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29 votes
6 answers
9k views

Nonseparable Hilbert spaces

Being nonseparable Banach space is in fact nothing special: one meets the first examples in the standard functional analysis course, when one learns about $\ell^p$ or $L^p[0,1]$ spaces-these spaces ...
truebaran's user avatar
  • 9,330
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
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
21 votes
0 answers
732 views

Closed connected additive subgroups of the Hilbert space

It is a classical result that a closed and connected additive subgroup of $\mathbb{R}^n$ is necessarily a linear subspace. However, this is no longer true in infinite dimension: a very easy example is ...
Pietro Majer's user avatar
  • 60.5k
20 votes
12 answers
9k views

The role of completeness in Hilbert Spaces

Why do Hilbert spaces have to be complete? I've been studying (teaching myself about) Hilbert spaces for a while now as they have a habit of popping up in many of the papers I'm come across (I'm a ...
Olumide's user avatar
  • 661
19 votes
2 answers
2k views

Can we take a supremum over all Hilbert spaces?

In my paper On the optimal error bound for the first step in the method of cyclic alternating projections, I defined functions $f_n:[0,1]\to\mathbb{R}$, $n\geqslant 2$, by $$ f_n(c)=\sup\{\|P_n\dotsm ...
Ivan Feshchenko's user avatar
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
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
16 votes
2 answers
731 views

A reference to a characterization of metric spaces admitting an isometric embedding into a Hilbert space

I am looking for a reference to the bipartite version of the Schoenberg's criterion of embeddability into a Hilbert space. The Schoenberg criterion is formulated as Proposition 8.5(ii) of the book &...
Taras Banakh's user avatar
  • 41.9k
15 votes
4 answers
2k views

Naive questions about "matrices" representing endomorphisms of Hilbert spaces.

This is a very basic question and might be way too easy for MO. I am learning analysis in a very backwards way. This is a question about complex Hilbert spaces but here's how I came to it: I have in ...
Kevin Buzzard's user avatar
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
15 votes
2 answers
2k views

What is a projective space?

Is there a "recognition principle" for projective spaces? What categories are there with projective spaces for objects? Background: Although the title is a nod to What is a metric space?, ...
Andrew Stacey's user avatar
13 votes
4 answers
5k views

What is known about the Gaussian measure of the unit ball in a Hilbert Space?

Let $X$ be an infinite dimensional separable Hilbert Space with norm $||\cdot||$ and let $\mu$ be a Gaussian measure on $X$ such that $\mu(X) = 1$. What do we know about $\mu(B(0,1))$, where $B(0,1)$ ...
RadonNikodym'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
2 answers
478 views

$x f'$ bounded by $x^2f $ and $f''$?

Consider the Hilbert space of functions $f \in L^2(\mathbb R)$ such that $x^2f \in L^2(\mathbb R) $ and $ f'' \in L^2(\mathbb R).$ I am wondering whether it is true that $xf'\in L^2(\mathbb R)$ as ...
Zorgo's user avatar
  • 177
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
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
2 answers
606 views

A characterization of metric spaces admitting a bi-Lipschitz embedding into a Hilbert space?

Theorem (??) derived in this MO-post from Schoenberg's theorem yeilds a "bipartite" characterization of metric spaces that admit an isometric embedding into a Hilbert space. This Theorem (??)...
Taras Banakh's user avatar
  • 41.9k
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
10 votes
1 answer
593 views

Density of smooth function in Hilbert spaces

I am looking for a simple reference to the following fact: If $f:\Omega\to\mathbb{R}$ is continuous, where $\Omega\subset H$ is an open subset of a separable Hilbert space $H$, then for any $\...
Piotr Hajlasz'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
10 votes
1 answer
900 views

Approximation of a compactly supported function by Gaussians

Let $f:\mathbb{R}\to\mathbb{R}$ be a smooth function whose support is a closed interval, e.g. $\text{supp}(f)=[a,b]$. Then $f$ can be approximated (e.g. in $L^2$) by a linear combination of Gaussian ...
JohnA's user avatar
  • 710
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
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
9 votes
2 answers
321 views

For which $f \in L^2([0,1])$ is $f^\perp \cap C^\infty$ dense in $f^\perp$?

