Tagged Questions
1
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1answer
41 views
About compact operator
When seeing a proof of Fredholm's alternative I don't get the following:
Let $T$ be compact from a Banach space $X$ to itself, and $\lambda \neq 0$. Define $S=I-T$, $S^k$ its $k$-th power and ...
1
vote
1answer
54 views
Is restriction of compact operator always compact?
Question 1: Is a compact operator $T:X\to Y$ 's restriction on a subspace $Z\subset X$ still compact? (I think I've got the answer)
I think the compact operator's restriction on any subspace must ...
1
vote
1answer
56 views
Showing that the inverse of the perturbation of a compact operator by a bounded operator remains compact.
The title says it all. If we have a Hilbert space $H$, then if $B\in \mathcal B(H)$, $L$ is a linear operator that is not necessarily bounded, $L^{-1}$ is compact, and $0\in \rho(L)\cap\rho(L+B)$, ...
1
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1answer
37 views
The set of compact linear operators is a subspace of the set of bounded linear operators
I know that a linear operator $T:X \to Y$ (where $X$ and $Y$ are normed vector spaces) is compact if for every sequence
$\left(x_{n}\right)\subseteq X$ s.t. $\left\Vert x_{n}\right\Vert \leq C$,
the ...
2
votes
1answer
62 views
Show for compact operator $K$, if $||Kf|| < ||f|| \forall f$, then $||K|| < 1$.
I wanted to check my reasoning on proving this statement, and see if anyone had suggestions for other proofs of this fact.
Again, the statement is, if $K$ is a compact operator on a Hilbert space ...
1
vote
1answer
31 views
Strong operator convergence and adjoint operator
Let $H$ be a Hilbert space and $(T_n)_{n \in \mathbb{N}}$ be a sequence of bounded linear operators on $H$.
The strong convergence of $T_n$ doesn't imply the strong convergence of $T_n^*$, i.e.
...
3
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1answer
58 views
Invertibility of compact operators
I'm a little confused about compact operators and whether or not they are invertible. Just hoping someone here can clear up my confusion:
Let $T$ be a compact operator on a Banach space $X$. Since ...
3
votes
1answer
64 views
$K$ is a linear compact operator on Hilbert space $H$. Will the image of $I-K$ on every closed subspace of $H$ be also closed?
Just as the title. We know the image of $I-K$ is closed, but if we restrict $H$ to a closed subspace $V$, will $(I-K)(V)$ be a closed subspace of $H$? Any hint is appreciated.
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0answers
46 views
Hilbert Schmidt decomposition
Usually, for example in Reed and Simon, the Hilbert Schmidt (singular value) decomposition of a compact operator $T$ on a Hilbert Space is written as
$$T = \sum_{n=1}^{N} \lambda_n ...
3
votes
1answer
114 views
Spectral theorem of compact operators in Hilbert space
I am reading the following theorem from my lecture notes (English translation of German text). But I don't understand exactly what is meant from this theorem and the proof.
Theorem.
Let $H$ ...
5
votes
1answer
175 views
Bounded operator and Compactness problem
Let $H$ be a Hilbert space with orthonormal basis $(e_{n})_{n\in\mathbb{N}}$. Furthermore, let $T\colon H\rightarrow C[a,b]$ be a bounded operator.
a) Let $x\in [a,b]$. Show that there is a ...
2
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0answers
20 views
Trying to show that operator $T((x_n))=(2^{-n}x_n)$ is compact. [duplicate]
Consider $T\colon\ell^2\to\ell^2$ an operator such that
$$T((x_n))=(2^{-n}x_n); \forall x=(x_n)\in \ell^2 $$
Does anyone know how to prove that it is compact?
I understand that a linear operator ...
2
votes
1answer
54 views
How to prove the compactness of this Sobolev embedding?
I have a question on compactness of the following Sobolev embedding.
Let $W^{1,p}([0,1],\mathbb{R}^n)$ be the Sobolev space of functions $u:[0,1]\rightarrow \mathbb{R}^n$ for $1<p<\infty$. How ...
