Hilbert–Schmidt operator

Suppose that ${\displaystyle \left(H,\|\cdot \|\right)}$ is a Hilbert space. If ${\displaystyle \{e_{i}:i\in I\}}$ is an orthonormal basis of H then for any linear operator A on H define:

${\displaystyle \left\|A\right\|_{\operatorname {HS} }^{2}:=\sum _{i\in I}\left\|Ae_{i}\right\|^{2}}$

where this sum may be finite or infinite. Note that this value actually independent of the orthonormal basis ${\displaystyle \{e_{i}:i\in I\}}$ of H that is chosen. Moreover, if the Hilbert-Schmidt norm is finite then the convergence of the sum converges necessitates that at most countably many of the terms ${\displaystyle \left\|Ae_{i}\right\|^{2}}$ are non-zero (even if I is uncountable). If A is a bounded linear operator then we have ${\displaystyle \left\|A\right\|\leq \left\|A\right\|_{\operatorname {HS} }}$.[4]

A bounded operator A on a Hilbert space ${\displaystyle \left(H,\|\cdot \|\right)}$ is a Hilbert-Schmidt operator if ${\displaystyle \left\|A\right\|_{\operatorname {HS} }^{2}}$ is finite. Equivalently, A is a Hilbert-Schmidt operator if the trace ${\displaystyle \operatorname {Tr} }$ of the nonnegative self-adjoint operator ${\displaystyle A^{*}A}$ is finite, in which case ${\displaystyle \|A\|_{\text{HS}}^{2}=\operatorname {Tr} (A^{*}A)}$.[5][6]

If A is an Hilbert–Schmidt operator on H then

${\displaystyle \|A\|_{\text{HS}}^{2}=\sum _{i,j}|A_{i,j}|^{2}=\|A\|_{2}^{2}}$

where ${\displaystyle \{e_{i}:i\in I\}}$ an orthonormal basis of H, ${\displaystyle A_{i,j}:=\langle e_{i},Ae_{j}\rangle }$, and ${\displaystyle \|A\|_{2}}$ is the Schatten norm of ${\displaystyle A}$ for p = 2. In Euclidean space, ${\displaystyle \|\cdot \|_{\text{HS}}}$ is also called Frobenius norm.

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 ${\displaystyle x\otimes y:H\to H}$ by (x ⊗ y)(z) = <z, y> 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), ${\displaystyle \operatorname {Tr} \left(A\left(x\otimes y\right)\right)=\left\langle Ax,y\right\rangle }$.[7]

If T : H → H is a bounded compact operator with eigenvalues l₁, l₂, ... where each eigenvalue is repeated as often as its multiplicity, then T is Hilbert-Schmidt if and only if ${\displaystyle \sum _{i=1}^{\infty }l_{i}^{2}<\infty }$, in which case the Hilbert-Schmidt norm of T is ${\displaystyle \left\|T\right\|_{\operatorname {HS} }={\sqrt {\sum _{i=1}^{\infty }l_{i}^{2}}}}$.[4]

If ${\displaystyle k\in L^{2}\left(\mu \times \mu \right)}$, where ${\displaystyle \left(X,\Omega ,\mu \right)}$ is a measure space, then the integral operator ${\displaystyle K:L^{2}\left(\mu \right)\to L^{2}\left(\mu \right)}$ with kernel k is a Hilbert-Schmidt operator and ${\displaystyle \left\|K\right\|_{\operatorname {HS} }=\left\|k\right\|_{2}}$.[4]

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

${\displaystyle \langle A,B\rangle _{\text{HS}}=\operatorname {Tr} (A^{*}B)=\sum _{i}\langle Ae_{i},Be_{i}\rangle .}$

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 B₂(H), which can be shown to be naturally isometrically isomorphic to the tensor product of Hilbert spaces

${\displaystyle H^{*}\otimes H,}$

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).[7] 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).[7]

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

• Every Hilbert-Schmidt operator T : H → H is a compact operator.[4]
• A bounded linear operator T : H → H is Hilbert-Schmidt if and only if the same is true of the operator ${\displaystyle \left|T\right|:={\sqrt {T^{*}T}}}$, in which case the Hilbert-Schmidt norms of T and |T| are equal.[4]
• Hilbert–Schmidt operators are nuclear operators of order 2, and are therefore compact operators.[4]
• If ${\displaystyle S:H_{1}\to H_{2}}$ and ${\displaystyle T:H_{2}\to H_{3}}$ are Hilbert-Schmidt operators between Hilbert spaces then the composition ${\displaystyle T\circ S:H_{1}\to H_{3}}$ is a nuclear operator.[3]
• If T : H → H is a bounded linear operator then we have ${\displaystyle \left\|T\right\|\leq \left\|T\right\|_{\operatorname {HS} }}$.[4]
• If T : H → H is a bounded linear operator on H and S : H → H is a Hilbert-Schmidt operator on H then ${\displaystyle \left\|S^{*}\right\|_{\operatorname {HS} }=\left\|S\right\|_{\operatorname {HS} }}$, ${\displaystyle \left\|TS\right\|_{\operatorname {HS} }\leq \left\|T\right\|\left\|S\right\|_{\operatorname {HS} }}$, and ${\displaystyle \left\|ST\right\|_{\operatorname {HS} }\leq \left\|S\right\|_{\operatorname {HS} }\left\|T\right\|}$.[4] In particular, the composition of two Hilbert-Schmidt operators is again Hilbert-Schmidt (and even a trace class operator).[4]
• The space of Hilbert-Schmidt operators on H is an ideal of the space of bounded operators ${\displaystyle B\left(H\right)}$ that contains the operators of finite-rank.[4]

• Frobenius inner product
• Trace class

• Conway, John (1990). A course in functional analysis. New York: Springer-Verlag. ISBN 978-0-387-97245-9. OCLC 21195908.
• Schaefer, Helmut H. (1999). Topological Vector Spaces. GTM. 3. New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.CS1 maint: ref=harv (link)