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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Closure of compact operators?

If $X,Y$ are normed spaces (not nec complete) and $A_n:X\rightarrow{Y}$ bounded linear operators of finite rank converging in the operator norm to $A$ is $A$ compact? A diagonal argument and cauchy ...
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Compact operator in Hilbert spaces reach the maximum in the sphere.

I found the following question in my textbook: (QUESTION) Let $\mathcal{H}$ a Hilbert space and $T: \mathcal{H} \rightarrow \mathcal{H}$ a compact operator. Show that exists $x \neq 0$ in ...
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the c*-algebra generated by the Volterra operator

Let V be the Volterra operator on $\mathscr{L^2(0,1)}$.$V(f)(x)=\int_{0}^{x}{f(y)dy}$. Show that $C^*(V)$, the smallest C* algebra generated with V, is $\mathbb{C}+\mathscr{B_0(L^2(0,1))}$ where ...
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Compact operator with no non-zero eigenvalues is zero?

Suppose we have a Hilbert space $H$ and a compact operator $T$ acting on $H$. If $T$ has no non-zero-eigenvalues, is it necessarily the zero operator? Secondly, if I decompose $H$ into eigenspaces of ...
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Spectrum of operator $T((x_n)_{n\in\mathbb{Z}})=\left(\frac{1}{n^2+1}(x_n-x_{-n})\right)_{n\in\mathbb{Z}}$

The eigenvalues should satisfy: $$T(x_n)=\lambda x_n$$ $$\frac{1}{n^2+1}(x_n-x_{-n})=\lambda x_n$$ $$\left[(n^2+1)\lambda+1\right]x_n=x_{-n}$$ I suppose that this should mean that ...
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Trace-Operator is compact?

Assume $\Omega$ is a smooth, compact riemannian manifold with non-empty smooth boundary $\partial \Omega$. Let $T$ be the Trace-Operator $T \colon H^1(\Omega) \to L^2(\partial \Omega), \ f \mapsto ...
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Bounds on eigenfunctions of integraloperator

Let $K: [0,1]\times[0,1] \to \mathbb{R}$ be a symmetric positive definit and continuous function. It is known my Mercer's theorem that $$ [T_K \varphi](x) =\int_0^1 K(x,s) \varphi(s)\, ds $$ is ...
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Dual operator of a compact operator

Why is the dual operator $A^{\ast}$ of a compact operator $A:X \rightarrow Y$, where $X,Y$ are two Banach spaces again compact?
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Different versions of Mercer's theorem

I am reviewing materials on reproducing kernel Hilbert space (RKHS) and I've found various versions of Mercer's theorem: About the positive-definiteness conditions. In the Wikipedia pages on RKHS ...
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Nontrivial closed ideal of $\mathbb{B(H)}$, $\mathbb{H}$ is a non-separable Hilbert space.

$\mathbb{H}$ is a non-separable Hilbert space. Give an example of nontrivial closed ideal $I$of $\mathbb{B(H)}$, that is different from $\mathbb{B_0(H)}$ which is the ideal of compact operators. Any ...
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Sequence of operators that commute imply the limit commutes?

Given a sequence of compact operators $A_n\to A$ as $n\ \to \infty$ and $B$ (which has finite rank). $\varphi \in L^2([a,b])$ If $A_nB\varphi = BA_n \varphi$ Am I able to say anything about $A$, ...
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$T:\ell^2 \to \ell^2$ is a compact operator

Any example in Kreyszig: Introductory Functional Analysis with Applications: Prove compactness of $T:\ell^2 \to \ell^2$ defined by $y=(\eta_j)=Tx$ where $\eta_k=\xi_j/j$ for $j=1,2,\dots$. Defined ...
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38 views

can a sum of two non-compact operators be compact?

I'm supposed to say, whether an operator $$Tf(t)=f(1-t)$$ can be expressed as $$T=\lambda I-K$$ where $K$ is a compact operator. As $T$ is not compact, I suppose that $K=\lambda I-T$ can't be compact ...
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Modified shift operator is compact.

For the operator $$T(\eta_j) = \frac{\eta_{j+1}}{j}$$ on Hilbert Space $H$ where $(\eta_j)$ is a basis. Show it is compact. Can this work? Define $$f = (\eta_j)_{j \geq 1}$$ $$T_N(f) = ...
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Weak convergence = norm convergence for trace class operators?

