A compact operator is an operator from normed space $X$ to a normed space $Y$, such that image of every bounded subset of $X$ is relatively compact in $Y$. It's used with (functional-analysis) and (operator-theory) tags.

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Show this integral operator is compact for various values of $\alpha$

I am having some problems evaluating a multivariable integral. This question is features in Stakgold's book Green's functions and boundary value problems. page 359. Consider the kernel for $a\leq ...
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54 views

If a series converges, does it converge with additional log term multiplied?

If $\sum_{n} |a_n| < \infty$, is it true that $\sum_{n} |a_n\log(a_n)| < \infty$ if $0 \leq a_n \leq 1$? I am trying to see if $A$ is trace class operator, then $A \log(A)$ is also trace class ...
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What is a predual of the Banach space of compact operators on $\ell^2$?

I am wondering if the space $K(\ell^2)$ of compact operators on $\ell^2$ can have a predual. Thank you in advance for your help.
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Cauchy singular integral operator

Help on proving the following equality: $$K(-sgn)=S$$ where $K$ is the operator defined by $K(f)=F^{−1}fF$ ($F$=fourier transform, $f$=any function), sgn is the signum function and S is the Cauchy ...
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28 views

Extending a compact operator to the entire Hilbert space

In a course I'm taking we defined compact operators as a linear mapping $H\rightarrow H$, where $H$ is a Hilbert space, that maps bounded sets to relative compact ones. The lecturer mentioned that the ...
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Normal Compact Operator: not diagonalizable!

To proposition 5.17 in Weidmann's 'Lineare Operatoren in Hilberträumen' (german version) it is noted that the expansion of compact operators that are normal rather than self adjoint doesn't apply in ...
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Sum of the Eigenvalues of a Compact Positive-Definite Linear Operator on a Hilbert Space

Let $ A $ be a compact positive-definite linear operator on a Hilbert space $ \mathcal{H} $. Let $ \{ v_{1},v_{2},\ldots,v_{n} \} $ be an orthonormal $ n $-subset of $ \mathcal{H} $. Let $ \lambda_{1} ...
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Question about compact operator

So here is my question, Let $H$ be a Hilbert-space and $K:H\rightarrow H$ a compact operator. I know that if $K$ is self adjoint, then it has one eigenvalue $\mu$ such that $|\mu|=||K||$. Can some ...
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35 views

Fredholm Index: Finite Corank $\Rightarrow$Closed Range [duplicate]

Obviously closed subspaces turn quotient spaces into normed spaces rather than just merely vector spaces. However the dimension involved in Freholm's index are purely algebraic. Why do we thus ...
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40 views

Showing that a certain operator is compact

So here is my problem, I try to show that following operator is compact, \begin{align} J: h_1 & \rightarrow\ell^1 \\ (x_n) & \mapsto(x_n) \end{align} where $$h_1:=\left\{x_n\in ...
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Question about a counterexample concerning compact operators

Does anybody know if the following is true, Let $H$ be an infinite dimensional Hilbert-space and $K:H\rightarrow H$ a compact operator. Then if $|\mathrm{spec}(K)|<\infty$ i.e the spectrum is ...
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Question if an operator is compact

So here is my problem, Let $$J_p:\ell^p\rightarrow c_o$$ be the canonical embedding where $c_0:=\{x_n\subseteq\mathbb C:x_n\rightarrow 0\quad n \rightarrow\infty\}$. I have to decide whether the ...
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Question about an integral operator

So here is my question, I know that the operator $$T:L^2[0,1]\rightarrow L^2[0,1]$$ $$f\mapsto(Kf)(x)=\int_{[0,1]}k(x,y)f(y)\;dy$$ for a function $k$ continuous on $[0,1]^2$ is compact. Is this also ...
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What makes compact operators special?

I would like to understand why compact operators are considered so special to consider them as an extra class of operators. Over Hilbert spaces these (as far as I know) these are the ones with ...
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*IF $X$ is a normed spaces, then $L_c(X)$ two sided ideal in normed algebra $L(X)=L(X,X)$.*

Yesterday assistant filed during exercises work, but not one of us was able to resolve, and then he proved himself and noticed that he had the solution when attempting to problems. Therefore, please ...
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46 views

Is operator $T_n$ a compact operator?

