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

=== Assistant: What you are missing is not a property of the heat semigroup, but a minimal integrability requirement needed to even de… ===
What you are missing is not a property of the heat semigroup, but a minimal integrability requirement needed to even define the Duhamel term as a Bochner integral and to extract smoothing from it.

I will give a clean, structured answer.

==== In the mild formulation, the forcing (or nonlinear term) appears as ====

TF(t)  :=  ∫0te(t−s)ΔF(s) ds,\mathcal{T}F(t)
\;:=\;
\int_0^t e^{(t-s)\Delta} F(s)\,ds,TF(t):=∫0te(t−s)ΔF(s)ds,
where F(s)F(s)F(s) takes values in some Banach space XXX (e.g. L2L^2L2, Hs−1H^{s-1}Hs−1, etc.).

To talk about smoothing, this integral must first be well-defined.

==== ### ====

If XXX is a Banach space, the integral

∫0tF(s) ds∈X\int_0^t F(s)\,ds \in X∫0tF(s)ds∈X
is defined if

F∈L1(0,t;X).F \in L^1(0,t;X).F∈L1(0,t;X).
This is not optional:
* without L1L^1L1-integrability, the integral may not exist as an XXX-valued object,
* weaker conditions (e.g. pointwise bounds) do not suffice.

So the first reason we require F∈L1F\in L^1F∈L1 is purely foundational.

==== Assume: ====

F∈L1(0,T;L2).F \in L^1(0,T;L^2).F∈L1(0,T;L2).
Use the standard heat semigroup estimate:

∥∇ke(t−s)Δg∥L2  ≤  C(t−s)−k/2∥g∥L2,t>s.\|\nabla^k e^{(t-s)\Delta} g\|_{L^2}
\;\le\;
C (t-s)^{-k/2}\|g\|_{L^2},
\quad t>s.∥∇ke(t−s)Δg∥L2≤C(t−s)−k/2∥g∥L2,t>s.
Apply this pointwise in time:

∥∇kTF(t)∥L2  ≤  ∫0t(t−s)−k/2∥F(s)∥L2 ds.\|\nabla^k \mathcal{T}F(t)\|_{L^2}
\;\le\;
\int_0^t (t-s)^{-k/2}\|F(s)\|_{L^2}\,ds.∥∇kTF(t)∥L2≤∫0t(t−s)−k/2∥F(s)∥L2ds.
This integral is finite provided:
* F(s)∈L1(0,t;L2)F(s)\in L^1(0,t;L^2)F(s)∈L1(0,t;L2),
* and the kernel (t−s)−k/2(t-s)^{-k/2}(t−s)−k/2 is locally integrable on (0,t)(0,t)(0,t) (true for fixed t>0t>0t>0).

Thus:

TF(t)∈Hkfor every k,  t>0.\mathcal{T}F(t)\in H^k \quad \text{for every } k,\; t>0.TF(t)∈Hkfor every k,t>0.
This is the precise mechanism of smoothing.

==== If F∉L1(0,t;L2)F\notin L^1(0,t;L^2)F∈/L1(0,t;L2): ====
* the Bochner integral may not exist;
* even if a distributional integral exists, the estimate above cannot be justified;
* the singular kernel (t−s)−k/2(t-s)^{-k/2}(t−s)−k/2 may amplify the non-integrability of FFF.

In particular:
* continuity of F(s)F(s)F(s) in time is not sufficient;
* boundedness of F(s)F(s)F(s) in space is not sufficient.

Time-integrability is essential.

==== In Navier–Stokes: ====

F(s)=−P∇⋅(u⊗u)(s).F(s) = -P\nabla\cdot(u\otimes u)(s).F(s)=−P∇⋅(u⊗u)(s).
For local mild solutions with

u∈C([0,T];Hs),s>32,u \in C([0,T];H^s), \quad s>\tfrac32,u∈C([0,T];Hs),s>23,
we have:
* HsH^sHs is a Banach algebra,
* hence u⊗u∈Hsu\otimes u \in H^su⊗u∈Hs,
* therefore F(s)∈Hs−1⊂L2F(s)\in H^{s-1}\subset L^2F(s)∈Hs−1⊂L2,
* and continuity in time implies F∈L1(0,T;L2).F \in L^1(0,T;L^2).F∈L1(0,T;L2).

This is why the local theory works.

==== Although: ====
* F∈L1(0,T;L2)F\in L^1(0,T;L^2)F∈L1(0,T;L2) holds locally,
* smoothing follows for each fixed t>0t>0t>0,

you do not get a uniform bound as t→T∗t\to T^*t→T∗, because:

∥F(s)∥L2∼∥u(s)∥Hs2\|F(s)\|_{L^2} \sim \|u(s)\|_{H^s}^2∥F(s)∥L2∼∥u(s)∥Hs2
may blow up, destroying L1L^1L1-integrability near T∗T^*T∗.

Thus:
* local smoothing ≠ global regularity,
* and this is exactly where the Clay problem lives.

==== The heat semigroup smooths the argument because: ====
# The Duhamel term is a Bochner integral.
# Bochner integrability requires F∈L1F\in L^1F∈L1 in time.
# Heat kernel estimates convert L1L^1L1-in-time forcing into spatial smoothing.
# Loss of L1L^1L1-integrability is precisely the mechanism by which blowup can occur.

If you want, next I can:
* prove the Bochner-integrability claim rigorously,
* show how the Picard iteration guarantees F∈L1F\in L^1F∈L1 locally,
'' or show explicitly how loss of L1L^1L1 near T∗T^''T∗ obstructs smoothing.

Just say which.