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Effective action of unbounded operators on subspaces outside their domains of definition

Consider a densely defined, self-adjoint operator $$ H: \mathcal{D} \rightarrow \mathscr{H}. $$ Assume for simplicity that $H$ is nonnegative. We want to effectively restrict this operator $H$ to a ...
Qualearn's user avatar
  • 133
1 vote
1 answer
151 views

Is smoothness preserved under an isometric isomorphism?

Let $(X, \|.\|_1)$ is isometrically isomorphic to $(X, \|.\|_2)$ and $\|.\|_2\leq \|.\|_1$. Assume that $x_0$ is a smooth point of $(X, \|.\|_1)$ and $\|x_0\|_2=1$. According to the definition of a ...
Tuh's user avatar
  • 113
0 votes
0 answers
90 views

How to show a point is a weak* -weak continuous for the identity map on $X_1^*$ or on $X_1^{**}$?

I am trying to understand the Remark 3.2 mentioned in the paper titled as "On Weak* -Extreme Points in Banach Spaces" written by S. Dutta and T. S. S. R. K. Rao (http://library.isical.ac.in:...
Tuh's user avatar
  • 113
0 votes
1 answer
169 views

Is the evolution family self-adjoint?

$ \newcommand{\bR}{\mathbb{R}} \newcommand{\diff}{\mathop{}\!\mathrm{d}} \newcommand{\qtext}[1]{\quad\text{#1}} \newcommand{\qtextq}[1]{\quad\text{#1}\quad} $ I am reading Roland Schnaubelt's survey ...
Akira's user avatar
  • 835
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 ...
Hugo's user avatar
  • 31
1 vote
0 answers
111 views

References on the partial trace

For the Hilbert space $H^N:=L((\mathbb R^{3})^N,\mathbb C)$, consider the projection operator $D: H^N\to H^N$ as follows : $$D(\Phi):=\left(\int_{(\mathbb R^{3})^N}\overline{\Psi(x_1,\ldots, x_N)}\Phi(...
Fawen90's user avatar
  • 1,389
1 vote
1 answer
117 views

Lower bound for a commutator trace

I have this Hilbert space of square-integrable complex-valued functions on a square, $\mathbb{L}^2([0,1]^2)$. And let $M_x$, $M_y$, and $M_{x+y} = M_x+M_y$ be the operators of multiplication by the ...
Chilperic's user avatar
  • 121
0 votes
0 answers
50 views

Self-adjoint operators and index of quadratic form associated to it

Let $B$ a bounded self-adjoint operator on a real Hilbert space $H$ with an associated inner product $(\cdot,\cdot).$ Take $V=\operatorname{span}\{f_1, f_2, \ldots, f_n\}$ a finite dimensional ...
Frank Zermelo's user avatar
0 votes
1 answer
139 views

Existence of infinite rank compact operator

Given any separable Banach space $X$, we know that always there exists a Banach space $Y$ such that there is an injective compact operator from $X$ to $Y$. Can we show that given any infinite ...
Anupam's user avatar
  • 585
1 vote
0 answers
127 views

Trace type convergence of the Laplacian on the box to the Laplacian on $\mathbb R^d$

Let $-\Delta \colon H^2(\mathbb R^d) \to \mathbb R^d$ be the (negative) Laplacian on the full space and $-\Delta_L$ the Laplacian acting on $L^2([-L,L]^d)$ with some boundary conditions making it self-...
lasik43's user avatar
  • 61
9 votes
0 answers
163 views

Moore-Penrose partial isometries and hermitian elements

Let $A$ be a unital Banach algebra. An element $a \in A$ is hermitian if $\|\mathrm{exp}(ita)\|=1$ for every $t \in \mathbb{R}$. An element $a \in A$ is Moore-Penrose invertible if there exists $b \in ...
Hannes Thiel's user avatar
  • 3,497
7 votes
1 answer
415 views

Is there a “Closure-of-Range Theorem” for Banach spaces?

The classic Closed Range theorem states that for a linear bounded operator $T:X\to Y$ between Banach spaces, and its transpose $T^*:Y^*\to X^*$, the four conditions: $T(X)$ is $s$-closed; $T(X)$ is $...
Pietro Majer's user avatar
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0 votes
0 answers
68 views

Family of separable Hilbert spaces over locally compact form a continuous field of Hilbert space?