Is operator $T_n$ a compact operator? $$T_n:l_2\rightarrow l_2$$ $$T_nx=(\underbrace{0,0,\ldots,0,}_{n\text{ zeros}}, x_1,x_2,x_3,\ldots)\text{ where }x=(x_1,x_2,x_3,\ldots)\in l_2,\ ...
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50 views

Two questions about a proof of the compactness of an operator

There are a few things that I don't understand about a proof and I'd appreciate any help. The theorem and its proof are the following: (1) Is the equality $$ \|v(\tau) -v(\tau_j)\| = \max_{1 \le ...
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33 views

Problems proving that a compact operator is completely continuous [duplicate]

I would like to prove that if $T:X\rightarrow Y$ is a compact operator, then for every weak convergent sequence $(x_n)_{n\in\mathbb N}$ with $x_n\rightharpoonup x$ for some $x\in X$ it follows that ...
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Compact operator as a limit of finite ranked operators

So here is my question, I had to show that the following operator is compact, $$T:C[0,1]\rightarrow C[0,1]$$ $$f\mapsto\int_0^tf(s)ds$$ with $||f||=\mathrm{sup}_{x\in[0,1]}|f(x)|$ I think I ...
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Question about compact operators

I would like to prove the following, Let $X$,$Y$ be infinite dimensional Banach-Spaces and $T$ a compact, linear and bounded operator. Then there exists a sequence $(x_n)_{n\in\mathbb N}$ with ...
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36 views

About the Volterra operator and the approximation property

I need some help with these questions. $\bullet\;$ First of all, if we define the Volterra operator $V:L^{1}[0,2\pi]\rightarrow L^{1}[0,2\pi]$ as $(Vf)(x)=\int_0^xf(t)dt$, Is this operator compact? ...
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Volterra operator with continuously differentiable Kernel has no Eigenvalue

First I'll describe the entire question, as it stated in the exercise: let $K(t,s)\in C([0,1]^2$), continuously differentiable in the first coordinate (meaning $K_t(t,s)\in C([0,1]^2$). And let ...
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is $T$ compact operator?

is $T$ compact operator? $T:C[0,1]\rightarrow C[0,1]$: $x(t)\mapsto x(t^2)$ where $t\in[0,1]$ with supremum norm Could you please help.
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How to figure whether it is a compact operator

How to figure whether it is a compact operator: $$T:C[0,1]\rightarrow C[0,1] $$ $C[0,1]$:the space of all continous function on [0,1] with supremum norm $$(Tx)(t)=\int^t_0 x(s)ds, \ \ \forall ...
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61 views

How to find if it is a compact operator

How to find if it is a compact operator: $F\colon C[0,1]\rightarrow C[0,1]$ : $x(t)\mapsto \int^1_0 \cos(t^2+s^2)x(s)ds$ Could you please help with this question.
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inequality for compact operator

Let $K(x)$, $x\ge0$ be a nonnegative-valued continuous function with support $(0,\infty)$ and such that $\int_0^\infty K(x)\,dx=1$. Let $\mathcal{K}$ be an integral operator given by $$ ...
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If an upper semicontinuous multivalued map is compact on a set, is it compact on the boundary as well?

I have stumbled upon the following problem during my research: Let $X$ and $Y$ be Banach spaces, $K\subset X$ nonempty, $F:\overline{K}\rightarrow 2^{Y}$ an upper semicontinuous multivalued map with ...
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84 views

Compact operator whose range is not closed

I am asked to find a compact operator (on a Hilbert space) whose range is not closed, but I am having trouble coming up with one. My guess is that you need to have some sequence in the range that ...
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Examples of spectrum of compact operators on the sequence space $l_2$

Suppose $T$ is a compact operator on the sequence space $l_2$, and let $\sigma(T)$ be its spectrum. Is it possible to find a $T \ne 0$ such that $\sigma(T) = \{0\}$? Also, is it possible to find $T$ ...
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If the $A^*A\leq BB^*$ and $B$ is a compact operator, then the operator $A$ is compact

I am a student of mathematics and professor left us today for homework this example: If the $A^*A\leq BB^*$ and $B$ is a compact operator, then the operator $A$ is compact. Prove this. ($A^*$ ...
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Sufficient condition for compactness of an operator [closed]

Let $A$ be a bounded linear operator from a Hilbert space $H$ to itself such that for each orthonormal basis $\{e_n\}$ of $H$ we have $\langle Ae_n , e_n\rangle \rightarrow 0$. Then show that $A$ is ...
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Is the Neumann series a compact operator?

Let $X$ be an infinite dimensional Banach space and $A:X\to X$ be a compact operator with the operator norm $\|A\|<1$. Then $I-A$ is invertible and the Neumann series $$ S_N = \sum_{k=0}^N A^k $$ ...
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Compact operator on $L^{2}$

Let $K(t,s)$ be a real-valued function of two real variables, and let $T: L^2(a,b) \to L^2(a,b)$ be defined by $(Tf)(t) = \int_{a}^{b} K(t,s) f(s) ds$ where $$K(t,s) = \sum_{j=1}^{n} \phi_{j}(t) ...
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Show compactness of an operator with Arzelà–Ascoli

