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.

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extending a bounded linear operator

So I have a homework question which I have no idea how to start. Let $E_0$ be a dense subspace of the normed space $E$. Let $T_0:E_0 \rightarrow F$ be a bounded linear operator into the Banach space ...
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11 views

What does the Stein–Weiss Interpolation Theorem say?

I was looking for the statement for the Stein–Weiss Interpolation Theorem, but I cant find it anywhere on internet.
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14 views

What is iterative integral operator?

I don't know what is iterated integral operator, and why $K^{(3)}$ is square-integrable implies $K$ is compact. Can anyone help?
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1answer
9 views

Showing T intertwines $D_T$ and $D_{T^*}$ using Spectral Theorem

Suppose $T$ is a contraction on a Hilbert space $H$ (separable, if you wish). $D_T=(I-T^*T)^{1/2}$ and $D_{T^*}=(I-TT^*)^{1/2}$. I want to show that $TD_T=D_{T^*}T$. I had done this before using a ...
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57 views

T is not compact operator

I want to show that if $T$ is a bounded operator between two Hilbert spaces and $T$ is not compact then there exists an orthonormal sequence $y_{n}$ and an $R>0$ such that $\forall n\in ...
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26 views

Self-adjoint operator and eigenbasis

Let us assume that we have a self-adjoint operator $A: D(A) \subset L^2 \rightarrow L^2$ and we know that $A$ has a purely discrete spectrum and the eigenvalues of $A$ are simple. Does that mean that ...
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52 views

Derivatives of Differential operators

Let $V=H^1(\Omega)$ and $U = L^{\infty}(\Omega)^2$. Take for example, a differential operator, $L:U\times V \rightarrow V^* $ such that: \begin{equation} L(u,y) = \Delta y + \vec{u}(x)\nabla y ...
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22 views

Equivalent definitions of the trace of a Hilbert-Schmidt operator

I am currently reading the book Spectral Methods in Automorphic Forms, and Iwaniec defines the trace operator in a different way than I am accustomed to. Throughout, assume that everything converges ...
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1answer
42 views

How to show a Borel Operator Measure dilates to a Spectral Measure?

Does anyone know a simple proof of the following theorem stating that a positive Borel operator measure $P$ on $\mathbb{R}$ can be written as $V^{\star}EV$ for a Borel spectral measure $E$? ...
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70 views

Composition of projections has a fixed point in a Hilbert space

Let Let H be a Hilbert space with an inner product ⟨⋅,⋅⟩ : H×H→R, and induced norm $∥⋅∥ : H→R_+$ Let $C_1$ and $C_2$ be closed, convex, nonempty, disjoint subsets of $H$ with at least one of ...
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35 views

Spectral Measures: Spectral Subspaces

Given a Hilbert space $\mathcal{H}$ and let the Lebesgue measure be $\lambda$. Consider a normal operator $N:\mathcal{D}\to\mathcal{H}$. Denote its associated Borel spectral measure by: ...
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Closure in a Hilbertspace

Define for a self-adjoint pure contraction $S$ (remember: $\|S\|\leq1$ and $\pm1\notin\sigma_p(S))$ on a Hilbert space $\mathcal{H}$ the following set: $C_c^*(S):=\{g(S):g\in C_c(\hat{\sigma}(S))\}$ ...
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1answer
17 views

Reducing Subspaces: Nonexample?

Given a Hilbert space $\mathcal{H}$. Consider an operator $T:\mathcal{D}(T)\to\mathcal{H}$. Suppose there exists a closed subspace $Z\leq\mathcal{H}$: $$TZ\subseteq Z,TZ^\perp\subseteq Z^\perp$$ ...
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29 views

a question about contractions on Hilbert spaces

Let $\cal{H}$ be a Hilbert space, $T_1,T_2\in\cal{B(H)}$, $\|T_1(h_1)+T_2(h_2)\|^2\leq\|h_1\|^2+\|h_2\|^2$ for all $h_1,h_2\in\cal{H}$. $T_1T^\ast_1+T_2T^\ast_2\leq I$. Then can we verify that 1 ...
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1answer
45 views

Finding the norm of integral operator

I have the following operator: $A: (C^1[0;1];|||\cdot|||)\rightarrow(C^1[0;1];||\cdot||_{\infty})$ $Af(x)=\int_0^x f(t)dt$ where $|||f|||= ||f||_\infty+||Af||_\infty$ What is $||A||?$ I wrote ...
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1answer
18 views

Isometric Operators: Common Core

Given a Hilbert or Banach space $\mathcal{H}$. Consider two closed operators $S:\mathcal{D}(S)\to\mathcal{H}$ and $T:\mathcal{D}(T)\to\mathcal{H}$. Suppose they're isometric on a common core ...
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1answer
77 views

$\langle Tx,x\rangle =0$ , then T is zero

I just wanted to be sure about something. The implication $\langle Tx,x\rangle =0$ , then T is zero , holds only if $T$ is self-adjoint right? If $T$ is an arbitrary operator, we need to have $\langle ...
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1answer
47 views

The set of analytic functions on unit circle is not a C*-algebra

Let $\mathbb{D}$ be the open unit disc on the complex plane and consider the set $$A=\{f\in C({\rm cl}\, {\Bbb D})\colon f \text{ is an analytic function on } {\Bbb D}\}.$$ It is certainly closed ...
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1answer
17 views

Is $C(\Omega)$ a C*-algebra if $\Omega$ is not locally compact, nor compact?

