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

=== Assistant: You asked for a proof that the heat semigroup GteνtΔGte^{\nu t\Delta}GteνtΔ is infinitely smoothing. ===
You asked for a proof that the heat semigroup Gt=eνtΔG_t=e^{\nu t\Delta}Gt=eνtΔ is infinitely smoothing. Below I give a compact, rigorous statement and two standard proofs (Fourier / multiplier and kernel / convolution). Both make the same point: for every t>0t>0t>0 the operator GtG_tGt maps very rough data to C∞C^\inftyC∞; moreover one has quantitative smoothing estimates (derivatives gain powers of t−1/2t^{-1/2}t−1/2).

I will work on R3\mathbb{R}^3R3. Fix ν>0\nu>0ν>0. Write Gtf=Kt∗fG_t f = K_t * fGtf=Kt∗f where

Kt(x)=(4πνt)−3/2exp⁡ ⁣(−∣x∣2/(4νt))K_t(x)=(4\pi\nu t)^{-3/2}\exp\!\big(-|x|^2/(4\nu t)\big)Kt(x)=(4πνt)−3/2exp(−∣x∣2/(4νt))
is the heat kernel.

==== For every real sss and every real r≥0r\ge0r≥0 there exists C=C(ν,r)C=C(\nu,r)C=C(ν,r) such that for all t>0t>0t>0 and all fff with finite HsH^{s}Hs-norm, ====

∥Gtf∥Hs+r  ≤  C t−r/2 ∥f∥Hs.\boxed{\qquad
\|G_t f\|_{H^{s+r}} \;\le\; C\,t^{-r/2}\,\|f\|_{H^s}.
\qquad}∥Gtf∥Hs+r≤Ct−r/2∥f∥Hs.
Consequences:
* If f∈Hsf\in H^sf∈Hs (or L2L^2L2, or a distribution), then for every t>0t>0t>0 the function GtfG_t fGtf is C∞C^\inftyC∞ in xxx.
* More concretely, for each multiindex α\alphaα there is CαC_\alphaCα with ∥∂xαGtf∥L2≤Cα t−∣α∣/2 ∥f∥L2.\|\partial_x^\alpha G_t f\|_{L^2} \le C_\alpha\, t^{-|\alpha|/2}\,\|f\|_{L^2}.∥∂xαGtf∥L2≤Cαt−∣α∣/2∥f∥L2.
'' Kernel form: for each multiindex α\alphaα there is CαC_\alphaCα such that ∣∂xαKt(x)∣≤Cα t−(3+∣α∣)/2exp⁡ ⁣(−∣x∣2/(8νt)).|\partial_x^\alpha K_t(x)| \le C_\alpha\, t^{-(3+|\alpha|)/2}\exp\!\big(-|x|^2/(8\nu t)\big).∣∂xαKt(x)∣≤Cαt−(3+∣α∣)/2exp(−∣x∣2/(8νt)). Hence Gtf=Kt∗fG_t f=K_t''fGtf=Kt∗f is C∞C^\inftyC∞ for all t>0t>0t>0 even if fff is merely a tempered distribution.

==== Recall Gtf^(ξ)=e−νt∣ξ∣2f^(ξ)\widehat{G_t f}(\xi)=e^{-\nu t|\xi|^2}\widehat f(\xi)Gtf(ξ)=e−νt∣ξ∣2f(ξ). Using the Sobolev norm definition ====
∥g∥Hm2=∫R3(1+∣ξ∣2)m∣g^(ξ)∣2 dξ\|g\|_{H^m}^2=\int_{\mathbb{R}^3}(1+|\xi|^2)^m|\widehat g(\xi)|^2\,d\xi∥g∥Hm2=∫R3(1+∣ξ∣2)m∣g(ξ)∣2dξ,

