Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

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

share|cite|improve this answer
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
@DavideGiraudo Can you explain, please, how a finite rank operator is written as an integral operator? Maybe there is a good reference. – Janko Bracic Jan 14 '15 at 8:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.