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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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Integral Equation and Fredholm Alternative

I am learning about functional analysis at the moment and I have difficulties grasping the connection to integral equations or differential equations. For simplicity, let us consider the following ...
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Compact embedding of the domain and compact inverse

I have several problems in showing this point of a problem: we consider $X$ Banach space and $T: D(T) \to X$ a closed operator with domain $D(T) \subseteq X$. Let be $T$ bounded, invertible and ...
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Compact open operator between Banach spaces

Let $X,Y$ be Banach space, $Y$ infinite dimensional. Show that no $T \in \mathcal{K}(X,Y)$ is open. By definition $T$ is open if and only if $\exists r >0$ such that $B_Y(0,r) \subset T(B_X(0,1))$ ...
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Compact operator on $L_2([0,1],m)$

Consider the Hilbert space $H=L_2([0,1],m)$ where $m$ is the Lebesgue measure on the interval $[0,1]$. Let $T \in \mathcal{L}(H,H)$ given by \begin{equation*} T\ f(x)=x \ f(x) \ \ \ \ f \in H,\ x \...
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Compact operator $L:\ell^2\to\ell^2$ with $\Vert L\Vert=1$ such that $\Vert L(x)\Vert<\Vert x\Vert$ for all $x$

Let $\ell^2$ denote the space of square summable sequences of complex numbers. Let $L:\ell^2\to\ell^2$ be a linear operator with $\Vert L\Vert=1$ such that for all $x\in\ell^2\setminus\{0\}$, $\Vert L(...
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Show that $(Tu)(x)=\int_{\alpha(x)}^{\beta(x)} u(t)dt$ is Compact linear operator on $C([0,1])$

Show that \begin{equation} (Tu)(x)=\int_{\alpha(x)}^{\beta(x)} u(t)dt \end{equation} is Compact linear operator on $C([0,1],R)$ where $\alpha, \beta:[0,1]\rightarrow [0,1]$ are continuous. My ...
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Compact operators and weak convergence

Let $X$ and $Y$ be Banach spaces. (a) Let $T \in \mathcal{L}(X, Y )$. For each sequence $(x_n)_{n \geq 1}$ in $X$ and each $x \in X$, show that $x_n →x$ weakly, as $n \rightarrow \infty$ ,implies ...
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Isometric isomorphism between $L^2$ and $\mathcal{L}^2$

I was reading and trying to understand the proof that the space $\mathcal{L}^2 (\mathcal{H})$ (Hilbert-Schmidt operators) is made by all the $T_K:L^2(X,\mu) \rightarrow L^2(X,\mu)$ with $K \in L^2(X \...
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Compact operator on $L^2[0,1]^2$

Let $K\in L^2([0,1]\times[0,1])$, and we define the operator $T_k$ on $L^2[0,1]$. $$(T_kf)(x)=\int_{0}^{1}K(x,y).f(y).dy \quad \quad \forall f\in L^2[0,1]$$ How to prove that $T_k$ is a compact ...
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$||T-T_n|| \rightarrow 0$ and $T_n$ are compact but $T$ is not a compact operator.

It is a result that if $||T-T_n|| \rightarrow 0$ in the norm operator an that the $T_n \in \mathcal{L}(X,Y)$ (were $Y$ is a Banach space) are compact operators, then $T$ is compact. I found from here ...
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Integral Operator in $L^2$

I was trying to do this exercise and I'm wondering if I figured it out well: I have $\mathcal{H} := L^2(0,1)$ and $T$ the operator with integral kernel $K(x,y) = \min\{x,y\}$, $x,y \in [0,1]$. I have ...
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Compact Integral Operators induced by positive Kernels

Let $K$ be a compact operator induced by the kernel $k(s,t)\in L^2([0,1])^2$ with $k(s,t)>0$. Prove that $\|K\|<1$ if and only if $(I-K)$ has a bounded inverse $(I-K)^{-1}$ which is induced by a ...
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Matrices of bounded linear operators