4
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0answers
36 views
Fredholm alternative for polynomially compact operators
Let $T \in \mathcal{L}(E)$ be a polynomially compact operator, i.e., there exists a polynomial $p$ such that $p(T)$ is a compact operator. Suppose $p(1) \neq 0$. I want to show that $N(I-T) = \{0\} ...
0
votes
3answers
94 views
Trying to prove that operator is compact
Consider $T\colon\ell^2\to\ell^2$ an operator such that
$$T((x_n))=(2^{-n}x_n); \forall x=(x_n)\in \ell^2 $$
Does anyone know how to prove that it is compact?
I understand that I have to find a ...
1
vote
0answers
77 views
Proof of compactness of bounded linear operator
Define $T: l^2 \to l^2$ by $Tx = y =(\eta_j)$, where $x = (\xi_j)$ and
$$
\eta_j = \sum_{k=1}^{\infty} \alpha_{jk}\xi_k, \quad \quad \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} |\alpha_{jk}|^2 < ...
2
votes
0answers
129 views
Fredholm and Compact Operators
Let $X$ and $Y$ be Banach spaces and $T\in B(X,Y)$ be Fredholm. Then there is $S\in B(Y,X)$ such that $ST=I+K_{1}$ and $TS=I+K_{2}$ where $K_{1},K_{2}$ are compact operators.Proof: Since $T$ is ...
1
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0answers
53 views
If limit of $f(n)$ is zero then the operator is compact
I want to prove the following:
Suppose $\mathfrak{H}$ is the Hilbertspace $l^2(\Bbb{N})$ and $T_f$ the multiplication operator on $\mathfrak{H}$, thus $T_f\psi=f\psi$ for ...
0
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0answers
54 views
Why is this a compact operator?
I would like to know a proof of the following result:
Let $K: \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R} $ be an integral kernel such that there is an $\epsilon >0$ which fulfills
...
1
vote
1answer
73 views
Spectral radius of an operator .
I would like to know the spectral radius of $$T_k x (t)= \int_0^t k(t,s) x(s) ds$$
where $T_k$ is a map from $C[0,1] \to C[0,1]$ and $k(x,y)\colon [0,1]^2 \to \mathbb C$ is continuous.
And also ...
3
votes
1answer
148 views
Eigenvalues integral operator - general case
Let $T$ be an integral operator on $L^2([0,1])$, such that:
$$
(Tf)(x) = \int_0^1K(x,y)f(y)dy,
$$
with $K(x,y): [0,1]^2 \rightarrow \mathbb{R}$ continuous and $K(x,y) = K(y,x)$, $K(x,y)\geq0$ $ ...
0
votes
0answers
59 views
Diagonal of Hilbert--Schmidt operator
Suppose $U\in HS$, Hilbert--Schmidt operators on $V = L^2(\mathbb{R}^n)$. There is a natural isomorphism between HS operators and elements in $W = L^2(\mathbb{R}^n\times\mathbb{R}^n)$, in particular, ...
3
votes
2answers
96 views
Is $\ell^p \mathbb N \subset \ell^q \mathbb N$ inclusion compact ?
This question just struck me, is it true that if $1 <p <q <\infty$ , is the inclusion map
$$\ell^p \mathbb N \subset \ell^q \mathbb N$$compact ?
Hรถlders inequality gives us that the ...
3
votes
1answer
235 views
Is the inclusion $C^1[0,1]\subset C[0,1]$ compact?
I am working on this problem but i couldn't succeed .
Consider the space $C^1[0,1]$ with the norm $$\|f\|=\max \{\|f\|_{C[0,1]}, \|f'\|_{C[0,1]}\},$$
I don't know if the inclusion map is compact, ...
5
votes
2answers
160 views
Compact operators: why is the image of the unit ball only assumed to be relatively compact?
Recall the definition of compact operators between Hilbert spaces:
An operator $A$ is called compact if the image $A(\mathcal U_H)$ of the unit ball is relatively compact (i.e. its closure is ...
