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 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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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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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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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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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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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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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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$L(\ell_{p})$ 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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Why $K(X,Y)$ is closed?

If $X$ and $Y$ are Banach space then $K(X Y)$ is a closed vector space of $B(X, Y)$. $B(X,Y)$ is the vector space of all bounded linear maps from $X$ to $Y$. $K(X,Y)$ is the set of all compact ...
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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 ...
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Frechet derivative of compact operator is compact

... this seems to be a well known fact as mentioned in this and in this manuscript. However, I was not able to find a proof or to prove it by myself. So my question is: How to prove this? Any hint ...
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51 views

Are $T,T^2$ compact operators?

$T:l_2\to l_2$ is defined by $T(x_1,x_2,\dots)=(0,x_1,0,x_3,0,x_5,\dots)$ we need to find whether $T, T^2$ are compact or not. I see here the definition of compact operator but I'm not able to apply. ...
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Transpose of the Hilbert-Schmidt operator

Let $X = L^2(\Omega)$, $\Omega \subset \mathbb{R}^N$ be an open set (or a $\sigma$-finite measure space), $B \in L^2( \Omega \times \Omega)$. Then the Hilbert-Schmidt operator $T \in \mathcal L(X)$ ...
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On Fredholm operator

Consider operator $T: l^2(\mathbb{N})\to l^2(\mathbb{N})$ given by $T(x_1,x_2,\cdots)=(\lambda_1x_1,\lambda_2x_2,\cdots)$, where $\{\lambda_n\}_{n\in \mathbb{N}}$ is nonzero bounded complex numbers. I ...
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Show that $f \in c_0^*$ and $||f||=\sum_{j=1}^{\infty} \frac{1}{j!}$

Let $$\begin{eqnarray} f: c_0 & \to & \mathbb{R}\\ (x_i)_1^{\infty} & \to & \displaystyle \sum_{j=1}^{\infty} \frac{x_j}{j!}\\ \end{eqnarray}$$ Show that $f \in c_0^*$ and ...
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Invertibility of a linear operator on a Hilbert space.

Let $H$ be an infinite dimensional Hilbert space over $\mathbb C$, $T$ be a continuous linear operator of $H$, $r(T)=\sup_{||x||=1}|(Tx|x)|$ be the numerical radius of $T$, and $z\in \mathbb C$, such ...
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Want to show an operator is compact

With $V=L^2(0,T;H^1(\Omega))$, let $A:V \to V^*$ with $$\langle Au,v \rangle = \int_0^T \int_{\Omega} \nabla u(t) \cdot \nabla v(t).$$ I want to show that $A$ is a compact operator. So, one way to ...
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Inequality for compact operator between Banach spaces

I've been pondering about the following Lemma for a while now, but can't think of a proof. In fact, I can't even think of a way to prove it. Let $E$, $F$ and $G$ be Banach spaces, $T \in ...
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Is the limit of compact operators again compact?

Let $(T_n)_{n \in \mathbb{N}} \subset \mathcal{L}(\mathcal{X}, \mathcal{Y})$ where $T_n$, $n \in \mathbb{N}$, is compact. Now, assuming that $(T_n)_{n \in \mathbb{N}}$ has a limit $T \in ...
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how to prove this property of compact operator? [duplicate]

I read about this property of compact operator from wikipedia $K(X, Y)$ is a closed subspace of $B(X, Y)$: Let $T_{n}, n \in N$, be a sequence of compact operators from one Banach space to the other, ...
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Is $A$ a compact operator or not?

Let $ A\colon L_1[0,1] \to C[0,1] $ $$ Af(t) = \int_0^t f(s)ds,\quad f \in L_1[0,1] $$ Is $A$ a compact operator or not?
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Is a Hilbert space $H$ compactly embedded in its dual?

Is a Hilbert space $H$ compactly embedded in its dual? Is it compactly embedded in itself? No idea how to think of this.
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Compact integral operator

I have this exercise and I don't know how to solve the last question. In the following $a,b$ are two real numbers such that $a<b$ ,$E=C([a,b],\mathbb{R})$ with the norm $||.||_0$ given by ...
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Compact operators

Let $AA^*$ be compact I need to show that A is compact, I am quite sure that it needs to be shown that it maps weakly convergent sequence to a sequence that converge in norm but I cant manage to do ...
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Is there a noncompact operator from $\ell^\infty$ to a reflexive space?

