21.3. OPERATOR VALUED FUNCTIONS 665
Proof of claim: From 21.3.16 and the definition of An and ek⊗ ek,
(An−1en+1,en+1) =
((A−
n−1
∑k=1
λ kek⊗ ek
)en+1,en+1
)= (Aen+1,en+1) = (Anen+1,en+1)
Thus,
λ n+1 = (Anen+1,en+1)
= (An−1en+1,en+1)−λ n |(en,en+1)|2
= (An−1en+1,en+1)
By the previous claim. Therefore,
|λ n+1|= |(An−1en+1,en+1)| ≤ |(An−1en,en)|= |λ n|
by the definition of |λ n|. (en makes |(An−1x,x)| as large as possible.)Claim 3: limn→∞ λ n = 0.Proof of claim: If for some n,λ n = 0, then λ k = 0 for all k > n by claim 2. Thus, for
some n,
A =n
∑k=1
λ kek⊗ ek
Assume then that λ k ̸= 0 for any k. Then if limk→∞ |λ k|= ε > 0, one contradicts, ∥ek∥= 1for all k because
∥Aen−Aem∥2 = ∥λ nen−λ mem∥2
= λ2n +λ
2m ≥ 2ε
2
which shows there is no Cauchy subsequence of {Aen}∞
n=1 , which contradicts the compact-ness of A. This proves the claim.
Claim 4: ∥An∥→ 0Proof of claim: Let x,y ∈ B
|λ n+1| ≥∣∣∣∣(An
x+ y2
,x+ y
2
)∣∣∣∣=
∣∣∣∣14 (Anx,x)+14(Any,y)+
12(Anx,y)
∣∣∣∣≥ 1
2|(Anx,y)|− 1
4|(Anx,x)+(Any,y)|
≥ 12|(Anx,y)|− 1
4(|(Anx,x)|+ |(Any,y)|)
≥ 12|(Anx,y)|− 1
2|λ n+1|
and so3 |λ n+1| ≥ |(Anx,y)| .
It follows ∥An∥ ≤ 3 |λ n+1| . By 21.3.16 this proves 21.3.15 and completes the proof.