All Questions
Tagged with unbounded-operators fa.functional-analysis
85 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
148
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
69
views
Perturbation of one-parameter groups of unitary operators
Let $H$ be a Hilbert space and let $h$ be a fixed, densely defined, possibly unbounded, self-adjoint operator on $H$. Letting $B(H)$ denote the space of all bounded operators on $H$, it is well ...
- 2,263
2
votes
0
answers
111
views
Everywhere-defined unbounded operators between Banach spaces
In this post, it is said that there are no constructive examples of everywhere-defined unbounded operators between Banach spaces; every example furnished must use the axiom of choice. This seems like ...
- 151
2
votes
0
answers
261
views
When is an unbounded averaging operator on $\mathbb{R}\to \mathbb{R}$ closed?
Let $\{a_n\}_{n=1}^\infty$, $a_n\in \mathbb{R}$. Consider the following linear operator $A$ on functions $f:\mathbb{R}\to \mathbb{R}$:
$$(Af)(x) = \sum_{n=1}^\infty a_n f(x+n)+ \sum_{n=1}^\infty a_n f(...
- 20.2k
2
votes
2
answers
155
views
"Completeness" for weak convergence of unbounded closed operators on a separable Hilbert space $H$
Let $H$ be a separable Hilbert space with the inner product $\langle, \rangle$ and $\{ T_n \}$ be a sequence of unbounded closed linear operators with a common dense domain $D \subset H$ such that $...
- 3,477
3
votes
2
answers
280
views
Question about the Bessel operator
For $\nu>-1$ denote by $\{\lambda_{k,\nu}\}_{k\in\mathbb{N}}$ the succesive positive zeros of the Bessel function of the first kind $J_{\nu}$. The Bessel operator is given by
\begin{equation*}
L_\...
- 421
2
votes
0
answers
91
views
How to show $ |(Bx,x)|\leq (Ax,x) $ for any $ x\in D(A) $ here?
On the Hilbert space $ H $, $ A $ is a non-negative self-adjoint operator and $ B $ is a symmetric operator. Let $ D(B)\supset D(A) $, where $ D(A) $ and $ D(B) $ are definite domain for $ A $ and $ B ...
- 1,219
2
votes
1
answer
137
views
Disturbance of self-adjoint operator
Assume that $ A $ is self-adjoint operator and $ B $ is a bounded self-adjoint operator. The definite domain of $ A,B $, denoted by $ D(A) $ and $ D(B) $ satisfies $ D(A)\subset D(B) $. Show that
\...
- 1,219
3
votes
0
answers
98
views
Algebra core for generator of Dirichlet form
This is a question about the existence of a core $C$ for the generator $A$ of a regular Dirichlet form $\mathcal{E}$ having a carré du champ $\Gamma$, so that $C$ is an algebra with respect to ...
- 143
4
votes
1
answer
228
views
Diagonalizing selfadjoint operator on core domain
Let $A$ be a densely-defined, positive, self-adjoint operator with compact resolvent on a Hilbert space $H$. Then, $\text{Range}(1+A)=H$ and there is a basis for $H$ consisting of eigenvectors of $1+A$...
- 143
3
votes
1
answer
226
views
$\tau$-measurable operator
Problem: Let $M$ be a semifinite von Neumann algebra with a faithful semifinite normal trace $\tau$. Let $m$ be a positive element in $M$ and let $e_{(0,\infty)}(m)$ be the spectral projection of $m$ ...
- 85
1
vote
0
answers
385
views
Densely defined and closed operator
Question: Let $N\subseteq B(H)$ be a finite von Neumann algebra with a faithful normal trace $\tau$ and let $n\in N$. Let $B$ be a von Neumann subalgebra of $N$. Let $\mathbb{E}_B: N\rightarrow B$ be ...
- 85
1
vote
0
answers
153
views
Maximal domain of an unbounded linear operator in a weighted Hilbert-space
Let's consider the following (unbounded) linear operator. (So called transport operator in some context.)
$$ \mathrm{T}: \mathcal{H} \supset \mathcal{D}(\mathrm{T}) \to \mathcal{H} , f \mapsto \mathrm{...
