1
vote
1answer
15 views

Show compactness of an operator with Arzelà–Ascoli

We have $K\colon L^{2}(a,b) \rightarrow L^{2}(a,b)$ such that $ Kf(t)=\sum_{j=1}^{n}\phi_{j}(t) \int_{a}^{b} \psi_{j}(S) f(s)ds$ where $\phi_{j} ,\psi_{j} \in L^{2}(a,b)$. We want to show that K is ...
1
vote
1answer
49 views

Show the Volterra Operator is compact using only the definition of compact

The Volterra operator $V:L^{2}[0,1]\rightarrow L^{2}[0,1]$ is defined by $(Vf)(x)=\int_0^xf(t)dt$. I am wondering if it can be shown that $V$ is compact by definition - that is, either that $V$ ...
0
votes
0answers
51 views

Show these operators converge to a particular limit

Let $H$ be a Hilbert space, and $T$ be a operator on $H$ of the form $T=\sum_{n=1}^{\infty}{\lambda}_{n}<x,e_{n}>e_{n}$ where $e_{n}$ are the eigenvectors of $T$ and an orthonormal basis of H ...
0
votes
1answer
54 views

The eigenvalues of a compact and self-adjoint operator on Hilbert space

Show that if $K$ is a compact self-adjoint operator on Hilbert space then it has either finitely many eigenvalues or a sequence of eigenvalues $\lambda_n\to 0$ as $n\to \infty$.
0
votes
0answers
33 views

Operators on a Hilbert space question

For a Borel measure $\mu$ define $<S_\mu x,y>=\int_H<x,z><y,z> \mu(z)$. An exercise in my book that I am reading says that I could find a $\mu$ s.t. $S_\mu$ is a bounded operator ...
0
votes
0answers
10 views

characterisation up to unitary equivalence

My book says that the spectral theorem for compact normal operators characterises compact normal operators up to unitary equivalence. It doesn't expand on this so I was wondering what does this mean ...
1
vote
1answer
41 views

The trace class operators are the dual of the compact operators

I know that the map from the trace class operators $L_1(H)$ to the dual of the compact operators $K'(H)$ given by $A \mapsto tr( \cdot A)$ is an isometric isomorphism. Linearity is obvious by the ...
-1
votes
2answers
74 views

Normal compact operator commute with bounded self adjoint operator in Hilbert space.

Suppose $H$ is a Hilbert space and $A:H\rightarrow H$ is a normal compact operator such that $\ker(A)=0$. show that if $B$ is a bounded self adjoint operator that commutes with $A$ then the spaces in ...
4
votes
3answers
76 views

$A^2$ self-adjoint and Compact, prove $A$ has an eigenvalue

Suppose $H$ is a Hilbert space and $A \in L(H)$ is such that $A^2$ is compact and self-adjoint. Prove that $A$ has an eigenvalue. (Here $L(H)$ is the set of bounded linear operators on a Hilbert ...
0
votes
0answers
45 views

Verify that the operator $T$ defined by $T( \varphi _{k})=\frac{1}{k}\varphi _{k+1}$ is compact, but has no eigenvectors.

Let $H$ be a Hilbert space with basis $\left \{ \varphi _{k} \right \}_{k=1}^{\infty }$ .Verify that the operator $T$ defined by $$T( \varphi _{k})=\frac{1}{k}\varphi _{k+1}$$ is compact, but has no ...
2
votes
1answer
79 views

If E is a Hilbert space and $T \in B(E)$ is compact, show $T(E)$ does not contain a closed infinite dimensional subspace

It's the problem from "Essential Results of Functional Analysis," R.J. Zimmer, Chapter 3, problem 3.1. I try to prove this problem and I am confused with the condition "closed infinite dimensional." ...
1
vote
1answer
147 views

Invertibility of a linear operator on a Hilbert space.

Let $H$ be an infinite dimensional Hilbert space over $\mathbb C$, $T$ be a continuous linear operator of $H$, $r(T)=\sup_{||x||=1}|(Tx|x)|$ be the numerical radius of $T$, and $z\in \mathbb C$, such ...
2
votes
1answer
107 views

Want to show an operator is compact

With $V=L^2(0,T;H^1(\Omega))$, let $A:V \to V^*$ with $$\langle Au,v \rangle = \int_0^T \int_{\Omega} \nabla u(t) \cdot \nabla v(t).$$ I want to show that $A$ is a compact operator. So, one way to ...
2
votes
1answer
94 views

Is the limit of compact operators again compact?

Let $(T_n)_{n \in \mathbb{N}} \subset \mathcal{L}(\mathcal{X}, \mathcal{Y})$ where $T_n$, $n \in \mathbb{N}$, is compact. Now, assuming that $(T_n)_{n \in \mathbb{N}}$ has a limit $T \in ...
3
votes
2answers
107 views

Is a Hilbert space $H$ compactly embedded in its dual?