We have $K\colon L^{2}(a,b) \rightarrow L^{2}(a,b)$ such that $ Kf(t)=\sum_{j=1}^{n}\phi_{j}(t) \int_{a}^{b} \psi_{j}(S) f(s)ds$ where $\phi_{j} ,\psi_{j} \in L^{2}(a,b)$. We want to show that K is ...
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Compact operators is a linear subspace of bounded operators

Let $X,Y$ be Banach spaces. Let $B(X,Y)$ be the set of bounded linear operators and let $K(X,Y)$ be the set of compact linear operators. I want to prove that $K(X,Y)$ is a vector subspace of ...
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Proof of equivalent characterizations of compact operators

As an exercise I tried to prove the following theorem: If $X,Y$ are Banach spaces and $u \in B(X,Y)$ is a bounded linear operator then the following are equivalent: (1) $u$ is compact ...
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Showing a particular integral operator is trace class

Let $f$ and $P$ be continuous, integrable functions $\mathbb{R} \to \mathbb{C}$ vanishing at $\pm \infty%$. Concisely, $f,P \in C_0(\mathbb{R}) \cap L^1(\mathbb{R})$. Also, assume that $P$ is ...
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Equivalent conditions for composition to be compact operator

I did some exercises in Conway's functional analysis book and found the following problem: Let $\tau:[0,1]\to [0,1]$ be continuous and define $A:C[0,1]\to C[0,1]$ by $Af:= f\circ \tau$. Give ...
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Schauder's theorem: consequences and applications

I am about to give an informal talk about Schauder's theorem ($T:X\to Y$ linear operator between Banach spaces is compact if and only if its adjoint is). Does anyone know any derived ...
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Why is this estimate using a compact embedding in a sobolev space true?

Let $\Omega\subset\mathbb{R}^3$ be a bounded Lipschitz-domain. We then have, for $s\in[1,6)$ the compact embedding $H^1(\Omega)\stackrel{c}{\hookrightarrow}L^s(\Omega)$ ensuring the existence of a ...
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Question about the Morse index of $p_0$ where $f''(p_0)=id-T'(p_0)$

I have this perturbation $f'(x)=x-T(x)$ where $T$ is compact, i have that $p_0$ is non degenerate and i want to see if it's Morse index (i.e. the suprimum of the dimensions of subspaces where ...
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I want to show that some subset of $C([0,1])$ is equicontinous

First why the problem appeard. I want to show that the linear and continuous operator $T:C([0,1])\rightarrow C([0,1])$ , $ (Tf)(t)=\int_{[0,1]}k(t,s)f(s)ds$ where $k:[0,1]^2\rightarrow\mathbb R$ is ...
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93 views

Show the Volterra Operator is compact using only the definition of compact

The Volterra operator $V:L^{2}[0,1]\rightarrow L^{2}[0,1]$ is defined by $(Vf)(x)=\int_0^xf(t)dt$. I am wondering if it can be shown that $V$ is compact by definition - that is, either that $V$ ...
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55 views

Show these operators converge to a particular limit

Let $H$ be a Hilbert space, and $T$ be a operator on $H$ of the form $T=\sum_{n=1}^{\infty}{\lambda}_{n}<x,e_{n}>e_{n}$ where $e_{n}$ are the eigenvectors of $T$ and an orthonormal basis of H ...
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Subspaces in the image of compact operator

Let $X$ and $Y$ be some infinite dimensional Banach spaces. Let $T:X\longrightarrow Y$ be some compact linear operator. It is easy to understand that $T$ cannot be surjective: the Open Mapping Theorem ...
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The dual of a dual space with the topology of uniform convergence on compact subsets?

$W$ is a Banach space. The topology of $W^*$ is the uniform convergence on the compact subsets of $W$. That is generated by the family of seminorms $$p_K(f)=\sup_{x\in K}|f(x)|,$$ for all compact ...
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The eigenvalues of a compact and self-adjoint operator on Hilbert space

Show that if $K$ is a compact self-adjoint operator on Hilbert space then it has either finitely many eigenvalues or a sequence of eigenvalues $\lambda_n\to 0$ as $n\to \infty$.
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Perturbation of eigenvalues

I am looking at a certain operator, that is a Hilbert-Schmidt integral operator from $L^2(X,d\mu)$ to $L^2(X,d\mu)$.I want to see how its eigenvalues or singular values change as its kernel is ...
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Finite dimension and total boundedness

Let $T:X\to Y$ be a bounded operator between Banach spaces $X$ and $Y$. Assume that for any $\epsilon >0$ there is a finite-dimensional subspace $Y_\epsilon\subset Y$ so that $\|Q_\epsilon ...
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1answer
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Sufficient condition for compact embedding between Banach spaces

Let $(X,\|\cdot\|_X)$ and $(Y, \|\cdot\|_Y)$ be two Banach spaces and $X\subset Y$. Then $X$ is compactly embedded in $Y$, if the unit ball $B_X(0,1)$ in $X$ is a relatively compact subset in $Y$. ...