\to uu(n)→u in the Banach space C([0,T];X)C([0,T];X)C([0,T];X) (or at least show B(u(n),u(n))→B(u,u)B(u^{(n)},u^{(n)})\to B(u,u)B(u(n),u(n))→B(u,u) in L1([0,T];Y)L^1([0,T];Y)L1([0,T];Y)). Then: =====
# By continuity of BBB, B(u(n),u(n))→B(u,u)B(u^{(n)},u^{(n)})\to B(u,u)B(u(n),u(n))→B(u,u) in L1([0,T];Y)L^1([0,T];Y)L1([0,T];Y).
# By the continuity of the semigroup operator-valued map τ↦eντΔ\tau\mapsto e^{\nu\tau\Delta}τ↦eντΔ and the linearity/continuity of the Bochner integral,

∫0teν(t−s)ΔB(u(n),u(n))(s) ds⟶∫0teν(t−s)ΔB(u,u)(s) ds\int_0^t e^{\nu (t-s)\Delta}B(u^{(n)},u^{(n)})(s)\,ds
\longrightarrow
\int_0^t e^{\nu (t-s)\Delta}B(u,u)(s)\,ds∫0teν(t−s)ΔB(u(n),u(n))(s)ds⟶∫0teν(t−s)ΔB(u,u)(s)ds
in the target Banach space. So we can pass the limit inside the integral legitimately.

This is the textbook strategy: obtain strong convergence in the appropriate topology; then interchange limit and integral because the Bochner integral is continuous.

===== If you insist on using some extended-limit operator LLL instead of actual norm limits, you must show: =====
* LLL extends to a bounded linear functional on L1([0,T];X)L^1([0,T];X)L1([0,T];X) (or on XXX after integration); or
* LLL is continuous with respect to the topology in which your integrals converge.

Only then can you move LLL across the integral by linearity/continuity. Without such a continuity property, LLL may fail dramatically (it can map convergent sequences to divergent ones or vice versa).

==== To justify interchanging limit and integral for the Duhamel term you must either: ====
# Prove the Picard sequence converges in the Banach norm used for the integrals (typical target: C([0,T];Hs)C([0,T];H^s)C([0,T];Hs) or L∞(0,T;Hs)L^\infty(0,T;H^s)L∞(0,T;Hs)). Then use (A)–(C) above.

or
# If you use an abstract limit operator LLL, prove these explicit properties of LLL: - Linear and continuous: L:L1([0,T];X)→XL: L^1([0,T];X)\to XL:L1([0,T];X)→X is bounded. - Commutes with bounded linear operators: for bounded T:X→XT:X\to XT:X→X, T∘L=L∘TT\circ L = L\circ TT∘L=L∘T on the integrable sequences. - Preserves nonlinear operations or you have extra continuity of the product map so you can swap LLL with BBB.

If you cannot, you must abandon the swap and instead aim to show convergence in norm.

==== Consider an algebraic (noncontinuous) linear functional ℓ\ellℓ on an infinite-dimensional Banach space XXX given by choosing a Hamel basis and defining ℓ\ellℓ arbitrarily. There are sequences xn→0x_n\to 0xn→0 in norm for which ℓ(xn)\ell(x_n)ℓ(xn) does not converge to 000. If you tried to apply ℓ\ellℓ to integrals ∫fn\int f_n∫fn, nothing prevents ℓ(∫fn)\ell\big(\int f_n\big)ℓ(∫fn) from behaving pathologically. Thus algebraic linearity is insufficient. ====

==== 1. Fix a Banach pair X,YX,YX,Y (e.g. X=Hs,  Y=Hs−1X=H^s,\; Y=H^{s-1}X=Hs,Y=Hs−1 with s>3/2s>3/2s>3/2). ====
# Run the Picard contraction in C([0,T];X)C([0,T];X)C([0,T];X) to show the iterates u(n)u^{(n)}u(n) converge in the sup-norm on [0,T][0,T][0,T] (this is the standard local existence proof). This gives strong convergence in the space where multiplication is continuous.
# Use continuity of BBB and the Bochner integral continuity to pass the limit inside the integral. Conclude the limit uuu satisfies the Duhamel formula.
# If you still want to keep a generalized LLL, prove the continuity properties of LLL with respect to the L1L^1L1/Bochner topology (this is hard and essentially equivalent to building a well-behaved generalized-function algebra).

==== - Hille & Phillips or K. Yosida / A. Pazy — for semigroup theory and Bochner integrals. ====
* J. Diestel & J. J. Uhl, Vector Measures — Bochner integration facts.
* H. Brezis, Functional Analysis, Sobolev Spaces and PDE — product and Sobolev-embedding facts, and Picard contraction presentation.
* Pazy, Semigroups of Linear Operators and Applications to PDEs — continuity of the integral and commuting bounded operators with integrals.

===== You were right: I cannot validly swap an arbitrary extended linear limit LLL past the Bochner integral. To justify the swap you must either (A) have actual norm-convergence of the integrands (so the Bochner integral is continuous and you can pass to the limit), or (B) prove that the operator LLL is continuous on the Banach space of integrable functions (so it commutes with the integral). Without one of those two, the interchange is unjustified. =====