Is a Hilbert space $H$ compactly embedded in its dual? Is it compactly embedded in itself? No idea how to think of this.
2
votes
1answer
46 views

Prove that there $B : H\to H $ bounded such $ B^n = A $.

Let $ A : H\to H $ a compact self-adjoint operator. Suppose $ A $ is positive. let $ n \geq 2 $. Prove that there is $B : H\to H $ bounded such $ B^n = A $.
9
votes
1answer
191 views

Criteria of compactness of an operator

Suppose $K$ is a linear operator in a separable Hilbert space $H$ such that for any Hilbert basis $\{e_i\}$ of $H$ we have $\lim_{i,j \to \infty} (Ke_i,e_j) = 0$. Is it true that $K$ is compact? ...
1
vote
1answer
130 views

Hilbert space proof

$X$ is a separable Hilbert space and $ A\in L(X,X)$ and compact. I need to prove that $A$ is approximately of finite dimension.
2
votes
1answer
112 views

Show for compact operator $K$, if $||Kf|| < ||f|| \forall f$, then $||K|| < 1$.

I wanted to check my reasoning on proving this statement, and see if anyone had suggestions for other proofs of this fact. Again, the statement is, if $K$ is a compact operator on a Hilbert space ...
5
votes
1answer
292 views

Bounded operator and Compactness problem

Let $H$ be a Hilbert space with orthonormal basis $(e_{n})_{n\in\mathbb{N}}$. Furthermore, let $T\colon H\rightarrow C[a,b]$ be a bounded operator. a) Let $x\in [a,b]$. Show that there is a ...
1
vote
2answers
388 views

compact and self adjoint square root of an operator

Let H a Hilbert space and $T:H\rightarrow H$ a linear bounded, self-adjoint, positive and compact operator. How can i prove that the square root of T, $\ T^{1/2}:H\rightarrow H$ is also compact and ...
3
votes
2answers
78 views

Given $F:X \to X$ on $X$:Hilbert space satisfying some properties, prove that $F$ is surjective.

Given a map $F:X \to X$ where $X$ is a Hilbert space, $F$ satisfying $f(x):=x-F(x)$ is a compact map. $\lim_{\|x\|\to \infty} \frac{(F(x),x)}{\|x\|} = \infty$ I'm seeking to prove that $F$ is ...
1
vote
0answers
61 views

If limit of $f(n)$ is zero then the operator is compact

I want to prove the following: Suppose $\mathfrak{H}$ is the Hilbertspace $l^2(\Bbb{N})$ and $T_f$ the multiplication operator on $\mathfrak{H}$, thus $T_f\psi=f\psi$ for ...
6
votes
1answer
247 views

Trace class for operators

Let $ \mathcal{H} $ be a Hilbert space and $ T: \mathcal{H} \to \mathcal{H} $ a bounded linear operator. The $ n $-th singular number $ {\mu_{n}}(T) $ of $ T $ is defined as the distance from $ T $ ...
1
vote
1answer
74 views

How do I prove that a particular linear operator has an orthonormal basis?

I have to show that if $T$ is a linear operator such that $T: L^2(\mathbb(R)^n) \to L^2(\mathbb(R)^n)$ and $T(f)(x) = \int_{R^n}f(y)g(x,y)dy$, where $g(x,y)$ is an $L^2$ function, that there is an ...
1
vote
1answer
113 views

Symmetric bounded linear maps can be approximated by compact symmetric linear maps.

Let $H$ be a separable Hilbert space and let $T:H \rightarrow H$ be a symmetric bound linear map. a) Show that for every orthogonal projection $P$ on $H$ ($P' = P$, $P^2 = P$) PTP is symmetric. b) ...
3
votes
1answer
143 views

Determine the operator T in a Hilbert space

Let $H$ be a Hilbert space and let $\{e_n, n \geq 1\}$ be an orthonormal basis for $H$. a) Determine the operator $T\in B(H)$ that satisfies $$ Te_1 = 0,\; Te_n = \frac{1}{n}e_{n-1}, n ...
4
votes
1answer
568 views

How to prove this integral operator is compact?

