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Spatial derivative estimates for the heat flow

Statement

Assume Countable Choice, let n≥1, 1≤p≤q≤∞, and let Q and the constants be as in Lp to Lq smoothing estimate for the heat flow. For every multi-index α, every f∈Lp(Rn) and every t>0, the function x↦∫RnΓ(x−y,t)f(y) dy is C∞ on Rn, with DαHtf=(DαΓ(⋅,t))∗f as an absolutely convergent integral, and ∥DαHtf∥q≤Cn,p,q,α t−∣α∣2−n2(1p−1q)∥f∥p,Cn,p,q,α:=∥DαΓ(⋅,1)∥Q<∞.

Facts & Assumptions

Given: Countable Choice, n≥1, 1≤p≤q≤∞ with exponent Q determined by 1/Q=1+1/q−1/p, a multi-index α, f∈Lp(Rn) and t>0.

[A1]

Countable Choice is the hypothesis carried by the convolution and integration suppliers below (The Axiom of Countable Choice (ACω)).

[F1]

For every s>0: Γ(⋅,s) is smooth, unit mass holds, Γ(λx,λ2s)=λ−nΓ(x,s) for every λ>0, and DβΓ(x,s)=s−(n+∣β∣)/2(DβΓ)(x/s,1) for every multi-index β; moreover ∣DβΓ(x,1)∣≤Cn,βe−∣x∣2/8 (Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel).

[F2]

Htf has the everywhere-defined absolutely convergent representative u(x,t)=∫RnΓ(x−y,t)f(y) dy, which is C∞ in x, and DαHtf=(DαΓ(⋅,t))∗f for every multi-index α (Spatial and time derivatives pass through heat convolution for positive time).

[F3]

For 1≤p,q,r≤∞ with 1/r=1/p+1/q−1 and f∈Lp, g∈Lq, ∥f∗g∥r≤∥f∥p∥g∥q (Young's convolution inequality under Countable Choice).

[F4]

With Q as in the statement, 1/Q=1+1/q−1/p and the kernel estimate ∥Htf∥q≤Cn,p,qt−n2(1p−1q)∥f∥p holds (Lp to Lq smoothing estimate for the heat flow); the exponent range 1≤p≤q≤∞ makes Q∈[1,∞].

Proof

technique · direct
1.1A1F1F2given

Smoothness and derivative identity: by [F2] the representative u is C∞ in x for t>0 and DαHtf is the class of (DαΓ(⋅,t))∗f, an absolutely convergent integral by the derivative bounds of [F1] together with Hölder and f∈Lp.

2.1step 1.1F1givenalgebra

Scaling of the derivative kernel norms: by the derivative scaling identity of [F1], DαΓ(x,t)=t−(n+∣α∣)/2(DαΓ)(x/t,1); for Q<∞, substituting x=t z gives ∥DαΓ(⋅,t)∥QQ=t−(n+∣α∣)Q/2tn/2∥DαΓ(⋅,1)∥QQ, hence ∥DαΓ(⋅,t)∥Q=t−∣α∣/2t−n2(1−1Q)∥DαΓ(⋅,1)∥Q. For Q=∞, taking essential suprema in the same scaling identity gives ∥DαΓ(⋅,t)∥∞=t−(n+∣α∣)/2∥DαΓ(⋅,1)∥∞, the same formula with 1/Q=0. The constant Cn,p,q,α=∥DαΓ(⋅,1)∥Q is finite because ∣DαΓ(x,1)∣≤Cn,αe−∣x∣2/8 by [F1] and the Gaussian e−∣⋅∣2/8 lies in every LQ, 1≤Q≤∞.

3.1step 1.1step 2.1F3F4givenalgebra

Young estimate for the derivative: applying Young's convolution inequality [F3] with the exponent triple (p,Q,q), admissible because 1/Q=1+1/q−1/p, gives ∥DαHtf∥q≤∥DαΓ(⋅,t)∥Q∥f∥p, and step 2.1 together with the identity 1−1Q=1p−1q from [F4] turns this into ∥DαHtf∥q≤Cn,p,q,αt−∣α∣2−n2(1p−1q)∥f∥p, which is the stated estimate.

4.1step 1.1step 3.1given∎

Steps 1.1, 2.1 and 3.1 prove the smoothness of the representative, the absolutely convergent convolution formula for DαHtf and the displayed derivative estimate with finite constant Cn,p,q,α=∥DαΓ(⋅,1)∥Q.

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