Stack Exchange Network

Stack Exchange network consists of 174 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Visit Stack Exchange

Questions tagged [operator-theory]

Operator theory is the branch of functional analysis that focuses on bounded linear operators, but it includes closed operators and nonlinear operators. Operator theory is also concerned with the study of algebras of operators.

0
votes
0answers
7 views

Laurent and Toeplitz operators properties

I would like to have an exact definition of Laurent and Toeplitz operators. I would also want to know if these operators are bounded and why? And how could we compute their norm and know if they are ...
1
vote
0answers
22 views

Right half-plane in resolvent set of $B$. Why is $B$ invertible?

Let $B$ be a densely defined operator on a Banach space $X$ and let $B$ generate an analytic semigroup on $X$. I know that the right half-plane satisfies $\{Re\ z>0\}\subset \rho(B)$ ($\rho(B)$ ...
1
vote
1answer
31 views

How can we use Hahn-Banach Theorem to prove a functional is linear and bounded?

Let $V$ and $W$ be Banach spaces, bounded linear operators $T: V \to W$ and $S: W' \to V'$ such that \begin{align*} f(T(v)) = S(f)(v), \qquad \forall v \in V, \quad \forall f \in W' \end{align*} ...
0
votes
1answer
16 views

To show an element in the sprectum of a matrix $M$ is negative in the spectrum of $-M$

I am trying to solve the following problem: Let $V$ be a Banach space , $\lambda \in \mathbb{C}$, an operator $H \in \mathbb{B}(V \oplus V)$ in a form of \begin{align} H = \begin{pmatrix} 0 &...
1
vote
0answers
22 views

Please check my proof of a bounded linear operator and to show it a finite rank operator

There is an exercise I am trying to solve, but I am not sure if my solution is 100% correct. Please help me to check it carefully. Let $A = (A_{n,m})_{n,m \in \mathbb{N}}$ with $A_{n,m} \in \...
2
votes
1answer
36 views

Trace of $x\otimes y\tilde\otimes x\otimes y$

Suppose that $x,y\in\mathbb H$, where $\mathbb H$ is a separable Hilbert space. Define the tensor $x\otimes y:\mathbb H\to\mathbb H$ by setting $(x\otimes y)(z)=\langle z,y\rangle x$ for $z\in\mathbb ...
2
votes
1answer
50 views

A property of the convolution of operator

For any $f\in C_c^\infty(\mathbb{R})$, let $T_f$ denote the bounded operator on $L^2(\mathbb{R})$ given by convolution by $f$: that is, $$(T_f g)(x):=\int f(y)g(x-y)\,dy,$$ for $g\in L^2(\mathbb{R})$...
0
votes
0answers
12 views

Infimal Dual Norms

If $\|A\|_p$ is the Schatten $p$-norm of a matrix $A$ and $p^{-1} + q^{-1}=1$, $$\|A\|_q = \sup\{|\langle A,B\rangle_\textsf{F}| : \|A\|_p=1\}$$ is the dual norm, which turns out to be the Schatten $q$...
2
votes
1answer
21 views

the commutant of $\mathcal{HS(H)}$

Let $\mathcal{HS}(H)$ be the set of Hilber Schmidt operators on a Hilbert space,it is a $C^*$ algebra.I wonder whether we have an explicit description of the commutant of $\mathcal{HS(H)}$.Is the ...
0
votes
1answer
21 views

construct a closed subspace of $\mathcal{HS}(H)$ such that all elements in the subspace are commutative.

Suppose $K=\mathcal{HS}(H)$,where$\mathcal{HS}(H)$ is the set of all Hilbert Schmidt operators on the Hilbert space $H$.I have two questions. 1.Can we construct a closed subspace $K_1$ of $K$ such ...
1
vote
0answers
44 views

Operator norm of $T: L^{p}(0,\infty) \rightarrow L^{p}(0,\infty)$

I hope you can help me with this exercise. Let $p\in (1,\infty)$. Also, \begin{align*} T: L^{p}(0,\infty) &\rightarrow L^{p}(0,\infty)\\ f&\rightarrow (Tf)(x)=\frac{1}{x}\int_{0}^{x} f(y)\, ...
2
votes
2answers
38 views

