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The zero-boundary W1p space is separable and reflexive

Statement

Assume the ultrafilter lemma, Dependent Choice and Hahn–Banach. Let Ω⊆Rd be bounded and open, d≥1, and let 1<p<∞. Then W01,p(Ω;R), with the norm inherited from W1,p(Ω;R), is a separable reflexive real Banach space.

Facts & Assumptions

Given: The ultrafilter lemma, Dependent Choice and Hahn–Banach; a bounded open Ω⊆Rd, d≥1; and 1<p<∞.

[F1]

The displayed W1,p formula is a norm, and weak differentiation is linear; hence Φ(u)=(u,∂1u,…,∂du) is a linear embedding of W1,p(Ω;R) into Y:=Lp(Ω;R)d+1 with its maximum product norm. Moreover, ∥Φu∥max⁡≤∥u∥W1,p≤(d+1)1/p∥Φu∥max⁡ (Integer-order Sobolev spaces and their norms, The Sobolev norm descends to equivalence classes, Linearity, locality, and commutation of weak derivatives).

[F2]

W01,p(Ω) is the norm closure of Cc∞(Ω) in W1,p(Ω), so it is a closed linear subspace (Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms).

[F3]

Under Hahn–Banach, a closed linear subspace of a reflexive Banach space is reflexive with its inherited norm; reflexive spaces are Banach spaces by definition (Closed subspaces of reflexive spaces are reflexive, Reflexivity is surjectivity of the canonical map).

[F4]

Under Countable Choice, rational box-step functions form a countable dense subset of Lp(Rd) for 1≤p<∞ (Rational box-step functions form a countable dense subset of Lp(Rn) for 1≤p<∞).

[F5]

A normed space is separable when it has a countable dense subset (Separability: the existence of an at most countable dense subset).

[F6]

Under Countable Choice, Lp(Ω;R) is complete for every 1≤p≤∞ (Riesz-Fischer completeness of Lp for 1≤p≤∞).

[F7]

A finite product of Banach spaces is Banach for its maximum product norm (Finite products of Banach spaces are Banach).

[F8]

Under Countable Choice, Lp(Ω;R) is reflexive when 1<p<∞ (Reflexivity of Lp for one less p less infinity).

[F9]

For conjugate exponents p,p′, Hölder's inequality bounds each test pairing by the product of the corresponding Lp and Lp′ norms (Holder's inequality for integrals, including the endpoint cases).

[F11]

Under the ultrafilter lemma, Dependent Choice and Hahn–Banach, a real Banach space is reflexive exactly when every bounded sequence has a weakly convergent subsequence (Reflexivity is equivalent to weak subsequential compactness of bounded sequences).

[F12]

Q is countable and dense in R, subsets and finite products of countable sets are countable, and in a metric space balls centered at a countable dense set with positive rational radii form a countable base. Indeed, for y∈B(x,r) let δ=r−d(x,y)>0, choose a dense center z with d(y,z)<δ/3, and choose rational q with d(y,z)<q<δ−d(y,z); then y∈B(z,q)⊆B(x,r) (Q is countably infinite, Both Q and R∖Q are dense in R, and every nonempty open subset of R is uncountable, Every subset of an at most countable set is at most countable, A product of two at most countable sets is at most countable, Finite, countably infinite, countable, uncountable, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

Proof

1.1A1F4F12givenconstructchoose

Let D be the countable rational box-step dense family in [F4]. Zero extension embeds Lp(Ω) isometrically in Lp(Rd). Given g∈Lp(Ω), choose s∈D approximating its zero extension g~; then ∥s∣Ω−g∥Lp(Ω)≤∥s−g~∥Lp(Rd). Thus the restrictions of D are countable and dense in Lp(Ω). Their finite products give a countable dense subset of Y; balls centered there with rational radii form a countable base by [F12].

1.2A1F1F6F9F10givenalgebra

Let (uj) be Cauchy in W1,p(Ω). By [F1], the coordinate sequences (uj) and (∂iuj) are Cauchy in Lp(Ω) for each i. Countable Choice is available by [A1], so [F6] gives limits u,gi∈Lp(Ω). Fix φ∈Cc∞(Ω) with compact support K. By [F10], K has finite measure and φ,∂iφ are bounded there; Hölder [F9] therefore lets both sides of ∫Ωuj∂iφ=−∫Ω(∂iuj)φ pass to the limits u and gi. Hence gi=∂iu weakly for each i, so u∈W1,p(Ω) and uj→u in its norm by coordinate convergence. Thus W1,p(Ω) is Banach.

2.1A1F1F3F6F7F8F11step 1.2givenconstruct

Each factor of Y is Banach by [F6], so the finite product is Banach for its maximum norm by [F7]. Let (yj) be a bounded sequence in Y. Its finitely many coordinate sequences are bounded in Lp(Ω); by [F8] and the bounded-sequence criterion [F11], successive subsequence extractions give one subsequence weakly convergent in every coordinate. Every bounded linear functional on Y is the sum of its restrictions to the coordinate factors, so this subsequence converges weakly in Y. Thus [F11] makes Y reflexive. By [F1], Φ is a bounded linear isomorphism onto its image S:=Φ(W1,p(Ω)). The image S is closed in Y: a convergent sequence in S has Cauchy preimages by the lower norm bound in [F1], and they converge by step 1.2. Hahn–Banach and [F3] make S reflexive. The inverse isomorphism Φ−1:S→W1,p(Ω) preserves weak convergence, so [F11] shows that W1,p(Ω) is reflexive.

3.1A1F2F3step 2.1

By [F2], W01,p(Ω) is a closed linear subspace of the reflexive Banach space W1,p(Ω) from step 2.1. Hahn–Banach and [F3] therefore make W01,p(Ω) reflexive and Banach with its inherited norm.

4.1A1F1F5F12step 1.1step 3.1choose

Let S0=Φ(W01,p(Ω))⊆Y. The countable base from step 1.1 restricts to a countable base of S0. By [A1], Countable Choice selects one point from each nonempty member of this relative base; the resulting countable set meets every nonempty relatively open set and is dense in S0. Since Φ is an isomorphism with equivalent norms by [F1], W01,p(Ω) is separable by [F5].

5.1step 3.1step 4.1∎

Steps 3.1 and 4.1 establish that W01,p(Ω;R) is a separable reflexive real Banach space with its inherited norm, as claimed.

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