All Questions
Tagged with hilbert-spaces operator-theory
161 questions
20
votes
0
answers
333
views
Existence of orthonormal basis for $L^2(G)$ in $C_c(G)$
Remark: I cross-posted this question on MSE and added a bounty to it.
Suppose that $G$ is a locally compact (Hausdorff) group endowed with the Haar measure. It is well-known that the compactly ...
15
votes
2
answers
660
views
Multiple of identity plus compact
Is there an example of a bounded operator $T\in\mathcal{B}(H)$, where $H$ is a separable complex Hilbert space, such that no restriction to an infinite dimensional closed subspace is multiple of ...
13
votes
3
answers
1k
views
Is the set of separable quantum states closed?
Let $\mathcal H,\mathcal H'$ be Hilbert spaces (not necessarily separable).
A "separable state" is a trace-class operator of the form $\sum_i \rho_i\otimes\rho_i'$ where $\rho_i,\rho_i'$ are positive ...
11
votes
0
answers
389
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 ...
11
votes
0
answers
529
views
Contraction semigroup on Hilbert space
I'd like to know whether a certain unbounded operator on a Hilbert space is the generator of a strongly continuous contraction semigroup.
(Such operators are known as maximally dissipative operators.)
...
10
votes
0
answers
225
views
Can the trace be computed in any Schauder basis?
I'm cross-posting this question from Math.SE, as it didn't get much attention there.
Let $H$ be a separable Hilbert space and $T \in L(H)$ a trace-class operator. It is well known that the trace of $T$...
9
votes
1
answer
669
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" ...
8
votes
1
answer
747
views
Strongly continuous semigroups that cannot be contractions
Let $X$ be a Banach space, and $(P_t)_{t \ge 0}$ a strongly continuous semigroup of bounded operators on $X$. Using the uniform boundedness principle, it's simple to prove that there are constants $M,...
8
votes
1
answer
844
views
A doubt about the parts of the spectrum of tensor products
Let $\mathcal{H}$ be any complex Hilbert space of infinite dimensional. By an operator $T$ I mean a linear bounded transformation from $\mathcal{H}$ into $\mathcal{H}$, i.e, $T:\mathcal{H}\rightarrow\...
8
votes
2
answers
640
views
Does a random sequence of vectors span a Hilbert space?
Let $\mathcal{H}$ be a separable Hilbert space. Let $v$ be a random variable taking values in $\mathcal{H}$ such that $P(v \perp h) < 1$ for all $h \in \mathcal{H}.$ Suppose we sample an infinite ...
8
votes
1
answer
359
views
Lax pairs in an abstract formalism
I am reading Integrals of Nonlinear Equations of
Evolution and Solitary Waves by Peter Lax and I'm having a hard time. The methods are pioneering, of course, but Lax does not bother much to provide ...
8
votes
0
answers
251
views
Struggling with a proof in the preprint 'Hermitian geometry on resolvent set'
I have been struggling for awhile with a particular argument in the paper below. I posted the question first on MathSE, but I got no answers. I understand however that MO might be an overreach for ...
7
votes
1
answer
959
views
a claim for a proof of the invariant subspace problem [closed]
Recently four mathematicians claimed to have proven the invariant subspace problem, which is the problem that states
Does every bounded operator on a separable Hilbert space have a non-trivial ...
7
votes
3
answers
881
views
Criterion for compactness
Let $T:H\to H$ be a continuous operator on a Hilbert space.
Assume there exists an orthonormal base $(e_j)_{j\in\mathbb N}$, such that the sequence $Te_j$ tends to zero.
Must $T$ be compact?
7
votes
3
answers
2k
views
Essential spectrum of multiplication operator
Let $a\in \mathcal{L}(L^2([0, 1], \mathbb{R}))$ be a multiplication operator. I wonder whether there is any work on calculating its essential spectrum. Is there any way to explicitly compute its ...
7
votes
1
answer
938
views
What is the idea behind interpolation spaces?
I am working through a text on Numerics for SPDEs and there the concept an interpolation (Hilbert-)space associated to an operator is used. To be specific:
Definition. Let $H$ be an $\mathbb{R}$-...
7
votes
1
answer
242
views
Is there a nice "minimum" of two symmetric operators?
Let $A$ and $B$ be two bounded symmetric positive operators in Hilbert space, such that $A-B$ is trace class. If needed, $A$ and $B$ may be assumed reasonably "small", let's say, Hilbert-Schmidt.
