0
votes
1answer
27 views

Show this integral operator is compact for various values of $\alpha$

I am having some problems evaluating a multivariable integral. This question is features in Stakgold's book Green's functions and boundary value problems. page 359. Consider the kernel for $a\leq ...
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.
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 ...
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 ...
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
36 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? ...
5
votes
4answers
225 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.
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
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 $T$ ...
3
votes
2answers
73 views

Compact operator on $L^{2}$

Let $K(t,s)$ be a real-valued function of two real variables, and let $T: L^2(a,b) \to L^2(a,b)$ be defined by $(Tf)(t) = \int_{a}^{b} K(t,s) f(s) ds$ where $$K(t,s) = \sum_{j=1}^{n} \phi_{j}(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
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 ...
2
votes
0answers
84 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 $L^2(X,d\mu)$.I want to see how its eigenvalues or singular values change as its kernel is ...
1
vote
0answers
41 views

Operators on a Hilbert space question

For a Borel measure $\mu$ define $\langle S_\mu x,y\rangle=\int_H\langle x,z\rangle \langle y,z\rangle \mu(z)$. An exercise in my book that I am reading says that I could find a $\mu$ s.t. $S_\mu$ ...
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 ...
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 ...
3
votes
2answers
54 views

Distance between unilateral shift and compact operator

We have $S\in\mathbb{B}(\mathcal{H})$ (where $\mathbb{B}(\mathcal{H})$ is algebra of bounded linear operators in Hilbert space) and $S$ is unilateral shift. Compute ...
1
vote
0answers
43 views

properties of integral operator $x^{-1}\int_0^xf(x,y)v(y)dy $

here we have two cases to study $(1)$ let us fix any $f \in C^{1}[ [0,1] \times [0,1]]$ ($k \neq 0$). Set $$[T(v)](x) := x^{-1}\int_0^xf(x,y)v(y)dy $$ for any $x \neq 0$ otherwise $[T(v)](0) := ...
2
votes
2answers
78 views

restriction a non compact operator to compact operator

If $T\in\mathcal{B}(X,Y)$ is not compact can the restriction of $T$ to an infinite dimensional subspace of $X$ be compact?
0
votes
2answers
57 views

Locally compact operators and their spectrum

At the moment, I'm studying the book "Introduction to Spectral Theory" from P.D. Hislop and I.M. Sigal, I arrived at chapter 10 and I'm stuck on two problems there. Problem 10.1: Let $A$ be a closed ...
-1
votes
2answers
82 views

Normal compact operator commute with bounded self adjoint operator in Hilbert space.

Suppose $H$ is a Hilbert space and $A:H\rightarrow H$ is a normal compact operator such that $\ker(A)=0$. show that if $B$ is a bounded self adjoint operator that commutes with $A$ then the spaces in ...
0
votes
1answer
59 views

Interpretation of Fredholm Alternative with respect to PDEs

I have studied the Fredholm Alternative, which states the following: Theorem: Let $K:H \rightarrow H$ be a compact linear operator and let $I$ be the identity operator on $H$. Then: 1.$N(I-K)$ is ...
3
votes
1answer
72 views

Fredholm alternative and orthonormal basis

The following question relates to the Fredholm alternative: Let $K:H \rightarrow H$ be a compact linear operator and let $I$ be the identity operator. Notation: $N$ is the nullspace and $R$ is the ...
3
votes
1answer
98 views

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 $T:C^0[0,1]\rightarrow C^1[0,1]$? ...
1
vote
0answers
46 views

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 $$\cal{K(\cal{H})}\otimes\cal{K}(\cal{H})\cong\cal{K}(\cal{H}\otimes H).$$ ...
2
votes
3answers
229 views

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 ...
0
votes
2answers
69 views

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 ...
0
votes
1answer
62 views

Why is the set of compact operators closed in the space of all bounded operators between Banach spaces?

Let $X$ and $Y$ be Banach space. $B(X,Y)$ is the vector space of all bounded linear maps from $X$ to $Y$. Also, $K(X,Y)$ is the set of all compact operators from $X$ to $Y$. Why is $K(X ,Y)$ ...
6
votes
1answer
161 views

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 ...
1
vote
1answer
54 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. ...
1
vote
0answers
63 views

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 $T \in \mathcal L(X)$ ...
3
votes
1answer
89 views

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 ...
1
vote
1answer
125 views

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, ...
1
vote
0answers
132 views

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 ...
2
votes
1answer
48 views

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 $.
5
votes
1answer
97 views

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) ...
1
vote
1answer
81 views

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{% ...
1
vote
1answer
401 views

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 ...
0
votes
2answers
52 views

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. ...