2
votes
2answers
216 views

A question on unbounded operators

Assume that $H$ is a separable Hilbert space. Is there a polynomial $p(z)\in \mathbb{C}[x]$ with $deg(p)>1$ with the following property?: Every densely defined operator $A:D(A)\to ...
0
votes
1answer
86 views

Densely-defined unbounded operators with large support

Most densely-defined unbounded linear operators on Hilbert spaces have a very large domain. In fact, for a lot of natural operators the intersection of their domains are still dense. Let us consider ...
1
vote
1answer
115 views

Laplacian on space of measures

Let $X$ be a compact Riemannian manifold and let $\mathcal{M}(X)$ be the space of regular finite Borel measures with the total variation as norm. The Laplace-Belrami-Operator $\Delta$ on $X$ with ...
1
vote
1answer
99 views

Special form of unbounded operators on $L_2(\mathbb{R}_+, \mathcal{H})$

I have the following problem; Fix a Hilbert space $\mathcal{H}$. Let $S \colon \mathrm{Dom}S \subset L_2(\mathbb{R}_+, \mathcal{H}) \rightarrow L_2(\mathbb{R}_+, \mathcal{H}) $ be a closed densely ...
9
votes
3answers
204 views

Does the generator of a 1-parameter group of Banach space isometries know which elements are entire?

Let $X$ be a complex Banach space. Let $(\sigma_t)_{t \in \mathbb{R}}$ be a 1-parameter group of linear isometries of $X$ which is strongly continuous i.e. $t \mapsto \sigma_t(x)$ is continuous for ...
7
votes
2answers
146 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 ...
2
votes
1answer
116 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 ...
6
votes
2answers
323 views

C*-algebraic representation of observables vs self-adjoint operators one

I am trying to reconcile the "physicist" definition of an observable: self-adjoint operator on a Hilbert space, and the operational one as given by Strocchi in "An introduction to the mathematical ...
13
votes
2answers
294 views

Perturbation of unbounded self-adjoint operators

In the paper "A CRITERION FOR THE NORMALITY OF UNBOUNDED OPERATORS AND APPLICATIONS TO SELF-ADJOINTNESS" by M. H MORTAD (http://arxiv.org/pdf/1301.0241.pdf), the author states the following theorem ...
7
votes
1answer
205 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 ...
3
votes
1answer
190 views

Hormander's bracket condition for the adjoint of an operator

Let $X_0, X_1, \dots, X_k$ be smooth vector fields over ${\mathbb R}^n$, and let us consider the operator $$ L = \sum_{i=1}^k X_i^2 + X_0~. $$ Here, I assume that Hörmander's bracket condition is ...
4
votes
3answers
312 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 $$ ...
2
votes
0answers
92 views

non-closed weak graph limit of symmetric operators

Hi Everyone, I was recently reading Reed & Simon's functional analysis textbook (the first volume), and it mentions casually on page 294 that weak graph limits of a sequence of symmetric ...
8
votes
2answers
961 views

What is a good reference that compact resolvent implies Fredholm operator?

Suppose $A \in \mathcal{L}(E_1, E_0)$ is a bounded linear operator between Banach spaces $E_1$ and $E_0$, and we also have that $E_1$ is densely, continuously embedded in $E_0$ (i.e. $A$ can be ...
4
votes
2answers
329 views

Transpose of unbounded operators between Banach spaces.

Let $X$ and $Y$ be Banach spaces, and let $L : X \rightarrow Y$ be a unbounded operator with dense domain $\operatorname{dom}(L)$. We can then talk about the transposed operator $L' : ...
8
votes
1answer
338 views

When the adjoint of a hypoelliptic operator hypoelliptic

Assume, $M$ is a smooth manifold with a measure $\mu$ and let $L^2(M, \mu)$ be a space of all square-integrable functions on $M$. Recall that $L$ is a hypoelliptic differential operator, if for ...
2
votes
0answers
180 views

Core of divergence form operator with unbounded coefficient

Consider the unbounded operator $L$ on $L^2(\mathbb{R^d})$ to be the self-adjoint extension of $$Lf := \nabla \cdot \left(a(x) \nabla f(x) \right)$$ on $C^2_c(\mathbb{R^d})$. I also assume that ...
2
votes
0answers
100 views

Invariant linear manifolds for multiplication by the independent variable in L^2 (R)

In general I am trying to determine when the self-adjoint operator $M$ of multiplication by the independent variable in $L^2 (\mathbb{R})$ has a symmetric restriction to a dense linear manifold ...
-2
votes
2answers
593 views

Question on Linear Operators

Let $V$ be a normed infinite dimensional vector space. Let $L: V \longrightarrow V$ be a bounded linear operator. Moreover assume that $L$ is 'locally nilpotent' that is: $$ \forall v \in V \quad ...
4
votes
2answers
650 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 ...