...
7
votes
1
answer
283
views
A characterization of Hilbert spaces by norm one projections
Suppose a (separable) Banach space $X$ has the following property: If $P:X\to X$ is a bounded projection different from $I$ such that $\|P\|=1$, then $\|I-P\|=1$. Does this imply that $X$ is a Hilbert ...
7
votes
1
answer
343
views
Is every $C_0$ semigroup on a Hilbert space automatically a $C_0$ group on a larger space?
Let $\{T(t),t\ge 0\}$ be a $C_0$ semigroup on a Hilbert space $X$, does that exist a larger Hilbert space $Y$ such that $X\subset Y$, and $T(t)$ extend to a $C_0$ group $T'(t)$ (so $t<0$ make ...
6
votes
2
answers
514
views
Convergence criterion in the domain of an unbounded operator
Cross-post from math.sx.
My question is somewhat close to this one, but the counterexamples given there do not apply here.
Setup. Given a Hilbert space $\mathcal H$, a closed operator $A$ and a ...
6
votes
2
answers
665
views
Unbounded Fredholms operators
Motivated by the situation of bounded Fredholm operators, I have the following question about "unbounded Fredholm operators".
Let $\mathcal{H}_1$ and $\mathcal{H}_2$ be two Hilbert spaces, and
$$
D: ...
6
votes
1
answer
765
views
An equivalence relation on the space of polynomials in one complex variable
Let $P(z)$ be a polynomial with complex variable $z$. We consider the following distribution for the roots of $P(z)=0$: the distribution is a triple $(n_{1},n_{2},n_{3})$
where these integers are ...
6
votes
1
answer
1k
views
Is the sum of spectral projections a projection?
Let $T$ be a closed operator on a Hilbert space with discrete spectrum. Then for $\{\lambda_1,...\lambda_n\}\in\sigma(T)$ one can define the spectral projections
$$P_{\{\lambda_1,...\lambda_n\}}=\frac{...
5
votes
2
answers
2k
views
On the domains and extensions of unbounded operators
I am not an expert in functional analysis but I was studying some, motivated from some mathematical physics considerations. I am not quite sure whether this is research-level, but let me state some ...
5
votes
2
answers
5k
views
When are two operators simultaneously diagonalisable?
I am reading a paper and they have diagonalised both operators in an equation, on a separable Hilbert space, with respect to the same basis. My question is, when can two operators be simultaneously ...
5
votes
1
answer
1k
views
Space of compact operators defined on separable Hilbert space
Let $X$ be a separable Banach space and consider $\mathcal{K}(X)$ the space of compact operators $K\colon X \rightarrow X$. Is it true that the space $\mathcal{K}(X)$ is separable? If yes, why? If no, ...
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^*...
5
votes
1
answer
465
views
Quasinilpotent , non-compact operators
If $X$ is a separable Banach space and $(\epsilon_n)\downarrow 0$, can we find a quasinilpotent, non-compact operator on $X$ such that $||T^n||^{1/n}<\epsilon_n$ for all $n$? I suspect the answer ...
5
votes
2
answers
276
views
Dilation of bounded linear operators
Let $H$ be a Hilbert space, and let $A$ be a contraction (bounded linear operator of norm $\leq 1$) on $H$. I heard in a recent talk that there is a (apparently famous) result due to Sz-Nagy which ...
5
votes
1
answer
388
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?
...
5
votes
2
answers
149
views
Showing an operator is (or not) closed on $L^2(\mathbb{R})$
I am linearizing nonlinear waves and get operators of the form below. Everything is considered in $L^2(\mathbb{R})$.
Consider the operator $L_1=\frac{d}{dx}$. The domain is $H^1(\mathbb{R})$ and it is ...
5
votes
1
answer
230
views
Canonical multiplication representation of self-adjoint operator in quantum chemistry and coding theory research
In my applied math research group, we are studying and going over functional analysis results from papers and theses from our institution to generalize their results and apply them in our discrete ...
5
votes
1
answer
201
views
$C(X)$-compact operators and families of compact operators
In this question Operators on Hilbert $C^*$-module and families of Fredholm operators I asked about the relation between being a family of compact operators $F:X \to K(H)$ on Hilbert space $H=\ell^2$ ...
5
votes
1
answer
179
views
An extension of Lomonosov Theorem
Let $H$ be a complex infinite dimensional separable Hilbert space. There are various extensions of the following well known result:
Theorem (Lomonosov): Every nonscalar $T \in B(H)$ which commutes ...
