All Questions
15 questions
3
votes
1
answer
226
views
$\tau$-measurable operator
Problem: Let $M$ be a semifinite von Neumann algebra with a faithful semifinite normal trace $\tau$. Let $m$ be a positive element in $M$ and let $e_{(0,\infty)}(m)$ be the spectral projection of $m$ ...
- 85
1
vote
0
answers
385
views
Densely defined and closed operator
Question: Let $N\subseteq B(H)$ be a finite von Neumann algebra with a faithful normal trace $\tau$ and let $n\in N$. Let $B$ be a von Neumann subalgebra of $N$. Let $\mathbb{E}_B: N\rightarrow B$ be ...
- 85
-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 ...
- 969
0
votes
1
answer
138
views
Antilinear unbounded operator has closed graph
Let $H$ and $K$ be Hilbert spaces and $D(T)$ a vector subspace of $H$. Let $T: D(T) \to K$ be a densely defined antilinear operator. Its adjoint $T^*: D(T^*)\to K$ is defined by the relation
$$\langle ...
- 175
2
votes
1
answer
917
views
Characterising closed range self-adjoint operators
Let $T:\mathrm{dom}(T) \subseteq H \to H$ be a densely defined, self-adjoint operator on a Hilbert spaces $𝐻$. In general the range of $T$ is not guaranteed to be closed. What tools are available to ...
- 969
1
vote
0
answers
277
views
Adjoint for a non-densely defined unbounded operator on a Hilbert space
Let $\mathbf{H}$ be a Hilbert space, and $D$ an unbounded densely-defined operator on $\mathbf{H}$. As is well-known, every such operator admits an adjoint, with domain possibly different from that ...
- 993
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 ...
- 969
6
votes
1
answer
1k
views
Unbounded version of continuous functional calculus
For a normal operator $T$ on a Hilbert space ${\cal H}$, it is well known that for any continuous complex valued function $f$ on the spectrum of $T$, we have a well-defined operator $f(T) \in B({\cal ...
- 969
2
votes
1
answer
134
views
Decomposition of the spectrum of an unbounded opeator [closed]
The Wikipedia article on spectral decomposition, see here
https://en.wikipedia.org/wiki/Self-adjoint_operator
says the following:
A self-adjoint operator A on $H$ has pure point spectrum if and ...
7
votes
2
answers
485
views
The von Neumann algebra generated by a non-closable operator
Let $H$ be a separable Hilbert space and let $M$ be a densely defined operator $\mathcal{D}(M) \subset H \to H$. It is closable iff its adjoint $M^{\star}$ is densely defined, and then its closure $\...
8
votes
1
answer
302
views
Does every integer map generate a von Neumann algebra of type I?
Consider a map $m: \mathbb{N} \to \mathbb{N}$ (we call it an integer map). Let $E_r$ be the set $m^{-1}(\{r\})$.
Let $H$ be the Hilbert space $\ell^2(\mathbb{N})$ and consider the densely defined ...
3
votes
1
answer
229
views
Symmetric diagonalizable operators and self-adjointness
Given a densely defined symmetric operator $L$ on a Hilbert space $H$, which is also assumed to be diagonalizable, will there always exist a unique extension of $L$ to a self-adjoint operator?
2
votes
1
answer
926
views
Eigenvalues and Compact Resolvent
For $A$ an unbounded (densely defined) operator on a separable Hilbert space, what conditions on its eigenvalues will show that, for $\lambda \notin $spec$(A)$, we have that $(A-\lambda)^{-1}$ is a ...
- 465
1
vote
2
answers
398
views
A question on unbounded operators
Assume that $H$ is a separable Hilbert space.
Is there a polynomial $p(z)\in \mathbb{C}[x]$ with $deg(p)>1$ with the following property?:
Every densely defined operator $A:D(A)\to D(A),\;D(A)\...
- 356
7
votes
2
answers
1k
views
C*-algebraic representation of observables vs self-adjoint operators one
I am trying to reconcile the "physicist" definition of an observable: self-adjoint operator on a Hilbert space, and the operational one as given by Strocchi in "An introduction to the mathematical ...
- 585