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The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction

Statement

Assume Countable Choice (The Axiom of Countable Choice (ACω)). Let Ω⊆Rn be open and bounded in one direction: there are a unit vector e and a<b with a<x⋅e<b for every x∈Ω. Let 1≤p<∞. Then ∥u∥Lp(Ω)≤C(p) (b−a) ∥Du∥Lp(Ω) for every u∈W01,p(Ω;K).

Facts & Assumptions

Given: Countable Choice; an open set Ω⊆Rn bounded in the direction e∈Sn−1 between a<b; an exponent 1≤p<∞; a field K∈{R,C}; and a class u∈W01,p(Ω;K).

[F1]

W01,p(Ω;K) is the closure in W1,p(Ω;K) of the compactly supported smooth functions Cc∞(Ω;K), and every such smooth function lies in W1,p with its classical derivatives as weak derivatives (Zero-boundary Sobolev space as a norm closure).

[F2]

Riemann vector-valued fundamental theorem: if f:[a,b]→Rm is differentiable with integrable derivative, then ∫abf′=f(b)−f(a) (If f:[a,b]→Rm is differentiable with integrable f′ then ∫abf′=f(b)−f(a); and a bounded derivative makes f Lipschitz).

[F4]

Tonelli's theorem on sigma-finite products, allowing iterated integrals of nonnegative measurable functions in either order (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F5]

Linear set-measure scaling: an invertible linear map T of Rn scales Lebesgue measure by ∣det⁡T∣ (A linear map T of Rn sends Lebesgue measurable sets to Lebesgue measurable sets, with λn(T[E])=∣det⁡T∣ λn(E) when T is invertible and T[E] Lebesgue null when it is not). This set formula is distinct from the integrand substitution in [F9].

[F6]

W1,p consists of Lp classes with weak derivatives in Lp and the norm is the ℓp norm of u and its coordinate weak derivatives (Integer-order Sobolev spaces and their norms); Lp classes are almost-everywhere classes (The space Lp(μ) as the quotient by null functions).

[F7]

Countable Choice, assumed for the measure and closure interfaces above (The Axiom of Countable Choice (ACω)).

[F8]

A continuous real function on a compact nondegenerate interval is bounded and Riemann integrable; under Countable Choice its Riemann and Lebesgue integrals agree (Every continuous function on a closed nondegenerate rectangle in Rm is Riemann integrable, A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral).

[F9]

Under Countable Choice, a C1 diffeomorphism T:U→V between open subsets of Rn satisfies ∫Vf(y) dy=∫Uf(T(z))∣det⁡DT(z)∣ dz for every nonnegative Lebesgue measurable f (A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions).

[F10]

The Lebesgue integral of a complex integrable function is the integral of its real part plus i times the integral of its imaginary part, and ∣∫h∣≤∫∣h∣ for real or complex integrable h (Integrable real and complex functions, and their integrals, The modulus of an integral is bounded by the integral of the modulus).

Proof

technique · direct
1.1F2F3F8F10givenalgebra

The smooth case: pointwise bound. Let φ∈Cc∞(Ω;K) and extend it by zero to Rn. Write x=x⊥+xne with x⊥⊥e. For each fixed x⊥, the profile g(s):=φ(x⊥+se) is smooth and supported in (a,b), since Ω⊂{a<y⋅e<b}. If x=x⊥+xne∈Ω, then a<xn<b and g(a)=0. On [a,xn], each real component of g′ is continuous, so [F8] supplies its Riemann integrability and converts its Riemann integral to its Lebesgue integral. Thus [F2] applied componentwise and the complex integral convention in [F10] give φ(x)=g(xn)−g(a)=∫axnDφ(x⊥+se)⋅e ds. The continuous derivative and its modulus are Lebesgue integrable on this compact interval by [F8]; [F10] and ∣Dφ⋅e∣≤∣Dφ∣ therefore give ∣φ(x)∣≤∫axn∣Dφ(x⊥+se)∣ ds. For p=1 the resulting integral bound is already the desired estimate; for p>1, Holder [F3] applied to the real nonnegative function ∣Dφ∣ and the constant 1, followed by xn−a≤b−a and enlargement of the integration interval, yields ∣φ(x)∣p≤(b−a)p−1∫ab∣Dφ(x⊥+se)∣p ds.

2.1F4F5F9step 1.1algebra

The smooth case: integration. If n=1, then ∣Ω∣≤b−a, so integrating step 1.1 gives ∫Ω∣φ∣p≤(b−a)p∫ab∣Dφ(se)∣p ds=(b−a)p∫Ω∣Dφ∣p by [F9] for the global linear diffeomorphism s↦se of R (where e=1 or e=−1 and the absolute Jacobian is 1), and the support of Dφ in Ω. Now suppose n≥2. Choose orthonormal coordinates with last vector e and write x=x⊥+xne; the corresponding orthogonal map T:Rn−1×R→Rn is a global C1 diffeomorphism with ∣det⁡DT∣=1. By [F9], nonnegative integrands have the same integral after composing with T; this applies both to 1Ω∣φ∣p and to ∣Dφ∣p, which are Borel measurable since the zero extension is smooth and Ω is open. For each x⊥, the slice Sx⊥:={xn:(x⊥,xn)∈Ω} is a measurable subset of (a,b), so ∣Sx⊥∣≤b−a. Integrating step 1.1 over Ω and applying Tonelli [F4] gives ∫Ω∣φ∣p dx≤(b−a)p−1∫Rn−1∫Sx⊥∫ab∣Dφ(x⊥+se)∣p ds dxn dx⊥≤(b−a)p∫Rn−1∫ab∣Dφ(x⊥+se)∣p ds dx⊥. The last integral equals ∫Rn∣Dφ∣p dx=∫Ω∣Dφ∣p, since φ and its gradient vanish outside Ω and Ω lies in the slab. Therefore ∥φ∥Lp(Ω)p≤(b−a)p∥Dφ∥Lp(Ω)p.

3.1F1F6F7step 2.1algebra∎

The general class. By [F1] there are φk∈Cc∞(Ω;K) with φk→u in W1,p(Ω;K); by step 2.1, ∥φk∥Lp≤(b−a)∥Dφk∥Lp for every k. Both sides are continuous in the W1,p norm: ∥φk∥p→∥u∥p and ∥Dφk∥p→∥Du∥p by [F6]. Passing to the limit gives ∥u∥Lp(Ω)≤(b−a)∥Du∥Lp(Ω), so the asserted inequality holds with C(p)=1.

Source notes

Kinnunen's Theorem 3.10 proves the estimate on bounded open sets by taking the primitive in one coordinate direction and applying Holder; the proof above runs the same argument along the unit vector e of the hypothesis, uses the orthonormal coordinate decomposition for the integration, and then extends from Cc∞(Ω) to W01,p(Ω) by the definition of the latter as a closure. The constant obtained is 1, independent of p; the statement permits a p-dependent constant.

Depends on

Dependency tree · 2 levels

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