Let $\{H_{x}\}_{x\in G^{0}}$ be a family of separable Hilbert spaces and $G^{0}$ be a locally compact second countable topological space. Let $\mathbb{B}_{x}$ be the orthonormal basis of $H_{x}$. If ...
K N SRIDHARAN NAMBOODIRI's user avatar
-1 votes
1 answer
98 views

Spectrum of sum of positive and negative operators

Let $(\mathscr{H}, \langle \cdot, \cdot \rangle)$ be a separable complex Hilbert space, and let $\mathscr{D}$ be a dense subset of $\mathscr{H}$. Let $P: \mathscr{D} \to \mathscr{H}$ and $N: \mathscr{...
d'Alembert's user avatar
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 ...
Gateau au fromage's user avatar
0 votes
0 answers
121 views

How to find the inverse of this linear integral operator?

Let $f(x): \mathbb{R}^d \rightarrow \mathbb{R}$ be a function that decays ``fastly enough'' at infinity. We can define the following linear operator $$L[f](x):= \int_{\mathbb{R}^d} d^d y \, \frac{f(y)}...
mnerone's user avatar
-6 votes
1 answer
180 views

An analog of Anderson's result in C* algebra setting [closed]

Let $\mathcal{A}$ be a unital $C^{*}$-algebra and $S(\mathcal{A})$ denote the states space of $\mathcal{A}$. For $a\in \mathcal{A}$ , define $W(a) =\{\phi(a):\phi\in S(\mathcal{A})\}$ It's known that $...
SoG's user avatar
  • 307
0 votes
0 answers
78 views

Definition of Moore-Penrose inverse for unbounded self-adjoint operators?

I know there is a concept of Moore-Penrose or pseudoinverse of a matrix. I would like to know if one can define it for densely defined unbounded self-adjoint operators on Hilbert spaces as well. ...
InMathweTrust's user avatar
3 votes
1 answer
116 views

Does a bounded positive modular sesquilinear form on a $C^\ast$-algebra induces an element of its multiplier algebra?

This is a question that originates from my attempt at this question. Specifically, for a $C^\ast$-algebra $A$, I am attempting to construct a map $\phi: A \times A \to A$ s.t., $\phi$ is sesquilinear,...
David Gao's user avatar
  • 2,830
0 votes
0 answers
141 views

The tensor product of two Fredholm operators

What can be said about the tensor product $T\otimes S$ of two Fredholm operators $T:X_1\to Y_1$ and $S:X_2 \to Y_2$ where $X_1,X_2,Y_1, Y_2$ are Banach spaces and tensor product of ...
Ali Taghavi's user avatar
1 vote
1 answer
122 views

distance in the matrix algebra w.r.t. the nuclear norm

Let $\varphi\in\mathcal{M}_n(\mathbb{C})$ and let $Z:=\mathbb{C}\cdot I=\{zI\colon\,z\in\mathbb{C}\}$ be the one-dimensional subspace spanned by the identity matrix $I$. Let moreover $\|\cdot\|_{\...
Krzysztof's user avatar
  • 375
3 votes
0 answers
108 views

A question on essentially self-adjoint differential operators of the type $\Delta=D^{\ast}D$

Let $(M,g)$ be a smooth (connected, complete, oriented) Riemannian manifold and let $D:C^{\infty}(M)\to C^{\infty}(M)$ be a linear partial differential operator, which I view as an operator in $L^{2}(...
B.Hueber's user avatar
  • 1,171
2 votes
0 answers
80 views

Surjectivity of kernel operator with kernel in $L^1(\nu \times \mu)$

Let $ \mu $ and $ \nu $ be two finite and non discrete measures. Let's begin with a well-known fact. Let $ k \in L^2(\nu \otimes \mu) $, then we can define an operator $ \tilde{T} $ as follows: $$ \...
thibault jeannin's user avatar
2 votes
1 answer
142 views

Bounded differentiation operator on compact intervals with $L^2$ norm

It is known that the differentiation operator $D$ is not bounded on $C^1([0,1])$ with $L^2$ norm (counterexample: $f(x)=x^n$). Now I am wondering whether there is an infinitely dimensional subspace ...
graham's user avatar
  • 153
4 votes
1 answer
52 views

Krein-Rutman for integral transforms: proof of convergence to leading eigenvector

Disclaimer: This is a question in functional analysis, on which I don't have much background. It arose from me trying to prove on my own a folklore result in probability theory. Consider an integral ...
Plemath's user avatar
  • 312
6 votes
1 answer
951 views

a claim for a proof of the invariant subspace problem [closed]

Recently four mathematicians claimed to have proven the invariant subspace problem, which is the problem that states Does every bounded operator on a separable Hilbert space have a non-trivial ...
euleroid's user avatar
1 vote
0 answers
87 views

Convergence and sequential compactness for nonlinear operators

I have a family of operators $T_n\colon X \to Y$ where $X,Y$ are Hilbert spaces. These operators are nonlinear. What kind of notions of convergence does one have for such operators? I'm specifically ...
C_Al's user avatar
  • 251
0 votes
0 answers
49 views

