All Questions
Tagged with sp.spectral-theory oa.operator-algebras
30
questions
2
votes
1
answer
148
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 ...
0
votes
0
answers
100
views
Given $\sigma(AB-BA) = \{0\}$, what can be said about $\sigma(A)$ and $\sigma(B)$?
Let $\mathcal H$ be a separable Hilbert space, and $\mathfrak B(\mathcal H)$ denote the algebra of bounded linear operators on $\mathcal H$. Furthermore, let $A,B \in \mathfrak B(\mathcal H)$ be two ...
3
votes
1
answer
163
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 $m$ ...
4
votes
1
answer
155
views
Characters of algebra of Schwartz functions
Consider the (non-unital) $\mathbb{C}$-algebra (point-wise multiplication) of $\mathcal{S}$ of Schwartz functions on $\mathbb{R}$.
Question: Does there exist some character (non-zero multiplicative ...
1
vote
0
answers
67
views
Spectral measure for a finite set of mutually commuting normal operators
The following question is from Exercise $\S 11.11$ in A Course in Operator Theory written by John B. Conway:
Suppose $\{N_1, \cdots, N_p\}$ is a finite set of mutually commuting normal operators in $...
-1
votes
1
answer
101
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 ...
3
votes
0
answers
139
views
Non-emptiness of spectrum $\sigma(a)$ in non-Archimedean Banach algebras
I'm trying to determine whether or not the standard proof that the spectrum of a point in a unital Banach algebra is non-empty can be adapted to prove the same thing over certain non-Archimedean ...
8
votes
1
answer
349
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
$$\...
4
votes
1
answer
303
views
When is rank-1 perturbation to a positive operator still positive?
Let $A : \mathcal{H} \to \mathcal{H}$ and $B : \mathcal{H} \to \mathcal{H}$ be trace-class (hence compact) Hermitian operators on a separable Hilbert space. Assume that $A$ is strictly positive and ...
22
votes
5
answers
1k
views
Rigorous justification for this formal solution to $f(x+1)+f(x)=g(x)$
Let $g\in C(\Bbb R)$ be given, we want to find a solution $f\in C(\Bbb R)$ of the equation
$$
f(x+1) + f(x) = g(x).
$$
We may rewrite the equation using the right-shift operator $(Tf)(x) = f(x+1)$...
-1
votes
2
answers
527
views
Invariance of spectrum under conjugation
Let $T$ be a self-adjoint invertible operator on $\mathcal{H}$ with a continuous spectrum, means the spectral measure is nonatomic. For which class of invertible operators $V$( with continuous ...
3
votes
2
answers
720
views
Is the ring of $p$-adic integers extremally disconnected?
We call a topological space $X$ extremally disconnected if the closures of its open sets remain open. Obviously, Hausdorff extremally disconnected spaces are totally disconnected in the sense that ...
0
votes
2
answers
400
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
180
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 ...
1
vote
0
answers
590
views
What is spectral multiplicity for multiplication operators in general von Neumann algebra set up?
When two multiplication operators $M_{f}$ and $M_{g}$ acting on $L^2(X,\mu) $and $L^2(Y,\nu)$ are unitary equivalent? How multiplicity function look like here? What is the spectral multiplicity in ...
9
votes
0
answers
220
views
Using Property (T) to approximate invertible matrices
In the wikipedia article for Kazhdan's Property (T), there's an intriguing application:
Similarly, groups with property (T) can be used to construct finite sets of invertible matrices which can ...
8
votes
0
answers
218
views
Regularilty of Commutative Spectral Triples
In Connes' approach to non-commutative geometry, the notion of a spectral triple is said to generalize compact Riemannian manifolds to the non-commutative setting. Motivating classical examples ...
8
votes
1
answer
1k
views
Spectral mapping theorem for polynomials in $z,\overline{z}$ and direct construction of the function calculus for a normal operator
Suppose that $A$ is an element in Banach algebra and $p$ is a polynomial. Then we have an equality $p(\sigma(A))=\sigma(p(A))$ where $p(A)$ has an elementary meaning. This theorem (the spectral ...
4
votes
1
answer
150
views
Commutator representation of certain smoothing operators
I have a question regarding the classical trace $\text{Tr} \colon \Psi^{-\infty}(S^1)\to \mathbb C$ on pseudodifferential operators of infinite negative order (i.e. smoothing operators), defined over ...
27
votes
0
answers
1k
views
Unital $C^{*}$ algebras whose all elements have path connected spectrum
A unital $C^{*}$ algebra is called a "Path connected algebra" if the spectrum of all its elements is a path connected subset of $\mathbb{C}$.
What is an example of a non commutative ...
5
votes
2
answers
463
views
When is the group algebra $L^1(G)$ semisimple?
Let $G$ be locally compact group. Define group algebra as
$$L^1(G)=\{f\colon G\to\Bbb{C}\mid\int\lvert f(x)\rvert\, dx<\infty\}$$
with convolution product. When is the group algebra $L^1(G)$ ...
3
votes
2
answers
534
views
A version of the spectral theorem for group actions
Suppose $G$ is a sufficiently nice (maybe locally compact and abelian) group which acts on the separable Hilbert space $\mathcal{H}$ by unitary transformations. Is there a generalization of the ...
2
votes
0
answers
467
views
Versions of the spectral theorem
Since any $C^*$-algebra can be represented as an algebra of bounded operators $\mathcal{B(H)}$ on a Hilbert space $\mathcal{H}$, the spectral theorem applies to all $C^*$-algebras:
($*$) $A=\int_{\...
0
votes
1
answer
218
views
Spectral decomposition function [closed]
Once I met a notation of "spectral decomposition function" (for a self-adjoint operator). No definition was given.
Could someone give me a clue what can that be, cause I can't find this exact phrase ...
6
votes
0
answers
367
views
Paving conjecture for Toeplitz matrices
Let me first recall what is the so-called paving conjecture:
for any $\epsilon >0$, there exists $r\in \mathbb N$ such that
for any bounded operator $A$ on $\ell^2(\mathbb Z)$, there exists a ...
5
votes
1
answer
248
views
Well defined Tensoring of spectral triples
Hi,
I have a misunderstanding that I am hoping is really quite trivial. I will give my question directly and context below for those that need/want it.
Question: In connes standard model he takes ...
1
vote
0
answers
836
views
Matrix conditions under which spectral radius is smaller than 1?
Hello everyone,
I would like to find out which conditions are necessary so that the spectral radius $\rho(M)<1$ where $M$ represents the following matrix:
$M = \left( \begin{array}{ccc}
W & 0 ...
5
votes
2
answers
481
views
Is independence meaningful for commutative $C^*$-algebras?
I don't know very much about spectral theory so probably the answer to my question has a basic reference which I would appreciate.
Let's say I have two self-adjoint operators on a Hilbert space and ...
11
votes
2
answers
2k
views
How "generalized eigenvalues" combine into producing the spectral measure?
Hi... I am wondering how 'eigenvalues' that don't lie in my Hilbert space combine into producing the spectral measure. I study probability and I am quite ignorant in the field of spectral analysis of ...
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" ...