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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Why's Daugavet equation important?

I've been recently studing Daugavet equation in $L^1[0,1]$ and $C[0,1]$. I understand most of the results I've found but I can't figure out why is it important to find operators that hold Daugavets ...
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Is $T$ a compact mapping from $W_{0}^{1,2}\left(\Omega\right)$ into itself? [closed]

Let $\Omega$ be an open bounded subset in $\mathbb{R}^{6}$ and $f$ be in $L^{8}\left(\Omega\right)$. For any $w$ in $W_{0}^{1,2}\left(\Omega\right)$, define $T\left(w\right)$ be in ...
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Show that $f \overset{T}{\rightarrow} \frac{1}{x} \int_0^x f(t) dt$ is Bounded, and is NOT Compact in $L^2(0, \infty)$

Can someone help me with this question? Let $f \in X = L^2(0, \infty)$, and define \begin{equation} (Tf)(x) = \frac{1}{x} \int_0^x f(t) dt \ . \end{equation} Show that $T$ is Bounded, and is NOT ...
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63 views

Show that $\{ x_n \} \overset{T}{\mapsto} \{ \sum_{k=1}^{\infty} a_{nk} x_k \}$ is compact

Can someone help me with this question? Let $\ell^2$ be the space of complex sequences $\{ x_1, x_2, \ldots \}$ that $\sum_{n=1}^{\infty} \lvert x_n \rvert ^2 < \infty$. If $\mu$ be Counting ...
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22 views

Adjoint of canonical expansion of compact operator

Lets say I have given a rank-$n$ operator $A = \sum^n_{k=1} \lambda_k \langle u_k, \cdot \rangle v_k$. Then it is straightforward to compute its adjoint as $A^\ast = \sum^n_{k=1} \lambda_k \langle ...
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Find eigenfunctions of the integral operator with kernel $\sum\limits_{n=1}^\infty \frac{1}{n^2} \sin((n+1)x)\sin(ny)$

Find the eigenvalue and eigenfunctions of the integral operator $Ku=\int_0^\pi k(x,y)u(y)dy$. $k( x,y) = \sum\limits_{n=1}^\infty \frac{1}{n^2} \sin\big((n+1)x\big)\sin(ny)$. This is how I ...
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44 views

functional analysis. Compact operator. Hilbert-Schmidt theorem.

I have the following problem: "Under which $ \alpha \in \mathbb{R}$ is the operator $ T: L_2 [1, + \infty) \to L_2 [1, + \infty) $: \begin{equation*} (Tf) (x) = x^{\alpha} \int_x^{\infty} ...
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52 views

Is there an example of a non compact operator whose square is compact?

Is there an example of a non compact linear operator T from a Banach space X to itself such that T^2 is compact? Of course the converse is true, as T ^2 is compact if T is. Here T^2 means T composite ...
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57 views

Powers of compact operators

Consider a Hilbert space $H$ and a compact self-adjoint operator $T : H \to H$. I want to prove that all positive powers (especially fractional powers) of $T$ are compact. From the spectral theorem, I ...
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29 views

Counter-examples of direct sum of compact operators on Banach spaces is compact

Given a Banach space $X$ which can be written a direct sum of two subspaces $Y\oplus Z$ and the $u\in B(X), w\in B(Y), v\in B(Z)$ and $u=w\oplus v$. where $B(\cdot)$ denotes the space of bounded ...
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31 views

Compact operators and functions

Assume we have a look to the space $L^2(S^1)$. An orthonormal basis of $L^2(S^1)$ is given by $p_n(x)=e^{2\pi in x}$ $(n\in\Bbb{Z})$. One can also have a look at the operator $S(p_n)=sgn(n)\cdot p_n$ ...
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20 views

Diagonal non-compact operator

Suppose we have an operator $I:l_2 \rightarrow l_2$ which is diagonal but not compact. Does that follow: there exists a constant $C$ such that infinite number of diagonal terms $>C$?
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81 views

Definition of a compact operator

Operator compactness is characterized by maps the send the unit ball to relatively compact sets. Does anyone have a good justification for why we call this property compactness? The best ...
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29 views

Convergence in strong operator topology and norm topology

Let $(T_n)\subset B(H)$ be a sequence of operators such that $T_n\to 0$ in strong operator topology. Show that $\|T_nK\|\to 0$ and $\|KT_n\|\to 0$ for every compact operator $K$. Let $f,g \neq 0 ...
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28 views

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 ...
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45 views

Prove an operator is compact and classify the spectral values

Let $T(\eta) = \{\frac{1}{2^n}\eta_n\}$ for $\eta = \{\eta_1, \eta_2, ...\} \in \ell_2$. I need to prove T is compact, and evaluate $\sigma_p(T), \sigma_c(T), \sigma_r(T)$. My feeling is that I need ...
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42 views

Product of compact, bounded and self adjoint operator.