0
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0answers
34 views
Spectrum of weighted shift operator [duplicate]
Possible Duplicate:
Compact operator? self adjoint operator? Stirlingโs formula
Let $H$ be a Hilbert space and let $\{e_n, n \geq\}$ be an orthonormal basis in H.
Let $T \in B(H)$ be the ...
3
votes
1answer
186 views
Compactness and spectrum of integral operator
Show that the operator $C: L^2([0,1]) \rightarrow L^2([0,1])$ defined by
$$Cf(x) = \int_0^x\int_1^tf(s)dsdt$$
is compact and determine its spectrum.
Im not sure how to find the spectrum when we are ...
6
votes
1answer
148 views
Trace class for operators
Let $ \mathcal{H} $ be a Hilbert space and $ T: \mathcal{H} \to \mathcal{H} $ a bounded linear operator. The $ n $-th singular number $ {\mu_{n}}(T) $ of $ T $ is defined as the distance from $ T $ ...
3
votes
1answer
83 views
Does there exist a diagonal dominance concept for integral kernels?
A self-adjoint diagonally dominant square matrix $M$ with nonnegative diagonal is positive semi-definite. Does there exist a similar concept for integration kernels that define compact operators over, ...
4
votes
0answers
71 views
The control of norm in quotient algebra
Let $B_1,B_2$ be two Banach spaces and $L(B_i,B_j),K(B_i,B_j)(i,j=1,2)$ spaces of bounded and compact linear operator between them respectively. If $T \in L(B_1,B_1)$, we have a $S \in K(B_1,B_2)$ and ...
3
votes
2answers
204 views
Show that a finite-dimensional Banach space has a bijective compact operator
It is clear that if $ T: X \rightarrow X $ is a bijective compact operator, where $ X $ is a Banach space, then $ \dim(\text{Range}(T)) = \dim(X) $, which implies that $ \dim(X) $ must be $ < ...
2
votes
3answers
138 views
Compact integral and multiplication operator in Banach spaces
Let $ A\colon C[0,1] \to C[0,1] $
$$ A(x)(t) = f(t)x(t) + \int_0^t x(s)ds,\quad f \in C[0,1]: f(1) \neq 0, \forall t \in [0,1] $$
Is $A$ a compact operator or not?
3
votes
1answer
201 views
Operators on $C([0,1])$ that is compact or not.
For $f\in C([0,1])$ set
$$Hf(x) = \frac{1}{x}\int_0^x f(t)dt.$$
a) Show that $H$ is a bounded operator from $C([0,1])$ into itself which is not compact.
b) From a) it follows that $H$ induces a ...
1
vote
1answer
78 views
Symmetric bounded linear maps can be approximated by compact symmetric linear maps.
Let $H$ be a separable Hilbert space and let $T:H \rightarrow H$ be a symmetric bound linear map.
a) Show that for every orthogonal projection $P$ on $H$ ($P' = P$, $P^2 = P$) PTP is symmetric.
b) ...
2
votes
1answer
100 views
Determine the operator T in a Hilbert space
Let $H$ be a Hilbert space and let $\{e_n, n \geq 1\}$ be an orthonormal basis for $H$.
a) Determine the operator $T\in B(H)$ that satisfies
$$ Te_1 = 0,\; Te_n = \frac{1}{n}e_{n-1}, n ...
3
votes
1answer
90 views
Eigenvalues of Hilbert-Schmith operator
I am having trouble determining the eigenvalues and eigenvectors of the operator $Kv(x)= \int_0^1((x+t)v(t)dt$, where the kernel is $k=x+t$. I have tried to solve the equation $Kv(x)=\lambda v(x)$, ...
5
votes
1answer
121 views
Show $T$ is compact
$H$ and $K$ are Hilbert Spaces, $(u_n)$ and $(v_n)$ are sequences in $H$ and $K$ respectively. $\sum_{n=1}^{n=\infty} \|u_n\|\|v_n\| $ converges.
