Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Hilbert-adjoint identities

Statement

Assume the Axiom of Countable Choice. Let H,K,L be real or complex Hilbert spaces and let R,SB(H,K) and TB(K,L) be bounded linear operators. Then the Hilbert adjoints satisfy:

  1. (aR+bS)=aR+bS for scalars a,b, and the adjoint of an operator is unique;
  2. (TS)=ST;
  3. T=T and T=T;
  4. TT=T2.

Facts & Assumptions

[A1]

The Hilbert adjoint of TB(K,L) is the unique map T:LK with Tx,yL=x,TyK for all x,y (The Hilbert-space adjoint of a bounded operator).

[A2]

The pairing is linear in the first argument, conjugate-linear in the second, conjugate symmetric, and v,v=0 implies v=0 (Real and complex inner-product spaces and their induced length).

[A3]

The operator norm satisfies TuTu and is the unit-ball supremum (The operator norm as the least bound and as the unit-sphere or unit-ball supremum), while composition obeys UVUV (Composition satisfies |ST|\le|S|,|T|); a linear map is bounded if it admits a finite constant C0 with UyCy for every y (A bounded linear operator between normed spaces).

[A4]

Cauchy–Schwarz gives u,vuv (Cauchy–Schwarz: x,yxy, with equality exactly for dependent pairs).

[A5]

Countable Choice is the hypothesis of the Riesz construction of adjoints (The Axiom of Countable Choice (ACω)).

Proof

technique · direct

Given: Countable Choice, Hilbert spaces H,K,L and bounded operators R,SB(H,K), TB(K,L).

1.1

First let U:XY be any bounded linear operator between Hilbert spaces over the same scalar field. The map U:YX exists by the adjoint construction. For y,zY, scalars a,b and xX, its identity gives x,U(ay+bz)=Ux,ay+bz=ax,Uy+bx,Uz=x,aUy+bUz. Positive definiteness applied to the difference proves that U is linear. Also Uy2=U(Uy),yU(Uy),yUUyy, so division when Uy0, and the trivial inequality otherwise, give UyUy. Thus U is bounded and its now-defined operator norm satisfies UU. This applies to each bounded operator used below, including an adjoint once its boundedness has been established.

A1A2A3A4A5algebra
2.1

For uniqueness of the adjoint and conjugate-linearity in the operator, let U,VB(K,H) both satisfy the defining adjoint identity for the same operator in B(H,K). For yK, one has x,(UV)yH=0 for every xH; taking x=(UV)y gives (UV)y=0. For scalars a,b, the identities (aR+bS)x,yK=aRx,yK+bSx,yK=x,aRy+bSyH hold for all xH and yK, so (aR+bS)=aR+bS.

step 1.1A1A2A5
3.1

Composition and involution: for xH and yL, x,STyH=Sx,TyK=TSx,yL, so (TS)=ST by uniqueness. Likewise, for xK and yL, the defining identity for T gives Ty,xK=y,TxL; conjugate symmetry and the defining identity for T give Tx,yL=x,TyK=Tx,yL, so T=T by uniqueness.

step 1.1step 2.1A1A2
4.1

Norms: Cauchy–Schwarz gives Ty2=Ty,TyK=T(Ty),yLTTyy, hence TyTy (trivially when Ty=0) and TT; applying this to T and using T=T gives T=T. Moreover TTTT=T2, while for x1 one has Tx2=x,TTxKxTTxTT, so T2TT and hence TT=T2.

step 1.1step 3.1A3A4algebra

Depends on

Used by

…and 6 more results.

Dependency tree · two levels

27 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