 About 9,330 results  books.google.com If dom(A') = H and if A is symmetric, then from the closed graph theorem it follows
that K is bounded and hence selfadjoint. So that a symmetric everywhere
defined operator is bounded and selfadjoint. A. 16. Essentially selfadjoint
operators A ... 

 books.google.com Show that B is a bounded linear operator on H. Show that B is selfadjoint and
that A = B~l exists. (e) Show that ... Let L be a densely defined bounded
symmetric operator on a Hilbert space H. Show that L is essentially selfadjoint. 9.
Let M be ... 

 books.google.com (b) Show that if A is selfadjoint, then so is A\j. (c) Assume that A is closed. Show
that Au and A have the same spectrum and the same eigenvalues. 2.22. Show
that a bounded symmetric operator is essentially selfadjoint. 2.23. (a) Let A be ... 

 books.google.com (A.2.1) For a symmetric operator A, its adjoint operator A∗ is defined on the
domain D(A∗) consisting of the points x ∈ H for ... Except for bounded operators
(bounded symmetric operators are selfadjoint), there is a huge difference
between symmetric and selfadjoint operators. ... a unique selfadjoint extension
to some larger domain D(L), then L is said to be essentially selfadjoint (this
notion refers both ... 

 books.google.com Then S is essentially selfadjoint if and only if T is; in this case S and T have the
same domain. In particular, S is selfadjoint if and only if T is. 2. The case of
relative bound 1 In the theorems proved in the preceding paragraph, the
assumption ... 

 books.google.com An operator a is said to be bounded on D(a) if there exists a real number r such
that ae<re, e G £>(o). If otherwise, it ... For bounded operators, the notions of
symmetric, selfadjoint and essentially selfadjoint operators coincide. Let fibe a ... 

 books.google.com C8 A^ symmetric operator Ajs essentially selfadjoint if and only if its closure A^ is
selfadjoint, i.e., A^ = A^. Given that ... Two bounded selfadjoint operators A\ and
An are said to commute if their commutator [AUA2] = A1A2A2A1 (2.85) vanishes. 

 books.google.com Hence, C1IB(AIi,51)—l is a bounded operator with domain D(C) I H and norm
less than 1. Therefore ... Proposition 8.6 Suppose that A is an essentially self
adjoint operator and B is a symmetric operator on H such that D(B) Q D(A).
Suppose ... 

 books.google.com But A* is densely defined and closed, by Theorem 1.5(2) we have A*** = A*, thus
A* = A**, i.e., A* is selfadjoint. ... theorem gives a useful characterization for a
symmetric, bounded below operator to be selfadjoint or essentially selfadjoint. 

 books.google.com Then, At±A2 is essentially selfadjoint. Proof: The proof follows directly from the
spectral theorem, y A useful criterion for essential selfadjointness is given in :
Lemma 3. Let A be symmetric operator and let (A+I)~l be bounded and densely ... 

 