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 that the span of eigenvalues of a compact operator is a closed

Let $(X, ||\cdot||)$ be a real Banach space and $T: X \to X$ a compact operator (so $\{x_n\}_{n=1}^\infty$ bounded implies that $\{Tx_n\}_{n=1}^\infty$ has a convergent subsequence). Let $x_1, \dots, ...
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85 views

Totally boundedness of a compact operator [closed]

Let $T:\ell_2(\mathbb N)\longrightarrow \ell_2(\mathbb N)$ bounded linear operator such that $$T(\{x_n\})=\{x_n/n\}.$$ I need to prove that $TB(\ell_2(\mathbb N))$, that is closed unit ball in ...
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33 views

Compact embedding

Prove that the embedding $j\colon (C^1[0,1],\|\cdot\|)\to(L^1[0,1],\|\cdot\|_{L^1})$ where $\|f\|=\max\{\|f\|_\infty,\|f'\|_\infty\}$ and $\|f\|_\infty$ denotes the supremum norm, ...
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41 views

Operators on a Hilbert space question

For a Borel measure $\mu$ define $\langle S_\mu x,y\rangle=\int_H\langle x,z\rangle \langle y,z\rangle \mu(z)$. An exercise in my book that I am reading says that I could find a $\mu$ s.t. $S_\mu$ ...
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characterisation up to unitary equivalence

My book says that the spectral theorem for compact normal operators characterises compact normal operators up to unitary equivalence. It doesn't expand on this so I was wondering what does this mean ...
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111 views

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 ...
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1answer
94 views

Direct sum of eigenspaces of a compact operator has finite codimension

In an infinite dimensional Hilbert space the orthogonal complement of the (closure) of the direct sum of eigenspaces of a compact normal operator is finite dimensional. Why is this the case? thanks.
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51 views

About a compact imbedding of Sobolev spaces

I am studying the Compactness lemma ( on page 570) of the article http://projecteuclid.org/euclid.cmp/1103922134. The lemma says (Compactness lemma ): for $0 < \sigma < \frac{2}{N-2}$, $(N ...
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48 views

Basic question about weak/strong convergence

Let $0<\sigma< \frac{2}{N-2}$ with $N \geq 3$. I know that $H^{1}_{\operatorname{rad}}(R^n)$ (radial functions of $H^{1}(R^n)$ ) is compactly embedded in $L^{2 \sigma +2}(R^N)$. Let $(\psi_v )$ ...
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30 views

2 questions on compact operators

Here are 2 facts I think are true and want to prove: 1) If T=LS (composition), S is compact and L commutes with T, then T is compact. My thoughts. Let $(x_{n})$ be a sequence with norm of $x_{n}$ ...
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99 views

Exercise about compact operator.

In $X=\ell^p$, $p\in[1,\infty]$ we consider: $$ T(x_1,x_2,x_3,\ldots)=(0,x_1,0,x_3,\ldots) $$ Prove that $T$ isn't a compact operator and that $T^2$ is a compact operator. I think I solved the second ...
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155 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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78 views

Compact kernel operator on $L^p$ space

Let $\displaystyle U_1 \subset \mathbb R^{n_1}$ and $\displaystyle U_2 \subset \mathbb R^{n_2} $ measurable sets, $\displaystyle 1 < p,q < \infty $ and consider the measurable function ...
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1answer
52 views

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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45 views

Show $T$ cannot be a compact operator

Let $(X,\lVert\cdot\Vert_x)$ and $(Y,\lVert\cdot\Vert_y)$ be normed spaces, X be infinite dimensional and $T\in\mathcal{L}(X,Y)$ Which has the property: there exists $m>0$ such that $ \Vert{T ...
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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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21 views

Pitt's theorem on automatic compactness of bounded operators between sequence spaces

Why is it called Pitt's theorem? I couldn't locate the origin of the statement.
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53 views

Compactness of the solution operator

Let $\Omega$ be a smooth open bounded subset of $\mathbb{R}^n$. The bilinear form $$a(u,v)=\int_{\Omega}\frac{\partial u}{\partial x}\frac{\partial v}{\partial x}dx$$ is elliptic on ...
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54 views

Distance between unilateral shift and compact operator

We have $S\in\mathbb{B}(\mathcal{H})$ (where $\mathbb{B}(\mathcal{H})$ is algebra of bounded linear operators in Hilbert space) and $S$ is unilateral shift. Compute ...
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1answer
48 views

compact operators and finite dimentional spaces

Let $Q_n$ a finite dimentional space. Since any finite rank operator is compact, it's true that any linear operator $K:Q_n\to Q_n$ is compact 'cause $\dim(R(K))<\infty$?
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Operator from $\ell_{4}$ to $\ell_{1}$ is compact, if it's continuous.