Given a (separable) Hilbertspace $H$, I look at the traceclass operators $\mathfrak{S}_1$. I recall the fact that the weak convergence implies norm convergence in the sequence space $\mathcal{l}^1$. ...
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Compact operator for which the image of the closed unit ball is not compact [duplicate]

Do you have an example of a compact operator for which the image of the closed unit ball is not compact?
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compactness in $\ell^2$

How can I show T is compact when T is defined as $$ \text{T :}\,\ell^2 \to\ell^2\,\text{by Tx=y where} \,y_j=\alpha_jx_j\text{and}\,\alpha_j\to0\,\text{as}\,n\to\infty$$
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An operator satisfying in a sequence of equations

Assume that $H$ is a non-separable Hilbert space. Let $\{\eta_n\}$ be an arbitrary sequence in $H$. Let $\{\zeta_n\}$ be a sequence in $H$ which forms a linearly independent set. Does there exist ...
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bounded operator whose image is contained in the image of a compact operator

Let $A,K$ be 2 linear operators from $X$ to $Y$ (Banach spaces). $A$ is bounded and $K$ is compact. If $A(X)\subset K(X)$, is it true that $A$ is alse compact? I know that if the rank of $K$ is ...
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A translation invariant sigma algebra in $B(H)$

Assume that $H$ is a non-separable Hilbert space. Let $s_0$ be the family of all basic neighborhoods in the strong operator topology. We denote $M_s$ by the sigma algebra generated by $s_0$. ...
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Eigenvalues of an integral operator

The following operator is defined on $L_2(0,1)$: $$Kf(t)=\int_0^1|s-t|f(s)ds$$ I am wondering how I can calculate the eigenvalues and eigenfunctions of such an operator. I start with ...
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27 views

Measure of non-compactness

Can someone give me some simple examples of measure of non-compactness of sets in Banach spaces or metric spaces, which are easy to understand.
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Compactness of a certain Integral Operator on $L^2$

Let $X\subset \Bbb R^n$ be a compact subset and $\alpha \in (0,n)$. Let $m:X\times X\to \Bbb R$ be a bounded measurable function. Consider the integral operator $$Tf(x)=\int_X ...
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Prove the following are equivalent.

Let $E$ and $F$ be Hilbert spaces. For, $T \in B(E,F)$, show that the following are equivalent: $(i)$ $T$ is compact $(ii)$ $T^*$ is compact $(iii)$ There exists a sequence of finite rank operators ...
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T compact if and only if T*T is compact.

I have an operator $T \in B(\mathcal{H})$. I need to prove that T is comapct if and only if $T^*T$ is compact. One way is ok, because if A or B is comapct then AB is compact, so I get at once that if ...
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Strong convergence due to Compact Operator [duplicate]

Given a sequence $u_{n}$ such that: $u_{n} \rightharpoonup 0$ in $L^{2}(\mathbb R^{n})$ & $A$ is a compact operator. The problem is to show that : $Au_{n} \rightarrow 0$ in $L^{2}(\mathbb R^{n})$ ...
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35 views

Are Hilbert-Schmidt operators in non-separable Hilbert spaces compact?

The definition of Hilbert-Schmidt operator should still be valid even when the Hilbert space is not separable: If $e_i$ for $i\in I$ is an orthonormal basis for a Hilbert space, and ...
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Compact operator in terms of exact sequences

I know a pretty equivalent definition for Fredholm operators in terms of exact sequences. Here is it: We called operator $S : E \to F$ between Banach spaces $E$ and $F$ as Fredholm iff exist exact ...
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trace norm and tensor product

Let $(M_n (\mathbb{C}), n\|.\|)$ , $(M_n (\mathbb{C}), n\|.\|)$ and $(M_{nm} (\mathbb{C}), nm\|.\|)$ be three Banach algebras. where $$\|A\| = \mathrm{tr}\sqrt{(A^* A)}. $$ What is the norm of $\phi$ ...
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Functional Analysis, infimum of $A+K$ where $K\in K(X)$

This is a problem from Martin Schechter's Book (Principles of Functional Analysis) If $X$ is a Banach Spaces and $A\in B(X)$, let $$|A|_K=\inf_{K\in K(X)}\|A+K\|.$$ Show that $|A|_K<1$ implies ...
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$T$ is a compact operator but not a finite rank operator

Prove that if $X$ is a Banach Space and if $T$ is a compact operator but is not a finite rank operator, then $0 \in \bar {T(S_X)}$(where $S_X$ is the unite sphere). Please provide some solution ...
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Prove compactness of an operator.

Suppose $$ X=\left\{x \in C^2(\Bbb R,\Bbb R):x(t+T)=x(t)\; \text{for all}\;t \in \Bbb R \right\}, $$ $$ Y=\left \{h \in C(\Bbb R,\Bbb R):h(t+T)=h(t)\;\text{for all}\;t \in \Bbb R \right \} ,$$ and ...
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C*-algebraic intrinsic definition for compactness of an operator?

