All Questions
Tagged with hilbert-spaces oa.operator-algebras
86 questions
6
votes
1
answer
170
views
Do projections in an $AW^\ast$-algebra form an orthomodular lattice?
I’m currently studying orthomodular lattices arising out of operator algebras. One of the most standard examples is the projection lattice of a von Neumann algebra - if $M$ acts on a Hilbert space $H$,...
- 151
0
votes
0
answers
35
views
Operator-form correspondence without lower semiboundedness
When dealing with a normal unbounded operator $A$, it is often useful to frame questions about the operator in terms of questions about the associated form $\omega,$ which has domain $D(|A|^{1/2})$ ...
- 109
2
votes
0
answers
316
views
What are alternative or equivalent definitions of a positive-definite function on a group?
The standard definition of a positive-definite function on a group goes as follows:
Let $\varphi : G \rightarrow L(H)$, where $G$ is a group (with an involution) and $H$ a Hilbert space. $L(H)$ is the ...
- 63
2
votes
0
answers
158
views
Question about the ergodic mean
This is a repost from this MathStackExchange question, where unfortunately I was not able to resolve this question.
I've read a thesis where there is an example on ergodic mean, where however there is ...
- 6,357
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 ...
- 4,351
1
vote
1
answer
142
views
An inner product and projection property in RKHS
I'm reading an article regarding the condition for the existence of a solution to a constrained Pick interpolation problem, and the author wrote the following equality - for background, we have the ...
- 2,830
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" ...
- 41.1k
9
votes
2
answers
516
views
Why operator systems?
A $\mathrm{C}^*$-algebra $\mathcal{A}\subset B(\mathsf{H})$ is a norm-closed, self-adjoint subalgebra of bounded operators on a Hilbert space. If we then take a unital self-adjoint (possibly closed) ...
- 42.8k
0
votes
1
answer
93
views
Does ${\rm Tr}(l(a)+l(b)) ={\rm Tr}(l(\alpha_1 a +\alpha_2 b ))$ imply that $l(a)l(b)=r(a)r(b)=0$?
Let $H$ be a Hilbert space and let $a,b\in B(H)$ be such that
$${\rm Tr}(l(a))={\rm Tr}(r(a))<\infty , {\rm Tr}(l(b))={\rm Tr}(r(b))<\infty,$$
$${\rm Tr}(l(a)+l(b)) ={\rm Tr}(l(\alpha_1 a +\...
- 617
3
votes
1
answer
332
views
Takesaki II Lemma 1.13: stuck in proof
Consider the following fragment from Takesaki's book "Theory of operator algebras II" (Lemma 1.13 on p8, in chapter VI "Left Hilbert algebras"):
Here, we associate with an ...
- 1,027
1
vote
0
answers
89
views
Let $T\in B(H)$. Then prove that $T$ is a contraction if and only if the closed unit disc is spectral set for $T$
Let's first recall the definition of Spectral set for a bounded operator $T$ on a hilbert space $H$. Let $X\subseteq \Bbb{C}$ be compact such that $\sigma(T)\subseteq X$. Define $$\mathcal{R}(X):=\...
- 111
4
votes
0
answers
152
views
Maximally fine topologies on $B(H)$ making the unit ball compact
Let $H$ be a Hilbert space, and $B(H)$ its algebra of bounded operators. One of the reasons the Ultraweak topology is (in a way) more useful than the weak operator topology is that the Ultraweak ...
- 190
-1
votes
1
answer
210
views
A commuting pair of isometries
Let $H$ be a Hilbert space and $B(H)$ be the space of all bounded operators on $H$.
The Wold decomposition says that: an operator $x$ in $B(H)$ is an isometry if and only if $x=x_u\oplus x_s$ where $...
- 66
4
votes
1
answer
371
views
Motivation for Heisenberg's modeling of observables
What's the motivation for observables to be modeled by self-adjoint operators? I can't seem to find any place where this is laid out clearly. Maybe von Neumann's book is decent, but it's not ...
- 433
2
votes
0
answers
56
views
Existence of a suitable smooth kernel
Denote by $H=L^2([0,1])$ the Hilbert space of square integrable real valued functions on the interval $[0,1]$. Does there exist a nontrivial smooth real valued function $k$ on the unit interval such ...
- 4,135
9
votes
1
answer
667
views
Reference for "Every compact quasinilpotent operator is the limit of nilpotent ones"
It was mentioned on Page 916 Problem 7 of Halmos's "Ten Problems in Hilbert space" that there is a proof for "Every compact quasinilpotent operator is the limit of nilpotent ones" ...
CommunityBot
- 1
3
votes
1
answer
475
views
Extension of a bounded linear functional
Let $\mathcal{H}$ be a finite-dimensional Hilbert space and $\mathcal{A}\subseteq\mathcal{B(\mathcal{H})}$ be an operator system. Suppose $T_n$ is a collection of all $n$-by-$n$ matrices equipped with ...
- 231
-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 ...
