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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Spectrum of a bilateral shift

Let $u$ be a bilateral shift on Hilbert space $\ell^2(\Bbb Z)$. As unilateral shifts, spectrum $u$ does not contain any eigenvalue. Also $u$ is unitary, so $\sigma(u) \subset \Bbb T$ ($\Bbb T$ means ...
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35 views

Wot convergence and sot convergence

Let $\{A_n\} $ be a sequence of bounded linear operators on Hilbert space $H$ and $\langle A_n\xi,\eta \rangle \to \langle A \xi,\eta\rangle$ for $\xi,\eta\in H$ with $\|\eta\|=1$. Show that $\|A_n\xi ...
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Showing that an operator generates a unitary group

Consider the following operator on $X=L^{2}(0,1)$: $\displaystyle Af=\frac{df}{d\zeta}$ with domain: $D(A)=\{f\in L^{2}(0,1)|f$ is absolutely continuous, $\frac{df}{d\zeta}\in L^{2}(0,1)$ and ...
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The $C_0-$group generated by the operator $(Af)(x)=f'(x)+a(x)f(x)$

Consider the Banach space $L^1(\mathbb{R})$ of integrable functions $f:\mathbb{R}\to \mathbb{R}$. Consider the unbounded operator $A$ defined by $$(Af)(x)=f'(x)+a(x)f(x), \ \ \ x\in \mathbb{R}$$ for ...
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Quadratic Functional Differentiability

I would like to solve the following: Let $T$ be a self-adjoint bounded operator on a Hilbert space $H$. Consider the quadratic functional $\Phi$ defined by: \begin{equation} ...
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Binomial-like expansion for non-commuting operators

I would like to evaluate the following and express it in a compact form: $(\hat{a}^\dagger(x)+\hat{a}(x))^n\,\vert0\rangle$, where the $n^{\text{th}}$ power of the sum of the annihilation and creation ...
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How to calculate the adjoint of an operator and its domain?

Let $A : D(A) \subset L^2(0,1) \to L^2(0, 1)$, $$D(A) = \{u \in H^2([0, 1]) : u(0) = u'(1) = 0\}$$ $$Au = u''.$$ Can someone explain how to calculate the adjoint of A, $A^*$, and the domain of $A^*$, ...
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Showing that a domain of an operator is dense in $L^2$

Let $A : D(A) \subset L^2(\Omega) \to L^2(\Omega)$, where $$D(A) = \{u \in H^2([0,1]) : u(0) = u_x (1) = 0\}.$$ Show that $D(A)$ is dense in $L^2((0, 1))$. $D(A)$ is dense in $L^2((0, 1))$ if ...
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an operator question

I know how the derivative operator $\Big(\frac{d}{dx}\Big)^n$ works. But then how does it work if I have $$\exp{\Big(a\frac{d}{dx}+b\frac{d^2}{dx^2}\Big)}f(x)$$ I thought to use $$\exp ...
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45 views

Operator matrix is invertible if and only if its determinant is invertible

Let $A,B,C,D$ are pairwise commutative operators on a Hilbert space $H$, then a necessary and sufficient condition that the operator matrix $$\begin{pmatrix} A&B\\C&D\end{pmatrix}$$ be ...
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Does this show that it is a bounded linear operator?

Let $X$ be a Hilbert space and $A\in\mathcal{L}(X)$. I want to show that $\displaystyle e^{At}:=\sum_{n=0}^{\infty}\frac{(At)^{n}}{n!}=T(t)$ defines a strongly continuous semigroup (i.e. a ...
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30 views

Differential of an operator $\phi: Mat_{2 \times 2}{\mathbb{R}} \rightarrow Mat_{2 \times 2}{\mathbb{R}}$

Let's consider an operator $ \phi: Mat_{2 \times 2}{\mathbb{R}} \rightarrow Mat_{2 \times 2}{\mathbb{R}}$ so that $A \rightarrow A^{-1}$. How to evaluate its differential? By the differential we ...
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The closed unit ball is not compact in infinite dimension spaces. Why?

We know that in finite dimension spaces the closed unit ball is compact, that is if H is a finite dimension space, then there exists an $u$ in the closed unit ball in H and $T \in \mathcal{L}(H, ...
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35 views

Calculating a norm of an operator

Let $T \in (C([a, b]))^*$, $$ T(u) = \underset{a}{\overset{(a+b)/2}\int} u(x) dx - \underset{(a+b)/2}{\overset{b}\int} u(x) dx. $$ Show that $ || T || = b - a $. We have that $$|| T || = ...
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Application of Uniform Bounded Principle (UBP)

Let $Y$ be a Banach space, and $Z$ be a n.v.s. If $(B_n)_n\in L(E,F)$ with the property that for all $(y_n)_n\in Y$, that $\|y_n\|\rightarrow 0$, we have $\|B_n(y_n)\|\rightarrow 0$. Prove that ...
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adjoint of an operator. on $L^2(0,1)$, $Bf(x)=\int_0^x f(t)dt$