It is well known that bounded operators from $c_{0}$ to a reflexive Banach space $X$ are in fact all compact. Indeed, since it can be shown that an operator is compact iff for any weakly convergent ...
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Prove that there $B : H\to H $ bounded such $ B^n = A $.

Let $ A : H\to H $ a compact self-adjoint operator. Suppose $ A $ is positive. let $ n \geq 2 $. Prove that there is $B : H\to H $ bounded such $ B^n = A $.
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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 ...
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Pitt's theorem and reflexivity

Does it follow from Pitt's theorem that the space of bounded operators from $\ell_2$ to $\ell_p$ ($p<2$) is actually reflexive? We have $$ \mathcal{B}(\ell_2, \ell_p) = \mathcal{K}(\ell_2, \ell_p) ...
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how to show that the norm-limit of compact operators 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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How to show that a sequence in $l^{2}$ has a convergent subseqence after the action of an operator. [duplicate]

Problem: Let $l^{2}=\{(x_{1},x_{2},...)| x_{n} \in \mathbb{R}, \forall n,(x_{1}^{2}+x_{2}^{2}+...)<\infty\}$, $\|x\|=(x_{1}^{2}+x_{2}^{2}+...)^{\frac{1}{2}}$. Given ...
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Norm operator and compactness

For the operator $U\colon \ell_{p}\to\ell_{p},\;\left( 1\leqslant p<\infty \right) :$ \begin{equation*} Ux=U\left( x_{1},x_{2},\dots \right) =\left( 0,x_{1},\frac{x_{2}}{2},\frac{% ...
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Invertible operator

Let $K:V\to W$ such that $Kf = k$, where $V,W$ are infinite-dimensional Banach spaces. Is it correct to say that in general $f = (K^*K)^{-1}k$, however, when $V=W$, then $f = K^{-1}k$. $T^*$ here ...
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Solution to operator equation, surjectivity

Suppose $T:V\to W$, where $V,W$ are banach spaces and $Tf = k$ (for instance $T$ might be an integral operator). They say that the equation has solution when $T$ is injective and so $T^{-1}$ exists. ...
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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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Hilbert space proof

$X$ is a separable Hilbert space and $ A\in L(X,X)$ and compact. I need to prove that $A$ is approximately of finite dimension.
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About compact operator

When seeing a proof of Fredholm's alternative I don't get the following: Let $T$ be compact from a Banach space $X$ to itself, and $\lambda \neq 0$. Define $S=I-T$, $S^k$ its $k$-th power and ...
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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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Is restriction of compact operator always compact?

Question 1: Is a compact operator $T:X\to Y$ 's restriction on a subspace $Z\subset X$ still compact? (I think I've got the answer) I think the compact operator's restriction on any subspace must ...
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Showing that the inverse of the perturbation of a compact operator by a bounded operator remains compact.

The title says it all. If we have a Hilbert space $H$, then if $B\in \mathcal B(H)$, $L$ is a linear operator that is not necessarily bounded, $L^{-1}$ is compact, and $0\in \rho(L)\cap\rho(L+B)$, ...
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The set of compact linear operators is a subspace of the set of bounded linear operators

I know that a linear operator $T:X \to Y$ (where $X$ and $Y$ are normed vector spaces) is compact if for every sequence $\left(x_{n}\right)\subseteq X$ s.t. $\left\Vert x_{n}\right\Vert \leq C$, the ...
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Show for compact operator $K$, if $||Kf|| < ||f|| \forall f$, then $||K|| < 1$.

I wanted to check my reasoning on proving this statement, and see if anyone had suggestions for other proofs of this fact. Again, the statement is, if $K$ is a compact operator on a Hilbert space ...
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Strong operator convergence and adjoint operator

Let $H$ be a Hilbert space and $(T_n)_{n \in \mathbb{N}}$ be a sequence of bounded linear operators on $H$. The strong convergence of $T_n$ doesn't imply the strong convergence of $T_n^*$, i.e. ...
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Invertibility of compact operators

I'm a little confused about compact operators and whether or not they are invertible. Just hoping someone here can clear up my confusion: Let $T$ be a compact operator on a Banach space $X$. Since ...