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Convex hulls of compact sets

Let $A$ be a compact set in a separable Hilbert space $H$, and let $\bar A$ denote its convex hull. Is $\bar A$ compact?
Tom LaGatta's user avatar
  • 8,512
2 votes
1 answer
257 views

Cardinality of the set of Boolean subalgebras of the lattice of projections on a Hilbert space

I have a simple question I've managed to get myself quite confused about. Given a Hilbert space H, what do we know about the cardinality of (a) the set $P(H)$ of projection operators onto $H$ (...
King Kong's user avatar
  • 631
18 votes
2 answers
1k views

compact-open topology on $B(H)$

In topology, it is common to use the compact-open topology on the set of continuous maps between two given topological spaces. Let now $H$ be a Hilbert space and $B(H)$ the set of continuous linear ...
André Henriques's user avatar
4 votes
0 answers
112 views

status of Invariant subspace problem on Krein Space

What is the status of Invariant subspace problem on Krein Space? What sort of developments have taken place in this area.
Koushik's user avatar
  • 2,106
2 votes
0 answers
86 views

Terminology and reference question

I am working on a problem involving bilinear forms over complex Hilbert spaces, and in my case it is not natural to make the forms sesquilinear, i.e., $a(u,v)$ is linear in both complex arguments. ...
Nemis L.'s user avatar
  • 143
0 votes
1 answer
265 views

find a weak solution in an intersection of Sobolev spaces

In using-lax-milgram-to-find-a-weak-solution-in-an-intersection-of-sobolev-spaces the weak solution for $$ -\Delta^2 u = f \in L^2(U)\\ \\ u|_{\partial U}=\Delta u|_{\partial U} = 0 $$ was discussed,...
All's user avatar
  • 111
9 votes
2 answers
485 views

why is this a sufficient condition for a domain to be a core of an unbounded operator?

Let $\alpha:\mathbb R\to U(H)$ be a strongly continuous action of the reals on some Hilbert space, and let $A=-i\frac d{dt}\alpha(t)|_{t=0}$ be its infinitesimal generator, so that $\alpha(t)=e^{itA}$....
André Henriques's user avatar
2 votes
1 answer
178 views

Self-adjointness of a perturbed quantum mechanical Hamiltonian specified in an infinite matrix form

Consider an operator $H$ on the Hilbert space $\ell_2$ given as an infinite matrix with two pieces, one diagonal and one arbitrary: $H_{ij}=E_i\delta_{ij}+V_{ij}$. This has a physical meaning in ...
Slava Rychkov's user avatar
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
8 votes
1 answer
548 views

Product of commuting nonnegative operators

Let $V$ be a real vector space with an inner product and $A,B : V \to V$ linear maps which are self-adjoint nonnegative-definite, i.e. $\langle Ax,y \rangle = \langle x,Ay \rangle$ and $\langle Ax,x \...
Martin Brandenburg's user avatar
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
3 votes
0 answers
168 views

Deleting "weak homeomorphism" in a Hilbert space

It is well-known that there exists a homeomorphism $h$ from an infinite-dimensional Hilbert space $H$ to $H\setminus\{0\}$. Does there exist a "weak homeomorphism" $g:H \to H\setminus\{0\}$, that is, $...
Ilnara's user avatar
  • 91
4 votes
1 answer
1k views

RKHSs containing constant functions

Suppose $H$ is the reproducing kernel Hilbert space on a space $X$ with reproducing kernel $K$. If, say, $K - c$ is a positive definite kernel for some $c>0$ then $H$ contains the constant ...
Mark Meckes's user avatar
  • 11.4k
1 vote
1 answer
184 views

Special kind of operators

Consider an operator $A: H \longrightarrow X$ ($H$ is a Hilbert space and $X$ is a Banach space) that has a representation $$ A = \sum_{j=0}^\infty a_j \langle \cdot, e_j\rangle \cdot x_j,$$ where $(...
Matthias Ludewig's user avatar
3 votes
1 answer
174 views

A space parameterizing the choices of orthonormal bases for a Hilbert space

Let $\mathcal{H}$ be an infinite dimensional separable (complex) Hilbert space. What is a natural space which parameterizes the choices of orthonormal bases for $\mathcal{H}$? It seems like one ...
Iian Smythe's user avatar
  • 3,115
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
2 votes
1 answer
959 views

Do kernels provide a basis for a RKHS?

Let $H$ be a Reproducing Kernel Hilbert Space with elements $f:X\rightarrow \mathbb{C}$, with kernel $K(x, y)$. My question is whether, for some choice of $x_i\in X$, it is the case that $u_i:=K(x_i, \...
gappy3000's user avatar
  • 461
2 votes
1 answer
1k views

Coercive Symmetric Bilinear form on a Hilbert space

I need to show one of the two following equivalent results. If true, it must be a simple proof but I do not seem to be able to make it work. Thank you in advance. 1) Consider a continuous symmetric ...
Gateau au fromage's user avatar
1 vote
1 answer
499 views

For a bounded sequence in a hilbert space, does $\|u_n\|^2 u_n \to \|u_0\|^2u_0$ ?

