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

=== Assistant: Good catch — you’re right that continuity of ututut alone does not imply continuity of ∇ ⁣⋅u⊗ut\nabla\!\cdotu\otimes ut… ===
Good catch — you’re right that continuity of u(t)u(t)u(t) alone does not imply continuity of ∇ ⁣⋅(u⊗u)(t)\nabla\!\cdot(u\otimes u)(t)∇⋅(u⊗u)(t) in the same norm, so you cannot simply assert the integrand

s↦Gt−s P∇ ⁣⋅(u⊗u)(s)s\mapsto G_{t-s}\,P\nabla\!\cdot\big(u\otimes u\big)(s)s↦Gt−sP∇⋅(u⊗u)(s)
is continuous just from u∈C([0,T];X)u\in C([0,T];X)u∈C([0,T];X). That objection is fixable in two standard ways; I’ll give the precise assumptions and a short, rigorous lemma you can use in the Picard / mild-solution argument.

Two ways to salvage the step
# Work in a space XXX with enough spatial regularity so that the nonlinear mapping u↦P∇ ⁣⋅(u⊗u)u\mapsto P\nabla\!\cdot(u\otimes u)u↦P∇⋅(u⊗u) is continuous C([0,T];X)→C([0,T];Y)C([0,T];X)\to C([0,T];Y)C([0,T];X)→C([0,T];Y) for a suitable YYY. For Navier–Stokes the classical choice is X=HsX=H^sX=Hs, s>32s>\tfrac32s>23. Then u∈C([0,T];Hs)u\in C([0,T];H^s)u∈C([0,T];Hs) implies u⊗u∈C([0,T];Hs)u\otimes u\in C([0,T];H^s)u⊗u∈C([0,T];Hs) (because HsH^sHs is an algebra when s>3/2s>3/2s>3/2), hence ∇ ⁣⋅(u⊗u)∈C([0,T];Hs−1)\nabla\!\cdot(u\otimes u)\in C([0,T];H^{s-1})∇⋅(u⊗u)∈C([0,T];Hs−1). With this you get time-continuity of the integrand away from the endpoint s=ts=ts=t; near s=ts=ts=t the semigroup factor Gt−sG_{t-s}Gt−s may create an integrable singularity but the Bochner integral still exists.
# Use Bochner integration instead of pointwise-in-time classical integrals. The Bochner integral requires only that the integrand be strongly (Bochner) measurable and norm-integrable on [0,t][0,t][0,t]. Continuity is not necessary. For the mild integral it is enough to show the integrand is in L1([0,t];X)L^1([0,t];X)L1([0,t];X) (as an XXX-valued function).

Below is a compact lemma combining both ideas and giving the exact estimates you need.

===== Let s>3/2s>3/2s>3/2. Suppose =====

u∈C([0,T];Hs(R3)).u\in C([0,T];H^{s}(\mathbb{R}^3)).u∈C([0,T];Hs(R3)).
Define the bilinear map B(u,v)=P∇ ⁣⋅(u⊗v)B(u,v)=P\nabla\!\cdot(u\otimes v)B(u,v)=P∇⋅(u⊗v). Then for every fixed t∈(0,T]t\in(0,T]t∈(0,T] the HsH^sHs-valued function

Φ(s):=Gt−s B(u(s),u(s)),s∈[0,t],\Phi(s):=G_{t-s}\,B\big(u(s),u(s)\big),\qquad s\in[0,t],Φ(s):=Gt−sB(u(s),u(s)),s∈[0,t],
is strongly measurable on [0,t][0,t][0,t] and belongs to L1([0,t];Hs(R3))L^1([0,t];H^s(\mathbb{R}^3))L1([0,t];Hs(R3)). Moreover there is a constant CCC (depending on ν,s\nu,sν,s) such that for a.e. s∈[0,t]s\in[0,t]s∈[0,t]

∥Φ(s)∥Hs≤C (t−s)−1/2 ∥u(s)∥Hs2.\|\Phi(s)\|_{H^s}
\le C\,(t-s)^{-1/2}\,\|u(s)\|_{H^s}^2.∥Φ(s)∥Hs≤C(t−s)−1/2∥u(s)∥Hs2.
In particular Φ\PhiΦ is Bochner-integrable on [0,t][0,t][0,t] since ∫0t(t−s)−1/2 ds<∞\int_0^t (t-s)^{-1/2}\,ds<\infty∫0t(t−s)−1/2ds<∞.

====== 1. Product estimate (algebra property). For s>3/2s>3/2s>3/2 we have the continuous product embedding Hs×Hs→HsH^s\times H^s\to H^sHs×Hs→Hs. Hence ======

∥u(s)⊗u(s)∥Hs≤C1∥u(s)∥Hs2.\|u(s)\otimes u(s)\|_{H^s} \le C_1 \|u(s)\|_{H^s}^2.∥u(s)⊗u(s)∥Hs≤C1∥u(s)∥Hs2.
Applying ∇ ⁣⋅\nabla\!\cdot∇⋅ reduces regularity by one:

∥B(u,u)(s)∥Hs−1=∥P∇ ⁣⋅(u⊗u)(s)∥Hs−1≤C2∥u(s)∥Hs2,\|B(u,u)(s)\|_{H^{s-1}} = \|P\nabla\!\cdot(u\otimes u)(s)\|_{H^{s-1}}
\le C_2 \|u(s)\|_{H^s}^2,∥B(u,u)(s)∥Hs−1=∥P∇⋅(u⊗u)(s)∥Hs−1≤C2∥u(s)∥Hs2,
because PPP is bounded on Sobolev spaces.
# Heat-semigroup gain estimate. The heat semigroup satisfies the smoothing bound