Adjoint of T $\in B(\ell^{2})$

I was wondering if you guys can help me with this one! Let $T \in B(\ell^{2})$ a continuous operator. Also, \begin{align*} (Ts)(n)=\frac{1}{n}\sum^{t}_{m=1} s(m),\hspace{1cm}n\in \mathbb{N}, s\in \...
1
vote
0answers
12 views

Constructing a product of commuting operators with certain properties

Motived by elementary examples I wanted to find an example of a pair of commuting operators $A$ and $B$ with the following properties. $A$ is left invertible, but not right invertible, and $B$ is ...
0
votes
0answers
7 views

Semi-Fredholm quasidiagonal operator has index zero

For a bounded linear operator $T$ on Hilbert space, it is known that if $T$ is left Fredholm and quasidiagonal, then it is Fredholm with index zero. This can be seen by writing $T=D+K$, where $D$ is ...
2
votes
1answer
40 views

How to make sense of operators' inverse and the derivative of operators themselves?

I'm a physics graduate student, it's when my professor for mathematical physics mentioned KdV equations that I encountered the following problem. This is what our professor wrote: $\mathbb{L}:=-d^2+...
0
votes
0answers
9 views

Do the poles of the resolvent of an operator correspond to eigenvectors?

Consider a matrix $M$ with resolvent $$R_M(z)=(z-M)^{-1}$$ The poles $\lambda_i$ of $R_M(z)$ are eigenvalues of $M$, and the eigenvectors $v_i$ of $R_M$ are also eigenvectors of $M$ such that $$...
1
vote
0answers
9 views

Are the poles of the resolvent of a Hermitian operator real?

The resolvent of a matrix $M$ is defined as $$R_M(z)=(z-M)^{-1}.$$ The spectrum of $M$ can then be defined as the set of points $\lambda$ for which $R_M(\lambda)$ is singular. Since all the ...
0
votes
0answers
10 views

Example: $F(t)$ is strongly continuous on $X$ but $F^*(t)$ is not

Let $X$ be a reflexive Banach space. I am looking for examples of the strongly continuous operators $F(t):t\mapsto \mathcal{L}(X)$ such their adjoint on $X$ for every $t$ is not a strongly continuous ...
0
votes
1answer
34 views

multiplication of Hilbert Schmidt operators

Suppose $H$ is an infinite dimensional Hilbert space,$B(H)$ is the set of bounded operators on $H$,$\mathcal{HS}(H)$ is the set of Hilbert-Schmidt operators on $H$. I have two questions: 1.If $T$ is ...
1
vote
0answers
17 views

bimodule over a non-unital $C^*$ algebra

Let $A$ be a $C*$ algebra,suppose $\mathcal{H}$ is a $M(A)–M(A)$ bimodule,where $M(A)$ is a multiplier algebra of $A$.Can we deduce that $\mathcal{H}$ is a $A-A$ bimodule? My thought:since $\mathcal{...
1
vote
1answer
19 views

Minimal Polynomial of null operator

I think that is a simple question, but there is some details confusing me. I need to calculate the minimal polynomial and the characteristic polynomial of the null operator in a $\mathbb{F}$-space. My ...
0
votes
0answers
29 views

How to compactly write “element-wise less than”

I'm trying to compactly notate the cumulative distribution function of a degenerate distribution; in other words, one of a continuous random variable of which one always observes the same outcome ($\...
0
votes
0answers
25 views

$i d/dx$ is symmetric on the domain

Let $\Omega$ be the domain $\Omega = \lbrace f : [0,1] \rightarrow \mathbb{C} | f \ \text{is abs. cont.}, f' \in L^2 ([0,1]), f(0)=0 \rbrace$. I am asked whether the operator $T= i \frac{d}{dx}$ is ...
1
vote
0answers
21 views

Finding the eigenfunctions of an operator with multiple derivatives

The differential operator is given by $L= x\cdot \frac{d^2}{dx^2} + \frac{d}{dx} -\frac{a}{x}$ Any advice or strategies to find the eigenfunctions of this operator would be greatly appreciated. I ...
1
vote
1answer
18 views

A question on hypercyclic operators and their adjoints.

