All Questions
Tagged with hilbert-spaces unbounded-operators
40 questions
-3
votes
0
answers
43
views
Do the domains of the two square roots of a positive (unbounded) operator coincide? [closed]
Let $H$ be a Hilbert space and $D:\mathrm{Dom}(D) \to H$ a densely defined operator on $H$. We further assume that $D$ is closed and self-adjoint. If we further assume that $D$ is positive, then we ...
- 415
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 ...
0
votes
0
answers
35
views
Operator-form correspondence without lower semiboundedness
When dealing with a normal unbounded operator $A$, it is often useful to frame questions about the operator in terms of questions about the associated form $\omega,$ which has domain $D(|A|^{1/2})$ ...
- 109
2
votes
1
answer
187
views
$(S\otimes T)^{it}= S^{it}\otimes T^{it}$ for unbounded operators
Let $S,T$ be unbounded, closed operators in Hilbert spaces $H,K$. In that case, we can form the tensor product operator $S\otimes T$ on the Hilbert space $H\otimes K$ which is the closure of the ...
- 175
2
votes
1
answer
245
views
If $A$ is a closed operator, is $A^k$ closed?
Let $A$ be a closed (densely defined) operator on a Hilbert space $H$.
We define for a natural number $k$, the operator $A^k$ with its natural domain.
Is $A^k$ closed?
- 175
2
votes
1
answer
233
views
Identity for spectral resolution: $dE_{\xi, \xi}= |g|^2 dE_{\eta, \eta}$
Let $(\Omega, \mathcal{F})$ be a measurable space. Let $E: \mathcal{F}\to B(H)$ be a regular resolution of the identity on the Hilbert space $H$, see e.g. Rudin's functional analysis book.
Suppose ...
- 175
3
votes
1
answer
332
views
Takesaki II Lemma 1.13: stuck in proof
Consider the following fragment from Takesaki's book "Theory of operator algebras II" (Lemma 1.13 on p8, in chapter VI "Left Hilbert algebras"):
Here, we associate with an ...
- 175
3
votes
2
answers
275
views
Unbounded positive self-adjoint without $0$ in its spectrum: can we construct its inverse using functional calculus?
Let $P$ be a positive, self-adjoint (unbounded) operator in a Hilbert space $H$ with $0\notin \sigma(P)$. Consider its spectral decomposition
$$P = \int_{\sigma(P)} t dE(t).$$
Since $0 \notin \sigma(P)...
- 175
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
335
views
When the adjoint of an unbounded operator on a Hilbert space coincides with the formal adjoint on its natural domain?
This is almost a copy of https://math.stackexchange.com/questions/3931318/when-the-adjoint-of-an-unbounded-operator-on-a-hilbert-space-coincides-with-the
I am trying to work with infinite matrices in ...
- 21
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
0
votes
1
answer
113
views
An adjoint characterization of (unbounded) Fredholm operators
Let $\mathcal{H}_i$, for $i=1,2$, be Hilbert spaces, and $T:{\frak Dom}(T) \subseteq \mathcal{H}_1 \to \mathcal{H}_2$ a densely-defined closed operator. If the kernel of $T$, and the kernel of its ...
- 993
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
0
votes
2
answers
465
views
Spectrum equals eigenvalues for unbounded operator
Let $D$ be an unbounded densely defined operator on a separable Hilbert space $H$. If $D$ is diagonalisable with all eigenvalues having finite multiplicity and growing towards infinity, does it follow ...
- 219
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
0
votes
1
answer
54
views
Simplify the expression of $ T^+$ for an unbounded operator $T$?
For a negative unbounded operator $T$, what equals the operator
$$ T^+ = \left[\frac{1}{2}(|T| + T) \right]^{**},$$
where $|T|= (T^2)^{1/2}$ and $A^{**} $ is the minimal closed extension of an ...
- 650
-3
votes
1
answer
320
views
Is :$\frac{\Bbb d}{\Bbb d x}$ a chaotic operator in infinite-dimensional Hilbert space? [closed]
I proposed this question in SE but no answer ,may I have a problem in my question, I would like to know when $\frac{\Bbb d}{\Bbb d x}$ does chaotic operator in Hilbert space ?
Let $H$=$L^2(\mathbb R)$...
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
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 $\alpha(t)=e^{itA}$....
- 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
3
votes
3
answers
3k
views
Countability of eigenvalues of a linear operator
Is it true that every closed operator on a separable Hilbert H space only has countably many eigenvalues?
Or put the other way around, if I want to ensure that a (not necessarily bounded) linear ...
- 9,853
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
3
votes
3
answers
3k
views
Infinite hermitian matrix
Suppose we have a finite square n x n matrix of complex numbers H that is Hermitian and skew-symmetric:
$H^\dagger = H$ and $H^T = -H$.
(T denotes transpose, $\dagger$ denote conjugate transpose. I ...
- 195