Tagged Questions

28 views

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 ...
23 views

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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Compact operator whose range is not closed

I am asked to find a compact operator (on a Hilbert space) whose range is not closed, but I am having trouble coming up with one. My guess is that you need to have some sequence in the range that ...
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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 ...
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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 $V$ ...
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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 ...
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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 $n\to \infty$.
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Operators on a Hilbert space question

For a Borel measure $\mu$ define $\langle S_\mu x,y\rangle=\int_H\langle x,z\rangle \langle y,z\rangle \mu(z)$. An exercise in my book that I am reading says that I could find a $\mu$ s.t. $S_\mu$ ...
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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 ...
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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 ...
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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 ...
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$A^2$ 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 ...
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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 ...
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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." ...
205 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 ...
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 ...