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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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847 views

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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265 views

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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100 views

trace $(ADA^{-1})=$ trace $(D)$ in infinite dimensions?

Let $X$ be a separable Hilbert space, $D$ nonnegative definite (by which I also mean self-adjoint) and trace class operator on $X$. Let $A$ be a compact and injective operator with dense range $R$, ...
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49 views

Linear bounded operator from $L^p[0,1]$ to itself whose range consists of continuous functions.

Let $T\colon \mathbb L^p[0,1]\to \mathbb L^p[0,1]$, $1<p<+\infty$, be a linear bounded operator such that $\operatorname{Im}(T)$ is contained in the space of continuous functions. It was shown ...
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51 views

convergence RATE of the square root of a self-adjoint operator.

I am assuming $T$ is a compact operator and $\{T_j\}$ is a sequence of compact operators such that $\|T-T_j\| < \epsilon_j$ where $\epsilon_j$ is a quantity that goes to zero as $j \to \infty$. It ...
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113 views

Why is this compact operator a Fredholm operator?

Let $X$ be a Banach space with the $L^{\infty}$ norm and let $A$: $X \rightarrow X$ be an integral operator of the following form, \begin{equation} Ax(s) = \lambda\int^{b}_{a}K(s,t)x(t)dt, \end{...
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82 views

Compact embedding between $H^{m+1}(\Omega)$ and $H^{m}(\Omega)$ for $\Omega$ bounded

I know that we have Rellich-Kondrachov Theorem that says that there is a compact embedding between $H^{1}(\Omega)$ and $H^{0}(\Omega)$, or more generally as Adams states (pag 168 theorem 6.3) we have ...
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110 views

Is this integral operator on $L_2(0, \infty)$ compact?

Let's define $T:L_2(0,\infty) \to L_2(0,\infty)$ as $$(Tf)(x) = \int_0^\infty \frac{f(y)\sqrt{xy}}{x^2y^2+1}dy.$$ I'm interested, if this operator is compact. $T$ is integral operator with kernel $K = ...
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93 views

I don't see why $W^{1, 2}(\partial D)$ being compactly embedded in $L^2(\partial D)$ lets us show an operator is Fredholm of index zero.

Let $D$ be a bounded Lipschitz domain. Let $A$ be the single layer potential which maps $L^2(\partial D)$ into $W^{1, 2}(\partial D)$ boundedly. $A$ is given by: $$ A_D[\phi] = \int_{\partial D}G(x-y)...
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100 views

Fredholm alternative for polynomially compact operators

Let $T \in \mathcal{L}(E)$ be a polynomially compact operator, i.e., there exists a polynomial $p$ such that $p(T)$ is a compact operator. Suppose $p(1) \neq 0$. I want to show that $N(I-T) = \{0\} \...
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37 views

Reference request on operators with compact powers

In this Wikipedia page about compact operators at the very bottom it says:"If $B$ is an operator on a Banach space X such that $B^n$ is compact for some $n$, then the theorem proven above also holds ...
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32 views

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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51 views

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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32 views

Optimal conditions for Schrödinger semigroup to be Hilbert-Schmidt

Let $V:\mathbb R\to[0,\infty)$ be (for simplicity) a smooth function. Consider the Schrödinger operator $$H:=-\Delta/2+V$$ acting on $L^2(\mathbb R)$. By the Feynman-Kac formula, the semigroup of this ...
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65 views

Compactness of operators and norming sets

Let $T$ be a linear map from a normed space $E$ into a Banach space $F$. Let $D\subset \overline{B}_{F^{*}}$ be norming, i.e. there is $r>0$ such that $\sup_D|\left<f,v\right>|\ge r\|f\|$, ...
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62 views

Show that the integral-operator compact?

We have the operator $T$ on $L_2[0,1]$ which is defined as $Tf = y$ where $y$ is the solution to the ODE $y^{\prime \prime} + y^\prime = f$ with boundary conditions $y(0)=0, y(1) = 1$ Show that $T$ ...
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87 views

Does there exist an infinite dimensional compact operator on any infinite dimensional Banach space?

Let $X$ be an infinite dimensional Banach space. Does there always exist a compact linear operator on $X$ with infinite dimensional range? I can see that the answer is yes if $X$ is a Hilbert space, ...
3
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203 views

Volterra Operator Property

It can be shown that if $V(f) = \int_0^x f(t)\,dt$ denotes the Volterra operator on $C([0,1])$, then the closed subspace $A_b = \{f : f|_{[0,b]} =0\}$ satisfies $ V(A_b) \subset A_b$, that is, $A_b$ ...
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988 views

Show that the inverse of the Laplacian is compact

Let $d\in\mathbb N$ $\Lambda\subseteq\mathbb R^d$ be bounded and open $\mathcal D(A):=\left\{u\in H_0^1(\Lambda):\Delta u\in L^2(\Lambda)\right\}$ and $$Au:=-\Delta u\;\;\;\text{for }u\in\mathcal D(A)...
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318 views

Characterization of compact operators by their spectra

In any functional analysis book there is usually a section devoted to the study of the properties of the spectrum of compact operators. Is there any spectral characterization of compact (self-adjoint)...
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459 views

How is the Point Spectrum of a Compact Operator Countable?

