| dbp:mathStatement
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- Let X be a Banach space, C be a compact operator acting on X, and σ be the spectrum of C.
Every nonzero λ ∈ σ is an eigenvalue of C.
For all nonzero λ ∈ σ, there exists m such that Ker = Ker, and this subspace is finite-dimensional. It is called the generalized eigenspace for λ.
The eigenvalues can only accumulate at 0.
σ is at most countably infinite.
If the dimension of X is not finite, then σ must contain 0.
Every nonzero λ ∈ σ is a pole of the resolvent function ζ → −1. (en)
- If C is compact, then Ran is closed. (en)
- 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 is the distance from x to Y. (en)
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| dbp:proof
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- i) Without loss of generality, assume λ = 1.
Assume is not an eigenvalue, then is injective. Since it is bounded, but has no bounded inverse, it is not surjective, by the bounded inverse theorem.
By Lemma 2, Y1 = Ran is a closed proper subspace of X.
Since is injective, Y2 = Y1 is again a closed proper subspace of Y1.
Define Yn = Ran'n. Consider the decreasing sequence of subspaces
:
where all inclusions are proper, since is injective. By Riesz's lemma, we can choose unit vectors yn ∈ Yn such that d > ½.
Compactness of C means {C yn} has a convergent subsequence. But for n < m
:
and notice that , thus
:
which implies Cyn − Cym > ½, so it cannot have a convergent subsequence. Contradiction. ▮
ii) Consider the sequence { Yn = Kern} of closed subspaces. It satisfies . If we ever have some , then the sequence stops increasing from there on.
The theorem claims it stops increasing after finitely many steps. Suppose it does not stop, i.e. the inclusion Kern ⊂ Kern+1 is proper for all n.
By Riesz's lemma, there exists a sequence {yn}n ≥ 2 of unit vectors such that yn ∈ Y'n and d > 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 = Kern} must terminate at some finite m.
Kern is compact by induction on . By Riesz's lemma, this means it is finite-dimensional.
For , given any sequence with norm bounded by 1, by compactness of , there exists a subsequence such that . Thus, the closed unit ball in is compact.
Induct. Let be a sequence in the unit ball with norm bounded by 1. Now, is contained within a ball in , which is compact by induction. We also use the compactness of the operator. So we take subsequences twice, to obtain some , such that and are both convergent, so is also convergent.
▮
iii) Suppose there exist infinite distinct {λn} in the spectrum, such that λn > ε for all n. We derive a contradiction, thus concluding that there are no nonzero accumulation points.
By part i, they are eigenvalues. Pick corresponding eigenvectors {xn}.
Define Yn = span{x1...xn}.
The sequence {Yn} is a strictly increasing sequence.
Choose unit vectors such that yn ∈ Y'n and d > ½. Then for n < m
:
Since , we have
:
therefore Cyn − Cym > ε/2, a contradiction. ▮
iv) By iii) and the Cantor–Bendixson theorem. ▮
v) If does not contain zero, then has a bounded inverse, so is compact, so is finite-dimensional. ▮
vi''') As in the matrix case, this is a direct application of the holomorphic functional calculus. ▮ (en)
- Let in norm.
If is bounded, then there exists a sequence such that is bounded, and we still have .
So WLOG, is bounded. Then compactness of implies that there exists a subsequence such that is norm convergent. So is norm convergent, to some . Thus .
Now we show that is bounded.
If not, then select a divergent subsequence , and define vectors .
Since is unbounded, we have . Further, we also have that is a sequence of unit vectors in .
So by the previous half of the proof, there exists a convergent subsequence , such that , so , so is a zero vector, contradiction. (en)
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