An operator $\( T \)$ is said to be Hilbert-Schmidt if there exists an orthonormal basis $\( \mathopen{}\left({e}_{n}\right)\mathclose{} \)$ for $\( H \)$ such that $$\[ \sum_{n}{} {\mathopen{}\left\lVert{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} \lt \infty \text{.} \]$$ Denote by $\( \mathcal{HS}\mathopen{}\left( H\right)\mathclose{} \)$ the set of all Hilbert-Schmidt operators on $\( H \)$.

1. $\( T \)$ is Hilbert-Schmidt if and only if $\( T^{*}\mathopen{}\left( T\right)\mathclose{}= {\mathopen{}\left\lvert{}T\right\rvert\mathclose{}}^{2}\in \mathcal{T}\mathopen{}\left( H\right)\mathclose{} \)$ since $$\[ \sum_{n}{} {\mathopen{}\left\lVert{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} = \sum_{n}{} \mathopen{}\left\langle{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}, T\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rangle\mathclose{} = \sum_{n}{} \mathopen{}\left\langle{} T^{*}\mathopen{}\left( T\mathopen{}\left( {e}_{n}\right)\mathclose{}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{} \text{.} \]$$
2. It follows from above and Proposition III.29 that $\( \mathcal{HS}\mathopen{}\left( H\right)\mathclose{} \)$ is an ideal of $\( \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \)$.
3. $\( T\in \mathcal{HS}\mathopen{}\left( H\right)\mathclose{} \)$ implies $\( \sum{} {\mathopen{}\left\lVert{}T\mathopen{}\left( {f}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} \lt \infty \)$ for any orthonormal basis $\( \mathopen{}\left({f}_{n}\right)\mathclose{} \)$ of $\( H \)$.

$\( \mathcal{T}\mathopen{}\left( H\right)\mathclose{}\subseteq \mathcal{HS}\mathopen{}\left( H\right)\mathclose{}\subseteq \mathcal{K}\mathopen{}\left( H\right)\mathclose{} \)$.

