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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
1 vote
0 answers
120 views

Injective differential of linear operators on a Hilbertspace

Given a complex Hilbertspace $\mathcal{H}$ of dimension $\dim(\mathcal{H}) = d$ and the set $$\mathcal{F} := \{q\in L(\mathcal{H})\vert\quad \text{rank}(q) = 4 \quad \wedge \lambda^q_{1,2} < 0\ \ \...
Alpha001's user avatar
  • 143
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
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,...
Kashayar's user avatar
  • 146
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
4 votes
1 answer
157 views

Geometric Motivation for Hilbert $C^*$-Bimodules

I'm trying to get an understanding of Hilbert $C^*$-bi-modules from a geometric point of view. As is well-known, we have that i) Commutative unital $C^*$-algebras correspond to compact Hausdorff ...
Ago Szekeres's user avatar
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
3 votes
0 answers
152 views

Whether a projection can "overlap" certain projections yet not commute with them

Question here about how the projections of a von Neumann algebra $\mathcal{R}$ might be arranged, relative to a projection that is not in $\mathcal{R}$. Stipulate the following:     $H$ is ...
Doug McLellan's user avatar
0 votes
1 answer
134 views

Bounded operators $T: B(H)\to H$ whose Kernel is a Lie algebra

Assume that $H$ is an infinite dimensional Hilbert space.The space of all bounded operators on $H$ is denoted by $B(H)$.We consider the Lie algebra structure $[T,S]=TS-ST$ on $B(H)$. Is there a ...
Ali Taghavi'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
3 votes
0 answers
144 views

Deformation and Representations

Let $\widetilde{U_q(sl_n)}$ denote a deformation of the algebra $U_q(sl_n)$. In particular, $\widetilde{U_q(sl_n)}$ is defined by the same generators and relations and $*$-operations as $U_q(sl_n)$ ...
Kendra Haynes's user avatar
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
5 votes
1 answer
516 views

Stinespring's dilation without $C^{\ast}$-algebras

Does Stinespring's dilation theorem hold if the algebra of interest is a topological $\ast$-algebra instead of the usual $C^{\ast}$-algebra? I will now state the version of Stinespring's dilation ...
sunspots's user avatar
  • 161
4 votes
1 answer
384 views

A Hilbert-space completion of a Hilbert $ C^{*} $-module over a separable $ C^{*} $-algebra

Let $ B $ be a separable $ C^{*} $-algebra and $ \mathcal{E} $ a Hilbert $ B $-module. We know that $ B $ has a faithful state $ \phi $. Using $ \phi $, we can construct a $ \mathbb{C} $-valued pre-...
Transcendental's user avatar
2 votes
1 answer
926 views

Eigenvalues and Compact Resolvent

For $A$ an unbounded (densely defined) operator on a separable Hilbert space, what conditions on its eigenvalues will show that, for $\lambda \notin $spec$(A)$, we have that $(A-\lambda)^{-1}$ is a ...
Juan Corrida's user avatar
0 votes
0 answers
255 views

Bounded operators with infinite matrix representations

I asked this question on StackExchange originally, but I'm giving it a go here as well. Suppose that $A$ is a unital $C^*$-algebra, $\varphi\colon A\to B(H)$ is a unital, completely positive map and ...
Bryder's user avatar
  • 123
4 votes
1 answer
386 views

Invertible unbounded linear maps defined on a Hilbert space

It is well-known that, assuming the axiom of choice, there are unbounded linear maps defined not only on a dense subset but on all of Hilbert space. Is it possible that such a map is invertible?
Arnold Neumaier's user avatar
1 vote
1 answer
176 views

Does a dense submodule of a free module always contain a basis?

Let $R$ be a completed normed ring, eg Banach algebra. Suppose that $F$ is a free $R$-module of infinite rank with a norm defined by the square root of sum of all norms of its components. If $F'$ is a ...
yeshengkui's user avatar
  • 1,373
3 votes
1 answer
1k views

Self-adjoint bounded operator, resolution of the identity, def. of the diagonal

Let $A$ be a self adjoint bounded linear operator with a continuous spectrum $\sigma(A)=[a,b]$ which acts on a separable Hilbert space. Let $E_\lambda$ be its resolution of the identity. For ...
Yakov Dymarskii's user avatar
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" ...
Mahdiyar's user avatar
  • 355
0 votes
2 answers
377 views

"Frobenius-finite" linear operators on a Hilbert Space

Let $H = L_2(S)$ be the complex Hilbert space over $S$ with the counting measure. (There might be another term for this concept, but) I define a continuous linear operator $L$ on $H$ with matrix ...
user avatar
2 votes
3 answers
1k views

Norm on quotient algebra of a tensor algebra

Suppose you have a finite dimensional real Hilbert space $V$ and you form the tensor algebra $$T(V) = \mathbb{R} \oplus V \oplus (V\otimes V) \oplus (V\otimes V \otimes V) \oplus \cdots$$ where the ...
Daoud's user avatar
  • 223
9 votes
5 answers
870 views

Abelianization of GL(H)

This is related to Theo's question about the abelianizations of finite dimensionsal Lie groups. I am interested in a specific (infinite-dimensional) case of the above question. Let H be an infinite-...
Mike Hartglass's user avatar

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