∥Gtf∥Hs+r2=∫R3(1+∣ξ∣2)s+r e−2νt∣ξ∣2 ∣f^(ξ)∣2 dξ=∫(1+∣ξ∣2)s [(1+∣ξ∣2)re−2νt∣ξ∣2] ∣f^(ξ)∣2 dξ≤sup⁡ξ∈R3[(1+∣ξ∣2)re−2νt∣ξ∣2]⋅∥f∥Hs2.\begin{aligned}
\|G_t f\|_{H^{s+r}}^2
&= \int_{\mathbb{R}^3} (1+|\xi|^2)^{s+r}\, e^{-2\nu t|\xi|^2}\,|\widehat f(\xi)|^2\,d\xi \\
&= \int (1+|\xi|^2)^{s}\,\big[(1+|\xi|^2)^{r}e^{-2\nu t|\xi|^2}\big]\,|\widehat f(\xi)|^2\,d\xi \\
&\le \sup_{\xi\in\mathbb{R}^3}\big[(1+|\xi|^2)^{r}e^{-2\nu t|\xi|^2}\big]\cdot \|f\|_{H^s}^2 .
\end{aligned}∥Gtf∥Hs+r2=∫R3(1+∣ξ∣2)s+re−2νt∣ξ∣2∣f(ξ)∣2dξ=∫(1+∣ξ∣2)s[(1+∣ξ∣2)re−2νt∣ξ∣2]∣f(ξ)∣2dξ≤ξ∈R3sup[(1+∣ξ∣2)re−2νt∣ξ∣2]⋅∥f∥Hs2.
Estimate the supremum: set ρ=∣ξ∣2\rho=|\xi|^2ρ=∣ξ∣2. The scalar function h(ρ)=(1+ρ)re−2νtρh(\rho)=(1+\rho)^{r}e^{-2\nu t\rho}h(ρ)=(1+ρ)re−2νtρ attains its maximum at some ρ≥0\rho\ge0ρ≥0; elementary calculus (or the bound (1+ρ)r≤Cr(1+ρr)(1+\rho)^r\le C_r(1+\rho^r)(1+ρ)r≤Cr(1+ρr) and the maximum of ρre−2νtρ\rho^r e^{-2\nu t\rho}ρre−2νtρ is attained at ρ∼1/t\rho\sim 1/tρ∼1/t) shows

sup⁡ξ(1+∣ξ∣2)re−2νt∣ξ∣2≤Cr t−r.\sup_{\xi}(1+|\xi|^2)^{r}e^{-2\nu t|\xi|^2} \le C_r\, t^{-r}.ξsup(1+∣ξ∣2)re−2νt∣ξ∣2≤Crt−r.
Therefore

∥Gtf∥Hs+r2≤Cr t−r ∥f∥Hs2,\|G_t f\|_{H^{s+r}}^2 \le C_r\, t^{-r}\, \|f\|_{H^s}^2,∥Gtf∥Hs+r2≤Crt−r∥f∥Hs2,
and taking square roots gives the desired inequality with t−r/2t^{-r/2}t−r/2 factor.

Specializing to s=0s=0s=0 and integer r=∣α∣r=|\alpha|r=∣α∣ gives the derivative estimate

∥∂αGtf∥L2≲t−∣α∣/2∥f∥L2.\|\partial^\alpha G_t f\|_{L^2} \lesssim t^{-|\alpha|/2}\|f\|_{L^2}.∥∂αGtf∥L2≲t−∣α∣/2∥f∥L2.
Since this holds for every rrr, Gtf∈⋂m≥0Hm⊂C∞G_t f\in\bigcap_{m\ge0}H^m\subset C^\inftyGtf∈⋂m≥0Hm⊂C∞ for any t>0t>0t>0.

==== We use explicit derivatives of the Gaussian kernel. ====

The heat kernel is

Kt(x)=(4πνt)−3/2e−∣x∣2/(4νt).K_t(x)=(4\pi\nu t)^{-3/2} e^{-|x|^2/(4\nu t)}.Kt(x)=(4πνt)−3/2e−∣x∣2/(4νt).
Differentiate under xxx. Each spatial derivative ∂xj\partial_{x_j}∂xj produces a polynomial factor in xxx multiplied by 1/(t)1/(t)1/(t) from differentiating the exponential; more precisely for any multiindex α\alphaα there are constants PαP_\alphaPα (polynomials) and CαC_\alphaCα with

∂xαKt(x)=t−(3+∣α∣)/2 Pα ⁣(x/t) e−∣x∣2/(4νt).\partial_x^\alpha K_t(x) = t^{-(3+|\alpha|)/2}\, P_\alpha\!\big(x/\sqrt{t}\big)\, e^{-|x|^2/(4\nu t)}.∂xαKt(x)=t−(3+∣α∣)/2Pα(x/t)e−∣x∣2/(4νt).
From that one obtains the standard Gaussian bound (adjust constants):

