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

=== Assistant: Short answer up front: you prove the mild Duhamel expression is well-defined by placing the extended space inside a Ban… ===
Short answer up front: you prove the mild (Duhamel) expression is well-defined by placing the extended space inside a Banach algebra (or giving it Banach-type axioms) so you can use Bochner integration + the heat-semigroup smoothing estimate. Below I state a minimal list of explicit, checkable hypotheses and then give the exact estimate/argument that shows the Duhamel integral

∫0teν(t−s)Δ P∇ ⁣⋅(u(s)⊗u(s)) ds\int_0^t e^{\nu (t-s)\Delta}\,P\nabla\!\cdot\big(u(s)\otimes u(s)\big)\,ds∫0teν(t−s)ΔP∇⋅(u(s)⊗u(s))ds
is a well-defined element of the space for every t>0t>0t>0 and every uuu in the solution class. I finish with the alternative conditional route if you insist on working in a nonnormed “extended” algebra with a linear limit functional LLL.

==== Fix a Banach space (X,∥⋅∥X)(X,\|\cdot\|_X)(X,∥⋅∥X) of vector fields on R3\mathbb{R}^3R3 with these properties: ====

(H1) Heat semigroup bounds. There exist constants CG>0C_G>0CG>0 and α∈[0,1)\alpha\in[0,1)α∈[0,1) such that for all τ∈(0,T]\tau\in(0,T]τ∈(0,T]

∥eντΔ∥X←Y≤CG τ−α,\|e^{\nu \tau\Delta}\|_{X\leftarrow Y} \le C_G\,\tau^{-\alpha},∥eντΔ∥X←Y≤CGτ−α,
for some Banach space YYY with a continuous embedding Y↪XY\hookrightarrow XY↪X. (Often Y=X−1Y=X_{-1}Y=X−1 or Hs−1H^{s-1}Hs−1 while X=HsX=H^sX=Hs; typical α=1/2\alpha=1/2α=1/2.)

(H2) Product estimate (algebraic control). There is a constant CBC_BCB such that the bilinear map

B(u,v):=P∇ ⁣⋅(u⊗v)B(u,v) := P\nabla\!\cdot(u\otimes v)B(u,v):=P∇⋅(u⊗v)
satisfies

∥B(u,v)∥Y≤CB∥u∥X∥v∥Xfor all u,v∈X.\|B(u,v)\|_{Y} \le C_B \|u\|_{X}\|v\|_{X}\qquad\text{for all }u,v\in X.∥B(u,v)∥Y≤CB∥u∥X∥v∥Xfor all u,v∈X.
(H3) XXX-continuity in time. We consider functions u∈C([0,T];X)u\in C([0,T];X)u∈C([0,T];X). (You can weaken this, but this is the standard mild-solution class.)

These three hypotheses are exactly the ones used in standard mild/Picard proofs (for example take X=HsX=H^sX=Hs, s>3/2s>3/2s>3/2, Y=Hs−1Y=H^{s-1}Y=Hs−1, then α=1/2\alpha=1/2α=1/2 and (H2) holds).

==== Under (H1)–(H3) show that for any u∈C([0,T];X)u\in C([0,T];X)u∈C([0,T];X) and any fixed t∈(0,T]t\in(0,T]t∈(0,T] the XXX-valued function ====

Φu(s):=eν(t−s)Δ B(u(s),u(s)),s∈[0,t],\Phi_u(s) := e^{\nu (t-s)\Delta}\,B\big(u(s),u(s)\big),\qquad s\in[0,t],Φu(s):=eν(t−s)ΔB(u(s),u(s)),s∈[0,t],
is Bochner integrable on [0,t][0,t][0,t] and therefore the Duhamel term

I(t):=∫0tΦu(s) ds\mathcal{I}(t):=\int_0^t \Phi_u(s)\,dsI(t):=∫0tΦu(s)ds
exists in XXX. Moreover t↦I(t)t\mapsto \mathcal I(t)t↦I(t) is continuous in XXX and hence the mild formula defines an element of C([0,T];X)C([0,T];X)C([0,T];X).

