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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If $H$ is a one-dimensional Hilbert space then the zero representation of a C*-algebra on $H$ is irreducible.

It says on page 143 of Murphy's $C^*$-algebras and operator theory that if $H$ is a one-dimensional Hilbert space then the zero representation of any C*-algebra on H is irreducible. What is the zero ...
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19 views

Proving that each compact operator is bounded

I'm trying to prove that each compact operator is bounded, from the definition: On the internet I find solutions for this problem, but they always start from a different definition. I would be glad ...
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$s \in L^{1}(H)$ $\iff$ $s=\sum_{i=0}^\infty x_{i} \otimes y_{i} $

Let $H$ be a separable Hilbert space, and let $L^1(H)$ be the space of trace-class operators on $H$. I'd like to prove that $s\in L^{1}(H)$ if and only if there exists $\{ x_{i} \} , \{ y_{i} \} ...
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29 views

The ideal generated by a non-compact operator

I wanted to find a quick proof of the following well-known fact. Since I couldn't easily find a reference, I provide a proof below. Let $H$ be a separable Hilbert space, and $J\subset B(H)$ be a ...
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28 views

Spectrum of a bounded operator on a (not necessarily Banach) normed vector space

It's well known that on a Banach space, the spectrum of each bounded operator is compact in $\mathbb C$. What about a general normed vector space? Is there a counterexample if we don't assume ...
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Compactness of solution space of semi-linear parabolic PDE

Under what conditions a closed and bounded subset of solution space of following parabolic PDE is compact? $$x_{t}=x_{zz}+f(x,z)$$ Thank you!
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26 views

Toeplitz Operator is compact if and only if it has finite rank

A referee has pointed out to me that it is "well known that a Toeplitz operator is compact if and only if it has finite rank" and pointed me to R. Douglas: Banach algebra techniques in the ...
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36 views

sup norm of operator

Let $T$ be a compact linear operator defined as $$ T\circ u = \int_a^b k(x,y)\,u(y)\,dy, $$ where $k(x,y)\in C([a,b]\times[a,b])$ and $k(x,y)\ge0$ for all $x,y$, and $u\in C([a,b])$. Suppose that the ...
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Green-Operator for Sturm-Liouville Differential equation compact on Sobolev space?

Let $g$ be Green's Function for a Sturm-Liouville differential equation. Is the operator $G: H_{0}^{1}(0,1) \rightarrow H_{0}^{1}(0,1)$ defined by $(Gf)(x) := \int_{0}^{1} g(x,y)f(y) dy, \quad f \in ...
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Product of two positive compact, self adjoint operators

If we have two positive compact , self adjoint operators; $A$, $B$. Is the product $AB$ a positive operator?
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Is $L^\infty$ compactly embedded in $L^1$?

I'm trying to find a contraction example to show that the space $L^\infty$ is not compactly embedded in $L^1$ with the Lebesgue measure. Please help me!
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Prove the operator on hilbert space is compact

My question is actually the same as the first part of this one, Prove that T is compact which has not been answered. I am thinking about two ways, 1) use a bounded sequence $\{g_n\}$, and try to ...
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1answer
20 views

Understanding connectedness argument in proof of Analytic Fredholm Theorem

Let $X$ be a complex Banach space, and let $D \subset \mathbb{C}$ be a domain. Let $\mathcal{L}(X)$ denote the Banach space of bounded linear transformations $X \to X$. The Analytic Fredholm Theorem ...
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46 views

Adjoint Operator of a Compact Operator

In the proof of the fact that the ad-joint operator $T^*$ of a compact operator $T$ defined on a separable, infinite dimensional Hilbert space $\mathcal H$ is also compact, I read that "$\|P_nT - T\| ...
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39 views

Equivalent formulation for compact operators

According to Wikipedia, an operator is compact if it can be written in the form $T(u)=\sum_{n=1}^\infty \lambda_n<f_n, u> g_n$, where $\{f_n\}$ and $\{g_n\}$ are orthonormal sets and ...
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25 views

spectrum of compact operators

Let $\phi\in\ell^\infty(\mathbb{N})$. For $p\in[1,\infty]$, define $$M_\phi:\ell^p\to\ell^p,\quad f\mapsto\phi f.$$ Use spectral theory to show that, if $M_\phi$ is compact, then $\phi\in c_0$. Here ...
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Possible flaw in “proof” that a sum of two compact operators is compact

If X and Y are Banach spaces, and $A: X \to Y$, $B: X \to Y$ are both compact operators, then $A + B$ is compact. A + B is compact if and only if for every bounded sequence $\lbrace x_n \rbrace$ ...
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38 views

If $A$ is a compact diagonal operator, with diagonal $\{\alpha_n\}$, then $\lim_{n\to\infty}\alpha_n=0$.