∥Gτf∥Hs≤C3 τ−1/2 ∥f∥Hs−1,τ>0,\|G_\tau f\|_{H^{s}} \le C_3\,\tau^{-1/2}\,\|f\|_{H^{s-1}},\qquad \tau>0,∥Gτf∥Hs≤C3τ−1/2∥f∥Hs−1,τ>0,
which is the r=1r=1r=1 case of the general estimate ∥Gτf∥Hs+r≤Cτ−r/2∥f∥Hs\|G_\tau f\|_{H^{s+r}}\le C\tau^{-r/2}\|f\|_{H^s}∥Gτf∥Hs+r≤Cτ−r/2∥f∥Hs.
# Combine the two with τ=t−s\tau=t-sτ=t−s:

∥Φ(s)∥Hs=∥Gt−sB(u,u)(s)∥Hs≤C3(t−s)−1/2∥B(u,u)(s)∥Hs−1≤C (t−s)−1/2∥u(s)∥Hs2.\|\Phi(s)\|_{H^s} = \|G_{t-s} B(u,u)(s)\|_{H^s}
\le C_3 (t-s)^{-1/2} \|B(u,u)(s)\|_{H^{s-1}}
\le C\,(t-s)^{-1/2}\|u(s)\|_{H^s}^2.∥Φ(s)∥Hs=∥Gt−sB(u,u)(s)∥Hs≤C3(t−s)−1/2∥B(u,u)(s)∥Hs−1≤C(t−s)−1/2∥u(s)∥Hs2.
# Since u∈C([0,T];Hs)u\in C([0,T];H^s)u∈C([0,T];Hs), the function s↦∥u(s)∥Hs2s\mapsto \|u(s)\|_{H^s}^2s↦∥u(s)∥Hs2 is bounded on [0,t][0,t][0,t]. Therefore the right-hand side is dominated by a constant times (t−s)−1/2(t-s)^{-1/2}(t−s)−1/2, which is integrable on [0,t][0,t][0,t]. This proves Bochner integrability.

Strong measurability follows from continuity of the composition s↦B(u,u)(s)s\mapsto B(u,u)(s)s↦B(u,u)(s) into Hs−1H^{s-1}Hs−1 and continuity of the map s↦Gt−ss\mapsto G_{t-s}s↦Gt−s into bounded operators on Hs−1H^{s-1}Hs−1 for s<ts<ts<t; standard measurability results then imply Φ\PhiΦ is strongly measurable on [0,t][0,t][0,t].

∎

===== - No need for pointwise continuity of ∇⋅(u⊗u)\nabla\cdot(u\otimes u)∇⋅(u⊗u) at s=ts=ts=t. The integrand may blow up in operator norm as s→t−s\to t^-s→t− (because of the (t−s)−1/2(t-s)^{-1/2}(t−s)−1/2 factor), but this is an integrable singularity; the Bochner integral still exists. Thus the Duhamel integral ∫0tGt−sB(u(s),u(s)) ds\int_0^t G_{t-s} B(u(s),u(s))\,ds∫0tGt−sB(u(s),u(s))ds is well-defined as a Bochner integral in HsH^sHs. =====
* If you prefer genuine continuity of the integrand: impose stronger time-regularity on uuu. For instance u∈Cα([0,T];Hs)u\in C^\alpha([0,T];H^s)u∈Cα([0,T];Hs) for some α>0\alpha>0α>0 often yields better behavior of B(u,u)B(u,u)B(u,u) in time; or if you require u∈C([0,T];Hs+1)u\in C([0,T];H^{s+1})u∈C([0,T];Hs+1) then ∇⋅(u⊗u)∈C([0,T];Hs)\nabla\cdot(u\otimes u)\in C([0,T];H^{s})∇⋅(u⊗u)∈C([0,T];Hs), and semigroup smoothing is no longer needed to obtain continuity in HsH^sHs.
* How this fits into Picard: the contraction estimates use exactly the integrability estimate above. You do not need continuity of the integrand; you only need the operator norm estimates to show T\mathcal TT maps a ball in C([0,T];Hs)C([0,T];H^s)C([0,T];Hs) into itself and is Lipschitz for small TTT. The Bochner-integral formulation + the estimate ∫0t(t−s)−1/2 ds=2t\int_0^t (t-s)^{-1/2}\,ds = 2\sqrt{t}∫0t(t−s)−1/2ds=2t provides the required smallness factor in time.

===== 1. Work in X=HsX=H^sX=Hs with s>3/2s>3/2s>3/2. =====
# Define mild solutions via the Bochner integral (Duhamel formula in C([0,T];Hs)C([0,T];H^s)C([0,T];Hs)).
# Use the estimate above to show the Duhamel integral is well-defined and to run the Picard contraction (small TTT gives small factor ∝T1/2\propto T^{1/2}∝T1/2).
# You do not need the integrand to be continuous in time everywhere — Bochner integrability and the smoothing estimate suffice.

If you want, I can now (a) write the Picard contraction proof in full using the precise constants from the lemma, or (b) show the variant when you work in critical Besov spaces (if you aim for a more delicate existence theory). Which do you prefer?