$T_kf=\int K(x,y)f(y)dy$ where $K(x,y)=\frac{\phi(x)\phi(y)}{|x-y|^{n-\alpha}}$ $\phi(x)$ is a smooth function on a compact support. $f$ is defined on $R^n$ and $K$ is defined on $R^n\times R^n$ ...
5
votes
1answer
144 views

Show $T$ is compact

$H$ and $K$ are Hilbert Spaces, $(u_n)$ and $(v_n)$ are sequences in $H$ and $K$ respectively. $\sum_{n=1}^{n=\infty} \|u_n\|\|v_n\| $ converges. $T\colon H\rightarrow K$ is defined by ...
3
votes
3answers
468 views

Showing that the orthogonal projection in a Hilbert space is compact iff the subspace is finite dimensional

Suppose that we have a Hilbert Space $H$ and $M$ is a closed subspace of $H$. Let $T\colon H\rightarrow M$ be the orthogonal projection onto $M$. I have to show that $T$ is compact iff $M$ is finite ...
4
votes
1answer
520 views

Hilbert-Schmidt Operator

We have just covered Hilbert-Schmidt operators in class (which I missed) and I am having a hard time getting my head around them. I know the definition: If $H$ is a Hilbert space and ...
0
votes
1answer
151 views

Hilbert space the trace

I need help from someone to solve this problem. Given a bounded sequence $(\lambda_n)$ in $\mathbb С$ define an operator $S$ in $B(\ell_2)$ by $S(x_1) = 0$ and $S(x_n) = \lambda_n x_{n-1}$ , ...
2
votes
2answers
462 views

No Nonzero multiplication operator is compact [duplicate]

Let $f,g \in L^2[0,1]$, multiplication operator $M_g:L^2[0,1] \rightarrow L^2[0,1]$ is defined by $M_g(f(x))=g(x)f(x)$. Would you help me to prove that no nonzero multiplication operator on $L^2[0,1]$ ...
3
votes
1answer
208 views

eigenvalue question

I think this question isn't that hard, but I am a bit confused. Define the linear operator $T_k:H\mapsto H$ by \begin{align} T_ku=\sum^\infty_{n=1}\frac{1}{n^3}\langle u,e_n\rangle e_n+k\langle ...
2
votes
1answer
176 views

Compact operator on $l^2$

Let A be a bounded linear operator on $l^2$ defined by A($a_n$)=($\frac{1}{n} a_n$). Would you help me to prove that A is compact operator. I guess the answer using an approximation by a sequences of ...
0
votes
1answer
82 views

Compact Operator defined by inner product

Let $H$ be a Hilbert space and $y,z \in H$. Define bounded linear operator $Ax=\langle x,y\rangle z$ where $\langle,\rangle$ is inner product. Would you help me to prove that $A$ is compact operator.
2
votes
3answers
418 views

Compact operators and uniform convergence

Suppose $T: H \rightarrow H$ is a compact operator, $H$ is a Hilbert space, and let $(A_n)$ be a sequence of bounded linear operators on $H$ converging strongly to $A$. Show that $A_nT$ converges in ...
1
vote
1answer
169 views

Direct sum of compact operators

I am having some trouble proving this: Let $T_1\in H_1$ and $T_2\in H_2$ where $H_1,H_2$ are Hilbert spaces. Let $T=T_1\oplus T_2$ on $H_1\oplus H_2$. I need to show $T$ is compact iff $T_1$ and $T_2$ ...
3
votes
1answer
175 views

Compact operators between Hilbert spaces

I have the suspect that the following statement is true, but I don't how to prove it. Any suggestion? Thanks to all! Let $X$, $Y$ be Hilbert spaces and let $T \colon X \to Y$ be a linear continuous ...
3
votes
2answers
282 views

The image of orthonormal basis under compact operator

I need a help to prove that statement: if $\{e_n\}$ an orthonormal basis in Hilbert space $H$ and $A$ is a compact operator from $H$ to $H$, then $Ae_n\rightarrow 0$. Thx for any help.
0
votes
1answer
352 views

Square root of compact operator

I'm trying to solve a functional analysis problem A self-adjoint non-negative operator $A$ on a Hilbert space $H$ is compact if and only if its $\sqrt{A}$ is compact.
12
votes
1answer
992 views

How to prove that an operator is compact?

Consider $T\colon\ell^2\to\ell^2$ an operator such that $Te_k=\lambda_k e_k$ with $\lambda_k\to 0$ as $k \to \infty$ how to prove that it is compact?
4
votes
1answer
191 views

Set of all compact operators $K(H)$ is the unique ideal in $B(H)$?

I want to show that the set of all compact operators $K(H)$ is the unique ideal in $B(H)$. Is there any relation between invertibility and compactness of an operator?
5
votes
2answers
381 views

On the isometry between bounded linear operators and the dual of nuclear linear operators

Let $H$ be a separable Hilbert space. Let $(e_i)_i$ be an orthonormal basis. For any bounded linear map $T$ we write, whenever possible $$\operatorname{tr} T := \sum_{i}^{\infty} \langle T e_i, e_i ...