Let $X,Y$ be Banach spaces and let $A=(A_{n,k})$ be an infinite matrix of bounded linear operators $A_{n,k}:X \to Y$. Suppose $\sup_n \sum_k \|A_{n,k}\|<\infty$. Property: For each sequence $x=(...
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Range of a compact operator [closed]

Let $T$ be a bounded operator and $K$ be a compact operator on a Banach space $X$ such that $TX \subseteq KX$. Is $T$ a compact operator?
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Non existence of solution of a special first kind Fredholm integral equation

Let $k \in {L^2}((0,4) \times (0,1))$, $g \in {L^2}(0,1)$. We consider the following first kind Fredholm equation $$\int\limits_0^4 {k(s,x)f(s)ds=g(x), x\in(0,1).} $$ Where $f$ is the unknown. How ...
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Isomorphisms in the proof of the Fredholm alternative/Theorem of Riesz-Schauder (for compact operators)

In a proof of the Fredholm alternative/Theorem of Riesz-Schauder, I came across the following: Let $X$ be a Banach space, $T:X \to X$ be a compact operator and $A:=T-I$. We proved that $\mbox{ker}(A)...
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1answer
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A linear operator is compact if and only if the image of any bounded set is relatively compact

Let $U$ and $V$ be normed spaces. Show that a linear operator $T:V \to U$ is compact if and only if the image of any bounded set is relatively compact. (Here $T$ is compact if the image of any ...
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A non compact operator on $L^2[0,1]$

Let $H$ be the Hilbert space $L^2[0,1]$. and the operator $T : H\rightarrow H$, such as $T(f)(x)=x.f(x)$ Why $T$ isn't compact ?
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Third kind Fredholm integral equation

Let us consider the following integral equation $$a(x)u(x) + \int\limits_0^1 {K(s,x)u(s)ds} = f(x)$$ Let f in $L^p(0,1)$ for some $p \in [1,\infty] and let $ $K \in L^q((0,1) \times (0,1))$. Assume ...
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Example of a non-trace class operator

Let $H$ be a complex Hilbert space. For an operator $T: H \rightarrow H$ fix an orthonormal basis and define the "absolute value" of $T$ as $$|T| = (TT^{*})^{\frac{1}{2}}$$ We say that the operator ...
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Given an averaging operator $A: l_p \rightarrow l_p$, why $A$ is not compact

Let $A$ be an operator $A: l_p \rightarrow l_p , 1 <p<\infty$ $$A(x_1, ..., x_n, ...)=\left(x_1, \frac{x_1+x_2}{2}, ..., \frac{x_1+...+x_n}{n}, ...\right)$$ I want to show that operator $A$ is ...
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Spectral representation of compact self-adjoint operators on a Hilbert space H

Theorem 9.9-1 of Kreyszig's "Introductory Functional Analysis with Applications" states that if $H$ is a complex Hilbert space and $T : H \longrightarrow H$ is a bounded self-adjoint operator, then $T$...
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1answer
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Bounded operator but not Compact

Let $T : (C([-1,1]),||.||_{\infty}) \rightarrow (C([-1,1]),||.||_{\infty}) $ Such as : $(Tf)(x)=\frac{1}{2}(f(x)+f(-x))$ . For all $f\in C([-1,1])$ Why $T$ isn't Compact ? I tried to use the ...
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1answer
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Kernel decomposition of a finite rank integral opeartor

Given a self-adjoint finite rank integral operator P on $L_2[0,1]$, it has the eigen-decomposition $P=\sum_{i=1}^k\lambda_i \langle u_i,\cdot\rangle u_i$ where $u_i$ are eigenfunctions and $\lambda_i$ ...
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How to show that $(\lambda_n x_n)$ is compact iff $(\lambda_n)\in c_0$? [closed]

Let $(\lambda_n) \in l_\infty$, and $A:l_2 \rightarrow l_2$ a linear operator defined as $$A(x_n)=(\lambda_n x_n), \,\, (x_n) \in l_2$$ How to show that $A$ is compact if and only if ($\lambda_n)\in ...
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Operator on $C[-1, 1]$ compact? [closed]

Let $A:C[-1, 1]\to C[-1, 1]$ with $Ax(t)=1/2 (x(t)+x(-t))$ and sup-norm? Is this operator compact? Thanks in advance!
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dim $ker(T-\lambda)\le1$ for all $\lambda$ in $\mathbb{C}$ $\iff$ $\exists h\in \mathbb{H} \ s.t. \ h$ is a cyclic vector for $T$.

