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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What role does the range of an operator play in showing the operator is compact?

To show an operator is compact I understand you have to show the operator is the limit of finite rank operators. However the proof I have doesn't do this. I have an operator $$k:C0[,\pi]) \to C([0, ...
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Proving a linear operator is compact: understanding the statement “norm limit of a sequence of finite rank operators”.

I am having serious trouble understanding the proof that an operator is compact. This is the original question I asked and the proof is included very helpfully in the answer. Show if $\lim_{n \to ...
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7 views

Integral operator with sub-exponential eigenvalue decay

I am looking for a positive-definite symmetric function $K: \Omega \times \Omega \to \mathbb{R}$ ($\Omega \subset \mathbb{R}$), such that $$ K(x,y) = \sum_{k=0}^\infty \exp(-\sqrt{k}) \phi_k(x) ...
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19 views

A question on Bounded Approximation property

Let $V$ be a Banach space and we say that $V$ has the $C$-BAP if there exists a net of bounded finite rank operators $T_\alpha$ in $B(V,V)$ and a constant $C$ such that $\|T_\alpha\| \leq C$ for each ...
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Showing a C* Algebra contains a compact operator

In my functional analysis class we are currently dealing with C* Algebras, and I just met this problem: Let $ \mathbb{H} $ be a separable Hilbert space, and suppose we have $ A \subset ...
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25 views

Problem about a compact operator $T:l^p\rightarrow l^p$

I have to solve this problem. Let $\{\lambda_n\}$ be a sequence of real number such that $\lim_{n\rightarrow\infty}\lambda_n=0$ and consider the operator $T:l^p\rightarrow l^p$, $1\leq p\leq ...
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31 views

Singular integral operator

i got the following problem to solve. Let $0 < \alpha < 1$, $L \in L_\infty([0,1]^2)$, $D = \{(x,y) \in \mathbb{R}^2: x = y\}$ the diaagonal of $\mathbb{R}^2$ and $k:[0,1]^2 \setminus D \to ...
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Hilbert space …

i got the following operator. Let H be a Hilbert space, $(\lambda_n)_{n \in \mathbb{N}}$ a bounded sequence in $\mathbb{K}$ and $T^: H \to H, x \to \sum_{n\in\mathbb{N}} \lambda_n \left<x, ...
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26 views

The Hilbert-Schmidt Theorem for Compact, Self-adjoint operators

Suppose $T$ is a compact, self-adjoint operator on a Hilbert space $\mathcal{H}$. Then there exists an orthonormal set $\{e_n\}_{n=1}^{\infty}$ of eigenvalues of $T$ such that every $x \in ...
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Spectrum of an unbounded operator

Consider a densely defined unbounded operator $A_0:D(A_0)(\subset{H})\to H$ which has the following properties: 1- Symmetric, $\langle A_0x,y\rangle=\langle x,A_0y \rangle$ 2- Positive, $\langle ...
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Show that there is an operator on $H^{2}$ and it's compact.

Let $H^{2}=W_{0}^{2,2}(\Omega)$. Define $(u,v)=\int_{\Omega} (\triangle u\triangle v+2v\triangle u)\mathrm{d}S$ as an inner product on $H^{2}$. Define $a(u;v)=\int_{\Omega} (\nabla u\cdot \nabla ...
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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, ...
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25 views

Problem understanding compact and Fredholm operators

I'm trying to understand the general interaction/duality between Fredholm and compact operators and I ran into the following: Let $L$ be the Laplacian or some elliptic operator on the Sobolev space ...
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9 views

If two compact, positive operators are close, are the projections onto subspaces also close?

Let $H$ be a Hilbert space. Let $a$ and $b$ be compact, positive operators acting on $H$. I wonder if the inequality $$\Vert \Pi_{\ker[a - \lambda_j(a)]}\, -\, \Pi_{\ker[b - \lambda_j(b)]}\Vert\leq ...
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44 views

Compact operators are orthogonally equivalent to a diagonal matrix?

On Brezis's Functional Analysis, the last question of Problem 44 (near the end of the book) reads (modified to include context) Assume that the Hilbert space $H$ is separable and $T\in\mathcal ...
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14 views

$A$ and $A^*$ dissipative implies $D(A) \subset H$ is compact embedding

For selfstudy purpose I want to show the following: $H$ Hilbertspace, $D(A)$ dense subspace of $H$, $A\colon H \supset D(A) \to H$ linear closed dense defined operator. If $A$ and $A^*$ are both ...
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18 views

how is a compact embedding of infinite dimensional Banach spaces possible?

