Questions tagged [unbounded-operators]

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Rudin functional analysis: theorem 13.27 questions

Consider the following fragment from Rudin's book "Functional analysis": Questions: (1) I'm a bit confused about the formulation of (a). If $E(\omega_\alpha)\ne 0$, isn't it automatically ...
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43 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 ...
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29 views

Unbounded operator with closed range

Consider the spaces $C_{2\pi}^m$ of smooth $2\pi$-periodic functions, and the unbounded operator $L:C_{2\pi}^m\to C_{2\pi}$, given by $L(u)(t)=P(\partial)u(t)+u(t-\tau)$ where $P(\partial)$ is a ...
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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 ...
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8 votes
1 answer
365 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 $\mathcal{H}$, ...
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2 answers
269 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 $...
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1 vote
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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 $\{a,b\}\cap \sigma(T)=\emptyset$, ...
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3 votes
3 answers
507 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 $...
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1 vote
1 answer
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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 $...
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6 votes
2 answers
322 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 ...
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2 votes
1 answer
184 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 ...
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10 votes
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395 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....
2 votes
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172 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)^{-...
1 vote
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89 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 ...
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4 votes
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153 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 ...
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2 votes
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93 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 ...
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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 ...
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1 answer
213 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 ...
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54 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 $f(A)=Q^{-1} M_{f\circ \alpha} Q$. (...
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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\...
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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 $V_{j}(t)=V_{j}(t,\xi_{j})$...
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1 answer
93 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 ...
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1 answer
205 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{...
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1 vote
1 answer
470 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 ...
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1 answer
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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
206 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 $\mathcal{H}_2$ ...
5 votes
2 answers
434 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: ...
1 vote
1 answer
225 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 ...
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1 vote
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86 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 ...
3 votes
1 answer
119 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 ...
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2 answers
310 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 ...
3 votes
1 answer
137 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 ...
5 votes
1 answer
611 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 ...
3 votes
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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
0 answers
196 views

Common eigenvector of commuting unbounded Operators

Let be T,S two self adjoint linear Operator on a Hilbert Space $\mathcal{H}$ with pure point spectrum. Then T,S commute if and only if they have a complete set of common eigenvectors. This is the ...
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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}_{...
4 votes
1 answer
498 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 ...
2 votes
1 answer
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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 ...
1 vote
1 answer
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Dirichlet fractional Laplacian and zero boundary conditions

Does there exists a non-zero function $$f\in C_0([0,1]):=\{f:[0,1]\to \mathbb R:\ f\text{ is continuous and } f(0)=f(1)=0\},$$ such that $(-\Delta)^{\frac\alpha 2}f\in C_0([0,1]) $, where $(-\Delta)^{\...
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7 votes
2 answers
436 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
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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
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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 $...
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1 answer
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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 ...
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4 votes
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118 views

Bounded self-adjoint perturbation of a p-summable spectral triple

I am new to the field of Noncommutative Geometry.I was reading the chapter on Spectral triple from the book 'Elements of Noncommutative Geometry' by Gracia-Bondía,Várilly and Figueroa.Now,after ...
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3 votes
1 answer
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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
326 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 ...
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4 votes
1 answer
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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 ...
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4 votes
2 answers
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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 ...
4 votes
1 answer
149 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 ...
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3 answers
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Notations for dual spaces and dual operators

I'm asking for opinions about the 'best' notations for: 1. the algebraic dual of a vector space $X$; 2. the continuous dual of a TVS; 3. the algebraic dual (transpose) of an operator $T$ between ...