==== Because u∈C([0,T];X)u\in C([0,T];X)u∈C([0,T];X), the real function m:=sup⁡0≤s≤t∥u(s)∥Xm:=\sup_{0\le s\le t}\|u(s)\|_Xm:=sup0≤s≤t∥u(s)∥X is finite. ====

Using (H2) and (H1) we have for each s∈[0,t)s\in[0,t)s∈[0,t)

∥Φu(s)∥X=∥eν(t−s)Δ B(u(s),u(s))∥X≤∥eν(t−s)Δ∥X←Y ∥B(u(s),u(s))∥Y≤CG (t−s)−α CB ∥u(s)∥X2≤C (t−s)−α,C:=CGCB m2.\begin{aligned}
\|\Phi_u(s)\|_X
&= \big\|e^{\nu (t-s)\Delta} \, B(u(s),u(s))\big\|_X\\[4pt]
&\le \|e^{\nu (t-s)\Delta}\|_{X\leftarrow Y}\,\|B(u(s),u(s))\|_Y\\[4pt]
&\le C_G\,(t-s)^{-\alpha}\,C_B\,\|u(s)\|_X^2\\[4pt]
&\le C\, (t-s)^{-\alpha},\qquad C:=C_G C_B\, m^2.
\end{aligned}∥Φu(s)∥X=eν(t−s)ΔB(u(s),u(s))X≤∥eν(t−s)Δ∥X←Y∥B(u(s),u(s))∥Y≤CG(t−s)−αCB∥u(s)∥X2≤C(t−s)−α,C:=CGCBm2.
Because α<1\alpha<1α<1 by hypothesis, the function s↦(t−s)−αs\mapsto (t-s)^{-\alpha}s↦(t−s)−α is integrable on [0,t][0,t][0,t]. Therefore Φu∈L1([0,t];X)\Phi_u\in L^1([0,t];X)Φu∈L1([0,t];X), so the Bochner integral I(t)\mathcal I(t)I(t) exists as an element of XXX.

==== Fix t∈(0,T]t\in(0,T]t∈(0,T]. Write ====

I(t)=∫0teν(t−s)ΔB(u(s),u(s)) ds.\mathcal I(t) = \int_0^t e^{\nu (t-s)\Delta} B(u(s),u(s))\,ds.I(t)=∫0teν(t−s)ΔB(u(s),u(s))ds.
Standard dominated-convergence arguments for Bochner integrals give continuity of t↦I(t)t\mapsto\mathcal I(t)t↦I(t) provided:
* the integrand is strongly measurable in sss, (true here because u(s)u(s)u(s) is continuous into XXX and the composition with bounded linear maps is continuous), and
* there is an integrable dominating function independent of small perturbations of ttt. The estimate above provides such a uniform integrable bound ∝(t−s)−α\propto(t-s)^{-\alpha}∝(t−s)−α when ttt ranges over a compact subinterval away from 000. One checks the contribution near the moving endpoint is small uniformly.

A standard calculation (write difference I(t+h)−I(t)\mathcal I(t+h)-\mathcal I(t)I(t+h)−I(t) and split the integral) shows I(⋅)\mathcal I(\cdot)I(⋅) is continuous on (0,T](0,T](0,T]; continuity at t=0t=0t=0 follows from small-time smoothing estimates and smallness of the integral on small intervals. Therefore I∈C([0,T];X)\mathcal I\in C([0,T];X)I∈C([0,T];X).

==== Define the mild operator ====

(Tu)(t):=eνtΔu0−∫0teν(t−s)ΔB(u(s),u(s)) ds.(\mathcal T u)(t) := e^{\nu t\Delta}u_0 - \int_0^t e^{\nu (t-s)\Delta} B(u(s),u(s))\,ds.(Tu)(t):=eνtΔu0−∫0teν(t−s)ΔB(u(s),u(s))ds.
From the previous steps we conclude that for every u∈C([0,T];X)u\in C([0,T];X)u∈C([0,T];X) the right-hand side is an element of C([0,T];X)C([0,T];X)C([0,T];X). Hence T\mathcal TT maps C([0,T];X)C([0,T];X)C([0,T];X) into itself; i.e. the Duhamel (mild) expression is well-defined in this Banach setting.

