# Tagged Questions

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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$X,Y$ 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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### 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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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 $x,y\in\... 3answers 1k views ### Compactness of a bounded operator$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 \...
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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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### 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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### 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. $$e^{a\partial}f(x)=f(a+x)$$ This can be easily verified from a Taylor series ...
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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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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: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\... 1answer 39 views ### Proving an operator is compact exercise Suppose (a_{ij})_{i,j\in \mathbb N} satisfy \sum_{i,j}|a_{ij}|^2<\infty and define A:\ell ^2 \rightarrow \ell ^2 by (Ax)_i)=\sum _j a_{ij}x_j. I need to prove A is compact. Unfortunately,... 1answer 305 views ### Finite dimensional C^*-algebras [closed] Show that if a C^*-algebra A is reflexive as a Banach space, then A must be finite dimentional. I tried to solve it; but, I could not. please help me for this exercise. Thanks a lot! 1answer 122 views ### Normal Operators: Transform Given a Hilbert space \mathcal{H}. Consider a normal operator:$$N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*$$Construct the operator:$$Q:=(1+N^*N)^{-1}:\quad Z:=N\sqrt{Q}$$Then it is ... 5answers 5k views ### How could we define the factorial of a matrix? Suppose I have a square matrix \mathsf{A} with \det \mathsf{A}\neq 0. How could we define the following operation?$$\mathsf{A}!$$Maybe we could make some simple example, admitted it makes any ... 3answers 6k views ### Differential equations and Fourier and Laplace transforms Why do both the Fourier transform and the Laplace transform appear in the study of differential equations? I've never understood why there are some situations where the Fourier transform is used and ... 2answers 2k views ### Gelfand-Naimark Theorem The Gelfandâ€“Naimark Theorem states that an arbitrary C*-algebra  A  is isometrically *-isomorphic to a C*-algebra of bounded operators on a Hilbert space. There is another version, which states that ... 2answers 510 views ### Selfadjoint compact operator with finite trace I have a compact selfadjoint operator T on a separable Hilbert space. For some fixed orthonormal basis, the operator's diagonal is in \ell^1(\mathbb{N}). Can we conclude that T is trace ... 1answer 450 views ### Graph of symmetric linear map is closed A homework problem: Let H be a Hilbert space. Let T:H\rightarrow H be a symmetric linear map (\langle Tx,y\rangle=\langle x,Ty\rangle). Show that S is bounded. My attempt: I'd ... 2answers 713 views ### Fourier transform as diagonalization of convolution I've read this in a lot of places but never quite got how this is true or meant. Let's say we have a convolution Operator$$ A_f(g) = \int f(\tau)g(t-\tau)d\tau  and apply it to $g(t)=e^{ikt}$. ...
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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: $\rho(AB) \leq \rho(A)\rho(B)$ ?
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### Square root is operator monotone

This is a fact I've used a lot, but how would one actually prove this statement? Paraphrased: given two positive operators $X, Y \geq 0$, how can you show that $X^2 \leq Y^2 \Rightarrow X \leq Y$ (or ...
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### Showing that the orthogonal projection in a Hilbert space is compact iff the subspace is finite dimensional

Suppose that we have a Hilbert Space $H$ and $M$ is a closed subspace of $H$. Let $T\colon H\rightarrow M$ be the orthogonal projection onto $M$. I have to show that $T$ is compact iff $M$ is finite ...
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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. ...
### $T$ 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 Tx,y\...