Self-Adjoint Extensions

  • Thread starter Nusc
  • Start date
754
2
Let A be a closed symmetric operator with deficiency indices n_+/-.

A has self-adjoint extension iff n_+=n_-. In this case the set of self-adjoint extensions is in natural correspondence with the set of isomorphisms of L_+ onto L_-.

Isn't this just von Neumann's theorem?

Let A be a symmetric operator and suppose that there exists a conjugation C with C: dom A-> dom A and AC = CA. Then A has equal deficiency indices and therefore has self-adjoint extensions.
 
I only know of a VN thm involving one-parameter unitary groups of operators. Your problem looks as if you have to prove closure and to look at the associated resolvent operator.
 

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