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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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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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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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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53 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
25 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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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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39 views

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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1answer
51 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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57 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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35 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

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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83 views

Maximal monotone operator without convex domain?

I'm looking for an example of a (multi-valued) maximal monotone operator $A$ mapping a Banach space $X$ into its dual $X^*$ such that the domain $D(A)=\{x\in X: Ax\neq\emptyset\}$ is not convex. ...
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1answer
81 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
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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37 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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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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1answer
28 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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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
38 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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27 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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36 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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1answer
30 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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1answer
32 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 ...
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1answer
34 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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22 views

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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32 views

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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1answer
29 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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118 views

Is an orthogonal projector bounded in $L_p$-spaces?

Let $P$ be an orthogonal projector on $C^\infty([0,1])$. For $0<p<\infty$, we define for $f \in C^\infty$ the norm (quasi-norm if $p<1$) $\lVert f \rVert_p$ in the usual way. Moreover, we ...
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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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21 views

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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34 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
38 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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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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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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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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32 views

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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25 views

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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42 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
38 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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1answer
19 views

A question about Invariant subspaces of an algebra.

I feel that this is a very simple problem, but somehow I don't see the solution. I want to show that if $A$ is a subalgebra of $B(H)$ containing $1$ then if $B\in SOTcl(A)$, for every n, ...
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14 views

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 ...
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1answer
20 views

Norm of $A$ is zer0 when $\Bbb H$ is Complex Hiblert Space

If $\Bbb H$ is a $\Bbb C$-Hilbert space and $A\in \Bbb B(\Bbb H),$i.e. bounded linear operator on $\Bbb H$ such that $\langle Ah,h\rangle=0$ for all h in $\Bbb H$, then $A=0$ For the proof of this ...
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When $\int_\Omega f(x)K(x,y)dx$ is injective

Let $T:X\to Y$ be a compact integral operator and consider $g=Tf$, i.e. $$g(y) = \int_\Omega f(x)K(x,y)dx$$ where $\Omega=[a,b]$ is some subset of $\mathbb{R}$. Are there some particular conditions on ...
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Why this operator is not fredholm?

Define $f:S^{2}\to \mathbb{R}$ by $f(x,y,z)=z$. Let $D:=D_{\nabla f}$. As I learned from the following post this operator is not counted as a fredholm operator.( I did not underestand, ...
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When a symmetric densely defined operator is an adjoint operator?

I am wondering if it is possible to say that if a symmetric differential operator is densely defined then the operator is self-adjoint indeed? More Precisely, Let $A:D(A)(\subset H)\to H$ a densely ...
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1answer
11 views

Calculating the form domain of an operator

I am reading the book "Mathematical Methods in Quantum Mechanics" by Gerald Teschl and just came across the concept of a form domain. It is defined for non-negative operators i.e $<\phi,A \phi> ...
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34 views

Approximate unit of a separable C*-algebra

The following is a corollary of Takesaki's Operator Theory: My question: I do not know why the author says"there exists an n such that $||x(1-v_n)^\frac{1}{2}||<\epsilon$" . Please help me to ...
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55 views

What is the image of operator exponential?

Given a Banach space $V$ and a bounded linear operator $A:V\to V$, the operator $e^A$ is bounded and invertible. When $V$ is finite dimensional, every invertible operator is of the form $e^B$ (one can ...
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35 views

An routine exercise about matrix norm

If $T_{n}\in M_{k(n)}(\mathbb{C})$ and $||T_{n}^{*}T_{n}-1_{k(n)}||\rightarrow0$, then $||T_{n}T_{n}^{*}-1_{k(n)}||\rightarrow 0$ too? (Here, $M_{k(n)}(\mathbb{C})$ denotes the $k(n) \times k(n)$ ...
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Norm of an element in a C*-algebra

The following is a part of a proof in Takasaki's Operator theory: Let $\epsilon>0$ and $A$ is a C*-algebra. For an $x\in A$, put $h=x^*x$ and $u_\epsilon=(h+\epsilon)^{-1}h$. We have then ...