All Questions
57 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
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
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
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
$${\rm Tr}(l(a))={\rm Tr}(r(a))<\infty , {\rm Tr}(l(b))={\rm Tr}(r(b))<\infty,$$
$${\rm Tr}(l(a)+l(b)) ={\rm Tr}(l(\alpha_1 a +\...
- 617
1
vote
0
answers
89
views
Let $T\in B(H)$. Then prove that $T$ is a contraction if and only if the closed unit disc is spectral set for $T$
Let's first recall the definition of Spectral set for a bounded operator $T$ on a hilbert space $H$. Let $X\subseteq \Bbb{C}$ be compact such that $\sigma(T)\subseteq X$. Define $$\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
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
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
$$\langle ...
- 18.7k
6
votes
1
answer
277
views
Characterize this subspace of the bounded operators on $ L^2(\mathbb{R}) $
I posted this on MSE a couple months ago and it got three upvotes but no answers or even comments so I decided to cross-post it here:
For every pair $ a,b $ of real numbers define the operator $ U_{a,...
- 42.8k
3
votes
1
answer
260
views
norm estimates for Schatten class
Let $C
_p$ be the Schatten-p-classes on a separable Hilbert spaces, $p\ge 1$.
Let ${\rm Tr}$ be the standard trace.
Let $y\in C_p$ be a self-adjoint operator (or even a positive operator) and let $...
- 9,486
0
votes
1
answer
152
views
Inequality between matrix elements of positive self-adjoint operators
We have three positive semi-definite self-adjoint operators $\hat{A}_-$, $\hat{B}$, $\hat{A}_+$ on the Hilbert space $\mathcal{H}$. They are unbounded operators and satisfy the following inequality
\...
- 456
8
votes
1
answer
393
views
A question about comparison of positive self-adjoint operators
I have the following question but have no idea on its proof (one direction is trivial):
Let $A$ and $B$ be (bounded) positive self-adjoint operators on a complex Hilbert space $H$. Prove that
$$\...
- 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
$$
E \otimes_{\phi}...
- 18.7k
2
votes
1
answer
916
views
Characterising closed range self-adjoint operators
Let $T:\mathrm{dom}(T) \subseteq H \to H$ be a densely defined, self-adjoint operator on a Hilbert spaces $𝐻$. In general the range of $T$ is not guaranteed to be closed. What tools are available to ...
- 18.7k
2
votes
1
answer
107
views
tensor stability of block-positive matrices
Let $X_{AB}$ be an operator acting on the tensor-product Hilbert space $\mathcal{H}_A \otimes \mathcal{H}_B$. Suppose that $X_{AB}$ is block positive, meaning that (in Dirac notation)
$\langle \psi |...
- 18.7k
5
votes
2
answers
642
views
Non-standard tensor products of inner product spaces
For two inner product spaces $(\mathcal{V}, (\cdot,\cdot)_V)$ and $(\mathcal{W}, (\cdot,\cdot)_W)$, we can put an inner product on their tensor product in the obvious way:
$$
(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
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
2
votes
1
answer
407
views
Strongly Continuous Group Actions on the $ C^{\ast} $-Algebra of Compact Operators on a Hilbert Space
Let $ \mathcal{H} $ be a not-necessarily-separable Hilbert space. Let $ G $ be a locally compact Hausdorff group. It is easy to see that if $ U: G \to \mathbb{U}(\mathcal{H}) $ is a norm-continuous ...
- 42.8k
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 ...
- 23k
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 ...
- 356
2
votes
1
answer
94
views
Fredholmness of formal selfadjoint operator $AA^*$ and Fredholmenss of $A$
Let $X$ and $Y$ be Hilbert spaces with respective inner products $\langle , \rangle_{X,Y}$. Let $A:X \rightarrow Y$ be a bounded linear operator. Assume there is a non-degenerate sesquilinear product $...
- 5,407
4
votes
1
answer
357
views
Extending maps from dense $*$-algebras of $C^*$-algebras
Given $\cal{A},\cal{B}$ two dense $*$-algebras of two $C^*$-algebras $A$ and $B$ respectively, together with a $*$-algebra homomorphism $f:\cal{A} \to \cal{B}$, is it clear that $f$ extends to a ...
- 18.7k
3
votes
1
answer
266
views
Convergence of nuclear operators
Let $H$ be a separable infinite-dimensional real Hilbert space. We consider operators in $H.$
Nuclear norm of a nuclear operator is the sum of its singular values.
A nuclear, positive and self-...
