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Questions tagged [unbounded-operators]

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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 $D(|A|^{1/2})$ ...
user_35's user avatar
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66 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 ...
Ruy's user avatar
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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 ...
Damalone's user avatar
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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(...
H A Helfgott's user avatar
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8 votes
1 answer
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On commutator of bounded operators

Let $\mathbb H$ be a Hilbert space and let $\mathcal B(\mathbb H)$ be the bounded operators on $\mathbb H$. Let $J,K\in \mathcal B(\mathbb H)$ such that $ J=J^*, K=-K^*. $ Then the commutator $[J,K]$ ...
Bazin's user avatar
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2 votes
2 answers
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"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 $...
Isaac's user avatar
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3 votes
2 answers
252 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_\...
Tony419's user avatar
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1 answer
178 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 ...
Andromeda's user avatar
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1 answer
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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\...
Bazin's user avatar
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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 ...
Luis Yanka Annalisc's user avatar
2 votes
1 answer
124 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 \...
Luis Yanka Annalisc's user avatar
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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 ...
Curious's user avatar
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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 $1+A$...
Curious's user avatar
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2 answers
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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 ...
Andromeda's user avatar
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1 answer
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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?
Andromeda's user avatar
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1 answer
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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. ...
Ali's user avatar
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2 votes
1 answer
232 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 ...
Andromeda's user avatar
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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 ...
Andromeda's user avatar
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2 answers
273 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 $0 \notin \sigma(P)...
Andromeda's user avatar
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1 answer
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$\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 $m$ ...
John's user avatar
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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 ...
John's user avatar
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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{...
kumquat's user avatar
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0 answers
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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)} $ ...
Luis Yanka Annalisc's user avatar
-1 votes
1 answer
146 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 ...
Dave Shulman's user avatar
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1 answer
123 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 ...
Andromeda's user avatar
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8 votes
2 answers
620 views

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 $\mathcal{H}$, ...
rpk's user avatar
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2 answers
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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 $...
Anupam's user avatar
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0 answers
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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 $\{a,b\}\cap \sigma(T)=\emptyset$, ...
Watanabe's user avatar
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3 votes
3 answers
993 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 $...
The Ju's user avatar
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1 vote
1 answer
320 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 $...
Guilherme's user avatar
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6 votes
2 answers
501 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 ...
Benjamin's user avatar
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2 votes
1 answer
215 views

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 ...
Migalobe's user avatar
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11 votes
1 answer
557 views

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....
Michael_1812's user avatar
2 votes
0 answers
576 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)^{-...
user270619's user avatar
1 vote
0 answers
155 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 ...
Isaac's user avatar
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4 votes
0 answers
214 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 ...
Isaac's user avatar
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2 votes
0 answers
140 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 ...
Isaac's user avatar
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3 votes
0 answers
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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 ...
Photographer's user avatar
2 votes
1 answer
324 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 ...
apyshkin's user avatar
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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 $f(A)=Q^{-1} M_{f\circ \alpha} Q$. (...
Ma Joad's user avatar
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2 votes
0 answers
382 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\...
Kris's user avatar
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0 answers
36 views

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 $V_{j}(t)=V_{j}(t,\xi_{j})$...
demolishka's user avatar
2 votes
1 answer
122 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 ...
Jacob Lu's user avatar
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5 votes
1 answer
249 views

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{...
Migalobe's user avatar
  • 395
2 votes
1 answer
849 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 ...
Dave Shulman's user avatar
0 votes
1 answer
104 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 ...
Max Schattman's user avatar
3 votes
1 answer
277 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 $\mathcal{H}_2$ ...
Dave Shulman's user avatar
6 votes
2 answers
631 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: ...
Dave Shulman's user avatar
1 vote
1 answer
248 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 ...
Ma Pa's user avatar
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1 vote
0 answers
261 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 ...
Max Schattman's user avatar