Hilbert–Schmidt operator and Hilbert–Schmidt norm

Throughout, $H$ and $K$ are real or complex Hilbert spaces with the pairing linear in the first argument and conjugate-linear in the second (Hilbert space), $T\in \mathcal{B}(H,K)$ is a bounded linear operator (A bounded linear operator between normed spaces, The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators) of operator norm $\mathrm{\parallel }T\mathrm{\parallel }$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum), and $E$ is a Hilbert basis of $H$, that is, a complete orthonormal subset of $H$ (Orthonormal families, complete orthonormal systems and Hilbert bases). The basis $E$ is supplied as data; the definitions below are stated relative to it.

The Hilbert–Schmidt square-sum. The family $(\mathrm{\parallel }Te{\mathrm{\parallel }}^2{)}_{e\in E}$ consists of nonnegative real numbers and is indexed by the arbitrary set $E$. Its sum is the supremum of the finite subsums,

$s_E(T):={\sum }_{e\in E}\mathrm{\parallel }Te{\mathrm{\parallel }}^2:=\sup \{{\sum }_{e\in F}\mathrm{\parallel }Te{\mathrm{\parallel }}^2:F\subseteq E\text{ finite}\}\in [0,+\mathrm{\infty }],$

in the finite-subset-supremum convention of Square-summable families on an arbitrary index set and the space ${\mathrm{\ell }}^2(I)$; the empty finite subset contributes the empty sum $0$, so the supremum is over a nonempty set and exists in $[0,+\mathrm{\infty }]$. No enumeration, ordering or countability of $E$ is used or asserted, and no choice is performed: the supremum ranges over the set of finite subsets of the fixed index set $E$. For finite $E$ the supremum is the ordinary finite sum over $E$, because all terms are nonnegative.

Hilbert–Schmidt relative to a basis. The operator $T$ is Hilbert–Schmidt relative to $E$ when $s_E(T)<+\mathrm{\infty }$, that is, when the finite subsums are bounded above in $\mathbb{R}$. In that case the Hilbert–Schmidt norm of $T$ relative to $E$ is the nonnegative square root

$\mathrm{\parallel }T{\mathrm{\parallel }}_{HS,E}:=({\sum }_{e\in E}\mathrm{\parallel }Te{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}\in [0,+\mathrm{\infty })\mathrm{.}$

When $s_E(T)=+\mathrm{\infty }$, no Hilbert–Schmidt norm relative to $E$ is defined, and $T$ is not Hilbert–Schmidt relative to $E$.

Degenerate and extreme cases. The zero operator satisfies $s_E(0)=0$ for every basis $E$, so it is Hilbert–Schmidt relative to every basis with norm $0$. If $H=\{0\}$ then the empty family is a Hilbert basis of $H$, the only finite subset is empty, and $s_{\mathrm{\varnothing }}(T)=0$ for the only linear operator $T:\{0\}\to K$; that operator is Hilbert–Schmidt relative to the empty basis with norm $0$. If $H$ is finite dimensional and $E$ is finite, $s_E(T)$ is an ordinary finite sum of the squared norms $\mathrm{\parallel }Te{\mathrm{\parallel }}^2$, and $T$ is automatically Hilbert–Schmidt relative to $E$.

The basis is part of the notation. The symbol $\mathrm{\parallel }T{\mathrm{\parallel }}_{HS,E}$ keeps the basis in the subscript on purpose. Until the next result is proved, the phrase "$T$ is Hilbert–Schmidt" is never used without a specified Hilbert basis, and nothing here asserts that a Hilbert basis of $H$ exists: the existence of $E$ is a hypothesis of the definition, and the question of whether the finiteness of $s_E(T)$ and the value $\mathrm{\parallel }T{\mathrm{\parallel }}_{HS,E}$ depend on $E$ is taken up in The Hilbert–Schmidt norm is basis independent. In particular this definition makes no basis-existence claim and no comparison with the operator norm $\mathrm{\parallel }T\mathrm{\parallel }$; every such statement is proved later, where its own hypotheses are displayed.

Notation. For a finite $F\subseteq E$ we write $s_F(T):={\sum }_{e\in F}\mathrm{\parallel }Te{\mathrm{\parallel }}^2$ for the finite subsum, so that $s_E(T)=\sup \{s_F(T):F\subseteq E\text{ finite}\}$.