Measure, Integration & Real Analysis, 2021a

316 Chapter 10 Linear Maps on Hilbert Spaces
Spectrum of Compact Operator and Fredholm Alternative
We noted earlier that the identity map on an infinite-dimensional Hilbert space is not
compact. The next result shows that much more is true.
10.74 no infinite-dimensional closed subspace in range of compact operator
The range of each compact operator on a Hilbert space contains no infinitedimensional
closed subspaces.
Proof Suppose T is a bounded operator on a Hilbert space V and U is an infinitedimensional
closed subspace contained in range T. We want to show that T is not
compact.
Because T is a continuous operator, T −1 (U) is a closed subspace of V. Let
S = T| T −1 (U). Thus S is a surjective bounded linear map from the Hilbert space
T −1 (U) onto the Hilbert space U [here T −1 (U) and U are Hilbert spaces by 6.16(b)].
The Open Mapping Theorem (6.81) implies S maps the open unit ball of T −1 (U) to
an open subset of U. Thus there exists r > 0 such that
10.75 {g ∈ U : ‖g‖ < r} ⊂{Tf : f ∈ T −1 (U) and ‖ f ‖ < 1}.
Because U is an infinite-dimensional Hilbert space, there exists an orthonormal
sequence e 1 , e 2 ,...in U, as can be seen by applying the Gram–Schmidt process (see
the proof of 8.67) to any linearly independent sequence in U. Each re n
is in the left
2
side of 10.75. Thus for each n ∈ Z + , there exists f n ∈ T −1 (U) such that ‖ f n ‖ < 1
and Tf n = re n
2 . The sequence f 1, f 2 ,...is bounded, but the sequence Tf 1 , Tf 2 ,...
has no convergent subsequence because ∥ re j
2 − re √
k
2r
∥ = for j ̸= k. Thus T is
2 2
not compact, as desired.
Suppose T is a compact operator on an infinite-dimensional Hilbert space. The
result above implies that T is not surjective. In particular, T is not invertible. Thus
we have the following result.
10.76 compact implies not invertible on infinite-dimensional Hilbert spaces
If T is a compact operator on an infinite-dimensional Hilbert space, then
0 ∈ sp(T).
Although 10.74 shows that if T is compact then range T contains no infinitedimensional
closed subspaces, the next result shows that the situation differs drastically
for T − αI if α ∈ F \{0}.
The proof of the next result makes use of the restriction of T − αI to the closed
subspace ( null(T − αI) ) ⊥ . As motivation for considering this restriction, recall that
each f ∈ V can be written uniquely as f = g + h, where g ∈ null(T − αI) and
h ∈ ( null(T − αI) ) ⊥ (see 8.43). Thus (T − αI) f =(T − αI)h, which implies that
range(T − αI) =(T − αI) (( null(T − αI) ) ⊥)
.
Measure, Integration & Real Analysis, by Sheldon Axler