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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949 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 ...
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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 ...
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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 ...
6
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501 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, ...
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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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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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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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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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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\}\\ ...
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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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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 ...
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686 views

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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315 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 ...
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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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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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141 views

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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4answers
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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 ...
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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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$\operatorname{Range}T$ 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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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 ...
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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 ...
3
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Normal Operators: Superalgebra (I)

Given a Hilbert space $\mathcal{H}$. Consider normal operators: $$N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*$$ Denote their calculus: ...
3
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1answer
191 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
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2answers
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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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73 views

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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1answer
75 views

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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138 views

Compact Operators: Trace

Given a Hilbert space $\mathcal{H}$. Consider a bounded operator: $$A:\mathcal{H}\to\mathcal{H}:\quad\|A\|<\infty$$ Regard ONB's: ...
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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 ...
0
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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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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 ...
8
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696 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
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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? ...
4
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Norm of a matrix equals greatest eigenvalue

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 ...
8
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1answer
581 views

Normal operator + only real eigenvalues implies self-adjoint operator?

Let say we are in a complex vector space, is there an example of a normal operator with only real eigenvalues(or without eigenvalues) that is not a self-adjoint operator? Cause of the spectral theorem ...
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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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Spectrum of shift-operator

Hoi, consider the Hilbertspace $l^2$ and the Left and Right-shift operator \begin{align*} L(x_1,x_2,\cdots) &= (x_2,x_3,\cdots)\\ R(x_1,x_2,\cdots) &= (0,x_1,x_2,\cdots ) \end{align*} I know ...
5
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1answer
733 views

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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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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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. ...
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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 ...
3
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1answer
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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 ...
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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. ...
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1answer
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Exponential of matrices and bounded operators

Let $A$ be a complex $n \times n$ matrix, such that the function $t\mapsto e^{tA}x$ is bounded on $\mathbb{R}$ and nonzero, for some vector $x\in \mathbb{C}$. How can we prove that $\inf_{t\in ...
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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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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 $x^{(N)}\in\ell^1$ ...