- 12.5k
6
votes
1
answer
788
views
A spectral description of Fredholm operators
Let $L:H \to H$ be a bounded operator on a Hilbert space $H$, with finite dimensional kernel, and whose adjoint also has finite dimensional kernel. Is it true that $L$ is Fredholm if and only if its ...
- 1
4
votes
1
answer
196
views
A kind of holomorphicity of maps on Hilbert space
Let $H$ be an infinite dimensional seperable Hilbert space. Is there an Irreducible involutive sub algebra $D$ of $B(H)$ with the following properties?:
1)For every open set $U\subset H$ and every ...
- 25.3k
3
votes
2
answers
135
views
Unicellular compact operators
An operator $T$ on a separable Hilbert space $H$ is called unicellular if any two closed invariant subspaces $M$ and $N$ are comparable; that is either $M\subseteq N$ or $N\subseteq M$. There are many ...
- 24.2k
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
$$\langle ...
- 18.7k
6
votes
0
answers
175
views
Functional calculus for the Dolbeault operator over Hilbert C*-modules
$\newcommand{\odd}{\mathrm{odd}}\newcommand{\even}{\mathrm{even}}$Let $X$ be a complex manifold, you can assume it's compact, if necessary. We have the Dolbeault complex $$0 \rightarrow \mathcal{A}^{0,...
- 63.9k
6
votes
1
answer
276
views
Characterize this subspace of the bounded operators on $ L^2(\mathbb{R}) $
I posted this on MSE a couple months ago and it got three upvotes but no answers or even comments so I decided to cross-post it here:
For every pair $ a,b $ of real numbers define the operator $ U_{a,...
- 42.8k
1
vote
1
answer
247
views
Linearity of the directional derivative of a convex functional at the minimum
Let $H$ be a Hilbert space, $T_+(H)$ the set of positive self-adjoint trace-class operators on $H$, and $f : T_+(H) \to [0,m]$ a non-negative, bounded, convex functional. I don't necessarily know that ...
- 6,357
3
votes
1
answer
260
views
norm estimates for Schatten class
Let $C
_p$ be the Schatten-p-classes on a separable Hilbert spaces, $p\ge 1$.
Let ${\rm Tr}$ be the standard trace.
Let $y\in C_p$ be a self-adjoint operator (or even a positive operator) and let $...
- 9,486
0
votes
1
answer
154
views
Why is $q(f,g) = (f-g,0)$ not adjointable?
Let $A= C([0,1])$ and $J= \{f \in A: f(0) = 0\}$. Consider the Hilbert $C^*$-module
$E:= A \oplus J$ (with the obvious right $A$-action and inner product). I want to prove that
$$q: E \to E: (f,g) \...
- 18.7k
0
votes
1
answer
152
views
Inequality between matrix elements of positive self-adjoint operators
We have three positive semi-definite self-adjoint operators $\hat{A}_-$, $\hat{B}$, $\hat{A}_+$ on the Hilbert space $\mathcal{H}$. They are unbounded operators and satisfy the following inequality
\...
- 456
2
votes
2
answers
235
views
Can a sum of commutators of selfadjoint bounded operators be a multiple of the identity?
Let $H$ be an infinite-dimensional Hilbert space, and $a_1,b_1,\ldots,a_k,b_k$ be bounded selfadjoint operators on $H$. Can the sum $\sum_{i=1}^k[a_i,b_i]$
be a (pure imaginary) multiple of the ...
- 9,486
2
votes
1
answer
335
views
When the adjoint of an unbounded operator on a Hilbert space coincides with the formal adjoint on its natural domain?
This is almost a copy of https://math.stackexchange.com/questions/3931318/when-the-adjoint-of-an-unbounded-operator-on-a-hilbert-space-coincides-with-the
I am trying to work with infinite matrices in ...
- 21
1
vote
0
answers
57
views
Continuity of a linear functional for sequences of projections
Let $T_+$ be the set of a positive trace-class operators over some separable Hilbert space and
$A: T_+ \to \mathbb{R}\cup \{\infty\}$ some linear functional.
In general, $A$ will not be continuous. ...
- 695
5
votes
1
answer
385
views
hereditary C*-subalgebra of a non-elementary simple C*-algebra
A is said to be elementary if A is isomorphic to some $K(H)$ or $M_n$.
A C*-subalgebra $B$ is said to be hereditary if for every $0≤a≤b∈B$ we have $a∈B$.
I wanted to know that is this statement true?
...
- 2,689
4
votes
0
answers
160
views
Solution without using any k-theory tools
Let $A$ be the UHF-algebra of type $2^{\infty}$. Suppose that $p$ and $q$ are two projections in $A$ and $\tau(p) = \tau(q)$, where $\tau$ is the unique normalized trace. Then there is a partail ...
- 581
1
vote
0
answers
70
views
Is $F: B(H)\to B(H\otimes K)$, $X\mapsto X\otimes \mathbb{1}_K$ completely contractive?
