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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 $H$,...
David Gao's user avatar
  • 2,830
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 ...
S-F's user avatar
  • 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 ...
B.Hueber's user avatar
  • 1,161
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 +\...
user92646's user avatar
  • 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):=\...
DeltaEpsilon's user avatar
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 ...
Aareyan Manzoor's user avatar
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) ...
JP McCarthy's user avatar
  • 1,027
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 ...
Ali's user avatar
  • 4,135
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 ...
Piku's user avatar
  • 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 ...
Dave Shulman's user avatar
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 ...
Markus's user avatar
  • 1,361
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 ...
Andromeda's user avatar
  • 175
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,...
Ian Gershon Teixeira's user avatar
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 $...
user92646's user avatar
  • 617
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 \...
Harry Song's user avatar
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" ...
Rye's user avatar
  • 191
-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 $...
ABB's user avatar
  • 4,058
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 ...
Ali Taghavi's user avatar
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 $$\...
Lao-tzu's user avatar
  • 1,906
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 ...
Leon Lang's user avatar
  • 253
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 ...
erz's user avatar
  • 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^*...
NewB's user avatar
  • 243
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{...
NewB's user avatar
  • 243
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 ...
Dave Shulman's user avatar
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}...
Dave Shulman's user avatar
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 ...
Dave Shulman's user avatar
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 \...
Pierre Dubois's user avatar
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 (...
Max Schattman's user avatar
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 ...
Max Schattman's user avatar
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 |...
Mark M. Wilde's user avatar
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 ...
Transcendental's user avatar
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 ...
Dave Shulman's user avatar
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 ...
Dave Shulman's user avatar
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 $...
inquisitor's user avatar
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 ...
Max Schattman's user avatar
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-...
Alexander Kukush's user avatar
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?
mathlover's user avatar
  • 227
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 ...
MSMalekan's user avatar
  • 2,118
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 ...
MSMalekan's user avatar
  • 2,118
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\...
0xbadf00d's user avatar
  • 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 ...
user92646's user avatar
  • 617
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 ...
Schüler's user avatar
  • 724
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 ...
geometricK's user avatar
  • 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 \...
Schüler's user avatar
  • 724
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 ...
heller's user avatar
  • 481
3 votes
1 answer
228 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?
Milan Bernolak's user avatar
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 ...
Don John Prep's user avatar
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: \...
ABB's user avatar
  • 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 ...
Ali Taghavi's user avatar
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 ...
David E Speyer's user avatar