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Zero mean of the divergence of a zero-boundary field

Statement

Assume Countable Choice. Let D⊂Rd be bounded and open. If v∈H01(D;Rd), then div⁡v∈L2(D) and ∫Ddiv⁡v=0.

Facts & Assumptions

Given: A bounded open D and a vector field in the H1-closure of Cc∞(D;Rd).

[A1]

Countable Choice permits selecting the H0 test-function approximation sequence used to pass the integral identity to the limit. (The Axiom of Countable Choice (ACω))

[F1]

H01(D) is the closure of compactly supported smooth functions in the W1,2 norm. (Zero-boundary Sobolev space as a norm closure)

[F2]

First weak derivatives of a W1,2 field lie in L2, and its W1,2 norm controls their L2 norms. (Integer-order Sobolev spaces and their norms)

[F3]

On a finite-measure set, ∣∫Dg∣≤∣D∣1/2∥g∥2. (Holder's inequality for integrals, including the endpoint cases)

[F4]

Fubini evaluates the integral of each compactly supported coordinate derivative by iterated one-dimensional integration. (Fubini's theorem for L^1 functions on a sigma-finite product)

Proof

1.1F2algebra

By [F2], each weak derivative ∂ivi is in L2, so their finite sum div⁡v is in L2(D).

1.2givenF1F4

For v∈Cc∞(D;Rd), each component derivative has integral zero by one-dimensional integration along its coordinate, hence ∫Ddiv⁡v=0.

2.1A1F1F2F3step 1.2∎

By [A1, F1], choose vn∈Cc∞(D;Rd) converging to v in H1. Then div⁡vn→div⁡v in L2 by [F2]. Applying [F3] to the difference shows their integrals converge; the zero integrals in step 1.2 therefore give ∫Ddiv⁡v=0.

Depends on

Used by

Dependency tree · two levels

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Sources