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.
Questions tagged [compact-operators]
1337 questions
5
votes
2 answers
What is the intuition/motivation behind compact linear operator.
Compact Linear Operator is defined such that the operator will map any bounded set into a relatively compact set. Why is this property so special that it can be named as "compact"? Does it share some similar properties as compact sets? What is the…
TH000
- 401
3
votes
1 answer
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) Calculate the Singular Value Decomposition of $R$.
EDIT: My…
user34632
3
votes
1 answer
Does this imply compactness
Suppose H is Hilbert space with a countable orthonormal basis $\{e_n\}$. Let T be a bounded operator on H such that $∥Te_n∥$ tends to 0. Can we conclude that T is compact?
Kavi Rama Murthy
- 311,013
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0 answers
Is $|x|^{-d+\alpha}$ square integrable in $\mathbb{R}^d$ given $\alpha>0$?
This is a problem in the S.-T. Yau College Student Mathematics Contests in 2013.
Suppose $H=L^2(B)$, $B$ is the unit ball in $\mathbb{R}^d$. Let $K(x,y)$ be a measurable function on $B\times B$ that satisfies $$|K(x,y)|\le A|x-y|^{-d+\alpha}$$ for…
Apocalypse
- 888
2
votes
2 answers
$K(X,Y)$ 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 subset of $B(X,Y)$ the set of bounded operators…
roo
- 5,598
1
vote
1 answer
*IF $X$ is a normed spaces, then $L_c(X)$ two sided ideal in normed algebra $L(X)=L(X,X)$.*
Yesterday assistant filed during exercises work, but not one of us was able to resolve, and then he proved himself and noticed that he had the solution when attempting to problems. Therefore, please know that there are many people who can solve this…
user145717
- 217
1
vote
1 answer
Dimmension of null spaces of a convergent sequence of operators
I am trying to solve the following problem:
Let $H$ be an infinite-dimensional Hilbert space. $A_\varepsilon, A$ are compact operators in $L(H,H)$ and $A_\varepsilon \to A$ in $L(H,H)$ as $\varepsilon\to 0^+$. We know that $\dim N(I-A_\varepsilon)$…
Jane T.
- 63
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0 answers
Compact operator and its norm attaining set.
Let $X=l_{2}$ and $T:X\to X$ be a compact operator. Define $M=\{x\in l_{2}: \|x\|=1 \ \text{and} \ \|Tx\|=\|T\| \}$. My question is that define $T$ such that $M$ is not compact.
We know that in the case of a bounded linear operator the identity…
akshay
- 19
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vote
0 answers
Closure of compact operators plus the identity
I have a question about compact operators.
Suppose $\mathcal{H}$ is an infinite-dimensional separable Hilbert space, and denote by $\mathcal{K}\left(H\right)$ the compact operators from $\mathcal{H}$ to $\mathcal{H}$. It is well known that $1$, the…
Diego MMG
- 11
0
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0 answers
If the $A^*A\leq BB^*$ and $B$ is a compact operator, then the operator $A$ is compact
I am a student of mathematics and professor left us today for homework this example:
If the $A^*A\leq BB^*$ and $B$ is a compact operator, then the operator $A$ is compact. Prove this. ($A^*$ hermitian adjoint operator)
Thanks very much for your…
user145717
- 217
0
votes
3 answers
Showing operator has no eigenvalues
Let $T:\ell_2 \to \ell_2$ be defined by $(T(x))_n = \dfrac{x_{n-1}}{n}$ for $n \gt 1$ and $(T(x))_1=0$.
I have showed that $T$ is a compact operator.
Now I need to show that $T$ has no eigenvalues.
So assume $\lambda \ne 0$ is such, we have $T(x) =…
user335501
0
votes
1 answer
Conditions under which a compact operator with finite-dimensional domain has a finite rank
Let $X$ be a separable Hilbert space, and denote by $\mathcal{L}\left(X,X\right)$ the usual space of bounded linear maps on $X$ (that is, taking $X$ into $X$). Let $n$ be some nonnegative integer and consider the operators…
Trevor3
- 147
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votes
1 answer
$\ell^2$ space, compute $\|L\|$.
Consider $\ell^2$ space, define $L\in\mathcal{L}(\ell^2):(x_1,x_2,\ldots)\mapsto(x_2,x_3,\ldots)$.
Compute $\|L\|$.
Show that $\{\lambda\in C:|\lambda|<1\}\subseteq\sigma_p(L)$, $\sigma_p(L)$ is the point spectrum.
Show that $\{\lambda\in…
lethecullen
- 43
0
votes
1 answer
Calculating the following Volterra operator
I have the following space $X =C([0,1],\mathbb{K}$ and let $T$ be the operator defined by
\begin{equation}
Tf(x)=\int^{x}_{0} (x-t)f(t)dt, \space\space\space f \in X
\end{equation}
Now I have showed the operator $T$ is bounded by $1/2$. I am left to…
johnd1992
- 41
0
votes
1 answer
Compact operators
From a textbook:
Suppose $X$ and $Y$ are Banach spaces and $T : X \to Y$ is linear. Then $T$ is a compact operator if one of the following holds:
(a) $\{Tx_n\}$ contains a convergent subsequence in $Y$ whenever $\{x_n\}$ is a bounded sequene in…