4
votes
3
answers
728
views
Inequality of von Neumann for more than two contractions
Good morning,
I'm doing the Master 2 Practice at the University of Toulouse 3, France, on the spectral Nevanlinna-Pick interpolation, via operator theory. This problem leads to study the symmetrized ...
4
votes
2
answers
353
views
Why $\lim_{n\to+\infty}\bigg(\bigg\|\sum_{f\in F(n,d)} A_{f}^* A_{f}\bigg\|^{\frac{1}{2n}} \bigg)\;\text{exists}?$
Let $E$ be a complex Hilbert space and $\mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$.
For $A= (A_1,\cdots,A_d)\in\mathcal{L}(E)^d$ (not necessary to be commuting). Why
$$...
4
votes
2
answers
244
views
Compact images of nowhere dense closed convex sets in a Hilbert space
Let $B=-B$ be a nowhere dense bounded closed convex set in the Hilbert space $\ell_2$ such that the linear hull of $B$ is dense in $\ell_2$.
Question. Is there a non-compact linear bounded operator ...
4
votes
2
answers
730
views
Finite dimensional approximations of operators on Hilbert spaces
Let $e_1,e_2,\dots$ be a Schauder basis for a Hilbert space $(V , \langle \cdot , \cdot \rangle)$. Let $A:V \to V$ be an operator. Finally, let $V_n = {\rm span}( e_1, \dots, e_n)$. Let $i_n : V_n \...
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?
4
votes
1
answer
461
views
On the self-adjoint part of a quasinilpotent operator
Disclaimer: this is not research-level, but I've read some non research-level questions/answers on quasinilpotent operators here, some of them involving renowned users. So I thought I'd give it a try. ...
4
votes
1
answer
301
views
Trying to recover a proof of the spectral mapping theorem from old thesis/paper with continuous functional calculus
In my research group in functional analysis and operator theory (where we do physics and computer science as well), we saw in an old Russian combination paper/PhD thesis in our library a nice claim ...
4
votes
1
answer
378
views
Closure of the space of Fredholm operators
Let $X,Y$ be two Banach spaces.
A bounded operator $A$ is Fredholm if $\ker A$ and $\mathrm{coker} A$ are finite dimensional. Denote by $Fred(X,Y) \subset \mathcal{L}(X,Y)$ the space of Fredholm ...
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 ...
4
votes
1
answer
92
views
Continuous section inside a family of rank-varying operators
Good morning everybody,
my question is as follows: let $K$ be a compact set and assume $F:K\to L(\mathcal H,\mathbb R^{m+1})$ be a continuous map from the compact set $K$ to the space of linear ...
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 ...
4
votes
0
answers
2k
views
Eigenvalues and spectrum of the adjoint
In a finite-dimensional Hilbert space, the eigenvalues of the adjoint $A^*$ of an operator $A$ are the complex conjugates of the eigenvalues of $A$.
But in infinite dimensions this need no longer be ...
3
votes
1
answer
555
views
Trace norm of operators obtained by restricting the matrix of a trace class operator
Suppose $H$ is a Hilbert space with orthonormal basis $\{e_i\}_{i\in \mathbb N}$. To every operator $T$, we associate a infinite matrix $[T_{ij}]$, where $T_{ij}=\left<Te_j,e_i\right>$. We know ...
3
votes
1
answer
157
views
Operator in the commutant which is small on a given vector
Suppose $x$ is a non-zero vector in a Banach space, and $T$ is a fixed operator. Is the following true:
For any $\varepsilon, \delta$, there exists $S$ in the commutant of $T$ such that $1\leq\|S\|<...
3
votes
1
answer
724
views
Is this consequence of the invariant subspace problem known?
An interesting fact popped out of a paper I'm writing: if the invariant subspace problem for Hilbert space operators has a positive solution, then every $A \in B(\mathcal{H})$ can be made "upper ...
3
votes
1
answer
261
views
Can the $L^{\infty}\to L^{\infty}$ norm be bounded by the trace norm?
Let $k\in C(\mathbb{R}^2; \mathbb{R})$ be a continuous function. Suppose that the operator $K\colon L^2(\mathbb{R}) \rightarrow L^2(\mathbb{R})$ defined by the formula
$$(Kf)(x)=\int_{\mathbb{R}} k(x,...