Lemma 1 (Riesz's lemma) — Let X be a Banach space and Y ⊂ X, Y ≠ X, be a closed subspace.
For all ε > 0, there exists x ∈ X such that x = 1 and
where d(x, Y) is the distance from x to Y.
This fact will be used repeatedly in the argument leading to the theorem.
Notice that when X is a Hilbert space, the lemma is trivial.
Proof
i) Without loss of generality, assume λ = 1. λ ∈ σ(C) not being an eigenvalue means (I − C) is injective but not surjective.
By Lemma 2, Y1 = Ran(I − C) is a closed proper subspace of X.
Since (I − C) is injective, Y2 = (I − C)Y1 is again a closed proper subspace of Y1.
Define Yn = Ran(I − C)n. Consider the decreasing sequence of subspaces
where all inclusions are proper. By lemma 1, we can choose unit vectors yn ∈ Yn such that d(yn, Yn+1) > ½.
Compactness of C means {C yn} must contain a norm convergent subsequence. But for n < m
and notice that
which implies Cyn − Cym > ½.
This is a contradiction, and so λ must be an eigenvalue.
ii) The sequence { Yn = Ker(λi − A)n} is an increasing sequence of closed subspaces.
The theorem claims it stops. Suppose it does not stop, i.e. the inclusion Ker(λi − A)n ⊂ Ker(λi − A)n+1 is proper for all n.
By lemma 1, there exists a sequence {yn}n ≥ 2 of unit vectors such that yn ∈ Yn and d(yn, Yn − 1) > ½.
As before, compactness of C means {C yn} must contain a norm convergent subsequence. But for n < m
and notice that
which implies Cyn − Cym > ½.
This is a contradiction, and so the sequence { Yn = Ker(λi − A)n} must terminate at some finite m.
Using the definition of the Kernel, we can show that the unit sphere of Ker(λi − C) is compact, so that Ker(λi − C) is finite-dimensional.
Ker(λi − C)n is finite-dimensional for the same reason.
iii) Suppose there exist infinite (at least countable) distinct eigenvalues {λn}, with corresponding eigenvectors {xn}, such that λn > ε for all n.
Define Yn = span{x1...xn}.
The sequence {Yn} is a strictly increasing sequence.
Choose unit vectors such that yn ∈ Yn and d(yn, Yn − 1) > ½. Then for n < m
But
therefore Cyn − Cym > ε/2, a contradiction.
So we have that there are only finite distinct eigenvalues outside any ball centered at zero.
This immediately gives us that zero is the only possible limit point of eigenvalues and there are at most countable distinct eigenvalues (see iv).
iv) This is an immediate consequence of iii). The set of eigenvalues {λ} is the union
Because σ(C) is a bounded set and the eigenvalues can only accumulate at 0, each Sn is finite, which gives the desired result.
v) As in the matrix case, this is a direct application of the holomorphic functional calculus.