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We have just covered Hilbert-Schmidt operators in class (which I missed) and I am having a hard time getting my head around them. I know the definition:

If $H$ is a Hilbert space and $T\in\mathcal{L}(H)$ then $T$ is Hilbert-Schmidt if there is some orthonormal basis of $H$ such that: $$\sum_n \lVert T(e_n)\rVert^2<\infty$$

So I am a bit unsure what this is saying about the operator? Is this saying that the operator is "bounded enough" in some sense, to do something?

Thanks very much for the help (sorry for my confusion and rubbish question)

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1 Answer 1

up vote 8 down vote accepted
  • Hilbert-Schmidt operators are more than bounded, as in the case of separable Hilbert space they are compact. Indeed, if $\{e_n\}$ is an orthonormal basis, and $P_N$ the projection over $\operatorname{Span}\{e_j,1\leqslant j\leqslant N\}$, then $\{TP_N\}$ converges in norm to $T$ and $TP_N$ is finite ranked.

  • The Hilbert-Schmidt operators form an ideal of the set of bounded operators.

  • An interest of the Hilbert-Schmidt operators is that it can be endowed with an inner product, defining $$\langle S,T\rangle_{HS}:=\sum_{j=1}^{+\infty}\langle Se_n,Te_n\rangle.$$

  • It can be shown with Bessel's equality that this doesn't depend on the choice of the Hilbert basis.

For applications, I don't know too much. In the context of probability measures on the Borel subsets of the topology induced by an inner product space (with a countable Hilbert basis), Hilbert-Schmidt operators help us to characterize the sets of probability measures which are tight. In Araujo and Giné's book Probability measures in Banach spaces, we encounter the following result:

Theorem 1.4.17. Let $H$ be a separable Hilbert space, and $\Gamma$ a set of probability measures on the Borel $\sigma$-algebra of $H$. The set $\Gamma$ has a compact closure for the weak-$^*$ topology if and only if for all $\varepsilon>0$, we can find a family of Hilbert-Schmidt operators on $H$ $\{A_{\mu}^\varepsilon\}_{\mu\in\Gamma}$ such that for a Hilbert basis $\{e_j\}$, the following properties hold:

  1. $\displaystyle\sup_{\mu\in\Gamma}\sum_{j=1}^{+\infty}\lVert A_\mu^\varepsilon(e_j)\rVert^2<\infty$;
  2. $\displaystyle\lim_{N\to +\infty}\sup_{\mu\in\Gamma}\sum_{j=N}^{+\infty}\lVert A_\mu^\varepsilon(e_j)\rVert^2=0$;
  3. for all $v\in H$, $\mu\in\Gamma$, $$\left|1-\int_He^{i\langle v,x\rangle_H}d\mu(x)\right|\leqslant \lVert A\mu^{\varepsilon}(v)\rVert+\varepsilon.$$
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