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

=== Assistant: This is a very good question, because it goes to the heart of what is actually proved by heat-kernel estimates and what… ===
This is a very good question, because it goes to the heart of what is actually proved by heat-kernel estimates and what is not.

I will answer in a strictly logical, step-by-step way and point out the exact misunderstanding.

==== You wrote (in essence): ====

∥∇kGtf∥L2  ≤  C t−k/2 ∥f∥L2,t>0,\|\nabla^k G_t f\|_{L^2} \;\le\; C\, t^{-k/2}\,\|f\|_{L^2},
\qquad t>0,∥∇kGtf∥L2≤Ct−k/2∥f∥L2,t>0,
where Gt=etΔG_t = e^{t\Delta}Gt=etΔ is the heat semigroup.

This estimate is true and standard.

==== ### ====

For any fixed t>0t>0t>0:
* ∇kGtf∈L2\nabla^k G_t f \in L^2∇kGtf∈L2 for every kkk,
* hence Gtf∈HkG_t f \in H^kGtf∈Hk for all kkk,
* therefore Gtf∈C∞G_t f \in C^\inftyGtf∈C∞.

So yes:

: 

No Fourier transform is logically necessary for this conclusion.

===== It does not give: =====
* uniform control as t→0+t\to 0^+t→0+,
* continuity at t=0t=0t=0 in higher norms,
* or boundedness of derivatives independent of time.

In fact,

∥∇kGtf∥L2∼t−k/2blows up as t↓0.\|\nabla^k G_t f\|_{L^2} \sim t^{-k/2} \quad \text{blows up as } t\downarrow 0.∥∇kGtf∥L2∼t−k/2blows up as t↓0.
So:
* smoothing is instantaneous,
* but singular at t=0t=0t=0.

This is crucial in Navier–Stokes.

==== You are correct that once the estimate is known, Fourier transform is unnecessary. ====

But Fourier analysis is used to:
# Prove the estimate ∥∇kGtf∥L2≤Ct−k/2∥f∥L2\|\nabla^k G_t f\|_{L^2} \le C t^{-k/2}\|f\|_{L^2}∥∇kGtf∥L2≤Ct−k/2∥f∥L2 in a clean, one-line way: Gtf^(ξ)=e−t∣ξ∣2f^(ξ),∇kGtf^=(iξ)ke−t∣ξ∣2f^,\widehat{G_t f}(\xi)=e^{-t|\xi|^2}\hat f(\xi), \quad \widehat{\nabla^k G_t f}=(i\xi)^k e^{-t|\xi|^2}\hat f,Gtf(ξ)=e−t∣ξ∣2f^(ξ),∇kGtf=(iξ)ke−t∣ξ∣2f^, then apply Plancherel.
# Show exact scaling t−k/2t^{-k/2}t−k/2, which is critical for: - Picard iteration, - Duhamel estimates, - blowup analysis.
# Control nonlinear interactions frequency-by-frequency, which physical-space estimates alone cannot do cleanly.

So Fourier transform is not philosophically essential, but technically indispensable for sharp results.

==== The misunderstanding is not about smoothing. ====

It is here:

: 

That statement is false without fixing time.

Correct version:
* For each fixed t>0t>0t>0, all derivatives are bounded.
* But as t→0t\to 0t→0, higher derivatives blow up like t−k/2t^{-k/2}t−k/2.

So you do not get:

sup⁡t∈(0,T)∥∇kGtf∥L2<∞\sup_{t\in(0,T)}\|\nabla^k G_t f\|_{L^2}<\inftyt∈(0,T)sup∥∇kGtf∥L2<∞
unless f∈Hkf\in H^kf∈Hk already.

This is why:
* smoothing does not prevent blowup,
* energy bounds do not control higher derivatives,
* Clay problem is not trivialized.

==== In Navier–Stokes: ====

u(t)=Gtu0−∫0tGt−sP∇⋅(u⊗u)(s) ds.u(t)=G_t u_0 - \int_0^t G_{t-s}P\nabla\cdot(u\otimes u)(s)\,ds.u(t)=Gtu0−∫0tGt−sP∇⋅(u⊗u)(s)ds.
Even though:
* each Gt−sG_{t-s}Gt−s smooths infinitely for s<ts<ts<t,

the singular factor:

(t−s)−k/2(t-s)^{-k/2}(t−s)−k/2
appears inside the integral.

To control higher derivatives globally, you must control:

∫0t(t−s)−k/2∥u⊗u∥L2(s) ds,\int_0^t (t-s)^{-k/2}\|u\otimes u\|_{L^2}(s)\,ds,∫0t(t−s)−k/2∥u⊗u∥L2(s)ds,
which fails without strong a-priori bounds.

This is exactly where blowup may occur.

==== - ✔ Your estimate is enough to prove infinite smoothing for t>0t>0t>0. ====
* ✘ It does not give uniform bounds near t=0t=0t=0.
* ✘ It does not control nonlinear growth.
* Fourier transform is mainly a tool to prove and sharpen these estimates, not a logical necessity once they are known.

If you want, next I can:
* show precisely how the nonlinear term breaks these bounds,
* or explain why smoothing + energy inequality still allows blowup,
* or rewrite the Picard iteration showing exactly where t−k/2t^{-k/2}t−k/2 kills uniform control.

Just say which one.