Hilbert–Schmidt operator

In mathematics, a Hilbert–Schmidt operator, named after David Hilbert and Erhard Schmidt, is a bounded operator $$ A \colon H \to H $$ that acts on a Hilbert space $$ H $$ and has finite Hilbert–Schmidt norm

$$\|A\|^2_{\operatorname{HS}} \ \stackrel{\text{def}}{=}\ \sum_{i \in I} \|Ae_i\|^2_H,$$

where $$\{e_i: i \in I\}$$ is an orthonormal basis. The index set $$ I $$ need not be countable. However, the sum on the right must contain at most countably many non-zero terms, to have meaning. This definition is independent of the choice of the orthonormal basis. In finite-dimensional Euclidean space, the Hilbert–Schmidt norm $$\|\cdot\|_\text{HS}$$ is identical to the Frobenius norm.

||·||$HS$ is well defined
The Hilbert–Schmidt norm does not depend on the choice of orthonormal basis. Indeed, if $$\{e_i\}_{i\in I}$$ and $$\{f_j\}_{j\in I}$$ are such bases, then $$ \sum_i \|Ae_i\|^2 = \sum_{i,j} \left| \langle Ae_i, f_j\rangle \right|^2 = \sum_{i,j} \left| \langle e_i, A^*f_j\rangle \right|^2 = \sum_j\|A^* f_j\|^2. $$ If $$e_i = f_i, $$ then $ \sum_i \|Ae_i\|^2 = \sum_i\|A^* e_i\|^2. $ As for any bounded operator, $$ A = A^{**}. $$ Replacing $$ A $$ with $$ A^* $$ in the first formula, obtain $ \sum_i \|A^* e_i\|^2 = \sum_j\|A f_j\|^2. $ The independence follows.

Examples
An important class of examples is provided by Hilbert–Schmidt integral operators. Every bounded operator with a finite-dimensional range (these are called operators of finite rank) is a Hilbert–Schmidt operator. The identity operator on a Hilbert space is a Hilbert–Schmidt operator if and only if the Hilbert space is finite-dimensional. Given any $$x$$ and $$y$$ in $$H$$, define $$x \otimes y : H \to H$$ by $$(x \otimes y)(z) = \langle z, y \rangle x$$, which is a continuous linear operator of rank 1 and thus a Hilbert–Schmidt operator; moreover, for any bounded linear operator $$A$$ on $$H$$ (and into $$H$$), $$\operatorname{Tr}\left( A\left( x \otimes y \right) \right) = \left\langle A x, y \right\rangle$$.

If $$T: H \to H$$ is a bounded compact operator with eigenvalues $$\ell_1, \ell_2, \dots$$ of $$|T| = \sqrt{T^*T}$$, where each eigenvalue is repeated as often as its multiplicity, then $$T$$ is Hilbert–Schmidt if and only if $\sum_{i=1}^{\infty} \ell_i^2 < \infty$, in which case the Hilbert–Schmidt norm of $$T$$ is $\left\| T \right\|_{\operatorname{HS}} = \sqrt{\sum_{i=1}^{\infty} \ell_i^2}$.

If $$k \in L^2\left( \mu \times \mu \right)$$, where $$\left( X, \Omega, \mu \right)$$ is a measure space, then the integral operator $$K : L^2\left( \mu \right) \to L^2\left( \mu \right)$$ with kernel $$k$$ is a Hilbert–Schmidt operator and $$\left\| K \right\|_{\operatorname{HS}} = \left\| k \right\|_2$$.

Space of Hilbert–Schmidt operators
The product of two Hilbert–Schmidt operators has finite trace-class norm; therefore, if A and B are two Hilbert–Schmidt operators, the Hilbert–Schmidt inner product can be defined as

$$\langle A, B \rangle_\text{HS} = \operatorname{Tr}(A^* B) = \sum_i \langle Ae_i, Be_i \rangle.$$

The Hilbert–Schmidt operators form a two-sided *-ideal in the Banach algebra of bounded operators on $H$. They also form a Hilbert space, denoted by $B_{HS}(H)$ or $B_{2}(H)$, which can be shown to be naturally isometrically isomorphic to the tensor product of Hilbert spaces

$$H^* \otimes H,$$

where $H^{∗}$ is the dual space of $H$. The norm induced by this inner product is the Hilbert–Schmidt norm under which the space of Hilbert–Schmidt operators is complete (thus making it into a Hilbert space). The space of all bounded linear operators of finite rank (i.e. that have a finite-dimensional range) is a dense subset of the space of Hilbert–Schmidt operators (with the Hilbert–Schmidt norm).

The set of Hilbert–Schmidt operators is closed in the norm topology if, and only if, $H$ is finite-dimensional.

Properties

 * Every Hilbert–Schmidt operator $T : H → H$ is a compact operator.
 * A bounded linear operator $T : H → H$ is Hilbert–Schmidt if and only if the same is true of the operator $\left| T \right| := \sqrt{T^* T}$, in which case the Hilbert–Schmidt norms of T and |T| are equal.
 * Hilbert–Schmidt operators are nuclear operators of order 2, and are therefore compact operators.
 * If $$S : H_1 \to H_2$$ and $$T : H_2 \to H_3$$ are Hilbert–Schmidt operators between Hilbert spaces then the composition $$T \circ S : H_1 \to H_3$$ is a nuclear operator.
 * If $T : H → H$ is a bounded linear operator then we have $$\left\| T \right\| \leq \left\| T \right\|_{\operatorname{HS}}$$.
 * $T$ is a Hilbert–Schmidt operator if and only if the trace $$\operatorname{Tr}$$ of the nonnegative self-adjoint operator $$T^{*} T$$ is finite, in which case $$\|T\|^2_\text{HS} = \operatorname{Tr}(T^* T)$$.
 * If $T : H → H$ is a bounded linear operator on $H$ and $S : H → H$ is a Hilbert–Schmidt operator on $H$ then $$\left\| S^* \right\|_{\operatorname{HS}} = \left\| S \right\|_{\operatorname{HS}}$$, $$\left\| T S \right\|_{\operatorname{HS}} \leq \left\| T \right\| \left\| S \right\|_{\operatorname{HS}}$$, and $$\left\| S T \right\|_{\operatorname{HS}} \leq \left\| S \right\|_{\operatorname{HS}} \left\| T \right\|$$. In particular, the composition of two Hilbert–Schmidt operators is again Hilbert–Schmidt (and even a trace class operator).
 * The space of Hilbert–Schmidt operators on $H$ is an ideal of the space of bounded operators $$B\left( H \right)$$ that contains the operators of finite-rank.
 * If $A$ is a Hilbert–Schmidt operator on $H$ then $$\|A\|^2_\text{HS} = \sum_{i,j} |\langle e_i, Ae_j \rangle|^2 = \|A\|^2_2$$ where $$\{e_i: i \in I\}$$ is an orthonormal basis of H, and $$\|A\|_2$$ is the Schatten norm of $$A$$ for $p = 2$. In Euclidean space, $$\|\cdot\|_\text{HS}$$ is also called the Frobenius norm.