Define $T: \ell_{4} \rightarrow \ell_{1}$ as $Tx=(a_1x_1, a_2x_2, \ldots)$. I showed that $T$ is continuous if and only if $\sum \left| a_i \right|^{\frac{4}{3}} < \infty$. How can I prove that if ...
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35 views

Is operator from $X^{*}$ to $c_{0}$ compact

$X$ - Banach space. Let $X \ni (x_n)$ be such that $x_n \stackrel{\omega}{\rightarrow}0 $ weakly. Define $T: X^{*} \rightarrow c_0$ by $T=\left(x^{*}(x_n) \right)_{n=1}^{\infty}$. Is operator $T$ ...
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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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properties of integral operator $x^{-1}\int_0^xf(x,y)v(y)dy $

here we have two cases to study $(1)$ let us fix any $f \in C^{1}[ [0,1] \times [0,1]]$ ($k \neq 0$). Set $$[T(v)](x) := x^{-1}\int_0^xf(x,y)v(y)dy $$ for any $x \neq 0$ otherwise $[T(v)](0) := ...
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1answer
47 views

Is the canonical injection between $ C^{k+1}[0, 1] $ and $ C^k [0, 1] $ compact?

If $ k=0 $ by the Ascoli - Arzelá theorem the answer is yes, but i don't know how to proceed in the general case ($ k> 0, k \in \mathbb{N} $). I tried to build a counter example using Riesz lemma ...
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restriction a non compact operator to compact operator

If $T\in\mathcal{B}(X,Y)$ is not compact can the restriction of $T$ to an infinite dimensional subspace of $X$ be compact?
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Locally compact operators and their spectrum

At the moment, I'm studying the book "Introduction to Spectral Theory" from P.D. Hislop and I.M. Sigal, I arrived at chapter 10 and I'm stuck on two problems there. Problem 10.1: Let $A$ be a closed ...
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Normal compact operator commute with bounded self adjoint operator in Hilbert space.

Suppose $H$ is a Hilbert space and $A:H\rightarrow H$ is a normal compact operator such that $\ker(A)=0$. show that if $B$ is a bounded self adjoint operator that commutes with $A$ then the spaces in ...
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Arzela-Ascoli and adjoint of compact operator compact

I have seen in this thread a nice answer where it is shown that Thread that the adjoint operator of a compact operator is compact by using the Arzela Ascoli theorem. Unfortunately, there is one thing ...
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$A^2$ self-adjoint and Compact, prove $A$ has an eigenvalue

Suppose $H$ is a Hilbert space and $A \in L(H)$ is such that $A^2$ is compact and self-adjoint. Prove that $A$ has an eigenvalue. (Here $L(H)$ is the set of bounded linear operators on a Hilbert ...
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1answer
59 views

Interpretation of Fredholm Alternative with respect to PDEs

I have studied the Fredholm Alternative, which states the following: Theorem: Let $K:H \rightarrow H$ be a compact linear operator and let $I$ be the identity operator on $H$. Then: 1.$N(I-K)$ is ...
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Verify that the operator $T$ defined by $T( \varphi _{k})=\frac{1}{k}\varphi _{k+1}$ is compact, but has no eigenvectors.

Let $H$ be a Hilbert space with basis $\left \{ \varphi _{k} \right \}_{k=1}^{\infty }$ .Verify that the operator $T$ defined by $$T( \varphi _{k})=\frac{1}{k}\varphi _{k+1}$$ is compact, but has no ...
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Fredholm alternative and orthonormal basis

The following question relates to the Fredholm alternative: Let $K:H \rightarrow H$ be a compact linear operator and let $I$ be the identity operator. Notation: $N$ is the nullspace and $R$ is the ...
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Sufficient and necessary conditions for a given operator $T$ to be $T \in \mathcal{B}$, $T \in \mathcal{K}$ or $T \in \mathcal{F}$