Here is my question: If $A\in \mathscr{B}(\mathscr{H})$ is a diagonal operator with diagonal $\{\alpha_n\}$, show that if $A$ is compact, then $\lim_{n\to\infty}\alpha_n=0$. Here is what I have: I ...
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Compact operators on a Banach space $X$ are closed in the bounded operators on $X$. - Proof correction help

I am given a proof of the following statement (see below). Compact operators on a Banach space $X$ are closed in the bounded operators on $X$. I was told that there is an error in this proof - I ...
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Prove $Tx=(r_1x_1, r_2x_2, r_3x_3,…)$ is compact, $T:l^2\to l^2$, $r\in l^2$

Here is my question: Fix $r=(r_1,r_2,...)\in l^2$. Define $T:l^2\to l^2$ by $$Tx=(r_1x_1, r_2x_2, r_3x_3,...)$$ Prove that $T$ is compact. Here is what I have, input would be appreciated: Let ...
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33 views

Prove an integral operator is compact

The statement is like this, $K\subset\mathbb{R}$ is compact, the operator $A:L^\infty(K)\mapsto L^\infty(K)$ is defined by $f(x)\mapsto\int_K k(x,y)f(y)dy$. For $x\neq y$, $|k(x,y)|\leq ...
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How do endomorphisms of Banach space modulo compact operators look like?

It is well-known that given a Banach space $X$, the set of compact operators (let's denote it by $K(X)$) on $X$ forms a both-sided ideal in $L(X)$, the ring of bounded linear operators on $X$. My ...
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adjoint operator

I have a bounded linear operator L on a Hilbert Space! L and adjoint operator L1 Null(L1)=Null(LL1); I want to verify it L1x = 0 in Null(L1) and LL1x = 0 in Null(LL1) then L1x - LL1x = 0 in ...
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canonical form of finite rank operators

Let $X,Y$ be banach spaces and let $T:X\rightarrow Y$ be an linear continuous operator with finite dimensional image $Im(T)\subset Y$. Now I want to prove that there exists continuous linear ...
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85 views

T is not compact operator

I want to show that if $T$ is a bounded operator between two Hilbert spaces and $T$ is not compact then there exists an orthonormal sequence $y_{n}$ and an $R>0$ such that $\forall n\in ...
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Equivalent definitions of the trace of a Hilbert-Schmidt operator

I am currently reading the book Spectral Methods in Automorphic Forms, and Iwaniec defines the trace operator in a different way than I am accustomed to. Throughout, assume that everything converges ...
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34 views

Isomorphism between compact operators and compact operators tensor matrices ($\mathbb{K}\otimes M_n(\mathbb{C})\cong \mathbb{K}$)

Let $\mathbb{K}$ be the compact operators and $M_n(\mathbb{C})$ the complex valued matrices. I have read the algebra $\mathbb{K}\otimes M_n(\mathbb{C})$ is isomorphic to $\mathbb{K}$. Could you tell ...
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Compact projection on Banach space has finite rank

Let $E$ be a Banach space. Show that every compact projection has finite rank. I have no idea where to start.
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Does a compact operator always have a kernel?

I am sorry if this question is stupid..... I raise it when I read Lax's book Functional Analysis. We know that some integral operators are compact, for example an integral operator from $L^2[Y]$ to ...
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Does $\|f(T^*T)T^*\|_\infty = \|f(T^*T)(T^*T)^{\frac{1}{2}}\|_\infty$?

If $T:X\to Y$ is a compact operator and $X,Y$ are some Hilbert spaces, can we say that $\|f(T^*T)T^*\|_\infty = \|f(T^*T)(T^*T)^{\frac{1}{2}}\|_\infty$, where $T^*$ is its adjoint and $f$ some ...
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Compact operators, space of sequences

Let $\phi\in\ell^\infty$. For $p\in[1,\infty]$, define $M_\phi:\ell^p\to\ell^p$ by $$M_\phi(f)=\phi f.$$ Show that $\Vert M_\phi\Vert=\Vert\phi\Vert_\infty$, and $M_\phi$ is compact if and only if ...
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Hilbert Spaces - an application of the minimax principle.

Let $A$ be a compact, self-adjoint operator, $A \geq 0$. We need to prove that for any orthonormal system $\{e_i\}_1^{\infty}$ and for any $N$, $$\sum_1^N \langle Ae_i,e_i \rangle \leq \sum_1^N ...
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Show that an operator is weakly compact

If $(X,\Omega,\mu)$ is a finite measure space, $k\in L^\infty(X\times X, \Omega\times \Omega,\mu \times \mu)$ , and $K:L^1(\mu)\to L^1(\mu)$ is defined by $$(Kf)(x)=\int k(x,y) f(y) d\mu(y)$$ show ...
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Weakly compact operator on $c_0$ is compact

Show that if $T\in {\cal B}(c_0)$ and $T$ is weakly compact, then $T$ is compact. My attempt: $T$ is weakly compact, so there is a reflexive space $X$ , and operators $A\in {\cal B}(X,c_0) $ and $B ...
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Compactness of $(x_1,x_2,…)\mapsto(0,x_1,x_2/2…)$

