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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 ...
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" ...
-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 ...