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.

learn more… | top users | synonyms

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 ...
0
votes
0answers
28 views

Question about the Morse index of $p_0$ where $f''(p_0)=id-T'(p_0)$

I have this perturbation $f'(x)=x-T(x)$ where $T$ is compact, i have that $p_0$ is non degenerate and i want to see if it's Morse index (i.e. the suprimum of the dimensions of subspaces where ...
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
106 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
56 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
81 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
34 views

The dual of a dual space with the topology of uniform convergence on compact subsets?

$W$ is a Banach space. The topology of $W^*$ is the uniform convergence on the compact subsets of $W$. That is generated by the family of seminorms $$p_K(f)=\sup_{x\in K}|f(x)|,$$ for all compact ...
0
votes
1answer
66 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 $n\to \infty$.
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 ...
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
32 views

Sufficient condition for compact embedding between Banach spaces

Let $(X,\|\cdot\|_X)$ and $(Y, \|\cdot\|_Y)$ be two Banach spaces and $X\subset Y$. Then $X$ is compactly embedded in $Y$, if the unit ball $B_X(0,1)$ in $X$ is a relatively compact subset in $Y$. ...
0
votes
1answer
73 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
50 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
86 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
35 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, ...
1
vote
0answers
42 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. $S_\mu$ ...
0
votes
0answers
16 views

characterisation up to unitary equivalence

My book says that the spectral theorem for compact normal operators characterises compact normal operators up to unitary equivalence. It doesn't expand on this so I was wondering what does this mean ...
7
votes
3answers
122 views

Spectrum of a compact operator

If the spectrum of a compact operator is finite, I don't understand why $0$ has to be a member. I have proved that for all $\epsilon > 0$, there is only a finite number of eigenvectors which have ...
2
votes
1answer
96 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
55 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 $(\psi_v )$ ...
0
votes
0answers
31 views

2 questions on compact operators

Here are 2 facts I think are true and want to prove: 1) If T=LS (composition), S is compact and L commutes with T, then T is compact. My thoughts. Let $(x_{n})$ be a sequence with norm of $x_{n}$ ...
1
vote
3answers
104 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
156 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
80 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
54 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
57 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
203 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
22 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
53 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 ...
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 ...
2
votes
1answer
49 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
46 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
35 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 $T$ ...
2
votes
3answers
339 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
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) := ...
1
vote
1answer
47 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
2answers
79 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?
1
vote
2answers
61 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
83 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
79 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
79 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
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
73 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 ...
2
votes
0answers
18 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
61 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
63 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
65 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
104 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 $T:C^0[0,1]\rightarrow C^1[0,1]$? ...
4
votes
0answers
119 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. ...