I'm looking at a dense defined closed operator $A\colon H \supset D(A) \to H$ with a Hilbertspace $H$ and $D(A)$ a dense subspace of $H$. In my notes there are some phrase like "if the embedding $D(A) ...
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1answer
32 views

Spectrum is finite or unbounded

Let $X$ be a Banach space and $A:D(A)\subset X\to X$ be a linear and closed operator with $\rho(A)\neq \emptyset$. Suppose that the map $$j:\left(D(A),\|\bullet\|_A\right)\hookrightarrow ...
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43 views

Algebra of compact operators on $\ell_p$

Are the algebras of compact operators $K(\ell_p)$ and $K(\ell_q)$ isomorphic as Banach algebras for $1\leq p<q<\infty$?
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28 views

Is range of completely continuous of bounded set finite dimensional set?

Define of completely continuous operator : $L$ is continuous operator and map bounded set to relatively compact set , then $L$ is completely continuous operator. Let $\Omega$ is a bounded set of ...
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26 views

Multiplication functional $M_f$ compact iff $f$ is in $c_0(\mathbb{N})$

Let $f:\mathbb{N} \to \mathbb{C}$ be a bounded function and let $$ M_f : l^2(\mathbb{N}) \to l^2(\mathbb{N}) \hspace{0.2in} (M_fu)(n):=f(n)u(n) $$ be the correspoding multiplication operator. Show ...
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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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Finite rank operators are dense in compact operators on $L^p(\mu)$

Let $1\le p \le \infty$ and let $(X,\Omega, \mu)$ be a $\sigma$-finite measure space.If $A\in \mathscr{B_0}(L^P(\mu))$, show that there is a sequence $\{A_n\}$ of finite rank operators such that ...
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42 views

necessary and sufficient conditions for certain operators on $C[0,1]$ to be compact

This is an excersice in Conway's 《A Course in Functional Analysis》. Let $\tau:[0,1]\rightarrow [0,1]$ be continuous and define $A:C[0,1]\rightarrow C[0,1]$ by $A(f)=f\circ \tau$. Give necessary and ...
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27 views

Compact operators form the only closed proper ideal of bounded linear operators

I am trying to understand the following proof in Trace Ideals and Their Applications by Barry Simon (Proposition 2.1): Let $\mathcal{J}$ be a two-sided ideal in $\mathcal{L}(\mathcal{H})$ containing ...
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35 views

Adjoint of Compact Operators in Normed spaces

I have a little trouble understanding adjoint operators in spaces without inner products. So the definition of the adjoint operator is the following: Definition (Adjoint operator $T^\times$): ...
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Closure of compact operators?

If $X,Y$ are normed spaces (not nec complete) and $A_n:X\rightarrow{Y}$ bounded linear operators of finite rank converging in the operator norm to $A$ is $A$ compact? A diagonal argument and cauchy ...
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Compact operator in Hilbert spaces reach the maximum in the sphere.

I found the following question in my textbook: (QUESTION) Let $\mathcal{H}$ a Hilbert space and $T: \mathcal{H} \rightarrow \mathcal{H}$ a compact operator. Show that exists $x \neq 0$ in ...
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the c*-algebra generated by the Volterra operator

Let V be the Volterra operator on $\mathscr{L^2(0,1)}$.$V(f)(x)=\int_{0}^{x}{f(y)dy}$. Show that $C^*(V)$, the smallest C* algebra generated with V, is $\mathbb{C}+\mathscr{B_0(L^2(0,1))}$ where ...
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Compact operator with no non-zero eigenvalues is zero?

Suppose we have a Hilbert space $H$ and a compact operator $T$ acting on $H$. If $T$ has no non-zero-eigenvalues, is it necessarily the zero operator? Secondly, if I decompose $H$ into eigenspaces of ...
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Spectrum of operator $T((x_n)_{n\in\mathbb{Z}})=\left(\frac{1}{n^2+1}(x_n-x_{-n})\right)_{n\in\mathbb{Z}}$

The eigenvalues should satisfy: $$T(x_n)=\lambda x_n$$ $$\frac{1}{n^2+1}(x_n-x_{-n})=\lambda x_n$$ $$\left[(n^2+1)\lambda+1\right]x_n=x_{-n}$$ I suppose that this should mean that ...
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Trace-Operator is compact?

