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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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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When $\int_\Omega f(x)K(x,y)dx$ is injective

Let $T:X\to Y$ be a compact integral operator and consider $g=Tf$, i.e. $$g(y) = \int_\Omega f(x)K(x,y)dx$$ where $\Omega=[a,b]$ is some subset of $\mathbb{R}$. Are there some particular conditions on ...
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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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$A^{\ast}$ 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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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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41 views

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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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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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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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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$\sup$ 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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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 ...
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Spectrum of $Tu=\int^1_{-1} (1-|x-y|)u(y)dy $

Consider the operator $$ Tu(x)=\int^1_{-1} (1-|x-y|)u(y)dy $$ We want to find the spectrum of $T$. The kernel is certainly bounded and so this operator is Hilbert-Schmidt, so $T$ is compact. We ...
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If a compact operator satisfies $T^nx\to0$ weakly for all $x$, then $\|T^n\|\to0$

Let $H$ be a real Hilbert space, $T:H\to H$ be a compact operator. Suppose that for every $x\in H$, sequence $(T^n x)_{n\in \mathbb{N}}$ converges weakly to $0$. How to prove that $ ...
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Compactness of the identity operator

As far as I know, by Rellich-Kondrachov theorem, we can say $I:H_{0}^{k}\to H_{0}^{m}$, for $m<k$ is a compact operator, where: $H_{0}^{k}=\{f\in H_{{}}^{k}|f(0)={f}'(0)=\cdots ={{f}^{(k)}}(0)=0\}$ ...
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Sufficient condition for an operator to be compact in Hilbert space of holomorphic function with respect to Gaussion weight (Fock space).

What I read in a book I could not understand, some one please help. Let $\mathcal{F}=\{f:\mathbb{C^n}\rightarrow\mathbb{C}: \text{$f$ is holomorphic and}\int_{\mathbb{C}^n}\lvert ...
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Prove that operator is completely continuous

Let's consider Banach space $\ell^\infty$ of bounded sequences $x = \{ \xi_n\}_{n=1}^\infty$: $$ ||x|| = \sup_{n \in \mathbb N} |\xi_n|. $$ Suppose matrix $||a_{i j}||_1^\infty$ specifies operator $A$ ...
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Rellich's theorem for Sobolev space on the torus

From John Roe: Elliptic operators, topology and asymptotic methods, page 73: Let $H^{k}$ be the Soblev space defined on the torus $\mathbb{T}^{n}$ with the discrete $k$-norm: $$ \langle f_{1}, ...
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Space of bounded functions vs. bounded space of functions.

Suppose I have a bounded set of functions, say $B\subset C[0,1]$. What exactly does this mean? I.e. is a bounded set of continuous functions equivalent to a set of continuous bounded functions? For ...
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Show $T: C([0,1]) \rightarrow C([0,1])$ is compact

Consider $T: C([0,1]) \rightarrow C([0,1])$ defined by $$(Tf)(t) := \int_0^1 \kappa_t(s)f(s)ds,$$ where $\kappa:[0,1]^2 \rightarrow \mathbb{R}$ satisfies the following properties: for all $t\in ...
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Proof of Weierstrass' second theorem using the Fejér operator

Weierstrass' second theorem states the following: Let $f$ be a real continuous $2\pi$-periodic function (write $f\in C_{2\pi}$). Then for all $\epsilon>0$ there exists a trigonometric polynomial ...
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Canonical inclusion $L^q(0,1) \to L^p(0,1)$ is compact?

Does there exist $q>p$ such that the canonical inclusion $L^q(0,1) \to L^p(0,1)$ is compact? My answer is no. Since we know that $L^\infty (0,1) \to L^p(0,1)$ is not compact, take $\{\sin(nx)\}$ ...
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Compactness of an operator involving the resolvent of laplacian

Let $w\in L^n(\mathbb{R}^n)$, ($n\geq 3$), and for $\tau\in\mathbb{C}$, $Im(\tau)\neq 0$, let $R_{\tau}=(-\Delta-\tau)^{-1}$ be the resolvent of the Laplacian. I need to show that $T:=wR_{\tau}w$ is a ...
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Compact operators on $L^2(G)$ as a reduced cross product of $C_0(G)$ and $G$.