I see that the above operator is bounded. I ended up with an argument to calculate the adjoint as follows, $$ <f,Bg>=\int_0^1\overline{f(x)} \int_0^xg(t)\,dt\,dx $$ I see $f(x)$ as the ...
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11 views

Eigenvalue dependent operator

Consider the wave equation in a Riemann metric $g^{\mu\nu}$ with spacetime off-diagonal components $g^{i0}$: ...
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21 views

Extension of a self adjoint Operator

Suppose we have a open (bounded) domain $\Omega$ in $\mathbb R^d$. And let a plane $\mathcal P$ in $\mathbb R^d$ divides the domain in two (disjoint) open sets. (say $\Omega_1$ and $\Omega_2$) Hence ...
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A sequence of strongly continuous one-parameter unitary groups

Suppose that for a sequence $\{A_n\}_n$ of bounded self-adjoint operators in a Hilbert space $\mathcal H$ we have $e^{itA_n} \to e^{itA}$ strongly, for all $t \in \mathbb R$, where $A$ is a (possibly ...
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Discrete and Essential spectrum of Laplacian in $\mathbb R_{+}$ (with weird boundary conditions)

I am given on Hilbert Space $\mathcal H=L^2(\mathbb R_{+})$ $$ Af(x)=-f''(x) $$ and Domain of A is $$ D(A)=\{f\in H_2(\mathbb R_{+})\;\;| \;\;f'(0)+\alpha f(0)=0\} $$ for some $\alpha \in ...
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Relating Fourier transform theory on two distinct subspaces

In Fourier transform theory (on $\mathbb{R}$), three vector spaces play a very important role: $L^1(\Bbb R)$, $L^2(\Bbb R)$ and the Schwartz space $\mathcal{S}(\Bbb R)$. Arguably the nicer spaces of ...
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how to prove this epsilon-delta property for continuous functional calculus with normal elements?

Let $ A$ be a C* algebra, $f\in C([-1,1])$. Prove that for every $\epsilon >0, \exists \delta >0,$ s.t. for $\forall x \in A, x=x^*, \| x \| \leq 1$ and $\forall y \in A, \|y\| \leq 1$, we have ...
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45 views

Compact operators on Hilbert Space

I m working on the following problem: Let $K:H\rightarrow H$ be a compact operator on a Hilbert space. Show that if there exists a sequence $(u_n)_n\in H$ such that $K(u_n)$ is orthonormal, then ...
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43 views

Resolvent: Decay Behavior [closed]

Given a Hilbert space $\mathcal{H}$. Consider a Hamiltonian: $$H:\mathcal{D}(H)\to\mathcal{H}:\quad H=H^*$$ Regard the resolvent set: ...
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34 views

Eigenvectors of operators on a tensor product Hilbert Space

Suppose I have finite dimensional Hilbert spaces $V$, $W$, and an operator $A$ acting on vectors in $V$ such that it has eigenvectors/values $Ax_a=\lambda_ax_a$. In the tensor product space I want to ...
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1answer
52 views

Spectral Measures: Core Lemma

Given a Hilbert space $\mathcal{H}$. Consider a Hamiltonian: $$H:\mathcal{D}(H)\to\mathcal{H}:\quad H=H^*$$ Regard a dense domain: ...
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Spectral Measures: Scaled Spaces

Problem Given a Hillbert space $\mathcal{H}$. Consider a Hamiltonian: $$H:\mathcal{D}(H)\to\mathcal{H}:\quad H=H^*$$ Denote its probability measures by: ...
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76 views

Convergence of the spectrum under norm resolvent convergence

Suppose $\{A_n\}$ is a sequence of self-adjoint operators in a Hilbert space $\mathcal H$, and $A$ is a self-adjoint operator, with $A_n \to A$ in norm resolvent sense. Since $A_n \to A$ in strong ...
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Orthonormal system of simultaneous eigenvectors

Suppose we have a commutative family of compact, self-adjoint operators on a Hilbert space. Prove that there is an orthonormal system of simultaneous eigenvectors for the family. I'm not sure how to ...
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1answer
33 views

Discrete Derivative: Closure?

Problem Given the Hilbert space $\ell^2(\mathbb{N})$. Consider the operators: $$T_0:\ell^2_0(\mathbb{N})\to\ell^2(\mathbb{N}):\quad T_0(a_k)_k:=(ka_k)_k$$ ...
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Assuming $A$ is a nonexpansion in some norm, in what norm is $A^\top$ a nonexpansion.

