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Questions tagged [compact-operators]

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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Is the Riemann Liouville fractional integral compact operator? [closed]

I am about to figured out that is the Riemann Liouville fractional integral compact operator or not? where f is continuous function in [0, b].
Jabar S. Hassan's user avatar
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Perturbation of semi-Fredholm operators in Frèchet spaces

It is well known that the index is a continuous function on the set of Semi-Fredholm operators on a Banach space, and even on quasi-Banach spaces. The result is unfortunately false in general Fréchet ...
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Divergent Tail Sums of Approximations of Non-trace Class Compact Operators

I'm working on approximations of compact operators that are not trace class, and I'm looking for ways to provide meaningful approximation error estimates for truncated eigenfunction expansions. I ...
user avatar
4 votes
1 answer
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Convolution preserve the boundary condition

Here, I want to know if convolution will preserve the Neumann condition or not. Suppose $K$ is a continuous function on some interval $[-L,L]$, and $u$ is some 'good enouth' function on $[0,L]$ that ...
SaltedFishKing's user avatar
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1 answer
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Compactness of an operator on l^2

I am working on some problems from Kreyzig's Functional Analysis and got stuck on this problem: Let $T: l^2 \rightarrow l^2$ be defined by $Tx = y = (\eta_j)$ where $x = (\xi_j) $ and $\eta_j = \sum_{...
Bureking's user avatar
2 votes
1 answer
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Is the notion of compactness of operators preserved under quasi-similarity?

Let $H$ be a Hilbert space and $T \in \mathcal L (H).$ Then $T$ is said to be quasi-invertible if it is one-one and has dense range. The operator $T$ is said to be quasi-similar to an operator $S \in \...
Anacardium's user avatar
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What is a conjugate unitary operator?

I'm studying operator theory and I have encountered the concept of a cojugate unitary operagtor several times. However, I cannot find any reliable references. There is one paper which claims that a ...
OSCAR's user avatar
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$u^*$ comapact implies $u$ compact proof verification

Suppose that $X, Y$ are Banach spaces and suppose that $u \in L(X, Y)$, with $u^*$ compact, where $u^*: Y^* \to X^*$, given by $u^*(\tau) = \tau \circ u$ is the adjoint of $u$. I am asked to prove ...
blomp's user avatar
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1 answer
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An interesting family of seminorms $\mathcal F$ and comparison between the topology generated by this seminorms and the Weak Operator Topology.

I am learning functional analysis and I am stuck with the following questions from Strong Operator Topology and Weak Operator topology on $\mathcal B(H)=\{T:H\to H:T$ is Op-Norm continuous,linear $\}$....
Kishalay Sarkar's user avatar
4 votes
1 answer
150 views

Proving that operator in $L^2[0,1]$ is compact

I need help with some functional analysis: Let $A$ be a continuous linear operator on $L^2[0,1]$ and for any $f \in L^2[0,1]$ the function $Af$ is Lipschitz continuous. Show that $A$ is compact. It is ...
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I am trying to show that $K: H_0^1(\Omega) \rightarrow H_0^1(\Omega)$ is a compact operator.

$\textbf{Suppose.}$ $\Omega$: bounded open in $\mathbb{R}^n$, ${W^{1,2}}(\Omega)$: the space of $u:\Omega \to\mathbb{R}$ in $L^2(\Omega)$ with a weak derivative $Du$ in $L^2(\Omega)$, and ${H_0}^1(\...
topst's user avatar
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2 votes
1 answer
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Showing $T$ is a compact operator

Let $\{\varphi_n\}$ be an orthonormal sequence in a Hilbert space $H$ and consider the operator $T:H \to H$ defined by $$T(f) = \sum_{n=1}^\infty a_n \color{blue}{\langle \varphi_n, f \rangle} \...
Grigor Hakobyan's user avatar
2 votes
1 answer
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Equivalence of Kraus operators on a single element

Let's say that I'm working in a Hilbert space $\mathcal{H}$,suppose I have two bounded operators $A.B\in B(\mathcal{H})$ and a positive semidefinite operator $x$, for example take $x$ to be a density ...
ana's user avatar
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1 answer
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Eventual equality of ranges of $(T-\lambda)^n$

Let $X$ be a Banach space, $T:X\to X$ a compact operator, and $\lambda$ a non-zero complex number. I've shown that $(\operatorname{ran}((T-\lambda)^n)_{n<\infty}$ is an $\supseteq$-decreasing ...
Good Morning Captain's user avatar
3 votes
1 answer
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Problem of Fourier transform and compact operator

