All Questions
26 questions
3
votes
0
answers
95
views
Commutator of $A\otimes I$ and $I \otimes B$ vanishes?
Consider two Hilbert spaces $H_1$ and $H_2$, and $A$, $B$ unbounded operators on $H_1$, $H_2$ respectively. $(A \otimes I)$ is classically defined as the closure of the operator defined on the set of ...
- 31
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 ...
3
votes
0
answers
390
views
How to prove the polar decomposition of unbounded operators?
Let $ T $ be a closed, densely defined operator on a Hilbert space $ H $. Then there exists a positive self-adjoint operator $ A $, $ D(A)=D(T) $ and a isometric operator $ V:R(A)\to \overline{R(T)} $ ...
- 1,219
-1
votes
1
answer
164
views
Closure of the point spectrum of an unbounded diagonalizable operator
Given a (separable) Hilbert space H and an unbounded densely defined linear operator $T:{\cal D}(T) \to $H such that ${\cal D}$ is diagonalizable (it means $\exists$ an O.N.B. of H such that all basis ...
- 969
0
votes
1
answer
138
views
Antilinear unbounded operator has closed graph
Let $H$ and $K$ be Hilbert spaces and $D(T)$ a vector subspace of $H$. Let $T: D(T) \to K$ be a densely defined antilinear operator. Its adjoint $T^*: D(T^*)\to K$ is defined by the relation
$$\langle ...
- 175
1
vote
1
answer
335
views
A consequence of the Min-Max Principle for self-adjoint operators
Let $H=(H, (\cdot, \cdot))$ be a Hilbert space. Let $T_1,T_2:D \subset H \longrightarrow H$ be a self-adjoint operators (not necessarily bounded). It's well-know that the spectrum $\sigma(T_i)$ of $...
- 205
6
votes
2
answers
514
views
Convergence criterion in the domain of an unbounded operator
Cross-post from math.sx.
My question is somewhat close to this one, but the counterexamples given there do not apply here.
Setup. Given a Hilbert space $\mathcal H$, a closed operator $A$ and a ...
- 245
2
votes
1
answer
917
views
Characterising closed range self-adjoint operators
Let $T:\mathrm{dom}(T) \subseteq H \to H$ be a densely defined, self-adjoint operator on a Hilbert spaces $𝐻$. In general the range of $T$ is not guaranteed to be closed. What tools are available to ...
- 969
3
votes
1
answer
285
views
Closable unbounded operators and Banach space adjoints
For an unbounded operator $T:\mathcal{H}_1 \to \mathcal{H}_2$, if its adjoint $T^*$ is densely defined, then we know that $T$ is closable. What happens if we replace $\mathcal{H}_1$ or $\mathcal{H}_2$ ...
- 969
6
votes
2
answers
665
views
Unbounded Fredholms operators
Motivated by the situation of bounded Fredholm operators, I have the following question about "unbounded Fredholm operators".
Let $\mathcal{H}_1$ and $\mathcal{H}_2$ be two Hilbert spaces, and
$$
D: ...
- 969
1
vote
0
answers
277
views
Adjoint for a non-densely defined unbounded operator on a Hilbert space
Let $\mathbf{H}$ be a Hilbert space, and $D$ an unbounded densely-defined operator on $\mathbf{H}$. As is well-known, every such operator admits an adjoint, with domain possibly different from that ...
- 993
3
votes
1
answer
214
views
Non-point spectrum for diagonalisable self-adjoint unbounded operator
Given a (separable) Hilbert space H and an unbounded densely defined linear operator $T:{\cal D}(T) \to $H such that ${\cal D}$ is diagonalizable (it means $\exists$ an O.N.B. of H such that all basis ...
- 969
6
votes
1
answer
1k
views
Unbounded version of continuous functional calculus
For a normal operator $T$ on a Hilbert space ${\cal H}$, it is well known that for any continuous complex valued function $f$ on the spectrum of $T$, we have a well-defined operator $f(T) \in B({\cal ...
- 969
3
votes
0
answers
74
views
A question about a theorem in 'Quantum dynamical semigroups generated by noncommutative unbounded elliptic operators'
I have asked this question on MathSE and someone advised me to ask it here. The link is .
I'm studying the paper Quantum dynamical semigroups generated by noncommutative unbounded elliptic operators ...
- 31
3
votes
1
answer
229
views
Symmetric diagonalizable operators and self-adjointness
Given a densely defined symmetric operator $L$ on a Hilbert space $H$, which is also assumed to be diagonalizable, will there always exist a unique extension of $L$ to a self-adjoint operator?
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 ...
- 7,746
4
votes
1
answer
165
views
Scattering of relativistic particle by long-range potential
Let
$\mathcal{H}=L^2(\mathbb{R}^3)$,
$H_0=\sqrt{-\Delta+M^2}$, ($M$ is a positive constant, $\Delta$ is the laplacian)
and
$H=H_0+V(\vec{x})$
(where $V(\vec{x})$ is the operator of ...
- 552
2
votes
1
answer
221
views
Selfadjointness of hamiltonian with 1/x potential
Let us consider the Hilbert space $L^2([0,\infty))$ and operator $H=-\frac{d^2}{dx^2} + \frac{1}{x}$ on the domain of $C^{\infty}_0((0,\infty))$ (smooth functions with compact support away from $0$).
...
- 552
1
vote
1
answer
2k
views
Operator theory of the Hessian
How can I learn more about the operator theory of the Hessian? The Hessian of a function $u : \Omega \rightarrow \mathbb R$ over a domain $\Omega \subseteq \mathbb R^n$ is the matrix of second ...
- 5,327
2
votes
1
answer
926
views
Eigenvalues and Compact Resolvent
For $A$ an unbounded (densely defined) operator on a separable Hilbert space, what conditions on its eigenvalues will show that, for $\lambda \notin $spec$(A)$, we have that $(A-\lambda)^{-1}$ is a ...
- 465
2
votes
1
answer
206
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 ...
- 23
9
votes
2
answers
483
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 $\alpha(t)=e^{itA}$....
- 43.2k
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 ...
- 679
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 \...
- 63.1k
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
$$
\...
- 43.2k
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 ...
- 1,411