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The partial-trace displacement space is a Hilbert space

Statement

Assume the geometry and trace conventions of Partial-trace displacement space on a polygon. Then the maps TD,TC:H1(Ω;R2)→L2(ΓD;R2),L2(ΓC;R2) are bounded and linear, and V=ker⁡TD is a Hilbert space with the inherited H1 inner product.

Facts & Assumptions

Given: The bounded connected polygon, its fixed bounded C1 collars UD,UC, and the trace maps defined in Partial-trace displacement space on a polygon.

[A1]

The Axiom of Choice is assumed by the bounded C1 trace theorem and its smooth-density interface. (The Axiom of Choice, The Lp trace operator on a bounded C1 domain)

[F1]

On a bounded C1 domain U, TU:H1(U;R2)→L2(∂U;R2) is bounded and linear, agrees with boundary restriction for Sobolev functions having a representative continuous on U‾, and satisfies ∥TUw∥L2(∂U)≤CU∥w∥H1(U). (The Lp trace operator on a bounded C1 domain)

[F2]

Restriction from H1(Ω) to H1(Uj) is well defined and does not increase the H1 norm when Uj⊂Ω. (Sobolev extension domains and extension operators, Integer-order Sobolev spaces and their norms)

[F3]

H1(Ω;R2) is complete, and a closed subspace of a complete inner-product space is complete. (Integer-order Sobolev spaces are Banach, Closed subspaces of complete metric spaces are complete; the converse under countable choice)

Proof

technique · direct

Given: The fixed collars and the trace maps TD,TC on the bounded connected polygon Ω.

1.1A1F1F2given

For j∈{D,C}, restriction [F2] followed by the bounded trace map [F1] and restriction from ∂Uj to its segment Γj gives ∥Tjv∥L2(Γj)≤Cj∥v∣Uj∥H1(Uj)≤Cj∥v∥H1(Ω) for every v∈H1(Ω;R2). Both maps are compositions of linear maps, so they are bounded and linear.

2.1F1F3step 1.1∎

The space V is the kernel of the bounded linear map TD, hence is closed in H1(Ω;R2); by [F3] the inherited inner product is complete on V, so V is Hilbert.

Depends on

Used by

Cited to discharge well-definedness by Partial-trace displacement space on a polygon.

Dependency tree · two levels

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Sources