Problem: Is the Fourier transform $F(f(x))=\int_{-\infty }^{\infty }f(y)e^{-ixy}dy$ a compact operator in the case of $F:L_1\left ( \mathbb {R} \right )\to \mathrm{BC}\left ( \mathbb{R}\right )$. $\...
Dmitry's user avatar
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3 answers
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Prove that the operator is not compact

Problem: Prove that the operator $A\in\mathcal{L}\left ( C[0,1] \right ):(Af)(x)=f\left ( x^2 \right )$ is not compact My attempt at a solution: It is necessary that the image of the unit ball be ...
Dmitry's user avatar
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1 answer
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continuouty of operators

I was given a task to understand, wheter operators $A$ and $B$ are compact, $$\displaystyle A:\ell_2 \rightarrow L_1(\mathbb{R}), (Ax)(t) = \sum\limits_{k=1}^{+\infty}\frac{x(k)}{\cosh^2(kt)},$$ $$B:...
GeoArt's user avatar
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Upper bound of operator norm of Hadamard (Schur) product

As in Extension of the Schur product theorem to operators, we can define two compact self-adjoint integral operators $A$ and $B$ on $L^2(\Omega)$, by $$ A \phi(x)=\int_{\Omega} a(x, y) \phi(y) d y ; \...
user1247096's user avatar
1 vote
1 answer
52 views

Hilbert-Schmidt integral operator without second integrable kernel

From this webpage, we know that if $X$ is a measurable space ($\sigma$-algebra is omitted) and $\mu$ is the measure, $K(x,y) \in L_2(X\times X,\mu\times \mu)$, we can define the following operator ...
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1 vote
1 answer
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compact and trace operator [closed]

I am study compact and trace classes of operators in Hilbert space. To clarify, I asked the following question, but have not been able to resolve it yet. Consider $L_p \left( \mathbb{R}^d \right) $ ...
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2 answers
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How do you prove that compact operators on a Hilbert space are weak to weak continuous?

I'm actually trying to understand the proof for the statement that Let $T \in K(H,H)$ for Hilbert space $H$, and $(x_n) \subset H, x_n > \rightharpoonup 0$ weakly. Then $\|Tx_n\| \to 0$ strongly. ...
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0 answers
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Compactness and boundedness of integral convolution operator

Doing my homework on functional analysis I faced with the following problem. Let $\displaystyle (Uf) (s)=\int\limits_{-1}^{1}\frac{f(t)}{|s-t|^{5/6}}dt$. Using Young inequality it can be shown that $U$...
Jackson Harris's user avatar
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1 answer
50 views

$T\in\mathcal K(H)$ , $(e_n)$ orthonormal sequence. Prove $Te_n \rightarrow 0$. [closed]

Let $H$ be a Hilbert space and $T:H\to H$ a compact operator. Let $(e_n)$ be an orthonormal sequence in H. Prove $Te_n \rightarrow 0$. Hint: what do you know about $\langle Te_n,f\rangle$ for $f\in H$...
Its me's user avatar
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Prove operator $(Tf)(x)=sin(x)\cdot f(x)$ is not compact

Given the following operator in $L_2[0,1]$ $$(Tf)(x):=sin(x)\cdot f(x)$$ Prove or Disprove that the opertor is Compact. I thought it is compact and used arzelà–Ascoli theorem, but apparently I am ...
Its me's user avatar
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0 answers
109 views

Spectrum of compact self-adjoint operator vs symmetric matrix

I know that compact self-adjoint operators have a countable, orthonornal eigendecomposition, and the spectrum is real and positive, and the only cluster point is zero. I know that self-adjoint ...
900edges's user avatar
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1 answer
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Compact embedding in $L^1$

Is it true that the space $H^1(\mathbb{R}^2)\cap L^2(\mathbb{R}^2,(1+|x|^2)^2dx)$ is compactly embedded in $L^1(\mathbb{R}^2)?$ I don't think that the dimension $2$ is important. Here $H^1$ is the ...
Frank Zermelo's user avatar
1 vote
1 answer
36 views

Range of $I-T$ is closed for $T$ being a compact operator in a Banach Space

I know from this question that this is true for Hilbert spaces. But is the result true for Banach spaces as well? So suppose $x_{n}$ is a bounded sequence and $(I-T)(x_{n})$ converges. Then upto a ...
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Sequence of range space of compact operator stabilizes. Sheldon Axler MIRA Ex 10C 11

Suppose $T$ is a compact operator from a Hilbert Space. Then for $a\neq 0$, we have that $\text{range}(T- aI)^{n-1}=\text{range}(T-aI)^{n}$ for some $n$ and $\ker(T-aI)^{m-1}=\ker(T-aI)^{m}$ for some $...
Dovahkiin's user avatar
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1 vote
2 answers
39 views

Compactness of $C[J,X]$?

