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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
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 $D(|A|^{1/2})$ ...
user_35's user avatar
  • 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 ...
S-F's user avatar
  • 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 ...
MBlrd's user avatar
  • 33
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
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 ...
GBA's user avatar
  • 167
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
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 ...
Andromeda's user avatar
  • 175
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 ...
MrPajeet's user avatar
  • 433
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
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) \...
Andromeda's user avatar
  • 175
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
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 ...
Fabien Besnard'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
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 ...
apyshkin's user avatar
-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
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. ...
Artemy's user avatar
  • 695
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 ...
Artemy's user avatar
  • 695
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 ...
Peg Leg Jonathan's user avatar
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? ...
Peg Leg Jonathan's user avatar
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
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 ...
Maria  Dmitrieva'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
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 ...
Max Schattman'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
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 ...
Max Schattman'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
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 ...
Max Schattman'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
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)=\...
Sergei Akbarov's user avatar
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 ...
Dave Shulman'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
0 votes
2 answers
465 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 ...
Bas Winkelman'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