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CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)
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A bounded map into H01 yields a compact L2 operator

Statement

Assume the Axiom of Choice. Let Ω⊂Rn be a bounded open set and T:L2(Ω)→H01(Ω) a bounded linear operator. Then ιT:L2(Ω)→L2(Ω) is compact, where ι:H01(Ω)↪L2(Ω) is the inclusion. No boundary regularity of Ω is needed.

Facts & Assumptions

Given: the Axiom of Choice, a bounded open set Ω⊆Rn, and a bounded linear operator T:L2(Ω)→H01(Ω), with ι:H01(Ω)→L2(Ω) the inclusion.

[F1]

Zero-boundary Rellich theorem. W01,2(Ω)=H01(Ω) is compactly embedded in L2(Ω): every sequence bounded in H01(Ω) has a subsequence converging in L2(Ω). (Compactness of W01,p(Ω)↪Lp(Ω) on bounded open sets, Compactly embedded normed spaces, The notation Hk and the reserved zero-boundary symbol)

[F2]

Bounded operators map bounded sequences to bounded sequences. If (fj) satisfies ∥fj∥L2≤M, then ∥Tfj∥H01≤∥T∥M for the operator norm of A bounded linear operator between normed spaces. (A bounded linear operator between normed spaces)

[F3]

Compactness is the sequential extraction criterion. A bounded operator S is compact exactly when the image of every bounded sequence has a convergent subsequence. (Compact linear operator, Compactly embedded normed spaces)

Proof

technique · direct
1.1F1F2given

Let (fj) be bounded in L2(Ω) with ∥fj∥L2≤M. By [F2], (Tfj) is bounded in H01(Ω), so [F1] supplies a subsequence with Tfjk→g in L2(Ω), that is, ιTfjk→g.

2.1F1F3step 1.1∎

Since every bounded sequence in L2(Ω) has an image under ιT with a convergent subsequence, [F3] makes ιT a compact operator. The inclusion ι is bounded because ∥u∥L2≤∥u∥H01, and the Axiom of Choice is inherited through the Rellich theorem [F1].

Depends on

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