Kadec-Klee property of an equivalent norm on a Hilbert space

Let us consider the space $\ell_2$ with the Hilbert norm $\Vert \cdot \Vert$ and consider the following eqivalent norm: $$ \Vert (r,x) \Vert_A^2 = \Vert (r, Tx)\Vert^2 + \max \{ \Vert x \Vert, \vert r ...
PPB's user avatar
  • 85
4 votes
1 answer
251 views

Show that $\Lambda_\varphi(x_n)\to \Lambda_\varphi(x)$ for an nsf weight $\varphi$ on a von Neumann algebra

Let $\varphi$ be an nsf weight on a von Neumann algebra $M$. Fix a square-integrable element $x\in \mathscr{N}_\varphi$. Put $$x_n := \sqrt{\frac{n}{\pi}}\int_{-\infty}^{+\infty} \exp(-nt^2) \sigma_t^\...
Andromeda's user avatar
  • 175
5 votes
1 answer
183 views

Question about modular group (Modular theory in operator algebras, section 2.14)

Consider the following fragment from Stratila's book "Modular theory in operator algebras", section 2.14, p20: I'm trying to understand the claim $(3)$ (see the red box). The main strategy ...
Andromeda's user avatar
  • 175
0 votes
0 answers
59 views

Decomposition of a contractive representation into an orthogonal sum for the $n$-dimensional case. Has this been done yet?

I know that it has been done for the two-dimensional case. Marek Kosiek showed it in his work "Decomposition of operator representations of the algebra $R(K_1 \times K_2)$" and "...
S-F's user avatar
  • 63
10 votes
0 answers
225 views

Can the trace be computed in any Schauder basis?

I'm cross-posting this question from Math.SE, as it didn't get much attention there. Let $H$ be a separable Hilbert space and $T \in L(H)$ a trace-class operator. It is well known that the trace of $T$...
WillG's user avatar
  • 233
0 votes
0 answers
29 views

On constructing the canonical boundary operator for a given differential operator

Given an $n\times n$ matrix $$X=\begin{pmatrix} x_{11} & x_{12} & \cdots & x_{1n} \\ x_{21} & x_{22} & \cdots & x_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ x_{n1}...
Ryan Hendricks's user avatar
6 votes
1 answer
255 views

Example/Existence of Positive Linear Functional which is NOT Hermitian

We know that if $\mathcal{A}$ is a unital $C^*$-algebra and if $f:\mathcal{A}\to\mathbb{C}$ is a positive linear functional then it is Hermitian. It simply follows from the fact that in $\mathcal{A}$ ...
UtsabrajSarkar's user avatar
1 vote
1 answer
151 views

Some operators on spheres

Let $S_2$ be the unit sphere in $\mathbb R^3$ equipped with normalized Haar measure. For a continuous function f and $\delta\in (-1,1)$ define $T_\delta f(x):=\int_{\{y:<x,y>=\delta\}}f(y)d_\...
A beginner mathmatician's user avatar
2 votes
0 answers
94 views

Nemytskij operator for Lebesgue variable UNBOUNDED exponent spaces

Let $f:\Omega\times\mathbb{R}\to\mathbb{R}$ be a Caratheodory function (i.e. $f(x,\cdot)$ is continuous for a.a. $x\in\Omega$ and $f(\cdot,t)$ is measurable for all $t\in\mathbb{R}$), where $\Omega\...
Bogdan's user avatar
  • 1,759
5 votes
1 answer
220 views

How big is the class of all closed range bounded linear operator?

Let $X$ and $Y$ be Banach spaces and let $CR(X,Y)$ denote the set $B(X,Y)$ of all bounded linear maps from $X$ to $Y$ with $T(X)$ closed in $Y$. Certainly $CR(X,Y)$ is not open in $B(X,Y)$ as given ...
Anupam's user avatar
  • 585
1 vote
0 answers
98 views

$(\lambda I-A)^{-1}-(\lambda I-B)^{-1}$ compact implies $\sigma_\text{ess}(A)=\sigma_\text{ess}(B)$

Suppose $H$ is a Hilbert space and $A$, $B$ are two adjoint operators on it (not necessarily bounded), satisfying $D(A)=D(B)$. Question: If $\exists \lambda\in \rho(A)\cap\rho(B)$ such that $(\lambda ...
Richard's user avatar
  • 775
5 votes
2 answers
432 views

Does closedness of the image of unit sphere imply the closed range of the operator

Let $X$ and $Y$ be Banach spaces and let $T:X\to Y$ be a bounded linear operator such that $T(S_X)$ is closed in $Y$. Does it imply that $T(X)$ is closed? Any hint is appreciated.
Anupam's user avatar
  • 585
2 votes
0 answers
96 views

Isometric Schröder-Bernstein theorem for injective Banach spaces?

