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The gradient norm dominates the H-one norm on zero-boundary Sobolev space

Statement

Assume the Axiom of Choice (The Axiom of Choice), used to invoke the published Sobolev completeness theorem through its Countable-Choice interface (The Axiom of Countable Choice (ACω)). Let Ω⊆(−L,L)3 be bounded and open for some L>0, and write H1(Ω)=W1,2(Ω;R) with the norm of Integer-order Sobolev spaces and their norms, and H01(Ω):=W01,2(Ω;R), the closure of Cc∞(Ω;R) in that norm (Zero-boundary Sobolev space as a norm closure). Then H1(Ω) is complete, H01(Ω) is a closed subspace and is complete, the W1,2 quadratic norm is a Hilbert norm (its coordinate L2 pairings satisfy the parallelogram law), and

∥v∥L22≤(2L)2∥∂1v∥L22≤(2L)2∥∇v∥L22for every v∈H01(Ω).

In particular ∥v∥L2≤2L∥∇v∥L2 and the gradient norm is equivalent to the full H1 norm on H01(Ω). The argument needs neither a smooth domain nor a general boundary-trace theorem, and no choice beyond the completeness interface is used.

Facts & Assumptions

Given: The Axiom of Choice AC (so Countable Choice ACω holds), a number L>0, and a bounded open Ω⊆(−L,L)3. Write H1=W1,2(Ω;R) and H01=W01,2(Ω;R), with ∥u∥W2=∥u∥22+∥∇u∥22=∥u∥22+∑j=13∥∂ju∥22.

[A1]

The Axiom of Choice implies the Axiom of Countable Choice (The Axiom of Choice, The Axiom of Countable Choice (ACω)).

[F1]

Sobolev spaces and norm. For open Ω⊆Rn, Wk,p(Ω;K) consists of the Lp classes whose weak derivatives up to order k are Lp classes, the Dαu are well-defined classes and ∥u∥Wk,p=(∑∣α∣≤k∥Dαu∥Lpp)1/p for p<∞; at k=1, p=2 this is ∥u∥W2=∥u∥22+∥∇u∥22 (Integer-order Sobolev spaces and their norms). The definition leaves the norm axioms separate; under AC they are supplied by the complete-normed-space assertion of [F3].

[F2]

Zero-boundary space as a closure. W0k,p(Ω;K) is the metric closure of Cc∞(Ω;K) in Wk,p, so u∈W0k,p exactly when for every δ>0 there is φ∈Cc∞(Ω;K) with ∥u−φ∥Wk,p<δ; no boundary trace is asserted (Zero-boundary Sobolev space as a norm closure).

[F3]

Completeness. Under the Axiom of Choice, Wk,p(Ω;K) with the displayed norm is a complete normed space; at k=1, p=2, H1 is complete (Integer-order Sobolev spaces are Banach).

[F4]

Hilbert structure of real L2. Under ACω, real L2(Ω) with ⟨f,g⟩=∫fg is a real Hilbert space and ⟨f,f⟩=∥f∥22 (L2 with the integral pairing is a Hilbert space); in every real inner-product space the parallelogram law ∥x+y∥2+∥x−y∥2=2∥x∥2+2∥y∥2 holds (The parallelogram law).

[F5]

Cauchy-Schwarz for L2. ∫∣fg∣≤∥f∥2∥g∥2 for f,g∈L2 (Cauchy-Schwarz inequality for L2); applied to g=1(−L,L) this gives (∫−LL∣f∣)2≤2L∫−LL∣f∣2.

[F6]

Newton-Leibniz with interior derivative. If G:[a,b]→R is continuous on [a,b], differentiable on (a,b), and f is Riemann integrable with f=G′ on (a,b), 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).

[F7]

Tonelli. A nonnegative measurable function on a σ-finite product may be integrated in either order (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F8]

Norm axioms. A norm satisfies ∥x+y∥≤∥x∥+∥y∥ and ∥x∥=0⇒x=0 (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms), so a limit of a sequence in a subspace that exists in an ambient complete space and is approached by elements of the subspace lies in its closure whenever the closure is closed for the norm topology.

