Questions tagged [unbounded-operators]
The unbounded-operators tag has no usage guidance.
111
questions
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$(S\otimes T)^{it}= S^{it}\otimes T^{it}$ 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 ...
0
votes
1
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80
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$L^p$ boundedness for pseudo-differential operators
Let $\rho, \delta, m$ be real parameters such that $0\le \delta\le \rho\le 1, \delta<1$. The set $S^m_{\rho, \delta}(\mathbb R^{2n})$ is defined as the set of smooth functions $a$ on $\mathbb R^n\...
2
votes
0
answers
84
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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 ...
2
votes
1
answer
104
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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
\...
3
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0
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44
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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 ...
4
votes
1
answer
216
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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 $1+A$...
2
votes
2
answers
405
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Takesaki II "Connes cocycle derivative"
Consider the following fragments from Takesaki's second volume "Theory of operator algebras" (chapter VIII $\oint 3$, Modular Automorphism groups, p107-108:
Why are the second and third ...
2
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1
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198
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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?
2
votes
1
answer
112
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On the Fractional Laplace-Beltrami operator
I would appreciate it if a reference could be given for the following claim.
Let $g$ be a Riemannian metric on $\mathbb R^n$, $n\geq 2$ that is equal to the Euclidean metric outside some compact set. ...
2
votes
1
answer
222
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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 ...
3
votes
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319
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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 ...
3
votes
2
answers
220
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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 $0 \notin \sigma(P)...
3
votes
1
answer
161
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$\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 $m$ ...
1
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0
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191
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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 ...
1
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0
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101
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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.)
$$ \mathrm{T}: \mathcal{H} \supset \mathcal{D}(\mathrm{T}) \to \mathcal{H} , f \mapsto \mathrm{...
3
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How to prove the polar decomposition of unbounded operators?
Let $ T $ be a closed, densely defined operator on a Hilbert space $ H $. Then there exists a positive self-adjoint operator $ A $, $ D(A)=D(T) $ and a isometric operator $ V:R(A)\to \overline{R(T)} $ ...
0
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1
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97
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Closure of the point spectrum of an unbounded diagonalizable operator
Given a (separable) Hilbert space H and an unbounded densely defined linear operator $T:{\cal D}(T) \to $H such that ${\cal D}$ is diagonalizable (it means $\exists$ an O.N.B. of H such that all basis ...
0
votes
1
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114
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Antilinear unbounded operator has closed graph
Let $H$ and $K$ be Hilbert spaces and $D(T)$ a vector subspace of $H$. Let $T: D(T) \to K$ be a densely defined antilinear operator. Its adjoint $T^*: D(T^*)\to K$ is defined by the relation
$$\langle ...
8
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2
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583
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An inverse to functional calculus
Given a Borel function $f:\mathbb{R}\rightarrow\mathbb{R}\cup\{\infty\}$, functional calculus allows to calculate $F(x)$ for any unbounded selfadjoint operator $x$ on a Hilbert space $\mathcal{H}$, ...
0
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2
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588
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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 $...
1
vote
0
answers
128
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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 $\{a,b\}\cap \sigma(T)=\emptyset$, ...
3
votes
3
answers
745
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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 $...
1
vote
1
answer
263
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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 $...
6
votes
2
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432
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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 ...
2
votes
1
answer
209
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Limit of $e^{-t(-\Delta)^\alpha}$ when $\alpha \to 1$
Let us consider the fractional heat semigroup $\left(e^{-t(-\Delta)^\alpha}\right)_{t\ge 0}$ for $\alpha\in (0,1)$ (the fractional power is taken in whole $\mathbb R^d$). Is there any result about the ...
10
votes
1
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469
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A detail in the proof of Schur's lemma: the closures of the $\mathcal{Ker}$ and $\mathcal{Im}$ of the intertwiner
$\renewcommand\Im{\operatorname{\mathcal{Im}}}\newcommand\Ker{\operatorname{\mathcal{Ker}}}$I was sure that this is a trivial question and placed it on Math Stackexchange
https://math.stackexchange....
2
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395
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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)^{-...
1
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0
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112
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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 ...
4
votes
0
answers
189
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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 ...
2
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0
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132
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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 ...
3
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147
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'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 ...
2
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1
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292
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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 ...
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0
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"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 $f(A)=Q^{-1} M_{f\circ \alpha} Q$. (...
2
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0
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313
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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\...
0
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0
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What is the analog of the symmetrized Jacobi matrix for delay equations?
For a linear system of ODEs in $\mathbb{R}^{n}$ (with the usual inner product), say $\dot{V}(t) = A(t) V(t)$, we know that if $\xi_{1},\ldots,\xi_{k} \in \mathbb{R}^{n}$ and $V_{j}(t)=V_{j}(t,\xi_{j})$...
2
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1
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113
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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 ...
5
votes
1
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224
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Domain of $(-\Delta)^\alpha$, $\alpha \in (0,1)$
Is there any simple way to characterize explicitly the domain of fractional powers for a given operator? For example, the domain of Dirichlet Laplacian on a bounded nice domain $\Omega \subset \mathbb{...
2
votes
1
answer
717
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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 ...
0
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95
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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 ...
2
votes
1
answer
256
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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 $\mathcal{H}_2$ ...
5
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2
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539
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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: ...
1
vote
1
answer
239
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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 ...
1
vote
0
answers
153
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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 ...
3
votes
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122
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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 ...
0
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2
answers
399
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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 ...
3
votes
1
answer
177
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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 ...
6
votes
1
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880
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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 ...
3
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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 ...
2
votes
1
answer
318
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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.
...
2
votes
0
answers
39
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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}_{...