∣∂xαKt(x)∣≤Cα t−(3+∣α∣)/2exp⁡ ⁣(−∣x∣2/(8νt)).|\partial_x^\alpha K_t(x)| \le C_\alpha\, t^{-(3+|\alpha|)/2}\exp\!\big(-|x|^2/(8\nu t)\big).∣∂xαKt(x)∣≤Cαt−(3+∣α∣)/2exp(−∣x∣2/(8νt)).
Now let fff be a tempered distribution or an LpL^pLp-function. For any fixed t>0t>0t>0 the function x↦∂xαKt(x)x\mapsto\partial_x^\alpha K_t(x)x↦∂xαKt(x) is in L1(R3)L^1(\mathbb{R}^3)L1(R3) because the Gaussian tail makes the integral finite and the prefactor has only a polynomial singularity in ttt. Thus the convolution ∂xαKt∗f\partial^\alpha_x K_t * f∂xαKt∗f is well-defined as a continuous function whenever fff is (say) L∞L^\inftyL∞ or as a function in appropriate Lebesgue spaces via Young's inequality. In particular if fff is a tempered distribution, convolution with the smooth rapidly decaying function KtK_tKt produces a C∞C^\inftyC∞ function (standard distribution convolution theorem).

Concretely, for f∈L∞f\in L^\inftyf∈L∞,

∥∂αGtf∥L∞=∥∂αKt∗f∥L∞≤∥∂αKt∥L1 ∥f∥L∞≲t−∣α∣/2∥f∥L∞.\|\partial^\alpha G_t f\|_{L^\infty} = \|\partial^\alpha K_t * f\|_{L^\infty}
\le \|\partial^\alpha K_t\|_{L^1}\,\|f\|_{L^\infty}
\lesssim t^{-|\alpha|/2}\|f\|_{L^\infty}.∥∂αGtf∥L∞=∥∂αKt∗f∥L∞≤∥∂αKt∥L1∥f∥L∞≲t−∣α∣/2∥f∥L∞.
For f∈Lpf\in L^pf∈Lp combine ∥∂αKt∥L1\|\partial^\alpha K_t\|_{L^1}∥∂αKt∥L1 or other Young-type bounds to get analogous estimates. Thus for every t>0t>0t>0 the function GtfG_t fGtf has derivatives of all orders; i.e. Gtf∈C∞(R3)G_t f\in C^\infty(\mathbb{R}^3)Gtf∈C∞(R3).

==== - The factor t−∣α∣/2t^{-|\alpha|/2}t−∣α∣/2 is natural: roughly speaking, differentiating a Gaussian of width ∼t\sim\sqrt{t}∼t produces a factor (t)−∣α∣(\sqrt{t})^{-|\alpha|}(t)−∣α∣. ====
* The above shows instant regularization: even if fff is only an L2L^2L2-function (or distribution), for any fixed t>0t>0t>0 the result GtfG_t fGtf is smooth in space. Time-regularity at t=0t=0t=0 is weaker (there is a singularity as t↓0t\downarrow0t↓0 in the derivative estimates).
* The Fourier and kernel proofs are two faces of the same phenomenon: the multiplier e−νt∣ξ∣2e^{-\nu t|\xi|^2}e−νt∣ξ∣2 decays rapidly for large ∣ξ∣|\xi|∣ξ∣ (cutting off high frequencies), and convolution with a Gaussian smooths and spreads singularities.

==== For the Navier–Stokes Picard setup you often need the particular case sss and r=1r=1r=1: ====

∥eνtΔf∥Hs+1≤C t−1/2 ∥f∥Hs,\|e^{\nu t\Delta} f\|_{H^{s+1}} \le C\, t^{-1/2}\,\|f\|_{H^s},∥eνtΔf∥Hs+1≤Ct−1/2∥f∥Hs,
which is exactly the heat-gain used to control the Duhamel integrand by a factor (t−s)−1/2(t-s)^{-1/2}(t−s)−1/2.

If you want, I can now:
* give a fully worked numeric constant for the HsH^sHs estimate (evaluate the sup⁡ρ\sup_\rhosupρ bound explicitly), or
* show how the same smoothing works on the torus T3\mathbb{T}^3T3 (Fourier series version), or
* derive the pointwise-in-space bound for ∂αKt\partial^\alpha K_t∂αKt with explicit polynomial PαP_\alphaPα. Which of these would you like?