Hilbert–Schmidt operator

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

$$\Vert A{\Vert }_{\mathrm{HS}}^2\stackrel{\text{def}}{=}\sum _{i\in I}\Vert A{e}_i{\Vert }_H^2,$$

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.Template:Sfn This definition is independent of the choice of the orthonormal basis. In finite-dimensional Euclidean space, the Hilbert–Schmidt norm $\Vert \cdot {\Vert }_{\text{HS}}$ is identical to the Frobenius norm.

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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 $$\sum _i\Vert A{e}_i{\Vert }^2=\sum _{i,j}{|⟨A{e}_i,f_j⟩|}^2=\sum _{i,j}{|⟨e_i,A^{\ast }{f}_j⟩|}^2=\sum _j\Vert A^{\ast }{f}_j{\Vert }^2.$$ If $e_i=f_i,$ then ${\sum }_i\Vert A{e}_i{\Vert }^2={\sum }_i\Vert A^{\ast }{e}_i{\Vert }^2.$ As for any bounded operator, $A=A^{\ast \ast }.$ Replacing $A$ with $A^{\ast }$ in the first formula, obtain ${\sum }_i\Vert A^{\ast }{e}_i{\Vert }^2={\sum }_j\Vert A{f}_j{\Vert }^2.$ 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 $(x\otimes y)(z)=⟨z,y⟩x$, 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 $H$), $\mathrm{tr}\left(A(x\otimes y)\right)=⟨Ax,y⟩$.Template:Sfn

If $T:H\to H$ is a bounded compact operator with eigenvalues ${\ell }_1,{\ell }_2,\dots$ of $|T|:=\sqrt{T^{\ast }T}$, where each eigenvalue is repeated as often as its multiplicity, then $T$ is Hilbert–Schmidt if and only if ${\sum }_{i=1}^{\mathrm{\infty }}{\ell }_i^2<\mathrm{\infty }$, in which case the Hilbert–Schmidt norm of $T$ is ${\Vert T\Vert }_{\mathrm{HS}}=\sqrt{\sum _{i=1}^{\mathrm{\infty }}{\ell }_i^2}$.Template:Sfn

If $k\in L^2(X\times X)$, where $(X,\mathrm{\Omega },\mu )$ is a measure space, then the integral operator $K:L^2\left(X\right)\to L^2\left(X\right)$ with kernel $k$ is a Hilbert–Schmidt operator and ${\Vert K\Vert }_{\mathrm{HS}}={\Vert k\Vert }_2$.Template:Sfn Conversely, if a bounded operator on $L^2\left(X\right)$ is Hilbert-Schmidt then it may be written as an integral operator of this form.Template:Sfn

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

$$⟨A,B{⟩}_{\text{HS}}=\mathrm{tr}(B^{\ast }A)=\sum _i⟨A{e}_i,B{e}_i⟩.$$

The Hilbert–Schmidt operators form a two-sided *-ideal in the Banach algebra of bounded operators on Template:Math. They also form a Hilbert space, denoted by Template:Math or Template:Math, which can be shown to be naturally isometrically isomorphic to the tensor product of Hilbert spaces

$$H^{\ast }\otimes H,$$

where Template:Math is the dual space of Template:Math. 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).Template:Sfn 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).Template:Sfn

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

Properties

• Every Hilbert–Schmidt operator Template:Math is a compact operator.Template:Sfn
• A bounded linear operator Template:Math is Hilbert–Schmidt if and only if the same is true of the operator $\left|T\right|:=\sqrt{T^{\ast }T}$, in which case the Hilbert–Schmidt norms of $T$ and $\left|T\right|$ are equal.Template:Sfn
• Hilbert–Schmidt operators are nuclear operators of order 2, and are therefore compact operators.Template:Sfn
• 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.Template:Sfn
• If Template:Math is a bounded linear operator then we have $\Vert T\Vert \le {\Vert T\Vert }_{\mathrm{HS}}$.Template:Sfn
• Template:Math is a Hilbert–Schmidt operator if and only if the trace $\mathrm{tr}$ of the nonnegative self-adjoint operator $T^{\ast }T$ is finite, in which case $\Vert T{\Vert }_{\text{HS}}^2=\mathrm{tr}(T^{\ast }T)$.
• If Template:Math is a bounded linear operator on Template:Math and Template:Math is a Hilbert–Schmidt operator on Template:Math then ${\Vert S^{\ast }\Vert }_{\mathrm{HS}}={\Vert S\Vert }_{\mathrm{HS}}$, ${\Vert TS\Vert }_{\mathrm{HS}}\le \Vert T\Vert {\Vert S\Vert }_{\mathrm{HS}}$, and ${\Vert ST\Vert }_{\mathrm{HS}}\le {\Vert S\Vert }_{\mathrm{HS}}\Vert T\Vert$.Template:Sfn In particular, the composition of two Hilbert–Schmidt operators is again Hilbert–Schmidt (and even a trace class operator).Template:Sfn
• The space of Hilbert–Schmidt operators on Template:Math is an ideal of the space of bounded operators $B\left(H\right)$ that contains the operators of finite-rank.Template:Sfn
• If Template:Math is a Hilbert–Schmidt operator on Template:Math then $$\Vert A{\Vert }_{\text{HS}}^2=\sum _{i,j}|⟨e_i,A{e}_j⟩{|}^2=\Vert A{\Vert }_2^2$$ where $\{e_i:i\in I\}$ is an orthonormal basis of H, and $\Vert A{\Vert }_2$ is the Schatten norm of $A$ for Template:Math. In Euclidean space, $\Vert \cdot {\Vert }_{\text{HS}}$ is also called the Frobenius norm.

See also

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References

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