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.
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Showing that smoothing operators are compact
Suppose I have a bounded, linear map $T: H^1(X) \to H^1(X)$ such that $T(H^1(X)) \subset C^\infty(X)$. Is $T$ a compact operator?
I'm guessing this depends on whether or not $X$ is (pre)compact, and ...
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1answer
52 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 ...
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1answer
55 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 $0\in \rho(L)\cap\rho(L+B)$, ...
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1answer
36 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 ...
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61 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 ...
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1answer
28 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.
...
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1answer
56 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 ...
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63 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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45 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
$$T = \sum_{n=1}^{N} \lambda_n ...
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109 views
Spectral theorem of compact operators in Hilbert space
I am reading the following theorem from my lecture notes (English translation of German text). But I don't understand exactly what is meant from this theorem and the proof.
Theorem.
Let $H$ ...
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172 views
Bounded operator and Compactness problem
Let $H$ be a Hilbert space with orthonormal basis $(e_{n})_{n\in\mathbb{N}}$. Furthermore, let $T\colon H\rightarrow C[a,b]$ be a bounded operator.
a) Let $x\in [a,b]$. Show that there is a ...
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Trying to show that operator $T((x_n))=(2^{-n}x_n)$ is compact. [duplicate]
Consider $T\colon\ell^2\to\ell^2$ an operator such that
$$T((x_n))=(2^{-n}x_n); \forall x=(x_n)\in \ell^2 $$
Does anyone know how to prove that it is compact?
I understand that a linear operator ...
3
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1answer
75 views
$W^{1,p}$ compact in $L^\infty$?
Is $W^{1,p}(0,1)$ compactly contained in $L^\infty(0,1)$? Can I use this to show that I can select a sequence $(u_{n_k})$ from every bounded sequence $(u_n)$ in $W^{1,p}(0,1)$ such that $\lVert ...
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1answer
52 views
How to prove the compactness of this Sobolev embedding?
I have a question on compactness of the following Sobolev embedding.
Let $W^{1,p}([0,1],\mathbb{R}^n)$ be the Sobolev space of functions $u:[0,1]\rightarrow \mathbb{R}^n$ for $1<p<\infty$. How ...
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35 views
Fredholm alternative for polynomially compact operators
Let $T \in \mathcal{L}(E)$ be a polynomially compact operator, i.e., there exists a polynomial $p$ such that $p(T)$ is a compact operator. Suppose $p(1) \neq 0$. I want to show that $N(I-T) = \{0\} ...
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3answers
93 views
Trying to prove that operator is compact
Consider $T\colon\ell^2\to\ell^2$ an operator such that
$$T((x_n))=(2^{-n}x_n); \forall x=(x_n)\in \ell^2 $$
Does anyone know how to prove that it is compact?
I understand that I have to find a ...
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2answers
127 views
compact and self adjoint square root of an operator
Let H a Hilbert space and $T:H\rightarrow H$ a linear bounded, self-adjoint, positive and compact operator. How can i prove that the square root of T, $\ T^{1/2}:H\rightarrow H$ is also compact and ...
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1answer
50 views
1.4.5 Theorem of Murphy's book
See 1.4.5 Theorem of Murphy's book : I want to prove that if $u$ be compact operator on $X$ which is Banach space and $\lambda\in \mathbb{C}\setminus\{0\}$, then ...
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54 views
Continuous, selfadjoint and compact?
Hell0 there!
I have to show whether the operator
$$
T\colon L^2(\mathbb{R})\to L^2(\mathbb{R}), f\mapsto\chi_{[0,1]}f
$$
is continuous, selfadjoint and compact.
I have problems to show the ...
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2answers
44 views
$T:\ell^{2} \rightarrow \ell^{2}$ defined by $T(\{x_{n}\})=\{2^{-n}x_{n}\}$ is compact
Please help me to proof of problem :
Show that the operator $T:\ell^{2} \rightarrow \ell^{2}$ defined by $T(\{x_{n}\})=\{2^{-n}x_{n}\}$ is compact.
Tanks for your hint.