Some properties of operators (normal, self adjoint, hermitian) have intrinsic definitions for any element of a $C^*$-algebra. Is there such definition for compact operators? Equivalently: Let ...
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$\mathcal{A}+K$ is norm-closed where $\mathcal{A}$ is a $C^*$-algebra and $K$ is the compact operators.

Let $\mathcal{A}\subset B(H)$ be a unital $C^*$-algebra and let $K$ be the closed ideal of compact operators. I need to show that $\mathcal{A}+K$ is also a $C^*$-subalgebra of $B(H)$. I am stuck at ...
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Compactness of Single Layer operator

The single layer operator defined by $S: H^{-1/2}(\Gamma) \rightarrow H^{1/2}(\Gamma), \, Sf(x) = \int_{\Gamma} G(x,x') f(x') dx'$ with $G(x,x')$ the Green's function and $H^s$ the Sobolev space of ...
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Compactness of the trace operator

Is it true that for a set $\Omega$ with Lipschitz boundary the trace operator $T : H^1(\Omega) \to L^2(\partial \Omega)$ is compact? Can you please give a reference? I found a theorem in Necas' ...
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Looking for a general approach to proving compactness of linear operators.

Preparing for my exam in functional analysis, I often have to prove that certain explicitly given operators are compact. I now have a decent amount of operators of which I can prove the compactness, ...
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How does the definition of compactness imply that all continuous operators are compact in finite dimensional spaces?

Let $S \subset X, Y$ be normed spaces over $K$. An operator $A:S \to Y$ is called compact if: $A$ is continuous $A$ transforms bounded set into relatively compact sets i.e. if $(c_n)$ ...
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Matrix representatin of a compact operator

Fact : I know for every compact operator $x : H \to H$, there are sequences $\{\xi_n\}$ and $\{\eta_n\}$ of orthonormal vectors of separable Hilbert space $H$, and sequence $\{\alpha_n\}$ in $\Bbb C$ ...
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Question about norm in trace class

I'm having troubles to proof the next inequality $$\|A\|\leq\|A\|_1$$ where A is an operator in the trace class and $\|A\|_1=Tr|A|$. And $\|A\|$ is the norm of operators. I just got this ...
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Proving that an operator $T$ on a Hilbert space is compact

Let $H$ be a Hilbert space, $T:H \to H$ be a bounded linear operator and $T^{*}$ be the Hilbert Adjoint operator of $T$. Show that $T$ is compact if and only if $T^{*}T$ is compact. My attempt: ...
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Existence of a surjective compact linear operator on an infinite dimensional Banach space

Does there exist a surjective compact operator $T:l^{\infty} \to l^{\infty}$ ? Even though this might be tagged as a repeat question, i still have some doubts that i would like to clarify. My ...
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Proving that an operator is Compact

I have to check that the following operator $T$ is compact: Define $T:l^{2} \to {l^2}$ by $Tx=y=(\eta_{j})$, where $x=(\zeta_{j})$ and $(\eta_{j})=\sum_{k=1}^\infty \alpha_{jk} \zeta_{j}$ and ...
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Compactness of a linear operator

The question is as follows: Show that a linear operator $T:X\to Y$ where $X$ and $Y$ are normed spaces is compact if and only if the image $T(M)$ of the unit ball $M\subset X$ is relatively compact ...
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Spectrum of compact operators on an infinite dimensional normed space

The question is as follows: Let $T:X \to X$ be a compact linear operator on a normed space. If the $dim X= \infty$ then show that $0 \in \sigma(T)$. My attempt: Suppose on the contrary that $0 ...
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How do you quickly show whether an operator is compact?

In my text on functional analysis, the author defines an operator on normed space as compact if it: continuous transforms bounded sets into relatively compact sets Okay, number 1 we can work with. ...
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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 ...
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Compact diagonal operator

Suppose $A : H \to H$, where $H$ is a Hilbert space, is bounded. Also, $A$ is a diagonal operator with diagonal $\{a_n\}$. Show: If $A$ is compact, then $a_n \to 0$ as $n \to \infty$. Should I prove ...
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Is there an error in the solution for this exercise?

I have this exercise: H is a complex hilbert space. And T is a compact operator on H. Show that if H is not separable, then 0 is an eigenvalue of T. Hint: Use lemma 1, and theorem 2. The ...
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Can a positive definite kernel expanded as the product form with an arbitrary orthonormal system?

Notations mostly follow https://en.wikipedia.org/wiki/Mercer%27s_theorem. Mercer's theorem uses the eigenfunctions $\{e_j\}$ of the integral operator as the expansion function. I wonder if we could ...