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Maximum principle on the whole space under Gaussian growth

Statement

Assume Countable Choice. Let T>0, a≥0, A<∞, and let u∈C([0,T]×Rn)∩C1,2((0,T]×Rn) satisfy ut−Δu≤0,u(t,x)≤Aea∣x∣2((t,x)∈[0,T]×Rn). Then sup⁡[0,T]×Rnu≤sup⁡Rnu(0,⋅). (The same statement holds for a finite union of consecutive strips of length less than 1/(4a) when a>0.)

Facts & Assumptions

Given: Countable Choice, T>0, a≥0, A<∞, and a continuous u on the closed strip, C1,2 in positive time, with ut−Δu≤0 and u(t,x)≤Aea∣x∣2.

[A1]

Countable Choice is the ambient hypothesis (The Axiom of Countable Choice (ACω)).

[F1]

The weak maximum principle on a bounded cylinder BR(0)×(0,T]: a C2,1 subsolution attains its maximum on the parabolic boundary (Weak parabolic maximum principle, Parabolic cylinder and parabolic boundary).

Proof

Given: Countable Choice, T>0, a≥0, A<∞, and u satisfying the subsolution inequality and the Gaussian growth bound.

1.1A1F2F3given

Put M:=sup⁡Rnu(0,⋅). If M=+∞ the conclusion is immediate, so assume M<∞ (it is greater than −∞ since the initial trace is real valued). Assume first aT<1/4; choose δ>0 with b:=14(T+δ)>a, which is possible under this assumption because 1/(4T)>a and δ↦1/(4(T+δ)) is continuous with value 1/(4T) at δ=0, and set B(t,x):=(T+δ−t)−n/2e∣x∣2/[4(T+δ−t)] for t≤T; writing τ=T+δ−t and differentiating, [F2] gives Bt=(n2τ+∣x∣24τ2)B=ΔB, so (∂t−Δ)B=0 and v:=u−εB satisfies (∂t−Δ)v≤0 for every ε>0.

2.1step 1.1F2F3given

For every ε>0 there is R with v(t,x)≤M for all ∣x∣≥R and all t∈[0,T]: indeed B(t,x)≥(T+δ)−n/2eb∣x∣2 and u≤Aea∣x∣2, so v≤Aea∣x∣2−ε(T+δ)−n/2eb∣x∣2, whose right-hand side tends to −∞ as ∣x∣→∞ because b>a and exponentials dominate constants and polynomials [F3]; hence it is at most the fixed value M for ∣x∣≥R. On the initial slice v(0,x)=u(0,x)−εB(0,x)≤u(0,x)≤M, since B(0,x)>0.

3.1step 1.1step 2.1F1F3given

Fix ε>0 and R as in step 2.1. For 0<h<T, v has all required derivatives continuous on B‾R×[h,T], so [F1] applies on this positive-time cylinder. Continuity on B‾R×[0,T] gives ηh:=max⁡(0,max⁡B‾Rv(h,⋅)−M)→0 as h↓0, because v(0,⋅)≤M. Its lateral values are at most M by step 2.1, hence [F1] gives v≤M+ηh for h≤t≤T in the ball. At each fixed positive time let h↓0; combining with step 2.1 outside the ball gives v≤M on the whole closed strip. Letting ε↓0 gives u≤M when aT<1/4.

4.1step 3.1F3given∎

If aT≥1/4, choose an integer N>4aT and divide [0,T] into the N equal intervals [kT/N,(k+1)T/N]. Each has positive length T/N<1/(4a) and inherits the same Gaussian bound. Apply the short-strip case to the time-translated function on each interval. On the first interval its supremum is at most M, and inductively the initial supremum of each subsequent strip is at most M. Thus u≤M throughout [0,T]×Rn, with no time-zero derivative assumption.

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