- 185
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
0
votes
2
answers
971
views
Example of a linear operator whose graph is not closed
I want an example of a linear operator $T:X\to Y$ such that graph of $T$ is not closed.
My thoughts: $T$ must be unbounded. Again by closed graph theorem any unbounded linear map from a Banach space $...
- 585
1
vote
0
answers
205
views
Convergence in the resolvent sense and spectral properties
Let $\{T_k\}_k, T$ be unbounded selfadjoint operators on a Hilbert space $H$. If $T_k\to T$ in the norm-resolvent sense, then for any $(a,b)\subset \mathbb R$ with $\{a,b\}\cap \sigma(T)=\emptyset$, ...
- 111
3
votes
3
answers
1k
views
Unbounded operators vs compact operators
The operator $L:\operatorname{dom}(L)\subset C[0,1]\to C[0,1]$ given by $Lx=x'$
a) is closed, unbounded and densely defined
b) also has a compact right inverse, namely $K:C[0,1] \to C[0,1]$ given by $...
- 31
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
0
answers
654
views
Convergence of operator in norm resolvent sense and their eigenvectors
Let $\{T_n\}_{n=1}^\infty$ and $T$ be (unbounded) self-adjoint operators and $T_n\to T$ in norm resolvent sense, that is, for some $z\in \mathbb{C} \setminus \mathbb{R}$, $\|(zI- T_n)^{-1}- (zI- T)^{-...
- 21
1
vote
0
answers
167
views
Common core for unbounded operators
Suppose that $\mathcal H$ is a Hilbert space representing some physical system, $H$ is the Hamiltonian for the system, and $A$ is some observable for the system, that is, some unbounded self-adjoint ...
- 771
4
votes
0
answers
230
views
Spectral theorem for unbounded operators
Part of the Spectral theorem for unbounded operators states that if $A$ is a self adjoint unbounded operator and $B$ is a bounded operator such that $BA$ is contained in $AB$, then $B$ commutes with ...
- 771
2
votes
0
answers
144
views
Extensions of symmetric unbounded operators
I saw it claimed that every symmetric operator on a Hilbert space $H$ can be extended to a self-adjoint operator on some potentially larger space K. But I seem to be able to prove from this that every ...
- 771
3
votes
0
answers
201
views
'Local' commutativity of self-adjoint operators
Preamble
Two (unbounded) self-adjoint operators $A, B$ on a Hilbert space $\mathcal{H}$ are said to (strongly) commute if the unitary groups they generate commute or equivilantely if all the ...
- 131
0
votes
0
answers
62
views
"Trade-off" between bound on the function and on the spectrum for functional calculus in spectral theory
Let $A$ be a self-adjoint (unbounded) operator on a separable Hilbert space $H$.
From the following form of spectral theorem, we may define a functional calculus by $f(A)=Q^{-1} M_{f\circ \alpha} Q$. (...
- 1,755
2
votes
0
answers
395
views
On strong resolvent convergence of unbounded operators
Given a sequence of unbounded self-adjoint operators $\Delta_N$ defined on some fixed domain of the Hilbert space $L^2(\mathbb{R})$ converging to i*identity in the sense that: for all $\phi\in\...
- 63
2
votes
1
answer
127
views
Spectral representation of closed operators with finite spectral bound
Assume $A$ is a closed linear operator on a Banach space $X$ and is densely defined. Assume the spectral bound $s(A) = \sup\{Re\lambda: \lambda\in \sigma(A)\}$ is finite. For example, if $A$ is the ...
- 903
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
1
answer
253
views
Power of an infinitesimal generator of a $C_0$-semigroup in a Banach space
Let $A$ be the infinitesimal generator of a $C_0$-semigroup of linear operators in a Banach space. Let $n$ be a positive integer, $n\geq2$. Is $A^n$ closed?