$T\colon H\rightarrow K$ is defined by ...
3
votes
2answers
116 views
Proof that certain operators are compact
I want to examine which of the following operators $T \colon C[0,1] \to C[0,1]$. are compact, by some I think I got the argument, but others I have no idea.
a) $Tx(t) = x(t^2)$
Guess it is ...
0
votes
1answer
137 views
Hilbert space the trace
I need help from someone to solve this problem.
Given a bounded sequence $(\lambda_n)$ in $\mathbb ะก$ define an operator $S$ in $B(\ell_2)$ by $S(x_1) = 0$ and
$S(x_n) = \lambda_n x_{n-1}$ , ...
1
vote
0answers
49 views
Containment of an element to an operator system
This question will probably appeal to people in operator systems theory as it is very much related. However, I'm interested in down-to-earth concrete systems with finite dimensional Hilbert space ...
1
vote
1answer
161 views
No Nonzero multiplication operator is compact
Let $f,g \in L^2[0,1]$, multiplication operator $M_g:L^2[0,1] \rightarrow L^2[0,1]$ is defined by $M_g(f(x))=g(x)f(x)$. Would you help me to prove that no nonzero multiplication operator on $L^2[0,1]$ ...
3
votes
1answer
196 views
eigenvalue question
I think this question isn't that hard, but I am a bit confused.
Define the linear operator $T_k:H\mapsto H$ by
\begin{align}
T_ku=\sum^\infty_{n=1}\frac{1}{n^3}\langle u,e_n\rangle e_n+k\langle ...
2
votes
1answer
82 views
Compact operator on $l^2$
Let A be a bounded linear operator on $l^2$ defined by A($a_n$)=($\frac{1}{n} a_n$). Would you help me to prove that A is compact operator. I guess the answer using an approximation by a sequences of ...
0
votes
1answer
61 views
Compact Operator defined by inner product
Let $H$ be a Hilbert space and $y,z \in H$. Define bounded linear operator $Ax=\langle x,y\rangle z$ where $\langle,\rangle$ is inner product. Would you help me to prove that $A$ is compact operator.
1
vote
2answers
91 views
Proof of the Pitt's theorem
I'm reading the book Topics in Banach Space Theory by Albiac F. Kalton N. J. I got stuck at the proof of the Pitt's theorem.
In the second paragraphs authors tries to prove ad absurdum that for ...
5
votes
2answers
114 views
Behaviour of the spectrum of a compact operator w.r.t. perturbations.
Suppose $A$ and $B$ are linear compact operators on a Hilbert space with $\sigma(A)$ and $\sigma(B)$ as their spectrum.
Is it possible to obtain some continuity result of $\sigma(A+\epsilon B)$ as ...
1
vote
1answer
131 views
Arzela-Ascoli on space $C_0(X)$ (vanishing at infinity), $X$ locally compact
Does Arzela-Ascoli hold in $C_0(X)$ (vanishing at infinity)?... namely a subset of $C_0(X)$ is relatively compact iff it is equicontinuous and bounded. In particular, if $C_0^1(X)$ is the Banach ...
5
votes
1answer
454 views
Compact operators and completely continuous operators
A compact operator between Banach spaces is an operator that maps bounded sets into relatively compact sets, while a completely continuous operator maps all weakly convergent sequences into convergent ...
3
votes
1answer
186 views
Compact operator and compact embeddings
Here, all spaces are Banach spaces.
Definition: A map $S:X \to X$ is compact if for every bounded sequence $\{u_n\}$, there exists a subsequence $\{u_{n_k}\}$ such that $\{S(u_{n_k})\}$ converges in ...
2
votes
3answers
241 views
Compact operators and uniform convergence
Suppose $T: H \rightarrow H$ is a compact operator, $H$ is a Hilbert space, and let $(A_n)$ be a sequence of bounded linear operators on $H$ converging strongly to $A$. Show that $A_nT$ converges in ...
