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

=== Assistant: Good — this is exactly the right point to question. ===
Good — this is exactly the right point to question. Let me account for the Leray projector PPP explicitly and rigorously, without hand-waving.

==== On R3\mathbb{R}^3R3 (or T3\mathbb{T}^3T3), the Leray projector is defined by ====

P:=I−∇Δ−1div⁡.P := I - \nabla \Delta^{-1}\operatorname{div}.P:=I−∇Δ−1div.
Equivalently, in Fourier variables,

(Pf)^i(ξ)=(δij−ξiξj∣ξ∣2)f^j(ξ).\widehat{(P f)}_i(\xi)
=
\left(\delta_{ij} - \frac{\xi_i\xi_j}{|\xi|^2}\right)\hat f_j(\xi).(Pf)i(ξ)=(δij−∣ξ∣2ξiξj)f^j(ξ).
This is a matrix-valued Fourier multiplier of order 0.

==== ### ====

For every s∈Rs\in\mathbb{R}s∈R,

P:Hs(R3;R3)→Hs(R3;R3)is bounded.P : H^s(\mathbb{R}^3;\mathbb{R}^3)\to H^s(\mathbb{R}^3;\mathbb{R}^3)
\quad\text{is bounded.}P:Hs(R3;R3)→Hs(R3;R3)is bounded.
That is,

∥Pf∥Hs≤Cs∥f∥Hs.\|P f\|_{H^s} \le C_s \|f\|_{H^s}.∥Pf∥Hs≤Cs∥f∥Hs.
===== - The symbol mij(ξ)=δij−ξiξj∣ξ∣2m_{ij}(\xi)=\delta_{ij}-\frac{\xi_i\xi_j}{|\xi|^2}mij(ξ)=δij−∣ξ∣2ξiξj is: - smooth away from ξ=0\xi=0ξ=0, - homogeneous of degree 0, - uniformly bounded with all derivatives. =====
* Such symbols define bounded operators on all HsH^sHs (this is a standard Fourier multiplier theorem).

No regularity is lost or gained by PPP.

==== The Navier–Stokes nonlinearity in mild form is ====

F(s)=−P ∇⋅(u⊗u)(s).F(s) = -P\,\nabla\cdot (u\otimes u)(s).F(s)=−P∇⋅(u⊗u)(s).
Let us track regularity step by step.

===== Assume =====

u(s)∈Hs,s>32.u(s)\in H^s,\qquad s>\tfrac32.u(s)∈Hs,s>23.
Then:
* HsH^sHs is a Banach algebra,
* hence u⊗u∈Hs,∥u⊗u∥Hs≲∥u∥Hs2.u\otimes u \in H^s, \qquad \|u\otimes u\|_{H^s} \lesssim \|u\|_{H^s}^2.u⊗u∈Hs,∥u⊗u∥Hs≲∥u∥Hs2.

===== ∇⋅(u⊗u)∈Hs−1.\nabla\cdot (u\otimes u)\in H^{s-1}.∇⋅(u⊗u)∈Hs−1. =====
===== Using boundedness of PPP: =====

P∇⋅(u⊗u)∈Hs−1,P\nabla\cdot (u\otimes u)\in H^{s-1},P∇⋅(u⊗u)∈Hs−1,
with

∥P∇⋅(u⊗u)∥Hs−1  ≤  C∥∇⋅(u⊗u)∥Hs−1.\|P\nabla\cdot (u\otimes u)\|_{H^{s-1}}
\;\le\;
C\|\nabla\cdot (u\otimes u)\|_{H^{s-1}}.∥P∇⋅(u⊗u)∥Hs−1≤C∥∇⋅(u⊗u)∥Hs−1.
So PPP does not change the Sobolev index.

==== This is trivial but worth stating cleanly: ====

If s>32s>\tfrac32s>23, then s−1>12>0s-1>\tfrac12>0s−1>21>0, and by definition

Hs−1⊂L2with continuous embedding.H^{s-1}\subset L^2
\quad\text{with continuous embedding.}Hs−1⊂L2with continuous embedding.
Thus:

F(s)=−P∇⋅(u⊗u)(s)∈L2.F(s)= -P\nabla\cdot(u\otimes u)(s)\in L^2.F(s)=−P∇⋅(u⊗u)(s)∈L2.

==== Since: ====
* u∈C([0,T];Hs)u\in C([0,T];H^s)u∈C([0,T];Hs),
* the map u↦P∇⋅(u⊗u)u\mapsto P\nabla\cdot(u\otimes u)u↦P∇⋅(u⊗u) is continuous Hs→Hs−1H^s\to H^{s-1}Hs→Hs−1,

we obtain:

F∈C([0,T];Hs−1)  ⊂  L1(0,T;L2).F\in C([0,T];H^{s-1})
\;\subset\;
L^1(0,T;L^2).F∈C([0,T];Hs−1)⊂L1(0,T;L2).
This is the precise justification needed for the Duhamel term.

==== : ====

Nothing is hidden here — PPP is completely harmless at this level.

==== - Temam, Navier–Stokes Equations, Ch. I–II ====
* Constantin & Foias, Navier–Stokes Equations, §2
* Taylor, Pseudodifferential Operators, multiplier theory
* Grafakos, Classical Fourier Analysis, Fourier multipliers

If you want, next I can:
* show the boundedness of PPP on HsH^sHs without Fourier transform, using singular integrals,
* or explain how PPP interacts with the heat semigroup etΔe^{t\Delta}etΔ,
* or track the same argument in Besov spaces (critical case).

Just tell me which.