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.

learn more… | top users | synonyms

13
votes
1answer
1k views

How to prove that an operator is compact?

Consider $T\colon\ell^2\to\ell^2$ an operator such that $Te_k=\lambda_k e_k$ with $\lambda_k\to 0$ as $k \to \infty$ how to prove that it is compact?
3
votes
1answer
894 views

Question about Fredholm operator

$X,Y$ are Banach spaces and $A\in B(X,Y)$ is a Fredholm operator (that is, the dimensions of ker($A$) and coker($A$) are both finite), then are closed linear subspaces ker($A$) and Im($A$) ...
20
votes
1answer
858 views

Why do zeta regularization and path integrals agree on functional determinants?

When looking up the functional determinant on Wikipedia, a reader is treated to two possible definitions of the functional determinant, and their agreement is trivial in finite dimensions. The first ...
14
votes
3answers
919 views

Compactness of a bounded operator $T\colon c_0 \to \ell^1$

Pitt Theorem says that any bounded linear operator $T\colon \ell^r \to \ell^p$, $1 \leq p < r < \infty$, or $T\colon c_0 \to \ell^p$ is compact. I know how to prove this in case $\ell^r \to ...
5
votes
1answer
2k views

Easy Proof Adjoint(Compact)=Compact

I am looking for an easy proof that the adjoint of a compact operator on a Hilbert space is again compact. This makes the big characterization theorem for compact operators (i.e. compact iff image of ...
8
votes
2answers
303 views

Question about Angle-Preserving Operators

This an exercise out of Spivak's "Calculus on Manifolds". Edit: There was a typo in the exercise as is noted below in the answers. The statement has been edited to reflect this. Given ...
5
votes
1answer
406 views

Sum of Closed Operators Closable?

Let $A$ and $B$ be closed operators on a (separable complex) Hilbert space with dense domains $D(A)$ and $D(B)$ respecitvely. Then, we may define the operator $A+B$ on $D(A)\cap D(B)$. In general, ...
18
votes
2answers
812 views

Compact sets as point spectrum of a bounded operator

It is well known that if $K$ is any compact set in $\mathbb{C}$, then there exist a bounded linear operator $T:l_2\to l_2$ such that $\sigma(T)=K$. My questions are: Q1) Does there exist $T$, a ...
8
votes
3answers
1k views

Is there a formula similar to $f(x+a) = e^{a\frac{d}{dx}}f(x)$ to express $f(\alpha\cdot x)$?

Using the Taylor expansion $$f(x+a) = \sum_{k=0}^\infty \frac{a^k}{k!}\frac{d^k }{dx^k}f(x)$$ one can formally express the sum as the linear operator $e^{a\frac{d}{dx}}$ to obtain $$f(x+a) = ...
9
votes
2answers
3k views

Equivalent Definitions of the Operator Norm

Would you give me a proof of the equivalence of these ones? $$\begin{align*} \lVert A\rVert_{\mathrm{op}} &= \inf\{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for all }v\in V\}\\ ...
3
votes
2answers
190 views

Resolvent Set: Definition

Given Banach spaces: $X,Y$ Consider a linear operator: $T:\mathcal{D}(T)\to Y$ (not necessarily bounded nor closed nor closable nor densely defined) Define for the shorthand the shifted operator: ...
3
votes
1answer
236 views

Graph of symmetric linear map is closed

A homework problem: Let $H$ be a Hilbert space. Let $T:H\rightarrow H$ be a symmetric linear map ($\langle Tx,y\rangle=\langle x,Ty\rangle$). Show that $S$ is bounded. My attempt: I'd ...
6
votes
1answer
481 views

Prove that $\sigma(AB) \backslash \{0\} = \sigma(BA)\backslash \{0\} $

Prove that $\sigma(AB) \backslash \{0\} = \sigma(BA)\backslash \{0\} $. Where $A,\ B$ are bounded operators in Banach space and $\sigma$ denotes spectrum.
2
votes
1answer
620 views

A compact operator is completely continuous.

I have a question. If $X$ and $Y$ are Banach spaces, we have to prove that a compact linear operator is completely continuous. A mapping $T \colon X \to Y$ is called completely continuous, if it maps ...
0
votes
2answers
76 views

Normal Operators: Spectrum vs. Numerical Range

Disclaimer: As I realized in the comments that this works for normal operators I decided to modify this question. Besides, I got the proof now - thanks to T.A.E.! Prove that for normal operators the ...
8
votes
4answers
341 views

Determining the action of the operator $D\left(z, \frac d{dz}\right)$

This question was motivated by a question by Tobias Kienzler and its wonderful answers. I begin as in the linked question... Using the Taylor expansion $$f(z+a) = \sum_{k=0}^\infty ...
3
votes
2answers
725 views

$\operatorname{Range}T$ is a closed subspace.