If $\{u_n\}$ is bounded in a real Hilbert space $H$, with inner product $(\cdot,\cdot)$, then ${\|u_n\|^2u_n}$ is also bounded. As there is a weakly converging sub-sequence, we can WLOG assume that $\...
jiahua's user avatar
  • 11
2 votes
1 answer
386 views

Decomposing bilinear forms in Hilbert spaces

You are given a complex Hilbert space $H$ with two equivalent Hilbert space structures $<,>$ and $<,>'$. Define $<,>''=<,> + <,>'$ to be the sum of our two scalar ...
Hammerhead's user avatar
  • 1,211
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
10 votes
3 answers
1k views

ordered exponential of unbounded operators

Let $H$ be a Hilbert space, and let $A_t$ be a family of unbounded positive (self-adjoint) operators on $H$ parametrized by $\mathbb t\in R_{\ge 0}$. Consider the ordinary differential equation $$ \...
André Henriques's user avatar
0 votes
2 answers
1k views

Weak versus strong convergence

This is my first time posting. I am well aware that an $L^2$ weakly converging sequence is not convergent in the corresponding strong topology. However, my question is as follows, do the sequence of ...
dcs24's user avatar
  • 213
0 votes
1 answer
795 views

Can we construct a Hilbert space where the operator following differencial operator is symmetric?

I'd like to know if one can define a pertinent Hilbert space where the operator $$A_p v := -\frac{1}{2} v" + (vF + v\int_\mathbb{R} Sp + p\int_\mathbb{R} Sv )'$$ is symmetric. Here, $p$ satisfies ...
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
3 votes
1 answer
3k views

Approximate point spectrum

I have a question concerning the relation between the approximate point spectrum and the spectrum of an operator. Let $T$ be a bounded linear operator of a complex Hilbert space $H$. The approximate ...
Đức Anh's user avatar
3 votes
1 answer
598 views

is a non-invertible operator a boundary point of the group of invertible operators?

Good evening, I have a question concerning non-invertible operators. Let $H$ be a Hilbert space and $T$ a non-invertible bounded operator on $H.$ Is it true that $T$ is the limit of some sequence ...
Đức Anh's user avatar
5 votes
1 answer
3k views

Inner product of linear bounded operators between Hilbert spaces

Let $X$ and $Y$ be Hilbert spaces, and let $L(X,Y)$ be the set of bounded linear operators between Hilbert spaces. Can we equip $L(X,Y)$ with a natural inner product? I think it should look like $\...
shuhalo's user avatar
  • 5,327
0 votes
1 answer
261 views

Norm functionals of $B(H)$ restricted to sub ven-Neumann algebras [closed]

Let $H$ be a Hilbert space, we know that weak topology over $B(H)$, operator algebra of bounded linear operators from $H$ into $H$, is the topology generated by $\{\langle \cdot \xi,\eta\rangle:\; \...
Mahmood Alaghmandan's user avatar
6 votes
3 answers
1k views

Topological vector spaces that are isomorphic to their duals

After reviewing the (locally convex) topological vector spaces that I know, the only examples I could find where there is an isomorphism from the space to its (anti)dual, are Hilbert spaces. So my ...
Oliver's user avatar
  • 357
1 vote
0 answers
149 views

Banach spaces with simple best approximate solutions

Let $\langle V,||.||\rangle$ be a Banach space such that: $\;\;$ for all continuous linear maps $\: L : V\to V \:$ and members $v$ of $V$, there exists a unqiue member $u$ of $V$ $\;\;$ that ...
user avatar
10 votes
1 answer
869 views

Complement of a subspace which is a cartesian product

Let $H$ be a Hilbert space and $U$ a closed subspace of $H\times H$ . Does then exist closed subspaces $V$ and $W$ of $H$ such that $H\times H = U \oplus (V\times W)$ ? See also Perturbations of an ...
jjcale's user avatar
  • 2,753
2 votes
1 answer
901 views

Geometry of the Hilbert sphere

Let $X$ be the unit sphere in $\ell^2$, i.e. $X=\{x\in\ell^2: \|x\|=1\}$. Let the metric on $X$ be the geodesic metric, i.e. $d(x,y)=\cos^{-1}\langle x,y\rangle$. Call a set a ball-intersection if ...
TCL's user avatar
  • 744
2 votes
1 answer
452 views

What do we get from an euclidian affine structure ?

Imagine you investigate a set of objects $\mathcal{E}$, and you just realize this that $\mathcal{E}$ possesses an affine structure with respect to some real vector space $\mathcal{V}$ having a scalar ...
Adrien Hardy's user avatar
  • 2,135
-1 votes
1 answer
934 views

Domain and exponential of self- adjoint operator

Let $A$ be a self - adjoint operator on a Hilbert space $\mathcal{H}$ and let $D(A)$ be its domain. If $\psi \in D(A)$ then $exp(-itA) \psi \in D(A)$ iff $A$ is bounded ? Thank ...
Physics  beginner's user avatar
28 votes
2 answers
1k views

Can an operator have Exp(z) as its characteristic "polynomial"?

