Alphabeta Math
DefinitionDefinition: AI-adaptedProof: Not applicablePipeline-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.

Symmetric, self-adjoint and essentially self-adjoint operators

Definition

Assume Countable Choice. Let T be a densely defined linear operator on H.

  • T is symmetric when TT, that is, D(T)D(T) and Tx=Tx for all xD(T); equivalently Tx,y=x,Ty for all x,yD(T).
  • T is self-adjoint when T=T, that is, the domains and the values agree.
  • T is essentially self-adjoint when its closure T is self-adjoint; by Closability is equivalent to density of the adjoint domain this presupposes that T is closable.

Consequences, with proofs. These are part of the content of the definition.

  1. A symmetric T is closable. TT and T is closed (The adjoint is well defined, closed, and reverses inclusions), so T is a closed extension of T. In particular the closure T exists and TT.
  2. The closure of a symmetric operator is symmetric. If TT, the inclusion reversal of The adjoint is well defined, closed, and reverses inclusions applied to TT gives (T)T, that is TT. Applying it once more to the inclusion TT gives (T)(T), that is TT. Since T=T and (T)=T (Closability is equivalent to density of the adjoint domain), this reads T(T): the closure is symmetric. (Each application is legitimate because the adjoint is defined once its operator is densely defined, and D(T) is dense because TT=T, so that D(T) contains the dense domain D(T).)
  3. A self-adjoint operator has no proper symmetric extension. If T=T and TSS, then ST by inclusion reversal, so TSST=T and all inclusions are equalities.
  4. A self-adjoint operator is closed, being equal to the adjoint T of the densely defined T, and an essentially self-adjoint operator has exactly one self-adjoint extension, namely T: a self-adjoint extension S of T is closed, hence contains the least closed extension T, and then SS(T)=T by item 2 and symmetry, so S=T.

Depends on

Used by

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Sources