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Questions tagged [compact-operators]

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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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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Proof of Pitt's theorem

I'm reading the book Topics in Banach Space Theory by Albiac F. Kalton N. J. I got stuck at the proof of Pitt's theorem. In the second paragraphs authors tries to prove ad absurdum that for weakly ...
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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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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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A Question on Compact Operators

Let $ \mathcal{H} $ and $ \mathcal{K} $ be Hilbert spaces, and let $ T: \mathcal{H} \to \mathcal{K} $ be a bounded linear operator. Show that if $ T $ is a compact operator, then $$ \lim_{n \to \infty}...
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Compact operators on an infinite dimensional Banach space cannot be surjective

I am reading a book about functional analysis and have a question: Let $X$ be a infinite-dimensional Banach-space and $A:X \rightarrow X$ a compact operator. How can one show that $A$ can not be ...
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How to show that the limit of compact operators in the operator norm topology is compact

When I read the item of compact operator on Wikipedia, it said that Let $T_{n}, n\in \mathbb{N}$, be a sequence of compact operators from one Banach space to the other, and suppose that $Tn$ ...
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Density of finite rank operator in compact operators on Hilbert spaces

Show the subspace of finite rank operators (defined on a Hilbert space) is dense in the space of all compact operators. I just started reading about compact operators and I saw this questions. So ...
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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 ...
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1answer
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Problem about a compact operator $T:l^p\rightarrow l^p$

I have to solve this problem. Let $\{\lambda_n\}$ be a sequence of real number such that $\lim_{n\rightarrow\infty}\lambda_n=0$ and consider the operator $T:l^p\rightarrow l^p$, $1\leq p\leq \...
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Compact operators and uniform convergence

Suppose $T: H \rightarrow H$ is a compact operator, $H$ is a Hilbert space, and let $(A_n)$ be a sequence of bounded linear operators on $H$ converging strongly to $A$. Show that $A_nT$ converges in ...
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1answer
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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,...
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1answer
427 views

a compact operator on $l^2$ defined by an infinite matrix

Let $A$ be an infinite matrix such that $\displaystyle \sum_{i,j}|a_{i,j}|^2<\infty$. Then $A$ defined a compact operator on $l^2$.
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No Nonzero multiplication operator is compact [duplicate]

Let $f,g \in L^2[0,1]$, multiplication operator $M_g:L^2[0,1] \rightarrow L^2[0,1]$ is defined by $M_g(f(x))=g(x)f(x)$. Would you help me to prove that no nonzero multiplication operator on $L^2[0,1]$ ...
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Criteria of compactness of an operator

Suppose $K$ is a linear operator in a separable Hilbert space $H$ such that for any Hilbert basis $\{e_i\}$ of $H$ we have $\lim_{i,j \to \infty} (Ke_i,e_j) = 0$. Is it true that $K$ is compact? ...
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Compact operator with closed range has finite dimensional range

Let $X,Y$ be Banach Spaces, and let $T\in K(X,Y)$ be a compact operator from $X$ to $Y$. I have to prove that $T(X)$ is closed in Y if, and only if, $\dim(T(X))<\infty$. Can anybody help me with ...
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Compact operators: why is the image of the unit ball only assumed to be relatively compact?

Recall the definition of compact operators between Hilbert spaces: An operator $A$ is called compact if the image $A(\mathcal U_H)$ of the unit ball is relatively compact (i.e. its closure is compact)...
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1answer
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The ideal generated by a non-compact operator

I wanted to find a quick proof of the following well-known fact. Since I couldn't easily find a reference, I provide a proof below. Let $H$ be a separable Hilbert space, and $J\subset B(H)$ be a (...
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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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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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show that every compact operator on Banach space is a norm-limit of finite rank operators (in a particular way, under the given hypotheses)

Let $X$ be a Banach space and suppose there is a net $\{F_i\}$ of finite-rank operators on $X$ such that $(a)$ $\sup_i||F_i||<\infty$ ; $(b)$ $||F_ix-x||\to 0$ for all $x$ in $X$. Show that if $A$ ...
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1answer
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There are compact operators that are not norm-limits of finite-rank operators

Given an example of a Banach space for which There are compact operators that are not norm-limits of finite-rank operators. Tanks for answer
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1answer
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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 T(x),...
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1answer
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Square root of compact operator

I'm trying to solve a functional analysis problem A self-adjoint non-negative operator $A$ on a Hilbert space $H$ is compact if and only if its $\sqrt{A}$ is compact.
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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 $x^{(N)}\in\ell^1$ one ...
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1answer
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Prove that an infinite matrix defines a compact operator on $l^2$.

Let $(a(i))_{i=1}^\infty$ be an absolutely summable sequence, i.e., $\sum_{i=1}^\infty |a(i)|<\infty$, and consider the infinite matrix $$A=\begin{bmatrix} a(1)&a(2)&a(3)&\cdots\\ a(2)&...
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1answer
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Algebra of compact operators on $\ell_p$

Are the algebras of compact operators $K(\ell_p)$ and $K(\ell_q)$ isomorphic as Banach algebras for $1\leq p<q<\infty$?
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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 ...
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compact and self adjoint square root of an operator

Let H a Hilbert space and $T:H\rightarrow H$ a linear bounded, self-adjoint, positive and compact operator. How can i prove that the square root of T, $\ T^{1/2}:H\rightarrow H$ is also compact and ...
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3answers
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Trying to prove that operator is compact

Consider $T\colon\ell^2\to\ell^2$ an operator such that $$T((x_n))=(2^{-n}x_n); \forall x=(x_n)\in \ell^2 $$ Does anyone know how to prove that it is compact? I understand that I have to find a ...
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How to prove that a bounded linear operator is compact?