Let $\mathcal{H}$ be a Hilbert space, and let $T: \mathcal{H} \rightarrow \mathcal{H}$ be a trace-class operator. Define $$ f_T(z) = \sum_{i=0}^\infty \mbox{Tr}(\wedge^k T) \cdot z^k, $$ the ...
John Wiltshire-Gordon's user avatar
0 votes
0 answers
155 views

General form of a symplectic map

A symplectic automorphism of a Hilbert space has the form $T=U(\cosh S+J\sinh S)$ for a unitary $U$, an antilinear involution $J$ and a positive operator $S$. In fact a version of this goes through in ...
Ollie's user avatar
  • 1,411
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
7 votes
2 answers
7k views

Dual operators between Hilbert spaces: with or without Riesz representation

Let $X$ and $Y$ be Hilbert spaces over the real numbers (so complex conjugation plays no role, and everything will be linear in the strict sense). Let $f : X \rightarrow Y$ be a linear continuous ...
shuhalo's user avatar
  • 5,327
4 votes
1 answer
1k views

Doubts on Reproducing Kernel Hilbert Spaces and orthogonal decomposition

I'm a CS student and I'm trying to learn RKHS theory to understand the passages made in this paper . Among the bibliography I'm using there are "On the mathematical fundamentals of learning" and "...
user16348's user avatar
  • 151
0 votes
1 answer
2k views

What is the orthonormal basis for the Bergman space on the disk?

[EDIT by YC: the original question's title asked about a basis for the Hardy space on the disk. It is clear from the actual question that what was meant was the Bergman space.] In arXiv:0310.5297, ...
john mangual's user avatar
  • 22.8k
5 votes
1 answer
578 views

Infimum over all vector-valued L^2 spaces

Suppose I have a Banach space $E$ (which may be finite dimensional if you wish), a Hilbert space $H$ and a tensor $\tau \in H\otimes E$ in the algebraic tensor product. There are lots of ways to ...
Matthew Daws's user avatar
  • 18.7k
3 votes
1 answer
588 views

orthonormal basis of eigenvectors for laplacian on a concave polygon

I am interested in the Laplace operator $\Delta$ on a concave polygon. When the polygon is convex, it is known that $\Delta: H^2(\Omega) \rightarrow L^2(\Omega)$ is boundedly invertible. In addition, ...
localizer's user avatar
0 votes
1 answer
474 views

Hilbert space having all norms (and seminorms) continous.

Suppose I have a Hilbert space $H$ such that every seminorm on $H$ is continuous with respect to the inner-product induced norm. Is $H$ necessarily finite-dimensional? If not, is there an easy ...
RadonNikodym's user avatar
2 votes
1 answer
205 views

Do unitary bijections act invariantly on irreducible representations?

Let $\mathcal{A}$ be a $C^*$ algebra. Let $(\pi, \mathcal{H})$ be a faithful, irreducible, unitary, Hilbert space representation of $\mathcal{A}$; i.e., $\pi:\mathcal{A}\rightarrow\mathcal{B}(\mathcal{...
soulphysics's user avatar
18 votes
2 answers
1k views

Complex structure on $L^2(\mathbb R)$ generalizing the Hilbert transform

The Hilbert transform on the real Hilbert space $L^2(\mathbb R)$ is the singular integral operator $$ \mathcal H(f)(x) := \frac{1}{\pi} \int_{-\infty}^\infty \frac{1}{x-y} f(y) dy. $$ It satisfies $\...
André Henriques's user avatar
22 votes
3 answers
7k views

Subspace of $L^2$ that lies in $L^\infty$

Let $E$ be a closed subspace of $L^2[0,1]$. Suppose that $E\subset{}L^\infty[0,1]$. Is it true that $E$ is finite dimensional? PS. This is actually a question from the real analysis qualifier. I came ...
Rostyslav Kravchenko's user avatar
9 votes
2 answers
2k views

Nice Classes of Non-Closable Operators

The only thing I know about non-closable operators can be summarised as "they exist, but they're nasty, so let's not talk about them!" This seems to be the case with everyone else I've talked to. I'd ...
Ollie's user avatar
  • 1,411
3 votes
1 answer
1k views

If $H$ is a separable Hilbert space, is its dual dense in $L^2(H)$?

Let $H$ be an infinite-dimensional, separable Hilbert space, and let $\gamma$ be a Radon probability measure on $H$ with mean zero and covariance operator the identity $I$. Let $H^*$ denote the space ...
Tom LaGatta's user avatar
  • 8,512
7 votes
1 answer
1k views

If $H$ is a separable Hilbert space, is $L^2(H)$ separable?

Let $H$ be a separable Hilbert space, and let $\gamma$ be a Radon probability measure on $H$ with mean zero and covariance operator the identity $I$. Is the Hilbert space $L^2(H,\gamma)$ separable?
Tom LaGatta's user avatar
  • 8,512

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