We always say if $\Omega$ is compact or locally compact, then C(\Omega) is a C*-algebra. Now is $C(\Omega)$ a C*-algebra if $\Omega$ is not compact nor locally compact? If not, I want to know which ...
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1answer
25 views

spectral projection of an element in a C*-algebra

I'm studying Takesaki's Operator theory and I preferred "spectral projection "in the page 43 of this book while he didn't speak about it before. I searched it, but I could not find it. Please explain ...
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22 views

Getting U.C.P map on group operator algebras using Fell's absorbtion principle.

I'm struggling a bit with this theorem: Let $\Gamma$ be a discrete group and $\mathbb{C}\Gamma$ be the group ring of $\Gamma$ i.e. the set of formal sums $\sum_{t \in \Gamma} \alpha_t t$. Furthermore ...
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1answer
16 views

Characterization of multipliers on the Dirichlet space of holomorphic functions

Can someone tell me whether there is a characterization of boundedness of multiplication operators on the Dirichlet space of the unit disc? These are holomorphic functions $f$ on the unit disk $D$ ...
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24 views

Operator for scaling a function?

Let $\mathbb{F}$ denote the set of functions of the form $f: \mathbb{R} \to \mathbb{R}$. I am interested to know whether there exists a well-known linear map $T_\alpha: \mathbb{F} \to \mathbb{F}$ ...
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Selfadjointness of the Dirac operator on the infinite-dimensional Hilbert space

I am a physicist, so my background in functional analysis is limited only to basics. However, I would like to prove that the free Dirac operator is selfadjoint (or Hermitian, or neither). The free ...
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1answer
22 views

norm of inverse of a bounded operator [duplicate]

Are there any conditions in which norm of inverse of a bounded operator T is equal to reciprocal of norm of the operator T.
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1answer
46 views

boundary conditions for operator

if you have a Schrödinger operator on a sphere ( $\mathbb{S}^2$) $-\Delta_{\theta,\phi} \psi(\theta,\phi) + V(\theta) \psi(\theta,\phi) = E\psi(\theta,\phi),$ where the potential does not depend on ...
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1answer
28 views

Positive linear functional on an involutive Banach algebra

Why is every positive linear functional on an involutive Banach algebra with a bounded approximate continuous?
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1answer
26 views

Approximate unit of an involutive Banach algebra

I know that every C*-algebra has an approximate unit. I have two questions: why we cannot show that every involutive Banach algebra has an approximate unit? I need an example of an involutive Banach ...
2
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1answer
25 views

The spectrum of $L:=-\Delta+V(x)$ on complex $L^2(\mathbb{R}^N)$ and real $L^2(\mathbb{R}^N)$

In general, when one talks about the spectrum of an self-adjoint operator, it is naturally considered in a complex Hilbert space (say $L^2(\mathbb{R}^N,\mathbb{C})$). Moreover, the spectral ...
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1answer
25 views

Positive elements in a Banach algebra

Let $A$ be a unital Banach algebra. If $a$ is an element of $A$ with $||1-a||_{sp}<1$, then there exists $b\in A$ such that $b^2=a$. Furthermore, if $A$ is an involutive Banach algebra and if $a$ ...
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On calculating spectral projections

Consider following operator from this paper; Let $h$ be any function in $L^1$ relative to the measure $g(w)dw$ and $K\in\mathbb{C}$ Consider the linear operator $B$ on $L^1$ defined by $$(Bh)(x) = ...
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Eigenvalues of a Self-Adjoint Operator

It is easy to see that eigenvectors corresponding to distinct eigenvalues of a self-adjoint operator $T$ are mutually orthogonal. However, from this, it is supposed to be easy to see that for a given ...
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30 views

Laplacian on $\mathbb{S}^2$ has a pure point spectrum

Consider an operator $T = -\Delta + V(\theta)$ where $V(\theta)$ is $C^{\infty}$ and $T : C^{\infty}(\mathbb{S}^2) \subset L^2(\mathbb{S}^2)\rightarrow C^{\infty}(\mathbb{S}^2).$ I was wondering why ...
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How to calculate Hill's discriminant?

