Measure, Integration & Real Analysis, 2021a

326 Chapter 10 Linear Maps on Hilbert Spaces
10D
Spectral Theorem for Compact
Operators
Orthonormal Bases Consisting of Eigenvectors
We begin this section with the following useful lemma.
10.96 T ∗ T −‖T‖ 2 I is not invertible
If T is a bounded operator on a nonzero Hilbert space, then ‖T‖ 2 ∈ sp(T ∗ T).
Proof Suppose T is a bounded operator on a nonzero Hilbert space V. Let f 1 , f 2 ,...
be a sequence in V such that ‖ f n ‖ = 1 for each n ∈ Z + and
10.97 lim n→∞
‖Tf n ‖ = ‖T‖.
Then
∥
∥T ∗ Tf n −‖T‖ 2 f n
∥ ∥
2 = ‖T ∗ Tf n ‖ 2 − 2‖T‖ 2 〈T ∗ Tf n , f n 〉 + ‖T‖ 4
= ‖T ∗ Tf n ‖ 2 − 2‖T‖ 2 ‖Tf n ‖ 2 + ‖T‖ 4
10.98
≤ 2‖T‖ 4 − 2‖T‖ 2 ‖Tf n ‖ 2 ,
where the last line holds because ‖T ∗ Tf n ‖≤‖T ∗ ‖‖Tf n ‖≤‖T‖ 2 .Now10.97
and 10.98 imply that
lim
n→∞ (T∗ T −‖T‖ 2 I) f n = 0.
Because ‖ f n ‖ = 1 for each n ∈ Z + , the equation above implies that T ∗ T −‖T‖ 2 I
is not invertible, as desired.
The next result indicates one way in which self-adjoint compact operators behave
like self-adjoint operators on finite-dimensional Hilbert spaces.
10.99 every self-adjoint compact operator has an eigenvalue.
Suppose T is a self-adjoint compact operator on a nonzero Hilbert space. Then
either ‖T‖ or −‖T‖ is an eigenvalue of T.
Proof
Because T is self-adjoint, 10.96 states that T 2 −‖T‖ 2 I is not invertible. Now
T 2 −‖T‖ 2 I = ( T −‖T‖I )( T + ‖T‖I ) .
Thus T −‖T‖I and T + ‖T‖I cannot both be invertible. Hence ‖T‖ ∈sp(T) or
−‖T‖ ∈sp(T). Because T is compact, 10.85 now implies that ‖T‖ or −‖T‖ is an
eigenvalue of T, as desired, or that ‖T‖ = 0, which means that T = 0, in which case
0 is an eigenvalue of T.
Measure, Integration & Real Analysis, by Sheldon Axler