Proof

technique · direct
1.1givenA1F1F3

By [F3] the space H1 is complete for ∥u∥W2=∥u∥22+∥∇u∥22; the Axiom of Choice assumed here is used only to supply Countable Choice for that theorem by [A1].

1.2givenF6

Let v∈Cc∞(Ω;R) and extend it by zero to R3; fix (x2,x3)∈R2 and let G(s)=v(s,x2,x3). Since Ω⊆(−L,L)3 and v vanishes outside Ω, G is continuous on [−L,L], differentiable on (−L,L) with G′=∂1v(⋅,x2,x3) Riemann integrable, and G(−L)=0; hence [F6] gives v(x1,x2,x3)=∫−Lx1∂1v(s,x2,x3) ds for every x1∈(−L,L).

2.1step 1.1F1F2F3F8algebra

H01 is a closed subspace of H1 and is complete: if (um)⊆H01 is ∥⋅∥W-Cauchy, [F3] gives u∈H1 with ∥um−u∥W→0; given δ>0 choose m with ∥um−u∥W<δ and then, because um∈H01, a test function φ∈Cc∞(Ω;R) with ∥um−φ∥W<δ by [F2]; the triangle inequality [F8] gives ∥u−φ∥W<2δ, so u∈H01 by the characterization [F2], and ∥um−u∥W→0 in the subspace norm; the identical argument with H01 in place of the arbitrary subspace shows closedness of H01 in H1.

2.2step 1.1F1F4algebra

The W1,2 norm is a Hilbert norm: writing ∥u∥W2=∑∣α∣≤1∥Dαu∥22 by [F1], each summand ∥Dαu∥22 is the squared norm of the real Hilbert space L2(Ω) by [F4], so the parallelogram law [F4] holds for each summand and, summed over the finite index set {∣α∣≤1}, for ∥⋅∥W2 itself; combining this with the completeness of step 1.1 exhibits H1 and its closed subspace H01 as real Hilbert spaces for the coordinate L2 pairings.

2.3step 1.2F5algebra

For such v and every x=(x1,x2,x3), step 1.2 and the Cauchy-Schwarz bound [F5] give ∣v(x)∣2≤(∫−LL∣∂1v(s,x2,x3)∣ds)2≤2L∫−LL∣∂1v(s,x2,x3)∣2ds.

3.1step 2.3F7algebra

Integrating the inequality of step 2.3 over (−L,L)3 and applying Tonelli [F7] to the nonnegative integrand gives ∥v∥L22≤2L∫(−L,L)3∫−LL∣∂1v(s,x2,x3)∣2ds dx=(2L)2∥∂1v∥L22, because the x1-integration contributes the length 2L and v vanishes outside the box.

4.1step 3.1F1F2F8algebra

Fix v∈H01 and δ>0. By [F2] choose one φ∈Cc∞(Ω) with ∥φ−v∥W<δ. The triangle inequality and step 3.1 give ∥v∥2≤δ+∥φ∥2≤δ+2L∥∂1φ∥2≤2L∥∂1v∥2+(1+2L)δ. Since this holds for every δ>0, ∥v∥2≤2L∥∂1v∥2; squaring and using ∥∂1v∥2≤∥∇v∥2 gives the displayed chain. This uses one approximant for each fixed tolerance, without selecting an approximation sequence.

5.1step 2.1step 2.2step 4.1F1algebra∎

Equivalence of norms on H01: ∥∇v∥2≤∥v∥W is immediate from ∥v∥W2=∥v∥22+∥∇v∥22 by [F1], while step 4.1 gives ∥v∥W2=∥v∥22+∥∇v∥22≤(1+(2L)2)∥∇v∥22; the gradient norm is therefore equivalent to the full W1,2 norm on H01, both operators being applied to the weak derivatives of [F1] and no smoothness of Ω being used. This proves all the claims of the Statement, and no choice beyond the completeness interface [F3]/[A1] has been used.

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