Consider a matrix $A \in \mathbb{R}^{n \times n}$. Consider the vector norm $\| \cdot \|_\triangle = \| F \cdot \|_1$, where $F \in \mathbb{R}^{n \times m}$ and we have $m < n$ and $F$ has ...
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Zero Tensor Product

Suppose we have a space $|\psi_1\rangle \otimes |\psi_2\rangle \otimes |\psi_3\rangle$, and operators (matrices) A ⊗ B ⊗ C acting on this Hilbert space (like in quantum mechanics). I'm trying to ...
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Riccati Finite Difference Equation:

Question: Consider the finite difference equation in Zeilberger notation: $$D_{1,x} [y] = a_0(x) + a_1 (x) y + a_2 (x)y^2 $$ Which in functional form is: $$ y(x+1) - y(x) = a_0(x) + a_1 (x) y + ...
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Spectral Measures: Analytic Elements

Given a Hilbert space $\mathcal{H}$. Consider a Hamiltonian: $$H:\mathcal{D}(H)\to\mathcal{H}:\quad H=H^*$$ Denote the convergence radius by: ...
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1answer
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Sot closure of the unit ball of a subalgebra of $B(H)$

Let $A$ be a $C^* -$ subalgebra of B(H) and $S$ be the closed unit ball of $A$. 1- $S$ is convex and bounded, so $ S = weak^* -cl ~S$. (Is it correct?) 2- By the Kapalansky density theorem, we have ...
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WOT convergence in the unit ball of B(X)

My questions is (probably) related to: On separable Hilbert space $H$, weak operator topology is metrizable on bounded parts of $B(H)$ Does the theorem quoted in the above question, together with ...
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1answer
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Spectral Measures: Nelson

Problem Given a Hilbert space $\mathcal{H}$. Consider a symmetric operator: $$T:\mathcal{D}(T)\to\mathcal{H}:\quad T\subseteq\overline{T}\subseteq T^*$$ Denote the convergence radius by: ...
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Compactness of unit ball in WOT of B(X), with $X$ reflexive [duplicate]

This is connected with the question below: Compactness of unit ball in WOT of B(X) It is known that the unit ball in $\mathcal{B}(H)$, where $H$ is a separable Hilbert space is compact in the weak ...
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Compactness of unit ball in WOT of B(X)

It is known that the unit ball in $\mathcal{B}(H)$, where $H$ is a separable Hilbert space is compact in the weak operator topology. Is it the same true if instead of $H$ we have any separable Banach ...
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Proof of von Neumann's Theorem about dense domain

The theorem due to von Neumann is the following: if $T$ is a closed densely defined Operator with domain $D(T)$, then also $D(T^*T)$ is also dense. I am searching for one proof of this non-trivial ...
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Reducing Subspaces: Characterization

Given a Hilbert space $\mathcal{H}$. Consider an operator: $$T:\mathcal{D}(T)\to\mathcal{H}:\quad\mathcal{D}:=\mathcal{D}(T)$$ Regard a subspace: ...
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Reducing Subspaces: Domain

Problem Given a Hilbert space $\mathcal{H}$. Consider a closed operator: $$T:\mathcal{D}(T)\to\mathcal{H}:\quad T=\overline{T}$$ Regard a closed subspace: ...
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Proving projection operators are bounded

Suppose $X$ is the direct sum of two normed linear spaces $X_1,X_2$, ($X=X_1 \dot{+} X_2$). Then I need to prove the projection operators $\Pi_1:X \to X_1$ and $\Pi_2:X \to X_2$ are bounded ...
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Singular values of compact operators

Let $T$ be a compact operator.Then show that $\sum_{n=1}^\infty$$M_n(T)=\sup\{||PT||_1:P\mbox{ is a rank } N\mbox{ projection}\}$ where $M_n$'s are the singular values of $T$ and ...
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1answer
55 views

Prove the operator is positive

I'm searching for an alternative proof of the following: Let $U$ be a self-adjoint operator on a Hilbert space $H$, define $m=\inf_{\|x\|=1}\langle Ux,x\rangle$ and $M=\sup_{\|x\|=1}\langle ...
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About closed linear operators defined on a subset of a Banach space into a Banach space?

I have this weird theorem here: If $A$ is a closed linear operator from Banach space $X$ into Banach space $Y$ i.e. $A:X\rightarrow Y$ (please take note that $A$ is closed but not necessarily ...
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Interpreting the lingo of a definition

The Terms I grew up with: A bounded linear operator $U$ on a Hilbert space $H$ is a partial isometry if there exists a subspace $M$ of $H$ such that $\|Ux\| = \|x\|$ for all $x\in M$, and $Ux = ...
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1answer
29 views

Heat operator formalism via spectral projections and Dirac measure

I am currently reading very helpful notes on the Heat kernel $p(t,x,y)$ on a Riemannian manifold $M$ -- there is one aspect though that I am not sure I understand notationally, it says that we can ...
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How to show Legendre Operator $L_{m}=-\frac{d}{dx}(1-x^{2})\frac{d}{dx}+\frac{m^{2}}{1-x^{2}}$ is Selfadjoint?

Let $m$ be a positive integer and define $$ Lf = -\frac{d}{dx}(1-x^{2})\frac{df}{dx}+\frac{m^{2}}{1-x^{2}}f $$ on the domain $\mathcal{D}(L)\subset L^{2}(-1,1)$ consisting of all twice ...
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Self adjoint operator with countable eigenvalues

It has been explained here that any self adjoint operator on a seperable Hilbert space inhabits at most countable many eigenvalues. Now I wonder, can one construct an operator $$ T : L^2([0,1]) \to ...