I encountered a homework problem that says: If $A$ is a bounded linear operator from $X$ to $Y$. And $K$ is a compact operator from $X$ to $Y$, where $X$ and $Y$ are both Banach spaces, and Ran$(A)\...
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Rellich–Kondrachov theorem for traces

Let $W^{1,p}(\Omega)$ be the Sobolev space of weakly differentiable functions whose weak derivatives are $p$-integrable, where $\Omega \subset \mathbb R^n$ is a domain with Lipschitz boundary. Let ...
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is $T:C[0,1]\rightarrow C[0,1]$: $x(t)\mapsto x(t^2)$ compact?

Is $T$ a compact operator? $T:C[0,1]\rightarrow C[0,1]$: $x(t)\mapsto x(t^2)$ where $t\in[0,1]$ with supremum norm.
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Compact operators and completely continuous operators

A compact operator between Banach spaces is an operator that maps bounded sets into relatively compact sets, while a completely continuous operator maps all weakly convergent sequences into convergent ...
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The trace class operators are the dual of the compact operators

I know that the map from the trace class operators $L_1(H)$ to the dual of the compact operators $K'(H)$ given by $A \mapsto tr( \cdot A)$ is an isometric isomorphism. Linearity is obvious by the ...
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finite dimensional range implies compact operator

Let $X,Y$ be normed spaces over $\mathbb C$. A linear map $T\colon X\to Y$ is compact if $T$ carries bounded sets into relatively compact sets (i.e sets with compact closure). Equivalently if $x_n\in ...
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1answer
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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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Is every bounded derivation from compact to finite rank operators inner?

This question is related to Derivation into dense ideal of Banach algebras. Depending on which way it goes, an answer to one might answer the other (this is elaborated below). Let $H$ be a Hilbert ...
3
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1answer
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Rellich's theorem for Sobolev space on the torus

From John Roe: Elliptic operators, topology and asymptotic methods, page 73: Let $H^{k}$ be the Soblev space defined on the torus $\mathbb{T}^{n}$ with the discrete $k$-norm: $$ \langle f_{1}, f_{2}\...
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1answer
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Inequalities on kernels of compact operators

Suppose we have a $\sigma$-finite positive measure $\mu(v)$ on $\Bbb R^d$ and we have two positive kernels on $\Bbb R^d\times \Bbb R^d$ $k_1(v,u)>0$, $k_2(v,u)>0$. We define integral operators $$...
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2answers
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Equivalent definitions of the trace of a Hilbert-Schmidt operator

I am currently reading the book Spectral Methods in Automorphic Forms, and Iwaniec defines the trace operator in a different way than I am accustomed to. Throughout, assume that everything converges ...
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1answer
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Compactness of an integral operator from $L^2$ to $L^2$

I want to prove that The operator (linear and bounded) $T: L^2(0,1) \rightarrow L^2(0,1)$, defined by: $Tu(x)=\int_0^1\sin(x^2+y^2)u(y)dy$, is compact. Just by using theory, it's an Hilbert ...
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1answer
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Compact operator on $l^2$

Let A be a bounded linear operator on $l^2$ defined by A($a_n$)=($\frac{1}{n} a_n$). Would you help me to prove that A is compact operator. I guess the answer using an approximation by a sequences of ...
3
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1answer
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Integral operator on $L^p$ is compact

Let $(X,\Omega,\mu)$ be an arbitrary measure space, $1<p<\infty$ , and $\frac{1}{p}+ \frac{1}{q} = 1$. If $k:X. X\to \Bbb C$ is an $\Omega.\Omega-$ measurable function such that $$M = [\int (\...
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3answers
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Spectrum of a compact operator

If the spectrum of a compact operator is finite, I don't understand why $0$ has to be a member. I have proved that for all $\epsilon > 0$, there is only a finite number of eigenvectors which have ...
7
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1answer
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True/False: Self-adjoint compact operator

Let $H$ be a hilbert space and $T$ a compact self-adjoint operator on it. T is also injective on a dense subspace $U \subset H$ and we also have that $T(H) \subset U$. Now I am asked whether it is ...
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Show that a finite-dimensional Banach space has a bijective compact operator

It is clear that if $ T: X \rightarrow X $ is a bijective compact operator, where $ X $ is a Banach space, then $ \dim(\text{Range}(T)) = \dim(X) $, which implies that $ \dim(X) $ must be $ < \...
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1answer
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Composition of bounded operator and compact operators

In Hilbert space $H$, is it true that the composition of operators $ST$ and $TS$ of the bounded operator $S$ and the compact operator $T$ are compact?
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Convergence of $A_nT$ to $AT$ in operator norm for compact $T$

$A_n:Y\rightarrow Z$ are operators that strongly converge to $A$. Also, $\|A_n\|_\text {op}\le c$ for $c>0$. Given a compact operator $T:X\rightarrow Y$, I need to show that $A_nT$ converges to $AT$...
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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 $V$ ...