Given $f \in L^2([0,1])$, $f \neq 0$, we can consider the orthogonal complement $f^\perp$ . The smooth functions $C^\infty([0,1])$ are dense in $L^2([0,1])$. Is the intersection $f^\perp \cap C^\infty(...
user40707's user avatar
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
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
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
  • 1,037
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
3 answers
1k views

When does a unitary Hilbert space rep of a reductive Lie group decompose into a direct sum of irreps with finite multiplicities?

I'm giving some lectures on the trace formula. Here's something I proved in the last lecture. Let $G$ be a locally compact Hausdorff unimodular topological group (e.g. a reductive Lie group), let $\...
Kevin Buzzard's user avatar
8 votes
1 answer
305 views

Subspaces isomorphic with quotients

Suppose $X$ is a Banach space not isomorphic to a Hilbert space. Can we always find a subspace of $X$ that is not isomorphic to a quotient of $X$?
Markus's user avatar
  • 1,361
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
1 answer
523 views

Are the following subsets of a Hilbert space always homeomorphic?

Let $F$ be a infinite-dimensional complex Hilbert space, with inner product $\langle\cdot\;| \;\cdot\rangle$, the norm $\|\cdot\|$, the 1-sphere $S(0,1)=\{x\in F;\;\|x\|=1\}$ and let $\mathcal{B}(F)$ ...
Schüler's user avatar
  • 724
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
576 views

On the definition of Hilbert spaces and real structures on Hilbert spaces

Let us consider the space $L^2:=L^2(\mathbb{R}^n,\mathbb{C})$ and the associated scalar product $S(f,g):=\int f \overline g$. In distribution theory, we have a situation where we have to deal with two ...
LCO's user avatar
  • 506
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
8 votes
1 answer
446 views

Parallelogram law for vectors of equal length

Does the parallelogram law for vectors of equal length imply the full parallelogram law? That is, if for all norm one vectors $x$ and $y$ in a Banach space $X$ it holds that $\lVert x-y\rVert^2+\lVert ...
Markus's user avatar
  • 1,361
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
2k views

Definitions of Hilbert Bundles

I have some doubts regarding definitions and conventions on Hilbert Bundles. Some authors like Peter Kuchment (Floquet Theory for Partial Differential Equations) and Serge Lang (Differential and ...
Thangachelli Debopritama's user avatar
8 votes
1 answer
392 views

Proving that a space is Hilbert

Let $H=H_0^1(0,\ell)\times H_*^1(0,\ell)\times H_*^1(0,\ell)$ be equipped with the norms \begin{align*} \|(\varphi,\psi,w)\|_1^2&=A\|\varphi_x+\psi+lw\|_{L^2}^2+B\|w_x-l\varphi\|_{L^2}^2+C\|\psi_x\...
Robert's user avatar
  • 171
8 votes
0 answers
192 views

Is $L^2(I,\mathbb Z)$ homeomorphic to the Hilbert space?

I am somehow puzzled by the subset $G:=L^2(I,\mathbb Z)$ of $H:=L^2(I,\mathbb R)$ of all integer valued functions on $I=[0,1]$ (in fact I mentioned as an example in this old MO question). Some simple ...
Pietro Majer's user avatar
  • 60.5k
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
8 votes
0 answers
6k views

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
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
1k views

Nice orthonormal basis for L^2(Cantor set)

Let X be the Cantor set, which we view as the space $2^\mathbb{N}$ (the set of all infinite binary sequences), equipped with the product topology. We can construct a Borel probability measure $\mu$ on ...
akerber47's user avatar
7 votes
2 answers
464 views

Representing an $L^2$-functional by a non-$L^2$-function on a dense subspace

Let $(X, \mu)$ be your favourite measure space (finite or $\sigma$-finite if you like), let $g \in L^2$ (say, the scalar field of $L^2$ is $\mathbb{R}$, though this probably doesn't matter). Let $\...
Jochen Glueck's user avatar
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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