Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12
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Hilbert cesaro averages converge to the fixed subspace

Statement

Assume AC. Let U be a linear isometry on a closed complex L2 subspace H, and put F=ker⁡(I−U). For each f∈H, ANf=1N∑n=0N−1Unf⟶PFfin norm as N→∞. Also R=U(H) is closed, and V=U−1PR is a linear contraction satisfying ⟨Uf,g⟩=⟨f,Vg⟩ and VU=I. Here U−1 means the inverse from R to H, not a surjectivity assumption on U.

Facts & Assumptions

[F1]

Closed subspaces have unique orthogonal projections and orthogonal decompositions under AC Closed l two subspaces have orthogonal projections.

[F2]

Isometries preserve the pairing and the norm; adjoint and invariant-subspace conventions are fixed locally L two operator conventions for weak mixing.

[F3]

The complex pairing is sesquilinear and satisfies Cauchy–Schwarz The complex L2 pairing is well-defined and satisfies Cauchy–Schwarz.

[F4]

Assume AC The Axiom of Choice, as required for the projections and completeness used in their proof.

[F5]

Under countable choice, complex L2 is complete Complex Lp completeness and almost-everywhere subsequences.

Proof

Given: H, U and AC as stated; N is a positive integer.

1.1F1F2F4F5

F5 and AC make the ambient complex L2 complete. Hence its closed subspace H is complete: an H-valued Cauchy sequence converges in the ambient space by F5, and closedness puts its limit in H. If Ufn converges in H, then ∥fn−fm∥=∥Ufn−Ufm∥ makes (fn) Cauchy. Its limit f∈H satisfies Ufn→Uf by isometry, so R is closed. Isometry makes U injective; its inverse on R is linear and isometric. The projection PR therefore defines the linear contraction V=U−1PR.

1.2F2F3

Let M=(I−U)H‾. This is a closed subspace: sums and scalar multiples of limits remain limits by the norm inequalities. If h⊥M, then ⟨h,h−Uh⟩=0, whence ⟨h,Uh⟩=∥h∥2. Expansion and isometry give ∥h−Uh∥2=2∥h∥2−2Re⁡⟨h,Uh⟩=0, so Uh=h. Conversely, if Uh=h, then for each g∈H, ⟨h,Ug⟩=⟨Uh,Ug⟩=⟨h,g⟩. Thus h is orthogonal to (I−U)H, and Cauchy–Schwarz extends orthogonality to its closure. Consequently M⊥=F.

2.1F1F2step 1.1

Write PRg=Uh. Orthogonality gives ⟨Uf,g⟩=⟨Uf,Uh⟩=⟨f,h⟩=⟨f,Vg⟩. Since PRUf=Uf, we have VUf=f. This proves the adjoint identity without a representation theorem or an inverse of U on all of H.

2.2F1step 1.2

By orthogonal decomposition, H=M⊕F. The subspace F is closed, either as M⊥ or directly by continuity of I−U. In the decomposition f=m+h, m∈M, h∈F, the vector m is orthogonal to F, so uniqueness of projection gives h=PFf.

2.3F2step 1.2

Isometry and the triangle inequality give ∥AN∥≤1. For g∈H, cancellation of the finite sum gives AN(I−U)g=(g−UNg)/N, of norm at most 2∥g∥/N. For m∈M and ε>0, choose one g with ∥m−(I−U)g∥<ε. Hence lim sup⁡N∥ANm∥≤ε. As ε is arbitrary, ANm→0. No sequence of such approximants is needed.

3.1F4step 1.1step 2.2step 2.3∎

For h∈F, every Unh=h, so ANh=h. Applying this and the previous limit to f=m+h gives ANf→h=PFf. For N=1 the average is the identity; zero vectors and H={0} obey every formula without division by a vector norm. AC is inherited from the projection/completeness argument in step 1.1 and the projections in step 2.2.

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Sources