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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 ...
Calculix's user avatar
  • 321
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 ...
user129564's user avatar
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 ...
Dominique Unruh's user avatar
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 ...
erz's user avatar
  • 5,529
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.) ...
André Henriques's user avatar
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 $T$...
WillG's user avatar
  • 233
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" ...
Rye's user avatar
  • 191
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,...
Nate Eldredge's user avatar
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\...
portella's user avatar
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 ...
J. E. Pascoe's user avatar
  • 1,429
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 ...
Delio Mugnolo's user avatar
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 ...
WeakMath's user avatar
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 ...
euleroid's user avatar
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?
user avatar
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 ...
potionowner's user avatar
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 $\mathbb{R}$-...
alexlo's user avatar
  • 131
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. ...
Alexander Shamov's user avatar
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 ...
Markus's user avatar
  • 1,361
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 ...
Tomas's user avatar
  • 879
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 ...
Benjamin's user avatar
  • 245
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: ...
Dave Shulman'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
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{...
Frank's user avatar
  • 241
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 ...
Konstantinos Kanakoglou's user avatar
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 ...
David's user avatar
  • 161
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, ...
Marco's user avatar
  • 51
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
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 ...
Markus's user avatar
  • 1,361
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 ...
SKNEE's user avatar
  • 51
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? ...
Peg Leg Jonathan's user avatar
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 ...
Gateau au fromage's user avatar
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 ...
groupoid's user avatar
  • 620
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 $H=\ell^2$ ...
truebaran's user avatar
  • 9,330
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 ...
Manuel Norman's user avatar
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 ...
Đức Anh's user avatar
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 $$...
Student's user avatar
  • 1,154
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 ...
Taras Banakh's user avatar
  • 41.9k
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 \...
hoj201's user avatar
  • 614
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
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. ...
Julien's user avatar
  • 660
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 ...
groupoid's user avatar
  • 620
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 ...
Overflowian's user avatar
  • 2,533
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
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 ...
guido giuliani's user avatar
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
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 ...
Arnold Neumaier's user avatar
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 ...
NewB's user avatar
  • 243
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\|<...
Markus's user avatar
  • 1,361
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 ...
Nik Weaver's user avatar
  • 42.8k
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,...
Iosif Pinelis's user avatar