I'm working on understanding a proof that if an operator $A$ on a Hilbert space $\mathcal{H}$ is compact, then show that $\sigma(A) - \{0\} \subseteq \sigma_p(A)$. If you're not familiar with this ...
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172 views

Trace class operator

Let $A\in B(H)$ and $\sum_{E}|\langle A e,e\rangle|< \infty$ for every orthonormal basis $E$. Show that $A$ is a trace class (means $\sum_E \langle |A|e,e\rangle < \infty$). I can not prove it. ...
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49 views

Compactness of solution space of semi-linear parabolic PDE

Under what conditions a closed and bounded subset of solution space of following parabolic PDE is compact? $$x_{t}=x_{zz}+f(x,z)$$ Thank you!
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150 views

Compact operators on a Banach space $X$ are closed in the bounded operators on $X$. - Proof correction help

I am given a proof of the following statement (see below). Compact operators on a Banach space $X$ are closed in the bounded operators on $X$. I was told that there is an error in this proof - I ...
3
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171 views

Sufficient condition for an operator to be compact in Hilbert space of holomorphic function with respect to Gaussion weight (Fock space).

What I read in a book I could not understand, some one please help. Let $\mathcal{F}=\{f:\mathbb{C^n}\rightarrow\mathbb{C}: \text{$f$ is holomorphic and}\int_{\mathbb{C}^n}\lvert f(z)\rvert^2e^{-\...
3
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959 views

Fredholm and Compact Operators

Let $X$ and $Y$ be Banach spaces and $T\in B(X,Y)$ be Fredholm. Then there is $S\in B(Y,X)$ such that $ST=I+K_{1}$ and $TS=I+K_{2}$ where $K_{1},K_{2}$ are compact operators.Proof: Since $T$ is ...
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17 views

Is $|x|^{-d+\alpha}$ square integrable in $\mathbb{R}^d$ given $\alpha>0$?

This is a problem in the S.-T. Yau College Student Mathematics Contests in 2013. Suppose $H=L^2(B)$, $B$ is the unit ball in $\mathbb{R}^d$. Let $K(x,y)$ be a measurable function on $B\times B$ that ...
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24 views

If $A \in \mathcal{L}_c(X)$ and $X$ is Banach, then $\dim \ker (\text{id}-A) < + \infty$.

Exercise : Let $X$ be a Banach space and $A \in \mathcal{L}_c(X)$. Show that $\dim \ker ( \text{id} - A) < + \infty$. Attempt/Thoughts : The kernel of the operator $(\text{id}-A) : X \to X$ ...
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62 views

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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55 views

What is known about the reciprocal $1/f$ of a holomorphic Banach-valued function $f$?

Let $U \subseteq \mathbb C$ be open, $A$ a (unital, associative) complex Banach algebra and $f : U \to A$ holomorphic and invertible in a punctured neighborhood of $0 \in U$, so that $0$ is an ...
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25 views

Existence of a finite rank operator $F$ such that $A+F-\lambda$ is invertible for all $\lambda$

This is an exercise in the book "A First Course in Functional Analysis" written by Conway, page 362: Let $A$ be a bounded linear operator on a Hilbert space and $G$ be an open connected subset of $ ...
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67 views

Compact or noncompact operator?

I found an example of operator: $$A(x_1, x_2, x_3, ...)=\left(x_1+ \frac{x_1+x_2}{4}+\frac{x_1+x_2+x_3}{9}+..., \frac{x_1+x_2}{4}+\frac{x_1+x_2+x_3}{9}+..., \frac{x_1+x_2+x_3}{9}+..., ..., \frac{x_1+...
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51 views

Expressing an operator as a sum of the identity and a compact operator

My problem concerns with the unique solvability of a linear system of integral equations. In my problem, I was able to write the system in matrix form: $$ M \begin{align} \begin{bmatrix} ...
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35 views

Square Root of Operator forms a two sided ideal

If $H$ is an infinite dimensional separable hilbert space, define $T(H)$ as $$A \in T(H) \qquad \text{iff} \qquad (A^*A)^{\frac{1}{2}} \, \, \text{bounded and compact} \, \, $$ and $$\sum_{i} \...
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51 views

Cuntz-Krieger algebra as crossed product

Let $A$ be a square matrix and let $\mathcal{O}_A$ be the corresponding Cuntz-Krieger algebra. Let furthermore $\gamma \colon \mathbb{T} \to \text{Aut}(\mathcal{O}_A)$ be the canonical guage-action ...
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86 views