I am currently reading this paper on Schrödinger operators see here. On page 6 and 7 they talk about Hill's discriminant and how this is connected with the spectral properties. They also show some ...
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31 views

Compact projection on Banach space has finite rank

Let $E$ be a Banach space. Show that every compact projection has finite rank. I have no idea where to start.
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14 views

Numerical range of closure of operator

Let $B$ be an unbounded densely defined and closable operator. If $\mathcal{N}(B)$ is the numerical range, what can be said about the numerical range of its closure $\overline{B}$?
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computation example of Maximal operator

how to compute the answer M(f(x))? can anyone show a detail step? thank you!!
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36 views

Fixed Point Theorem in finite dimensional Euclidean space

A fixed point theorem says that: "any continuous mapping of $\mathbb{R}^n$ into a bounded subset of $\mathbb{R}^n$ has a fixed point". So consider $f: \mathbb{R}^n \rightarrow X \subset ...
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1answer
32 views

Multiplication operator on $L^2$ and spectral theorem.

Let's consider the multiplication operator by the independent variable in $L^2(\mu)$, where $\mu$ is a borel regular measure on $\mathbb{C}$: $Mf(z)=zf(z)$. I want to show that if $\phi$ is a borel ...
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1answer
27 views

Spectrum of multiplication operator by the independent variable in $L^2$

If $\mu$ is a regular Borel measure on $\mathbb{C}$ with compact support $K$, define $N_\mu$ on $L^2(\mu)$ by $N_\mu f=zf$ (the multiplication by the indipendent variable). An exercise in "Conway" ...
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1answer
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Simple proof that $\|p(A)\|\le \sup_{|z|\le 1}|p(z)|$ for polynomials $p$ and $\|A\| \le 1$.

Let $\mathcal{H}$ be a complex Hilbert space, and let $A$ be a bounded operator linear operator on $\mathcal{H}$ with $\|A\| \le 1$. It is known that $\|p(A)\|\le \sup_{|z|=1}|p(z)|$ for all complex ...
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19 views

Nondegenerate representation

By the definition, we say a representation $(\pi,H)$ is nondegenerate if $cl[\pi(A)H ]= H$. Below I have two theorem, the first from Conway's Functional analysis and the second from Takesaki's ...
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1answer
27 views

Spectral measure and commutativity.

I want to prove that if $A\in B(H)$ and $N\in B(H)$ is a normal operator, and $AE(\Delta)=E(\Delta)A$, where $E$ is the spectral measure given by $N$ and $\Delta$ is a Borel subset of $\sigma(N)$, ...
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Some doubts concerning spectral theory.

Probably I'm saying something wrong (that's why the conclusions are strange) so please correct me! There is the continuous functional calculus for a normal element $N$ of a C*-Algebra. This means ...
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1answer
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If $N$ is normal, $W^\ast(N)=\{N\}''$

I want to prove that if $N$ is normal, then $W^\ast(N)=\{N\}''$, where $W^\ast(N)$ is the Von Neumann algebra generated by $N$ and $\{N\}''$ is the bicommutant of $N$. for the inclusion $W^\ast(N) ...
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1answer
40 views

Why is $\int_{\mathbb{R}^3} |p\rangle \langle p| d\lambda(p)=id$?

As I have written in the headline, I am curious how the relation $\int_{\mathbb{R}^3} |p \rangle \langle p| d\lambda(p)=id$ that physicists use, where $|p\rangle$ is the eigenfunction to the ...
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1answer
34 views

Cyclic representation on $L^2(\mu)$

Show that if $(X,\Omega,\mu)$ is a $\sigma-$ finite measure space and $H=L^2(\mu)$, then $\pi:L^\infty(\mu)\to B(H)$ defined by $\pi(\phi)=M_\phi$ is a cyclic representation and find all the cyclic ...
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1answer
91 views

The spectrum of a product of operators

Suppose $A,B\in\mathcal{B}(\mathcal{H})$, where $\mathcal{H}$ is an infinite dimensional Hilbert space. In general, we know that there is no relationship between $\sigma(AB)$ and $\sigma(A)$ and ...
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37 views

Showing $||L|| = \sup_{x,\, y\, \in H,\, x,\, y\, \neq 0} \frac{|\langle Ax,y\rangle|}{||x||\cdot ||y||}.$

Prove that, for a Hilbert space $H$ and a linear bounded operator $L:H \to H$ such that the domain of $L$ is $H$, $$||L|| = \sup_{x,\, y\, \in H}_{ x,\, y\, \neq 0} \frac{|\langle ...
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spectrum of a convolution operator is connected

Let $k \in L^1(\mathbb{R}; \mathbb{C})$ and define a linear operator $T$ acting on $L^2(\mathbb{R}; \mathbb{C})$ by $$ Tf(x) := [k \ast f] (x) \ldotp$$ Linearity is obvious, while the fact that $T$ is ...