All Questions
6 questions
2
votes
1
answer
237
views
On spectral calculus and commutation of operators
Let $\mathcal{H}$ be a Hilbert space, $B\in\mathcal{B}(\mathcal{H})$ be bounded and self-adjoint and $A:\mathcal{D}(A)\to\mathcal{H}$ closed (but not necessarily self-adjoint or bounded). The ...
-1
votes
1
answer
164
views
Closure of the point spectrum of an unbounded diagonalizable operator
Given a (separable) Hilbert space H and an unbounded densely defined linear operator $T:{\cal D}(T) \to $H such that ${\cal D}$ is diagonalizable (it means $\exists$ an O.N.B. of H such that all basis ...
8
votes
1
answer
393
views
A question about comparison of positive self-adjoint operators
I have the following question but have no idea on its proof (one direction is trivial):
Let $A$ and $B$ be (bounded) positive self-adjoint operators on a complex Hilbert space $H$. Prove that
$$\...
0
votes
2
answers
465
views
Spectrum equals eigenvalues for unbounded operator
Let $D$ be an unbounded densely defined operator on a separable Hilbert space $H$. If $D$ is diagonalisable with all eigenvalues having finite multiplicity and growing towards infinity, does it follow ...
3
votes
1
answer
214
views
Non-point spectrum for diagonalisable self-adjoint unbounded operator
Given a (separable) Hilbert space H and an unbounded densely defined linear operator $T:{\cal D}(T) \to $H such that ${\cal D}$ is diagonalizable (it means $\exists$ an O.N.B. of H such that all basis ...
20
votes
8
answers
12k
views
Can a self-adjoint operator have a continuous set of eigenvalues?
This should be a trivial question for mathematicians but not for typical physicists.
I know that the spectrum of a linear operator on a Banach space splits into the so-called "point," "continuous" ...