Let $T$ be a compact normal operator on a $\mathbb{C}$-Hilbert space,$\mathbb{H}$. dim $ker(T-\lambda)\le1$ for all $\lambda$ in $\mathbb{C}$ $\iff$ $\exists h\in \mathbb{H} \ s.t. \ h$ is a cyclic ...
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Unbounded self-adjoint operator with compact resolvent: Expansion

Given an unbounded self-adjoint operator $T\colon H \supset \mathrm{dom}(T) \to H$ on a Hilbert space $H$ with compact resolvent, i.e. $T$ has a purely discrete spectrum, the eigenfunctions $e_n$ form ...
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How to show the continuity of the following operator?

Let $T:\mathcal{l}^2\to\mathcal{l}^2$ be an operator with $$(Tu)_n:= \frac{1}{2}u_n + \frac{1}{n^2}u_n.$$ I've found that the operator is not compact, since $$(Tu)_n = \left(\frac{1}{2}+\frac{1}{n^...
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1answer
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Centralizer of projections

Let $H$ be a Hilbert space and $p, q$ self-adjoint projectors in $B(H)$, i.e. $$p^2=p=p^* \space \text{ and } \space q^2=q=q^*.$$ Suppose they have the same centralizers $C(p)=C(q)$. Is it true that ...
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Compact operators on $\ell^1$

Let $T\in \ell^1$, $Tx = (\lambda_1x_1,\dots,\lambda_nx_n,\dots)$. Want to show that if $T$ is compact, then $\lambda_n\to0$. I know for $p\in(1,\infty]$, canonical basis $e_n \rightharpoonup 0$ (so ...
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Range-Kernel Uncomplementation of Fredholm Operators

I need to find a normed space $X$ and a compact operator $K:X\to X$ such that $$X\neq N(I-K)\oplus R(I-K).$$ However, all my examples failed so far.
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If a bounded operator has finitely many eigenvalues above zero, is it compact?

In order to give you some context: In the paper "Magnetic Lieb-Thirring estimates" by Laszlo Erdös he claims that since the number of eigenvalues of a bounded operator $K$ acting on $L^2(\mathbb{R}^3)$...
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Limit of an operator and eigenvalues

can you help me how to solve this exercice? Define in $L^2(-1,1)$ $$ T_k[f](x):= \int_{-1}^1 \frac{(-1)^k}{(2k+1)!}(xy)^{2k+1} f(y)\,\,dy $$ i) show that $||T_k||\to 0$; ii) find eigenvalues of $T_0$....
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Kernel of an operator

can you help me to solve this exercice? Show that the kernel of the operator $M:L^2((-1,1))\to L^2((-1,1))$ has infinite dimension. The operator is: $$ M[f](x):= \int_{(-1,1)} \sin(xy)f(y)\,\,\,dy. $$ ...
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Spectrum of two operators on Hilbert Space whose difference is compact

Let $\mathcal{H}$ be a Hilbert space, and $v$ and $w$ two elements of $\mathcal{L}(\mathcal{H}),$ the space of the continuous linear operators from $\mathcal{H}$ to $\mathcal{H}.$ All we know is that ...
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Showing compact operator

Let $X,Y,Z$ be Banach spaces and let $T:X\to Y$ be a compact linear operator, $S:Z\to Y$ be a bounded linear operator such that $S(Z)\subset T(X)$. I have to show that $S$ is a compact linear operator....
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multiplication of Hilbert Schmidt operators and bounded operators