I read that the linear operator in the Hilbert space $\ell_2$ defined by $(x_1,x_2,...,x_n,...)\mapsto(0,x_1,x_2/2,...,x_n/n,...)$ is compact. I wanted to prove it by proving that the image of the ...
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43 views

Prove that T is compact

If $H$ is a Hilbert space with basis $\{\varphi_{k}\}^{\infty}_{k=1}$, how do I show that the operator $T$ defined by $T(\varphi_{k})=\frac{1}{k}\varphi_{k+1}$ is compact and has no eigenvectors? ...
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60 views

$A^{\ast}$ compact $\Rightarrow A$ compact

I read that if $A:E\to E$ is a bounded linear operator where $E$ is a Banach space and $A^{\ast}$ is a compact operator, then $A$ is a compact operator. I know that the converse is true (th. 4 here), ...
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28 views

Compactness of the image of $S^{\ast}$ throw compact operator

I read, on an Italian language version of Kolmogorov-Fomin's Introductory Real Analysis, that, for any Banach space $E$, the unit closed sphere $S^{\ast}$ of $E^{\ast}$ is compact in the $\ast$-weak ...
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Show an operator is compact if $\sum \|Te_n\| < \infty$

Let $H$ be a separable Hilbert space, define a bounded linear operator $T:H \rightarrow H$, show it is compact if $\sum \|Te_n\|_H < \infty$. My attempt: We show that $T(B)$ is totally ...
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Bounded linear operator commuting with every compact operators

Let $A$ be a bounded linear operator on the Banach space $X$. Assuming that $AK = KA$ for every compact operator $K$, how do I show that $A$ must be a scalar multiple of the identity, i.e., we have $A ...
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show that every compact operator on Banach space is a norm-limit of finite rank operators (in a particular way, under the given hypotheses)

Let $X$ be a Banach space and suppose there is a net $\{F_i\}$ of finite-rank operators on $X$ such that $(a)$ $\sup_i||F_i||<\infty$ ; $(b)$ $||F_ix-x||\to 0$ for all $x$ in $X$. Show that if $A$ ...
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An example of a non-compact operator which is equal to (norm) limit of compact operators

For a normed linear space $X$ and a Banach space $Y$, the set of all compact operator from $X$ to $Y$, which is denoted by $K(X,Y)$, is normed closed in $B(X,Y)$. Is there a counterexample which ...
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Perturbation of operators and eigenvalues

Suppose $P\in\mathcal{B}(\mathcal{H})$ is a self-adjoint compact operator. Lets perturb $P$ by multiplying it by a bounded operator $S$ and set $T=PS.$ Then what can be said for the spectrum of $T?$ ...
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Compactness of an operator on $c_0$ in terms of its infinite matrix representation

Let $A\in {\cal B}(c_0)$ (${\cal B}(c_0)$ is linear bounded operators on $c_0$) and for $n\geq 1$, define $e_n \in c_0$ by $e_n(m)=\delta_{nm}$. Put $\alpha_{nm}=(Ae_n)(m)$ for $n,m\geq 1$. we have $M ...
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Proving $u$ is compact whenever $u^\ast$ is

Let $X,Y$ be Banach spaces and let $u: X \to Y$ be a linear operator. Let $u^\ast: Y^\ast \to X^\ast$ denote its transpose and assume that $u^\ast$ is compact. I am trying to prove that $u$ is ...
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60 views

Compactness of the Volterra opelator

The Volterra operator is given as \begin{eqnarray} (Vf)(x)=\int_0^xK(x,y)f(y)\,{\rm d}y. \end{eqnarray} By the Arzelà–Ascoli theorem, $V\colon C^0[0,1]\rightarrow C^0[0,1]$ is compact operator. But, ...
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116 views

Spectrum of an integral operator.

For any $f\in C([0,1],\mathbb{R})$ set $$ Tf(x) = \int_0^1 [\min\{x,y\}\cdot f(y)]dy. $$ I have just proved that $T$ is a compact operator from $C([0,1],\mathbb{R})$ into itself. I would like to know ...
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1answer
25 views

An abstract a priori estimate in finite element method

Let $V$ and $K$ be Banach spaces (with norms $\|\cdot\|_V$ and $\|\cdot\|_K$ resp.) and suppose that there is a compact linear embedding $K\hookrightarrow V$. Furthermore, let $P_n$ be a family of ...
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1answer
26 views

$\sup$ norm of a function

The following is an example of Murphy's C*-algebras and operator theory: I do not know how he concludes $$\int_0^1 |k(s,t) - k(s',t)||f(t)| dt \leq \sup|k(s,t) - k(s',t)|||f||_\infty$$ Please help ...
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1answer
74 views

Integral operator on $L^p$ is compact

Let $(X,\Omega,\mu)$ be an arbitrary measure space, $1<p<\infty$ , and $\frac{1}{p}+ \frac{1}{q} = 1$. If $k:X. X\to \Bbb C$ is an $\Omega.\Omega-$ measurable function such that $$M = [\int ...