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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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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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441 views

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 $$...
12
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1answer
680 views

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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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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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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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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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 ...
10
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1answer
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Trace class for operators

Let $ \mathcal{H} $ be a Hilbert space and $ T: \mathcal{H} \to \mathcal{H} $ a bounded linear operator. The $ n $-th singular number $ {\mu_{n}}(T) $ of $ T $ is defined as the distance from $ T $ to ...
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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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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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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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A compact operator in $L^2(\mathbb R)$

Let $g \in L^{\infty}(\mathbb R)$. Consider the operator $$ \begin{split} T_g\colon & L^2(\mathbb R)\to L^2(\mathbb R) \\ & f \mapsto gf \end{split} $$ Prove that $T_g$ is compact (i.e., ...
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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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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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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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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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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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Compactness criterion for operator between reflexive Banach spaces

I found (without any proof) the following proposition: Let $T \in \mathcal{L}(X,Y)$ be a linear continuous operator between two reflexive Banach spaces $X,Y$, then $T$ is compact if and only if for ...
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Showing that smoothing operators are compact

Suppose I have a bounded, linear map $T: H^1(X) \to H^1(X)$ such that $T(H^1(X)) \subset C^\infty(X)$. Is $T$ a compact operator? I'm guessing this depends on whether or not $X$ is (pre)compact, and ...
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Show that $f \overset{T}{\rightarrow} \frac{1}{x} \int_0^x f(t) dt$ is Bounded, and is NOT Compact in $L^2(0, \infty)$

Can someone help me with this question? Let $f \in X = L^2(0, \infty)$, and define \begin{equation} (Tf)(x) = \frac{1}{x} \int_0^x f(t) dt \ . \end{equation} Show that $T$ is Bounded, and is NOT ...
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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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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 $L^2[0,1]$ ...
7
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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 (...
7
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1answer
325 views

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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2answers
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On the isometry between bounded linear operators and the dual of nuclear linear operators

Let $H$ be a separable Hilbert space. Let $(e_i)_i$ be an orthonormal basis. For any bounded linear map $T$ we write, whenever possible $$\operatorname{tr} T := \sum_{i}^{\infty} \langle T e_i, e_i \...
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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 ...
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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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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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T compact if and only if T*T is compact.

I have an operator $T \in B(\mathcal{H})$. I need to prove that T is comapct if and only if $T^*T$ is compact. One way is ok, because if A or B is comapct then AB is compact, so I get at once that if ...
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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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1answer
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Hilbert-Schmidt Operator

We have just covered Hilbert-Schmidt operators in class (which I missed) and I am having a hard time getting my head around them. I know the definition: If $H$ is a Hilbert space and $T\in\mathcal{...
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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 t\in[0,...
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Is the inclusion $C^1[0,1]\subset C[0,1]$ compact?

I am working on this problem but i couldn't succeed . Consider the space $C^1[0,1]$ with the norm $$\|f\|=\max \{\|f\|_{C[0,1]}, \|f'\|_{C[0,1]}\},$$ I don't know if the inclusion map is compact, ...
6
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1answer
287 views

Compactness of the Volterra opelator

The Volterra operator is given as \begin{eqnarray} (Vf)(x)=\int_0^xK(x,y)f(y)\,{\rm d}y. \end{eqnarray} By the Arzelà–Ascoli theorem, $V\colon C^0[0,1]\rightarrow C^0[0,1]$ is compact operator. But, ...
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Convergence of eigenvalues for sequence of compressions of a compact operator

Suppose $H$ is a separable Hilbert space, $A$ is a Hilbert Schmidt operator on $H$, and $P_n$ is an increasing sequence of finite rank orthogonal projections of $H$ (so $P_nx\rightarrow x$ for all $x\...
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1answer
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Set of all compact operators $K(H)$ is the unique ideal in $B(H)$?

I want to show that the set of all compact operators $K(H)$ is the unique ideal in $B(H)$. Is there any relation between invertibility and compactness of an operator?
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1answer
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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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1answer
248 views

Show $T$ is compact

$H$ and $K$ are Hilbert Spaces, $(u_n)$ and $(v_n)$ are sequences in $H$ and $K$ respectively. $\sum_{n=1}^{n=\infty} \|u_n\|\|v_n\| $ converges. $T\colon H\rightarrow K$ is defined by $Tx=\sum_{n=1}^{...
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Criterion for a bounded linear operator to be compact

Let $X,Y$ be Banach spaces. $T\colon X\to Y$ be a bounded linear operator. How can I prove that $T$ is compact if and only if there is $\lbrace x_n^*\rbrace\subset X^*$ such that $\|x_n^*\|\to 0$ and ...
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Proof of the Riesz-Schauder Theorem (for compact operators) using the Analytical Fredholm Theorem

First of all sorry for my bad English, I'm an Italian student, hope to let you understand! I'm having a little troubles with the proof of the Riesz-Schauder theorem for Compact Operators. Some infos ...
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5answers
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Compact operator on a Hilbert space

This is a homework problem that I'm having some trouble with, so any hints would be appreciated. Let $H$ be a Hilbert space with an orthonormal basis $\{e_n\}$. Consider the operator $F : H\rightarrow ...
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6answers
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Possible flaw in “proof” that a sum of two compact operators is compact

If X and Y are Banach spaces, and $A: X \to Y$, $B: X \to Y$ are both compact operators, then $A + B$ is compact. A + B is compact if and only if for every bounded sequence $\lbrace x_n \rbrace$ in X,...
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2answers
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Is convolution operator compact?

I know convolution is not a Hilbert–Schmidt integral operator, but it needs more to tell if convolution is compact or not.
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Why is $A$ a compact operator?

Let $X$ be a compact space and let $\mu$ be a positive Borel measure on X. Let $T\in \mathscr{B}(L^p(\mu),C(X))$ where $1\lt p \lt \infty$. Show that if $A:L^p(\mu)\rightarrow L^p(\mu)$ defined by $...
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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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If $T^2$ is a compact operator on a Banach space then is $T$ also compact?

If $T^2$ is a compact operator on a Banach space $X$ then is it true that $T$ is also compact operator on X? Here $T:X\to X.$ First I was trying to prove(assuming it is true). Since $X$ is Banach ...
5
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1answer
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Does a compact operator always have a kernel?

I am sorry if this question is stupid..... I raise it when I read Lax's book Functional Analysis. We know that some integral operators are compact, for example an integral operator from $L^2[Y]$ to $L^...
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How to prove this integral operator is compact?

$T_kf=\int K(x,y)f(y)dy$ where $K(x,y)=\frac{\phi(x)\phi(y)}{|x-y|^{n-\alpha}}$ $\phi(x)$ is a smooth function on a compact support. $f$ is defined on $R^n$ and $K$ is defined on $R^n\times R^n$ ...
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Example of a noncompact operator on $L^2([0,1])$

For $f \in L^2([0,1])$, define operator $Tf: x \mapsto \frac{1}{x}\int_0^x f(y)dy$. Show that $T$ is not a compact operator on $L^2([0,1])$ and that $T$ is bounded. For the second part, I can show $...