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Let $(X,\mu)$ be a $\sigma$-finite measure space. If $K\in\mathcal{L}^2(X\times X,\mu\times\mu)$ then the map $A_K:\mathcal{L}^2(X,\mu)\to\mathcal{L}^2(X,\mu)$ defined by\begin{equation} A_Kf(x)=\int_XK(x,y)f(y)d\mu(y) \end{equation} is Hilbert-Schmidt.

But Arveson (Proposition 2.8.6) says this $K\mapsto A_K$ is an isomorphism from $\mathcal{L}^2(X\times X,\mu\times\mu)$ to the space of Hilbert-Schmidt operators on $\mathcal{L}^2(X,\mu)$.

So in particular this map is onto. I do not know how to prove this. I tried to focus on the easiest case $X=[0,1]$ but still got no progress.

Can someone give a hint? Thanks!

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@JonasMeyer It is Proposition 2.8.6 as in my book. Thanks! –  Hui Yu Dec 14 '12 at 16:45

1 Answer 1

up vote 3 down vote accepted

A Hilbert-Schmidt operator is compact (see proposition 2.8.4), and in a Hilbert space a compact operator is a norm limit of finite ranked operators. So let $T$ a Hilbert-Schmidt operator on $L^2(X,\mu)$; and $T_n$ a sequence of finite-ranked operators which converge in norm to $T$.

Each finite ranked operator can be written as an integral operator, so write $T_n$ as $A_{K_n}$. As $K\to A_K$ is an isometry, the sequence $\{K_n\}$ is Cauchy in $L^2(X\times X,\mu\otimes \mu)$. And the limit does the job.

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But if this was true, then it would imply every compact operator is Hilbert-Schmidt. –  Hui Yu Dec 15 '12 at 14:50
    
Since it would imply that every compact operator is of an integration against a kernel in $\mathcal{L}^2(X\times X,\mu\times\mu)$ and such kind of operators are Hilbert-Schmidt. –  Hui Yu Dec 15 '12 at 14:51
    
I meant an isometry when $HS(L^2)$ is endowed with the Hilbet-Schmidt norm. That's what makes all work. –  Davide Giraudo Dec 15 '12 at 15:40
    
Get it! Thanks! –  Hui Yu Dec 16 '12 at 6:57

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