468 CHAPTER 17. BANACH SPACES

Lemma 17.6.4 Let Y be an infinite dimensional Banach space. Then there exists a se-quence {xn} in the unit sphere S, ∥xn∥= 1, such that ∥xn− xm∥ ≥ 1

2 whenever n ̸= m.

Proof: Pick x1 ∈ S. Now the span of x1 is not everything and so there exists u2 /∈span(x1) . Let w2 be a point of span(x1) such that ∥u2−w2∥ ≤ 2dist(u2,span(x1)) . Thenx2 =

u2−w2∥u2−w2∥

. Then

∥x1− x2∥=∥∥∥∥∥u2−w2∥x1− (u2−w2)

∥u2−w2∥

∥∥∥∥≥ dist(u2,span(x1))

2dist(u2,span(x1))=

12

Now repeat the argument with span(x1,x2) in place of span(x1) and continue to get thedesired sequence.

Lemma 17.6.5 Let L be a compact linear map. Then the eigenspace of L is finite dimen-sional for each eigenvalue λ ̸= 0.

Proof: Consider (L−λ I)−1 (0)∩S where S is the unit sphere. The eigenspace is just(L−λ I)−1 (0) . Let Y be this inverse image. If Y is infinite dimensional, then the aboveLemma 17.6.4 applies. There exists {xn} ⊆ (L−λ I)−1 (0)∩S where ∥xn− xm∥ ≥ 1/2 forall n ̸= m. Then there is a subsequence, still denoted with subscript n such that {Lxn} is aCauchy sequence. Thus Lxn = λxn and so, since λ ̸= 0, it follows that {xn} is also a Cauchysequence and converges to some x. But this is impossible because of the construction of the{xn} which prevents there being any Cauchy sequence. Thus Y must be finite dimensional.

This lemma is useful in proving the following major spectral theorem about the eigen-values of a compact operator. I found this theorem in Deimling [38].

Theorem 17.6.6 Let L ∈L (X ,X) with L compact. Let Λ be the eigenvalues of L. That isλ ∈ Λ means there exists x ̸= 0 such that Lx = λx. It is assumed the field of scalars is R orC. Let Rλ ≡ L−λ I. Then the following hold.

1. If µ ∈ Λ then |µ| ≤ ∥L∥ ,Λ is at most countable and has no limit points other thanpossibly 0.

2. Rλ is a homeomorphism onto X whenever λ /∈ Λ∪{0} .

3. For all λ ∈ Λ\{0} , there exists a smallest k = k (λ ) ,

(a) Rkλ

X ⊕N(Rk

λ

)= X where N

(Rk

λ

)is the vectors x such that Rk

λx = 0. Rk

λX is

closed, dim(N(Rk

λ

))< ∞.

(b) Rkλ

X and N(Rk

λ

)are invariant under L and Rλ |Rk

kX is a homeomorphism onto

Rkλ

X.

(c) N(Rk

µ

)⊆ Rk

λX for all λ ,µ ∈ Λ\{0} where λ ̸= µ .

468 CHAPTER 17. BANACH SPACESLemma 17.6.4 Let Y be an infinite dimensional Banach space. Then there exists a se-quence {Xp} in the unit sphere S, ||xp|| = 1, such that ||x» —xXm|| > 5 whenever n # m.Proof: Pick x; € S. Now the span of x; is not everything and so there exists uz ¢span (x;). Let w2 be a point of span (x;) such that ||u2 —wa|| < 2dist (uz, span (x;)). ThenXo = 72”. ThenIlu2—wa|||v2 — wa||1 — (uz — w2)I|u2 — wa|~ 2dist(u2,span(x;)) 2dist (uz, span (x 11 al =| E (us,span(x1)) _ 1Now repeat the argument with span (x1,x2) in place of span(x;) and continue to get thedesired sequence. §fLemma 17.6.5 Let L be a compact linear map. Then the eigenspace of L is finite dimen-sional for each eigenvalue 0 # 0.Proof: Consider (L—AJ)~' (0) MS where S is the unit sphere. The eigenspace is just(L—A1)~'(0). Let Y be this inverse image. If Y is infinite dimensional, then the aboveLemma 17.6.4 applies. There exists {x,,} C (L—AD)~' (0) MS where ||xy —xm|| > 1/2 forall n 4m. Then there is a subsequence, still denoted with subscript n such that {Lx,} is aCauchy sequence. Thus Lx, = Ax, and so, since A 4 0, it follows that {x,} is also a Cauchysequence and converges to some x. But this is impossible because of the construction of the{xn} which prevents there being any Cauchy sequence. Thus Y must be finite dimensional.|This lemma is useful in proving the following major spectral theorem about the eigen-values of a compact operator. I found this theorem in Deimling [38].Theorem 17.6.6 Let L € &(X,X) with L compact. Let A be the eigenvalues of L. That isA € Ameans there exists x #0 such that Lx = Ax. It is assumed the field of scalars is R orC. Let Ry =L—AI. Then the following hold.1. If w € A then |p| < ||L||, A is at most countable and has no limit points other thanpossibly 0.2. Ry is a homeomorphism onto X whenever A ¢ AU {0}.3. Forall A € A\ {0}, there exists a smallestk=k(A),(a) RX @N (Ri ) = X where N (Ri) is the vectors x such that Rix =0. RX isclosed, dim (N (Ri )) <_00,(b) RUX and N (Ri) are invariant under L and R)| REX is a homeomorphism ontoJkRyX.(c) N (Ri) C RKX for all A, € A\ {0} where AF .