operator theory
Hilbert–Schmidt Operators
You should know: compact operators fa, lp spaces fa
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
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.
Notation
| Notation | Meaning |
|---|---|
| Hilbert space of Hilbert–Schmidt operators | |
| Hilbert–Schmidt norm | |
| Hilbert–Schmidt inner product tr(S*T) | |
| Integral kernel of a Hilbert–Schmidt operator |
Theorems
Worked Examples
- 1
The kernel is K(x,y) = 1, which is in L^2([0,1]^2) with:
- 2
By Theorem 8.2, T is Hilbert–Schmidt with ||T||_2 = ||K||_{L^2} = 1.
- 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:
- 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
Show that the Hilbert–Schmidt norm ||T||_2 is independent of the choice of orthonormal basis.
Is the identity operator I on an infinite-dimensional Hilbert space Hilbert–Schmidt? Justify your answer.
Quiz
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
- BookSimon, B. — Trace Ideals and Their Applications (2nd ed.), AMS, 2005
- BookConway, J. — A Course in Functional Analysis (2nd ed.), Chapter II
- Websiteen.wikipedia.org
Mathematics