Norm of translation operator on $\mathbb{R}^2$

I'm interested in calculating an operator norm of $T : \mathbb{R}^2 \to \mathbb{R}^2$, where $T(x,y)=(x+1, y)$, i.e. a translation by $1$ in direction $x$. From wiki for bounded operator $$\left\...
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46 views

Show that a $N$-body Schrodinger operator has compact resolvent

I have a $N$-body Sch\"odinger operator of the following form \begin{equation} H := \sum_{j=1}^{N} (-\Delta_{x_j}+U_{trap}(x_j))+\sum_{i<j} V(x_i-x_j), \end{equation} where $U_{trap}$ is a positive ...
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137 views

Fixed point of compact operator in Banach space

Let $M:=\overline{B(o,1)} $ be a closed unit ball of Banach space X. Assume that $F: M \rightarrow X$ is a nonlinear compact operator and $F(\partial M) \subset M$. Then F has a fixed point in M. ...
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78 views

How to show if a nonlinear operator is Ill-Posed, in the concrete gravemetry problem.

I'm considering a nonlinear integral equation $F:U \subset \mathcal{L}^2(a,b) \to \mathcal{L}^2(a,b)$, where $a,b \in \mathbb{R}$ and $H$ is a constant. The operator is defined as $$ F[x](t)=\int_a^b \...
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0answers
111 views

Difference of Orthogonal Projections is Compact

Let $M$ and $N$ be closed subspaces of a Hilbert space $H$ and let $P_M$ and $P_N$ denote the orthogonal projections onto $M$ and $N$, respectively. I have proved so far the following equivalences: (...
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0answers
52 views

Convergence of Compact Linear Opeators in $L^2([0,1])$

Let $A_n: L^2([0,1]) \to L^2([0,1])$ be $(A_nf)(x)= \int_0^1 sin(n\pi(x-y))f(y)dy$. Are they compact operators? Is there any kind of convergence? They are continous since $\|A_nf\| \leq \|f\| $. ...
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49 views

Injectivity of index map for $K_1(S^1)$

This example/problem is from Valette's notes on the Baum-Connes conjecture (p. 45). The exercise is to prove that the (trivially equivariant) $K$-homology group $K_1(S^1)$ is $\mathbb{Z}$. For this, ...
2
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0answers
328 views

Different versions of Mercer's theorem

I am reviewing materials on reproducing kernel Hilbert space (RKHS) and I've found various versions of Mercer's theorem: About the positive-definiteness conditions. In the Wikipedia pages on RKHS ...
2
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0answers
387 views

Weakly compact operator on $c_0$ is compact

Show that if $T\in {\cal B}(c_0)$ and $T$ is weakly compact, then $T$ is compact. My attempt: $T$ is weakly compact, so there is a reflexive space $X$ , and operators $A\in {\cal B}(X,c_0) $ and $B \...
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113 views

Spectrum of $Tu=\int^1_{-1} (1-|x-y|)u(y)dy $

Consider the operator $$ Tu(x)=\int^1_{-1} (1-|x-y|)u(y)dy $$ We want to find the spectrum of $T$. The kernel is certainly bounded and so this operator is Hilbert-Schmidt, so $T$ is compact. We ...
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707 views

Compact operators is a linear subspace of bounded operators

Let $X,Y$ be Banach spaces. Let $B(X,Y)$ be the set of bounded linear operators and let $K(X,Y)$ be the set of compact linear operators. I want to prove that $K(X,Y)$ is a vector subspace of $B(X,Y)$...
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0answers
98 views

Showing a particular integral operator is trace class

Let $f$ and $P$ be continuous, integrable functions $\mathbb{R} \to \mathbb{C}$ vanishing at $\pm \infty%$. Concisely, $f,P \in C_0(\mathbb{R}) \cap L^1(\mathbb{R})$. Also, assume that $P$ is real-...
2
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0answers
145 views

Perturbation of eigenvalues

I am looking at a certain operator, that is a Hilbert-Schmidt integral operator from $L^2(X,d\mu)$ to $L^2(X,d\mu)$. The question is how eigenvalues, or singular values, change as it's kernel is ...
2
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319 views

Proof of compactness of bounded linear operator

Define $T: l^2 \to l^2$ by $Tx = y =(\eta_j)$, where $x = (\xi_j)$ and $$ \eta_j = \sum_{k=1}^{\infty} \alpha_{jk}\xi_k, \quad \quad \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} |\alpha_{jk}|^2 < \infty....
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48 views

Every Dunford-Pettis operator is compact

I was trying this problem. Let $X$ be a Banach space that does not contain a copy of $l^1$. Show that every Dunford-Pettis operator $T: X \rightarrow Y$, with $Y$ any Banach space, is compact. I ...