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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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?
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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($A$) ...
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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 ...
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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 ...
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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 ...
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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 ...
6
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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, ...
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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 ...
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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) = ...
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Norm of a symmetric matrix equals spectral radius

How do I prove that the norm of a matrix equals the absolutely largest eigenvalue of the matrix? This is the precise question: Let $A$ be a symmetric $n \times n$ matrix. Consider $A$ as an ...
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$\operatorname{Range}T$ 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 :)
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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\}\\ ...
12
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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 $\mathcal I$ ...
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50 views

Spectral Measures: Pushforward

This thread is Q&A. Problem Given a Hilbert space $\mathcal{H}$. Consider a normal operator: $$N:\mathcal{D}(N)\subseteq\mathcal{H}\to\mathcal{H}:\quad N^*N=NN^*$$ And its spectral measure: ...
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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 ...
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Exponential of a function times derivative

Exponential of a derivative $e^{a\partial}$ is simply a shift operator, i.e. \begin{equation} e^{a\partial}f(x)=f(a+x) \end{equation} This can be easily verified from a Taylor series \begin{equation} ...
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Resolvent: Definition

Given a Banach space. Consider linear operators: $$T:\mathcal{D}(T)\to E:\quad T(\kappa x+\lambda y)=\kappa T(x)+\lambda T(y)$$ (No other assumptions on the operator!) Denote for shorthand: ...
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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 ...
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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 ...
11
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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 ...
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For positive invertible operators $C\leq T$ on a Hilbert space, does it follow that $T^{-1}\leq C^{-1}$?

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 ...
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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.
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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 ...
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Summary: Spectrum vs. Numerical Range

This thread is only Q&A! Given a Hilbert space $\mathcal{H}$. Consider operators: $$T:\mathcal{D}(T)\to\mathcal{H}$$ Denote for shorthand: ...
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322 views

Complementability of von Neumann algebras

Is every von Neumann algebra complemented in its bidual? It is certainly true for commutative von Neumann algebras as their spectrum is hyperstonian. Is it 1-complemented?
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If space of bounded operators L(V,W) is Banach, V nonzero, then W is Banach (note direction of implication)

Let $V,W$ be normed vector spaces, and $L(V,W)$ be the space of bounded linear operators. Usually I would only see the statement "If $W$ is Banach, then $L(V,W)$ is Banach.". But Wikipedia writes that ...
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Normal Operators: 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 ...
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Selfadjoint compact operator with finite trace

I have a compact selfadjoint operator $T$ on a separable Hilbert space. For some fixed orthonormal basis, the operator's diagonal is in $\ell^1(\mathbb{N})$. Can we conclude that $T$ is trace ...
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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 ...
8
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$\exp(A+B)$ and Baker-Campbell-Hausdorff

A few years ago, I did research in quantum mechanics, specifically dealing with generalized displacement operators. In such musings, BCH lights (or gets in, depending on your viewpoint) the way. A ...
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Proof Complex positive definite => self-adjoint

I am looking for a proof of the theorem that says: A is a complex positive definite endomorphism and therefore is A self-adjoint. Does anybody of you know how to do this?
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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 ...
4
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1answer
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Proving Stone's Formula for Constructively obtaining the Spectral Measure for $A=A^\star$

Let $A$ be a bounded or unbounded selfadjoint linear operator on a complex Hilbert space $H$ with spectral representation $A=\int_{\sigma}\lambda \, dE(\lambda)$ given by the Spectral Theorem for ...
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$T$ surjective iff $T^*$ injective in infinite-dimensional Hilbert space?

Let $T:H_1\rightarrow H_2$ be a bounded linear operator where $H_1$ and $H_2$ are Hilbert spaces. The Hilbert-adjoint is defined to the the operator $T^*:H_2\rightarrow H_1$ such that $\langle ...
5
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1answer
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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 ...
4
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Normal Operators: Meet

Given a Hilbert space $\mathcal{H}$. Normal Operators: $$\mathcal{N}(\mathcal{H}):=\{N:N^*N=NN^*\}$$ Borel Calculus: $$\mathcal{B}(N):=\{\eta({N}):\eta\in\mathcal{B}(\mathbb{C},\mathbb{C})\}$$ ...
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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
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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 ...
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1answer
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Operator: not closable!

Is there an operator between Banach spaces with the following properties: $$T:\mathcal{D}(T)\subseteq X\to Y:\text{ injective, dense range, continuously invertible, not closable!}$$ (Note that the ...
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generalized functions & operators

I am dealing with a function $f(r) $that behaves like ~ $\frac{1}{r}$ when approaching zero. When I take the Laplacian of this guy and then integrate the result ([0,$\infty$]) I get some additional ...
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1answer
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Spectral Measures: Reducibility

Given a Hilbert space $\mathcal{H}$. Consider a normal operator: $$N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*$$ And its spectral measure: ...
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Trace Class: Relativeness

Given a Hilbert space $\mathcal{H}$. Consider a selfadjoint: $$H\in\mathcal{B}(\mathcal{H}):\quad H=H^*$$ Denote trace class: ...
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Finite dimensional $C^*$-algebras [closed]

Show that if a $C^*$-algebra $A$ is reflexive as a Banach space, then $A$ must be finite dimentional. I tried to solve it; but, I could not. please help me for this exercise. Thanks a lot!
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Trace Class (Obsolete!) [duplicate]

Given a Hilbert space $\mathcal{H}$. Consider ONB's $\mathcal{S}$ and $\mathcal{S}'$. For trace class it is: $$\operatorname{Tr}A<\infty:\quad\sum_{\sigma\in\mathcal{S}}\langle ...
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Normal Operators: Transform

Given a Hilbert space $\mathcal{H}$. Consider a normal operator: $$N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*$$ Construct the operator: $$Q:=(1+N^*N)^{-1}:\quad Z:=N\sqrt{Q}$$ Then it is ...
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1answer
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Spectral Measures: Lebesgue

Preface Dominated convergence: $$f_n(\omega)\to f(\omega)\quad(\omega\in\Omega)\implies f_n(E)\to f(E)$$ (This gives a tool for analysis of operators.) Problem Given a Borel space $\Omega$ and a ...
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5answers
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How could we define the factorial of a matrix?

Suppose I have a square matrix $\mathsf{A}$ with $\det \mathsf{A}\neq 0$. How could we define the following operation? $$\mathsf{A}!$$ Maybe we could make some simple example, admitted it makes any ...
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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 ...
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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 ...
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Fourier transform as diagonalization of convolution

I've read this in a lot of places but never quite got how this is true or meant. Let's say we have a convolution Operator $$ A_f(g) = \int f(\tau)g(t-\tau)d\tau $$ and apply it to $g(t)=e^{ikt}$. ...