LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)

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Nonzero compact kernel operators yield nonzero positive compact k star k

Statement

Assume AC. For a nonzero compact kernel operator $K$ with $K 1 = K^{*} 1 = 0$ , $S = K^{*} K$ on $H_{0}$ is bounded, compact, self-adjoint, positive and nonzero. If $K$ and $K^{*}$ commute with a Koopman isometry $U$ , then $S U = U S$ on $H_{0}$ .

Facts & Assumptions

[F1]

Kernel adjoints are bounded, satisfy the pairing identity, and have double adjoint $K$ Conjugate transpose kernels give adjoints.

[F2]

Positivity, self-adjointness and compactness use the local operator conventions L two operator conventions for weak mixing.

[F3]

The complex pairing is positive definite The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz.

[F4]

Assume AC The Axiom of Choice.

Proof

Given: $K$ and its two zero images of $1$ , under AC.

1.1

If $h \in H_{0}$ , the adjoint identities and the two zero images give $⟨ K h, 1 ⟩ = ⟨ h, K^{*} 1 ⟩ = 0$ and $⟨ K^{*} h, 1 ⟩ = ⟨ h, K 1 ⟩ = 0$ . Thus both operators preserve $H_{0}$ . Their restrictions include a nonzero K: choose f with $K f \neq 0$ and write $f = (\int f) 1 + f_{0}$ ; then $f_{0} \in H_{0}$ and $K f_{0} = K f \neq 0$ . Thus $S$ is a well-defined bounded endomorphism of $H_{0}$ , with $∥ S ∥ \leq ∥ K^{*} ∥ ∥ K ∥$ . AC supplies the inherited kernel results.

F1F3F4

2.1

For $f, g \in H_{0}$ , the two adjoint identities give $⟨ S f, g ⟩ = ⟨ K f, K g ⟩ = ⟨ f, S g ⟩$ . Also $⟨ S f, f ⟩ = ∥ K f ∥^{2} \geq 0$ . Therefore $S$ is self-adjoint and positive. For the $f_{0}$ in step 1.1, this quantity is strictly positive; hence $S f_{0} \neq 0$ and $S \neq 0$ .

F1F2F3step 1.1

3.1

For any bounded sequence in $H_{0}$ , compactness of $K$ gives a subsequence of its images convergent in $L^{2}$ . The limit lies in $H_{0}$ because it is closed. Applying the bounded, hence continuous, operator $K^{*}$ shows the corresponding $S$ images converge in $H_{0}$ . This is compactness of $S$ . A Koopman operator fixes $1$ ; since the isometry U preserves the pairing, $⟨ U h, 1 ⟩ = ⟨ U h, U 1 ⟩ = ⟨ h, 1 ⟩$ , so U preserves $H_{0}$ . If both factors commute with U, then $S U = K^{*} K U = K^{*} U K = U K^{*} K = U S$ on $H_{0}$ . No eigenvalue of K itself has been asserted.

F1F2step 2.1∎

Depends on

Used by

Compact intertwiners produce finite dimensional invariant subspaces Lemma

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Axler Example 10.5; 10.69(b) p.313 and 10.96 p.326; direct K-star-K argument (standard reference, not scraped)