If $J=[0,T]\subset \mathbb{R}$, and $X$ is a Banach space? Then, is $C[J,X]$ (Banach space of all continuous functions from $J$ to $X$) a compact metric space with respect to supnorm defined as $\|x\|...
Pratima Tiwari's user avatar
1 vote
1 answer
33 views

Convergence in trace

I am having difficulties with this problem. Given a non-negative compact self-adjoint operator $\gamma$ i.e. $\langle \gamma u, u\rangle \geq 0$ for all $u$. Denote the cut-off function $\chi \in C^\...
Larry Baynes's user avatar
2 votes
1 answer
105 views

Convergent sum of operator norms of iterates of kernel operator corresponding to stepping of graphon $W$

Suppose that we are given a possibly non-symmetric graphon $W:[0,1]^2\to\mathbb [0,K]$, where $K>0$ is some fixed, known constant. Given such a graphon, we consider the kernel operator $T_W:L^1[0,1]...
Václav Mordvinov's user avatar
0 votes
1 answer
61 views

if $S \in \mathscr{H}$ and $T \in \mathscr{K}$ then ST is compact. [duplicate]

I understand that $\mathscr{K}(\mathscr{H})$ is a closed subspace of $\mathscr{B}(\mathscr{H})$. Where $\mathscr{K}$ is the compact operators and $\mathscr{H}$ is the bounded operators. I need to ...
Sarah W's user avatar
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1 answer
40 views

Can both of these linear operators be unbounded?

Suppose $A$, $B$, $C$, and $D$ are compact and injective linear operators in $L_2[0,1]$ and that $AB^{*}=CD^{*}$. Is it possible that both $D^{-1}B$ and $C^{-1}A$ are unbounded?
SecretlyAnEconomist's user avatar
1 vote
1 answer
56 views

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\...
Jane T.'s user avatar
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2 votes
3 answers
125 views

Prove $T$ with bounded basis sum in Hilbert space is compact.

Let $T:\mathcal{H}\rightarrow \mathcal{H}$ be a linear continuous operator between Hilbert spaces and $\{b_i\: |\: i \in I\}$ an orthonormal basis. Prove that if: $$\sum_{i\in I}\lVert T b_i\rVert^2&...
Kadmos's user avatar
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2 votes
0 answers
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Is this integral operator a compact operator?

Let $D$ be the closed unit ball in $\mathbb{R}^n$. Define $$(T x)(t)=\int_{D}\frac{x(s)}{|t-s|^\alpha}{\rm d}s$$ Where $0<\alpha<n$. Then $T$ is a continuous operator on $C(D)$. Is it compact? I'...
meow_w's user avatar
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1 answer
104 views

Showing that if $\lim_{i\to\infty}\lambda_i=0$ and $Ae_i=\lambda_ie_i$ for $A\in\mathcal{B}(H)$, then $A$ is a compact operator

Let $\{e_i\}$ be a Hilbert basis for a Hilbert space $H$ over $\mathbb{C}$ and $\{\lambda_i\}$ be a sequence of complex numbers tending to zero. Suppose that $A\in\mathcal{B}(H)$ such that $Ae_i=\...
Epsilon Away's user avatar
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Space of compact operators defined on separable Hilbert space

Let $H$ be a separable Hilbert space and $K(H)$ the Banach space of all compact linear maps from $H$ into itself (with the operator norm). Show that $K(H)$ is separable. There is a hint that states ...
BasicUser's user avatar
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3 votes
2 answers
71 views

Linear positive definite bounded operator $T:H\to H$

I was trying to solve this problem: Let $H$ be a Hilbert space and $T:H\to H$ a bounded linear operator such that $$ (Tx,x)\geq ||x||^2 \quad \forall x\in H $$ where $(\cdot,\cdot)$ and $||\cdot||$ ...
Gio5520's user avatar
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1 answer
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Why does this prove that the span of the eigenvectors is dense in $\text{Im } T$?