To run Picard you further show T\mathcal TT is a contraction on a small time interval (use the same convolution estimate with the factor t1−αt^{1-\alpha}t1−α coming from the time integral); that gives existence and uniqueness of a fixed point in C([0,T];X)C([0,T];X)C([0,T];X).

==== Take X=Hs(R3)X=H^s(\mathbb{R}^3)X=Hs(R3) with s>3/2s>3/2s>3/2, Y=Hs−1Y=H^{s-1}Y=Hs−1. Then: ====
* The heat semigroup bound (H1) holds with α=1/2\alpha=1/2α=1/2: ∥eντΔ∥Hs←Hs−1≲τ−1/2\|e^{\nu \tau\Delta}\|_{H^s\leftarrow H^{s-1}}\lesssim \tau^{-1/2}∥eντΔ∥Hs←Hs−1≲τ−1/2.
* The product estimate (H2) holds because HsH^sHs is an algebra for s>3/2s>3/2s>3/2: ∥u⊗v∥Hs≲∥u∥Hs∥v∥Hs\|u\otimes v\|_{H^{s}} \lesssim \|u\|_{H^s}\|v\|_{H^s}∥u⊗v∥Hs≲∥u∥Hs∥v∥Hs, so ∥∇⋅(u⊗v)∥Hs−1≲∥u∥Hs∥v∥Hs\|\nabla\cdot(u\otimes v)\|_{H^{s-1}}\lesssim\|u\|_{H^s}\|v\|_{H^s}∥∇⋅(u⊗v)∥Hs−1≲∥u∥Hs∥v∥Hs.

Plugging these into the argument above yields the familiar integrability factor (t−s)−1/2(t-s)^{-1/2}(t−s)−1/2, integrable on [0,t][0,t][0,t]. This is the standard textbook setting.

==== If your space is the nonnormed extended algebra EEE and you want to assert the mild expression exists there, you must add precise axioms for EEE. At minimum you need: ====
* A topology (or net/sequence notion) and a Banach-type seminorm ∥⋅∥X\|\cdot\|_X∥⋅∥X on the subspace where the Picard iterates live, or an explicit statement that the semigroup operators eνtΔe^{\nu t\Delta}eνtΔ act and satisfy a bound of the form in (H1).
* Closure of EEE under multiplication with a bilinear estimate like (H2).
* The ability to form Bochner (or primitive-based) integrals in EEE and that your linear limit LLL commutes with these integrals.

Without explicit axioms of this type you cannot rigorously show existence of the Bochner integral in EEE; the claim “the mild equation is well-defined in the extended space” is therefore conditional on supplying those axioms.

==== 1. Find a norm or seminorm ∥⋅∥X\|\cdot\|_X∥⋅∥X making the Picard iterates elements of C([0,T];X)C([0,T];X)C([0,T];X). ====
# Prove product estimate ∥B(u,v)∥Y≲∥u∥X∥v∥X\|B(u,v)\|_{Y}\lesssim\|u\|_X\|v\|_X∥B(u,v)∥Y≲∥u∥X∥v∥X and semigroup bound ∥eντΔ∥X←Y≲τ−α\|e^{\nu\tau\Delta}\|_{X\leftarrow Y}\lesssim\tau^{-\alpha}∥eντΔ∥X←Y≲τ−α with α<1\alpha<1α<1.
# Apply the estimate ∥eν(t−s)ΔB(u,u)(s)∥X≲(t−s)−α∥u(s)∥X2\|e^{\nu(t-s)\Delta}B(u,u)(s)\|_X\lesssim (t-s)^{-\alpha}\|u(s)\|_X^2∥eν(t−s)ΔB(u,u)(s)∥X≲(t−s)−α∥u(s)∥X2 to dominate and integrate.
# Conclude Bochner integrability and continuity in ttt.

If you want, I can now (a) instantiate the whole argument with explicit constants in the HsH^sHs case, or (b) write down the minimal axioms for an extended algebra EEE that would make the above proof go through and examine whether any known generalized-function framework satisfies them. Which do you prefer?