0
votes
1answer
32 views

Is the operator compact? [closed]

Suppose the the space C([0,$\pi$]) is endowed with the norm $\|u\| = \sup \|u(x)\|$. Consider the linear operator $K : C([0,\pi]) \rightarrow C([0,\pi])$ given by $$(Ku)(x) = \int_0^\pi(\sin t + ...
1
vote
0answers
12 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
32 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
16 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
33 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
29 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
49 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
51 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
52 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
53 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
22 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
48 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
39 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
79 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
28 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
82 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
40 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
33 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
60 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
138 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
57 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
41 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
36 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
169 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
18 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.
1
vote
2answers
52 views

Compactness of the solution operator

Let $\Omega$ be a smooth open bounded subset of $\mathbb{R}^n$. The bilinear form $$a(u,v)=\int_{\Omega}\frac{\partial u}{\partial x}\frac{\partial v}{\partial x}dx$$ is elliptic on ...
2
votes
1answer
43 views

compact operators and finite dimentional spaces

Let $Q_n$ a finite dimentional space. Since any finite rank operator is compact, it's true that any linear operator $K:Q_n\to Q_n$ is compact 'cause $\dim(R(K))<\infty$?
0
votes
1answer
39 views

Operator from $\ell_{4}$ to $\ell_{1}$ is compact, if it's continuous.

Define $T: \ell_{4} \rightarrow \ell_{1}$ as $Tx=(a_1x_1, a_2x_2, \ldots)$. I showed that $T$ is continuous if and only if $\sum \left| a_i \right|^{\frac{4}{3}} < \infty$. How can I prove that if ...
1
vote
1answer
32 views

Is operator from $X^{*}$ to $c_{0}$ compact

$X$ - Banach space. Let $X \ni (x_n)$ be such that $x_n \stackrel{\omega}{\rightarrow}0 $ weakly. Define $T: X^{*} \rightarrow c_0$ by $T=\left(x^{*}(x_n) \right)_{n=1}^{\infty}$. Is operator $T$ ...
1
vote
3answers
157 views

Compact operator with closed range has finite dimensional range

Let $X,Y$ be Banach Spaces, and let $T\in K(X,Y)$ be a compact operator from $X$ to $Y$. I have to prove that $T(X)$ is closed in Y if, and only if, $\dim(T(X))<\infty$. Can anybody help me with ...
1
vote
0answers
34 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) := ...
1
vote
1answer
43 views

Is the canonical injection between $ C^{k+1}[0, 1] $ and $ C^k [0, 1] $ compact?

If $ k=0 $ by the Ascoli - Arzelá theorem the answer is yes, but i don't know how to proceed in the general case ($ k> 0, k \in \mathbb{N} $). I tried to build a counter example using Riesz lemma ...
2
votes
1answer
69 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
46 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
72 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 ...
2
votes
1answer
63 views

Arzela-Ascoli and adjoint of compact operator compact

I have seen in this thread a nice answer where it is shown that Thread that the adjoint operator of a compact operator is compact by using the Arzela Ascoli theorem. Unfortunately, there is one thing ...
4
votes
3answers
75 views

$A^2$ self-adjoint and Compact, prove $A$ has an eigenvalue

Suppose $H$ is a Hilbert space and $A \in L(H)$ is such that $A^2$ is compact and self-adjoint. Prove that $A$ has an eigenvalue. (Here $L(H)$ is the set of bounded linear operators on a Hilbert ...
0
votes
1answer
49 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
56 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 ...
1
vote
0answers
17 views

Sufficient and necessary conditions for a given operator $T$ to be $T \in \mathcal{B}$, $T \in \mathcal{K}$ or $T \in \mathcal{F}$

I was rewiewing for an upcoming exam and found this problem. It is quite basic and straightforward but still causing me some confusion mainly since I would have to justify my answer. Let a linear ...
1
vote
1answer
56 views

How do I prove that this operator is compact. Functional analysis

We have $$K:l_2\to l_2$$ $$Kx=\left(x_1, \frac{1}{2}x_2,\frac{1}{3}x_3,...\right)$$ where $x=(x_1,x_2,x_3,...)\in l_2$. I want to use the definition of a compact operator: $K$ is compact if ...
2
votes
2answers
61 views

Functional analysis, help to show a short result

The following problem is from the theory of compact operators: Suppose $X,Y$ are normed spaces and $T:X\to Y$ is linear. Show that if $T$ is compact and invertible then $\mbox{dim}(X)$ and ...
3
votes
2answers
52 views

Is this operator compact and how do I prove it? [duplicate]

I have a very big problem with the following question: Is the operator $T$ defined by $(Tx)t=tx(t)$, $(0<t<1)$ compact in $L_2(0,1)$? My guess is no and I've tried 3 different approaches to ...
3
votes
1answer
69 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 $T:C^0[0,1]\rightarrow C^1[0,1]$? ...
3
votes
0answers
111 views

Non linear compact map

Suppose to have two Banach spaces $E$ and $F$, with $E$ reflexive. Suppose to have a continuous map $T:E \to F$ which maps bounded subsets into precompact subsets. $T$ is not assumed to be linear. ...
3
votes
1answer
89 views

Completely continuous implies compact under separability assumption

My professor left for self-convincing the following statement: "If $T: X \to Y$ with $X$ Banach is completely continuous (that is, takes $w$-convergent sequences to norm convergent sequences) and $X$ ...
0
votes
1answer
51 views

Compact Operators and Complete Metrics Spaces

I have a couple of questions about compact operators and compactness in complete metric spaces: 1.I have the following implications: Let $Y$ be a metric space with $A$ a subset of $Y$. $A$ is ...
1
vote
0answers
43 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
1answer
79 views

If E is a Hilbert space and $T \in B(E)$ is compact, show $T(E)$ does not contain a closed infinite dimensional subspace

It's the problem from "Essential Results of Functional Analysis," R.J. Zimmer, Chapter 3, problem 3.1. I try to prove this problem and I am confused with the condition "closed infinite dimensional." ...
1
vote
1answer
41 views

a compact operator on $l^2$ defined by an infinite matrix

Let $A$ be an infinite matrix such that $\displaystyle \sum_{i,j}|a_{i,j}|^2<\infty$. Then $A$ defined a compact operator on $l^2$.