Let $X,Y$ two Banach spaces. If $T \in \mathcal{B}(X,Y)$ study if $\operatorname{Range}T$ is a closed subspaces. How can I prove this fact ? What theorems can I use ? thanks :)
11
votes
1answer
740 views

Closure of the invertible operators on a Banach space

Let $E$ be a Banach space, $\mathcal B(E)$ the Banach space of linear bounded operators and $\mathcal I$ the set of all invertible linear bounded operators from $E$ to $E$. We know that $\mathcal I$ ...
7
votes
2answers
2k views

What is operator calculus?

I watched the excellent interview with Richard Feynman: http://www.youtube.com/watch?v=PsgBtOVzHKI In the interview Feynman mention that he at young age re-invented operator calculus. I have searched ...
5
votes
2answers
315 views

If $(I-T)^{-1}$ exists, can it always be written in a series representation?

If $X$ is a Banach space, and $T:X \to X$ is a bounded linear operator with norm < $1$, then $I-T$ has a bounded inverse defined by $(I-T)^{-1} = \sum_{n=0}^\infty T^n$. Thinking in terms of a ...
3
votes
1answer
183 views

Hilbert's Inequality

Could you help me to show the following: The operator $$ T(f)(x) = \int _0^\infty \frac{f(y)}{x+y}dy $$ satisfies $$\Vert T(f)\Vert_p \leq C_p \Vert f\Vert_p $$ for $1 <p< \infty$ where ...
3
votes
2answers
558 views

Hellinger-Toeplitz theorem use principle of uniform boundedness

Suppose $T$ is an everywhere defined linear map from a Hilbert space $\mathcal{H}$ to itself. Suppose $T$ is also symmetric so that $\langle Tx,y\rangle=\langle x,Ty\rangle$ for all ...
16
votes
3answers
4k views

Differential equations and Fourier and Laplace transforms

Why do both the Fourier transform and the Laplace transform appear in the study of differential equations? I've never understood why there are some situations where the Fourier transform is used and ...
11
votes
2answers
1k views

Gelfand-Naimark Theorem

The Gelfand–Naimark Theorem states that an arbitrary C*-algebra $ A $ is isometrically *-isomorphic to a C*-algebra of bounded operators on a Hilbert space. There is another version, which states that ...
8
votes
3answers
564 views

Spectral radius inequality

Suppose $A,B \in M(n \times n, \mathbb{C})$ or $ A,B \in M(n \times n, \mathbb{R}) $. Under wich hypothesis can I state that: $\rho(AB) \leq \rho(A)\rho(B)$ ?
6
votes
1answer
102 views

local convexity of $L_p$ spaces

wiki says The spaces $L_p([0, 1])$ for $0 < p < 1$ are equipped with the F-norm they are not locally convex, since the only convex neighborhood of zero is the whole space Why is this so? ...
3
votes
1answer
113 views

Positive operator is bounded

For a real Banach space $X$ let $A:X\rightarrow X^*$ be a positive operator in the sense that $(Ax)(x)\geq 0$ for all $x\in X$. Show that $A$ is bounded. I don't know how to do that, maybe it's ...
4
votes
1answer
737 views

Compact multiplication operators

In class, we started talking about operators on Banach spaces after covering the Arzela-Ascoli Theorem. We defined a continuous operator $T\colon X \to Y$ to be compact if $\overline{T(B_X)}^{Y}$ is ...
4
votes
1answer
566 views

positive invertible operators

I need the following result. I think it's quite obvious but I don't know how to prove that: Let $C, T : \mathcal{H} \rightarrow \mathcal{H}$ be two positive, bounded, self-adjoint, invertible ...
3
votes
1answer
1k views

Prove that $T$ is an orthogonal projection

Let $T$ be a linear operator on a finite-dimensional inner product space $V$. Suppose that $T$ is a projection such that $\|T(x)\| \le \|x\|$ for $x \in V$. Prove that $T$ is an orthogonal projection. ...
10
votes
1answer
347 views

Every Hilbert space operator is a combination of projections

I am reading a paper on Hilbert space operators, in which the authors used a surprising result Every $X\in\mathcal{B}(\mathcal{H})$ is a finite linear combination of orthogonal projections. The ...
7
votes
2answers
378 views

Proof that operator is compact

Prove that the operator $T:\ell^1\rightarrow\ell^1$ which maps $x=(x_1,x_2,\dots)$ to $\left(x_1,\frac{x_2}{2},\frac{x_3}{3},\dots\right)$ is compact. For an arbitrary sequence $x^{(N)}\in\ell^1$ ...
6
votes
1answer
208 views

Weak* operator topology and finite rank operators

We will say that ${T_i}\subset B(X,Y^*)$ converges to $T$ in W*-operator topology if $T_i(x)\rightarrow T(x)$ in W*-topology of $Y^*$( $\forall y\in Y \langle T_i(x),y\rangle \rightarrow \langle ...
6
votes
1answer
175 views

Questions about a PDE: $-u_{txx}+u_{t}=u_{x}$, $u(x,0)=u_0(x)$, $x,t\in{\bf R}$

Consider the BBM equation: $-u_{txx}+u_{t}=u_{x}$, $u(x,0)=u_0(x)$, $x,t\in{\bf R}$. One may rewrite this equation as following $u_t=((I-A)^{-1}\partial_x)u$ where $Au=u_{xx}$ if $(I-A)^{-1}$ ...
4
votes
1answer
253 views

Norms involving positive operators

Let's say we have $A \leq B$. Is it then true that $||Ax|| \leq ||Bx||$ (where $x, A, B$ all belong to the same finite-dimensional Hilbert space $H$)?
3
votes
1answer
135 views

Alternative definition of strong/weak operator topology.