Here (setting $A^1$ $:=$ $A$, and ...
- 31
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
126
views
The imaginary exponential of a tangent field on a manifold
If $M$ is a compact Riemannian manifold and $X$ is a tangent field, I am seeking to define the object $\exp {\mathrm i t X}$ for $t \in \mathbb R$, and I do not know how to do it.
One option was to ...
- 5,407
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
2
votes
1
answer
486
views
Common eigenvector of commuting unbounded operators
Let $T$, $S$ be two self-adjoint linear operators on a Hilbert space $\mathcal{H}$ with pure point spectrum.
Then $T$ and $S$ commute if and only if they have a complete set of common eigenvectors.
...
- 21
2
votes
0
answers
40
views
Invariance of simple functions
Let $(A(s),D(A(s)))_{s\in\mathbb{R}}$ be a family of unbounded operators on a Banach space $X$ and $g:\mathbb{R}\rightarrow X$ be a simple function, i.e.,
\begin{align*}
g=\sum_{i=1}^n{x_i\textbf{1}_{...
4
votes
1
answer
878
views
Commuting with an unbounded operator
Let $H$ be a Hilbert space. Let $A$ be a closed unbounded operator, and let $B\in B(H)$ be a bounded operator.
Definition:
$A$ and $B$ strong-commute if the partial isometry in the polar ...
- 43.2k
2
votes
1
answer
134
views
Decomposition of the spectrum of an unbounded opeator [closed]
The Wikipedia article on spectral decomposition, see here
https://en.wikipedia.org/wiki/Self-adjoint_operator
says the following:
A self-adjoint operator A on $H$ has pure point spectrum if and ...
7
votes
2
answers
485
views
The von Neumann algebra generated by a non-closable operator
Let $H$ be a separable Hilbert space and let $M$ be a densely defined operator $\mathcal{D}(M) \subset H \to H$. It is closable iff its adjoint $M^{\star}$ is densely defined, and then its closure $\...
8
votes
1
answer
302
views
Does every integer map generate a von Neumann algebra of type I?
Consider a map $m: \mathbb{N} \to \mathbb{N}$ (we call it an integer map). Let $E_r$ be the set $m^{-1}(\{r\})$.
Let $H$ be the Hilbert space $\ell^2(\mathbb{N})$ and consider the densely defined ...
1
vote
1
answer
394
views
Are $\| \Delta u \|_{L^p(\Bbb R^d)} + \| u \|_{L^p(\Bbb R^d)}$ and $\| u \|_{W^{2,p}(\Bbb R^d)}$ equivalent norms?
Is it true that $$\| \Delta u \|_{L^p(\Bbb R^d)} + \| u \|_{L^p(\Bbb R^d)}\quad\text{and}\quad\| u \|_{W^{2,p}(\Bbb R^d)}$$ are equivalent norms?
This results is pretty easy and straightforward for $...
- 1,101
5
votes
1
answer
2k
views
Commuting with self-adjoint operator
Let $T$ be an (unbounded) self-adjoint operator. Assume that there is a bounded operator $S$ such that $TS=ST.$ For which kind of $f$ do we have that $f(T)S=Sf(T)?$
My thought was that using a ...
- 501
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?
3
votes
2
answers
441
views
Minimum eigenvalue of One-dimensional Schrodinger Operator
Consider the One dimensional Schrodinger Operator
$$
-\frac{d^2}{dx^2} + V(x)
$$
Where the Potential Function $V$ is of the form $V(x) = x^4 - a^2x^2$ , $a \in \mathbb{R} $.
Now of course,the ...
- 73
4
votes
1
answer
280
views
Spectral growth of One dimensional Schrodinger Operator
Conside the One dimentional Schrodinger Operator
$$
-\frac{d^2}{dx^2} + ( V(x) + E )
$$
Where the Potential Function $V$ is of the form $V(x) = ax^2 + b^2x^4$ , $a,b \in \mathbb{R} $.
What is known ...
- 73