7
$\begingroup$

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!

1 Answers 1

7

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.

  • 0
    That's what I thought too. But thinking of that, I got the answer. For that you have to write down $\alpha(x)$ with $x \in L^2$ and use $x = \sum_{j=1}^\infty\langle x, e_j\rangle_{L^2}e_j$ where $(e_j)$ is an orthonormal basis of $L^2$ (ideally you should use the orthonormal system $(e_1, ..., e_N)$ of $\alpha$ with $N \in \mathbb{N}$ [since $\alpha$ is finite rank]). Then you should get your representation (otherwise feel free to ask).2017-02-07