I was rewiewing for an upcoming exam and found this problem. It is quite basic and straightforward but still causing me some confusion mainly since I would have to justify my answer. Let a linear ...
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1answer
59 views

How do I prove that this operator is compact. Functional analysis

We have $$K:l_2\to l_2$$ $$Kx=\left(x_1, \frac{1}{2}x_2,\frac{1}{3}x_3,...\right)$$ where $x=(x_1,x_2,x_3,...)\in l_2$. I want to use the definition of a compact operator: $K$ is compact if ...
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62 views

Functional analysis, help to show a short result

The following problem is from the theory of compact operators: Suppose $X,Y$ are normed spaces and $T:X\to Y$ is linear. Show that if $T$ is compact and invertible then $\mbox{dim}(X)$ and ...
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Is this operator compact and how do I prove it? [duplicate]

I have a very big problem with the following question: Is the operator $T$ defined by $(Tx)t=tx(t)$, $(0<t<1)$ compact in $L_2(0,1)$? My guess is no and I've tried 3 different approaches to ...
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1answer
98 views

Compactness of integral operator

I need some help with this exercise. Let $f\in C^0_b(R^2)$ and consider the operator $[T(v)](x)=\int_0^x f(x,y)v(y)dy$ for every $x\in R$. Is this a compact operator $T:C^0[0,1]\rightarrow C^1[0,1]$? ...
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Non linear compact map

Suppose to have two Banach spaces $E$ and $F$, with $E$ reflexive. Suppose to have a continuous map $T:E \to F$ which maps bounded subsets into precompact subsets. $T$ is not assumed to be linear. ...
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Completely continuous implies compact under separability assumption

My professor left for self-convincing the following statement: "If $T: X \to Y$ with $X$ Banach is completely continuous (that is, takes $w$-convergent sequences to norm convergent sequences) and $X$ ...
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Compact Operators and Complete Metrics Spaces

I have a couple of questions about compact operators and compactness in complete metric spaces: 1.I have the following implications: Let $Y$ be a metric space with $A$ a subset of $Y$. $A$ is ...
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A question about tensor product of algebras of compact operators. [duplicate]

Let $\cal{H}$ be a separable Hilbert space and $\cal{K(\cal{H})}$ the algebra of compact operators acting on $\cal{H}$. Then $$\cal{K(\cal{H})}\otimes\cal{K}(\cal{H})\cong\cal{K}(\cal{H}\otimes H).$$ ...
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37 views

Determine when operator is compact

Let $B$ be the Banach space of bounded complex functions on $[0,1]$ with sup-norm. For $q \in B$, define the (multiplication) operator $T_q : B\rightarrow B$ by $(M_q f)(t) = q(t)f(t)$. Which $q$ ...
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1answer
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If E is a Hilbert space and $T \in B(E)$ is compact, show $T(E)$ does not contain a closed infinite dimensional subspace

It's the problem from "Essential Results of Functional Analysis," R.J. Zimmer, Chapter 3, problem 3.1. I try to prove this problem and I am confused with the condition "closed infinite dimensional." ...
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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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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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69 views

Multiplication operator with a function non-vanishing on the cantor set

Let $M_f$ be the multiplication operator, which acts on bounded functions $g$ on the unit interval as $g\mapsto fg$, with $f:[0,1]\rightarrow \mathbb{C}$ such that $f$ is nonzero only on the Cantor ...
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1answer
247 views

Closure of the range of a compact operator

Let $X$ be an infinite-dimensional Banach space, and let $Y$ be a banach. Let $T$ be a compact operator from $X$ to $Y$, ie. if $(x_n)$ is a sequence in $X$ then there is a subsequence s.t. ...
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70 views

$L(\ell_{p})$ contains only one proper closed ideal

I am trying to solve the following problem: Show that if $1<p<\infty$ and $T:\ell_{p}\rightarrow\ell_{p}$ is not compact then there is a complemented infinite dimensional subspace $E$ of ...
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1answer
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Why is the set of compact operators closed in the space of all bounded operators between Banach spaces?

Let $X$ and $Y$ be Banach space. $B(X,Y)$ is the vector space of all bounded linear maps from $X$ to $Y$. Also, $K(X,Y)$ is the set of all compact operators from $X$ to $Y$. Why is $K(X ,Y)$ ...