588 CHAPTER 22. HILBERT SPACES

As noted, this orthonormal set A∞ is countable. Let it be {uk}Nk=1 where N ≤ ∞. Thus

for x ∈ H,Ax ∈ A(H)⊆ A(H) and so, by Corollary 22.4.4,

Ax =N

∑k=1

(Ax,uk)uk =N

∑k=1

(x,Auk)uk =N

∑k=1

λ k (x,uk)uk (22.17)

and the series converges. Also, the formula implies directly that Aum = λ mum so |λ m| ≤∥A∥.

I claim limsupn→∞ |λ n|= 0. If this were not so, then for some ε > 0, 0 < ε ≤ |λ n| fora subsequence still denoted as λ n but then

∥Aun−Aum∥2 = ∥λ nun−λ mum∥2 = |λ n|2 + |λ m|2 ≥ 2ε2

and so there could not exist a convergent subsequence of {Auk}∞

k=1 contrary to the assump-tion that A is compact. This verifies the claim that limn→∞ λ n = 0. Also, since |λ m| ≤ ∥A∥ ,if S ⊆ N, supm∈S |λ m| = maxm∈S |λ m| and so, we can re-number the uk if necessary suchthat the eigenvalues satisfy |λ k| ≥ |λ k+1| for all k. Thus, if λ m = 0 for some m, it followsfrom 22.17 that A(H) is contained in the span of finitely many of the vectors {uk}. ■

Define v⊗u ∈L (H,H) by v⊗u(x) = (x,u)v, then 22.17 is of the form

A =N

∑k=1

λ kuk⊗uk

This is the content of the following corollary.

Corollary 22.5.4 The main conclusion of the above theorem can be written as A =

∑Nk=1 λ kuk⊗uk where the convergence of the partial sums takes place in the operator norm.

Proof: Without loss of generality, assume N = ∞.∣∣∣∣∣((

A−n

∑k=1

λ kuk⊗uk

)x,y

)∣∣∣∣∣=∣∣∣∣∣(

Ax−n

∑k=1

λ k (x,uk)uk,y

)∣∣∣∣∣=

∣∣∣∣∣(

∞

∑k=n

λ k (x,uk)uk,y

)∣∣∣∣∣=∣∣∣∣∣ ∞

∑k=n

λ k (x,uk)(uk,y)

∣∣∣∣∣≤ |λ n|

(∞

∑k=n|(x,uk)|2

)1/2(∞

∑k=n|(y,uk)|2

)1/2

≤ |λ n|∥x∥∥y∥

It follows ∥(A−∑nk=1 λ kuk⊗uk)(x)∥ ≤ |λ n|∥x∥ ■

Corollary 22.5.5 Let A be a compact self adjoint operator and

A =N

∑k=1

λ kuk⊗uk

where Auk = λ kuk with ∥uk∥= 1, the |λ k| decreasing. Then |λ k+1|= ∥Ak∥ where Ak is therestriction of A to {u1, ...,uk}⊥.