4
votes
3answers
83 views

What is a predual of the Banach space of compact operators on $\ell^2$?

I am wondering if the space $K(\ell^2)$ of compact operators on $\ell^2$ can have a predual. Thank you in advance for your help.
0
votes
1answer
23 views

Normal Compact Operator: not diagonalizable!

To proposition 5.17 in Weidmann's 'Lineare Operatoren in Hilberträumen' (german version) it is noted that the expansion of compact operators that are normal rather than self adjoint doesn't apply in ...
2
votes
1answer
59 views

Sum of the Eigenvalues of a Compact Positive-Definite Linear Operator on a Hilbert Space

Let $ A $ be a compact positive-definite linear operator on a Hilbert space $ \mathcal{H} $. Let $ \{ v_{1},v_{2},\ldots,v_{n} \} $ be an orthonormal $ n $-subset of $ \mathcal{H} $. Let $ \lambda_{1} ...
2
votes
1answer
42 views

Question about compact operator

So here is my question, Let $H$ be a Hilbert-space and $K:H\rightarrow H$ a compact operator. I know that if $K$ is self adjoint, then it has one eigenvalue $\mu$ such that $|\mu|=||K||$. Can some ...
2
votes
2answers
35 views

Fredholm Index: Finite Corank $\Rightarrow$Closed Range [duplicate]

Obviously closed subspaces turn quotient spaces into normed spaces rather than just merely vector spaces. However the dimension involved in Freholm's index are purely algebraic. Why do we thus ...
1
vote
1answer
40 views

Showing that a certain operator is compact

So here is my problem, I try to show that following operator is compact, \begin{align} J: h_1 & \rightarrow\ell^1 \\ (x_n) & \mapsto(x_n) \end{align} where $$h_1:=\left\{x_n\in ...
4
votes
2answers
61 views

Question about a counterexample concerning compact operators

Does anybody know if the following is true, Let $H$ be an infinite dimensional Hilbert-space and $K:H\rightarrow H$ a compact operator. Then if $|\mathrm{spec}(K)|<\infty$ i.e the spectrum is ...
0
votes
1answer
32 views

Question if an operator is compact

So here is my problem, Let $$J_p:\ell^p\rightarrow c_o$$ be the canonical embedding where $c_0:=\{x_n\subseteq\mathbb C:x_n\rightarrow 0\quad n \rightarrow\infty\}$. I have to decide whether the ...
3
votes
1answer
23 views

Question about an integral operator

So here is my question, I know that the operator $$T:L^2[0,1]\rightarrow L^2[0,1]$$ $$f\mapsto(Kf)(x)=\int_{[0,1]}k(x,y)f(y)\;dy$$ for a function $k$ continuous on $[0,1]^2$ is compact. Is this also ...
4
votes
1answer
78 views

What makes compact operators special?

I would like to understand why compact operators are considered so special to consider them as an extra class of operators. Over Hilbert spaces these (as far as I know) these are the ones with ...
0
votes
1answer
46 views

Is operator $T_n$ a compact operator?

Is operator $T_n$ a compact operator? $$T_n:l_2\rightarrow l_2$$ $$T_nx=(\underbrace{0,0,\ldots,0,}_{n\text{ zeros}}, x_1,x_2,x_3,\ldots)\text{ where }x=(x_1,x_2,x_3,\ldots)\in l_2,\ ...
2
votes
1answer
50 views

Two questions about a proof of the compactness of an operator

There are a few things that I don't understand about a proof and I'd appreciate any help. The theorem and its proof are the following: (1) Is the equality $$ \|v(\tau) -v(\tau_j)\| = \max_{1 \le ...
1
vote
1answer
33 views

Problems proving that a compact operator is completely continuous [duplicate]

I would like to prove that if $T:X\rightarrow Y$ is a compact operator, then for every weak convergent sequence $(x_n)_{n\in\mathbb N}$ with $x_n\rightharpoonup x$ for some $x\in X$ it follows that ...
3
votes
1answer
52 views

Compact operator as a limit of finite ranked operators

So here is my question, I had to show that the following operator is compact, $$T:C[0,1]\rightarrow C[0,1]$$ $$f\mapsto\int_0^tf(s)ds$$ with $||f||=\mathrm{sup}_{x\in[0,1]}|f(x)|$ I think I ...
1
vote
0answers
46 views

Question about compact operators

I would like to prove the following, Let $X$,$Y$ be infinite dimensional Banach-Spaces and $T$ a compact, linear and bounded operator. Then there exists a sequence $(x_n)_{n\in\mathbb N}$ with ...
0
votes
1answer
37 views

About the Volterra operator and the approximation property

I need some help with these questions. $\bullet\;$ First of all, if we define the Volterra operator $V:L^{1}[0,2\pi]\rightarrow L^{1}[0,2\pi]$ as $(Vf)(x)=\int_0^xf(t)dt$, Is this operator compact? ...
0
votes
0answers
53 views

Volterra operator with continuously differentiable Kernel has no Eigenvalue

First I'll describe the entire question, as it stated in the exercise: let $K(t,s)\in C([0,1]^2$), continuously differentiable in the first coordinate (meaning $K_t(t,s)\in C([0,1]^2$). And let ...
5
votes
4answers
226 views

is $T$ compact operator?

is $T$ compact operator? $T:C[0,1]\rightarrow C[0,1]$: $x(t)\mapsto x(t^2)$ where $t\in[0,1]$ with supremum norm Could you please help.
6
votes
3answers
152 views

How to figure whether it is a compact operator

How to figure whether it is a compact operator: $$T:C[0,1]\rightarrow C[0,1] $$ $C[0,1]$:the space of all continous function on [0,1] with supremum norm $$(Tx)(t)=\int^t_0 x(s)ds, \ \ \forall ...
2
votes
1answer
63 views

How to find if it is a compact operator

How to find if it is a compact operator: $F\colon C[0,1]\rightarrow C[0,1]$ : $x(t)\mapsto \int^1_0 \cos(t^2+s^2)x(s)ds$ Could you please help with this question.
1
vote
0answers
35 views

If an upper semicontinuous multivalued map is compact on a set, is it compact on the boundary as well?