I'm reading Linear Chaos by Karl-G. Grosse-Erdmann and Alfred Peris Manguillot. I'm and having trouble connecting the dots in the "equivalently" part of proof. $\textbf{Theorem}$ Let $T$ be a ...
1
vote
1answer
22 views

A condition to ensure that adjoint and inverse commute

Let $A$ be a densely defined operator on a Hilbert space $H$. Let us assume that $A$ is injective and $Ran(A)$ is dense. Then I want to show that $$ (A^*)^{-1} = (A^{-1})^*. $$ So far, I was able to ...
0
votes
1answer
37 views

A question about self-adjoint operators

Let $A$ be a densely defined symmetric operator on some Hilbert space $\mathcal{H}$. Let's say that we know that there exists $(A + 1)^{-1}$ and that this is a self-adjoint operator. How can I prove ...
1
vote
1answer
24 views

tracial states on $C^*$ subalgebra

Let $A$ be a $C^*$ algebra,$A=A_1\oplus A_2$.If $A$ has tracial state $\tau$,I want to show $A_1$ also has tracial state,say $\tau_1$. My thought: let $\tau_1(a_1)=\tau(a_1,0)$,where $a_1 \in A_1$,...
2
votes
0answers
58 views

Proving a Certain Inequality Without Utilizing the Full Theory of $ C^{\ast} $-Tensor Products

Suppose that we have the following objects: $ X $ — a locally compact Hausdorff space. $ A $ — a $ C^{\ast} $-algebra. $ \pi $ and $ \rho $ — commuting representations of, respectively, $ {C_{0}}(X) $...
1
vote
0answers
31 views

What are the properties of this differential-like operator?

So I've recently came across this question, which is about defining the differential operator $D_x$ in terms of axioms rather than the traditional "taking the limit"-way. This made me think about a ...
1
vote
1answer
40 views

tracial states on corona algebra

Let $A$ be the $c_0$ direct sum of $M_{n}(\mathbb{C})$,I know the fact that the multiplier algebra of $A$ ,M($A$) is $\prod M_n(\mathbb{C})$. Does the corona algebra $M(A)/A$ have uncountable tracial ...
2
votes
1answer
25 views

On the image of two bounded operators

Let $A, B \in B(\mathcal{H})$ such that $Im(A) = Im(B)$, where $Im(A)$ denotes the image of $A$. Is there any $\lambda>0$ such that $AA^*=\lambda BB^*$?
0
votes
1answer
21 views

existence of finite irreducible reprentation of a nonunital $C^*$ algebra

Suppose $A$ is a non-unital $C^*$ algebra,can we conclude that there must exist a nonzero finite irreducible representation of $A$.
2
votes
1answer
33 views

Is the spectral decomposition of self-adjoint compact operators unique?

Given a compact, self-adjoint operator $T$ on a Hilbert space $H$, then there is the Spectral Theorem which says that $T=\sum_i \lambda_i P_i$ where the sum is over the number of eigenvalues of $T$, ...
0
votes
1answer
24 views

Pseudo Differential Operator example

I was reading a introduction about Pseudo differential operators, and the definition of that type of operator was: $$|\partial_x^\alpha \partial_\xi^\beta f(x,\xi) | \le C_{\alpha,\beta}(1 + |\xi|)^{...
3
votes
1answer
44 views

Question regarding kernels in Hilbert-Schmidt operators

I'm studying some Hilbert-Schmidt integral operator theory. I read that a Hilbert-Schmidt integral operator is uniquely determined by its kernel, and I wonder how exactly that is. I guess what I'm ...
1
vote
1answer
30 views

Operator valued analytic functions on an annulus

Let $\mathscr{L, M}$ be two Hilbert spaces (not necessarily finite dimensional) and let $$\mathbb{A}=\{z\in \mathbb{C}:0<q<|z|<1\}$$ be an annulus. I was trying to learn about $B(\mathscr{L, ...
0
votes
2answers
43 views

$T$ is an isometry if and only if $\langle Tx, Ty \rangle = \langle x, y \rangle$

I want to prove: A linear mapping $T:X \to Y$ between two pre-Hilbert spaces is an isometry if and only if the inner products $\langle Tx, Ty \rangle = \langle x, y \rangle$ for all $x, y \in ...
3
votes
1answer
48 views

Continuous Functional Calculus Argument in Fell's Paper

I am reading an old paper of Fell's and I am having some problems sorting out one of his continuous functional calculus arguments. The essential problem is this. Let $A$ be a C$^{*}$-algebra and let $...
2
votes
1answer
72 views

Under what conditions we have $\|Tx\| = \|T\| \|x\|$?