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77 views
Proof of compactness of bounded linear operator
Define $T: l^2 \to l^2$ by $Tx = y =(\eta_j)$, where $x = (\xi_j)$ and
$$
\eta_j = \sum_{k=1}^{\infty} \alpha_{jk}\xi_k, \quad \quad \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} |\alpha_{jk}|^2 < ...
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2answers
66 views
Compactness of multiplication operator on $C[0,1]$
Find a condition on function $a\in C[0,1]$ such that the operator $A:C[0,1]\rightarrow C[0,1]$ $$(Ax)(t) = a(t)x(t)$$ is compact? We are taking uniform norm on $C[0,1]$.
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128 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 ...
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1answer
66 views
Show compactness of an operator and calculate its SVD
Consider
$$T\colon\ell^2\to\ell^2, (s_1,s_2,\ldots)\mapsto (s_2,s_3,\ldots)$$
$$S\colon\ell^2\to\ell^2, (s_n)\mapsto (s_n/n)$$
$$R:=TS.$$
1) Show that $R$ is a compact operator.
2) ...
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1answer
142 views
There are compact operators that are not norm-limits of finite-rank operators
Given an example of a Banach space for which There are compact operators that are not norm-limits of finite-rank operators.
Tanks for answer
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2answers
63 views
Given $F:X \to X$ on $X$:Hilbert space satisfying some properties, prove that $F$ is surjective.
Given a map $F:X \to X$ where $X$ is a Hilbert space, $F$ satisfying
$f(x):=x-F(x)$ is a compact map.
$\lim_{\|x\|\to \infty} \frac{(F(x),x)}{\|x\|} = \infty$
I'm seeking to prove that $F$ is ...
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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 ...
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41 views
u is compact if and only if $\lim_{n \to \infty}{\lambda_n}=0$ [duplicate]
Please help me for solve of the following problem :
Let $H$ be a Hilbert space with an orthonormal basis $(e_n)_{n=1}^{\infty}$, and let $u$ be an operator in $B(H)$ diagonal with respect to $(e_n)$ ...
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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
...
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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 ...
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1answer
145 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 $K(x,y) = K(y,x)$, $K(x,y)\geq0$ $ ...
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1answer
35 views
A Linear map $u : X \longrightarrow Y$ is not bounded below iff there is …
Do you help me to:
checking that a linear map $u : X \longrightarrow Y$ between Banach spaces is not bounded below if and only if there is a sequence of unit vector ...
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2answers
74 views
$K(X,Y)$ a closed subset of $B(X,Y)$ for normed spaces $X,Y$
Note this is a homework problem so I am looking for a hint not a solution:
For normed linear spaces $X$ and $Y$, I'm trying to show that $K(X,Y)$, the set of compact operators $X\to Y$ is a closed ...
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57 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, ...
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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 ?
Hölders inequality gives us that the ...
3
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1answer
234 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, ...
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2answers
159 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 ...
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34 views
Spectrum of weighted shift operator [duplicate]
Possible Duplicate:
Compact operator? self adjoint operator? Stirling’s 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 ...
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1answer
185 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 ...
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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 $ ...
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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, ...
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70 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 ...
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199 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 $ < ...
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1answer
195 views
Rellich–Kondrachov theorem for traces
Let $W^{1,p}(\Omega)$ be the Sobolev space of weakly differentiable functions whose weak derivatives are $p$-integrable, where $\Omega \subset \mathbb R^n$ is a domain with Lipschitz boundary. Let ...
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3answers
137 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?
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1answer
199 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 ...
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1answer
53 views
How do I prove that a particular linear operator has an orthonormal basis?
I have to show that if $T$ is a linear operator such that $T: L^2(\mathbb(R)^n) \to L^2(\mathbb(R)^n)$ and $T(f)(x) = \int_{R^n}f(y)g(x,y)dy$, where $g(x,y)$ is an $L^2$ function, that there is an ...
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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
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1answer
99 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
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1answer
338 views
How to prove this integral operator is compact?
$T_kf=\int K(x,y)f(y)dy$
where $K(x,y)=\frac{\phi(x)\phi(y)}{|x-y|^{n-\alpha}}$
$\phi(x)$ is a smooth function on a compact support. $f$ is defined on $R^n$ and $K$ is defined on $R^n\times R^n$
...