$T \in B(H)$, and $T = S^2$ for some self adjoint operator $S \in B(H)$. I need to prove that T is compact if and only if S is compact. If S is compact, it is easy to show that T is compact since S ...
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Proof Check for Compactness of Integral Operator

Above is my question. I have completed the question, but I'm not 100% about my proof for the final part - it seems like I haven't done enough. I've shown that if $U$ and $V$ are compact, then so is ...
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39 views

Prove that an infinite matrix defines a compact operator on $l^2$.

Let $(a(i))_{i=1}^\infty$ be an absolutely summable sequence, i.e., $\sum_{i=1}^\infty |a(i)|<\infty$, and consider the infinite matrix $$A=\begin{bmatrix} a(1)&a(2)&a(3)&\cdots\\ ...
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80 views

Sufficient conditions on integration kernel for continuity of the integral operator

Suppose that we have a measure $d\mu(v)=e^{-|v|^2}dv$ on $\Bbb R^d$. We define a linear operator $$T[f](u)=\int_{\Bbb R^d} |u-v|^\beta f(v) d\mu(v).$$ I want to establish conditons on $\beta\in\Bbb ...
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On closed ranges and sequences which converge to zero

I'm reading a proof of the Fredholm alternative, and there is a claim that goes like this: Let $K:X\rightarrow X$ be a compact linear map. Define $T=I-K$, then $Y=\ker(T)$ is a finite dimensional ...
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Hilbert-Schmidt and compact operators

I am new to this site and i dont really know how to ask questions properly, so i am really sorry if i did something wrong. My question is if there is a way to prove that a Hilbert-Schmidt operator is ...
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How is the Point Spectrum of a Compact Operator Countable?

I'm working on understanding a proof that if an operator $A$ on a Hilbert space $\mathcal{H}$ is compact, then show that $\sigma(A) - \{0\} \subseteq \sigma_p(A)$. If you're not familiar with this ...
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29 views

Show that a set is totally bounded

Let Y = $L^1 $($\mu$) where $\mu$ is counting measure on N. Let X = {$f$ $\in$ Y : $\sum_{n=1}^{\infty}$ n|$f(n)$|<$\infty$} Define T : X -> Y by $Tf(n)=nf(n)$ Equip X with the graph norm ...
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About the largest eigenvalue of Hilbert-Schmidt integral operators

Let $\Omega$ be an open set of $\mathbb{R}^d$ and $K \in L^2(\Omega\times \Omega)$ such that for almost all $x,y \in \Omega$ : $K(x,y)=K(y,x)$ $K(x,y)>0$ One can show that under these ...
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A sequence $L_n$ of compact bounded linear transformations on a hilbert space defines a convergent subsequence in each $L_n$ for a bounded sequence?

Let $L_n:\mathcal{H}\to\mathcal{H}$ be a sequence of compact bounded linear transformation on a Hilbert space $\mathcal{H}$, and $h_m$ be a sequence in $\mathcal{H}$. Since each $L_n$ is compact, ...
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Singular values of compact operators

Let $T$ be a compact operator.Then show that $\sum_{n=1}^\infty$$M_n(T)=\sup\{||PT||_1:P\mbox{ is a rank } N\mbox{ projection}\}$ where $M_n$'s are the singular values of $T$ and ...
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Do $L^p$ spaces have the approximation property?

A Banach space $X$ has the approximation property if every compact operator $T:X \to X$ is the norm-limit of a sequence of finite-rank operators. My question is if there is a simple proof that the ...
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34 views

Problem 8.11 in Young's book (An introduction to Hilbert space)

I've tried to prove this problem which appears in Young's Book: Let $K$ be a compact Hermitian operator on a Hilbert space $H$ and let the kernel of $K$ be $\{0\}$. Show that there is a sequence ...
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Unclear passage of a theorem concerning compact operators (Schauder fixed point theorem)

I'm looking at this proof of Schauder theorem and I am struggling with a passage. This is my problem: Let $X$ be a Banach space, $K \subset X$ a convex, close and bounded set and $F:K \rightarrow ...
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Compactness of translation operator in weighted spaces

Let $x,v\in\Bbb R^d$, $t\in \Bbb R$ and $m(x,v)$ be a smooth strictly positive function rapidly decaying on infinity - think $m(x,v) = \exp(-|x|^2-|v|^2)$. Define Banach spaces $X$ and $Y$ by ...
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Norm, adjoint operator and compactness os some operators

Let $A_i:\ell^2\rightarrow \ell^2$ be two operators given as follows: $A_1x=(0,x_1,0,\frac{x_2}{2},0,\frac{x_3}{3},...)$ and $A_2x=(x_1,x_1,x_2,x_2,x_3,x_3,...)$ Compute the norm and the adjoint ...
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compact projection

Show that the operator T: l^2→l^2 defined by T({x_n })={( x_n)/2^n } is compact. How one can show that the sequence{ََ T(x_n)} is contain convergent subsequence if {x_n } is bounded? I know that ...
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40 views

norms of operator

I am stuck on a question on the operator norm. If $L$ is a bounded linear operator $L:H\rightarrow H$ where $H$ is a Hilbert space. How would you show that $$\|L\|\leq \sup_{\|u\|=\|v\|=1}(Lu,v)$$ any ...
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Question about the Image of a compact transformation of a Hilbert space