For bounded linear operator we have $\|Tx\| \le \|T\| \|x\|$ but when the equality takes, i.e., $\|Tx\| = \|T\| \|x\|$? PS. If $T$ is an isometry then $\|Tx\|=\|x\|$ and $\|T\|=1$, therefore the ...
2
votes
1answer
35 views

Spectrum included in spectral measure support

Let $\mathcal{H}$ be an Hilbert space and $A=\displaystyle{\int} \lambda \,dE(\lambda)$ a spectral operator on $\mathcal{H}$ associated to the spectral measure $E$. I am interested in the very basic ...
1
vote
2answers
41 views

A linear bounded operator has to be continuous?

I just found on the wiki https://en.wikipedia.org/wiki/Continuous_linear_operator stating that An operator between two normed spaces is a bounded linear operator if and only if it is a continuous ...
1
vote
1answer
29 views

Unbounded inverse of a bounded operator

Suppose that $T\in B(H)$. As we know $T^{-1}$ is bounded if and only if $T$ is bijective. Also, $T^{-1}T=TT^{-1}=I$. In the other word, $T^{-1}:\mathcal{H}\to \mathcal{H}$ such that $T^{-1}T\xi=\xi$ ...
1
vote
1answer
19 views

Pure isometries are unitarily equivalent to a shift (Wold decomposition). Do the corresponding intertwining relations also hold for the adjoints?

Let $T \in L(H)$ be an isometry on a Hilbert space. Further assume that $T$ is pure, i.e. $T^{*m} \xrightarrow {m \rightarrow \infty} 0$ in the strong operator topology, or equivalently, there is no ...
1
vote
2answers
62 views

Example of measures being ergodic but not invariant

Let $\mu$ be a measure, $f:X\rightarrow X$ being a map and $\Sigma$ is a $\sigma$-algebra. $\mu$ is $f$-invariant if $\mu(E) = \mu(f^{-1}(E)), \forall E\in \Sigma$. $\mu$ is ergodic with ...
2
votes
1answer
17 views

About definition of Ergodic theorem

Let $(X,\Sigma, \mu)$ be a probability space, and $T:X\rightarrow X$ be a measure-preserving transformation. We say $\mu$ is ergodic with respect to $T$ if for every $E\in \Sigma $ with $T^{-1}(...
-1
votes
1answer
26 views

Image of a dense set [closed]

Is the image of a dense set under an isometric operator is again dense set? i.e., Given two Hilbert spaces $X, Y$. If $T:X\to Y$ is an isometric operator and $S$ is a dense span subset of $X$, is ...
3
votes
1answer
31 views

Proof attempt at “Adjoint of compact operator is compact” on Banach spaces

I was trying to prove this result and this is my attempt. Theorem [Schauder] Let $E,F$ be Banach spaces, and let $T:E\to F$ a compact operator. Then its adjoint $T^*:F^*\to E^*$ is compact. ...
9
votes
0answers
85 views

Trace on $\mathcal{S}(\mathbb{R}^k) \mathbin{\hat{\otimes}_\pi} \mathcal{S}'(\mathbb{R}^k)$

$\newcommand{\Tr}{\operatorname{Tr}}$Let $\mathcal{S}(\mathbb{R}^k)$ denote the $k$-dimensional Schwartz space with the usual topology, and let $\mathcal{S}'(\mathbb{R}^{k}))$ denote its strong dual (...
3
votes
1answer
58 views

Is this operator on $L^\infty$ injective / surjective?

$$ f \in L^\infty (0,1) \\ Tf(x) = \int_0^x e^{y-x}f(y)dy, x\ge0 $$ I've shown that T is a bounded linear operator from $L^\infty(0,\infty)$ into itself. I've computed its norm (it should be $\|T\| = ...