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Sobolev level-set step: energy decay with explicit level gap and radius loss

Statement

Assume Countable Choice and the Axiom of Choice. Let n≥3, let BR⊂Rn be a ball, and let u∈H1(BR;R). Suppose there is C0≥1 such that for every 0<ρ<R and every level k ∫Bρ∣D(u−k)+∣2dx≤C0 (R−ρ)−2∫BR(u−k)+2dx, i.e. the truncated Caccioppoli estimate of Caccioppoli inequality for truncated subsolutions holds with f=0 on BR. Then there is C=C(n,C0) such that for all 0<r<R and all h<k: ∫Br(u−k)+2dx≤C (R−r)−2(k−h)−4/n(∫BR(u−h)+2dx)1+2/n, and consequently, for u≥0 and h>0, ∣{u>k}∩Br∣≤C (k−h)−2(R−r)−2h−4/n(∫BRu2 dx)1+2/n. For n=2 and each 0<δ<1, the same conclusions hold with 1+δ in place of 1+2/n, (k−h)−2δ and h−2δ in place of the powers −4/n, and the common factor (R−r)−2 replaced by R2−2δ(R−r)−2. Here the constant may also depend on δ. Indeed, the critical Sobolev inequality on BR has the scaled form ∥v∥Lκ(BR)≤SκR2/κ∥Dv∥L2(BR) for finite κ>2, and choosing κ=2/(1−δ) gives δ=1−2/κ. Thus the open range 0<δ<1 is exactly the range supplied by finite κ, and the radius factor is the one dictated by dilation (The critical Sobolev embedding into every finite Lq, The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction).

Facts & Assumptions

Given: Countable Choice and the Axiom of Choice; n≥2; a ball BR; a real class u∈H1(BR;R); a constant C0≥1 with ∫Bρ∣D(u−k)+∣2≤C0(R−ρ)−2∫BR(u−k)+2 for every 0<ρ<R and every level k; radii 0<r<R and levels h<k.

[F1]

Assume Countable Choice. For k∈R the class uk=(u−k)+ lies in H1(BR;R), and for η∈Cc∞(BR) the product ηuk lies in H01(BR) with D(ηuk)=ηDuk+ukDη almost everywhere (Positive-part truncation calculus and admissible cut-off weak tests, Integer-order Sobolev spaces and their norms).

[F2]

Assume the Axiom of Choice. Sobolev inequality: for n≥3 there is S=S(n)<∞ with ∥v∥L2∗(BR)≤S∥Dv∥L2(BR) for every v∈H01(BR), where 2∗=2n/(n−2) (The Sobolev inequality for zero-boundary Sobolev closures on open sets, The Sobolev conjugate exponent and the scaling identity).

[F3]

Assume the Axiom of Choice. On the unit ball in R2, the critical embedding into every finite Lκ, combined with Poincaré's inequality for H01, gives ∥v∥Lκ(B1)≤Sκ∥Dv∥L2(B1) for finite κ>2. Dilation therefore gives ∥v∥Lκ(BR)≤SκR2/κ∥Dv∥L2(BR) for v∈H01(BR) (The critical Sobolev embedding into every finite Lq, The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction).

[F4]

Chebyshev's inequality: for a nonnegative measurable v and t>0, ∣{v>t}∣≤t−2∫v2; and Hölder's inequality gives ∫Ef2≤∣E∣2/n∥f∥L2∗2 for measurable E of finite measure (Chebyshev-Markov inequality for the integral, Holder's inequality for integrals, including the endpoint cases, The space Lp(μ) as the quotient by null functions).

[F5]

The radial cutoff used in the ball-form Caccioppoli estimate has a universal gradient constant: for 0<r<ρ, take s=(r+ρ)/2 and η(x)=σ((s2−∣x∣2)/(s2−r2)). Then η∈Cc∞(Bρ), 0≤η≤1, η=1 on Br, and ∣Dη∣≤CU/(ρ−r) with CU=4∥σ′∥∞, since on the support ∣x∣≤s and s−r=(ρ−r)/2; this is the explicit cutoff calculation in Caccioppoli inequality for truncated subsolutions.

