Measure, Integration & Real Analysis, 2021a

Section 10C Compact Operators 315
As a special case of the previous result, we can now see that the Volterra operator
V : L 2 ([0, 1]) → L 2 ([0, 1]) defined by
(V f )(x) =
is compact. This holds because, as shown in Example 10.15, the Volterra operator is
an integral operator of the type considered in the previous result.
∫ The Volterra operator is injective [because differentiating both sides of the equation
x
0
f = 0 with respect to x and using the Lebesgue Differentiation Theorem (4.19)
shows that f = 0]. Thus the Volterra operator is an example of a compact operator
with infinite-dimensional range. The next example provides another class of compact
operators that do not necessarily have finite-dimensional range.
∫ x
10.72 Example compact multiplication operators on l 2
Suppose b 1 , b 2 ,...is a sequence in F such that lim n→∞ b n = 0. Define a bounded
linear map T : l 2 → l 2 by
T(a 1 , a 2 ,...)=(a 1 b 1 , a 2 b 2 ,...)
and for n ∈ Z + , define a bounded linear map T n : l 2 → l 2 by
T n (a 1 , a 2 ,...)=(a 1 b 1 , a 2 b 2 ,...,a n b n ,0,0,...).
Note that each T n is a bounded operator with finite-dimensional range and thus is
compact (by 10.67). The condition lim n→∞ b n = 0 implies that lim n→∞ T n = T.
Thus T is compact because C(V) is a closed subset of B(V) [by 10.69(a)].
The next result states that an operator is compact if and only if its adjoint is
compact.
10.73 T compact ⇐⇒ T ∗ compact
Suppose T is a bounded operator on a Hilbert space. Then T is compact if and
only if T ∗ is compact.
Proof First suppose T is compact. We want to prove that T ∗ is compact. To do
this, suppose f 1 , f 2 ,...is a bounded sequence in V. Because TT ∗ is compact [by
10.69(b)], some subsequence TT ∗ f n1 , TT ∗ f n2 ,...converges. Now
‖T ∗ f nj − T ∗ f nk ‖ 2 = 〈 T ∗ ( f nj − f nk ), T ∗ ( f nj − f nk ) 〉
0
f
= 〈 TT ∗ ( f nj − f nk ), f nj − f nk
〉
≤‖TT ∗ ( f nj − f nk )‖‖f nj − f nk ‖.
The inequality above implies that T ∗ f n1 , T ∗ f n2 ,...is a Cauchy sequence and hence
converges. Thus T ∗ is a compact operator, completing the proof that if T is compact,
then T ∗ is compact.
Now suppose T ∗ is compact. By the result proved in the paragraph above, (T ∗ ) ∗
is compact. Because (T ∗ ) ∗ = T (see 10.11), we conclude that T is compact.
Measure, Integration & Real Analysis, by Sheldon Axler