Given two normed spaces $(X, ||\cdot||_X)$ and $(Y,||\cdot||_Y)$ the space of bounded linear maps $\mathcal{B}(X,Y)$ can be equipped with the strong operator topology (SOT) as follows: The ...
3
votes
2answers
58 views

Projection of the third dual of a Banach space onto the first dual

Let $j_X:X\rightarrow X^{**}$ denote the canonical embedding. I've read several articles where it is assumed that the reader is familiar with the idea that there is a norm one projection from ...
3
votes
1answer
433 views

Operators on $C([0,1])$ that is compact or not.

For $f\in C([0,1])$ set $$Hf(x) = \frac{1}{x}\int_0^x f(t)dt.$$ a) Show that $H$ is a bounded operator from $C([0,1])$ into itself which is not compact. b) From a) it follows that $H$ induces a ...
3
votes
1answer
417 views

Weakly compact operators on $\ell_1$

Is the following assertion true/known? Let $V$ be a Banach space and let $T\colon \ell_1\to V$ be a bounded linear operator. Is it true that $T$ is not weakly compact if and only if there is a ...
2
votes
1answer
83 views

Limit point / limit circle and self-adjointness

I have a collection of questions about the limit point/circle concept and self-adjointness that are kind of connected, so I would like to ask them in a row. Apparently, an operator that is limit ...
2
votes
1answer
70 views

Exercise 31 from chapter 4 (“Hilbert Spaces: An Introduction”) of Stein & Shakarchi's “Real Analysis”

The following problem shows that $\{e^{inx}\}_{n \in \mathbb{Z}}$ is an orthonormal basis of $L^2([-\pi, \pi])$. It is taken from [1] (exercise 31 of chapter 4: "Hilbert Spaces: An Introduction", pp. ...
2
votes
1answer
120 views

Prove that the limit exist II

First question was here. I add one new condition. Let $T : H \rightarrow H$ is a linear continuous unitary ($T^*=T^{-1}$) operator, $H$ is a Hilbert space (not necessary). Suppose that $\forall h ...
1
vote
1answer
49 views

Spectral Measures: Domain Criterion

Given a topological space $\Omega$ and a Hilbert space $\mathcal{H}$. Let $\mathcal{B}(\Omega)$ be its Borel algebra and $\mathcal{B}(\mathcal{H})$ its bounded operators. Moreover, given a spectral ...
1
vote
1answer
71 views

Proving that $AB-BA=cI$ for nontrivial $c \in \mathbb{C}$

I have a homework question I can`t solve: Let $X$ be a normed linear space, $A,B \in B(X)$. Show that there exists no nontrivial $c \in \mathbb{C} $ such that $AB-BA=cI$. Thanks alot already guys! I ...
1
vote
1answer
185 views

Normal operators in Hilbert spaces

Let $H$ be a separable Hilbert space and let $T:H\to H$ be a continues linear map such that there exists an orthonormal basis of $H$ that consists of the eigenvectors of $T$. Show that $T$ is normal. ...
1
vote
1answer
316 views

Compact operator? self adjoint operator? Stirling's formula

Define a pair of operators $S\colon\ell^2 \rightarrow \ell^2$ and $M\colon\ell^2 \to\ell^2$ as follows: $$S(x_1,x_2,x_3,x_4,\ldots)=(0,x_1,x_2,x_3,\ldots) $$ and ...
6
votes
3answers
159 views

If a linear operator has an adjoint operator, it is bounded

This is a question I'm struggling with for a while: Let $H$ be a Hilber space. Let $T,S: H\rightarrow H$ be linear operators (not neccessarily bounded) such that for every $x,y\in H$: $\langle ...
5
votes
1answer
366 views

Bounded operator and Compactness problem

Let $H$ be a Hilbert space with orthonormal basis $(e_{n})_{n\in\mathbb{N}}$. Furthermore, let $T\colon H\rightarrow C[a,b]$ be a bounded operator. a) Let $x\in [a,b]$. Show that there is a ...
4
votes
1answer
78 views

Prove that the limit exist

Let $T : H \rightarrow H$ is a linear continuous unitary ($T^*=T^{-1}$) operator, $H$ is a Hilbert space (not necessary). Suppose that $\forall h \in H \Rightarrow Th=h$ $T_n$ - a sequence of ...
3
votes
1answer
171 views

Does an irreducible operator generate an exact $C^{*}$-algebra?

Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators. Definition : An operator $T \in B(H)$ is irreducible if $W^{*}(T)=B(H)$. Definition : A ...