Hellinger-Toeplitz theorem
[10/60%]Prove that if is an operator on a Hilbert space such that for all , then is a bounded operator.
Let be linear operators in the Hilbert space such that for all . Prove that .
Let be linear operators in Hilbert space such that . Prove that and .
(E. Hellinger, O. Toeplitz, 1910). Let be a linear operator in a Hilbert space . Prove that if for all the equality holds, then the operator is bounded. 1
Definition 8.3: An operator is called self-adjoint if .
Footnotes
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Operators satisfying such a relation are called self-adjoint (see Definition 8.3 below). ↩
Is a symmetric linear transformation on an inner-product space necessarily bounded?
What if is a Hilbert space?
A linear transformation (not necessarily bounded) on an inner-product space (not necessarily complete) is called symmetric if for all and in . (This neutral term is used rather than "Hermitian" or "self-adjoint," because in the customary approach to Hermitian operators, , boundedness is an assumption necessary for the very formulation of the definition.)
Let be a Hilbert space and a linear operator which is symmetric, i.e.,
Prove that is continuous.
Let be Hilbert space and let be linear operators satisfying for all . Prove that and are both bounded.
Let be Hilbert space, let be a linear operator that is symmetric, i.e., , for all . Prove that is bounded. Give an example to show that the assertion is not true if is a dense subspace of and is symmetric.
Let be Hilbert space, let be a linear operator such that , for all . Prove that is bounded.
Let be a Hilbert space and the linear operator have the property that for all . Show that belongs to , that is, is bounded.