Suppose $T_1,T_2 \in B(H)$,where $H$ is an infinite dimensional Hilbert space. $S\in \mathcal{HS}(H)$,$\mathcal{HS}(H)$ is the set of Hilbert Schmidt operators on $H$. Does there exist nonzero ...
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1answer
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Definition of a compact linear operator

We know that if $X$ and $Y$ are Banach spaces, then a linear $T:X\to Y$ is compact if $T(B_X)$ is relatively compact, i.e. $\overline{T(B_X)}$ is compact. My question is why it is not defined as $T(...
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1answer
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Compact Operator and Unit Ball in C([0,1])

My professor mentioned that for $V=C([0,1])$, equipped with the norm $||*||_{\infty}$, the function $T:V \rightarrow V$ given as $T(f(t))=\int_0^t f(s) ds$ is a "compact operator". I'm not too ...
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compactness of weighted resolvent

In several papers I've seen the following argument and I don't understand it. Let $H$ be a self-adjoint defined on the sobolev space $H^2(\mathbb R^d)$ with absolutely continuos spectrum. Moreover, ...
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Banach spaces and compactness

Let X, Y be Banach spaces and $ X \hookrightarrow Y$ (which would imply $Y' \hookrightarrow X'$). Let the bilinear form $a_0: X \times X \to \mathbb R$ be continuous, symmetric and coercive. Moreover, ...
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The operator $T+i Id$

I have the operator $$T:\mathcal{H}\rightarrow\mathcal{H}$$ the operator $T$ is linear, symmetric and compact, $\mathcal{H}$ is a Hilbert space my problem is Prove that ($\,T+i\,$Id) is $\,\bf{...
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1answer
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Please check my proof of a bounded linear operator and to show it a finite rank operator

There is an exercise I am trying to solve, but I am not sure if my solution is 100% correct. Please help me to check it carefully. Let $A = (A_{n,m})_{n,m \in \mathbb{N}}$ with $A_{n,m} \in \...
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1answer
76 views

A property of the convolution of operator

For any $f\in C_c^\infty(\mathbb{R})$, let $T_f$ denote the bounded operator on $L^2(\mathbb{R})$ given by convolution by $f$: that is, $$(T_f g)(x):=\int f(y)g(x-y)\,dy,$$ for $g\in L^2(\mathbb{R})$...
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1answer
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Is the spectral decomposition of self-adjoint compact operators unique?

Given a compact, self-adjoint operator $T$ on a Hilbert space $H$, then there is the Spectral Theorem which says that $T=\sum_i \lambda_i P_i$ where the sum is over the number of eigenvalues of $T$, ...
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1answer
48 views

Hilbert-Schmidt operator defined by non-orthogonal basis

I have the following operator on $L^2(0,1)$ $$ Tf = \sum_{n \geq 0} 2^{-n}\langle f,v_n\rangle v_n$$ Where $v_n(t) = t^n$. I was able to prove that it is Hilbert-Schmidt, but now I need to ...
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Why are nonzero eigenvalues of a compact operator poles of its resolvent?

Let $X$ be a complex Banach space and $T$ a compact operator on $X$. I read on Wikipedia that nonzero elements of the spectrum of $T$ are poles of its resolvent. It says "by functional calculus", but ...
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References for “Compactness Operator in set theory and topology” - why antispace is kind a topological space?

I find here that Antispace is a kind of topological space I try to find this little book (70p) to read online 1968, Evert Wattel, The Compactness Operator in Set Theory and Topology: There ...
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Continuity of a bivariate series.

Assume that $\lambda_i = \mathcal{O} (i^{-r})$ with $r > 1/2$. If $$ K (t,s) = \sum_{i=1}^\infty \lambda_i \phi_i (t) \phi_i(s) $$ and $\{ \phi_i (t): i \geq 1\}$ are continuous (functions) on $[0,...