Mathematics.

operator theory

Hilbert–Schmidt Operators

Functional Analysis55 minDifficulty8 out of 10

Overview

A bounded operator T on a Hilbert space H is Hilbert–Schmidt if the sum of squares of its singular values is finite: sum sigma_n(T)^2 < infinity. Equivalently, for any orthonormal basis {e_n}, sum ||Te_n||^2 < infinity. Hilbert–Schmidt operators form a Hilbert space B_2(H) under the inner product <S,T>_HS = tr(S*T), and every Hilbert–Schmidt operator is compact. The integral operators with square-integrable kernels are the prototypical Hilbert–Schmidt operators.

Intuition

Hilbert–Schmidt operators are the 'L^2 version' of compact operators, just as trace-class operators are the 'L^1 version'. An integral operator (Tf)(x) = integral K(x,y)f(y) dy on L^2 is Hilbert–Schmidt precisely when the kernel K is in L^2(X x X). This makes Hilbert–Schmidt operators the natural setting for Fredholm integral equations and many problems in mathematical physics.

Formal Definition

Definition

T in B(H) is Hilbert–Schmidt if ||T||_2^2 = sum_{n=1}^inf ||Te_n||^2 < infinity for some (equivalently, any) orthonormal basis {e_n}. The Hilbert–Schmidt norm is ||T||_2 = (sum ||Te_n||^2)^{1/2} = (sum sigma_n^2)^{1/2} = (tr(T*T))^{1/2}. The space B_2(H) with inner product <S,T>_2 = tr(S*T) = sum_n <Se_n, Te_n> is a Hilbert space.

T22=n=1Ten2=n=1σn(T)2=tr(TT)\|T\|_2^2 = \sum_{n=1}^\infty \|T e_n\|^2 = \sum_{n=1}^\infty \sigma_n(T)^2 = \operatorname{tr}(T^*T)
Hilbert–Schmidt norm squared
S,THS=tr(ST)=nSen,Ten\langle S, T \rangle_{\mathrm{HS}} = \operatorname{tr}(S^*T) = \sum_n \langle Se_n, Te_n \rangle
Hilbert–Schmidt inner product
T1T22/σ1(T),T2T11/2T1/2\|T\|_1 \le \|T\|_2^2 / \sigma_1(T),\quad \|T\|_2 \le \|T\|_1^{1/2}\|T\|^{1/2}
Relations between norms

Notation

NotationMeaning
B2(H)\mathcal{B}_2(H)Hilbert space of Hilbert–Schmidt operators
T2\|T\|_2Hilbert–Schmidt norm
S,THS\langle S,T\rangle_{\mathrm{HS}}Hilbert–Schmidt inner product tr(S*T)
K(x,y)K(x,y)Integral kernel of a Hilbert–Schmidt operator

Theorems

Theorem 8.1: Theorem 8.1
B2(H)isaHilbertspacewithinnerproduct<S,T>2=tr(ST),andeveryHilbertSchmidtoperatoriscompact.ThecontainmentB1(H)subsetB2(H)subsetK(H)holds,andB1(H)=B2(H)B2(H)(productoftwoHSoperatorsistraceclass).B_2(H) is a Hilbert space with inner product <S,T>_2 = tr(S*T), and every Hilbert–Schmidt operator is compact. The containment B_1(H) subset B_2(H) subset K(H) holds, and B_1(H) = B_2(H) * B_2(H) (product of two HS operators is trace-class).
Theorem 8.2: Theorem 8.2 (Integral operators)
AnintegraloperatorTonL2(X,mu)definedby(Tf)(x)=integralK(x,y)f(y)dmu(y)isHilbertSchmidtifandonlyifKinL2(XxX,muxmu),andthenT2=KL2(XxX).An integral operator T on L^2(X, mu) defined by (Tf)(x) = integral K(x,y)f(y)dmu(y) is Hilbert–Schmidt if and only if K in L^2(X x X, mu x mu), and then ||T||_2 = ||K||_{L^2(X x X)}.
Theorem 8.3: Theorem 8.3
EveryHilbertSchmidtoperatorTadmitstheSchmidtdecompositionT=sumnsigman<,en>fnwhere(en)and(fn)areorthonormalsequencesandthesumconvergesintheHilbertSchmidtnorm.Every Hilbert–Schmidt operator T admits the Schmidt decomposition T = sum_n sigma_n <·, e_n> f_n where (e_n) and (f_n) are orthonormal sequences and the sum converges in the Hilbert–Schmidt norm.

Worked Examples

  1. 1

    The kernel is K(x,y) = 1, which is in L^2([0,1]^2) with:

    KL2([0,1]2)2=010112dxdy=1\|K\|_{L^2([0,1]^2)}^2 = \int_0^1\int_0^1 1^2\,dx\,dy = 1
  2. 2

    By Theorem 8.2, T is Hilbert–Schmidt with ||T||_2 = ||K||_{L^2} = 1.

  3. 3

    Direct check: take the ONB phi_0 = 1, phi_n(x) = sqrt(2)cos(n pi x). We compute ||T phi_n||^2 directly. T(1)(x) = integral_0^1 1 dy = 1, so T phi_0 = 1. For n >= 1:

    (Tϕn)(x)=01ϕn(y)dy=0 (since cos integrates to 0 over [0,1])(T\phi_n)(x) = \int_0^1 \phi_n(y)\,dy = 0 \text{ (since } \cos \text{ integrates to 0 over } [0,1])
  4. 4

    So sum ||T phi_n||^2 = ||T phi_0||^2 + sum_{n>=1} 0 = 1. This confirms ||T||_2 = 1.

✓ Answer

The constant kernel operator has ||T||_2 = 1, consistent with ||K||_{L^2([0,1]^2)} = 1.

Practice Problems

Mediumproof writing

Show that the Hilbert–Schmidt norm ||T||_2 is independent of the choice of orthonormal basis.

Mediumfree response

Is the identity operator I on an infinite-dimensional Hilbert space Hilbert–Schmidt? Justify your answer.

Quiz

T in B(H) is Hilbert–Schmidt if:
An integral operator on L^2(X) with kernel K is Hilbert–Schmidt iff:
The product of two Hilbert–Schmidt operators ST belongs to:

Summary

  • T is Hilbert–Schmidt if sum sigma_n(T)^2 < infinity, equivalently sum ||Te_n||^2 < infinity for any ONB.
  • The HS inner product <S,T>_2 = tr(S*T) makes B_2(H) a Hilbert space.
  • Integral operators on L^2(X) are HS iff the kernel is in L^2(X x X); ||T||_2 = ||K||_{L^2}.
  • Containment: B_1(H) subset B_2(H) subset K(H); B_2 * B_2 subset B_1.
  • The identity on an infinite-dimensional Hilbert space is not Hilbert–Schmidt.

References

  1. BookSimon, B. — Trace Ideals and Their Applications (2nd ed.), AMS, 2005
  2. BookConway, J. — A Course in Functional Analysis (2nd ed.), Chapter II