Measure, Integration & Real Analysis, 2021a

312 Chapter 10 Linear Maps on Hilbert Spaces
10C
Compact Operators
The Ideal of Compact Operators
A rich theory describes the behavior of compact operators, which we now define.
10.66 Definition compact operator; C(V)
• An operator T on a Hilbert space V is called compact if for every bounded
sequence f 1 , f 2 ,... in V, the sequence Tf 1 , Tf 2 ,... has a convergent
subsequence.
• The collection of compact operators on V is denoted by C(V).
The next result provides a large class of examples of compact operators. We will
see more examples after proving a few more results.
10.67 bounded operators with finite-dimensional range are compact
If T is a bounded operator on a Hilbert space and range T is finite-dimensional,
then T is compact.
Proof Suppose T is a bounded operator on a Hilbert space V and range T is
finite-dimensional. Suppose e 1 ,...,e m is an orthonormal basis of range T (a finite
orthonormal basis of range T exists because the Gram–Schmidt process applied to
any basis of range T produces an orthonormal basis; see the proof of 8.67).
Now suppose f 1 , f 2 ,...is a bounded sequence in V. For each n ∈ Z + ,wehave
Tf n = 〈Tf n , e 1 〉e 1 + ···+ 〈Tf n , e m 〉e m .
The Cauchy–Schwarz inequality shows that |〈Tf n , e j 〉|≤‖T‖ sup
k∈Z + ‖ f k ‖ for every
n ∈ Z + and j ∈{1, . . . , m}. Thus there exists a subsequence f n1 , f n2 ,...such that
lim k→∞ 〈Tf nk , e j 〉 exists in F for each j ∈{1,...,m}. The equation displayed above
now implies that lim k→∞ Tf nk exists in V. Thus T is compact.
Not every bounded operator is compact. For example, the identity map on an
infinite-dimensional Hilbert space is not compact (to see this, consider an orthonormal
sequence, which does not have a convergent subsequence because the distance
between any two distinct elements of the orthonormal sequence is √ 2).
10.68 compact operators are bounded
Every compact operator on a Hilbert space is a bounded operator.
Proof We show that if T is an operator that is not bounded, then T is not compact.
To do this, suppose V is a Hilbert space and T is an operator on V that is not bounded.
Thus there exists a bounded sequence f 1 , f 2 ,...in V such that lim n→∞ ‖Tf n ‖ = ∞.
Hence no subsequence of Tf 1 , Tf 2 ,...converges, which means T is not compact.
Measure, Integration & Real Analysis, by Sheldon Axler