I have stumbled upon the following problem during my research: Let $X$ and $Y$ be Banach spaces, $K\subset X$ nonempty, $F:\overline{K}\rightarrow 2^{Y}$ an upper semicontinuous multivalued map with ...
0
votes
2answers
85 views

Compact operator whose range is not closed

I am asked to find a compact operator (on a Hilbert space) whose range is not closed, but I am having trouble coming up with one. My guess is that you need to have some sequence in the range that ...
1
vote
3answers
39 views

Examples of spectrum of compact operators on the sequence space $l_2$

Suppose $T$ is a compact operator on the sequence space $l_2$, and let $\sigma(T)$ be its spectrum. Is it possible to find a $T \ne 0$ such that $\sigma(T) = \{0\}$? Also, is it possible to find $T$ ...
1
vote
1answer
36 views

Show compactness of an operator with Arzelà–Ascoli

We have $K\colon L^{2}(a,b) \rightarrow L^{2}(a,b)$ such that $ Kf(t)=\sum_{j=1}^{n}\phi_{j}(t) \int_{a}^{b} \psi_{j}(S) f(s)ds$ where $\phi_{j} ,\psi_{j} \in L^{2}(a,b)$. We want to show that K is ...
1
vote
0answers
76 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 ...
2
votes
2answers
103 views

Proof of equivalent characterizations of compact operators

As an exercise I tried to prove the following theorem: If $X,Y$ are Banach spaces and $u \in B(X,Y)$ is a bounded linear operator then the following are equivalent: (1) $u$ is compact ...
1
vote
0answers
23 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 ...
2
votes
1answer
45 views

Equivalent conditions for composition to be compact operator

I did some exercises in Conway's functional analysis book and found the following problem: Let $\tau:[0,1]\to [0,1]$ be continuous and define $A:C[0,1]\to C[0,1]$ by $Af:= f\circ \tau$. Give ...
1
vote
0answers
22 views

Schauder's theorem: consequences and applications

I am about to give an informal talk about Schauder's theorem ($T:X\to Y$ linear operator between Banach spaces is compact if and only if its adjoint is). Does anyone know any derived ...
3
votes
1answer
43 views

Why is this estimate using a compact embedding in a sobolev space true?

Let $\Omega\subset\mathbb{R}^3$ be a bounded Lipschitz-domain. We then have, for $s\in[1,6)$ the compact embedding $H^1(\Omega)\stackrel{c}{\hookrightarrow}L^s(\Omega)$ ensuring the existence of a ...
1
vote
1answer
36 views

I want to show that some subset of $C([0,1])$ is equicontinous

First why the problem appeard. I want to show that the linear and continuous operator $T:C([0,1])\rightarrow C([0,1])$ , $ (Tf)(t)=\int_{[0,1]}k(t,s)f(s)ds$ where $k:[0,1]^2\rightarrow\mathbb R$ is ...
1
vote
1answer
94 views

Show the Volterra Operator is compact using only the definition of compact

The Volterra operator $V:L^{2}[0,1]\rightarrow L^{2}[0,1]$ is defined by $(Vf)(x)=\int_0^xf(t)dt$. I am wondering if it can be shown that $V$ is compact by definition - that is, either that $V$ ...
0
votes
0answers
55 views

Show these operators converge to a particular limit

Let $H$ be a Hilbert space, and $T$ be a operator on $H$ of the form $T=\sum_{n=1}^{\infty}{\lambda}_{n}<x,e_{n}>e_{n}$ where $e_{n}$ are the eigenvectors of $T$ and an orthonormal basis of H ...
3
votes
2answers
79 views

Subspaces in the image of compact operator

Let $X$ and $Y$ be some infinite dimensional Banach spaces. Let $T:X\longrightarrow Y$ be some compact linear operator. It is easy to understand that $T$ cannot be surjective: the Open Mapping Theorem ...
0
votes
1answer
64 views

The eigenvalues of a compact and self-adjoint operator on Hilbert space

Show that if $K$ is a compact self-adjoint operator on Hilbert space then it has either finitely many eigenvalues or a sequence of eigenvalues $\lambda_n\to 0$ as $n\to \infty$.
0
votes
1answer
23 views

Finite dimension and total boundedness

Let $T:X\to Y$ be a bounded operator between Banach spaces $X$ and $Y$. Assume that for any $\epsilon >0$ there is a finite-dimensional subspace $Y_\epsilon\subset Y$ so that $\|Q_\epsilon ...
0
votes
1answer
66 views

Does Hilbert Transform commute with Function Multiplication modulo Compact on $L^p(R)$?

Define Hilbert Transform (HT) as the convolution with the function $1/x$. E. Stein proves in his book Singular Integrals and Differentiability Properties of Functions that HT, when understood as a ...
2
votes
1answer
49 views

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, ...
0
votes
2answers
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 ...
1
vote
1answer
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, ...
2
votes
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.
0
votes
1answer
52 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 ...
0
votes
1answer
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 $(\psi_v )$ ...
1
vote
3answers
101 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 ...
7
votes
1answer
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 ...
3
votes
1answer
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 ...
1
vote
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 ...
0
votes
1answer
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 ...
10
votes
1answer
201 views

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 ...
0
votes
1answer
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.