It's known that every injective Banach space is of the form $C(M)$ where $M$ is a compact, Hausdorff, extremally disconnected topological space. Let $X$, $Y$ be two injective Banach spaces such that, ...
Onur Oktay's user avatar
  • 2,605
2 votes
0 answers
71 views

How to naturally define an output space with certain properties

Consider the following regression problem $v=A(u) + \varepsilon$ for some operator $A:\mathcal{U} \rightarrow \mathcal{V}$ and some function spaces $\mathcal{U},\mathcal{V}$, mapping from $\mathcal{X}$...
emma bernd's user avatar
2 votes
0 answers
66 views

interchange of integrals and semigroup without the semigroup being an integral operator

In Cazenave's book: BREZIS, HAIM.; CAZENAVE, T. Nonlinear evolution equations. IM-UFRJ, Rio, v. 1, p. 994, 1994. The following corollary appears The formula (1.5.2) is Duhamel formula: $$u(t) = T(t)u(...
Ilovemath's user avatar
  • 677
0 votes
0 answers
72 views

Domain of a Jacobi operator with unbounded coefficients

Is it possible to describe the domain of a Jacobi operator explicitly? Let $J$ be the linear operator acting on a real sequence $(u_{n})_{n\in\mathbb{N}}$ by $$ J(u_{n}) = a_{n+1} u_{n+1} + a_{n} u_{n-...
Bastien's user avatar
  • 23
0 votes
0 answers
49 views

Existence of sequence of regular projections

Reading the book :Krasnosel'skii, M.A.; Pustylnik, E.I.; Sobolevskii, P.E.; Zabreiko, P.P. (1976), Integral Operators in Spaces of Summable Functions, Leyden: Noordhoff International Publishing, 520 p....
Guillermo García Sáez's user avatar
2 votes
1 answer
149 views

Show that $\|P(f\circ\varphi_{\lambda})-\widetilde{f}(\lambda)\|_p=\|P(f\circ\varphi_{\lambda}-\overline{P(\overline{f}\circ\varphi_{\lambda}}))\|_p.$

Let $\Omega = \mathbb B_n,$ the unit ball in $\mathbb C^n$ and $L^2_a(\Omega)$ be the Bergman space endowed with the normalized volume measure on $\Omega.$ Let $k_{\lambda}$ be the associated Bergman ...
Anacardium's user avatar
1 vote
0 answers
176 views

Unitary operators in $\mathcal{U}(L^2(X,\mu))$ not coming from $\operatorname{Aut}(X,\mu)$

$\DeclareMathOperator\Aut{Aut}$Let $(X,\mu)$ be a measured space with probability measure $\mu$ and $\Aut(X,\mu)$ denote the group of measure preserving bijections of $(X,\mu)$. It is easy to see ...
Kajal Das's user avatar
  • 105
2 votes
1 answer
214 views

Forming real positive semidefinite matrices from complex matrices

I have asked this question on the Mathematics Stack Exchange: https://math.stackexchange.com/questions/4924554/forming-real-symmetric-positive-semidefinite-matrices-from-complex-matrices. Let $Q \in \...
Mthpd's user avatar
  • 31
0 votes
0 answers
66 views

Equality between operators, on dense subspace, from a quadratic form point of view

Let $L \ge 1$ and consider a finite box $\Omega = [0,L]^{d} \subset \mathbb{R}^{d}$. The set of functions: $$\psi_{p}(x) = \frac{1}{L^{d/2}}e^{i\langle p,x\rangle} \quad p\in \frac{2\pi}{L}\mathbb{Z}^{...
MathMath's user avatar
  • 1,305
1 vote
1 answer
232 views

When is this operator positive semi-definite?

I have the following operator $$\Phi(\chi_A)=\int \text{d}\eta\, \text{d}\zeta\,\chi_A(\eta,\zeta)\,e^{i(\eta \hat{P}+\zeta\hat{Q})}.$$ With $\chi_A$ the indicator function associated to a set $A\...
Nicolas Medina Sanchez's user avatar
2 votes
1 answer
139 views

Domain of the infinitesimal generator of a composition $C_0$-semigroup

In the paper [1] the following $C_0$-group is presented, $$ T(t)f(x) = f(e^{-t} x) , \quad x \in (0,\infty) \quad f \in E $$ where $E$ is an ($L^1,L^\infty$)-interpolation space. In mi case, I'm just ...
Scottish Questions's user avatar

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