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Heat generator at zero on compactly supported smooth data

Statement

Assume Countable Choice. For n≥1, φ∈Cc∞(Rn) and 1≤p<∞, ∥(Htφ−φ)/t−Δφ∥p→0 as t↓0.

Facts & Assumptions

Given: Countable Choice, n≥1, φ∈Cc∞(Rn), 1≤p<∞, and 0<ε<t.

[A1]

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

[F1]

Cc∞(Rn) is the test-function space of Test function space d of an open set; φ and every derivative of it is smooth with compact support, so Δφ∈Cc∞(Rn) and in particular Δφ∈Lp(Rn). In this item Hsψ denotes the evolution of the class ψ as in The heat evolution Ht of initial data.

[F2]

For s>0 the function Hsφ is C∞ on Rn and every spatial and time derivative passes through the convolution, Dα∂skHsφ=(Dα∂skΓs)∗φ (Spatial and time derivatives pass through heat convolution for positive time).

[F3]

The kernel satisfies ∂sΓ=ΔxΓ on Rn×(0,∞) (Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel).

[F4]

For continuous F,G on [a,b] differentiable on (a,b) with F′=f, G′=g Riemann integrable there, ∫abFg+∫abfG=F(b)G(b)−F(a)G(a) (Integration by parts for continuous factors with Riemann-integrable extensions of their interior derivatives).

[F5]

If G is continuous on [a,b], differentiable on (a,b) and f=G′ is Riemann integrable, then ∫abf=G(b)−G(a) (Newton–Leibniz needs only continuity on [a,b], differentiability on (a,b), and a Riemann-integrable extension of the interior derivative).

[F6]

For 1≤p<∞ and ψ∈Lp(Rn), ∥Hsψ−ψ∥p→0 as s↓0+, and ∥Hsψ∥p≤∥ψ∥p (The heat Cauchy problem for Lp data).

[F7]

Let 1≤p<∞ and let F(x,s) be measurable with ∫0t∥F(⋅,s)∥Lp ds<∞; then ∥∫0t∣F(x,s)∣ ds∥Lp≤∫0t∥F(⋅,s)∥Lp ds (Minkowski's integral inequality).

[F8]

Bounded uniformly continuous data are recovered locally uniformly at t=0 by their heat convolution (The heat Cauchy problem for bounded uniformly continuous data). This applies to both φ and Δφ, which are smooth with compact support.

Proof

technique · direct
1.1A1F1F2F3F4given

Derivative identity: fix x∈Rn and s>0. By [F2] with α=0, k=1 and f=φ the function s↦Hsφ(x) is differentiable with ∂sHsφ(x)=∫∂sΓ(x−y,s)φ(y) dy; by [F3] and ∂xi∂xiΓ(x−y,s)=∂yi∂yiΓ(x−y,s), applying the scalar integration by parts [F4] twice in each coordinate (the boundary terms vanish because φ has compact support, and this works for every n≥1 including n=1) turns the last integral into ∫Γ(x−y,s)Δφ(y) dy=HsΔφ(x); hence ∂sHsφ(x)=HsΔφ(x) for every x and s>0.

2.1step 1.1F5given

Newton–Leibniz: by step 1.1 the map s↦Hsφ(x) is continuous on [ε,t] with derivative HsΔφ(x) on (ε,t), for its real and imaginary parts separately; the fundamental theorem [F5] applied on [ε,t] gives Htφ(x)−Hεφ(x)=∫εtHsΔφ(x) ds for every x.

3.1step 2.1F1F2F7F8given

Both Hεφ(x)→φ(x) and HsΔφ(x)→Δφ(x) hold at every x by [F8]. Thus the integrand in step 2.1 extends continuously to s=0, and letting ε↓0 gives Htφ(x)−φ(x)=∫0tHsΔφ(x) ds. Dividing by t and subtracting Δφ(x) yields Htφ(x)−φ(x)t−Δφ(x)=1t∫0t(HsΔφ(x)−Δφ(x)) ds. The integrand is jointly continuous for s>0 by [F2] applied to Δφ, hence measurable as required for [F7].

4.1step 3.1F6F7given

Lp bound and limit: applying the Minkowski integral inequality [F7] to F(x,s):=HsΔφ(x)−Δφ(x) on Rn×(0,t) — whose hypothesis holds because ∥F(⋅,s)∥p≤2∥Δφ∥p by the contraction clause of [F6] — gives ∥Htφ−φt−Δφ∥p≤1t∫0t∥HsΔφ−Δφ∥p ds. Given η>0, the strong convergence clause of [F6] applied to Δφ∈Lp supplies δ>0 with ∥HsΔφ−Δφ∥p<η for every 0<s<δ; for every 0<t<δ the right-hand side is then at most η. Hence ∥(Htφ−φ)/t−Δφ∥p→0 as t↓0+.

5.1step 1.1step 4.1given∎

Steps 1.1, 2.1, 3.1 and 4.1 prove the stated generator limit for compactly supported smooth data in every Lp, 1≤p<∞.

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