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Let

  • $H$ be a $\mathbb R$-Hilbert space
  • $A$ be a compact and self-adjoint bounded linear operator on $H$
  • $I:=\left\{n\in\mathbb N:n\le\operatorname{rank}A\right\}$

By the Hilbert-Schmidt theorem, there is a $(\lambda_i)_{i\in I}\subseteq\mathbb R\setminus\left\{0\right\}$ with $$Ae_i=\lambda_ie_i\;\;\;\text{for all }i\in I$$ for some orthonormal basis $(e_i)_{i\in I}$ of $\overline{AH}$.

Now, if $A$ is nonnegative (i.e. $\langle Ax,x\rangle_H\ge0$ for all $x\in H$) and has finite trace, why can we conclude that $\overline{AH}=H$?

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1 Answer 1

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Your claim is incorrect, even for finite dimensional spaces: let $H=\mathbb{R}^2$ and pick $A$ to be the operator corresponding to the matrix $\left(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}\right)$. Then $A$ is selfadjoint (since the matrix is symmetrical), nonnegative (since the eigenvalues $0$, $1$ are both nonnegative), and, trivially, has finite trace. However, $\overline{AH}=AH=\text{span}\{e_1\}\neq\mathbb{R}^2=H$.

Even simpler (but perhaps a little too trivial) is the operator $Ax:=0_H$ for any Hilbert space $H$.

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  • $\begingroup$ I've made a thought in the wrong direction. It is the orthonormal basis which can be enriched (by an orthonormal basis of $\ker A$) to an orthonormal basis of $H$. $\endgroup$
    – 0xbadf00d
    Commented Aug 15, 2017 at 23:20
  • $\begingroup$ @0xbadf00d Oh alright...btw, that's how I found my counter example: I took an easy operator (the identity), enlarged $H$ and threw the additional vector into $\text{ker} A$. $\endgroup$
    – haemi
    Commented Aug 16, 2017 at 21:38

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