If any of the terminology is unclear then please don't hesitate to point it out. My question is: is it true that when $G$ is a locally compact second countable group then: \begin{equation*} C_0(G) ...
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Show this integral operator is compact for various values of $\alpha$

I am having some problems evaluating a multivariable integral. This question is features in Stakgold's book Green's functions and boundary value problems. page 359. Consider the kernel for $a\leq ...
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If a series converges, does it converge with additional log term multiplied?

If $\sum_{n} |a_n| < \infty$, is it true that $\sum_{n} |a_n\log(a_n)| < \infty$ if $0 \leq a_n \leq 1$? I am trying to see if $A$ is trace class operator, then $A \log(A)$ is also trace class ...
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What is a predual of the Banach space of compact operators on $\ell^2$?

I am wondering if the space $K(\ell^2)$ of compact operators on $\ell^2$ can have a predual. Thank you in advance for your help.
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Cauchy singular integral operator

Help on proving the following equality: $$K(-sgn)=S$$ where $K$ is the operator defined by $K(f)=F^{−1}fF$ ($F$=fourier transform, $f$=any function), sgn is the signum function and S is the Cauchy ...
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Extending a compact operator to the entire Hilbert space

In a course I'm taking we defined compact operators as a linear mapping $H\rightarrow H$, where $H$ is a Hilbert space, that maps bounded sets to relative compact ones. The lecturer mentioned that the ...
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Normal Compact Operator: not diagonalizable!

To proposition 5.17 in Weidmann's 'Lineare Operatoren in Hilberträumen' (german version) it is noted that the expansion of compact operators that are normal rather than self adjoint doesn't apply in ...
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Sum of the Eigenvalues of a Compact Positive-Definite Linear Operator on a Hilbert Space

Let $ A $ be a compact positive-definite linear operator on a Hilbert space $ \mathcal{H} $. Let $ \{ v_{1},v_{2},\ldots,v_{n} \} $ be an orthonormal $ n $-subset of $ \mathcal{H} $. Let $ \lambda_{1} ...
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Question about compact operator

So here is my question, Let $H$ be a Hilbert-space and $K:H\rightarrow H$ a compact operator. I know that if $K$ is self adjoint, then it has one eigenvalue $\mu$ such that $|\mu|=||K||$. Can some ...
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Fredholm Index: Finite Corank $\Rightarrow$Closed Range [duplicate]

Obviously closed subspaces turn quotient spaces into normed spaces rather than just merely vector spaces. However the dimension involved in Freholm's index are purely algebraic. Why do we thus ...
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Showing that a certain operator is compact

So here is my problem, I try to show that following operator is compact, \begin{align} J: h_1 & \rightarrow\ell^1 \\ (x_n) & \mapsto(x_n) \end{align} where $$h_1:=\left\{x_n\in ...
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Question about a counterexample concerning compact operators

Does anybody know if the following is true, Let $H$ be an infinite dimensional Hilbert-space and $K:H\rightarrow H$ a compact operator. Then if $|\mathrm{spec}(K)|<\infty$ i.e the spectrum is ...
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37 views

Question if an operator is compact

So here is my problem, Let $$J_p:\ell^p\rightarrow c_o$$ be the canonical embedding where $c_0:=\{x_n\subseteq\mathbb C:x_n\rightarrow 0\quad n \rightarrow\infty\}$. I have to decide whether the ...
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33 views

Question about an integral operator

So here is my question, I know that the operator $$T:L^2[0,1]\rightarrow L^2[0,1]$$ $$f\mapsto(Kf)(x)=\int_{[0,1]}k(x,y)f(y)\;dy$$ for a function $k$ continuous on $[0,1]^2$ is compact. Is this also ...
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What makes compact operators special?

I would like to understand why compact operators are considered so special to consider them as an extra class of operators. Over Hilbert spaces these (as far as I know) these are the ones with ...
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*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 ...