Proof

technique · direct; combine the truncated Caccioppoli estimate with the Sobolev embedding and Chebyshev's inequality, then iterate the resulting measure–energy inequality once
1.1givenF4

Put w:=(u−k)+ and v:=(u−h)+. Since h<k one has 0≤w≤v, and {w>0}={u>k}={v>k−h}⊆{v>0}. Applying Chebyshev's inequality to the nonnegative function v at level k−h>0 gives ∣{w>0}∩Br∣≤∣{v>k−h}∣≤(k−h)−2∫BRv2dx.

1.2givenF1F5algebra

Choose ρ:=(R+r)/2∈(r,R) and the bump η of [F5] with 0≤η≤1, η=1 on Br, supp⁡η⊆Bρ and ∣Dη∣≤CU/(ρ−r)=2CU/(R−r). By [F1], ηw∈H01(BR), and D(ηw)=ηDw+wDη almost everywhere, so the Caccioppoli hypothesis at radius ρ and level k, together with the elementary bound (a+b)2≤2a2+2b2, gives ∫BR∣D(ηw)∣2dx≤2∫Bρη2∣Dw∣2dx+2∫Bρw2∣Dη∣2dx≤8(C0+CU2)(R−r)2∫BRw2dx.

2.1step 1.1F2F4algebra

Assume n≥3 and let 2∗=2n/(n−2). Applying the Sobolev inequality [F2] to ηw∈H01(BR) and then Hölder's inequality [F4] on the support of ηw, which is contained in {w>0}∩Bρ up to a null set, gives ∫Brw2dx≤∫BR(ηw)2dx≤∣{w>0}∩Bρ∣2/nS2∫BR∣D(ηw)∣2dx.

3.1step 1.1step 1.2step 2.1algebra

Substituting the bound of step 1.2 into step 2.1 and then the Chebyshev bound of step 1.1, and using ∫BRw2≤∫BRv2, yields ∫Br(u−k)+2dx≤C(n,C0)(R−r)−2(k−h)−4/n(∫BR(u−h)+2dx)1+2/n, which is the first displayed estimate.

3.2step 1.1step 1.2step 2.1F3F4algebra

Assume n=2 and fix a finite exponent κ>2, put δ:=1−2/κ∈(0,1), and fix η and w as in steps 1.1-1.2. Replacing 2∗ by κ in step 2.1, using the scaled inequality [F3] and Hölder in the form ∫Ef2≤∣E∣1−2/κ∥f∥Lκ2, and inserting steps 1.1 and 1.2 gives ∫Br(u−k)+2dx≤C(κ,C0)R2−2δ(R−r)−2(k−h)−2δ(∫BR(u−h)+2dx)1+δ. As finite κ>2 varies, δ=1−2/κ ranges over exactly (0,1); the factor R2−2δ is precisely the dilation factor from [F3].

4.1step 3.1step 3.2F4algebra

For the measure clause assume u≥0 and h>0. On {u>k}∩Br one has (u−h)+>k−h>0, so Chebyshev's inequality gives ∣{u>k}∩Br∣≤(k−h)−2∫Br(u−h)+2dx, and the first estimate applied with levels 0<h bounds the integral by the displayed energy expression, since u≥0; multiplying the two bounds gives the displayed measure estimate. For n=2 the same argument carries the factor R2−2δ(R−r)−2h−2δ from step 3.2.

5.1step 3.1step 3.2step 4.1∎

Both displayed estimates follow from steps 3.1-4.1 with constants depending only on n, the Sobolev constants and C0; the hypothesis list uses the Caccioppoli estimate of Caccioppoli inequality for truncated subsolutions and the declared Countable Choice and Axiom of Choice only, so no further choice principle is used.

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