Here is a rich source of examples of Hilbert-Schmidt operators. Let $\( X \)$ be a measure space with measure $\( μ \)$. (Let us assume, for convenience, that $\( H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( X\right)\mathclose{} \)$ is separable.) Fix $\( k\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( X\times X, μ\times μ\right)\mathclose{} \)$. Think $\( k= k\mathopen{}\left( t, s\right)\mathclose{} \)$. For $\( t\in X \)$, define the slice $\( {k}_{t} \)$ of $\( k \)$ by $\( {k}_{t}\mathopen{}\left( s\right)\mathclose{}= k\mathopen{}\left( t, s\right)\mathclose{} \)$. Then by Fubini's theorem each $\( {k}_{t}\in H \)$ for almost every $\( t \)$ and $$\[ \int _{X}{} {\mathopen{}\left\lVert{}{k}_{t}\right\rVert\mathclose{}}^{2} \,\mathrm{d}μ\mathopen{}\left( t\right)\mathclose{}= \int _{X\times X}{} {\mathopen{}\left\lvert{}k\right\rvert\mathclose{}}^{2} \,\mathrm{d} \mathopen{}\left(μ\times μ\right)\mathclose{} \lt \infty \text{.} \]$$ For $\( f\in H \)$, define the function $\( K\mathopen{}\left( f\right)\mathclose{} \)$ almost everywhere by $$\[ \mathopen{}\left(K\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}= \int _{X}{} k\mathopen{}\left( t, s\right)\mathclose{}f\mathopen{}\left( s\right)\mathclose{} \,\mathrm{d}μ\mathopen{}\left( s\right)\mathclose{}= \mathopen{}\left\langle{}f, \overline{{k}_{t}}\right\rangle\mathclose{} \text{.} \]$$ Then $\( K\mathopen{}\left( f\right)\mathclose{} \)$ is measurable (by Fubini). Further, $$\[ \int _{X}{} {\mathopen{}\left\lvert{}\mathopen{}\left(K\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}\right\rvert\mathclose{}}^{2} \leq \int _{X}{} {\mathopen{}\left\lVert{}f\right\rVert\mathclose{}}^{2}{\mathopen{}\left\lvert{}\overline{{k}_{t}}\right\rvert\mathclose{}}^{2} \,\mathrm{d}μ\mathopen{}\left( t\right)\mathclose{}= {\mathopen{}\left\lVert{}f\right\rVert\mathclose{}}^{2}\int _{X\times X}{} {\mathopen{}\left\lvert{}k\right\rvert\mathclose{}}^{2} \,\mathrm{d} \mathopen{}\left(μ\times μ\right)\mathclose{} \text{,} \]$$ so $\( K \)$ maps $\( H \)$ boundedly into itself, with $\( \mathopen{}\left\lVert{}K\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}k\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}\mathopen{}\left( X\times X\right)\mathclose{}} \)$. Let $\( \mathopen{}\left\{\, {e}_{1}, {e}_{2}, \dotsc\,\right\}\mathclose{} \)$ be an orthonormal basis for $\( H \)$. Then by the calculations above $$\[ \sum_{n}{} {\mathopen{}\left\lVert{}K\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} = \sum_{n}{} \int _{X}{} {\mathopen{}\left\lvert{}\mathopen{}\left\langle{}{e}_{n}, \overline{{k}_{t}}\right\rangle\mathclose{}\right\rvert\mathclose{}}^{2} \,\mathrm{d}μ\mathopen{}\left( t\right)\mathclose{} = \int _{X}{} \sum_{n}{} {\mathopen{}\left\lvert{}\mathopen{}\left\langle{}{e}_{n}, \overline{{k}_{t}}\right\rangle\mathclose{}\right\rvert\mathclose{}}^{2} \,\mathrm{d}μ\mathopen{}\left( t\right)\mathclose{}= \int _{X}{} {\mathopen{}\left\lVert{}{k}_{t}\right\rVert\mathclose{}}^{2} \,\mathrm{d}μ\mathopen{}\left( t\right)\mathclose{}= {\mathopen{}\left(\mathopen{}\left\lVert{}k\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}\mathopen{}\left( X\times X\right)\mathclose{}}\right)\mathclose{}}^{2} \text{.} \]$$ Thus $\( K \)$ is a Hilbert-Schmidt operator on $\( H \)$, and its Hilbert-Schmidt norm (to be defined shortly) coincides with the $\( \mathrm{L}^{\mathrm{2}} \)$ norm of the kernel function $\( k \)$.

For $\( {S}_{1} \)$ and $\( {S}_{2} \)$ in $\( \mathcal{HS}\mathopen{}\left( H\right)\mathclose{} \)$, define $$\[ \mathopen{}\left\langle{}{S}_{1}, {S}_{2}\right\rangle\mathclose{}_{\mathcal{HS}}= \sum{} \mathopen{}\left\langle{}{S}_{1}\mathopen{}\left( {e}_{n}\right)\mathclose{}, {S}_{2}\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rangle\mathclose{} = \operatorname{Tr}\mathopen{}\left( {S}_{2}^{*}{S}_{1}\right)\mathclose{} \text{.} \]$$

This is an inner product. The algebraic properties are obvious, and $\( \mathopen{}\left\langle{}S, S\right\rangle\mathclose{}_{\mathcal{HS}}= 0 \)$ implies $\( S\mathopen{}\left( {e}_{n}\right)\mathclose{}= 0 \)$ for all $\( n \)$ implies $\( S= 0 \)$. We define the Hilbert-Schmidt norm on $\( \mathcal{HS}\mathopen{}\left( H\right)\mathclose{} \)$ by $\( \mathopen{}\left\lVert{}S\right\rVert\mathclose{}_{\mathcal{HS}}= {\mathopen{}\left(\mathopen{}\left\langle{}S, S\right\rangle\mathclose{}_{\mathcal{HS}}\right)\mathclose{}}^{\frac{1}{2}} \)$.

$\( \mathcal{HS}\mathopen{}\left( H\right)\mathclose{} \)$ with $\( \mathopen{}\left\lVert{}\cdot\right\rVert\mathclose{}_{\mathcal{HS}} \)$ is complete (that is, a Hilbert space).

It turns out that $\( \mathcal{HS}\mathopen{}\left( H\right)\mathclose{}\simeq H\otimes \overline{H} \)$, where $\( \overline{H} \)$ is $\( H \)$ with conjugate scalar multiplication.