$T$ is a compact operator on a Hilbert space. Show that $\operatorname{im}(T)$ does not contain a closed infinite dimensional subspace. Here is my attempt at the problem: Suppose that ...
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64 views

Inverse of laplacian operator

I recently read a paper, the author treats $$\int_{\mathbb{R}^d}f(y)\cdot \frac{1}{|x-y|^{d-2}}\,dx = (- \Delta)^{-1} f(y)$$ up to a constant in $\mathbb{R}^d$. I am not familiar with unbounded ...
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Question on the range of values of compact operator

Suppose $X$ is a Banach space and $T$ is a compact operator from $X$ to $X$. Let $B_1$ is closed unit ball in $X$. Then, can we get a resual that $\overline{T(B_1)} \subseteq T(X)$ ? It is easy to ...
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42 views

Approximation Property: Decomposition

This is a real question of me. Given a Banach space $E$. Consider a finite rank operator $F\in\mathcal{F}(X,E)$. Introduce a basis on the finite dimensional range: ...
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spectrum of convolution integral operator

Let $A f(x)= \int_{-\pi}^{\pi} h(x-y) f(y) dy$ operator $L^2( {-\pi},{\pi})->L^2( {-\pi},{\pi}), h$ is continuous, periodic with period $2\pi$ and $h(x)=h(-x)$ on $ [ {-\pi},{\pi}] $. How can I ...
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Adjoint of Integral Operator in $L^p$

Let $E = L^p(0,1)$ with $1 ≤ p < ∞$. Given $u ∈ E$, set $$Tu(x):=\int_0^x u(t)dt$$ Find the adjoint of $T$. I know how to this in the case $p=2$ as shown here. But in general $L^p$ is not an ...
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32 views

Bound for Integrator Operator

Let $E = L^p(0,1)$ with $1 ≤ p < ∞$. Given $u ∈ E$, set $$Tu(x):=\int_0^x u(t)dt$$ Prove that $T$ is compact on $E$. I would like to use Ascoli-Arzela', but I need to prove: $$|T u(x) − T u(y)| ...
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Hilbert- Schmidt class is an ideal

Definitions: 1 - An operator $y\in B(H)$ is said to be of trace class if $y$ is compact, and also $\sum|\alpha_n| <\infty$ where $\alpha_n \in \sigma(y)$ and $y$ has a representation $\sum ...
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Compact operator space is the greatest ideal of $B(H)$

Suppose $H$ is a separable infinite dimensional Hilbert space. Show that if $A\in B(H)$ is noncompact, then there exist two operators $B,C$ such that $BAC=1$. Clearly if $A$ is invertible it holds, ...
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Schrödinger Semigroup is compact if potential goes to infinity

Several papers (e.g. this one: arXiv:0810.3275v1 [math.SP] 17 Oct 2008 ) claim that if $H=-\Delta+V$ and $V(x)\to\infty$ if $|x|\to\infty$, then the semigroup $e^{-tH}$ is eventually compact. Does ...
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Trace class operator

Let $A\in B(H)$ and $\sum_{E}|\langle A e,e\rangle|< \infty$ for every orthonormal basis $E$. Show that $A$ is a trace class (means $\sum_E \langle |A|e,e\rangle < \infty$). I can not prove it. ...
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Compact operator and a sot convergent sequence of operators

The following is an exercise of Conway's operator theory: I proved all parts of this exercise except $\|KT_n\| \to 0$. I can easily prove $\|KT_n^*\|\to 0$, but do not have any idea to prove ...
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122 views

Show that the operator $(x_n)_n\mapsto (\frac{x_n}{n}) $ is compact

I want to show that the following operator is compact: $$T:\mathbb \ell^p\rightarrow \mathbb \ell^p, \text{ }(x_n)_n\mapsto(\frac{x_n}{n})_n \text{ } 1\leq p<\infty$$ Its the first time that ...
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47 views

parameter operator $A_a$ is compact??

I need some help in this exercise. Let define operator on $ L^2[0,1]$: $$ A_af(x)=\int_{0}^{1}{|x-y|}^{a-1} f(y)dy $$ for f $\in L^2[0,1] $. Prove that A_a is compact for all $a>0$. I see that ...
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Compactness of the fractional integral operator.

Is the fractional integral operator $J_a^{\alpha} :L^2[a,b] \rightarrow L^2[a,b]$ ( or $J_a^{\alpha}:C[a,b] \rightarrow C[a,b]$) which is defined by $$J_a^{\alpha} f(x) = {1 \over ...
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1answer
79 views

Finite rank volterra operator

I am wondering when a Volterra integral operator $V_K:L_2(0,1)\to L_2(0,1)$ is a finite rank operator: $$V_Kf=\int_0^xK(x,y)f(y)dy$$ thanks in advance for your help