Let $H$ be a Hilbert space with the inner product $(\cdot,\cdot)_H$ and let $T:H\to H$ be a bounded compact and self-adjoint operator on $H$. In this case there exists an $H$-orthonormalsystem $(\...
Max Stuthmann's user avatar
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0 answers
60 views

Show that the image of S under the map $ f \mapsto f * g $ is a compact set in $ C_0([-2, 2])$.

Given $\frac{1}{p} + \frac{1}{q} = 1 $, let $S = f \in L^p(\mathbb{R})$ $spt(f) \subset [-1,1]$, and $\|f\|_p \leq 1$ , and let $ g$ be a fixed but arbitrary function in $ L^1(\mathbb{R})$, with spt(...
Mr. Proof's user avatar
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2 votes
1 answer
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Clarify proof of theorem 6.6 Fredholm alternative in Haim Brezis book (Funtional Analysis)

I'm currently reading proof of this theorem. My question is in part b), it says that: I have already wrote But i'm not sure why $w_{n_k}-Tw_{n_k}$ corverges to $0$, which is implies from $(4)$. Can ...
Nguyen Thy's user avatar
2 votes
0 answers
47 views

Dual of space of compact linear operators on a Hilbert space [duplicate]

For a Hilbert space $H$, consider the space of compact linear operators on this Hilbert space, $\mathcal{K}(\mathcal{H})$. Is it correct to say that the dual space of $\mathcal{K}(H)$ is the space of ...
Ronan's user avatar
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0 answers
39 views

Using Finite Rank Operators to Prove that Compact Operators form an Ideal

This is from Analytic K-Homology by Higson and Roe. Show that if $S$ is a rank one operator and $T$ is any operator, then $ST$ and $TS$ are rank one operators. Deduce that $\mathcal{K(H)}$ forms a ...
Vinay Deshpande's user avatar
1 vote
0 answers
72 views

Solving an Integral Equation Using Neumann Series in $C([0,1])$

Question: Let $U, V \subset \mathbb{R}^d$ be compact, $K \in C(U \times V)$ and $T_K: C(V) \rightarrow C(U)$ given by $$ T_K(u)(x)=\int_V K(x, y) u(y) d y $$ Let $U=V=[0,1] \subset \mathbb{R}$. Show ...
CanDoMajoringMath's user avatar
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0 answers
22 views

How to Prove $T_K\left(B_1(0)\right)$ is Precompact in $C(U)$ for a Kernel Operator $K$ in $\mathbb{R}^d$

Question: Let $U, V \subset \mathbb{R}^d$ be compact, $K \in C(U \times V)$ and $T_K: C(V) \rightarrow C(U)$ given by $$ T_K(u)(x)=\int_V K(x, y) u(y) d y . $$ Prove that $T\left(B_1(0)\right)$ is ...
CanDoMajoringMath's user avatar
0 votes
1 answer
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Brezis' exercise 8.28.2: prove that $T$ is a compact operator

Let $I$ be the open interval $(0, 1)$ and $H := L^2 (I)$ equipped with the usual inner product. Consider the linear map $T: H \to H$ defined by $$ (Tf) (x) = \int_0^x t f(t) dt + x \int_x^1 f(t)dt, \...
Akira's user avatar
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1 vote
0 answers
68 views

T is Compact iff the image of the unit ball $B1(0)$ is relatively compact (the closure is compact)

Hey all I have this question: Let $X, Y$ be normed spaces, $T \in \mathrm{L}(X, Y)$. Prove the following. $T$ is compact if and only if $\overline{T\left(B_1(0)\right)}$ compact. Proof: Let $X, Y$ be ...
CanDoMajoringMath's user avatar
2 votes
1 answer
131 views

How do I show that the kernel of $\mathrm{Id}-T$ is finite-dimensional for a compact operator $T$ in a Banach space.

My attempted Solution: We need to show that for a compact operator $T$ in a Banach space, the kernel of Id $T$, denoted as $\operatorname{ker}(\mathrm{Id}-T)$, is finite-dimensional. The kernel of $\...
CanDoMajoringMath's user avatar
1 vote
0 answers
422 views

Is $L$ compact if $\| Lf_k \|\rightarrow 0$ for Orthonormal Basis $f_k$?

Suppose $f_k$ is an orthonormal basis of a separable Hilbert space. $T$ is bounded. $\| (T-\lambda_k)f_k \| \rightarrow 0$, $\| (T-\lambda_k)^*f_k \| \rightarrow 0$. $R$ is the diagonal operator with ...
Vinay Deshpande's user avatar

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