- 61.9k
1
vote
1
answer
171
views
On projection theory for inseparable Hilbert spaces
How can one see that $I$ is an infinite projection in $B(\mathcal{H})$, where $\mathcal{H}$ is an inseparable Hilbert space?
- 3,574
2
votes
2
answers
178
views
Point spectrum of a positive invertible operator
Let $G$ be a l.c. group and $f$ belong to $C_c(G)$, the space of continuous functions with compact support. Define an operator$T_f$ on $L^2(G)$ by $T_f(g)=f*g$ (the convolution product). If $T_f$ is ...
- 12.5k
1
vote
1
answer
133
views
Does the image of $f$ contain a positive number?
Let $H$ be a Hilbert space and $T$ be a bounded and positive operator on $H$. Define a real function $f$ on positive real numbers by
$$f(r):=\|(r+T)^{-1}\|^{-1}-r\quad(r\in\mathbb R_+).$$
Does the ...
- 60.5k
1
vote
0
answers
74
views
If $f$ takes values in $L(H,L(H,\Bbb R))$ and $μ$ is a $H\hat ⊗_πH$-valued measure, how are $\int f\:dμ$ and $\int f⊗_π\text{id}_Hdμ$ related?
Let
$H$ be a separable $\mathbb R$-Hilbert space
$H\:\hat\otimes_\pi\:H$ denote the projective tensor product of $H$ and $H$
$(\Omega,\mathcal A)$ be a measurable space
$\mu$ be a $H\:\hat\otimes_\pi\...
- 167
0
votes
1
answer
203
views
For $B=\int \lambda d E_\lambda $ and $X$ commutes with every $E_\lambda $, why $BX$ is positive and self-adjoint?
Let $B$ be an unbounded closed operator on a Hilbert space $H$. If $B=\int \lambda d E_\lambda $ is positive self-adjoint and a positive bounded operator $X$ commutes with every $E_\lambda $, then why ...
- 16.2k
0
votes
1
answer
365
views
When $\lambda$-commutativity implies commutativity?
Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite-dimensional complex Hilbert $F$.
Let $T,S\in\mathcal{B}(F)$. The pair $(T,S)$ is said to $\lambda$-commute if there ...
- 2,670
1
vote
0
answers
233
views
Bochner integrals with values in a Hilbert $A$-module
I'm wondering whether there exists a generalisation of Bochner integration with values in a Hilbert $A$-module $M$, where $A$ is a general $C^*$-algebra rather than $\mathbb{C}$ (and whether there are ...
- 1,903
2
votes
1
answer
411
views
Problem of convergence of the following sequence
Let $E$ be a complex Hilbert space, with inner product $\langle\cdot\;, \;\cdot\rangle$ and the norm $\|\cdot\|$. Let $T\in \mathcal{L}(E)$
be bounded linear operators from $E$ to $E$ and $M\in \...
- 3,984
2
votes
2
answers
260
views
Bounded operators leaving dense subspace invariant
Let $A$ be a C$^*$-algebra. A pre-Hilbert $A$-module $H$ is a right $A$ module with a $A$-valued inner product (which is linear in the second variable and conjugate linear in the first variable) such ...
- 481
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?
- 24.2k
2
votes
3
answers
865
views
The multiplier algebra of a Reproducing Kernel Hilbert Space and its commutant
In my research in the theory of Reproducing Kernel Hilbert Spaces I was concerned with this topic which came up but I could not find a reference on:
If $ \mathbb{H} $ is an RKHS and we denote the ...
- 550
4
votes
0
answers
185
views
A strongly open set which is not measurable in the weak operator topology
Let $H$ be a non-separable Hilbert space and $\{e_i\}_{i\in I}$ be an orthonormal basis for $H$. Let $J$ be a uncountable proper subset in $I$.
Let us put $$E=\{x\in B(H): \lVert xe_j\rVert <1: \...
- 4,058
6
votes
1
answer
765
views
An equivalence relation on the space of polynomials in one complex variable
Let $P(z)$ be a polynomial with complex variable $z$. We consider the following distribution for the roots of $P(z)=0$: the distribution is a triple $(n_{1},n_{2},n_{3})$
where these integers are ...
CommunityBot
- 1
11
votes
2
answers
2k
views
Schur's Lemma for Hilbert spaces
Let $H$ be a complex Hilbert space and let a group $G$ act on $H$ such that there are no invariant closed subspaces besides $H$ and $(0)$. Let $D$ be the ring of bounded operators which commute with ...
CommunityBot
- 1