Measure, Integration & Real Analysis, 2021a

Section 10C Compact Operators 313
If V is a Hilbert space, then a twosided
ideal of B(V) is a subspace of
B(V) that is closed under multiplication
on either side by bounded operators on V.
The next result states that the set of compact
operators on V is a two-sided ideal
of B(V) that is closed in the topology on
B(V) that comes from the norm.
If V is finite-dimensional, then the
only two-sided ideals of B(V) are
{0} and B(V). In contrast, if V is
infinite-dimensional, then the next
result shows that B(V) has a closed
two-sided ideal that is neither {0}
nor B(V).
10.69 C(V) is a closed two-sided ideal of B(V)
Suppose V is a Hilbert space.
(a) C(V) is a closed subspace of B(V).
(b) If T ∈C(V) and S ∈B(V), then ST ∈C(V) and TS ∈C(V).
Proof Suppose f 1 , f 2 ,...is a bounded sequence in V.
To prove that C(V) is closed under addition, suppose S, T ∈C(V). Because
S is compact, Sf 1 , Sf 2 ,...has a convergent subsequence Sf n1 , Sf n2 ,.... Because
T is compact, some subsequence of Tf n1 , Tf n2 ,... converges. Thus we have a
subsequence of (S + T) f 1 , (S + T) f 2 ,...that converges. Hence S + T ∈C(V).
The proof that C(V) is closed under scalar multiplication is easier and is left to
the reader. Thus we now know that C(V) is a subspace of B(V).
To show that C(V) is closed in B(V), suppose T ∈B(V) and there is a sequence
T 1 , T 2 ,...in C(V) such that lim m→∞ ‖T − T m ‖ = 0. To show that T is compact, we
need to show that Tf n1 , Tf n2 ,...is a Cauchy sequence for some increasing sequence
of positive integers n 1 < n 2 < ···.
Because T 1 is compact, there is an infinite set Z 1 ⊂ Z + with ‖T 1 f j − T 1 f k ‖ < 1
for all j, k ∈ Z 1 . Let n 1 be the smallest element of Z 1 .
Now suppose m ∈ Z + with m > 1 and an infinite set Z m−1 ⊂ Z + and
n m−1 ∈ Z m−1 have been chosen. Because T m is compact, there is an infinite set
Z m ⊂ Z m−1 with
‖T m f j − T m f k ‖ <
m
1
for all j, k ∈ Z m . Let n m be the smallest element of Z m such that n m > n m−1 .
Thus we produce an increasing sequence n 1 < n 2 < ··· of positive integers and
a decreasing sequence Z 1 ⊃ Z 2 ⊃··· of infinite subsets of Z + .
If m ∈ Z + and j, k ≥ m, then
‖Tf nj − Tf nk ‖≤‖Tf nj − T m f nj ‖ + ‖T m f nj − T m f nk ‖ + ‖T m f nk − Tf nk ‖
≤‖T − T m ‖(‖ f nj ‖ + ‖ f nk ‖)+ 1 m .
We can make the first term on the last line above as small as we want by choosing
m large (because lim m→∞ ‖T − T m ‖ = 0 and the sequence f 1 , f 2 ,...is bounded).
Thus Tf n1 , Tf n2 ,...is a Cauchy sequence, as desired, completing the proof of (a).
To prove (b), suppose T ∈C(V) and S ∈B(V). Hence some subsequence of
Tf 1 , Tf 2 ,...converges, and applying S to that subsequence gives another convergent
sequence. Thus ST ∈C(V). Similarly, Sf 1 , Sf 2 ,...is a bounded sequence, and thus
T(Sf 1 ), T(Sf 2 ),...has a convergent subsequence; thus TS ∈C(V).
Measure, Integration & Real Analysis, by Sheldon Axler