Assume $\Omega$ is a smooth, compact riemannian manifold with non-empty smooth boundary $\partial \Omega$. Let $T$ be the Trace-Operator $T \colon H^1(\Omega) \to L^2(\partial \Omega), \ f \mapsto ...
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Bounds on eigenfunctions of integraloperator

Let $K: [0,1]\times[0,1] \to \mathbb{R}$ be a symmetric positive definit and continuous function. It is known my Mercer's theorem that $$ [T_K \varphi](x) =\int_0^1 K(x,s) \varphi(s)\, ds $$ is ...
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Dual operator of a compact operator

Why is the dual operator $A^{\ast}$ of a compact operator $A:X \rightarrow Y$, where $X,Y$ are two Banach spaces again compact?
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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 ...
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Nontrivial closed ideal of $\mathbb{B(H)}$, $\mathbb{H}$ is a non-separable Hilbert space.

$\mathbb{H}$ is a non-separable Hilbert space. Give an example of nontrivial closed ideal $I$of $\mathbb{B(H)}$, that is different from $\mathbb{B_0(H)}$ which is the ideal of compact operators. Any ...
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1answer
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Sequence of operators that commute imply the limit commutes?

Given a sequence of compact operators $A_n\to A$ as $n\ \to \infty$ and $B$ (which has finite rank). $\varphi \in L^2([a,b])$ If $A_nB\varphi = BA_n \varphi$ Am I able to say anything about $A$, ...
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40 views

$T:\ell^2 \to \ell^2$ is a compact operator

Any example in Kreyszig: Introductory Functional Analysis with Applications: Prove compactness of $T:\ell^2 \to \ell^2$ defined by $y=(\eta_j)=Tx$ where $\eta_k=\xi_j/j$ for $j=1,2,\dots$. Defined ...
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1answer
41 views

can a sum of two non-compact operators be compact?

I'm supposed to say, whether an operator $$Tf(t)=f(1-t)$$ can be expressed as $$T=\lambda I-K$$ where $K$ is a compact operator. As $T$ is not compact, I suppose that $K=\lambda I-T$ can't be compact ...
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Modified shift operator is compact.

For the operator $$T(\eta_j) = \frac{\eta_{j+1}}{j}$$ on Hilbert Space $H$ where $(\eta_j)$ is a basis. Show it is compact. Can this work? Define $$f = (\eta_j)_{j \geq 1}$$ $$T_N(f) = ...
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Weak convergence = norm convergence for trace class operators?

Given a (separable) Hilbertspace $H$, I look at the traceclass operators $\mathfrak{S}_1$. I recall the fact that the weak convergence implies norm convergence in the sequence space $\mathcal{l}^1$. ...
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Compact operator for which the image of the closed unit ball is not compact [duplicate]

Do you have an example of a compact operator for which the image of the closed unit ball is not compact?
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27 views

compactness in $\ell^2$

How can I show T is compact when T is defined as $$ \text{T :}\,\ell^2 \to\ell^2\,\text{by Tx=y where} \,y_j=\alpha_jx_j\text{and}\,\alpha_j\to0\,\text{as}\,n\to\infty$$
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An operator satisfying in a sequence of equations

Assume that $H$ is a non-separable Hilbert space. Let $\{\eta_n\}$ be an arbitrary sequence in $H$. Let $\{\zeta_n\}$ be a sequence in $H$ which forms a linearly independent set. Does there exist ...
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2answers
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bounded operator whose image is contained in the image of a compact operator

Let $A,K$ be 2 linear operators from $X$ to $Y$ (Banach spaces). $A$ is bounded and $K$ is compact. If $A(X)\subset K(X)$, is it true that $A$ is alse compact? I know that if the rank of $K$ is ...
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A translation invariant sigma algebra in $B(H)$

Assume that $H$ is a non-separable Hilbert space. Let $s_0$ be the family of all basic neighborhoods in the strong operator topology. We denote $M_s$ by the sigma algebra generated by $s_0$. ...
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54 views

Eigenvalues of an integral operator

The following operator is defined on $L_2(0,1)$: $$Kf(t)=\int_0^1|s-t|f(s)ds$$ I am wondering how I can calculate the eigenvalues and eigenfunctions of such an operator. I start with ...
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27 views

Measure of non-compactness

Can someone give me some simple examples of measure of non-compactness of sets in Banach spaces or metric spaces, which are easy to understand.
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48 views

Compactness of a certain Integral Operator on $L^2$

Let $X\subset \Bbb R^n$ be a compact subset and $\alpha \in (0,n)$. Let $m:X\times X\to \Bbb R$ be a bounded measurable function. Consider the integral operator $$Tf(x)=\int_X ...
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Prove the following are equivalent.

Let $E$ and $F$ be Hilbert spaces. For, $T \in B(E,F)$, show that the following are equivalent: $(i)$ $T$ is compact $(ii)$ $T^*$ is compact $(iii)$ There exists a sequence of finite rank operators ...