Take Hilbert spaces $H$ and $K$. Consider a linear map $F: B(H)\to B(H\otimes K)$, $X\mapsto X\otimes \mathbb{1}_K$.
Is it true that $F$ is completely contractive? If it is, I would be very grateful ...
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
$$\...
- 42.8k
0
votes
0
answers
132
views
Spectral Theorem for compact self-adjoint operators on real Hilbert spaces [duplicate]
Is the spectral theorem for self-adjoint compact operators on a Hilbert space also true if the Hilbert space is real (instead of complex)?
Wikipedia says this is true.
However, it seems to me that ...
- 63.9k
11
votes
0
answers
388
views
Von Neumann Inequality in Banach spaces
It is known that the only Banach space that satisfies the von-Neumann inequality is the Hilbert space:
Theorem (see e.g. Pisier, "Similarity Problems and Completely Bounded Maps", p 27) For a Banach ...
- 5,529
5
votes
1
answer
334
views
Iterated limits equal?
Consider the Banach algebra $B_2(H)$ of Hilbert Schmidt operators on a Hilbert space $H$ with Hilbert Schmidt norm. We know that $B_2(H)$ is a Hilbert space as well with $\left<A,B\right>=tr(B^*...
- 18.7k
0
votes
1
answer
673
views
Weak convergence of Hilbert Schmidt operators
So I am stuck at this situation. Let $\{A_n\}$ be a weakly convergent sequence in $B_2(H)$ converging to $0$ in the weak topology on $B_2(H)$. Which means that $\left<A_n,D\right>=\operatorname{...
4
votes
1
answer
110
views
Graded adjointable operators on a graded Hilbert space
Given a graded Hilbert space $\mathbf{H} = \bigoplus_{k \in \mathbb{N}} \mathbf{H}_k$, one might extend the notion of adjoint to a "graded adjoint" defined as follows: $L \in B(\mathbf{H})$ is said to ...
- 18.7k
6
votes
1
answer
119
views
Tensoring adjointable maps on Hilbert modules
Given a right Hilbert $A$-module $E$, and a right Hilbert $B$-module $F$, together with non-degenerate $*$-homomorphism $\phi:A \to \mathcal{L}_B(F)$, we can form the tensor product
$$
E \otimes_{\phi}...
- 18.7k
6
votes
2
answers
297
views
Finite-dimensional Hilbert $C^*$-modules
Does there exist a classification, or characterization, of finite-dimensional Hilbert $C^*$-modules? More generally, does there exist a characterization of countable direct sums of finite-dimensional ...
- 42.8k
2
votes
1
answer
916
views
Characterising closed range self-adjoint operators
Let $T:\mathrm{dom}(T) \subseteq H \to H$ be a densely defined, self-adjoint operator on a Hilbert spaces $𝐻$. In general the range of $T$ is not guaranteed to be closed. What tools are available to ...
- 18.7k
0
votes
1
answer
113
views
An adjoint characterization of (unbounded) Fredholm operators
Let $\mathcal{H}_i$, for $i=1,2$, be Hilbert spaces, and $T:{\frak Dom}(T) \subseteq \mathcal{H}_1 \to \mathcal{H}_2$ a densely-defined closed operator. If the kernel of $T$, and the kernel of its ...
- 18.7k
2
votes
1
answer
107
views
tensor stability of block-positive matrices
Let $X_{AB}$ be an operator acting on the tensor-product Hilbert space $\mathcal{H}_A \otimes \mathcal{H}_B$. Suppose that $X_{AB}$ is block positive, meaning that (in Dirac notation)
$\langle \psi |...
- 18.7k
5
votes
2
answers
642
views
Non-standard tensor products of inner product spaces
For two inner product spaces $(\mathcal{V}, (\cdot,\cdot)_V)$ and $(\mathcal{W}, (\cdot,\cdot)_W)$, we can put an inner product on their tensor product in the obvious way:
$$
(1) ~~~~ \langle v \...
- 643
2
votes
0
answers
116
views
Closable operators on Hilbert modules
For $T:{\frak{Dom}}(T) \to H$ a densely defined operator, with $H$ a (separable) Hilbert space, we know that $T$ is closable if its adjoint $T^*$ has dense domain in $H$.
Does this extend to the (...
- 95
5
votes
2
answers
273
views
Is a set of nuclear functionals equicontinuous in compact-open topology if it is equicontinuous on each compact set?
Let $H$ be a Hilbert space and $B(H)$ be its space of all (bounded) operators. A nuclear functional on $B(H)$ is a linear functional $f:B(H)\to{\mathbb C}$ that can be represented in the form
$$
f(A)=\...
- 41
8
votes
1
answer
570
views
Comparing the definitions of $K$-theory and $K$-homology for $C^*$-algebras
In Higson and Roe's Analytic K-homology, for a unital $C*$-algebra $A$, the definitions of K-theory and K-homology have quite a similar flavor.
Roughly, the group $K_0(A)$ is given by the ...
- 29.2k
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